\documentstyle[twoside]{article}
\input amssym.def     % used for R in Real numbers
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\markboth{\hfil Radial Symmetric Solutions \hfil EJDE--1996/07}%
{EJDE--1996/07\hfil C.O. Alves, D.C.de Morais \& M.A.S. Souto \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.\ {\bf 1996}(1996), No.\ 07, pp. 1--12. \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} \\
Radially Symmetric Solutions for a Class of Critical Exponent Elliptic
Problems in $\Bbb R^N$
\thanks{ {\em 1991 Mathematics Subject Classifications:}
35A05, 35A15 and 35J20. \newline\indent
{\em Key words and phrases:} Radial solutions, critical Sobolev  exponents,
 \newline\indent Palais-Smale condition, Mountain Pass Theorem.\newline\indent
\copyright 1996 Southwest Texas State University  and University of
North Texas.\newline\indent
Submitted July 04, 1996. Published: August 30, 1996.\newline\indent
Second and third author are partially supported by  CNPq/Brazil } }
\date{}

\author{C. O. Alves, D. C. de Morais Filho
 \& M. A. S. Souto}
\date{}
\maketitle

\newtheorem{theo}{Theorem}
\newtheorem{remark}{Remark}
\newtheorem{lemma}{Lemma}

\begin{abstract} We give a method for obtaining radially
symmetric solutions for the critical exponent problem
\[
\left\{
\begin{array}{c}
-\Delta u+a(x)u=\lambda u^q+u^{2^{*}-1}\mbox{ in }\Bbb R^N \\
u>0\mbox{ and }\int_{\Bbb R^N}|\nabla u|^2<\infty 
\end{array}
\right.
\]
where, outside a ball centered at the origin, the non-negative function $a$ 
is bounded from below by a positive constant $a_o>0$. We remark that, 
differently from the literature, we do not require any
conditions on $a$ at infinity.
\end{abstract}
\section{Introduction}

Our purpose in this paper is to solutions for the semi-linear
elliptic problem:
\begin{equation}
\left\{
\begin{array}{c}
-\Delta u+a(x)u=\lambda u^q+u^{2^{*}-1}\mbox{ in }\Bbb R^N \\
u>0\mbox{ and }\int_{\Bbb R^N}|\nabla u|^2<\infty .
\end{array}
\right.  \label{global}
\end{equation}
where $a:\Bbb R^N\to \Bbb R$ is a non-negative radially symmetric $C^1$
function, \linebreak $2*=2N/(N-2)$; $1<q<2^*-1$, $\lambda >0$,
and $N\geq 3$.

Several researchers have studied variants of
problem (\ref{global}). Among others, we can cite the article by
Br\`ezis \& Nirenberg \cite{BN1} which treats the case $a\equiv 0$ in 
bounded domains. 
 Azorero \& Alonzo in \cite{AA1} and \cite{AA3}
generalize some similar results for the $p-$Laplacian operator in bounded
domains. Egnell \cite{E1} also generalizes some results in \cite{BN}. In 
the case of unbounded domains, Rabinowitz \cite{R} considers
a more general non-linearity, but he does not treat the Sobolev critical
exponent case. Benci \& Cerami \cite{BC} consider the problem (\ref{global})
when $\lambda =0,$ and \cite{AGM} deals with the case where $\lambda $ 
is
replaced by an integrable function. In \cite{CMP}, a variation of this 
problem, with $a$ constant, was solved for the biharmonic operator. To 
finish  citations we list the
following works: \cite{GA} Alves
\& Gon\c calves, \cite{GM} Gon\c calves \& Miyagaki, \cite{J} Jianfu, 
\cite{JX} Jianfu \& Xiping. All the last results
in unbounded domains are obtained under the crucial hypothesis that $a$ 
is
a coercive function or that $\lim\limits_{|x|\rightarrow +\infty }a(x)$
exists.

We improve their results, relaxing the coerciveness of $a$ and the existence
of the above limit.
As in \cite{R}, we shall use variational method to solve problem (\ref
{global}).
To describe  precisely our results, we present below the hypotheses on the
function $a$:

$(A_o)$ $a\in C^1(\Bbb R^N)$ is a radially symmetric function and there are 
$a_o,$ $R>0$ such that $a(x)\geq a_o,$ for all $|x|\geq R$ .

Let us consider the following $W^{1,2}(\Bbb R^N)$ Hilbert subspace:
\[
H_{\rm rad}^1(\Bbb R^N)=\{u\in W^{1,2}(\Bbb R^N):u\mbox{ radially symmetric
}\}.
\]

Our main result is the following.

\begin{theo}
If ($A_o$)  is satisfied, then problem (\ref{global}) possesses a
nontrivial classical solution $u\in H_{\rm rad}^1(\Bbb R^N),$ for all $\lambda >0
$ and $1<q<2^{*}-1$ when $N\geq 4.$ In the case $N=3$ the same result is
valid if $3<q<6$ .
\end{theo}

\begin{remark}
When $\lambda $ is large enough, (\ref{global}) possesses a nontrivial
classical solution. Later we shall justify this remark.
\end{remark}

Employing the same techniques used to prove the above theorem, we improve
the results obtained in the subcritical exponent case due to Rabinowitz (see
\cite{R}), where he considers the problem
\begin{equation}
-\Delta u+a(x)u=f(x,u)\mbox{ in }\Bbb R^N  \label{global2}
\end{equation}
for a given $C^1$-function $f:\Bbb R^N\times \Bbb R\to \Bbb R$ with $a$
coercive.

Results related to this kind of problem can be found in \cite{BL}, \cite{R},
among others.

In \cite{BL}, H. Berestycki and P. L. Lions obtained positive solution of
problem (\ref{global}) when the non-linearity $f$ does not depend on $x.$
They obtained the solution as a limit of positive solutions of the problem
restricted to bounded domains. In their paper they basically made use of 
$H^1 $-estimates.

Our second result is a global version on $\Bbb R^N$ of a well known result 
for bounded domain due to Rabinowitz (theorem 2.15 in \cite{R1}):

\begin{theo}
Suppose that $a\in C^1(\Bbb R^N)$ satisfies $(A_o)$ and $f$ satisfies:

\begin{description}
\item{$(f_o)$} The function $f$ is a $C^1$, radially symmetric
function in $x$, i.e.,
$f(x,s)=f(r,s)$ where $r=|x|$, for all $x\in \Bbb R^N$, 
$s\in \Bbb R$.
\item{$(f_1)$} For each $\varepsilon >0,$ there is a constant $a_1>0,$ such
that
\[
|f(x,s)|\leq \varepsilon |s|+a_1|s|^p,\mbox{ for all }x\in \Bbb R^N,
\quad s\in \Bbb R,
\]
where $1\leq p<2^{*}-1.$

\item{$(f_2)$} There is $\mu >2,$ such that
\[
0<\mu F(x,s)\dot \leq sf(x,s),\mbox{ for all }x\in \Bbb R^N,\mbox{ }s\in
\Bbb R\backslash \{0\},
\]
where $F(x,s)=\int_o^sf(x,t)dt.$
\end{description}

Then (\ref{global2}) possesses a nontrivial classical solution $u\in
W^{1,2}(\Bbb R^N).$
\end{theo}

\section{Proof of Theorem 1}
First, let us formulate a proper framework to solve problem (\ref{global}).
Define the Hilbert space
\[
E=\{u\in H_{\rm rad}^1(\Bbb R^N):\int_{\Bbb R^N}a(x)u^2<\infty \},
\]
endowed with the inner product $\langle u,v\rangle=:\int_{\Bbb R^N}(\nabla u\nabla
v+a(x)uv)$ and the norm $||u||^2=:\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2$

Now we present two lemmas that will be used in the proof of the Theorem~1.

\begin{lemma}
Let $w$ be a $W_{\rm loc}^{1,s}(\Bbb R^N)$ function satisfying
\begin{equation}
-\Delta w=h,\mbox{ }  \label{Lwh}
\end{equation}
in $\Bbb R^N\backslash \{0\}$ in the weak sense, where $h$ is a $%
L_{\rm loc}^1(\Bbb R^N)$ function and $s\geq \frac N{N-1}$. Then (\ref{Lwh})
is weakly satisfied in the whole $\Bbb R^N.$
\end{lemma}

\paragraph{Proof:} In order to prove this result, consider $\varphi
\in C^\infty (\Bbb R^N)$ such that $\varphi (x)=0$ in $|x|\leq 1$ and $%
\varphi (x)=1$ in $|x|\geq 2.$ For each $\varepsilon >0,$ define $\psi
_\varepsilon (x)=\varphi (\frac x\varepsilon ).$ Fix a function $\phi \in
C_c^\infty (\Bbb R^N).$ As $\psi _\varepsilon \phi \in C_c^\infty
(\Bbb R^N\backslash \{0\})$ we have that
\[
\int_{\Bbb R^N}\nabla w\nabla (\psi _\varepsilon \phi
)=\int_{\Bbb R^N}h(x)(\psi _\varepsilon \phi ),
\]
and then
\begin{equation}
\int_{\Bbb R^N}\psi _\varepsilon \nabla w\nabla \phi +\int_{\Bbb R^N}\phi
\nabla w\nabla \psi _\varepsilon =\int_{\Bbb R^N}h(x)(\psi _\varepsilon
 \phi
).  \label{epsil}
\end{equation}
Using the dominated convergence theorem, we obtain the limits
\begin{eqnarray}
\lim_{\varepsilon \rightarrow 0}\int_{\Bbb R^N}\psi _\varepsilon \nabla
w\nabla \phi &=&\int_{\Bbb R^N}\nabla w\nabla \phi \mbox{ }  
\label{limites}
\\
\lim_{\varepsilon \rightarrow 0}\int_{\Bbb R^N}h(x)(\psi _\varepsilon \phi )
&=&\int_{\Bbb R^N}h(x)\phi .  \nonumber
\end{eqnarray}
We claim that the limit of the second term on the left side of (\ref{epsil}) is
zero. In fact,
\[
\left| \int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon \right| \leq
||\phi ||_{L^\infty (\Bbb R^N)}\int_{|x|\leq 2\varepsilon }|\nabla w||\nabla
\psi _\varepsilon |.
\]
Using H\"older's inequality in the above inequality with $\frac 1s+\frac 1q
=1,$
we obtain that
\[
\left| \int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon \right| \leq
||\phi ||_{L^\infty (\Bbb R^N)}\left( \int_{|x|\leq 2\varepsilon }|\nabla
w|^s\right) ^{1/s}\left( \int_{|x|\leq 2\varepsilon }|\nabla \psi
_\varepsilon |^q\right) ^{1/q}
\]
and then
\[
\left| \int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon \right| \leq
||\phi ||_{L^\infty (\Bbb R^N)}||\nabla \varphi ||_{L^q(\Bbb R^N)}\left(
\int_{|x|\leq 2\varepsilon }|\nabla w|^s\right) ^{1/s}\varepsilon ^{
\frac{N-q}q}.
\]
Observe that $N\geq q$ and passing to the limit in this last inequality we
prove the claim.

Finally using the claim and the limits (\ref{limites}) in (\ref{epsil}) we
have that
\[
\int_{\Bbb R^N}\nabla w\nabla \phi =\int_{\Bbb R^N}h(x)\phi .
\]

\begin{remark}
The above result is not valid for $W_{\rm loc}^{1,1}(\Bbb R^N)$ functions.
The
function $w=|x|^{2-N}$(if $N\geq 3,$ or $w=\log |x|,$ if $N=2$) belongs to $%
W_{\rm loc}^{1,1}(\Bbb R^N),$ satisfies $-\Delta w=0$ in $\Bbb R^N\backslash
\{0\},$ but if $v$ is a radially symmetric function in $C_c^\infty \left(
\Bbb R^N\right) $ such that $v(0)\neq 0,$ we have that
\[
\int_{\Bbb R^N}\nabla w\nabla v=\frac{\omega _N}{2-N}\int_o^\infty
r^{N-1}r^{1-N}v^{\prime }(r)dr=\frac{\omega _N}{2-N}v(0)\neq 0,\mbox{ if }%
N\geq 3
\]
or
\[
\int_{\Bbb R^N}\nabla w\nabla v=2\pi v(0)\neq 0,\mbox{ if }N=2.
\]
\end{remark}

\begin{lemma}
Let $f:\Bbb R^N\times \Bbb R\rightarrow \Bbb R$ be a $C^1$ function
satisfying $(f_o)$ such that
\[
|f(x,s)|\leq c|s|+|s|^{2^{*}-1}\mbox{ for all }x\in \Bbb R^N,\mbox{ }s\in
\Bbb R;
\]
and let $a$ be  a radially symmetric function. Suppose that $u\in E$
satisfies
\[
\int_{\Bbb R^N}(\nabla u\nabla v+a(x)uv)=\int_{\Bbb R^N}f(x,u)v,\mbox{ 
for
all }v\in E\,.
\]
Then $u\in C^2(\Bbb R^N)$ and $-\Delta u(x)+a(x)u(x)=f(x,u(x))$ for all 
$x\in \Bbb R^N$.
\end{lemma}

\paragraph{ Proof:} Since $a$ and $f$ are radially symmetric we rewrite
the above expression as
\begin{equation}
\int_o^\infty r^{N-1}(u^{\prime }v^{\prime }+a(r)uv)dr=\int_o^\infty
r^{N-1}f(r,u)vdr,  \label{EDOinte}
\end{equation}
for all $v \in E$. We have that $h(r):=-a(r)u(r)+f(r,u(r))$ is in $C^{o
,\alpha
}(\Bbb R^N\backslash \{0\}),$ since $H_{\rm rad}^1(\Bbb R^N)$ is contained
in  $C^{o,\alpha }(\Bbb R^N\backslash \{0\})$. Hence
\[
\int_o^\infty r^{N-1}u^{\prime }\psi ^{\prime }dr=\int_o^\infty
r^{N-1}h(r)\psi dr,\mbox{ for all }\psi \in C_c^\infty (0,+\infty ),
\]
and
\begin{equation}
\int_o^\infty u^{\prime }(r^{N-1}\psi )^{\prime }dr=\int_o^\infty [\frac{N-1}%
ru^{\prime }+h(r)](r^{N-1}\psi )dr,\mbox{ }  \label{phipsi}
\end{equation}
for all $\psi \in C_c^\infty (0,+\infty ).$ For each $\varphi \in C_c^\infty
(0,+\infty ),$ considering $\psi =r^{1-N}\varphi $ in (\ref{phipsi}) we
conclude that $u$ is a weak solution of
\[
-u^{\prime \prime }=\frac{N-1}ru^{\prime }+h(r),\mbox{ for }r>0.
\]
Since $u^{\prime }\in L_{\rm loc}^2(0,+\infty ),$ it follows that $u\in
H_{\rm loc}^2(0,+\infty )$, $u^{\prime }\in H_{\rm loc}^1(0,+\infty )$, 
and 
$$u\in H_{\rm rad}^1(\Bbb R^N)\cap C^2(\Bbb R^N\backslash \{0\})\,.$$ Moreover for $|x|>0$
, the function $u$ satisfies (\ref{global}) in the classical sense.
\hfill $\diamondsuit$

\paragraph{Proof of Theorem 1}. This proof consists of using variational 
methods to get critical points of the Euler-Lagrange functional associated
 to (1) and defined on $E$:
\[
I(u)=\frac 12\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2)-\frac \lambda
{q+1}\int_{\Bbb R^N}\left( u^{+}\right) ^{q+1}-\frac
1{2^{*}}\int_{\Bbb R^N}\left( u^{+}\right) ^{2^{*}}
\]
where $u^+(x)=\max \{u(x),0 \}$ and $u^+(x)=\min \{-u(x),0 \}$.

The critical points of $I$ are precisely the weak solutions of (1). These
solutions may be regularized.

The Hilbert space $E$ is immersed continuously in $W^{1,2}(\Bbb R^N).$ This
assertion comes from $(A_o)$ and the following inequalities
\begin{eqnarray*}
\left( \int_{|x|\leq R}u^2\right) ^{1/2}& \leq&  c_1\left( \int_{|x|\leq
R}\left| u\right| ^{2^{*}}\right) ^{1/2^*}\\ 
& \leq& c_1\left(\int_{\Bbb R^N}\left| u\right| ^{2^{*}}\right) ^{1/2^*}\\
&\leq&
c_2\left( \int_{\Bbb R^N}|\nabla u|^2\right) ^{1/2}\,.
\end{eqnarray*}

We also have that $H_{\rm rad}^1(\Bbb R^N)\subset L^p(\Bbb R^N)$ continuously
 if
$2 \leq p\leq 2^{*}$ and compactly if $2 \leq p<2^{*}$(see \cite{K}). Using these
results one has the following lemma:

\begin{lemma}
The Banach space $E$ is continuously immersed in $L^p(\Bbb R^N)$ if %
\linebreak $2 \leq p\leq 2^{*}$ and compactly if $2 \leq p<2^{*}$.
\end{lemma}

Using lemma 3  we verify that $I$ is a well-defined $C^1(E)$ functional -
see \cite{RZ}. It is easy to verify that
\begin{equation}
\frac \lambda {q+1}\int_{\Bbb R^N}(u^+)^{q+1}+\frac
1{2^{*}}\int_{\Bbb R^N}(u^+)^{2^{*}}=o(||u||^2)\mbox{ as }u\rightarrow 
0,
\label{opeqeno}
\end{equation}
and hence that $I$ has a local minimum at the origin. This is not a global
minimum. If $u\in E\backslash \{0\}$ , $u\geq 0,$ we have that
\[
I(tu)=\frac{t^2}2\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2)-\frac{\lambda t^{q+1}%
}{q+1}\int_{\Bbb R^N}(u^+)^{q+1}-\frac{t^{2^{*}}}{2^{*}}\int_{\Bbb R^N}
(u^+)^{2^{*}}.\]
Since $\int_{\Bbb R^N}(u^+)^{2^{*}}\neq 0,$ we conclude that $
I(tu)\rightarrow -\infty $ as $t\rightarrow \infty $.
So, we have just seen that $I$ has the\textit{\ Mountain Pass Theorem
Geometry.}

Let $e\in E$ such that $I(e)<0,$ and
\[
\Gamma =\{g:[0,1]\rightarrow E:g(0)=0\mbox{ and }g(1)=e\}
\]
and
\[
c=\inf_{g\in \Gamma }\max_{0\leq t\leq 1}I(g(t)).
\]
Thus $c\ $ is the mountain pass minimax value associated to $I$. At this
moment, it is important to notice that $c$ is not the minimax value
associated to the Euler Lagrange functional of problem (\ref{global})
defined in the whole $W^{1,2}(\Bbb R^N).$ Assertion (\ref{opeqeno}) implies 
$c>0.$ Using an application of the Ekeland Variational Principle (Theorem 
4.3
of \cite{MW}), there exists a sequence $\{u_m\}\subset E$ such that
\begin{equation}
I(u_m)\rightarrow c,\mbox{ }I^{\prime }(u_m)\rightarrow 0.  \label{PSseq}
\end{equation}

\begin{lemma}
The above sequence $\{u_m\}$ is bounded.
\end{lemma}

\paragraph{Proof:} Notice that
\[
I(u_m)-\frac 1{q+1}I^{\prime }(u_m)u_m=\left( \frac 12-\frac 1{q+1}\right)
||u_m||^2+\left( \frac 1{q+1}-\frac 1{2^{*}}\right) \int_{\Bbb R^N}\left(
u_m^{+}\right) ^{2^{*}},
\]
then
\[
I(u_m)-\frac 1{q+1}I^{\prime }(u_m) u_m \geq \left( \frac 12-\frac 1{q+1}
\right)
||u_m||^2.
\]
Combining this last inequality with
\[
I(u_m)-\frac 1{q+1}I^{\prime }(u_m)u_m \leq 1+c+||u_m||
\]
for large $m$, we conclude the proof.
\hfill $\diamondsuit $

The following lemma shows that we can choose a vector $e\in E\backslash
\{0\} $ in the definition of $\Gamma ,$ such that $I(e)<0$ and
\begin{equation}
0<c<\frac 1NS^{N/2},  \label{cns}
\end{equation}
where $S$ is the best constant of the Sobolev immersion $W^{1,2}(\Bbb R^N
)\subset
L^{2^{*}}(\Bbb R^N),$ this is
\[
S=\inf \{\int_{\Bbb R^N}|\nabla u|^2;\mbox{ }u\in W^{1,2}(\Bbb R^N)\mbox{
and }\int_{\Bbb R^N}|u|^{2^{*}}=1\}.
\]
Using the above facts and arguments due to Br\`ezis \& Nirenberg \cite{BN},
 we
will show that the choice in (\ref{cns}) applies in obtaining a non-trivial
solution of (\ref{global}).

\begin{lemma}
Suppose that $\lambda >0$ and one of the following conditions is satisfied:
\begin{description}
\item{(i)} $N\geq 4;$
\item{(ii)} $N=3$ and $3<q<6.$
\end{description}
Then, there is a vector $e\in E\backslash \{0\},$ $e\geq 0,$ $I(e)<0$ such
that
\begin{equation}
\sup_{t\geq 0}I(te)<\frac 1NS^{N/2},  \label{ite}
\end{equation}
\end{lemma}

\paragraph{Proof:} For each $\varepsilon >0,$ consider the function
\[
\phi _\varepsilon (x)=\frac{\left[ N(N-2)\varepsilon \right] ^{(N-2)/4}}{%
\left( \varepsilon +|x|^2\right) ^{(N-2)/2}}.
\]
The functions $\phi _\varepsilon $ satisfy the problem
\[
\left\{
\begin{array}{c}
-\Delta u=u^{2^{*}-1},\mbox{ in }\Bbb R^N \\
u>0,\mbox{ }\int_{\Bbb R^N}|\nabla u|^2<\infty
\end{array}
\right.
\]
and
\[
\int_{\Bbb R^N}|\nabla \phi _\varepsilon |^2=\int_{\Bbb R^N}|\phi
_\varepsilon |^{2^{*}}=S^{N/2}
\]
(see \cite{Ta}, lemma 2 - pp. 364). Now, consider $v_\varepsilon =\varphi
\phi _\varepsilon $ where $\varphi \in C_o^\infty (\Bbb R^N),$ $0\leq
\varphi (x)\leq 1$ and
\[
\varphi (x)=\left\{
\begin{array}{ll}
1&\mbox{ if }x\in B_1 \\
0&\mbox{ if }x\notin B_1
\end{array}
.\right.
\]
Using arguments due to \cite{Mi} there is $\varepsilon >0$ such that
\[
\sup_{t\geq 0}I(tv_\varepsilon )<\frac 1NS^{N/2}.
\]
If $t_\varepsilon >0$ is such that $I(t_\varepsilon v_\varepsilon )<0,$ we
choose $e=t_\varepsilon v_\varepsilon $ and the proof is
complete.~$\diamondsuit $  


In order to complete the proof of Theorem 1, let us consider $e\in
E\backslash \{0\}$ given by lemma 5. Let $\{u_m\}$ be the sequence in $E$
satisfying (\ref{PSseq}). From Lemmas 3 and 4, we may assume that
\begin{eqnarray*}
&u_m\rightharpoonup u\mbox{ in }E&\\
&u_m\rightarrow u\mbox{ in }L^s(\Bbb R^N),\, 2 \leq s<2^* &\\
&u_m(x)\rightarrow u(x)\mbox{ a.e. in }\Bbb R^N\,.&
\end{eqnarray*}
The above limits with an observation in Br\`ezis \& Lieb \cite{BLieb} 
yield that $u$ must be a critical point of $I$ in $E,$ that is,
\[
I^{\prime }(u)=0.
\]

We claim that $u\neq 0$. In fact, if $u\equiv 0$ and taking $l\geq 0$ such
that
\[
\int_{\Bbb R^N}|\nabla u_m|^2\rightarrow l,
\]
then
\[
\int_{\Bbb R^N}\left( u_m^{+}\right) ^{2^{*}}\rightarrow l
\]
for the reason that $I^{\prime }(u_m)\rightarrow 0$ and $E\subset
L^{q+1}(\Bbb R^N)$ compactly. Since $I(u_m)\rightarrow c,$ we get
\begin{equation}
Nc=l.  \label{ncl}
\end{equation}
From the definition of $S$,
\[
\int_{\Bbb R^N}|\nabla u_m|^2\geq S\left( \int_{\Bbb R^N}\left| u_m\right|
^{2^{*}}\right) ^{\frac 2{2^{*}}}\geq S\left( \int_{\Bbb R^N}\left(
u_m^{+}\right) ^{2^{*}}\right) ^{\frac 2{2^{*}}}.
\]
Taking the limit in the last inequalities, we achieve that
\[
l\geq Sl^{ 2/ 2^*}
\]
and by (\ref{ncl}) that
\[
c\geq \frac 1NS^{N/2}>c
\]
which contradicts the above choice of $e$, and thus the claim is proved.

Observe that $I^{\prime }(u)u^{-}=0$ implies $\int_{\Bbb R^N}|\nabla
u^{-}|^2+a(x)\left( u^{-}\right) ^2=0$ and then $u^{-}\equiv 0$ which
implies $u\geq 0.$
Notice that at this moment we do not know if $u$ satisfies (\ref{global})
 in the
$W^{1,2}(\Bbb R^N)$ sense but, thanks to lemma 2,  $u$ is a nontrivial 
classical solution of (\ref{global}) with $u\geq 0$. The Hopf maximum 
principle assures that $u>0$. Theorem 1 is
proved.
\hfill $\diamondsuit $

We conclude this section by justifying Remark 1 in the beginning of 
Section~1. The argument we are going to use is due to Azorero \& Alonzo \cite{AA3}.

\paragraph{Justification of Remark 1.} Fix $\varphi \in C_o^\infty
(\Bbb R^N\backslash \{0\})$, $\varphi (x)\geq 0$. Notice that the real
function $I(t\varphi )$ possesses a positive maximum value . Suppose that
this maximum value is assumed for $t=t_\lambda .$ Thus
\[
\frac d{dt}\left. I(t\varphi )\right| _{t=t_\lambda }=0
\]
then
\[
||\varphi ||^2=t_\lambda ^{q-1}\lambda \int_{\Bbb R^N}\varphi
^{q+1}+t_\lambda ^{2^{*}-2}\int_{\Bbb R^N}\varphi ^{2^{*}}\geq t_\lambda
^{q-1}\lambda \int_{\Bbb R^N}\varphi ^{q+1}.
\]
From the last inequality we have that $t_\lambda \rightarrow 0$ as $\lambda
\rightarrow \infty .$ On the other hand
\[
\sup_{t\geq 0}I(t\varphi )\leq \frac{t_\lambda }2\int_{\Bbb R^N}|\nabla
\varphi |^2
\]
and for large enough $\lambda >0$ we get
\[
\sup_{t\geq 0}I(t\varphi )<\frac 1NS^{\frac N2}.
\]
Using the same arguments employed in the proof of Theorem 1 we
conclude the justification.

We have just finished the proof of Theorem 1. Our next step is the proof
of Theorem 2

\section{Proof of Theorem 2}
Let $\left( E,||\cdot ||\right) $ be the same
defined in the proof of Theorem 1 and consider
\begin{equation}
I(u)=\frac 12\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2)-\int_{\Bbb R^N}F(x,u)
\label{Euler2}
\end{equation}
defined in $E$, as the associated Euler-Lagrange functional to problem (\ref
{global2}), which is $C^1-$ see \cite{RZ} . Under hypothesis $(f_1)$, it 
is
easy to verify that
\begin{equation}
\int_{\Bbb R^N}F(x,u)=o(||u||^2)\mbox{ as }u\rightarrow 0,  \label{opeqeno2}
\end{equation}
and hence that $I$ has a local minimum at the origin. Hypothesis $(f_2)$
implies that
\begin{equation}
F(x,s)\geq a_2|s|^\mu  \label{Fmu}
\end{equation}
for large $|s|$ . Then, by (\ref{opeqeno2}) and (\ref{Fmu}), $I$ has the%
\textit{\ Mountain Pass Theorem Geometry. }Let
\[
\Gamma =\{g:[0,1]\rightarrow E:g(0)=0\mbox{ and }I(g(1))\leq 0\}
\]
and
\[
c=\inf_{g\in \Gamma }\max_{0\leq t\leq 1}I(g(t)).
\]
As in the proof of Theorem 1, $c>0$ and there is a sequence $\{u_m\}\subset
E $ satisfying (\ref{PSseq}). Using standard arguments, $\left( f_2\right) $
implies that $||u_m||$ is a bounded sequence. Therefore, along a
subsequence, $u_m$ converges weakly in $E$ and strongly in $L^p(\Bbb R^N)$, 
$2\leq p<\frac{2N}{N-2},$ to a function $u\in E$ which is a weak solution 
of (%
\ref{global2}). We claim that $u\neq 0.$ In fact, for large $m$,
\[
\frac c2\leq I(u_m)-\frac 12I^{\prime }(u_m)u_m=\int_{\Bbb R^N}[\frac
12f(x,u_m)u_m-F(x,u_m)].
\]

Taking $m\rightarrow \infty ,$ in the above expression we obtain that
\[
\int_{\Bbb R^N}[\frac 12f(x,u)u-F(x,u)]\geq \frac c2
\]
contradicting a possible vanishing of $u.$ Then the claim is proved.

We have that $u\in E\subset H_{\rm rad}^1(\Bbb R^N)\ $is a non-zero function
satisfying
\[
\int_{\Bbb R^N}(\nabla u\nabla v+b(x)uv)=\int_{\Bbb R^N}f(x,u)v,\mbox{ for
all }v\in E\,.
\]
As in the proof of Theorem 1, using Lemma 2 we have $u$ is a classical
solution of (\ref{global2}).
\hfill $\diamondsuit $

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\noindent
{\sc Claudianor O. Alves} (coalves@dme.ufpb.br) \newline
{\sc Daniel C. de Morais Filho} (daniel@dme.ufpb.br) \newline
{\sc Marco Aurelio S. Souto} (marco@dme.ufpb.br)\newline

\noindent
Departamento de Matem\'atica e Estat\'\i stica \newline
Centro de Ci\^encias e Tecnologia \newline
Universidade Federal da Para\'\i ba \newline
Caixa Postal 10044 \newline
CEP 58.109-970 - Campina Grande - Para\'\i ba - Brazil\newline


\end{document}