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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 133, pp. 1--10.\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-2017/133\hfil Existence and nonexistence of solutions]
{Existence and nonexistence of solutions for sublinear
problems with prescribed number of zeros on exterior domains}

\author[J. Joshi \hfil EJDE-2017/133\hfilneg]
{Janak Joshi}

\address{Janak Joshi \newline
Department of Mathematics,
University of North Texas, 
Denton, TX 76203, USA}
\email{janakrajjoshi@my.unt.edu}

\dedicatory{Communicated by Jerome A. Goldstein}

\thanks{Submitted March 6, 2017. Published May 16, 2017.}
\subjclass[2010]{34B40, 35B05}
\keywords{Exterior domain; sublinear; radial solution}

\begin{abstract}
 We prove existence of radial solutions of $\Delta u+K(r)f(u)=0$ on the
 exterior of the ball, of radius $R$, centered at the origin in $\mathbb{R}^{N}$
 such that  $\lim_{r \to \infty} u(r)=0$ if $R>0$ is sufficiently small.
 We assume $f:\mathbb{R} \to \mathbb{R}$ is odd and there exists a $\beta>0$
 with $f<0$ on $(0,\beta)$, $f>0$ on $(\beta,\infty)$ with $f$ sublinear
 for large $u$, and $K(r) \sim r^{-\alpha}$ for large $r$ with $ \alpha >2(N-1)$.
 We also prove nonexistence if $R>0$ is sufficiently large.
\end{abstract}

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

\section{Introduction}

In this article we study radial solutions to
 \begin{equation} \label{maineq}
\begin{gathered}
 \Delta u + K(|x|)f(u) = 0 \quad \text{for } R<|x|<\infty,  \\
  u(x)=0 \text{ when } |x|=R, \quad \lim_{|x| \to \infty} u(x) = 0,
\end{gathered}
 \end{equation}
 where  $u:\mathbb{R}^N \to \mathbb{R} $  with $N\geq 2$, $R>0$, $f$
 is odd and locally Lipschitz.
In addition we have the following:
 \begin{itemize} 
\item[(H1)] 
$f'(0)<0$, there exists  $\beta> 0$  such that
$f(u)< 0$  on $(0, \beta)$, $f(u)>0$  on $(\beta, \infty)$.   

\item[(H2)]  $f(u) = |u|^{p-1}u + g(u)$ with  
$0<p<1$ , and  $\lim_{u \to \infty} \frac{|g(u)|}{|u|^p} = 0$. 
 
\item[(H3)] Denoting $F(u) \equiv  \int_{0}^{u} f(t) \, dt$ we assume that
there exists   $\gamma$ with $0< \beta < \gamma$ such that 
$F<0$ on $(0, \gamma)$  and $F>0$  on $(\gamma, \infty)$.  

\item[(H4)] We  assume $K$ and $K'$ are continuous on $[R, \infty)$,
$K(r)> 0$, and there exists $\alpha> 2(N-1)$ such that 
$\lim_{r \to \infty} \frac{rK'}{K} = -\alpha$,
$2(N-1)+\frac{rK'}{K}<0$.


\item[(H5)] We assume that there exists  positive   $d_1, d_2$ such that 
$d_1 r^{-\alpha}\leq K(r) \leq d_2 r^{-\alpha}$  for  
$r\geq{R}$.
\end{itemize}

\begin{theorem} \label{thm1} 
Assume {\rm (H1)--(H5)}, $N > 2$, and $\alpha>2(N-1)$. Then for each 
nonnegative integer $n$ there exists a radial solution, $u_{n}$, of \eqref{maineq}
 such that $u_{n}$ has exactly $n$ zeros on $(R, \infty)$ if $R$ is positive and
sufficiently small.
\end{theorem}

\begin{theorem} \label{thm2} 
 Let $N\geq{2}$ and $\alpha>2(N-1)$. If $R$ is positive and sufficiently large,
 then  there are no nontrivial radial solutions of \eqref{maineq}.
\end{theorem}

The radial solutions of \eqref{maineq} on $\mathbb{R}^{N}$ with $f$ superlinear
for large $u$ and $K(r) \equiv 1$ have been well-studied; 
see for example \cite{BL,BL2,JK,M}. 
Recently there has been  interest in studying these problems on 
$\mathbb{R}^{N} \backslash B_{R}(0)$,
with various types of non linearities, see \cite{C,JI,I,I2,JU,C2,S}.
 Interest in the topic for this paper comes from the papers 
\cite{JI,I,I2} by  Iaia where he studied  the solutions of the 
differential equations 
with superlinear and hilltop type of nonlinearity. 
The type of nonlinearity  addressed here has not been studied as 
extensively as other cases, see \cite{C,JU}.
In  \cite{JU} the same nonlinearity as here was studied but in the case $\alpha<2$. 
In this article we use different methods to study the case $\alpha>2(N-1)$.
 We use a scaling argument as in \cite{M} to prove existence of 
solutions for \eqref{maineq}, \eqref{e3} with large number of zeros.

For the proofs in sections 2 and 3, we will need 
to temporarily extend $K$ and $K'$ continuously to $(0,\infty)$ so that
 (H4) and (H5) continue to hold on $(0,\infty)$.
 We define
\begin{equation} \label{e3}
   \tilde{K}(r)= \begin{cases}
      K(R)-\frac{K'(R) R^{\alpha+1}}{\alpha}
[\frac{1}{r^{\alpha}}-\frac{1}{R^{\alpha}}] & r\leq R \\
      K(r) & r\geq{R}. 
      \end{cases}
\end{equation}
It follows that
$$
\lim_{r\to{R^-}}\tilde{K}(r)=K(R)\quad \text{and}\quad 
\lim_{r\to{R^-}}\tilde{K'}(r)=K'(R).
$$
It is straightforward to verify that $\tilde{K}$ and $\tilde{K'}$ extend
 $K$ and $K'$ continuously on $(0, \infty)$ and $\tilde{K}$ 
satisfies (H4) on $(0, \infty)$.
 In addition 
\[
\lim_{r\to{0^+}}{r^{\alpha}\tilde{K}(r)}=-\frac{K'(R)R^{\alpha+1}}{\alpha}>0.
\]
 Therefore with perhaps a smaller positive number $d_1$ and larger positive 
number $d_2$ we may ensure (H5) on $(0,\infty)$.

 \section{Preliminaries}

Since we are interested in radial solutions of \eqref{maineq} we denote 
$r = |x|$ and consider $u(x)= u(|x|)$ where $u$  satisfies
 \begin{gather} 
u'' + \frac{N-1}{r} u' + K(r) f(u) = 0 \quad \text{for } R<r<\infty, \label{DE} \\ 
u(R) = 0, \quad u'(R) = b>0. \label{IC}
 \end{gather}
 We will occasionally write $u(r,b)$ to emphasize the dependence of the solution 
on $b$.   By the standard existence-uniqueness theorem \cite{BR} there 
is a unique solution of \eqref{DE}-\eqref{IC} on $[R, R+\epsilon)$ for some $\epsilon > 0$.

 We next consider:
\begin{equation} 
E(r) = \frac{1}{2} \frac{u'^2}{K(r)} + F(u). \label{E} 
\end{equation}
It is straightforward using \eqref{DE} and (H4) to show that
\begin{equation} 
E'(r) = -\frac{u'^2}{2rK}[2(N-1) + \frac{rK'}{K}]  \geq 0. \label{E'} 
\end{equation}
Thus $E$ is non-decreasing.
Therefore
\begin{equation}
 \frac{1}{2} \frac{u'^2}{K(r)} + F(u) = E(r) \geq E(R) 
= \frac{1}{2}\frac{b^{2}}{K(R)}
\quad \text{for } r \geq R. \label{E'2} 
\end{equation}


\begin{proof}[Proof of Theorem \ref{thm2}]
  Suppose there is a nontrivial solution $u(r)$ of \eqref{DE}-\eqref{IC}
 with $\lim_{r\to{\infty}}{u(r)=0}$. Then $u$ has a local maximum at some 
$M_b>R$ and $u'>0 \text{ on }[R, M_b)$.
Evaluating \eqref{E'2} at $r=M_b$ gives
$$ 
F(u(M_b))\geq{\frac{1}{2}\frac{b^2}{K(R)}}>0.
$$  
So by (H3) we see
 \begin{equation} \label{e9}
u(M_b)>\gamma.
\end{equation}
Now by \eqref{E'} we have
$$
E(r)\leq{E(M_b)}\quad \text{for } R\leq{r}\leq{M_b}
$$ 
so
\begin{equation} \label{e10}
\frac{1}{2}\frac{u'^2}{K}+F(u)\leq{F(u(M_b))}\quad \text{for } 
R\leq{r}\leq{M_b}.
\end{equation}
Rewriting this expression, integrating over $[R, M_b]$ and applying (H5) yields
\begin{equation} \label{e11}
\int_R^{M_b}{\frac{|u'|\,dr}{\sqrt{2}\sqrt{F(u(M_b))-F(u)}}}
\leq{\int_R^{M_b}{\sqrt{K}}}\leq{\sqrt{d_2}
\int_R^{M_b}{r^{-{\frac{\alpha}{2}}}\,dr}}.
\end{equation}
Setting $u(r)=t$ we have
\begin{equation} \label{e12}
\int_0^{u(M_b)}{\frac{dt}{\sqrt{2}\sqrt{F(u(M_b))-F(t)}}}
\leq{\frac{2\sqrt{d_2}}{\alpha-2}
\big[R^{\frac{2-\alpha}{2}}-M_b^{\frac{2-\alpha}{2}}\big]}
\leq{\frac{2\sqrt{d_2}}{\alpha-2}R^{\frac{2-\alpha}{2}}}.
\end{equation}

Next since $f(0)=0$, $f'(0)<0$ and $0<p<1$ it follows that  
$\lim_{u\to{0}} \frac{F(u)}{|u|^{p+1}}=0$.
Also from (H2) we see that
 $\lim_{u\to{\infty}} \frac{F(u)}{|u|^{p+1}}=\frac{1}{p+1}$.
Then there exists $c_1>0$ such that
\begin{equation} \label{e13}
|F(u)|\leq{c_1|u|^{p+1}}\quad \forall u.
\end{equation}   
Moreover, by (H1) and (H3) we have
 \begin{equation} \label{e14}
F(u)\geq{-F_0 }\quad \text{for some }F_0>0
\end{equation}
 and so it follows from \eqref{e13} and \eqref{e14} that 
$$
F(u(M_b))-F(u)\leq{c_1 |u(M_b)|^{p+1}+ F_0}
$$ 
for all $u$.
This along with \eqref{e12}  implies
\begin{equation} \label{e15}
\frac{2\sqrt{d_2}}{\alpha-2}R^{\frac{2-\alpha}{2}}
\geq{\int_0^{u(M_b)}{\frac{dt}{\sqrt{F(u(M_b))-F(t)}}}}
\geq{\frac{u(M_b)}{\sqrt{c_1|u(M_b)|^{p+1}+F_0}}}.
\end{equation}
Since $u(M_b)>{\gamma}$ by \eqref{e9}  and $0<p<1$, we have
\begin{equation} \label{e16}
\frac{u(M_b)}{\sqrt{c_1|u(M_b)|^{p+1}+F_0}}
=\frac{|u(M_b)|^{\frac{1-p}{2}}}{\sqrt{c_1+\frac{F_0}{|(u(M_b)|^{p+1}}}}
\geq{\frac{\gamma^{\frac{1-p}{2}}}{\sqrt{c_1 +\frac{F_0}{\gamma^{p+1}}}}}.
\end{equation}
Thus by \eqref{e15} and \eqref{e16} it follows that
 $$
\frac{2\sqrt{d_2}}{\alpha-2}R^{\frac{2-\alpha}{2}}
\geq{\frac{\gamma^{\frac{1-p}{2}}}{\sqrt{c_1+\frac{F_0}{\gamma^{p+1}}}}} 
$$
which implies
 \begin{equation} \label{e17}
R^{\frac{\alpha-2}{2}}\leq{\frac{2\sqrt{d_2}}{\alpha-2 }
\frac{ \sqrt{c_1+\frac{F_0}{\gamma^{p+1}}}}{\gamma^{\frac{1-p}{2}}}}.
\end{equation}
Since $\alpha>2$ we see that (17) is violated if $R$ is sufficiently large. 
Hence there are no solutions of \eqref{maineq} such that 
$\lim_{r\to{\infty}}{u(r)}=0$ if $R$ is sufficiently large.
This completes the proof.
\end{proof}

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

To prove Theorem \ref{thm1} we first make the following change of variables.
Let 
\begin{gather}
    u(r)=u_1(r^{2-N}) \label{D'}, \\
R^*=R^{2-N},\quad  b^*=\frac{b R^{N-1}}{N-2}.\label{D1}
\end{gather}
This transforms \eqref{DE}-\eqref{IC} into
\begin{equation}
u_1''(t)+h(t)f(u_1(t))=0, \quad \text{with}\quad
t=r^{2-N} \text{ for } 0<t<R^*,\label{D3}
\end{equation}
where
\begin{gather}
u_1(R^*)=0,\quad u_1'(R^*)=-b^*<0,\label{D4} \\
h(t)=\frac{1}{(N-2)^2}t^{\frac{2(N-1)}{2-N}}K(t^{\frac{1}{2-N}}).\label{D41}
\end{gather}
Since $(r^{2(N-1)}K)'<0$ and $N>2$ it follows that
 \begin{equation}
h'(t)>0 \quad \text{for } 0<t<R^*.\label{D5}
\end{equation}
In addition, from (H4)-(H5) we see that
\begin{equation}
\frac{d_1}{(N-2)^2}\leq{\frac{h(t)}{t^q}}
\leq{\frac{d_2}{(N-2)^2}}\quad\text{for } 0<t\leq{R^*},
\label{D6}
\end{equation}
where 
\[
q=\frac{\alpha-2(N-1)}{N-2}>0.
\]
 Moreover by using \eqref{e3} and \eqref{D41},
 we can extend $h$ on $(0, \infty)$ such that 
\eqref{D5}-\eqref{D6} hold on $(0,\infty)$
 and that
 \begin{equation} \label{e25}
\lim_{t\to{\infty}}{\frac{h(t)}{t^q}
=-\frac{K'(R)R^{\alpha+1}}{(N-2)^2\alpha}}\equiv L>0.
\end{equation}
Now instead of considering \eqref{D3}-\eqref{D4} we look at the  
initial value problem
\begin{equation}
u_1''(t)+ h(t) f(u_1(t))=0,\quad t>0\label{D7}
\end{equation}
with:\begin{equation}
u_1(0)=0, \quad u_1'(0)=a>0.\label{D8}
\end{equation}
From \eqref{D41} and \eqref{D6} it follows that $h(t)$ can be 
extended to be continuous at $t=0$ with $h(0)=0$ and so by the standard
 existence-uniqueness theorem there is a unique solution of 
\eqref{D7}-\eqref{D8} on $[0,2\epsilon]$ for some $\epsilon>0$.
Let
 $$
E_1(t)=\frac{1}{2}\frac{u_1'^2}{h(t)}+F(u_1).
$$
Then using \eqref{D5} and \eqref{D7} we see that
 $$
E_1'(t)=-\frac{u_1'^2\,h'}{2h^2}\leq{0}.
$$
Thus
 \begin{equation} \label{e28}
\frac{1}{2}\frac{u_1'^2}{h(t)}+F(u_1)=E_1(t)
\leq{E_1(\epsilon)}=\frac{1}{2}\frac{u_1'^2(\epsilon)}{h(\epsilon)}
+F(u_1(\epsilon))\quad \text{for } t>\epsilon.
\end{equation}
It follows from \eqref{e28} that $u_1$ and $u_1'$ are uniformly bounded
 on $[\epsilon,t]$ from which it follows that the solution of 
\eqref{D7}-\eqref{D8} exists on $[0, t]$. Since $t$ is arbitrary,
 we see the solution of \eqref{D7}-\eqref{D8} exists on $[0,\infty)$.

 \begin{lemma} \label{lem1} 
 Suppose {\rm (H1)--(H5)} hold and  $u_1$ solves \eqref{D7}-\eqref{D8}. Then
 there exists $t_a>0$ such that $u(t_a)=\beta$.
  Moreover $\lim_{a\to{0^+}}t_a=\infty$.
\end{lemma}

\begin{proof} 
  From \eqref{D8} we see that $u_1$ increases initially and $u_1>0$ for small 
$t>0$. If $u_1$ has a first local maximum $M$ then $u_1'(M)=0$ and 
$ u_1''(M)\leq0$. By uniqueness of solutions of initial value problems it 
follows that $u''(M)<0$. Then since $h>0$ it follows from \eqref{D7} that 
$f(u_1(M))>0$ which in turn from (H1) implies $u_1(M)>\beta$. 
Since $u_1(0)=0$ it then follows by the Intermediate Value Theorem that the 
first part of the lemma holds. Otherwise suppose $0<u_1<\beta$ and 
$u_1'>0$ for $t>0 $.
Then by (H1) we have $f(u_1(t))<0$ for all $t>0$.
Thus it follows from \eqref{D7} that $u_1''(t)>0$  for $t>0$
 so $u_1'(t)>a$ and hence $u_1(t)>a t$  for $t>0$.
 This implies $ u_1(t)\to{\infty}$ as $t\to{\infty}$ contradicting that 
$0<u_1(t)<\beta$ for all $t>0$. Hence there exists $t_a>0$ such that 
$u_1(t_a)=\beta$ and $0<u_1<\beta$ on $(0,t_a)$.
  This proves the first part of Lemma \ref{lem1}.

Now we let
 $$
E_2(t)=\frac{1}{2}u_1'^2+h(t)F(u_1).
$$
 By (H3) and \eqref{D5} we see
 $$ 
E_2'(t)=\big[\frac{1}{2}u_1'^2+h(t)F(u_1)\big]'
=h'(t)F(u_1)<0\quad \text{if }u_1<\gamma
$$
 and since $0\leq u_1\leq \beta<\gamma \text{ on }[0, t_a]$
this implies
\begin{equation} \label{e29}
\frac{1}{2}u_1'^2+h(t)F(u_1)\leq{\frac{1}{2}a^2} \quad\text{on }[0,t_a].
\end{equation}
 Also by (H2)-(H3) there exists $c_2>0$ such that
 \begin{equation} \label{e30}
F(u_1)\geq-c_2 u_1^2 \quad \text{on }[0, \gamma].
\end{equation}
It then follows from \eqref{e29} and \eqref{e30} that
$$
\frac{1}{2}u_1'^2- c_2\, h(t)\,u_1^2
\leq\frac{1}{2}u_1'^2+h(t)F(u_1)\leq{\frac{1}{2}a^2}\quad \text{on }[0,t_a]
$$
which on rewriting and using \eqref{D6} gives
 $$
u_1'^2\leq{a^2+ 2 c_2 h(t)u_1^2}\leq{a^2
+\frac{2 c_2 d_2}{(N-2)^2}t^q u_1^2}\quad \text{on }[0,t_a]
$$
which implies
\begin{equation} \label{e31}
u_1'\leq{a+ c_3 t^{q/2} u_1}\quad \text{on }[0,t_a] 
\end{equation}
where $c_3=\frac{\sqrt{2 c_2 d_2}}{N-2}$.
  After rewriting \eqref{e31}, multiplying by $e^{-y(s)}$ where  
$y(t)=\frac{c_3\,t^{\frac{q}{2}+1}}{\frac{q}{2}+1}$, and integrating on 
$(0,t)$ we see 
$$
u_1 e^{-y(t)}=\int_0^t{\left(u_1e^{-y(s)}\,ds\right)}'
\leq{\int_0^t{a\,e^{-y(s)}\,ds}}\text{ on }[0,t_a].
$$
This implies
$$
u_1\leq{a\,e^{y(t)}\int_0^t{e^{-y(s)}\,ds}}\text{ on }[0,t_a]
$$
and since $e^{-y(t)}\leq{1}$ it then follows that
 \begin{equation} \label{e32}
\beta=u_1(t_{a})\leq{a\,e^{y(t_a)}\int_0^{t_a}{e^{-y(s)}\,ds}}
\leq{a\,t_a\,e^{y(t_a)}}.
\end{equation}
Now suppose $|t_a|\leq{S}$ for some constant $S$. Since $y(t)$ is continuous 
on $[0,M]$ then   $y(t_a)$ is bounded and thus the right side of 
\eqref{e32} goes to $0$ as $a\to{0^+}$. This is a contradiction as the left 
side is $\beta>0$.
Hence $t_a\to{\infty}$ as $a\to{0^+}$. This completes the proof 
\end{proof}


 We now use a rescaling argument to show $u_1$ has a large number of zeros 
if $a>0$ is sufficiently large.
 For this we let
 \begin{equation} \label{e33}
v_\lambda(t)=\lambda^{-\frac{2+q}{1-p}}u_1(\lambda t),
\end{equation}
where $q=\frac{\alpha-2(N-1)}{N-2}>0 $  (from \eqref{D6}). 
Also from \eqref{D3} and (H2) we have
 $$
u_1''(\lambda\,t)+h(\lambda t )[u_1^p(\lambda t)+g(u_1(\lambda t)) ]=0.
$$ 
It now follows using \eqref{e33} that
\begin{equation} \label{e34}
v_\lambda''+\frac{h(\lambda t)}{(\lambda t)^q}t^q
\left[v_\lambda^p+\lambda^{-\frac{p(q+2)}{1-p}} 
g(\lambda^{\frac{2+q}{1-p}} v_\lambda)\right]=0
\end{equation}
with
$$
v_\lambda(0)=0,\quad v_\lambda'(0)=\lambda^{-\frac{p+q+1}{1-p}}a.
$$
On setting $a=\lambda^{\frac{1+p+q}{1-p}}$ we have
\begin{equation} \label{e35}
v_\lambda(0)=0,\quad v_\lambda'(0)=1.
\end{equation}

Integrating \eqref{e34} on $(0,t)$ and using \eqref{e35} gives
 \begin{equation} \label{e36}
v_\lambda'(t)=1-\int_0^t{\frac{h(\lambda s)}{(\lambda s)^q}}s^q 
\left[v_\lambda^p(s)+\lambda^{-\frac{p(q+2)}{1-p}}g
\left(\lambda^{\frac{2+q}{1-p}}v_\lambda(s)\right)\right]\,ds.
\end{equation}
Integrating  on $(0,t)$ by using \eqref{e35} gives
\begin{equation} \label{e37}
v_\lambda(t)= t-\int_0^t
\int_0^s \frac{h(\lambda x)}{(\lambda x)^q}x^q\left[v_\lambda^p(x)
+\lambda^{-\frac{p(q+2)}{1-p}}g\left(\lambda^{\frac{2+q}{1-p}}v_\lambda(x)\right)
\right]\,dx\,ds.
\end{equation}
Therefore
$$
|v_\lambda(t)|\leq t+\int_0^t
\int_0^s \big|\frac{h(\lambda x)}{(\lambda x)^q}\big|x^q
{\big| v_\lambda^p(x)+\lambda^{-\frac{p(q+2)}{1-p}}g
\big(\lambda^{\frac{2+q}{1-p}}v_\lambda(x)\big)\big|}\,dx\,ds. 
$$
Using \eqref{D6} yields
\begin{equation} \label{e38}
|v_\lambda(t)|\leq{t+\frac{d_2}{(N-2)^2}
\int_0^t s^q {\int_0^s{\big| v_\lambda^p(x)+\lambda^{-\frac{p(q+2)}{1-p}}
g\big(\lambda^{\frac{2+q}{1-p}}v_\lambda(x)\big)\big|}\,dx\,ds}}.
\end{equation}
Since from (H2) we have $|\frac{g(u)}{u^p}|\to{0}$ as $u\to{\infty} $, 
it follows that given $\epsilon>0$ there exists $u_0$ such that 
$|g(u)|\leq{\epsilon|u|^p}$ for $|u|\geq{u_0}$
 and also the continuity of $g$ implies $|g(u)|\leq{c_4}$ for $u\leq{u_0}$ 
where $c_4$ is some positive constant.
 Thus
 \begin{equation} \label{e39}
|g(u)|\leq{c_4+\epsilon\,|u|^p}\quad \text{for all } u.
\end{equation}
Now rewriting \eqref{e38} and using \eqref{e39} we see that
 \begin{equation} \label{e40}
 |v_\lambda(t)|\leq{t+\frac{d_2}{(N-2)^2}\int_0^t s^q
\int_0^t\big[{(1+\epsilon)|v_\lambda^p(x)|
+\frac{c_4}{\lambda^{\frac{(2+q)p}{1-p}}}}\big]\,dx\,ds}.
\end{equation}
 Now choose $\lambda>0$ large enough so that 
$\frac{c_4}{\lambda^{\frac{(2+q)p}{1-p}}}\leq{1}$. Since  $0<p<1$ it 
follows that $|v_\lambda|^p\leq{1+|v_\lambda|}$.
 Hence it follows from \eqref{e40} that for large $\lambda$,
 $$
|v_\lambda(t)|\leq t+ \Big[\frac{d_2}{(q+1)(N-2)^2}t^{q+2}\Big]
+\Big[\frac{d_2(1+\epsilon)}{(q+1)(N-2)^2} t^{q+1}
\int_0^t{ |v_\lambda(x)|}\,dx\Big] 
$$
 so for large $\lambda$,
  \begin{equation} \label{e41}
|v_\lambda(t)|\leq{t+ B t^{q+2}
+\Big[A t^{q+1}\int_0^t{ |v_\lambda(x)|}\,dx\Big]}
\end{equation}
where  
\[
A=\frac{d_2(1+\epsilon)}{(q+1)(N-2)^2}, \quad  
B=\frac{d_2}{(q+1)(N-2)^2}.
\]
Now let $w(t)=\int_0^t{|v_\lambda(x)|}\,dx$.
Then $w'(t)=|v_\lambda(t)|$ and hence from (41) we have
 \begin{equation} \label{e42}
|v_\lambda(t)|=w'(t)\leq{t+B\,t^{q+2}+ A\,t^{q+1}w(t)}.
\end{equation}
Thus
$$
w'(t)-A\,t^{q+1}w(t)\leq{t+B\,t^{q+2}}
$$
and therefore
$$
\int_0^t{\big(w\, e^{-(\frac{A}{q+2})t^{q+2}}\big)'
\leq{[t+B\,t^{q+2}}] e^{-(\frac{A}{q+2})t^{q+2}}}.
$$
which implies
$$
w(t)\leq{e^{(\frac{A}{q+2})t^{q+2}}}\int_0^t{[s+ 2A\,s^{q+2}]
e^{-(\frac{A}{q+2})t^{q+2}}}\,ds.
$$
Since $e^{-(\frac{A}{q+2})t^{q+2}}\leq{1}$ it follows that
 $$
w(t)\leq{e^{(\frac{A}{q+2})t^{q+2}}}
\big[\frac{t^2}{2}+\frac{B\,t^{q+3}}{q+3}\big].
$$ 
Thus for any fixed $0<T<\infty$ we have
 \begin{equation} \label{e43}
 w(t)\leq{e^{(\frac{A}{q+2})T^{q+2}}}
\big[\frac{T^2}{2}+\frac{B\,T^{q+3}}{q+3}\big]
 = C_T \quad \text{on } [0, T]
.\end{equation}
Now from \eqref{e42} and \eqref{e43} we have
 \begin{equation} \label{e44}
|v_\lambda(t)|\leq{T+B\,T^{q+2}+A\,T^{q+1}C_T}=D_T \quad\text{on } [0, T] 
\end{equation} 
and using \eqref{D6}, \eqref{e39} and \eqref{e44} in \eqref{e36} it follows that
\begin{equation} \label{e45}
|v_\lambda'(t)|\leq{1+\frac{d_2(1+ D_T)(2+\epsilon)}{(q+1)(N-2)^2}}T^{q+1}
=Q_T \quad \text{on }[0, T].
\end{equation} 
Using this inequality along with \eqref{D7}, \eqref{e39} and \eqref{e44} 
in \eqref{e34} it follows, for sufficiently large $\lambda$,  that
$$
|v_\lambda''(t)|\leq{\frac{d_2(2+\epsilon)(1+D_T)}{(N-2)^2}T^q}=J_T \quad
\text{on }  [0,T].
$$ 
where $C_T, D_T, Q_T,J_T$ are constants for the fixed $T<{\infty}$.
Hence by the Arzela-Ascoli theorem $v_\lambda\to{v}$ and $v_\lambda'\to{v'}$ 
uniformly on $[0,T]$ as $\lambda\to{\infty}$ for some subsequence still 
denoted by $v_\lambda$.
Since $T$ is arbitrary we see that $v$ and $v'$ are continuous on $[0,\infty)$.

Now by \eqref{e36} we have
$$
\lim_{\lambda\to{\infty}}{v_\lambda'(t)}
=1-\lim_{\lambda\to{\infty}} \Big(\int_0^t{\frac{h(\lambda s)}{(\lambda s)^q}s^q 
\big[v_\lambda^p(s)+\lambda^{-\frac{(2+q)p}{1-p}} 
g\big(\lambda^{\frac{2+q}{1-p}}v_\lambda(s)\big)}\big]\,ds\Big).
$$  
But we know that $\lim_{\lambda\to{\infty}}{v_\lambda'(t)}=v'(t)$
 so
\begin{equation} \label{e46}
v'(t)=1-\lim_{\lambda\to{\infty}}
\Big(\int_0^t{\frac{h(\lambda s)}{(\lambda s)^q}}s^q 
\big[v_\lambda^p(s)+\lambda^{-\frac{(2+q)p}{1-p}} 
g\big(\lambda^{\frac{2+q}{1-p}}v_\lambda(s)\big)\big]\,ds\Big).
 \end{equation}
Next we show that 
\[
\lim_{\lambda\to{\infty}}{\int_0^t{\Big(\frac{h(\lambda\,s)}{(\lambda s)^q}
\,\lambda^{-\frac{(2+q)p}{1-p}}g(\lambda^\frac{2+q}{1-p}v_\lambda)\Big)}\,ds}=0.
\]
 From \eqref{D6} and \eqref{e39} it follows that
\begin{align*}
&\big|\int_0^t{\frac{h(\lambda\,s)}{(\lambda s)^q} s^q
 \lambda^{-\frac{(2+q)p}{1-p}} g\big(\lambda^{\frac{2+q}{1-p}}v_\lambda(s)\big)}\,ds
\big| \\
&\leq{\frac{d_2}{(N-2)^2}\int_0^t{s^q\left(c_4\lambda^{-\frac{(2+q)p}{1-p}}
+\epsilon\,v_\lambda^p(s)\right)\,ds}} \\
&\leq{\frac{d_2\,c_4}{(q+1)(N-2)^2}\frac{T^{q+1}}{\lambda^{\frac{(2+q)p}{1-p}}}}
+\Big( \frac{d_2\,D_T^p\,T^{q+1}}{(q+1)(N-2)^2}\Big)\epsilon\quad \text{on }[0, T].
\end{align*}
Since $\frac{T^{q+1}}{\lambda^{\frac{(2+q)p}{1-p}}}\to{\infty}$  as 
$\lambda\to{\infty}$ and $\epsilon>0$ is arbitrary it follows that
 $$
\lim_{\lambda\to{\infty}}
\int_0^t\Big(\frac{h(\lambda\,s)}{(\lambda s)^q}\,\lambda^{-\frac{(2+q)p}{1-p}}
g(\lambda^\frac{2+q}{1-p}v_\lambda)\Big)\,ds=0.
$$
Therefore from \eqref{e44} it follows that
 \begin{equation} \label{e47}
v'(t)=1-\lim_{\lambda\to{\infty}}
\int_0^t \Big(\frac{h(\lambda\,s)}{(\lambda s)^q} s^q
v_\lambda^p(s)\Big)\,ds
\end{equation}
and since from \eqref{e25} we have $0<\lim_{t\to{\infty}}{\frac{h(t)}{t^q}}=L$ 
then it follows from \eqref{e44} that
$$
v'(t)=1-L\int_0^t{s^qv^p(s)}\,ds
$$
therefore $v''(t)=-L\,t^q\,v^p(t)$.
Hence $v$ satisfies:
\begin{gather}
v''(t)+Lt^qv^p(t)=0, \label{e48}\\
 v(0)=0,\quad v'(0)=1. \label{e49}
\end{gather}

\begin{lemma} \label{lem2} 
Suppose $v$ satisfies \eqref{e47}-\eqref{e48}. Then $v(t)$ has an infinite number
 of zeros on $(0,\infty)$.
\end{lemma}

\begin{proof}  
We first show $v$ has a local maximum. If not $v'>0$ and 
$v(t)\geq{c_0}$ for some $c_0>0$ on $(t_0,\infty)$ for some $t_0>0$.
Then from \eqref{e48} we see
 \begin{equation} \label{e50}
-v''\geq{L\,c_0\,t^q}
\end{equation}
Integrating  on $[t_0,t]$ gives
 $$
-v'(t)\geq{-v'(t_0)+\frac{L c_0}{q+1}[t^{q+1}-t_0^{q+1}]}.
$$
This implies
$v'(t)\to{-\infty}$ as $t\to{\infty}$ which contradicts that $v'>0$. 
Thus there exists an $M>0$ such that $v'(M)=0$ and $v(M)>{0}$. 
Thus by \eqref{e48} $v''(M)<0$. So $v$ has a local maximum at $M$.
Integrating \eqref{e48} on $[M,t]$ gives
\begin{equation} \label{e51}
 -v'(t)=L\int_{M}^t{s^q\,v^p}\,ds.
\end{equation}
Now let us assume that $v>0$ for $t>M$ then \eqref{e51} implies that 
$v'<0$ for $t>M$ hence estimating \eqref{e51} on $[M,t]$ gives
$$
-v'(t)=L\int_{M}^t{s^q v^p\,ds\geq{L v^p(t)
\big[\frac{t^{q+1}-M^{q+1}}{q+1}\big]}}
$$
which on rewriting it yields
$$
-v^{-p}\,v'\geq L\big[{\frac{t^{q+1}-M^{q+1}}{q+1}}\big].
$$
 Integrating this on $[M,t]$ gives
 $$
\frac{v^{1-p}(M)}{1-p}-\frac{v^{1-p}(t)}{1-p}
\geq{L\int_{M}^t{\frac{s^{q+1}-M^{q+1}}{q+1}}\,ds}
$$
Thus
$$
\frac{v^{1-p}(M)}{1-p}-\frac{L}{(q+1)(q+2)}
\left[t^{q+2}-(q+2)M^{q+1}t+(q+1)M^{q+2}\right]
\geq{\frac{v^{1-p}(t)}{1-p}}
$$
the left side of which goes to $-\infty$ as $t\to{\infty}$ since
 $L>0$ and $q>0$ hence
$$
\frac{v^{1-p}(t)}{1-p}\to{-\infty}\quad \text{as }t\to{\infty}.
$$
This is a contradiction since we assumed $v>0$ for $t>M$. 
Hence there exists $z_1>0$ such that $v(z_1)=0$ and $v>0$ on $[M,z_1]$ 
so $v'(z_1)\leq{0}$.
In addition by uniqueness of solution of initial value problems it 
follows that $v'(z_1)<0$. Similarly we can then show that $v$ has a local minimum, 
$m>z_1$ and that $v$ has a second zero $z_2>m$.
Proceeding similarly we can show that $v$ has infinitely many zeros on 
$[0,\infty)$.
This completes the proof.
\end{proof}

Since $v_\lambda\to{v}$ uniformly on compact sets and since $v$ has an infinite 
number of zeros, so   $v_\lambda$ has large number of zeros for large 
values of $\lambda$ and hence $u_1$ satisfying \eqref{D7}-\eqref{D8} 
has a large number of zeros for large $a$.

 
\begin{proof}[Proof of Theorem \ref{thm1}]
From Lemma \ref{lem2} we see that $u_1$ has a first zero, $z_1(a)$, if $a$ is 
sufficiently large. Note by continuous dependence on initial conditions 
$z_1(a)$ is a continuous function of $a$. Now choose $R>0$ small enough 
so that $z_1(a)<R^{2-N}$. Then by Lemma \ref{lem1} if $a$ is sufficiently 
small $t_a>R^{2-N}$ and so $u_1(t,a)>0$ on $(0, R^{2-N})$ if $a$ 
is sufficiently small.
Thus $\{a>0: z_1(a)<R^{2-N}\}$ is nonempty and bounded from below.
Then let:  
$$ 
a_{0} = \inf \{a: z_1(a)<R^{2-N}\}.
$$
Since $z_1(a)$ is continuous it follows that   $z_1(a_0)\leq{R^{2-N}}$.

 Now suppose $z_1(a_0)< R^{2-N}$.
Then by the continuous dependency of the solutions  on initial conditions we have 
$z_1(a)<R^{2-N}$  for $a< a_0$ with $a$ sufficiently close to $a_0$ which
contradicts the definition of $a_0$. So $z_1(a_0)=R^{2-N}$ and hence 
$u_1(z_1(a_0))=u_1(R^{2-N})=0$.
Now let $U_0(r)=u_1(r^{2-N},a_0)$. Then $U_0$ satisfies \eqref{DE} and \eqref{IC}
 with $b=u_1'(R^{2-N})(2-N)R^{1-N}>0$, $\lim_{r\to{\infty}}U_0(r)=0$ and
 $U_0>0$ on $(R,\infty)$.

Next using Lemma \ref{lem2} we see that if $a$ is sufficiently large then $u_1$ has 
two zeros $z_1(a)$ and $z_2(a)$ with  $u_1>0$ on $(0,z_1(a))$ and 
$u_1<0$ on $(z_1(a),z_2(a))$.
Choose $R>0$ small enough so that $z_2(a)<R^{2-N}$.
Let 
$$ 
a_1 = \inf \{a|\,\, z_2(a)<R^{2-N}\}.
$$
As above we can show that $z_2(a_1)=R^{2-N}$.
Let $U_1(r)=-u_1(r^{2-N},a_1)$.
Then $U_1'(R)=(N-2)R^{1-N}u_1(R^{2-N})>0$, $U_1$ has one zero on $(R,\infty)$, 
satisfies \eqref{DE} and \eqref{IC} and $\lim_{r\to{\infty}}U_1(r)=0$.
Proceeding inductively, we obtain $a_2, a_3, a_4, \dots,a_n$ for any non negative 
integer $n$ such that $u_1(R^{2-N},a_n)=0$ by choosing $R$ sufficiently small.
Define $U_n(r)=(-1)^n u_1(r^{2-N},a_n)$. 
Then $U_n'(R)=(-1)^n (2-N)R^{1-N} u_1'(R^{2-N},a_n)>0$ and as above we can 
show that $U_n$ has $n$ zeros on $(R,\infty)$, satisfies \eqref{DE} and 
\eqref{IC} and $\lim_{r\to{\infty}}U_n(r)=0$. This completes the proof.
\end{proof}

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\end{document}