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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 114, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/114\hfil Superlinear Dirichlet problem]
{Radial solutions with a prescribed number of zeros for a superlinear
Dirichlet problem in annular domain}

\author[B. Azeroual, A. Zertiti \hfil EJDE-2016/114\hfilneg]
{Boubker Azeroual, Abderrahim Zertiti}

\address{Azeroual Boubker \newline
Universit\'e Abdelmalek Essaadi\\
Facult\'e des sciences \\
D\'epartement de Math\'ematiques \\
BP 2121, Tetouan, Morocco}
\email{boubker\_azeroual@yahoo.fr}

\address{Abderrahim Zertiti \newline
Universit\'e Abdelmalek Essaadi\\
Facult\'e des sciences \\
D\'epartement de Math\'ematiques \\
BP 2121, Tetouan, Morocco}
\email{abdzertiti@hotmail.fr}

\thanks{Submitted February 4, 2016. Published May 3, 2016.}
\subjclass[2010]{35J25, 35B05, 35A24}
\keywords{Superlinear; radial solution; Bessel's equation}

\begin{abstract}
 In this article we study the existence of radially symmetric solutions
 to a superlinear Dirichlet problem in annular domain in $\mathbb{R}^{N}$.
 Using fairly straightforward tools of the theory of ordinary differential
 equations, we show that if $k$ is a sufficiently large nonnegative integer,
 there is a solution $u$ which has exactly $(k-1)$ interior zeros.
\end{abstract}

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

\section{Introduction}

The main goal in this article is to study the existence of radially
symmetric solutions $ u:\mathbb{R}^{N} \to \mathbb{R}$
to the superlinear boundary-value problem
 \begin{equation}\label{eq00}
\begin{gathered}
-\Delta u(x) =f(u)+g(| x |)  \quad\text{if } x\in \Omega\\
 u=0 \quad \text{if } x\in \partial\Omega,
\end{gathered}
\end{equation}
where $| x |$ denotes the standard norm of $x$ in $\mathbb{R}^{N}$,
$N\geq 3 $ and  $\Omega $ is the annulus of  $\mathbb{R}^{N}$ defined
by
\[
\Omega=\textbf{C}(0,R,T)=[ x\in \mathbb{R}^{N}: R<| x |<T ]
 \]
where $ R$ and $ T $ are two real numbers such that $ 0<R<T$,
$f: \mathbb{R}\to \mathbb{R}$ is a nonlinear function and
$g\in C^{1}([R,T],\mathbb{R})$.

We will focus on studying the problem  \eqref{eq00} with the following hypotheses:
\begin{itemize}
\item[(H1)] $f$ is locally Lipschitzian,

\item[(H2)] $f$ is superlinear i.e.,
\[
\lim _{| u | \to \infty}\frac{f(u)}{u}=+\infty  ,
\]

\item[(H3)] there exists $m>0$ such that
\[
NF(u)-\frac{N-2}{2} u f(u)-\frac{N+2}{2}\| g \|\,| u |-T\| g'\|\,| u|\geq -m
 \]
where $ F(u)= \int_0^u {f(s)}\mathrm{d}s $ and
$\| g \| = \sup_{R\leq t\leq T}| g(t)| $,

\item[(H4)] $u\to f(u)$ is increasing for $| u|$ large.

\end{itemize}

 From (H2) and L'Hopital's Rule it follows that
\begin{equation} \label{eq01}
\lim _{| u | \to \infty}\frac{F(u)}{u^2}=+\infty.
\end{equation}

In recent decades, the existence of solutions to the superlinear
Dirichlet problem \eqref{eq00} in general domains has been widely studied.
 Most of these results are based on variational methods.
This requires finding a critical point of some energy functional in
Sobolev spaces, by assuming that $f$ is locally Lipschitz and satisfies
a growth condition. A standard way to do this is to apply the
mountain pass theorem. In this context, we mention as examples the authors
 Berestyki,  Bahri and  Struwe.
When the growth of the nonlinearity surpasses the critical exponent of
the Sobolev embedding theorem and the domain is the ball,
 Castro and Kurepa \cite{c1} proved the superlinear Dirichlet problem  \eqref{eq00}
 has infinitely many radially symmetric solutions by offering sufficient
condition and using the ``shooting method'' and ``phase-plane angle analysis''.
However, these arguments are quite difficult and provide no specific
information about the solution. In particular we ask whether radial
solutions exist with prescribed numbers of zeros.  Mcleod, Troy and  Weissler
in \cite{m1} studied this question for the following problem
\begin{gather*}
\Delta u(x) +f(u)=0 \quad \text{if } x\in\mathbb{R}^{N} \\
u(x)\to 0 \quad \text{as } | x | \to +\infty.
\end{gather*}
Thereafter, in the case of the ball,  Iaia and  Pudipeddi \cite{i1}
answered the question above and give an easy proof by using Bessel
functions and proved the problem \eqref{eq00} has infinitely many
radially symmetric solutions with (H1)--(H4) and adding the additional condition
\begin{itemize}
\item[(H5)] There exists a $0< k\leq 1$, such that
\[
\lim _{u \to \infty }\Big( \frac{u}{f(u)}\Big) ^{N/2}
\Big( NF(ku)-\frac{N-2}{2} u f(u)-\frac{N+2}{2}\,\| g \|\,| u |
-T\| g'\|\,| u| \Big) =+\infty .
\]
\end{itemize}
An important contribution was made by  Gidas, Li and Nirenberg \cite{g1}
who showed that if $\Omega $ is a ball, then all positive solutions of
the problem
 \begin{gather*}
\Delta u(x) +f(u)=0 \quad \text{if } x\in \Omega\\
u =0 \quad \text{if } x\in \partial\Omega
\end{gather*}
are radially symmetric. This is not the case in the annulus domain.
The difficulty resides with the fact that a positive radial solution in
annular domain is not monotonic in the radial direction.
Our aim here is to extend the results in  \cite{c1,i1} to the case
in an annular domain, by assuming  (H1)--(H4) without adding
(H5). Our method is based on the same approach used by  Iaia and
Pudipeddi \cite{i1, p1}; by approximating the solution of\eqref{eq00}
with an appropriate linear equation. At last, we note that by (H2)
 the assumption (H3) is more general than (H5)

Our paper is organized as follows:
in Section 2 we begin to establish some preliminary results concerning
the existence of radial solutions and by analysing the energy we show
that the energy function converges uniformly to infinity.
In Section 3 we obtain to localize the zeros of the solution and finally,
we shall prove the following theorem.

\begin{theorem} \label{theo1}
 If {\rm (H1)--(H4)} are satisfied then   \eqref{eq00} has infinitely
many radially symmetric solutions $u$ with $u'(R)\neq 0$.
 For $k\in \mathbb{N^{*}}$ sufficiently large there exist two radially
symmetric solutions $u_{k}$ and $w_{k}$ of problem  \eqref{eq00}
which have exactly $(k-1)$ zeros on $(R,T) $ such that
$ w'_{k}(R)<0< u'_{k}(R)$.
\end{theorem}

\section{Preliminaries}

The existence of radially symmetric solution $ u (x) = u (r) $ with
$r =| x| $  of  \eqref{eq00} is equivalent to the existence of a solution
$u$ of the nonlinear ordinary differential equation
\begin{gather} \label{eq02}
u''(r)+\frac{N-1}{r}u'(r)+f(u)+ g(r)=0 \quad \mbox {if } R<r<T  ,\\
\label{eq03}
u(R)=u(T)=0.
\end{gather}
To solve  \eqref{eq02}-\eqref{eq03}, we apply the shooting method,
 by considering the  initial value problem
\begin{equation} \label{eq04}
\begin{gathered}
u''(r)+\frac{N-1}{r}u'(r)+f(u)+g(r)=0 \quad \text{if } R<r<T, \\
u(R)=0\quad\text{and}\quad u'(R)=d
\end{gathered}
\end{equation}
with $d$ an arbitrary nonzero real number.
Denote $u(r,d)$ as the solution of  \eqref{eq04} which depends on parameter $d$.
By varying $d$, we shall attempt to choose the parameter appropriately
to have  \eqref{eq03} and if $k$ is a sufficiently large nonnegative integer
then $u(r,d)$ has exactly $(k-1)$ zeros on $( R,T)$.

\begin{lemma} \label{lem1}
 Let $d>0$, assume {\rm (H1)} and {\rm (H2)} hold. Then   \eqref{eq04}
has a unique solution $u(r,d)$ defined on interval $[R,T]$.
\end{lemma}

\begin{proof} The proof is divided into two steps.
First we show the existence and uniqueness of the local solution of \eqref{eq04}.
 In the second step we prove that a unique solution can be extended to
a maximal interval $[R,T]$.
\smallskip

\noindent\textbf{Step 1.}
 We consider the initial value problem
\begin{equation}\label{eq05}
\begin{gathered}
u''(r)+\frac{N-1}{r}u'(r)+f(u)+g(r)=0\quad \text{if }\rho<r<T \\
u(\rho)=a,\quad u'(\rho)=b
\end{gathered}
\end{equation}
with $R\leq\rho <T $ and $ (a,b)\in\mathbb{R}^2$. Let $u(r)$ be a solution 
of  \eqref{eq05}.
 Multiplying  \eqref{eq02} by $r^{N-1}$ and by integrating on $(\rho,r)$ 
 with the initial condition gives
\begin{equation}
\label{eq06}
u'(r)=\frac{1}{r^{N-1}}\Big( b\,\rho^{N-1}-\int_\rho^r {t^{N-1} (f(u)+g(t))}
\,\mathrm{d}t\Big),
\end{equation}
Integrating this, we obtain 
\begin{equation}\label{eq07}
u(r)=a+\frac{b\,\rho^{N-1}}{N-2}\Big( \frac{1}{\rho^{N-2}}-\frac{1}{r^{N-2}}\Big)
-\int_\rho^r {\frac{1}{t^{N-1}}}
\Big(\int_\rho^t {s^{N-1} (f(u)+g(s))}\,\mathrm{d}s \Big) \,\mathrm{d}t.
\end{equation}
Conversely, if $u(r)$ is a continuous function and satisfies \eqref{eq07} then 
$u$ is a solution of  \eqref{eq05}. 
Let $\varepsilon>0 $ and $\Psi(u)$ be equal to the right hand side of  
\eqref{eq07} where $X=C( [\rho,\rho+\varepsilon],\mathbb{R}) $ 
the Banach space of real continuous functions on $[\rho,\rho+\varepsilon]$
 with uniform norm. By $(H1)$ we can choose $\varepsilon$ sufficiently 
 small such that $\Psi$ is a contraction mapping. This enables us to conclude 
 that the problem  \eqref{eq04} has a unique solution $u(r,d)$ defined on 
 $[R,R+\varepsilon]$ for $\varepsilon$ sufficiently small (we take $a=0$, $b=d$ and 
 $\rho=R$ in  \eqref{eq05}).
\smallskip

\noindent\textbf{Step 2.}
 Let $u(r,d)=u(r)$ be the unique solution of  \eqref{eq04} and denote by 
 $[R,R_1[ $ its maximal domain. We will show that $R_1=T$.
Otherwise, we suppose that $ R_1<T$. Then we claim that $u$ is bounded
on $[R,R_1[ $. 
We define the energy function of a solution of  \eqref{eq04} as
\begin{equation} \label{eq08}
E(r,d)=E(r)=\frac{u'^2(r)}{2}+F(u(r))\quad \forall r\in [R,R_1).
\end{equation}
Then we see from \eqref{eq01} that $ F(u)>0$ for $u$ large enough so there
 exists a $J>0$ such that 
 \begin{equation} \label{eq09}
F(u)>-J \quad \forall u\in\mathbb{R}.
\end{equation}
It follows from  \eqref{eq01}, \eqref{eq08} and  \eqref{eq09} that
\[
E'(r)= -u' g(r)-\frac{N-1}{r}u'^2\leq \| g\| | u'|\leq \| g\| \sqrt{2(E+J)}.
\]
Dividing by $\sqrt{2(E+J)}$ and integrating this on $(R,r)$ we obtain 
\begin{gather*}
\sqrt{2(E+J)}-\sqrt{2(E(R)+J)}\leq \| g\| (r-R),\\
 | u'|\leq\sqrt{2(E(r)+J)}\leq \| g\|(R_1-R)+\sqrt{d^2+2J} .
\end{gather*}
 It follows that $u'$ is bounded on $[ R,R_1[$. Therefore, by the mean value 
theorem and since $u(R)=0$ we see that $u$ is bounded on $[ R,R_1[$.
By using this, \eqref{eq06} and  \eqref{eq07} (we take $a=0,b=d$ and $\rho=R$) 
we deduce that $(u(r_n))$ and $(u'(r_n))$ are Cauchy sequences for all
sequence $(r_n)$ on $[R,R_1)$ increasing and converging to $R_1$ which implies
the existence of the finite limits 
\[ 
\lim_{r\to R_1^-}u(r)=a,\,\lim_{r\to R_1^-}u'(r)=b .
\]
Now we consider the  initial value problem 
\begin{gather*}
 v''(r)+\frac{N-1}{r}v'(r)+f(v)+ g(r)=0 \quad \text{if } r> R_1 \\ 
v(R_1)=a,\quad v'(R_1)=b. 
\end{gather*}
By step 1, there exists a $\varepsilon >0 $ and a solution $v(r)$ defined 
on $[R_1,R_1+\varepsilon]$. Then it is easy to see that 
\[ 
\widetilde u(r) = \begin{cases} 
u(r) &\text{if }  R<r<R_1 \\ 
v(r) & \text{if } R_1<r<R_1+\varepsilon 
\end{cases}
\] 
is a solution of   \eqref{eq04} on the interval $[R,R_1+\varepsilon] $ 
which contains the maximal domain. This is a contradiction. Hence $R_1=T$. 
\end{proof}

\begin{remark}\label{rem1} \rm
 Using the Arzela-Ascoli theorem the solution $u(r,d)$ of  \eqref{eq04} 
depends continuously on $d$ in the sense that
if the sequence $(d_n)$ converges to $d$, then the sequence of functions
$u(.,d_n)$ converges uniformly to $u(\cdot,d)$ on any bounded interval.
 A similar property is also true for $u'(\cdot,d_n)$.
\end{remark}

\begin{remark}\label{rem2} \rm
We can use the standard ODE existence-uniqueness theorem to obtain a 
local solution of  \eqref{eq04} on $[R,R+\varepsilon]$ for some 
$\varepsilon>0$.
\end{remark}

As $ u'(R,d)=d >0$ and by continuity then, there exists 
$r>R$ such that $u'>0$ on $(R,r)$. 
Denote $r_0(d)$ as the largest $r\in (R,T)$ such that $ u'>0 $ on $(R,r) $.

\begin{lemma} \label{lem2}
Assume {\rm (H1)} and {\rm (H2)} hold. Then 
\begin{enumerate}
\item $\lim_{d\to +\infty}r_0(d)=R$.
\item $\lim_{d\to +\infty}u(r_0(d),d)=+\infty$.
\end{enumerate}
\end{lemma}

\begin{proof}
For (1), we argue by contradiction. Suppose that 
there exists $\varepsilon >0$ such that for all $\gamma>0$ there exists $d>\gamma$
for which 
\[
R+\varepsilon \leq r_0(d).
\]
Denote $R_0=R+\varepsilon$. Then there exists a sequence $d_n\to +\infty$
such that
\begin{equation} \label{eq010}
\begin{gathered}
r_0(d_n)\geq R_0 \\
u(r,d_n)>0,\quad  u'(r,d_n)\geq 0 \quad \forall r\in(R,R_0) ,
\forall n\in\mathbb{N}.
\end{gathered}
\end{equation} 
We set $\overline{r}=(R+R_0)/2$ and $u(\overline{r},d_n)=u_n(\overline{r}) $.
We now show that the sequence $( u_n(\overline{r})) $ is unbounded.
Again by contradiction we suppose that there exists $M>0$ such that for all 
$n\in \mathbb{N}$, $ 0<u_n(\overline{r})\leq M$. 
By  \eqref{eq07} (with $a=0$, $b=d_n $ and $\rho=R $) and $u_n$ is increasing 
on $[R,R_0]$ we obtain 
\begin{align*}
\frac{d_nR^{N-1}}{N-2}\Big( \frac{1}{R^{N-2}}-\frac{1}{\overline{r}^{N-2}}\Big) 
&= u_n(\overline{r})+\int_R^{\overline{r}} {\frac{1}{t^{N-1}}}
\Big( \int_R^t {s^{N-1} (f(u)+g(s))}\,\mathrm{d}s \Big) \,\mathrm{d}t \\ 
& \leq M+ \frac{T^2}{N}\sup_{0\leq \zeta\leq M}
\big(| f(\zeta)| +\| g\|\big)< \infty
\end{align*}
which is a contradiction to $d_n\to +\infty$. Hence, the sequence 
$(u_n(\overline{r})) $ is unbounded and passing to subsequence we can 
suppose that 
\[
\lim_{n \to +\infty} u_n(\overline{r})=+\infty.
\]
Now, for all $n\in \mathbb{N}$, we denote
\[
 M_n=\inf_{\overline{r}\leq r\leq R_0} \big\{ \frac{f(u_n)}{u_n}+\frac{g(r)}{u_n}\big\}.
\]
Since,  $ 0< u_n(\overline{r})\leq u_n(r)$ for all 
$r\in [\overline{r},R_0]$ we see that 
\begin{align*}
M_n \geq \inf_{u_n(\overline{r})\leq u \leq u_n(R_0)}
\{ \frac{f(u)}{u}\}-\frac{\| g\|}{u_n(\overline{r})}.
\end{align*} 
On the other hand, from (H2) and $\lim_{n \to +\infty} u_n(\overline{r})=+\infty $ 
we have $\lim_{n \to +\infty} M_n=+\infty $. 
Thus, there exists $n_0\in\mathbb{N}$ such that $ M_{n_0}> \mu_2$ where
$\mu_2>0$ is the second eigenvalue of
 $-[\frac{d^2}{dr^2}+\frac{N-1}{r}\frac{d}{dr}]$ in
$(\overline{r},R_0)$ with Dirichlet boundary conditions. 
It is known that the first eigenfunction of this operator can be chosen 
to be positive. Then since the second eigenfunction is orthogonal to the 
first eigenfunction then necessarily the second $\Phi_2$ eigenfunction
must be zero somewhere on $(\overline{r},R_0)$. 
Then by Sturm comparison theorem since $\mu_2< M_{n_0}$ it follows that
$u_{n_0}$ has at least one zero in $( \overline{r},R_0)$. 
This is a contradiction with  \eqref{eq010} and finally we deduce that
 $\lim_{d\to +\infty}r_0(d)=R$.

For (2), since $\lim_{d\to +\infty}r_0(d)=R$ then for $d>0$ sufficiently 
large we have $ R<r_0(d)<T $.
On the other hand, $u$ has a local maximum at $r_0(d)$ then, there exists 
$r^{*}\in (r_0(d),T)$ such that $u$ is decreasing and nonnegative on 
$(r_0(d),r^{*})$. Now, we will show that 
\[
\lim_{d\to +\infty}u(r_0(d),d)=+\infty .
\]
Suppose that there exists a sequence $d_n\to +\infty$ such that 
$(u(r_0(d_n),d_n))$ is bounded by $M$. From  \eqref{eq06} we obtain that
for all $n\in\mathbb{N} $ and for all $r\in( r_0(d_n),r^{*})$
\begin{gather*}
r^{N-1}u'(r)= d_n\,R^{N-1}-\int_ R^r {t^{N-1}(f(u)+g(t))}\,\mathrm{d}t\leq0,\\ 
\begin{aligned} 
d_n R^{N-1}&\leq \int_R^r {t^{N-1}( f(u)+g(t))}\,\mathrm{d}t \quad 
(0\leq u \leq M)\\
 &\leq \sup_{0\leq \zeta\leq M}(| f(\zeta)| +\| g\|)\frac{T^{N}}{N} <\infty.
\end{aligned}
\end{gather*} 
It follows that $(d_n)$ is bounded which is a contradiction to $d_n\to +\infty$.
\end{proof}

\begin{lemma} \label{lem3}
Assume {\rm (H1)--(H3)} hold. Then 
\begin{equation*}
\lim_{d\to +\infty}\inf_{r\in[R,T]}E(r,d)=+\infty.
\end{equation*}
\end{lemma}

\begin{proof}
Let $r\in[R,T]$. We consider the Pohozaev-type identity
\begin{align*}
&\Big( r^{N}E+r^{N}\,g(r) u+\frac{N-2}{2}r^{N-1}u u'\Big)'\\
&=r^{N-1}\Big( NF(u)-\frac{N-2}{2}u f(u)+\frac{N+2}{2}g(r) u+rg'(r) u\Big).
\end{align*}
From (H3), we have
 \[
N F(u)-\frac{N-2}{2} u f(u)-\frac{N+2}{2}\| g \|\,| u |-T\| g'\|\,| u|\geq -m .
\]
Integrating Pohozaev's identity on $(R,r)$ with the  initial conditions, gives
\begin{equation} \label{eq011}
 r^{N}E+r^{N}\,g(r) u+\frac{N-2}{2}r^{N-1}u u'
\geq \frac{R^{N}d^2}{2}-\frac{m}{N}(T^{N}-R^{N}).
\end{equation}
Now from  \eqref{eq01} we deduce there exists $B>0$ such that 
for all $| u|>B$,
\[
0<u^2<F(u)<F(u)+J .
\]
If $| u|\leq B$ then from  \eqref{eq09} we see that
\begin{equation}\label{eq012}
u^2\leq F(u)+J+B^2\leq E+J+B^2\quad \forall u\in\mathbb{R}.
\end{equation}
Using Young's inequality we have 
\[
| uu'| \leq \frac{u^2}{2}+\frac{u'^2}{2}\leq F(u)+J+B^2+\frac{u'^2}{2},
\]
We deduce that 
\begin{equation} \label{eq013}
| uu'|\leq E+J+B^2.
\end{equation} 
Hence using  \eqref{eq012} and  \eqref{eq013},
\begin{align*}
&r^{N}E+r^{N}g(r) u+\frac{N-2}{2}r^{N-1}u u'\\
&\leq T^{N}E+T^{N}\| g\|| u|+\frac{N-2}{2}T^{N-1}| u u'| \\ 
&\leq T^{N}E+T^{N}(\| g\|^2+u^2)+\frac{N-2}{2}T^{N-1}(E+J+B^2)\\ 
& \leq T^{N}E+T^{N}\| g\|^2+(T^{N}+\frac{N-2}{2}T^{N-1})(E+J+B^2)\\
& \leq \Big(2T^N+\frac{N-2}{2}T^{N-1}\Big)E+T^{N}\| g\|^2
+(T^N+\frac{N-2}{2}T^{N-1})(J+B^2)\\
& \leq C_1\,E+C_2
\end{align*} 
with $ C_1$ and $C_2$ two positive real numbers depending only on $N,T,J$ and $g$.
From  \eqref{eq011}, then we have 
\[
\inf_{r\in[R,T] }E \geq \frac{R^{N}d^2}{2C_1}-\frac{C_2}{C_1}
-\frac{m}{NC_1}(T^{N}-R^{N}).
\]
 Finally we deduce that $\lim_{d\to +\infty}\inf_{r\in[R,T]}E(r,d)=+\infty$.
\end{proof}

\begin{lemma} \label{lem4}
If $d$ is sufficiently large, then 
\begin{enumerate}
\item all the zeros of $u(r,d)$ are simple on $[R,T]$.
\item $u(r,d)$ has a finite number of zeros on $[R,T]$.
\end{enumerate}
\end{lemma}

\begin{proof}
(1) From Lemma \ref{lem3}, for $d$ sufficiently large and all 
$r\in [R,T]$, we have $E(r,d)>0$. 
If $t_0$ is a zero of $u(r,d)$, then  $E(t_0,d)=\frac{u'^2(t_0,d)}{2}>0$;
 thus $ u'(t_0,d)\neq 0$. Then  $t_0$ is a simple zero of $ u(r,d)$.

For (2), we argue by contradiction. Suppose if $d$ is sufficiently large 
there exists $R<t_1<\dots .<t_n<t_{n+1}\leq T$ and $u(t_n)=0$ for all
$n\in \mathbb{N}$. Using the mean value theorem, there exists 
$ z_n\in(t_n,t_{n+1})$ such that $u'(z_n,d)=0 $ for all $n\in \mathbb{N}$. 
So $(t_n)$ converges to $t\leq T$ and by continuity of $u$ and $u'$ we 
deduce that $u(t,d)=u'(t,d)=0$. This is a contradiction to (1). 
Thus for $d$ sufficiently large $u$ has a finite number of zeros on $[R,T]$.
\end{proof}


\section{Solution with a prescribed number of zeros}


In this section we show the solution $u(r,d)$ has a large number of zeros for 
$d$ sufficiently large. For this we study the behavior  of zeros of $u(r,d)$ 
for $d$ large enough. Also, assuming (H1)--(H4) hold,  it is obvious that 
the first zero of $ u(r,d)$ is $z_0(d)=R$. In the following we focus 
on finding the zeros of $u(r,d)$ on interval $] R,T] $.
From (H2), the mapping $u\mapsto F(u)$ is increasing for large $u$ and 
decreasing when  $u$ is a large negative number. By \eqref{eq01}, we have $F(u)>0$ 
for sufficiently large $| u|$ and from Lemma \ref{lem3} we deduce that for 
$d$ sufficiently large the equation $F(u)=\frac{1}{2}\inf_{r\in[R,T]}E(r,d)$
 has exactly two solutions, which we denote $h_1(d)$ and $h_2(d)$ such that
\begin{gather*}
h_2(d)<0<h_1(d),\\
 F(h_{i}(d))=\frac{1}{2}\inf_{r\in[R,T]}E(r,d)\quad \text{for } i=1,2.
\end{gather*}
From  \eqref{eq01} and Lemma \ref{lem3} we see that 
\begin{equation} \label{eq014}
\lim_{d\to +\infty}h_1(d)=+\infty.
\end{equation} 
Also, $\lim_{d\to +\infty}h_2(d)=-\infty$.

On the other hand by (H2), for $d$ large enough,
$ u''(r_0(d))=-f(u(r_0(d))-g(r_0(d)) <0$. As $u'(r_0(d))=0$ so  
$u$ is decreasing on $(r_0(d),r)$ for $r$ close enough to $r_0(d)$. 
Denote for $d$ sufficiently large 
\[
r^{*}(d)=\sup\big\{ r\in(r_0(d),T) :  u \text{ is decreasing on } (r_0(d),r) \big\} .
\]
There are two cases  $r^{*}(d)=T$ and $r^{*}(d)<T$.

\begin{lemma} \label{lem31}
If {\rm (H1)--(H4)} are satisfied, then for $d$ sufficiently large there 
exist $r_1\in (r_0(d),T)$ such that
$u(r_1)=h_1(d)$ and $h_1(d)<u\leq u(r_0(d))$ on $[r_0(d),r_1)$.
\end{lemma}

\begin{proof}
Suppose by contradiction there exists a sequence $d_n\to\infty$ such 
that  for all $n\in\mathbb{N})$,
\[
u(r,d_n)= u_n(r)> h_1(d_n)\quad \text{on } (r_0(d_n),T).
\]
If $r^{*}(d_n)=T$ then $u_n$ is decreasing on $[r_0(d_n),T]$ for $n$ large enough. 
From  \eqref{eq014}, (H2) and (H4) we obtain for $n$ large enough and for all 
$r\geq r_0(d_n)$
\begin{equation} \label{eq015}
u_n(r)>h_1(d_n)\quad\text{and}\quad f(u_n(r))>f(h_1(d_n))> \| g\|.
\end{equation} 
Let $n$ be large enough and $s\geq r_0(d_n)=r_{0,n}$. From  \eqref{eq06} we have
\begin{align*}
-u'_n(s)=\frac{1}{s^{N-1}} \int_{r_{0,n}}^s {t^{N-1}(f(u_n)+g(t))}\,\mathrm{d}t.
\end{align*}
Integrating on $(r_{0,n}+\frac{r}{2},r_{0,n}+r)$ with $r\in(0,T-r_{0,n})$ 
gives 
\[
u_n(r_{0,n}+\frac{r}{2})= u_n(r_{0,n}+r)
+\int_{r_{0,n}+\frac{r}{2}}^{r_{0,n}+r} {\frac{1}{s^{N-1}}}
\Big( \int_{r_{0,n}}^s {t^{N-1} (f(u_n)+g(t))}\,\mathrm{d}t \Big) \,\mathrm{d}s.
\] 
As $u_n$ is decreasing and by  \eqref{eq015} we have
\[ 
u_n(r_{0,n}+\frac{r}{2})\geq \frac{f(u_n(r_{0,n}+\frac{r}{2}))-\| g\| }{2N T^{N-1}}
\Big([ r_{0,n}+\frac{r}{2}]^{N}-r_{0,n}^{N}\Big)r.
\]
Taking $r=T-r_{0,n}$ by  \eqref{eq014}, \eqref{eq015} and (H2) we see that 
\[ 
\Big([ \frac{r_{0,n}+T}{2}] ^{N}-r_{0,n}^{N}\Big)
\frac{(T-r_{0,n})}{2N T^{N-1}}\leq \frac{u_n\big( \frac{r_{0,n}+T}{2}\big) }
{f\big( u_n\big( \frac{r_{0,n}+T}{2}\big)\big) -\| g\|} \to 0.
\]
Since $r_{0,n}\to R $, it follows that
 \[
\Big([ \frac{R+T}{2}] ^{N}-R^{N}\Big)\frac{(T-R)}{2N T^{N-1}}=0 
\] 
which implies $T=R$ which is impossible. Thus it must be that $r^{*}(d_n)<T$.

For $r^{*}(d_n)= r^{*}<T$, we haven $u'_n(r^{*})=0$ and
 $\int_{r_{0,n}}^{r^{*}} {t^{N-1}(f(u_n)+g(t))}\,\mathrm{d}t=0$.
 However by  \eqref{eq015} we deduce that $f(u_n(t))-g(t)> f(u_n(t))-\| g\| >0 $ 
on $[r_{0,n},r^*]$ and so
 $ \int_{r_{0,n}}^{r^{*}} {t^{N-1}(f(u_n)+g(t))}\,\mathrm{d}t>0$. 
This is impossible. End of the proof.
\end{proof}

Thus, for $d$ sufficiently large we denote by $r_1(d)$ the smallest 
$r\in (r_0(d),T)$ such that
\begin{equation} \label{eq016}
u(r_1(d))=h_1(d),\quad h_1(d)<u\leq u(r_0(d))\quad \text{on }[r_0(d),r_1(d)).
\end{equation}

\begin{lemma} \label{lem32} 
If {\rm (H1)--(H4)} are satisfied, then
\begin{enumerate}
\item $\lim_{d\to+\infty}r_1(d)=R$.
\item For $d$ sufficiently large,  $u(r,d)$ has a first zero $z_1(d)$ in the interval 
$(R,T) $, and  $\lim_{d\to +\infty}z_1(d)=R$.
\end{enumerate}
\end{lemma}

\begin{proof}
For (1), let  
\[
C(d)=\frac{1}{2}\min_{r\in[r_0(d),r_1(d)]}\frac{f(u)}{u}
=\frac{1}{2}\min_{r\in[h_1(d),u(r_0(d))]}\frac{f(s)}{s}.
\]
It follows from  \eqref{eq014} and (H2) that 
\begin{equation} \label{eq017}
\lim_{d\to +\infty}C(d)=+\infty.
\end{equation}
We  now compare the problem 
\begin{equation}\label{eq018}
u''(r)+\frac{N-1}{r}u'(r)+\frac{f(u)}{u}u+ g(r)=0
\end{equation} 
with 
\begin{equation}\label{eq019}
v''(r)+\frac{N-1}{r}v'(r)+C(d)v =0
\end{equation} 
with the initial conditions 
\begin{equation}\label{eq020}
u(r_0(d))=v(r_0(d))\quad\text{and}\quad u'(r_0(d))=v'(r_0(d))=0.
\end{equation} 
Then by  \eqref{eq017} we see that for $d$ sufficiently large 
and all $r\in[r_0(d),r_1(d)]$, we have
\begin{equation}\label{eq021}
 \frac{f(u)}{u}\geq 2C(d)>C(d).
\end{equation} 
\smallskip

\noindent\textbf{Claim:}
for $d$ sufficiently large, $ u<v$ on $(r_0(d),r_1(d)]$. 

Indeed, multiplying  \eqref{eq018} by $r^{N-1}v$ and 
 \eqref{eq019} by $r^{N-1}u$ and subtracting, gives
\[
\big( r^{N-1}(u'v-uv')\big)'+r^{N-1}uv
\Big( \frac{f(u)}{u}+\frac{g(r)}{u}-C(d)\Big) =0.
\]
Integrating this on $(r_0(d),r)$ and using the initial conditions, gives 
\begin{equation}\label{eq022}
r^{N-1}(u'v-uv')=-\int_{r_0(d)}^r {t^{N-1}uv
\Big( \frac{f(u)}{u}+\frac{g(t)}{u}-C(d)\Big) }\,\mathrm{d}t.
\end{equation} 
From  \eqref{eq014}, \eqref{eq017} and  \eqref{eq021} we see that for $d$ 
sufficiently large,
 \begin{equation} \label{eq023}
\frac{f(u)}{u}+\frac{g(r)}{u}-C(d) \geq C(d)-\frac{\| g\|}{h_1(d)}>0.
\end{equation}
For $d$ sufficiently large, let
$\mathscr{F}=\{ r\in (r_0(d),r_1(d)) :u<v \text{ on }(r_0(d),r)\}$. Then
\begin{align*}
 u''(r_0(d))&=-g(r_0(d))-f(u(r_0(d)))\\
& =u(r_0(d))\Big( -\frac{g(r_0(d))}{u(r_0(d))}
 -\frac{f(u(r_0(d)))}{u(r_0(d))}+C(d)\Big) -C(d) u(r_0(d)).
\end{align*}
From (H2) and Lemma \ref{lem2} it follows that for $d$ sufficiently large
\[
u(r_0(d))>0 \quad\text{and}\quad
 -\frac{g(r_0(d))}{u(r_0(d))}-\frac{f(u(r_0(d)))}{u(r_0(d))}+C(d)<0 .
\]
Then, for $d$ sufficiently large we have
\[
u''(r_0(d))<-C(d) u(r_0(d))=v''(r_0(d)).
\]
By continuity there exists $ \varepsilon>0 $ such that $(u-v)''(r)<0 $
on $(r_0(d),r_0(d)+\varepsilon)$. Using the initial conditions  \eqref{eq020}
we deduce that $ u<v $ on $(r_0(d),r_0(d)+\varepsilon)$. Thus
$\mathscr{F}\neq\emptyset $. We denote $\overline{r}=\sup\mathscr{F}$.
Now we will show that $ \overline{r}=r_1(d)$. Otherwise, suppose that
\[
u<v\quad\text{on }(r_0(d),\overline{r}) \;\quad\text{and}\quad
u(\overline{r})=v(\overline{r}).
\]
Since $0<h_1(d)<u<v$ on $(r_0(d),\overline{r})$ and by  \eqref{eq023} we see that,
for $d$ sufficiently large then
\[
 r^{N-1}uv\Big( \frac{f(u)}{u}+\frac{g(r)}{u}-C(d)\Big) >0 .
\]
Therefore, by  \eqref{eq022} $u'(r)v(r)-u(r)v'(r)<0$ on $(r_0(d),\overline{r}]$.
Thus, $ u'(\overline{r})<v'(\overline{r})$. On the other hand, as
$ u(r)<v(r)$ for $ r<\overline{r}$ we have
\[
\frac{u(r)-u(\overline{r})}{r-\overline{r}}
> \frac{v(r)-v(\overline{r})}{r-\overline{r}}.
\]
Hence $u'(\overline{r})\geq v'(\overline{r})$. This is a contradiction.
It follows that $ \overline{r}=r_1(d)$ which completes the proof of the claim.
\smallskip

 Now, we set 
\[
z(r)=\left(r/\sqrt{C(d)}\right)^\frac{N-2}{2} v\left( r/\sqrt{C(d)}\right) .
\]
 It is easy to verify that $z(r)$ is a solution of Bessel's equation of 
order $\nu =\frac{N-2}{2}>0$. i.e., 
\[ 
z''+\frac{z'}{r}+\big( 1-\frac{\nu^2}{r^2}\big) z =0 .
\] 
Then there exists a constant $K>0$ such that every interval of length $K$ 
has at least one zero of $z(r)$ (see \cite{s1}). It follows that every interval 
of length $K/\sqrt{C(d)}$ contains at least one zero of $v(r)$. 
Hence by claim for $d$ sufficiently large we have \begin{equation*}
r_0(d)<r_1(d)<r_0(d)+\frac{K}{\sqrt{C(d)}}.
\end{equation*}
Now (1) of this lemma is a consequence of Lemma \ref{lem2} and  \eqref{eq017}.

For (2), suppose not, which means $u>0$ on $(R,T]$ and consider $r>r_1(d)$. 
Then $0<u<u(r_1(d))$. Also as $F(h_1(d))=\frac{1}{2}\inf_{r\in[R,T]}E(r,d)$ 
for large d, thus  
\[
2F(h_1(d))\leq \frac{u'^2}{2}+F(u)\leq \frac{u'^2}{2}+F(h_1(d)).
\] 
Therefore 
\[
-u'=| u'|\geq \sqrt{2F(h_1(d))} \quad \text{for }r_1(d)\leq r\leq T. 
\] 
Integrating on $(r_1(d),r)$ and by  \eqref{eq016} we obtain 
\[
h_1(d)-u(r)=u(r_1(d))-u(r)\geq \sqrt{2F(h_1(d))}(r-r_1(d)),
\] 
so that 
\[ 
h_1(d)-\sqrt{2F(h_1(d))}(r-r_1(d))\geq u(r)>0,
\]
thus 
\begin{equation}\label{eq024}
 r-r_1(d)\leq\frac{h_1(d)}{\sqrt{2F(h_1(d))}}
 \end{equation} 
for large $d$.

 Taking $r=T$ and taking the limit as $d\to \infty$ in  \eqref{eq024} as 
well as using  \eqref{eq01}, \eqref{eq014} and $r_1(d)\to R $ we see 
that \[0<T-R\leq\frac{h_1(d)}{\sqrt{2F(h_1(d))}} \to 0 \]as $d\to \infty$. 
This is impossible since $T>R$. Thus $u$ has a first zero $z_1(d)$. 
Then using a similar argument on $[r_1(d),z_1(d)]$ and letting $r=r_1(d)$ 
in  \eqref{eq024} we obtain $\lim_{d\to +\infty}z_1(d)=R$. 
The proof  is complete.
\end{proof}

\begin{lemma} \label{lem33} 
Let {\rm (H1)--(H4)} be satisfied. Then for $d$ sufficiently large the 
solution $u(r,d)$ attains a local minimum at $r_3(d)\in (r_2(d) ,T)$
and moreover $\lim_{d\to\infty}r_3(d)=R$.
\end{lemma}

\begin{proof}
We begin to establish the following claim.
\smallskip

\noindent\textbf{Claim:} for $d$ sufficiently large $u(r,d)$ attains the value
 $h_2(d)$ on $(z_1(d),T) $.

Otherwise, there exists a sequence $d_n\to\infty$ such that for all 
$n\in\mathbb{N}$, $ u_n(r)> h_2(d_n)$ on $(z_1(d_n),T)$. 
By Lemma \ref{lem4} we have $u'_n(z_1(d_n))\neq 0$ for $n$ large enough. 
As $u'_n<0$ on $]r_1(d_n),z_1(d_n)[$ therefore $u'_n(z_1(d_n)) <0 $. 
Then by continuity we see that $ u'_n<0 $ on some maximal interval $[z_1(d_n),r^*[$ 
for $n$ large enough, therefore $h_2(d_n)<u_n$. 
Thus $F(u_n)<F(h_2(d_n))$ on $[z_1(d_n),r^*[$. Hence by the definition of
 $h_2(d)$ at the beginning of section 3 we have 
\[
2F(h_2(d_n))\leq E(r,d_n)<\frac{u_n'^2}{2}+F(h_2(d_n)).
\]
Therefore 
\[
0<\sqrt{2F(h_2(d_n))}\leq| u_n'|=-u_n'\quad \forall r\in[z_1(d_n),r^*].
\] 
In particular $u_n'(r^*)<0$. This implies $r^*=T$ for if $r^*<T$ then by 
definition of $r^*$ we wold have $u_n'(r^*)=0$. Now integrating this 
inequality on $(z_1(d_n)),r)$ we obtain, for $n$ large enough 
\begin{equation}\label{eq025}
h_2(d_n)<u_n(r)\leq -\sqrt{2F(h_2(d_n))}(r-z_1(d_n))\quad\forall r\in[z_1(d_n),T].
\end{equation} 
Taking $r=T$ we have 
\[ 
T-z_1(d_n)\leq \frac{-h_2(d_n)}{\sqrt{2F(h_2(d_n))}} .
\] 
Since $\lim_{n\to\infty}h_2(d_n)=-\infty $, by  \eqref{eq01} we deduce that 
$\lim_{n\to\infty}\frac{-h_2(d_n)}{\sqrt{2F(h_2(d_n))}}=0$. As
 $\lim_{n\to\infty}z_1(d_n)=R$ (by Lemma \ref{lem32}) then $T=R$. 
This is a contradiction. End of proof of claim.
\smallskip

We denote by $r_2(d)$ the smallest $r\in( z_1(d),T)$ such that $u(r_2(d))=h_2(d)$ 
and $h_2(d)<u(r,d)$ on $[ z_1(d),r_2(d)[ $. By  \eqref{eq025} taking $r=r_2(d)$ 
we see that 
\begin{equation}\label{eq*}
\lim_{d\to\infty}r_2(d)=R.
\end{equation}
Now, suppose by contradiction that $u$ is decreasing on $(r_2(d) ,T)$. 
Then $u<h_2(d)<0$ on $(r_2(d) ,T)$. 
We set
\[
C(d)=\frac{1}{2}\;\min_{u\leq h_2(d)}\frac{f(u)}{u}.
\]
 By (H2), we see that
\begin{equation}\label{eq026}
\lim_{d\to+\infty}C(d)=+\infty.
 \end{equation}
Now, we compare the problem 
\begin{equation}\label{eq027}
u''(r)+\frac{N-1}{r}u'(r)+\frac{f(u)}{u}u+g(r)=0
\end{equation} 
with  
\begin{equation}\label{eq028}
v''(r)+\frac{N-1}{r}v'(r)+C(d)v = 0
\end{equation} 
and with the initial conditions  
\begin{equation}\label{eq029}
v(r_2(d))=u(r_2(d))=h_2(d)\;\text{and}\:v'(r_2(d))=u'(r_2(d)).
\end{equation}
As in the proof of Lemma \ref{lem32} we see that $ u>v$ on $(r_2(d) ,T)$, 
for $d$ large enough. We saw that 
\[
z(r)=\Big( r/\sqrt{C(d)}\Big) ^\frac{N-2}{2}v\Big( r/\sqrt{C(d)}\Big) 
\]
 is a solution of the Bessel's equation of order $\nu=\frac{N-2}{2}$. 
Then, there exists $K>0$ such every interval of length $K$ has at least one 
zero of $z(r)$. We deduce that for large $d$, $v$ must have a zero on 
$(r_2(d),T)$ and since $u>v$ we see that $u$ gets positive which contradicts 
that $u$ is decreasing on $(r_2(d),T)$. It follows that $u$ has a local
 minimum at $r_3(d)\in (r_2(d),T)$. Also , for $d$ sufficiently large we 
have 
\[
r_2(d)<r_3(d)\leq r_2(d)+\frac{K}{\sqrt{C(d)}}.
\] 
It follows from  \eqref{eq026} and  \eqref{eq*} as $d\to\infty$ that $r_3(d)\to R$.
This completes the proof.
\end{proof}

 As $F(u(r_3(d)))= E(r_3(d))\to\infty$ as $d\to \infty$ 
(by Lemma  \ref{lem3}), in similar way we can show that for $d$ large enough, 
$u(r,d)$ has a second zero $z_2(d)$ with $r_3(d)<z_2(d)<T$ and moreover
 $\lim_{d\to +\infty}z_2(d)=R $. Proceeding in the same way, we can show 
that for $d$ sufficiently large, $u (r,d)$ has a second local maximum at
 $ r_{4}(d)\in(z_2(d),T) $ with $\lim_{d \to+\infty}u(r_{4}(d))=+\infty$ 
and therefore, there exists $z_3(d) $ the third zero of 
$u(r,d)$ on $(R,T)$ with $\lim_{d \to+\infty}z_3(d)=R$.

\begin{remark}\label{rem3} \rm
Continuing in the same way we can obtain as many zeros of $u(r,d)$ as desired 
on $(R,T)$ for $d$ large enough.
\end{remark}


\section{Proof of main result}

For $d>0$, let us denote by $N_{d}\operatorname{card}\{\text{zeros zeros of
$u(r,d)$ on $(R,T)$}\}$.
For $k\geq 1$  defined by set
\[S_{k}=\{ d : N_{d}=k-1 \text{ and }\inf_{r\in[R,T]}E(r,d)>0\}.
\]
By Lemma \ref{lem3} and remark \ref{rem3}, we see that for $d$ sufficiently large,
 $S_{k}$ is not empty for some $k$ and $\inf_{r\in [R,T]}E(r,d)>0$ and we denote
 $ k_0= \min \{ k\in\mathbb{N^*}\:| \;S_{k}\neq\emptyset\} $. It follows 
that $ S_{k_0}$ is not empty and is bounded above. 
Let $d_{k_0}=\sup S_{k_0} $.

\begin{lemma}\label{lem41}
$N_{d_{k_0}}=k_0-1$.
\end{lemma}

\begin{proof} 
By definition of $k_0$ we have $N_{d_{k_0}}\geq k_0-1$. Suppose now that 
$N_{d_{k_0}}\geq k_0$. Then for $d $ close to $d_{k_0} $ and $d\leq d_{k_0}$ 
by remark \ref{rem1} with respect to initial conditions and by Lemma \ref{lem4} 
we see that $N_{d}\geq k_0$. However, if $ d\in S_{k_0}$ and is close 
to $d_{k_0} $ and $d<d_{k_0}$ then $N_{d}=k_0-1 $. This is a contradiction 
to the definition of $d_{k_0}$. Hence $N_{d_{k_0}}=k_0-1$.
\end{proof}

\begin{lemma}\label{lem42}
$u(T,d_{k_0})=0$.
\end{lemma}

\begin{proof} 
We argue by contradiction and assume that $u(T,d_{k_0})\neq 0$, then by remark 
\ref{rem1} with respect to initial conditions and by Lemma \ref{lem4}, 
we deduce that if $d$ is close to $d_{k_0}$ then $N_{d}=N_{d_{k_0}}$
Now, for $d$ close to $d_{k_0}$ and $d>d_{k_0}$ then $d \notin S_{k_0} $ 
therefore, $N_{d}\neq k_0-1$.  This is a contradiction with Lemma \ref{lem41}. 
Hence $u(T,d_{k_0})=0$.
\end{proof}

We denote  $ S_{k_0+1}=\{ d> d_{k_0} : N_{d}=k_0 \text{ and }
\inf_{r\in[R,T]}E(r,d)>0\}$.

\begin{lemma}\label{lem43}
$ S_{k_0+1}\neq \emptyset$.
\end{lemma}

\begin{proof} 
We want to show the following result first.
\smallskip

\noindent\textbf{Claim:}
If $d$ close to $d_{k_0}$ and $ d>d_{k_0}$ then $N_{d}\leq k_0$.

Suppose by contradiction that there exists a sequence $q_n\to d_{k_0}$ such that 
$ N_{q_n}\geq k_0+1$.
For all $ 1\leq i\leq k_0$ let us denote $z_{i}^n$ the $i$th zero of 
$u(r,q_n)$ on $(R,T)$ such that 
\[
 R<z_1^n<z_2^n<\dots <z_{k_0}^n<z_{k_0+1}^n<T.
\] 
For every $1\leq i\leq k_0+1$ the sequence $(z_{i}^n)$ is bounded and converges 
to $z_{i}$ thus, we see that 
\[
R<z_1<z_2<\dots <z_{k_0}<z_{k_0+1}<T .
\] 
It follows that $N_{d_{k_0}}\geq k_0$, which contradicts Lemma \ref{lem41}. 
Thus the claim is proven.
\smallskip

Finally, if $d>d_{k_0}$ then $ N_{d}\leq k_0$ and $N_{d}\neq k_0-1 $ thus, 
$N_{d}= k_0$ and $ S_{k_0+1}\neq \emptyset$ which completes the proof. 
\end{proof}

By remark \ref{rem3} it follows that $ S_{k_0+1}$ is not empty and bounded 
above thus, we denote $d_{k_0+1}=\sup S_{k_0+1}$. We show in a similar way as 
Lemmas \ref{lem41} and  \ref{lem42} that $N_{d_{k_0+1}}=k_0$ and $u(T,d_{k_0+1})=0$.
Proceeding inductively we can show, for all $k \geq k_0 $ there exists a 
solution $ u_{k}(r)=u(r,d_{k}) $ of  \eqref{eq02}-\eqref{eq03} which 
has exactly $ (k-1) $ zeros on $(R,T)$ with $ u'_{k}(R)=d_{k}>0 $.

Now, in the case $ d <0 $ we consider the  problem
\begin{equation} \label{e4.1}
\begin{gathered}
u''(r)+\frac{N-1}{r}u'(r)+f(u)+g(r)= 0 \quad \text{if } R<r<T\\ 
u(R)=0,\quad u'(R)= d <0.
\end{gathered}
\end{equation} 
We denote $ v(r)=-u(r)$ and $g_1(r)=-g(r)$ on $[R,T]$ and $f_1(s)=-f(-s)$ on 
$\mathbb{R}$ then the problem \eqref{e4.1} is equivalent to
\begin{equation} \label{e4.2}
\begin{gathered}
v''(r)+\frac{N-1}{r}v'(r)+f_1(v)+g_1(r)= 0 ,\quad \text{if } R<r<T\\ 
v(R)=0,\quad v'(R)=-d > 0.
\end{gathered}
\end{equation} 
Then $g_1$ is  $ C^{1}([R,T],\mathbb{R})$. It is clear that the assumptions 
(H1), (H2) and (H4) are satisfied. 

It remains to prove (H3).
We set $F_1(v)=\int_0^v {f_1(s)}\mathrm{d}s$. Then 
$F_1(v)=F(-v)$ for all $v\in \mathbb{R}$; thus
\begin{align*}
&NF_1(v)-\frac{N-2}{2}v f_1(v)-\frac{N+2}{2}\| g_1 \|\,| v |-T\| g'_1\|\,| v| \\
& = NF(u)-\frac{N-2}{2}u f(u)-\frac{N+2}{2}\| g \|\,| u |-T\| g'\|\,| u|
 >-m .
\end{align*} 
Next, according to the case  $d>0$ we deduce that, for $k$ sufficiently large,
\eqref{eq02}-\eqref{eq03} has a solution $v_{k}$ which has exactly $(k-1)$ 
zeros on $(R,T)$ with $v'_{k}(R)>0$.
Finally, for $k$ sufficiently large, \eqref{eq02}-\eqref{eq03} has a solution 
$w_{k}=-v_{k}$ which has $(k-1)$ zeros on $(R,T)$ and $w'_{k}(R)<0$. 
End of proof of the main Theorem \ref{theo1}.

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