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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 214, pp. 1--13.\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-2017/214\hfil Existence and nonexistence of solutions]
{Existence and nonexistence of solutions
 for sublinear equations on exterior domains}

\author[J. A. Iaia \hfil EJDE-2017/214\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 Zhaosheng Feng}

\thanks{Submitted December 29, 2016. Published September 13, 2017.}
\subjclass[2010]{34B40, 35B05}
\keywords{Exterior domains; semilinear; sublinear; radial}

\begin{abstract}
 In this article we study radial solutions of $\Delta u + K(r)f(u)= 0$ on
 the exterior of the ball of radius $R>0$, $B_{R}$, centered at the origin
 in ${\mathbb R}^{N}$ with $u=0$ on $\partial B_{R}$ where $f$ 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 $K(r) \sim r^{-\alpha}$ for large $r$.
 We prove that if $N>2$ and   $K(r)\sim r^{-\alpha}$ with $2< \alpha < 2(N-1)$
 then there are no solutions with $\lim_{r \to \infty} u(r)=0$ for sufficiently
 large $R>0$. On the other hand, if $2<N-p(N-2) < \alpha< 2(N-1)$ and $k, n$
 are nonnegative integers with $0 \leq k \leq n$ then there exist solutions,
 $u_{k}$, with $k$ zeros on $(R, \infty)$ and $\lim_{r \to \infty} u_{k}(r)=0$
 if $R>0$ is sufficiently small.
\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 $(\beta , \infty)$,  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 $(\gamma, \infty)$.
\end{itemize}

 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}-\cite{B}, \cite{JK,M,ST}.

Interest in the topic for this paper comes from recent papers
\cite{C,C2,S} about solutions of differential equations on exterior domains.
In \cite{I5}-\cite{I7} we studied \eqref{1}-\eqref{3} with
$K(r) \sim r^{-\alpha}$, $f$ superlinear, and
$\Omega = {\mathbb R}^{N} \backslash B_{R}$ 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
$u \to 0$  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 paper 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 $0<p<1$.

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
\begin{itemize}
\item[(H4)] there exist constants $k_{1}>0$, $k_{2}>0$,
 and $\alpha$ with $0< \alpha < 2(N-1) $ such that
\begin{equation}
 k_{1} r^{-\alpha} \leq K(r) \leq k_{2} r^{-\alpha}  \quad\text{on } [R, \infty).
       \label{K}
\end{equation}
\item[(H5)] $K$  is differentiable, on $[R, \infty)$,
 $\lim_{r \to \infty} \frac{rK'}{K} = -\alpha$,  and
$\frac{rK'}{K} + 2(N-1)>0$.   %\label{K2}
\end{itemize}
Note that (H5) implies $r^{2(N-1)}K(r)$ is increasing.
In this article we prove the following result.

\begin{theorem} \label{thm1}
Let $N > 2$, $0<p<1$,  and $ 2<  N-p(N-2) < \alpha < 2(N-1)$.
 Assuming {\rm (H1)--(H5)} then
given nonnegative integers $k, n$  with $0\leq k \leq n$ then there exist
 solutions, $u_{k}$, of \eqref{DE}-\eqref{DE2} with $k$ zeros on
$(R, \infty)$ and $\lim_{r \to \infty} u_{k}(r) = 0$ if $R>0$ is
sufficiently small.
\end{theorem}

In addition we also prove:

\begin{theorem} \label{thm2}
Let $N > 2$, $0<p<1$ and  $  2< \alpha < 2(N-1)$.  Assuming {\rm (H1)--(H5)},
there are no solutions of \eqref{DE}-\eqref{DE2} such that
$\lim_{r \to \infty} u(r) = 0$  if $R>0$ is sufficiently large.
\end{theorem}


Note that for the superlinear problems studied in \cite{I5}-\cite{I7}
we were able to prove existence for {\it any} $R>0$ whereas in the
sublinear case and in \cite{I4} we only get solutions if $R$ is sufficiently small.


\section{Preliminaries and proof of Theorem \ref{thm2}}

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)}
\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 $N>2$, $0<p<1$, and $0 < \alpha < 2(N-1)$. Assume
{\rm (H1)--(H5)} and suppose $u$ satisfies
\eqref{DE}-\eqref{DE2} with $b>0$. If $u$ has a zero, $z_{b}$,
with $u>0$ on $(R, z_{b})$ or if $u>0$ for $r>R$ and
$\lim_{r \to \infty} u=0$ then
$u$ has a local maximum, $M_{b}$, with $R<M_{b}$, $u'>0$ on
$(R, M_{b})$, $M_{b} \to \infty$ as $b \to \infty$, and
$u(M_{b})\to \infty$ as $b \to \infty$.
\end{lemma}

\begin{proof}
Since $u(R)=0$ and $u'(R) = b >0$ we see that $u$ gets positive for $r>R$
and if $u$ has a zero, $z_{b}$, or if $u>0$ and
$\lim_{r \to \infty} u(r) =0$  then $u$ has a critical point, $M_{b}$,
such that $u'>0$ on $(R, M_{b})$. Then $u'(M_{b})=0$ and $u''(M_{b})\leq 0$.
By uniqueness of solutions of initial value problems it follows that
$u''(M_{b})< 0$ and thus $M_{b}$ is a local maximum. Next suppose there
exists $M_{0}>R$ such that $M_{b} \leq M_{0}$ for all $b>0$.
Letting $v_{b}(r) = \frac{u(r)}{b} $ then from \eqref{DE2} we have
$v_{b}(R)=0$, $v_{b}'(R) = 1$ and
\begin{equation}
v_{b}''(r) + \frac{N-1}{r} v_{b}'(r) + K(r)\frac{f(b v_{b}(r))}{b} = 0 \quad
\text{for } r \geq R. \label{frank}
\end{equation}
It follows from \eqref{energy}-\eqref{energy2} that
 $$
\Big(\frac{1}{2} \frac{v_b'^{2}}{K} + \frac{F(bv_b)}{b^2} \Big)' \leq 0 \quad
 \text{for } r\geq R
$$
and thus
 \begin{equation}
\frac{1}{2} \frac{v_b'^{2}}{K} + \frac{F(bv_b)}{b^2}
 \leq \frac{1}{2K(R)}\quad \text{for } r\geq R. \label{energy3}
\end{equation}
It then follows from \eqref{energy3} and (H2)--(H4) that  $|v_{b}'|$ is
 uniformly bounded for large $b>0$ on $[R, \infty)$.
So there is a constant $C_{1}>0$ such that
\begin{equation}
 |v_{b}'| \leq C_{1} \text{ for large } b>0 \quad \text{and all } r \geq R.
  \label{taft}
\end{equation}
We now fix a compact set $[R, R_{0}]$. Then on $[R, R_0]$ we have by
\eqref{taft}
$$
|v_{b}| = |(r-R) + \int_{R}^{r} v_{b}'(t) \, dt| \leq ( 1 + C_{1})(R_{0}-R)
$$
so we see that $|v_{b}|$ is uniformly bounded for large $b$ on $[R, R_{0}]$.

In addition from (H1)--(H2) it follows there is a constant $C_{2}>0$ such that
\begin{equation}
|f(u)| \leq C_{2}|u|^{p} \quad \text{for all } u \label{waylon}
\end{equation}
and therefore since the $v_{b}$ are uniformly bounded on $[R, R_{0}]$
and $0<p<1$ it follows that
\begin{equation}
|  \frac{f(b v_{b})}{b} | \leq \frac{C_{2}|v_{b}|^p}{b^{1-p}} \to 0 \quad
\text{as } b \to \infty. \label{estreet}
\end{equation}

Then from \eqref{frank} and \eqref{estreet} we see that $|v_{b}''|$
is uniformly bounded on $[R, R_{0}]$. So by the Arzela-Ascoli theorem
there is a subsequence of  $v_{b}$ (still denoted $v_{b}$) such that
$v_{b} \to v_{0}$ and $v_{b}'\to v_{0}'$ uniformly on $[R, R_{0}]$ as
$b\to \infty$. It then follows from \eqref{frank} that $v_{b}''$ converges
uniformly to $v_{0}''$ on $[R, R_{0}]$ and  $v_{0}'' + \frac{N-1}{r}v_{0}' =0$.
Since $R_{0}$ is arbitrary we see that $v_{0}'' + \frac{N-1}{r}v_{0}' =0$
on $[R, \infty)$.
Thus, $r^{N-1}v_{0}' = R^{N-1}$ and
$v_{0} = \frac{R^{N-1} [R^{2-N} - r^{2-N}]}{N-2}$. Now since $M_{b}\leq M_{0}$
 for all $b>0$ then a subsequence
of $M_{b}$ converges to some $M$ and since $v_{b}'(M_b)=0$ it follows that
$v_{0}'(M)=0$. However this contradicts that $v_{0}'= \frac{R^{N-1}}{r^{N-1}}>0$.
 Therefore our assumption that the $M_{b}$ are bounded is false and so we
see $M_{b} \to \infty$ as $b \to \infty$.

Next we see that since $M_{b} \to \infty$ then $M_{b}> 2R$ if $b$ is
sufficiently large and since $u$ is increasing on $[R, M_{b}]$ then
$\frac{u(M_{b})}{b} \geq \frac{u(2R)}{b} =  v_{b}(2R) \to v_{0}(2R)>0$
for sufficiently large $b$. Thus $u(M_{b})> \frac{v_{0}(2R)}{2}b$
for sufficiently large $b$ and so we see  that $u(M_{b}) \to \infty$
as $b \to \infty$. This completes the proof.
\end{proof}

\begin{lemma} \label{lem2}
Let $N>2$, $0<p<1$, $2<\alpha<2(N-1)$, and  assume {\rm (H1)--(H5)}.
If $u(z_b)=0$ with $u>0$ on $(R, z_{b})$ or
$u>0$ on $(R, \infty)$  with $\lim_{r \to \infty} u=0$  then
\begin{equation}
[u(M_{b})]^{\frac{1-p}{2}} M_{b}^{\frac{\alpha}{2}-1}
\leq \frac{k_{2}}{\frac{\alpha}{2}-1}
\sqrt{\frac{1}{p+1} + \frac{F_{0}}{\gamma^{p+1}}}.  \label{tonyb}
 \end{equation}
\end{lemma}

\begin{proof}
We first show that if $u(z_{b})=0$ with $u>0$ on $(M_{b}, z_{b})$ then
$u'<0$ on $(M_{b}, z_{b})$ and if $u>0$ on $(M_{b}, \infty)$ with
$\lim_{r \to \infty} u(r)=0$ then $u'<0$ on $(M_{b}, \infty)$.
In the first case, if $u$ has a positive local minimum, $m_{b}$, with
$M_{b} < m_{b} < z_{b}$ then
$u'(m_{b})=0$, $u''(m_{b})\leq 0$, so $f(u(m_{b}))\geq 0$ which implies
$0< u(m_{b}) \leq \beta$. On the other hand, since $E$ is nonincreasing
$0 > F(u(m_{b})) = E(m_{b}) \geq E(z_{b})
= \frac{1}{2} \frac{u'^2(z_{b})}{K(z_{b})} \geq 0$
 which is impossible.
Secondly, suppose $u>0$ on $(R, \infty)$ and $\lim_{r \to \infty} u(r)=0$.
Since $E$ is nonincreasing it follows that $\lim_{r \to \infty} E(r)$
exists and since $\frac{1}{2}\frac{u'^2}{K} \geq 0$ and $F(u(r))\to 0$ as
$r \to \infty$ we see that $\lim_{r \to \infty} E(r) \geq 0$.
Thus $E(r) \geq 0$ for all $r \geq R$. On the other hand,
if $u$ has a positive local minimum, $m_{b}$, then $0< u(m_{b}) \leq \beta$
and $E(m_{b}) = F(u(m_{b})) <0$ again yielding a contradiction.

Next, it follows from \eqref{energy}-\eqref{energy2} that $E(t) \leq E(M_{b})$
 for $t \geq M_{b}$. Rewriting this inequality we obtain
\begin{equation}
\frac{|u'(t)|}{\sqrt{2}\sqrt{F(u(M_b)) - F(u(t))}}
\leq \sqrt{K} \text{ for } t \geq M_{b}.  \label{energy5}
\end{equation}
If $u(z_{b})=0$ then integrating \eqref{energy5} on $(M_b,z_b)$ and using that
$u'<0$ on $(M_{b}, z_{b})$ gives
\begin{equation}
\begin{aligned}
\int_{0}^{u(M_{b})} \frac{dt}{\sqrt{F(u(M_b))-F(t)}}
&=  \int_{M_{b}}^{z_{b}} \frac{-u'(t)}{\sqrt{2}\sqrt{F(u(M_b)) - F(u(t))}} \, dt \\
& \leq \int_{M_{b}}^{z_{b}}\sqrt{K} \, dt \\
& \leq   \frac{k_{2}}{\frac{\alpha}{2}-1} ( M_{b}^{1 - \frac{\alpha}{2}}
 -z_{b}^{1 - \frac{\alpha}{2}})  \\
&\leq \frac{k_{2}}{\frac{\alpha}{2}-1}  M_{b}^{1 - \frac{\alpha}{2}}.
\end{aligned} \label{auh2o}
\end{equation}
Similarly if $u(r)>0$ and $\lim_{r \to \infty} u=0$ then integrating
\eqref{energy5} on $(M_{b}, \infty)$ and using that $u'<0$ on $(M_{b}, \infty)$
we again obtain
$$
\int_{0}^{u(M_{b})} \frac{dt}{\sqrt{F(u(M_b))-F(t)}}
\leq  \frac{k_{2}}{\frac{\alpha}{2}-1}  M_{b}^{1 -\frac{\alpha}{2}}.
$$
Next from (H2), (H3) and \eqref{waylon} it follows that
$-F_{0} \leq F(u) \leq  \frac{C_{2}|u|^{p+1}}{p+1}$ for all $u$.
Therefore estimating the left-hand side of \eqref{auh2o} gives
\begin{equation}
\int_{0}^{u(M_{b})} \frac{\, dt}{ \sqrt{F(u(M_{b})) - F(t)}}
\geq \frac{u(M_{b})}{\sqrt{\frac{C_2[u(M_{b})]^{p+1}}{p+1} + F_{0}}}
= \frac{[u(M_{b})]^{\frac{1-p}{2}}}{ \sqrt{\frac{C_{2}}{p+1}
+ \frac{F_{0}}{[u(M_{b})]^{p+1}}}}. \label{JFK}
 \end{equation}
Also from \eqref{energy}-\eqref{energy2} if $u(z_{b})=0$ then we have
$F(u(M_{b})) = E(M_{b}) \geq E(z_{b}) = \frac{1}{2}
\frac{u'^2(z_{b})}{K(z_{b})} \geq 0$ and so $u(M_{b}) \geq \gamma$.
On the other hand, if $u>0$ and $\lim_{r \to \infty} u=0$ then as we saw
earlier $E(r) \geq 0$ for all $r \geq R$.
Thus $F(u(M_{b})) =E(M_{b})\geq 0$ and again we see
$u(M_{b}) \geq \gamma$.
Now using \eqref{JFK} in \eqref{auh2o} and rewriting gives
\begin{equation}
\begin{aligned}
[u(M_{b})]^{\frac{1-p}{2}} M_{b}^{\frac{\alpha}{2}-1}
&\leq \frac{k_{2}}{\frac{\alpha}{2}-1}
\sqrt{\frac{C_2}{p+1} + \frac{F_{0}}{[u(M_{b})]^{p+1}}} \\
&\leq \frac{k_{2}}{\frac{\alpha}{2}-1}\sqrt{\frac{C_{2}}{p+1}
+ \frac{F_{0}}{\gamma^{p+1}}}.     \label{LBJ}
\end{aligned}
 \end{equation}
This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 If $u$ has a zero, $z_{b}$, with $u>0$ on $(R, z_{b})$ or $u>$ on
 $(R, \infty)$ with $\lim_{r \to \infty} u(r) =0$  then by Lemmas 
\ref{lem1} and \ref{lem2}
we know that $u$ has a local maximum, $M_{b}$, with $R< M_{b}$ and $u'>0$
on $(R, M_{b})$. In addition, from the proof of Lemma \ref{lem2} we have
$u(M_{b}) \geq \gamma$. Combining this with \eqref{LBJ} and the fact that
$\alpha >2$  and $0<p<1$ we obtain
\begin{equation}
\gamma^{\frac{1-p}{2}} R^{\frac{\alpha}{2}-1}
\leq [u(M_{b})]^{\frac{1-p}{2}} M_{b}^{\frac{\alpha}{2}-1}
\leq \frac{k_{2}}{\frac{\alpha}{2}-1}\sqrt{\frac{1}{p+1}
+ \frac{F_{0}}{\gamma^{p+1}}}. \label{auh2o2}
\end{equation}
Thus we see that if $R$ is sufficiently large then \eqref{auh2o2}
is violated and so we obtain a contradiction.
This completes the proof of Theorem \ref{thm2}.
\end{proof}

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

We now turn to the proof of existence for $N>2$, $0<p<1$,
$ 2< N-p(N-2) < \alpha < 2(N-1)$ and $R>0$ sufficiently small.
First we make the change of variables:
$$
u(r) = u_{1}(r^{2-N})\,.
$$
Using \eqref{DE} we see that $u_{1}$ satisfies
\begin{equation}
 u_{1}'' + h(t) f(u_{1}) =0  \label{DE4}
\end{equation}
where it follows from (H4)--(H5) that:
\begin{gather}
0< h(t) = \frac{t^{\frac{2(N-1)}{2-N}}K(t^{\frac{1}{2-N}})}{(N-2)^2} \quad
\text{and}\quad   h'(t)<0  \text{ for } t > 0, \label{h equation} \\
u_{1}(R^{2-N})=0 \quad \text{and}\quad
 u_{1}'(R^{2-N}) = -\frac{bR^{N-1}}{N-2}<0.  \label{johnpaul}
\end{gather}
In addition, from (H4) we have
\begin{equation}
\frac{k_{1}}{(N-2)^2 t^q} \leq h(t) \leq \frac{k_{2}}{(N-2)^2t^{q}} \quad
\text{for all } t>0,
\quad \text{where } q = \frac{2(N-1)-\alpha}{N-2}. \label{ringo}
\end{equation}

\textbf{Note:} Since $2 < \alpha< 2(N-1)$, $N>2$, and $q= \frac{2(N-1)-\alpha}{N-2}$
it follows that $0<q<2$.

Now instead of considering \eqref{DE4} with \eqref{johnpaul} we consider
\eqref{DE4} with
\begin{equation}
u_{1}(0) =0, \quad u_{1}'(0) = b_{1}>0.  \label{initial}
 \end{equation}
Integrating \eqref{DE4} twice on $(0,t)$ and using \eqref{initial} we see
that a solution of \eqref{DE4}, \eqref{initial} is equivalent to a solution of:
\begin{equation}
u_{1} = b_{1}t -\int_{0}^{t} \int_{0}^{s} h(x) f(u_{1}) \, dx \, ds. \label{george}
 \end{equation}
Letting $u_{1}=tv_{1}$ we see that a solution of \eqref{george} is equivalent
to a solution of
\begin{equation}
 v_{1} = b_{1} - \frac{1}{t} \int_{0}^{t} \int_{0}^{s} h(x) f(xv_{1}) \, dx \, ds.
\label{joe}
\end{equation}
Now we define
\begin{equation}
Tv_{1} = b_{1} - \frac{1}{t} \int_{0}^{t} \int_{0}^{s} h(x) f(xv_{1}) \, dx \, ds.
\label{jimi}
\end{equation}
Let $0<\epsilon < 1$. Denoting $\|w\| = \sup_{[0,\epsilon]} |w(x)|$ we let
$$
B = \{ v \in C[0,\epsilon] \ | \ \|v-b_{1}\| \leq 1 \}
$$
where $C[0, \epsilon]$ is the set of continuous functions on $[0,\epsilon]$.
It follows from (H1)--(H2) that there exists $L>0$ such that
\begin{equation}
|f(u)|\leq L|u| \quad \text{for all } u.  \label{zappa}
\end{equation}
Then by \eqref{ringo}, \eqref{jimi}-\eqref{zappa}, and since $q<2$
as well as $|v_{1}| \leq 1 + b_{1}$:
\begin{align*}
| Tv_{1} - b_{1} |
&\leq \frac{L k_{2}}{(N-2)^2 t} \int_{0}^{t} \int_{0}^{s} x^{-q} x |v_{1}|\, dx \, ds\\
&\leq  \frac{L k_2 (1 + b_1)t^{2-q}}{(2-q)(3-q)(N-2)^2} \\
&\leq \frac{L k_2 (1 + b_1)\epsilon^{2-q}}{(2-q)(3-q)(N-2)^2}.
\end{align*}
Thus for sufficiently small $\epsilon>0$ we have $T:B \to B$.
Next we see by the mean value theorem, \eqref{ringo}, and \eqref{zappa} that we have
\begin{align*}
|Tv_{1}- T v_{2}| 
&= | \frac{1}{t} \int_{0}^{t} \int_{0}^{s} h(x) [f(xv_{1}) - f(xv_{2})] \, dx \, ds|\\
&\leq \frac{L}{t} \int_{0}^{t} \int_{0}^{s} xh(x) |v_{1} - v_{2}|  \, dx \, ds \\
&\leq \frac{L k_{2}}{(N-2)^2} \|v_{1} - v_{2}\| \frac{1}{t} \int_{0}^{t}
  \int_{0}^{s} x x^{-q} \, dx \, ds \\
&\leq \frac{L k_{2} \epsilon^{2-q}}{(2-q)(3-q)(N-2)^2} \|v_{1} - v_{2}\|. 
\end{align*}
Thus for small enough $\epsilon>0$ we see that $T$ is a contraction for any $b_{1}>0$ 
and so by the contraction mapping principle there is a solution of  \eqref{joe} 
and hence of \eqref{DE4}, \eqref{initial} on $[0, \epsilon]$ for some $\epsilon>0$.


Next from \eqref{joe} and \eqref{zappa} we have
\begin{align}  
|\frac{u_{1}}{t}| 
&= |v_{1}| \leq b_1 + \frac{L}{t}\int_{0}^{t} \int_{0}^{s} xh(x)|v_{1}(x)| \, dx \, ds
  \label{lennon} \\
& \leq b_1 + \frac{Lk_2}{(N-2)^2 t} \int_{0}^{t} \int_{0}^{s} x^{1-q}|v_{1}(x)|
  \, dx \, ds \nonumber \\
&\leq b_1 + \frac{k_2 L}{(N-2)^2} \int_{0}^{t} x^{1-q}|v_{1}(x)| \, dx.
  \label{neptune}  
\end{align}
Now let $ w_{1} = \int_{0}^{t} s^{1-q}|v_{1}(s)| \, ds$.
Then
 \begin{equation} 
w_{1}' = t^{1-q}|v_{1}(t)|= t^{-q} |u_1(t)|  \label{hank} 
\end{equation}
and from \eqref{lennon}-\eqref{hank} we obtain
 \begin{equation} 
w_{1}' - \frac{k_2L}{(N-2)^2} t^{1-q} w_{1} \leq b_1 t^{1-q}. \label{pete}
 \end{equation}
Multiplying \eqref{pete} by  $\mu(t) =e^{-\frac{k_2L t^{2-q}}{(2-q)(N-2)^2}} \leq 1$, 
integrating on $[0, t]$, and rewriting gives
\begin{equation} 
w_{1} \leq \frac{b_1}{\mu(t)} \int_{0}^{t} s^{1-q} \mu(s)  \, ds  
\leq  \frac{b_{1}}{(2-q)} \frac{t^{2-q}}{\mu(t)}.   \label{roger}  
\end{equation}
Then from \eqref{hank}-\eqref{roger} we obtain
\begin{equation} 
u_{1} \leq  \Big(\frac{k_{2}L}{(2-q)(N-2)^2} \Big) \frac{b_1t^{3-q}}{\mu(t)}  
+ b_{1} t  =
b_{1}\left(t + B(t) t^{3-q} \right) \label{mercury}
 \end{equation} 
where
\begin{equation}  
B(t) = \Big(\frac{k_{2}L}{(2-q)(N-2)^2} \Big) \frac{1}{\mu(t)}. \label{albert}  
\end{equation}
Note that $\mu(t)$ is decreasing and continuous hence $B(t)$ is increasing 
and continuous.

Next it follows from \eqref{george} that
\begin{equation} 
u_{1}' = b_{1} -\int_{0}^{t}  h(x) f(u_{1}) \, dx  \label{saturn} 
\end{equation}
and thus from \eqref{ringo}, \eqref{mercury}, \eqref{saturn}, and since $B(t)$ 
is increasing:
\begin{equation} 
\begin{aligned}
|u_{1}'| &\leq b_{1} + \frac{k_{2}L}{(N-2)^2}  \int_{0}^{t} x^{-q}  
b_{1}\left( x + B(x) x^{3-q} \right) \, dx \\
&\leq  b_{1} + \frac{k_{2}Lb_{1}}{2(N-2)^2(2-q)}  \left(2t^{2-q} + B(t) t^{4-2q} 
\right). 
\end{aligned} \label{venus} 
\end{equation}

Thus from \eqref{mercury} and \eqref{venus} we see that $u_{1}$ and $u_{1}'$ 
are bounded on $[0,t]$ and so it follows that the solution of 
\eqref{DE4}, \eqref{initial} exists
on $[0,t]$. Since $t$ is arbitrary it follows that the solution of 
\eqref{DE4}, \eqref{initial} exists on $[0, \infty)$.

\begin{lemma} \label{lem3}  
Let $N>2$, $0<p<1$, and $2 < \alpha <2(N-1)$. Assuming {\rm (H1)--(H5)}
 and that $u_{1}$ solves
\eqref{DE4}, \eqref{initial} then there exists $t_{b_1}>0$ such that 
$u_{1}(t_{b_1}) =\beta$ and  $0< u_{1}<\beta$ on $(0, t_{b_1})$. 
In addition,  $u_{1}'(t)>0$ on $[0, t_{b_1}]$.
\end{lemma}

\begin{proof} 
Since $u_{1}'(0)=b_{1}>0$ we see that $u_{1}$ is initially increasing, 
positive, and less than $\beta$.
On this set $f(u_{1})< 0$ and so by \eqref{DE4} we have
$u_{1}''> 0$. Thus by \eqref{initial} we have $u_{1}' > b_1>0$ when 
$0<u_{1}<\beta$ and so on this set we have $u_{1} >  b_1t$.
Since $b_1 t$ exceeds $\beta$ for sufficiently large $t$ we see then
that there exists $t_{b_1}>0$ such that $u_{1}(t_{b_1}) =\beta$ and  
$0< u_{1}<\beta$ on $(0, t_{b_1})$.
This completes the proof. 
\end{proof}

\begin{lemma} \label{lem4}  
Let $N>2$, $0<p<1$, and $2 < \alpha < 2(N-1)$. Assuming {\rm (H1)--(H5)}
 and that $u_{1}$ solves
\eqref{DE4}, \eqref{initial} then $t_{b_1} \to \infty$ as $b_1 \to 0^{+}$.
\end{lemma}

\begin{proof} 
Evaluating  \eqref{mercury} at $t = t_{b_1}$ gives:
\begin{equation} \beta = u_{1}(t_{b_1})
\leq b_1 ( t_{b_1}  + B(t_{b_1}) t_{b_1}^{3-q}).  \label{peace} 
\end{equation}
Since $2 < \alpha < 2(N-1)$ it then follows from the note after 
\eqref{ringo} that $0<q<2$. Now if $t_{b_{1}}$ is bounded as $b_{1} \to 0^{+}$ 
then the right-hand side of \eqref{peace}  goes to $0$ as $b_{1} \to 0^{+}$ 
which violates \eqref{peace}. Thus we obtain a contradiction and so we see 
that $t_{b_1} \to \infty$ as $b_1 \to 0^{+}$. This completes the proof.
\end{proof}

\begin{lemma} \label{lem5}  
Let $N>2$, $0<p<1$, and $N-p(N-2)< \alpha < 2(N-1)$. Assuming {\rm (H1)--(H5)}
 and that $u_{1}$ solves \eqref{DE4}, \eqref{initial} then  $u_{1}$ has 
a local maximum, $M_{b_1}$, on $(0, \infty)$.
\end{lemma}

\begin{proof} 
From Lemma \ref{lem3} it follows that there exists $t_{b_1}>0$ such that 
$u_{1}(t_{b_1}) = \beta$ and $u_{1}'>0$ on
$[0, t_{b_1}]$. Now if $u_{1}$ does not have a local maximum then 
$u_{1}'\geq 0$ for $t >t_{b_1}$ and so
$u_{1}\geq u_{1}(t_{b_1} + \delta) > \beta>0$ for $t> t_{b_1} + \delta$ 
and some $\delta >0$.
Then from (H2) we see that there is a $C_3>0$ such that $f(u_1) \geq C_3$ on 
$[ t_{b_1} + \delta, \infty)$.
Thus
\begin{equation}
 -u_{1}'' = h(t)f(u_{1})\geq C_3h(t) \text{ for } t > t_{b_1} + \delta. \label{star} 
\end{equation}

We now divide the rest of the proof into 3 cases.
\smallskip

\noindent\textbf{Case 1:} $N< \alpha < 2(N-1)$
In this case we see from \eqref{ringo} that $0<q<1$ so integrating 
\eqref{star} on $(t_{b_1} + \delta, t)$ and using
\eqref{ringo} gives
\[
 u_{1}'\leq u_{1}'(t_{b_1}+\delta)-\frac{k_1 C_3}{(1-q)(N-2)^2}
 \left( t^{1-q}-(t_{b_1}+\delta)^{1-q} \right) \to -\infty
\quad\text{as } t \to \infty.
\]  
Thus $u_{1}'$ gets negative which contradicts that $u_{1}' \geq 0$ for $t>0$ 
and so $u_{1}$ must have a local maximum.
\smallskip

\noindent\textbf{Case 2:} $ \alpha = N$
In this case we have $q=1$ by \eqref{ringo}
and so again integrating \eqref{star} on $(t_{b_1}+\delta,t)$ we obtain
$$ 
u_{1}' \leq u_{1}'(t_{b_1}+\delta) - \frac{k_1C_3}{(N-2)^2}
 \left( \ln(t) -\ln(t_{b_1}+\delta) \right) \to -\infty \text{ as } t \to \infty 
$$
which again contradicts that $u_{1}'\geq 0$ for $t>0$. Thus $u_{1}$ must 
have a local maximum.
\smallskip

\noindent\textbf{Case 3:} $ N-p(N-2)<\alpha < N$
We denote
\begin{equation}
E_{1} = \frac{1}{2} \frac{	u_{1}'^{2}}{h(t)} + F(u_{1})  \label{munson}
\end{equation}
and observe from \eqref{DE4}-\eqref{h equation} that
\begin{equation} E_{1}'= \Big( \frac{1}{2} 	\frac{	u_{1}'^{2}}{h(t)}
 + F(u_{1}) \Big)' =   -\frac{u_{1}'^{2}h'}{2h^2}\geq 0.
\label{polaris}  \end{equation}
In addition we see from \eqref{ringo} that $E_{1}(0)=0$ and so 
$E_{1}(t)\geq 0$ for $t \geq 0$.

We suppose now that $u_{1}$ is increasing for $t> t_{b_1}$.
We first show that there exists $t_{b_2}> t_{b_1}$ such that 
$u(t_{b_2})= \gamma$.
So we suppose by the way of contradiction that $0< u_1< \gamma$ and 
$u_{1}'\geq 0$ for $t > t_{b_1}$.

Then from \eqref{DE4}-\eqref{h equation} and (H3) we have
\begin{equation} 
\Big(\frac{1}{2} u_{1}'^2 + h(t) F(u_{1}) \Big)' 
= h'(t) F(u_{1}) \geq 0  \quad  \text{when } 0 \leq u_{1} \leq \gamma.  
\label{paul} 
\end{equation}

Now we recall from (H1) that 
$ \lim_{u_{1} \to 0} \frac{F(u_{1}) }{u_{1}^2} = \frac{f'(0)}{2}$.
Also since $u_{1}(0)=0$ and $u_{1}'(0)=b_{1}$ then
 $\lim_{t \to 0^{+}} \frac{u_{1}}{t} = b_{1}$.
Therefore for small positive $t$ and \eqref{ringo} we have
\begin{equation} 
0 \leq  h(t)|F(u_{1})| = t^2h(t)\frac{|F(u_{1})|}{u_{1}^2} \frac{u_{1}^2}{t^2}
 \leq  \frac{|f'(0)| \, k_2 \, b_{1}^2 \, t^{2-q}}{(N-2)^2}  \to 0 
\label{mccartney} \end{equation}
as $t \to 0^{+}$  since $q<2$. Therefore, integrating \eqref{paul} on $(0,t)$ 
and using \eqref{mccartney} we obtain
\begin{equation} 
\frac{1}{2} u_{1}'^2 + h(t) F(u_{1}) \geq \frac{1}{2} b_1^2 \quad
\text{when } 0 \leq u_{1} \leq \gamma. \label{starr} 
\end{equation}
In addition, since $ 0 \leq u_{1} \leq \gamma$ it follows that
 $h(t) F(u_{1}) \leq 0$ and thus from \eqref{starr},
\begin{equation} 
u_{1}' \geq b_{1} \quad \text{when } 0\leq u_{1}\leq \gamma. \label{hilbert} 
 \end{equation}
Integrating on $(0,t)$ we obtain
$$ 
u_{1} \geq  b_1 t  \to \infty  \text{ as } t \to \infty
$$
- a contradiction since we assumed $u_{1} < \gamma$.
Thus there exists $t_{b_2}>t_{b_1}$ such that $u(t_{b_2})= \gamma$ and 
 $u_{1}'\geq b_1>0$ on $[0, t_{b_2}]$ by \eqref{hilbert}.

We show now that $u_{1}(t) \to \infty$ as $t \to \infty$. 
If not then $u_{1}$ is bounded from above and so there exists $Q> \gamma$ such that
$\lim_{t \to \infty} u_{1}(t) = Q$. Returning to \eqref{DE4} we see that this implies:
\begin{equation} 
\lim_{t \to \infty} \frac{u_{1}''}{h(t)} = -f(Q)<0.  \label{euler} 
\end{equation}

In particular, $u_{1}''<0$ for large $t$ and so $u_{1}'$ is decreasing for large $t$.
Since $u_{1}'>0$ for large $t$ it follows that $\lim_{t \to \infty} u_{1}' $ exists. 
This limit must be zero otherwise this would imply $u_{1} \to \infty$
as $ t \to \infty$ contradicting the assumption that $u_{1}$ is bounded. 
Thus $\lim_{t \to \infty} u_{1}'=0$.
Next denoting $H(t) = \int_{t}^{\infty} h(s) \, ds$ we see that since
 $N-p(N-2) < \alpha < N$ and
$q= \frac{2(N-1)-\alpha}{N-2}$ this implies:
\begin{equation}
 1 <q < 1+p<2. \label{mars}
\end{equation}
Therefore by \eqref{ringo} we see that $h(t)$ is integrable at infinity so
$H(t)$ is defined. Then by \eqref{euler} and L'H\^opital's rule we see that
\begin{equation} 
\lim_{t \to \infty} \frac{u_{1}'}{H(t)} 
= \lim_{t \to \infty} -\frac{u_{1}''}{h(t)} = f(Q) >0. \label{orion} 
\end{equation}
Then from \eqref{ringo} and \eqref{mars}-\eqref{orion} we see
\begin{equation} 
u_{1}' \geq \frac{f(Q)}{2} H(t) \geq \frac{k_1 f(Q)}{2(q-1)(N-2)^2}t^{1-q} \quad
\text{for large } t. \label{laplace}
 \end{equation}
Now integrating \eqref{laplace} on $(t_{0},t)$ where $t_{0}$ and $t$ are 
sufficiently large gives
$$ 
u_{1} \geq u_{1}(t_{0}) +\frac{k_1 f(Q)}{2(q-1)}\frac{t^{2-q}}{(2-q)(N-2)^2} 
\to \infty \quad \text{as } t \to \infty \text{ since } q<2
$$
- a contradiction since we assumed $u_{1}$ was bounded.
Thus if $u_{1}'>0$ for $t>0$ then it must be that $u_{1} \to \infty$ as 
$ t \to \infty$.

Next recalling \eqref{paul} we have
\begin{equation} 
\Big( \frac{1}{2} u_{1}'^2 + h(t) F(u_{1}) \Big)' 
= h'(t) F(u_{1}) <0  \quad  \text{when }  u_{1}> \gamma. \label{nobel}
 \end{equation}
Integrating this on $(t_{b_2},t)$ gives
\begin{equation} 
\frac{1}{2} u_{1}'^2 + h(t) F(u_{1}) \leq \frac{1}{2} u_{1}'^2(t_{b_{2}})\quad
 \text{for } t > t_{b_2}.  \label{harrison}
 \end{equation}
On $(t_{b_2},t)$ we have $h(t) F(u_{1}) > 0$ and thus from \eqref{harrison}:
\begin{equation}  
|u_{1}'| < |u_{1}'(t_{b_{2}})| \text{ for } t > t_{b_2}. \label{mick}
 \end{equation}
We claim now that
\begin{equation} 
\lim_{ t \to \infty} \frac{t^2 h(t) f(u_{1})}{u_{1}} =\infty.  \label{keith} 
\end{equation}
Integrating \eqref{mick} on $(t_{b_2},t)$ gives
\begin{equation}
u_{1} < \gamma + (t - t_{b_2})|u_{1}'(t_{b_{2}})| \leq C_{4}t \quad
 \text{ for some } C_{4}>0 \text{ for large } t. \label{ferry} 
\end{equation}

Next from (H2) we have
$$ 
\frac{f(u_{1})}{u_{1}^p} \geq 1-\epsilon \text{ for large } u_{1}.  
$$
Thus by \eqref{ferry},
\begin{equation} 
\frac{f(u_{1})}{u_{1}}  \geq \frac{(1-\epsilon)u_{1}^p}{u_{1}} 
=  \frac{(1-\epsilon)}{u_{1}^{1-p}} \geq
\frac{(1-\epsilon)}{C_4^{1-p}t^{1-p}}\quad \text{for large } t. \label{sirius} 
\end{equation}
Therefore by \eqref{ringo}, \eqref{mars}, and \eqref{sirius}:
$$  
\frac{t^2 h(t) f(u_{1})}{u_{1}} \geq \frac{k_{1}(1-\epsilon)}{C_4^{1-p}(N-2)^2} 
\frac{t^{2-q}}{t^{1-p}} = \frac{k_{1}(1-\epsilon)}{C_4^{1-p}(N-2)^2} t^{1+p-q} 
\to \infty, 
$$
since $1+p > q$.
This establishes \eqref{keith}.

Next we rewrite \eqref{DE4} as
\begin{equation} 
u_{1}'' + \frac{t^2h(t)f(u_{1})}{u_{1}} \frac{u_{1}}{t^2} = 0. \label{helium} 
\end{equation}
Now it follows from \eqref{keith} that we may choose $t_{0}$ sufficiently 
large so that
$$ 
\frac{t^2 h(t) f(u_{1})}{u_{1}} \geq A > \frac{1}{4} \quad\text{on } [t_{0}, \infty).
$$
Next let $y_{1}$ be the solution of
\begin{equation} 
y_{1}'' + A \frac{y_{1}}{t^2} = 0 \label{hydrogen}
\end{equation}
with $y_{1}(t_0)= u_{1}(t_{0})= \gamma$ and $y_{1}'(t_{0})= u_{1}'(t_{0}) >0$.
It follows then for some constants $d_{1}\neq 0$ and $d_{2}$ that
$$ 
y_{1} = d_{1}\sqrt{t}\Big( \sin\big(\ln\big(t\sqrt{A- \frac{1}{4}} \big) + d_{2}\big)\Big)
$$
and so clearly $y_{1}$ has an infinite number of local extrema on $[t_{0}, \infty)$.
Consider now the interval $[t_{0}, M]$ such that $y_{1}>0$, $y_{1}'>0$ on 
$[t_{0}, M]$ and $y_{1}'(M) =0$.
We claim now that $u_{1}'$ must get negative on $[t_{0}, M]$. 
So suppose not. Then $u_{1}'\geq 0$ on $[t_{0}, M]$.
Then multiplying \eqref{helium} by $y_{1}$, multiplying \eqref{hydrogen} 
by $u_{1}$, and subtracting we obtain
$$ 
(y_{1}u_{1}' - y_{1}'u_{1})' + \Big(\frac{t^2h(t)f(u_{1})}{u_{1}} -A\Big)
\frac{y_{1}u_{1} }{t^2} = 0. 
$$
Integrating this on $[t_{0}, M]$ gives
\begin{equation}  
y_{1}(M) u_{1}'(M) + \int_{t_{0}}^{M} \Big(\frac{t^2h(t)f(u_{1})}{u_{1}} -A\Big)
\frac{y_{1}u_{1} }{t^2}  \, dt = 0.\label{rome} 
\end{equation}
The integral term in \eqref{rome} is positive by \eqref{keith} and also 
$y_{1}(M) u_{1}'(M) \geq 0$ yielding a contradiction.
Therefore we see that $u_{1}$ must have a maximum, $M_{b_1}>0$, and
$u_{1}'>0$ on $[0, M_{b_{1}})$. This completes the proof.
\end{proof}

\begin{lemma} \label{lem6} 
Let $N>2$, $0<p<1$, and $N-p(N-2)< \alpha < 2(N-1)$. Assuming {\rm (H1)--(H5)}
 and that $u_{1}$ solves
\eqref{DE4}, \eqref{initial} then there exists $t_{b_{3}}> M_{b_1}$ such that
$u_{1}(t_{b_3}) =  \frac{\beta + \gamma}{2}$ and $u_{1}'<0$ on $(M_{b_1}, t_{b_3}]$.
\end{lemma}

\begin{proof} 
If $u_{1} \geq \frac{\beta + \gamma}{2}$ for all $t \geq M_{b_1}$, 
then $f(u_{1})>0$ for $t \geq M_{b}$.
Then from \eqref{DE4} it follows that
$ u_{1}'' < 0$ and thus $u_{1}'(t) \leq u_{1}'(t_{0}) < 0$ for $t > t_{0}> M_{b_1}$.
Integrating this inequality on $(t_{0}, t)$ gives
 $$ 
u_{1}(t) \leq u_{1}(t_{0}) + u_{1}'(t_{0})(t-t_{0}) \to -\infty \quad
\text{as } t \to \infty 
$$
which gives a contradiction since we assumed 
$u_{1} \geq  \frac{\beta + \gamma}{2}$ for all $t \geq M_{b_1}$.
Thus there exists $t_{b_3}> M_{b_{1}}$ such that
$u_{1}(t_{b_3}) =  \frac{\beta + \gamma}{2}$, $u_{1}> \frac{\beta + \gamma}{2}$, 
and $u_{1}'<0$ on $(M_{b_{1}}, t_{b_3}]$.
\end{proof}

\begin{lemma} \label{lem7} 
Let $N>2$, $0<p<1$, and $N-p(N-2)< \alpha < 2(N-1)$. Assuming {\rm (H1)--(H5)}
 and that $u_{1}$ solves
\eqref{DE4}, \eqref{initial} then there exists $z_{1,b_1}> M_{b_1}$ such that 
$u_{1}(z_{1, b_1}) =0$.
In fact, $u_{1}$ has an infinite number of zeros on $(0, \infty)$.
\end{lemma}

\begin{proof} 
Suppose now by the way of contradiction that $0< u_{1} < \gamma$ and thus  
$F(u_{1})< 0$  for $t > t_{b_{3}}$.
Then from \eqref{munson}-\eqref{polaris} we have
\begin{equation} 
\frac{1}{2} 	\frac{	u_{1}'^{2}}{h(t)} + F(u_{1}) 
\geq F(u_{1}(M_{b_1}))>0 \text{ for } t \geq M_{b_1}. \label{gauss} 
\end{equation}
  Therefore by \eqref{ringo} and \eqref{gauss} we have
\[
u_{1}'^2 \geq 2h(t)F(u_1(M_{b_1})) \geq \frac{2k_{1}F(u_1(M_{b_1}))}{(N-2)^2t^q}
\]
 for $t > t_{b_{3}}$. Thus:
\begin{equation}
-u_{1}' \geq C_{5}t^{-q/2}\quad \text{where }  
C_{5} = \frac{\sqrt{2k_{1}F(u_1(M_{b_1}))}}{N-2}>0 
\text{ for } t > t_{b_3}. \label{naples} 
\end{equation}
Integrating \eqref{naples} on $(t_{b_3},t)$ gives
$$
u_{1} \leq \frac{\beta + \gamma}{2} - C_{5}
\Big(  \frac{t^{1-\frac{q}{2}} - t_{b_{3}}^{1-\frac{q}{2}} }{1- \frac{q}{2}} \Big)
\to -\infty \quad \text{as } t \to \infty \text{ since } q<2.
$$
Thus $u_{1}$ gets negative contradicting that $u_{1}>0$ on $(0, \infty)$.
 Hence there exists $z_{1, b_1}> M_{b_1}$ such that
$u_{1}(z_{1,b_1})=0$ and $u_{1}' < 0$ on $(M_{b_{1}}, z_{1,b_{1}}]$.

In a similar way to Lemma \ref{lem5} we can show that $u_{1}$ has a negative local minimum,
 $m_{b_1}>z_{1,b_1}$, and similar to Lemma \ref{lem7}
we can show that $u_{1}$ has a second zero $z_{2, b_{1}} > m_{b_{1}}$. 
It then in fact follows that $u_{1}$ has an infinite number of zeros 
$z_{n, b_{1}}$. This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm1}]
  By continuous dependence on initial conditions it follows that $z_{1, b_{1}}$ 
is a continuous function of $b_{1}$. In addition, by Lemma \ref{lem4} it follows that
$t_{b_1} \to \infty$ as $b_{1} \to 0^{+}$ and since  $z_{1, b_{1}} > t_{b_{1}}$ 
it follows that $z_{1,b_{1}} \to \infty$ as $b_{1} \to 0^{+}$.

So now  let $k, n$ be nonnegative integers with $0 \leq k \leq n$. 
Choose $R>0$ sufficiently small so that
$ z_{1, b_{1}} < \cdots < z_{n,b_{1}} < R^{2-N}$. 
Then by the intermediate value theorem there exists a smallest value of 
$b_{1}>0$, say $b_{1,k}$,  such that $z_{k,b_{1,k}} =R^{2-N}$. 
Then $u_{1}(t, b_{1,k})$ is a solution of \eqref{DE4} and  \eqref{initial} such that
$u_{1}(t, b_{1,k})$ has $k$ zeros on $(0, R^{2-N})$.

Finally  defining
$$
 U_{k}(r) = (-1)^k u_{1}(r^{2-N}, b_{1,k})
$$ 
we see that $U_{k}$ solves \eqref{DE}, $U_{k}$ has $k$ zeros on $(R, \infty)$, and
$\lim_{r \to \infty} U_{k}(r) = 0.  $ This completes the proof.
\end{proof}

\noindent\textbf{Note:} A crucial step in proving Theorem \ref{thm1} is 
Lemma \ref{lem5}  which says that if $N - p(N-2) < \alpha < 2(N-1)$ then every solution 
of \eqref{DE4}, \eqref{initial} must have a local maximum. We conjecture 
that a similar lemma does not hold for $2< \alpha< N-p(N-2)$ because for 
an appropriate constant $c>0$ the function $ c t^{\frac{\alpha-2}{(N-2)(1-p)}} $
 is a monotonically increasing solution of the  model equation
 $$ 
u'' + \frac{1}{t^q} u^p = 0
$$ 
with  $q = \frac{2(N-1)-\alpha}{N-2}$ and $0<p<1$.

\begin{thebibliography}{00}

\bibitem{BL} H. Berestycki, P. L. Lions;
Non-linear scalar field equations I, {\it Arch. Rational Mech. Anal.}, 
Volume 82, 313-347, 1983.

\bibitem{BL2} H. Berestycki, P.L. Lions;
Non-linear scalar field equations II,
 {\it Arch. Rational Mech. Anal.}, Volume 82, 347-375, 1983.

\bibitem{B} M. Berger;
 {\it Nonlinearity and functional analysis,} Academic Free Press, New York, 1977.

\bibitem{BR} G. Birkhoff, G. C. Rota;
{\it Ordinary differential equations}, Ginn and Company, 1962.

\bibitem{C} A. Castro, L. Sankar, R. Shivaji;
Uniqueness of nonnegative solutions for semipositone problems on exterior domains,
 {\it Journal of Mathematical Analysis and Applications}, Volume 394, 
Issue 1, 432-437, 2012.

\bibitem{I4} J. Iaia;
Existence and nonexistence for semilinear equations on exterior domains,
 submitted to the {\it Journal of Partial Differential Equations}, 2016.

\bibitem{I5} J. Iaia;
Existence for semilinear equations on exterior domains,
  {\it Electronic Journal of the Qualitative Theory of Differential Equations}, 
No. 108, 1-12, 2016.

\bibitem{I6} J. Iaia; 
Existence of solutions for semilinear problems on exterior domains,
 submitted to {\it Journal of Mathematical Analysis and Applications}, 2016.

\bibitem{I7} J. Iaia; 
Existence of solutions for semilinear problems with prescribed number of 
zeros on exterior domains, to appear in {\it Journal of Mathematical Analysis and Applications},  446, 591-604, 2017.

\bibitem{JK} C. K. R. T. Jones, T. Kupper;
On the infinitely many solutions of a semi-linear equation,
 {\it SIAM J. Math. Anal.}, Volume 17, 803-835, 1986.

\bibitem{C2} E. Lee, L. Sankar, R. Shivaji;
Positive solutions for infinite semipositone problems on exterior domains,
 {\it Differential and Integral Equations}, Volume 24, Number 9/10, 861-875, 2011.

\bibitem{M} K. McLeod, W.C.  Troy, F.B. Weissler;
Radial solutions of $\Delta u + f(u) = 0$ with prescribed numbers of zeros;
 {\it Journal of Differential Equations}, Volume 83, Issue 2, 368-373, 1990.

\bibitem{S} L. Sankar, S. Sasi, R. Shivaji;
Semipositone problems with falling zeros on exterior domains,
 {\it Journal of Mathematical Analysis and Applications}, Volume 401,
 Issue 1, 146-153, 2013.

\bibitem{ST} W. Strauss;
Existence of solitary waves in higher dimensions,
 {\it Comm. Math. Phys.}, Volume 55, 149-162, 1977.

\end{thebibliography}








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