\documentstyle[twoside]{article} \input amssym.def % used for R in Real numbers \pagestyle{myheadings} \nofiles \markboth{\hfil Elliptic Equations in $\Bbb R^N$ \hfil EJDE--1996/09}% {EJDE--1996/09\hfil C.O. Alves, J.V. Goncalves, \& O.H. Miyagaki\hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol.\ {\bf 1996}(1996), No.\ 09, pp. 1--11. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113} \vspace{\bigskipamount} \\ ON ELLIPTIC EQUATIONS IN $R^N$ WITH CRITICAL EXPONENTS \thanks{ {\em 1991 Mathematics Subject Classifications:} 35J20, 35K20.\newline\indent {\em Key words and phrases:} Elliptic equations, unbounded domains, critical exponents,\newline\indent variational methods. \newline\indent \copyright 1996 Southwest Texas State University and University of North Texas.\newline\indent Submitted August 7, 1996. Published October 22, 1996. \newline\indent Partially supported by CNPq/Brasil.}} \date{} \author{C.O. Alves, J.V. Goncalves, \& O.H. Miyagaki} \maketitle \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \begin{abstract} In this note we use variational arguments --namely Ekeland's Principle and the Mountain Pass Theorem-- to study the equation $$-\Delta u + a(x)u = \lambda u^q + u^{2^*-1} \mbox{ in }\Bbb R^N\,.$$ The main concern is overcoming compactness difficulties due both to the unboundedness of the domain $\Bbb R^N$, and the presence of the critical exponent $2^*= 2N/(N-2)$. \end{abstract} \section{Introduction} In this note we use variational methods to explore existence of weak solutions for the problem $$ \left\{ \begin{array}{ll} -\Delta u + a(x)u = \lambda u^{q} + u^{2^*-1} \mbox{ in } \Bbb R^N \\ \int a(x)u^{2} < \infty,~~ \int \vert \nabla u \vert^{2} < \infty \\ u \geq 0, \;\; u \not\equiv 0 \end{array} \right. \leqno(*) $$ where $a$ is a nonnegative $L^\infty _{\mbox{\small loc}}$ function, $\lambda \geq 0$, $0 < q \leq 1$ and $2^*$ is the critical exponent, $2^* = 2N/(N-2)$, for $N \geq 3$. This problem has been explored by many authors including Br\'ezis \& Nirenberg [6], Ambrosetti-Br\'ezis \& Cerami [1], Guedda \& Veron [9] (see also their references) for the case of elliptic equations in bounded domains. As far as unbounded domains are concerned we recall the work by Benci \& Cerami [12], Noussair-Swanson \& Jianfu [3], Jianfu \& Xiping [14], Egnell [7], Azorero \& Alonso [4], Miyagaki [10]. In this work we shall assume the following condition on $a$. $$ a(x) > 0, \ x \in B_{R_{o}}^{c} \mbox{ and } \int_{B_{R_{o}}^{c}} \frac{1}{a}~ < \infty\,. \leqno(a_{1}) $$ Our main results are the following: \begin{theorem} Let $0 < q < 1$ and assume $(a_{1})$. Then there exists $\lambda^* > 0$ such that $(*)$ has a solution for $0 < \lambda < \lambda^*$, $N \geq 3$. \end{theorem} \begin{theorem} Let $N \geq 4$ and $q = 1$. Assume $(a_{1})$ and $a(x) = 0, ~~x \in B_{2 r_0}$ for some $r_0 \in (0,R_0/2)$. Then there is $\lambda^* > 0$ such that $(*)$ has a solution for $0 < \lambda < \lambda^*$. \end{theorem} These theorems complement the results in [10] in the sense that here the function $a$ is allowed to vanish on a ball $B_{R_o}$. Actually in Theorem 2, we require $a$ to vanish on $B_{2r_0}$. In addition we consider the case $0 < q \leq 1$ while in [10], $a>0$ is continuous and $q \in (1, 2^*)$. \section{Preliminaries} Let $$ E = \left \{ u \in {\cal D}^{1,2}~ \vert~ \int a u^{2} < \infty \right \} $$ with inner product and norm given by $$ \left = \int \left (\nabla u \nabla v + a u v \right ),~~ \Vert u \Vert^{2} = \int \left ( \vert \nabla u \vert^{2} + a u ^{2} \right ). $$ Recall that ${\cal D}^{1,2}$ is the closure of ${\cal C}_{o}^{\infty}$ with respect to the gradient norm $\Vert u \Vert^{2}_{1} = \int \vert \nabla u \vert^{2}$. Moreover $$ {\cal D}^{1,2}= \left \{u \in L^{2^*}~ \vert~ \partial_{i} u \in L^{2} \right \} $$ and the norm $$ \Vert u \Vert^{'} \equiv \vert u \vert_{L^{2^*}} + \vert \nabla u \vert_{L^{2}} $$ is equivalent to the ${\cal D}^{1,2}$ norm. In addition ${\cal D}^{1,2} \rightarrow L^{2^*}$. The following lemma is a variant of a result by Willem \& Omana [13] and by Costa [2]. \begin{lemma} Assume $(a_{1})$. Then $E \rightarrow L^{s}$ for $1 \leq s \leq 2^*$ and $E \hookrightarrow L^{s}$ for $1 \leq s < 2^*$. \end{lemma} We shall look for the critical points of the functional $$ I(u) = \frac{1}{2} \int \left ( \vert \nabla u \vert^{2} + a \vert u \vert^{2} \right ) - \frac{1}{q + 1} \int \lambda u_{+}^{q + 1} - \frac{1}{2^*} \int u_{+}^{2^*} $$ in the Hilbert space $E$. Using standard techniques we can show that $I \in C^{1}(E, \Bbb R)$, and that its derivative is given by $$ \left = \int \left (\nabla u \nabla v + a u v \right ) - \lambda \int u_{+}^{q} v - \int u_{+}^{2^* - 1} v\,. $$ Therefore, the critical points of $I$ are the weak solutions of $(*)$. The following auxiliary result concerns the geometry of $I$. \begin{lemma} If $a$ satisfies $(a_{1})$ and if $0 < q \leq 1$ then there exists $\lambda^* > 0$ such that if $0 < \lambda < \lambda^*$ then $$ I(u) \geq r,~~ \Vert u \Vert = \rho,~ \mbox{ for some }~ r, \rho > 0 \leqno(i) $$ If in addition $\phi \geq 0,~ \phi \not\equiv 0, \mbox{ and } \phi \in E$ then $$ I(t \phi) \rightarrow -\infty~~ \mbox{as}~~ t \rightarrow \infty \leqno(ii) $$ $$ I(t \phi) < 0,~ \mbox{ for small } t > 0,\mbox{ and } 0 < q < 1. \leqno(iii) $$ \end{lemma} \section{Proofs} For the sake of completeness, we present a proof of Lemma 3, which is based on the proof in [2]. \paragraph{Proof of Lemma 3.} At first let $R > R_{o}$. Then we have $$ \int_{B_{R}^{c}} \vert u \vert = \int_{B_{R}^{c}} \frac{a ^{1/2} \vert u \vert}{a^{1/2}} \leq \left (\int_{B_{R}^{c}}\frac{1}{a} \right )^ {1/2} \left (\int_{B_{R}^{c}} a \vert u \vert^{2} \right )^{1/2} \leq C \Vert u \Vert $$ which shows that $$ \vert u \vert_{L^{1}} \leq C \Vert u \Vert,~~ u \in E. $$ Now using the interpolation inequality $$ \vert u \vert_{s} \leq \vert u \vert_{1}^{\alpha} \vert u \vert_{r}^{1 - \alpha},~~ \alpha + \frac{1 - \alpha}{r} = \frac{1}{s},~ 1 \leq s \leq r \leq 2^*,~~ 0 \leq \alpha \leq 1 $$ and the embedding $E \rightarrow L^{2^*}$, we infer that $E \rightarrow L^{s},~~ 1 \leq s \leq 2^*$. On the other hand, for sufficiently large $R > 0$ we have $$ \int_{B_{R}^{c}} \frac{1}{a} < \epsilon\,. $$ So if $u_{n} \rightharpoonup 0$ in $E$, then for large $n$ $$ \int_{B_{R}^{c}} \vert u_{n} \vert \leq C \int_{B_{R}^{c}} \frac{1}{a} \leq \epsilon\,. $$ Using compact Sobolev embeddings we also have $$ u_{n} \rightarrow 0\mbox{ in } L^{1}(B_{R}) $$ so that $u_{n} \rightarrow 0$ in $L^{1}$. Using again the interpolation inequality stated above, one concludes the proof of Lemma 3. \paragraph{Proof of Lemma 4.} \paragraph{Verification of $(i)$.} From the continuous embedding in Lemma 3, we have $$ \begin{array}{lcl} I(u) & \geq & \frac{1}{2} \Vert u \Vert^{2} - \frac{\lambda c_{q + 1}}{q + 1} \Vert u \Vert^{q + 1} - \frac{S^{-2^*/2}}{2^*} \Vert u \Vert^{2^*} \\ & \geq & \Vert u \Vert^{q + 1} \left (\frac{1}{2} \Vert u \Vert^{2 - (q + 1)} - \frac{\lambda c_{q + 1}}{q + 1} - \frac{S^{-2^*/2}} {2^*} \Vert u \Vert^{2^* - (q + 1)} \right ) \end{array} $$ where $S$ is the best constant for the embedding ${\cal D}^{1,2} \rightarrow L^{2^*}$, that is $$ S = \inf \left \{ \frac{\int \vert \nabla u \vert^{2}}{\left (\int \vert u \vert^{2^{\ast}} \right)^{2/2^*} }~~ \vert~~ u \in {\cal D}^{1,2},\ u \not\equiv 0 \right \}. $$ Letting $$ Q(t) \equiv \frac{1}{2} t^{{2 - (q + 1)}} - \frac{S^{-2^*/2}}{2^*} t^{2^* - (q + 1)},~~ t \geq 0\,, $$ there is $\rho > 0$ such that $$ \max_{t \geq 0} Q(t) = Q(\rho) > 0. $$ Taking $\Vert u \Vert = \rho$ and $\lambda^* = \frac{q+1}{c_{q +1}} Q(\rho)$ we get $(i)$. \paragraph{Verification of $(ii)$.} Taking $\phi \not\equiv\ 0$, $\phi \geq 0,$ $\phi \in E$ we have $$ I(t \phi) = \frac{t^{2}}{2} \Vert \phi \Vert^{2} - \frac{t^{q + 1}}{q + 1} \lambda \int \phi^{q + 1} - \frac{t^{2^*}}{2^*} \int \phi^{2^*} $$ which gives $(ii)$. \paragraph{Verification of $(iii)$.} It is clear from the expression of $I(t \phi)$ above taking into account that $0 < q < 1$. \paragraph{Proof of Theorem 1.} By the proof of lemma 4, $I$ is bounded from below on $\overline{B_\rho}$. By the Ekeland Principle [8], there exists $u_{\epsilon} \in \overline{B_\rho}$ such that $$ I(u_{\epsilon}) \leq \inf_{\bar{B _{\rho}}} I + \epsilon $$ and $$ I(u_{\epsilon}) < I(u) + \epsilon \Vert u - u_{\epsilon} \Vert,~~ u \not\equiv\ u_{\epsilon}. $$ Now since $0 < q < 1$ it follows that $$ I(t \phi) < 0, \mbox{ for small } t > 0,~~ \phi \not\equiv\ 0, \mbox{ and } \phi \in {\cal C}_{o}^\infty \,. $$ Again by Lemma 4 $$ \inf_{\partial B_{\rho}} I \geq r > 0~~ \mbox{and}~~ \inf_{\overline{B_\rho}} I < 0. $$ Choose $\epsilon > 0$ such that $$ 0 < \epsilon < \inf_{\partial B_{\rho}} I - \inf_{\overline{B_\rho}} I . $$ Hence $$ I(u_{\epsilon}) < \inf_{\partial B_{\rho}} I $$ so that $$ u_{\epsilon} \in B_{\rho}. $$ Hence letting $$ F(u) \equiv I(u) + \epsilon \Vert u - u_{\epsilon} \Vert $$ we notice that $u_{\epsilon}$ is a point of minimum of $F$ on $\overline{B_\rho}$ and so $$ \frac{I(u_{\epsilon} + \delta v) - I(u_{\epsilon})}{\delta} + \epsilon \Vert v \Vert \;\; \geq \;\; 0 $$ which by passing to the limit as $\delta \rightarrow 0$ gives that $$ \langle I'(u_{\epsilon}),v\rangle + \epsilon \Vert v \Vert \geq 0 $$ and hence $\Vert I'(u_{\epsilon}) \Vert \leq \epsilon$. Therefore, there is a sequence $u_{n} \in \overline{B_\rho}$ such that $$ I(u_{n}) \rightarrow c^* \equiv \inf_{\overline{B_\rho}} I < 0 ~~ \mbox{and}~~ I'(u_{n}) \rightarrow 0. $$ Since of course $u_{n}$ is bounded, $$ u_{n} \rightharpoonup u^*~~ \mbox{in}~~ E $$ and $$ u_{n} \rightarrow u^*~~ \mbox{a.e. in}~~ \Bbb R^N. $$ Now passing to the limit in $$ o(1) = \int \left ( \nabla u_{n} \nabla \phi + a u_{n} \phi \right ) - \lambda \int u_{n+}^{q} \phi - \int u_{n+}^{2^* - 1} \phi,~~ \phi \in E $$ we infer that $I'(u^*) = 0$ showing that $u^*$ is a solution of problem $(*)$. In order to show that $u^* \not\equiv\ 0$, we follow the arguments in [6]. Assume that $u^* \equiv 0$ and that $$ \Vert u_{n} \Vert^{2} \rightarrow \ell \geq 0. $$ Using $I'(u_{n}) \rightarrow 0$ we have $$ \Vert u_{n} \Vert^{2} - \int u_{n+}^{2^*} = o(1) $$ so that $\int u_{n+}^{2^*} \rightarrow \ell$ and from the expression $$ c^* + o(1) = \lambda \left ( \frac{1}{2} - \frac{1}{q + 1} \right ) \int u_{n+}^{q + 1} + \frac{1}{N} \int u_{n+}^{2^*} $$ we infer that $$ c^* = \frac{\ell}{N} $$ which is impossible. \paragraph{Proof of Theorem 2.} By Lemma 4 and the Mountain Pass Theorem, there exists a sequence $u_{n}~ in~ E$ such that $$ I(u_{n}) \rightarrow c~~ \mbox{and}~~ I'(u_{n}) \rightarrow 0 $$ where $$ c = \inf_{\gamma \in \Gamma} \max_{0 \leq t \leq 1} I(\gamma(t)),~~ c \geq r $$ and $$ \Gamma = \left \{ \gamma \in C([0,1], X)~~ \vert~~ \gamma(0) = 0,~ \gamma(1) = e \right \} $$ where $e \in E$ satisfies $I(e) \leq 0$. \paragraph{Claim.} {\it There is $e \equiv e_{\lambda}$ such that $0 < c < \frac{1}{N} S^{\frac{N}{2}}$, $0 < \lambda < \lambda^*$}. From the expression $$ \left < I'(u_{n}), u_{n} \right > - 2^* I(u_{n}) = \left (1 - \frac{2^*}{2} \right) \Vert u_{n} \Vert^{2} + \lambda \left (\frac{2^*}{2} - 1 \right) \int u_{n+}^{2} $$ one shows, by taking $\lambda^* > 0$ smaller than the one found in lemma 4, that $u_{n}$ is bounded. So that, passing to subsequences, $$ u_{n} \rightharpoonup u\mbox{ in }E \mbox{ and } u_{n} \rightarrow u \mbox{ a.e. in } \Bbb R^N $$ for some $u \in~ E$. \paragraph{Remark.} $I(u_{n+}) \rightarrow c$ and $I'(u_{n+}) \rightarrow 0$. Indeed, $u_{n-}$ is also bounded so that $$ o(1) = \left = \int (\vert \nabla u_{n -} \vert^{2} + au_{n -}^{2})\,. $$ Moreover, if $\phi \in E$ then \begin{eqnarray*} \left &=& \left - \lambda \int u_{n +} \phi - \int u_{n +}^{2^* -1} \phi\\ &=&\left - \left \end{eqnarray*} so that, $I'(u_{n+}) \rightarrow 0.$ On the other hand, \begin{eqnarray*} \lefteqn{I(u_{n}) - \frac{1}{2} \int (\vert \nabla u_{n -} \vert^{2} + au_{n -}^{2}) } \\ &=&\frac{1}{2} \int(\vert \nabla u_{n +} \vert^{2} + au_{n +}^{2}) - \frac{\lambda}{2} \int u_{n +}^{2} - \frac{1}{2^*} \int u_{n +}^{2^*} \\ &=&I(u_{n +})\,, \end{eqnarray*} which gives $I(u_{n +}) \rightarrow c$. So we may assume that $u_{n} \geq 0$ and thus $u \geq 0$. Now as in the proof of Theorem 1 one shows that $u$ satisfies the equation in $(*)$. Again arguing as in [6] we assume that $u \equiv 0$. Then $$ \Vert u_{n} \Vert^{2} \rightarrow \ell~~ \mbox{for some}~ \ell \geq 0 $$ and using the facts that $$ I(u_{n}) \rightarrow c,~~ I'(u_{n}) \rightarrow 0 $$ we infer that $c \geq \frac{1}{N} S^{N/2}$, contradicting $0 < c < \frac{1}{N} S^{N/2}$ given by the Claim. \paragraph{Proof of the Claim.} (Arguments adapted from [6].) Consider the cut-off function $\phi \in {\cal{C}}_{o}^{\infty}$ such that $$ \phi \equiv 1~~ \mbox{on}~~ B_{r_0},~~\phi \equiv 0~~ \mbox{on}~~ \Bbb R^N \backslash B_{2r_0}. $$ Now consider the function $$ w_{\epsilon}(x) = \frac{\left[N (N - 2) \epsilon \right]^{(N - 2)/4}} {\left (\epsilon + \vert x \vert^{2} \right )^{(N - 2)/2}},~~~~ x \in \Bbb R^N,~ \epsilon > 0 $$ which satisfies $$ -\Delta w_{\epsilon} = w_{\epsilon}^{2^* - 1}~~ \mbox{in}~ \Bbb R^N. $$ It is well known (see e.g. Talenti [5], Aubin [11] ) that $$ \Vert w_{\epsilon} \Vert^{2}_{1}~ = \vert w_{\epsilon} \vert_{2^*}^{2^*}~ =~ S^{N/2}. $$ Let $$ \psi_{\epsilon} = \phi w_{\epsilon} $$ and let $v_{\epsilon} \in {\cal C}_{o}^{\infty}$ given by $$ v_{\epsilon} = \frac{\psi_{\epsilon}}{\left ( \int {\psi_{\epsilon}}^{2{*}} \right)^{1/2^*}}. $$ Now it can be shown (see e.g. [6], [10]) that $X_{\epsilon} \equiv \int \vert \nabla v_{\epsilon} \vert^{2}$ satisfies $$ X_{\epsilon} \leq S + O(\epsilon^{(N - 2)/2}). $$ Moreover there is some $t_{\epsilon} > 0$ such that $$ \max_{t \geq 0} I(t v_{\epsilon}) = I(t_{\epsilon} v_{\epsilon} ) $$ and $$ \frac{d}{dt} I(t v_{\epsilon}) \vert_{ {t = t_{\epsilon}}} = 0. $$ which gives $$ 0 < t_{\epsilon} < X_\epsilon^{1/(2^* - 2)} \equiv t_{0}. $$ Notice that $a = 0~ \mbox{on}~ B_{2r_0}$ and $v_\epsilon = 0~ \mbox{on}~ \Bbb R^N \backslash B_{2r_0}$. Moreover $t_\epsilon \geq d_0 \equiv d_0({r_0})$ for some $d_0 > 0$. Otherwise since $X_\epsilon$ is bounded, if $t_\epsilon \rightarrow 0$, then $I(t_\epsilon v_{\epsilon}) \rightarrow 0$ contradicting $$ I(t_\epsilon v_{\epsilon}) = \max_{t \geq 0} I(t v_{\epsilon}) \geq r > 0 $$ given by lemma 4 (i). On the other hand \begin{eqnarray*} I(t v_{\epsilon}) & = & \frac{t^2}{2} \int \vert \nabla v_\epsilon \vert^{2} - \frac{t^{2^*}}{2^*} - \frac{\lambda t^2}{2} \int v_\epsilon^{2} \\ & \leq & \left (\frac{t^2}{2} t_0^{2^* - 2} - \frac{t^{2^*}}{2^*} \right ) - \frac{\lambda t^2}{2} \int_{B_{2r_0}} v_\epsilon^{2}\,. \end{eqnarray*} Now recalling that as a function of $t$, $$ \left (\frac{t^2}{2} t_0^{2^* - 2} - \frac{t^{2^*}}{2^*} \right ) $$ increases on the interval $\left (0, t_0 \right )$ we get \begin{eqnarray*} I(t_\epsilon v_{\epsilon}) &\leq & t_0^{2^*} \left (\frac{1}{2} - \frac{1}{2^*} \right ) - \frac{\lambda t_{\epsilon}^2}{2} \int_{B_{2r_0}} v_\epsilon^{2}\\ & \leq & \frac{1}{N} t_0^{2^*} - \frac{\lambda d_0^{2}}{2} \int_{B_{2r_0}} v_\epsilon^{2}\\ & \leq & \frac{1}{N} \left [ S + O \left (\epsilon^{(N - 2)/2} \right ) \right ]^{2^*/(2^* - 2)} - \frac{\lambda d_0^{2}}{2} \int_{B_{2r_0}} v_\epsilon^{2}\\ & = & \frac{1}{N} \left [ S + O \left (\epsilon^{(N - 2)/2} \right ) \right ]^{N/2} - \frac{\lambda d_0^{2}}{2} \int_{B_{2r_0}} v_\epsilon^{2}. \end{eqnarray*} Using the inequality $$ (b + c)^{\alpha} \leq b^{\alpha} + \alpha (b + c)^{\alpha - 1}c~~~ b, c \geq 0, \alpha \geq 1 $$ with $b = S$, $c = O \left (\epsilon^{(N - 2)/2} \right )$ and $\alpha = N/2$ we get $$ I(t_{\epsilon} v_{\epsilon}) \leq \frac{1}{N} S^{N/2} + O(\epsilon^{(N - 2)/2}) - c_{0} \lambda \int_{B_{2r_0}} v_{\epsilon}^{2}. $$ Therefore, $$ I(t_{\epsilon} v_{\epsilon}) \leq \frac{1}{N} S^{N/2} + \epsilon^{(N - 2)/2} \left \{M - c_0 \lambda \epsilon^{(2 - N)/2} \int_{B_{2r_0}} v_{\epsilon}^{2} \right \}, $$ where $c_0 = d_{0}^{2}/2$ and $M$ is a positive constant. We shall show that $$ \epsilon^{(N - 2)/2} \left \{M - c_0 \lambda \epsilon^{(N - 2)/2} \int_{B_{2r_0}} v_{\epsilon}^{2} \right \} < 0,~~ \mbox{ for small }\epsilon > 0\,. $$ So that $$ I(t_{\epsilon} v_{\epsilon}) < \frac{1}{N} S^{N/2} $$ and hence $$ 0 < c < \frac{1}{N} S^{N/2}. $$ Noticing that $$ d_1 \leq \int_{B_{2r_0}} \psi_{\epsilon}^{2^*} \leq d_2,~~ \mbox{for some}~~ d_1, d_2 > 0, $$ (see [6]), it follows by a change of variables that $$ I(t_{\epsilon} v_{\epsilon}) \leq \frac{1}{N} S^{N/2} + \epsilon^{(N - 2)/2} \left \{M - c_0 \lambda \epsilon^{(4 - N)/2} \int_{0}^{r_0 \epsilon^{-1/2}} \frac{s^{N - 1} ds}{(1 + s^{2})^{N - 2}} \right \}. $$ We are going to consider separately the cases $N = 4$ and $N \geq 5$. \paragraph{Case $N = 4$.} We have $$ \begin{array}{lcl} I(t_{\epsilon} v_{\epsilon}) & \leq & \displaystyle \frac{1}{4} S^{2} + \epsilon \left \{M - c_0 \lambda \int_{0}^{r_0 \epsilon^{-1/2}} \frac{s^{3} ds}{(1 + s^{2})^{2}} \right \}\\ & \leq & \displaystyle \frac{1}{4} S^{2} + \epsilon \left \{M - c_0 \lambda \ln (r_0 \epsilon^{-1/2}) \right \}. \end{array} $$ Now since $$ c_0 \lambda \ln (r_0 \epsilon^{-1/2}) \rightarrow \infty~ \mbox{as}~ \epsilon \rightarrow 0 $$ we infer that $$ I(t_{\epsilon} v_{\epsilon}) < \frac{1}{4} S^{2},~ \mbox{for small }\epsilon > 0. $$ \paragraph{Case $N \geq 5$.} Noticing that $$ c_0 \lambda \epsilon^{(4 - N)/2} \int_{0}^{r_0 \epsilon^{-1/2}} \frac{s^{N - 1} ds}{(1 + s^{2})^{N - 2}} \rightarrow \infty \mbox{ as } \epsilon \rightarrow 0 $$ we infer that $$ I(t_{\epsilon} v_{\epsilon}) < \frac{1}{N} S^{N/2}\mbox{ for small } \epsilon > 0\,, $$ which concludes the proof of this claim. \begin{thebibliography}{99} \bibitem{} A. 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Xiping, {\it On the existence of nontrivial solution of a quasilinear elliptic boundary value problem for unbounded domains}, Acta Math. Sci. 7(1987) 341-359. \end{thebibliography} \bigskip {\small C.O. Alves \\ Dep. Mat. Univ. Fed. Paraiba\\ 58109-970 - Campina Grande(PB), Brasil} \\ E-mail coalves@dme.ufpb.br \medskip \noindent{\small J.V. Goncalves \\ Dep. Mat. Univ. Bras\'{\i}lia\\ 70.910-900 Brasilia(DF), Brasil} \\ E-mail jv@mat.unb.br \medskip \noindent{\small O.H. Miyagaki \\ Dep. Mat. Univ. Fed. Vi\c cosa\\ 36570-000 Vi\c cosa(MG), Brasil} \\ E-mail olimpio@mail.ufv.br \end{document}