\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 253, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/253\hfil Sublinear equations on exterior domains] {Existence of solutions for sublinear equations on exterior domains} \author[J. A. Iaia \hfil EJDE-2017/253\hfilneg] {Joseph A. Iaia} \address{Joseph A. Iaia \newline Department of Mathematics University of North Texas P.O. Box 311430 Denton, TX 76203-1430, USA} \email{iaia@unt.edu} \dedicatory{Communicated by Ira Herbst} \thanks{Submitted February 12, 2017. Published October 10, 2017.} \subjclass{34B40, 35B05} \keywords{Exterior domains; semilinear; sublinear; radial} \begin{abstract} In this article we prove the existence of an infinite number of radial solutions of $\Delta u + K(r)f(u)= 0$, one with exactly $n$ zeros for each nonnegative integer $n$ on the exterior of the ball of radius $R>0$, $B_{R}$, centered at the origin in ${\mathbb R}^{N}$ with $u=0$ on $\partial B_{R}$ and $\lim_{r \to \infty} u(r)=0$ where $N>2$, $f$ is odd with $f<0$ on $(0, \beta)$, $f>0$ on $(\beta, \infty)$, $f(u)\sim u^p$ with $00$ centered at the origin in ${\mathbb R}^{N}$ and $K(r)>0$. We assume: \begin{itemize} \item[(H1)] $f$ is odd and locally Lipschitz, $f<0$ on $(0, \beta)$, $f>0$ on $(\beta , \infty)$, $f'(\beta)>0$, and $f'(0)<0$. %\label{f} \tag{H1} \item[(H2)] there exists $p$ with $00$ \ on $(\gamma, \infty)$. %\label{F2} \tag{H3} \end{itemize} Interest in the topic for this paper comes from recent papers \cite{C,CSS,LSS,C2,S} about solutions of differential equations on exterior domains. When $f$ grows superlinearly at infinity - i.e. $\lim_{u \to \infty} \frac{f(u)}{u} = \infty$, $\Omega = {\mathbb R}^{N}$, and $K(r)\equiv 1$ then the problem \eqref{1}, \eqref{3} has been extensively studied \cite{BL,BL2,B,cdgm,cgy,JK,M,serrtang,ST}. In \cite{I5,I7} equations \eqref{1}-\eqref{3} were studied with $K(r) \sim r^{-\alpha}$, $f$ superlinear, and $\Omega = {\mathbb R}^{N} \backslash B_{R}$ with $R>0$ with various values for $\alpha$. In those papers we proved existence of an infinite number of solutions - one with exactly $n$ zeros for each nonnegative integer $n$ such that $u \to 0 \text{ as } |x| \to \infty$ for all $R>0$. In \cite{I4} we studied \eqref{1}-\eqref{3} with $K(r) \sim r^{-\alpha}$, $f$ bounded, and $\Omega = {\mathbb R}^{N} \backslash B_{R}$. In this article we consider the case where $f$ grows sublinearly at infinity - i.e. $\lim_{u \to \infty} \frac{f(u)}{u^p} = c_{0}>0$ with $0 N - p(N-2)$ was investigated. Since we are interested in radial solutions of \eqref{1}-\eqref{3} we assume that $u(x) = u(|x|) = u(r)$ where $x \in {\mathbb R}^{N}$ and $r=|x|$=$\sqrt{x_{1}^2 + \cdots + x_{N}^2}$ so that $u$ solves \begin{gather} u''(r) + \frac{N-1}{r} u'(r) + K(r)f(u(r)) = 0 \quad\text{on } (R, \infty), \text{ where } R > 0, \label{DE} \\ u(R) = 0, u'(R) = b \in {\mathbb R}. \label{DE2} \end{gather} We will also assume that there exist constants $k_{1}>0$, $k_{2}>0$, and $\alpha$ with $0< \alpha < 2$ such that \begin{itemize} \item[(H4)] $k_{1} r^{-\alpha} \leq K(r) \leq k_{2} r^{-\alpha}$ on $[R, \infty)$. % \label{K} \tag{H4} \item[(H5)] $K$ is differentiable, on $[R, \infty)$, $\lim_{r \to \infty} \frac{rK'}{K} = -\alpha$ where $0< \alpha <2$, and $\frac{rK'}{K} + 2(N-1)>0$ on $[R, \infty)$. %\label{K2} \tag{H5} \end{itemize} Note that (H5) implies $r^{2(N-1)}K(r)$ is increasing. In this paper we prove the following result. \begin{theorem} \label{thm1} Let $N > 2$, $0 2$. When $\alpha> 2$ and $R$ is sufficiently large then it was shown in \cite{I8,JJ} that there are {\it no} solutions of \eqref{1}-\eqref{3} with $\lim_{r \to \infty} u(r)=0$. On the other hand, it was also shown in \cite{I8,JJ} that if $R>0$ is sufficiently small then solutions of \eqref{1}-\eqref{3} exist for $\alpha > N - p(N-2)$. We note in Theorem \ref{thm1} that existence of solutions is established {\it for all} $R>0$. Also to the best of our knowledge existence of solutions of \eqref{1}-\eqref{3} is still unknown when $2< \alpha < N-p(N-2)$, $00$ sufficiently small. \section{Preliminaries} From the standard existence-uniqueness theorem for ordinary differential equations \cite{BR} it follows there is a unique solution of \eqref{DE}-\eqref{DE2} on $[R, R+\epsilon)$ for some $\epsilon>0$. We then define \begin{equation} E = \frac{1}{2} \frac{u'^{2}}{K} + F(u). \label{energy} \end{equation} Using (H5) we see that \begin{equation} E' = -\frac{u'^2}{2rK}\Big(2(N-1) + \frac{rK'}{K}\Big) \leq 0 \quad \text{for } 0 < \alpha < 2(N-1). \label{energy2} \end{equation} Thus $E$ is nonincreasing. Hence it follows that \begin{equation} \frac{1}{2} \frac{u'^{2}}{K} + F(u)= E(r) \leq E(R)=\frac{1}{2} \frac{b^2}{K(R)} \quad \text{for } r\geq R \label{energy4} \end{equation} and so we see from (H2)--(H4) that $u$ and $u'$ are uniformly bounded wherever they are defined from which it follows that the solution of \eqref{DE}-\eqref{DE2} is defined on $[R, \infty)$. \begin{lemma} \label{lem1} Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold. Let $N>2$, $0R$ gives $$r^{N-1}u' = r_{0}^{N-1} u'(r_{0}) - \int_{r_{0}}^{r} t^{N-1}K f(u) \, dt.$$ Dividing by $r^{N}K$ gives \begin{equation} \frac{u'}{rK} = \frac{r_{0}^{N-1} u'(r_{0})}{r^{N}K} - \frac{\int_{r_{0}}^{r} t^{N-1}K f(u) \, dt}{r^{N} K}. \label{walrus} \end{equation} Using (H4) and that $0<\alpha<2R$ such that \begin{equation} |u'| \geq \frac{|f(L)|k_{1}}{2(N-\alpha)}r^{1-\alpha} >0 \quad\text{for } r > r_{0}. \label{yellow sub} \end{equation} Integrating \eqref{yellow sub} on $(r_{0},r)$ then gives \begin{equation} |u(r)-u(r_{0})| \geq \frac{|f(L)|k_{1}}{2(N-\alpha)(2 - \alpha)}(r^{2-\alpha} - r_{0}^{2- \alpha}). \label{64} \end{equation} Since $0 < \alpha <2$ we see the right-hand side of \eqref{64} goes to $+\infty$ but the left-hand side goes to $|L-u(r_{0})|$ - a contradiction. Thus it must be that $f(L)=0$. \end{proof} \begin{lemma} \label{lem2} Let $u$ satisfy \eqref{DE}-\eqref{DE2} with $b>0$ and suppose {\rm (H1)--(H5)} hold. Let $N>2$, $0R$ such that $u(t_{\epsilon,b}) = \beta - \epsilon$ and $u'>0$ on $[R, t_{\epsilon,b}]$. \end{lemma} \begin{proof} From \eqref{DE2} and since $b>0$ by assumption we see that $u$ is initially increasing and positive. Now if $u$ has a first critical point, $M$, with $u'>0$ on $[R, M)$ then $u'(M)=0$ and $u''(M) \leq 0$ from which it follows that $f(u(M)) \geq 0$. In addition, by uniqueness of solutions of initial value problems it follows that $u''(M)<0$ and so $f(u(M))>0$ and thus $u(M)> \beta$. Since $u(R)=0$ the lemma then follows in this case by the intermediate value theorem. Otherwise suppose the lemma does not hold. Then $u'>0$ and $0< u < \beta - \epsilon$ for all $r> R$ for some $\epsilon>0$ and so by (H1) there exists a constant $\epsilon_{1}>0$ and $r_{0}>R$ such that $f(u) \leq - \epsilon_{1}<0$ for $r > r_{0}>R$. Next multiplying \eqref{DE} by $r^{N-1}$, integrating on $(r_{0},r)$, and using (H4) gives \begin{align*} -r^{N-1}u' &= -r_{0}^{N-1}u'(r_{0}) + \int_{r_{0}}^{r} t^{N-1}K f(u) \, dt\\ &\leq -r_{0}^{N-1}u'(r_{0}) - \frac{\epsilon_{1} k_{1}}{N-\alpha} (r^{N-\alpha} - r_{0}^{N-\alpha}). \end{align*} Thus for some constant $C_{1}$, \begin{equation} u' \geq C_{1} r^{1-N} + \frac{\epsilon_{1} k_{1}}{N-\alpha} r^{1-\alpha}. \label{hole in pocket} \end{equation} Integrating on $(r_{0}, r)$ gives: \begin{equation} u(r) \geq u(r_{0}) + \frac{C_{1}}{2-N} (r^{2-N}- r_{0}^{2-N}) + \frac{\epsilon_{1}k_{1}}{(N-\alpha)(2-\alpha)} (r^{2-\alpha} - r_{0}^{2 - \alpha}). \label{george} \end{equation} Now the left-hand side of \eqref{george} is bounded above by $\beta$ but the right-hand side goes to $+\infty$ as $r \to \infty$ since $0< \alpha <22$, $0R$ such that $u(t_{b}) = \beta$ and $u'>0$ on $[R,t_{b}]$. \end{lemma} \begin{proof} We rewrite \eqref{DE} as $$u'' + \frac{N-1}{r} u' + K(r)\frac{f(u)}{u-\beta} (u- \beta) = 0$$ and then make the change of variables \begin{equation} u - \beta = r^{\frac{1-N}{2}} v. \label{pennsylvania} \end{equation} Thus $v$ satisfies $$v'' + \Big(K(r)\frac{f(u)}{u-\beta} - \frac{(N-1)(N-3)}{4r^2}\Big)v = 0.$$ Suppose now that the lemma does not hold. Then by Lemma \ref{lem2} we see for some sufficiently small $\epsilon >0$ we have $u'>0$, $\beta - \epsilon < u < \beta$, and $\frac{f(u)}{u-\beta} > \frac{f'(\beta)}{2}$ (by (H1)) for $r > t_{\epsilon,b}$. Also for some $r_{0}>R$ sufficiently large then by (H4) and since $0< \alpha < 2$, $K(r)\frac{f(u)}{u-\beta} - \frac{(N-1)(N-3)}{4r^2} \geq \frac{k_{1}f'(\beta)}{2r^{\alpha} }- \frac{(N-1)(N-3)}{4r^2} \geq \frac{1}{r^2} \text{ for } r > r_{0}. %\label{ringo}$ Next we consider a nontrivial solution $w$ of $$w'' + \frac{1}{r^2} w = 0 \text{ for } r >r_{0}.$$ It is straightforward to show $w = C_{2}e^{r/2}\sin\big(\frac{\sqrt{3}}{2} \ln(r) + C_3\big)$ for constants $C_{2}\neq 0$ and $C_3$. Hence $w$ has an infinite number of zeros on $(r_{0}, \infty)$. It follows by the Sturm comparison theorem \cite{BR} that between any two zeros of $w$ then $v$ must have a zero and from \eqref{pennsylvania} we see that $u$ must equal $\beta.$ Hence there exists a smallest value of $r$, denoted $t_{b}$, such that $u(t_{b}) = \beta$ and $00$. Also from Lemma \ref{lem2} we have $u'>0$ on $[R, t_{\epsilon,b}]$ for all $\epsilon > 0$ and since $u'(t_{b})>0$ it follows that $u'>0$ on $[R, t_{b}]$. This completes the proof. \end{proof} \begin{lemma} \label{lem4} Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold. Let $N>2$, $00$ such that $-f(u) \leq \epsilon_{2}u$ on } [0, \beta/2]. \label{kansas} \end{equation} Next we define $t_{b_0}0$ such that $u(t_{b_0}) = \frac{\beta}{2}$. The existence of $t_{b_0}$ follows from Lemma \ref{lem3}, since $u(R)=0$, and the intermediate value theorem. Combining \eqref{kansas} with (H4) gives \begin{equation} -r^{N-1}K f(u) \leq \epsilon_{2} k_{2} r^{N-1-\alpha}u \text{ on } [R, t_{b_0}]. \label{mckinley} \end{equation} Integrating \eqref{pierce} on $[R, t_{b_0}]$, using \eqref{mckinley} and that $u$ is increasing on $[R, t_{b_0}]$ (by Lemma \ref{lem3}) gives \begin{equation} \begin{aligned} r^{N-1}u' &\leq R^{N-1}b + \int_{R}^{r} \epsilon_{2} k_{2} t^{N-1-\alpha}u(t) \, dt \\ &\leq R^{N-1}b + \epsilon_{2} k_{2} u(r) \int_{R}^{r} t^{N-1-\alpha} \, dt \\ &\leq R^{N-1}b + \frac{\epsilon_{2} k_{2} }{N-\alpha}r^{N-\alpha}u. \end{aligned} \label{buchanan} \end{equation} Rewriting this inequality gives \begin{equation} u' - \frac{\epsilon_{2} k_{2}}{N-\alpha} r^{1-\alpha}u \leq R^{N-1}br^{1-N}. \label{gehrig} \end{equation} Now let $\epsilon_{3} = \frac{\epsilon_2 k_{2}}{(2-\alpha)(N-\alpha)}>0$ and denote \begin{equation} \mu(r) = e^{-\epsilon_{3}(r^{2-\alpha}- R^{2-\alpha})}\leq 1 \quad \text{for } R\leq r \leq t_{b_0}. \label{maryland} \end{equation} Multiplying \eqref{gehrig} by $\mu(r)$, using \eqref{maryland}, and integrating on $[R, r]\subset[R, t_{b_0}]$ gives \begin{equation} u \leq \frac{R^{N-1}b}{N-2}(R^{2-N} - r^{2-N})e^{\epsilon_{3}(r^{2-\alpha}- R^{2-\alpha})}. \label{rhizutto} \end{equation} Now evaluating \eqref{rhizutto} at $t_{b_0}$ gives \begin{equation} \frac{\beta}{2} \leq \frac{R^{N-1}b}{N-2}(R^{2-N} - t_{b_0}^{2-N}) e^{\epsilon_3(t_{b_0}^{2-\alpha}- R^{2-\alpha})}. \label{dimaggio} \end{equation} Since $0 < \alpha < 2$ it follows from \eqref{dimaggio} that $\lim_{ b \to 0^{+}} t_{b_0} = \infty$ and since $t_{b_0}< t _{b}$ it follows that $$\lim_{b \to 0^{+}} t_{b} = \infty.$$ This completes the proof. \end{proof} \begin{lemma} \label{lem5} Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {\rm (H1)--(H5)} hold. Let $N>2$, $00$ on $[R, M_{b})$. \end{lemma} \begin{proof} From Lemma \ref{lem3} we know $u(t_{b})= \beta$ and $u'(t_{b})>0$ so if the lemma does not hold then it follows from Lemma \ref{lem3} that $u'>0$ for $r\geq R$. Since $u$ is bounded by \eqref{energy4} then it follows from (H2) and (H3) that there exists an $L$ such that $u\to L> \beta$ with $L$ finite. We see then by Lemma \ref{lem1} that $f(L)=0$ and so (H1) implies $|L| \leq \beta$ contradicting that $L> \beta$. Thus $u$ has a local maximum and so there is a smallest value of $t$, denoted $M_{b}$, such that $u'(M_{b})=0$ and $u'>0$ on $[R, M_{b})$. This completes the proof. \end{proof} \begin{lemma} \label{lem6} Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose [(H1)--(H5)] hold. Let $N>2$, $0 0$ if $b>0$ is sufficiently small. \end{lemma} \begin{proof} We use a similar argument as in \cite{I7}. First, if $u'>0$ for $r\geq R$ then $u>0$ for all $r >R$ and so we are done in this case. Thus we suppose that $u$ has a first critical point $M_{b}$. Then $u'(M_{b})=0$, $u''(M_{b}) \leq 0$, and $u'>0$ on $[R, M_{b})$. By uniqueness of solutions of initial value problems it follows that $u''(M_{b})<0$ and thus $M_{b}$ is a local maximum. Now if $0< u(M_{b})< \gamma$ then it follows that $E(M_{b})=F(u(M_{b}))<0$ (by (H3)). Since $E$ is nonincreasing by \eqref{energy2} it follows that $u$ cannot be zero for $r> M_{b}$ for at such a zero, $z_{b}$, of $u$ we would have $0 \leq \frac{1}{2} \frac{u'^{2}(z_{b})}{K(z_{b})} = E(z_{b}) \leq E(M_{b}) < 0$ a contradiction. So we suppose now that $u(M_{b}) \geq \gamma$. Then there exists $t_{b_1}$ with $t_{b} < t_{b_1} < M_{b}$ so that $u(t_{b_1}) = \frac{\beta + \gamma}{2}$ and $u'>0$ on $[R, M_{b})$. Next we have the following identity which follows from \eqref{DE} and \eqref{energy2}, \begin{equation} (r^{2(N-1)}KE)' = (r^{2(N-1)}K)' F(u). \label{virginia} \end{equation} Integrating this on $[R,r]$ gives \begin{equation} r^{2(N-1)}KE = \frac{1}{2} R^{2(N-1)}b^2 + \int_{R}^{r} (t^{2(N-1)}K)' F(u) \, dt. \label{wilson} \end{equation} By (H3) we have $F(u) \leq 0$ on $[R, t_{b}]$ and by (H5) we have $(r^{2(N-1)}K)'>0$ so for $R < t_{b} < r$ we have \begin{equation} \int_{R}^{r} (t^{2(N-1)}K)' F(u) \, dt \leq \int_{t_{b}}^{r} (t^{2(N-1)}K)' F(u) \, dt. \label{grant} \end{equation} Next on $[\beta, \frac{ \beta+\gamma}{2}]$ it follows that there exists an $\epsilon_{4}>0$ such that $F(u) \leq -\epsilon_{4} < 0$. Also from (H5) we see there is a $k_{0}>0$ such that \begin{equation} 2(N-1) + \frac{rK'}{K} \geq k_{0} \text{ for } r\geq R. \label{fillmore} \end{equation} Then it follows from \eqref{fillmore} and (H4) that \begin{equation} (t^{2(N-1)}K)' = t^{2N-3}K[2(N-1) + \frac{rK'}{K}] \geq k_{0}k_{1} t^{2N-3-\alpha} \text{ for } t\geq R. \label{van buren} \end{equation} Thus from \eqref{wilson}-\eqref{van buren} we see \begin{equation} t_{b_1}^{2(N-1)}K(t_{b_1}) E(t_{b_1}) \leq \frac{1}{2} R^{2(N-1)}b^2 - \frac{\epsilon_{4}k_{0}k_{1}}{2N-2-\alpha}[ t_{b_1}^{2N-2-\alpha} - t_{b}^{2N-2-\alpha} ]. \label{hayes} \end{equation} Next solving \eqref{energy4} for $u'$, using (H4), and integrating on $[t_{b}, t_{b_{1}}]$ where $u'>0$ gives \begin{equation} \begin{aligned} \int_{\beta}^{\frac{\beta + \gamma}{2}} \frac{ dt}{\sqrt{\frac{b^{2}}{K(R)} - 2F(t)}} &=\int_{t_{b}}^{t_{b_1}} \frac{ u'(r) \, dr}{\sqrt{\frac{b^{2}}{K(R)} - 2F(u(r))}} \leq \int_{t_{b}}^{t_{b_1}} \sqrt{K} \, dr \\ &= \frac{\sqrt{k_{2}}}{1- \frac{\alpha}{2}}( t_{b_{1}}^{1- \frac{\alpha}{2}} - t_{b}^{1- \frac{\alpha}{2}}) \end{aligned}\label{maine} \end{equation} and so by (H4) we see from \eqref{maine} that for small $b>0$ we have \begin{equation} 0< \frac{1}{2}\int_{\beta}^{\frac{\beta + \gamma}{2}} \frac{ \, dt}{\sqrt{ - 2F(t)}} \leq \int_{\beta}^{\frac{\beta + \gamma}{2}} \frac{ dt}{\sqrt{\frac{b^{2}}{K(R)} - 2F(t)}} \leq \frac{\sqrt{k_{2}}}{1- \frac{\alpha}{2}} \big( t_{b_{1}}^{1- \frac{\alpha}{2}} - t_{b}^{1- \frac{\alpha}{2}}\big). \label{cleveland} \end{equation} It follows then from \eqref{cleveland} and since $0 < \alpha < 2$ that there exists an $\epsilon_5>0$ such that \begin{equation} t_{b_{1}}^{1- \frac{\alpha}{2}} \geq t_{b}^{1- \frac{\alpha}{2}} + \epsilon_{5}. \label{bruce} \end{equation} From the inequality \begin{equation} (x+y)^{l} \geq x^l + y^l \label{clarence} \end{equation} which holds if $l\geq 1$, $x\geq 0$, and $y\geq 0$, it follows from \eqref{bruce} and since $\frac{2}{2-\alpha} \geq 1$ that \begin{equation} t_{b_{1}} \geq t_{b} + \epsilon_6 \label{vermont} \end{equation} where $\epsilon_6 = \epsilon_5^{\frac{2}{2-\alpha}}$. Next from \eqref{bruce}-\eqref{vermont} we see \begin{equation} \begin{aligned} t_{b_1}^{2N-2-\alpha} - t_{b}^{2N-2-\alpha} &= [t_{b_1}^{N-1-\frac{\alpha}{2}} - t_{b}^{N-1-\frac{\alpha}{2}}] [t_{b_1}^{N-1-\frac{\alpha}{2}} + t_{b}^{N-1-\frac{\alpha}{2}}] \\ &\geq [ (t_{b}+\epsilon_6)^{N-1-\frac{\alpha}{2}} - t_{b}^{N-1-\frac{\alpha}{2}}]t_{b}^{N-1-\frac{\alpha}{2}} \\ &\geq \epsilon_7 t_{b}^{N-1-\frac{\alpha}{2}} \end{aligned}\label{little steve} \end{equation} where $\epsilon_7 = \epsilon_6^{N-1-\frac{\alpha}{2}} > 0$ and since $N-1-\frac{\alpha}{2}\geq 1$ by (H5). Thus we see it follows from \eqref{hayes}, \eqref{little steve}, and Lemma \ref{lem4} that $$t_{b_{1}}^{2(N-1)}K(t_{b_1})E(t_{b_1}) \leq \frac{1}{2} R^{2N-2}b^2 - \frac{\epsilon_{4}\epsilon_{7}k_{0}k_{1}}{2N-2-\alpha}t_{b}^{N-1-\frac{\alpha}{2}} \to -\infty \text{ as } b\to 0^{+}.$$ Therefore, $E(t_{b_{1}})<0$ if $b>0$ is sufficiently small. It then follows that $u(t)> 0$ for $t > t_{b_{1}}$ for if there were a $z_{b} > t_{b_1}$ such that $u(z_{b})=0$ then since $E$ is nonincreasing we would have $0 \leq E(z_{b}) \leq E(t_{b_1})<0$ - a contradiction. In addition we know from earlier that $u>0$ on $(R, M_{b}]$ and $R< t_{b_1}0$ on $(R, \infty)$. This completes the proof. \end{proof} \begin{lemma} \label{lem7} Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {(H1)--(H5)} hold. Let $N>2$, $0 R$ such that $M_{b} \leq M_{0}$ for all large $b$. Now let $v_{b} = \frac{u}{b}$. Then $v_{b}(R)=0$, $v_{b}'(R) =1$ and $v_{b}$ satisfies \begin{equation} v_{b}'' + \frac{N-1}{r} v_{b}' + \frac{K(r) f(bv_{b})}{b} = 0 \quad\text{for } r \geq R. \label{ike} \end{equation} As in \eqref{energy}-\eqref{energy2}, $$\Big( \frac{1}{2} \frac{v_{b}'^{2}}{K(r)} + \frac{F(bv_{b})}{b^2} \Big)' \leq 0 \quad \text{for } r \geq R$$ and therefore $$\frac{1}{2} \frac{v_{b}'^{2}}{K(r)} + \frac{F(bv_{b})}{b^2} \leq \frac{1}{2K(R)} \quad\text{for } r \geq R.$$ It follows from this that the $v_{b}'$ are uniformly bounded on $[R, \infty)$ and since $|v_{b}(r)|\leq \int_{R}^{r} |v_{b}'(s)| \, ds$ it follows that the $v_{b}$ are uniformly bounded on $[R, M_{0}+1]$. Since $f$ is sublinear we now show that $\frac{K(r) f(bv_{b})}{b} \to 0$ on $[R, M_{0}+1]$ as $b \to \infty$. To see this note that from (H2) we have $\frac{|g(u)|}{|u|^p} \leq 1$ if $|u|\geq u_{0}>0$ and since $g$ is continuous on $[0,u_{0}]$ then $|g(u)|\leq C_{4}$ for $|u| \leq u_{0}$ for some constant $C_4$. Combining these we see: \begin{equation} |g(u)| \leq C_{4} + |u|^{p} \text{ for all } u. \label{georgia} \end{equation} Therefore since the $v_{b}$ are uniformly bounded on $[R, M_{0}+1]$ and $00$. On the other hand, $v_{b}'(M_{b}) =0$ and since the $M_{b}$ are bounded by $M_{0}$ then there is a subsequence (still labeled $M_{b}$) such that $M_{b} \to M$ and since $v_{b}' \to v'$ uniformly on $[R, M_{0}+1]$ then $02$, $00$ such that $$[u(M_{b})]^{\frac{1-p}{2}} \geq \epsilon_{5} M_{b}^{1 - \frac{\alpha}{2}}.$$ \end{lemma} \begin{proof} It follows from Lemma \ref{lem6} that $$\frac{u}{b} \to \frac{R}{N-2}\Big(1 - \Big(\frac{R}{r}\Big)^{N-2}\Big)\quad \text{uniformly on } [R, 2R].$$ Hence $$\frac{u(2R)}{b} \to \frac{R}{N-2}\left(1 - 2^{2-N} \right) \quad \text{as } b \to \infty.$$ Thus $u(2R) \geq \frac{R}{2(N-2)}\left(1 - 2^{2-N} \right)b$ for sufficiently large $b$, and therefore $u(2R) \to \infty$ as $b \to \infty$. Since $M_{b} \to \infty$ as $b \to \infty$ (by Lemma \ref{lem7}), it follows that $M_{b} > 2R$ for large $b$, and since $u$ is increasing on $[R, M_{b})$ it follows that $u(M_{b}) \geq u(2R)\to \infty$ from which it follows that $u(M_{b}) \to \infty$ as $b \to \infty$. This completes the first part of the proof. Next, from \eqref{energy}-\eqref{energy2} we have $$\frac{1}{2} \frac{u'^2}{K} + F(u) \geq F(u(M_{b})) \quad\text{on } [R, M_{b}].$$ Rewriting this, integrating on $[R, M_{b}]$ and using (H4) gives \begin{equation} \begin{aligned} \int_{R}^{M_{b}} \frac{u'}{\sqrt{2}\sqrt{F(u(M_{b})) - F(u(t))}} & \geq \int_{R}^{M_{b}} \sqrt{K} \, dr \\ &\geq \int_{R}^{M_{b}} \sqrt{k_{1}} r^{-\frac{\alpha}{2}} \, dr \\ &= \frac{\sqrt{k_{1}}(M_{b}^{1 - \frac{\alpha}{2}} - R^{1 - \frac{\alpha}{2}})}{1 - \frac{\alpha}{2}}. \end{aligned} \label{taft} \end{equation} Changing variables on the left-hand side, rewriting, and changing variables again gives \begin{equation} \begin{aligned} \int_{R}^{M_{b}} \frac{u'}{\sqrt{2}\sqrt{F(u(M_{b})) - F(u(t))}} &= \int_{0}^{u(M_{b})} \frac{dt}{\sqrt{2}\sqrt{F(u(M_{b})) - F(t)}} \\ &= \frac{u(M_{b})}{\sqrt{2}\sqrt{F(u(M_{b}))}} \int_{0}^{1} \frac{ds}{\sqrt{1 - \frac{F(u(M_{b})s)}{F(u(M_{b}))}}}. \end{aligned}\label{coolidge} \end{equation} From the first part of the theorem we know that $u(M_{b})\to \infty$ as $b \to \infty$. Then from (H2) it follows that $F(u) = \frac{u^{p+1}}{p+1} + G(u)$ where $G(u) = \int_{0}^{u} g(s) \, ds$. In a similar way to \eqref{georgia} it follows that \begin{equation} |G(u)| \leq C_5 +\frac{1}{2(p+1)} |u|^{p+1} \quad \text{for all $u$ for some constant } C_5. \label{iowa} \end{equation} This along with (H2) and that $00$. This completes the proof. \end{proof} \begin{lemma} \label{lem9} Let $u$ satisfy \eqref{DE}-\eqref{DE2} and suppose {(H1)--(H5)} hold. Let $N>2$, $0M_{b}$ such that $u'<0$ on $(M_{b}, z_{b}]$ and $u(z_{b})=0$. In addition, given a positive integer $n$ then if $b$ is sufficiently large then $u$ has $n$ zeros on $(R, \infty)$. \end{lemma} \begin{proof} First let $v(r) = u(r+M_{b})$. Then $v(0)= u(M_{b})$, $v'(0)=u'(M_{b})=0$, and \eqref{DE} becomes \begin{equation} v'' + \frac{N-1}{r + M_{b}}v' + K(r+M_{b})\left( |v|^{p-1}v + g(v) \right) = 0. \label{eeyore} \end{equation} Next let \begin{equation} w_{\lambda}(r) = \lambda^{-\frac{2-\alpha}{1-p}} v(\lambda r) = \lambda^{-\frac{2-\alpha}{1-p}} u(\lambda r + M_{b}) \quad\text{where } \lambda^{\frac{2-\alpha}{1-p}} = u(M_{b}). \label{harrison} \end{equation} Then $w_{\lambda}(0) = \lambda^{-\frac{2-\alpha}{1-p}}u(M_{b}) = 1$ and $w_{\lambda}'(0)=0$. Now recall from Lemmas \ref{lem7} and \ref{lem8} that $M_{b}\to \infty$ and $u(M_{b}) \to \infty$ as $b \to \infty$. Thus $[u(M_{b})]^{\frac{2-\alpha}{1-p}} = \lambda \to \infty$ as $b \to \infty$. In addition we see from \eqref{eeyore}-\eqref{harrison} that $w_{\lambda}$ solves \begin{equation} w_{\lambda}'' + \frac{N-1}{r + \frac{M_{b}}{\lambda}} w_{\lambda}' + \lambda^{\alpha}K(\lambda r + M_{b}) \left[ |w_{\lambda}|^{p-1}w_{\lambda} + \lambda^{-\frac{(2-\alpha)p}{1-p}}g(\lambda^{\frac{2-\alpha}{1-p}}w_{\lambda} ) \right] = 0. \label{hoover} \end{equation} We now define \begin{equation} E_{\lambda} = \frac{1}{2} \frac{w_{\lambda}'^2}{\lambda^{\alpha}K(\lambda r + M_{b})} + \frac{1}{p+1} |w_{\lambda}|^{p+1} + \lambda^{\frac{-(2-\alpha)(1+p)}{1-p}} G(\lambda^\frac{2-\alpha}{1-p}w_{\lambda}). \label{jqa} \end{equation} Using \eqref{hoover} and (H5) we obtain \begin{equation} \begin{aligned} E_{\lambda}' &= -\frac{\lambda^{1-\alpha} w_{\lambda}'^2}{2(\lambda r + M_{b}) K(\lambda r + M_{b})} \Big( 2(N-1) + \frac{(\lambda r + M_{b})K'(\lambda r + M_{b})}{{K(\lambda r + M_{b})}} \Big) \\ &\leq 0 \quad \text{for } r \geq 0. \end{aligned} \end{equation} Thus \begin{equation} E_{\lambda}(r) \leq E_{\lambda}(0) = \frac{1}{p+1} + \lambda^{\frac{-(2-\alpha)(1+p)}{1-p}} G(\lambda^\frac{2-\alpha}{1-p}). \label{adams} \end{equation} From (H2) it follows that $\lambda^{\frac{-(2-\alpha)(1+p)}{1-p}} G(\lambda^\frac{2-\alpha}{1-p}) \to 0$ as $\lambda \to \infty$. In addition it follows from \eqref{iowa}, \eqref{jqa}, and \eqref{adams} that for large $\lambda$, \begin{equation} \frac{1}{2} \frac{w_{\lambda}'^2}{\lambda^{\alpha}K(\lambda r + M_{b})} + \frac{|w_{\lambda}|^{p+1}}{2(p+1)} \leq \frac{2}{p+1}. \label{truman} \end{equation} Hence the $w_{\lambda}$ are uniformly bounded on $[0, \infty)$. In addition, it follows from (H4) and \eqref{truman} that the $w_{\lambda}'$ are uniformly bounded by $\sqrt{\frac{4k_2}{p+1}} r^{-\alpha/2}$ on $(r, \infty)$. Then from \eqref{hoover} it follows that the $w_{\lambda}''$ are uniformly bounded by $C_{6} r^{-(\frac{\alpha}{2}+1)}$ for some constant $C_{6}$. Thus $w_{\lambda}$, $w_{\lambda}'$, and $w_{\lambda}''$ are uniformly bounded on compact subsets of $(0, \infty)$ and so by the Arzela-Ascoli theorem a subsequence (still labeled $w_{\lambda}$ and $w_{\lambda}'$) converges uniformly on compact subsets of $(0, \infty)$ to some $w$ and $w'$. In addition, by the fundamental theorem of calculus with $0\leq r_{1} < r_{2}$ we have \begin{equation} \begin{aligned} |w_{\lambda}(r_{1}) - w_{\lambda}(r_{2})| &\leq \int_{r_{1}}^{r_{2}} |w_{\lambda}'(s)| \, ds \\ &\leq \int_{r_{1}}^{r_{2}}\sqrt{\frac{4k_2}{p+1}} s^{-\alpha/2} \, ds \\ &= \sqrt{\frac{4k_2}{p+1}} [ r_2^{1-\alpha/2} - r_{1}^{1-\alpha/2} ] \end{aligned} \label{donovan} \end{equation} and so we see from \eqref{donovan} and since $0 < \alpha <2$ that the $w_{\lambda}$ are equicontinuous on compact subsets of $[0,\infty)$. Thus it follows that $w(r)$ is continuous on $[0,\infty)$ and in particular $w(0)=1$. Next we show that $w_{\lambda}$ has a large number of zeros for large $\lambda$ and hence $u$ has a large number of zeros for large $b$. So suppose $w>0$ on $[0,\infty)$. We see then from \eqref{hoover} and (H4) that \begin{equation} \begin{aligned} & -(r + \frac{M_{b}}{\lambda})^{N-1} w_{\lambda}' \\ &=\int_{0}^{r} \lambda^{\alpha}K(\lambda(r + \frac{M_{b}}{\lambda})) (r + \frac{M_{b}}{\lambda})^{N-1} \Big( |w_{\lambda}|^{p-1}w_{\lambda} + \lambda^{-\frac{(2-\alpha)p}{1-p}} g(\lambda^\frac{2-\alpha}{1-p} w_{\lambda}) \Big). \end{aligned}\label{harding} \end{equation} We claim now that \begin{equation} \lim_{\lambda \to \infty} \int_{0}^{r} \lambda^{\alpha} K(\lambda(r + \frac{M_{b}}{\lambda}))(r + \frac{M_{b}}{\lambda})^{N-1} \Big( \lambda^{-\frac{(2-\alpha)p}{1-p}} g(\lambda^\frac{2-\alpha}{1-p} w_{\lambda}) \Big) =0 \label{new york} \end{equation} on any fixed compact subset of $[0, \infty)$. To see this note as in \eqref{georgia} we can similarly obtain the inequality $|g(u)| \leq C_7 + \epsilon |u|^{p}$ for all $u$ for some constant $C_{7}$. Therefore using this and (H4) in \eqref{harding} we see that \begin{equation} \begin{aligned} &\Big|\int_{0}^{r} \lambda^{\alpha}K(\lambda(t + \frac{M_{b}}{\lambda})) (t + \frac{M_{b}}{\lambda})^{N-1} \left( \lambda^{-\frac{(2-\alpha)p}{1-p}} g(\lambda^\frac{2-\alpha}{1-p} w_{\lambda}) \right) \, dt \Big| \\ &\leq \int_{0}^{r} k_2 (t + \frac{M_{b}}{\lambda})^{N-1-\alpha} \left( C_7 \lambda^{-\frac{(2-\alpha)p}{1-p}} + \epsilon |w_{\lambda}|^{p} \right) \, dt. \end{aligned} \label{bitten} \end{equation} Now it follows from \eqref{bbragg} and \eqref{harrison} that $M_{b} \leq \epsilon_{6} [u(M_{b})]^{\frac{1-p}{2-\alpha}} = \epsilon_6 \lambda$ where $\epsilon_{6} = \epsilon_{5}^{-\frac{2}{2-\alpha}}$ so that for some subsequence $\frac{M_{b}}{\lambda} \to A$ with $0\leq A \leq \epsilon_{6}$ and thus for large $\lambda$ we obtain from \eqref{bitten}, \begin{align*} &\int_{0}^{r} k_2 (t + \frac{M_{b}}{\lambda})^{N-1-\alpha} \left( C_7 \lambda^{-\frac{(2-\alpha)p}{1-p}} + \epsilon |w_{\lambda}|^{p} \right) \, dt \\ &\leq C_7 k_2\lambda^{-\frac{(2-\alpha)p}{1-p}} \int_{0}^{r} (t + 2\epsilon_{6})^{N-1-\alpha} +\epsilon k_2 \int_{0}^{r}(t + 2\epsilon_{6})^{N-1-\alpha}|w_{\lambda}|^{p}. \end{align*} Both of these terms are small on any compact subset of $[0, \infty)$ (the first since $\lambda \to \infty$ and the second term by \eqref{truman}) and so both of these limit to zero as $\lambda \to \infty$. This establishes \eqref{new york}. Therefore we see by using (H4) and taking limits in \eqref{harding} we obtain \begin{equation} -(r + A)^{N-1} w' \geq k_{1}\int_{0}^{r} (t+A)^{N-1-\alpha} w^{p} \, dt \quad \text{on } (0, \infty). \label{max} \end{equation} Since $w> 0$ on $[0, \infty)$ it follows from \eqref{max} that $w$ is decreasing so that \begin{equation} -(r + A)^{N-1} w' \geq k_{1} w^{p} \frac{(r+A)^{N-\alpha} - A^{N-\alpha}}{N-\alpha} \quad\text{on } (0, \infty). \label{delaware} \end{equation} Rewriting \eqref{delaware} gives \begin{equation} -w' w^{-p} \geq \frac{k_{1}}{N-\alpha}(r+A)^{1-\alpha} - \frac{k_{1} A^{N-\alpha}}{N-\alpha}(r+A)^{1-N} \text{ on } (0,\infty). \label{usg} \end{equation} Next we analyze the two cases $A=0$ and $A\neq 0$ separately. \smallskip \noindent\textbf{Case 1:} $A\neq 0$. Integrating \eqref{usg} on $(0,r)$ gives \begin{align*} & -\Big( \frac{w^{1-p} - 1}{1-p} \Big) \\ &\geq \frac{k_{1}}{N-\alpha} \Big( \frac{(r+A)^{2-\alpha}}{2-\alpha} - \frac{ A^{2 -\alpha}}{2-\alpha} \Big) - \frac{k_{1}A^{N-\alpha}}{N-\alpha} \Big( \frac{(r+A)^{2-N}}{2-N} - \frac{A^{2 -N}}{2-N} \Big). \end{align*} Thus for some constant $C_{8}$ we obtain \begin{equation} \frac{w^{1-p} - 1}{1-p} \leq -\frac{k_{1}(r+A)^{2-\alpha}}{(N-\alpha)(2-\alpha)} - \frac{k_1(r+A)^{2-N}A^{N-\alpha}}{(N-2)(N-\alpha)} + C_{8}. \label{nixon} \end{equation} The right-hand side of \eqref{nixon} goes to $-\infty$ as $r \to \infty$ since $0 < \alpha < 2$, $0 2$ and so we see that $w$ becomes negative which is a contradiction because we assumed $w>0$ and so $w$ and hence $u$ must have a zero for sufficiently large $b$. \smallskip \vskip .1 in \noindent\textbf{Case 2:} $A=0$. In this case we see that \eqref{usg} becomes \begin{equation} -w' w^{-p} \geq \frac{k_{1}}{N-\alpha}r^{1-\alpha} \quad\text{on } (0,\infty). \label{ohio} \end{equation} Integrating \eqref{ohio} on $(0,r]$ gives $$\frac{w^{1-p} - 1}{1-p} \leq -\frac{k_{1}r^{2-\alpha}}{(N-\alpha)(2-\alpha)}$$ and therefore we see that $w$ becomes negative. Thus we again obtain a contradiction and so $w$ and hence $u$ has a zero if $b$ is sufficiently large. Thus there exists a $z_{b}>R$ such that $u(z_{b})=0$ and $u>0$ on $(R, z_{b})$. In addition by uniqueness of solutions of initial value problems it follows that $u'(z_{b})<0$ and then we can similarly show as in Lemma \ref{lem5} that $u$ has a local minimum $m_{b}> z_{b}$ for large enough $b>0$ and also that $w$ has a second zero $z_{2,b}$ (and hence $u$ has a second zero) if $b$ is sufficiently large. In a similar way, given any positive integer $n$ we can show for large enough $b$ that $u$ has $n$ zeros on $(R, \infty)$. Since $w_{\lambda} \to w$ uniformly on compact sets it follows then that if $\lambda$ is sufficiently large then $w_{\lambda}$ will have $n$ zeros on $(0, \infty)$ and hence $u(r,b)$ will have $n$ zeros on $(R, \infty)$ if $b>0$ is sufficiently large. This completes the proof. \end{proof} \section{proof of Theorem \ref{thm1}} We consider the set $$\{ b>0 | \ u(r,b) > 0 \text{ for all } r>R \}.$$ This set is nonempty by Lemma \ref{lem6} and is bounded from above by Lemma \ref{lem9} so there exists a $b_{0}>0$ such that $$b_{0} = \sup \{ b>0 | \ u(r,b) > 0 \text{ for all } r>R \}.$$ We show now that $u(r, b_{0}) > 0$ for $r>R$. If $u(r_{0}, b_{0})=0$ and $u(r,b_{0})>0$ on $(R, r_{0})$ then $u'(r_{0}, b_{0})\leq 0$. By uniqueness of solutions of initial value problems it follows that $u'(r_{0}, b_{0})<0$. Thus for $r_1> r_{0}$ and $r_{1}$ sufficiently close to $r_0$ we have $u(r_1, b_{0})<0$. Then for $b$ close to $b_{0}$ with $b < b_{0}$ then $u(r_{1}, b)<0$ contradicting the definition of $b_{0}$. Hence $u(r, b_{0})>0$ for $r>R$. Now by Lemma \ref{lem3} we know that $u(r,b_{0})$ must get larger than $\beta$. If $u'>0$ for all $r\geq R$ then since $u$ is bounded it follows that $u$ would has a limit which by Lemma \ref{lem1} would have to be less than or equal to $\beta$. Thus we see that $u(r, b_{0})$ must have a local maximum $M_{b_{0}}>R$ and $u'>0$ on $[R, M_{b_0})$. Next we show $E(r, b_{0}) \geq 0$ for all $r\geq R$. If $E(r_{0}, b_{0})<0$ then $E(r_{0}, b)<0$ for $b > b_{0}$ and $b$ close to $b_{0}$. On the other hand, since $b>b_{0}$ it follows that there exists a $z_{b}$ such that $u(z_{b}, b)=0$. Thus $E(z_{b}, b)\geq 0$. Since $E$ is nonincreasing this implies $z_{b} < r_{0}$ for all $b > b_{0}$. However $z_{b} \to \infty$ as $b \to b_{0}^{+}$ for if the $z_{b}$ were bounded then this would force a subsequence of the $z_{b}$ to converge to some $z_{0}$ and then $u(z_{0}, b_{0})=0$ contradicting that $u(r,b_{0})>0$. Thus $E(r, b_{0}) \geq 0$ for all $r\geq R$. It now follows that $u(r, b_{0})$ cannot have a positive local minimum, $m_{b_0} > M_{b_{0}}$ for at such a point $u'(m,b_{0})=0$, $u''(m,b_{0}) \geq 0$ and so $f(u(m,b_{0}))\leq 0$. Since $u(m,b_{0})>0$ this then forces $0M_{b_0}$ and so $\lim_{r \to \infty} u(r,b_{0})$ exists. Denoting this limit as $L$ then $L \geq 0$ since $u(r, b_{0})>0$ for $r>R$ and by Lemma \ref{lem1} we have $f(L)=0$ so that $L=0$ or $L = \beta$. Then a similar argument as in Lemma \ref{lem2} shows that $u(r,b_{0})$ gets less than $\beta$ and so it follows that $L=0$ and thus $\lim_{r \to \infty} u(r, b_{0})=0$. Hence $u(r, b_{0})$ is a positive solution on $(R, \infty)$ and $\lim_{r \to \infty} u(r, b_{0})=0$. Next from a lemma in \cite{I4} if $b> b_{n}$ is sufficiently close to $b_{n}$ where $u(r, b_{n})$ has $n$ zeros on $(R, \infty)$ and $$\lim_{r \to \infty} u(r,b) =0$$ then $u(r, b)$ has at most $n+1$ zeros on $(R, \infty)$. From this lemma it then follows that $$\{ b>b_0 | u(r,b) \text{ has exactly one zero on } (R, \infty) \}$$ is nonempty and again from Lemma \ref{lem9} this set is bounded from above. Thus there exists a $b_{1}> b_{0}$ such that $$b_{1} = \sup\{ b>b_{0} | u(r,b) \text{ has exactly one zero on } (R, \infty) \}.$$ As above we can show $u(r, b_{1})$ has exactly one zero on $(R, \infty)$ and $$\lim_{r \to \infty} u(r,b_{1}) =0.$$ Similarly we can find $b_{n}>b_{n-1}$ such that $u(r, b_{n})$ has exactly $n$ zeros on $(R, \infty)$ and $$\lim_{r \to \infty} u(r,b_{n}) =0.$$ This completes the proof of Theorem \ref{thm1}. \begin{thebibliography}{00} \bibitem{BL} H. Berestycki, P.L. Lions; Non-linear scalar field equations I, {\it Arch. Rational Mech. 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