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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 111, 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  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/111\hfil A sub-super critical Dirichlet problem]
{Infinitely many radial solutions for a sub-super critical
Dirichlet boundary value problem in a ball}

\author[A. Castro, J. Kwon, C. M. Tan\hfil EJDE-2007/111\hfilneg]
{Alfonso Castro, John Kwon, Chee Meng Tan}  % in alphabetical order

\address{Departament of Mathematics\\
Harvey Mudd College \\
Claremont, CA 91711, USA}
\email[A. Castro]{castro@math.hmc.edu}
\email[J. Kwon]{kwonjy@math.uci.edu}
\email[C. M. Tan]{ctan@hmc.edu}

\thanks{Submitted February 4, 2007. Published August 14, 2007.}
\subjclass[2000]{35J65, 34B16}
\keywords{Sub-super critical;  radial solutions;
 nonlinear elliptic equation; \hfill\break\indent Pohozaev identity}

\begin{abstract}
 We prove the existence of infinitely many solutions to a
 semilinear  Dirichlet boundary value problem in a ball
 for a nonlinearity $g(u)$ that grows  subcritically for $u$
 positive and  supercritically for $u$ negative.
\end{abstract}

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

\begin{section}{Introduction}

In this paper we consider the {\it sub-super critical}
boundary-value problem
\begin{equation}\label{bvp}
 \begin{gathered}
\Delta u+g(u(x))=0, \quad  x \in \mathbb{R}^N, \; \|x\| \leq 1\\
 u(x) =0 \quad \hbox{for }   \|x\| = 1,
 \end{gathered}
\end{equation}
where
\begin{equation}\label{defg}
g(u) =  \begin{cases}
   u^{p},   & u \geq 0\\
   |u|^{q-1}u,   & u < 0,
 \end{cases}
\end{equation}
with
\begin{equation}\label{defpq}
    1 < p < \frac{N+2}{N-2} < q < \infty,
\end{equation}
that is, $g$ has subcritical growth for $u >0$ and supercritical growth for
$u < 0$. Our results hold for more general nonlinearities.
For example, it is easy
to see that (\ref{defg}) may be replaced by
$\lim_{u \to +\infty} g(r,u)/u^p \in (0, \infty)$
and $\lim_{u \to -\infty} g(r,u)/(|u|^{q-1}u) \in (0, \infty)$,
 uniformly for $r \in [0,1]$.

Our main result is as follows.

\begin{theorem} \label{main}
Problem \textup{(\ref{bvp})} has infinitely many radial solutions.
\end{theorem}

This theorem extends the results of \cite{ca-ka1} where it was
established that if
$1 < p < (N+1)/(N-1)$ and $q >1$,  or
$p,q \in (1, (N+2)/(N-2))$, or $p \in (1, (N+2)/(N-2))$ and
$q = (N+2)/(N-2)$,  then
(\ref{bvp}) has infinitely many radial solutions.
This result is optimal in the
sense that if $p,q \in [(N+2)/(N-2), \infty)$ then $u =0$ is the only solution
to (\ref{bvp}) (see \cite{ps}).
For related results for quasilinear equations the reader is referred
to \cite{eha-dtf} and \cite{gm-mr-zf}. Studies on positive solutions
for sub-super critical problems may be found in \cite{el-tm}.
For other studies  on the critical case,  $p = q = (N+2)/(N-2)$
and $\lim_{|u| \to \infty}/(u|u|^{p-1}) \in \mathbb{R}$,
 see  \cite{af-bh-pl,bh-nl,ca-ka1,ca-ka3,cg-ss-sm}.
 In  \cite{br-dj-em} the reader can find a complete classification
of the radial solutions to (\ref{bvp}) for $1< p=q < (N+2)/(N-2)$.
For a recent survey of radial solutions for elliptic
boundary-value problems that includes the case where the Laplacian
operator is replaced by the more general
{\it $k$-Hessian} operator, see \cite{jj-sk}.

Radial solutions to \eqref{bvp} are the solutions to the singular
ordinary differential equation
\begin{equation}\label{odebvp}
 \begin{gathered}
    u''+\frac{n}{t} u'+g(u(t))=0\\
         u'(0)= u(1)  = 0,
 \end{gathered}
\end{equation}
where, and henceforth, $n= N-1$.


For $d >0$ let $u(t,d)$ be the solution to the initial-value problem
\begin{equation}\label{ivp}
 \begin{gathered}
    u''+\frac{n}{t} u'+g(u(t))=0\\
         u(0)= d, \ u'(0)  = 0.
 \end{gathered}
\end{equation}
We define the {\it energy } function
\begin{equation}\label{defE}
E(t,d) \equiv \frac{(u'(t,d))^2}{2} + G(u(t,d)),
\end{equation}
where $G(u) = \int_0^u g(s) ds$. For future reference we note
that
\begin{equation}\label{derE}
\frac{dE}{dt} (t) = -\frac{n}{t}(u'(t))^2 \leq 0.
\end{equation}
The proof of Theorem \ref{main} is based on the properties of the
energy and the  {\it argument} function defined below
(see (\ref{deftheta})).

 \begin{theorem} \label{enertheo}
There exists $D>0$ such that if $d \geq D$, then
 \begin{equation}\label{defxi}
 t^{N-1}\Big(tE(t) + \frac{N-2}{2}u(t) u'(t)\Big) \geq cd^{\xi}
\quad \hbox{for all }   t \geq \sqrt{N}d^{(1-p)/2},
 \end{equation}
 where $\xi = \frac{N+2 -p(N-2)}{2}$. Also $u(t) \geq d/2$ for
$t \in [0,   \sqrt{N}d^{(1-p)/2}]$.
 \end{theorem}

As a consequence of Theorem \ref{enertheo} we see that, for $d \geq D$,
 $\rho(t,d) \equiv u^2(t,d) + (u'(t,d))^2 > 0$ for all $t  \in [0,1]$.
Hence  there exists
 a continuous function $\theta:[0,1] \times [D, \infty) \to \mathbb{R}$ such that
 \begin{equation} \label{deftheta}
 u(t,d) = \rho(t,d) \cos(\theta(t,d))  \quad \hbox{and}  \quad
 u'(t,d) = -\rho(t,d) \sin(\theta(t,d)).
 \end{equation}
In section 7 we prove that
\begin{equation}\label{argtheo}
  \lim_{d \to \infty} \theta(1,d) = +\infty,
  \end{equation}
see (\ref{thetaprove}) below.
\end{section}

\begin{section}{First zero}

Let $d>0$ and $t_0 >0$ be such that $u(t_0) = d/2$, and $u(t) > d/2$ for
$t \in (0, t_0)$. Following the arguments in \cite{ca-ka1}, based on
\begin{equation}\label{u'}
-u'(t) = t^{-n}\int_0^t s^n g(u(s)) ds,
\end{equation}
 it is easily seen that
\begin{equation}\label{estt0}
\sqrt{N} d^{(1-p)/2} \leq t_{0} \leq \sqrt{2^p N} d^{(1-p)/2}.
\end{equation}

Multiplying (\ref{ivp}) by $r^{N-1}u$ and integrating on $[s, t]$, then
multiplying the same equation by $r^N u'$ and integrating also on $[s,t]$ one
has the following identity, known as Pohozaev's identity,
\begin{equation}\label{poho}
    t^{n}H(t)  = s^{n}H(s)  + \int_{s}^{t}r^{n}\Big(NG(u(r))
-\frac{N-2}{2}u(r)g(u(r)) \Big) \,dr,
\end{equation}
where $H(x) \equiv xE(x)\  +
    \frac{N-2}{2}\, \,u'(x)\,u(x)$. In particular, taking $s=0$ and
    $t=t_0$ we have
    \begin{equation}
\begin{aligned}\label{pohot0}
t_0^nH(t_0) & \geq \frac{t_0^N \gamma d^{p+1}}{2^{p+1} N}
&\geq   \frac{N^{(N-2)/2} \gamma }{2^{p+1}} d^{\xi}
& \equiv c_1d^{\xi},
\end{aligned}
\end{equation}
where
$\gamma = N/(p+1) - (N-2)/2$ and $\xi$ is as in  (\ref{defxi}).
Also, from \eqref{poho},  if $u(r) \geq 0$ for all  $r \in [0,t]$ we have
\begin{equation}\label{pohot1}
    t^{N}(u'(t))^{2} = - {(N-2)u}  \cdot t^{n}u' -2 t^{N} \frac{u^{p+1}}{p+1} +
    2 \int_{0}^{t}\gamma r^{n}u^{p+1}\,dr.
\end{equation}
Thus from \eqref{pohot1} and the fact that
$t^{-n} \int_{0}^{t} s^{n} u^{p+1} \,ds \geq -u(t)u'(t)$ we have
\begin{equation}\label{esttu'overu}
\begin{aligned}
    \big(\frac{-tu'}{u}\big)'
 &= \frac{(-tu''-u')u+t(u')^{2}}{u^{2}}\\
 &= \frac{-t(-\frac{n}{t}u'-u^{p})u-uu'+t(u')^{2}}{u^{2}}\\
 &= \frac{(\overbrace{n-1}^{N-2})uu'+tu^{p+1}+t(u')^{2}}{u^{2}}\\
 &= \frac{2t^{-n} \int_{0}^{t} s^{n} \gamma u^{p+1} \,ds -2 \frac{tu^{p+1}}{p+1}
     +tu^{p+1}}{u^{2}}\\
 &= \frac{2t^{-n} \int_{0}^{t} s^{n} \gamma u^{p+1} \,ds
    + t \big(\frac{p-1}{p+1}\big)  u^{p+1}}{u^{2}}\\
& \geq \frac{2\gamma}{t}\big( \frac{-tu'(t)}{u(t)}\big),
\end{aligned}
\end{equation}
provided $u(s) > 0$ for $s \in (0, t)$.
Integrating (\ref{esttu'overu}) on $[t_0, t)$ we have
\begin{equation*}
 \ln\left(\frac{{-tu'(t)}/{u(t)}}{{-t_0u'(t_0)}/{u(t_0)}}\right)
\geq \ln\big(\frac{t}{t_0}\big)^{\gamma}.
\end{equation*}
Letting $\Gamma = -t_0u'(t_0)/u(t_0)$  we conclude that
\begin{equation*}
\frac{-tu'(t)}{u(t)} \geq \Gamma \big(\frac{t}{t_0}\big)^{\gamma}.
\end{equation*}
For future reference we note that
\begin{equation}\label{estGamma}
\Gamma \geq 2^{1-p},
\end{equation}
where we have used (\ref{estt0}), and $-u'(t_0) \geq t_0d^p/(2^pN)$
(see (\ref{u'})).
Integrating again in $[t_0, t]$ yields
\begin{equation*}
\ln\big(\frac{u(t_0)}{u(t)}\big) \geq
\frac{\Gamma}{ \gamma t_0^{\gamma}}[t^{\gamma}-t_0^{\gamma}].
\end{equation*}
Assuming that $u(t) \geq 0$ for all $t \in [t_0, t_0 \ln^{1/\gamma}(d)\equiv T]$
we have
\begin{equation*}
u(T) \leq u(t_0) (ed^{-1})^{\Gamma/\gamma}
 = \frac{e^{\Gamma/\gamma}}{2}d^{1-\Gamma/\gamma}.
\end{equation*}
Now we estimate $E$ for $t \geq t_0$ with $u(s) \geq 0$ for $s \in (t_0, t]$.
Since $E'(t) \geq -2nE(t)/t$,
\begin{equation}\label{endec}
E(t) \geq E(s)(s/t)^{2n} \quad  \hbox{for any } \quad 0 \leq s \leq t \leq 1.
\end{equation}
 Thus
\begin{equation}\label{u'T}
\begin{aligned}
\frac{(u'(T))^2}{2}
& \geq E(t_0)\big(\frac{t_0}{T}\big)^{2n} - \frac{u^{p+1}(T)}{p+1} \\
& \geq \frac{d^{p+1}}{(p+1)2^{p+1}\ln^{2n/\gamma}(d)}
  - \frac{1}{p+1}\Big( \frac{e^{\Gamma/\gamma}}{2}d^{1-\Gamma/\gamma}\Big)^{p+1} \\
& \geq \frac{d^{p+1}}{(p+1)2^{p+2}\ln^{2n/\gamma}(d)},
\end{aligned}
\end{equation}
for $d$ sufficiently large. 

\medskip
Let us suppose now that that $u(t) > 0$,
for any $t\in [T, 2T]$. Arguing as in (\ref{u'T}) we have
\begin{equation}\label{u't}
\begin{aligned}
\frac{(u'(t))^2}{2} & \geq E(T)\left(\frac{T}{t}\right)^{2n} - \frac{u^{p+1}(T)}{p+1} \\
& \geq \frac{d^{p+1}}{(p+1)2^{p+1+2n}\ln^{2n/\gamma}(d)} - \frac{1}{p+1}\left( \frac{e^{\Gamma/\gamma}}{2}d^{1-\Gamma/\gamma}\right)^{p+1} \\
& \geq \frac{d^{p+1}}{(p+1)2^{p+2+2n}\ln^{2n/\gamma}(d)},
\end{aligned}
\end{equation}
for $d$ large. Integrating on $[T,t]$ we have
\begin{equation}
\begin{aligned}
0 \leq u(t) & = u(T) + \int_T^t u'(s) ds \\
& \leq \frac{e^{\Gamma/\gamma}}{2}d^{1-\Gamma/\gamma}
- (t-T)\frac{\sqrt{2} d^{(p+1)/2}}{2^{1 + n + (p/2)}\ln^{n/\gamma}(d)\sqrt{p+1}}.
\end{aligned}
\end{equation}
Hence $u$ has a zero in $[d^{(1-p)/2},T
 +  {e^{\Gamma/\gamma}}d^{(1-p)/2-\Gamma/\gamma}
{2^{n + (p/2)}\ln^{n/\gamma}(d)\sqrt{p+1}}]$.
We summarize the above in the following lemma.

\begin{lemma} \label{lemmat1} For
$d>0$ sufficiently large, there  exists
\begin{equation}\label{estt1}
t_1 \in (\sqrt{N} d^{(1-p)/2},  2\sqrt{N}d^{(1-p)/2}\ln^{1/\gamma}(d))
\end{equation}
such that $u(t_1) =0$, $u(s) >0$ for $s \in [0, t_1)$,
and
\begin{equation}\label{Et1}
\frac{d^{p+1}}{(p+1)2^{p+2+2n}\ln^{2n}(d)}  \leq E(t_1)
\leq \frac{d^{p+1}}{p+1}
\end{equation}
\end{lemma}
\end{section}

\begin{section}{First local minimum}

Let $t \in (t_1, t_1 + (1/2)(2/(q+1))^{q/(q+1)}|u'(t_1)|^{(1-q)/(1+q)}
 \equiv t_1+\tau)$. From (\ref{estt1})  and (\ref{Et1}),
\begin{equation}\label{t1/t}
\begin{aligned}
\frac{t_1}{t} & \geq 1 - \frac{\tau}{t_1 + \tau} \geq 1 - \frac{\tau}{t_1} \\
& \geq 1 - \frac{(1/2)(2/(q+1))^{q/(q+1)}|u'(t_1)|^{(1-q)/(1+q)} }{\sqrt{N} d^{(1-p)/2}} \\
& \geq 1 - \frac{(1/2)(2/(q+1))^{q/(q+1)}(d^{(p+1)/2}/\sqrt{p+1})^{(1-q)/(1+q)} }{\sqrt{N} d^{(1-p)/2}} \\
& \equiv 1 - md^{(p-q)/(1+q)}
 \geq 0.9^{1/n}.
\end{aligned}
\end{equation}
for $d$ large. Hence for $d$ positive and large
\begin{equation}\label{u't>t1}
\begin{aligned}
u'(t)
& = t^{-n} \Big[t_1^nu'(t_1) - \int_{t_1}^t s^n |u(s)|^{q-1}u(s) ds \Big]\\
& \leq 0.9 u'(t_1) +  (t-t_1)\big( \frac{q+1}{2}\big)^{q/(q+1)}
  |u'(t_1)|^{(2q)/(q+1)} \\
&  \leq 0.4 u'(t_1),
\end{aligned}
\end{equation}
where we have used that, since $E' \leq 0$,
$|u(t)|^{q+1}\leq (q+1)(u'(t_1))^2/2$ for
$t \geq t_1$ with $u(t) \leq 0$. This and (\ref{u't>t1}) yield
\begin{equation}\label{utau1}
u(t_1+ \tau )
 \leq 0.4u'(t_1) \tau
\leq -0.2(2/(q+1))^{q/(q+1)} |u'(t_1)|^{2/(1+q)}.
\end{equation}
Now for $t \geq t_1+\tau$ with $u(s)
\leq  -0.2(2/(q+1))^{q/(q+1)}  |u'(t_1)|^{2/(1+q)}$
for all $s \in (t_1+\tau, t)$ we have
\begin{equation}\label{u't>t1a}
\begin{aligned}
&u'(t)\\
& = t^{-n} \Big[t_1^n u'(t_1) - \int_{t_1}^t s^n |u(s)|^{q-1}u(s) ds\Big]\\
& \geq u'(t_1) + t^{-n} ( 0.2(2/(q+1))^{q/(q+1)} )^q |u'(t_1)|^{2q/(1+q)}
  \int_{t_1+\tau }^t s^n   ds \\
&  \geq -u'(t_1)\Big[-1 +  t^{-n} ( 0.2(2/(q+1))^{q/(q+1)} )^q
   |u'(t_1)|^{(q-1)/(1+q)}\frac{t^N - (t_1+\tau)^N}{N}\Big] \\
&  \geq -u'(t_1)\Big[-1 +  \frac{ ( 0.2(2/(q+1))^{q/(q+1)} )^q}N
   |u'(t_1)|^{(q-1)/(1+q)}(t - (t_1+\tau))\Big].
\end{aligned}
\end{equation}
This and the definition of $\tau$ imply the following lemma.

\begin{lemma}\label{lemmatau1} There exists $\tau_1 $  in
\begin{align*} %32
& \Big(t_1, t_1+\big\{(1/2)(2/(q+1))^{q/(q+1)}
   + \frac{N}{( .2(2/(q+1))^{q/(q+1)} )^q}\big\}|u'(t_1)|^{(1-q)/(1+q)}\Big] \\
& \equiv \big(t_1, t_1 + \kappa_1 |u'(t_1)|^{(1-q)/(1+q)}\big]
\end{align*}
such that $u'(\tau_1) = 0$.
\end{lemma}
\end{section}

\begin{section}{Second zero}

Let $\tau_0 > \tau_1$ be such that $u(s) \leq 0.5 u(\tau_1)$ for all
$s \in [\tau_1, \tau_0]$. Imitating the arguments leading to
(\ref{estt0}) we see that
\begin{equation}
\tau_1+ |u(\tau_1)|^{(1-q)/2} \leq \tau_0
\leq \tau_1 + \sqrt{2^qN}|u(\tau_1)|^{(1-q)/2}.
\end{equation}
Hence
\begin{equation}\label{estu'tau0}
\begin{aligned}
u'(\tau_0) & = \tau_0^{-n}\int_{\tau_1}^{\tau_0}s^n  |u(s)|^q ds \\
& \geq \frac{|u(\tau_1)|^q(\tau_0^N - \tau_1^N)}{N2^q\tau_0^n} \\
& \geq \frac{|u(\tau_1)|^q(\tau_0 - \tau_1)}{ 2^qN}
\\
& \geq \frac{|u(\tau_1)|^{(1+q)/2}}{2^qN},
\end{aligned}
\end{equation}
and
\begin{equation}\label{tau0/t}
\tau_0^n \geq .9 s^n \quad \hbox{for any }
s \in  (\tau_0, \tau_0 + 2^{q+2}N|u(\tau_1)|^{(1-q)/2}],
\end{equation}
for $d>0$ sufficiently large.


Suppose now that for all
$s \in [\tau_0, r \equiv  \tau_0 + 2^{q+1}N|u(\tau_1)|^{(1-q)/2}]$
we have
$u(s) \leq 0$. Then
\begin{equation}\label{u's,s>tau0}
u'(s) \geq 0.9 u'(\tau_0) \quad  \hbox{for all }  s \in [\tau_0, r].
\end{equation}
This and and the definition of $r$  give
\begin{align*}
0 \geq u(r)
& \geq \frac{u(\tau_1)}{2} + .9(r-\tau_0)u'(r) \\
&  \geq \frac{u(\tau_1)}{2} + .9(2^{q+1})N|u(\tau_1)|^{(1-q)/2}
   \frac{|u(\tau_1)|^{(1+q)/2}}{2^qN} \\
&  = 1.3|u(\tau_1)|,
\end{align*}
which is a contradiction.
 From \eqref{utau1}, $|u(\tau_0)| \geq 0.2(2/(q+1))^{q/(q+1)}
|u'(t_1)|^{2/(1+q)}$. Since also
$\tau_0 \leq t_1 + (\kappa_1 +0.2\sqrt{2^qN}(2/(q+1))^{q/(q+1)})
|u'(t_1)|^{(1-q)/(1+q)}$ (see Lemma \ref{lemmatau1} and \eqref{estu'tau0}).
Thus
\begin{align*}
&\tau_0+ 2^{q+1}N|u(\tau_1)|^{(1-q)/2} \\
& \leq t_1 + (\kappa_1 +.2(  2^{q+2}N)(2/(q+1))^{q/(q+1)})|u'(t_1)|^{(1-q)/(1+q)} \\
& \equiv t_1 + k_2 |u'(t_1)|^{(1-q)/(1+q)}.
\end{align*}
Thus we have proven the following lemma.

\begin{lemma} \label{lemmat2}
There exists $t_2 \in [t_1, t_1 + k_2|u'(t_1)|^{(1-q)/(1+q)}]$ such that
$u(t_2) = 0$ and $u(s) < 0$ in $(t_1, t_2)$.
\end{lemma}

\end{section}

\begin{section}{First positive maximum}

Let $t >t_2$ be such that $u'(s)>0$ on $[t_2, t]$. Thus $u'' \leq 0$ in
$[t_2,t]$. Hence  $u(s) \leq u'(t_2)(s - t_2)$ for all  $s \in [t_2, t]$. Integrating (\ref{ivp}) on $[t_2, s]$, we have
\begin{equation} \label{u's>t_2}
\begin{aligned}
s^nu'(s)
& =  t_2^{n}u'(t_2) -   \int_{t_2}^s r^n |u(r)|^{p-1}u(r) dr\\
& \geq t_2^n u'(t_2) - s^n \frac{|u'(t_2)|^p(s-t_2)^{p+1}}{p+1} \\
&  \geq u'(t_2)\big( t_2^n   -  \frac{s^n}{p+1}\big),
\end{aligned}
\end{equation}
for $s \leq t_2 + u'(t_2)^{(1-p)/(1+p)}$.
Since $t_2^N |u'(t_2)|^2 \geq 2c_1 d^{\xi}$ (see (\ref{pohot0})) and
$(u'(t_2))^2 \leq 2d^{p+1}/(p+1)$, we have
\begin{equation}\label{estt_2}
t_2^N \geq 2c_1\big(\frac{p+1}2\big)^{\xi/(p+1)}|u'(t_2)|^{N(1-p)/(1+p)}.
 \end{equation}
Now for
$$
s \in [t_2,  \min\big\{2^{1/n},  1 + (2c_1)^{-1/N}
\big(\frac{2}{p+1}\big)^{\frac{\xi}{N(p+1)}}\big\}t_2\equiv \alpha t_2],
$$
 from (\ref{u's>t_2}) and (\ref{estt_2}), we have
\begin{equation}
u'(s) \geq u'(t_2) \big( \frac{t_2^n}{s^n} - \frac{1}{p+1}\big)
\geq u'(t_2)\frac{p-1}{p+1}.
\end{equation}
Integration on $[t_2, \alpha t_2]$ yields
\begin{equation}
u(\alpha t_2) \geq  \frac{p-1}{p+1}\alpha t_2 u'(t_2).
\end{equation}
Therefore, assuming again that $u'>0$ on $[t_2, t]$, we have
 \begin{equation} \label{u's>t2A}
\begin{aligned}
t^nu'(t) & \leq  t_2^{n}u'(t_2) -   \int_{\alpha t_2}^t r^n |u(r)|^{p-1}u(r) dr\\
& \leq t_2^n u'(t_2) - t_2^n (t - \alpha t_2)
 \Big(\frac{p-1}{p+1}\alpha t_2 u'(t_2)\Big)^p.
\end{aligned}
\end{equation}
This and (\ref{estt_2}) imply
\begin{equation}
\begin{aligned}
t - \alpha t_2 & \leq \big(\frac{p-1}{p+1}\alpha  \big)^{-p}t_2^{-p}|u'(t_2)|^{1-p} \\
& \leq \big(\frac{p-1}{p+1}\alpha  \big)^{-p}(2c_1)^{-p/N}
\big(\frac{p+1}{2}\big)^{-p\xi/(N(p+1))}|u'(t_2)|^{(1-p)/(p+1)} \\
&  \equiv \kappa_2|u'(t_2)|^{(1-p)/(p+1)}.
\end{aligned}
\end{equation}
This proves the following result.

\begin{lemma}\label{lemmatau2}
There exists $\tau_2 \in [t_2, \alpha t_2 +\kappa_2|u'(t_2)|^{(1-p)/(p+1)}]$
such that
$u'(\tau_2) = 0$ and $u'(s) > 0$ on $[t_2, \tau_2)$.
\end{lemma}
\end{section}

\begin{section}{Energy on the interval $[t_0, \tau_2]$}

Now we estimate the energy on $[t_0, \tau_2]$.

\begin{lemma}\label{enerlemma}
For $t\in [t_0,  \tau_2]$,
\begin{equation} t^nH(t)  \geq c_1d^{\xi}.
\end{equation}
\end{lemma}

\begin{proof} Let us prove first that
\begin{equation}\label{Et2}
\int_{t_0}^{t_1}t^n \gamma u^{p+1}(t) dt \geq \int_{t_1}^{t_2}
t^n\gamma_1 |u(t)^{q+1}| dt,
\end{equation}
where $\gamma_1 = ((q+1)(N-2)-2N)/(2(q+1))$.
Let $\hat t_0 \in [t_0, t_1]$ be such that $u(\hat t_0) = d/4$.
Then, for $t \in [t_0,\hat t_0]$, we have
\begin{equation}
-u'(t) = t^{-n}\int_{0}^{t}s^n u^p(s)\, ds \leq \frac{td^p}{N}.
\end{equation}
Integrating on $[t_0, \hat t_0]$ we have
$(d/4) \leq (\hat t_0^2 - t_0^2)d^p/(2N)$. This and (\ref{estt0}) yield
\begin{equation}
\hat t_0 \geq \sqrt{\frac{Nd^{1-p}}{2}+t_0^2} = t_0\sqrt{1+\frac{Nd^{1-p}}{2t_0^2}} \geq t_0\sqrt{1 + \frac{1}{2^{p+1}}},
\end{equation}
which combined with  (\ref{estt0}) gives
\begin{equation}\label{Et-hat}
\begin{aligned}
\int_{t_0}^{t_1} t^n \gamma u^{p+1}(t) dt & \geq \int_{t_0}^{\hat t_0}t^n \gamma u^{p+1}(t) dt \\
& \geq \gamma (d/4)^{p+1}  \frac{\hat t_0^N -t_0^N}{N}  \\
& \geq \frac{\gamma}{4^{p+1}N}
t_0^N\Big(\big(1+\frac{1}{2^{p+1}} \big)^{N/2}-1\Big) d^{p+1}\\
& \geq \frac{\gamma}{4^{p+1}N}
\Big(\big(1+\frac{1}{2^{p+1}}\big)^{N/2}-1\Big) N^{N/2} d^{\xi}.
\end{aligned}
\end{equation}
Using (\ref{derE}), we have $|u(t)|^{q+1} \leq (q+1)d^{p+1}/(p+1)$.
Also from (\ref{poho}) and (\ref{pohot0}),
 we have $t_1^N |u'(t_1)|^2/2 = t_1H(t_1) \geq c_1 d^{\xi}$.
This implies that $k_2|u'(t_1)|^{(1-q)/(1+q)}< t_1$ for $d > 0$ large.
These inequalities and  Lemma \ref{lemmat2} imply
\begin{equation}\label{intt1t2}
\begin{aligned}
\int_{t_1}^{t_2} t^n |u(t)|^{q+1} dt
& \leq \big(\frac{q+1}{p+1}d^{p+1} \big) \frac{t_2^N -t_1^N}{N} \\
& \leq \big(\frac{q+1}{p+1}d^{p+1} \big) \frac{(t_1+ k_2|u'(t_1)|^{(1-q)/(1+q)})^N   -t_1^N}{N} \\
& \leq \big(\frac{q+1}{p+1}d^{p+1} \big) t_1^n \frac{(2^N-1) k_2|u'(t_1)|^{(1-q)/(1+q)} }{N} \\
& \leq \big(\frac{q+1}{p+1}d^{p+1} \big)  \frac{(2^N-1) k_2 }{N} \,
t_1^n \left(d^{\xi/2}t_1^{-N/2}\right)^{(1-q)/(1+q)} \\
& = \big(\frac{q+1}{p+1} \big)  \frac{(2^N-1) k_2 }{N} \,
d^{p+1+(\xi(1-q)/(2(1+q))} t_1^{N-1-N(1-q)/(2(1+q))} \\
& \leq \big(\frac{q+1}{p+1} \big)  \frac{(2^N-1) k_2 }{N}
\ln^{M/\gamma}(d) \, d^\eta,
\end{aligned}
\end{equation}
where
\begin{gather*}
\eta  = {p+1 + \frac{\xi(1-q)  +
(1-p)(2(N-1)(1+q) -N(1-q))}{2(1+q)}}, \\
M = (2(N-1)(1+q)-N(1-q))/(2(1+q)).
\end{gather*}
An elementary calculation shows that  $\xi > \eta$.
Thus from (\ref{Et-hat}) and
(\ref{intt1t2}), (\ref{Et2}) follows.
Thus for $t \in [t_1, \tau_2]$,
\begin{equation}\label{pohotau2}
\begin{aligned}
t^nH(t) & = t_0^nH(t_0)  + \int_{t_0}^t s^n\left(NG(u(s))
  -\frac{N-2}2 u(s)g(u(s))\right)ds \\
& \geq t_0^nH(t_0)  + \int_{t_0}^{t_2} s^n\left(NG(u(s))
  -\frac{N-2}2 u(s)g(u(s))\right)ds \\
& \geq t_0^nH(t_0) \\
& \geq c_1d^{\xi},
\end{aligned}
\end{equation}
which proves the lemma.
\end{proof}
\end{section}


\begin{section}{Proof of Theorem \ref{main}}

Arguing as in Lemmas \ref{lemmat1} and \ref{lemmat2}, we see that for $d>0$
sufficiently large there exist
numbers
\begin{equation}\label{deftk}
t_3 < \dots <t_k \leq 1
\end{equation}
such that
\begin{equation}
u(t) <0 \quad  \hbox{in }  (t_{2i-1}, t_{2i}), \quad\hbox{and}\quad
u(t) >0  \quad \hbox{in }  (t_{2i}, t_{2i+1}), \;
 i = 1, \dots \min\{\frac{k}{2}, \frac{k+1}{2}\}.
\end{equation}
Imitating the arguments leading to (\ref{Et2}), one sees that
\begin{equation}\label{Et2i}
\int_{t_{2i}}^{t_{2i+1}}t^n \gamma u^{p+1}(t) dt \geq
\int_{t_{2i+1}}^{t_{2i+2}} t^n\gamma_1 |u(t)^{q+1}| dt.
\end{equation}
This in turn (see (\ref{pohotau2})) leads to
\begin{equation}\label{poho1}
t^nH(t) \geq c_1d^{\xi} \quad \hbox{for all }  t \in [t_0, 1].
\end{equation}
This, together with Lemma \ref{lemmat1},  proves Theorem \ref{enertheo}.
From (\ref{poho1}) we see that
\begin{equation}\label{defro(t)}
\rho^2(t) \equiv u^2(t) + (u'(t))^2 \to \infty  \quad
 \hbox{as } d \to + \infty,
\end{equation}
uniformly for $t \in [0,1]$. Therefore, there exists
a continuous  {\it argument} function $\theta(t,d) \equiv \theta(t)$
such that
\begin{equation}\label{deftheta(t)}
u(t) = \rho(t)\cos(\theta(t)) \quad \hbox{and} \quad
u'(t) =  - \rho(t)\sin(\theta(t)).
\end{equation}
 From this we see that
$\theta'(t) = \{((n/t)u'(t) + g(u(t)))u(t) + (u'(t))^2\}/\rho^2(t)$.
Thus $\theta'(t) > 0$ for $\theta(t) = j \pi/2$  with $j= 1, \dots $,
which implies that if $\theta(t) = j \pi/2$ then
$\theta(s) >  j \pi/2$ for all $s \in (t, 1]$.

Imitating the arguments of Lemmas \ref{lemmat1} and \ref{lemmat2},
we see that
$t_{2i} - t_{2(i-1)} \leq c_3 \ln^{1/\gamma}(d) d^{(1-p)/2}$. Thus
$k \geq c_4 \ln^{-1/\gamma}(d)d^{(p-1)/2}$ (see (\ref{deftk})),
which implies that
\begin{equation}\label{thetaprove}
\lim_{d \to + \infty} \theta(1,d) = + \infty.
\end{equation}
By the continuity of $\theta$ and the intermediate value theorem  we see that there exists a sequence
$d_1 < \dots < d_j  < \dots \to \infty$ such that
$\theta(1,d_j) = j \pi + (\pi/2)$. Hence
$u(t, d_j)$ is a solution to (\ref{bvp}) having exactly $j$ zeroes in $(0,1)$,
which proves Theorem \ref{main}.
\end{section}

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