\documentclass [11pt,reqno]{article} \usepackage{graphicx,amssymb,amsfonts,latexsym,amsmath,amsthm,times} \usepackage{epsfig} \usepackage{color} \usepackage[english]{babel} \usepackage[latin1]{inputenc} \usepackage[T1]{fontenc} \usepackage{hyperref} \newtheorem{rien}{} \newtheorem{thm}{Theorem}[section] \newtheorem{pro}{Proposition}[section] \newtheorem{rem}{Remark}[section] \newtheorem{defn}{Definition}[section] \newtheorem{preuve}{proof}[section] \newtheorem{cor}{Corollary}[section] \setlength{\textwidth}{6.5in} \textheight=8.5in \oddsidemargin 0in \topmargin -0.5 in \renewcommand{\theequation}{\thesection.\arabic{equation}} \numberwithin{equation}{section} %******************************************************************* \usepackage{fancyhdr} \rhead{\bfseries {\it }} \lhead{\bfseries {\it A. El Amrouss and A.Ourraoui.}} \cfoot{\footnotesize \it{ Existence of solutions of a boundary problem involving p(x)-biharmonic operator.}} \rfoot{\thepage} \renewcommand{\headrulewidth}{0.4pt} \renewcommand{\footrulewidth}{0.4pt} \begin{document} \vspace*{2cm} \normalsize \centerline{\Large \bf Existence of solutions for a boundary problem } \normalsize \centerline{\Large \bf involving p(x)-biharmonic operator} \vspace*{1cm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%% \centerline{\bf Abdel Rachid El Amrouss$^1$ and Anass Ourraoui$^2$ } \vspace*{0.5cm} %******************************************************************* %ADDRESS OF THE AUTHORS %******************************************************************* \centerline{$^1$ Faculty of Sciences, Department of Mathematics,Oujda, Morocco.} \centerline{$^2$ $~$ Faculty of Sciences, Department of Mathematics,Oujda, Morocco. } \centerline{$^1$ \footnotesize \it{ E-mail:a.elamrouss@fso.ump.ma, elamrouss@hotmail.com}} \centerline{$^2$ \footnotesize \it{ E-mail:anas.our@hotmail.com}} \vspace*{0.5cm} %******************************************************************* \noindent {\bf Abstract.} {\it In this paper, we establish the existence of at least three solutions to a boundary problem involving the p(x)-biharmonic operator. Our technical approach is based on theorem obtained by B. Ricceri's variational principale and local mountain pass theorem without (Palais.Smale) condition.} \vspace*{0.5cm} \noindent {\bf Key words:}{\it p(x)-biharmonic, neumann problem, variational methods.} \vspace*{1cm} \setcounter{equation}{0} %%%%%%%%%%%%%%%%%%%%%%========================================= \section{\hspace*{-.5cm}-Introduction} The study of various mathematical problems with variable exponent have received a lot of attention in recent years \cite{1,14}. Fourth order equations appears in many contexts. Some of these problems come from different areas of applied mathematics and physics such as Micro Electro-Mechanical systems, surface diffusion on solids, flow in Hele-Shaw cells (see \cite{9}). In addition, this type of equation can describe the static from change of beam or the sport of rigid body, there are many authors have pointed out that type of non linearity furnishes a model to study traveling waves in suspension bridges (see \cite{5,11}).\\ \hspace*{-.59cm}In this paper, we consider the following $p(x)-$biharmonic problem with a boundary condition, \vspace*{.6cm} $(\mathcal{P})~~\left\{\begin{array}{rl} & \mbox{$\Delta_{p(x)}^{{2}} u+a(x)\mid u\mid^{p(x)-2}u=f(x,u)+\lambda g(x,u)~~in~~\Omega ,$} \\ & \mbox{$ Bu=Tu=0~~on~~\partial\Omega.$}\end{array}\right.$ \vspace*{.6cm} \hspace*{-.62cm} Here $\Omega$ is a bounded open domain in $\mathbb{R}^N$ with smooth boundary $\partial\Omega$, $\Delta_{p(x)}^{{2}} u=\Delta(\mid \Delta\mid^{p(x)-2}\Delta u)$ is the $p(x)-$biharmonic with $p\in C(\overline{\Omega}),~ p(x)>1,$ $\lambda\in \mathbb{R}$ , $a\in L^{\infty}(\Omega)$ such that $ \displaystyle ess\inf_{x\in \Omega}a(x)=a^- >0$. \\ $Bu=Tu=0$ denotes the following boundary conditions:\\ $(1)$ $B=B_1,T=T_1$, Navier boundary condition, i.e. $$B_1u=\Delta u=0~~and ~~T_1u=u=0~~on~~\partial\Omega.$$\\ $(2)$ $B=B_2,T=T_2$, Neumann boundary condition, i.e \\$$B_2u=\frac{\partial u}{\partial\nu}=0~~and ~~T_2u=\frac{\partial}{\partial\nu}(\mid \Delta u\mid^{p(x)-2}\Delta u)=0~~on~~\partial\Omega,$$ \\ where $\nu$ is the outward unit normal to $\partial\Omega.$ \\ We define \\ $F(x,t)=\displaystyle\int_0^t f(x,s)ds $ and $G(x,t)=\displaystyle\int_0^t g(x,s)ds.$ \\ \hspace*{.0cm}Throughout this paper, we suppose the following assumptions:\begin{equation}\label{H} \hspace*{-0.15cm}(F)~~f,g \in C(\overline{\Omega}\times\mathbb{R},\mathbb{R}) ~~such~~ that~~\mid f(x,t)\mid,\mid g(x,t)\mid\leq C_1+C_2 \mid t\mid^{q(x)-1}~~ \forall (x,t)\in \overline{\Omega}\times \mathbb{R}), \end{equation} \hspace*{0cm} where $q\in C(\overline{\Omega}),C_1,C_2>0$ and $1\leq q(x)< p^*(x)~~\forall x\in\overline{\Omega}.$\\ With \\ $~~p^*(x)=\left\{\begin{array}{rl} & \mbox{$\frac{N-p(x)}{N-2 p(x)}~~if~~ p(x)< \frac{N}{2}$} \\ & \mbox{$+\infty~~~~~~~if~~ p(x)\geq \frac{N}{2}$}\end{array}\right.$ \\ \hspace*{-0.62cm} $(F_1)~~$ $\displaystyle\lim_{\mid t\mid\rightarrow\infty} [F(x,t)-\frac{\lambda_1}{p(x)}\mid t\mid^{p^-}]=-\infty$ uniformly for almost every $x\in\overline{\Omega}.$ \\ \hspace*{-0.62cm} $(F_2)~~$ There exist $x_0\in\Omega,r_0\in]0,1[$ and $t_0>1$ with $B(x_0,2r_0)\subset\Omega$ such that \\ \hspace*{.5cm}$F(x,t)~\geq~0~~~~ ~~for ~~~~x\in B(x_0,2r_0)\subset\Omega~~and~~t\in]0,t_0],$\\ \hspace*{1.08cm}$F(x,t_0)\geq C_0~~~for~~~~x\in B(x_0,r_0).$\\ Where \\ \hspace*{1.1cm}$C_0=[(\frac{2}{r_0})^{p^+(B(x_0,2r_0))}(2^N-1)+\mid a\mid_\infty 2^N]\frac{\mid t_0\mid^{p^+(B(x_0,2r_0))}}{p^-(B(x_0,2r_0)}.$\\ \hspace*{-.66cm} $(F'_2)~~$ There exist $\xi\in \mathbb{R}$ such that \\ $$\int_\Omega F(x,\xi)dx>\int_\Omega \frac{a(x)}{p(x)}\mid\xi\mid^{p(x)}dx.$$ \\ \hspace*{-.62cm} $(F_3)~~$ There exist \hspace*{0cm} $b_0>0,\delta>0$ such that \\ \hspace*{1cm}$F(x,t)\leq b_0\mid t\mid^{q_0(x)},\forall x\in\Omega,\mid t\mid<\delta,$ where $q_0\in C(\overline{\Omega})$ with $p^+0$ and $\gamma>0$ such that $$G(x,t)\geq b\mid t\mid^{\beta^+(B(x_1,r_1))} ~~for~~ all~~x\in(B(x_1,r_1))~~and~~\mid t\mid<\gamma.$$ \\ With \\ $$\beta^+(B(x_1,r_1))=\displaystyle\sup_{x\in B(x_1,r_1)}\beta(x) ~~,~~\beta^-(B(x_1,r_1)=\displaystyle\inf_{x\in B(x_1,r_1)}\beta(x).$$ \hspace*{-0.086cm} $ (G_2)~~ \displaystyle\limsup_{t\rightarrow 0}\frac{\inf_{x\in\Omega}G(x,t)}{\mid t\mid^{p-}}=+\infty.$ \\ \hspace*{.1cm}In the case $B=B_1$ and $T=T_1$, we claim the following theorem. \begin{thm} Suppose that assumptions $(F_1)$, $(F_2)$, $(F_3)$, $(G_1)$ and $(F)$ hold. Then the problem $(\mathcal{P})$ admits at least three weak solutions. \end{thm} The case $B=B_2,T=T_2$, we have the following result: \begin{thm} Under the assumptions $~(F_1)$ ,$(F'_2)$ ,$(F_3)$ $(G_2)$ and $(F)$ ,\\ the problem $(\mathcal{P})$ has at least three weak solutions. \end{thm} Many authors consider the existence of multiple nontrivial solutions for some fourth order problems \cite{11,16}. In particular, Li and Tang \cite{10} consider the p-biharmonic equation. Using the modified three critical points theorem of B. Ricceri they get at least three solutions. The $p(x)-$biharmonic operator possesses more complicated nonlinearities than $p-$biharmonic, for example, it is inhomogeneous. Recently, in\cite{4} A. Ayoujil and A. R. El Amrouss interested to the spectrum of a fourth order elliptic equation with variable exponent. They proved the existence of infinitely many eigenvalue sequences and $sup\Lambda = +\infty$, where $\Lambda$ is the set of all eigenvalues. Moreover, they present some sufficient conditions for $inf \Lambda = 0$. The technical approach is based on the Ricceri's variational principal and local mountain pass theorem\cite{3}, without Palais- Smale condition. One of the first result in this direction was obtained by Shao-Gao Deng \cite{6} for the $p(x)-laplacien$, here, we borrow some ideas from that work. The purpose of this work is to improve the results of \cite{6} and extend them to the case of p(x)-biharmonic equation with Navier and Neumann condition. \\ \hspace*{.5cm} This article consists of three sections. In section 2, we start with some preliminary basic results on theory of Lesbegue-Sobolev spaces with variables exponent( we refer to the book of Musielak \cite{13} , ${\rm Mih\breve{a}ilescu}$ and ${\rm R\breve{a}dulescu}$ \cite{12}), we recall Ricceri's variational principal with some results which are needed later. In section 3, we give the proof of the main result. \section{\hspace*{-.5cm} -Preliminaries} \subsection{\hspace*{-.5cm} -Variable exponent space and Sobolev Spaces} In order to deal with the problem ($\mathcal{P}$), we need some theory of variable exponent Sobolev Space. For convenience, we only recall some basic facts which will be used later.\\ \hspace*{-.57cm}Suppose that $\Omega \subset \mathbb{R^N}$ be a bounded domain with smooth boundary $\partial\Omega.$ \\ Let $C_+(\overline{\Omega})=\{p\in C(\overline{\Omega})~~and~~\displaystyle ess\inf_{x\in \overline{\Omega}}p(x)>1\}.$ \\ For any $p(x)\in C_+(\overline{\Omega})$, denote by $p^-=\displaystyle\min_{x\in \overline{\Omega}}p(x)$, $~p^+=\displaystyle\max_{x\in \overline{\Omega}}p(x)$ \\ and $~~p^*_k(x)=\left\{\begin{array}{rl} & \mbox{$\frac{N-p(x)}{N-k p(x)}~~if~~k p(x)< N$} \\ & \mbox{$+\infty~~~~~~~if~~k p(x)\geq N$}\end{array}\right.$ \\ Define the variable exponent Lebesgue space $L^{p(x)}(\Omega)$\\ $L^{p(x)}(\Omega)=\{u\in L^1(\Omega):~\displaystyle\int_{\Omega}\mid u\mid^{p(x)}dx < \infty\}$ \vspace*{.3cm} \hspace{-.65cm} then $L^{p(x)}(\Omega)$ endowed with the norm \\ $\parallel u\parallel_{p(x)}=\inf\{\lambda>0:~ \displaystyle\int_{\Omega}\mid \frac{u}{\lambda}\mid^{p(x)}dx \leq1\}$ \\ becomes a Banach space separable and reflexive space. \begin{pro} Set, $\rho(u)=\int_{\Omega} \mid u\mid^{p(x)} dx,$ if $u\in L^{p(x)}(\Omega)$ we have : \\ (1) $\parallel u\parallel_{p(x)}\geq 1 \Rightarrow \parallel u\parallel_{p(x)}^{p^-}\leq \rho(u)\leq \parallel u\parallel_{p(x)}^{p^+}.$ \\ (2) $\parallel u\parallel_{p(x)}\leq 1 \Rightarrow \parallel u\parallel_{p(x)}^{p^-}\geq \rho(u)\geq \parallel u\parallel_{p(x)}^{p^+}.$ \end{pro} \vspace*{.5cm} \hspace*{-.6cm} Define the variable exponent Sobolev space $W^{k,p(x)}(\Omega)$ by \\ $$W^{k,p(x)}(\Omega)=\{u\in L^{p(x)}(\Omega):D^{\alpha}u\in L^{p(x)}(\Omega),\mid\alpha\mid\leq k\}$$ \\ \hspace*{0.cm}where $~~D^{\alpha}u=\frac{\partial^{\mid\alpha\mid}}{\partial^{\alpha_{1}} x_1...\partial^{\alpha_{N}} x_N}$ \ \hspace*{-.7cm} (with $\alpha=(\alpha_1,\alpha_2,...,\alpha_N)$ is a multi-index and $\mid\alpha\mid=\displaystyle\Sigma_{i=1}^{N}\alpha_i.)$ \\\\ The space $W^{k,p(x)}(\Omega)$ with the norm $\parallel u\parallel=\displaystyle\Sigma_{\mid\alpha\leq k\mid} \mid D^{\alpha}u\mid_{p(x)}$ is a Banach separable \\ and reflexive space. \begin{pro} \cite{2,7,13} For $p,r\in C_+(\overline{\Omega})$ such that $r(x)\leq p^*_k(x)$ for all $x\in\overline{\Omega}$, \\ there is a continuous and compact embedding \\ $$W^{k,p(x)}(\Omega)\hookrightarrow L^{r(x)}(\Omega).$$ We denote by $W^{k,p(x)}_0(\Omega)$ the closure of $C^{\infty}_0(\Omega)$ in $W^{k,p(x)}(\Omega).$ \end{pro} \begin{rem} $\vspace*{0.5cm}\\ ~~1) (W^{2,p(x)}(\Omega)\cap W^{1,p(x)}_0(\Omega),\parallel . \parallel)$ is a Banach space separable and reflexive space. \hspace{-.58cm}2) Define $\parallel u\parallel_a=\inf\{\lambda>0:~\int_\Omega [\mid \frac{\Delta u}{\lambda}\mid^{p(x)}+a(x)\mid\frac{u}{\lambda}\mid^{p(x)}] dx \leq1 \}$ \\ for all $u\in W^{2,p(x)}(\Omega)~or~W^{2,p(x)}(\Omega)\cap W^{1,p(x)}_0(\Omega)$ then , $\parallel u\parallel_a$ is a norm on $W^{2,p(x)}(\Omega)~and~W^{2,p(x)}(\Omega)\cap W^{1,p(x)}_0(\Omega)$ and equivalent the usual one.\\ 3) By the above remark and proposition 2.2 there is a continuous and compact embedding of $W^{2,p(x)}(\Omega)\cap W^{1,p(x)}_0(\Omega)$ into $L^{r(x)}(\Omega)$, where $r(x)0,$ which means that $L'$ is strictly \hspace*{.55cm} monotone. \\ \hspace*{.62cm}Let $(u_n)_n$ be a sequence of $X$ such that \\ $$u_n\rightharpoonup u ~~{\rm in}~~X$$ and $$\displaystyle\limsup_{n\rightarrow\infty}\langle L'(u_n),u_n-u\rangle\leq0.$$ It suffices to show that \begin{equation} \label{eq3.1} \int_\Omega (\mid \Delta u_n-\Delta u\mid^{p(x)}+a(x) \mid u_n-u\mid^{p(x)})\rightarrow 0 \end{equation} by the monotonicity of $L'$ , we claim that\\ $$ \langle L'(u_n)-L'(u),u_n-u \rangle \geq 0 .$$ \hspace*{.6cm}Since $ u_n\rightharpoonup u$ in $X$ then \begin{equation} \label{eq3.2} \displaystyle\limsup_{n\rightarrow\infty}\langle L(u_n)-L(u),u_n-u\rangle=0. \end{equation} \hspace*{-.1cm}Put \\ \hspace*{0cm}$$\chi_n(x)=(\mid \Delta u\mid^{p(x)-2}\Delta u_n-\mid \Delta u\mid^{p(x)-2}\Delta u).(\Delta u_n-\Delta u),$$\\ \hspace*{0cm}$$\xi_n(x)=(\mid u_n\mid^{p(x)-2}u_n-\mid u\mid^{p(x)-2}u).(u_n-u).$$\\ \hspace*{.5cm}By the compact embedding of $X$ into $L^{p(x)}(\Omega),$ \\ $$ u_n\rightarrow u~~ in~~L^{p(x)}(\Omega) $$ \\ $$ \mid u_n\mid^{p(x)-2}u_n\rightarrow \mid u\mid^{p(x)-2}u ~~in~~L^{q(x)}(\Omega)$$ where $\frac{1}{p(x)}+\frac{1}{q(x)}=1$ for all $x\in\Omega.$ It follows that, \begin{equation} \label{eq3.3} \int_\Omega \xi_n(x)dx\rightarrow 0 \end{equation} using \eqref{eq3.1} and \eqref{eq3.2}, then we have \\ \begin{equation} \label{eq3.4} \displaystyle\limsup_{n\rightarrow\infty}\int_\Omega \chi_n(x)dx =0. \end{equation} Thanks to the above inequalities, \\ $$ \int_{U_p}\mid \Delta u_n-\Delta u_k\mid^{p(x)}dx\leq 2^{p^+}\int_{U_p}\chi_n(x)dx,$$ $$\int_{U_p}\mid u_n- u_k\mid^{p(x)}dx\leq 2^{p^+}\int_{U_p}\xi_n(x)dx.$$ It results that\\ \begin{equation} \label{eq3.5} \int_{U_p}(\mid \Delta u_n-\Delta u\mid)\mid^{p(x)}+a(x)\mid u_n-u\mid^{p(x)}dx\rightarrow 0 ~~ as~~ n\rightarrow\infty. \end{equation} \hspace*{.6cm}Besides, in $V_p,$ put $\delta_n=\mid \Delta u_n\mid+\mid \Delta u\mid,$ we have \\ \hspace*{.5cm}$$\int_{V_p}\mid \Delta u_n-\Delta u\mid^{p(x)}dx\leq \frac{1}{p^{-1}-1}\int_{V_p}(\chi_n)^{\frac{p(x)}{2}}(\delta_n)^{\frac{p(x)}{2}(2-p(x))}dx.$$ \hspace*{.5cm} By Young's inequality, \begin{eqnarray} \label{eq3.6} d\int_{V_p}\mid \Delta u_n-\Delta u\mid^{p(x)}dx &\leq& \int_{V_p}(d\chi_n)^{\frac{p(x)}{2}}(\delta_n)^{\frac{p(x)}{2}(2-p(x))}dx\nonumber \\ &\leq &\int_{V_p}\chi_n(d)^{\frac{2}{p(x)}}dx+\int_{V_p}(\delta_n)^{p(x)}dx. \end{eqnarray} \hspace*{.65cm} From \eqref{eq3.4} and since $\chi_n\geq0,$ one consider that \\ $$ 0\leq\int_{V_p}\chi_n dx<1.$$ \hspace*{.55cm} If $\int_{V_p}\chi_n dx=0$ then $\int_{V_p}\mid \Delta u_n-\Delta u\mid^{p(x)}dx=0.$ Or else, we take $d=(\int_{V_p}\chi_n dx)^{\frac{-1}{2}}>1,$ \\ \hspace*{.65cm}and the fact that $\frac{2}{p(x)}<2$, inequality \eqref{eq3.6} becomes \\ \begin{eqnarray} \int_{V_p}\mid \Delta u_n-\Delta u\mid^{p(x)}dx\leq \frac{1}{d}(\int_{V_p}\chi_n d^2 dx+\int_\Omega \delta^{p(x)}_n dx), \\ \hspace*{4.5cm} \leq \int_{V_p}(\chi_n dx)^{\frac{1}{2}}(1+\int_{\Omega}\delta^{p(x)}_n dx). \end{eqnarray} Note that, $\int_\Omega \delta^{p(x)}_n dx$ is bounded, which implies \\ $$ \int_{V_p}\mid \Delta u_n -\Delta u\mid^{p(x)}dx\rightarrow 0~~as~~n\rightarrow\infty. $$ Similarly we can have $$\int_{V_p}\mid u_n-u\mid^{p(x)}dx\rightarrow 0~~as ~~n\rightarrow\infty.$$ Hence, it result that \begin{equation} \label{eq3.7} \int_{V_p}(\mid \Delta u_n-\Delta u\mid^{p(x)}+a(x)\mid u_n-u\mid^{p(x)})dx\rightarrow 0~~as~~n\rightarrow\infty. \end{equation} Finally, \eqref{eq3.1} is given by combining \eqref{eq3.5} and \eqref{eq3.7}. \\It remains to show that $L'$ is is a homeomorphism.\\ In view of strict monotonicity of $L'$ which implies the injectivity of $L'.$Moreover, $L'$ is a coercive. Indeed, since $p^{-1}>1$, for each $u\in X$ such that $u\geq1$ we have \\\\ $$\frac{\langle L'(u),u \rangle}{\parallel u\parallel}=\frac{\rho_a(u)}{\parallel u\parallel}\geq \parallel u\parallel^{p^-1}\rightarrow\infty~~ as~~ n\rightarrow\infty.$$ \\\\ Consequently, thanks to a Minty-Browder \cite{15}, $L'$ is surjective and admits an inverse mapping. It suffices to show the continuity of $(L')^{-1}.$ Let $(f_n)_n$ be a sequence of $X'$ such that $f_n\rightarrow f$ in $X'.$Let $u_n$ and $u$ in $X$ such that $$ (L')^{-1}=u_n ~~and~~(L')^{-1}(f)=u. $$ By the coercivity of $L'$, one deducts that the sequence $(u_n)$ is bounded in the reflexive space $X.$ For a subsequence, we have $u_n\rightharpoonup \widehat{u}$ in $X,$ which implies \\ $$ \displaystyle\lim_{n\rightarrow\infty}\langle L'(u_n)-L'(u),u_n-\widehat{u} \rangle=\displaystyle\lim_{n\rightarrow\infty}\langle f_n-f,u_n-\widehat{u}\rangle=0.$$ Since $L'$ is of $(S_+)$ type and continuous, it follows that \\\\ $$u_n\rightarrow \widehat{u}~~ in ~~X~~ and~~L'(u_n)\rightarrow L'(\widehat{u})=L(u)~~ in ~~X'.$$\\ Moreover, since $L'$ is an injection, we deduce that $u=\widehat{u}.$ This completes the proof.$\centerdot$ \begin{pro} let $\sigma(u)=\int_{\Omega}G(x,u)dx$, then $\sigma$ is a $C^1$ in $L^{q(x)}(\Omega)$ and $\sigma'$ are weakly-strongly continuous, i.e $u_n\rightharpoonup u$ implies $\sigma'(u_n)\rightarrow \sigma'(u).$ \end{pro} \hspace*{-.6cm}\textbf{proof} : by \eqref{H} we have, $G(x,t)\leq A(x)+B\mid t\mid^{q(x)}.$ \\ where $A\in L^{1}(\Omega)~,A(x)\geq 0,~B\geq0.$ \\Then Nemytskii operator properties implies that $\sigma$ is a $C^1$ in $L^{q(x)}(\Omega).$\\By the continuous embedding of $X$ into $L^{q(x)}(\Omega),$ we have $\sigma$ is also $C^{1}$ in $X$ and for $u,v\in X$ and $$\sigma'(u)v=\int_{\Omega}g(x,u)vdx.$$ Let $(u_n)_n\subset X$ be a sequence such that $u_n\rightharpoonup u.$ Since there is a compact embedding of $X$ into $L^{q(x)}(\Omega),$ there is a subsequence, noted also $(u_n)_n$, such that $u_n\rightarrow u$ in $L^{q(x)}(\Omega).$ According to the Krasnoselki's theorem, the Nemytskii operator $$~~\begin{array}{rl} & \mbox{$ N_g:~L^{q(x)}(\Omega) ~\rightarrow L^{\frac{q(x)}{q(x)-1}}(\Omega)$} \\ & \mbox{~~~$~~~~~~~~~~~~u~~~~~~~\mapsto ~~f(., ~u)$}\end{array}$$ \hspace*{-.62cm}is continuous. Hence, $N_g(u_n)\rightarrow N_g(u)$ in $L^{\frac{q(x)}{q(x)-1}}(\Omega).$ Using Holder's inequality and the continuous embedding of $X$ into $L^{q(x)}(\Omega),$ we obtain \\ $\mid < \sigma'(u_n)-\sigma'(u), v > \mid=\mid \int_{\Omega} (g(x, u_n)-g(x,u))v(x)dx\mid$ \\ \hspace*{3.8cm}$\leq 2\parallel N_g(u_n)-N_g(u) \parallel_{\frac{q(x)}{q(x)-1}} \parallel v(x)\parallel_{q(x)}$ \\ \hspace*{4.4cm}$\leq C\parallel\parallel N_g(u_n)-N_g(u)\parallel_{\frac{q(x)}{q(x)-1}} \parallel v\parallel_a. $ \\ Thus, $\sigma'(u_n)\rightarrow \sigma'(u)\centerdot$ \subsection{\hspace*{-.5cm} -Ricceri's variational principal} \begin{defn} Let $G$ a bounded subset of $X$ and $\rho\in \mathbb{R}$. $G$ is called a \textbf{block} of I \\ with type $\rho$ if $$I(u)<\rho \ \forall x\in G~~ and ~~I(x)=\rho ~\forall x\in \overline{\partial G},$$ Where $\overline{\partial G}=\overline{G}^{ W}\backslash G$ and $\overline{G}^{\small{W}}$ is the closure of $G$ in $X$ in the weak topology. \end{defn} \begin{defn} Let D a bounded open subset of $X$ and $c0$ small enough, we have \\ $\displaystyle\inf_{\partial B(u,\epsilon)}I>I(u_*)$ i.e $B(u_*,\epsilon)$ is a Recceri box of $I$. \\ ii) By proposition 2.6) in \cite{8}, $I$ and $J:~~X\rightarrow \mathbb{R}$ are sequentially weakly continuous. \end{rem} \vspace*{.2cm} \begin{pro}\cite{8} Suppose that $G$ is a Ricceri box of $I$ with type $(c, b)$ and $I:~X\rightarrow \mathbb{R}$ continuous.\\ Then $\forall \rho\in ]c, b]$ we have $I^{-1}(]-\infty, \rho[)\cap G$ is a Ricceri block de $I$ with type $\rho.$ \end{pro} By Proposition \ref{pro}, Remark \ref{rem} and Theorem \ref{thm} we obtain the following result: \begin{pro}\label{proimp}\cite{6} Suppose that $I,J:~~X\rightarrow \mathbb{R}$ are continuous. \\ For some $r>0$, $u_1\in B(u_0,r),~I(u_0)=\inf_{B(u_0,r)}I=c_0;~\displaystyle\inf_{\partial B(x_0,r)}I=b>c_0$ and $u_1$ is a strictly local minimizer of $I$ and $I(u_1)=c_1>c_0.$ \\ Then for $\epsilon>0$ small enough and $\rho_1>c_1,\rho_0\in]c_0,\min\{b,c_1\}[$ and $\forall\lambda\in ]0, \lambda_*[,~I+\lambda J$ has at least two local minima $u_0^*,u_1^*$ in $B(u_0,r).$ where $u_0^*\in I^{-1}(]-\infty,\rho_0[)\cap B(u_0,r), u_0^*\not\in \overline{B(u_1,\epsilon)}$ and \\ $u_1^*\in I^{-1}(]-\infty,\rho_1[)\cap B(u_1,r).$ \end{pro} \begin{thm} Let $Y$ be a reflexive Banach space.Assume that \\ $1)\phi\in C^1 (Y,\mathbb{R})$,the mapping $\phi':Y\rightarrow Y^*$ is of type $S_+.$\\ $2)(G_0,G)$ is a valley box of $\phi$ with $G_0,G$ being connected and $0\in G_0.$ \\ $3)there~~exist~~e\in G_0$ and $r>0$ such that $\parallel e\parallel>r,$ \\ $\displaystyle\inf_{\partial B(0,r)}\phi>\max\{\phi(0),\phi(e)\}$ \\ Then, the functional $\phi$ has at least a critical point $u_0\in \overline{G}$ with $\phi(u_0)=c$ \\ where $c=\inf_{\gamma\in\Gamma}\sup_{t\in[0,1]}\phi(\gamma(t)),$ \\ $\Gamma=\{\gamma\in C([0,1],G):\gamma(0)=0;\gamma(1)=e \}.$ \end{thm} \hspace*{-.6cm}\textbf{proof} : see Theoreme 3.1 \cite{6} \begin{cor} Under the same assumption as in previous theorem, furthermore, if $J: Y\rightarrow \mathbb{R}\in C^1$ and $J':Y\rightarrow Y^{'}$ are weakly strongly continuous. Then, $\forall \lambda\in]0,\lambda_*[~~I+\lambda J$ has still a mountain pass type critical point $u_2\in \overline{G}.$ \end{cor} \section{\hspace*{-.5cm}-Proof of the main result} The critical point of the integral functional $\varphi=I+\lambda J$ is solution of the problem $\mathcal{(P)}.$ \vspace*{.4cm} Define $$\lambda_1=\displaystyle\inf_{u\in X\setminus\{0\}}\frac{\int_\Omega \frac{1}{p(x)}[\mid \Delta u\mid^{p(x)}+a(x)\mid u\mid^{p(x)}]dx}{\int_\Omega\frac{1}{p(x)}\mid u\mid^{p(x)}dx},$$ \hspace*{-.6cm}\textbf{proof of Theorem 1.1} : The proof is divided into four steps, \texttt{step(1)}:\\ We show that $v_0=0$ is strictly local minimizer of $I.$ \\ By \eqref{H} and assumption $(F_3)$, we may find $q_1\in C(\overline{\Omega})$ with \\ $p^+p^+$ then there exists $\epsilon >0$ such that $\forall u\in \overline{B(0,\epsilon)}\setminus{0}$ we have $I(u)>0=I(v_0).$ \\ \texttt{step(2)}:\\ \hspace*{-.1cm}We show that the functional $I$ has a global minimizer $v_1\neq 0.$ \\ \hspace*{-.1cm}Set $H(x,t)=F(x,t)-\frac{\lambda_1}{p(x)}\mid t\mid^{p^-}.$ \\ \hspace*{-.09cm}Then, from $(F_1)$ we conclude that, for every $M>0$, there is $R_M>0$ such that \\ \begin{equation}\label{eq.3.1} H(x,t)\leq -M,~~\forall \mid t\mid\geq R_M,~~{\rm almost~~ every~~ x\in\Omega.} \end{equation} We have $I$ is coercive, or else there exist $K\in \mathbb{R}$ and $(u)_n\subset X$ such that \\ $$\parallel u_n\parallel_a\rightarrow\infty~~ and~~ I(u_n)\leq K.$$ \hspace*{-.65cm} Putting $v_n=\frac{u_n}{\parallel u_n\parallel_a}$ i.e $\parallel v_n\parallel_a=1$. \\ Then for subsequence, we may assume that for $v\in X$, we have $v_n\rightharpoonup v $ in $X$, $v_n\rightarrow v$ strongly in $L^{p(x)}(\Omega)$, $v_n(x)\rightarrow v(x)$ for almost every $x\in \Omega.$ \\ \hspace*{-.62cm} Now, using \ref{eq.3.1}, we obtain \begin{eqnarray}\label{eq3.2} K\geq I(u_n)&=&\int_\Omega \frac{1}{p(x)}[ \mid \Delta u\mid^{p(x)}+ a(x)\mid u_n\mid^{p(x)}] dx-\int_\Omega F(x,u_n)dx.\nonumber\\ &\geq&\int_\Omega \frac{1}{p(x)}[ \mid \Delta u_n\mid^{p(x)}+a(x)\mid u_n \mid^{p(x)}] dx-\lambda_1\int_\Omega\frac{1}{p(x)} \mid u_n\mid^{p^-}dx-\int_\Omega H(x,u_n)dx \nonumber\\ &\geq & \frac{1}{p^+} \parallel u_n \parallel_a^{p^-}-\lambda_1\int_\Omega \frac{1}{p(x)} \mid u_n\mid^{p^-}dx+M_1, \end{eqnarray} where $M_1\in \mathbb{R}.$\\ Dividing \ref{eq3.2} by $\parallel u_n\parallel_a^{p^-}$ and passing to the limit, we conclude \\ $$\frac{1}{p^+}-\lambda_1\int_\Omega \mid v\mid^{p^{-}}dx\leq 0,$$ \\ hence, $v\not\equiv0$. Therfore $\mid\Omega^\backprime\mid> 0$ such that $\Omega^\backprime=\{ x\in\Omega ~\setminus~v_0(x)\neq 0\}$, \\ then $\mid u_n(x)\mid\rightarrow+\infty$ for almost every $x\in\Omega^\backprime.$ \hspace*{-.6cm}On the other hand, \begin{eqnarray} \label{equat.above} \lambda_1\int_{[\mid u\mid\geq1]}\frac{\mid u\mid^{p^-}}{p(x)}dx&\leq& \lambda_1\int_{[\mid u\mid\geq1]}\mid u\mid^{p(x)}dx \nonumber\\&\leq& \lambda_1\int_{\Omega}\mid u\mid^{p(x)}dx\nonumber\\&\leq&\int_\Omega\frac{1}{p(x)}[\mid \Delta u\mid^{p(x)}+a(x)\mid u\mid^{p(x)}]dx. \end{eqnarray} \\ Where, \\ $$[\mid u\mid\geq1]=\{x\in \Omega\setminus \mid u\mid\geq1\}~~;~~[\mid u\mid<1]=\{x\in \Omega\setminus \mid u\mid<1\}.$$ \\ It is clear that $\int_{[\mid u_n\mid<1]}\frac{1}{p(x)}\mid u_n \mid^{p(x)}dx $ is bounded.\\ From $(F_1)$ and the above inequalities \ref{equat.above} we deduce \\ \begin{eqnarray} K&\geq&\textstyle\int_\Omega\frac{1}{p(x)}[\mid\Delta u\mid^{p(x)}+a(x)\mid u\mid^{p(x)}]dx -\int_\Omega F(x,u_n)dx\nonumber\\ &=&\displaystyle\textstyle\int_\Omega\frac{1}{p(x)}[\mid\Delta u_n\mid^{p(x)}+a(x)\mid u_n\mid^{p(x)}]dx -\lambda_1\int_\Omega \frac{\mid u_n\mid^{p^-}}{p(x)} dx-\int_\Omega \small H(x,u_n)dx\nonumber\\ &=&\displaystyle\textstyle\int_\Omega\frac{1}{p(x)}[\mid\Delta u_n\mid^{p(x)}+a(x)\mid u_n\mid^{p(x)}]dx -\lambda_1\int_{[\mid u_n\mid\geq1]} \frac{\mid u_n\mid^{p^-}}{p(x)} dx-\lambda_1\int_{[\mid u_n\mid<1]} \frac{\mid u_n\mid^{p^-}}{p(x)} dx\nonumber \\&&\displaystyle\textstyle-\int_\Omega H(x,u_n)dx\nonumber\\ &\geq&\displaystyle\textstyle-\lambda_1\int_{[\mid u_n\mid<1]} \frac{\mid u_n\mid^{p^-}}{p(x)} dx-\int_{\Omega^{\backprime}} H(x,u_n)dx+\int_{\Omega\setminus\Omega^\backprime} H(x,u_n)dx\rightarrow +\infty,\nonumber \end{eqnarray} \hspace*{-.7cm} which is a contradiction. \\ \hspace*{-.15cm} Hence $I$ is coercive and has a global minimizer $v_1.$ \\ When the assumption $(F_2)$ holds, taking $\omega_1\in C^\infty_0 (B(x_0,2r_0))$ such that $0\leq \omega_1\leq t_0$ for all $x\in B(x_0,2r_0)$, $\omega_1(x)\equiv t_0$ for $x\in B(x_0,r_0)$ and $\mid \Delta \omega_1(x)\mid\leq\frac{2t_0}{r_0}$ then $\omega_1\in X.$ \\ On the other hand, \begin{eqnarray} I(\omega_1)& \leq & \int_{B(x_0,2r_0)\backslash B(x_0,r_0)}\mid \frac{2t_0}{r_0}\mid dx+\mid a\mid_\infty\int_{B(x_0,2r_0)}\frac{1}{p(x)}\mid t_0\mid^{p(x)}dx -\int_\Omega F(x,\omega_1)dx\nonumber \\ & \leq & C_0 \mid B(x_0,r_0)\mid -\int_\Omega F(x,\omega_1)dx \nonumber \end{eqnarray} Since $$ \int_\Omega F(x,\omega_1)dx> \int_{B(x_0,r_0)}F(x,\omega_1)dx\geq C_0 \mid B(x_0,r_0)\mid, $$ we obtain $I(\omega_1)<0$ then $I(v_1)<0=I(v_0).$ So $v_1\neq 0.$ \\ \texttt{step(3)}:\\ We show that $\varphi$ has two local minima.\\ Since $I$ is coercive there is $r_0>0$ large enough such that \\ $v_0, v_1\in B(0,r_0)~~$ and $~~\displaystyle\inf_{\partial B(0,r_0)} I>I(v_0)>I(v_1).$ \\ By proposition \ref{proimp}, given any $\epsilon> 0,~\rho_1\in]I(v_1),0[$ and $\rho_2> 0$ then $\forall \lambda\in]0,\lambda_*[$ \\ $\varphi$ has at least two local minima $u_0\in B(0,\epsilon))\cap I^{-1}(]-\infty, \rho_2[),~u_1\in I^{-1}]-\infty, \rho_1[$\\ and $u_1 \notin \overline{B(0,\epsilon)}.$ \\ The minimizer $u_0\neq 0$. In fact, when $(G_1)$ holds, taking $\omega\in C_0^\infty(B(x_1,r_1))$ such that $0\leq\omega\leq 1$ and $\omega(x)\equiv1$ for $x\in B(x_1,\frac{r_1}{2})$, then, it is easy to see that for $\lambda\in ]0,\lambda_*[$, when $t>0$ is small enough, we get $t\omega\in B(0,\epsilon)\cap I^{-}(]-\infty,\rho_2[)$ and $I(u_0)+\lambda J(u_0)\leq I(t\omega)+\lambda J(t\omega)<0.$ In particular, $u_0\neq 0$ $\centerdot$\\ \texttt{step(4)}:\\ \hspace*{-0cm}$\varphi$ has a mountain pass type critical point $\forall \lambda\in]0,\lambda_*[$.\\ \hspace*{-0.6cm}We take $r_1>0$ such that $B(0,r_1)\subset X$ and $B(0,r_1)\supset I^{-1}(]-\infty,\rho_1[)\cup B(0,\epsilon).$ \\ Since $I$ is coercive, there exists $r_2>r_1$ such that \\ $\displaystyle\inf_{\partial B(0,r_2)}I> \displaystyle\sup_{B(0,r_1)} I$ then $(B(0, r_1), B(0,r_2))$ is a valley box of $I.$ Since $I(v_1)<0=I(v_0)$ and by step 1), we have that for some $\epsilon_0>0$ with $\epsilon_0> \parallel v_1\parallel_a$ , $\displaystyle\inf_{\partial B(\epsilon,0)}I>0.$ We apply the Corollary 3.1, then $\varphi$ admits a mountain pass point $u_2$. \\ \hspace*{-.0cm}Consequently, $u_0,~u_1$ and $u_2$ are at least three nontrivial solutions of the problem $(\mathcal{P}).$\\ \hspace*{.0cm}\textbf{proof of Theorem 1.2} : It was the same steps of the previous proof.\\ \texttt{step(1)}:\\ To show that $v_0=0$ is strictly local minimizer of $I$, we follow the same procedure as in step (1) in the previous proof. \texttt{step(2)}:\\ We show that the functional $I$ has a global minimizer $v_1\neq 0.$ \\ Similarly in step(2) in the last proof of Theorem 1.2, one shows the coercivity of I and then $I$ has a global minimizer $v_1.$ \\ We use the condition $(F'_2)$, we obtain $I(\xi)<0$ and then $I(v_1)<0=I(v_0)$. So $v_1\neq0.$\\ \texttt{step(3)}:\\ We show that $\varphi$ has two local minima.\\ \hspace*{-.13cm} The same way as In step 3 in the last proof of Theorem 1.2, $\varphi$ has at least two local minima $u_0$ and $u_1\neq 0.$\\ \hspace*{-.7cm} Moreover, the minimizer $u_0\neq 0$. Indeed, \\ \hspace*{-.7cm} by assumption $(G_2)$, taking $t_n\rightarrow0$ such that \\ $$\frac{\displaystyle\inf_{x\in \Omega}G(x,t_n)}{\mid t_n\mid^{p^-}}\rightarrow +\infty.$$ \\ Let $w_n=t_n$( i.e $w_n(x)\in X).$ \\ \hspace*{-.65cm}For all $\lambda\in ]0, \lambda_*[$, by \ref{eq 4.1} we get\\ $\hspace*{.9cm}\varphi(w_n)\leq \mid t_n\mid^{p^-}\int_\Omega \frac{a(x)}{p(x)}dx-\int_\Omega F(x,t_n)dx-\lambda\int_\Omega G(x,t_n)dx$\\ $ \hspace*{2cm}\leq \mid t_n\mid^{p^-}K_1+c_3\int_\Omega\mid t_n\mid^{q_{1}(x)}-\lambda \mid t_n\mid^{p^-}\int_\Omega \frac{G(x,t_n)}{\mid t_n\mid^{p^-}}dx $ \\ $ \hspace*{2cm}\leq \mid t_n\mid^{p^-}K_1+\mid t_n\mid^{q_{1}^{-}}K_2-\lambda \mid t_n\mid^{p^-}\int_\Omega \frac{G(x,t_n)}{\mid t_n\mid^{p^-}}dx < 0 $ \\ and \\ $$w_n\in B(0, \epsilon)\cap I^{-}(]-\infty,\rho_2[)$$ \\\\ Thus, $$\varphi(u_0)\leq\varphi(w_n)<0=\varphi(0), ~~so~~u_0\neq0\centerdot$$ \\ \texttt{step(4)}:\\ As in step(4) of the theorem 1.1), $\varphi$ has a mountain pass type critical point $u_2$.\\ \hspace*{-.0cm}Consequently, $u_0,~u_1$ and $u_2$ are at least three nontrivial solutions of the problem $(\mathcal{P}).$ \\ \addcontentsline{toc}{chapter}{Bibliographie} \begin{thebibliography}{99} \bibitem{1} E. 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