\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 29, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/29\hfil An Orlicz-Sobolev space setting]
{An Orlicz-Sobolev space setting for quasilinear elliptic problems }
\author[N. Halidias\hfil EJDE-2005/29\hfilneg]
{Nikolaos Halidias}

\address{Nikolaos Halidias \hfill\break
University of the Aegean \\
Department of Statistics and Actuarial Science\\
Karlovassi, 83200, Samos, Greece}
\email{nick@aegean.gr}

\date{}
\thanks{Submitted October 14, 2004. Published March 8, 2005.}
\subjclass[2000]{32J15, 34J89, 35J60}
\keywords{Landesman-Laser conditions; critical point theory;
 nontrivial solution; \hfill\break\indent
  Cerami (PS) condition; Mountain-Pass Theorem;
 interpolation inequality}

\begin{abstract}
 In this paper we give two existence theorems  for a class of
 elliptic problems in an Orlicz-Sobolev space setting concerning
 both the sublinear and the superlinear case with Neumann boundary
 conditions. We use the classical critical point theory with
 the Cerami (PS)-condition.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In this paper we consider the following elliptic problem with
Neumann boundary conditions,
\begin{equation} \label{e1}
\begin{gathered}
-\mathop{\rm div}(\alpha (|\nabla u(x)|)\nabla u(x)) = g(x,u)
\quad \mbox{ a.e. on }\Omega \\
\frac{\partial u}{\partial v} = 0, \mbox{ a.e. on } \partial \Omega.
\end{gathered}
\end{equation}
 We assume that $\Omega$ is a bounded domain with smooth boundary
$\partial \Omega$.  By $\frac{\partial}{\partial v}$ we denote the
outward normal derivative. As in \cite{Clement} we assume that the
function $\alpha$ is such that $\phi:\mathbb{R} \to \mathbb{R}$ defined by
$\phi(s) = \alpha (|s|)s$ if $s \neq 0$ and $0$ otherwise, is an
increasing homeomorphism from $\mathbb{R}$ to $\mathbb{R}$.

In \cite{Clement}, the authors study a Dirichlet problem when
the right-hand side is superlinear. They show the existence of a
nontrivial solution and show that it is important to
use an Orlicz-Sobolev space setting. Here, we consider a Neumann
problem  when the right-hand side is sublinear. Also we
consider the superlinear case using the ideas in \cite{CostaMag}.
Assuming Landesman-Laser conditions for the sublinear case  and
using the interpolation inequality for the superlinear case, we
prove the existence of a nontrivial solution.


Let us recall the Cerami (PS) condition  \cite{BBF}. Let $X$
be a Banach space. We say that a functional $I:X \to \mathbb{R}$
satisfies the $(PS)_c$ condition if for any sequence such that
$|I(u_n)|\leq M$ and $(1+\| u_n \|)\langle I'(u_n),\phi\rangle \to 0$ for
all $\phi \in X$ we can show that there exists a convergent
subsequence.


Let
\[
\Phi (s) = \int_0^s \phi(t)dt, \quad \Phi^* (s)  =\int_0^s
\phi^{-1} (t) dt, \quad s \in \mathbb{R},
\]
it is well-known that $\Phi$ and $\Phi^*$ are complementary $N$
functions which define the Orlicz spaces $L_{\Phi}, L_{\Phi^*}$
respectively. We  use the well-known Luxenburg norm,
\[
 \| u \|_{\Phi} = \inf \{ k> 0: \int_{\Omega} \Phi
 (\frac{|u(x)|}{k}) dx \leq 1 \}.
\]
 As in \cite{Clement} we denote by $W^1L_{\Phi}$ the corresponding
 Orlicz-Sobolev space with the norm
 $\| u \|_{1,\Phi} = \| u \|_{\Phi} + \| |\nabla u| \|_{\Phi}$.

Now we introduce the Orlicz-Sobolev conjugate $\Phi_*$ of $\Phi$,
defined as
\[
\Phi_*^{-1}  (t) = \int_0^t \frac{\Phi^{-1}
(\tau)}{\tau^{\frac{N+1}{N}}},d \tau,
\]
and as in \cite{Clement}, we suppose that
\[
\lim_{t \to 0}\int_t^1 \frac{\Phi^{-1}
(\tau)}{\tau^{\frac{N+1}{N}}},d \tau < +\infty\,,  \quad
\lim_{t \to \infty} \int_1^t \frac{\Phi^{-1} (\tau)}{\tau^{\frac{N+1}{N}}},d
\tau = + \infty.
\]
To state our hypotheses on $\phi,g$, we need the following three numbers,
\[
p^1 = \inf_{t > 0} \frac{t \phi(t)}{\Phi(t)}, \quad p_{\Phi} =
\liminf_{t \to \infty} \frac{t\phi(t)}{\Phi(t)}, \quad p^0 =
\sup_{t>0} \frac{t \phi(t)}{\Phi(t)} \,.
\]
\begin{itemize}
\item[(H1)] The function $\phi$ is such that
\begin{enumerate}
\item[(i)]
For  every $\varepsilon>0$, there is  $k_{\varepsilon} > 1$ such that
$\Phi'((1+\varepsilon)x) \geq k_{\varepsilon} \Phi'(x)$, $x \geq
x_o(\varepsilon) \geq 0$ and that $\Phi$ is strictly convex.
\item[(ii)] Both $\Phi,\Phi^*$ satisfy a $\Delta_2$
condition, namely
\[
1 < \liminf_{s \to \infty} \frac{s \phi (s)}{\Phi (s)} \leq
\limsup_{s \to \infty} \frac{s \phi (s)}{\Phi (s)} < + \infty.
\]
\end{enumerate}
\end{itemize}


\begin{remark} \label{rmk1} \rm
Under hypotheses (H1), $L_{\Phi}$
is uniformly convex  \cite[p.288]{Rao}.
\end{remark}

We assume the following conditions on $g$.
\begin{itemize}
\item[(H2)] The function $g:\Omega \times \mathbb{R} \to \mathbb{R}$ is a
continuous and satisfies the following hypotheses:
\begin{enumerate}
\item[(i)] There exists nonnegative constants $a_1,a_2$ such that $|g(x,s)| \leq
a_1 + a_2 |s|^{a-1}$, for all $(x,s) \in \Omega \times \mathbb{R}$, with
$p^0\leq a < \frac{N p^1}{N-p^1}$.

\item[(ii)] For all $x \in \Omega$,
\[
\limsup_{u \to 0} \frac{G(x,u)}{\Phi(u)} \leq -\mu < 0,
\quad \lim_{u \to \infty} \frac{G(x,u)}{|u|^{p^1}} = 0\,.
\]
\item[(iii)] There is a function $h : \mathbb{R}^{+} \to \mathbb{R}^{+}$ with
the property $\liminf \frac{h(a_nb_n)}{h(b_n)} > 0$,
$h(b_n) \to\infty$ when $a_n \to a>0$ and $b_n \to +\infty$ such that
\[
\liminf_{|u| \to \infty} \frac{p^1 G(x,u)-g(x,u)u}{h(|u|)}
\geq k (x)>0,
\]
with $k \in L^{1}(\Omega)$,
\end{enumerate}
with $G(x,u) = \int_0^u g(x,r)dr$.
\end{itemize}

\begin{remark} \label{rmk2} \rm
Using the definition of $p^1$ we can prove that
$\Phi (t) \geq ct^{p^1}$ for $t\geq 1$. From this we obtain that
$W^1L_{\Phi} \hookrightarrow L^{\frac{Np^1}{N-p^1}}$ (see \cite{Clement}).
\end{remark}

Our energy functional $I:W^1L_{\Phi} \to \mathbb{R}$ is defined as
\[
I(u) = \int_{\Omega} \Phi(|\nabla u(x)|)dx - \int_{\Omega}
G(x,u(x))dx.
\]
 From the arguments of \cite{Clement,Garcia} we know that
this functional is well defined and $C^1$.

\begin{lemma} \label{lem1}
If (H1), (H2) hold,  then the energy functional
satisfies the $(PS)_c$ condition.
\end{lemma}

\begin{proof} Let $X = W^1L_{\Phi}(\Omega)$.  Suppose that there
exists a sequence $\{ u_n \} \subseteq X$ such that $|I(u_n)| \leq
M$ and
\begin{equation} \label{e2}
|\langle I'(u_n),\phi\rangle |
 \leq \varepsilon_n \frac{\| \phi \|_{1,\Phi}}{1 +\| u_n \|_{1,\Phi}}.
\end{equation}
Suppose that $\| u_n \|_{1,\Phi} \to \infty$. Let $y_n(x) =
\frac{u_n(x)}{\| u_n \|_{1,\Phi}}$. It is easy to see that $y_n
\to y$ weakly in $X$ and $y_n \to y$ strongly in
$L_{\Phi}(\Omega)$.
 From the first inequality we have
\begin{equation} \label{e3}
\big| \int_{\Omega} \Phi(|\nabla u_n(x)|)dx  - \int_{\Omega}
G(x,u_n(x))dx \big| \leq M.
\end{equation}
We can prove that $\Phi(t) \geq \rho^{p^1} \Phi(\frac{t}{\rho})$.
Indeed, we have that $\Phi(t) p^1 \leq t \phi(t)$ for $t>0$. Then
we obtain
\[
\int_{t/\rho}^{t} \frac{p^1}{s}ds \leq \int_{t/\rho}^{t}
\frac{\phi(s)}{\Phi(s)}ds,
\]
for all $t>0$ and for $\rho > 1$. Calculating the above integrals
we arrive at the fact that $\Phi(t) \geq \rho^{p^1}
\Phi(\frac{t}{\rho})$ for all $t>0$ and all $\rho > 1$.
When we divide the above inequality by
$\| u_n\|_{1,\Phi}^{p^1} > 1$, we obtain
\[
\int_{\Omega}\Phi(|\nabla y_n(x)|dx \leq
\int_{\Omega}\frac{G(x,u_n(x))}{\| u_n \|_{1,\Phi}^{p^1}}dx\,.
\]
Next, we prove that $\int_{\Omega}\frac{G(x,u_n(x))}{\| u_n
\|_{1,\Phi}^{p^1}}dx \to 0$. Indeed, from $(H2)(ii)$ we have that
for every $\varepsilon>0$ there exists some $M>0$ such that for
$|u| > M$ we have $\frac{G(x,u)}{|u|^{p^1}} \leq \varepsilon$ for
all $x \in \Omega$. Thus,
\begin{align*}
&\int_{\Omega}\frac{G(x,u_n(x))}{\| u_n \|_{1,\Phi}^{p^1}}dx \\
&\leq \int_{\{x \in \Omega:|u_n(x)| \leq M\}}\frac{G(x,u_n(x))}{\| u_n
\|_{1,\Phi}^{p^1}}dx + \int_{\{x \in \Omega:|u_n(x)| \geq M\}}
\varepsilon |y_n(x)|^{p^1}dx.
\end{align*}
Note that $p^1 \leq p^0 \leq a$ so we have that $W^1L_{\Phi}
\hookrightarrow L^{p^1}$. From that we obtain
\[
\int_{\Omega}\frac{G(x,u_n(x))}{\| u_n \|_{1,\Phi}^{p^1}}dx \leq
\int_{\{x \in \Omega:|u_n(x)| \leq M\}}\frac{G(x,u_n(x))}{\| u_n
\|_{1,\Phi}^{p^1}}dx +\varepsilon c \| y_n \|_{1,\Phi}^{p^1}.
\]
Finally, note that $\| y_n \|_{1,\Phi} =1$ so we have proved our
claim.

 Now $\int_{\Omega} \Phi(|\nabla y_n(x)|dx
\to 0$ thus, $\| \nabla y_n \|_{\Phi} \to 0$. Since
\[
\| \nabla y \|_{\Phi} \leq \liminf_{n \to \infty} \| \nabla y_n \|_{\Phi}
\to 0,
\]
 so $\| \nabla y_n \|_{\Phi} \to \| \nabla y \|_{\Phi}$
and moreover $y_n \to y$ weakly in $X$, thus from the uniform
convexity of $X$ we deduce that $y_n \to y$ strongly in $X$. Note
that $\| y_n \|_{1,\Phi} =1$ so, $y \neq 0$ and from the fact
that $\| \nabla y \|_{\Phi} = 0$ we have that $y = c \in \mathbb{R}$
with $c \neq 0$. From this we obtain that $|u_n(x)| \to \infty$.

Choosing now $\phi = u_n$ in \eqref{e2} and substituting with
\eqref{e3},  we arrive at
\begin{align*}
&\int_{\Omega} p^1 G(x,u_n(x)) - g(x,u_n(x))u_n(x)dx
+ \int_{\Omega}\phi(|\nabla u_n|)|\nabla u_n|-p^1 \Phi(|\nabla
u_n|)dx \\
&\leq M +\varepsilon_n \frac{\| u_n \|_{1,\Phi}}{1 + \| u_n \|_{1,\Phi}}.
\end{align*}
 From the definition of $p^1$ we have  $p^{1} \Phi(t) \leq t
\phi(t)$. Using this fact and dividing  the last inequality with
$h(\| u_n \|_{1,\Phi})$ we obtain
\begin{align*}
& \int_{\Omega} \frac{p^1
G(x,u_n(x)) - g(x,u_n(x))u_n(x)}{h(|u_n(x)|)} \frac{h(|y_n(x)| \|
u_n \|_{1,\Phi})}{h(\| u_n \|_{1,\Phi})} dx \\
&\leq   \frac{M +\varepsilon_n \frac{\| u_n \|_{1,\Phi}}{1 + \| u_n
\|_{1,\Phi}}}{h(\| u_n \|_{1,\Phi})}.
\end{align*}
 From this we can see that
\[
\liminf_{n \to \infty}\int_{\Omega} \frac{p^1 G(x,u_n(x)) -
g(x,u_n(x))u_n(x)}{h(|u_n(x)|)} \frac{h(|y_n(x)| \| u_n
\|_{1,\Phi})}{h(\| u_n \|_{1,\Phi})} dx \leq 0.
\]
Using  Fatou's lemma and  $(H2)(iii)$ we obtain the contradiction.
That is $u_n$ is bounded. So, we can say, at least for a
subsequence, that $u_n \to u$ weakly in $X$ and $u_n \to u$
strongly in $L_a(\Omega)$.

To show the strong convergence we going back to \eqref{e2}
and choose $\phi = u_n-u$. Thus, we obtain
\begin{align*}
&\big|\int_{\Omega} {\Big ( } \alpha (| \nabla u_n|)\nabla u_n -
\alpha(| \nabla u|)\nabla u {\Big ) }{\Big (} \nabla u_n-\nabla
u{\Big )} dx\big| \\
&\leq  \int_{\Omega} g(x,u_n)(u_n-u)dx +\varepsilon_n \|
u_n -u \|_{1,\Phi} -\int_{\Omega} \alpha(| \nabla u|)\nabla u
(\nabla u_n-\nabla u)dx  .
\end{align*}
 Using the compact imbedding $X \hookrightarrow L^{a}(\Omega)$ and
the fact that $u_n \to u$ weakly in $X$ we arrive at
 $\int_{\Omega} \big( a(|\nabla u_n|)\nabla u_n - a(| \nabla u|)\nabla u \big)
 \big(\nabla u_n-\nabla u\big ) dx \to 0$ and using
\cite[Theorem 4]{Le1} we obtain the strong convergence of $u_n$.
\end{proof}

\begin{lemma} \label{lem2}
If hypotheses $(H1)(ii),(H2)$ holds, then there exists some $e \in
X$ with $I(e) \leq 0$.
\end{lemma}

\begin{proof} We will show that there exists some $a \in \mathbb{R}$ such
that $I(a) \leq 0$. Suppose that this is not the case. Then there
exists a sequence $a_n \in \mathbb{R}$ with $a_n \to \infty$ and $I(a_n)
\geq c >0$.
We can easily see that
\begin{align*}
(-\frac{G(x,u)}{u^{p^1}})'
&= \frac{p^1 G(x,u)- g(x,u)u}{u^{p^1+1}} \\
&= \frac{p^1 G(x,u) - g(x,u) u}{h(|u|)}\frac{h(|u|)}{u^{p^1+1}} \\
&\geq (k (x)-\varepsilon) \frac{1}{u^{p^1+1}} = \frac{k(x)-\varepsilon}{p^1}
 (-\frac{1}{u^{p^1}})',
\end{align*}
for a large enough $u \in \mathbb{R}$.
We can say then
\[
\int_t^s \big(-\frac{G(x,u)}{u^{p^1}}\big)'du \geq \int_t^s \frac{k
(x)-\varepsilon}{p^1}
 \big(-\frac{1}{u^{p^1}}\big)'du.
 \]
Take now $s \to \infty$ and using (H2)(iii), we obtain
\[
G(x,t) \geq \frac{k (x)}{p^1},
\]
for large enough $t \in \mathbb{R}$.  From this we obtain
\[
\limsup_{a_n \to \infty} I(a_n) \geq \liminf_{a_n \to \infty}
I(a_n) \geq 0
\]
implies
\[
\limsup_{a_n \to \infty}\int_{\Omega} -G(x,a_n)d x \geq 0
\]
which implies $\int_{\Omega}\frac{-k (x)}{p^1}d x \geq 0$.
 Then using (H2)(iii) we obtain the contradiction.
\end{proof}


\begin{lemma} \label{lem3}
If (H1)(ii) and (H2) hold, then there exists some $\rho > 0$
such that for all $u \in X$ with $\| u \|_{\Phi} = \rho$ we have
that $I(u) > \eta >0$.
\end{lemma}

\begin{proof}
>From (H2)(ii) we have that for every $\varepsilon>0$ there
exists some $u^* \leq 1$ such that for every $|u| \leq u^*$ we
have $G(x,u) \leq (-\mu +\varepsilon)\Phi (|u|) \leq k(-\mu+\varepsilon)|u|^{p^0}$
with $k > 0$.
 On the other hand there exists
$c_1,c_2>0$ such that $|G(x,u)| \leq c_1
|u|^{\frac{Np^1}{N-p^1}}+c_2$ for every $u \in \mathbb{R}$. Recall that
$p^0 < \frac{Np^1}{N-p^1}$ so we can find some $\gamma > 0$ such
that $G(x,u) \leq k (-\mu +\varepsilon)|u|^{p^0} + \gamma
|u|^{\frac{Np^1}{N-p^1}}$. Indeed,  we can choose
\[
\gamma \geq c_1+ \frac{c_2}{|u^*|^{\frac{Np^1}{N-p^1}}} + k(\mu - \varepsilon)
\frac{|u^*|^{p^0}}{|u^*|^{\frac{Np^1}{N-p^1}}}.
\]
Take now a sequence $\{ u_n \} \in X$ such that $\| u_n
\|_{1,\Phi} \to 0$. Thus, we can see that
\[
I(u_n) \geq \int_{\Omega} \Phi(| \nabla u_n |)dx  +k (\mu - \varepsilon)
\| u_n \|_{p^0}^{p^0} - \gamma \| u_n
\|_{\frac{Np^1}{N-p^1}}^{\frac{Np^1}{N-p^1}}
\]
implies
\[
I(u_n) \geq c \| | \nabla u_n |\|^{p^0}_{\Phi}+k (\mu - \varepsilon)
\| u_n \|_{\Phi}^{p^0} - \gamma \| u_n
\|_\frac{Np^1}{N-p^1}^{\frac{Np^1}{N-p^1}}
\]
which implies
\[
I(u_n) \geq C \| u_n \|_{1,\Phi}^{p^0} - \gamma \| u_n
\|_{1,\Phi}^{\frac{Np^1}{N-p^1}}.
\]
Here we have used the fact that $L^{p^0}(\Omega)$ imbeds
continuously in $L_{\Phi}(\Omega)$ and the fact that
$L^{Np^1/(N-p^1)}$ imbeds continuously in $W^1L_{\Phi}$.
Finally we have $C = \min \{c,k(\mu-\varepsilon)\}$. Thus, for big enough
$n \in \mathbb{N}$ and noting that $p^0 < \frac{Np^1}{N-p^1}$ we deduce
that there exists some $\rho > 0$ such that  for all $u \in X$
with $\| u \|_{\Phi} = \rho$ we have that $I(u) > \eta >0$.
The  Lemma is proved.
 \end{proof}

The existence theorem follows from the Mountain-Pass theorem. Note
that we also extend the recently results of Tang \cite{Tang} for
Neumann problems because the author there needs $h(u) = u$.


\section{Superlinear Case}

In this section we consider problem \eqref{e1} with  a superlinear right
hand side.
We assume the following conditions on $g$,
\begin{itemize}
\item[(H3)] The funciton $g:\Omega \times \mathbb{R} \to \mathbb{R}$ is
a continuous function satisfying the following hypotheses:
\begin{enumerate}

\item[(i)] There
exists nonnegative constants $a_1,a_2$ such that $|g(x,s)| \leq
a_1 + a_2 |s|^{a-1}$, for all $(x,s) \in \Omega \times \mathbb{R}$, with
$p^0\leq a < \frac{N p^1}{N-p^1}$, .

\item[(ii)] There exists some $q > 0$ such that for all $x \in \Omega$,
\[
\limsup_{u \to 0} \frac{G(x,u)}{\Phi(|u|)} < -k <0\,\quad
\lim_{u \to \infty} \frac{G(x,u)}{|u|^q} = 0, \quad
0< \beta \leq \liminf_{|s| \to \infty} \frac{G(x,s)}{\Phi(s)}
\]
\item[(iii)]
There exists  $\mu > N/p^1 (q-p^1)$ such that
\[
 \liminf_{|u| \to \infty} \frac{g(x,u)u-p^1 G(x,u)}{|u|^{\mu}}
\geq m >0.
\]
\end{enumerate}
with $G(x,u) = \int_0^u g(x,r)dr$.
\end{itemize}


\begin{theorem} \label{thm1}
If hypotheses (H1)(ii) and (H3) hold,
then problem \eqref{e1}  has a nontrivial solution $u \in X$.
\end{theorem}


\begin{proof}
 Let us denote first  by $N(u) = \int_{\Omega} G(x,u)dx$.
 Suppose that there exists a sequence $\{ u_n \} \subseteq X$
such that
 $I(u_n) \to c$ and $|<I'(u_n),y>| \leq \varepsilon_n \frac{\| y
 \|_{1,\Phi}}{1+\| u_n \|_{1,\Phi}}$ for all $y \in X$. We are
 going to show that $\| u_n \|_{1,\Phi}$ is bounded in $X$.
 Suppose not. Then there exists a subsequence such that $ \| u_n
 \|_{1,\Phi} \to \infty$.

Using the definition of $p^1$ it is easy to see that
$|\langle I'(u),u \rangle- p^1 I(u)| \geq |\langle N'(u),u\rangle -p^1N(u)|$
and using (H3)(iii), we arrive at $\| u_n \|_{\mu}^{\mu} \leq C$.

Next, we use the interpolation inequality, namely
\[
 \| u \|_q \leq \| u \|_{\mu}^{1-t} \| u
 \|_{\frac{Np^1}{N-p^1}}^t,
 \]
 where $0 < \mu \leq q \leq \frac{Np^1}{N-p^1}$, $t \in [0,1]$.
Using the fact that $X$ imbeds continuously in
 $L^{\frac{Np^1}{N-p^1}}$ we have
 \begin{equation} \label{e4}
 \begin{aligned}
 \int_{\Omega} \Phi (| \nabla u_n |)dx
& = I(u_n) + N(u_n) \\
&\leq c_1  \| u_n \|_q^q +c_2 \\
&\leq \| u_n \|_{\mu}^{(1-t)q} \| u_n  \|^{qt}_{\frac{Np^1}{N-p^1}}\\
&\leq c_1 \| u_n \|^{qt}_{1,\Phi}  + c_2,
\end{aligned}
\end{equation}
here we have used the second assertion of (H3)(ii).
 From the relation $|I(u_n)|\leq M$ we obtain
\[
\int_{\Omega} G(x,u_n)dx \leq \int_{\Omega} \Phi(|\nabla u_n|)dx
+M
\]
and
\[
 \beta \int_{\Omega} \Phi(u_n)dx \leq
\int_{\Omega} \Phi(|\nabla u_n|)dx +M\,.
\]
We have used here the third assertion of (H3)(ii).
Adding $\beta \int_{\Omega} \Phi(|\nabla u_n|)dx$ to the last
inequality, we obtain
\begin{equation} \label{e5}
\beta ( \int_{\Omega} \Phi(u_n)dx + \int_{\Omega} \Phi(| \nabla
u_n |)dx) \leq C \int_{\Omega} \Phi(|\nabla u_n|)dx +M.
\end{equation}
 We can prove that $\Phi(t) \geq \rho^{p^1} \Phi(t/\rho)$ for
 $\rho \geq 1$ and combining \eqref{e4} and \eqref{e5},  we arrive at
 \[
 c_1\|  u_n  \|_{1,\Phi}^{p^1} -c_2 \leq \int_{\Omega}
 \Phi(| \nabla u_n|)dx \leq c_1 \| u_n \|_{1,\Phi}^{qt} +c_2.
 \]
 for some $c_1,c_2>0$.
 Choosing $qt < p^1$ (or equivalently $\mu > N/p^1(q-p^1)$) we
 obtain a contradiction. Thus, $\{ u_n \} \subseteq X$ is bounded
 and using the same arguments as in Lemma \ref{lem1} we can prove
 that in fact $\{ u_n \}$ has a strongly convergent subsequence in
 $X$.

Next we prove that there exists some $e \in X$ such that $I(e)
\leq 0$. Indeed, take a sequence $t_n \to \infty$, then
\[
I(t_n) = -\int_{\Omega} G(x,t_n)dx \leq -\beta \int_{\Omega}
\Phi(t_n)dx +C.
\]
It is clear now that for big enough $n \in \mathbb{N}$ we have
$I(t_n) \leq 0$.
Using Lemma \ref{lem3} and the Mountain-Pass theorem, we obtain a nontrivial
solution.
 \end{proof}

As an example of functions that satisfy the above hypotheses,
we have $\Phi(u) = \log(1+|u|)|u|^2$ and
$G(u) = \log(1+|u|) \Phi(u)$.

\subsection*{Acknowledgement}
The author wishes to thank Professor Vy
Khoi Le for his helpful suggestions and remarks.


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