\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 119, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/119\hfil Radial positive solutions]
{Radial positive solutions for a nonpositone problem in an annulus}

\author[S. Hakimi, A. Zertiti \hfil EJDE-2014/119\hfilneg]
{Said Hakimi, Abderrahim Zertiti}  % in alphabetical order

\address{Said Hakimi \newline
Universit\'e Sultan Moulay Slimane\\
Facult\'e polydisciplinaire \\
D\'epartement de Math\'ematiques \\
B\'eni Mellal, Morocco}
\email{h\_saidhakimi@yahoo.fr}

\address{Abderrahim Zertiti \newline
Universit\'e Abdelmalek Essaadi\\
Facult\'e des sciences \\
D\'epartement de Math\'ematiques \\
BP 2121, T\'etouan, Morocco}
\email{zertitia@hotmail.com}

\thanks{Submitted April 11, 2013. Published April 25, 2014.}
\subjclass[2000]{35J25, 34B18}
\keywords{Nonpositone problem; radial positive solutions}

\begin{abstract}
 The main purpose of this article is to prove the existence of radial
 positive solutions for a nonpositone problem in an annulus when the
 nonlinearity is superlinear and has more than one zero.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

In this article we study the existence of radial positive solutions
for the boundary-value problem
\begin{equation}
\begin{gathered}
-\Delta u(x)=\lambda f(u(x))\quad x\in \Omega, \\
u(x)=0\quad  x\in \partial \Omega,
\end{gathered}  \label{eq1}
\end{equation}
 where $\lambda >0$, $f:[0,+\infty ) \to \mathbb{R}$ is a continuous nonlinear 
function that has more than one zero, and $\Omega \subset \mathbb{R}^N$
is the annulus: $\Omega =C(0,R,\widehat{R})
 =\{ x\in \mathbb{R}^N :R<| x| <\widehat{R}\} $ 
($N>2$, $0<R<\widehat{R}$).

When $f$ is a nondecreasing nonlinearity satisfying $f(0)<0$ 
(the nonpositone case) and has only one zero, problem \eqref{eq1} has
been studied by Arcoya and Zertiti \cite{a1} and by Hakimi and
Zertiti in a ball when $f$ has more than one zero \cite{h1}.

We observe that the existence of radial positive solutions of
\eqref{eq1} is equivalent to the existence of positive solutions of
the problem
\begin{equation}
\begin{gathered}
-u''(r)-\frac{N-1}ru'(r)=\lambda f(u(r)) \quad R<r<\widehat{R} \\
u(R)=u(\widehat{R})=0.
\end{gathered} \label{eq2}
\end{equation}
Our main objective in this article is to prove that the result of
existence of radial positive solutions of the problem \eqref{eq1}
remains valid when $f$ has more than one zero and is not increasing
entirely on $[0,+\infty )$; see \cite[Theorem 2.4]{a1}.

\begin{remark} \label{rmk1} \rm 
In this article, we assume (without loss of generality) that $f$ 
has exactly three zeros.
\end{remark}

We assume that the map $f:[0,+\infty ) \to \mathbb{R}$ satisfies the following
hypotheses:
\begin{itemize}
\item[(F1)] $f$\ $\in C^1([0,+\infty ),\mathbb{R})$ such that $f$
 has three zeros $\beta _1<\beta_2<\beta_3$, with
$ f'(\beta _i) \neq 0$ for all $i\in \{ 1,2,3\}$.
 Moreover, $f'\geq 0$ on $[\beta _3,+\infty )$.

\item[(F2)] $f(0)<0$.

\item[(F3)] $\lim_{u\to +\infty } \frac{f(u)}u=+\infty$.

\item[(F4)] The function $h(u) =NF(u) -\frac{N-2}2f(u) u$
 is bounded from below in $[0,+\infty )$, where
$F(x) =\int_0^xf(r)dr$.
\end{itemize}

\begin{remark} \label{rmk2} \rm 
We observe that our arguments also work in the case $\Omega =B(O,R)$, 
improving slightly the results in  \cite{h1}. In fact in \cite{h1},
besides imposing that $f$ is increasing, we need (F1), (F2), (F3) and that
For some $k\in (0,1) $, 
\[
\lim_{d \to +\infty } \big(\frac {d}{f(d)}\big)^{N/2}\big(F(kd)
-\frac{N-2}{2N}\,df(d)\big)
=+\infty.
\]
On the other hand, it is clear that our hypothesis (F4) is more general 
than this assumption.

For a nonexistence result of positive solutions for superlinearities 
satisfying (F1), (F2) and (F3) see \cite{h2}. 
Also see \cite{c2} for existence and nonexistence of positive solutions 
for a class of superlinear semipositone systems, and \cite{c3} for existence 
and multiplicity results for semipositone problems.
\end{remark}

\section{Main Result}

In this section, we give the main result in this work. More
precisely we shall prove the following theorem.

\begin{theorem} \label{thm2.1}
Assume that the hypotheses {\rm (F1)--(F4)} are
satisfied. Then there exists a positive real number $\lambda_*$ such that if 
$\lambda <\lambda_{*}$, problem \eqref{eq1} has at least one radial
positive solution.
\end{theorem}

 To prove Theorem \ref{thm2.1}, we need the next four
technical lemmas. The first lemma assures the existence of a unique
solution $u(.,d,\lambda)$ of \eqref{eq2} in $[R,+\infty)$ for all $\lambda, d>0$.
The three last lemmas concern the behaviour of the solution of \eqref{eq2}.

\begin{remark} \label{rmk3} \rm
In this article we follow the work of Arcoya and Zertiti \cite{a1}, 
and we note that the proofs of Lemmas \ref{lem2.4} and \ref{lem2.7}
are analogous with  those of \cite[Lemmas 1.1 and 2.3]{a1}. 
On the other hand, the proofs of the second and third lemmas are different
from that of \cite[Lemma 2.1 and 2.2]{a1}.
This is so because our $f$ has more than one zero. So we apply the 
Shooting method. For this we consider the auxiliary boundary-value problem
\begin{equation}
\begin{gathered}
-u''(r)-\frac{N-1}ru'(r)=\lambda f(u(r)), \quad r>R \\
u(R)=0, \quad u'(R)=d, \label{eq3}
\end{gathered}
\end{equation}
where $d$ is the parameter of Shooting method.
\end{remark}

\begin{remark} \label{rmk4} \rm 
For suitable $d$, problem \eqref{eq3} has a
solution $u:=u(.,d,\lambda)$ such that $u>0$ on $(R,\widehat{R})$
and $u(\widehat{R})=0$. So, such solution $u$ of \eqref{eq3} is
also a positive solution of \eqref{eq2}.
\end{remark}

In this sequel, we suppose that the nonlinearity 
$f\in C^1([0,+\infty ))$ is always extended to $\mathbb{R}$ by 
$f|_{(-\infty ,0)}\equiv f(0)$.

\begin{lemma} \label{lem2.4} 
Let $\lambda,d>0$ and $f\in C^1([0,+\infty ))$ a function which is bounded 
from below.
Then problem \eqref{eq3} has a unique solution
$u(.,d,\lambda)$ defined in $[R,+\infty )$,
In addition, for every $d>0$ there exist $M=M(d)>0$ and 
$\lambda =\lambda(d)>0$ such that
$$
\max_{r\in [R,\widehat{R}]} | u(r,d,\lambda )|
\leq M,\quad \forall \lambda \in (0,\lambda(d) ).
$$
\end{lemma}

\begin{proof} 
The proof of the existence is given in two steps.
In first, we show the existence and uniqueness of a local solution
of \eqref{eq3}; i.e, the existence a $\varepsilon =\varepsilon
(d,\lambda )>0$ such that \eqref{eq3} has a unique solution
on $[R,R+\varepsilon ]$.
In the second step we prove that this unique solution can be
extended to $[R,+\infty )$.
\smallskip

\noindent\textbf{Step 1:} (Local solution).
 Consider the problem
\begin{equation}
\begin{gathered}
-u''(r) -\frac{N-1}ru'(r) =\lambda f(u(r)),\quad r>R_1\\
u(R_1)=a,\quad u'(R_1)=b,
\end{gathered} \label{eq4}
\end{equation}
where $R_1\geq R$.
Let $u$ be a solution of \eqref{eq4}. Multiplying the equation
 by $r^{N-1}$ and using the initial conditions, we obtain
\begin{equation}
u'(r)=\frac 1{r^{N-1}}\big\{ R_1^{N-1}b-\lambda
\int_{R_1}^rs^{N-1}f(u(s))ds\big\}.  \label{eq5}
\end{equation}
from which $u$ satisfies
\begin{equation}
u(r)=a+\frac{bR_1^{N-1}}{N-2}\Big(\frac 1{R_1^{N-2}}-\frac
1{r^{N-2}}\Big) -\lambda \int_{R_1}^r\frac 1{t^{N-1}}\Big[
\int_{R_1}^ts^{N-1}f(u(s))ds\Big] dt.  \label{eq6}
\end{equation}
Conversely, if $u$ is a continuous function satisfying
\eqref{eq6}, then $u$ is a solution of \eqref{eq4}.

Hence, to prove the existence and uniqueness of a solution $u$ of
\eqref{eq4} defined in some interval $[R_1,R_1+\varepsilon] $, it is 
sufficient to show the existence of a unique fixed
point of the operator $T$ defined on $X$ (the Banach space of the
real continuous functions on $[R_1,R_1+\varepsilon ]$
with the uniform norm),
\begin{align*}
T:X=C([R_1,R_1+\varepsilon ] ,\mathbb{R})
&\to X \\
v &\mapsto Tv,
\end{align*}
where
\begin{equation}
(Tv)(r)=a+\frac{bR_1^{N-1}}{N-2}\Big(\frac 1{R_1^{N-2}}-\frac
1{r^{N-2}}\Big) -\lambda \int_{R_1}^r\frac 1{t^{N-1}}\Big[
\int_{R_1}^ts^{N-1}f(v(s))ds\Big]dt,  \label{eq7}
\end{equation}
for all $r\in [R_1,R_1+\varepsilon ]$ and $v\in X$.
 To check this, Let $\delta >0$ such that $\delta >| a| $ and 
$\overline{B} (0,\delta )=\{ u\in X:\|u\| \leq \delta \}$. 
For all $u,v\in\overline{B}(0,\delta )$, we have
$$
(Tu-Tv)(r)=\lambda \int_{R_1}^r\frac 1{t^{N-1}}\Big[
\int_{R_1}^ts^{N-1}\{ f(v(s))-f(u(s))\} ds\Big] dt,
$$
then
\begin{align*}
| (Tu-Tv)(r)| 
&\leq \lambda \int_{R_1}^r\frac 1{t^{N-1}}[\int_{R_1}^ts^{N-1}
 \sup_{\zeta \in (0,\delta]} |f'(\zeta )| \, | v(s)-u(s)| ds] dt \\
&\leq \lambda \int_{R_1}^r\frac 1{t^{N-1}}\Big[
\int_{R_1}^ts^{N-1}ds\Big] dt\sup_{\zeta \in (0,\delta] } 
| f'(\zeta )| \,\|u-v\|.
\end{align*}
However,
\begin{align*}
\int_{R_1}^r\frac 1{t^{N-1}}[\int_{R_1}^ts^{N-1}ds] dt
&= \int_{R_1}^r\frac 1{t^{N-1}}[\frac{t^N}N-\frac{R_1^N}N] dt \\
&\leq  \int_{R_1}^r\frac tNdt-\frac{R_1^N}N\int_{R_1}^r\frac{dt}{t^{N-1}} \\
&= \frac 1{2N}(r^2-R_1^2)_{-}\frac{R_1^N}N\Big(\frac
1{(2-N)r^{N-2}}-\frac 1{(2-N)R_1^{N-2}}\Big) \\
&= \frac{r^2-R_1^2}{2N}+\frac 1{N(N-2)}.\frac{R_1^N}{r^{N-2}}-\frac{R_1^2}{
N(N-2)} \\
&\leq  \frac{(R_1+\varepsilon ) ^2-R_1^2}{2N},\quad
\text{because $r\in [R_1,R_1+\varepsilon ]$} \\
&= \frac{\varepsilon (2R_1+\varepsilon ) }{2N};
\end{align*}
therefore,
\begin{align*}
\|Tu-Tv\| 
&\leq  \frac{\varepsilon ( 2R_1+\varepsilon ) }{2N}
\lambda \sup_{\zeta \in [0,\delta]} |f'(\zeta )| \|u-v\|  \\
&\leq  \frac{\varepsilon (R_1+\varepsilon ) }N\lambda
\sup_{\zeta \in [0,\delta] } | f'(\zeta )| \|u-v\|.
\end{align*}
Hence
\begin{equation}
\|Tu-Tv\|\leq \frac \lambda N\sup_{\zeta \in (0,\delta] } 
| f'(\zeta )| \varepsilon (R_1+\varepsilon ) \|u-v\|. \label{eq8}
\end{equation}
Similarly,
\begin{equation}
\|Tu\| \leq | a| +\frac{| b| R_1^{N-1}}{N-2
}\Big(\frac 1{R_1^{N-2}}-\frac 1{(R_1+\varepsilon )^{N-2}}\Big)
+\frac \lambda N\sup_{\zeta \in [0,\delta] } |f(\zeta )| 
\varepsilon (R_1+\varepsilon ).
\label{eq9}
\end{equation}
Now, by \eqref{eq8} and \eqref{eq9}, we can choose 
$\varepsilon =\varepsilon (\delta )>0$ (depending on $\delta $) sufficiently
small such that $T$ is a contraction from $\overline{B}(0,\delta )$ to 
$\overline{B} (0,\delta )$.
Consequently, $T$ has a fixed point $u$ in $\overline{B}(0,\delta)$.
 The fixed point $u$ is unique in $X\;$for a $\delta $\ as
large as we wanted.
\smallskip

\noindent\textbf{Step 2:} Let $u(.)=u(.,d,\lambda )$ be the unique solution 
of \eqref{eq3} (we take $a=0$, $b=d$ and $R_1=R$ in \eqref{eq4}), and  
denote by $[R,R(d,\lambda ))$ its maximal domain.
We shall prove by contradiction that $R(d,\lambda )=+\infty$. 
For it, assume $R^{*}:=R(d,\lambda )<+\infty $.
$u$ is bounded on $[R,R^{*})$. In fact, using \eqref{eq6} and that $f$
is bounded from below, we have
\begin{align*}
\frac{dR}{N-2} 
&\geq \frac{dR^{N-1}}{N-2}\Big(\frac
1{R^{N-2}}-\frac 1{r^{N-2}}\Big) \\
&= u(r)+\lambda \int_R^r\frac 1{t^{N-1}}\Big[
\int_R^ts^{N-1}\;f(u(s))ds\Big] dt \\
&\geq u(r)+\lambda \inf_{\xi \in [0,+\infty )} 
f(\xi )\int_R^{R^{*}}\frac 1{t^{N-1}}\Big[\int_R^ts^{N-1}ds\Big]
dt,\quad \forall r\in [R,R^{*}),
\end{align*}
then, there exists $K_1>0$ such that $u(r)\leq K_1$ for all
$r\in[R,R^{*})$.

On the other hand, using again \eqref{eq6}, we obtain
\begin{align*}
u(r) &\geq \frac{dR^{N-1}}{N-2}\Big(\frac 1{R^{N-2}}-\frac
1{r^{N-2}}\Big) 
-\lambda \max_{\xi \in [0,K_1 ]} f(\xi )\int_R^{R^{*}}
\frac 1{t^{N-1}}\Big[\int_R^ts^{N-1}ds\Big] dt \\
&\geq -K_2,\quad \forall r\in [R,R^{*}),
\end{align*}
for convenient $K_2>0$. Hence $u$ is bounded.

By using this and \eqref{eq5} and \eqref{eq6}, we deduce that
$\{u(r_n)\}$ and $\{u'(r_n)\}$ are the Cauchy sequence for all sequence 
$(r_n)\subset [R,R^{*})$ converging to $R^{*}$. This is equivalent 
to the existence of the finite limits
\begin{align*}
\lim_{r\to R^{*-}} u(r)=a \quad \text{and} \quad
\lim_{r\to R^{*-}} u'(r)=b.
\end{align*}
Now, consider the  problem
\begin{equation}
 \begin{gathered} 
-v''(r) -\frac{N-1}rv'(r) =\lambda f( v( r)),\quad R^{*}<r\\
v(R^{*})=a,\quad v'(R^{*})=b
\end{gathered} \label{eq10}
\end{equation}
and by step 1, we deduce the existence of a positive number
$\varepsilon >0$ and a solution $v$ of this problem in
 $ [R^{*},R^{*}+\varepsilon ]$.
It is easy to see that
\[
w(r)= \begin{cases}
u(r),&\text{if }R\leq r<R^{*}\\
v(r),&\text{if }R^{*}\leq r\leq R^{*}+\varepsilon,
\end{cases}
\]
is a solution of \eqref{eq3} in $[R,R^{*}+\varepsilon]$ which is a
contradiction, so $R^{*}=+\infty $.

To prove the second part of the lemma, we consider the operator
$T$ defined by \eqref{eq7} on  $X_0=C([R,\widehat{R}],\mathbb{R})$ with
$ R_1=R$, $a=0$ and $b=d$.
Taking $M=\delta >\frac{2dR}{N-2}$ and
\[
\lambda (d)=\min \big\{ \frac M{2M_1\max_{\xi \in [0,M]} | f(\xi )| },
\frac 1{M_1 \max_{\xi \in [0,M]} | f'(\xi )| }\}
\]
with
$M_1=\int_R^{\widetilde{R}}\frac 1{t^{N-1}}\big[\int_R^{t}s^{N-1}ds\big]dt$.

By \eqref{eq8} and \eqref{eq9}, we deduce that $T$ is a contraction from
 $\overline{B}(0,M,X_0)$ into $\overline{B}(0,M,X_0)$, where 
\[
 \overline{B}(0,M,X_0)=\{ u\in X_0:\max_{r\in [R,\widehat{R}]}| u(r)| \leq M\}.
\]
So, the unique fixed point of $T$ belongs to $\overline{B}(0,M,X_0)$.
The lemma is proved.
\end{proof}

\begin{lemma}\label{lem2.5} 
Assume {\rm (F1), (F2)} and let $d_0>0$. Then
there exists
$\lambda _1=\lambda _1(d_0) >0$ such that the unique solution $
u(r,d_0,\lambda ) $ of \eqref{eq3} satisfies
\[
u(r,d_0,\lambda ) >0,\quad \forall r\in (R,\widehat{R}], \forall
\lambda \in (0,\lambda_1).
\]
\end{lemma}

\begin{proof} 
For $\lambda>0$, we consider the set 
\[
\Psi =\{ r\in (R,\widehat{R}) : u(.)=u(.,d_0,\lambda ) \text{is nondecreasing in
}(R,r) \}.
\]
Since $u'(R)=d_0>0$, $\Psi$ is nonempty, and clearly bounded from above.
Let $r_1=\sup \Psi $ (which depends on $\lambda $). We have two cases:

\noindent\textbf{Case 1.} If $r_1=\widehat{R}$, the proof is complete.

\noindent\textbf{Case 2.} If $r_1<\widehat{R}$, we shall prove
 $u(.)=u(.,d_0,\lambda )>0$,
 for all $r\in (R,\widehat{R}]$ for all $\lambda $ sufficiently small.
In order to show it, assume that $r_1<\widehat{R}$. Then $u'(r_1) =0$, and since
\[
u'(r)=\frac 1{r^{N-1}}\Big[R^{N-1}d_0-\lambda\int_R^rs^{N-1}f(u(s))ds\Big],
\]
then $u(r_1)>\beta _1$.
Hence the set  $\Gamma =\{r\in [r_1,\widehat{R}] : u(t)
\geq \beta_1\text{ and }u'(t) \leq 0,\,\forall t\in [r_1,r] \} $
 is nonempty and bounded from above.
Let $r_2=\sup \Gamma >r_1$.
We shall prove that for $\lambda $ sufficiently small
$r_2=\widehat{R}$.
We observe that  $u'(r)\leq0$ for all $r\in \Gamma $, then 
$u(r) \leq u(r_1)$, for all $r\in [R,r_2]$.
Therefore, by the mean value theorem, there exists $c\in (r_1,r_2)$
such that
\[
u( r_2) =u(r_1) +u'(c)(r_2-r_1),
\]
but
\[
u'(c)=-\frac\lambda{c^{N-1}}\int_{r_1}^ct^{N-1}f(u(t))dt,
\]
then
\[
u( r_2) >u(r_1) -\frac{\lambda
\widehat{R}}N\sup_{[\beta_1,u(r_1) ]} | f(\zeta ) |(\widehat{R}-R).
\]
If $M=M(d_0) >0$ and $\lambda (d_0)>0$ (defined in Lemma \ref{lem2.4}), then
$$
\beta_1 <u(r_1) \leq M, \quad  \forall \lambda \in (0,\lambda (d_0)).
$$
Let $K=K(d_0) >0$ such that 
$| f(\zeta ) |<K(\zeta-\beta_1 )$  for all $\zeta \in (\beta_1,M] $.
We deduce that
\[
u(r_2) >u(r_1) -\frac{\lambda K\widehat{R}}N(\widehat{R}-R) (u(r_1)
-\beta_1 ) ,\quad \forall \lambda \in (0,\lambda (d_0)),
\]
Thus, if $\lambda \in (0,\lambda _1)$ with 
$\lambda _1=\min\{\lambda (d_0),\frac N{K\widehat{R}(\widehat{R}
-R) }\}$ we have $u(r_2) >\beta_1$,
which implies that $r_2=\widehat{R}$.
\end{proof}

\begin{lemma}\label{lem2.6} 
Assume {\rm (F1)--(F3)}. Let $\lambda >0$. Then
\begin{itemize}
\item[(i)]  $\lim_{d\to +\infty } r_1(d,\lambda )=R$

\item[(ii)] $\lim_{d\to +\infty } u(r_1,d,\lambda) =+\infty $
\end{itemize}
\end{lemma}

\begin{proof}  If (i) is not true, then there exists $\varepsilon >0$ so that
for all $n$ there exists $d_n$ such that
\[
| r_1(d_n,\lambda ) -R| \geq \varepsilon,
\]
from which
\[
r_1(d_n,\lambda ) \geq R+\varepsilon \quad (\text{because } r_1(d_n,\lambda ) 
\geq R),
\]
then there exists $R_0\in (R,\widehat{R}) $ and a sequence 
$(d_n) \subset (0,+\infty ) $ converging to $\infty$
such that $u_n:=u(.,d_n,\lambda ) $ satisfies
\[
u_n(r) >0,\quad u_n'(r) \geq 0,\quad \forall
r\in (R,R_0] ,\quad \forall n\in \mathbb{N}.
\]
Let $\overline{r}=\frac{R+R_0}2$. By the  equality
\[
u_n(\overline{r})
=\frac{d_nR^{N-1}}{N-2}\Big(\frac 1{R^{N-2}}-\frac 1{^{
\overline{r}^{N-2}}}\Big) -\lambda \int_R^{\overline{r}}\frac
1{t^{N-1}}\Big[\int_R^ts^{N-1}f(u_{n}(s))ds\Big]dt,
\]
we observe that $(u_n(\overline{r}))$ is unbounded.
Passing to a subsequence of $(d_n)$, if it is necessary, we can suppose 
$\lim_{n\to +\infty } u_n(\overline{r}) =+\infty $.
Now, consider
\[
M_n=\inf \big\{ \frac{f(u_n(r) )}{u_n(r) }:r\in (\overline{r},R_0)\big\}.
\]
By (F3), $\lim_{n\to +\infty} M_n=+\infty $.
Let $n_0\in \mathbb{N}$ such that $\lambda M_{n_0}>\mu _3$
 where $\mu _3$ is the third eigenvalue of 
$-[\frac{d^2}{d^2r}+\frac{N-1}r\frac d{dr}] $\ in $(
\overline{r},R_0) $ with Dirichlet boundary conditions.

We take a nonzero eigenfunction $\phi _3$ associated to $\mu _3$; i.e.,
\begin{gather*}
\phi _3''(r) +\frac{N-1}r\phi _3'(r)  +\mu _3\phi _3(r) =0,\quad 
\overline{r}<r<R_0 \\
\phi _3(\overline{r}) =0=\phi _3(R_0). 
\end{gather*}
Since $\phi _3$ has two zeros in $(\overline{r},R_0)$, we deduce from 
the Sturm comparison Theorem \cite{h3} that $u_{n_0}$ has at least one zero 
in $(\overline{r},R_0)$.
Which is a contradiction 
(because $u_n(r)>0$   for all $r\in (R,R_0]$ and all $n\in \mathbb{N}$).

(ii) Let $r_1$ be the same number as in the proof of lemma \ref{lem2.5}.
 we have $u'(r_1)=0$.
However,
\[
u'(r_1) =\frac 1{r_1^{N-1}}\Big[dR^{N-1}-\lambda
\int_R^{r_1}t^{N-1}f(u(t) ) dt\Big],
\]
then
\[
dR^{N-1}=\lambda \int_R^{r_1}t^{N-1}f(u(t) )dt.
\]
Hence
\[
\underset{d\to +\infty }{\lim }u(r_1,d,\lambda )=+\infty.
\]
\end{proof}

\begin{lemma}\label{lem2.7} 
Assume {\rm (F1)--(F4)} and let $\gamma _1$ be
a positive number. Then there exists a $\lambda _2>0$ such that:
\begin{itemize}
\item[(a)] For all $\lambda \in (0,\lambda _2) $ the
unique solution $u(r,d,\lambda )$ of \eqref{eq3}
satisfies
$$
u^2(r,d,\lambda ) +u'^2(r,d,\lambda )>0,\quad \forall r\in [R,\widehat{R}] ,\;
\forall d\geq \gamma _1.
$$

\item[(b)] For all $\lambda \in (0,\lambda _2)$, there exists 
$d>\gamma _1$ such that $u(r,d,\lambda ) <0$ for some
$r\in (R,\widehat{R}]$.
\end{itemize}
\end{lemma}

\begin{proof} 
(a) Let $\lambda, d>0$ and $u(.)=u(.,d,\lambda ) $ the unique solution of
\eqref{eq3}.
We define the auxiliary function $H$ on $[R,+\infty ) $
by setting
\[
H(r) =r\frac{u'^2(r)}2+\lambda rF(u(r) )
+\frac{N-2}2u(r) u'(r) ,\quad \forall r\in [R,+\infty ).
\]
We can prove, as in \cite{c1, h1} the next identity of Pohozaev-type:
\[
r^{N-1}H(r) =t^{N-1}H(t) +\lambda\int_t^rs^{N-1}[NF(u(s) )-\frac{N-2}2f(u(s) ) u(s)
] ds,\quad \forall t\in [R,r].
\]
Taking $t=R$, in this identity we obtain
\[
r^{N-1}H(r)=\frac{R^Nd^2}2+\lambda \int_R^rs^{N-1}\Big[NF(
u(s) )-\frac{N-2}2f(u(s))u(s) \Big] ds,
\]
hence
\begin{equation}
r^{N-1}H(r)\geq \frac{R^Nd^2}2+\lambda m\Big(
\frac{r^N}N-\frac{R^N}N\Big), \label{eq11}
\end{equation}
where $m$ is a strictly negative real such that 
$NF(u) -\frac{N-2}2f(u) u\geq m$ for all $u\in \mathbb{R}$,
so
\[
r^{N-1}H(r)\geq \frac{R^N\gamma _1^2}2+\lambda 
m\Big(\frac{\widehat{R}^N}N-\frac{R^N }N\Big) ,\quad \forall r\in [ R,\widehat{R}],
\;\forall d\geq \gamma _1.
\]
We note that $m$ exists by $(f_4)$.
 Hence there exists $\lambda _2>0$ such that
\begin{equation}
H(r) >0,\quad \forall r\in [R,\widehat{R}] ,\;\forall
d\geq \gamma _1,\;\forall \lambda \in (0,\lambda _2).  \label{eq12}
\end{equation}
Therefore,
\[
u^2(r,d,\lambda ) +u'^2(r,d,\lambda ) >0,\quad \forall r\in [R,\widehat{R}] ,
\quad \forall d\geq \gamma _1,\;\forall \lambda \in (0,\lambda _2).
\]

(b) We argue by contradiction: fix $\lambda \in (0,\lambda _2) $ and suppose that
\[
u(r,d,\lambda ) \geq 0,\quad \forall r\in [R,\widehat{R}] ,\;\forall 
d\geq \gamma _1.
\]
Choose $\varrho >0$ such that there exists a solution of
$\omega''+\frac{N-1}r\omega'+\varrho \omega =0$, where
$$
\omega(0)=1,\quad \omega'(0)=0, \quad
\frac{\widehat{R}-R}{4}\text{ is the first zero of } \omega.
$$
We note (see \cite{g1}) that $\omega(r)\geq0$ and $\omega'(r)<0$ , for all 
$r\in (0,\frac{\widehat{R}-R}4]$.

By (F3), there exists $d_0=d_0(\lambda )>\gamma _1$ such that
\begin{equation}
\frac{f(u)}u\geq \frac \varrho \lambda, \quad 
\forall u\geq d_0.
\label{eq13}
\end{equation}
On the other hand, let $r_1=r_1(d,\lambda ) $ and
$r_2=r_2(d,\lambda ) $ be the same numbers as in the
 proof of Lemma \ref{lem2.5}.
By Lemma \ref{lem2.6}, we can assume that
\[
r_1=r_1(d,\lambda )<R+\frac{\widehat{R}-R}4<\widehat{R}\quad \text{and}\quad
u( r_1,d,\lambda ) >d_0,\quad \forall d\geq d_0,
\]
the definitions of $r_1$ and $r_2$ imply
\begin{equation}
u'(r,d,\lambda )\leq0,\quad \forall r\in [r_1,\widehat{R}] ,\quad
\forall d\geq d_0. \label{eq14}
\end{equation}
Define  $v(r)=u(r_1) \omega (r-r_1)$,
hence $v''(r) +\frac{N-1}{r-r_1}v'(r) +\varrho v(r)=0$,
 for all $r\in (r_1,r_1+\frac{\widehat{R}-R}4)$ with
$u(r_1)=v(r_1)$, $v'(r_1)=0$, $v(r_1+\frac{\widehat{R}-R}4)=0$, 
$v(r)>0$ and $v'(r)\leq 0$, for all $r\in(r_1,r_1+\frac{\widehat{R}-R}4)$,
thus
\[
v''(r)+\frac{N-1}rv'(r)+\varrho v(r)\geq 0,\quad\forall 
r\in (r_1,r_1+\frac{\widehat{R}-R}4),
\]
if $u(r)\geq d_0$, for all $r\in(r_1,r_1+\frac{\widehat{R}-R}4) $,
hence by \eqref{eq13} and the Sturm comparison theorem (see \cite{h3}),
 $u$ have a zero in $(r_1,r_1+\frac{\widehat{R}-R}4) $. Which is a contradiction.
Hence, there exists $r^{*}\in (r_1,r_1+\frac{\widehat{R}-R}4)$ such that 
$u(r^{*},d,\lambda)=d_0$. 

Now, consider the energy function
\[
E(r,d,\lambda ) =\frac{u^{^{\prime }2}(r,d,\lambda) }2+\lambda F(u(r,d,\lambda ) )
,\quad \forall r\geq R.
\]
By \eqref{eq11}, \eqref{eq14} and the equality
$H(r)=rE(r)+\frac{N-2}{2}u(r)u'(r)$, we obtain
\begin{align*}
r^NE(r,d,\lambda ) 
&\geq r^{N-1}H(r,d,\lambda ) \\
&\geq \frac{R^Nd^2}2+\lambda m\Big(\frac{\widehat{R}^N}N-\frac
{R^N}N\Big) ,\quad \forall r\in [r_1,\widehat{R}] ,
\end{align*}
hence, there exists $d_1=d_1(\lambda) \geq d_0$ such that
\[
E(r,d,\lambda) \geq \lambda F(d_0) +\frac
2{(\widehat{R}-R)^2}d_0^2,\quad \forall r\in
[r_1,\widehat{R}],\quad \forall d\geq d_1.
\]
However,
\[
E'(r) =-\frac{N-1}ru'(r) ^2\leq 0,\quad \forall r\in[R,\widehat{R}],
\]
hence
\[
E(r^{*})\geq E(r),\quad \forall r\in[ r^{*},\widehat{R}],
\]
thus
\[
\frac {u'(r)^2}2\geq \frac{2d_0^2}{(\widehat{R}-R)
^2},\quad  \forall r\in[r^{*},\widehat{R}]
,\;\forall d\geq d_1,
\]
and by \eqref{eq14}, we deduce
\[
u'(r)\leq -\frac{2d_0}{\widehat{R}-R}
\;,\quad \forall r\in [
r^{*},\widehat{R}] ,\;\forall d\geq d_1.
\]
The mean value theorem implies  that there exists a 
$c\in( r^{*},r^{*}+\frac{\widehat{R}-R}2)$ such that
\[
u\Big(r^{*}+\frac{\widehat{R}-R}2\Big)-u(r^{*})=\frac{\widehat{R}-R}2u'(c).
\]
Hence
$$
u\Big(r^{*}+\frac{\widehat{R}-R}2\Big)\leq 0.
$$
Which is a contradiction (because
$u'(r^{*}+\frac{\widehat{R}-R}2)<0$).
\end{proof}

\begin{proof}[Proof of theorem \ref{thm2.1}]
Let $d_0>0$. By Lemmas \ref{lem2.5} and \ref{lem2.7}, there exists
$\lambda _{*}>0$ such that, if $\lambda \in (0,\lambda _{*})$ then
\begin{itemize}
\item[(i)] $u(r,d_0,\lambda )>0$ for all $r\in (R,\widehat{R}]$

\item[(ii)] $u'(r,d,\lambda )^2+u(r,d,\lambda)^2>0$ for all
$r\in [R,\widehat{R}]$ and all $d\geq d_0$,

\item[(iii)] there exist $d_1>d_0$ and $r\in (R,\widehat{R}] $ such that 
$u(r,d_1,\lambda )<0$.
\end{itemize}
Define $\Gamma =\{ d\geq d_0\;\mid u(r,\overline{d},\lambda)>0,\;\forall r\in (
R,\widehat{R}),\;\forall \overline{d}\in [d_0,d]\}$.
By (i), $d_0\in \Gamma $ then $\Gamma $ is nonempty. 
In addition, by (iii) $\Gamma $ is bounded from above by $d_1$.
Take $d^{*}=\sup \Gamma $. it is clear that
$$
u(r,d^{*},\lambda )\geq 0,\quad \forall r\in [R,\widehat{R}].
$$
Since $d^{*}<d_1$, we deduce (using (ii)) that
\begin{equation}
u(r,d^{*},\lambda )>0,\quad \forall r\in (R,\widehat{R}). \label{eq15}
\end{equation}
$u(.,d^{*},\lambda )$ will be a solution searching, if we prove 
$u (\widehat{R},d^{*},\lambda )=0$. Assume that 
$u(\widehat{R},d^{*},\lambda)>0$. Then by \eqref{eq15} and the fact that
 $u'(R,d^{*},\lambda)=d^{*}>0$, we have that
$$
u(r,d,\lambda )>0,\quad \forall r\in (R,\widehat{R}],\forall 
d\in [d^{*},d^{*}+\delta ],
$$
where $\delta$ is sufficiently small.
Hence $d^{*}+\delta \in \Gamma $,\;which is a contradiction.
Therefore, $u(\widehat{R},d^{*},\lambda )=0$.
\end{proof}

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\end{document}