\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{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 $02$ 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 $20$ 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 $00$ 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 $0 0, \label{DE} \\ u(R) = 0, u'(R) = b \in {\mathbb R}. \label{DE2} \end{gather} We will also assume that \begin{itemize} \item[(H4)] there exist constants$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$,$00$is sufficiently small. \end{theorem} In addition we also prove: \begin{theorem} \label{thm2} Let$N > 2$,$00$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$N>2$,$00$. 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$R0$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$00$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$N>2$,$00$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$(M_{b}, z_{b})gives \begin{equation} \begin{aligned} \int_{0}^{u(M_{b})} \frac{dt}{\sqrt{F(u(M_b))-F(t)}} &= \int_{M_{b}}^{z_{b}} \frac{-u'(t)}{\sqrt{2}\sqrt{F(u(M_b)) - F(u(t))}} \, dt \\ & \leq \int_{M_{b}}^{z_{b}}\sqrt{K} \, dt \\ & \leq \frac{k_{2}}{\frac{\alpha}{2}-1} ( M_{b}^{1 - \frac{\alpha}{2}} -z_{b}^{1 - \frac{\alpha}{2}}) \\ &\leq \frac{k_{2}}{\frac{\alpha}{2}-1} M_{b}^{1 - \frac{\alpha}{2}}. \end{aligned} \label{auh2o} \end{equation} Similarly ifu(r)>0$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$u(M_{b}) \geq \gamma. Now using \eqref{JFK} in \eqref{auh2o} and rewriting gives \begin{equation} \begin{aligned} [u(M_{b})]^{\frac{1-p}{2}} M_{b}^{\frac{\alpha}{2}-1} &\leq \frac{k_{2}}{\frac{\alpha}{2}-1} \sqrt{\frac{C_2}{p+1} + \frac{F_{0}}{[u(M_{b})]^{p+1}}} \\ &\leq \frac{k_{2}}{\frac{\alpha}{2}-1}\sqrt{\frac{C_{2}}{p+1} + \frac{F_{0}}{\gamma^{p+1}}}. \label{LBJ} \end{aligned} \end{equation} This completes the proof. \end{proof} \begin{proof}[Proof of Theorem \ref{thm2}] Ifu$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$02$,$00$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$00. \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 $N>2$, $00$ 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 $0 b_1t$. 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 $N>2$, $02$, $00$ 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 $00$ 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 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 $N>2$, $0 M_{b_1}$ 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 $N>2$, $0 M_{b_1}$ 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