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\begin{document}

\title[Multiple solutions for two classes of quasilinear problems defined on .... ]
{Multiple solutions for two classes of quasilinear problems defined on a nonreflexive Orlicz-Sobolev space}
\author{Claudianor O. Alves, Sabri Bahrouni and Marcos L. M. Carvalho}
\address[Claudianor O. Alves ]
{\newline\indent Unidade Acad\^emica de Matem\'atica
	\newline\indent
	Universidade Federal de Campina Grande
\newline\indent
e-mail:coalves@mat.ufcg.edu.br
\newline\indent
58429-970, Campina Grande - PB, Brazil}

\address[Sabri Bahrouni]
{\newline\indent Mathematics Department, Faculty of Sciences
	\newline\indent
	University of Monastir
	\newline\indent
	e-mail:sabribahrouni@gmail.com
	\newline\indent
	5019 Monastir, Tunisia}

\address[ Marcos Carvalho]
{\newline\indent Instituto de Matem\'atica e Estat\'istica
	\newline\indent
	Universidade Federal de Goias
\newline\indent
e-mail: marcos$\_$leandro$\_$carvalho@ufg.br
\newline\indent 74001-970, Goi\^ania, GO, Brazil }

\pretolerance10000

\begin{abstract}
	In this paper we prove  the existence and multiplicity of solutions for a large class of quasilinear problems on a nonreflexive  Orlicz-Sobolev space. Here, we use the variational methods developed by Szulkin \cite{Szulkin} combined with some properties of the weak$^*$ topology.
\end{abstract}

\thanks{ C. O. Alves was partially
	supported by  CNPq/Brazil 304804/2017-7 }
\subjclass[2010]{35A15, 35J62, 46E30 }
\keywords{Orlicz-Sobolev Spaces, Variational Methods, Quasilinear Elliptic Problems, $\Delta_{2}$-condition, Modular}

\maketitle

\section{Introduction}
This paper concerns the existence and multiplicity of weak solutions for a class of quasilinear elliptic problem of the type
$$
\left\{
\begin{array}{ll}
\displaystyle-\Delta_{\Phi}{u}=\lambda f(x,u), \quad \mbox{in} \quad \Omega, \\
u =0,\quad\mbox{on} \quad \partial \Omega,
\end{array}
\right.
\leqno{(P)}
$$
where $\Omega \subset \mathbb{R}^N$  is a smooth bounded domain, $N \geq 1$, $\lambda$ is a positive parameter and $f:\Omega\times\mathbb{R}\to\mathbb{R}$ is a Carath\'eodory function verifying some conditions that will be mentioned later on. It is important to recall that
$$
\Delta_\Phi u = \mbox{div}(\phi(|\nabla u|)\nabla u),
$$
where $\Phi:\mathbb{R} \to \mathbb{R}$ is a N-function of the form
$$
\Phi(t)=\int_{0}^{|t|}s\phi(s) \,ds
$$
and $\phi:(0, +\infty) \to (0, +\infty) $ is a $C^{1}$ function verifying some technical assumptions.

We would like to detach that this type of operator appears in a lot of physical applications, such as: Nonlinear Elasticity,  Plasticity, Generalized Newtonian Fluid, Non-Newtonian Fluid and Plasma Physics. For more details about the physical applications we cite \cite{Db}, \cite{FN2} and their references.

Motivated by above applications, many authors have studied problems involving quasilinear problem driven by a N-function $\Phi$, we would like to cite  Bonanno, Bisci and Radulescu \cite{BBR, BBR2}, Cerny \cite{Cerny}, Cl\'ement, Garcia-Huidobro and Man\'asevich \cite{VGMS}, Donaldson \cite{donaldson}, Fuchs and Li \cite{Fuchs1},  Fuchs and Osmolovski \cite{Fuchs2}, Fukagai, Ito and Narukawa \cite{FN}, Gossez \cite{gossez2}, Le and Schmitt \cite{LK}, Mihailescu and Radulescu \cite{MR1, MR2}, Mihailescu and Repovs \cite{MD}, Mihailescu, Radulescu and Repovs \cite{MRR}, Mustonen and Tienari \cite{MT}, Orlicz \cite{O} and their references, where quasilinear problems like $(P)$ have been considered in bounded and unbounded domains of $\mathbb{R}^{N}$.

 In all of these works the so called $\Delta_{2}$-condition has been assumed on $\Phi$ and $\widetilde{\Phi}$, which ensures that the Orlicz-Sobolev space $W^{1,\Phi}(\Omega)$ is a reflexive Banach space. This assertion is used several times in order to get a nontrivial solution for elliptic problems taking into account the weak topology and the classical variational methods to $C^1$ functionals.

In recent years many researchers have studied the nonreflexive case, which is more subtle from a mathematical point of view, because in general the energy functional associated with these problems are in general only continuous and the classical variational methods to $C^1$ cannot be used. For example, in \cite{GKMS}, Garc\'ia-Huidobro, Khoi, Man\'asevich and Schmitt considered the existence of solution for the following nonlinear eigenvalue problem
\begin{equation} \label{P1}
	\left\{
	\begin{array}{ll}
		-\Delta_{\Phi}{u}=\lambda \Psi (u), \quad \mbox{in} \quad \Omega \\
		u=0, \quad \mbox{on} \quad \partial \Omega,
	\end{array}
	\right.
\end{equation}
where $\Omega$ is a bounded domain, $\Phi:\mathbb{R} \to \mathbb{R}$ is a N-function and $\Psi:\mathbb{R} \to \mathbb{R}$ is a continuous function verifying some others technical conditions. In that paper, the authors studied the situation where $\Phi$ does not satisfy the well known $\Delta_2$-condition. More precisely, in the first part of that paper the authors consider the function
\begin{equation} \label{phi*}
	\Phi(t)= (e^{t^{2}}-1)/2, \quad \forall t \in \mathbb{R}.
\end{equation}
More recently, Bocea and  Mih\u{a}ilescu \cite{BM}  made a careful study about the eigenvalues of the problem
\begin{equation} \label{P2}
	\left\{
	\begin{array}{ll}
		-div(e^{|\nabla u|^{2}}\nabla u)-\Delta u=\lambda u, \quad \mbox{in} \quad \Omega \\
		u=0, \quad \mbox{on} \quad \partial \Omega.
	\end{array}
	\right.
\end{equation}
After that, Silva, Gon\c calves and Silva \cite{EGS} considered existence of multiple solutions for a class of problem like (\ref{P1}). In that paper the $\Delta_2$-condition is not also assumed and the main tool used was the truncation of the nonlinearity together with a minimization procedure for the energy functional associated to the quasilinear elliptic problem (\ref{P1}).

In \cite{DMKV}, Silva, Carvalho, Silva and Gon\c calves study a class of problem (\ref{P1}) where the energy functional satisfies the mountain pass geometry and the N-function $\widetilde{\Phi}$ does not satisfies the $\Delta_2$-condition and has a polynomial growth. Still related to the mountain pass geometry, in \cite{AEM}, Alves, Silva and Pimenta also considered the problem (\ref{P1}) for a large class of function $\Psi$, but supposing that $\Phi$ has an exponential growth like (\ref{phi*}).

Motivated by above study involving nonreflexive Banach spaces, we intend to consider two new classes of problem $(P)$ where $W_0^{1,\Phi}(\Omega)$ can be nonreflexive. The plan of the paper is as follows: In Section 2 we done a review about the main properties involving the Orlicz-Sobolev spaces that will be used in our approach. In Section 3 we consider our first class of problem  assuming the following conditions:
$$
t \mapsto t\phi(t); \quad t>0 \;\; \quad \mbox{increasing;}\leqno{(\phi_1)
$$
$$
\lim_{t \to 0}t\phi(t)=0, \quad  \lim_{t \to +\infty}t\phi(t)=+\infty. \leqno{(\phi_2)}
$$
$$
t\mapsto t^2\phi(t)\qquad\mbox{is convex and}\qquad\frac{t^{2}{\phi(t)}}{\Phi(t)} \geq l >1, \quad \forall t >0. \leqno{(\phi_3)}
$$

	 The Carath\'eodory function $f:\Omega\times\mathbb{R}\to\mathbb{R}$ satisfies:
		\begin{itemize}
			\item[$(f_0)$] There exist a constant $C>0$ and a function $a:[0, +\infty) \to (0, +\infty) $ such that
							$$|f(x,t)|\leq C(a(t)t+1),~\mbox{a.e. }\ x\in\Omega,~t\in[0,\infty),$$
							where
							$$A(t)=\int_0^ta(s)sds$$
							is a N-function satisfying $1<\displaystyle m_A:=\sup_{t>0}\frac{a(t)t^2}{A(t)}<l.$
			\item[$(f_1)$] There exits $\delta>0$ such that $t\mapsto F(x,t):=\int_0^{t}f(x,s)ds$ is decreasing in $[0,\delta)$ a.e. in $\Omega$;
			\item[$(f_2)$] There exits $t_1>0$ such that
			$$F(x,t_1)>0,~\mbox{a.e. in }\Omega.$$
		\end{itemize}

		 A model of nonlinearity satisfying $(f_0)-(f_2)$ is
			$$
			f(x,t)=t_+^{p-1}-t_+^{q-1}, \quad t \in \mathbb{R},
			$$
			where $t_+=\max\{t,0\}$, $1<q<p$, which was considered in \cite{MRep}.
			However, our hypothesis are more general than those in \cite{MRep}, because in our case, $\Phi$ does not satisfies the $\Delta_2$-condition and $t\mapsto\Phi(\sqrt{t})$ does not need be a convex function. Another model of nonlinearity satisfying $(f_0)-(f_2)$, which was not treated in \cite{MRep}, is
			$$
			f_1(x,t)=pt_+^{p-1}\ln(1+t_+)-\frac{t^p}{\ln(1+t_+)}-qt_+^{q-1}, \quad t \in \mathbb{R},
			$$
			where $1<q<p$. Related to the conditions $(\phi_1)-(\phi_3)$, it is possible to show that the functions below  satisfy these conditions: \\
			\noindent $(i)$ \, $ \Phi(t) = (1+|t|^{2})^{\alpha}-1, \alpha \in (1, \frac{N}{N-2})$, \\
			\noindent $(ii)$ $ \Phi(t) = t^{p}\ln(1+|t|), 1< \frac{-1+\sqrt{1+4N}}{2}<p<N-1, N\geq 3$, \\
			\noindent  $(iii)$ $\Phi(t) = \int_{0}^{|t|}s^{1-\alpha}(\sinh^{-1}s)^{\beta}ds, 0\leq \alpha \leq 1$ and $\beta > 0$, \\
			\noindent $(iv)$ $\Phi(t) = \frac{1}{p}|t|^{p}$ for $p>1$,\\
			\noindent $(v)$ $\Phi(t) = \frac{1}{p}|t|^{p} + \frac{1}{q}|t|^{q}$ where $1<p<q<N$ with $q \in (p, p^{*}),$ \\
			\noindent and \\
			\noindent  $(vi)$ $\Phi(t)=(e^{|t|^{2}}-1)/2$. \\

		From now on, we say that $u \in W_0^{1,\Phi}(\Omega)$ is a weak solution of $(P)$ whenever
$$
\int_{\Omega}\phi(|\nabla u|)\nabla u \nabla v\,dx=\lambda\int_{\Omega}f(x,u)v\,dx, \quad \forall v \in W_0^{1,\Phi}(\Omega).
$$

Under these assumptions the main result in this section can be stated as follows:

\begin{theorem} \label{T1} Assume $(f_0)-(f_2)$ and $(\phi_1) - (\phi_3)$. Then there exists $\lambda^*>0$ such that problem $(P)$ has at least two nontrivial weak solutions for all $\lambda>\lambda^*$.
\end{theorem}

In Section 4, we study a second class of problem, where we require the following structural assumptions on $\Phi$ and $f$:
\begin{itemize}
    \item[$(\phi_4)$ ] $\displaystyle 0\leq\ell -1 =\inf_{t>0}\frac {(t\phi(t))^\prime}{\phi(t)}\leq \frac {(t\phi(t))^\prime}{\phi(t)}\leq m-1,~t>0$
	\item[$(f_3)$] There exist $C>0$ and $0<\alpha<1$ such that
$$
f(x,0)\in L^\infty(\Omega)\quad\text{and}\quad |F(x,t)|\leq C {\Phi(t)^\alpha}, ~t\in\mathbb{R} \setminus \{0\}.
$$
\end{itemize}
The condition $(\phi_4)$ does not guarantee that $W_0^{1,\Phi}(\Omega)$ is reflexive, because it permits to work with the case $\ell=1$, where we have a loss of reflexivity, because in this situation $\widetilde{\Phi}$ does not satisfies the $\Delta_2$-condition. For $\ell=1$, we have as model $\Phi(t)=|t|\log(1+|t|)$. In this section our main result is the following
\begin{theorem}\label{T2}
	Assume $(\phi_1),(\phi_2),(\phi_4)$, $(f_{1})- (f_{3})$. Then there exist $\lambda^*>0$ such that problem $(P)$ has at least two solutions $u_1,u_2\in  W_0^{1,\Phi}(\Omega)\cap L^{\infty}(\Omega) \setminus\{0\}$
	for all $\lambda>\lambda^*$.
\end{theorem}

Before concluding this introduction, we would like to point out that Theorems \ref{T1} and \ref{T2} complement the study made in \cite{MRep}, in the sense that we are considering new classes of N-functions that were not considered in that reference. Moreover, the above theorems are the first results in the literature involving multiplicity of solutions for a class of quasilinear problems driven by a N-function $\Phi$ whose the Orlicz-Sobolev space $W_0^{1,\Phi}(\Omega)$ can be nonreflexive.

\section{Basics on Orlicz-Sobolev spaces}

In this section we recall some properties of Orlicz and Orlicz-Sobolev spaces, which can be found in \cite{Adams, RR}. First of all, we recall that a continuous function $\Phi : \mathbb{R} \rightarrow [0,+\infty)$ is a
N-function if:
\begin{itemize}
	\item[$(i)$] $\Phi$ is convex.
	\item[$(ii)$] $\Phi(t) = 0 \Leftrightarrow t = 0 $.
	\item[$(iii)$] $\displaystyle\lim_{t\rightarrow0}\frac{\Phi(t)}{t}=0$ and $\displaystyle\lim_{t\rightarrow+\infty}\frac{\Phi(t)}{t}= +\infty$ .
	\item[$(iv)$] $\Phi$ is even.
\end{itemize}
We say that a N-function $\Phi$ verifies the $\Delta_{2}$-condition, if
\[
\Phi(2t) \leq K\Phi(t),\quad \forall t\geq 0,
\]
for some constant $K > 0$. For instance, it can be shown that functions  $\Phi$ given in $(i)-(v)$ satisfy the $\Delta_2$-condition, while $\Phi(t)=(e^{t^{2}}-1)/2$ does not verify it.

In what follows, fixed an open set $\Omega \subset \mathbb{R}^{N}$ and a N-function $\Phi$, we define the Orlicz space associated with $\Phi$ as
\[
L^{\Phi}(\Omega) = \left\{  u \in L_{loc}^{1}(\Omega) \colon \ \int_{\Omega} \Phi\left(\frac{|u|}{\lambda}\right)dx < + \infty \ \ \mbox{for some}\ \ \lambda >0 \right\}.
\]
The space $L^{\Phi}(\Omega)$ is a Banach space endowed with the Luxemburg norm given by
\[
\Vert u \Vert_{\Phi} = \inf\left\{  \lambda > 0 : \int_{\Omega}\Phi\Big(\frac{|u|}{\lambda}\Big)dx \leq1\right\}.
\]
The complementary function $\widetilde{\Phi}$ associated with $\Phi$ is given
by its Legendre's transformation, that is,
\[
\widetilde{\Phi}(s) = \max_{t\geq 0}\{ st - \Phi(t)\}, \quad  \mbox{for} \quad s\geq0.
\]
The functions $\Phi$ and $\widetilde{\Phi}$ are complementary each other. Moreover, we also have a Young type inequality given by
\[
st \leq \Phi(t) + \widetilde{\Phi}(s), \quad \forall s, t\geq0.
\]
Using the above inequality, it is possible to prove a H\"older type inequality, that is,
\[
\Big| \int_{\Omega}uvdx \Big| \leq 2 \Vert u \Vert_{\Phi}\Vert v \Vert_{ \widetilde{\Phi}},\quad \forall u \in L^{\Phi}(\Omega) \quad \mbox{and} \quad \forall v \in L^{\widetilde{\Phi}}(\Omega).
\]
The corresponding Orlicz-Sobolev space is defined by
\[
W^{1, \Phi}(\Omega) = \Big\{ u \in L^{\Phi}(\Omega) \ :\ \frac{\partial u}{\partial x_{i}} \in L^{\Phi}(\Omega), \quad i = 1, ..., N\Big\},
\]
endowed with the norm
\[
\Vert u \Vert_{1,\Phi} = \Vert \nabla u \Vert_{\Phi} + \Vert u \Vert_{\Phi}.
\]

The space $W_0^{1,\Phi}(\Omega)$ is defined as the weak$^*$ closure of $C_0^{\infty}(\Omega)$ in $W^{1,\Phi}(\Omega)$. Moreover, by the Modular Poincar\'e's inequality
$$
\int_{\Omega}\Phi(|u|)\,dx\leq \int_{\Omega}\Phi(d|\nabla u|)\,dx, \quad \forall u \in W_0^{1,\Phi}(\Omega),
$$
where $d=diam(\Omega)$, and it follows that
$$
\|u\|_\Phi\leq 2d\|\nabla u\|_\Phi,\qquad u\in W_0^{1,\Phi}(\Omega).
$$
The last inequality yields that  the functional $\|\cdot\|:=\|\nabla \cdot\|_\Phi$ defines an equivalent norm in $W_0^{1,\Phi}(\Omega)$. Here we refer the readers to the important works \cite{gossez,gossez2}. The spaces $L^{\Phi}(\Omega)$, $W^{1, \Phi}(\Omega)$ and $W_0^{1, \Phi}(\Omega)$ are separable and reflexive, when $\Phi$ and $\widetilde{\Phi}$ satisfy  $\Delta_{2}$-condition.

If $E^{\Phi}(\Omega)$ denotes the closure of $L^{\infty}(\Omega)$ in $L^{\Phi}(\Omega)$ with respect to the norm $\|\,\,\|_{\Phi}$, then  $L^{\Phi}(\Omega)$ is the dual space of $E^{\widetilde{\Phi}}(\Omega)$, while $L^{\widetilde{\Phi}}(\Omega)$ is the dual space of $E^{\Phi}(\Omega)$. Moreover, $E^{\Phi}(\Omega)$ and $E^{\widetilde{\Phi}}(\Omega)$ are separable spaces and any continuous linear functional $M:E^{\Phi}(\Omega) \to \mathbb{R}$ is of the form
$$
M(v)=\int_{\Omega}v(x)g(x)\,dx \quad \mbox{for some} \quad g \in L^{\widetilde{\Phi}}(\Omega).
$$
We recall that if $\Phi$ verifies the $\Delta_2$-condition, we then have $E^{\Phi}(\Omega)=L^{\Phi}(\Omega)$.

The next result is crucial in the approach explored in Section 3, and its proof follows directly from a result by Donaldson \cite[Proposition 1.1]{donaldson}.

\begin{lemma}  \label{Estrela} Assume that $\Phi$ is a N-function and $\widetilde{\Phi}$ verifies the $\Delta_2$-condition. If $(u_n) \subset W^{1,\Phi}_0(\Omega) $ is a bounded sequence, then there are a subsequence of $(u_n)$, still denoted by itself, and $u \in  W^{1,\Phi}_0(\Omega)$ such that
	$$
	u_n \stackrel{*}{\rightharpoonup} u \quad \mbox{in} \quad W^{1,\Phi}_0(\Omega)
	$$
	and
	$$
	\int_{\Omega}u_n v \,dx  \to  \int_{\Omega}u v \,dx, \quad  \int_{\Omega}\frac{\partial u_n}{\partial x_i} w \,dx  \to  \int_{\Omega}\frac{\partial u}{\partial x_i} w \,dx, \quad \forall v,w \in  E^{\tilde{\Phi}}(\Omega)=L^{\tilde{\Phi}}(\Omega).
	$$
\end{lemma}

The lemma just above is crucial when the space $W^{1,\Phi}_0(\Omega)$ is not reflexive, for example if $\Phi(t)=(e^{t^2}-1)/2$. However, if $\Phi$ is one of the functions given in $(i)-(v)$, the above lemma is not necessary since $\Phi$ and $\widetilde{\Phi}$ satisfy the $\Delta_2$-condition, and so, $W^{1,\Phi}_0(\Omega)$ is reflexive. Here we would like to point out that $(\phi_3)$ ensures that $\widetilde{\Phi}$ verifies the $\Delta_2$-condition, for more details see Fukagai and Narukawa \cite{FN}.

\begin{lemma}\label{ed1}
	Suppose $(\phi_{1}),~(\phi_{2})$ and either $(\phi_{3})$ or $(\phi_{4})$. Let $(u_{n}) \subset W_0^{1,\Phi}(\Omega)$ be a fixed sequence such that $\|u_{n}\| \rightarrow \infty$. Then there exists $n_0 \in \mathbb{N}$ such that
	$$
	\int_{\Omega} \Phi(|\nabla u_n|) dx \geq \|u_n\|^l, \quad \forall n \geq n_0.
	$$
	Hence
	$$
	\|u_n\| \to +\infty \,\, \mbox{implies that} \,\, \int_{\Omega}  \Phi(|\nabla u_n|)\, dx \to +\infty.
	$$
\end{lemma}
\begin{proof}
	The proof is similar to that given in \cite[Lemma 2.1]{FN}.
\end{proof}

\section{Proof of Theorem \ref{T1}}
Note that under hypotheses $(\phi_1)-(\phi_3)$, it is well known that $\Phi$ does not satisfy the $\Delta_2$-condition, and as a consequence, $ W^{1,\Phi}_0(\Omega)$ is nonreflexive anymore. Under these conditions,  it is also well known that there exists $u \in W_0^{1,\Phi}(\Omega)$ such that
$$
\int_{\Omega}\Phi(|\nabla u|)\,dx=+ \infty.
$$
However, independent of the $\Delta_2$-condition, $(f_0)$ guarantees that the embedding \linebreak $W_0^{1,\Phi}(\Omega)\hookrightarrow L^A(\Omega)$ is continuous. Having this in mind, the energy functional \linebreak $I: W_0^{1,\Phi}(\Omega) \to \mathbb{R}\cup \{+\infty\}$ associated with $(P)$ given by
\begin{equation} \label{I}
I(u)=\int_{\Omega}\Phi(|\nabla u|)dx -\lambda\int_{\Omega}F(x,u) dx,~u\in  W_0^{1,\Phi}(\Omega)
\end{equation}
is well defined.  Hereafter, we denote by $D_\Phi \subset  W_0^{1,\Phi}(\Omega)$ the  set
$$
D_\Phi=\left\{u \in W_0^{1,\Phi}(\Omega)\,:\, \int_{\Omega}\Phi(|\nabla u|)\,dx<+\infty\right\}.
$$
The reader is invited to observe that $D_\Phi= W_0^{1,\Phi}(\Omega)$ when $\Phi$ satisfies the $\Delta_2$-condition.

As an immediate consequence of the above remarks, we cannot guarantee that $I$ belongs to $C^{1}(W_0^{1,\Phi}(\Omega),\mathbb{R})$. However, the functional $\mathcal{F}: W_0^{1,\Phi}(\Omega) \to \mathbb{R} $ given by
$$
\mathcal{F}(u)=\int_{\Omega}F(x,u) dx
$$
belongs to $C^{1}( W_0^{1,\Phi}(\Omega),\mathbb{R})$ and its derivative is given by
$$
\mathcal{F}'(u)v=\int_{\mathbb{R}^N}f(x,u)v \,dx, \quad \forall u,v \in  W_0^{1,\Phi}(\Omega).
$$

Related to the functional $Q: W_0^{1,\Phi}(\Omega) \to \mathbb{R} \cup\{+\infty\}$ given by
\begin{equation} \label{Q}
Q(u)=\int_{\Omega}\Phi(|\nabla u|)dx,
\end{equation}
we know that it is strictly convex   and  l.s.c. with respect to the weak$^*$ topology. Furthermore, $Q \in C^{1}(W_0^{1,\Phi}(\Omega),\mathbb{R})$ when $\Phi$ and $\widetilde\Phi$ satisfy the $\Delta_2$-condition.

From the above commentaries, in the present paper we will use a minimax method developed by Szulkin  \cite{Szulkin}. In this sense, we will say that $u \in D_\Phi$ is a critical point for $I$ if $0 \in \partial I(u) = \partial Q(u) - \mathcal{F}'(u)$, since $\mathcal{F} \in C^{1}( W_0^{1,\Phi}(\Omega),\mathbb{R})$. Then $u \in D_\Phi$ is a critical point of $I$ if, and only if, $\mathcal{F}'(u) \in \partial Q(u)$, what, since $Q$ is convex, is equivalent to
\begin{equation} \label{E1}
Q(v)-Q(u)\geq \lambda\int_{\Omega}f(x,u)(v-u)\,dx, \quad \forall v \in   W_0^{1,\Phi}(\Omega).
\end{equation}

In the case where  $\Phi$ and $\widetilde\Phi$ satisfy the $\Delta_2$-condition, we would like to point out that the energy functional $I \in C^{1}(  W_0^{1,\Phi}(\Omega),\mathbb{R})$, and the last inequality is equivalent to
\begin{equation} \label{E1.0}
I'(u)v=0, \quad \forall v \in   W_0^{1,\Phi}(\Omega),
\end{equation}
that is,
$$
\int_{\Omega}\phi(|\nabla u|)\nabla u \nabla v\,dx=\lambda\int_{\Omega}f(x,u)v\,dx, \quad \forall v \in  W_0^{1,\Phi}(\Omega).
$$
The last identity ensures that $u$ is a weak solution of $(P)$. However, when $\Phi$ does not satisfy the $\Delta_2$-condition, the above conclusion is not immediate, and a careful analysis must be done. For more details see Lemma \ref{Lema0} below.

From now on, let us denote by $\| \cdot\|$ the usual norm in  $ W_0^{1,\Phi}(\Omega)$ given by
$$
\|u\|=\|\nabla u\|_{\Phi}
$$
where
$$
\|\nabla u\|_{\Phi}=\inf\left\{\lambda >0\,:\, \int_{\Omega}\Phi\left(\frac{| \nabla u|}{\lambda}\right)\,dx \leq 1\right\}.
$$
Moreover, we also denote by $\text{dom}(\phi(t)t) \subset  W_0^{1,\Phi}(\Omega)$ the following set
$$
\text{dom}(\phi(t)t)=\left\{ u \in  W_0^{1,\Phi}(\Omega)\,:\, \int_{\Omega}\widetilde{\Phi}(\phi(|\nabla u|)|\nabla u|)\,dx<\infty \right\}.
$$
As $\widetilde{\Phi}$ verifies the $\Delta_2$-condition, the above set can be written of the form
$$
\text{dom}(\phi(t)t)=\left\{ u \in   W_0^{1,\Phi}(\Omega)\,:\, \phi(|\nabla u|)|\nabla u|  \in L^{\widetilde{\Phi}}(\Omega) \right\}.
$$
The set $\text{dom}(\phi(t)t)$ is not empty, since it is easy to see that $C_0^{\infty}(\Omega) \subset \text{dom}(\phi(t)t)$.

\begin{lemma}\label{dominio} Suppose $(\phi_{1})-(\phi_{2})$. For each $u \in D_\Phi$, there is a sequence $(u_n) \subset  \text{dom}(\phi(t)t)$ such that
	$$
|u_n|\leq |u|,\qquad\int_{\Omega}\Phi(|\nabla u_n|)\,dx \leq \int_{\Omega}\Phi(|\nabla u|)\,dx  \quad \mbox{and} \quad \|u-u_n\| \leq {1}/{n}.
	$$
\end{lemma}
\begin{proof}
	See \cite[Lemma 3.2]{AEM}.
\end{proof}

\begin{lemma}\label{I-coercive}
	Assume that $(\phi_1)-(\phi_3)$ and $(f_0)$ hold. Then, functional $I$ is coercive.
\end{lemma}
\begin{proof}
	Indeed, supposing  $\|u\|\geq 1$, using $(f_0)$ and the embeddings $W_0^{1,\Phi}(\Omega)\hookrightarrow L_A(\Omega)$ and $W_0^{1,\Phi}(\Omega) \hookrightarrow L^1(\Omega)$, we conclude that there exist positive constants $C_1$ and $C_2$ satisfying
	$$
	I(u)\geq \|u\|^l- \lambda C_1\|u\|^{m_A}-\lambda C_2\|u\|.
	$$
Recalling the $l>m_A>1$, we get the desired result.
\end{proof}

\begin{lemma}\label{negative}
	Assume that $(\phi_1)-(\phi_3)$ and $(f_0)$, $(f_2)$ hold. Then, there exists $\lambda_*>0$ such that $I$ is bounded from below in  $W^{1,\Phi}_0(\Omega)$ and  $\displaystyle \inf_{u \in W_0^{1,\Phi}(\Omega)} I(u)<0$ for all $\lambda>\lambda_*$.
\end{lemma}
\begin{proof} First of all, the fact that $I$ is coercive yields that there is $R>0$ such that
$$
I(u) \geq 1, \quad \mbox{for} \quad \|u\| \geq R.
$$
On the other hand, by definition of $I$, a simple computation gives
$$
|I(u)| \leq K, \quad \mbox{for} \quad \|u\| \leq R.
$$
From this, there is $M>0$ such that
$$
I(u) \geq -M, \quad \mbox{for all} \quad u \in W_0^{1,\Phi}(\Omega),
$$
showing the boundedness of $I$ from below in $W_0^{1,\Phi}(\Omega)$.

Next, we will show that $\displaystyle \inf_{u \in W_0^{1,\Phi}(\Omega)} I(u)<0$. By $(f_2)$, there exists $t_1>0$ such that $F(x,t_1)>0$. Let $\Omega_1\subset \Omega$ be a compact subset large enough and $u_0\in W_0^{1,\Phi}(\Omega)$ such that $u_0(x)=t_1$ in $\Omega_1$ and $0\leq u_0(x)\leq t_1$ in $\Omega\setminus\Omega_1$. Note that $\{x\in\Omega:F(x,u_0(x))<0\}\subset \Omega\setminus\Omega_1$. So, using $(f_0)$,
\begin{eqnarray}\label{Negative}
	\int_{\Omega}F(x,u_0)d
	&\geq & \int_{\Omega_1}F(x,t_1)dx-C\int_{\Omega\setminus\Omega_1}(A(u_0)+|u_0|)dx\nonumber\\
	&\geq & \int_{\Omega_1}F(x,t_1)dx-C|\Omega\setminus\Omega_1|(A(t_1)+|t_1|)>0
\end{eqnarray}
provided that $|\Omega\setminus\Omega_1|$ is small enough. Thus $I(u_0) < 0$ for $\lambda > 0$ large
enough. This proves the lemma.
\end{proof}
From Lemmas \ref{I-coercive} and \ref{negative},  $I$ is bounded from below in  $W^{1,\Phi}_0(\Omega)$. Thereby, there is $(u_n) \subset W^{1,\Phi}_0(\Omega)$ such that
$$
I(u_n) \to I_{\infty}=\inf_{u \in W^{1,\Phi}_0(\Omega)}I(u) \quad \mbox{as} \quad n \to +\infty.
$$
Consequently, taking into account that $I$ is coercive, the sequence $(u_n)$ must be bounded in $W^{1,\Phi}_0(\Omega)$. Therefore, by Lemma \ref{Estrela}, for some subsequence denoted by itself, we obtain
$$
u_n \stackrel{*}{\rightharpoonup} u_1 \quad \mbox{in} \quad  W^{1,\Phi}_0(\Omega).
$$
Now, applying \cite[Lemma 3.2]{GKMS} and \cite{ET}, it follows that $I$ is weak$^*$ lower semicontinuous. As a consequence,
$$
\liminf_{n \to +\infty}I(u_n) \geq I(u_1).
$$
The last estimate implies that
$$
I(u_1)=I_{\infty}.
$$
From this, $u_1\in W_0^{1,\Phi}(\Omega)\setminus\{0\}$, $I(u_1)<0$ and
\begin{equation} \label{sol-u_1}
Q(v) -Q(u_1)\geq \lambda\int_{\Omega}f(x,u_1)(v-u_1)\,dx, \quad \forall v \in   W_0^{1,\Phi}(\Omega).
\end{equation}
\begin{lemma}\label{Domin}
	$u_1\in D_\Phi\cap dom(\phi(t)t)$.
\end{lemma}
\begin{proof}
	Making $v=0$ in \eqref{sol-u_1}, using $(f_0)$ and H\"{o}lder's inequality we find
	\begin{eqnarray}
		\int_{\Omega}\Phi(|\nabla u_1|)\,dx = Q(u_1)  \leq \lambda\int_{\Omega} f(x,u_1)u_1\,dx<\infty.	\nonumber
	\end{eqnarray}
	This proves that $u_1\in D_\Phi$.

	In the sequel, we will show that $u_1 \in \text{dom}(\phi(t)t)$.
	Setting $v=\left(1-\frac{1}{n}\right)u_1$ in \eqref{sol-u_1}, we get
	$$
	\begin{array}{l}
	\displaystyle \int_{\Omega}\left[\Phi(|\nabla (1-{1}/{n})u_1|)-\Phi(|\nabla u_1|)\right]\,dx=Q\left((1-{1}/{n})u_1\right)-Q(u_1)  \leq 	\displaystyle-\frac{1}{n}\lambda \int_{\Omega}f(x,u_1)u_1\,dx.
	\end{array}
	$$
	Since $\Phi$ is $C^{1}$, there exists $\theta_n(x) \in [0,1]$ such that
	$$
	\frac{\Phi(|\nabla u_1-\frac{1}{n}\nabla u_1|)-\Phi(|\nabla u_1|)}{-\frac{1}{n}}=\phi(|(1-{\theta_n(x)}/{n})\nabla u_1|)(1-{\theta_n(x)}/{n})|\nabla u_1|^{2}.
	$$
	Recalling that $0< 1-{\theta_n(x)}/{n}\leq 1$, we know that
	$$
	1-{\theta_n(x)}/{n} \geq (1-{\theta_n(x)}/{n})^{2}:=g_n(x),
	$$
	which leads to
	$$
	\int_{\Omega}\phi(|g_n(x)\nabla u_1|)|g_n(x)\nabla u_1|^{2}\,dx \leq \lambda\int_{\Omega}f(x,u_1)u_1\,dx, \quad \forall n \in \mathbb{N}.
	$$
Letting  $n \to +\infty$, we derive that

	$$
	\int_{\Omega}\phi(|\nabla u_1|)|\nabla u_1|^{2}\,dx\leq\lambda\int_{\Omega}f(x,u_1)u_1\,dx.
	$$
	Recalling that
	$$
	\phi(t)t^{2}=\Phi(t)+\widetilde{\Phi}(\phi(t)t), \quad \forall t \in \mathbb{R}
	$$
	so, we have
	$$
	\phi(|\nabla u_1|)|\nabla u_1|^{2}=\Phi(|\nabla u_1|)+\widetilde{\Phi}(\phi(|\nabla u_1|)|\nabla u_1|),
	$$
	which leads to
	$$
	\int_{\Omega}\phi(|\nabla u_1|)|\nabla u_1|^{2}\,dx=\int_{\Omega}\Phi(|\nabla u_1|)\,dx+\int_{\Omega}\widetilde{\Phi}(\phi(|\nabla u_1|)|\nabla u_1|)\,dx.
	$$
As $\displaystyle \int_{\Omega}\phi(|\nabla u_1|)|\nabla u_1|^{2}\,dx$ and $\displaystyle \int_{\Omega}\Phi(|\nabla u_1|)\,dx$ are finite, we infer that $\displaystyle \int_{\R^N}\widetilde{\Phi}(\phi(|\nabla u_1|)|\nabla u_1|)\,dx$ is also finite, then $u_1 \in \text{dom}(\phi(t)t)$. This finishes the proof.
\end{proof}

\begin{lemma} \label{Lema0}
	Suppose $(\phi_{1})-(\phi_{2})$ and let $u \in  D_\Phi $ be a critical point of $I$. If $u \in \text{dom}(\phi(t)t)$, then it is a weak solution for $(P)$, that is,
	$$
	\int_{\mathbb{R}^N}\phi(|\nabla u|)\nabla u \nabla v\,dx =\lambda\int_{\mathbb{R}^N}f(x,u)v\,dx,\quad \forall v \in W_0^{1,\Phi}(\Omega).
	$$
\end{lemma}
\begin{proof}
	Hereafter, we adapt the arguments found in  \cite[Lemma 4.5]{EGS}. Given $\epsilon \in (0,\frac{1}{2})$ and $v \in C_0^{\infty}(\R^N)$, we set the function
	$$
	v_\epsilon =\frac{1}{1-\frac{\epsilon}{2}}((1-\epsilon)u+\epsilon v).
	$$
	Hence, as $u$ is a critical point of $I$,
	$$
	\int_{\mathbb{R}^N}\Phi(|\nabla v_\epsilon|)\,dx - \int_{\R^N}\Phi(|\nabla u|)\,dx \geq \lambda\int_{\R^N}f(x,u)(v_\epsilon-u)\,dx, \forall \epsilon \in (0,{1}/{2}),
	$$
	and so,
	$$
	\frac{\int_{\mathbb{R}^N}\Phi(|\nabla v_\epsilon|)\,dx- \int_{\R^N}\Phi(|\nabla u|)\,dx}{\epsilon}  \geq \lambda\int_{\R^N}f(x,u)\left(\frac{v_\epsilon-u}{\epsilon}\right)\,dx.
	$$
	 Taking the limit as $\epsilon \to 0$, we get
	$$
	\int_{\R^N}\phi(|\nabla u|)\nabla u (\nabla v -\nabla u/2) \, dx  \geq \lambda\int_{\R^N}f(x,u)(v-u/2)\,dx
	$$
	or equivalently
	$$
	\int_{\R^N}\phi(|\nabla u|)\nabla u \nabla v  \, dx -\lambda\int_{\R^N}f(x,u)v\,dx \geq A, \quad \forall v \in C_0^{\infty}(\R^N),
	$$
	where
	$$
	A=\frac{1}{2}\int_{\R^N}\phi(|\nabla u|)|\nabla u|^2   \, dx -\frac{\lambda}{2}\int_{\R^N}f(x,u)u\,dx.
	$$
	As $C_0^{\infty}(\R^N)$ is a vector space, the last inequality gives
	$$
	\int_{\R^N}\phi(|\nabla u|)\nabla u \nabla v  \, dx -\lambda\int_{\R^N}f(x,u)v\,dx=0, \quad \forall v \in C_0^{\infty}(\R^N).
	$$

	Now  the result follows using the weak$^*$ density of $C_0^{\infty}(\R^N)$ in $W^{1,\Phi}(\R^N)$ together with the fact that $\phi(|\nabla u|)|\nabla u|  \in L^{\widetilde{\Phi}}(\R^N)$.
\end{proof}

As a byproduct of the last lemma is the following corollary.

\begin{corollary}\label{sol1}
	Suppose that $(\phi_{1})-(\phi_{3})$, $(f_0)$ and $(f_2)$ hold. Then $u_1$ is a solution of problem $(P)$ with $I(u_1)<0$ for $\lambda \geq \lambda^*$.
\end{corollary}

We are going to use the Mountain Pass Theorem to find a second
critical point of $I$. To this end, we define
$$
g(x,t):=
\left\{
\begin{array}{ll}
	0 & \mbox{if } t<0;\\
	f(x,t) & \mbox{if }0\leq t\leq u_1(x);\\
	f(x,u_1) & \mbox{if } t>u_1(x).
\end{array}
\right.
$$
Now, let us consider the functional  $J:W_0^{1,\Phi}(\Omega) \to \mathbb{R}\cup\{+\infty\}$ defined by
\begin{eqnarray}\label{J}
	J(u)=\int_{\Omega}\Phi(|\nabla u|)dx-\lambda\int_{\Omega}G(x,u)dx,
\end{eqnarray}
where $G(x,t):=\int_0^tg(x,s)ds$. Due to Lemma \ref{Lema0}, we can follow the same ideas found in \cite[Lemma 2]{MRep} to prove the lemma below

	\begin{lemma}\label{Auxiliar}
	If $u$ is a solution of problem
	$$
	\left\{
	\begin{array}{ll}
	\displaystyle-\Delta_{\Phi}{u}=\lambda g(x,u), \quad \mbox{in} \quad \Omega, \\
	u =0,\qquad\mbox{on}\qquad\partial\Omega,
	\end{array}
	\right.
	\leqno{(PA)}
	$$
	then $u\leq u_1$.
\end{lemma}

The next lemma establishes the first mountain pass geometry.
\begin{lemma}\label{MPG}
	Suppose that $(\phi_{1})-(\phi_{3})$, $(f_0)$ and $(f_1)$ hold. There exist $r,\rho>0$ such that $J(u)\geq \rho$ for all $u\in W_0^{1,\Phi}(\Omega)$ with $\|u\|=r$.
\end{lemma}
\begin{proof}
	Condition $(f_1)$ implies that there exists $\delta>0$ such that
	$$F(x,t)\leq 0,~0\leq t\leq \delta.$$
	Define $\Omega_u:=[u>\min\{\delta,u_1\}]$. From the last inequality,
	$$G(x,u(x))=F(x,u(x))\leq 0,~\mbox{a.e. in }x\in \Omega\setminus\Omega_u.$$
	On the other hand,
	$$G(x,u(x))\leq 0,~\mbox{a.e. in }\Omega_u\cap[u_1<u<\delta].$$
	 Let  $\Omega_{u,\delta}:=\Omega_u\setminus [u_1<u<\delta]$, $\Omega_{u,\delta}^-:=\Omega_{u,\delta}\cap [u\leq u_1]$ and  $\Omega_{u,\delta}^+:=\Omega_{u,\delta}\cap [u> u_1]$. From $(f_0)$,  $W^{1,\Phi}_0(\Omega)\hookrightarrow L^\Phi(\Omega)\hookrightarrow L^l(\Omega)\hookrightarrow L^s(\Omega)$ for  $s\in(m_A,l)$. Then,
	\begin{eqnarray}\label{des}
		\lambda\int_{\Omega_{u,\delta}}G(x,u)dx &\leq& \lambda C\int_{\Omega_{u,\delta}^-}(A(|u|)+|u|)\,dx+\lambda \int_{\Omega_{u,\delta}^+}(F(x,u_1)+f(x,u_1)(u-u_1))\,dx\nonumber\\
		&\leq&\lambda \overline{C}\int_{\Omega_{u,\delta}}(A(|u|)+|u|)\,dx \nonumber\\
		&\leq&\lambda C_\delta\int_{\Omega_{u,\delta}}\max\left\{\left(\frac{u}{\delta}\right),\left(\frac{u}{\delta}\right)^{m_A}\right\} \,dx\nonumber\\
		&\leq&\lambda \frac{C_\delta}{\delta^s} \int_{\Omega_{u,\delta}} |u|^s\,dx \leq \overline{C}_\delta\|u\|^s,
	\end{eqnarray}
	where $\overline{C},~C_\delta$ and $\overline{C}_\delta$ are positive constants. Therefore, considering $\|u\|\geq 1$, by Lemma \ref{ed1} and \eqref{des},
	$$J(u)\geq \|u\|^l-\lambda\int_{\Omega_{u,\delta}}G(x,u)dx\geq \|u\|^l(1-\lambda\overline{C}_\delta \|u\|^{s-l}).$$
	This proves the lemma.
\end{proof}

\begin{lemma}\label{jc}
	$J$ is coercive.
\end{lemma}
\begin{proof}
By $(f_0)$
$$|F(x,t)|\leq C_A(A(t)+|t|),~t\in\mathbb{R}$$
and
$$f(x,u_1)(t-u_1)\leq 2C_A(A(t)+|t|),~\forall t\geq u_1. $$
Thus, for $\|u\|\geq 1$,

$\displaystyle\int_\Omega\Phi(|\nabla u|)dx-\lambda\int_\Omega G(x,u)dx=$
\begin{eqnarray}
&=  & \int_\Omega\Phi(|\nabla u|)dx-\lambda\int_{u\leq u_1}F(x,u)dx-\lambda\int_{u>u_1}F(x,u_1)+f(x,u_1)(u-u_1)dx\nonumber \\
&\ge& \int_\Omega\Phi(|\nabla u|)dx-3\lambda C_A\int_{\Omega}(A(u)+|u|)dx\nonumber\\
&\geq  & \|u\|^l-\lambda c_1\|u\|^{m_A}-\lambda c_2\|u\|,
\end{eqnarray}
 where $c_1,c_2$ are positive constants. As $1<m_A<l$, we get the desired result.
\end{proof}

\noindent{\bf Proof of Theorem \ref*{T1}: } Gathering Lemma \ref{MPG} with the fact that $J(u_1)=I(u_1)<0$ for $\lambda \geq \lambda^*$, we can apply the Mountain Pass Theorem found in \cite[Theorem 3.1]{AlvesdeMorais} to guarantee the existence of a $(PS)$ sequence $(u_n)\subset W_0^{1,\Phi}(\Omega)$ associated with the mountain pass level of $J$, that is, $J(u_n)\to c\geq \rho$ and $\tau_n\to 0$ in $\R$ such that
\begin{equation} \label{sequencia2}
Q(v)-Q(u_n) \geq \lambda\int_{\Omega}g(x,u_n)(v-u_n)\,dx- \tau_n\|v-u_n\|,
\end{equation}
for all $v \in W_0^{1,\Phi}(\Omega)$ holds true for  all $n \in\mathbb{N}$,
where
$$c:=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}J(\gamma(t))$$
and
$$\Gamma=\{\gamma\in C[0,1]|~\gamma(0)=0,~\gamma(1)=u_1\}.$$
By Lemma \ref{jc}, $(u_n)$ is bounded. Hence, we can assume without loss of generality that $u_n\stackrel{*}\rightharpoonup u_2$ in $W_0^{1,\Phi}(\Omega)$.

In the sequel, we will show that $u_2 \in \text{dom}(\phi(t)t)$. By Lemma \ref{dominio}, there is $(v_n) \subset \text{dom}(\phi(t)t)$ such that
$$
|v_n|\leq |u_n|,\qquad\|v_n-u_n\| \leq 1/n \quad \mbox{and} \quad \int_{\Omega}\Phi(|\nabla v_n|)\,dx \leq \int_{\Omega}\Phi(|\nabla u_n|)\,dx, \quad \forall n \in \mathbb{N}.
$$
Consequently,
$$
Q(v)-Q(v_n) \geq \lambda\int_{\Omega}g(x,u_n)(v-u_n)\,dx- |\tau_n|\, \|v-u_n\|,
$$
for all $v \in W_0^{1,\Phi}(\Omega)$. Setting $v=v_n-\frac{1}{n}v_n$, we get
$$
Q(v_n-\frac{1}{n}v_n)-Q(v_n)
\geq \lambda\int_{\Omega}g(x,u_n)(v_n-\frac{1}{n}v_n-u_n)\,dx- |\tau_n| \, \|v_n-\frac{1}{n}v_n-u_n\|,
$$
that is,
$$
\begin{array}{l}
\displaystyle		\int_{\Omega}\frac{(\Phi(|\nabla v_n-\frac{1}{n}\nabla v_n|)-\Phi(|\nabla v_n|))}{-\frac{1}{n}}\,dx \leq  \\
\mbox{}\\
\displaystyle  -n\lambda\int_{\Omega}g(x,u_n)(v_n-u_n)\,dx+ \lambda\int_{\Omega}g(x,u_n)v_n \,dx \nonumber + n|\tau_n|\,\|v_n-u_n\|+|\tau_n|\,\|v_n\|. \nonumber
\end{array}
$$
As $(u_n)$ is bounded in $W_0^{1,\Phi}(\Omega)$, $(g(x,u_n))$ is bounded in $L_{\widetilde{A}}(\Omega)$, $(\tau_n)$ is bounded in $\mathbb{R}$ and $\|v_n-u_n\| \leq \frac{1}{n}$, it follows that the right side of the above inequality is bounded. Therefore, there is $M>0$ such that
$$
\int_{\Omega}\frac{(\Phi(|\nabla v_n-\frac{1}{n}\nabla v_n|)-\Phi(|\nabla v_n|))}{-\frac{1}{n}}\, dx  \leq M, \quad \forall n \in \mathbb{N}.
$$
Using again that $\Phi$ is $C^{1}$, there exists $\theta_n(x) \in [0,1]$ such that
$$
\frac{\Phi(|\nabla v_n-\frac{1}{n}\nabla v_n|)-\Phi(|\nabla v_n|)}{-\frac{1}{n}}=\phi(|(1-{\theta_n(x)}/{n})\nabla v_n|)(1-{\theta_n(x)}/{n})|\nabla v_n|^{2}.
$$
Recalling that $0< 1-{\theta_n(x)}/{n}\leq 1$, we know that
$$
1-{\theta_n(x)}/{n} \geq (1-{\theta_n(x)}/{n})^{2},
$$
which leads to
$$
\int_{\Omega}\phi(|(1-{\theta_n(x)}/{n})\nabla v_n|)(1-{\theta_n(x)}/{n})^{2}|\nabla v_n|^{2}\,dx \leq M \quad \forall n \in \mathbb{N}.
$$
As $u_n \stackrel{*}{\rightharpoonup} u_2$ in $W_0^{1,\Phi}(\Omega)$, we also have $(1-{\theta_n(x)}/{n})v_n \stackrel{*}{\rightharpoonup} u_2$ in $W_0^{1,\Phi}(\Omega).$ Therefore,  using the fact that $\phi(t)t^{2}$ is convex, we can apply \cite[Lemma 3.2]{GKMS} to obtain
$$
\liminf_{n \to +\infty}\int_{\Omega}\phi(|(1-{\theta_n(x)}/{n})\nabla v_n|)(1-{\theta_n(x)}/{n})^{2}|\nabla v_n|^{2}\,dx  \geq \int_{\Omega}\phi(|\nabla u_2|)|\nabla u_2|^{2}\,dx,
$$
and so,
$$
\int_{\Omega}\phi(|\nabla u_2|)|\nabla u_2|^{2}\,dx \leq M.
$$
Recalling that
$$
\phi(t)t^{2}=\Phi(t)+\widetilde{\Phi}(\phi(t)t), \quad \forall t \in \mathbb{R},
$$
we have
$$
\phi(|\nabla u_2|)|\nabla u_2|^{2}=\Phi(|\nabla u_2|)+\widetilde{\Phi}(\phi(|\nabla u_2|)|\nabla u_2|),
$$
which leads to
$$
\int_{\Omega}\phi(|\nabla u_2|)|\nabla u_2|^{2}\,dx=\int_{\Omega}\Phi(|\nabla u_2|)\,dx+\int_{\Omega}\widetilde{\Phi}(\phi(|\nabla u_2|)|\nabla u_2|)\,dx.
$$
Since $\displaystyle \int_{\Omega}\phi(|\nabla u_2|)|\nabla u_2|^{2}\,dx$ and $\displaystyle \int_{\Omega}\Phi(|\nabla u_2|)\,dx$ are finite, we see that $\displaystyle \int_{\Omega}\widetilde{\Phi}(\phi(|\nabla u_2|)|\nabla u_2|)\,dx$ is also finite. \\
{\bf Claim:} $J(u_n)\to J(u_2)$ as $n \to +\infty$. \\

In fact, from \eqref{sequencia2} and $W_0^{1,\Phi}(\Omega)\stackrel{comp}\hookrightarrow L_A(\Omega)$,
$$\int_{\Omega}G(x,u_n) dx\to \int_{\Omega}G(x,u_2) dx\qquad\mbox{and}\qquad \int_{\Omega}g(x,u_n)(v-u_n) dx\to \int_{\Omega}g(x,u_2)(v-u_2) dx.$$
Since $(J(u_n))$ is bounded sequence, for some a subsequence of $(u_n)$, still denoted by itself, we can assume that
$$\lim_{n\to \infty}Q(u_n)= L.$$
Using the fact  that $Q$ is lower semicontinuous with respect to the weak* topology, we derive that
$$Q(u_2)\leq \liminf_{n\to\infty} Q(u_n)=L.$$
On the other hand, making $v=u_2$ in \eqref{sequencia2}, we conclude that
$$
Q(u_2)\geq  \liminf_{n\to\infty} Q(u_n)=L,
$$
and so,
$$\lim_{n\to \infty} Q(u_n)=Q(u_2).$$
Thus, $J(u_n)\to J(u_2)=c$. Now, we can use the same ideas of Lemma \ref{Domin} to prove that $u_2\in D_\Phi\cap dom(\phi(t)t)$. Letting $n \to +\infty$  in \eqref{sequencia2}, we conclude that $u_2$ is a critical point of $J$. Using similar idea explored in the proof of Lemma \ref{Lema0}, $u_2$ is a weak solution of $(PA)$.  From Lemma \ref{Auxiliar} we obtain that $u_2\leq u_1$, then $g(x,u_2(x))=f(x,u_2(x))$ for all $x \in \Omega$, from where it follows that $u_2$ is a weak solution of
$(P)$ with $J(u_2)=I(u_2)$. Moreover, by Lemma \ref{Lema0}, $u_2$ is a weak solution of
$(P)$. But, from Lemma \ref{negative}, $I(u_1)<0<c=I(u_2)$, from where it follows that $u_1\neq u_2$. This finishes the proof.

\section{Proof of Theorem \ref{T2}}
In this section we will use the same approach of the last section. In order to avoid some repetitions, we are going to show only the different accounts. For example, it is important to point out that if $(\phi_4)$ holds, then $D_\Phi=dom(t\phi(t))=W_0^{1,\Phi}(\Omega)$.
\begin{lemma}\label{I-coercive-1}
	Assume that $(\phi_1),~(\phi_2),~(\phi_4)$ and $(f_3)$ hold. Then the functional $I$ is coercive.
\end{lemma}
\begin{proof}
	Initially, from  $(f_3)$,
	\begin{eqnarray}\label{cons-f_3}
		|F(x,t)|\leq \frac{C}{\alpha}\Phi(t)^\alpha.
	\end{eqnarray}
Using \eqref{cons-f_3}, H\"{o}lder's and Poincar\'e's Inequalities, there exist positive constants $C_1$ and $C_2$ satisfying
	\begin{eqnarray}
	I(u)&\geq&\int_{\Omega}\Phi(|\nabla u|)dx - \lambda \frac{C}{\alpha}\int_{\Omega}[\Phi(u)]^\alpha dx\nonumber\\
        &\geq& \int_{\Omega}\Phi(|\nabla u|)dx - \lambda C_1 \left(\int_{\Omega}\Phi(|\nabla u|) dx\right)^{\alpha}\nonumber\\

		&\geq&\int_{\Omega}\Phi(|\nabla u|)dx\left[1 - \lambda C_2\left(\int_{\Omega}\Phi(|\nabla u|) dx\right)^{\alpha-1}\right].\nonumber
	\end{eqnarray}
As $\alpha \in (0,1)$, the Lemma \ref{ed1} ensures that $I$ is coercive functional. This finishes the proof.
\end{proof}

\begin{lemma}\label{negative-1}
	Assume that $(\phi_1),~(\phi_2),~(\phi_4),~(f_2)$ and $(f_3)$ hold. Then, there exist $\lambda_*>0$ such that $I$ is bounded from below in  $W^{1,\Phi}_0(\Omega)$ and  $\displaystyle \inf_{u \in W_0^{1,\Phi}(\Omega)} I(u)<0$ for all $\lambda>\lambda_*$.
\end{lemma}
\begin{proof}
Since the boundedness of $I$ from below follows as in Lemma \ref{negative}, we will omit this part. The rest of the proof follows as the same ideas of Lemma \ref{negative}. But, we need to change the inequality \eqref{Negative} by
	\begin{eqnarray}
		\int_{\Omega}F(x,u_0)d
		&\geq & \int_{\Omega_1}F(x,t_1)dx-C\int_{\Omega\setminus\Omega_1}\Phi(u_0)^\alpha dx\nonumber\\
		&\geq & \int_{\Omega_1}F(x,t_1)dx-C|\Omega\setminus\Omega_1|\Phi(t_1)^\alpha >0.
	\end{eqnarray}
	Here, we point out that in the first inequality we have used \eqref{cons-f_3}.
\end{proof}

\begin{corollary}\label{sol12}
	Suppose that $(\phi_{1}),~(\phi_{2}),~(\phi_{4})$, $(f_2)$ and $(f_3)$ hold. Then, there exists a solution $u_1$ of problem $(P)$ such that $I(u_1)<0$.
\end{corollary}
\begin{proof}
	Since $I$ is lower semicontinuous in weak$^*$ topology, we can use Lemmas \ref{I-coercive-1} and \ref{negative-1} to obtain $u_1\in W_0^{1,\Phi}(\Omega)$ such  that
	$$I(u_1):=\inf_{u\in W_0^{1,\Phi}(\Omega)}I(u)<0.$$
	Since $I$ is G\^ateaux differentiable, it follows from \cite[Prop. 1.1]{Szulkin} that $u_1$ is a critical point of I. Moreover, using the fact that $\Phi$ satisfies the $\Delta_2$-condition, we obtain that $D_\Phi=dom(\phi(t)t)=W_0^{1,\Phi}(\Omega)$. From Lemma \ref{Lema0}, we conclude that $u_1$ is a solution of problem $(P)$.
\end{proof}

Now, we shall consider the functional $J$ defined by \eqref{J}.

\begin{lemma}\label{MPG2}
	Assume $(\phi_1),~(\phi_2),~(\phi_4),~(f_2)$, $(f_3)$ and $m<l^*$. Then there exist $r,\rho>0$ such that $I(u)\geq \rho$ for all $u\in W_0^{1,\Phi}(\Omega)$ with $\|u\|=r$.
\end{lemma}

\begin{proof}
	We shall use the same ideas of Lemma \ref{MPG}, but we need to do some adjusts. Assume that $\|u\|\leq 1$. We will change the inequality \eqref{des}. For this end, by $(f_3)$, \eqref{cons-f_3} and choosing $s\in(m, l^*)$,
	\begin{eqnarray}\label{des1}
	\lambda\int_{\Omega_{u,\delta}}G(x,u)dx &\leq& \lambda C\int_{\Omega_{u,\delta}^-}\Phi(u)^\alpha\,dx+\lambda C\int_{\Omega_{u,\delta}^+}(F(x,u_1)+f(x,u_1)(u-u_1))\,dx\nonumber\\
	&\leq&\lambda \overline{C}\int_{\Omega_{u,\delta}}\Phi(u)^\alpha\,dx \nonumber\\
	&\leq&\lambda C_\delta\int_{\Omega_{u,\delta}}\max\left\{\left(\frac{u}{\delta}\right)^\alpha,\left(\frac{u}{\delta}\right)^{m\alpha}\right\} \,dx\nonumber\\
	&\leq&\lambda \frac{C_\delta}{\delta^s} \int_{\Omega_{u,\delta}} |u|^s\,dx \leq \overline{C}_\delta\|u\|^s.
	\end{eqnarray}
	Thus, from \eqref{des1},
	$$J(u)\geq \|u\|^m-\lambda\int_{\Omega_{u,\delta}}G(x,u)dx\geq \|u\|^m(1-\lambda\overline{C}_\delta \|u\|^{s-m}).$$

\end{proof}

The Lemma \ref{MPG2} combined with the equality $J(u_1) =I(u_1)<0$ permit to use again the Mountain Pass Theorem to obtain
a sequence $(u_n)\subset W_0^{1,\Phi}(\Omega)$ such that $J(u_n)\to c\geq \rho>0$ and \eqref{sequencia2} holds.

\begin{lemma}\label{Bounded}
	Assume $(\phi_1),(\phi_2),(\phi_4)$ and $(f_{1})- (f_{3})$. Then $(u_n)$ is bounded.
\end{lemma}
\begin{proof}
	Using similar idea of Lemmas \ref{I-coercive-1} and \ref{jc}, we can prove that $J$ is coercive, and so, $(u_n)$ must be bounded.
\end{proof}

	\noindent {\bf Proof of Theorem \ref{T2}: } Initially, by Lemma \ref{Bounded}, $u_n\stackrel{*}\rightharpoonup u_2$, for some $u_2\in W_0^{1,\Phi}(\Omega)$. Moreover, $u_n\to u_2$ a.e. in $\Omega$ and there exists $h\in L^\Phi(\Omega)$ such that $|u_n|\leq h$.\\
	{\bf Claim:} $J(u_n)\to J(u_2)$ as $n \to +\infty$.

	Firstly, we will prove that
	\begin{eqnarray}\label{conv}
		\int_{\Omega}G(x,u_n) dx\to \int_{\Omega}G(x,u_2) dx.
	\end{eqnarray}
	In fact, by \eqref{cons-f_3},
	$$|G(x,t)|=|F(x,t)|\leq C\Phi^\alpha(t),~0<t<u_1.$$
	Now, using \cite[Theorem 1.6]{SCA}, we have $u_1\in L^{\infty}(\Omega)$. From this,
$$|G(x,t)| \leq |F(x,u_1)|+ |f(x,u_1)||t-u_1|\leq C_1+C_2t, \quad u_1<t.$$
	Consequently,
$$|G(x,u_n)|\leq C_1 + C\Phi^\alpha(h)+C_2h \in L^{1}(\Omega).$$
Now, (\ref{conv}) follows from Lebesgue dominated convergence theorem. A similar argument works to prove that
	$$\int_{\Omega}g(x,u_n)(v-u_n) dx\to \int_{\Omega}g(x,u_2)(v-u_2) dx,~v\in W_0^{1,\Phi}(\Omega)$$
As in the proof of Theorem \ref{T1},
	$$\lim_{n\to \infty} Q(u_n)=Q(u_2).$$
	Thus, $J(u_n)\to J(u_2)=c$. Letting $n \to +\infty$ in \eqref{sequencia2}, we conclude that $u_2$ is a critical point of $J$. Using the same ideas explored in the proof of Lemma \ref{Lema0}, it is easy to see that $u_2$ is a weak solution of $(PA)$.  Arguing as in the proof of Theorem \ref{T1}, $u_2$ is a weak solution of
	$(P)$ with $J(u_2)=I(u_2)$. But, from Lemma \ref{negative}, $I(u_1)<0<c=I(u_2)$, this implies that $u_1\neq u_2$. Moreover, From \cite[Theorem 1.6]{SCA},  $u_2\in L^{\infty}(\Omega)$, which finishes the proof.

redpajama_set_name: RedPajamaArXiv


text:
music sa appoints new general manager
by Skye Walter | Nov 27, 2014 | News
The MUSIC SA Board is delighted to announce the appointment of Lisa Bishop to the role of General Manager.
A highly qualified executive with extensive experience in the private, public & not-for-profit sectors, Lisa brings a broad range of skills and expertise in management, HR, governance and business development to the organisation. Lisa has worked in diverse fields including the arts, tourism, retail and health, most recently working in film & TV, in addition to over 20 years as a company director.
"There is currently unprecedented support for contemporary music in South Australia from policy makers across all levels of government" she said, "I'm thrilled to be working with the Music SA board and staff to drive positive change that capitalises on the current local and state government support, the recommendations of the Thinker's Report and the outstanding achievements of Music SA to date."
Lisa will commence the role of General Manager from January 2015.
The Music SA Board would like to acknowledge the outstanding work of PJ Murton over the past 5 months in the caretaker role of Managing Director. PJ's passion and commitment to Music SA has ensured the smooth continuation of the organization during the recruitment and transition period to a new General Manager. PJ will resume his position on the Music SA Board from the end of the caretaker period in January.
For enquiries, please contact:
Anne Wiberg
Music SA
annewiberg@internode.on.net

redpajama_set_name: RedPajamaCommonCrawl


How much is Walter Isaacson worth? – Wondering how wealthy & rich is Walter Isaacson? Or maybe you're just curious about Walter Isaacson's age, body measurements, height, weight, hair color, eye color, bra & waist size, bio, wiki, wealth and salary?
Walter Isaacson (New Orleans, May 20, 1952), American journalist, writer and biographer.
A native of New Orleans, Isaacson earned his B.A. in history and literature at Harvard University in 1974. He was subsequently a benefactor of a Rhodes Scholarship at Pembroke College, Oxford University.
He began his journalistic career in the Sunday Times of London. Later he was president and CEO of CNN and managing editor of Time Magazine.
He also held public office: in 2005, in his home state he worked at the Louisiana Recovery Authority, in charge of recovery from the Katrina disaster. Two years later he joined the friendship committee EE. UU – Palestine: Hillary Clinton appointed him vice president of Partners for a New Beginning, in pursuit of public-private partnerships in the Muslim world, and co-chairs the EE dialogue board. UU – Vietnam: In 2009, Obama named him president of the Broadcasting Board of Governors, which manages La Voz de America, Radio Europa Libre and other international radio stations.
He is currently president and CEO of the Aspen Institute, a center for political and educational studies in Washington, D.C.
He has written the biographies of Henry Kissinger, Benjamin Franklin, Albert Einstein and Steve Jobs. In 2012 he was selected to be part of the Time 100 list of the most influential people in the world.
The Walter Isaacson's statistics like age, body measurements, height, weight, bio, wiki, net worth posted above have been gathered from a lot of credible websites and online sources. But, there are a few factors that will affect the statistics, so, the above figures may not be 100% accurate.


text:
SINCE IT WAS FRIDAY
I HAD THE PHONE
As soon as I logged on, the owner transferred the phone to me. The only scheduled pickup was an 8:30p from the Kea Lani to OGG. I was also the only cab available, anywhere, and was "first up" in Wailea. About 5 minutes later, I get a call from some people at the Maui Oceanfront Inn (MOF) inquiring about the cab they had scheduled for 7:00p. Dropped call #1. I roll on it. Two minutes after that, the valet desk at The Grand calls, wanting to know where the cab was for the 7:00p pickup to OGG. Dropped call #2. Cab #27 had cleared in Kihei and was returning to Wailea. I told him to "beat feet" to The Grand. He finally made the pickup at 7:30p. The passengers were not "happy campers". Luckily for #27, this was being billed to a master account at The Grand, which includes an automatic 15% gratuity for the driver.
It is extremely frustrating to be "blindsided", twice, at the beginning of the shift.
The first couple of hours stayed very busy and I had to decline about 8 calls, since every cab at my disposal was tied up on runs and I was backlogged about 30-45 minutes.
About 10:00p, the rush ended and became just the routine of a Friday night. #05 had called off and myself, with #15, #25 and #27, trudged through the tedium of the night. Bar close was the usual Keystone Kop dog and pony show and then the phone never rang again until 4:20a. (one exception below)
That call was from MPD. They were requesting a cab in the middle of Kihei, on the highway, near Safeway. Some people needing to go to Kahului. I started to roll on it when the little bulb went on above my head. Flor, one of my dayshift reliefs, picks his van up at the far north end of Kihei but usually doesn't check-in until he reaches Wailea. I called him on the phone and, sure enough, he was just 1/2 mile from where the police had the people's vehicle stopped. I gave him the run. He had driven less than 2 miles when they demanded to be let out. They paid Flor and started walking back to their SUV. He radioed me and I called the non-emergency number for MPD and told them what had happened. They relayed this to the patrol officers. It must have been interesting when MPD stopped the vehicle again, after having given all the occupants a very big break. Can you spell S-T-U-P-I-D?
Speaking of stupid. During the bar close rush, I get a call from a guy in Mau'i Meadows who want to go to The Shops At Wailea. I dispatch #27. A short while later, #27 advises that there is no such address. I remembered how many calls back this was on the phone and do a call-back. I confirmed the address and then told the man that there was no such house number. The number he had given me was 679. After a brief discussion, he insisted that he was at 978. Whoa! Thats another mile further up into the hills. I told him to forget it. He was wasting our time and gas and sent #27 on to another call in Wailea. 45 minutes later the guy calls back and wants to know when his cab is going to arrive. I ask him for the pickup address and he tells me 679 again. I gently and kindly (Yeah, right!) informed him that no cab was coming and that I was the only cab on the southside at that time of the morning and he was probably going to have to walk the 4 niles to The Shops. Unlike Brian, over at The Blank Top Chronicles, who chews stupid callers up with the finesse of a scalpel, my response is more sledgehammer and chainsaw. I know that the use of rude, coarse and vulgar language is the sign of a limited vocabulary but I have found that it is the only thing that penetrates what few brain cells that most drunks have left functioning in their pea-sized cranial cavity.
Turned the phone and reservations over to Flor and then hung around until a second cab came on duty.
Overall, an average Friday, moneywise.
This morning's sunrise at Mauna Kea:
Here's another on-line pumpkin carving site. This one is legit, not like the one I posted before.
Pretty kewl.
Shaws Reef
Polo Beach
Po'olenalena
Wailea Beach
Ulua Beach
Keawakapu Beach
Mokapu Beach
Kihei Boat Ramp
Kamaole Beach
Butterfly Fish
Pukalani Sunset


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information management strategy in the lean production system, in the transportation sector in England. Lean construction and VM have recently gained momentum in England's transportation construction supply chain with high efficiency targets. VM in construction is a scarcely researched topic, particularly outside the building construction context with its quantitative and qualitative benefits. The recorded benefits of four different visual tools/systems (visual workplace structuring or the 5S, visual performance system, visual specification/indicator and visual control) from one highways construction and one metro station upgrade project in England were presented through an action and case study research effort. The findings show that VM systems can contribute to (i) increased self-management, (ii) increased coordination, (iii) increased PPC through better promises, (iv) easier control and (v) improved site conditions in transportation projects. Although both qualitative and quantitative data were collected to triangulate the findings, the main limitation of the research is in abstracting the benefits or contributions of a particular VM tool from an overall improvement in the projects' performance. Some future research opportunities for VM in the transportation sector were also discussed in the paper.
Tezel, A. , Tezel, T. , Koskela, L. & Koskela, K. 2016, 'Benefits of Visual Management in the Transportation Sector' In:, 24th Annual Conference of the International Group for Lean Construction. Boston, USA, 20-22 Jul 2016.


Wild Night -- As Brian Rolston correctly pointed out after Monday's 3-1 win against the Colorado Avalanche, every game the Minnesota Wild play the rest of the way will feel like a postseason contest.
And as we all know, goaltending is the biggest factor in the Stanley Cup Playoffs. That's why the Minnesota Wild should feel encouraged following Niklas Backstrom's 29-save performance against the Avs -- a performance that gave the Wild a two-point lead in the Northwest Division.
The victory was also key when you consider the fact that Minnesota begins a crucial four-game road trip Wednesday night at San Jose. The Wild will also visit Vancouver, Calgary and Edmonton.
"We know that we're leaving tomorrow for four games on the road," Backstrom said after Monday's victory. "We needed this win. We needed the two points. It's good for us. We got it in regulation. It's good to move on and get ready for Wednesday. It's good for our confidence to show we can play against these clubs. We know how we should play against them."
Colorado, meanwhile, is still very much alive in the division race despite the loss. The Avalanche return home to host the Flames on Thursday night in yet another "playoff" game.
Let the best part of the year begin.
"It's exactly like a playoff game," Rolston said. "The good thing about all of us being around the 84(-point) mark is we'll all be playoff ready. These last nine games will be a playoff game every night because we're all fighting, scratching and clawing to get into the playoffs."
The Canucks' Taylor Pyatt is congratulated by Daniel Sedin after scoring against the Phoenix Coyotes last night.
Twin Peaks -- Henrik and Daniel Sedin have been separated as of late by Vancouver Canucks coach Alain Vigneault, but the twins found enough time to play a role in a pair of goals in Monday night's 3-1 win over the Phoenix Coyotes at GM Place.
Henrik scored twice for his first non-empty net goals in 17 games, while Daniel had two assists for his first points of any kind in the last six contests. Their play helped the Canucks pull into a three-way tie for second place with Calgary and Colorado in the Northwest Division.
"It's great to see Hank score, he was right in front of the net on both goals and he stood up and played a good game tonight," Daniel Sedin said. "I think you have to shoot the puck and look for rebounds and then you can make plays. Maybe we got way from that when we were in this slump."
It's the perfect time for the Sedins to snap out of it, considering the Canucks will play four of their next five games on the road. They will visit the Edmonton Oilers Thursday night.
"We're trying to catch the teams in front of us," Henrik Sedin said, as Vancouver is just two points behind Minnesota for the division lead. "We have a really tough stretch here with a lot of games on the road. It was a huge win for us."
Here Comes Okposo -- While the New York Islanders are all but out of the Eastern Conference playoff race, the franchise is still doing its best to keep its fans excited.
Kyle Okposo, the team's first-round draft pick in 2006, with make his NHL debut Tuesday night when the Isles host the Toronto Maple Leafs at Nassau Coliseum. The 19-year-old has eight goals and 16 assists in 29 games with the Bridgeport Sound Tigers in the American Hockey League.
Listed at six feet and 195 pounds, Okposo will skate on a line Tuesday with Richard Park and Blake Comeau.
"I think I fit in pretty well out there, playing with Richard Park and Blake Comeau," Okposo said after Monday's practice. "Parkie's pretty responsible defensively, so that's going to help me out there. I feel like I'm ready."
The Islanders are certainly ready for Okposo. Entering Tuesday's action, New York is tied with the St. Louis Blues for the fewest number of goals scored in the NHL (176), a paltry average of 2.4 per game.
"I just look at is as a challenge and kind of try to ignore all the hype," Okposo said. "I'm just going to go out there and try to play my game and do the best I can to help the New York Islanders win."
That would be a nice change of pace, as the Isles have lost six straight games to basically fall out of playoff contention. If anything, though, these remaining nine games in the regular season will help coach Ted Nolan and GM Garth Snow gauge Okposo's development heading into the 2008-09 campaign.
"This is not the American Hockey League or college hockey," Nolan said. "Things happen a little bit quicker. It's the best League in the world. We're trying to put him with one of the most reliable guys that we have in Richard Park. He's probably going to make a few mistakes out there, but when you have Richard out there to compensate for that, that's a big thing. Blake Comeau has some good speed and good hockey sense. The three of those guys together should form a pretty good line."


Science Central provides a platform for the researchers and scientists to expose their ideas to the global community. It is the portal for scientific field which includes conferences, journals, researches, current issues and upcoming events related to Science. Research in science education is important, as it leads to progress in science. Exchange of thoughts and the novel ideas with individuals will lead in accelerating scientific discovery. e-Science Central will act as a bridge between researchers/scientists and the global community for human welfare. e-Science Central is a part of OMICS International. A Rebirth of scientific discovery has begun and OMICS International is proud to be the one accelerating it.


text:
Mandy Moore To Share Charity Hosting Duties With Hilary Duff
By Tim Saunders on December 10, 2008
Hilary Duff and Mandy Moore will co-host a special charity gala later tonight, December 10.
The event is being held at the Beverly Hills boutique Kitson to launch a collection of holiday merchandise to benefit Stand Up To Cancer. Items in the collection include a star necklace designed by Jennifer Meyer, Love Zooey Tees by Alice Heller, a set of Graf & Lantz felt star ornaments, and a Kumi KooKoon cashmere and silk throw blanket.
"Today's cancer scientists are on the verge of life-saving discoveries, but they desperately need the funds to mount an all-out assault," said Laura Ziskin, executive producer and founding member of Stand Up To Cancer. "The money raised from the sale of these beautiful items will go directly to funding these important research programs. Together we will defeat this insidious disease."
The collection can be viewed at the Stand Up To Cancer website, and all items will be available at Kitson stores across the US.
Copyright © 2008 Look to the Stars
Stand Up To Cancer
Stand Up To Cancer and Major League Baseball Launch Star-Studded New PSAs Jul 10, 2019
Stand Up To Cancer (SU2C) is set to launch a brand new public service announcement (PSA) campaign in collaboration with founding donor, Major League Baseball (MLB). More →
Tim McGraw Joins SU2C and AA to Launch Life's Journeys PSA Jul 4, 2019
Katherine Schwarzenegger Launches New Podcast - The Dog That Changed Me May 20, 2019
Avengers: Endgame Stands Up To Cancer Apr 23, 2019
Stand Up To Cancer and American Airlines: Life's Journeys
Stand Up To Cancer's 2018 Show Announcement with Bradley Cooper and Mitch Carbon
Rinse, Recycle, Repeat with Mandy Moore
Treasured Chest


In group "(CMIst) Affiliated event"
In group "(SOSS) Public events"
The course covers basic concepts in longitudinal design and analysis.
The morning session focuses on the strengths and methodological difficulties of the longitudinal approach such as defining longitudinal populations and target samples; levels and dimensions of change; age, period and cohort effects. There will be one small group discussion.
After lunch, we start with an overview session on the sources and causes of missing data (attrition, etc), and how to adjust for missingness; this will be followed by another group exercise.
confidence to address questions about longitudinal design and missing data.


The skin undergoes changes with time. Sun exposure, health habits, environmental pollutants and the pull of gravity, if allowed free rein on the skin, will wear it down. Within the body, the aging process also exacts its toll with reduced collagen production and slower cell turnover. Understanding skincare at every age and taking measures to slow the aging effects will help your skin stay resilient longer.
The skin is at its peak in the 20's with strong collagen production and oodles of elastin, making it plump and firm. However, take nothing for granted. Dermatologists agree that prevention is key in the 20's. Develop a good skin care routine: use sunscreen (at least SPF 30), moisturize and keep it clean. Maintain a healthy lifestyle with adequate sleep and healthy eating, while keeping stress levels down.
Imperfections start to creep in. You may notice first signs of aging such as expression lines around the mouth, eyes and on the forehead. All these due to slower cell renewal and reduced sebum production. Continue to be vigilant in your skin care routine and consider moisturizers that offer multiple benefits such as hydration, sun protection and skin repair. Skin products with antioxidants will help to encourage collagen production.
Skin shows more drastic changes as cell turnover slows down further and collagen, elastin and moisture level in the skin drop. Actual wrinkles and deeper expression lines develop as a result. This is the age where you may need to be more militant in your skincare. Continue to use sunscreen, daily moisturizing and more advanced skin care products. Look for skin products with vitamin C and peptides to help the connective tissue under the skin to produce collagen. You may need to exfoliate more frequently to get rid of dull skin. Salicylic and glycolic acids and retinoic acids make good exfoliating agents and help to promote skin radiance. Eye cream is now a definite must.
As hormones level drop, collagen production slows drastically. Sagging, jowling and hollowing may become the landscape on your skin. Fight back with collagen-boosting peptides and vitamin C serum. Encourage skin to retain moisture by using hyaluronic acid and ceramides. Moisturize well and hydrate continually to help skin retain its elasticity, bounce and radiance.
Knowing how your skin age and taking good care of it will give you the best chance of aging beautifully. Then, there is always a team of skin experts who can advise you on your skin care and take care of skin needs—whether it be moisturizing, hydrating, firming or renewing.
Call today for a free skin analysis and allow our experts to take care of your skin at every age.


text:
Home > Authors > BookLife
How to Network Better by Saying Less
Knowing what to say and when to say it can help writers make professional connections
By Jane Friedman |
When I was growing up, my mother often repeated the adage "Better to remain silent and be thought a fool than to speak and remove all doubt." It's rare that you hear that advice today (we live in very different times), but, for better or worse, I took it to heart and soon uncovered the strange power of silence. Far from preventing people from passing judgment on me, it did the opposite. By remaining quiet in a room full of people talking, and then offering a sharp observation, it increased my presence and influence. Sometimes people listen more carefully when you do finally speak, or they attribute meaning that isn't there to acts of silence.
Silence—or, in fact, just saying less—is an art that can be strategically practiced by authors trying to garner attention for their work. It is an especially critical strategy when trying to influence people who hear about new books all day, every day—such as editors, agents, booksellers and retailers, mainstream media, and reviewers. Such people, approached constantly by authors who want to talk about their books, feel immense relief when engaging with someone who knows how to give just enough information that allows for a conversation and a natural connection. None of them want a protracted speech that boils down to, "Please pay attention to me and my book."
Recently I attended an educational session by the publishing industry vet Amy Collins of New Shelves Books on how authors can effectively pitch their work to retail outlets. She asked an author in the audience to stand up and try pitching her published book as if she had met a bookseller at a party. The writer's response wasn't bad: it was essentially the meaty hook that would be appropriate in a query letter to an agent or editor. But, for a networking context, or even for a brief sales conversation, it was too detailed.
Collins emphasized that first the author should have stated that her work is a novel and then mentioned a couple of details that help relate the novel to something that the bookseller likely already knows. For example, let's say an author has written a novel based on historical events about a submarine that sinks off the coast of England during World War II. It appeals to men who love Michener. End there. The bookseller doesn't need any more to understand the readership and whether it's the sort of thing for her store. Maybe she'll say, "Oh, I sell tons of Michener—tell me more about the historical events you're writing about." Or maybe she'll respond, "Oh, my bookstore is focused on women's fiction, but you should talk to so-and-so librarian who's looking for that kind of thing."
What kind of details the author shares with someone is, of course, dependent on context. If speaking with someone who specializes in World War II historical novels, the author may be more specific about the historical events or the name of the submarine in the book. If the author is talking to someone who may not understand the appeal or readership of Michener, a different comparison would be better—something like, "The guys who love to watch the History Channel have been my most devoted readers."
Part of being a good salesperson for your work is developing a rapport with others and understanding their interests and needs. The conversation can't be focused on you. You're not sitting down with Terry Gross and divulging your origin story as a writer, how you were inspired to write the book, the twists and turns the story takes, and the research surprises along the way. Instead, you're focused on how your product (book) might fill a need for someone else or on looking for points of commonality. You have to put aside any impulse to digress about the content of the book, and instead be curious about and interested in exploring a business connection.
There may not be a match, but it's okay. There may still be a wonderful relationship to develop, and you won't become a person to be avoided later on, along the lines of, "Oh, there's that author who cornered me for 15 minutes and wouldn't stop talking about his book."
You've probably been at a reading or author event where people in the audience stand up to ask questions, but their questions are really a way for them to talk about their own work. You can feel people shifting in their seats, giving each other sidelong glances. The scenario often ends with the moderator awkwardly interrupting, "Sorry, but what's the question?"—that is, what do we have in common that we can talk about?
That urge to divulge detail, though, can be irresistible. When you feel that urge, identify it and ask yourself, can I gain more by saying less? What can I ask or share that will spark a connection with this person rather than a desire to escape? As an author, you don't want to follow my mother's advice and be entirely silent, but you can allow silence to spark a more useful and engaging conversation.
Jane Friedman is the former publisher of Writer's Digest; her newest book is The Business of Being a Writer from the University of Chicago Press.
A version of this article appeared in the 05/28/2018 issue of Publishers Weekly under the headline: How to Sell Better By Saying Less


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The need for a federal privacy law governing Facebook and other tech companies has felt like a necessity for a while now. However, Facebook isn't about to let the government oversee it without heavily influencing the process; according to a report last weekend in The New York Times, major tech companies are lobbying to basically create their own regulations.
While it sounds like these companies are being helpful on paper, in reality they want to create a national framework that could supersede a strict California privacy law that passed earlier this summer. So their commitment to being a "constructive part of the process," as tech lobbyist Dean Garfield said, is really just a way to protect themselves and neuter a law they don't like at the same time. Facebook and its cohort don't want stricter standards; they want to avoid them.
"The idea that the companies that violated our privacy for more than a decade will suddenly have a self-regulatory blueprint is ridiculous," Jim Steyer, president of privacy advocacy group Common Sense Media, told the Times.
Facebook almost certainly requires government regulation at this point in its history; it's simply too big. However, if the catch is Facebook will help create the regulations itself, then maybe that isn't worth it after all.


Transit police are asking for the public's help to identify a suspect following reports that a seven-year-old girl was groped on a SkyTrain.
The alleged incident took place shortly after 7 p.m. on Saturday, Feb. 2 as the young victim rode the train with her mother from Commercial-Broadway Station.
According to authorities, the suspect was already seated on the train across the aisle from the girl, and allegedly began directing his attention towards the victim.
"The suspect allegedly made several comments to the girl that were extremely vulgar and sexual in nature, including explicitly detailing sexual acts he wanted to engage in with the seven-year-old," police said.
"He then allegedly made several attempts to touch the girl and, at one point, allegedly touched the girl's buttocks, causing her mother to protectively pull her daughter onto her lap."
Investigators said the suspect left the train at Production Way Station when a bystander confronted him over the alleged inappropriate remarks.
Police received a text message report from another passenger about the incident. Officers met the train in question at Lougheed Station.
The suspect is described as a white man between 50 and 60 years old. He is about six feet tall, has grey curly hair and was wearing a grey sweater with white stripes on the sleeves, a black baseball cap, grey pants, black shoes and was carrying a Nike duffle bag at the time of the alleged incident.
Transit police released surveillance video of the suspect entering Coquitlam Station on Feb. 2, prior to the alleged incident.
Anyone with information that could advance the investigation is asked to contact authorities at 604-516-7419 or by texting 87.77.77 and referencing file number 2019-1962.


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GeoVax Awarded NIH Grant to Advance COVID-19 Vaccine Development
SmallCaps Daily 1 week ago
Company Focused on Single-dose Vaccine Against Multiple COVID Strains
Atlanta, GA – (NewMediaWire) – January 11, 2021 – GeoVax Labs, Inc. (NasdaqCM: GOVX), a biotechnology company developing immunotherapies and vaccines against cancers and infectious diseases, announced today that the National Institute of Allergy and Infectious Diseases (NIAID), part of the National Institutes of Health (NIH), has awarded the Company a Small Business Innovative Research (SBIR) grant in support of its development of a vaccine against SARS-CoV-2, the virus that causes COVID-19.
The Phase 1 grant, titled, "Preclinical Development of GV-MVA-VLP Vaccines Against COVID-19," will support the ongoing design, construction and preclinical testing of GeoVax's vaccine candidates in preparation for human clinical trials. The efficacy testing will be performed in collaboration with the University of Texas Medical Branch (UTMB).
GeoVax is leveraging its GV-MVA-VLP™ platform to address the global need for an effective and safe SARS-CoV-2 vaccine. Unique among other vaccines under development, the experimental GeoVax candidates are specifically designed to provide a broader and more long-lived level of protective immunity against SARS-CoV-2 while avoiding the potential side effects that can limit vaccine utility and acceptance. GeoVax's vaccine candidates will be tested for antigen expression and genetic stability under conditions designed to simulate those in manufacturing, which will demonstrate the likely suitability of each vaccine construct as a candidate for full-scale production and clinical testing.
Mark Newman, Ph.D., GeoVax's Chief Scientific Officer, commented, "The first generation of SARS-CoV-2 vaccines are based on the 'Spike (S)' protein and are designed to induce antibodies that block infection of human cells, an effect referred to as virus neutralization. The GV-MVA-VLP platform provides the opportunity to design and test vaccine candidates that differ significantly through the inclusion of multiple SARS-CoV-2 proteins that are presented to the immune system as virus-like particles (VLPs). Our goal is to safely increase vaccine potency and efficacy by inducing both neutralizing antibody and cellular immune responses to optimize the level of protection against existing and potential new variants of COVID as well as establish immunological memory to provide multi-year protection. Vaccines using the GV-MVA-VLP platform developed for other pathogens have proven to be efficacious with a single dose, having strong durability which would be a significant advantage for SARS-CoV-2 global vaccination campaigns."
David Dodd, GeoVax's Chairman & CEO, added, "We are pleased to receive this Phase 1 SBIR funding award, which will supplement the internal resources allocated to our COVID-19 vaccine program and accelerate our progress toward human clinical trials. We appreciate and welcome this support from NIH/NIAID, which follows our signing of a Patent and Biological Materials License Agreement providing GeoVax with access to key NIAID patent rights, and whichprovides GeoVax with nonexclusive rights to develop, manufacture and commercialize our COVID-19 vaccine. While we continue to advance our COVID-19 vaccine program, we remain in discussions and negotiations related to additional funding support that will further accelerate our progress into clinical testing and supply chain preparation. We anticipate that additional vaccines, such as ours will be necessary against COVID-19, as well as potentially new strains and variants, requiring broader immune response, strong durability, exquisite safety within various cohort populations and minimal refrigeration for distribution and supply throughout the world. These attributes represent the focus of our overall 'COVID-X' vaccine program (think 'COVID-20, -21', etc.)."
Coronaviruses are common in many species of animals, including bats and birds. These viruses can evolve to cross the animal species and infect humans and quickly spread from person to person resulting in lethal but rare respiratory infections. Recent epidemics with SARS and MERS coronaviruses resulted in 774 and 858 deaths, respectively. Since 2015 there have not been any cases of SARS and MERS reported, but in January 2020, World Health Organization (WHO) identified a novel coronavirus, recently named SARS-CoV-2, in the city of Wuhan, China. On January 31, 2020, the WHO declared the novel coronavirus to be a global health emergency, and on March 11, 2020 the WHO declared a global pandemic. Worldwide, more than 85 million people have been infected (over 20 million in the U.S.), with nearly 2 million deaths (over 350,000 in the U.S). The situation is fluid, with the infection and death statistics changing significantly on a regular basis.
About GeoVax
GeoVax Labs, Inc. is a clinical-stage biotechnology company developing human vaccines against infectious diseases and cancers using a novel patented Modified Vaccinia Ankara-Virus Like Particle (MVA-VLP) based vaccine platform. On this platform, MVA, a large virus capable of carrying multiple vaccine antigens, expresses proteins that assemble into VLP immunogens within (in vivo) the person receiving the vaccine. The production of VLPs in the person being vaccinated mimics virus production in a natural infection, stimulating both the humoral and cellular arms of the immune system to recognize, prevent, and control the target infection. The MVA-VLP derived vaccines have been shown to elicit durable immune responses in the host similar to a live-attenuated virus, while providing the safety characteristics of a replication-defective vector. The goal of the GeoVax technology (GV-MVA-VLP™) is to provide high efficacy in a single dose, without the need for additional immune response components ("adjuvants"), strong durability ("minimizing the need for boosters"), the highly-validated safety of MVA, minimal refrigeration requirements, and cost-effective manufacturing.
GeoVax's current development programs are focused on preventive vaccines against COVID-19, Zika Virus, hemorrhagic fever viruses (Ebola, Lassa, Marburg and Sudan), HIV and malaria, as well as therapeutic vaccines against multiple cancers. The Company has successfully completed six human clinical trials related to HIV, having designed a preventive HIV vaccine candidate to fight against the subtype of HIV prevalent in the geographic markets of the Americas, Western Europe, Japan, and Australia. Human clinical trials for this program are managed by the HIV Vaccine Trials Network (HVTN) with the support of the National Institutes of Health (NIH). GeoVax's HIV vaccine is also part of two separate collaborative efforts to apply its innovative approach toward a functional cure for HIV, addressing those already infected with HIV.
This press release contains forward-looking statements regarding GeoVax's business. The words "believe," "may," "estimate," "continue," "anticipate," "intend," "should," "plan," "could," "target," "potential," "is likely," "will," "expect" and similar expressions, as they relate to us, are intended to identify forward-looking statements. We have based these forward-looking statements largely on our current plans and expectations about our product development, future events and trends that we believe may affect our financial condition, results of operations, business strategy and financial needs. Actual results may differ materially from those included in these statements due to a variety of factors, including whether: GeoVax can develop and manufacture its vaccines with the desired characteristics in a timely manner, GeoVax's vaccines will be safe for human use, GeoVax's vaccines will effectively prevent targeted infections in humans, GeoVax's vaccines will receive regulatory approvals necessary to be licensed and marketed, GeoVax raises required capital to complete vaccine development, there is development of competitive products that may be more effective or easier to use than GeoVax's products, GeoVax will be able to enter into favorable manufacturing and distribution agreements, the impact of the COVID-19 pandemic continues, and other factors, over which GeoVax has no control.
Further information on our risk factors is contained in our most recent Quarterly Report on Form 10-Q that we have filed with the SEC. Any forward-looking statement made by us herein speaks only as of the date on which it is made. Factors or events that could cause our actual results to differ may emerge from time to time, and it is not possible for us to predict all of them. We undertake no obligation to publicly update any forward-looking statement, whether as a result of new information, future developments or otherwise, except as may be required by law.
GeoVax Labs, Inc.
investor@geovax.com
Advance Awarded covid19 development GeoVax grant NIH Vaccine
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Margaret started her accountancy training in local government but following the death of her husband when her two children were toddlers her career changed direction. Together with two colleagues Margaret owned and ran two small businesses before her retirement in 1998.
After reading an article on the work of Nepacs at Durham Visitors' Centre she became a volunteer over 21 years ago. Margaret continues to volunteer at the visitors' centre and in her spare time enjoys watching most sports, particularly her local football team, Darlington FC.


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Hetty's Secret War by Rosie Clark: A Review and Extract
Greetings! Rosie Clarke is happily married and lives in a quiet village in East Anglia. Writing books is her passion. Rosie also likes to read, watch good films, shoes, holidays in the sunshine, and animals (especially dogs and squirrels). Rosie Clarke also writes The Mulberry Lane series (I just love it). Readers can follow Ms. Clarke on Twitter (@AnneHerries) and Facebook (@RosieClarke).
Hetty's Secret War by Rosie Clarke is the final installment in Women at War Trilogy. It is early summer in 1939 in Yorkshire when Georgie Bridges receives the news that her husband, Arthur is ill and will soon pass away. Georgie has had a good life with Arthur, and she will miss him when he is gone. However, she has never stopped loving Ben Tarleton who is training for a special war mission. Georgie is happy to reconnect with Ben, but their time is short. Ben is soon sent overseas, and Georgie may never see him again.
Beth Rawlings is nineteen and has just finished her secretarial courses. Annabel offers her a job at Rowntree House Hotel, but Beth would like to spread her wings and help the war efforts. She meets Captain Drew Bryant on the train and falls head over heels in love. When they receive word, that Drew is being deployed, they quickly marry. Beth keeps busy at her job with Arnold Pearson, but she soon discovers that she is with child. Just before Beth is due, she receives a telegram that Drew is missing in action. Will Drew be found, or will Beth become a war widow?
Hetty Tarleton has been living in Paris for ten years when war breaks out. She is a free spirit and artist. Her friends encourage her to return to England, but Hetty does not wish to leave. With the Germans about to enter France, Hetty departs in her car where she navigates roads clogged with people trying to get out of France before it is too late. After a horrific incident, Hetty meets Pierre de Faubourg who takes her to his family's home, Chateau de Faubourg. Hetty becomes close with Pierre's mother, Adele. She joins the resistance where she meets the enigmatic Stefan Lefarge. On one mission Hetty encounters the last person she expected to find in France. Hetty takes great risks fighting the German invaders. Will she make it through the war alive? Join Hetty, Beth, and Georgie as they struggle to survive World War II in Hetty's Secret War.
Hetty's Secret War can be read as a standalone, but you will find it confusing in the beginning. I suggest you read Jessie's Promise and The Runaway Wife before embarking on Hetty's Secret War. You will have a better understanding of the characters and their situations. I have always found World War II a fascinating time period (I would not have wanted to experience it though). Women were encouraged to work outside the home and take on tasks that were previously done by men. It was a difficult time, but people rallied together for the war effort. This is what we see in Hetty's Secret War. I found the story nicely written with a variety of characters in different situations. There is foul language and intimate scenes included in the book (fair warning). While the book focused on Hetty, Beth and Georgie, we also catch up with Annabel. I like how the separate characters stories intertwined to create one charming book. They face losses and experience heartache, but there is joy as well. Hetty experiences danger and life threatening situations. Despite the hardships, they continued to have hope and faith as well as the comfort of family and friends. The ending had me smiling. Hetty's Secret War is an emotional novel that will touch your heart.
Here is an extract from Hetty's Secret War to entice you:
'But it makes no sense for you to stay here in Paris, ma chérie,' Madame Arnoud said and spread her hands in an expression of disbelief. 'You are English not French. You should go home, get away from this madness before it is too late. Believe me, I am old enough to remember the last time the Germans paid us a visit. It was not pleasant.' Although well into her middle years the Frenchwoman's clothes, make-up and dark brown hair followed the latest mode and she looked both stylish and attractive.
'But I feel more French than English these days,' Hetty replied, wrinkling her nose at the older woman's comment. 'I have so many friends here and I love the life I'm leading – why should I give it all up?'
'Because the Germans will make you suffer if they catch you out,' Madame Arnoud said. 'You will probably be sent to an internment camp, that's if you're not shot as a spy.' 'Perhaps they won't invade…'
'Pouff!' the Frenchwoman snorted her disbelief. 'It is more likely that pigs will sprout wings. They will come, Hetty, believe me – it is merely a question of when.'
They were sitting in Madame's private parlour drinking wine, something they often did in the evenings when Hetty called to discuss her latest designs or simply to talk about what she had seen or done. They were good friends and had been for some years, since Hetty had first approached her rather tentatively with a design for an evening gown.
'Yes, perhaps,' Hetty agreed. 'But there's time yet, madame. I shall think about leaving when it becomes inevitable. Not that I've any idea of what I'll do when I get back to England. It will be difficult to settle anywhere else but Paris. Oh, it's such a shame that wars have to happen! Why must the Germans be so awful? Why can't they just leave us alone?'
'If we knew the answer to that the world would be a different place,' Madame Arnoud said and offered a world-weary smile. 'It is men who make wars, ma chérie, and we all know about them, do we not?'
Hetty laughed. At twenty-six years of age she had matured into a woman of some style, her hair a rich honey blonde that she wore long and in soft waves rather like Marlene Dietrich, the German film star with the gravelly voice, who had first made her name in the 1930 film The Blue Angel.
When at the age of seventeen she'd eloped to France to be with Henri, Hetty had been pretty rather than beautiful, but now she was stunning. Many of the artists she knew begged to paint her portrait, but these days she preferred to use the brush herself and earned a precarious living drawing quick sketches of the tourists, supplementing her meagre income with the work she did for Madame Arnoud.
'Yes, of that there is no doubt,' Hetty agreed. She had learned how selfish a man could be the hard way, weeping bitter tears the first time she'd discovered her lover, Henri, had been unfaithful to her with his latest model. She'd given up everything to come to Paris with him – her family home, the chance of marriage and a normal life – but she'd adored the fascinating artist who had challenged her to be bold. His betrayal had almost torn her in two that first time, making her weep into her pillow. He had told her he was sorry afterwards, swearing that the girl meant nothing and that it was her he loved. Hetty had forgiven him, but it had happened again, and again, until she woke up one day to discover that he no longer meant anything to her. It was over – the passion and love she'd had for him gone, destroyed by his lack of care for her. In the end, he was the one who had wept when she walked out on him, begging her to reconsider.
It had been hard at first without Henri, difficult to find work, her income barely enough to keep body and soul together, and lonely too. She had thought about going home to England, but something inside her had refused to give in – just as she had refused all the offers from Henri's friends to take his place in her life. Whether that had been from pride, a lack of interest sexually in the men themselves or her fierce independence, she had never been sure, but she had remained alone. And gradually she had found a new life and new friends; she had won respect for her own work, both as an artist on the Left Bank and as a dress designer for Madame Arnoud.
She could have worked full-time for the woman who had become both a friend and almost the mother Hetty felt she'd never had, if she'd wanted to be a model or a vendeuse, but neither of those things appealed to her. Besides, she now earned enough to pay the rent of her little apartment and to be able to buy food and clothes. She had no interest in more and found the relaxed, pleasant way of living suited her nature.
She might not always have been happy, but her life was busy, interesting, and she made sure it stayed that way. Love was something she'd learned to do without when she was a small child. Her father had been a kindly but remote figure, her mother cold and severe; both Ben and Annabel had been generous and kind, but they were twins and closer to each other. In the early years she had wept bitter tears over her mother's lack of affection, but then she had come to realise that it was something lacking in Lady Tarleton: she was a woman incapable of loving anyone other than herself and treated Annabel even worse than Hetty. Becoming independent and resourceful beyond her years, Hetty had found the best times were when she was in the kitchen with Cook and the maids who always had a soft word and a smile for her. And then Henri had come into her life and she'd given her innocent heart to him – a gift he'd taken and discarded without thought.
But Madame Arnoud was talking to her, scolding her, giving her the advice she knew she ought to take but was stubbornly resisting. Here in France she had a life, but there was nothing waiting for her in England.
Hetty's Secret War is available at Amazon, Kobo, Google Play, and iBooks. I hope you take the time to visit the other stops on the tour for Hetty's Secret War (see below). Thank you for joining me today. I will showcase The Last Year of the War by Susan Meissner next time. I hope I have helped you to find a new book and author. Take care and Happy Reading!
Labels: historical fiction, Historical romance, Rosie Clarke, women at war
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