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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 181, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/181\hfil Sublinear equations on exterior domains]
{Existence of solutions for sublinear equations \\ on exterior domains}

\author[J. A. Iaia \hfil EJDE-2018/181\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}

\thanks{Submitted  May 10, 2018. Published November 6, 2018}
\subjclass[2010]{34B40, 35B05}
\keywords{Exterior domains; semilinear; sublinear; radial solution}

\begin{abstract}
 In this article we consider the 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 $\partial B_{R}$
 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
 $\frac{(N+2)-p(N-2)}{2} \leq \alpha < N-p(N-2)$ for large $r$.
 We prove existence of $n$ solutions - one with exactly $n$ zeros on
 $[R, \infty)$ -  if $R>0$ is sufficiently small.
 If $R>0$ is sufficiently large then there are no solutions
 with $\lim_{r \to \infty} u(r)=0$.
\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} 
\end{itemize}
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[(H2)] $-F_0 <F < 0$  on  $(0, \gamma)$, $F>0$  on
$(\gamma, \infty)$. %  \label{F2} 
\end{itemize}
We also assume:
\begin{itemize}
\item[(H3)] 
 There exists $p$ with $0<p<1$ such that  $f(u) = |u|^{p-1}u + g(u)$
 where $\lim_{u \to \infty} \frac{|g(u)|}{|u|^p} = 0$
 and $g(u) \geq 0$ for $u \geq \gamma$.  % \label{f1}
\end{itemize}
In addition, we assume $K$ is differentiable on $(R, \infty)$ and that 
there exist constants $k_2>k_1>0$, and $\alpha$ with
$\frac{(N+2)-p(N-2)}{2}\leq \alpha < N-p(N-2) $ such that
\begin{itemize}
\item[(H4)] $k_1 r^{-\alpha} < K(r)  < k_2 r^{-\alpha}$  on $[R, \infty)$,
% \label{K}  
and:
\item[(H5)]
 $\lim_{r \to \infty} \frac{rK'}{K} = -\alpha$  for 
$\frac{(N+2)-p(N-2)}{2}\leq \alpha <N-p(N-2)$,  and 
$\frac{rK'}{K} >- 2(N-1)$  on $[R, \infty)$.   %\label{K2} 
\end{itemize}
Note that (H5) implies $r^{2(N-1)}K(r)$ is increasing.


Interest in the topic for this article 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.\
 $\lim_{u \to \infty} f(u)/u= \infty$, $\Omega = {\mathbb R}^N$, and
 $K(r)\equiv 1$ then the problem \eqref{1}, \eqref{3} has been extensively studied
\cite{BL,BL2,B,JK, M, ST}.

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 $\frac{(N+2)-p(N-2)}{2}\leq \alpha < N-p(N-2)$.
 In earlier papers \cite{I8,I9,JJ} the cases where $f$ is sublinear with 
 $0 < \alpha < 2$,  $  N - p(N-2) < \alpha < 2(N-1)$, and $\alpha > 2(N-1)$ 
were investigated.

In \cite{S} existence of  a positive solution if $R>0$ is sufficiently 
small was proved using sub and super solutions when $f$ is semipositone.
 Here we prove the existence of $n$ solutions if $R>0$ is sufficiently small.

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 
 $(R, \infty)$  where } R > 0,  \label{DE} \\
 u(R) = 0, \quad u'(R) = b \in {\mathbb R}. \label{DE2}
\end{gather}

In this article we prove the following result.

\begin{theorem} \label{mainthm}
Assume {\rm(H1)--(H5)} hold. Then given a nonnegative integer  $n$ there exists 
a solution  of \eqref{DE}-\eqref{DE2} with $n$ zeros on 
$(R, \infty)$ and $\lim_{r \to \infty} u(r) = 0$
if $R>0$ is sufficiently small.
\end{theorem}

We note that if $ 0 < \alpha < 2$ then it was shown in \cite{I8} that 
there are solutions for \emph{all} $R>0$ whereas when $ \alpha> 2$ and $R$
is sufficiently large then it was shown in \cite{I9} that there 
are \emph{no} solutions of \eqref{1}-\eqref{3} with $\lim_{r \to \infty} u(r)=0$.
In this paper we prove existence of solutions of \eqref{1}-\eqref{3} for 
$  \frac{(N+2)-p(N-2)}{2} \leq \alpha < N - p(N-2)$ and $R>0$ sufficiently small.
A similar result was proved in \cite{I9}, \cite{JJ} for
 $N-p(N-2) < \alpha < 2(N-1)$ and for $\alpha > 2(N-1)$.

\section{Behavior for small $b>0$}

We suppose that $U(r)$ solves \eqref{DE}
and then make the change of variables:
$$ 
U(r) = u(r^{2-N}).  
$$
Then for $0<t< \infty$ we see that $u$ satisfies
\begin{equation} 
 u'' + h(t)f(u) = 0,  \label{DE3}  
\end{equation}
where
 $$
h(t) = \frac{t^{\frac{2(N-1)}{2-N}}K(t^{\frac{1}{2-N}})}{(N-2)^2}. 
$$  
It follows from (H4), (H5) that
\begin{equation} 
h(t)>0, \quad  h'(t)<0,  \quad  h_1 t^{-q} < h(t) < h_2 t^{-q} \label{h eqn} 
\end{equation}
 for $t>0$  where $q = \frac{2(N-1)-\alpha}{N-2}$,
$h_i = \frac{k_i}{(N-2)^2}$. 
In addition it follows from (H3), (H5),  and \eqref{h eqn} that
\begin{equation} 
1< p+1<q<2. \label{ineq} 
\end{equation}
We also assume that
 \begin{equation}  
u(0)=0, \quad u'(0)=b>0.  \label{DE3IC} 
\end{equation}

For the rest of this article we will focus on finding solutions of 
 \eqref{DE3}, \eqref{DE3IC} such that $u(R^{2-N})=0$. If such a solution exists then
$U(r)= u(r^{2-N})$ will be a solution of \eqref{1}-\eqref{3}.

We first examine  \eqref{DE3}, \eqref{DE3IC}  assuming (H1)--(H5). 
It is straightforward to show that there is a unique solution
$u \in C^{1}([0, \epsilon))\cap C^{2}(0,\epsilon) $ of 
\eqref{DE3} for some $\epsilon >0$ such that  \eqref{DE3IC} holds.
 A proof is provided in the appendix.
Then from (H3) since $|f(u)| \leq C_1 |u|$ for some constant 
$C_1>0$ it follows that the solution  of \eqref{DE3}, \eqref{DE3IC} 
 exists on $(0, \infty)$.

Since $u>0$ for small positive $t$  (by \eqref{DE3IC}) and since $f(u) < 0$ 
for $0 < u < \beta$  by (H1) it follows from \eqref{DE3} that $u''> 0$ when 
$0<u<\beta$. Therefore $u' > u'(0)=b$  when $0<u< \beta$.  
Integrating this on $(0,t)$  and using \eqref{DE3IC} we obtain
\begin{equation}
 u> bt  \quad  \text{when  } 0< u <  \beta. \label{kubek}
 \end{equation}
It then follows that there is a smallest positive value of $t, t_{b}$, 
such that $u(t_{b}) = \beta$ and $0<u<\beta$ on $(0, t_{b})$.

We next show that $u$ gets larger than $\gamma$. So suppose not. 
That is suppose $0 < u < \gamma$ for all $t> 0$ and consider
\begin{equation} 
 E =\frac{1}{2} \frac{u'^2}{h(t)} + F(u)  \quad \text{for } t> 0.   
 \label{energy 0} 
\end{equation}
Then from \eqref{DE3} and \eqref{h eqn} we have
\begin{equation} 
E'= \Big( \frac{1}{2} \frac{u'^2}{h(t)} + F(u)\Big)' 
= -\frac{h' u'^2}{h^2}\geq 0   \label{energy}
 \end{equation}  
and so by the initial conditions \eqref{DE3IC} we see that
\begin{equation}  
E =\frac{1}{2} \frac{u'^2}{h(t)} + F(u)   > 0  \quad \text{for } t > 0. 
 \label{energy 1} 
\end{equation}

Now while $0< u <  \gamma$ we have $F(u) < 0$
and since $F(\gamma)=0$ with $F'(\gamma)=f(\gamma)>0$ 
(since $\gamma > \beta$) then we find that $-F(u) > a_0(\gamma-u) $
for $\beta <  u < \gamma$ for some $a_0>0$.
Using this in  \eqref{energy 1}  we obtain
\begin{equation} 
u'^2 \geq -2hF(u) > 2a_0 h (\gamma -u)  \quad\text{for } t > t_{b}.  \label{ineq2}
\end{equation}
Then by \eqref{h eqn} and \eqref{ineq2} we have
\begin{equation}  
\frac{u' }{ \sqrt{\gamma -u} } > \sqrt{2a_0h_1} t^{-\frac{q}{2}}  \quad
\text{for }  t >  t_{b}. \label{joe}  
 \end{equation}
Integrating this on $(t_{b},t)$  gives
\begin{equation} 
\sqrt{\gamma -u(t)} < \sqrt{\gamma - \beta} - \sqrt{2a_0h_1}
\Big(\frac{t^{1-\frac{q}{2}} - t_{b}^{1-\frac{q}{2}}}{1-\frac{q}{2}}\Big)
\quad \text{for }  t>t_{b}. \label{phil}  
\end{equation}
Since $1<q<2$ it follows that the right-hand side of  \eqref{phil}  goes to 
$-\infty$ as $t \to \infty$ contradicting that $0< u < \gamma$. 
Thus $u$ eventually gets larger than $\gamma$ and so there exists 
$t_{b,\gamma}> t_{b}$ such that $u(t_{b,\gamma})= \gamma$
and $0<u<\gamma$ on $(0, t_{b,\gamma})$.

Next we denote $t_{b,2}$ as the smallest positive value of $t$ 
such that $u(t_{b,2}) = \frac{\beta}{2}$. 
Note that $t_{b,2} < t_{b} < t_{b, \gamma}$.

\begin{lemma} \label{lem2.1} 
Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}. 
Then $\lim_{b \to 0^{+}} t_{b,2} = \infty$.
\end{lemma}

\begin{proof}
 From the above arguments we know $u>0$ and $u'>0$ on $(0, t_{b,\gamma})$. 
Also it follows from (H1) that
\begin{equation} 
f(u) \geq -a_1u \quad \text{for $u>0$  for some } a_1>0 \label{f lower}
 \end{equation} so that integrating \eqref{DE3} twice on $(0,t)$
gives
\begin{equation} 
0< u \leq  bt + a_1\int_0^{t} \int_0^{s} h(x) u(x) \, dx \, ds
 \leq     bt + a_1t \int_0^{t} h(x) u(x) \, dx. \label{IE}
\end{equation}
Now let
\begin{equation} 
y =  b +  a_1\int_0^{t} h(x) u(x) \, dx.  \label{yogi}
\end{equation}
Then from \eqref{IE}-\eqref{yogi} we have
 \begin{equation}
0<  u \leq ty \quad\text{and}\quad   y' =a_1 hu  \label{berra}  
\end{equation} 
and from \eqref{h eqn},  \eqref{berra} we obtain
\begin{equation} 
 y' = a_1 hu \leq a_1 th y \leq  a_1h_2t^{1-q} y. \label{IE2}
 \end{equation}
Dividing \eqref{IE2} by $y$, integrating on $(0,t)$, exponentiating, 
and recalling that $q<2$ gives
\begin{equation} 
u \leq b t \exp\Big({\frac{a_1 h_2 t^{2-q}}{2-q}}\Big) \quad \text{for } t > 0 
\label{mercer} 
\end{equation} 
so that evaluating \eqref{mercer} at $t_{b,2}$ gives
\begin{equation} 
\frac{\beta}{2} \leq   b t_{b,2}  \exp
\Big({\frac{ a_1 h_2 t_{b,2}^{2-q}}{2-q}}\Big). \label{rhizutto} 
\end{equation}
Since $q<2$ it then follows from \eqref{rhizutto} that $ t_{b,2} \to \infty$ 
as $b \to 0^{+}$. This completes the proof. 
\end{proof}

Now let $ c = \frac{2-q }{1-p}$ and note from \eqref{ineq} and (H5) 
that $\frac{1}{2}\leq c <1$.
We also note that $y = c_0 t^{c}$  with $c_0 = c(1-c)^{-\frac{1}{1-p}}>0$
is a solution of  $y'' + \frac{1}{t^q}y^p=0 $ for $t>0$.


\begin{lemma} \label{lem2.2} 
Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}. 
Then 
\[
\lim_{b \to 0^{+}}  t_{b,2}^{1-c} u'(t_{b,2}) = 0.
\]
\end{lemma}

\begin{proof}
 Using \eqref{h eqn} and \eqref{f lower} in \eqref{DE3} gives
\begin{equation} 
 u'' - \frac{ a_1 h_2}{t^{q}}u \leq 0.   \label{star}  
\end{equation}
Letting
\begin{equation} 
u(t) = \sqrt{t} \, w\Big( \frac{2 \sqrt{a_1h_2 } }{2-q} t^\frac{2-q}{2} \Big) 
\label{smile}  
\end{equation} 
then using \eqref{DE3IC} and   \eqref{star} it follows that  $w$ satisfies
\begin{equation} 
 w'' + \frac{1}{t} w' - \Big( 1 + \frac{1}{(2-q)^2 t^2}\Big) w \leq 0 , \quad
\lim_{t \to 0^{+}} \frac{w(t)}{t^{\frac{1}{2-q}}} 
=  \Big(\frac{2-q}{2 \sqrt{a_1h_2 } }\Big)^{\frac{1}{2-q}} b.   \label{antigua} 
 \end{equation}
Now let $w_1$ satisfy
\begin{equation} 
w_1'' + \frac{1}{t} w_1' - \Big( 1 + \frac{1}{(2-q)^2 t^2}\Big) w_1 =0 , \quad
\lim_{t \to 0^{+}} \frac{w_1(t)}{t^{\frac{1}{2-q}}} 
= \Big(\frac{2-q}{2 \sqrt{a_1h_2 } }\Big)^{\frac{1}{2-q}}  b. \label{barbados} 
 \end{equation}
That is, $w_1$ satisfies the modified Bessel equation and in fact
\begin{equation} 
w_1(t) = \Big(\frac{2-q}{2 \sqrt{a_1 h_2} }\Big)^{\frac{1}{2-q}} 
\frac{ b}{a} I_{\frac{1}{2-q}}(t)    \label{st kitts}
 \end{equation}
where $I_{\frac{1}{2-q}}(t)$ is the modified Bessel function of order
 $\frac{1}{2-q}$ that is finite at the origin.
It is well-known \cite{BR} that
\begin{equation}
\begin{gathered}
I_{\frac{1}{2-q}}(t)>0, \quad I'_{\frac{1}{2-q}}(t)>0 \quad 
\text{for } t>0, \lim_{t \to \infty}  I_{\frac{1}{2-q}}(t)= \infty,\\
\text{there exists  $a >0$  such that }  \lim_{t \to 0^{+}}
\frac{I_{\frac{1}{2-q}}(t)}{t^{\frac{1}{2-q}}} = a.
\end{gathered}  \label{bessel}   
\end{equation}
It is also true that
\begin{equation} 
\Big| \frac{I'_{\frac{1}{2-q}}(t)}{I_{\frac{1}{2-q}}(t)} \Big| 
\leq a_2  \quad \text{for $t>1$ for some } a_2>0. \label{bessel bound}  
\end{equation}
A proof of \eqref{bessel bound} is provided in the appendix.

It then follows from \eqref{st kitts}, \eqref{bessel bound} that
\begin{equation} 
\frac{w_1'}{w_1}=\frac{I_{\frac{1}{2-q}}'}{I_{\frac{1}{2-q}}} 
\leq a_2  \quad \text{for } t>1.   \label{bound}  
\end{equation}

Next multiplying \eqref{antigua} by $tw_1$, \eqref{barbados} by $tw$, and 
subtracting gives
\begin{equation} 
[t(w'w_1 - w_1'w)]' \leq 0  \quad \text{for }  t > 0. \label{vic} 
 \end{equation}
Since
 $$
\lim_{t \to 0^{+}}\frac{w(t)}{t^{\frac{1}{2-q}}} 
=\lim_{t \to 0^{+}} \frac{w_1(t)}{t^{\frac{1}{2-q}}} 
= \Big(\frac{2-q}{2 \sqrt{a_1h_2 } }\Big)^{\frac{1}{2-q}} b
$$
and
$$
\lim_{t \to 0^{+}}\frac{w'(t)}{t^{\frac{q-1}{2-q}}} 
=\lim_{t \to 0^{+}}\frac{w_1'(t)}{t^{\frac{q-1}{2-q}}}
= \frac{1}{2-q}\Big(\frac{2-q}{2 \sqrt{a_1h_2 } }\Big)^{\frac{1}{2-q}} b, 
$$
then  integrating \eqref{vic} on $(0,t)$ gives
\begin{equation} 
w'w_1 - ww_1' \leq 0 \quad \text{for } t> 0. \label{ringo} 
\end{equation}

Since $u>0$ on $(0, t_{b,2})$ it follows from \eqref{smile} that $w>0$ on  
$(0, t_{b,2})$.
It also follows that  $w_1>0$  since $I_{\frac{1}{2-q}} >0$ by \eqref{bessel}.  
Now dividing \eqref{ringo}  by $w w_1, $ using \eqref{bound}, and recalling 
$t_{b,2} \to \infty$ as $b \to 0^{+}$  from Lemma \ref{lem2.1} gives
\begin{equation}  
\frac{w'}{w} \leq  \frac{w_1'}{w_1} \leq a_2 \quad
\text{ for $t > t_{b,2}$  if $b>0$  is sufficiently small}. \label{guad}  
\end{equation}

Next we return to \eqref{smile} and differentiate to obtain
\begin{equation}  
u' = \frac{t^{-1}u}{2} +\sqrt{a_1 h_2} t^{-\frac{q}{2}}u \frac{w'}{w}.  
\label{costa rica}  
\end{equation}

Using \eqref{guad}-\eqref{costa rica}, Lemma \ref{lem2.1}, and assuming $b>0$ is 
sufficiently small we obtain
\begin{equation} 
t^{1-c}u' 
= \frac{t^{-c}u}{2} +\sqrt{ a_1 h_2} \,  t^{1-c-\frac{q}{2}}u \frac{w'}{w}
\leq \frac{t^{-c}u}{2} + a_2 \sqrt{ a_1 h_2}  t^{1-c-\frac{q}{2}}u 
  \label{pinky}
 \end{equation}
for $t>t_{b,2}$.
Since $1-c-\frac{q}{2} = - \frac{(2-q)(1+p)}{2(1-p)}<0$ and 
$\frac{1}{2} \leq c<1$, evaluating \eqref{pinky} at $t_{b, 2}$  along with 
Lemma \ref{lem2.1} gives
$$ 
0< t_{b,2}^{1-c}u'(t_{b,2})   \leq \frac{\beta t_{b,2}^{-c}}{4} 
+ \frac{\beta a_2}{2}\sqrt{ a_1 h_2} \, t_{b,2}^{1-c-\frac{q}{2}} \to 0
\quad \text{as } b \to 0^{+}.
$$ 
This completes the proof.
\end{proof}

\begin{lemma} \label{lem2.3} 
Assume {\rm (H1)-(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}.  
Then 
\[
\lim_{b \to 0^{+}}  t_{b,\gamma}^{1-c} u'(t_{b,\gamma}) = 0.
\]
\end{lemma}

\begin{proof}
Rewriting \eqref{DE3} we see that
\begin{equation}  
(t^{1-c}u')' = (1-c)t^{-c}u'  - t^{1-c}h(t)f(u).  \label{george}  
\end{equation}
Integrating this on $(t_{b,2}, t_{b,\gamma})$ gives
\begin{equation}
 t_{b,\gamma}^{1-c} u'(t_{b,\gamma}) 
= t_{b,2}^{1-c} u'(t_{b,2})  + \int_{t_{b,2}}^{t_{b,\gamma}}  (1-c)s^{-c}u' \, ds  -
\int_{t_{b,2}}^{t_{b,\gamma}}  s^{1-c}h(s)f(u) \, ds. \label{brain}  
\end{equation}
The first term on the right-hand side of \eqref{brain} goes to $0$ as 
$b \to 0^{+}$ by Lemma \ref{lem2.2}.
Integrating the first integral in \eqref{brain} by parts gives
\begin{equation}  
\int_{t_{b,2}}^{t_{b,\gamma}}  (1-c)s^{-c}u' \, ds 
= (1-c)t_{b,\gamma}^{-c}\gamma - (1-c) t_{b,2}^{-c}\frac{\beta}{2}    
+ \int_{t_{b,2}}^{t_{b,\gamma}}   c(1-c) s^{-c-1} u \, ds.  \label{rashi}  
\end{equation}
The first two terms on the right-hand side of this go to $0$ as $b \to 0^{+}$ 
by Lemma \ref{lem2.1}. Since $\frac{\beta}{2} < u < \gamma$ on $(t_{b,2}, t_{b, \gamma})$ 
estimating the third term on the right-hand side of \eqref{rashi} and using 
Lemma \ref{lem2.1} we obtain
\begin{equation}  
\big|\int_{t_{b,2}}^{t_{b,\gamma}}   c(1-c) s^{-c-1} u \, ds\big| 
\leq (1-c)\gamma (t_{b,2}^{-c} - t_{b,\gamma}^{-c}) \to 0 \quad\text{as } 
b \to 0^{+}. \label{blind willie} 
 \end{equation}
Finally, on $[\beta/2, \gamma]$ we have $|f| \leq a_3$ for some constant
 $a_3>0$ and thus by \eqref{h eqn},
\[
\big| \int_{t_{b,2}}^{t_{b,\gamma}}  s^{1-c}h(s)f(u) \, ds \big| 
\leq \frac{a_3h_2( t_{b,\gamma}^{2-q-c} - t_{b,2}^{2-q-c})}{2-q-c} \to 0
\quad\text{as } b \to 0^{+} 
\]
  since $2-q-c = -\frac{p(2-q)}{1-p}<0$.
Thus it follows from Lemma \ref{lem2.2} and \eqref{rashi}-\eqref{blind willie} that 
the right-hand side of  \eqref{brain} goes to 0 as $b \to 0^{+}$
and so the lemma follows.  This completes the proof. 
\end{proof}

\begin{lemma} \label{lem2.4} 
Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}.  
Let $u = t^c v$. Then $\lim_{b \to 0^{+}}  t_{b,\gamma} v'(t_{b,\gamma}) = 0$.
\end{lemma}

\begin{proof}
 We make the change of variables
$u= t^{c} v$ and after differentiating we obtain
\begin{equation}  
t^{1-c} u' = cv + tv'.  \label{paul}  
\end{equation}
Evaluating \eqref{paul} at $t_{b,\gamma}$ we see by Lemma \ref{lem2.3} that the 
left-hand side of \eqref{paul} goes to 0 as $b \to 0^{+}$.
Also by Lemma \ref{lem2.1} since $t_{b,\gamma} > t_{b,2} \to \infty$ as $b \to 0^{+}$ and
since $v(t_{b,\gamma}) = \frac{u(t_{b,\gamma})}{t_{b,\gamma}^c} 
= \frac{\gamma}{t_{b,\gamma}^c} \to 0$ as $b \to 0^{+}$ we see that the
 first term on the right-hand side of \eqref{paul}  goes to 0 as 
$b \to 0^{+}$ when evaluated at $t_{b, \gamma}$ and thus  it follows that
$$ 
t_{b,\gamma} v'(t_{b,\gamma}) \to 0 \text{  as } b \to 0^{+}.
$$
This completes the proof.  
\end{proof}


Next substituting $u=t^cv$ into \eqref{DE3} and using (H3) gives
 \begin{equation}  
t^2v'' + 2ctv' + c(c-1)v + t^q h(t) v^{p} +  t^{2-c}h(t)  g(t^c v)   =0 
\quad  \text{for }  t > t_{b,\gamma}.   \label{DE3v}
 \end{equation}
Now for $t > t_{b, \gamma}$ we have $u>\gamma>0$ (and hence $v>0$) and then  
by (H3) we have $g(t^{c}v)=g(u)\geq 0$. Then by \eqref{h eqn}  we have
\begin{equation}  
t^2v'' + 2ctv' + d(v) < 0  \quad \text{for }  t > t_{b,\gamma}   \label{DE3v2} 
\end{equation}
where $d(v)=  c(c-1)v + h_1 |v|^{p-1}v$.
Next we let $v_1$ be the solution of
\begin{gather}  
t^2v_1'' + 2ctv_1' + d(v_1)     =0,  \label{DE3v_1} \\
v_1(t_{b,\gamma}) =  v(t_{b,\gamma}),  \quad v_1'(t_{b,\gamma}) =  v'(t_{b,\gamma}). 
 \label{maris} 
 \end{gather}

Now let
\begin{equation}  
E_1 = \frac{1}{2} t^2 v_1'^2 + D(v_1) \label{brian}  
\end{equation}
where $D(v_1) \equiv \int_0^{v_1} d_1(s) \, ds
= \frac{c(c-1)}{2} v_1^2 + \frac{h_1}{p+1}|v_1|^{p+1}$.
It follows from \eqref{DE3v_1} that
\begin{equation} 
E_1'= (1-2c)tv_1'^2 \leq 0 \quad \text{since } \frac{1}{2} \leq c < 1.  
 \label{john}  
\end{equation}

Note $d(0)=0$ and that $d(v_1)$ is  increasing near $v_1=0$ since $0<p<1$. 
Also notice $d(v_1)$ has a unique positive zero at $\alpha_1>0$ and $d(v_1) < 0$ 
for $v > \alpha_1$.
Similarly $D(0)=0$ and $D(v_1)$ is increasing near $v_1=0$. 
In addition, $D$ has a local maximum at $\alpha_1$ and a unique positive 
zero at $\alpha_2 > \alpha_1 >0$. Also $D(v_1) <0$ for $v_1 > \alpha_2$.

Next it follows from Lemma \ref{lem2.4}  and since 
$v_1(t_{b,\gamma}) = v(t_{b,\gamma}) = \frac{\gamma}{t_{b,\gamma}^{c}} \to 0$ 
as $b \to 0^{+}$ (by Lemma \ref{lem2.1}) that
\begin{equation} 
E_1(t_{b,\gamma}) = \frac{1}{2} t_{b,\gamma}^2 v_1'^2(t_{b,\gamma}) 
+ D(v_1(t_{b,\gamma})) = \frac{1}{2} t_{b,\gamma}^2 v'^2(t_{b,\gamma}) 
+ D(v(t_{b,\gamma})) \to 0   \label{copa}  
\end{equation}
 as  $b \to 0^{+}$, 
and since $E_1$ is non-increasing for $\frac{1}{2} \leq c < 1$ (by \eqref{john}) 
it follows that
\begin{equation}
\frac{1}{2} t^2 v_1'^2 + D(v_1) = E_1(t) 
\leq  E_1(t_{b,\gamma}) <\epsilon \label{series of dreams} 
\end{equation}
 for $t > t_{b,\gamma}$  and $b$ sufficiently small. 
 
\begin{lemma} \label{lem2.5} 
Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}.  
If $b$ is sufficiently small then $v_1$ has a local maximum, $M_{1,b}$, 
and $v_1'> 0 $ on $(t_{b, \gamma}, M_{1,b})$.
\end{lemma}

\begin{proof} 
We suppose by way of contradiction that  $v_1'>0$ for all $t>t_{b,\gamma}$. 
Then $v_1(t) > v_1(t_{b,\gamma})= \frac{\gamma}{t_{b, \gamma}^c}>0$
and by \eqref{series of dreams} if $b$ is sufficiently small then 
$D(v_1)\leq E_1< \epsilon$ if $t>t_{b,\gamma}$ and so it follows that
$0<v_1< \frac{1}{2} \alpha_1$  if $b$ is sufficiently small.
In particular,  $v_1$ is bounded and since $v_1$ is increasing then 
$v_1\to L$ as $t \to \infty$ where $0<L< \alpha_1$.
Since  $D(v_1) \to D(L)$  as $t \to \infty$ and since $E_1'\leq 0$ 
it follows that $E_1$ has a finite limit as $ t \to \infty$. 
Thus from \eqref{brian} we have $t^{2}v_1'^{2} \to A \geq 0$ for some $A \geq 0$.  
If $A>0$ then $v_1' > \frac{A}{2t}$ for large $t$ implying 
$v_1(t) > v_1(t_0) + A \ln(t/t_0) \to \infty$ as
$t \to \infty$ contradicting that $v_1$ is bounded.
It follows therefore that $A=0$ and thus $tv_1' \to 0$ as $t \to \infty$. 
Using this and taking limits in \eqref{DE3v_1} we see that
$$ 
\lim_{t \to \infty} t^{2}v_1'' = - d(L).
$$ 
If $d(L)\neq 0$ then using a similar argument as we just showed with 
$v_1$ would imply that $v_1'$ is unbounded contradicting $tv_1' \to 0$
so it must be that $d(L)=0$ but this  is impossible since  
$0<L< \alpha_1$. Therefore from this contradiction we see that $v_1$ 
has a first local maximum, $M_{1,b}$, and  $v_1'>0$ on $(t_{b,\gamma}, M_{1,b})$   
if $b>0$ is sufficiently small. This completes the proof. 
\end{proof}

\begin{lemma} \label{lem2.6} 
 Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}.  
Then $v$ has a local maximum on $(t_{b,\gamma}, M_{1,b})$  if $b$ 
is sufficiently small.
\end{lemma}

\begin{proof}
 We assume $v'> 0$ on $(t_{b, \gamma}, M_{1,b})$  otherwise we are done. 
Since $v'>0$ it follows that $v> v(t_{b,\gamma})=\gamma>0$ on 
$(t_{b, \gamma}, M_{1,b})$. Multiplying \eqref{DE3v2} by $v_1$,  \eqref{DE3v_1}
 by $v$, and subtracting gives
$$ 
t^{2}(v_1 v' - v v_1')' + 2ct(v_1 v' - v v_1') + h_1(v^p v_1 - v_1^p v) \leq 0. 
$$
Multiplying this by $t^{2c-2}$ gives
\begin{equation}  
\left(t^{2c}(v_1 v' - v v_1')\right)' + h_1 t^{2c-2} v v_1 (v^{p-1} - v_1^{p-1}) 
\leq 0 \quad \text{for } t> t_{b,\gamma}.  \label{v_1} 
\end{equation}

Notice it follows from \eqref{DE3v2}--\eqref{maris} that
 $$  
t_{b, \gamma}^2 (v''(t_{b, \gamma})-v_1''(t_{b, \gamma})) < 0
$$ 
and so $v < v_1$ on $(t_{b, \gamma}, t_{b, \gamma}+ \epsilon_0)$
for some $\epsilon_0 > 0$.
We next show that $v$ and $v_1$ do not intersect on $(t_{b,\gamma}, M_{1,b})$.
Suppose they did and so there is a $t_0$ with $t_{b,\gamma} < t_0 <M_{1,b}$
such that $ v< v_1$ on $(t_{b, \gamma}, t_0)$ and
$v(t_0) = v_1(t_0)$.  Integrating \eqref{v_1}  on $(t_{b,\gamma}, t_0)$
and using \eqref{maris} gives
\begin{equation} 
t_0^{2c} \left(v_1(t_0) v'(t_0)  - v(t_0)v_1'(t_0)\right)  
+ h_1  \int_{t_{b, \gamma}}^{t_0} t^{2c-2} v v_1 (v^{p-1} - v_1^{p-1})  \, dt  
\leq 0. \label{cayman}  
\end{equation}
Since $v<v_1$ on $(t_{b,\gamma}, t_0)$ and $0<p<1$ it follows that 
$v^{p-1} > v_1^{p-1}$ on $(t_{b,\gamma}, t_0)$ and so the integral in 
\eqref{cayman} is positive.
So from \eqref{cayman} we see 
$(v_1(t_0) v'(t_0)  - v(t_0)v_1'(t_0)) = v(t_0)(   v'(t_0)  - v_1'(t_0)) < 0$ 
and since $v(t_0)>0$ we see that
\begin{equation}     
v'(t_0) < v_1'(t_0). \label{TR} 
\end{equation}
On the other hand, since $v<v_1$ on $(t_{b,\gamma}, t_0)$ and $v(t_0)=v_1(t_0)$ 
then $v(t) - v(t_0) < v_1(t) - v_1(t_0)$ and so
$\frac{v(t) - v(t_0)}{t-t_0} > \frac{v_1(t) - v_1(t_0)}{t-t_0} $ for 
$t_{b,\gamma}<t<t_0$. Therefore taking limits as $t \to t_0^{-}$ gives
\begin{equation} 
v'(t_0) \geq v_1'(t_0)    \label{RMN} 
\end{equation}
which contradicts \eqref{TR}.
Thus we see that $v<v_1$ on $(t_{b,\gamma}, M_{1,b})$.
Now we integrate \eqref{v_1} on $(t_{b,\gamma}, M_{1,b})$ and obtain
\begin{equation} 
 M_{1,b}^{2c} v_1(M_{1,b}) v'(M_{1,b}) + h_1 \int_{t_{b,\gamma}}^{M_{1,b}} 
t^{2c-2} v v_1 (v^{p-1} - v_1^{p-1})  \, dt  \leq 0. \label{M}
 \end{equation}
Then as earlier it follows that $v^{p-1} > v_1^{p-1}$  on $(t_{b,\gamma}, M_{1,b})$  
thus the integral term in \eqref{M} is positive.
But also $v_1(M_{1,b}) v'(M_{1,b})\geq  0$ since $v'>0$ on $(t_{b,\gamma}, M_{1,b})$ 
and so we get a contradiction to \eqref{M}.
Thus $v$ must have a local maximum, $M_{v, b}$, on $(t_{b,\gamma}, M_{1,b})$ 
if $b>0$ is sufficiently small. This completes the proof. 
\end{proof}

\begin{lemma} \label{lem2.7}
 Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}-\eqref{DE3IC}. 
Then $u$ has a local max if $b$ is sufficiently small.
\end{lemma}

\begin{proof} 
Since $u = t^c v$, it follows that $u' = t^{c-1}(tv' + cv)$ and so in 
order to show $u'<0$ somewhere we want to show that $tv' + cv$ gets negative 
for some  $t>t_{b,\gamma}$. So by the way of contradiction
let us suppose $u'>0$ for $t> t_{b,\gamma}$. Thus it follows that $u>0$ and 
hence $v>0$ for $t > t_{b,\gamma}$.
We next show that $v<v_1$ for  $t> t_{b, \gamma}$.
Integrating \eqref{v_1} on $(t_{b,\gamma}, t)$ gives
\begin{equation}   
t^{2c}(v_1 v' - v v_1')   + h_1 \int_{t_{b,\gamma}}^{t} t^ {2c-2} v v_1 (v^{p-1} 
- v_1^{p-1}) \, dt \leq 0  \quad \text{for } t > t_{b,\gamma}.\label{ramones}  
\end{equation}

Now using an identical argument as in Lemma \ref{lem2.6} it follows that  $v$ and $v_1$ 
do not intersect for $t> t_{b,\gamma}$ and thus $0< v<v_1$ for $t> t_{b, \gamma}$.
It follows then from \eqref{ramones} that $v_1 v' - v v_1' <0$ for 
$t> t_{b,\gamma}$.
And since $v_1>v>0$ we then have $\frac{v' }{v} < \frac{v_1' }{v_1}$.
From this it follows that 
$tv' + cv < t\frac{v_1'}{v_1} v + cv = \frac{v}{v_1}(tv_1' + cv_1)$ 
for $t>t_{b,\gamma}$. Thus to show $u'$ gets negative it suffices to show 
$tv_1' + cv_1$ gets negative for some $t> t_{b,\gamma}$.

Now recall from Lemma \ref{lem2.5} that $v_1$ has a local maximum at $M_{1,b}$. 
Thus $v_1'(M_{1,b})=0$ and  $v_1''(M_{1,b}) \leq 0$.
In fact by the uniqueness of solutions of initial value problems 
$v_1''(M_{1,b}) < 0$ and so $d(v_1(M_{1,b}))> 0$.
Thus
\begin{equation}
 0 < v_1(M_{1,b}) < \alpha_1.  \label{tom joad} 
\end{equation}
Now $v_1$ cannot have a positive local minimum at $m_{1,b}> M_{1,b}$ with 
$v_1'<0$ on $(M_{1,b}, m_{1,b})$
for at such a point we would have $v_1'(m_{1,b})=0$ and 
$v_1''(m_{1,b})\geq 0$ implying  $d(v_1(m_{1,b}))\leq 0$ which would imply
$v_1(m_{1,b})\leq 0$ contradicting $0<v<v_1$. Thus
\begin{equation} 
v_1'<0 \text{ for } t> M_{1,b}.   \label{gehrig}  
\end{equation}

Next we observe that if $v_1''<0$ for $t> M_{1,b}$  then $v_1$ has a zero for 
$t> M_{1,b}$ which implies $v$ has a zero since $v < v_1$ but this contradicts 
that $v>0$.  Thus   $v_1$ has an inflection point $t_2>M_{1,b}$ with $v_1''<0$ 
on $(M_{1,b}, t_2)$. In addition, $v_1$ cannot have another inflection point 
$t_3>t_2$ with
$v''>0$ on $(t_2, t_3)$ for at such a point  $t_3$ then  $v_1'$  would have 
a local maximum  so $v_1''(t_3)=0$ and $v_1'''(t_3)\leq 0$. Using this  and 
then differentiating  \eqref{DE3v_1} this implies
$ (c^2 + c + p h_1v_1^{p-1}(t_3)) v_1'(t_3) \geq 0$ and so 
$v_1'(t_3)\geq 0$ but this contradicts \eqref{gehrig}.
Thus $v_1''>0$ for $t > t_2$.

Now substituting that $v_1''>0$ for $t > t_2$ into \eqref{DE3v_1} gives:
$$ 
2c tv_1' + d(v_1)<0    \text{ for } t > t_2 
$$ 
and so rewriting
\begin{equation}  
tv_1' + cv_1 < \frac{1+c}{2}v_1  - \frac{h_1v_1^p}{2c}
 = v_1^p\Big( \frac{1+c}{2} v_1^{1-p} - \frac{h_1}{2c}\Big)
\quad  \text{ for } t > t_2.  \label{ny} 
\end{equation}

Now for $b>0$ sufficiently small we know $D(v_1)<\epsilon$ for $t>t_{b,\gamma}$  
by \eqref{series of dreams}. In particular,  $D(v_1(M_{1,b}))<\epsilon$
and since we also know  \eqref{tom joad}  this implies that for $b>0$ 
sufficiently small then $v_1$ is also arbitrarily small. In particular, for 
sufficiently small $b>0$ the right-hand side of  \eqref{ny} is
  negative which implies $tv_1' + cv_1$ and hence $u'$ gets negative. 
Therefore $u$ has a local maximum.
This completes the proof. 
\end{proof}

\begin{lemma} \label{lem2.8} 
Assume {\rm (H1)--(H5)} and let $u$ solve \eqref{DE3}, \eqref{DE3IC}. 
Then $u$ has a zero if $b$ sufficiently small.
\end{lemma}

\begin{proof} 
From Lemma \ref{lem2.8} we know that $u$ has a local maximum, $M_{u,b}$, with 
$M_{u,b}>t_{b}$, if $b>0$ is sufficiently small and so from \eqref{DE3} 
we see that $u''<0$ while $f(u)>0$. It follows then that $u$ must become 
less than or equal to $\beta$ and so there exists $s_{b}> M_{u,b}$ such that 
$u(s_{b})= \beta$. Returning to \eqref{energy 1}  we see then that
$\frac{1}{2} \frac{u'^2}{h} + F(u) = E\geq E_0>0$ for $t>s_{b}$  and since
 $F(u)< 0$ for $0< u < \beta$ it follows from \eqref{h eqn} that 
$-u' \geq \sqrt{2E_0} h_1 t^{-\frac{q}{2}}$ for $t> s_{b}$. 
Integrating this on $(s_{b},t)$ it then follows that
$u(t) \leq u(s_{b}) - \sqrt{2E_0h_1} 
\big( \frac{t^{1-\frac{q}{2}} - s_{b}^{1-\frac{q}{2}}}{1-\frac{q}{2}} \big) 
\to -\infty$
as $t \to \infty$ since $1<q<2$ and so $u$ has a zero.  
This completes the proof. 
\end{proof}


\section{Behavior for large $b>0$}

We now examine \eqref{DE3}, \eqref{DE3IC} for large $b>0$.
We want to show that either $u_{b}>0$ on $(0, \infty)$ or if there exists 
$z_{u,b}>0$ such that $u_{b}(z_{u,b})=0$ with $u>0$ on $(0, z_{u,b})$ 
then $z_{u,b} \to \infty$ as $b \to \infty$.
So let us suppose that $z_{u,b}$ is finite and suppose there exists $S$ 
such that $0< z_{u,b} \leq S$ for all large $b$.

Integrating \eqref{DE3} twice on $(0,t)$ gives
$$
 u_b = bt -\int_0^{t} \int_0^{s} h(x) f(u_b) \, dx \, ds. 
$$
Letting $\psi_b = \frac{u_b}{b}$ gives
\begin{equation} 
\psi_b = t - \int_0^t \int_0^s\frac{h(x) f(b\psi_b)}{b} \, dx \,ds \label{app} 
\end{equation}
and then letting $\psi_b=ty_b$ gives
$$ 
y_b = 1 - \frac{1}{t}\int_0^t \int_0^s\frac{h(x) f(bxy_b)}{b} \, dx \,ds. 
$$
It follows from (H1) and  (H3)  that $|f(u)| \leq C_1|u|$ for some $C_1>0$. Thus
\begin{equation} 
|y_b| \leq 1 + C_1 \int_0^{t} xh(x) |y_b(x)| \, dx. \label{app2} 
\end{equation}
We now denote the right-hand side of \eqref{app2}  as  $\phi_{b}$. 
Then we see $\phi_b' =C_1th(t)|y_b|$
and so  \eqref{app2} reads  $ \frac{\phi'_b}{C_1th} = |y_b|\leq \phi_b$.
Rewriting and integrating this on $(0,t)$ gives
$$ 
\frac{|\psi_b|}{t} = |y_{b}| \leq \phi_{b} 
\leq  \exp\Big(\frac{C_1h_2}{2-q} t^{2-q}\Big). 
$$
Thus
\begin{equation} 
|\psi_b| \leq t \exp\Big(\frac{C_1h_2}{2-q} t^{2-q}\Big)  
\leq C_2 \quad \text{on $[0,S]$  for some constant } C_2>0.   \label{southhampton} 
 \end{equation}
Also from \eqref{app} we have
\begin{equation} 
\psi_b' = 1 -  \int_0^t\frac{h(x) f(b\psi_b)}{b} \, dx.  \label{holland} 
\end{equation}
It follows then from \eqref{southhampton}-\eqref{holland},  
since $|f(u)| \leq C_1|u|$, and since $0< th(t)< h_2t^{1-q}$ is integrable 
near $t=0$ (because $1<q<2$) that there is a constant $C_3>0$ such that
\begin{equation} 
|\psi_b'| \leq 1 +  C_1\int_0^t h(x) |\psi_b| \, dx 
\leq 1 + C_1 \int_0^{t} xh(x) \exp\Big(\frac{C_1h_2}{2-q} x^{2-q}\Big) \, dx
\leq C_3   \label{pennylane} 
\end{equation}
 on $[0,S]$.
Thus from \eqref{southhampton} and \eqref{pennylane} we have  $|\psi_{b}|$
 and $|\psi_{b}'|$ are uniformly bounded on $[0, S]$ and so the 
$\{ \psi_{b} \}$ are equicontinuous on $[0,S]$.
In addition, differentiating \eqref{holland} and using \eqref{southhampton} 
we see there is some constant $C_{4}>0$ such that
$$ 
|\psi_{b}''| \leq \Big|\frac{h(t) f(b\psi_b)}{b}\Big|  
\leq C_1 h(t) |\psi_{b}| 
\leq C_1th(t)\exp\Big(\frac{C_1h_2}{2-q} t^{2-q}\Big)
\leq C_4  t^{1-q}
$$
 on $[0,S]$. 
It then follows that
\begin{equation} |\psi_{b}'(t_1) - \psi_{b}'(t_2)|
 \leq  \int_{t_1}^{t_2} |\psi_{b}''| 
\leq  \frac{C_{4}}{2-q}|t_2^{2-q} - t_1^{2-q}| \quad \text{on } [0,S].
\label{reprise}  \end{equation}
Thus since $q<2$ we see from \eqref{reprise} that $\{\psi_{b}' \}$ are
 also equicontinuous on $[0,S]$.
It then follows by the Arzela-Ascoli theorem that there is a subsequence
 (still denoted $\psi_{b}$)  such that
\begin{equation}
 \psi_{b} \to \psi  \text{  and } \psi_{b}' \to \psi' 
\text{  uniformly on $[0,S]$ as  } b \to \infty.  \label{youngstown} 
\end{equation}

Now since  $u_{b}$ (and hence $\psi_{b}$) has a  zero $z_{u,b}$ it follows that 
$u_{b}$ has a local maximum, $M_{u,b}$, with  $0< M_{u,b} < z_{u,b} \leq S$. 
Then since $E$ is nondecreasing (by \eqref{energy}) we have
$$ 
\frac{1}{2} \frac{u_b'^2}{h} + F(u_b) \leq F(u_b(M_{u,b})) \quad
\text{for } 0 \leq t \leq M_{u,b}.
$$
Rewriting and integrating this on $[0, M_{u,b}]$ using \eqref{h eqn} gives
\begin{align*}
\int_0^{u_b(M_{u,b})}   \frac{  dt}{\sqrt{2}\sqrt{F(u_{b}(M_{u,b}))- F(t) }} \, dt   
&= \int_0^{M_{u,b}}     \frac{ |u_{b}'|}{\sqrt{2}\sqrt{F(u_{b}(M_{u,b}))- F(u_{b}) }} 
\, dt \\
&\leq      \int_0^{M_{u,b}}  \sqrt{h}     \, dt  \\
&\leq \frac{\sqrt{h_2} M_{u,b}^{1-\frac{q}{2}} }{1-\frac{q}{2}}.  
\end{align*}
Since $F(t) \geq -F_0$ where $F_0>0$ it follows that the above inequality and (H3) 
 imply
$$  
\frac{ u_{b}(M_{u,b})}{\sqrt{2}\sqrt{C_{5}[u_{b}(M_{u,b})]^{p+1} + F_0}}   
\leq \frac{\sqrt{h_2} M_{u,b}^{1-\frac{q}{2}} }{1-\frac{q}{2}} 
\leq \frac{\sqrt{h_2} S^{1-\frac{q}{2}} }{1-\frac{q}{2}} 
$$
for some $C_{5}>0$.
It follows from this and since $0<p<1$ that $u_b(M_{u,b})$ must be bounded and 
thus $u_{b}$ is bounded.
Then since $\psi_{b} = \frac{u}{b}$ it follows that $\psi_{b} \to 0$ on $[0,S]$. 
Thus \eqref{youngstown} implies $\psi \equiv 0$. In addition,
\eqref{youngstown} also implies $\psi_{b}' \to \psi' \equiv 0$ but 
from \eqref{holland} it follows that $ 0= \psi'(0) \leftarrow \psi_b'(0)=1\neq 0$ 
yielding a contradiction. Thus the assumption that the $z_{u,b}$ are 
bounded must be false. Therefore $z_{u,b} \to \infty$ as $b \to \infty$.

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

 We now consider the  set
$$ 
S_0 = \{ b>0: u_{b}>0 \text{ on }   (0, R^{2-N})\}.
$$
From section three it follows that if $b>0$ is sufficiently large then 
$u_{b}>0$ on $(0, R^{2-N})$ and so $S_0$ is nonempty. 
From section two we know that if $b>0$ is sufficiently small then 
$u_{b}$ will have a zero on $(0, \infty)$ and so then if $R>0$ is sufficiently 
small then since $N>2$ it follows that $R^{2-N}$ is sufficiently large 
and thus $u_{b}$ will have a zero on $(0, R^{2-N})$. Thus $S_0$ is bounded 
from below by a positive quantity. Then we let
$$
b_0 = \inf\{ b>0 : u_{b} \text{ solves  \eqref{DE3}, \eqref{DE3IC}  and } u_{b}>0  
\text{ on }   (0, R^{2-N})\}.
$$
Then $b_0>0$ and a straightforward argument as in \cite{M}  shows $u_{b_0}>0$ 
on $(0, R^{2-N})$ and $u_{b_0}(R^{2-N})=0$.
We then define
$$ 
S_1 = \{ b>0| u_{b} \text{ solves  \eqref{DE3}, \eqref{DE3IC}  
and has exactly one zero on }   (0, R^{2-N})\}.
$$
Choosing $b$ slightly smaller than $b_0$ it follows then by continuity with 
respect to initial conditions that $u_{b}$ will have at least one zero on 
$(0, R^{2-N})$.
And as in \cite{M} it follows that if $b$ is sufficiently close to $b_0$ then 
$u_{b}$ has at most one zero on $(0, R^{2-N})$
and so $S_1$ is nonempty.  Then letting
$$ 
b_1 = \inf S_1
$$ 
a similar argument shows $u_{b_1}$ has one zero on
 $(0, R^{2-N})$ and $u_{b_1}(R^{2-N})=0$.
In a similar fashion we can show that given any $n$ then if $R>0$ is 
sufficiently small then there exists $b_0, b_1, \cdots, b_{n}$ such that
$u_{b_{k}}(t)$ has $k$ zeros on $(0, R^{2-N})$ and $u_{b_{k}}(R^{2-N}) = 0$. 
Finally, letting $U_{b_{k}}(t) = u_{b_{k}}(t^{2-N})$ it follows that
$U_{b_{k}}(t)$ satisfies \eqref{1}-\eqref{3} and $U_{b_{k}}$ has $k$ 
zeros on $(R, \infty)$. This completes the proof of the main theorem.

\section{Appendix}

\begin{lemma} \label{lem5.1} 
Assume {\rm (H1)--(H5)}. Then for any $b>0$ there is a solution 
$u \in C^{1}[0, \epsilon) \cap C^{2}(0,\epsilon) $   of \eqref{DE3} on 
$(0, \epsilon)$ for some $\epsilon > 0$ and such that \eqref{DE3IC} holds.
\end{lemma}

\begin{proof}
Integrating \eqref{DE3} twice on $(0,t)$ and using  \eqref{DE3IC}  gives
$$
 u=bt -\int_0^{t} \int_0^{s} h(x) f(u) \, dx \, ds. 
$$
Substituting $u=tv$ gives
\begin{equation} 
v=b - \frac{1}{t}\int_0^{t} \int_0^{s} h(x) f(xv) \, dx \, ds.  \label{RWR} 
\end{equation}
Defining the right-hand side of \eqref{RWR} as $Tv$ then we see that
$T: C[0, \epsilon] \to C[0, \epsilon]$  where  $C[0, \epsilon]$ is 
the set of continuous functions on $[0, \epsilon]$ such that $v(0)=b$ with the
 supremum norm, $\| \cdot \|$.
Then
$$ 
Tv_1 - Tv_2 = -\frac{1}{t}\int_0^{t} \int_0^{s} h(x) [f(xv_1)   
- f(xv_2)] \, dx \, ds 
$$
and so
$$ 
|Tv_1 - Tv_2| \leq \frac{L_1}{t}\int_0^{t} \int_0^{s}xh(x) |v_1   - v_2| \, dx \, ds 
$$ 
where $L_1$ is the Lipschitz constant for $f$  near $v= b$.
Then 
$$ 
|Tv_1 - Tv_2| \leq    \frac{L_1 h_2\epsilon^{2-q}}{(2-q)(3-q)}\|v_1 - v_2\| 
$$ 
and so $T$ is a contraction if $\epsilon>0$ is sufficiently small.
It then follows from the contraction mapping principle that $v(t)$ is 
a solution of  \eqref{RWR} and therefore
$u(t) = t v(t)$ satisfies  \eqref{DE3}, \eqref{DE3IC} on $[0, \epsilon]$. 
This completes the proof. 
\end{proof}


\begin{lemma} \label{lem5.2} 
Let $I$ be the solution of
\begin{equation} 
I'' + \frac{1}{t} I' - \Big( 1 + \frac{1}{(2-q)^2 t^2}\Big) I =0  \quad
\text{and}\quad
\lim_{t \to 0^{+}} \frac{I(t)}{t^{\frac{1}{2-q}}} = a>0. \label{revolution}
 \end{equation}
Then there exists a constant $C_6>0$ such that
$$ 
|\frac{I'}{I}| \leq C_6 \quad \text{for } t \geq 1\,.
$$
\end{lemma}

\begin{proof} 
A straightforward computation using \eqref{revolution} shows that
$$  
\Big(t^2 I'^2 -  t^2 I - \frac{1}{(2-q)^2} I\Big)' =   -tI^2 \leq 0\,.
$$
Thus since $I(0)=0$, it follows that
 $$ 
 t^2 I'^2 -  t^2 I^2 - \frac{1}{(2-q)^2} I^2 \leq 0 \quad \text{for } t>0.
$$
Therefore
$$ 
\frac{I'^2}{I^2}   \leq 1 + \frac{1}{(2-q)^2t^2}\,.  
$$
So we see that there exists  a $C_6>0$ such that
$$ 
\big|\frac{I'}{I}\big| \leq C_6  \text{ for } t > 1\,.
$$
This completes the proof. 
\end{proof}

\begin{thebibliography}{00}

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

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

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

\bibitem{BR} G. Birkhoff, G. C. Rota;
 \emph{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,
 \emph{Journal of Mathematical Analysis and Applications}, 
Volume 394, Issue 1, 432-437, 2012.

\bibitem{CSS} M. Chhetri, L. Sankar, R. Shivaji;
Positive solutions for a class of superlinear semipositone systems 
on exterior domains, \emph{Boundary Value Problems,} 198-207, 2014.

\bibitem{I8} J. Iaia;
Existence of solutions for sublinear equations on exterior domains,
 \emph{Electronic Journal of Differential Equations}, 2017 No. 253, pp. 1-14, 2017.

\bibitem{I9} J. Iaia;
Existence and nonexistence of solutions for sublinear equations on exterior domains,
 \emph{Electronic Journal of Differential Equations}, 2917 No. 214, 1-13,  2017.

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

\bibitem{JJ} J. Joshi;
Existence and nonexistence of solutions of sublinear problems with prescribed 
number of zeros on exterior domains,
 \emph{Electronic Journal of Differential Equations}, 2017 No. 133, 1-10, 2017.

\bibitem{LSS} E. K. Lee, R. Shivaji, B. Son;
Positive radial solutions to classes of singular problems on the exterior of
 a ball, \emph{Journal of Mathematical Analysis and Applications,}
 434, No. 2, 1597-1611, 2016.

\bibitem{C2} E. Lee, L. Sankar, R. Shivaji;
Positive solutions for infinite semipositone problems on exterior domains,
 \emph{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,
 \emph{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,
 \emph{Journal of Mathematical Analysis and Applications}, Volume 401, 
Issue 1, 146-153, 2013.

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

\end{thebibliography}

\end{document}