\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 108, pp. 1--10.\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-2018/108\hfil Infinitely many solutions] {Infinitely many solutions for a semilinear problem on exterior domains with nonlinear boundary condition} \author[J. Joshi, J. A. Iaia \hfil EJDE-2018/108\hfilneg] {Janak Joshi, Joseph A. Iaia} \address{Janak Joshi \newline Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-1430, USA} \email{JanakrajJoshi@my.unt.edu} \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 Jerome A. Goldstein} \thanks{Submitted July 8, 2017. Published May 8, 2018.} \subjclass[2010]{34B40, 35B05} \keywords{Exterior domain; superlinear; radial solution} \begin{abstract} In this article we prove the existence of an infinite number of radial solutions to $\Delta u+K(r)f(u)=0$ with a nonlinear boundary condition on the exterior of the ball of radius $R$ centered at the origin in $\mathbb{R}^{N}$ such that $\lim_{r \to \infty} u(r)=0$ with any given number of zeros where $f:{\mathbb R} \to {\mathbb R}$ is odd and there exists a $\beta>0$ with $f<0$ on $(0,\beta)$, $f>0$ on $(\beta,\infty)$ with $f$ superlinear for large $u$, and $K(r) \sim r^{-\alpha}$ with $0< \alpha < 2(N-1)$. \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{gather} \Delta u + K(|x|)f(u) = 0 \quad \text{for } R<|x|<\infty, \label{e1}\\ \frac{\partial u}{\partial{\eta}}+c(u)u=0 \quad\text{when } |x|=R \text{ and } \lim_{|x| \to \infty} u(x) = 0, \label{e2} \end{gather} where $u:{\mathbb R}^N \to {\mathbb R} $ with $N\geq 2$, $R>0$, $c:[0,\infty) \to (0,\infty)$ is continuous, $\frac{\partial}{\partial{\eta}}$ is the outward normal derivative, $f$ is odd and locally Lipschitz. We assume: \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)] $f(u) = |u|^{p-1}u + g(u)$ where $p>$ 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 there exists $\gamma$ with $0< \beta < \gamma$ such that $F<0$ on $(0, \gamma)$ and $F>0$ on $(\gamma, \infty)$. \item[(H4)] $K$ and $K'$ are continuous on $[R, \infty)$ with $K(r)> 0$, $2(N-1) + \frac{rK'}{K} >0$ and there exists $\alpha\in (0,2(N-1))$ such that $\lim_{r \to \infty} \frac{rK'}{K} = -\alpha$. \item[(H5)] There exist positive $d_1, d_2$ such that $d_1r^{-\alpha} \leq K(r) \leq d_2r^{-\alpha}$ for $r \geq R$. \end{itemize} Note that (H4) implies $r^{2(N-1)}K$ is increasing. Our main result reads as follows. \begin{theorem} \label{thm1} Assume {\rm (H1)--(H5)}, $N \geq 2$, and $0<\alpha<2(N-1)$. Then for each nonnegative integer $n$ there exists a radial solution, $u_{n}$, of \eqref{e1}-\eqref{e2} such that $u_{n}$ has exactly $n$ zeros on $(R, \infty)$. \end{theorem} The radial solutions of \eqref{e1} on ${\mathbb R}^{N}$ and $K(r) \equiv 1$ have been well-studied. These include \cite{BL,BL2,B,JK,M,ST}. Recently there has been an interest in studying these problems on ${\mathbb R}^{N} \backslash B_{R}(0)$. These include \cite{C,D,I,C2,S}. In \cite{D} positive solutions of a similar problem were studied for $ N <\alpha < 2(N-1)$. There the authors use the mountain pass lemma to prove existence of positive solutions. Here we use a scaling argument as in \cite{JI,M} to prove the existence of infinitely many solutions. \section{Preliminaries} Since we are interested in radial solutions of \eqref{e1}-\eqref{e2}, we denote $r = |x|$ and write $u(x)= u(|x|)$ where $u$ satisfies \begin{gather} u'' + \frac{N-1}{r} u' + K(r) f(u) = 0 \quad\text{for } R0, \quad u'(R) = b c(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}] \leq 0. \label{E'} \end{equation} Thus $E$ is non-increasing. Therefore \begin{equation} \frac{1}{2} \frac{u'^2}{K(r)} + F(u) = E(r) \leq E(R) = \frac{1}{2}\frac{b^{2}c^2(b)}{K(R)}+F(b) \text{ for } r \geq R. \label{E'2} \end{equation} Since $F$ is bounded from below by (H3), it follows from \eqref{E'2} that $u$ and $u'$ are uniformly bounded wherever they are defined from which it follows that the solution of \eqref{DE}-\eqref{IC} is defined on $[R, \infty)$. \begin{lemma} \label{lem2.1} Assume {\rm (H1)--(H5)} and $N \geq 2$. Let $u(r,b)$ be the solution of \eqref{DE}-\eqref{IC} and suppose $0< \alpha < 2(N-1)$. If $b>0$ and $b$ is sufficiently small then $u(r,b) > 0$ for all $r>R$. \end{lemma} \begin{proof} We proceed as in \cite{JI}. Since $u(R,b)=b>0$ and $u'(R,b)=bc(b)>0$ we see that $u(r,b)>0$ on $(R, R+\delta)$ for some $\delta >0$. If $u'(r,b)>0$ for all $r \geq R$ then $u(r,b)>0$ for all $r > R$ and so we are done in this case. If $u$ is not increasing for all $r>R$ then there exists a local maximum at some $M_{b}$ with $M_{b} > R$ and $u'(r,b)>0$ on $[R, M_{b})$. If $u(M_{b},b) < \gamma$ then $E(r) \leq E(M_{b}) < 0$ for $r > M_{b}$ since $E$ is non-increasing. It follows then that $u(r,b)$ cannot be zero for any $r > M_{b}$ for if there were such a $z_{b}>M_{b}$ then $ 0 \leq \frac{1}{2}\frac{u'^2(z_{b})}{K(z_{b})} = E(z_{b}) \leq E(M_{b}) < 0$ which is impossible. Also, since $u'(r,b)>0$ on $[R,M_{b})$ it follows then that $u(r,b)>0$ on $(R, \infty)$ if $u(M_{b},b) < \gamma$. So if $u(r,b)$ has a local maximum at $M_{b}$ with $u(M_{b},b)< \gamma$ then we are done in this case as well. In addition, if $E(R) = \frac{1}{2}\frac{b^{2}c^2(b)}{K(R)}+F(b) \leq 0$ then $E(t) <0$ for $t>R$ and a similar argument shows that $u(r,b)$ cannot be zero for $t>R$. So for the rest of this proof we assume that $u(r,b)$ has a local maximum at $M_{b}$, $u(M_{b},b) \geq \gamma$, $u'(r,b)>0$ on $[R, M_{b})$, and $E(R) = \frac{1}{2}\frac{b^{2}c^2(b)}{K(R)}+F(b) > 0$ for all sufficiently small $b>0$. From this it then follows from (H1) and (H3) that there exists $r_{b}$ and $r_{b_1}$ with $R0$ and so integrating \eqref{u'} on $[R, r_{b}]$ when $\alpha \neq 2$ gives \begin{equation} \begin{aligned} \int_0^{\beta} \frac{dt}{\sqrt{{\frac{b^2 c^2(b)}{K(R)}+2F(b) - 2F(t)}} } &= \int_{R}^{r_{b}} \frac{u'(r) \, dr}{\sqrt{{\frac{b^2 c^2(b)}{K(R)}+2F(b) - 2F(u(r))}}} \\ &\leq \frac{\sqrt{d_2}}{{\frac{\alpha}{2}-1}} \left(R^{1-\frac{\alpha}{2}}-{r_{b}}^{1-\frac{\alpha}{2}}\right). \end{aligned}\label{R r_b} \end{equation} In the case $\alpha =2$ the right-hand side of \eqref{R r_b} is replaced by: \begin{equation} \sqrt{d_2} \ln(r_{b}/R). \label{log} \end{equation} As $b \to 0^{+}$ the left-hand side of \eqref{R r_b} goes to $+\infty$ since by (H1) and the definition of $F$, \[ \sqrt{\frac{b^2c^2(b)}{K(R)}+2F(b) - 2F(t)} \leq \sqrt{\frac{b^2 c^2(b)}{K(R)}+2F(b) + 2|f'(0)|t^2} \] for small positive $t$ thus \begin{equation} \int_0^{\epsilon} \frac{dt}{\sqrt{\frac{b^2 c^2(b)}{K(R)}+2F(b) - 2F(t)}} \geq \int_0^{\epsilon} \frac{dt}{\sqrt{\frac{b^2 c^2(b)}{K(R)}+2F(b) + 2|f'(0)|t^2}} \to \infty \label{nixon} \end{equation} as $b \to 0^{+}$. Combining \eqref{R r_b} and \eqref{nixon} we see that if $2< \alpha < 2(N-1)$ then $$ \frac{\sqrt{d_2}}{{\frac{\alpha}{2}-1}}R^{1-\frac{\alpha}{2}} \geq \frac{\sqrt{d_2}}{{\frac{\alpha}{2}-1}} \left(R^{1-\frac{\alpha}{2}}-{r_b}^{1-\frac{\alpha}{2}}\right) \to \infty \quad\text{as } b \to 0^{+} $$ which is impossible since $R$ is fixed. Thus it follows that $u(M_{b},b) < \gamma$ if $b>0$ is sufficiently small and as indicated earlier in this lemma it then follows that $u(r,b)>0$ for $r>R$ if $b>0$ is sufficiently small. For the case $0 < \alpha \leq 2$ a lengthier argument is required and the details are carried out in \cite{JI}. There it is shown that $E(r_{b_1})<0$ for sufficiently small $b>0$ and therefore $u(r,b)$ cannot be zero for any $z_{b} > r_{b_1}$ as indicated earlier in this lemma. This completes the proof. \end{proof} \begin{lemma} \label{lem2.2} Assume {\rm (H1)--(H5)} and $N \geq 2$. Let $u(r,b)$ be the solution of \eqref{DE}-\eqref{IC} and suppose $0 < \alpha < 2(N-1)$. Given a positive integer $n$ then $u(r,b)$ has at least $n$ zeros on $(0, \infty)$ if $b>0$ is chosen sufficiently large. \end{lemma} \begin{proof} Let $v(r)=u(r+R)$. Then $v$ satisfies, \begin{gather} v''(r)+\frac{N-1}{R+r}v'(r)+K(R+r)f(v)=0, \label{DE2}\\ v(0)=b, v'(0)=b c(b).\label{DE3} \end{gather} Now let \begin{equation} v_{\lambda}(r)=\lambda^{-\frac{2}{p-1}}v\left(\frac{r}{\lambda}\right) \quad \text{for } \lambda>0. \label{rescale} \end{equation} Then \begin{gather*} v_\lambda'(r)=\lambda^{-\frac{2}{p-1}-1}v'\left(\frac{r}{\lambda}\right),\\ v_\lambda''(r)=\lambda^{-\frac{2}{p-1}-2}v''\left(\frac{r}{\lambda}\right). \end{gather*} Thus $$ v''\left(\frac{r}{\lambda}\right)+\frac{N-1}{R+\frac{r}{\lambda}} v'\left(\frac{r}{\lambda}\right) +K(\frac{r}{\lambda}+R)f\left(v\left(\frac{r}{\lambda}\right)\right)=0 $$ and so it then follows that \begin{equation} v_\lambda''+\frac{N-1}{(R\lambda+r)}v_\lambda'+ \frac{K(\frac{r}{\lambda}+R)}{\lambda^{\frac{2p}{p-1}}}f(\lambda^{\frac{2}{p-1}}v_{\lambda})=0. \label{JFK} \end{equation} From (H2) we have $f(u)=|u|^{p-1}u+g(u)$ and $\lim_{u \to \infty} \frac{|g(u)|}{|u|^p} = 0$ so rewriting \eqref{JFK} gives \begin{equation} v_\lambda''+\frac{N-1}{(R\lambda+r)}v_\lambda' +\frac{K(\frac{r}{\lambda}+R)}{\lambda^{\frac{2p}{p-1}}} \big[\lambda^{\frac{2p}{p-1}}|v_\lambda|^{p-1} v_{\lambda} +g(\lambda^{\frac{2}{p-1}}v_\lambda)\big]=0. \label{LBJ} \end{equation} Thus \begin{gather} v_\lambda''+\frac{N-1}{(R\lambda+r)}v_\lambda'+K(\frac{r}{\lambda}+R) \big[|v_\lambda|^{p-1}v_{\lambda} +\frac{g(\lambda^{\frac{2}{p-1} }v_\lambda)} {\lambda^{\frac{2p}{p-1}}}\big]=0, \label{DE4} \\ v_\lambda(0)=\lambda^{\frac{-2}{p-1}}b, \label{DE5}\\ v_\lambda'(0)= \lambda^{\frac{-2}{p-1}-1}bc(b) =\lambda^{-\frac{p+1}{p-1}}b c(b). \label{DE6} \end{gather} Now let \begin{equation} E_\lambda(r)=\frac{v_\lambda'^2}{2K(\frac{r}{\lambda}+R)} +\frac{F(\lambda^{\frac{2}{p-1}}v_\lambda)}{\lambda^{\frac{2p}{p-1}}}. \label{energy} \end{equation} A straightforward calculation using (H4) and \eqref{JFK} gives \[ E_\lambda'(r)=-\frac{v_\lambda'^2}{2(\frac{r}{\lambda}+R)K(\frac{r}{\lambda}+R)} \Big[\frac{(\frac{r}{\lambda}+R)K'(\frac{r}{\lambda}+R)} {K(\frac{r}{\lambda}+R)}+2(N-1)\Big]\leq{0} \] for $0<\alpha<2(N-1)$. Thus for $r\geq{0}$, \begin{equation} \frac{v_\lambda'^2}{2K(\frac{r}{\lambda}+R)}+\frac{F(v_\lambda)}{\lambda^{\frac{2p}{p-1}}} =E_\lambda(r)\leq{E_\lambda(0)} = {\frac{b^2c^2(b)}{2\lambda^\frac{2(p+1)}{p-1} K(R)}} + \frac{F(\lambda^{\frac{-2}{p-1}}b)}{\lambda^{\frac{2p}{p-1}} }. \label{FDR} \end{equation} We now divide the rest of the proof into two cases. \smallskip \noindent\textbf{Case 1:} $\frac{c(b)}{b^{\frac{p-1}{2}}}\leq{C_0}$ for all sufficiently large $b$ for some constant $C_0$. In this case we choose $b=\lambda^{\frac{2}{p-1}}$ so that \eqref{DE5}-\eqref{DE6} become $v_\lambda(0)=1$ and $$ v_\lambda'(0)=\lambda^{\frac{-2}{p-1}-1}b c(b)=\frac{c(b)}{\lambda} =\frac{c(b)}{b^{\frac{p-1}{2}}}\leq{C_0}. $$ Next using (H2)-(H3) it follows that \begin{equation} F(u)=\frac{|u|^{p+1}}{p+1}+G(u) \label{smile} \end{equation} where $G(u)=\int_0^{u} g(s)\,ds$ and from L'H\^opital's rule it follows that $ \frac{G(u)}{|u|^{p+1}} \to 0$ as $u\to{\infty}$. So from \eqref{rescale}, \eqref{FDR}-\eqref{smile} and since $b=\lambda^{\frac{2}{p-1}}$ we obtain \begin{gather} \frac{v_\lambda'^2}{2K(\frac{r}{\lambda}+R)}+\frac{|v_\lambda|^{p+1}}{p+1} +\frac{G(\lambda^{\frac{2}{p-1}}v_\lambda)}{\lambda^{\frac{2(p+1)}{p-1}}} \leq {\frac{b^2c^2(b)}{2\lambda^\frac{2(p+1)}{p-1} K(R)}} + \frac{F(1)}{{\lambda^{\frac{2p}{p-1}}}} \label{DE7} \\ =\frac{1}{2K(R)} \Big(\frac{c(b)}{b^{\frac{p-1}{2}}}\Big)^2 + \frac{F(1)}{\lambda{\frac{2p}{p-1}}} \leq{\frac{C_0^2}{2K(R)}}+\frac{F(1)}{\lambda{\frac{2p}{p-1}}}. \label{DE8} \end{gather} So since $\frac{G(u)}{|u|^{p+1}} \to 0$ as $u\to{\infty}$ it follows that $\frac{|G(u)|}{|u|^{p+1}}\leq{\frac{1}{2(p+1)}}$ for say $u>T$. Also, $|G(u)|\leq{G_0}$ for $|u|\leq{T}$ since $G$ is continuous on the compact set $[0,T]$ and thus $|G(u)|\leq{\frac{1}{2(p+1)}|u|^{p+1}+G_0}$ for all $u$. Similarly using (H2) it follows that $|g(u)| \leq \frac{1}{2}|u|^p + g_0$ for all $u$ for some constant $g_0$ where $|g(u)| \leq g_0$ on $[0, T]$. Therefore for $\lambda>0$ it follows from \eqref{DE7}-\eqref{DE8} that $$ \frac{v_\lambda'^2}{2K(\frac{r}{\lambda}+R)} +\frac{|v_\lambda|^{p+1}}{2(p+1)}\leq{\frac{C_0^2}{2K(R)}} +\frac{F(1)}{\lambda^{\frac{2p}{p-1}}} +\lambda^{\frac{-2(p+1)}{p-1}}G_0 \leq {\frac{C_0^2}{2K(R)}}+F(1) + G_0 \text{ for } \lambda > 1 . $$ It follows from this that $v_\lambda(r)$ and $v_\lambda'(r)$ are uniformly bounded on $[0, \infty)$ for large $\lambda$. It then follows that $\big(\frac{N-1}{R\lambda+r}\big) v_\lambda' $ is uniformly bounded on $[0, \infty)$ and also $K(\frac{r}{\lambda}+R)\big[|v_\lambda|^{p-1}v_{\lambda} +\frac{g(\lambda^{\frac{2}{p-1} }v_\lambda)}{\lambda^{\frac{2p}{p-1}}}\big] $ is uniformly bounded on $[0, \infty)$. Then from \eqref{DE4} we see that $v_\lambda''$ is uniformly bounded on $[0,\infty)$ for large $\lambda$. Therefore by the Arzela-Ascoli theorem it follows that there is a subsequence (still denoted $v_\lambda$) and continuous functions $v_0$ and $v_0'$ such that $v_\lambda\to{v_0}$ and $v_\lambda'\to{v_0'}$ uniformly on compact subsets of $[0, \infty)$ to a solution of \begin{equation} \label{DE9} \begin{gathered} v_0''+K(R) v_0^p =0, \\ v_0(0)=1, \quad v_0'(0)=d_0=\lim_{b\to{\infty}}{\frac{c(b)}{b^{\frac{p-1}{2}}}}\leq C_0. \end{gathered} \end{equation} It is now straightforward to show that $v_0$ has infinitely many zeros on $[0,\infty)$. Thus $v_\lambda$ has at least $n$ zeros for sufficiently large $\lambda$ and so $u(r,b)$ has at least $n$ zeros for sufficiently large $b$. This concludes the proof in Case 1. \smallskip \noindent\textbf{Case 2:} $\frac{c(b)}{b^{\frac{p-1}{2}}}\to{\infty}$ for some subsequence as $b\to{\infty}$. Then for these $b$ we let \begin{equation} \lambda = (bc(b))^{\frac{p-1}{p+1}} \quad\text{that is } b c(b)=\lambda^{\frac{p+1}{p-1}}. \label{calvin} \end{equation} From \eqref{DE6} and \eqref{calvin} we see that $$ v_\lambda(0)= \lambda^{-\frac{2}{p-1}}b =\Big[\frac{b^{\frac{p-1}{2}}}{c(b)}\Big]^{\frac{2}{p+1}}\to{0}\quad \text{as } b\to{\infty} \text{ and } v_{\lambda}'(0)=1. $$ As in case (1) we can show there exist continuous functions $v_0$ and $v_0'$ such that for some subsequence $v_\lambda\to{v_0}$ and $v_\lambda'\to{v_0'}$ as $\lambda \to{\infty}$ uniformly on compact subsets of $[0, \infty)$ and $v_0$ is a solution of \begin{equation} \label{DE10} \begin{gathered} v_0''+K(R) v_0^p =0, \\ v_0(0)=0, \quad v_0'(0)=1. \end{gathered} \end{equation} And again it is easy to show that $v_0$ has infinitely many zeros on $[0,\infty)$. Thus it follows that $v_{\lambda}(r)$ and hence $u(r,b)$ has at least $n$ zeros on $[0, \infty)$ when $b$ is sufficiently large. This completes the proof. \end{proof} \section{Proof of the main theorem} \begin{proof} We proceed as we did in \cite{JI}. It follows from Lemma \ref{lem2.1} that $$ \{ b>0 : u(r,b)>0 \text{ on } (R, \infty) \} $$ is nonempty and from Lemma \ref{lem2.2} it follows that this set is bounded from above. Hence we set $$ b_0 = \sup \{ b | u(r,b)>0 \text{ on } (R, \infty) \}. $$ We next show that $u(r, b_0)>0$ on $(R, \infty)$. This follows because if there is a $z>R$ such that $u(z,b_0)=0$ then $u'(z,b_0)<0$ (by uniqueness of solutions of initial value problems) and so $u(r,b_0)$ becomes negative for $r$ slightly larger than $z$. By continuity with respect to initial conditions it follows that $u(r, b)$ becomes negative for $b$ slightly smaller than $b_0$ contradicting the definition of $b_0$. Thus $u(r, b_0)>0$ on $(R, \infty)$. Next it follows by the definition of $b_0$ that if $b > b_0$ then $u(r,b)$ must have a zero, $z_{b}$, where $z_{b}>R$. We now show that $z_{b} \to \infty$ as $b \to b_0^{+}$. If not then the $z_{b}$ are uniformly bounded and so a subsequence of them (still denoted $z_{b}$) converges to some $z_0\geq R$. Then since $E'\leq 0$: \begin{equation} \frac{1}{2} \frac{u'^{2}(r,b)}{K(r)} + F(u(r,b)) \leq \frac{1}{2}\frac{ b^2 c^2(b)}{K(R)} \quad\text{for } r \geq R \label{cadillac} \end{equation} and since $F$ is bounded from below (by (H3)) it follows that $u(r,b)$ and $u'(r,b)$ are uniformly bounded on $[R, \infty)$ for $b$ near $b_0$. In addition it follows from \eqref{DE} that $u''(r,b)$ is also uniformly bounded on $[R, \infty)$ for $b$ near $b_0$. Then by the Arzela-Ascoli theorem a subsequence (still denoted $u(r,b)$ and $u'(r,b)$) converges uniformly to $u(r, b_0)$ and $u'(r, b_0)$ and so we obtain $u(z_0, b_0)=0$. But we know $u(r, b_0)>0$ for $r>R$ and so we get a contradiction. Thus $z_{b} \to \infty$ as $b \to b_0^{+}$. We now show that $E(r,b_0)\geq 0$ on $[R, \infty)$. If not then there is an $r_0>R$ such that $E(r_0, b_0)<0$. By continuity $E(r_0,b)<0$ for $b$ slightly larger than $b_0$. Also for $b > b_0$ the function $u(r,b)$ has a zero, $z_{b}$, (by definition of $b_0$) and $E(z_{b}) = \frac{1}{2}\frac{u'^2(z_{b},b)}{K(z_{b})} \geq 0$. But $E$ is non-increasing so $ z_{b}< r_0$ which contradicts $z_{b} \to \infty$ as $b \to b_0^{+}$. Thus, $E(r,b_0)\geq 0$ on $[R, \infty)$. Next either: (i) $u(r,b_0)$ has a local maximum at some $M_{b_0}>R$, or (ii) $u'(r,b_0)>0$ for $r> R$ and since $u(r, b_0)$ is bounded by \eqref{cadillac} then there is an $L>0$ such that $u(r,b_0) \to L$ as $r \to \infty$. We show now that (ii) is not possible. Suppose therefore that (ii) occurs. We divide this into three cases. \smallskip \noindent\textbf{Case 1:} $ 0 < \alpha < N$. Multiplying \eqref{DE} by $r^{N-1}$ and integrating on $(R,r)$ gives \begin{equation} -r^{N-1}u' = -R^{N-1}b_0 + \int_{R}^{r} t^{N-1} K(t) f(u) \, dt. \label{on a raft with taft} \end{equation} Dividing \eqref{on a raft with taft} by $r^{N} K \to \infty$ as $r \to \infty$ since $0 < \alpha < N$ and taking limits using L'H\^opital's rule and (H4) gives \begin{equation} -\frac{u'}{rK} = \lim_{r \to \infty} \frac{\int_{R}^{r} t^{N-1} K(t) f(u) \, dt}{r^{N}K} =\lim_{r \to \infty} \frac{f(u)}{N + \frac{rK'}{K}} = \frac{f(L)}{N-\alpha}. \label{keep cool} \end{equation} Thus since $0< \alpha0$, it follows that $f(L)\leq 0$ so that \begin{equation} 0< L \leq \beta< \gamma. \label{DDR} \end{equation} On the other hand integrating the identity $$ (r^{2(N-1}KE)' = (r^{2(N-1}K)'F $$ on $(R,r)$ and using L'H\^opital's rule gives \begin{align*} \lim_{r \to \infty} E(r,b_0) & = \lim_{r \to \infty}\frac{1}{2} \frac{u'^{2}}{K} +F(u) \\ &= \lim_{r \to \infty}\frac{1}{2} \frac{R^{2(N-1)}b_0^2}{r^{2(N-1)} K} + \frac{\int_{R}^{r} (t^{2(N-1)}K)' F(u(t,b_0)) \, dt }{r^{2(N-1)}K} = F(L). \end{align*} Since we showed earlier that $E(r,b_0)\geq 0$ we see then that \begin{equation} 0\leq \lim_{r \to \infty} E(r, b_0) = F(L). \label{bb} \end{equation} Thus $L \geq \gamma$ which contradicts \eqref{DDR}. Therefore it must be the case that $u(r,b_0)$ has a local maximum at some $M_{b_0}$. This completes Case 1. \smallskip \noindent\textbf{Case 2:} $ \alpha = N$. In this case as well it follows that $f(L)\leq 0$ for suppose $f(L)>0$. Then by (H5) the integral on the right-hand side of \eqref{on a raft with taft} grows like $f(L)\ln(r) \to \infty$ as $ r \to \infty$ and thus the right-hand side of \eqref{on a raft with taft} becomes arbitrarily large but the left hand side is negative. Thus it must be that $f(L)\leq 0$ and as in Case 1 we get a contradiction. \smallskip \noindent\textbf{Case 3:} $ N < \alpha < 2(N-1)$. For $b > b_0$ we know that there is an $z_{b}> R$ such that $u(z_{b},b) =0$ so there is an $M_{b}$ with $R< M_{b} < z_{b}$ such that $u(r,b)$ has a local maximum at $M_{b}$. If the $M_{b}$ are bounded as $b \to b_0^{+}$ then a subsequence of the $M_{b}$ converge to some $M_{b_0}<\infty$ and then $u(r, b_0)$ has a local maximum at $M_{b_0}$ contradicting our assumption that $u'(r, b_0)>0$ for $r > R$. So let us assume that $M_{b} \to \infty$ as $b \to b_0^{+}$. Since $E$ is non-increasing, it follows that $E(r) \leq E(M_{b})$ for $r \geq M_{b}$. Thus \begin{equation} \frac{1}{2} \frac{u'^2}{K} + F(u) \leq F(u(M_{b})) \text{ for } r \geq M_{b}. \label{hoover} \end{equation} Rewriting and integrating \eqref{hoover} on $[M_{b}, z_{b}]$ (using (H5)) gives \begin{equation} \begin{aligned} 0&\leq \int_0^{u(M_{b})}\frac{1}{\sqrt{2}{\sqrt{F(u(M_{b})) - F(t)}}} \, dt \\ & = \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{\sqrt{d_2}(M_{b}^{1 - \frac{\alpha}{2}} - z_{b}^{1 - \frac{\alpha}{2}})}{ \frac{\alpha}{2} -1 }. \end{aligned} \label{CC} \end{equation} Since $\alpha > N \geq 2$ and $M_{b} \to \infty$ as $b \to b_0^{+}$ (thus $z_{b} \to \infty$) we see that the right-hand side of \eqref{CC} goes to 0 as $b \to b_0^{+}$. On the other hand, since $u(r,b) \to u(r, b_0)$ uniformly on compact subsets of $[R, \infty)$ we see then that $u(M_{b}) \to L$ as $b \to b_0^{+}$. Taking limits in \eqref{CC} then gives: $$ \int_0^{L}\frac{1}{\sqrt{2}{\sqrt{F(L) - F(t)}}} \, dt = 0 $$ which is impossible. Thus the $M_{b}$ must be bounded as $b \to b_0^{+}$ which contradicts our assumption that $M_{b} \to \infty$. Thus $u(r,b_0)$ must have a local maximum $M_{b_0}$. This completes Case 3. Since we know $u(r, b_0)>0$ for $r>R$ and $u(r, b_0)$ has a local maximum $M_{b_0}$ it follows that $u(r, b_0)$ cannot have a local minimum at $m_{b_0}$ with $m_{b_0} > M_{b_0}$ for at such a point we would have $u(m_{b_0},b_0)>0$, $u'(m_{b_0},b_0)=0$, and $u''(m_{b_0})\geq 0$. Thus from \eqref{DE} we see that $f(u(m_{b_0},b_0)) \leq 0$ which implies $0< u(m_{b_0},b_0) \leq \beta$. On the other hand since $E(r, b_0) \geq 0$ for all $r\geq R$ then $E(m_{b_0}, b_0) = F(u(m_{b_0}, b_0))\geq 0$ and so $\beta \geq u(m_{b_0},b_0) \geq \gamma>\beta$ which is impossible. Thus it must be that $u'(r, b_0)<0$ for $r> M_{b_0}$ and hence there is an $L \geq 0$ such that $u(r,b_0) \to L$ as $r \to \infty$. Recalling \eqref{bb} we have $E(r, b_0) \to F(L) \geq 0$ as $r \to \infty$. Thus $L=0$ or $L \geq \gamma$. Finally we want to show $L=0$. There are again three cases to consider. \smallskip \noindent\textbf{Case 1:} $ 0 < \alpha < 2$. First suppose $f(L)\neq 0$. Recalling \eqref{keep cool} it then follows that $ \frac{u'}{rK} \to -\frac{f(L)}{N-\alpha}$. Thus for large $r$ we have $u' \sim -\frac{f(L)}{N-\alpha} rK$ and from (H5) we have $rK \sim r^{1-\alpha}$ so $$ |u(r) - u(r_0)| \sim |\frac{f(L)}{N-\alpha} [\frac{r^{2-\alpha} -r_0^{2 - \alpha}}{2-\alpha}]| \to \infty \quad \text{as } r \to \infty \text{ since } 0 < \alpha < 2 $$ contradicting that $u$ is bounded. Thus $f(L)=0$ so $L=0$ or $L=\beta$. But we also know from \eqref{bb} that $F(L)\geq 0$ so $L=0$ or $L \geq \gamma>\beta$. Thus we see that $L \neq \beta$ and so we must have $L=0$. \smallskip \noindent\textbf{Case 2:} $\alpha = 2$. Suppose again $f(L)\neq 0$. This is similar to case 1 but now we have $|u(r) - u(r_0)| \sim |\frac{f(L)}{N-\alpha}\ln(r/r_0)| \to \infty$ contradicting that $u$ is bounded. Thus $f(L)=0$ so $L=0$ or $L=\beta$. Since we also know $F(L)\geq 0$ so $L=0$ or $L \geq \gamma>\beta$. So again we see that $L \neq \beta$ and thus $L=0$. \smallskip \noindent\textbf{Case 3:} $ 2 < \alpha < 2(N-1)$. Here we let $$ u(r) = u_1(r^{2-N}). $$ This transforms \eqref{DE} to \begin{equation} u_1''(t) + h(t) f(u_1(t)) = 0 \quad \text{for } 0 \beta$ then $u_1''<0$ and when $00$. Now for $b> b_0$ we know that $u(r,b)$ has a zero (by definition of $b_0$) and thus $u_1(t,b)$ has a zero, $z_{1,b}$, with $0 < z_{1,b} < R^{2-N}$ for $b>b_0$. Therefore $u_1$ has a local maximum at some $M_{1,b}$ and an inflection point at some $t_{1,b}$ with $0< z_{1,b} < t_{1,b}< M_{1,b} < R^{2-N}$. Since $E_1(z_{1,b}) >0$ and $E_1$ is non-decreasing then it follows that $F(u_1(M_{1,b},b)) = E_1(M_{1,b}) \geq E_1(z_{1,b}) > 0$ and so $u_1(M_{1,b},b) > \gamma$. Note also that $u_1(t_{1,b},b) = \beta$. Since $u_1(t,b)$ is concave up on $(z_{1,b}, t_{1,b})$ we see then that $u_1(t,b)$ lies above the line through $(t_{1,b}, \beta)$ with slope $u_1'(t_{1,b},b)>0$. Thus: $$ u_1(t,b) \geq \beta + u_1'(t_{1,b},b)(t-t_{1,b}) \quad \text{on } [z_{1,b}, t_{1,b}]. $$ Evaluating this at $t=z_{1,b}$ and rewriting yields \begin{equation} t_{1,b} \geq t_{1,b} - z_{1,b} \geq \frac{\beta}{u'(t_{1,b},b)}. \label{fork} \end{equation} In addition, $E_1(t_{1,b}) \leq E_1(M_{1,b})$ so that there is a constant $c_1$ such that for $b$ close to $b_0$, $$ \frac{1}{2}\frac{u_1'^2(t_{1,b},b)}{h(t_{1,b})} + F(\beta) \leq F(u_1(M_{1,b}),b) \leq c_1 $$ and thus \begin{equation} 0< u_1'(t_{1,b}) \leq c_2 \sqrt{h(t_{1,b})} \label{fork2} \end{equation} where $c_2 = \sqrt{2[c_1 + |F(\beta)|]}$. Combining \eqref{fork}-\eqref{fork2} gives \begin{equation} \beta \leq t_{1,b}u_1'(t_{1,b},b) \leq c_2 t_{1,b} \sqrt{h(t_{1,b})} \leq c_3 t_{1,b}^{{\frac{2-q}{2}}} \label{joltinjoe} \end{equation} for some constant $c_3$ for $b$ close to $b_0$. Since $0 0$ for if $u_1'(t_{1,b_0})= 0$ then since $f(u_1(t_{1,b_0})) = f(\beta) = 0$ then $u_1''(t_{1,b_0},b_0)= 0$ implying by uniqueness of solutions of initial value problems that $u_1(t, b_0) \equiv \beta$ contradicting that $u_1'(R^{2-N}, b_0) = -\frac{b_0R^{N-1}}{N-2}>0$. Thus $u_1'(t_{1,b_0})> 0$ and this implies $u_1(t,b_0)< \beta$ for $00$ then in fact $\beta \geq L \geq \gamma > \beta$ which is impossible so we see it must be the case that $L=0$. Thus $\lim_{t \to 0^{+}} u_1(t, b_0)=0$ and therefore $\lim_{r \to \infty} u(r, b_0)=0$. Next, \cite[Lemma 4]{M} states that if $u(r, b_{k})$ is a solution of \eqref{DE}-\eqref{IC} with $k$ zeros on $(0, \infty)$ then if $b$ is sufficiently close to $b_{k}$ then $u(r,b)$ has at most $k+1$ zeros on $(0, \infty)$. Also \cite[Lemma 2.7]{I2} proves a similar result on $(R, \infty)$. Applying this lemma with $b=b_0$ we see that $u(r,b)$ has \emph{at most} one zero on $(R, \infty)$ for $b$ close to $b_0$. On the other hand, by the definition of $b_0$ if $b > b_0$ then $u(r,b)$ has \emph{at least} one zero on $(R, \infty)$. Therefore: $\{ b > b_0 | u(r,b) \text{ has \emph{exactly one zero} on } (R, \infty) \} $ is nonempty and by Lemma \ref{lem2.2} this set is bounded above. Then we let: $$ b_1 = \sup \{ b > b_0 | u(r,b) \text{ has exactly one zero on } (R, \infty) \}. $$ In a similar fashion we can show that $u(r,b_1)$ has exactly one zero on $(R, \infty)$ and $u(r,b_1) \to 0$ as $r \to \infty$. Similarly we can find $u(r,b_{n})$ which has exactly $n$ zeros on $(R, \infty)$ and $u(r,b_{n}) \to 0$ as $r \to \infty$. 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{D} R. Dhanya, Q. Morris, R. Shivaji; Existence of positive radial solutions for superlinear, semipositone problems on the exterior of a ball, \emph{Journal of Mathematical Analysis and Applications,} Volume 434, Issue 2, 1533-1548, 2016. \bibitem{I} J. Iaia; Loitering at the hilltop on exterior domains, \emph{Electronic Journal of the Qualitative Theory of Differential Equations}, Vol. 2015 (2015), No. 82, 1-11. \bibitem{I2} J. Iaia; Existence and nonexistence for semilinear equations on exterior domains, submitted to \emph{Journal of Partial Differential Equations}, Vol. 30 No. 4, 2017, pp. 1-17. \bibitem{JI} J. Iaia; Existence of solutions for semilinear problems with prescribed number of zeros on exterior domains, \emph{Journal of Mathematical Analysis and Applications}, 446, 591-604, 2017. \bibitem{JK} C. K. R. T. Jones, T. Kupper; On the infinitely many solutions of a semilinear equation, \emph{SIAM J. Math. Anal.}, Volume 17, 803-835, 1986. \bibitem{C2} E. Lee, L. Sankar, R. Shivaji; Positive solutions for infinite semipositone problems on exterior domains, \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}