\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 112, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/112\hfil Existence of solutions for semilinear problems] {Existence of solutions for semilinear problems with prescribed number of zeros on exterior domains} \author[J. Joshi, J. Iaia \hfil EJDE-2016/112\hfilneg] {Janak Joshi, Joseph 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 Iaia \newline Department of Mathematics, University of North Texas, P.O. Box 311430, Denton, TX 76203-1430, USA} \email{iaia@unt.edu} \thanks{Submitted December 31, 2015. Published May 3, 2016.} \subjclass[2010]{34B40, 35B05} \keywords{Exterior domains; semilinear; superlinear; radial} \begin{abstract} In this article we prove the existence of an infinite number of radial solutions of $\Delta(u)+f(u)=0$ with prescribed number of zeros on the exterior of the ball of radius $R>0$ centered at the origin in ${\mathbb R}^{N}$ where $f$ is odd with $f<0$ on $(0,\beta)$, $f>0$ on $(\beta,\infty)$ where $\beta>0$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} In this article we study radial solutions of \begin{gather} \Delta(u)+f(u)=0\quad\text{in }\Omega,\label{e1}\\ u=0 \quad\text{on }\partial{\Omega}, \label{e2}\\ u\to {0}\quad\text{as }|x|\to {\infty} \label{e3} \end{gather} where $x\in\Omega={\mathbb R}^{N} \backslash B_{R}(0)$ is the complement of the ball of radius $R>0$ centered at the origin.\ The function $f$ is odd, locally Lipschitz and is defined by \begin{equation} f(u)=|u|^{p-1}u+g(u)\quad\text{with $p>1$, $f'(0)<0$ and } \lim_{u\to{\infty}} \frac{g(u)}{u^p}=0. \label{e4} \end{equation} We assume that there exists $\beta>0$ such that $f(0)=f(\beta)=0$ and $F(u)=\int_0^{u}{f(s)}\,ds $ where \begin{equation} \text{$f<0$ on $(0,\beta)$, $f>0$ on $(\beta,\infty)$ } \label{e5} \end{equation} As $f$ is odd, it follows that $F(u)=\int_0^{u}{f(s)}\,ds $ is even. Also $F$ has a unique positive zero, $\gamma$, with $\beta<\gamma<\infty$ and $F$ is bounded below by some $-F_0<0$ so that \begin{equation} \text{$F<0$ on $(0,\gamma)$, $F>0$ on $(\gamma,\infty)$, and $ F \geq -F_0$ on $(0, \infty)$}. \label{e6} \end{equation} Since we are interested in radial solutions of \eqref{e1}--\eqref{e3} we assume that $u(x)=u(|x|)=u(r)$, where $r=|x|=\sqrt{x_1^2+x_2^2+\dots +x_N^2} $ so that $u$ solves \begin{gather} u''(r)+\frac{N-1}{r}u'(r)+f(u(r))=0\quad\text{on $(R,\infty)$ where $R>0$}, \label{e7} \\ u(R)=0,\quad u'(R)=a>0.\label{e8} \end{gather} \\[.25cm] We will occasionally denote the solution of the above by $u_a(r)$, to emphasize the dependence on the initial parameter $a$. \begin{theorem}\label{main-thm} For each nonnegative integer $n$, there exists a solution $u(r)$ of \eqref{e7}--\eqref{e8} on $[R,\infty)$ such that $\lim_{r\to{\infty}}u(r)=0$ and $u(r)$ has exactly $n$ zeros on $(R, \infty)$. \end{theorem} The radial solutions of \eqref{e1}, \eqref{e3} have been well-studied when $\Omega = {\mathbb R}^{N}$. These include \cite{BL,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,I,C2,S}. Here we use a scaling argument as in \cite{M} to prove existence of solutions. \section{Preliminaries} For $R>0$ existence and uniqueness of solutions of \eqref{e7}-\eqref{e8} on $[R,R+\epsilon)$ for some $\epsilon>0$ and continuous dependence of solutions with respect to $a$ follows from the standard existence-uniqueness theorem for ordinary differential equations \cite{BR}. For existence on $[R,\infty)$ we consider \begin{equation} E_a(r)= \frac{1}{2}u_a'^2+F(u_a). \label{e9} \end{equation} Using \eqref{e7} we see that \begin{equation} E_a'(r)=-\frac{N-1}{r}u_a'^2\leq{0} \label{e10} \end{equation} so $E_a$ is non-increasing on $[R,\infty)$. Therefore \begin{equation} \frac{1}{2}u_a'^2+F(u_a)=E_a(r)\leq{E_a(R)} =\frac{1}{2}a^2\quad\text{for }r\geq{R}. \label{e11} \end{equation} Therefore by \eqref{e6}, \[ \frac{1}{2}u_a'^2\leq{\frac{1}{2}a^2 + F_0}. \] So for a fixed $a$ we see that $u_a'$ is uniformly bounded and hence existence on all of $[R,\infty)$ follows. \begin{lemma} \label{lem1} Let $u_a(r)$ be the solution of \eqref{e7}-\eqref{e8}. If $a$ is sufficiently large then there exists $r>R$ such that $u_a(r)>\beta$. In particular, there exists $r_a> R$ such that $u_a(r_a) = \beta$. \end{lemma} \begin{proof} Since $u_a'(R)=a>0$ we see that $u_a(r)$ is increasing on $[R,R+\delta)$ for some $\delta>0$. If $u_a(r)$ has a first critical point $M_a>R$ with $u'_a(r)>0$ on $[R,M_a)$ then we must have $u'_a(M_a)=0,u''_a(M_a)\leq{0}$. In fact $u''_a(M_a)<0 $ (by uniqueness of solutions of initial value problems). Therefore from \eqref{e7} it follows that $f(u_a(M_a))>0 $ and using (5) we see that $u_a(M_a)>\beta$. On the other hand, if $u_a(r)$ has no critical point then $u_a'(r)>0$ for each $r\geq{R}$. Suppose now by the way of contradiction that $u_a(r)\leq{\beta}$ for each $r\geq{R} $. Since $u_a(r)$ is increasing and bounded above then $\lim_{r\to\infty}{u_a(r)}$ exists. Thus there exists $L>0$, $L\leq{\beta}$ such that \begin{equation} \lim_{r\to\infty}{u_a(r)} =L. \label{e12} \end{equation} Since $E_a(r)$ is non-increasing and bounded below, it follows that $\lim_{r\to\infty}E_a(r)$ exists. This implies $\lim_{r\to\infty}u'_a(r)$ exists and in fact $\lim_{r\to\infty}u'_a(r)=0$ since otherwise $u_a$ would become unbounded contradicting \eqref{e12}. Hence by \eqref{e7}, $\lim_{r\to\infty}u_a''(r) $ exists and as with $u_a'(r)$ we see that $\lim_{r\to\infty}u_a''(r)=0 $. Taking limits in \eqref{e7} we see that $f(L)=0$. Since $L>0$ it follows that $L=\beta$. Suppose now that this is true for all values of $a>0$. We then let $y_a(r)=\frac{u_a(r)}{a}$ and we see that: \begin{gather} y_a''+\frac{N-1}{r}y_a'+\frac{f(ay_a)}{a}=0. \label{e13} \\ y_a(R)=0 , \quad y_a'(R)=1.\label{e14} \end{gather} Since \[ \Big(\frac{y_a'^2}{2}+\frac{F(ay_a)}{a^2}\Big)' =y_a'y_a''+\frac{f(ay_a)}{a}y_a'=-\frac{(N-1)}{r}y_a'^2\leq{0}, \] it follows that \[ \frac{y_a'^2}{2}+\frac{F(ay_a)}{a^2}\leq{\frac{1}{2}} \quad \forall r\geq{R}. \] In addition, from \eqref{e6} it follows that \[ \frac{y_a'^2}{2}-\frac{F_0}{a^2}\leq{\frac{1}{2}}. \] Hence \[ \frac{y_a'^2}{2}\leq{\frac{1}{2}+\frac{F_0}{a^2}}\leq{1} \] if $a$ is sufficiently large. Therefore $|y_a'|$ is uniformly bounded if $a$ is sufficiently large. Also $0\leq{u_a}\leq{\beta}$ implies $0\leq{y_a}\leq{\frac{\beta}{a}}\leq{1}$ if $a$ is large so $y_a$ is uniformly bounded. And since $a y_a$ is bounded it follows that $\frac{f(a y_a)}{a} \to 0 $ as $a \to \infty$. Thus it follows from \eqref{e13} that $|y_a''|$ is uniformly bounded for sufficiently large $a$. Hence by the Arzela-Ascoli theorem $y_a\to{y}$ and $y_a'\to{y'}$ uniformly on the compact subsets of $[R,\infty)$ as $a\to{\infty}$ for some subsequence still denoted by $y_a$. Moreover from \eqref{e14} we see $y(R)=0$ and $y'(R)=1$. On the other hand, $0\leq{y_a}\leq{\frac{\beta}{a}}$ so it follows that $y_a\to{0}$ as $a\to{\infty}$. So $y\equiv{0}$ and therefore $y'\equiv{0}$ which is a contradiction to $y'(R)=1$. Hence there exists $r_a>R$ such that $u_a(r_a) =\beta$ and $00$. Thus $u_a'(r_a)>0$. Hence $u_a(r)$ must get larger than $\beta$. Thus there exists $r_a>R$ such that $u_a(r_a)=\beta, u_a'(r_a)>0$ and $u_a<\beta$ on $[R,r_a)$. This completes the proof. \end{proof} \begin{lemma} \label{lem2} If $a$ is sufficiently large then $u_a(r)$ has a maximum at $M_a>r_a$. In addition, $|u_a|$ has a global maximum at $M_a$ and $u_a(M_a)\to{\infty}$ as $a\to{\infty}$. \end{lemma} \begin{proof} Suppose by the way of contradiction that $u_a'(r)>0 $ for each $r>R$. Then $u_a(r)>\beta$ for $r>{r_a}$ as we saw in the proof of the Lemma \ref{lem1}. Also as in Lemma \ref{lem1}, $u_a'(r_a)>0$ thus $\exists\,r_{a_1}>r_a$ such that $u(r_{a_1})>\beta+\epsilon $ for some $\epsilon>0$ and since $u_a'>0$, for $r>r_{a_1}$ we have $f(u_a)\geq{f(\beta+\epsilon)}>0$. Therefore, \[ u_a''+\frac{N-1}{r}u_a'+f(\beta+\epsilon)\leq{u_a'' +\frac{N-1}{r}u_a'+f(u_a)}=0 \quad \text{for } r>r_{a_1}. \] This implies $$ \big(r^{N-1}u_a'(r)\big)'\leq{-f(\beta+\epsilon)r^{N-1}} \quad \text{for } r>r_{a_1}. $$ Hence for $r>r_{a_1}$ we have $$ r^{N-1}u_a'(r)0$ for $r>R$. So $\exists\,M_a>r_a$ such that $u_a'(M_a)=0$ and $u''_a(M_a)\leq{0}$. By uniqueness of solutions of initial value problems it follows that $u''_a(M_a)<0$ so $M_a$ is a local maximum. Thus $f(u_a(M_a))>0$ and therefore $u_a(M_a)> \beta$. To see this is a global maximum for $|u_a|$ suppose there exists $M_{a_2}>M_a$ with $|u_a(M_{a_2})|>u_a(M_a)>\beta$. Then since $F$ is even and increasing for $u> \beta$ it follows that \[ F(u_a(M_{a_2})) = F(|u_a(M_{a_2})|) < F(u_a(M_a)). \] On the other hand, $E_a$ is nonincreasing so \[ F(u_a(M_{a_2}))= E_a(M_{a_2})\leq{E_a(M_a)}=F(u_a(M_a)), \] a contradiction. Hence $M_a$ is the global maximum for $|u_a|$. We now show that $u_a(M_a)\to{\infty}$ as $a\to{\infty}$. Suppose not. Then $|u_a(r)|\leq{C}$ where $C$ is a constant independent of $a$. As in Lemma \ref{lem1}, let $y_a(r)=\frac{u_a(r)}{a}$. Then as in Lemma \ref{lem1}, $y_a\to{y}$ with $y\equiv{0}$ and $y'(R)=1$, a contradiction. Hence $u_a(M_a)\to{\infty}$ as $a\to{\infty}$. This proves the lemma. \end{proof} Next we proceed to show that $u_a(r)$ has zeros on $(R,\infty)$ and the number of zeros increases as $a\to{\infty}$. First we let $v_a(r)=u_a(M_a+r)$. It follows that $v_a$ satisfies \begin{gather} v_a''(r)+\frac{N-1}{M_a+r}v_a'(r)+f(v_a(r))=0 \quad\text{on } [R,\infty), \label{e15}\\ v_a(0) ={u_a(M_a)}\equiv\lambda_a^{\frac{2}{p-1}} \quad \text{and}\quad v_a'(0)=0. \label{e16} \end{gather} By Lemma \ref{lem2}, $\lim_{a\to{\infty}}u_a(M_a)=\infty$ and thus $\lambda_a\to{\infty}$ as $a\to{\infty}$. Next we let $w_{\lambda_a}(r)=\lambda_a^{-\frac{2}{p-1}}v_a (\frac{r}{\lambda_a})$ as in \cite{M}. Then using \eqref{e4} and \eqref{e15}--\eqref{e16} we see that \begin{gather} w_{\lambda_a}''(r)+\frac{N-1}{\lambda_aM_a+r}w_{\lambda_a}'(r) + |w_{\lambda_a}|^{p-1} w_{\lambda_a} +\frac{g(\lambda_a^\frac{2}{p-1}w_{\lambda_a})}{\lambda_a^\frac{2p}{p-1}}=0, \label{e17}\\ w_{\lambda_a}(0)=1, \quad w_{\lambda_a}'(0)=0. \label{e18} \end{gather} \begin{lemma} \label{lem3} $w_{\lambda_a}\to{w}$ uniformly on compact subsets of $[0,\infty)$ as $a\to{\infty}$ and $w$ satisfies $w''+|w|^{p-1}w=0$. \end{lemma} \begin{proof} From \eqref{e4} we know that $f(u)= |u|^{p-1}u+g(u)$ with $p>1$ where $\frac{g(u)}{u^p}\to{0}$ as $u\to{\infty}$. Letting $G(u)=\int_0^{u}g(s)\,ds $ then it follows that $\frac{G(u)}{u^{p+1}}\to{0}$ as $u\to{\infty}$. Let $w_{\lambda_a}(r)$ be the solution of the system \eqref{e17}--\eqref{e18} and $E_{\lambda_a}(r)$ be the energy associated with $w_{\lambda_a}(r)$ defined by \begin{equation} E_{\lambda_a} = \frac{w_{\lambda_a}'^2}{2} +\frac{|w_{\lambda_a}|^{p+1}}{p+1} +\frac{1}{\lambda_a^\frac{2(p+1)}{p-1}}G(\lambda_a^\frac{2}{p-1}w_{\lambda_a}). \label{e19} \end{equation} Then $E_{\lambda_a}'(r)=\frac{-(N-1)}{\lambda_a M_a +r} w_{\lambda_a}'^2\leq{0}$ which implies $E_{\lambda_a}(r)$ is a non-increasing function of $r$. Therefore, $$ E_{\lambda_a}(r)\leq{E_{\lambda_a}(0)} = \frac{1}{p+1}+\frac{1}{\lambda_a^\frac{2(p+1)}{p-1}}G(\lambda_a^\frac{2}{p-1}). $$ Since $\frac{G(u)}{u^{p+1}}\to{0}$ as $u\to{\infty}$ it follows for $a$ sufficiently large that $$ E_{\lambda_a}(r)\leq{E_{\lambda_a}(0)}\leq{\frac{1}{p+1}}+1< 2. $$ Also it follows that $|G(u)|\leq{\frac{1}{2(p+1)}|u|^{p+1}}$ if $|u|\geq{T_1}$ for some $T_1>0$. And since $G$ is continuous on the compact set $|u|\leq{T_1}$, there exists a constant $C_G>0$ such that $|G(u)|\leq{C_G}$ if $|u|\leq{T_1}$. Thus $$ |G(u)|\leq{C_G+\frac{1}{2(p+1)}|u|^{p+1}} \text{ for all } u. $$ Therefore if $a$ is sufficiently large we see from this upper bound for $G$ and \eqref{e19} that \[ \frac{w_{\lambda_a}'^2}{2}+\frac{|w_{\lambda_a}|^{p+1}}{p+1} \leq{2-\frac{1}{\lambda_a^\frac{2(p+1)}{p-1}} G(\lambda_a^\frac{2}{p-1}w_{\lambda_a}) \leq{2+\frac{C_G}{\lambda_a^\frac{2(p+1)}{p-1}}} +\frac{|w_{\lambda_a}|^{p+1}}{2(p+1)}}. \] Thus if $a$ is sufficiently large we have \begin{equation} \frac{w_{\lambda_a}'^2}{2}+\frac{|w_{\lambda_a}|^{p+1}}{2(p+1)} \leq{2+\frac{C_G}{\lambda_a^\frac{2(p+1)}{p-1}}}\leq{3}. \label{e20} \end{equation} Therefore $w_{\lambda_a}$ and $w_{\lambda_a}'$ are uniformly bounded for large $a$. So by the Arzela-Ascoli theorem $w_{\lambda_a}\to{w}$ uniformly on compact subsets of $[0,\infty)$ for some subsequence still labeled $w_{\lambda_a}$. Now using the definition of $f$ from \eqref{e4} we have: \begin{gather*} w_{\lambda_a}''+\frac{N-1}{\lambda_a M_a+r}w_{\lambda_a}' +|w_{\lambda_a}|^{p-1} w_{\lambda_a}+\lambda_a^\frac{-2p}{p-1} g(\lambda_a^\frac{2}{p-1}w_{\lambda_a}) =0, \label{e21}\\ w_{\lambda_a}(0)=1, w_{\lambda_a}'(0)=0. \label{e22} \end{gather*} Since $\lim_{u\to{\infty}}{\frac{g(u)}{u^p}} =0$, it follows that for all $\epsilon > 0$ there exists a $T_2>0$ such that $|g(u)|\leq{\epsilon|u|^{p}} $ if $|u|>T_2$ and the continuity of $g$ on the compact set $|u|\leq{T_2}$ implies $|g(u)|\leq{C_g}$ for some $C_g>0$ if $|u|\leq{T_2}$. Thus, $$ |g(u)|\leq{C_g+\epsilon|u|^{p}} \quad \text{for all } u $$ and hence $$ |g(\lambda_a^{\frac{2}{p-1}}w_{\lambda_a})| \leq{C_g+\epsilon\lambda_a^{\frac{2p}{p-1}}|w_{\lambda_a}|^p}. $$ Recall from \eqref{e20} that $|w_{\lambda_a}|\leq [6(p+1)]^\frac{1}{p+1}< 4$ for $p>1$. So: $$ \frac{|g(\lambda_a^{\frac{2}{p-1}}w_{\lambda_a})|}{\lambda_a^\frac{2p}{p-1}} \leq{\frac{C_g+\epsilon\lambda_a^\frac{2p}{p-1}4^p}{\lambda_a^\frac{2p}{p-1}}} = \frac{C_g}{\lambda_a^\frac{2p}{p-1}}+ \epsilon 4^p. $$ This implies $$ 0\leq{\limsup_{a\to{\infty}}{\frac{|g(\lambda_a^{\frac{2}{p-1}}w_{\lambda_a})|} {\lambda_a^\frac{2p}{p-1}}}} \leq{ \limsup_{a\to{\infty}}\frac{C_g}{\lambda_a^\frac{2p}{p-1}}+ \epsilon 4^p} = \epsilon 4^p. $$ This is true for each $\epsilon>0$. Hence \begin{equation} \lim_{a\to{\infty}}{\frac{|g(\lambda_a^{\frac{2}{p-1}} w_{\lambda_a})|}{\lambda_a^\frac{2p}{p-1}}} = 0 . \label{e23} \end{equation} In addition, recall that $M_a\geq R$ and so for $r\geq{R}$ we have \[ \frac{1}{\lambda_a M_a+r}\leq{\frac{1}{(\lambda_a+1)R}} \] and since $|w_{\lambda_a}'|$ is uniformly bounded (by \eqref{e20}) we see that $\frac{N-1}{\lambda_a M_a+r}w_{\lambda_a}' \to 0$ as $a \to \infty$. From this and \eqref{e23} we see that the second and fourth terms on the left-hand side of \eqref{e21} go to $0$ as $a \to \infty$. In addition, $w_{\lambda_a}$ is bounded by \eqref{e20} and therefore it follows from \eqref{e21} that $|w_{\lambda_a}''|$ is uniformly bounded. Therefore by the Arzela-Ascoli theorem for some subsequence still labeled $w_{\lambda_a}$ we have $w_{\lambda_a}\to{w}$ and $w_{\lambda_a}'\to{w'}$ uniformly on compact subsets of $[0,\infty)$ and from \eqref{e21} we have $\lim_{a\to\infty}w_{\lambda_a}''+|w|^{p-1}w=0$. Thus $\lim_{a\to{\infty}}w_{\lambda_a} ''$ exists and in fact $\lim_{a\to{\infty}}w_{\lambda_a}''=w''$. Hence \begin{gather} w''+|w|^{p-1}w=0 \label{e24}\\ w(0)=1,\quad w'(0)=0. \label{e25} \end{gather} Therefore $\frac{1}{2}w'^2+\frac{1}{p+1}|w|^{p+1}=\frac{1}{p+1}$. It is straightforward to show that solutions of \eqref{e24}--\eqref{e25} are periodic with period $ \sqrt{2(p+1)} \int _0^{1} \frac{dt}{\sqrt{1-t^{p+1}}}$ and they have an infinite number of zeros on $[0,\infty)$. Since $w_{\lambda_a}\to{w}$ uniformly on compact subsets of $[0,\infty)$ as $a\to{\infty}$ it follows that $w_{\lambda_a}$ has zeros on $(0,\infty)$ and the number of zeros of $w_{\lambda_a}$ gets arbitrarily large by taking $a$ sufficiently large. Recalling that \[ w_{\lambda_a}(r) = \lambda^{-\frac{2}{p-1}} u_a(M_a + \frac{r}{\lambda_a}) \] we see that $u_a(r)$ has zeros $(R,\infty)$ for large $a$ and the number of zeros of $u_a(r)$ increases as $a$ increases. \end{proof} Next we examine \eqref{e7}-\eqref{e8} when $a>0$ is small. \begin{lemma} \label{lem4} $r_a\to{\infty}$ as $a\to{0^+}$ where $r_a$ is defined in Lemma \ref{lem1}. \end{lemma} \begin{proof} From \eqref{e11} we have $\frac{1}{2}u_a'^2+F(u_a)\leq{\frac{1}{2}a^2}$ for $r\geq{R}$, and from Lemma \ref{lem2} we have $u_a'>0$ on $[R,r_a]$. So rewriting this inequality and integrating on $(R,r_a)$ gives \[ \int_{R}^{r_a}\frac{u_a'}{\sqrt{a^2-2F(u_a)}}\leq{\int_{R}^{r_a}}1\,dr = r_a -R. \] Letting $s=u_a(r) $ we see that \begin{equation} \int_0^{\beta}\frac{ds}{\sqrt{a^2-2F(s)}} =\int_{R}^{r_a}\frac{u_a'\,dr}{\sqrt{a^2-2F(u_a)}}\leq{r_a-R}. \label{e26} \end{equation} From \eqref{e4} we have $f'(0)<0$, thus $f(u)\geq{-\frac{3}{2}}|f'(0)|u $ for small $ u $. So $ a^2-2F(u)\leq{\frac{3}{2}}|f'(0)|u^2 +a^2$ for small $u$ and so \[ \sqrt{a^2-2F(u)}\leq{\sqrt{ a^2+\frac{3}{2}|f'(0)|u^2 }} \leq{a+\sqrt{\frac{3}{2}|f'(0)|}u} \text{ for small } u. \] Therefore, $$ \frac{1}{\sqrt{a^2-2F(u)}} \geq \frac{1}{{a+\sqrt{\frac{3}{2}|f'(0)|}u}} \quad \text{for small } u. $$ So for some $\epsilon$ with $0<\epsilon< \beta$ we have \begin{align*} \int_0^{\epsilon}{\frac{ds}{\sqrt{a^2-2F(s)}}} &\geq \int_0^{\epsilon} {\frac{ds}{a+\sqrt{\frac{3}{2}|f'(0)|}s}}\ &= \sqrt{\frac{2}{3 |f'(0)|}} \ln\Big({1+\sqrt{\frac{3}{2}|f'(0)|}\frac{\epsilon}{a}}\Big) \to \infty \end{align*} as $a \to 0^{+}$. Therefore from \eqref{e26} and the above computation we see that $$ r_a-R \geq \int_0^{\beta}{\frac{ds}{\sqrt{a^2-2F(s)}}} \geq{\int_0^{\epsilon}{\frac{ds}{\sqrt{a^2-2F(s)}}}} \to \infty \quad\text{as } a\to{0^+} $$ thus $r_a\to{\infty}$ as $ a\to{0^+}$. Hence the lemma is proved. \end{proof} Note that if $E(r_0)<0$, then \begin{equation} u(r)>0\quad\text{for each }r>r_0. \label{e27} \end{equation} Suppose not. Then there exists $z>r_0$ such that $u(z)=0$ and so $F(u(z))=0$. By \eqref{e10}, $E(r)$ is non-increasing so $E(z)\leq{E(r_0)}<0$. Therefore $$ 0\leq{\frac{u'(z)^2}{2}}=\frac{u'(z)^2}{2}+F(u(z))=E(z)<0 $$ which is impossible. Hence $u(r)>0$ for all $r>r_0$. \begin{lemma} \label{lem5} If $a>0$ and $a$ is sufficiently small then $u_a(r)>0$ for each $r>R$. \end{lemma} \begin{proof} Assume by the way of contradiction that $u_a(z_a)=0$ for some $z_a>R$. Since $u_a(R)=0$ and $u_a'(R)=a >0$ we see that $u_a(r)$ has a positive local maximum, $M_a$, with $R\gamma $ and so in particular there exist $p_a, q_a$ with $R 0 $ on $[R,q_a)$ by \eqref{e7}. Therefore $\int_{R}^{r} (r^{N-1}u_a')' \, dr > 0$ from which it follows that \[ r^{N-1}u_a'> R^{N-1}u_a'(R) > 0 \quad\text{on }[R,q_a). \] Thus $u_a(r)$ is increasing on $[R,q_a)$. In addition, $p_a\to{\infty}$ as $a\to{0^+}$ for if the ${p_a}$ were bounded then a subsequence would converge to say some finite $p_0$ as $a\to{0^+}$. Since $E_a(r)$ is non-increasing this would imply ${u_a(r)}$ and ${u_a'(r)}$ would be uniformly bounded on $[R,p_0+1]$ and so by the Arzela-Ascoli theorem for a subsequence $u_a(r)\to{u_0(r)}\equiv{0}$ as $a\to{0^+}$. On the other hand, $\frac{\beta}{2}=u_a(p_a)\to{u_0(p_0)} =0 $ as $a\to{0^+} $ which is a contradiction. Thus we see that $p_a\to{\infty}$ as $a\to{0^+}$. Next we return to \eqref{e11} and after rewriting we have $$ \frac{u_a'}{\sqrt{a^2-2F(u_a)}}\leq{1} \quad \text{for each } r\geq{R}. $$ Integrating on $[p_a, q_a]$ and setting $u_a(r)=t$ we obtain \begin{equation} \int_{\frac{\beta}{2}}^{\beta}\frac{dt}{\sqrt{a^2-2F(t)}} =\int_{p_a}^{q_a}{\frac{u_a'} {\sqrt{a^2-2F(u_a)}}}\,dr \leq{\int_{p_a}^{q_a}}1\, dr =q_a-p_a. \label{e28} \end{equation} Now on $[\frac{\beta}{2},\beta]$ we have $ 0 0 $$ for some constant $c>0$ and sufficiently small $a$. Combining this with \eqref{e28} we see that \begin{equation} q_a-p_a\geq{c} \quad \text{if $a$ is sufficiently small}. \label{e29} \end{equation} Now by the definition of $E_a(r)$ it is straightforward to show that $$ (r^{2(N-1)}E_a(r))'=(r^{2(N-1)})'F(u_a). $$ Integrating on $[p_a,q_a]$ gives $$ q_a^{2(N-1)}E_a(q_a)=p_a^{2(N-1)}E_a(p_a) +\int_{p_a}^{q_a}[r^{2(N-1)}]'F(u_a)\,dr. $$ Since $F(u_a)\leq{F(\frac{\beta}{2})}<0$ on $[p_a,q_a]$ we have \begin{align*} & p_a^{2(N-1)}E_a(p_a)+\int_{p_a}^{q_a}(r^{2(N-1)})'F(u_a)\,dr\\ &\leq{p_a^{2(N-1)}E_a(p_a)}-|F(\frac{\beta}{2})|[q_a^{2(N-1)}-p_a^{2(N-1)}]. \end{align*} But \[ p_a^{2(N-1)}E_a(p_a)=R^{2(N-1)}E_a(R)+\int_{R}^{p_a}[r^{2(N-1)}]'F(u_a)\,dr \] and $$ \int_{R}^{p_a}[r^{2(N-1)}]'F(u_a)\,dr\leq{0} $$ as $F(u_a)\leq{0}$ on $[R,p_a]$. Thus $$ p_a^{2(N-1)}E_a(p_a)\leq{R^{2(N-1)}}E_a(R)=\frac{1}{2}a^2\,R^{2(N-1)}. $$ Therefore, $$ q_a^{2(N-1)}E_a(q_a)\leq{\frac{1}{2}a^2\,R^{2(N-1)}} -|F(\frac{\beta}{2})|\left[q_a^{2(N-1)}-p_a^{2(N-1)}\right]. $$ So \begin{equation} q_a^{2(N-1)}E_a(q_a)\leq{\frac{a^2 R^{2(N-1)}}{2} -|F(\frac{\beta}{2})|(q_a^{2(N-1)}-p_a^{2(N-1)})} \label{e30} \end{equation} Now by \eqref{e29} we have \[ q_a^{2(N-1)}-p_a^{2(N-1)}\geq{(q_a-p_a)p_a^{2N-3}}\geq{c\,p_a^{2N-3}}, \] and from earlier in the proof of this lemma we saw $\lim_{a\to{0^+}}p_a^{2N-3}={\infty}$. Thus $q_a^{2(N-1)}-p_a^{2(N-1)}\to{\infty}$ as $a\to{0^+}$. It follows then from \eqref{e30} that $q_a^{2(N-1)}E_a(q_a)$ is negative if $a$ is sufficiently small. Thus by \eqref{e27} it follows that $u_a(r)>0$ for $r\geq{q_a}$. Also, since we have $u_a'>0$ on $[R,q_a]$ and $u_a(R)=0$ we see that $u_a(r)>0$ on $(R,\infty)$ if $a$ is sufficiently small. This completes the proof. \end{proof} \section{Proof of Theorem \ref{main-thm}} Let $$ S_0 = \{ a>0 | u_a(r)>0\, \forall r>R \}. $$ By Lemma \ref{lem5} we know that for $a>0$ and $a$ sufficiently small that $u_a(r)>0 $ so $S_0$ is nonempty. Also from Lemma \ref{lem3} we know that if $a$ is sufficiently large then $u_a(r)$ has zeros. Hence $S_0$ is bounded above and so the supremum of $S_0$ exists. Let $a_0=\sup (S_0)$. \begin{lemma} \label{lem6} $u_{a_0}(r)>0 $ on $(R,\infty)$. \end{lemma} \begin{proof} Suppose by the way of contradiction that there exists $z_0$ such that $u_{a_0}(z_0) = 0$ and $u_a(r) > 0$ on $[R, z_0)$. Then $u'_{a_0}(z_0)\leq{0}$ and by uniqueness in fact $u_{a_0}'(z_0)<0$. Thus $u_{a_0}(r)<0$ for $z_00 $ on $(R,\infty)$. This completes the lemma. \end{proof} \begin{lemma} \label{lem7} $u_{a_0}(r)$ has a local maximum, $M_{a_0}> R$. \end{lemma} \begin{proof} Suppose not. Then $u'_{a_0}(r)>0$ for all $r\geq{R}$. Since $E_{a_0}(r)\leq{E_{a_0}(R)}$ for all $r\geq{R}$, we have $$ \frac{u_{a_0}'^2(r)}{2} + F(u_{a_0}(r)) \leq{\frac{a_0^2}{2}}. $$ This implies $F(u_{a_0}(r))\leq{\frac{a_0^2}{2}}$ and hence $u_{a_0}(r)$ is bounded. Since we are also assuming $u_{a_0}'(r)>0$ it follows that $\lim_{r\to{\infty}}u_{a_0}(r)$ exists. Let us denote $\lim_{r\to{\infty}} u_{a_0}(r) = L$. Since $E_{a_0}(r)$ is a non-increasing function which is bounded below, it follows that $\lim_{r\to{\infty}} E_{a_0}(r) = \lim_{r\to{\infty}}[\frac{u_{a_0}'^2}{2}+F(u_{a_0})]$ exists. Since we also know that $\lim_{r\to{\infty}}u_{a_0}(r)$ exists it follows that $\lim_{r\to{\infty}}u_{a_0}'(r)$ exists and in fact $\lim_{r\to{\infty}}u_{a_0}'(r)=0$ (since otherwise $u_{a_0}(r)$ would be unbounded). Therefore from \eqref{e7} it follows that $\lim_{r\to{\infty}}u_{a_0}''(r)= -f(L)$ and in fact $f(L)=0$. (Otherwise, $u_{a_0}'$ would be unbounded but we know $u_{a_0}'\to 0$). So $ L=-\beta, 0$, or $\beta$. Since $u_{a_0}(r)>0 $ and $u_{a_0}'(r)>0$ thus $L=\beta$. Now by the definition of $a_0$ we know $u_a(r)$ has a zero if $a>a_0$, say $u_a(z_a)=0$. Next we show that \begin{equation} \lim_{a\to{a_0^+}} z_a =\infty. \label{e31} \end{equation} Suppose not. Then $|z_a|\leq{K} $ for some constant $K$ and so there is a subsequence of $z_a$ still denoted $z_a$ such that $z_a\to{z_0}$ as $a\to{a_0^+}$. But $u_a(r){\to{u_{a_0}(r)}}$ uniformly on the compact subset $[R,z_0+1]$ as $a\to{a_0^+}$ so $0=\lim_{a\to{a_0^+}}{u_a(z_a)}=u_{a_0}(z_0)$ which contradicts that $u_{a_0}(r)>0$ from Lemma \ref{lem6}. Thus $\lim_{a\to{a_0^+}}z_a=\infty$. In addition, $E_a(z_a)=\frac{u_a'^2(z_a)}{2}\geq 0$. Also: \[ \lim_{r\to{\infty}} E_{a_0}(r)=\lim_{r\to{\infty}} [\frac{u_{a_0}'^2(r)}{2}+F(u_{a_0}(r))]=F(\beta)<0. \] So there exists $R_0>R$ such that $E_{a_0}(R_0)<0$. Since $\lim_{a\to{a_0}} u_a(r) = u_{a_0}(r)$ uniformly on the compact set $[R, R_0+1]$, it follows that $\lim_{a\to{a_0}} E_a(R_0) = E_{a_0}(R_0)<0$. Since $E_a(R_0)<0R$. This completes the proof. \end{proof} \begin{lemma} \label{lem8} $u_{a_0}'(r)<0 $ if $r>M_{a_0}$. \end{lemma} \begin{proof} Suppose $u_{a_0}'(m_{a_0})=0 $ for some $m_{a_0}>M_{a_0}$. Then $u_{a_0}''(m_{a_0})>0$ and so $f(u(m_{a_0}))<0$. Since we also know that $u_{a_0}(r)>0$ (by Lemma \ref{lem6}) it follows that $0a_0$ then $u_a(r)$ has a zero, $z_a$, with $E_a(z_a)\geq{0}$ and by (31) we have seen that $\lim_{a\to{a_0}}z_a=\infty$. Since $E_a$ is non-increasing we therefore have $z_a0$ so $\lim_{r\to{\infty}}u_{a_0}(r)=L$ with $ L \geq 0$. Since $E_{a_0}(r)$ is non-increasing, we see as we did earlier that $f(L)=0$. Thus $L=0$ or $ \beta$. We now show $E_{a_0}(r)\geq 0$ for all $r\geq R$. So suppose there is an $r_0>R$ such that $E_{a_0}(r_0)<0$. Then $E_a(r_0)<0$ for $a$ close to $a_0$ and in particular if $a>a_0$. But then we know that $z_a$ exists and since $E_a(z_a)\geq{0}$ it follows that $z_a0 | u_a(r) \text{ has one zero on } (R, \infty) \}. $$ \cite[Lemma 4]{M} states that if $u_{a_{k}}(r)$ is a bounded solution of \eqref{e7} on $(0, \infty)$ with $k$ zeros and $\lim_{r \to \infty} u_{a_{k}}(r) =0$ then if $a$ is sufficiently close to $a_{k}$ then $u_a$ has at most $k+1$ zeros on $[0, \infty)$. A nearly identical lemma holds for solutions of \eqref{e7} on $(R, \infty)$. Applying this lemma with $a_0$ we see that $u_a$ on $(R, \infty)$ has at most one zero if $a$ is sufficiently close to $a_0$. On the other hand, for $a>a_0$ we know that $u_a(r)$ has at least one zero on $(R, \infty)$ by the definition of $a_0$. Thus if $a>a_0$ and $a$ is sufficiently close to $a_0$ then $u_a$ has exactly one zero and so we see that $S_{1}$ is nonempty. 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