\documentclass[twoside]{article} \usepackage{amssymb} % font used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Spike solutions concentrating at local minima \hfil EJDE--2000/32} {EJDE--2000/32\hfil Gregory S. Spradlin \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol.~{\bf 2000}(2000), No.~32, pp.~1--14. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % An elliptic equation with spike solutions concentrating at local minima of the \\ Laplacian of the potential \thanks{ {\em Mathematics Subject Classifications:} 35J50. \hfil\break\indent {\em Key words and phrases:} Nonlinear Schr\"odinger Equation, variational methods, \hfil\break\indent singularly perturbed elliptic equation, mountain-pass theorem, concentration compactness, \hfil\break\indent degenerate critical points. \hfil\break\indent \copyright 2000 Southwest Texas State University and University of North Texas. \hfil\break\indent Submitted February 4, 2000. Published May 2, 2000.} } \date{} % \author{ Gregory S. Spradlin } \maketitle \begin{abstract} We consider the equation $-\epsilon^2 \Delta u + V(z)u = f(u)$ which arises in the study of nonlinear Schr\"odinger equations. We seek solutions that are positive on ${\mathbb R}^N$ and that vanish at infinity. Under the assumption that $f$ satisfies super-linear and sub-critical growth conditions, we show that for small $\epsilon$ there exist solutions that concentrate near local minima of $V$. The local minima may occur in unbounded components, as long as the Laplacian of $V$ achieves a strict local minimum along such a component. Our proofs employ variational mountain-pass and concentration compactness arguments. A penalization technique developed by Felmer and del~Pino is used to handle the lack of compactness and the absence of the Palais-Smale condition in the variational framework. \end{abstract} \renewcommand{\theequation}{\thesection.\arabic{equation}} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \section{Introduction} This paper concerns the equation $$\label{1.0} -\epsilon^2 \Delta u + V(z)u = f(u)$$ on ${\mathbb R}^N$ with $N \geq 1$, where $f(u)$ is a superlinear'' type function such as $f(u) = u^p$, $p > 1$. Such an equation arises when searching for standing wave solutions of the nonlinear Schr\"odinger equation (see \cite{f1}). For small positive $\epsilon$, we seek ground states,'' that is, positive solutions $u$ with $u(z) \to 0$ as $|z| \to \infty$. Floer and Weinstein (\cite{f4}) examined the case $N=1$, $f(u) = u^3$ and found that for small $\epsilon$, a ground state $u_\epsilon$ exists which concentrates near a non-degenerate critical point of $V$. Similar results for $N > 1$ were obtained by Oh in \cite{o1}-\cite{o3}. In \cite{f1}, del~Pino and Felmer found that if $V$ has a strict local minimum, then for small $\epsilon$, (\ref{1.0}) has a ground state concentrating near that minimum. A strict local minimum occurs when there exists a bounded, open set $\Lambda \subset {\mathbb R}^N$ with $\inf_\Lambda V < \inf_{\partial \Lambda} V$. They extended their results in \cite{f2} to the more general case where $V$ has a topologically stable'' critical point, that is, a critical point obtained via a topological linking argument (see \cite{f2} for a precise formulation). Such a critical point persists under small perturbations of $V$. Examples are a strict local extremum and a saddle point. This very strong result is notable because the critical points of $V$ in question need not be non-degenerate or even isolated. Similar results have been obtained by Li \cite{l1}, and earlier work of Rabinowitz \cite{r1} is also interesting. The recent results of \cite{a1} and \cite{l2} also permit $V$ to have degenerate critical points. A common feature of all the papers above is that $V$ must have a non-degenerate, or at least topologically stable, set of critical points. Therefore it is natural to try to remove this requirement. That we must assume some conditions on $V$ is shown by Wang's counterexample \cite{w1} - if $V$ is nondecreasing and nonconstant in one variable (e.g. $V(x_1, x_2, x_3) = 2 + \tan^{-1}(x_1)$), then no ground states exist. In \cite{s1} the author showed that ground states to (\ref{1.0}) exist under the assumption that $V$ is almost periodic, together with another mild assumption. Those assumptions did not guarantee that $V$ had a topologically stable critical point. Aside from periodicity or recurrence properties of $V$, another approach is to impose conditions on the derivatives of $V$. That is the approach taken here. We will assume that $V$ has a (perhaps unbounded) component of local minima, along which $\Delta V$ achieves a strict local minimum. More specifically, assume $f$ satisfies the following: \begin{description} \item{(F1)} $f \in C^1({\mathbb R}^+, {\mathbb R})$ \item{(F2)} $f'(0) = 0 = f(0)$. \item{(F3)} $\lim_{q \to \infty} f(q)/q^s = 0$ for some $s > 1$, with $s < (N+2)/(N-2)$ if $N \geq 3$. \item{(F4)} For some $\theta > 2$, $0 < \theta F(q) \leq f(q)q$ for all $q > 0$, where $F(\xi) \equiv \int_0^\xi f(t)\,dt$. \item{(F5)} The function $q \mapsto f(q)/q$ is increasing on $(0, \infty)$. \end{description} Assumptions (F1)-(F5) are the same as in \cite{f1} and are satisfied by $f(q) = q^{s}$, for example, if $1 < s < (N+2)/(N-2)$. Assume that $V$ satisfies the following: \begin{description} \item{(V1)} $V \in C^2({\mathbb R}^n,{\mathbb R})$ \item{(V2)} $D^\alpha V$ is bounded and Lipschitz continuous for $|\alpha| = 2$. \item{(V3)} $0 < V_- \equiv \inf_{{\mathbb R}^N} V < \sup_{{\mathbb R}^N} V \equiv V^+ < \infty$ \item{(V4)} There exists a bounded, nonempty open set $\Lambda \subset {\mathbb R}^N$ and a point $z_0 \in \Lambda$ with $V(z_0) = \inf_\Lambda V \equiv V_0$, and $$\Delta_0 \equiv \inf\{\Delta V(z) \mid z \in \Lambda, \ V(z) = V_0\} < \inf\{\Delta V(z) \mid z \in \partial \Lambda, \ V(z) = V_0\}$$ \end{description} \paragraph{Note:} A special case of (V4) occurs when $\Lambda$ is bounded and $V(z_0) < \inf_{\partial \Lambda} V$; this case is treated, under weaker hypotheses, in \cite{f1}. A specific example of (V4) is if $N = 2$ and $V$ satisfies (V1)-(V4), with $V(z_1, z_2) = 1 + (z_1^2 - z_2)^2$ for $z_1^2 + z_2^2 \leq 1$. Then $\Delta V(z_1, z_1^2) = 8z_1^2 + 2$ for $z_1^2 + z_2^2 \leq 1$, so we may take $\Lambda = B_1(0,0) \subset {\mathbb R}^2$ and $z_0 = (0,0)$. Then $V$ has a component of local minima that includes the parabolic arc $\{z_2 = z_1^2\} \cap B_1(0,0)$, along which $\Delta V$ has a minimum of $2$ at $(0,0)$, with $\Delta V > 2$ at the two endpoints of the arc. We prove the following: \begin{theorem} \label{thm1.1} Let $V$ and $f$ satisfy {\rm (V1)-(V4)} and {\rm (F1)-(F5)}. Then there exists $\epsilon_0 > 0$ such that if $\epsilon \leq \epsilon_0$, then (1.0) has a positive solution $u_\epsilon$ with $u_\epsilon(z) \to 0$ as $|z| \to \infty$. $u_\epsilon$ has exactly one local maximum (hence, global maximum) point $z_\epsilon \in \Lambda$, where $\Lambda$ is as in {\rm (V4)}. There exist $\alpha, \beta > 0$ with $u_\epsilon (z) \leq \alpha \exp(- {\beta \over \epsilon} |z-z_\epsilon|)$ for $\epsilon \leq \epsilon_0$. Furthermore, with $V_0$ and $\Delta_0$ as in (V4), $V(z_\epsilon) \to V_0$ and $\Delta V(z_\epsilon) \to \Delta_0$ as $\epsilon \to 0$. \end{theorem} For small $\epsilon$, $u_\epsilon$ resembles a spike,'' which is sharper for smaller $\epsilon$. The spike concentrates near a local minimum of $V$ where $\Delta V$ has a strict local minimum. The proof of Theorem~1.1 employs the techniques of \cite{f1}, with some refinements necessary because $V$ does not necessarily achieve a {\it strict} local minimum. Section~2 introduces the penalization scheme developed by Felmer and del~Pino, and continues with the beginning of the proof of Theorem~1.1. These beginning arguments are taken practically verbatim from \cite{f1}, but are included, since the machinery of the penalization technique is used in the remainder of the proof. The reader is invited to consult \cite{f1} for more complete proofs. Section~3 contains the completion of the proof, which is original. This part contains delicate computations involving $\Delta V$. \section{The penalization scheme} \setcounter{equation}{0} Extend $f$ to the negative reals by defining $f(q) = 0$ for $q < 0$. Let $F$ be the primitive of $f$, that is, $F(q) = \int_0^q f(t)\,dt$. Define the functional $I_\epsilon$ on $W^{1,2}({\mathbb R}^N)$ by $$I_\epsilon(u) = \int_{{\mathbb R}^N} {1 \over 2} (\epsilon^2 |\nabla u|^2 + V(z)u^2) - F(u)\,dz.$$% $I_\epsilon$ is a $C^1$ functional, and there is a one-to-one correspondence between positive critical points of $I_\epsilon$ and ground states of (\ref{1.0}). It is well known that $I_\epsilon$ and similar functionals in related problems fail the Palais-Smale condition. That is, a Palais-Smale sequence,'' defined as a sequence $(u_m)$ with $I_\epsilon(u_m)$ convergent and $I'_\epsilon(u_m) \to 0$ as $m \to \infty$, need not have a convergent subsequence. To get around this difficulty, we formulate a penalized'' problem, with a corresponding penalized'' functional satisfying the Palais-Smale condition, by altering $f$ outside of $\Lambda$. Let $\theta$ be as in (F4). Choose $k$ so $k > \theta/(\theta - 2)$. Let $V_-$ be as in (V3) and $a > 0$ be the value at which $f(a)/a = V_-/k$. Define ${\tilde f}$ by $$\label{2.1} {\tilde f}(s) = \cases{f(s) &s \leq a;\cr s{V_- / k} &s > a,\cr}$$ $g( \cdot \, , s) = \chi_\Lambda f(s) + (1 - \chi_\Lambda){\tilde f}(s)$, and $G(z, \xi) = \int_0^\xi g(z, \tau)\,d\tau$. Although not continuous, $g$ is a Carath\'eodory function. For $\epsilon > 0$, define the penalized functional $J_\epsilon$ on $W^{1,2}({\mathbb R}^N)$ by $$\label{2.2} J_\epsilon(u) = \int_{{\mathbb R}^N} {1 \over 2} (\epsilon^2 |\nabla u|^2 + V(z)u^2) - G(z,u)\,dz.$$ A positive critical point of $J_\epsilon$ is a weak solution of the penalized equation'' $$\label{2.3} -\epsilon^2 \Delta u + V(z)u = g(z,u),$$ that is, a $C^1$ function satisfying (\ref{2.3}) wherever $g$ is continuous. It is proven in \cite{f1} that $J_\epsilon$ satisfies all the hypotheses of the Mountain Pass Theorem of Ambrosetti and Rabinowitz (\cite{a2}), including the Palais-Smale condition. Therefore $J_\epsilon$ has a critical point $u_\epsilon$, with the mountain pass critical level $c(\epsilon) = J_\epsilon(u_\epsilon)$. $c(\epsilon)$ is defined by the following minimax: let the set of paths $\Gamma_\epsilon = \{\gamma \in C([0,1], W^{1,2}({\mathbb R}^N)) \mid \gamma(0) = 0,\ J_\epsilon(\gamma(1)) < 0\}$, and $$c(\epsilon) = \inf_{\gamma \in \Gamma_\epsilon} \max_{\theta \in [0,1]} J_\epsilon(\gamma(\theta)).$$ As shown in (\cite{f1}), because of (F4), $c(\epsilon)$ can be characterized more simply as $$c(\epsilon) = \inf_{u \in W^{1,2}({\mathbb R}^N) \setminus \{0\} } \sup_{\tau > 0} J_\epsilon(\tau u).$$ The functions $g(z,q)$ and $f(q)$ agree whenever $z \in \Lambda$ or $q < a$. Therefore if $u$ is a weak solution of (\ref{2.3}) with $u < a$ on $\Lambda^{\bf C} \equiv {\mathbb R}^N \setminus \Lambda$, then $u$ solves (\ref{1.0}). Our plan is to find a positive critical point $u_\epsilon$ of $J_\epsilon$, which is a weak solution of (\ref{2.3}), then show that $u_\epsilon(z) < a$ for all $z \in \Lambda^{\bf C}$. For $\epsilon > 0$, let $u_\epsilon$ be a critical point of $J_\epsilon$ with $J_\epsilon(u_\epsilon) = c(\epsilon)$. Maximum principle arguments show that $u_\epsilon$ must be positive. Define the limiting functional'' $I_0$ by $$\label{2.6} I_0 (u) = \int_{{\mathbb R}^N} {1 \over 2} ( |\nabla u|^2 + V_0 u^2) - F(u)$$ and $$\label{2.7} \underline {c} = \inf_{u \in W^{1,2}({\mathbb R}^N) \setminus \{0\} } \sup_{\tau > 0} I_0(\tau u).$$ The equation corresponding to (\ref{2.6}) is $$\label{2.8} -\Delta u + V_0 u = f(u)$$ We will prove Theorem~1.1 by proving the following proposition: \begin{proposition} \label{pr2.9} Let $\epsilon > 0$. If $u_\epsilon$ is a positive solution of (\ref{2.3}) satisfying $J_\epsilon(u_\epsilon) = c(\epsilon)$, then \begin{description} \item{\rm (i)} $\lim_{\epsilon \to 0} \max_{z \in \partial \Lambda} u_\epsilon = 0$. \item{\rm (ii)} For all $\epsilon$ sufficiently small, $u_\epsilon$ has only one local maximum point in $\Lambda$ (call it $z_\epsilon$), with $\lim_{\epsilon \to 0} V(z_\epsilon) = V_0$ \item{\rm (iii)} $\lim_{\epsilon \to 0} \Delta V(z_\epsilon) =\Delta_0$. \end{description} \end{proposition} \paragraph{Proof of Theorem~\ref{thm1.1}:} Assuming Proposition~\ref{pr2.9}, there exists $\epsilon_0 > 0$ such that for $\epsilon < \epsilon_0$, $u_\epsilon < a$ on $\partial \Lambda$. In \cite{f1} it is shown that if we multiply (\ref{2.3}) by $(u_\epsilon - a)_+$ and integrate by parts, it follows that $u_\epsilon < a$ on $\Lambda^{\bf C}$, so $u_\epsilon$ solves (\ref{1.0}). By the definition of $a$ in (\ref{2.1}), and the maximum principle, $u_\epsilon$ has no local maxima outside of $\Lambda$, so $u_\epsilon$ has exactly one local maximum point $z_\epsilon$, which occurs in $\Lambda$. Define $v_\epsilon$ by translating $u_\epsilon$ from $z_\epsilon$ to zero and dilating it by $\epsilon$, that is, $$v_\epsilon (z) = u_\epsilon (z_\epsilon + \epsilon z).$$ % 2.10 Then $v_\epsilon$ is a weak ($C^1$) solution of the translated and dilated'' equation $$-\Delta v_\epsilon + V(z_\epsilon + \epsilon z)v_\epsilon = g(z_\epsilon + \epsilon z, v_\epsilon ).$$ % 2.11 Let $\epsilon _j \to 0$. Along a subsequence (called $(z_{\epsilon _j})$), $z_{\epsilon _j} \to {\bar z} \in {\overline{\Lambda}}$, with $V({\bar z} ) = V_0$ and $\Delta V({\bar z} ) = \Delta_0$. Along a subsequence, $v_{\epsilon _j}$ converges locally uniformly to a function $v^0$. Pick $R > 0$ so $v^0 < a$ on ${\mathbb R}^N \setminus B_R(0)$. For large enough $\epsilon$, $v_\epsilon < a$ on $\partial B_R(0)$. By the maximum principle arguments of \cite{f1}, for small $\epsilon$, $v_\epsilon$ decays exponentially, uniformly in $\epsilon$. \hfill$\diamondsuit$\smallskip The proof of Proposition~\ref{pr2.9} will follow if we can prove the following statement. \begin{proposition} \label{pr2.12} If $\epsilon _n \to 0$ and $(z_n) \subset {\overline{\Lambda}}$ with $u_{\epsilon_n}(z_n) \geq b > 0$, then \begin{description} \item{\rm (i)} $\lim_{n \to \infty} V(z_n) = V_0$. \item{\rm (ii)} $\lim_{n \to \infty}\Delta V(z_n) = \Delta_0$. \end{description} \end{proposition} It is proven in \cite{f1} that $u_\epsilon$ has exactly one local maximum point $z_\epsilon$ for small $\epsilon$. Since $u_\epsilon$ solves (\ref{2.3}), the maximum principle implies that $u_\epsilon(z_\epsilon)$ is bounded away from zero. Thus Proposition~\ref{pr2.12} and (V4) give Proposition~\ref{pr2.9}(ii)-(iii). To prove Proposition~\ref{pr2.12}, let $b$ and $(z_n)$ be as above. First we repeat the argument of \cite{f1} to show that $V(z_n) \to V_0$: suppose this does not happen. Then, along a subsequence, $z_n \to {\bar z} \in {\overline{\Lambda}}$ with $V({\bar z} ) > V_0$. Define $v_n$ by translating $u_{\epsilon_n}$ from $z_n$ to $0$ and dilating by $\epsilon_n$; that is, $$\label{2.13} v_n(z) = u_{\epsilon_n}(z_n + \epsilon_n z).$$ $v_n$ solves the translated and dilated'' penalized equation $$\label{2.14} - \Delta v_n + V(z_n + \epsilon_n z)v_n = g(z_n +\epsilon_n z, v_n)$$ on ${\mathbb R}^N$, with $v_n(z) \to 0$ and $\nabla v_n(z) \to 0$ as $|z| \to \infty$. As shown in \cite{f1}, $(v_n)$ is bounded in $W^{1,2}({\mathbb R}^N)$, so by elliptic estimates, $(v_n)$ converges locally along a subsequence (also denoted $(v_n)$) to $v^0 \in W^{1,2}({\mathbb R}^N)$. Define $\chi_n$ by $\chi_n(z) = \chi_\Lambda (z_n + \epsilon_n z)$, where $\chi_\Lambda$ is the characteristic function of $\Lambda$. $\chi_n$ converges weakly in $L^p$ over compact sets to a function $\chi$, for any $p > 1$, with $0 \leq \chi \leq 1$. Define $${\bar g}(z,s) = \chi(z)f(s) + (1- \chi(z)){\tilde f}(s)$$ Then $v^0$ satisfies $$\label{2.16} - \Delta v + V({\bar z} )v = {\bar g}(z,v)$$ on ${\mathbb R}^N$. Define ${\bar G}(z,s) = \int_0^s {\bar g}(z,t)\,dt$. Associated with (\ref{2.16}) we have the limiting functional ${\bar J}(u) = \int_{{\mathbb R}^N} {1 \over 2}(|\nabla u|^2 + V({\bar z} )u^2) - {\bar G}(z,u)\,dz$. $v^0$ is a positive critical point of ${\bar J}$. Define $J_n$ to be the translated and dilated'' penalized functional corresponding to (\ref{2.14}), that is, $$J_n(u) = \int_{{\mathbb R}^N} {1 \over 2}(|\nabla u|^2 + V(z_n + \epsilon_n z)u^2) - G(z_n + \epsilon_n z,u)\,dz.$$ Clearly $J_n(v_n) = \epsilon_n^{-N} J_{\epsilon_n}(u_{\epsilon_n})$. In \cite{f1} it is proven that $$\label{2.18} \lim\inf_{n \to \infty} J_n(v_n) \geq {\bar J}(v^0).$$ Also, by letting $w$ be a ground state for (\ref{2.8}) with $I_0(w) = {\underline {c}}$ (the mountain pass value for $I_0$, defined in (\ref{2.7}) and using $w$ as a test function for $J_n$, it is proven that ${\underline {c}} \geq \lim\inf_{n \to \infty} J_n(v_n)$. Thus ${\bar J}(v^0) \leq {\underline {c}}$. Therefore, as shown in \cite{f1}, $V({\bar z} ) \leq V_0$. This contradicts our assumption. Thus $V(z_n) \to V_0$. All the above is the same as was proven in \cite{f1}. Next, we must show that $\Delta V(z_n) \to \Delta_0$. That is the focus of the next section. \section{The effect of the Laplacian } \setcounter{equation}{0} Proving $\Delta V(z_n) \to \Delta_0$ is a subtle and delicate problem. Making $\epsilon_n$ approach $0$ is equivalent to dilating $V$, which has the effect of making local minima of $V$ behave more like global minima. This assists in finding solutions to (\ref{1.0}). However, making $\epsilon_n$ small {\it reduces} the effect of differences in $\Delta V$. For this reason, Theorem~\ref{thm1.1} is not only difficult to prove, but is not intuitively compelling, either. It is known (\cite{g1}) that a least energy solution'' of (\ref{2.8}), that is, a solution $w$ with $I_0(w) = {\underline {c}}$, must be radially symmetric. We will need to exploit this fact. In order to do this, we will need to work with the maximum points of $u_{\epsilon_n}$ instead of merely the $(z_n)$ as given in Proposition~\ref{pr2.12}. We need the following concentration-compactness result, which states that the sequence $(u_{\epsilon_n})$ of mountain-pass type solutions'' of (\ref{2.3}) does not split'': \begin{lemma} \label{lm3.0} If $(z_n) \subset {\overline \Lambda}$, $(y_n) \subset {\mathbb R}^N$, and $b > 0$ with $u_{\epsilon_n}(z_n) > b$ and $u_{\epsilon_n}(y_n) > b$ for all $n$, then $((z_n - y_n)/\epsilon_n)$ is bounded. \end{lemma} \paragraph{Proof:} define $v_n(z) = u_{\epsilon_n}(z_n + \epsilon_n z)$ as in (\ref{2.13}). Suppose the lemma is false. Then, along a subsequence, $|y_n - z_n|/\epsilon_n \to \infty$. Let $x_n = (y_n - z_n)/\epsilon_n$. $(\|v_n\|)$ is bounded in $W^{1,2}({\mathbb R}^N)$ and $|x_n| \to \infty$, so we may pick a sequence $(R_n) \subset {\mathbb N}$ with $R_n \to \infty$, $|x_n| - R_n \to \infty$, and $\|v_n\|_{W^{1,2} ( B_{R_n + 1}(0) \setminus B_{R_n - 1}(0) ) } \to 0$ as $n \to \infty$. Define cutoff functions $\varphi_n^{1,2} \in C^\infty({\mathbb R}^N, [0,1])$ satisfying $\varphi_1 \equiv 1$ on $B_{R_n - 1}(0)$, $\varphi_1 \equiv 0$ on $B_{R_n}(0)^{\bf C}$, $\varphi_2 \equiv 1$ on $B_{R_n + 1}(0)^{\bf C}$, $\varphi_2 \equiv 0$ on $B_{R_n}(0)$, and $\|\nabla \varphi_1\|_{L^\infty({\mathbb R}^N)} < 2$, $\|\nabla \varphi_2\|_{L^\infty({\mathbb R}^N)} < 2$. Set $v_n^1 = \varphi_n^1 v_n$ and $v_n^2 = \varphi_n^2 v_n$, and ${\bar v}_n = v_n^1 + v_n^2 = (\varphi_n^1 + \varphi_n^2)v_n$. Choose $T_n > 0$ so $J_n(T_n {\bar v}_n) = 0$. We claim that $T_n$ is well-defined, and bounded in $n$. Note that the existence of $T_n$ must be checked for the penalized functional $J_n$, because of the replacement of $F$ with $G$. By elliptic estimates, there exists an open set $U \subset {\mathbb R}^N$ such that along a subsequence, $v_n^1 > b/2$ on $U$ and $U \subset (\Lambda - z_n)/\epsilon_n \equiv \{z \in {\mathbb R}^N \mid z_n + \epsilon_n z \in \Lambda\}$. Let $a$ be as in (\ref{2.1}). For $t > 2a/b$ and $z \in U$, $t{\bar v}_n(z) > tb/2 > a$, so $G(z_n + \epsilon_n z, t{\bar v}_n) = F(t{\bar v}_n) > F(bt/2)$. Therefore, for $t > 2a/b$, \begin{eqnarray*} J_n(t{\bar v}_n) &=& t^2\int_{{\mathbb R}^N} {1 \over 2}(|\nabla {\bar v}_n|^2 + V(z_n + \epsilon_n z){\bar v}_n^2)\,dz - \int_{{\mathbb R}^N} G(z_n + \epsilon_n z,t{\bar v}_n)\,dz \\ &\leq& {t^2 \over 2}(1 + V^+) \|{\bar v}_n\|^2_{W^{1,2}({\mathbb R}^N)} -\int_U F(t{\bar v}_n) \\ &\leq& {t^2 \over 2}(1 + V^+) \|{\bar v}_n\|^2_{W^{1,2}({\mathbb R}^N)} - \lambda(U) F(tb/2), \end{eqnarray*} where $\lambda$ indicates the Lebesgue measure. By (F4), there exists $C>0$ such that for $t > 2a/b$, $F(tb/2) > Ct^\theta$. Therefore, for $t > 2a/b$, $$\label{3.2} J_n(t{\bar v}_n) \leq {t^2 \over 2}(1 + V^+) \|{\bar v}_n\|^2_{W^{1,2}({\mathbb R}^N)} - Ct^\theta.$$ Since $({\bar v}_n)$ is bounded in $W^{1,2}({\mathbb R}^N)$, this gives the existence and boundedness of $(T_n)$. Since $J_n(T_n {\bar v}_n) = J_n(T_n v_n^1) + J_n(T_n v_n^2) = 0$, we may pick $i_n \in \{1, 2\}$ with $J_n(T_n v_n^{i_n}) \leq 0$. By (F5) and (\ref{2.1}), the map $t \mapsto J_n(tv_n^{i_n})$ increases from zero at $t = 0$, achieves a positive maximum, then decreases to $- \infty$. We will see more of this in a moment. Thus there exists a unique $t_n \in (0, T_n)$ with $J_n(t_n v_n^{i_n}) = \max_{t>0} J_n(t v_n^{i_n})$. We claim that $t_n$ and $T_n - t_n$ are both bounded away from zero for large $n$: by $(f_1)-(f_4)$ and (\ref{2.1}), $J_n(w) \geq {1 \over \theta} \min (1, V_-)\|w\|^2_{W^{1,2}({\mathbb R}^N)}- o(\|w\|^2_{W^{1,2}({\mathbb R}^N)})$ uniformly in $n$, so $\max_{t>0} J_n(t v_n^{i_n})$ is bounded away from zero, uniformly in $n$. It is easy to show that $J_n$ is Lipschitz on bounded subsets of $W^{1,2}({\mathbb R}^N)$, uniformly in $n$. Since $(T_n)$ is bounded, this implies that $t_n$ and $T_n - t_n$ are both bounded away from zero for large $n$. By definition of $v_n$ as a mountain-pass type critical point'' of $J_n$, we have $$\max_{t>0} J_n(t v_n^{i_n}) \geq \max_{t>0} J_n(t v_n).$$ Using the facts that $\|v_n - {\bar v}_n\|_{W^{1,2}({\mathbb R}^N)} \to 0$ as $n \to \infty$, and $(T_n)$ is bounded, we have \begin{eqnarray} \lim\inf_{n \to \infty} J_n(t_n v_n^{i_n}) &=& \lim\inf_{n \to \infty} \max_{t>0}J_n(t v_n^{i_n})\nonumber\\ & \geq& \lim\inf_{n \to \infty} \max_{t>0}J_n(t v_n) \nonumber \\ &=& \lim\inf_{n \to \infty} \max_{t>0}J_n(t {\bar v}_n) \label{3.4} \\ &=& \lim\inf_{n \to \infty} J_n(t_n {\bar v}_n) \nonumber \\ &=& \lim\inf_{n \to \infty}(J_n(t_n v_n^{i_n}) +J_n(t_n v_n^{3 - i_n}) ) \nonumber \\ &\geq& \lim\inf_{n \to \infty}J_n(t_n v_n^{i_n}) +\lim\inf_{n \to \infty}J_n(t_n v_n^{3-i_n}). \nonumber \end{eqnarray} Now $J_n(T_n v_n^{3-i_n}) = - J_n(T_n v_n^{i_n}) \geq 0$ and $t_n < T_n$, so $J_n(t_n v_n^{3-i_n}) \geq 0$. By (\ref{3.4}), $\lim\inf_{n \to \infty}J_n(t_n v_n^{3-i_n}) \leq 0$. Therefore $J_n(t_n v_n^{3-i_n}) \to 0$ as $n \to \infty$. Since $J_n(w) \geq {1 \over \theta} \min (1, V_-)\|w\|^2_{W^{1,2}({\mathbb R}^N)} - o(\|w\|^2_{W^{1,2}({\mathbb R}^N)})$ uniformly in $n$, there exists $d \in (0, \lim\inf_{n \to \infty} t_n )$ such that $\lim\inf_{n \to \infty} J_n(d v_n^{3-i_n}) > 0$. Since $d < t_n$ and $J_n(d v_n^{3-i_n}) > J_n(t_n v_n^{3-i_n})$ for large $n$, the map $t \mapsto J_n(t v_n^{3-i_n})$ achieves a maximum at some $t'_n \in (0, t_n)$, and that maximum is bounded away from zero. Summarizing the important facts about the mapping $t \mapsto J_n(t v_n^{3-i_n})$, we have shown that there exists $\rho > 0$ such that for large $n$, \begin{description} \item{(i)} $0 < t'_n < t_n < T_n$ \item{(ii)} $(T_n)$ is bounded. \item{(iii)} $(T_n - t_n)$ is bounded away from zero. \item{(iv)} $J_n(t'_n v_n^{3-i_n}) > \rho > 0$ \item{(v)} $J_n(t_n v_n^{3-i_n}) \to 0$ \item{(vi)} $J_n(T_n v_n^{3-i_n}) \geq 0$ \end{description} From (i)-(vi) it is apparent that at some $t^*_n > t'_n$, the mapping $t \mapsto J_n(t v_n^{3-i_n})$ is at once decreasing and concave upward. But this is impossible: let $n \in {\mathbb N}$ and $w \in W^{1,2}({\mathbb R}^N) \setminus \{0\}$. Define $\psi(t) = J_n(tw)$ for $t > 0$. Then \begin{eqnarray*} \psi'(t) &=& t\int_{{\mathbb R}^N} |\nabla w|^2 + V(z_n + \epsilon_n z) w^2 \,dz - \int_{{\mathbb R}^N} g(z_n + \epsilon_n z, tw)w\,dz \\ &=& t \left[\int_{{\mathbb R}^N} |\nabla w|^2 + V(z_n + \epsilon_n z) w^2 \,dz - \int_{\{w \neq 0\} } {g(z_n + \epsilon_n z, tw) \over{tw} } w^2\,dz \right]. \end{eqnarray*} By (F5) and (\ref{2.1}), $t \mapsto g(z_n + \epsilon_n z, tw)/(tw)$ is nondecreasing, so if $\psi'(t)$ ever becomes negative, $\psi'$ is increasing for all time $t$ after that, and the graph of $\psi$ is concave down. Therefore the behavior of $J_n(t v_n^{3-i_n})$ as described in (i)-(vi) is impossible, and Lemma~\ref{lm3.0} is proven. \hfill$\diamondsuit$\smallskip As mentioned before, it will be advantageous to work with the maxima of $(u_{\epsilon_n})$. Choose $(y_n) \subset {\mathbb R}^N$ with $$u_{\epsilon_n}(y_n) = \max_{{\mathbb R}^N} u_{\epsilon_n}.$$ We will prove $$\label{3.8} \Delta V(y_n) \to \Delta_0.$$ By Lemma~3.0, $((y_n - z_n)/\epsilon_n)$ is bounded, so $y_n - z_n \to 0$. Thus (\ref{3.8}) gives Proposition~\ref{pr2.12}(ii), completing the proof of Theorem~\ref{thm1.1}. \hfill$\diamondsuit$\smallskip Along a subsequence, $y_n \to {\bar y} \in {\overline \Lambda}$. By Proposition~\ref{pr2.12}(i), $V({\bar y} ) = V_0$. Since is not apparent that ${\bar y} \in \Lambda$, we must proceed carefully. We will redefine the $v_n$'s like in (\ref{2.13}), by translating $u_{\epsilon_n}$ to $0$ and dilating it. That is, $$\label{3.9} v_n(z) = u_{\epsilon_n}(y_n +\epsilon_n z).$$ Then $v_n$ is a positive weak solution, vanishing at infinity, of the penalized, dilated, and translated'' PDE $$-\Delta v + V(y_n + \epsilon_n z)v = g(y_n + \epsilon_n z, v).$$ % 3.10 Like before, $(v_n)$ converges locally uniformly to a function $v_0$. We claim that $v_0$ is actually a ground state maximizing at $0$ of the autonomous limiting equation (\ref{2.8}). Proof: As before, define $\chi_n$ by $\chi_n(z) = \chi(y_n + \epsilon z)$. As before, along a subsequence, $\chi_n$ converges weakly in $L^p$, for any $p > 1$, on compact subsets of ${\mathbb R}^N$ to a function $\chi$ with $0 \leq \chi \leq 1$. Define ${\bar g}$ by $${\bar g}(z,s) = \chi(z)f(s) + (1-\chi(z)){\tilde f} (s).$$ % 3.11 By the argument of Proposition~\ref{pr2.12}, taken from \cite{f1}, $(v_n)$ converges locally along a subsequence to $v_0$, a ground state of $-\Delta v + V_0 v = {\bar g}(z,v)$. The functional corresponding to this equation is ${\bar J}(u) = \int_{{\mathbb R}^N} {1 \over 2}(|\nabla u|^2 + V_0 u^2) - {\bar G}(z,u)\,dz$, where ${\bar G}(z,s) = \int_0^s {\bar g}(z,t)\,dt$. As before, in (\ref{2.18}), ${\underline {c}} \geq \lim\inf_{n \to \infty}J_n(v_n) \geq {\bar J}(v_0)$, where ${\underline {c}}$ is from (\ref{2.7}). ${\bar J} \geq I_0$, where $I_0$ is the \lq\lq autonomous" limiting functional from (\ref{2.6}), so $${\underline {c}} \leq \max_{t>0} I_0(t v_0) \leq \max_{t>0} {\bar J}(t v_0) \leq {\underline {c}} ,$$ % 3.12 and $v_0$ is actually a ground state of (\ref{2.8}). \hfill$\diamondsuit$ \medskip Not only does $(v_n)$ converge locally to $v_0$, but it satisfies the following lemma. \begin{lemma} \label{lm3.13} With $(v_n)$ as in (\ref{3.9}), for any subsequence of $(v_n)$ there is a radially symmetric ground state $v_0$ of (\ref{2.8}) such that $v_n \to v_0$ uniformly along a subsequence and the $v_n$'s decay exponentially, uniformly in $n$. \end{lemma} \paragraph{Proof:} If one establishes uniform convergence, the uniform exponential decay follows readily, using a standard maximum principle argument found in \cite{f1}. Suppose the convergence is not uniform. Then there exist a subsequence of $(v_n)$ (denoted ($v_n$)) and a sequence $(x_n) \subset {\mathbb R}^N$ with $|x_n| \to \infty$ and $\lim_{n \to \infty} v_n(x_n) > 0$. Let $d > 0$ with $d < v_0(0)$ and $d < \lim_{n \to \infty} v_n(x_n)$. For large $n$, $d < v_n(0) = u_{\epsilon_n}(z_n)$ and $d < v_n(x_n) = u_{\epsilon_n}(z_n + \epsilon_n x_n)$. Letting $w_n = z_n + \epsilon_n x_n$, we obtain $((w_n - z_n)/\epsilon_n) = (x_n)$, which is unbounded, violating Lemma~\ref{lm3.0}. To show $\Delta V(y_n) \to \Delta_0$, we again argue indirectly. Suppose otherwise. Then, along a subsequence, $y_n \to {\bar y} \in {\overline \Lambda}$ with $$\label{3.14} \Delta V({\bar y} ) > \Delta_0\,.$$ For $x \in {\mathbb R}^N$, define the translation operator $\tau_x$ by $\tau_x u(z) = u(z-x)$, that is, $\tau_x u$ is $u$ translated by $x$. Assume for convenience, and without loss of generality, that $$0 \in \Lambda,\ V(0) = V_0, \hbox{ and } \Delta V(0) = \Delta_0.$$ We will prove that for large $n$, $$\label{3.16} \sup_{t > 0} J_{\epsilon_n}(t\tau_{-y_n/\epsilon_n}u_{\epsilon_n}) < J_{\epsilon_n}(u_{\epsilon_n}) = \sup_{t>0}J_{\epsilon_n}(tu_{\epsilon_n}),$$ recalling the definition of $J_\epsilon$ in (\ref{2.2}), and how $v_n$ is defined from $u_{\epsilon_n}$ in (\ref{3.9}). That is, translating $t u_{\epsilon_n}$ back to the origin reduces the value of $J_{\epsilon_n}(t v_n)$ because $V$ has lesser concavity at the origin. This occurs even though shrinking $\epsilon$ reduces the difference in concavity. (\ref{3.16}) contradicts the definition of $u_{\epsilon_n}$. Pick $T > 1$ large enough so that for large $n$, $J_n(Tv_n) = \epsilon_n^{-N} J_{\epsilon_n}(T u_{\epsilon_n}) < 0$. This is possible by the argument of (\ref{3.2}). Now (\ref{3.16}) is equivalent to $$\sup_{0 \leq t \leq T} J_{\epsilon_n}(t \tau_{-y_n}u_{\epsilon_n}) < \sup_{0 \leq t \leq T} J_{\epsilon_n}(t u_{\epsilon_n}).$$ % 3.17 To prove the above, it will suffice to prove the stronger fact that for large $n$, for all $t \in (0, T)$, $$J_{\epsilon_n}(t u_{\epsilon_n}) > J_{\epsilon_n}(t \tau_{-y_n}u_{\epsilon_n}).$$ % 3.18 Now, along a subsequence, $v_n \to v_0$ uniformly, so by the definition of $v_n$ as a dilation of $\tau_{-y_n}u_{\epsilon_n}$ ((\ref{3.9})), $u_{\epsilon_n} \to 0$ uniformly on ${\mathbb R}^N \setminus \Lambda$ as $n \to \infty$. Thus for large $n$ and $0 \leq t \leq T$, the definition of $G$ gives $G(z, t\tau_{-y_n}u_{\epsilon_n}(z)) = F(t\tau_{-y_n}u_{\epsilon_n}(z)$ for all $z \in {\mathbb R}^N$, so \begin{eqnarray*} \lefteqn{ J_{\epsilon_n}(t u_{\epsilon_n}) - J_{\epsilon_n}(t \tau_{-y_n}u_{\epsilon_n}) }\\ &=& \int_{{\mathbb R}^N} {1 \over 2} t^2 \left( |\nabla u_{\epsilon_n}(z)|^2 + V(z)u_{\epsilon_n}(z)^2\right) - G(z, tu_{\epsilon_n}(z))\,dz \\ &&-\big[ \int_{{\mathbb R}^N} {1 \over 2} t^2 \left( |\nabla \tau_{-y_n}u_{\epsilon_n}(z)|^2 + V(z)\tau_{-y_n}u_{\epsilon_n}(z)^2 \right) - F(t\tau_{-y_n}u_{\epsilon_n}(z))\,dz \big] \\ &\geq& {1 \over 2} t^2 \int_{{\mathbb R}^N} V(z)(u_{\epsilon_n}(z)^2 - u_{\epsilon_n}(z + y_n)^2)\,dz \\ && +\int_{{\mathbb R}^N} F(tu_{\epsilon_n}(z+y_n)- F(tu_{\epsilon_n}(z))\,dz \\ &=&{1 \over 2} t^2 \int_{{\mathbb R}^N} (V(z+y_n)- V(z)) u_{\epsilon_n}(z+y_n)^2\,dz \\ &=&{1 \over 2} t^2 \epsilon_n^{N} \int_{{\mathbb R}^N} (V(y_n + \epsilon_nz) - V(\epsilon_n z)) u_{\epsilon_n}(\epsilon_n z + y_n)^2\,dz \\ &=&{1 \over 2} t^2 \epsilon_n^{N}\int_{{\mathbb R}^N} (V(y_n + \epsilon_nz) - V(\epsilon_n z)) v_n(z)^2\,dz\,. \end{eqnarray*} For $n = 1, 2, \ldots$, define $h_n: {\mathbb R} \to {\mathbb R}$ by $$h_n(t) = \int_{{\mathbb R}^N} (V(y_n + tz) - V(t z))v_n^2\,dz.$$ Since $h_n(\epsilon_n) = \int_{{\mathbb R}^N} (V(y_n + \epsilon_n z) - V(\epsilon_n z))v_n^2$, we must prove that for large $n$, $$h_n(\epsilon_n) > 0\,. \label{3.21}$$ Assume without loss of generality that $\Lambda$ was chosen so that there exists $\rho > 0$ with $$\inf_{N_\rho(\Lambda)} V = V_0, \label{3.22}$$ where $N_\rho(\Lambda) = \{x \in {\mathbb R}^N \mid \exists y \in \Lambda \hbox{ with } |y - x| < \rho\}$. We will prove the following facts about $h_n$: \begin{lemma} \label{lm3.23} For some $\beta > 0$, for large $n$, \begin{description} \item{\rm (i)} $h_n \in C^2({\mathbb R}^+, {\mathbb R})$ \item{\rm (ii)} $h_n(0) \geq 0$ \item{\rm (iii)} $|h'_n(0)|^2 \leq o(1)h_n(0)$ \item{\rm (iv)} $h''_n(0) > \beta$ \item{\rm (v)} $h''_n$ is locally Lipschitz on ${\mathbb R}^+$, uniformly in $n$. \end{description} \end{lemma} Here $o(1) \to 0$ as $n \to \infty$. Before proving Lemma~\ref{lm3.23}, let us prove how it gives (\ref{3.21}). By (iv)-(v), there exists $d > 0$ such that for large $n$ and $0 \leq t \leq d$, $h''_n(t) > \beta/2$. For $t \in [0, d]$, a Taylor's series expansion shows that for large $n$, $$\label{3.24} h_n(t) \geq h_n(0) + h'_n(0)t + {\beta \over 4}t^2 \equiv l_n(t).$$ If $h_n(0) = 0$, then by Lemma~\ref{lm3.23}(iii), $h'_n(0) = 0$, so (\ref{3.24}) implies that $h_n(t) > 0$ for all $t \in (0, d)$, giving (\ref{3.21}) if $n$ is large enough that $\epsilon_n < d$. If $h_n(0) > 0$, then by elementary calculus, $l_n$ attains a minimum value at $t = -2h'_n(0)/\beta$, and the minimum value is $$\min_{\mathbb R} l_n = l_n(-2h'_n(0)/\beta) = h_n(0) - h'_n(0)^2/\beta \geq (1 - o(1))h_n(0),$$ % 3.25 where $o(1) \to 0$ as $n \to \infty$. For large $n$, if $h_n(0) > 0$ then $l_n(t) > 0$ for all $t \in {\mathbb R}$, so $h_n(t) > 0$ for all $t \in (0,d)$ for large $n$, implying (\ref{3.21}) if $n$ is large enough so that $\epsilon_n < d$. \paragraph{Proof of Lemma~\ref{lm3.23}} Statement (ii) is trivial, since $h_n(0) = (V(y_n) - V_0)\int_{{\mathbb R}^N} v_n^2$, and since $z_n \in {\overline \Lambda}$ and $y_n - z_n \to 0$, (\ref{3.22}) implies $V(y_n) \geq V_0$ for large $n$. (i) and (v) follow from Leibniz's Rule, $(V_1)-(V_2)$, and the fact that the $v_n$'s decay exponentially, uniformly in $n$. For $j = 1, 2$, $$h_n^{(j)}(t) = \int_{{\mathbb R}^N} \sum_{|\alpha| = j} (D^\alpha V(y_n + tz) - D^\alpha V(tz))z^\alpha v_n(z)^2\,dz.$$ % 3.26 Since (V2) holds, $v_n$ decays exponentially, uniformly in $n$, $y_n \to {\bar y}$, and $v_0$ is radially symmetric, we have \begin{eqnarray*} h''_n(0) &=& \int_{{\mathbb R}^N} \sum_{|\alpha| = 2} (D^\alpha V(y_n) - D^\alpha V(0))z^\alpha v_n(z)^2\,dz \\ &\to& \int_{{\mathbb R}^N} \sum_{|\alpha| = 2} (D^\alpha V({\bar y} ) - D^\alpha V(0))z^\alpha v_0(z)^2\,dz \\ &=& \int_{{\mathbb R}^N} \sum_{i=1}^N (D^{ii} V({\bar y} ) - D^{ii} V(0))z_i^2 v_0(z)^2\,dz \\ &=& \int_{{\mathbb R}^N} \sum_{i=1}^N (D^{ii} V({\bar y} ) - D^{ii} V(0)){1 \over N} |z|^2 v_0(z)^2\,dz \\ &=& {1 \over N}(\Delta V({\bar y} ) - \Delta V(0)) \int_{{\mathbb R}^N} |z|^2 v_0(z)^2\,dz > 0 \end{eqnarray*} by assumption (\ref{3.14}). Since Lemma~\ref{lm3.23}(v) holds, we have Lemma~\ref{lm3.23}(iv). To prove Lemma~\ref{lm3.23}(iii), we will need the following calculus lemma: \begin{lemma} \label{lm3.28} Let $U \subset {\mathbb R}^N$ and $r > 0$. Let $V \in C^2(N_r(U), {\mathbb R})$ with $\inf_{N_r(U)} V \equiv V_0 > - \infty$, $|\nabla V|$ bounded on $N_r(U)$, and $D^2 V$ Lipschitz on $N_r(U)$. Then there exists $C > 0$ with $$|\nabla V(z)|^2 \leq C(V(z) - V_0) \label{3.29}$$ for all $z \in U$. \end{lemma} \paragraph{Proof:} let $B > 0$ with $|D^2 V(z)\xi \cdot \xi| \leq B$ for all $\xi \in {\mathbb R}^N$ with $|\xi| = 1$. Also let $B$ be big enough so $$B > |\nabla V(z)|/r$$ % 3.30 for all $z \in U$. Pick $z \in U$. If $|\nabla V(z)| = 0$, then (\ref{3.29}) is obvious. Otherwise, let $d = |\nabla V(z)|/B < r$. Define $\varphi(t) = V(z - t\nabla V(z)/|\nabla V(z)|)$ for $t \in [0, d]$. $\varphi$ is $C^2$, $\varphi(0) = V(z)$, and $\varphi'(0) = -|\nabla V(z)|$. By choice of $B$ and the fact that $B_d(z) \subset N_r(U)$, $|\varphi''(t)| \leq B$ for all $t \in [0,d]$. Taylor's theorem gives $$\varphi(d) - \varphi(0) = \varphi'(0)d + \varphi''(\xi){d^2 \over 2} \leq -|\nabla V(z)|d + Bd^2/2 = -{|{\nabla V(z)}|^2 \over {2B} }.$$ % 3.31 Also $\varphi(d) \geq V_0$ because $B_d(z) \subset N_r(U)$. Therefore, $${|{\nabla V(z)}|^2 \over {2B} } \leq \varphi(0) - \varphi(d) \leq V(z) - V_0\,.$$ %3.32 Lemma~\ref{lm3.28} is proven. \hfill$\diamondsuit$ \medskip To prove Lemma~\ref{lm3.23}(iii), first note that, by the radial symmetry of $v_0$, the uniform exponential decay of $v_n$, and the uniform convergence $v_n \to v_0$, \begin{eqnarray*} |h'_n(0)| &=& |(\nabla V(y_n) - \nabla V(0)) \cdot \int_{{\mathbb R}^N} zv_n^2\,dz | \\ &=&|\nabla V(y_n) \cdot \int_{{\mathbb R}^N} zv_n^2\,dz | \\ &=& |\nabla V(y_n) \cdot \int_{{\mathbb R}^N} zv_0^2\,dz + \ \ \nabla V(y_n) \cdot \int_{{\mathbb R}^N} z(v_n^2 - v_0^2)\,dz| \\ &=& |\nabla V(y_n) \cdot \int_{{\mathbb R}^N} z(v_n^2 - v_0^2)\,dz| \\ &\leq& |\nabla V(y_n)| \, |\int_{{\mathbb R}^N} z(v_n^2 - v_0^2)\,dz| \\ &\leq& o(1)|\nabla V(y_n)|, \end{eqnarray*} so Lemma~\ref{lm3.28} implies \begin{eqnarray*} |h'_n(0)|^2 &\leq& o(1)|\nabla V(y_n)|^2 \leq o(1)(V(y_n)-V_0) \\ &\leq& o(1)(V(y_n)-V_0)\int_{{\mathbb R}^N} v_n^2\\ & =& o(1)h_n(0), \end{eqnarray*} since $\int_{{\mathbb R}^N} v_n^2$ is bounded away from zero. Lemma~\ref{lm3.23}(iii) is proven. Thence follow (\ref{3.21}), (\ref{3.8}), Proposition~\ref{pr2.12}, and Theorem~\ref{thm1.1}. \paragraph{Remarks:} Besides the results cited in the introduction, many important results for equations of type (\ref{1.0}) have been found recently. For instance, the work in \cite{f1}-\cite{f3} suggests that Theorem~\ref{thm1.1} could be strengthened by working on a smaller domain than ${\mathbb R}^N$, or by weakening the hypotheses on $V$. It is natural to try to extend Theorem~\ref{thm1.1} to cases where $V$ is not $C^2$, or to the case where the second derivatives of $V$ do not provide a condition like (V4), but higher-order derivatives do. \begin{thebibliography}{19} \bibitem{a1} Ambrosetti, A., Badiale, M., Cingolani, S.: Semiclassical states of nonlinear Schr\"odinger equations. Arch.~Rat.~Mech.~Anal. (1997) {\bf 140} no.~3, 285-300. \bibitem{a2} Ambrosetti, A., Rabinowitz, P.: Dual variational methods in critical point theory and applications. 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H.: On a class of nonlinear Schr\"odinger equations. Z.~angew Math.~Phys.~(1992) {\bf 43}, 270-291. \bibitem{s1} Spradlin, G.: A Singularly Perturbed Elliptic Partial Differential Equation with an Almost Periodic Term. Calc.~Var.~and PDE (1999) {\bf 9}, 207-232. \bibitem{w1} Wang, X.: On Concentration of positive bound states of nonlinear Schr\"odinger equations. Comm.~Math.~Phys.~(1993) {\bf 153}, 229-244. \end{thebibliography} \medskip \noindent{\sc Gregory S. Spradlin} \\ Department of Mathematical Sciences \\ United States Military Academy \\ West Point, New York 10996, USA \\ e-mail: gregory-spradlin@usma.edu \end{document}