\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx,epic}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 119, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/119\hfil
 Existence of infinitely many sign-changing solutions]
{Existence of infinitely many sign-changing solutions for
 elliptic problems with critical exponential growth}

\author[D. S. Pereira \hfil EJDE-2015/119\hfilneg]
{Denilson S. Pereira}

\address{Denilson S. Pereira \newline
 Universidade Federal de Campina Grande\\
 Unidade Acad\^emica de Matem\'atica - UAMat\\
 CEP: 58.429-900 - Campina Grande - PB,  Brazil}
 \email{denilsonsp@dme.ufcg.edu.br}

\thanks{Submitted January 29, 2015. Published April 30, 2015.}
\subjclass[2010]{35A15, 35J15}
\keywords{Variational method; critical exponential growth;
\hfill\break\indent sign-changing solution}

\begin{abstract}
 In this work we prove the existence of infinitely many nonradial solutions,
 that change sign, to the problem
 \begin{gather*}
 -\Delta u=f(u)\quad\text{in }B\\
 u=0\quad\text{on }\partial B,
 \end{gather*}
 where $B$ is the unit ball in $\mathbb{R}^2$ and $f$
 is a continuous and odd function with critical exponential growth.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{claim}[theorem]{Claim}
\allowdisplaybreaks

\section{Introduction}

Let $\Omega\subset\mathbb{R}^N$ be a bounded domain with smooth boundary and
$f:\mathbb{R}\to\mathbb{R}$ be a $C^1$ function with $f(-t)=-f(t)$. Consider the problem
\begin{equation} \label{eP}
\begin{gathered}
-\Delta u=f(u),\quad  \text{in } \Omega, \\
\mathcal{B}u=0,\quad \text{on } \partial \Omega,
\end{gathered}
\end{equation}
when $N\geq 4$, $\mathcal{B}u=u$ and $f(t)=|t|^{\frac{4}{N-2}}+\lambda t$,
Br\'ezis-Niremberg \cite{BN} proved that \eqref{eP} admits a non-trivial
positive solution, provided $0<f'(0)<\lambda_1(\Omega)$, where
$\lambda_1(\Omega)$ is the first eigenvalue of $(-\Delta,H_0^1(\Omega))$.
 Cerami-Solimini-Struwe \cite{CSS} proved that if $N\geq 6$, problem \eqref{eP}
admits a solution with changes sign. Using this, they also proved that
 when $N\geq 7$ and $\Omega$ is a ball, \eqref{eP} admits infinitely
many radial solution which change sign.

Comte and Knaap \cite{CK} obtained infinitely many non-radial solutions that
 change sign for \eqref{eP} on a ball with Neumann boundary condition
$\mathcal{B}u=\frac{\partial u}{\partial\nu}$, for every $\lambda\in\mathbb{R}$.
They obtained such solutions by cutting the unit ball into angular sectors.
This approach was used by Cao-Han \cite{CH}, where the authors dealt with
the scalar problem \eqref{eP} involving lower-order perturbation and by
de Morais Filho et al. \cite{MMF}  to obtain multiplicity results for a
class of critical elliptic systems
related to the Br\'ezis-Nirenberg problem with the Neumann boundary
condition on a ball.

When $N=2$, the notion of ``critical growth'' is not given by the Sobolev
imbedding, but by the \emph{Trudinger-Moser inequality} (see \cite{T,M}),
which claims that for any  $ u \in H^{1}_0(\Omega)$,
\begin{equation} \label{X0}
\int_{\Omega}
e^{\alpha u^2}dx
< +\infty,\quad\text{for every }\alpha >0.
\end{equation}
Moreover, there exists a positive constant $C=C(\alpha,|\Omega|)$ such that
\begin{equation} \label{X1}
\sup_{||u||_{H_0^{1}(\Omega)} \leq 1} \int_{\Omega} e^{\alpha u^2} dx \leq C ,
\quad \text{for all } \alpha  \leq 4 \pi .
\end{equation}

Motivated by inequality in \eqref{X1}, we say that the nonlinearity $f$
 has critical exponential growth if $f$ behaves like $e^{\alpha_0s^2}$, as
 $|s|\to\infty$, for some $\alpha_0>0$. More precisely,
$$
\lim_{|s|\to\infty}\frac{|f(s)|}{e^{\alpha s^2}}=0,\;
\forall\alpha>\alpha_0\quad \text{and}\quad
\lim_{|s|\to\infty}\frac{|f(s)|}{e^{\alpha s^2}}=+\infty,\; \forall \alpha<\alpha_0.
$$
In this case, Adimurthi \cite{Ad} proved that \eqref{eP} admits a positive
solution, provided that $\lim_{t\to\infty}tf(t)e^{\alpha t^2}=\infty$.
Using a more weaker condition (see \cite[Remark 4.2]{Ad}), Adimurthi
in \cite{ad1} also proved the exitence of many solutions for the Dirichlet
problem with critical exponential growth for the $N$-Laplacian.
 Adimurthi-Yadava  \cite{AY}, proved that \eqref{eP} has a solution that changes
sign and, when $\Omega$ is a ball in $\mathbb{R}^2$, \eqref{eP} has infinitely many
radial solutions that change sign. Inspired in \cite{CK}, this paper is concerned
 with the existence of infinitely many non-radial sign changing solutions
for \eqref{eP} when $f$ has critical exponential growth and $\Omega$
is a ball in $\mathbb{R}^2$. Our main result complements the studies made in
\cite{CK,MMF}, because we consider the case where $f$ has
critical exponential growth in $\mathbb{R}^2$.
It is important to notice that in both  studies mentioned above was
considered the Neumann boundary condition in order that the Pohozaev identity
(see \cite{P}) ensures that the problem \eqref{eP} with the
Dirichlet boundary condition, has no solutions for $\lambda<0$ and $N\geq 3$.
Since the Pohozaev identity is not available in dimension two,
in our case we can use the Dirichlet boundary condition.

Here we use the following assumptions
\begin{enumerate}

\item[(F1)] There is $C>0$ such that
$$
|f(s)|\leq Ce^{4\pi |s|^2},\quad \text{for all } s\in\mathbb{R};
$$

\item[(F2)] $\lim_{s\to 0} f(s)/s=0$;

\item[(H1)] There are $s_0>0$ and $M>0$ such that
$$
0< F(s):=\int_0^{s}f(t)dt\leq M|f(s)|\quad  \text{for all } |s|\geq s_0.
$$

\item[(H2)] $0<F(s)\leq\frac{1}{2}f(s)s$ for all $s\in\mathbb{R}\setminus\{0\}$.

\item[(H3)] $\lim_{s\to\infty} sf(s)e^{-4\pi s^2}=+\infty$
\end{enumerate}

Our main  result reads as follows.

\begin{theorem}\label{sb}
Let $f$ be an odd and continuous function satisfying {\rm (F1)--(F2)}
 and {\rm (H1)--(H3)}. Then \eqref{eP} has infinitely many sign-changing solutions.
\end{theorem}

\section{Notation and auxiliary results}

For each $m\in\mathbb{N}$, we define
$$
A_m=\{x=(x_1,x_2)\in B:
\cos\big(\frac{\pi}{2^m}\big)|x_1|<\sin\big(\frac{\pi}{2^m}\big)x_2\}.
$$
So $A_1$ is a half-ball, $A_2$ an angular sector of angle $\pi/2$, and
$A_3$ an angular sector of angle $\pi/4$ and so on; see Figure \ref{dddd}.

\begin{figure}[htb]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(35,28)(15,5)
\put(30,5){\line(-1,1){14.14}}
\put(30,5){\line(1,1){14.14}}
\qbezier(15.86, 19.14)(30,31)(44.14,19.14)
\put(31,17){$A_m$}
\dashline{1}(30,5)(50,5) 
\put(50,4){$\rightarrow$}
\put(50,2){\scriptsize $x_1$}
\dashline{1}(30,5)(30,29)
\put(29.1,29.5){$\uparrow$}
\put(26,30){\scriptsize $x_2$}
\end{picture} 
\end{center}
\caption{Angular sector $A_m$.}
\label{dddd}
\end{figure}

Using the above notation, we consider the auxiliary Dirichlet problem
\begin{equation} \label{ePm}
\begin{gathered}
-\Delta u = f(u), \quad  \text{in } A_m, \\
u=0,\quad \text{on } \partial A_m.
\end{gathered}
\end{equation}
We will use the Mountain Pass Theorem to obtain a positive solution of
\eqref{ePm}. Using this solution together with an anti-symmetric principle,
we construct a sign-changing solution of problem \eqref{eP}.

According to Figueiredo, Miyagaki and Ruf \cite{DOR}, to obtain a positive
solution of \eqref{ePm} it is sufficient to assume that the limit in (H3) satisfies
\begin{itemize}
\item[(H3')]
$\lim_{s\to+\infty}sf(s)e^{-4\pi s^2}\geq\beta>\frac{1}{2\pi d_m^2}$,
where $d_m$ is the radius of the largest open ball contained in $A_m$.
\end{itemize}
Hypothesis (H3) was initially considered in Adimurthi \cite{Ad}.
This hypothesis will be fundamental to ensure not only the existence but
also the multiplicity of sign-changing solutions. As we will see bellow,
assuming (H3) in place of (H3'), we have the existence of positive
solution of \eqref{ePm}, for every $m\in\mathbb{N}$. This is the content
of the next result.

\begin{theorem}\label{th1}
Under the assumptions {\rm (F1)--(F2)} and {\rm (H1)--(H3)},
problem \eqref{ePm} has a positive solution, for every $m\in\mathbb{N}$.
\end{theorem}

\section{Proof of Theorem \ref{th1}}

In what follows, for an open set $\Theta\subset\mathbb{R}^2$ we denote
 $L^q(\Theta)$ and  $H_0^1(\Theta)$ norms  by
$$
|u|_{q,\Theta}=\Big(\int_{\Theta}|u|^q\Big)^{1/q}, \quad
\|u\|_{\Theta}=\Big(\int_{\Theta}|\nabla u|^2\Big)^{1/2},
$$
respectively.
Since we are interested in positive solutions to  \eqref{ePm}, we assume that
$$
f(s)=0,\quad \text{for all } s\leq 0.
$$
Associated with problem \eqref{ePm}, we have the functional
$I: H_0^1(A_m)\to\mathbb{R}$ defined by
$$
I(u)=\frac{1}{2}\int_{A_m}|\nabla u|^2-\int_{A_m}F(u).
$$
In our case, $\partial A_m$ is not of class $C^1$. However, the functional
$I$ is well defined. In fact, for each $u\in H_0^1(A_m)$, let us consider
$u^{*}\in H_0^1(B)$ the zero extension of $u$ in $B$ defined by
$$
u^{*}(x)=\begin{cases}
u(x),& \text{if } x\in A_m,\\
0,& \text{if } x\in B\setminus A_m.
\end{cases}
$$
Clearly,
$$
\|u\|_{A_m}=\|u^{*}\|_{B}.
$$
Then, from (F1) and the Trudinger-Moser inequality \eqref{X0}
$$
\big|\int_{A_m}F(u)\big|=\big|\int_{B}F(u^{*})\big|
\leq \int_{B}|F(u^{*})|\leq C\int_{B}e^{4\pi|u^{*}|^2}<\infty.
$$
Moreover, using a standard argument we can prove that the functional $I$
 is of class $C^1$ with
$$
I'(u)v=\int_{A_m}\nabla u\nabla v-\int_{A_m}f(u)v,\quad \text{for all } u,v\in
H_0^1(A_m).
$$
Therefore, critical points of $I$ are precisely the weak solutions of \eqref{ePm}.
The next lemma ensures that the functional $I$ has the mountain pass geometry.

\begin{lemma}\label{mpg}
\begin{itemize}
\item[(a)] There are $r,\rho>0$ such that
$I(u)\geq\rho>0$ for all $\|u\|_{A_m}=r$.

\item[(b)] There is $e\in H_0^1(A_m)$ such that
$\|e\|_{A_m}>r$ and $ I(e)<0$.
\end{itemize}
\end{lemma}

\begin{proof}
 Using the definition of $I$ and the growth of $f$, we obtain
$$
I(u)\geq\frac{1}{2}\int_{A_m}|\nabla
u|^2-\frac{\epsilon}{2}\int_{A_m}|u|^2-C\int_{A_m}|u|^qe^{\beta|u|^2},
$$
or equivalently,
$$
I(u)\geq\frac{1}{2}\int_{B}|\nabla
u^{*}|^2-\frac{\epsilon}{2}\int_{B}|u^{*}|^2-C\int_{B}|u^{*}|^qe^{\beta|u^{*}|^2}.
$$
By the Poincar\'e's inequality,
$$
I(u)\geq\frac{1}{2}\int_{B}|\nabla
u^{*}|^2-\frac{\epsilon}{2\lambda_1}\int_{B}|\nabla
u^{*}|^2-C\int_{B}|u^{*}|^qe^{\beta|u^{*}|^2},
$$
where $\lambda_1$ is the first eigenvalue of $(-\Delta, H_0^1(B))$.
Fixing $\epsilon>0$ sufficiently small, we have
$C_1:=\frac{1}{2}-\frac{\epsilon}{2\lambda_1}>0$, from where it follows that
$$
I(u)\geq C_1\int_{B}|\nabla
u^{*}|^2-C\int_B|u^{*}|^qe^{\beta|u^{*}|^2}.
$$
Notice that, from Trudinger-Moser inequality \eqref{X1},
$e^{\beta|u^{*}|^2}\in L^2(B)$
and by continuous embedding
$|u^{*}|^q\in L^2(B)$.
Since $H_0^1(B)\hookrightarrow L^{2q}(B)$ for all $q\geq 1$, by H\"older's inequality
\begin{align*}
\int_B|u^{*}|^qe^{\beta|u^{*}|^2}
&\leq\Big(\int_B|u^{*}|^{2q}\Big)^{1/2}
\Big(e^{2\beta|u^{*}|^2}\Big)^{1/2}\\
&\leq |u^{*}|_{2q,B}^q\Big(\int_B e^{2\beta|u^{*}|^2}\Big)^{1/2}\\
&\leq C\|u^{*}\|_B^q\Big(\int_B e^{2\beta|u^{*}|^2}\Big)^{1/2}.
\end{align*}
We claim that for $r>0$ small enough, we have
$$
\sup_{\|u^{*}\|_B=r}\int_Be^{2\beta|u^{*}|^2}<\infty.
$$
In fact, note that
$$
\int_Be^{2\beta|u^{*}|^2}=\int_Be^{2\beta\|u^{*}\|_B^2
(\frac{|u^{*}|}{\|u^{*}\|_B})^2}.
$$
Choosing $r>0$ small enough such that $\alpha:=2\beta r^2<4\pi$ and
using the Trudinger-Moser inequality \eqref{X1},
$$
\sup_{\|u^{*}\|_B=r}\int_Be^{2\beta |u^{*}|^2}\leq\sup_{\|v\|_B\leq
1}\int_Be^{\alpha|v|^2}<\infty.
$$
Thus,
$$
I(u)\geq C_1\|u^{*}\|_B^2-C_2\|u^{*}\|_B^q.
$$
Fixing $q>2$, we derive
$$
I(u)\geq C_1r^2-C_2r^q:=\rho>0,
$$
for $r=\|u\|_{A_m}=\|u^{*}\|_B$ small enough, which shows that the item
(a) holds.

To prove (b), first notice that

\noindent{\bf Claim 1.} For each $\epsilon>0$, there exists
$\overline{s}_\epsilon>0$ such that
$$
F(s)\leq\epsilon f(s)s,\quad \text{for all }x\in A_m,~|s|\geq \overline{s}_\epsilon.
$$
In fact, from hypothesis (H1)
$$
\big|\frac{F(s)}{sf(s)}\big|\leq\frac{M}{|s|},\quad \text{for all } |s|\geq s_0.
$$
For $p>2$,  claim 1 with $\epsilon=1/p>0$, guarantees the existence
of $\overline{s}_\epsilon>0$ such that
$$
pF(s)\leq f(s)s,\quad \text{for all } s\geq \overline{s}_\epsilon,
$$
which implies the existence of constants $C_1,C_2>0$ satisfying
$$
F(s)\geq C_1|s|^{p}-C_2,\quad \text{for all } s\geq 0.
$$
Thus, fixing $\varphi\in C_0^\infty(A_m)$ with $\varphi\geq 0$ and
$\varphi\neq 0$. For $t\geq 0$, we have
$$
\int_{A_m}F(t\varphi)
\geq\int_{A_m}(C_1|t\varphi|^{p}-C_2)
\geq C_1|t|^{p}\int_{A_m}|\varphi|-C_2|A_m|,
$$
from where it follows that
\begin{equation}\label{4.1}
\int_{A_m}F(t\varphi)\geq C_3|t|^{p}-C_4.
\end{equation}
From \eqref{4.1}, if $t\geq 0$,
$$
I(t\varphi)\leq \frac{t^2}{2}\|\varphi\|_{A_m}^2-C_3|t|^{p}+C_4.
$$
Since $p>2$,
$I(t\varphi)\to-\infty$  as $t\to +\infty$.
Fixing $t_0$ large enough and let $e=t_0\varphi$, we obtain
$$
\|e\|_{A_m}\geq r\quad\text{and}\quad I(e)<0.
$$
\end{proof}


The next lemma is crucial for proving that the energy functional $I$ 
satisfies the Palais-Smale condition and its proof can be found in \cite{DOR}.

\begin{lemma}\label{1.6}
Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $(u_n)$ be a sequence of 
functions in $L^1(\Omega)$ such that $u_n$ converging to $u\in L^1(\Omega)$ 
in $L^1(\Omega)$. Assume that $f(u_n(x))$ and $f(u(x))$ are also 
$L^1(\Omega)$ functions. If
$$
\int_\Omega|f(u_n)u_n|\leq C,\quad \text{for all }n\in\mathbb{N},
$$
then $f(u_n)$ converges in $L^1(\Omega)$ to $f(u)$.
\end{lemma}

\begin{lemma}\label{ld}
The functional $I$ satisfies the $(PS)_d$ condition, for all
$d\in(0,1/2)$.
\end{lemma}

\begin{proof}
 Let $d<1/2$ and $(u_n)$ be a $(PS)_d$ sequence for the functional $I$;
 i.e., $$
I(u_n)\to d\quad \text{and}\quad
I'(u_n)\to 0, \quad\text{as } n\to+\infty.
$$
For each $n\in\mathbb{N}$, let us define 
$\epsilon_n=\sup_{\|v\|_{A_m}\leq 1}\{|I'(u_n)v|\}$, then
$$
|I'(u_n)v|\leq \epsilon_n\|v\|_{A_m},
$$
for all $v\in H_0^1(A_m)$, where $\epsilon_n=o_n(1)$. Thus
\begin{gather}\label{d1}
\frac{1}{2}\int_{A_m}|\nabla u_n|^2-\int_{A_m}F(u_n)=d+o_n(1),
\quad \forall n\in\mathbb{N}, \\
\label{d2}
\Big|\int_{A_m}\nabla u_n\nabla v-\int_{A_m}f(u_n)v\Big|
\leq\epsilon_n\|v\|_{A_m},\quad \text{for all } n\in\mathbb{N},\;
v\in H_0^1(A_m).
\end{gather}
From \eqref{d1} and Claim $1$, for any $\epsilon>0$, there is 
$n_0\in\mathbb{N}$ such that
$$
\frac{1}{2}\|u_n\|_{A_m}^2=\frac{1}{2}\int_{A_m}|\nabla u_n|^2
\leq\epsilon+d+\int_{A_m}F(u_n)\leq C_\epsilon+\epsilon\int_{A_m}f(u_n)u_n,
$$
for all $n\geq n_0$. Using \eqref{d2} with $v=u_n$, we obtain
$$
\big(\frac{1}{2}-\epsilon\big)\|u_n\|^2_{A_m}\leq C_\epsilon+\epsilon\|u_n\|_{A_m},
\quad \text{for all } n\geq n_0.
$$
Thus, the sequence $(u_n)$ is bounded. Since $H_0^1(A_m)$ is a reflexive
 Banach space, there exits $u\in H_0^1(A_m)$ such that, for some subsequence,
$$
u_n\rightharpoonup u\quad \text{in } H_0^1(A_m).
$$
Furthermore, from compact embedding,
\begin{gather*}
u_n\to u\quad \text{in } L^q(A_m),~q\geq 1, \\
u_n(x)\to u(x)\quad\text{a.e. in } A_m.
\end{gather*}
On the other hand, using \eqref{d2} with $v=u_n$, we obtain
$$
-\epsilon_n\|u_n\|_{A_m}\leq\int_{A_m}|\nabla u_n|^2-\int_{A_m}f(u_n)u_n,
$$
which implies
$$
\int_{A_m}f(u_n)u_n\leq\|u_n\|_{A_m}^2-\epsilon_n\|u_n\|_{A_m}\leq C,
\quad \text{for all } n\in\mathbb{N}.
$$
From Lemma \ref{1.6}, $f(u_n)\to f(u)$ in $L^1(A_m)$. Then, there is 
$h\in L^1(A_m)$ such that
$$
|f(u_n(x))|\leq h(x),\quad \text{a.e. in } A_m,
$$
and from (H1),
$|F(u_n)|\leq M h(x)$, a.e. in $A_m$.
Furthermore,
$$
F(u_n(x))\to F(u(x))\quad \text{a.e. in } A_m.
$$
Consequently, by the Lebesgue's dominated convergence,
$$
\int_{A_m}F(u_n)-\int_{A_m}F(u)=o_n(1).
$$
Thus, from \eqref{d1},
$$
\frac{1}{2}\|u_n\|_{A_m}^2-\int_{A_m}F(u)-d=o_n(1),
$$
which implies
\begin{equation}\label{d3}
\lim_{n\to\infty}\|u_n\|_{A_m}^2=2\Big(d+\int_{A_m}F(u)\Big).
\end{equation}
Using again \eqref{d2} with $v=u_n$, we obtain
$$
\Big|\|u_n\|_{A_m}^2-\int_{A_m}f(u_n)u_n\Big|\leq o_n(1),
$$
from where we derive
\begin{align*}
\Big|\int_{A_m}f(u_n)u_n-2\Big(d+\int_{A_m}F(u)\Big)\Big|
&\leq \Big|\|u_n\|_{A_m}^2-\int_{A_m}f(u_n)u_n\Big|\\
&\quad +\Big|\|u_n\|_{A_m}^2-2\Big(d+\int_{A_m}F(u)\Big)\Big|.
\end{align*}
Then
$$
\lim_{n\to\infty}\int_{A_m}f(u_n)u_n=2\Big(d+\int_{A_m}F(u)\Big).
$$
Furthermore, from (H2),
$$
2\int_{A_m}F(u)
\leq 2\lim_{n\to\infty}\int_{A_m}F(u_n)
\leq\lim_{n\to\infty}\int_{A_m}f(u_n)u_n=2d+2\int_{A_m}F(u),
$$
which implies that $d\geq 0$.
\smallskip

\noindent{\bf Claim 2.} For any $v\in H_0^{1}(A_m)$,
$$
\int_{A_m}\nabla u\nabla v=\int_{A_m}f(u)v.
$$
In fact, let us fix $v\in H_0^{1}(A_m)$ and notice that
\begin{align*}
&\Big|\int_{A_m}\nabla u\nabla v-\int_{A_m}f(u)v\Big|\\
&\leq\Big|\int_{A_m}\nabla u_n\nabla v-\int_{A_m}\nabla u\nabla v\Big|
+\Big|\int_{A_m}f(u_n)v-\int_{A_m}f(u)v\Big|\\
&\quad +\Big|\int_{A_m}\nabla u_n\nabla v-\int_{A_m}f(u_n)v\Big|.
\end{align*}
Using Lemma \ref{1.6}, the weak convergence $u_n\rightharpoonup u$ in 
$H_0^1(A_m)$
and the estimate in \eqref{d2}, we derive
$$
\Big|\int_{A_m}\nabla u\nabla v-\int_{A_m}f(u) v\Big|
\leq o_n(1)+\|v\|_{A_m}o_n(1),
$$
and the proof of Claim 2 is complete.

Note that from (H2) and Claim 2,
$$
J(u)\geq\frac{1}{2}\int_{A_m}|\nabla u|^2-\frac{1}{2}\int_{A_m}f(u)u=0.
$$

Now, We split the proof into three cases:
\smallskip

\noindent{\bf Case 1.} The level $d=0$. By the lower semicontinuity of the norm,
$$
\|u\|_{A_m}\leq\liminf_{n\to\infty}\|u_n\|_{A_m},
$$
then
$$
\frac{1}{2}\|u\|_{A_m}^2\leq\frac{1}{2}\|u_n\|_{A_m}^2.
$$
Using \eqref{d3},
$$
0\leq I(u)\leq\frac{1}{2}\liminf\|u_n\|_{A_m}^2-\int_{A_m}F(u),
$$
which implies
$$
0\leq I(u)\leq\int_{A_m}F(u)-\int_{A_m}F(u)=0,
$$
from where $I(u)=0$, or equivalently,
$$
\|u\|_{A_m}^2=2\int_{A_m}F(u).
$$
Using again \eqref{d3}, we derive
$$
\|u_n\|_{A_m}^2-\|u\|_{A_m}^2=o_n(1),
$$
since $H_0^1(A_m)$ be a Hilbert space,
$u_n\to u$ in $H_0^1(A_m)$.
Therefore, $I$ satisfies the Palais-Smale at the level $d=0$.
\smallskip

\noindent{\bf Case 2.} 
The level $d\neq 0$ and the weak limit $u\equiv 0$.
We will show that this can not occur for a Palais-Smale sequence.
\smallskip

\noindent{\bf Claim 3.} There are $q>1$ and a constant $C>0$ such that
$$
\int_{A_m}|f(u_n)|^q<C,\quad \text{for all } n\in\mathbb{N}.
$$
In fact, from \eqref{d3}, for each $\epsilon>0$,
$$
\|u_n\|_{A_m}^2\leq 2d+\epsilon, \quad \text{for all } n\geq n_0,
$$
for some $n_0\in\mathbb{N}$. Furthermore, from (F1),
$$
\int_{A_m}|f(u_n)|^q\leq C\int_{A_m}e^{4\pi qu_n^2}
=C\int_{B}e^{4\pi\|u_n^{*}\|_B^2(\frac{u_n}{\|u_n^{*}\|_B})^2}.
$$
By the Trudinger-Moser inequality \eqref{X1}, the last integral in the 
equality above is bounded if $4\pi q\|u_n^{*}\|_B^2<4\pi$ and this occur 
if we take $q>1$ sufficiently close to $1$ and $\epsilon$ small enough, 
because $d<1/2$, which proves the claim.

Then, using \eqref{d2} with $v=u_n$, we obtain
$$
\Big|\int_{A_m}|\nabla u_n|^2-\int_{A_m}f(u_n)u_n \Big|
\leq\epsilon_n\|u_n\|_{A_m}\leq \epsilon_nC,\quad \text{for all }n\in\mathbb{N}.
$$
Thus,
\begin{equation}\label{d4}
\|u_n\|_{A_m}^2\leq o_n(1)+\int_{A_m}f(u_n)u_n,\quad
\text{for all } n\in\mathbb{N}.
\end{equation}
Furthermore, from H\"older inequality, we can estimate the integral above 
as follows
$$
\int_{A_m}f(u_n)u_n\leq\Big(\int_{A_m}|f(u_n)|^q\Big)^{1/q}
\Big(\int_{A_m}|u_n|^{q'}\Big)^{1/q'},\quad \text{for all } n\in\mathbb{N},
$$
and since $u_n\to 0$ in $L^{q'}(A_m)$,
$\int_{A_m}f(u_n)u_n=o_n(1)$.
Then, from \eqref{d4},
\begin{equation}\label{vivo}
\|u_n\|_{A_m}^2\to 0,\quad \text{as }n\to\infty,
\end{equation}
which contradicts \eqref{d3}, because
$$
\|u_n\|_{A_m}^2\to 2d\neq 0,\quad \text{as } n\to\infty,
$$
proving that $d\neq 0$ and $u=0$ does not occur.
\smallskip

\noindent{\bf Case 3.} The level $d\neq 0$ and the weak limit $u\neq 0$. 
Since
$$
I(u)=\frac{1}{2}\|u\|_{A_m}^2-\int_{A_m}F(u)
\leq\liminf_{n}\Big(\frac{1}{2}\|u_n\|_{A_m}^2-\int_{A_m}F(u_n)\Big)=d,
$$
we have $I(u)\leq d$.
\smallskip

\noindent{\bf Claim 4.} $I(u)=d$.
Suppose by contradiction that $I(u)<d$, from definition of $I$,
\begin{equation}\label{d5}
\|u\|v^2<2\Big(d+\int_{A_m}F(u)\Big).
\end{equation}
On the other hand, if we consider the functions
\begin{gather*}
v_n=\frac{u_n^{*}}{\|u_n^{*}\|_B},~ n\in\mathbb{N},\\
v=u^{*}\Big[2\Big(d+\int_{B}F(u^{*})\Big)\Big]^{-1/2},
\end{gather*}
we have  $\|v_n\|_B=1$ e $\|v\|_B<1$. Furthermore, since
\begin{align*}
\int_B\nabla v_n\nabla \varphi
&=\|u_n\|_B^{-1}\int_{A_m}\nabla u_n\nabla\varphi\\
&\to \Big[2\Big(d+\int_BF(u^{*})\Big)\Big]^{-1/2}
\int_B\nabla u\nabla \varphi=\int_B\nabla v\nabla\varphi,
\end{align*}
for every $\varphi\in C_0^\infty(B)$, i.e.,
$$
\int_B\nabla v_n\nabla\varphi-\int_B\nabla v\nabla\varphi=o_n(1),
$$
we have
$v_n\rightharpoonup v$ in $H_0^1(B)$.


\begin{claim}\label{viv35} 
There are $q>1$ and $n_0\in\mathbb{N}$ such that
$$
\int_{A_m}|f(u_n)|^q<C,\quad \text{for all }n\geq n_0.
$$
\end{claim}

To prove this claim, we need the following result due to  Lions \cite{L}.

\begin{proposition}\label{c2}
Let $(u_n)$ be a sequence in $H_0^1(\Omega)$ such that 
$|\nabla u_n|_{2,\Omega}=1$ for all $n\in\mathbb{N}$. Furthermore, suppose 
that $u_n\rightharpoonup u$ in $H_0^1(\Omega)$ with $|\nabla u|_{2,\Omega}<1$. 
If $u\neq 0$, then for each $1<p<\frac{1}{1-|\nabla u|_{2,\Omega}^2}$, we have
$$
\sup_{n\in\mathbb{N}}\int_{\Omega}e^{4\pi pu_n^2}<\infty.
$$
\end{proposition}

From hypothesis (F1),
\begin{equation}\label{viv31}
\int_{A_m}|f(u_n)|^q\leq C\int_{A_m}e^{4\pi q u_n^2}
=C\int_Be^{4\pi q\|u_n^{*}\|_B^2v_n^2}.
\end{equation}
The last integral in the above expression  is bounded. 
In fact, by Proposition \ref{c2}, it is suffices to prove that there are 
$q,p>1$ and $n_0\in\mathbb{N}$ such that
\begin{equation}\label{viv30}
 q\|u_n^{*}\|_B^2\leq p<\frac{1}{1-\|v\|_B^2},\quad \text{for all }n\geq n_0.
\end{equation}
To prove that \eqref{viv30} occur, notice that $I(u)\geq 0$ and $d<1/2$,
 which implies that
$$
2<\frac{1}{d-I(u)},
$$
from where it follows that
$$
2\Big(d+\int_BF(u^{*})\Big)<\frac{d+\int_BF(u^{*})}{d-I(u)}
=\frac{1}{1-\|v\|_B^2}.
$$
Thus, for $q>1$ sufficiently close to $1$,
$$
2q\Big(d+\int_BF(u)\Big)<\frac{1}{1-\|v\|_B^2}.
$$
From \eqref{d3}, there are $p>1$ and $n_0\in\mathbb{N}$ such that
$$
q\|u_n^{*}\|_B^2\leq p<\frac{1}{1-\|v\|_B^2},
$$
for all $n\geq n_0$ which implies that \eqref{viv30} occur. 
Therefore, Claim \ref{viv35} holds.

Now, we will show that $u_n\to u$ in $H_0^1(A_m)$. First, notice that 
from H\"older inequality and \eqref{viv35},
$$
\int_{A_m}f(u_n)(u_n-u)
\leq\int_{A_m}(|f(u_n)|^q)^{1/q}\Big(\int_{A_m}|u_n-u|^{q'}\Big)^{1/q'}
\leq C|u_n-u|_{q',A_m},
$$
where $1/q+1/q'=1$. Since $u_n\to u$ in $L^{q'}(A_m)$,
\begin{equation}\label{viv40}
\int_{A_m}f(u_n)(u_n-u)=o_n(1).
\end{equation}
Using \eqref{d2} with $v=u_n-u$ and \eqref{viv40}, we obtain 
$\langle u_n-u,u_n\rangle=o_n(1)$, and since $u_n\rightharpoonup u$ 
in $H_0^1(A_m)$,
$$
\|u_n-u\|_{A_m}^2=\langle u_n-u,u_n\rangle-\langle u_n-u,u\rangle=o_n(1).
$$
Then $\|u_n\|_{A_m}^2\to\|u\|_{A_m}^2$ and this together with \eqref{d3}
 contradicts \eqref{d5}. Which proves that $I(u)=d$, i.e.,
$$
\|u\|_{A_m}^2=2\Big(d+\int_{A_m}F(u)\Big).
$$
Furthermore, from \eqref{d3}, $\|u_n\|_{A_m}\to\|u\|_{A_m}$ as $n\to \infty$. Therefore
$$
u_n\to u~~\text{em}~~ H_0^1(A_m).
$$
\end{proof}

From Lemma \ref{mpg} and the Mountain pass Theorem without compactness conditions 
(see \cite{W}), there is a $(PS)_{c_m}$ sequence $(u_n)\subset
H_0^1(A_m)$ such that
$$
I(u_n)\to c_m\quad text{and}\quad I'(u_n)\to 0,
$$
where
\begin{gather*}
c_m=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]} I(\gamma(t)), \\
\Gamma=\{\gamma\in C([0,1],H_0^1(A_m)): \gamma(0)=0 \text{ and } I(\gamma(1))<0\}.
\end{gather*}
To conclude the proof of existence of positive solution for \eqref{ePm}, 
it remains to show that $c_m\in(-\infty,1/2)$. For this, we introduce 
the following Moser's functions (see \cite{M}):
$$
\overline{w}_n(x)=\frac{1}{\sqrt{2\pi}}
\begin{cases}
(\ln(n))^{1/2},& 0\leq|x|\leq 1/n\\
\frac{\ln(1/|x|)}{(\ln(n))^{1/2}}, & 1/n\leq|x|\leq 1\\
0,&|x|\geq 1
\end{cases}
$$
Let $d_m>0$ and $x_m\in A_m$ such that $B_{d_m}(x_m)\subset A_m$ and define
$$
w_n(x)=\overline{w}_n\Big(\frac{x-x_m}{d_m}\Big),
$$
we have $w_n\in H_0^1(A_m)$, $\|w_n\|_{A_m}=1$ and $\operatorname{supp} w_n\subset
B_{d_m}(x_m)$.

We claim that the exists $n\in\mathbb{N}$ such that
$\max_{t\geq 0}I(tw_n)<\frac{1}{2}$.
In fact, suppose by contradiction that this is not the case. 
Then, there exist $t_n>0$ such that
\begin{equation}\label{e1}
\max_{t\geq 0}I(tw_n)=I(t_nw_n)\geq \frac{1}{2}.
\end{equation}
It follows from \eqref{e1} and (H1) that
\begin{equation}\label{e2}
t_n^2\geq 1.
\end{equation}
Furthermore, $\frac{d}{dt}I(tw_n)\left|_{t=t_n}\right.=0$, i.e.,
\begin{equation}\label{e5}
t_n^2=\int_{A_m}f(t_nw_n)t_nw_n,
\end{equation}
which implies that
\begin{equation}\label{e3}
t_n^2\geq\int_{B_{d_m/n}(x_m)}f(t_nw_n)t_nw_n.
\end{equation}
In what follows, we fix a positive constant $\beta_m$ satisfying
\begin{equation}\label{eq7}
\beta_m>\frac{1}{2\pi d_m^2}.
\end{equation}
From (H3), there exists $s_m=s_m(\beta_m)>0$ such that
\begin{equation}\label{bm}
f(s)s\geq\beta_m e^{4\pi s^2},~\text{for all}~ s\geq s_m.
\end{equation}
Using \eqref{bm} in \eqref{e3} and the definition of $w_n$ 
in $B_{d_m/n}(0)$, we obtain
\begin{equation}
t_n^2\geq \beta_m\pi \frac{d_m^2}{n^2}e^{2t_n^2\ln(n)}
\end{equation}
for $n$ large enough, or equivalently,
\begin{equation}\label{1}
t_n^2\geq \beta_m\pi d_m^2e^{2\ln(n)(t_n^2-1)},
\end{equation}
it implies that the sequence $(t_n)$ is bounded. Moreover, from \eqref{1} and
\eqref{e2},
$t_n^2\to 1$ as $n\to \infty$.
Now, let us define
$$
C_n=\{x\in B_{d_m}(x_m):t_nw_n(x)\geq s_m\},\quad 
D_n=B_{d_m}(x_m)\setminus C_n.
$$
With the above notations and using \eqref{e5},
$$
t_n^2\geq\int_{B_{d_m/n}(x_m)}f(t_nw_n)t_nw_n=\int_{C_n}f(t_nw_n)t_nw_n+\int_{D_n}f(t_nw_n)t_nw_n
$$
and by \eqref{bm},
$$
t_n^2\geq\int_{D_n}f(t_nw_n)t_nw_n+\beta_m\int_{C_n}e^{4\pi
t_n^2w_n^2}
$$
or equivalently,
\begin{equation}\label{e2.2}
t_n^2\geq\int_{D_n}f(t_nw_n)t_nw_n+\beta_m\int_{B_{d_m}(x_m)}e^{4\pi
t_n^2w_n^2}-\beta_m\int_{D_n}e^{4\pi t_n^2w_n^2}.
\end{equation}

Notice that
\begin{gather*}
w_n(x)\to 0 \quad \text{a.e. in } B_{d_m}(x_m),\\
\chi_{D_n}(x)\to 1~~\text{a.e. in}~~ B_{d_m}(x_m), \\
e^{4\pi t_n^2w_n^2}\chi_{D_n}\leq e^{4\pi t_n^2s_m^2}\in L^1(B_{d_m}(x_m)).
\end{gather*}
Then, by Lebesgue's dominated convergence
\begin{equation}\label{e2.1}
\lim_n\int_{D_n}e^{4\pi t_n^2w_n^2}=\lim_n\int_{B_{d_m}(x_m)}e^{4\pi
t_n^2w_n^2}\chi_{D_n}=\int_{B_{d_m}(x_m)}1=\pi d_m^2.
\end{equation}
Furthermore,
\begin{gather*}
f(t_nw_n)t_nw_n\chi_{D_n}\leq Ct_nw_ne^{4\pi t_n^2w_n^2}\leq Cs_m
e^{4\pi s_m^2}\in L^1(B_{d_m}(x_m)),\\
f(t_nw_n(x))t_nw_n(x)\chi_{D_n}(x)\to 0~~\text{a.e. in}~~B_{d_m}(x_m).
\end{gather*}
Thus, using again Lebesgue's dominated convergence,
\begin{equation}\label{e1.1}
\lim_n\int_{D_n}f(t_nw_n)t_nw_n=0
\end{equation}
Passing to the limit $n\to\infty$ in \eqref{e2.2} and using
\eqref{e2.1} and \eqref{e1.1},
$$
1\geq\beta_m\lim_n\int_{B_{d_m}(x_m)}e^{4\pi t_n^2w_n^2}-\beta_m\pi d_m^2.
$$
Since $t_n^2\geq 1$, we obtain
\begin{equation}\label{e2.3}
1\geq\beta_m\lim_n\Big[\int_{B_{d_m}(x_m)}e^{4\pi w_n^2}\Big]-\beta_m\pi
d_m^2.
\end{equation}
On the other hand, since
$$
\int_{B_{d_m}(x_m)}e^{4\pi w_n^2}
=d_m^2\int_{B_1(0)}e^{4\pi\overline{w}_n^2}
=d_m^2\big\{\frac{\pi}{n^2}e^{4\pi\frac{1}{2\pi}\ln(n)}
+2\pi\int_{1/n}^1e^{4\pi\frac{1}{2\pi}\frac{[\ln(1/r)]^2}{\ln(n)}}rdr\big\},
$$
making a changing of variables $s=\ln(1/r)/\ln(n)$,
$$
\int_{B_{d_m}(x_m)}e^{4\pi
w_n^2}=\pi d_m^2+2\pi
d_m^2\ln(n)\int_0^1e^{2s^2\ln(n)-2s\ln(n)},
$$
and since
$$
\lim_{n\to\infty}\left[ 2\ln(n)\int_0^1e^{2\ln(n)(s^2-s)}ds\right]=2,
$$
we have
$$
\lim_{n\to\infty}\int_{B_{d_m}(x_m)}e^{4\pi w_n^2}=\pi d_m^2+2\pi d_m^2=3\pi d_m^2.
$$
Using the last limit in \eqref{e2.3}, we obtain
$$
1\geq 3\beta_m\pi d_m^2-\beta\pi d_m^2=2\beta\pi d_m^2,
$$
from where we derive
$$
\beta_m\leq\frac{1}{2\pi d_m^2},
$$
which contradicts the choice of $\beta_m$ in \eqref{eq7}. Then,
$$
\max_{t\geq 0}I(tw_n)<\frac{1}{2},
$$
proving that $c_m<1/2$, for any $m\in\mathbb{N}$ fixed arbitrarily. 
\

\section{Proof of Theorem \ref{sb}}

We shall use the following proposition.

\begin{proposition}\label{p4u}
Let $A$ be an angular sector contained on the positive half plane of $\mathbb{R}^2$ 
such that one of its boundary lies in $x_1$ axis, and denote such boundary 
of $A$ by $B_0=\{x=(x_1,x_2)\in A:~ x_2=0\}$. Consider $A'$ the reflection $A$ 
with respect to $x_1$ axis. Suppose that $u$ is a solution of the  problem
\begin{equation} \label{eP2}
\begin{gathered}
-\Delta u=f(u),\quad \text{in } A, \\
u=0,\quad \text{on } B_0,
\end{gathered}
\end{equation}
where $f$ is a real, continuous and odd function. Then, the function $\tilde{u}$
such that $\tilde{u}= u$ in $A$  and $\tilde{u}$ is antisymmetric
with respect to $x_1$ axis,
$$
\tilde{u}(x_1,x_2)=\begin{cases}
u(x_1,x_2),&\text{in } A\\
-u(x_1,-x_2),&\text{in } A'\\
0,&\text{on }B_0
\end{cases}
$$
satisfies
$$
-\Delta \tilde{u}=f(\tilde{u})\quad \text{in } A\cup A'.
$$
\end{proposition}

\begin{proof} 
Since $u$ is a solution of \eqref{eP2}, we have
$$
\int_{A}\nabla u\nabla
\varphi=\int_{A}f(u)\varphi,\quad \text{for all } \varphi\in
C^\infty_c(A).
$$
We want to prove that
$$
\int_{A\cup A'}\nabla \tilde{u}\nabla
\phi=\int_{A\cup A'}f(\tilde{u})\phi,\quad \text{for all } \phi\in
C^\infty_0(A\cup A').
$$
For any $\phi\in C^\infty_0(A\cup A')$,
$$
\int_{A\cup A'}f(\tilde{u})\phi=\int_{A}f(u(x_1,x_2))\phi(x_1,x_2)
+\int_{A'}f(-u(x_1,-x_2))\phi(x_1,x_2).
$$
Since $f$ is an odd function,
\begin{align*}
 \int_{A\cup A'}f(\tilde{u})\phi
&=\int_{A}f(u(x_1,x_2))\phi(x_1,x_2)+\int_{A'}f(-u(x_1,-x_2))\phi(x_1,x_2)\\
&=\int_{A}f(u(x_1,x_2))\phi(x_1,x_2)-\int_{A'}f(u(x_1,-x_2))\phi(x_1,x_2)\\
&=\int_{A}f(u(x_1,x_2))\phi(x_1,x_2)-\int_{A}f(u(x_1,x_2))\phi(x_1,-x_2).
\end{align*}
Thus
\begin{equation}\label{e11}
\int_{A\cup A'}f(\tilde{u})\phi=\int_{A}f(u)\psi,
\end{equation}
where $\psi(x_1,x_2)=\phi(x_1,x_2)-\phi(x_1,-x_2)$. On the other hand,
\begin{align*}
\int_{A\cup A'}\nabla\tilde{u}\nabla\phi
&=\int_{A}\nabla u(x_1,x_2)\nabla\phi(x_1,x_2)
 -\int_{A'}\nabla u(x_1,-x_2)\nabla\phi(x_1,x_2)\\
&=\int_{A}\nabla u(x_1,x_2)\nabla\phi(x_1,x_2)
 -\int_{A}\nabla u(x_1,x_2)\nabla (\phi(x_1,-x_2))\\
&=\int_{A}\nabla u(x_1,x_2)\nabla(\phi(x_1,x_2)-\phi(x_1,-x_2)).
\end{align*}
Then
\begin{equation}\label{e12}
\int_{A\cup A' }\nabla\tilde{u}\nabla\phi=\int_{A}\nabla
u\nabla\psi.
\end{equation}
The function $\psi$ does not in general belong to $C^\infty_0(A)$. 
Therefore, $\psi$ can not be used as a function
test (in the definition of weak solution on $H^1(A)$). 
On the other hand, if we consider the sequence of functions $(\eta_k)$ 
in $C^{\infty}(\mathbb{R})$, defined by
$$
\eta_k(t)=\eta(kt),\quad t\in\mathbb{R},\; k\in\mathbb{N},
$$
where $\eta\in C^\infty(\mathbb{R})$ is a function such that
$$
\eta(t)=\begin{cases}
0, & \text{if } t<1/2,\\
1, & \text{if } t>1.
\end{cases}
$$
Then
$$
\varphi_k(x_1,x_2):=\eta_k(x_2)\psi(x_1,x_2)\in C^\infty_0(A),
$$
which implies that
\begin{equation}\label{e14}
\int_{A}\nabla
u\nabla\varphi_k=\int_{A}f(u)\varphi_k,\quad k\in\mathbb{N}.
\end{equation}
From \eqref{e11}, \eqref{e12} and \eqref{e14}, we can conclude the proof, 
in view of the following limits
\begin{gather}
\label{(I)} \int_{A}\nabla u\nabla\varphi_k\to
\int_{A}\nabla u\nabla\psi, \\
\label{(II)} \int_{A}f(u)\varphi_k\to \int_{A}f(u)\psi,
\end{gather}
as $k\to \infty$. To see that \eqref{(I)} occur, notice that
$$
\int_{A}\nabla u\nabla\varphi_k=\int_{A}\eta_k\nabla u\nabla \psi
+\int_{A}\frac{\partial u}{\partial x_2}k\eta'(kx_2)\psi.
$$
Clearly,
$$
\int_{A}\eta_k\nabla u\nabla \psi\to\int_{A}\nabla u\nabla \psi,
\quad \text{as } k\to\infty.
$$
Therefore, it remains to show that
\begin{equation}\label{stR}
\int_{A}\frac{\partial u}{\partial x_2}k\eta'(kx_2)\psi\to 0\quad
\text{as } k\to\infty.
\end{equation}
In fact this occurs,
$$
\big|\int_{A}\frac{\partial u}{\partial x_2}k\eta'(kx_2)\psi\big|
\leq kMC\int_{0<x_2<1/k}|\frac{\partial u}{\partial x_2}|x_2
\leq MC\int_{0<x_2<1/k}|\frac{\partial u}{\partial x_2}|,
$$
where $C=\sup_{t\in[0,1]}|\eta'(t)|$ and $M>0$ is such that
$$
|\psi(x_1,x_2)|\leq M|x_2|,\quad \text{for all } (x_1,x_2)\in A\cup A',
$$
and since
$$
\int_{0<x_2<1/k}|\frac{\partial u}{\partial x_2}|\to 0,\quad \text{as }k\to\infty,
$$
the limit in \eqref{stR} occur. The item \eqref{(II)} is an immediately consequence 
of the Lebesgue's dominated convergence.

Now, for each $m\in\mathbb{N}$, we apply the Proposition \ref{p4u} to the
 solution $u$ of problem \eqref{ePm}. Let $A_m'$ be the reflection of $A_m$ 
in one of its sides. On $A_m\cup A_m'$, wecan the define the function $\tilde{u}$ 
such that $\tilde{u}=u$ on $A_m$, and $\tilde{u}$ is antisymmetric with respect 
to the side of reflection. Now, let $A_m''$ be the reflection of $A_m\cup A_m'$ 
in one of its sides and
$\tilde{\tilde{u}}$ the function defined on $A_m\cup A_m'\cup A_m''$
such that $\tilde{\tilde{u}}=\tilde{u}$ on $A_m\cup A_m'$ and
$\tilde{\tilde{u}}$ is antisymmetric with respect to the side of reflection. 
Repeating this procedure, after finite steps, we finally obtain a function 
defined on the whole unit ball $B$,  denoted by $u_m$. Clearly, $u_m$ 
satisfies the Dirichlet condition on the boundary $\partial B$. 
That is, $u_m$ is a sign-changing solution of problem \eqref{eP}. 
Since for every $m\in\mathbb{N}$, problem \eqref{ePm} admits a positive solution, 
we conclude that there exist infinitely many  sign-changing solutions,
and the proof of Theorem  \ref{sb} is complete. 
\end{proof}


 In Figure $\ref{ddd}$, we represent the signal of three solutions, 
corresponding to the cases $m=1$, $m=2$, and $m=3$, respectively. 
The blue color represents the regions where the solutions are negative and 
the red color, the regions where the solutions are positive.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig2} %ddd
\end{center}
\caption{Signal of solutions}
\label{ddd}
\end{figure}

We show in Figure $3$ the profile of solution for the case $m=2$.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig3} %a22
\end{center}
\caption{Case $m=2$} \label{fig3}
\end{figure}

\begin{remark} \rm
It is possible to make a version of Theorem \ref{sb} with Neumann boundary 
condition using the same arguments that we used here, but we have to work 
with another version of Trudinger-Moser inequality in $H^1(\Omega)$ due 
to Adimurthi-Yadava \cite{AYa}, which says that if $\Omega$ is a bounded 
domain with smooth boundary, then for any $u\in H^1(\Omega)$,
\begin{equation} \label{X02}
\int_{\Omega} e^{\alpha u^2}< +\infty, \quad \text{for all }\alpha >0.
\end{equation}
Furthermore, there exists a positive constant $C=C(\alpha,|\Omega|)$ such that
\begin{equation} \label{X11}
\sup_{||u||_{H^{1}(\Omega)} \leq 1} \int_{\Omega} e^{\alpha u^2} \leq C ,
\quad \text{for all } \alpha  \leq 2 \pi .
\end{equation}
\end{remark}


\subsection*{Acknowledgments}  
The author would like to thank professor Claudianor O. Alves for his excellent 
advice during the graduate school. The author also thanks professor Adi Adimurthi 
for his helpful observations.


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\end{document}