\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small 
\emph{Electronic Journal of Differential Equations},
 Vol. 2011 (2011), No. 98, pp. 1--13.\newline 
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/98\hfil Existence and multiplicity of solutions]
{Existence and multiplicity of solutions for a singular
 semilinear elliptic problem in $\mathbb{R}^2$}

\author[Manass\'es de Souza \hfil EJDE-2011/98\hfilneg]
{Manass\'es de Souza}

\address{Manass\'es de Souza \newline
Departamento de Matem\'atica, Universidade Federal da Para\'iba\\
58.051-900 Jo\~ao Pessoa, PB, Brazil} \email{manasses@mat.ufpb.br}

\thanks{Submitted May 2, 2011. Published August 3, 2011.}
\thanks{Supported by the National Institute of Science
and Technology of Mathematics INCT-Mat.} 
\subjclass[2000]{35J60, 35J20, 35B33} 
\keywords{Variational methods; Trudinger-Moser
inequality; critical points; \hfill\break\indent 
critical exponents}

\begin{abstract}
 Using minimax methods we study the existence and multiplicity of
 nontrivial solutions for a singular class of semilinear elliptic
 nonhomogeneous equation where the potentials can change sign and
 the nonlinearities may be unbounded in $x$ and behaves like
 $\exp(\alpha s^2)$ when $|s|\to+\infty$. We establish the
 existence of two distinct solutions when the perturbation is
 suitable small.
\end{abstract}

\maketitle \numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

In this article, we consider the semilinear elliptic equation
\begin{equation}\label{P}
-\Delta u + V(x) u  = \frac{g(x)f(u)}{|x|^a}+h(x)\quad\text{in
}\mathbb{R}^2,
\end{equation}
where $a\in [0,2)$, the functions $V,g:\mathbb{R}^2\to \mathbb
{R}$ and $f:\mathbb{R}\to\mathbb{R}$ are continuous with $f(0)=0$
and $h\in(H^1(\mathbb{R}^2))^*\equiv H^{-1}$ is a small
perturbation, $h \not\equiv0$. We are interested in finding
nontrivial solutions of \eqref{P} when the nonlinearity $f(s)$ has
the maximal growth which allows to treat \eqref{P}  variationally
in the Sobolev space $H^1(\mathbb{R}^2)$.

On the potentials we assume the  hypothesis
\begin{itemize}
\item[(V1)] There exist $D>0$ such that $V(x)\geq
-D$, for all $x\in\mathbb{R}^2$;
\item[(V2)] $\lambda_{1}=\inf_{u\in E\backslash\{0\}}
\|u\|^2_E/\|u\|_2^2>0$;
\end{itemize}
where $E$ is the following subspace of $H^1(\mathbb{R}^2)$
\[
E=\big\{u\in
H^1(\mathbb{R}^2):\int_{\mathbb{R}^2}V(x)u^2\,\mathrm{d}
x<\infty\big\},
\]
which is a Hilbert space endowed with the scalar product
\[
\langle u,v\rangle_E=\int_{\mathbb{R}^2}[\nabla u\cdot\nabla
v+V(x)uv]\mathrm{d}x
\]
to which corresponds the norm $\|u\|_E=\langle u,u\rangle^{1/2}_E$
(see \cite[Lemma 2.1 and Proposition 3.1]{SIRA00}). Here, as
usual, $H^1(\mathbb{R}^2)$ denotes the Sobolev spaces modelled in
$L^2(\mathbb{R}^2)$ with norm
\[
\| u \|_{1,2}=\Big(\int_{\mathbb{R}^2}(|\nabla
u|^2+|u|^2)\,\mathrm{d} x\Big)^{1/2}.
\]
To ensure the continuous imbedding of $E$ into
$H^1(\mathbb{R}^2)$, we assume the condition (V2) on the first
eigenvalue of the operator $A=-\Delta+V(x)$ (see \cite[Proposition
2.2]{SIRA00}).

We use the following notation: if $\Omega\subset\mathbb {R}^2$ is
open and $s\geq 2$, we set
\[
\nu_{s}(\Omega)=\inf_{u\in
H^1_0(\Omega)\backslash\{0\}}\frac{\int_{\Omega}[|\nabla u|^2
+V(x)u^2]\, \mathrm{d}x}{\big(\int_\Omega|u|^s\,\mathrm{d}
x\big)^{2/s}},
\]
and we put $\nu_s(\emptyset)=\infty$. To obtain a compactness
result, we shall consider the following assumptions:
\begin{itemize}
\item[(V3)] $\lim_{R\to\infty}{\nu_{s}(\mathbb{R}^2\backslash
\overline{B}_R)}=\infty$.

\item[(V4)] There exist a function
$K(x)\in L^{\infty}_{\rm loc}(\mathbb{R}^2)$, with $K(x)\geq 1$,
 and constants
$\alpha>1$, $c_0, R_0>0$ such that
\[
K(x)\leq c_{0}[1+(V^+(x))^{1/\alpha}],
\]
for all $|x|\geq R_{0}$, where
$V^+(x)=\max_{x\in\mathbb{R}^2}\{0,V(x)\}$.
\end{itemize}
It is also well known that assumptions (V3)--(V4) imply that the
imbeddings of $E$ into $L^q(\mathbb{R}^2)$ are compact for all
$2\leq q<\infty$ (see \cite[Proposition 3.1]{SIRA00}).

Concerning the function $g$, we assume that it is strictly
positive and does not have to be bounded in $x$ provided that the
growth of $g$ is controlled by the growth of $V(x)$. More
precisely:
\begin{itemize}
\item[(H1)] There exists $a_0,b_0>0$ such that
$a_0\leq g(x)\leq b_0K(x)$ for all $x\in\mathbb{R}^2$.
\end{itemize}
Moreover, we suppose that $f(s)$ satisfies the following
conditions:
\begin{itemize}
\item[(H2)] $\lim_{s\to 0}\frac{f(s)}{s}=0$.
\item[(H3)] There is a number $\mu>2$ such that for all
 $s\in\mathbb{R}\backslash\{0\}$
\[
0<\mu F(s):= \mu \int_0^sf(t)\mathrm{d}t \leq sf(s).
\]
\end{itemize}
Motivated by Trudinger-Moser inequality (see \cite{MOSE71,TRUD67})
and by pioneer works of Adimurthi \cite{AD90} and de Figueiredo et
al. \cite{FMRU95} we treat the so-called subcritical case, which
we define next. We say that a function $f(s)$ has
\emph{subcritical growth} at $+\infty$ if for all $\beta>0$
\begin{equation}\label{cres-sub}
\lim_{|s|\to +\infty}\frac{|f(s)|}{e^{\beta s^2}}=0.
\end{equation}
Throughout this paper, we denote by $H^{-1}$ the dual space of
$H^1(\mathbb{R}^2)$ with the usual norm $\|\cdot\|_{H^{-1}}$.

Next we state our existence result.

\begin{theorem}\label{teoremaS1}
If $f(s)$ has subcritical growth at $+\infty$ and {\rm (V1)--(V4),
(H1)--(H3)} are satisfied then problem \eqref{P} has a weak
solution with positive energy if $h\equiv0$. Moreover, if
$h\not\equiv0$, there exists $\delta>0$ such that if
$\|h\|_{H^{-1}}<\delta$, problem \eqref{P} has at least two weak
solutions. One of them with positive energy, while the other one
with negative energy.
\end{theorem}

The results in this paper were in part motivated by several recent
papers on elliptic problems involving exponential growth. See for
example  de Souza \cite{DEDO2011} for the singular and homogeneous
case, Giacomoni-Sreenadh \cite{GEA} for the singular and
nonhomogeneous case,  do \'O et al. \cite{JMMU08} and Tonkes
\cite{TONK99} for the nonsingular and nonhomogeneous case, Cao
\cite{CAO92}, de Figueiredo et al. \cite{FMRU95} and  do \'O
\cite{JMAR97} for the nonsingular and homogeneous case. Our paper
is closely related to the recent works of  do \'O et al.
\cite{JMMU08} and  Rabelo \cite{PRabelo}. Indeed, we improve and
complement the results in  do \'O et al. \cite{JMMU08} for the
subcritical case in the sense that we use nonlinearities unbounded
in $x$ and potentials which can change sign. Moreover in
\cite{JMMU08} was studied the existence and multiplicity of weak
solutions of \eqref{P} in terms of the Trudinger-Moser inequality
for the nonsingular case. We point out that ours results are
closely related with results in
\cite{ADYANG2010,DEDO2011,MAN12011,MAN22011,MAN32011}.

The proofs of our existence results rely on minimization methods
in combination with the mountain-pass theorem. In the subcritical
case we are able to prove that the associated functional satisfies
the Palais-Smale compactness condition which allow us to obtain
critical points for the functional. As a consequence we can
distinguish the local minimum solution from the mountain-pass
solution.


\begin{remark} \label{rmk1.1} \rm
The study of such a class of problem has been motivated in part by
the search for standing waves for the nonlinear Schr\"odinger
equation (see for instance \cite{BELI83-I} and \cite{Rabinowitz2})
\[
 i\frac{\partial \psi }{\partial
t}=-\Delta \psi +W(x)\psi -G(|\psi |)\psi-e^{i\lambda t}L(x),\quad
x\in\mathbb{R}^2,
\]
where $\psi=\psi(t,x)$, $\psi:\mathbb{R}\times
\mathbb{R}^2\to\mathbb{C}$, $\lambda$ is a positive constant,
$W:\mathbb{R}^2\to \mathbb{R}$ is a given potential and for
suitable functions $G:\mathbb{R}^+\to \mathbb{R},\;
L:\mathbb{R}^2\to\mathbb{R}$.
\end{remark}

This article is organized as follows. Section \ref{prelim}
contains some preliminary results including a singular
Trudinger-Moser inequality. In Section \ref{geo}, contains the
variational framework and we also check the geometric conditions
of the associated functional. In Section \ref{palais-smale-seq},
we prove some properties of the Palais-Smale sequences. Finally,
in section \ref{proof-main-results} we complete the proofs of our
main results.

\section{Preliminary results}\label{prelim}

Let $\Omega$ be a bounded domain in $\mathbb{R}^2$, we know by the
Trudinger-Moser inequality that for all $\beta>0$ and $u\in
H_0^1(\Omega)$, $e^{\beta u^2}\in L^1(\Omega )$ (see
\cite{MOSE71,TRUD67}). Moreover, there exists a positive constant
$C$ such that
\[
\sup_{u\in H_0^1(\Omega) \; : \; \|\nabla u\|_2\leq
1}\int_{\Omega} e^{\beta
 u^2}\mathrm{d}x\leq C|\Omega| \quad\text{if }\beta \leq 4\pi,
\]
where $|\Omega|$ denotes Lebesgue measure of $\Omega$. This
inequality is optimal, in the sense that for any growth $e^{\beta
u^2}$ with $\beta > 4\pi$ the correspondent supremum is infinite.
 Adimurthi-Sandeep  \cite{ADSA07} proved a singular
Trudinger-Moser inequality, which in the case $N=2$ reads:
\[
\int_{\Omega}\frac{e^{\beta u^2}}{|x|^a} \mathrm{d}x <\infty \quad
\text{ for all } u \in H^1_0(\Omega),\;\beta >0,
\]
where $\Omega$ is a smooth bounded domain in $\mathbb{R}^2$
containing the origin and $a \in [0,2)$. Moreover, there exists a
positive constant $C(\beta,a)$ such that
\begin{equation}\label{adsa-singular}
\sup_{u\in H_0^1(\Omega) : \|\nabla u\|_2\leq
1}\int_{\Omega}\frac{e^{\beta u^2}}{|x|^a} \mathrm{d}x \leq
C(\beta,a) |\Omega|  \quad\text{if and only if}\quad \beta/4\pi +
a/2 \leq 1.
\end{equation}
Here we shall use the following extension of these results for the
whole space $\mathbb{R}^2$ obtained by  Giacomoni and Sreenadh in
\cite{GEA} (see also \cite{DEDO2011}):

\begin{lemma}\label{Trudinger-Moser}
If $\beta >0$, $a\in [0,2)$ and $u\in H^1( {\mathbb{R}}^2)$ then
\begin{equation}\label{TM1}
\int_{\mathbb{R}^2}\frac{(e^{\beta u^2} -1)}{|x|^a}\mathrm{d}x
<\infty.
\end{equation}
Moreover, if $\beta/4\pi + a/2 < 1$ and $\|u\|_2 \leq M$, then
there exists a positive constant $C=C(\beta,M)$ such that
\begin{equation}
\sup_{\|\nabla u\|_{2}\leq 1}\int_{\mathbb{R}^2}\frac{(e^{\beta
u^2} -1)}{|x|^a}\mathrm{d}x \leq C(\beta,M). \label{TM2}
\end{equation}
\end{lemma}

Our choice of the variational setting $E$ ensures that the
imbedding is continuous in $H^1(\mathbb{R}^2)$ and compact in
$L^s(\mathbb{R}^2)$, for $s\geq 2$ (see \cite[Lemma 2.1 and
Proposition 3.1]{SIRA00}). This lemma in \cite{SIRA00} provides a
inequality which will be needed throughout the paper:
\begin{equation}\label{eq.emb}
\|u\|_E^2\geq\zeta\int_{\mathbb{R}^2}|\nabla u|^{2}\,\mathrm{d}x,
\end{equation}
for some $\zeta>0$ and for all $u\in E$.

\begin{lemma}\label{estim-import}
Let $\beta>0$ and $r\geq1$. Then for each $\theta>r$ there exists
a positive constant $C=C(\theta)$ such that for all $s\in
\mathbb{R}$
\[
(e^{\beta s^2}-1)^r\leq C(e^{\theta\beta s^2}-1).
\]
In particular, for $r\in[1,\alpha)$, we have that
$K(x)^r\dfrac{(e^{\beta u^2}-1)^r}{|x|^a}$ belongs to
$L^1(\mathbb{R}^2)$ for all $u\in H^1(\mathbb{R}^2)$.
\end{lemma}

\begin{proof}
The proof of the  inequality above is a consequence of L'Hospital
Rule (see \cite[Lemma 2.2]{JMMU08} for a proof). Now, as $K(x)\in
L^\infty_{\rm loc}(\mathbb{R}^2)$, for $R>1$ we have that
%
\begin{align*}
&\int_{\mathbb{R}^2}K(x)^r\frac{(e^{\beta u^2}-1)^r}{|x|^a}\,
 \mathrm{d}x\\
&\leq  C_1\int_{|x|\leq R}\frac{(e^{\beta
u^2}-1)^r}{|x|^a}\,\mathrm{d}x
 +\int_{|x|>R}K(x)^r(e^{\beta u^2}-1)^r\,\mathrm{d}x\\
&\leq  C_2\int_{|x|\leq R}\frac{(e^{\theta\beta
u^2}-1)}{|x|^a}\,\mathrm{d}x+C_3\int_{|x|>R}K(x)^r(e^{\theta\beta
u^2}-1)\,\mathrm{d}x.
\end{align*}
From Lemma \ref{Trudinger-Moser} it follows that the first term is
integrable. To estimate the other term, we note that
\begin{align*}
\int_{|x|>R}K(x)^r(e^{\theta\beta u^2}-1)\,\mathrm{d}x
=\sum_{m=1}^{\infty}\frac{(\theta\beta)^m}{m!} \int_{|x|>R}
K(x)^r|u|^{2m}\,\mathrm{d}x.
\end{align*}
By (V4) and H\"older inequality, we have
\begin{align*}
&\int_{\mathbb{R}^2}K(x)^r|u|^{2m}\,\mathrm{d}x\\
&\leq  C_4\|u\|^{2m}_{2m}+C_5\int_{|x|>R_0}\big(V^+(x)\big)
 ^{r/\alpha}|u|^{2m}\,\mathrm{d}x\\
&\leq
C_4\|u\|^{2m}_{2m}+C_5\Big[\int_{|x|>R_0}V^+(x)|u|^2\,\mathrm{d}
x\Big]^{r/\alpha} \Big[\int_{|x|>R_0}|u|
^{2(m\alpha-r)/(\alpha-r)} \Big]^{(\alpha-r)/\alpha}.
\end{align*}
By (V2) and the continuous imbedding $E\hookrightarrow
L^s(\mathbb{R}^2)$, for all $s\geq 2$, we can conclude that
\[
\int_{\mathbb{R}^2}K(x)^r|u|^{2m}\,\mathrm{d}x \leq
C\|u\|^{2m}_E.
\]
Thus, we obtain
\begin{equation}\label{6001}
\begin{aligned}
&\int_{\mathbb{R}^2}K(x)^r\frac{(e^{\beta u^2}-1)^r}{|x|^a}
 \,\mathrm{d}x \\
& \leq  C_1\int_{|x|\leq R}\frac{(e^{\theta\beta
u^2}-1)}{|x|^a}\,\mathrm{d}x+
C\sum_{m=1}^{\infty}\frac{1}{m!}\left(\theta\beta\|u\|_E^2\right)^m\\
& \leq  C_1\int_{|x|\leq R}\frac{(e^{\theta\beta
u^2}-1)}{|x|^a}\,\mathrm{d}x+
C\left[\exp\left(\theta\beta\|u\|_E^2\right)-1\right]<\infty,
\end{aligned}
\end{equation}
which completes the proof.
\end{proof}


\begin{corollary}\label{desigualdade importante}
If $v\in E$, $\beta>0$, $q>0$ and $\|v\|_E\leq M$ with
$\frac{\beta M^2}{4\pi\zeta} + \frac{a}{2} <1$, then there exists
$C=C(\beta,M,q,\zeta)>0$ such that
\[
\int_{\mathbb{R}^2}K(x)|v|^q\frac{(e^{\beta
v^2}-1)}{|x|^a}\,\mathrm{d} x\leq C\|v\|_E^q.
\]
\end{corollary}

\begin{proof}
By H\"older inequality,
%
\begin{equation} \label{1001}
\int_{\mathbb{R}^2}K(x)|v|^{q}\frac{(e^{\beta
v^2}-1)}{|x|^a}\,\mathrm{d} x
\leq\|v\|^q_{qs}\Big[\int_{\mathbb{R}^2}K(x)^r \frac{(e^{\beta
v^2}-1)^r}{|x|^{ar}}\mathrm{d}x\Big]^{1/r},
\end{equation}
where $r>1$ is close to $1$ and $s=r/(r-1)$. Now, we consider
$\theta>r$ close to $r$ such that $\frac{\theta\beta
M^2}{4\pi\zeta}+\frac{ar}{2} <1$. By $(\ref{6001})$ and Lemma
\ref{Trudinger-Moser}, we have that
%
\begin{equation}\label{1002}
\begin{aligned}
&\int_{\mathbb{R}^2}K(x)  |v|^q\frac{(e^{\beta v^2}-1)}{|x|^a}\,\mathrm{d}x\\
&\leq  \Big\{C_1\int_{|x|\leq R}\frac{[e^{\frac{\theta\beta
M^2}{\zeta}\left(\frac{v}{\|\nabla v\|_2}\right)^2}-1]}{|x|^{ar}}
\,\mathrm{d}x+C_2\left[\exp\left(\theta\beta M^2\right)-1\right]\Big\}^{1/r}\|v\|^q_{qs}\\
&\leq  C_3\|v\|^q_E.
\end{aligned}
\end{equation}
\end{proof}


To show that the weak limit of a Palais-Smale sequence in $E$ is a
weak solution of \eqref{P} we will use the following convergence
result, which is a version of Lemma~2.1 in \cite{FMRU95}.


\begin{lemma}\label{Djairo}
Let $\Omega\subset\mathbb{R}^2$ be a bounded domain and $f:
\mathbb{R}\to\mathbb{R}$ a continuous function. Then for any
sequence $(u_n)$ in $L^1(\Omega)$ such that $u_n\to u$ in
$L^1(\Omega)$,
\[
\frac{g(x)f(u_n)}{|x|^a}\in L^1(\Omega)\quad \text{and}\quad
\int_\Omega\frac{g(x)|f(u_n)u_n|}{|x|^a}\,\mathrm{d} x\leq C_1,
\]
up to a subsequence we have
\[
\frac{g(x)f(u_n)}{|x|^a}\to \frac{g(x)f(u)}{|x|^a}\quad \text{in }
L^1(\Omega).
\]
\end{lemma}

\begin{proof}
It suffices to prove
\[
\int_{\Omega}\frac{|g(x)f(u_n)|}{|x|^a}\,\mathrm{d}x \to
\int_{\Omega}\frac{|g(x)f(u)|}{|x|^a} \,\mathrm{d}x.
\]
Since $u, {g(x)f(u)}/{|x|^a}\in L^1(\Omega)$, for each $\epsilon
>0$ there is a $\delta >0$ such that for any measurable subset
$A\subset \Omega$,
\begin{equation}\label{dj214}
\int_{A}|u| \,\mathrm{d}x <\epsilon  \quad \text{and}\quad
\int_{A}\frac{|g(x)f(u)|}{|x|^a} \, \mathrm{d}x < \epsilon
\quad\text{if } |A|\leq \delta.
\end{equation}
Next using the fact that $u \in L^1(\Omega)$ we find $M_1 > 0$
such that
\begin{equation}\label{dj215}
|\{x \in \Omega:|u(x)| \geq M_1\}|\leq \delta.
\end{equation}
Let $M=\max \{M_1, C_1/\epsilon\}$. We write
\[
\Big|\int_{\Omega}\frac{|g(x)f(u_n)|}{|x|^a}\,\mathrm{d}x
 - \int_{\Omega}\frac{|g(x)f(u)|}{|x|^a} \,\mathrm{d}x\Big| \leq I_{1,n}
+ I_{2,n} + I_{3,n},
\]
where
\begin{gather*}
I_{1,n} = \int_{[|u_n| \geq M]}\frac{|g(x)f(u_n)|}{|x|^a}\,\mathrm{d}x,\\
I_{2,n} = \Big|\int_{[|u_n| <
M]}\frac{|g(x)f(u_n)|}{|x|^a}\,\mathrm{d}
x - \int_{[|u| < M]}\frac{|g(x)f(u)|}{|x|^a}\,\mathrm{d}x\Big| ,\\
I_{3,n} = \int_{[|u| \geq
M]}\frac{|g(x)f(u)|}{|x|^a}\,\mathrm{d}x.
\end{gather*}
Now we estimate each integral separately.
\[
I_{1,n} = \int_{[|u_n| \geq
M]}\frac{|g(x)f(u_n)|}{|x|^a}\,\mathrm{d}x = \int_{[|u_n| \geq
M]}\frac{|g(x)f(u_n)u_n|}{|u_n||x|^a}\,\mathrm{d}x \leq
\frac{C_1}{M}\leq \epsilon.
\]
From \eqref{dj214} and \eqref{dj215}, we have $I_{3,n} \leq
\epsilon$.

Next we claim $I_{2,n} \to 0$ as $n\to +\infty$. Indeed,
\begin{align*}
I_{2,n} &\leq \Big|\int_{\Omega}\frac{\mathcal{X}_{[|u_n| <
M]}(|g(x)f(u_n)|
 - |g(x)f(u)|)}{|x|^a}\,\mathrm{d}x \Big| \\
&\quad + \Big|\int_{\Omega}\frac{(\mathcal{X}_{[|u_n| <M]}-
\mathcal{X}_{[|u|<M]})|g(x)f(u)|}{|x|^a}\,\mathrm{d}x\Big|
\end{align*}
and $g_n(x)=\mathcal{X}_{[|u_n| < M]}(|g(x)f(u_n)|
 - |g(x)f(u)|) \to 0$ almost everywhere in $\Omega$.
Moreover,
\[
|g_n(x)| \leq \begin{cases}
  |g(x)f(u)| &\text{if } |u_n(x)|\geq M,\\
 C + |g(x)f(u)| &\text{if }|u_n(x)|<M,
\end{cases}
\]
where $C=\sup\{|g(x)f(t)|: (x,t) \in
\overline{\Omega}\times[-M,M]\}$. So, by the Lebesgue dominated
convergence theorem, we obtain
\[
\Big|\int_{\Omega}\frac{\mathcal{X}_{[|u_n| < M]}(|g(x)f(u_n)|
 - |g(x)f(u)|)}{|x|^a}\,\mathrm{d}x \Big|
  \to 0 \quad\text{as } n \to \infty.
\]
Moreover,
\[
\{x\in \Omega: |u_n(x)| <M\} \setminus \{x\in \Omega: |u(x)|<M\}
\subset \{x\in \Omega:|u(x)| \geq M\}.
\]
Hence  by \eqref{dj214},
\[
\Big|\int_{\Omega}\frac{(\mathcal{X}_{[|u_n| <M]}-
\mathcal{X}_{[|u|<M]})|g(x)f(u)|}{|x|^a}\,\mathrm{d}x\Big| \leq
\int_{[|u|\geq M]} \frac{|g(x)f(u)|}{|x|^a}\,\mathrm{d}x <
\epsilon,
\]
which completes the proof.
\end{proof}

\section{The variational framework}\label{geo}

We now consider the functional $I$ given by
\begin{equation}\label{funcional}
I(u)=\dfrac{1}{2}\int_{\mathbb{R}^2}\left[|\nabla
u|^2+V(x)u^2\right]\mathrm{d}x
-\int_{\mathbb{R}^2}\frac{g(x)F(u)}{|x|^a}\mathrm{d}x
-\int_{\mathbb{R}^2}h(x)u\mathrm{d}x.
\end{equation}
Under our assumptions we have that $I$ is well-defined and is
$C^1$ on $E$. Indeed, by $(H_2)$, given $\varepsilon>0$ there
exists $\delta>0$ such that $|f(s)|\leq\varepsilon |s|$ always
that $|s|<\delta$. On the other hand, for $\beta>0$ we have that
there exists $C>0$ such that $|f(s)|\leq C(e^{\beta s^2}-1)$ for
all $s\geq\delta$. Thus
\begin{equation}\label{eq.bd1}
|f(s)|\leq\varepsilon|s|+C_1(e^{\beta s^2}-1),
\end{equation}
for all $s\in\mathbb{R}$. By (H1), (H3), (V4) and H\"older
inequality, we obtain
\begin{align*}
\int_{\mathbb{R}^2}\frac{g(x)F(u)}{|x|^a}\,\mathrm{d}x&\leq
\varepsilon\int_{\mathbb{R}^2}\frac{K(x)u^2}{|x|^a}\,\mathrm{d}x
+C_2\int_{\mathbb{R}^2}\frac{K(x)|u|
 (e^{\beta u^2}-1)}{|x|^a}\,\mathrm{d}x\\
&\leq  C_1 \int_{|x|\leq 1}\frac{u^2}{|x|^a}\,\mathrm{d}x +
\varepsilon\int_{|x|>1}K(x)u^2\,\mathrm{d}x\\
&\quad +C_2\|u\|_s \Big[\int_{\mathbb{R}^2}K(x)^r\frac{(e^{\beta
u^2}-1)^r}{|x|^{ar}}\,\mathrm{d}x\Big]^{1/r},
\end{align*}
%
where $r\in[1,\alpha)$ and $s=r/(r-1)$, with $ar<2$. Considering
the continuous imbedding $E\hookrightarrow
L^s_{K(x)}(\mathbb{R}^2)$ for $s\geq 2$, $a\in [0,2)$ and Lemma
\ref{estim-import}, it follows that ${g(x)F(u)}/{|x|^a}\in
L^1(\mathbb{R}^2)$ which implies that $I$ is well defined.

Next, we show that $I$ is in $C^1$ on $E$. Indeed, letting $
N(u)=\int_{\mathbb{R}^2}{g(x)F(u)}/{|x|^a}\,\mathrm{d}x$, we have
by dominated convergence theorem that
\begin{align*}
\langle I'(u),\phi\rangle &= \langle u,\phi\rangle_E-\lim_{t\to
0}\frac{1}{t}
[N(u+t\phi)-N(u)]-\int_{\mathbb{R}^2}h(x)\phi\,\mathrm{d}x\\
&= \langle u,\phi\rangle_E-\int_{\mathbb{R}^2}
 \frac{g(x)f(u)\phi}{|x|^a}\,\mathrm{d}x
 -\int_{\mathbb{R}^2}h(x)\phi\,\mathrm{d}x,
\end{align*}
%
for all $\phi\in E$. As $I'(u)$ is linear and bounded, it suffices
to show that the Gateaux derivative of $I$ is continuous. It is
clear that the first and last term are $C^1$. Hence, it remains to
prove that $N$ is $C^1$. Let $u_n\to u$ in $E$. By Proposition~2.7
in \cite{JMMU08}, there exists a subsequence $(u_{n_k})$ in $E$
and $\ell(x)\in H^1(\mathbb{R}^2)$ such that $u_{n_k}(x)\to u(x)$
and $|u_{n_k}(x)|\leq \ell(x)$ almost everywhere in
$\mathbb{R}^2$. Given $\xi\in E$, we define
\[
H_{n_k}(x)=\frac{g(x)f(u_{n_k}(x))\xi(x)}{|x|^a}.
\]
Then
\[
H_{n_k}(x)\to H(x)=\frac{g(x)f(u(x))\xi(x)}{|x|^a}\quad
\text{almost everywhere in } \mathbb{R}^2.
\]
Using $(\ref{eq.bd1})$ and Lemma \ref{Trudinger-Moser}, we obtain
that $H_{n_k}(x)$ is integrable, it follows by dominated
convergence theorem that
\[
\lim_{k\to\infty}\int_{\mathbb{R}^2}H_{n_k}(x)\,\mathrm{d}
x=\int_{\mathbb{R}^2}H(x)\,\mathrm{d}x.
\]
Thus, for each $\xi\in E$ with $\|\xi\|_E=1$, we obtain
\begin{align*}
\lim_{k\to\infty}\|N'(u_{n_k})-N'(u)\|_{E^*} &=
\lim_{k\to\infty}\sup_{\|\xi\|_E=1}|
 \langle N'(u_{n_k})-N'(u),\xi\rangle|\\
&=
\sup_{\|\xi\|_E=1}\lim_{k\to\infty}\int_{\mathbb{R}^2}\frac{g(x)
[f(u_{n_k})-f(u)]\xi}{|x|^a}\,\mathrm{d} x = 0
\end{align*}
and the proof is complete.

The geometric conditions of the mountain-pass theorem for the
functional $I$ are established by our next two lemmas.

\begin{lemma}\label{lema31}
Suppose that (V1)-(V2), (V4), (H1)-(H3) and \eqref{cres-sub} are
satisfied. Then there exists $\delta>0$ such that for each $h\in
H^1(\mathbb{R}^2)$ with $\|h\|_{H^{-1}}<\delta$, there exists
$\rho_h>0$ such that
\[
I(u)>0\quad whenever\quad\|u\|_E=\rho_h.
\]
\end{lemma}

\begin{proof}
In the same manner that \eqref{eq.bd1} was obtained, we can see
that
\begin{equation} \label{555}
|f(s)|\leq\varepsilon|s|+C_1|s|^q(e^{\beta s^2}-1),
\end{equation}
with $q>2$. Thus, considering the continuous imbedding
$E\hookrightarrow L^s_{K(x)}(\mathbb{R}^2)$ for $s\geq 2$ (see
\cite[Proposition 3.1]{SIRA00}), we obtain for $\varepsilon>0$
sufficiently small
\begin{align*}
I(u)& \geq\frac{1}{2}\|u\|^2_E -
\varepsilon\int_{\mathbb{R}^2}\frac{K(x)u^2}{|x|^a}\,\mathrm{d}x-
C_1\int_{\mathbb{R}^2}\frac{K(x)|u|^{q+1}(e^{\beta u^2}-1)}{|x|^a}
 \,\mathrm{d}x-\int_{\mathbb{R}^2}h(x)u\,\mathrm{d}x\\
& \geq \big(\frac{1}{2}-\varepsilon\big)\|u\|^2
_E-C_1\int_{\mathbb{R}^2}\frac{K(x)|u|^{q+1}(e^{\beta
u^2}-1)}{|x|^a}\,\mathrm{d}x
-\int_{\mathbb{R}^2}h(x)u\,\mathrm{d}x
\end{align*}
and since $\frac{\beta\sigma^2}{4\pi\zeta}+\frac{a}{2}<1$ if
$\|u\|_E<\sigma$ is sufficiently small, we can apply Corollary
\ref{desigualdade importante} to conclude that
\[
I(u)\geq \big(\frac{1}{2}-\varepsilon\big)\|u\|^2_E
-C\|u\|_E^{q+1}-\|h\|_{H^{-1}}\|u\|_E.
\]
Thus there exists $\rho_h>0$ such that $I(u)>0$ whenever
$\|u\|_E=\rho_h$ and $\|h\|_{H^{-1}}$ is sufficiently small.
Indeed, for $\varepsilon >0$ sufficiently small and $q>2$, we may
choose $\rho_h >0$ such that
\[
\big(\frac{1}{2}-\varepsilon\big)\rho_h-C_1\rho_h^{q}>0.
\]
Thus, for $\|h\|_{H^{-1}}$ sufficiently small there exists
$\rho_h>0$ such that $I(u)>0$ if $\|u\|_E=\rho_h$.
\end{proof}

\begin{lemma}
Assume that {\rm (H1), (H3)} and \eqref{cres-sub} are satisfied.
Then there exists $e\in E$ with $\|e\|_E>\rho_h$ such that
\[
I(e)<\inf_{\|u\|=\rho_h}I(u).
\]
\end{lemma}

\begin{proof}
Let $u\in E\backslash\{0\}$ with compact support and $u\geq 0$.
Integrating (H3) we obtain that there exist $c,d>0$ such that
\[
F(s)\geq cs^\mu-d
\]
for all $s\in\mathbb{R}$. Thus, denoting $K=supp(u)$ and using
(H1), we have that
\begin{align*}
I(tu) &\leq
\frac{t^2}{2}\|u\|^2_E-ct^\mu\int_{K}\frac{g(x)u^\mu}{|x|^a}
 \,\mathrm{d}x
+d\int_{K}\frac{g(x)}{|x|^a}\,\mathrm{d}x-t\int_{\mathbb{R}^2}h(x)u\,
 \mathrm{d}x\\
&\leq
\frac{t^2}{2}\|u\|^2_E-C_1t^\mu\int_{K}\frac{u^\mu}{|x|^a}\,\mathrm{d}x
 +C_2(|K|)-t\int_{\mathbb{R}^2}h(x)u\,\mathrm{d}x,
\end{align*}
for all $t>0$, which implies that $I(tu)\to-\infty$ as
$t\to\infty$. Setting $e=tu$ with $t$ large enough, the proof is
complete.
\end{proof}

To find an appropriate ball to use a minimization argument we need
the following result.

\begin{lemma}
If $f(s)$ satisfies $(\ref{cres-sub})$ and $h\neq0$, there exist
$\eta>0$ and $v\in E$ with $\|v\|_E=1$ such that $I(tv)<0$ for all
$0<t<\eta$. In particular,
\[
\inf_{\|u\|\leq\eta}I(u)<0.
\]
\end{lemma}

\begin{proof}
For each $h\in H^{-1}$, by applying the Riesz representation
theorem in the space $E$, the problem
\[
-\Delta v+V(x)v=h, \quad x\in\mathbb{R}^2
\]
has a unique weak solution  $v$ in $E$. Thus,
\[
\int_{\mathbb{R}^2} h(x)v\,\mathrm{d}x=\|v\|^2_E>0\quad\text{for
each } h\neq 0.
\]
Since $f(0)=0$, by continuity, it follows that there exists
$\eta>0$ such that
\[
\frac{\mathrm{d}}{\mathrm{d}t}I(tv)=t\|v\|^2_E-\int_{\mathbb{R}^2}
\frac{g(x)f(tv)v}{|x|^a}\,\mathrm{d}x-\int_{\mathbb{R}^2}
h(x)v\,\mathrm{d} x<0,
\]
for all $0<t<\eta$. Using that $I(0)=0$, it must hold that
$I(tv)<0$ for all $0<t<\eta$.
\end{proof}

\section{Palais-Smale sequences}\label{palais-smale-seq}

To prove that a Palais-Smale sequence converges to a solution of
problem \eqref{P} we need to establish the following lemma.

\begin{lemma}\label{solucao PM}
Assume {\rm (H3)} and that $f(s)$ satisfies \eqref{cres-sub}.
 Let $(u_n)$ in $E$ such that $I(u_n)\to c$
and $I'(u_n)\to 0$. Then $\|u_n\|_E\leq C$,
\[
\int_{\mathbb{R}^2}\frac{g(x)|f(u_n)u_n|}{|x|^a}\,\mathrm{d}x\leq
C\quad \text{and}\quad
\int_{\mathbb{R}^2}\frac{g(x)F(u_n)}{|x|^a}\,\mathrm{d} x\leq C.
\]
\end{lemma}

\begin{proof}
Let $(u_n)\subset E$ be a sequence such that $I(u_n)\to c$ and
$I'(u_n)\to 0$, that is, for any $\varphi\in E$,
\begin{equation} \label{026}
\frac{1}{2}\|u_n\|^2_E-\int_{\mathbb{R}^2}\frac{g(x)F(u_n)}{|x|^a}\,\mathrm{d}
x-\int_{\mathbb{R}^2}h(x)u_n\,\mathrm{d}x=c+\delta_n
\end{equation}
and
\begin{equation} \label{027}
\Big|\int_{\mathbb{R}^2}[\nabla
u_n\nabla\varphi+V(x)u_n\varphi]\,\mathrm{d}x
-\int_{\mathbb{R}^2}\frac{g(x)f(u_n)\varphi}{|x|^a}\,\mathrm{d}
x-\int_{\mathbb{R}^2}h(x)\varphi\,\mathrm{d}
x\Big|\leq\varepsilon_n\|\varphi\|_E,
\end{equation}
where $\delta_n\to 0$ and $\varepsilon_n\to 0$ as $n\to\infty$.
Taking $\varphi=u_n$ in \eqref{027} and using (H3), we have
\begin{align*}
&\mu(c+\delta_n)+\varepsilon_n\|u_n\|_E
+(\mu-1)\int_{\mathbb{R}^2}h(x)u_n\,\mathrm{d}x\\
&\geq (\frac{\mu}{2}-1)\|u_n\|^2_E
 -\int_{\mathbb{R}^2}\frac{g(x)[\mu F(u_n)-f(u_n)u_n]}{|x|^a}\,\mathrm{d}x\\
&\geq  (\frac{\mu}{2}-1)\|u_n\|^2_E.
\end{align*}
Consequently, $\|u_n\|_E\leq C$ and by \eqref{026} and
\eqref{027}, we obtain
\[
\int_{\mathbb{R}^2}\frac{g(x)F(u_n)}{|x|^a}\,\mathrm{d} x\leq C
\quad\text{and}\quad
\int_{\mathbb{R}^2}\frac{g(x)|f(u_n)u_n|}{|x|^a}\,\mathrm{d} x\leq
C.
\]
\end{proof}

\begin{corollary}\label{codaexistencia}
Let $(u_n)$ a Palais-Smale sequence for $I$. Then $(u_n)$ has a
subsequence, still denoted by $(u_n)$, which is weakly convergent
to a weak solution of \eqref{P}.
\end{corollary}


\begin{proof}
Using Lemma~\ref{solucao PM}, up to a subsequence, we can assume
that $u_n\rightharpoonup u$ weakly in $E$. Now, from \eqref{027},
taking the limit and using Lemma \ref{Djairo}, we have
\[
\int_{\mathbb{R}^2} (\nabla u \nabla \varphi+V(x)u
\varphi)\mathrm{d}x - \int_{\mathbb{R}^2}
\frac{g(x)f(u)}{|x|^a}\varphi \mathrm{d}x -\int_{\mathbb{R}^2}
h(x)\varphi \mathrm{d}x =0,
\]
for all $\varphi \in C^\infty_0(\mathbb{R}^2)$. Since
$C^\infty_0(\mathbb{R}^2)$ is dense in $E$, we conclude that $u$
is a weak solution of \eqref{P}.
\end{proof}

\section{Proof of Theorem 1.1}\label{proof-main-results}

Let $(u_n)$ in $E$ such that $I(u_n)\to c$ and $I'(u_n)\to 0$. We
will use the Mountain-Pass Theorem to obtain a nontrivial solution
of  \eqref{P}. Since
\[
\|u_n-u\|^2_E=\langle
I'(u_n)-I'(u),u_n-u\rangle+\int_{\mathbb{R}^2}\frac{g(x)
[f(u_n)-f(u)]}{|x|^a}(u_n-u)\,\mathrm{d}x,
\]
we have that the Palais-Smale condition is satisfied if
\[
\lim_{n\to\infty}\int_{\mathbb{R}^2}\frac{g(x)
[f(u_n)-f(u)]}{|x|^a}(u_n-u)\,\mathrm{d}x=0.
\]
By \eqref{eq.bd1} and H\"older inequality, we conclude that
\begin{align*}
&\int_{\mathbb{R}^2}\frac{g(x) [f(u_n)-f(u)]}{|x|^a}
 |u_n-u|\,\mathrm{d}x\\
&\leq  C_1\int_{\mathbb{R}^2}K(x)|x|^{-a}(|u_n|+|u|)|u_n-u|\,\mathrm{d}x\\
&\quad + C_2\int_{\mathbb{R}^2}K(x) \big[\frac{(e^{\beta
u_n^2}-1)}{|x|^a}
 +\frac{(e^{\beta u^2}-1)}{|x|^a}\big]|u_n-u|\,\mathrm{d}x\\
&\leq  C_1\int_{\mathbb{R}^2}K(x)|x|^{-a}(|u_n|+|u|)|u_n-u|\,\mathrm{d}x\\
&\quad + C_2\|u_n-u\|_s\Big\{\int_{\mathbb{R}^2}K(x)^r
\big[\frac{(e^{\beta u_n^2}-1)^r}{|x|^{ar}}
 +\frac{(e^{\beta u^2}-1)^r}{|x|^{ar}}\big]\,\mathrm{d}x\Big\}^{1/r},
\end{align*}
with $r>1$ close to 1 such that $ar<2$ and $s=r/(r-1)$. Since
$f(s)$ has subcritical growth and $E\hookrightarrow
L^s(\mathbb{R}^2)$ is compact for $s\geq 2$, the second term
converges to zero.

Now, to estimate the other term we will use H\"older inequality,
Young inequality and that $\|u_n\|_E \leq C$, thus we obtain
\begin{equation}\label{14001}
\begin{aligned}
&\int_{\mathbb{R}^2}K(x)|x|^{-a}(|u_n|+|u|)|u_n-u|\,\mathrm{d}x\\
&\leq
\sqrt{2}\Big(\int_{\mathbb{R}^2}\frac{K(x)|u_n|^2}{|x|^a}\,\mathrm{d}
x+\int_{\mathbb{R}^2}\frac{K(x)|u|^2}{|x|^a}\,\mathrm{d}
x\Big)^{1/2}\Big(\int_{\mathbb{R}^2}\frac{K(x)|u_n-u|^2}{|x|^a}\,\mathrm{d}
x\Big)^{1/2}\\
&\leq  C_1\left\{C_2 \|u_n-u\|_s^2 + \int_{\mathbb
{R}^2}K(x)|u_n-u|^{2}\,\mathrm{d}x \right\}^{1/2}.
\end{aligned}
\end{equation}
Using (V4), we have
\begin{equation}\label{1400}
\begin{aligned}
&\int_{\mathbb {R}^2}K(x)|u_n-u|^{2}\,\mathrm{d}x\\
&=\int_{|x|\leq R_0}K(x)|u_n-u|^{2}\,\mathrm{d}x
 +\int_{|x|>R_{0}}K(x)|u_n-u|^{2}\,\mathrm{d}x\\
&\leq \max_{|x|\leq R_{0}}\{K(x)\}\int_{|x|
 \leq R_{0}}|u_n-u|^{2}\,\mathrm{d}x\\
&\quad +\int_{|x|>R_0}c_{0}[1+(V^+(x))^{1/\alpha}]
|u_n-u|^{2}\,\mathrm{d}x\\
&\leq C\big\{\|u_n-u\|^2_{2}+
\int_{|x|>R_0}V^+(x)^{1/\alpha}|u_n-u|^2\,\mathrm{d}x\big\}.
\end{aligned}
\end{equation}
By H\"older inequality, we obtain
\begin{equation}\label{1500}
\begin{aligned}
& \int_{|x|>R_0}V^+(x)^{1/\alpha} |u_n-u|^2\,\mathrm{d}x\\
& \leq \Big[\int_{|x|>R_0}V^+(x)|u_n-u|^{2}\,\mathrm{d}
x\Big]^{1/\alpha}\Big[\int_{|x|>R_0}|u_n-u|^{(2\alpha-2)/(\alpha-1)}
\,\mathrm{d}x\Big]^{(\alpha-1)/\alpha}
\end{aligned}
\end{equation}
and by (V1), we have
\begin{equation}\label{1600}
\begin{aligned}
&\int_{|x|>R_0}V^+(x)|u_n-u|^{2}\,\mathrm{d}x\\
&= \int_{\mathbb{R}^2}V(x)|u_n-u|^{2}\,\mathrm{d}x-\int_{|x|\leq
R_0}
V(x)|u_n-u|^{2}\,\mathrm{d}x-\int_{|x|>R_0}V^{-}(x)|u_n-u|^2\,\mathrm{d}x\\
&\leq\int_{\mathbb {R}^2}\left[|\nabla (u_n-u)|^2+
V(x)|u_n-u|^2\right]\mathrm{d}x.
\end{aligned}
\end{equation}
From \eqref{1500}, \eqref{1600} in \eqref{1400} and using (V3), we
obtain
\begin{equation}\label{1700}
\begin{aligned}
&\int_{\mathbb {R}^2}K(x)|u_n-u|^{2}\,\mathrm{d}x\\
&\leq C\big\{\|u_n-u\|^2_{2}+\left(\|u_n-u\|^2_E
 +D\|u_n-u\|^2_{2}\right)^{1/\alpha}\|u_n-u\|^{(2\alpha-2)
 /\alpha}_{(2\alpha-2)/(\alpha-1)}\big\}\\
&\leq C\big\{\|u_n-u\|^2_{2}
+\big(1+\frac{D}{\lambda_1}\big)^{1/\alpha}\|u_n-u\|^{2/\alpha}_E
 \|u_n-u\|^{(2\alpha-2)/\alpha}_{(2\alpha-2)/(\alpha-1)}\big\}.
\end{aligned}
\end{equation}
Thus, by \eqref{14001},
\begin{align*}
&\int_{\mathbb{R}^2}K(x)|x|^{-a}(|u_n|+|u|)|u_n-u|\,\mathrm{d}x\\
&\leq   C_1\big\{C_2\|u_n-u\|^2_s +
\|u_n-u\|^2_2+C_3\|u_n-u\|^{2/\alpha}_E\|u_n-u\|_2^{2(\alpha-1)
 /\alpha}\big\}^{1/2}.
\end{align*}
By  compact embedding of $E$ in $L^s(\mathbb{R}^2)$ for any
$s\geq2$, we obtain
\[
\int_{\mathbb{R}^2}K(x)|x|^{-a}(|u_n|+|u|)|u_n-u|\,\mathrm{d} x\to
0\quad\text{as } n\to+\infty.
\]
Hence the Palais-Smale condition is satisfied. Therefore, the
functional $I$ has a critical point $u_M$ at minimax level
\begin{gather*}
 c_M=\inf_{\gamma\in \Gamma}\max_{t\in[0,1]}I(\gamma(t))>0, \\
\Gamma=\{\gamma\in C(E,\mathbb{R}) : \gamma(0)=0,\; \gamma(1)=e\}.
\end{gather*}
On the other hand, if $h\not\equiv 0$, then we obtain a second
solution of \eqref{P} with negative energy. Indeed, let $\rho_h$
be as in Lemma~\ref{lema31}. Since $\overline{B}_{\rho_h}$ is a
complete metric space with the metric given by norm of $E$, convex
and the functional $I$ is of class $C^1$ and bounded below on
$\overline{B}_{\rho_h}$, it follows by Ekeland variational
principle that there exists a sequence $(u_n)$ in
$\overline{B}_{\rho_h}$ such that
\begin{equation} \label{4141}
I(u_n)\to c_0=\inf_{\|u\|_E\leq\rho_h}I(u)\quad \text{and}\quad
\|I'(u_n)\|_{E'}\to 0.
\end{equation}
We now apply the argument above again to conclude that \eqref{P}
possesses a solution $u_0$ such that $I(u_0)=c_0<0$.


\subsection*{Acknowledgments}
 The author would like to thank the anonymous referees for their
valuable comments and suggestions which improved this article.

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\end{document}