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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 253, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/253\hfil Sublinear equations on exterior domains]
{Existence of solutions for sublinear equations on exterior domains}

\author[J. A. Iaia \hfil EJDE-2017/253\hfilneg]
{Joseph A. Iaia}

\address{Joseph A. Iaia \newline
Department of Mathematics
University of North Texas
P.O. Box 311430
Denton, TX 76203-1430, USA}
\email{iaia@unt.edu}

\dedicatory{Communicated by Ira Herbst}

\thanks{Submitted February 12, 2017. Published October 10, 2017.}
\subjclass[2010]{34B40, 35B05}
\keywords{Exterior domains; semilinear; sublinear; radial}

\begin{abstract}
 In this article we prove the existence of an infinite number of radial
 solutions of $\Delta u + K(r)f(u)= 0$, one with exactly $n$ zeros for each
 nonnegative integer $n$ on the exterior of the ball of radius $R>0$, $B_{R}$,
 centered at the origin in ${\mathbb R}^{N}$ with $u=0$ on $\partial B_{R}$
 and $\lim_{r \to \infty} u(r)=0$ where $N>2$, $f$ is odd with $f<0$ on
 $(0, \beta) $, $f>0$ on $(\beta, \infty)$, $f(u)\sim u^p$ with $0<p<1$ for
 large $u$ and $K(r) \sim r^{-\alpha}$ with $0 < \alpha < 2$ for large $r$.
\end{abstract}

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

\section{Introduction}

In this article we study radial solutions of
\begin{gather}
\Delta u + K(r)f(u) = 0 \quad \text{in } {\mathbb R}^{N} \backslash B_{R}, \label{1}\\
 u = 0 \quad \text{on } \partial B_{R}, \label{2} \\
 u \to 0 \quad \text{as } |x| \to \infty \label{3}
\end{gather}
where $B_{R}$ is the ball of radius $R>0$ centered at the origin in
${\mathbb R}^{N}$ and $K(r)>0$.
We assume:
\begin{itemize}
\item[(H1)] $ f$ is odd and locally Lipschitz,
$f<0$ on $(0, \beta)$, $f>0$ on $(\beta , \infty)$, $f'(\beta)>0$, and
$f'(0)<0$. %\label{f} \tag{H1}

\item[(H2)] there exists $p$ with $0<p<1$ such that
$f(u) = |u|^{p-1}u + g(u)$ where
$\lim_{u \to \infty} \frac{|g(u)|}{|u|^p} = 0$. %\label{f1} \tag{H2}
\end{itemize}
We let $F(u) = \int_{0}^{u} f(s) \, ds$. Since $f$ is odd it follows that $F$
is even and from (H1) it follows that $F$ is bounded below by $-F_{0}<0$,
$F$ has a unique positive zero, $\gamma$, with $0< \beta < \gamma$, and
\begin{itemize}
\item[(H3)] $-F_{0} <F < 0$ on $(0, \gamma)$, $F>0$ \ on
$(\gamma, \infty)$. %\label{F2} \tag{H3}
\end{itemize}

Interest in the topic for this paper comes from recent papers
\cite{C,CSS,LSS,C2,S} about solutions of
differential equations on exterior domains.
When $f$ grows superlinearly at infinity - i.e.
$\lim_{u \to \infty} \frac{f(u)}{u} = \infty$, $\Omega = {\mathbb R}^{N}$,
and $K(r)\equiv 1$ then the problem \eqref{1}, \eqref{3} has been
extensively studied
\cite{BL,BL2,B,cdgm,cgy,JK,M,serrtang,ST}.
In \cite{I5,I7} equations \eqref{1}-\eqref{3} were studied with
$K(r) \sim r^{-\alpha}$, $f$ superlinear, and
$\Omega = {\mathbb R}^{N} \backslash B_{R}$ with $R>0$ with various values for
$\alpha$. In those papers we proved existence of an infinite number of solutions
- one with exactly $n$ zeros for each nonnegative integer $n$ such that
$u \to 0 \text{ as } |x| \to \infty$ for all $R>0$.
In \cite{I4} we studied \eqref{1}-\eqref{3} with $K(r) \sim r^{-\alpha}$,
$f$ bounded, and $\Omega = {\mathbb R}^{N} \backslash B_{R}$.

In this article we consider the case where $f$ grows sublinearly at infinity
- i.e. $\lim_{u \to \infty} \frac{f(u)}{u^p} = c_{0}>0$ with $0<p<1$ and
$K(r)\sim r^{-\alpha}$ with $0 < \alpha < 2$.
In earlier papers \cite{I8,JJ} the case where $f$ is sublinear and
$ \alpha> N - p(N-2)$ was investigated.

Since we are interested in radial solutions of \eqref{1}-\eqref{3} we
assume that $u(x) = u(|x|) = u(r)$ where $x \in {\mathbb R}^{N} $ and
$r=|x|$=$\sqrt{x_{1}^2 + \cdots + x_{N}^2}$ so that $u$ solves
\begin{gather}
 u''(r) + \frac{N-1}{r} u'(r) + K(r)f(u(r)) = 0 \quad\text{on }
(R, \infty), \text{ where } R > 0, \label{DE} \\
 u(R) = 0, u'(R) = b \in {\mathbb R}. \label{DE2}
\end{gather}

We will also assume that there exist constants $k_{1}>0$, $k_{2}>0$,
and $\alpha$ with $0< \alpha < 2 $ such that
\begin{itemize}
\item[(H4)] $k_{1} r^{-\alpha} \leq K(r) \leq k_{2} r^{-\alpha}$ on
$[R, \infty)$. % \label{K} \tag{H4}

\item[(H5)] $K$ is differentiable, on $[R, \infty)$,
$\lim_{r \to \infty} \frac{rK'}{K} = -\alpha$ where
$ 0< \alpha <2$, and $\frac{rK'}{K} + 2(N-1)>0$ on
$[R, \infty)$. %\label{K2} \tag{H5}
\end{itemize}
Note that (H5) implies $r^{2(N-1)}K(r)$ is increasing.

In this paper we prove the following result.


\begin{theorem} \label{thm1}
 Let $N > 2$, $0<p<1$, and $ 0 < \alpha < 2$.
Assuming {\rm (H1)--(H5)} then given a nonnegative integer $n$ then there
exists a solution of \eqref{DE}-\eqref{DE2} with $n$ zeros on
$(R, \infty)$ and $\lim_{r \to \infty} u(r) = 0$.
\end{theorem}

It is interesting to compare this theorem with the case $\alpha > 2$.
 When $ \alpha> 2$ and $R$ is sufficiently large
then it was shown in \cite{I8,JJ} that there are {\it no} solutions of
\eqref{1}-\eqref{3} with $\lim_{r \to \infty} u(r)=0$.
On the other hand, it was also shown in \cite{I8,JJ} that if $R>0$ is
sufficiently small then solutions
of \eqref{1}-\eqref{3} exist for $\alpha > N - p(N-2)$.
We note in Theorem \ref{thm1} that existence of solutions is established
{\it for all} $R>0$. Also to the best of our knowledge existence of
solutions of \eqref{1}-\eqref{3} is still unknown when
$2< \alpha < N-p(N-2)$, $0<p<1$, and $R>0$ sufficiently small.

\section{Preliminaries}

From the standard existence-uniqueness theorem for ordinary differential
equations \cite{BR} it follows there is a unique solution of
\eqref{DE}-\eqref{DE2} on $[R, R+\epsilon)$ for some $\epsilon>0$. We then define
\begin{equation}
E = \frac{1}{2} \frac{u'^{2}}{K} + F(u). \label{energy}
\end{equation}
Using (H5) we see that
\begin{equation}
E' = -\frac{u'^2}{2rK}\Big(2(N-1) + \frac{rK'}{K}\Big) \leq 0 \quad
\text{for } 0 < \alpha < 2(N-1).
\label{energy2}
 \end{equation}
Thus $E$ is nonincreasing.
Hence it follows that
\begin{equation}
\frac{1}{2} \frac{u'^{2}}{K} + F(u)= E(r) \leq E(R)=\frac{1}{2} \frac{b^2}{K(R)}
\quad \text{for } r\geq R \label{energy4}
\end{equation}
and so we see from (H2)--(H4) that $u$ and $u'$ are uniformly bounded
wherever they are defined from which it follows that the solution of
\eqref{DE}-\eqref{DE2} is defined on $[R, \infty)$.

\begin{lemma} \label{lem1}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. If
$\lim_{r \to \infty} u(r) =L$ then $f(L)=0$.
\end{lemma}


\begin{proof}
 Multiplying \eqref{DE} by $r^{N-1}$ and integrating on $(r_{0}, r)$ where
$r_{0}>R$ gives
$$
r^{N-1}u' = r_{0}^{N-1} u'(r_{0}) - \int_{r_{0}}^{r} t^{N-1}K f(u) \, dt.
$$
Dividing by $r^{N}K$ gives
\begin{equation}
\frac{u'}{rK} = \frac{r_{0}^{N-1} u'(r_{0})}{r^{N}K}
- \frac{\int_{r_{0}}^{r} t^{N-1}K f(u) \, dt}{r^{N} K}.
\label{walrus}
\end{equation}
Using (H4) and that $0<\alpha<2<N$ then
 $r^{N}K\geq k_{1}r^{N-\alpha} \to \infty$ as $r \to \infty$.
Also if $f(L)\neq 0$ and $r_{0}, r$ are sufficiently large then it
 follows from (H4) that
$|\int_{r_{0}}^{r} t^{N-1}K f(u) \, dt|
\geq \frac{|f(L)|k_{1}}{2(N-\alpha)}(r^{N-\alpha} -r_{0}^{N - \alpha}) \to \infty$
as $r \to \infty$ and so by L'H\^opital's rule and \eqref{walrus} we see
\begin{equation}
\lim_{r \to \infty} \frac{u'}{rK} = - \lim_{r \to \infty}
\frac{\int_{r_{0}}^{r} t^{N-1}K f(u) \, dt}{r^{N} K}
= - \lim_{r \to \infty} \frac{f(u)}{N + \frac{r K'}{K}}
 = - \frac{f(L)}{N-\alpha}. \label{blue meany}
 \end{equation}
 Thus by (H4) and \eqref{blue meany} there exists an $r_{0}>R$ such that
 \begin{equation}
|u'| \geq \frac{|f(L)|k_{1}}{2(N-\alpha)}r^{1-\alpha} >0 \quad\text{for }
r > r_{0}. \label{yellow sub}
\end{equation}
 Integrating \eqref{yellow sub} on $(r_{0},r)$ then gives
 \begin{equation}
 |u(r)-u(r_{0})| \geq \frac{|f(L)|k_{1}}{2(N-\alpha)(2 - \alpha)}(r^{2-\alpha}
 - r_{0}^{2- \alpha}). \label{64}
 \end{equation}
 Since $0 < \alpha <2$ we see the right-hand side of \eqref{64} goes to
 $+\infty$ but the left-hand side goes to $|L-u(r_{0})|$ - a contradiction.
 Thus it must be that $f(L)=0$.
\end{proof}

\begin{lemma} \label{lem2}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} with $b>0$ and suppose {\rm (H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Let $0<\epsilon< \beta$.
Then there exists $t_{\epsilon,b}>R$ such that
$u(t_{\epsilon,b}) = \beta - \epsilon $ and $u'>0$ on $[R, t_{\epsilon,b}]$.
\end{lemma}

\begin{proof}
 From \eqref{DE2} and since $b>0$ by assumption we see that $u$ is initially
increasing and positive. Now if $u$ has a first critical point, $M$,
with $u'>0$ on $[R, M)$ then $u'(M)=0$ and $u''(M) \leq 0$ from which it follows
that $f(u(M)) \geq 0$.
In addition, by uniqueness of solutions of initial value problems it follows
that $u''(M)<0$ and so $f(u(M))>0$ and thus $u(M)> \beta$.
Since $u(R)=0$ the lemma then follows in this case by the intermediate value
theorem. Otherwise suppose the lemma does not hold.
Then $u'>0$ and $0< u < \beta - \epsilon$ for all $r> R$ for some $\epsilon>0$ and so
by (H1) there exists a constant
$\epsilon_{1}>0$ and $r_{0}>R$ such that $f(u) \leq - \epsilon_{1}<0$ for $r > r_{0}>R$.
Next multiplying \eqref{DE} by $r^{N-1}$, integrating on $(r_{0},r)$, and using
(H4) gives
\begin{align*}
 -r^{N-1}u'
&= -r_{0}^{N-1}u'(r_{0}) + \int_{r_{0}}^{r} t^{N-1}K f(u) \, dt\\
&\leq -r_{0}^{N-1}u'(r_{0}) - \frac{\epsilon_{1} k_{1}}{N-\alpha} (r^{N-\alpha}
 - r_{0}^{N-\alpha}).
\end{align*}
 Thus for some constant $C_{1}$,
\begin{equation}
u' \geq C_{1} r^{1-N} + \frac{\epsilon_{1} k_{1}}{N-\alpha} r^{1-\alpha}.
\label{hole in pocket}
\end{equation}
Integrating  on $(r_{0}, r)$ gives:
\begin{equation}
 u(r) \geq u(r_{0}) + \frac{C_{1}}{2-N} (r^{2-N}- r_{0}^{2-N})
+ \frac{\epsilon_{1}k_{1}}{(N-\alpha)(2-\alpha)} (r^{2-\alpha}
- r_{0}^{2 - \alpha}). \label{george}
\end{equation}
Now the left-hand side of \eqref{george} is bounded above by $\beta$
but the right-hand side goes to $+\infty$ as $ r \to \infty$ since
$0< \alpha <2<N$ - a contradiction. Hence the lemma holds.
\end{proof}

\begin{lemma} \label{lem3}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Then there exists a $t_{b}>R$
such that $u(t_{b}) = \beta$ and $u'>0$ on $[R,t_{b}]$.
\end{lemma}

\begin{proof}
 We rewrite \eqref{DE} as
$$
u'' + \frac{N-1}{r} u' + K(r)\frac{f(u)}{u-\beta} (u- \beta) = 0
$$
and then make the change of variables
\begin{equation}
u - \beta = r^{\frac{1-N}{2}} v. \label{pennsylvania}
\end{equation}
Thus $v$ satisfies
$$
v'' + \Big(K(r)\frac{f(u)}{u-\beta} - \frac{(N-1)(N-3)}{4r^2}\Big)v = 0.
$$
Suppose now that the lemma does not hold. Then by Lemma \ref{lem2} we see
for some sufficiently small $\epsilon >0$ we have $u'>0$,
$\beta - \epsilon < u < \beta$, and
$\frac{f(u)}{u-\beta} > \frac{f'(\beta)}{2}$ (by (H1)) for $r > t_{\epsilon,b}$.
Also for some $r_{0}>R$ sufficiently large then by (H4) and since $0< \alpha < 2$,
\[
K(r)\frac{f(u)}{u-\beta} - \frac{(N-1)(N-3)}{4r^2}
\geq \frac{k_{1}f'(\beta)}{2r^{\alpha} }- \frac{(N-1)(N-3)}{4r^2}
\geq \frac{1}{r^2} \text{ for } r > r_{0}. %\label{ringo}
 \]
Next we consider a nontrivial solution $w$ of
$$
w'' + \frac{1}{r^2} w = 0 \text{ for } r >r_{0}.
$$
It is straightforward to show
$w = C_{2}e^{r/2}\sin\big(\frac{\sqrt{3}}{2} \ln(r) + C_3\big)$
for constants $C_{2}\neq 0$ and $C_3$.
Hence $w$ has an infinite number of zeros on $(r_{0}, \infty)$.
It follows by the Sturm comparison theorem \cite{BR} that between any two
zeros of $w$ then $v$ must have a zero and from \eqref{pennsylvania} we
see that $u$ must equal $\beta. $ Hence there exists a smallest value
of $r$, denoted $t_{b}$, such that $u(t_{b}) = \beta$ and $0<u<\beta$ on
$(R, t_{b})$.
Thus $u'(t_{b}) \geq 0$ and by uniqueness of solutions of initial value
problems $u'(t_{b})>0$. Also from Lemma \ref{lem2} we have $u'>0$ on $[R, t_{\epsilon,b}]$
for all $\epsilon > 0$ and since $u'(t_{b})>0$ it follows that $u'>0$ on $[R, t_{b}]$.
This completes the proof.
\end{proof}

\begin{lemma} \label{lem4}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Then
$\lim_{b \to 0^{+}} t_{b} = \infty$.
\end{lemma}

\begin{proof}
First we rewrite \eqref{DE} as
\begin{equation} (r^{N-1}u')' = -r^{N-1}K f(u). \label{pierce}
 \end{equation}
From (H1) we have
\begin{equation}
\text{ there exists an $\epsilon_{2}>0$  such that $-f(u) \leq \epsilon_{2}u$  on } [0, \beta/2].
\label{kansas}
\end{equation}
Next we define $t_{b_0}<t_{b}$ to be the smallest value of $t>0$ such that
$u(t_{b_0}) = \frac{\beta}{2}$.
The existence of $t_{b_0}$ follows from Lemma \ref{lem3}, since $u(R)=0$, and the
intermediate value theorem.
Combining \eqref{kansas} with (H4) gives
\begin{equation}
-r^{N-1}K f(u)
\leq \epsilon_{2} k_{2} r^{N-1-\alpha}u \text{ on } [R, t_{b_0}]. \label{mckinley}
\end{equation}
Integrating \eqref{pierce} on $[R, t_{b_0}]$, using \eqref{mckinley} and that $u$
is increasing on $[R, t_{b_0}]$ (by Lemma \ref{lem3}) gives
\begin{equation}
\begin{aligned}
r^{N-1}u' &\leq R^{N-1}b + \int_{R}^{r} \epsilon_{2} k_{2} t^{N-1-\alpha}u(t) \, dt \\
&\leq  R^{N-1}b + \epsilon_{2} k_{2} u(r) \int_{R}^{r} t^{N-1-\alpha} \, dt  \\
&\leq R^{N-1}b + \frac{\epsilon_{2} k_{2} }{N-\alpha}r^{N-\alpha}u.
\end{aligned} \label{buchanan}
 \end{equation}
Rewriting this inequality gives
\begin{equation}
u' - \frac{\epsilon_{2} k_{2}}{N-\alpha} r^{1-\alpha}u
\leq R^{N-1}br^{1-N}. \label{gehrig}
 \end{equation}
Now let $\epsilon_{3} = \frac{\epsilon_2 k_{2}}{(2-\alpha)(N-\alpha)}>0$ and denote
\begin{equation}
 \mu(r) = e^{-\epsilon_{3}(r^{2-\alpha}- R^{2-\alpha})}\leq 1 \quad
\text{for } R\leq r \leq t_{b_0}. \label{maryland}
\end{equation}
Multiplying \eqref{gehrig} by $\mu(r)$, using \eqref{maryland}, and
 integrating on $[R, r]\subset[R, t_{b_0}]$ gives
\begin{equation}
u \leq \frac{R^{N-1}b}{N-2}(R^{2-N} - r^{2-N})e^{\epsilon_{3}(r^{2-\alpha}- R^{2-\alpha})}.
\label{rhizutto}
\end{equation}
Now evaluating \eqref{rhizutto} at $t_{b_0}$ gives
\begin{equation}
\frac{\beta}{2} \leq \frac{R^{N-1}b}{N-2}(R^{2-N} - t_{b_0}^{2-N})
 e^{\epsilon_3(t_{b_0}^{2-\alpha}- R^{2-\alpha})}. \label{dimaggio}
\end{equation}
Since $0 < \alpha < 2$ it follows from \eqref{dimaggio} that
$\lim_{ b \to 0^{+}} t_{b_0} = \infty$ and since $t_{b_0}< t _{b}$ it follows that
$$
\lim_{b \to 0^{+}} t_{b} = \infty.
$$
This completes the proof.
\end{proof}


\begin{lemma} \label{lem5}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Then $u$ has a local maximum,
$M_{b}$, and $u'>0$ on $[R, M_{b})$.
\end{lemma}

\begin{proof}
From Lemma \ref{lem3} we know $u(t_{b})= \beta$ and $u'(t_{b})>0$ so if the lemma
does not hold then it follows from Lemma \ref{lem3} that $u'>0$ for $r\geq R$.
Since $u$ is bounded by \eqref{energy4} then it follows from (H2) and (H3)
 that there exists an $L$ such that $u\to L> \beta$ with $L$ finite.
We see then by Lemma \ref{lem1} that $f(L)=0$ and so (H1) implies $|L| \leq \beta$
contradicting that $L> \beta$. Thus $u$ has a local maximum and so there
 is a smallest value of $t$, denoted $M_{b}$, such that $u'(M_{b})=0$ and
$u'>0$ on $[R, M_{b})$. This completes the proof.
\end{proof}

\begin{lemma} \label{lem6}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose [(H1)--(H5)] hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Then $u(r)> 0$ if $b>0$ is
sufficiently small.
\end{lemma}

\begin{proof}
 We use a similar argument as in \cite{I7}.
First, if $u'>0$ for $r\geq R$ then $u>0$ for all $r >R$ and so we are done
in this case. Thus we suppose that $u$ has a first critical point $M_{b}$.
Then $u'(M_{b})=0$, $u''(M_{b}) \leq 0$, and $u'>0$ on $[R, M_{b})$.
 By uniqueness of solutions of initial value problems it follows that
$u''(M_{b})<0$ and thus $M_{b}$ is a local maximum. Now if
$0< u(M_{b})< \gamma$ then it follows that $E(M_{b})=F(u(M_{b}))<0$ (by (H3)).
Since $E$ is nonincreasing by \eqref{energy2} it follows that $u$ cannot be
zero for $r> M_{b}$ for at such a zero, $z_{b}$, of $u$ we would have
 $0 \leq \frac{1}{2} \frac{u'^{2}(z_{b})}{K(z_{b})} = E(z_{b}) \leq E(M_{b}) < 0$
a contradiction. So we suppose now that $u(M_{b}) \geq \gamma$.
Then there exists $t_{b_1}$ with $t_{b} < t_{b_1} < M_{b}$ so that
$u(t_{b_1}) = \frac{\beta + \gamma}{2}$ and $u'>0$ on $[R, M_{b})$.

Next we have the following identity which follows from \eqref{DE} and \eqref{energy2},
\begin{equation}
(r^{2(N-1)}KE)' = (r^{2(N-1)}K)' F(u). \label{virginia}
\end{equation}
Integrating this on $[R,r]$ gives
\begin{equation}
r^{2(N-1)}KE = \frac{1}{2} R^{2(N-1)}b^2
+ \int_{R}^{r} (t^{2(N-1)}K)' F(u) \, dt. \label{wilson}
\end{equation}
By (H3) we have $F(u) \leq 0$ on $[R, t_{b}]$ and by (H5) we have
$(r^{2(N-1)}K)'>0$ so for $ R < t_{b} < r$ we have
\begin{equation}
\int_{R}^{r} (t^{2(N-1)}K)' F(u) \, dt
\leq \int_{t_{b}}^{r} (t^{2(N-1)}K)' F(u) \, dt. \label{grant}
 \end{equation}
Next on $[\beta, \frac{ \beta+\gamma}{2}]$ it follows that there exists an
$\epsilon_{4}>0$ such that $F(u) \leq -\epsilon_{4} < 0$.
Also from (H5) we see there is a $k_{0}>0$ such that
\begin{equation}
2(N-1) + \frac{rK'}{K} \geq k_{0} \text{ for } r\geq R. \label{fillmore}
 \end{equation}
Then it follows from \eqref{fillmore} and (H4) that
\begin{equation}
(t^{2(N-1)}K)' = t^{2N-3}K[2(N-1) + \frac{rK'}{K}]
\geq k_{0}k_{1} t^{2N-3-\alpha} \text{ for } t\geq R. \label{van buren}
\end{equation}
Thus from \eqref{wilson}-\eqref{van buren} we see
\begin{equation}
t_{b_1}^{2(N-1)}K(t_{b_1}) E(t_{b_1}) \leq \frac{1}{2} R^{2(N-1)}b^2
 - \frac{\epsilon_{4}k_{0}k_{1}}{2N-2-\alpha}[ t_{b_1}^{2N-2-\alpha}
- t_{b}^{2N-2-\alpha} ]. \label{hayes}
\end{equation}
Next solving \eqref{energy4} for $u'$, using (H4), and integrating on
$[t_{b}, t_{b_{1}}]$ where $u'>0$ gives
\begin{equation}
\begin{aligned}
\int_{\beta}^{\frac{\beta + \gamma}{2}} \frac{ dt}{\sqrt{\frac{b^{2}}{K(R)}
 - 2F(t)}}
&=\int_{t_{b}}^{t_{b_1}} \frac{ u'(r) \, dr}{\sqrt{\frac{b^{2}}{K(R)}
  - 2F(u(r))}} \leq \int_{t_{b}}^{t_{b_1}} \sqrt{K} \, dr \\
&= \frac{\sqrt{k_{2}}}{1- \frac{\alpha}{2}}( t_{b_{1}}^{1- \frac{\alpha}{2}}
- t_{b}^{1- \frac{\alpha}{2}})
\end{aligned}\label{maine}
\end{equation}
and so by (H4) we see from \eqref{maine} that for small $b>0$ we have
\begin{equation}
0< \frac{1}{2}\int_{\beta}^{\frac{\beta + \gamma}{2}} \frac{ \, dt}{\sqrt{ - 2F(t)}}
\leq \int_{\beta}^{\frac{\beta + \gamma}{2}}
 \frac{ dt}{\sqrt{\frac{b^{2}}{K(R)} - 2F(t)}}
\leq \frac{\sqrt{k_{2}}}{1- \frac{\alpha}{2}}
\big( t_{b_{1}}^{1- \frac{\alpha}{2}} - t_{b}^{1- \frac{\alpha}{2}}\big).
\label{cleveland}
\end{equation}

It follows then from \eqref{cleveland} and since $0 < \alpha < 2$ that there
exists an $\epsilon_5>0$ such that
\begin{equation}
t_{b_{1}}^{1- \frac{\alpha}{2}} \geq t_{b}^{1- \frac{\alpha}{2}} + \epsilon_{5}.
\label{bruce}
\end{equation}
From the inequality
\begin{equation}
(x+y)^{l} \geq x^l + y^l
\label{clarence} \end{equation}
which holds if $l\geq 1$, $x\geq 0$,  and $y\geq 0 $,
 it follows from \eqref{bruce} and since $\frac{2}{2-\alpha} \geq 1$ that
\begin{equation}
t_{b_{1}} \geq t_{b} + \epsilon_6 \label{vermont}
\end{equation}
where $\epsilon_6 = \epsilon_5^{\frac{2}{2-\alpha}}$.
Next from \eqref{bruce}-\eqref{vermont} we see
\begin{equation}
\begin{aligned}
t_{b_1}^{2N-2-\alpha} - t_{b}^{2N-2-\alpha}
&= [t_{b_1}^{N-1-\frac{\alpha}{2}} - t_{b}^{N-1-\frac{\alpha}{2}}]
  [t_{b_1}^{N-1-\frac{\alpha}{2}} + t_{b}^{N-1-\frac{\alpha}{2}}] \\
&\geq [ (t_{b}+\epsilon_6)^{N-1-\frac{\alpha}{2}}
  - t_{b}^{N-1-\frac{\alpha}{2}}]t_{b}^{N-1-\frac{\alpha}{2}} \\
&\geq \epsilon_7 t_{b}^{N-1-\frac{\alpha}{2}}
\end{aligned}\label{little steve}
\end{equation}
where $\epsilon_7 = \epsilon_6^{N-1-\frac{\alpha}{2}} > 0$ and since
$N-1-\frac{\alpha}{2}\geq 1$ by (H5).

Thus we see it follows from \eqref{hayes}, \eqref{little steve}, 
and Lemma \ref{lem4} that
$$
t_{b_{1}}^{2(N-1)}K(t_{b_1})E(t_{b_1})
\leq \frac{1}{2} R^{2N-2}b^2
 - \frac{\epsilon_{4}\epsilon_{7}k_{0}k_{1}}{2N-2-\alpha}t_{b}^{N-1-\frac{\alpha}{2}}
\to -\infty \text{ as } b\to 0^{+}.
$$
Therefore, $E(t_{b_{1}})<0$ if $b>0$ is sufficiently small.
It then follows that $u(t)> 0$ for $t > t_{b_{1}}$ for if there were a
$z_{b} > t_{b_1}$ such that $u(z_{b})=0$ then since $E$ is nonincreasing
 we would have $0 \leq E(z_{b}) \leq E(t_{b_1})<0$ - a contradiction.
In addition we know from earlier that $u>0$ on $(R, M_{b}]$
and $R< t_{b_1}<M_{b}$. Thus we see $u>0$ on $(R, \infty)$.
This completes the proof.
\end{proof}

\begin{lemma} \label{lem7}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {(H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Then $M_{b} \to \infty$ as $b \to \infty$.
\end{lemma}

\begin{proof}
If the $M_{b}$ are bounded then there exists $M_{0}> R$ such that
$M_{b} \leq M_{0}$ for all large $b$. Now let $v_{b} = \frac{u}{b}$.
Then $v_{b}(R)=0$, $v_{b}'(R) =1$ and $v_{b}$ satisfies
\begin{equation}
v_{b}'' + \frac{N-1}{r} v_{b}' + \frac{K(r) f(bv_{b})}{b} = 0 \quad\text{for }
 r \geq R. \label{ike}
\end{equation}
As in \eqref{energy}-\eqref{energy2},
$$
\Big( \frac{1}{2} \frac{v_{b}'^{2}}{K(r)} + \frac{F(bv_{b})}{b^2} \Big)'
\leq 0 \quad \text{for } r \geq R
$$
and therefore
$$
\frac{1}{2} \frac{v_{b}'^{2}}{K(r)} + \frac{F(bv_{b})}{b^2}
\leq \frac{1}{2K(R)} \quad\text{for } r \geq R.
 $$
It follows from this that the $v_{b}'$ are uniformly bounded on
$[R, \infty)$ and since $|v_{b}(r)|\leq \int_{R}^{r} |v_{b}'(s)| \, ds $
it follows that the $v_{b}$ are uniformly bounded on
$[R, M_{0}+1]$. Since $f$ is sublinear we now show that
$\frac{K(r) f(bv_{b})}{b} \to 0$ on $[R, M_{0}+1]$ as $ b \to \infty$.
To see this note that from (H2) we have $\frac{|g(u)|}{|u|^p} \leq 1$
if $|u|\geq u_{0}>0$ and since $g$ is continuous on $[0,u_{0}]$ then
$|g(u)|\leq C_{4}$ for $|u| \leq u_{0}$ for some constant $C_4$.
 Combining these we see:
\begin{equation}
|g(u)| \leq C_{4} + |u|^{p} \text{ for all } u. \label{georgia}
 \end{equation}

Therefore since the $v_{b}$ are uniformly bounded on $[R, M_{0}+1]$ and
$0<p<1$ then
\begin{equation}
\begin{aligned}
\big|\frac{K(r) f(bv_{b})}{b} \big|
&= K(r)\big|\frac{|v_{b}|^{p-1}v_{b}}{b^{1-p}} + \frac{g(bv_{b})}{b}\big| \\
&\leq K(r) \Big( \frac{|v_{b}|^{p}}{b^{1-p}} + \frac{C_4}{b}
 + \frac{|v_{b}|^p}{b^{1-p}} \Big) \to 0 \quad\text{as } b \to \infty.
\end{aligned} \label{lbj}
\end{equation}
Thus from \eqref{ike}, \eqref{lbj}, and since the $v_{b}'$ are uniformly
bounded it follows that the $v_{b}''$ are also uniformly bounded on $[R, M_{0}+1]$.
Then by the Arzela-Ascoli theorem there exists a subsequence of the $v_{b}$
and $v_{b}'$ (still denoted $v_{b}$ and $v_{b}'$) such that
$v_{b} \to v$ uniformly and $v_{b}' \to v'$ uniformly on $[R, M_{0}+1]$.
In addition, $v'' + \frac{N-1}{r} v' = 0$, $v(R)=0$, and $v'(R) = 1$.
Thus $v = \frac{R}{N-2}(1 - (\frac{R}{r})^{N-2})$. In particular $v'>0$.
On the other hand, $v_{b}'(M_{b}) =0$ and since the $M_{b}$ are bounded
by $M_{0}$ then there is a subsequence (still labeled $M_{b}$)
such that $M_{b} \to M$ and since $v_{b}' \to v'$ uniformly on $[R, M_{0}+1]$
then $0<v'(M)=\lim_{b \to \infty} v_{b}'(M_b)=0$ - a contradiction.
Thus it must be that $M_{b} \to \infty$ as $b \to \infty$.
This completes the proof.
\end{proof}

\begin{lemma} \label{lem8}
- Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)-(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$. Then $u(M_{b}) \to \infty$ as
$b \to \infty$. In addition, there exists a constant $\epsilon_{5}>0$ such that
$$
[u(M_{b})]^{\frac{1-p}{2}} \geq \epsilon_{5} M_{b}^{1 - \frac{\alpha}{2}}.
$$
\end{lemma}

\begin{proof}
It follows from Lemma \ref{lem6} that
$$
\frac{u}{b} \to \frac{R}{N-2}\Big(1 - \Big(\frac{R}{r}\Big)^{N-2}\Big)\quad
\text{uniformly on } [R, 2R].
$$
Hence
$$
\frac{u(2R)}{b} \to \frac{R}{N-2}\left(1 - 2^{2-N} \right) \quad
\text{as } b \to \infty.
$$
Thus
$u(2R) \geq \frac{R}{2(N-2)}\left(1 - 2^{2-N} \right)b$
 for sufficiently large  $b$,
and therefore
$ u(2R) \to \infty$  as $b \to \infty$.
Since $M_{b} \to \infty$ as $b \to \infty$ (by Lemma \ref{lem7}), it follows that
$M_{b} > 2R$ for large $b$, and since $u$ is increasing on $[R, M_{b})$
it follows that $u(M_{b}) \geq u(2R)\to \infty $ from which it follows that
$u(M_{b}) \to \infty$ as $b \to \infty$. This completes the first part of the proof.

Next, from \eqref{energy}-\eqref{energy2} we have
$$
\frac{1}{2} \frac{u'^2}{K} + F(u) \geq F(u(M_{b})) \quad\text{on } [R, M_{b}].
$$
Rewriting this, integrating on $[R, M_{b}]$ and using (H4) gives
\begin{equation}
\begin{aligned}
\int_{R}^{M_{b}} \frac{u'}{\sqrt{2}\sqrt{F(u(M_{b})) - F(u(t))}}
& \geq \int_{R}^{M_{b}} \sqrt{K} \, dr \\
&\geq \int_{R}^{M_{b}} \sqrt{k_{1}} r^{-\frac{\alpha}{2}} \, dr \\
&= \frac{\sqrt{k_{1}}(M_{b}^{1 - \frac{\alpha}{2}}
 - R^{1 - \frac{\alpha}{2}})}{1 - \frac{\alpha}{2}}.
\end{aligned} \label{taft}
\end{equation}
Changing variables on the left-hand side, rewriting, and changing
variables again gives
\begin{equation}
\begin{aligned}
\int_{R}^{M_{b}} \frac{u'}{\sqrt{2}\sqrt{F(u(M_{b})) - F(u(t))}}
&= \int_{0}^{u(M_{b})} \frac{dt}{\sqrt{2}\sqrt{F(u(M_{b})) - F(t)}} \\
&= \frac{u(M_{b})}{\sqrt{2}\sqrt{F(u(M_{b}))}} \int_{0}^{1}
\frac{ds}{\sqrt{1 - \frac{F(u(M_{b})s)}{F(u(M_{b}))}}}.
\end{aligned}\label{coolidge}
\end{equation}

From the first part of the theorem we know that $u(M_{b})\to \infty$ as
 $b \to \infty$.
Then from (H2) it follows that $F(u) = \frac{u^{p+1}}{p+1} + G(u)$ where
$G(u) = \int_{0}^{u} g(s) \, ds$. In a similar way to \eqref{georgia}
it follows that
\begin{equation}
|G(u)| \leq C_5 +\frac{1}{2(p+1)} |u|^{p+1} \quad
\text{for all $u$  for some constant } C_5. \label{iowa}
\end{equation}
This along with (H2) and that $0<p<1$ gives
\begin{equation}
\lim_{b \to \infty} \int_{0}^{1} \frac{ds}
{\sqrt{1 - \frac{F(u(M_{b})s)}{F(u(M_{b}))}}}
=\int_{0}^{1} \frac{ds}{\sqrt{1-s^{p+1}}}
< \int_{0}^{1} \frac{ds}{\sqrt{1-s}} = 2. \label{TR}
\end{equation}
In addition we see that
\begin{equation}
\sqrt{F(u(M_{b}))}
= [u(M_{b})]^{\frac{p+1}{2}}\sqrt{ \frac{1}{p+1}
 + \frac{G(u(M_{b}))}{[u(M_{b})]^{p+1}}}. \label{new hampshire}
 \end{equation}
Since $u(M_{b})\to \infty$ as $b \to \infty$ and $\frac{G(u)}{|u|^{p+1}} \to 0$
as $u \to \infty$ it follows from \eqref{new hampshire} that
\begin{equation}
\lim_{b \to \infty} \frac{ [u(M_{b})]^{\frac{p+1}{2}}}{\sqrt{F(u(M_{b}))}}
 = \sqrt{p+1}. \label{nils}
 \end{equation}
Therefore from \eqref{nils} for large $b$ we have
\begin{equation}
\frac{u(M_{b})}{\sqrt{F(u(M_{b}))}}
\leq 2\sqrt{p+1} \ [u(M_{b})]^{\frac{1-p}{2}}. \label{abe}
\end{equation}
Combining \eqref{taft}-\eqref{abe} we obtain for large $b$,
\begin{equation}
[u(M_{b})]^{\frac{1-p}{2}}
\geq \frac{\sqrt{k_1}}{2(2-\alpha){\sqrt{p+1}}}
 \left( M_{b}^{1 - \frac{\alpha}{2}} - R^{1 - \frac{\alpha}{2}} \right).
 \label{bragg}
\end{equation}
Finally since $M_{b} \to \infty$ as $b \to \infty$ (by Lemma \ref{lem7}) we obtain
\begin{equation}
[u(M_{b})]^{\frac{1-p}{2}} \geq
\epsilon_{5} M_{b}^{1 - \frac{\alpha}{2}} \label{bbragg}
\end{equation}
with $\epsilon_{5} = \frac{\sqrt{k_1}}{4(2-\alpha){\sqrt{p+1}}}>0$.
This completes the proof.
\end{proof}

\begin{lemma} \label{lem9}
Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {(H1)--(H5)} hold.
Let $N>2$, $0<p<1$, and $0 < \alpha < 2$.
Then for sufficiently large $b$ there exists a $z_{b}>M_{b}$ such that
$u'<0$ on $(M_{b}, z_{b}]$ and $u(z_{b})=0$.
In addition, given a positive integer $n$ then if $b$ is sufficiently large then
 $u$ has $n$ zeros on $(R, \infty)$.
\end{lemma}

\begin{proof}
 First let $v(r) = u(r+M_{b})$. Then $v(0)= u(M_{b})$, $v'(0)=u'(M_{b})=0$,
and \eqref{DE} becomes
\begin{equation}
 v'' + \frac{N-1}{r + M_{b}}v' + K(r+M_{b})\left( |v|^{p-1}v + g(v) \right) = 0.
 \label{eeyore} \end{equation}
Next let
\begin{equation}
w_{\lambda}(r) = \lambda^{-\frac{2-\alpha}{1-p}} v(\lambda r)
= \lambda^{-\frac{2-\alpha}{1-p}} u(\lambda r + M_{b}) \quad\text{where }
\lambda^{\frac{2-\alpha}{1-p}} = u(M_{b}). \label{harrison}
\end{equation}
Then $w_{\lambda}(0) = \lambda^{-\frac{2-\alpha}{1-p}}u(M_{b}) = 1$ and
$w_{\lambda}'(0)=0$.
Now recall from Lemmas \ref{lem7} and \ref{lem8} that $M_{b}\to \infty$ and
 $u(M_{b}) \to \infty$ as $b \to \infty$. Thus
$ [u(M_{b})]^{\frac{2-\alpha}{1-p}} = \lambda \to \infty$ as $b \to \infty$.
In addition we see from \eqref{eeyore}-\eqref{harrison} that $w_{\lambda}$ solves
\begin{equation}
w_{\lambda}'' + \frac{N-1}{r + \frac{M_{b}}{\lambda}} w_{\lambda}'
+ \lambda^{\alpha}K(\lambda r + M_{b})
\left[ |w_{\lambda}|^{p-1}w_{\lambda}
+ \lambda^{-\frac{(2-\alpha)p}{1-p}}g(\lambda^{\frac{2-\alpha}{1-p}}w_{\lambda} )
 \right] = 0. \label{hoover}
\end{equation}
We now define
\begin{equation}
E_{\lambda} = \frac{1}{2} \frac{w_{\lambda}'^2}{\lambda^{\alpha}K(\lambda r + M_{b})}
 + \frac{1}{p+1} |w_{\lambda}|^{p+1}
+ \lambda^{\frac{-(2-\alpha)(1+p)}{1-p}} G(\lambda^\frac{2-\alpha}{1-p}w_{\lambda}).
\label{jqa}
\end{equation}
Using \eqref{hoover} and (H5) we obtain
\begin{equation}
\begin{aligned}
E_{\lambda}' &= -\frac{\lambda^{1-\alpha} w_{\lambda}'^2}{2(\lambda r + M_{b})
 K(\lambda r + M_{b})}
\Big( 2(N-1) + \frac{(\lambda r + M_{b})K'(\lambda r + M_{b})}{{K(\lambda r
 + M_{b})}} \Big) \\
&\leq 0 \quad \text{for } r \geq 0.
\end{aligned}
 \end{equation}
Thus
\begin{equation}
E_{\lambda}(r)
\leq E_{\lambda}(0) = \frac{1}{p+1} +
\lambda^{\frac{-(2-\alpha)(1+p)}{1-p}} G(\lambda^\frac{2-\alpha}{1-p}). \label{adams}
\end{equation}

From (H2) it follows that
$\lambda^{\frac{-(2-\alpha)(1+p)}{1-p}} G(\lambda^\frac{2-\alpha}{1-p}) \to 0$ as
$\lambda \to \infty$.
In addition it follows from \eqref{iowa}, \eqref{jqa}, and \eqref{adams}
that for large $\lambda$,
\begin{equation}
\frac{1}{2} \frac{w_{\lambda}'^2}{\lambda^{\alpha}K(\lambda r + M_{b})}
+ \frac{|w_{\lambda}|^{p+1}}{2(p+1)} \leq \frac{2}{p+1}. \label{truman}
\end{equation}
Hence the $w_{\lambda}$ are uniformly bounded on $[0, \infty)$.
In addition, it follows from (H4) and \eqref{truman} that the
$w_{\lambda}'$ are uniformly bounded by $ \sqrt{\frac{4k_2}{p+1}} r^{-\alpha/2}$
 on $(r, \infty)$. Then from \eqref{hoover} it follows that the
$w_{\lambda}''$ are uniformly bounded by $C_{6} r^{-(\frac{\alpha}{2}+1)}$
for some constant $C_{6}$.
Thus $w_{\lambda}$, $w_{\lambda}'$, and $w_{\lambda}''$ are uniformly bounded
on compact subsets of $(0, \infty)$
and so by the Arzela-Ascoli theorem a subsequence
(still labeled $w_{\lambda}$ and $w_{\lambda}'$) converges uniformly on
 compact subsets of $(0, \infty)$ to some $w$ and $w'$.
In addition, by the fundamental theorem of calculus with
$0\leq r_{1} < r_{2}$ we have
\begin{equation}
\begin{aligned}
|w_{\lambda}(r_{1}) - w_{\lambda}(r_{2})|
&\leq \int_{r_{1}}^{r_{2}} |w_{\lambda}'(s)| \, ds \\
&\leq \int_{r_{1}}^{r_{2}}\sqrt{\frac{4k_2}{p+1}} s^{-\alpha/2} \, ds \\
&= \sqrt{\frac{4k_2}{p+1}} [ r_2^{1-\alpha/2} - r_{1}^{1-\alpha/2} ]
\end{aligned} \label{donovan}
\end{equation}
and so we see from \eqref{donovan} and since $0 < \alpha <2$ that the
$w_{\lambda}$ are equicontinuous on compact subsets of $[0,\infty)$.
Thus it follows that $w(r)$ is continuous on $[0,\infty)$ and in particular
 $w(0)=1$.

Next we show that $w_{\lambda}$ has a large number of zeros for large
$\lambda$ and hence $u$ has a large number of zeros for large $b$.

So suppose $w>0$ on $[0,\infty)$. We see then from \eqref{hoover} and (H4) that
\begin{equation}
\begin{aligned}
& -(r + \frac{M_{b}}{\lambda})^{N-1} w_{\lambda}' \\
&=\int_{0}^{r} \lambda^{\alpha}K(\lambda(r + \frac{M_{b}}{\lambda}))
 (r + \frac{M_{b}}{\lambda})^{N-1}
\Big( |w_{\lambda}|^{p-1}w_{\lambda} 
 + \lambda^{-\frac{(2-\alpha)p}{1-p}} g(\lambda^\frac{2-\alpha}{1-p} 
w_{\lambda}) \Big). 
\end{aligned}\label{harding} 
\end{equation}
We claim now that
\begin{equation} 
\lim_{\lambda \to \infty} \int_{0}^{r} \lambda^{\alpha}
K(\lambda(r + \frac{M_{b}}{\lambda}))(r + \frac{M_{b}}{\lambda})^{N-1}
\Big( \lambda^{-\frac{(2-\alpha)p}{1-p}} g(\lambda^\frac{2-\alpha}{1-p} 
w_{\lambda}) \Big) =0 \label{new york} 
\end{equation}
on any fixed compact subset of $[0, \infty)$.

To see this note as in \eqref{georgia} we can similarly obtain the 
inequality
$|g(u)| \leq C_7 + \epsilon |u|^{p}$ for all $u$ for some constant $C_{7}$.
Therefore using this and (H4) in \eqref{harding} we see that
\begin{equation} 
\begin{aligned}
&\Big|\int_{0}^{r} \lambda^{\alpha}K(\lambda(t + \frac{M_{b}}{\lambda}))
(t + \frac{M_{b}}{\lambda})^{N-1}
\left( \lambda^{-\frac{(2-\alpha)p}{1-p}} g(\lambda^\frac{2-\alpha}{1-p} 
w_{\lambda}) \right) \, dt \Big| \\
&\leq \int_{0}^{r} k_2 (t + \frac{M_{b}}{\lambda})^{N-1-\alpha}
\left( C_7 \lambda^{-\frac{(2-\alpha)p}{1-p}}
+ \epsilon |w_{\lambda}|^{p} \right) \, dt.
\end{aligned} \label{bitten} 
\end{equation}

Now it follows from \eqref{bbragg} and \eqref{harrison} that
 $M_{b} \leq \epsilon_{6} [u(M_{b})]^{\frac{1-p}{2-\alpha}} = \epsilon_6 \lambda$ 
where $\epsilon_{6} = \epsilon_{5}^{-\frac{2}{2-\alpha}} $ so that for some subsequence
$\frac{M_{b}}{\lambda} \to A$ with $0\leq A \leq \epsilon_{6}$ and thus for large
 $\lambda$ we obtain from \eqref{bitten},
\begin{align*}
&\int_{0}^{r} k_2 (t + \frac{M_{b}}{\lambda})^{N-1-\alpha}
\left( C_7 \lambda^{-\frac{(2-\alpha)p}{1-p}}
 + \epsilon |w_{\lambda}|^{p} \right) \, dt \\
&\leq  C_7 k_2\lambda^{-\frac{(2-\alpha)p}{1-p}} 
\int_{0}^{r} (t + 2\epsilon_{6})^{N-1-\alpha}
+\epsilon k_2 \int_{0}^{r}(t + 2\epsilon_{6})^{N-1-\alpha}|w_{\lambda}|^{p}.
\end{align*}
Both of these terms are small on any compact subset of $[0, \infty)$ 
(the first since $\lambda \to \infty$ and the second term by \eqref{truman}) 
and so both of these limit to zero as $\lambda \to \infty$.
 This establishes \eqref{new york}.

Therefore we see by using (H4) and taking limits in \eqref{harding} we obtain
\begin{equation}
 -(r + A)^{N-1} w' \geq k_{1}\int_{0}^{r} (t+A)^{N-1-\alpha} w^{p} \, dt \quad
\text{on } (0, \infty). \label{max}
 \end{equation}
Since $w> 0$ on $[0, \infty)$ it follows from \eqref{max} that $w$ is 
decreasing so that
\begin{equation}
-(r + A)^{N-1} w' \geq k_{1} w^{p} \frac{(r+A)^{N-\alpha} - A^{N-\alpha}}{N-\alpha} 
\quad\text{on } (0, \infty). \label{delaware}
\end{equation}
Rewriting \eqref{delaware} gives
\begin{equation} 
-w' w^{-p} \geq \frac{k_{1}}{N-\alpha}(r+A)^{1-\alpha} 
- \frac{k_{1} A^{N-\alpha}}{N-\alpha}(r+A)^{1-N}
\text{ on } (0,\infty). \label{usg}
 \end{equation}
Next we analyze the two cases $A=0$ and $A\neq 0$ separately.
\smallskip

\noindent\textbf{Case 1:} $A\neq 0$.
Integrating \eqref{usg} on $(0,r)$ gives
\begin{align*}
& -\Big( \frac{w^{1-p} - 1}{1-p} \Big) \\
&\geq \frac{k_{1}}{N-\alpha}
\Big( \frac{(r+A)^{2-\alpha}}{2-\alpha} 
- \frac{ A^{2 -\alpha}}{2-\alpha} \Big) 
- \frac{k_{1}A^{N-\alpha}}{N-\alpha} 
\Big( \frac{(r+A)^{2-N}}{2-N} - \frac{A^{2 -N}}{2-N} \Big). 
\end{align*}
Thus for some constant $C_{8}$ we obtain
\begin{equation} 
\frac{w^{1-p} - 1}{1-p} \leq -\frac{k_{1}(r+A)^{2-\alpha}}{(N-\alpha)(2-\alpha)} 
- \frac{k_1(r+A)^{2-N}A^{N-\alpha}}{(N-2)(N-\alpha)} + C_{8}. \label{nixon} 
\end{equation}
The right-hand side of \eqref{nixon} goes to $-\infty$ as $r \to \infty$ since 
$0 < \alpha < 2$, $0<p<1$, $N > 2$ and so we see that $w$ becomes negative 
which is a contradiction because we assumed $w>0$ and so $w$ and hence $u$ 
must have a zero for sufficiently large $b$.
\smallskip
\vskip .1 in
\noindent\textbf{Case 2:} $A=0$.
In this case we see that \eqref{usg} becomes
\begin{equation} 
-w' w^{-p} \geq \frac{k_{1}}{N-\alpha}r^{1-\alpha} \quad\text{on } 
(0,\infty). \label{ohio} 
\end{equation}
Integrating \eqref{ohio} on $(0,r]$ gives
$$ 
\frac{w^{1-p} - 1}{1-p} \leq -\frac{k_{1}r^{2-\alpha}}{(N-\alpha)(2-\alpha)} 
$$ 
and therefore we see that $w$ becomes negative.
Thus we again obtain a contradiction and so $w$ and hence $u$ has a 
zero if $b$ is sufficiently large.

Thus there exists a $z_{b}>R$ such that $u(z_{b})=0$ and $u>0$ on $(R, z_{b})$. 
In addition by uniqueness of solutions of initial value problems it follows
 that $u'(z_{b})<0$ and then we can similarly show as in Lemma \ref{lem5} that $u$ 
has a local minimum $m_{b}> z_{b}$ for large enough $b>0$
and also that $w$ has a second zero $z_{2,b}$ (and hence $u$ has a second zero) 
if $b$ is sufficiently large. In a similar way, given any positive integer 
$n$ we can show for large enough $b$ that $u$ has $n$ zeros on $(R, \infty)$.
Since $w_{\lambda} \to w$ uniformly on compact sets it follows then that if 
$\lambda$ is sufficiently large then
$w_{\lambda}$ will have $n$ zeros on $(0, \infty)$ and hence $u(r,b)$ 
will have $n$ zeros on $(R, \infty)$ if $b>0$ is sufficiently large. 
This completes the proof. 
\end{proof}

\section{proof of Theorem \ref{thm1}}

We consider the set
$$
\{ b>0 | \ u(r,b) > 0 \text{ for all } r>R \}. 
$$
This set is nonempty by Lemma \ref{lem6} and is bounded from above by Lemma \ref{lem9}
 so there exists a $b_{0}>0$ such that
$$ 
b_{0} = \sup \{ b>0 | \ u(r,b) > 0 \text{ for all } r>R \}. 
$$
We show now that $u(r, b_{0}) > 0$ for $r>R$. If $u(r_{0}, b_{0})=0$ 
and $u(r,b_{0})>0$ on $(R, r_{0})$ then $u'(r_{0}, b_{0})\leq 0$.
By uniqueness of solutions of initial value problems it follows that 
$u'(r_{0}, b_{0})<0$.
Thus for $ r_1> r_{0}$ and $r_{1}$ sufficiently close to $r_0$ we have 
$u(r_1, b_{0})<0$. Then for $b$ close to $b_{0}$ with $b < b_{0}$ then 
$u(r_{1}, b)<0$ contradicting the definition of $b_{0}$.
 Hence $u(r, b_{0})>0$ for $r>R$. Now by Lemma \ref{lem3} we know that $u(r,b_{0})$ 
must get larger than $\beta$. If $u'>0$ for all $r\geq R$ then since $u$ 
is bounded it follows that $u$ would has a limit which by Lemma \ref{lem1} would 
have to be less than or equal to $\beta$. Thus we see that $u(r, b_{0})$ 
must have a local maximum $M_{b_{0}}>R$ and $u'>0$ on $[R, M_{b_0})$.
Next we show $E(r, b_{0}) \geq 0$ for all $r\geq R$. If $E(r_{0}, b_{0})<0$ 
then $E(r_{0}, b)<0$ for $b > b_{0}$ and $b$ close to $b_{0}$. On the other hand,
 since $b>b_{0}$ it follows that there exists a $z_{b}$ such that $u(z_{b}, b)=0$.
 Thus $E(z_{b}, b)\geq 0$. Since $E$ is nonincreasing this implies $z_{b} < r_{0}$ 
for all $b > b_{0}$. However $z_{b} \to \infty$ as $b \to b_{0}^{+}$ for if the 
$z_{b}$ were bounded then this would force a subsequence of the $z_{b}$ to converge
 to some $z_{0}$ and then $u(z_{0}, b_{0})=0$ contradicting that $u(r,b_{0})>0$. 
Thus $E(r, b_{0}) \geq 0$ for all $r\geq R$.
It now follows that $u(r, b_{0})$ cannot have a positive local minimum, 
$m_{b_0} > M_{b_{0}}$ for at such a point $u'(m,b_{0})=0$, $u''(m,b_{0}) \geq 0$ 
and so $f(u(m,b_{0}))\leq 0$. Since $u(m,b_{0})>0$ this then forces 
$0<u(m,b_{0})\leq \beta$ and thus $E(m,b_{0}) = F(u(m,b_{0}))<0$ contradicting 
that $E(r,b_{0})\geq 0$. Thus
$u'(r, b_{0}) < 0$ for $r>M_{b_0}$ and so $\lim_{r \to \infty} u(r,b_{0})$ exists. 
Denoting this limit as $L$ then $L \geq 0$ since $u(r, b_{0})>0$ for $r>R$ and by 
Lemma \ref{lem1} we have $f(L)=0$ so that $L=0$ or $L = \beta$. Then a similar argument 
as in Lemma \ref{lem2} shows that $u(r,b_{0})$ gets less than $\beta$ and so it follows 
that $L=0$ and thus $\lim_{r \to \infty} u(r, b_{0})=0$.
Hence $u(r, b_{0})$ is a positive solution on $(R, \infty)$ and 
$\lim_{r \to \infty} u(r, b_{0})=0$.

Next from a lemma in \cite{I4} if $b> b_{n}$ is sufficiently close to $b_{n}$ 
where $u(r, b_{n})$ has $n$ zeros on $(R, \infty)$ and
$$
\lim_{r \to \infty} u(r,b) =0 
$$ 
then $u(r, b)$ has at most $n+1$ zeros on $(R, \infty)$.
From this lemma it then follows that
$$
\{ b>b_0 | u(r,b) \text{ has exactly one zero on } (R, \infty) \} 
$$ 
is nonempty and again from Lemma \ref{lem9} this set is bounded from above.
Thus there exists a $b_{1}> b_{0}$ such that
$$
b_{1} = \sup\{ b>b_{0} | u(r,b) \text{ has exactly one zero on } (R, \infty) \}. 
$$
As above we can show $u(r, b_{1}) $ has exactly one zero on $(R, \infty)$ and
$$
\lim_{r \to \infty} u(r,b_{1}) =0. 
$$
Similarly we can find $b_{n}>b_{n-1}$ such that $u(r, b_{n})$ has exactly $n$ 
zeros on $(R, \infty)$ and
$$
\lim_{r \to \infty} u(r,b_{n}) =0. 
$$ 
This completes the proof of Theorem \ref{thm1}. 

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\end{document}