\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 93, pp. 1--47.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/93\hfil Duality mappings on Orlicz-Sobolev spaces] {Variational and topological methods for operator equations involving duality mappings on Orlicz-Sobolev spaces} \author[G. Dinca, P. Matei\hfil EJDE-2007/93\hfilneg] {George Dinca, Pavel Matei} % in alphabetical order \address{George Dinca \newline Faculty of Mathematics and Computer Science, University of Bucharest, 14, Academiei Str., 010014 Bucharest, Romania} \email{dinca@fmi.unibuc.ro} \address{Pavel Matei \newline Department of Mathematics, Technical University of Civil Engineering, 124, Lacul Tei Blvd., 020396 Bucharest, Romania} \email{pavel.matei@gmail.com} \thanks{Submitted June 4, 2007. Published June 21, 2007.} \thanks{G. Dinca was supported by Contract CERES 12/25.07.2006 from the CEEX programm, \hfill\break\indent Romanian Ministry of Education and Research} \subjclass[2000]{35B38, 35B45, 47J30, 47H11} \keywords{A priori estimate; critical points; Orlicz-Sobolev spaces; \hfill\break\indent Leray-Schauder topological degree; Duality mapping; Nemytskij operator; \hfill\break\indent Mountain Pass Theorem} \begin{abstract} Let $a:\mathbb{R}\to \mathbb{R}$ be a strictly increasing odd continuous function with $\lim_{t\to +\infty }a(t)=+\infty $ and $A(t)=\int_{0}^{t}a(s)\,ds$, $t\in \mathbb{R}$, the $N$-function generated by $a$. Let $\Omega $ be a bounded open subset of $\mathbb{R}^{N}$, $N\geq 2$, $T[u,u]$ a nonnegative quadratic form involving the only generalized derivatives of order $m$ of the function $u\in W_{0}^{m}E_{A}(\Omega )$ and $g_{\alpha }:\Omega\times\mathbb{R}\to\mathbb{R}$, $| \alpha | 0$ and non-decreasing which satisfies the conditions $a(0)=0$, $\lim_{t\to \infty }a(t)=\infty $. \end{definition} It is assumed everywhere below that the function $a$ is continuous. \begin{remark}\label{R1}\rm In many applications, it will be convenient to extend the function $a$ for negative values of the argument. Thus, let $\widetilde{a}:\mathbb{R} \to \mathbb{R}_{+}$ be the function given by \begin{equation*} \widetilde{a}(s)=\begin{cases} a(t), & \text{if }t\geq 0 \\ -a(-t), &\text{if }t<0. \end{cases} \end{equation*} Then, the function $A:\mathbb{R}\to \mathbb{R}_{+}$, \begin{equation*} A(t)=\int_{0}^{t}\widetilde{a}(s)\,ds, \end{equation*} is an $N$-function. Obviously, the function $\widetilde{a}$ is continuous and odd. \end{remark} Throughout this paper, we suppose that $a:\mathbb{R}\to\mathbb{R}$ is a strictly increasing odd continuous function with $\lim_{t \to +\infty }a(t)=+\infty $ and $A$ is the $N-$function given by (\ref{ec1.1}). Let us consider the \textit{Orlicz class} \begin{equation*} K_{A}(\Omega )=\{u:\Omega \to \mathbb{R}\text{ measurable; } \int_{\Omega }A(u(x))dx<\infty \}. \end{equation*} The \textit{Orlicz space} $L_{A}(\Omega )$ is defined as the linear hull of $K_{A}(\Omega) $ and it is a Banach space with respect to the \textit{Luxemburg norm} \begin{equation*} \| u\| _{(A)}=\inf \{k>0\text{; }\int_{\Omega }A\big( \frac{u(x)}{k}\big) \,dx\leq 1\}. \end{equation*} \begin{remark} \label{R2} \rm If $a(t)=| t| ^{p-2}\cdot t$, $10$ and $t_{0}>0$ such that \begin{equation} A(2t)\leq kA(t),\quad\text{for all }t\geq t_{0}. \label{699} \end{equation} \begin{theorem}[{\cite[p. 24]{[KR]}}] \label{T1} A necessary and sufficient condition for the $N$- function $A$ to satisfy the $\Delta _{2}$-condition is that there exists a constant $\alpha $ such that, for $u>0$, \begin{equation} \frac{ua(u)}{A(u)}<\alpha . \label{ec2.0} \end{equation} \end{theorem} The $N$-function given by \begin{equation*} \overline{A}(u)=\int_{0}^{| u| }a^{-1}(s)\,ds, \end{equation*} is called the \textit{complementary }$N$\textit{-function} to $A$. \begin{remark} \label{R3} \rm Let $p$, $q$ be such that $p>1$ and $p^{-1}+q^{-1}=1$. If $A(t)= \frac{| t| ^{p}}{p}$, then $\overline{A}(t)=\frac{ | t| ^{q}}{q}$. Consequently $K_{\overline{A}}(\Omega )=L_{\overline{A}}(\Omega )=L^{q}(\Omega )$. \end{remark} We recall \textit{Young's inequality } \begin{equation*} uv\leq A(u)+\overline{A}(v), \quad \forall u,v\in \mathbb{R} \end{equation*} with equality if and only if $u=a^{-1}(| v| )\cdot \mathop{\rm sign}v $ or $v=a(| u| )\cdot \mathop{\rm sign}u$. The space $L_{A}(\Omega )$ is also a Banach space with respect to the \textit{Orlicz norm} \begin{equation*} \| u\| _{A}=\sup \big\{ \big| \int_{\Omega }u(x)v(x)\,dx\big| ; v\in K_{\overline{A}}(\Omega) , \int_{\Omega }\overline{A}(v(x))\,dx\leq 1\big\} . \end{equation*} Moreover \cite[p. 80]{[KR]}, \begin{equation*} \| u\| _{(A)}\leq \| u\| _{A}\leq 2\| u\| _{(A)}, \quad \forall u\in L_{A}(\Omega ). \end{equation*} One also has a \textit{H\"{o}lder's type inequality}: if $u\in L_{A}(\Omega) $ and $v\in L_{\overline{A}}(\Omega )$, then $uv\in L^{1}(\Omega )$ and \begin{equation} \big| \int_{\Omega }u(x)v(x)\,dx\big| \leq 2\| u\| _{(A)}\| v\| _{(\overline{A})}. \label{2.50} \end{equation} We shall denote the closure of $L^{\infty }(\Omega )$ in $L_{A}(\Omega )$ by $E_{A}(\Omega )$. One has $E_{A}(\Omega )\subset K_{A}(\Omega )$ and $E_{A}(\Omega )=K_{A}(\Omega )$ if and only if $A$ satisfies the $\Delta _{2}$ -condition. We shall denote by $\prod \big( E_{A}(\Omega ),r\big) $ the set of those $u$ from $L_{A}(\Omega )$ whose distance (with respect to the Orlicz norm) to $E_{A}(\Omega )$ is strictly less than $r$. If the $N$ -function $A$ does not satisfy the $\Delta _{2}$-condition, then \begin{equation*} \prod \left( E_{A}(\Omega ),r\right) \subset K_{A}(\Omega )\subset \overline{ \prod \left( E_{A}(\Omega ),r\right) }, \end{equation*} the inclusions being proper. \begin{theorem}[{\cite[p. 79]{[KR]}}] \label{T2} If $u\in L_{A}(\Omega )$ and $\|u\| _{(A)}\leq 1$, then $u\in K_{A}(\Omega )$ and $\rho (u;A)=\int_{\Omega }A(u(x))\,dx\leq \| u\| _{(A)}$. If $u\in L_{A}(\Omega )$ and $\| u\| _{(A)}>1$, then $\rho (u;A)\geq \| u\|_{(A)}$. \end{theorem} \begin{lemma}[\cite{[G74]}] \label{L1} If $u\in E_{A}(\Omega )$, then $a(| u|)\in K_{A}(\Omega )$. \end{lemma} The Orlicz-Sobolev space $W^{m}L_{A}(\Omega )$ $\big(W^{m}E_{A}(\Omega )\big) $ is the space of all $u\in L_{A}(\Omega )$ whose distributional derivatives $D^{\alpha }u$ are in $L_{A}(\Omega )$ $( E_{A}(\Omega )) $ for any $\alpha $, with $| \alpha |\leq m$; The spaces $W^{m}L_{A}(\Omega )$ and $W^{m}E_{A}(\Omega )$ are Banach spaces with respect to the norm \begin{equation} \| u\| _{W^{m}L_{A}(\Omega )}=\Big( \sum_{| \alpha | \leq m}\| D^{\alpha }u\| _{(A)}^{2}\Big) ^{1/2}. \label{ec2.2} \end{equation} If $\Omega $ has the segment property, then $\mathcal{C}^{\infty }(\overline{ \Omega })$ is dense in $W^{m}E_{A}(\Omega )$ \cite[Theorem 8.28]{[Ad]}. The space $W_{0}^{m}E_{A}(\Omega )$ is defined as the norm-closure of $\mathcal{D}(\Omega )$ in $W^{m}E_{A}(\Omega )$. Now, let us suppose that the boundary $\partial \Omega $ of $\Omega $ is $\mathcal{C}^{1}$. Consider the ``restriction to $\partial \Omega $'' mapping $\widetilde{\gamma }:\mathcal{C}^{\infty }(\overline{\Omega })\to \mathcal{C}(\partial \Omega )$, $\widetilde{\gamma }(u)= u|_{\partial \Omega }$. This mapping is continuous from $\big( \mathcal{C} ^{\infty }(\overline{\Omega }),\| \cdot \| _{W^{1}L_{A}(\Omega )}\big) $ to $\big( \mathcal{C}(\partial \Omega ),\| \cdot \| _{L_{A}(\partial \Omega )}\big) $ \cite[p. 69]{[G79]}. Consequently, the mapping $\widetilde{\gamma }$ can be extended into a continuous mapping, denoted $\gamma $ and called the "trace mapping", from $\big( W^{1}E_{A}(\Omega ),\| \cdot \| _{W^{1}L_{A}(\Omega )}\big) $ to $\big( E_{A}(\partial\Omega ),\| \cdot \| _{E_{A}(\partial \Omega )}\big) $. \begin{theorem}[{\cite[Proposition 2.3]{[G79]}}] \label{T3} The kernel of the trace mapping \\ $\gamma :W^{1}E_{A}(\Omega )\to E_{A}(\partial \Omega )$ is $W_{0}^{1}E_{A}(\Omega )$. \end{theorem} The following results are useful. \begin{theorem}[\cite{[CHZ]}]\label{T4} $W^{m}L_{A}(\Omega )$ is reflexive if and only if the $N$-functions $A$ and $\overline{A}$ satisfy the $\Delta _{2}$-condition. \end{theorem} \begin{proposition}[\cite{[G74]}]\label{P1} There exist constants $c_{m}$ and $c_{m,\Omega }$ such that \begin{equation*} \int_{\Omega }\sum_{| \alpha | 0$ \begin{equation} \lim_{t\to \infty }\frac{A(kt)}{B(t)}=0. \label{ec2.22} \end{equation} If the $N$-functions $A$ and $B$ are equivalent near infinity, then $A$ and $B$ define the same Orlicz space \cite[p. 234]{[Ad]}. Let us now introduce the Orlicz-Sobolev conjugate $A_{\ast }$ of the $N$-function $A$. We shall always suppose that \begin{equation} \lim_{t\to 0}\int_{t}^{1}\frac{A^{-1}(\tau )}{\tau ^{ \frac{N+1}{N}}}\,d\tau <\infty , \label{ec2.3} \end{equation} replacing, if necessary, $A$ by another $N$-function equivalent to $A$ near infinity (which determines the same Orlicz space). Suppose also that \begin{equation} \lim_{t\to \infty }\int_{1}^{t}\frac{A^{-1}(\tau )}{ \tau ^{\frac{N+1}{N}}}\,d\tau =\infty . \label{ec2.4} \end{equation} With \eqref{ec2.4} satisfied, we define the \textit{Sobolev conjugate} $A_{\ast }$ of $A$ by setting \begin{equation} A_{\ast }^{-1}(t)=\int_{0}^{t}\frac{A^{-1}(\tau )}{\tau ^{\frac{N+1 }{N}}}\,d\tau , t\geq 0. \label{ec2.5} \end{equation} \begin{theorem}[\cite{[Ad]}] \label{T5} If the $N$-function $A$ satisfies \eqref{ec2.3} and \eqref{ec2.4}, then \begin{equation*} W_{0}^{1}L_{A}(\Omega )\to L_{A_{\ast }}(\Omega ). \end{equation*} Moreover, if $\Omega _{0}$ is a bounded subdomain of $\Omega $, then the imbeddings \begin{equation*} W_{0}^{1}L_{A}(\Omega )\to L_{B}(\Omega _{0}) \end{equation*} exist and are compact for any $N$-function $B$ increasing essentially more slowly than $A_{\ast }$ near infinity. \end{theorem} \begin{theorem}[{\cite[Theorem 2.7]{[Ti]}}]\label{T6} The compact imbedding \begin{equation*} W_{0}^{1}L_{A}(\Omega )\to E_{A}(\Omega ) \end{equation*} holds. \end{theorem} \section{Geometry and smoothness of the space $(W_{0}^{m}E_{A}(\Omega ),\| \cdot \| _{m,A})$} \begin{definition}\label{D2} \rm The space $X$ is said to be smooth, if for each $x\in X$, $x\neq 0_{X}$, there exists a unique functional $x^{\ast }\in X^{\ast }$, such that $\| x^{\ast }\| =1$ and $\langle x^{\ast },x\rangle =\| x\| $. \end{definition} The following results will be useful. \begin{theorem}[\cite{[Di]}]\label{T7} Let $( X,\| \|) $ be a real Banach space. The norm of $X$ is G\^{a}teaux differentiable if and only if $X$ is smooth. \end{theorem} In order to study the smoothness of the space $W_{0}^{m}E_{A}(\Omega )$, we recall a result concerning the differentiability of the norm on Orlicz spaces. \begin{theorem}[\cite{[KR]}]\label{T8} The Luxemburg norm $\| \cdot \|_{(A)}$ is G\^{a}teaux-differentiable on $E_{A}(\Omega )$. For $u\neq 0$, we have \begin{equation} \langle \| \cdot \| _{(A)}'(u),h\rangle =\frac{\int_{\Omega }a\big( \frac{u(x)}{\| u\| _{(A)}}\big) h(x)\,dx}{\int_{\Omega }a\big( \frac{u(x)}{ \| u\| _{(A)}}\big) \frac{u(x)}{\| u\| _{(A)}}\,dx},\quad \text{for all }h\in E_{A}(\Omega ). \label{7} \end{equation} Moreover, if the $N$-function $\overline{A}$ satisfies the $\Delta _{2}$-condition, then the norm $\|\cdot\| _{(A)}$ is Fr\'{e}chet-differentiable on $E_{A}(\Omega )$. \end{theorem} The following results will be also useful. \begin{lemma}[{\cite[Lemma 2.5]{[Ti]}}] \label{L2} If $(u_{n})_{n}\subset E_{A}(\Omega )$ with $u_{n}\to u$ in $E_{A}(\Omega )$, then there exists $h\in K_{A}(\Omega )\subset L_{A}(\Omega )$ and a subsequence $(u_{n_{k}})_{n_{k}}$ such that $| u_{n_{k}}(x)| \leq h(x)$ a.e. and $u_{n_{k}}(x)\to u(x)$ a.e. \end{lemma} \begin{lemma}[{\cite[Lemma 18.2]{[KR]}}] \label{L3} Let $A$ and $\overline{A}$ be mutually complementary $N$ - functions the second of which satisfies the $\Delta _{2}$-condition. Suppose that the derivative $a$ of $A$ is continuous. Then, the operator $N_{a}$, defined by means of the equality $N_{a}u(x)=a(|u(x)| )$, acts from $\prod ( E_{A}(\Omega ),1) $ into $K_{\overline{A}}(\Omega )=L_{\overline{A}}(\Omega )=E_{\overline{A}}(\Omega ) $ and is continuous. \end{lemma} Now, let $T[u,v]$ be a nonnegative symmetric bilinear form involving the only generalized derivatives of order $m$ of the functions $u,v\in W_{0}^{m}E_{A}(\Omega )$, satisfying the inequalities (\ref{ec1.5}). From these inequalities and taking into account Corollary \ref{C1}, we obtain that $W_{0}^{m}E_{A}(\Omega )$ may be (equivalent) renormed by using the norm \begin{equation} \| u\| _{m,A}=\| \sqrt{T[u,u]}\| _{(A)}. \label{8} \end{equation} \begin{theorem}\label{T9} The space $\big( W_{0}^{m}E_{A}(\Omega ),\| \cdot \|_{m,A}\big) $ is smooth. Thus, the norm $\| \cdot \| _{m,A}$ is G\^{a}teaux-differentiable on $W_{0}^{m}E_{A}(\Omega) $. For $u\neq 0_{W_{0}^{m}E_{A}(\Omega )}$, we have \begin{equation} \langle \| \cdot \| _{m,A}'(u),h\rangle =\frac{\int_{\Omega }a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{T[u,h](x)}{\sqrt{T[u,u](x)}}\,dx}{ \int_{\Omega }a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\,dx},\quad \text{for all }h\in W_{0}^{m}E_{A}(\Omega ). \label{5} \end{equation} Moreover, if the $N$-function $\overline{A}$ satisfies the $\Delta _{2}$-condition, then $u\to \|\cdot \| _{m,A}'(u)$ is continuous thus $\| \cdot \|_{m,A}$ is Fr\'{e}chet-differentiable. \end{theorem} \begin{proof} Let $u\neq 0$ be in $W_{0}^{m}E_{A}(\Omega )$, that is $\sqrt{T[u,u]}\neq 0_{E_{A}(\Omega )}$. Let us denote $\psi (u)=\| \sqrt{T[u,u]}\| _{(A)}$. It is obvious that $\psi $ can be written as a product $\psi =QP$, where $Q:E_{A}(\Omega )\to \mathbb{R}$ is given by $Q(v)=\| v\| _{(A)}$ and $P:W_{0}^{m}E_{A}(\Omega )\to E_{A}(\Omega )$ is given by $P(u)=\sqrt{T[u,u]}$. The functional $Q$ is G\^{a}teaux differentiable (see Theorem \ref{T8}) and \begin{equation} \langle Q'(v),h\rangle =\| v\| _{(A)}'(h), \label{9} \end{equation} for all $v,h\in E_{A}(\Omega )$, $v\neq 0_{E_{A}(\Omega )}$. Simple computations show that the operator $P$ is G\^{a}teaux differentiable at $u$ and \begin{equation} P'(u)(h)=\frac{T[u,h]}{\sqrt{T[u,u]}}, \label{10} \end{equation} for all $u,h\in W_{0}^{m}E_{A}(\Omega )$, $u\neq 0_{W_{0}^{m}E_{A}(\Omega )}$. Combining (\ref{9}) and (\ref{10}), we obtain that $\psi $ is G\^{a}teaux differentiable at $u$ and \begin{align*} \langle \psi '(u),h\rangle &=\langle Q'(Pu),P'(u)(h)\rangle\\ &=\langle \|\cdot \| _{(A)}'(Pu),\frac{T[u,h]}{\sqrt{T[u,u]}}\rangle \\ &= \frac{\int_{\Omega }a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{T[u,h](x)}{\sqrt{T[u,u](x)}}\,dx}{ \int_{\Omega }a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\,dx}. \end{align*} Now, we will show that the mapping $u\mapsto \psi '(u)$ is continuous. In order to do that it is sufficient to show that any sequence $(u_{n})_{n}\subset W_{0}^{m}E_{A}(\Omega )$ converging to $u\in W_{0}^{m}E_{A}(\Omega )$ contains a subsequence $(u_{n_{k}})_{k}\subset (u_{n})_{n}$ such that $\psi '(u_{n_{k}})\to \psi '(u)$, as $k\to \infty $, in $\big(W_{0}^{m}E_{A}(\Omega )\big)^{\ast }$. We set \begin{equation*} \langle \psi '(u),h\rangle =\frac{\langle \varphi (u),h\rangle }{q(u)}, \quad \forall h\in W_{0}^{m}E_{A}(\Omega ), \end{equation*} where $\varphi :W_{0}^{m}E_{A}(\Omega )\to W_{0}^{m}E_{A}(\Omega)$ is defined by \begin{equation*} \langle \varphi (u),h\rangle =\int_{\Omega }a\big( \frac{ \sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{T[u,h](x)}{ \sqrt{T[u,u](x)}}\,dx \end{equation*} and $q:W_{0}^{m}E_{A}(\Omega )\to \mathbb{R}$ is given by \begin{equation*} q(u)=\int_{\Omega }a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\,dx. \end{equation*} First, we show that if $u_{n}\to u$ in $W_{0}^{m}E_{A}(\Omega )$, then the sequence $(u_{n})_{n}$ contains a subsequence $(u_{n_{k}})_{k}\subset (u_{n})_{n}$ such that $q(u_{n_{k}})\to q(u)$ as $k\to \infty $. Since \begin{equation} | \sqrt{T[u_{n},u_{n}]}-\sqrt{T[u,u]}| \leq \sqrt{T[u_{n}-u,u_{n}-u]}, \label{11} \end{equation} it follows from \begin{equation} \| u_{n}-u\| _{m,A}=\| \sqrt{T[u_{n}-u,u_{n}-u]} \| _{(A)}\to 0\quad \text{as }n\to \infty , \label{12} \end{equation} that \begin{equation} \sqrt{T[u_{n},u_{n}]}\to \sqrt{T[u,u]} \quad \text{as }n\to \infty, \text{ in }E_{A}(\Omega ); \label{17} \end{equation} therefore \begin{equation*} \frac{\sqrt{T[u_{n},u_{n}]}}{\| u_{n}\| _{m,A}}\to \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\quad \text{ as }n\to \infty, \text{ in }E_{A}(\Omega ). \end{equation*} By applying Lemma \ref{L3}, and obtain \begin{equation*} a\big( \frac{\sqrt{T[u_{n},u_{n}]}}{\| u_{n}\| _{m,A}} \big) \to a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \quad \text{as }n\to \infty, \text{ in } E_{\overline{A}}(\Omega ). \end{equation*} Then, from Lemma \ref{L2}, it follows that there exists a subsequence $(u_{n_{k}})_{k}\subset (u_{n})_{n}$ and $w\in K_{\overline{A}}(\Omega )=E_{\overline{A}}(\Omega )$, such that \begin{equation} a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \to a\big( \frac{\sqrt{ T[u,u](x)}}{\| u\| _{m,A}}\big) \quad \text{as }k\to \infty, \text{ for a.e. }x\in \Omega \label{16} \end{equation} and \begin{equation} a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \leq w(x),\quad \text{for a.e. }x\in \Omega . \label{116} \end{equation} Taking into account (\ref{12}), written for $(u_{n_{k}})_{k}$, and applying again Lemma \ref{L2}, it follows that there exists a subsequence (also denoted $(u_{n_{k}})_{k}$), and $w_{1}\in K_{A}(\Omega )$ such that \begin{equation} \sqrt{T[u_{n_{k}}-u,u_{n_{k}}-u](x)}\to 0\quad \text{as } k\to \infty, \text{ for a.e. }x\in \Omega . \label{13} \end{equation} and \begin{equation} \sqrt{T[u_{n_{k}},u_{n_{k}}](x)}\leq w_{1}(x),\quad \text{for a.e. }x\in \Omega . \label{15} \end{equation} Out of (\ref{13}) and (\ref{11}), we obtain \begin{equation} \sqrt{T[u_{n_{k}},u_{n_{k}}](x)}\to \sqrt{T[u,u](x)}\quad \text{as } k\to \infty , \text{ for a.e. }x\in \Omega . \label{14} \end{equation} Consequently \begin{align*} &a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \sqrt{T[u_{n_{k}},u_{n_{k}}](x)}\\ &\to a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}} \big) \sqrt{T[u,u](x)}\quad \text{as }k\to \infty, \text{ for a.e. } x\in \Omega \end{align*} and \begin{equation*} a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \sqrt{T[u_{n_{k}},u_{n_{k}}](x)}\leq w(x)\cdot w_{1}(x),\quad \text{for a.e. }x\in \Omega . \end{equation*} Since $w\cdot w_{1}\in L^{1}(\Omega )$, by using (\ref{17}) and Lebesgue's dominated convergence theorem, it follows that \begin{align*} &\int_{\Omega }a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{ \| u_{n_{k}}\| _{m,A}}\big) \frac{\sqrt{ T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\,dx \\ &\to \int_{\Omega }a\big( \frac{\sqrt{T[u,u](x)}}{ \| u\| _{m,A}}\big) \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}dx,\quad \text{as }k\to \infty , \end{align*} which is $q(u_{n_{k}})\to q(u)$ as $k\to \infty $. For the $(u_{n_{k}})_{k}$ obtained above, we shall show that \begin{equation*} \varphi ( u_{n_k}) \to \varphi (u), \quad \text{as }k\to \infty ,\text{ in }\big( W_{0}^{m}E_{A}(\Omega )\big) ^{\ast }. \end{equation*} But \begin{equation*} T[u,v]=\sum_{| \alpha | =| \beta | =m}c_{\alpha \beta }(x)D^{\alpha }uD^{\beta }v, \end{equation*} where $c_{\alpha \beta }\in \mathcal{C}( \overline{\Omega }) $, therefore they are bounded. First let us remark that, for arbitrary $h$, one has \begin{equation} \begin{aligned} &| \left( \varphi (u_{n_{k}})-\varphi (u)\right) (h)|\\ & =\Big| \sum_{| \alpha | =| \beta | =m}\int_{\Omega }c_{\alpha \beta }\Big[ a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{\sqrt{ T[u_{n_{k}},u_{n_{k}}](x)}}\\ &\quad -a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{T[u,u](x)}}\Big] D^{\beta }hdx\Big| \\ &\leq M\sum_{| \alpha | =| \beta | =m}\Big| \int_{\Omega }\Big[a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{\sqrt{ T[u_{n_{k}},u_{n_{k}}](x)}}\\ &\quad -a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{T[u,u](x)}}\Big] D^{\beta }hdx\Big| . \end{aligned} \label{111} \end{equation} We intend to apply H\"{o}lder's inequality (\ref{2.50}) in (\ref{111}). Since $D^{\beta }h\in E_{A}(\Omega )$, for all $\beta $ with $| \beta | =m$, it is sufficient to show that \begin{equation*} a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}]}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{ \sqrt{T[u_{n_{k}},u_{n_{k}}]}}-a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{T[u,u]}}\in L_{ \overline{A}}(\Omega ). \end{equation*} Moreover, we will show that \begin{equation} a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}]}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{ \sqrt{T[u_{n_{k}},u_{n_{k}}]}}-a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{T[u,u]}}\in E_{ \overline{A}}(\Omega )=K_{\overline{A}}(\Omega ). \label{112} \end{equation} Indeed, $a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}} \big) \frac{D^{\alpha }u}{\sqrt{T[u,u]}}\in K_{\overline{A} }(\Omega )$, because $\frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\in E_{A}(\Omega )$, by Lemma \ref{L1}, we obtain $a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \in K_{\overline{A} }(\Omega )$. On the other hand, since $T$ satisfies inequalities (\ref{ec1.5}), we have \begin{equation*} \frac{D^{\alpha }u}{\sqrt{T[u,u]}.}\leq \frac{1}{\sqrt{c_{1}}}; \end{equation*} therefore \begin{equation*} a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{T[u,u]}}\leq \frac{1}{\sqrt{c_{1}}} a\big(\frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \in K_{\overline{A}}(\Omega )=E_{\overline{A}}(\Omega ) \end{equation*} (the $N$-function $\overline{A}$ satisfies the $\Delta _{2}$-condition). Consequently, \begin{equation} a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{T[u,u]}}\in K_{\overline{A}}(\Omega )=E_{ \overline{A}}(\Omega ). \label{115} \end{equation} Now, using the same technique, we obtain \begin{equation*} a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}]}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{ \sqrt{T[u_{n_{k}},u_{n_{k}}]}}\in K_{\overline{A}}(\Omega )=E_{\overline{A} }(\Omega ); \end{equation*} therefore we have (\ref{112}). Applying H\"{o}lder's inequality in (\ref{111}), we obtain \begin{align*} | \left( \varphi (u_{n_{k}})-\varphi (u)\right) (h)| & \leq M_{1}\sum_{| \alpha | =| \beta | =m}\Big\| a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{ \| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}\\ &\quad -a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}} \big) \frac{D^{\alpha }u}{\sqrt{T[u,u](x)}}\| _{( \overline{A})}\Big\| h\| _{m,A}. \end{align*} Consequently, \begin{align*} \| \varphi (u_{n_{k}})-\varphi (u)\| &\leq M_{1}\sum_{| \alpha | =| \beta | =m}\Big\| a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{ \| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}\\ &\quad -a\big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}} \big) \frac{D^{\alpha }u}{\sqrt{T[u,u](x)}}\| _{(\overline{A})}. \end{align*} Finally, we show that \begin{equation} \big\| a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{ \sqrt{T[u_{n_{k}},u_{n_{k}}](x)}}-a\big( \frac{\sqrt{T[u,u](x)}}{ \| u\| _{m,A}}\big) \frac{D^{\alpha }u}{\sqrt{ T[u,u](x)}}\big\| _{(\overline{A})}\to 0, \label{114} \end{equation} as $k\to \infty $. We will use the following result \cite[Theorem 14, p. 84]{[RR]}. An element $f\in L_{A}(\Omega )$ has an absolutely continuous norm if and only if for each measurable $f_{n}$ such that $f_{n}\to \widetilde{f}$ a.e. and $| f_{n}| \leq | f| $, a.e., we have $\| f_{n}-\widetilde{f}\| _{(A)}\to 0$ as $n\to \infty $. The fact that $f\in L_{A}(\Omega )$ has an absolutely continuous norm means that for every $\varepsilon >0$ there exists a $\delta>0$ such that $\| f\cdot \chi _{E }\|_{A}<\varepsilon $ provided $\mathop{\rm mes}(E)<\delta $ ($E\subset \Omega $). Moreover, any function from $E_{A}(\Omega )$ has an absolutely continuous norm \cite[Theorem 10.3]{[KR]}. Then, (\ref{114}) follows from the above result with the following choices: \begin{gather*} f_{k}=a\big( \frac{\sqrt{T[u_{n_{k}},u_{n_{k}}]}}{\| u_{n_{k}}\| _{m,A}}\big) \frac{D^{\alpha }u_{n_{k}}}{ \sqrt{T[u_{n_{k}},u_{n_{k}}]}}\in E_{\overline{A}}(\Omega ), \\ \widetilde{f}=a\big( \frac{\sqrt{T[u,u]}}{\| u\| _{m,A}} \big) \frac{D^{\alpha }u}{\sqrt{T[u,u]}}\in E_{\overline{A} }(\Omega ) \end{gather*} From (\ref{ec1.5}) and (\ref{13}), it follows \begin{equation*} D^{\alpha }u_{n_{k}}(x)\to D^{\alpha }u(x), \quad \text{for a.e. }x\in \Omega ; \end{equation*} therefore, taking into account (\ref{16}), (\ref{14}), we obtain \begin{equation*} f_{k}(x)\to \widetilde{f}(x),\quad \text{as }k\to \infty, \text{ for a.e. }x\in \Omega . \end{equation*} On the other hand, from (\ref{ec1.5}) and (\ref{116}), we have \begin{equation*} | f_{k}(x)| \leq \frac{w(x)}{\sqrt{c_{1}}}, \quad \text{for a.e. }x\in \Omega , \end{equation*} with $w\in K_{\overline{A}}(\Omega )=E_{\overline{A}}(\Omega )$. Setting \begin{equation*} f=\frac{w}{\sqrt{c_{1}}}\in E_{\overline{A}}(\Omega ), \end{equation*} it follows (\ref{114}). It follows that $\| \varphi (u_{n_{k}})-\varphi (u)\| \to 0$ as $k\to \infty$. \end{proof} Now, we will study the uniform convexity of the space $ \big( W_{0}^{m}E_{A}(\Omega ),\| \cdot \| _{m,A}\big) $. To do it, we still need some prerequisites. We begin with a technical result due to Gr\"{o}ger (\cite{[Gr]}) (see, also \cite[p. 153]{[La]}). \begin{lemma} \label{L4} Let $A(u)=\int_{0}^{| u| }p(t)\,dt$ and $A_{1}(u)=\int_{0}^{| u| }p_{1}(t)\,dt$ be two $N$-functions, such that the functions $p$ and $p_{1}$ should satisfy the conditions \begin{gather} \frac{p(\tau )}{\tau }\geq \frac{p(t)}{t}, \quad \tau \geq t>0, \label{1} \\ p(t+\tau )-p(\tau )\geq p_{1}(t),\quad \tau \geq t>0. \label{2} \end{gather} Then \begin{equation} \frac{1}{2}A(a)+\frac{1}{2}A(b)-A(c)\geq A_{1}(c_{\ast }), \label{3} \end{equation} where \begin{equation} a\geq b\geq 0,\quad \frac{a-b}{2}\leq c\leq \frac{a+b}{2}, \quad c_{\ast}=\sqrt{\frac{a^{2}+b^{2}}{2}-c^{2}}. \label{4} \end{equation} \end{lemma} The next corollary is a direct consequence of the preceding lemma. \begin{corollary}\label{C2} Let $A(u)=\int_{0}^{| u| }p(t)\,dt$ be an $N$-function. Suppose that the function $p(t)/t$ is nondecreasing on $(0,\infty )$. Then \begin{equation*} \frac{1}{2}A(a)+\frac{1}{2}A(b)-A(c)\geq A(c_{\ast }), \end{equation*} where $a$, $b$, $c$ and $c_{\ast }$ are as in \eqref{4}. \end{corollary} \begin{proposition}\label{P2} Let $A(u)=\int_{0}^{| u| }p(t)dt$ be an $N$-function. Suppose that the function $\frac{p(t)}{t}$ is nondecreasing on $(0,\infty )$. Then \begin{equation*} \frac{1}{2}A\big(\sqrt{T[u,u]}\big) +\frac{1}{2}A\big(\sqrt{T[v,v]}\big) -A\big(\sqrt{T[\frac{u+v}{2},\frac{u+v}{2}]}\big) \geq A\big(\sqrt{T[\frac{u-v}{2},\frac{u-v}{2}]}\big). \end{equation*} \end{proposition} \begin{proof} We apply Corollary \ref{C2} with $a=\sqrt{T[u,u]}$, $b=\sqrt{T[v,v]}$, \[ c=\sqrt{T[\frac{u+v}{2},\frac{u+v}{2}]},\quad c_{\ast }=\sqrt{T[\frac{u-v}{2},\frac{u-v}{2}]}. \] \end{proof} \begin{proposition}\label{P3} Let $A$ be an $N$-function. If the $N$-function $A$ satisfies the $\Delta _{2}$-condition, then \begin{equation*} \rho :L_{A}(\Omega )=E_{A}(\Omega )=K_{A}(\Omega )\to \mathbb{R}, \quad \rho (u)=\int_{\Omega }A(u(x))dx, \end{equation*} is continuous. \end{proposition} \begin{proof} Obviously, $\rho $ is convex, therefore it suffices to show that $\rho $ is upper bounded on a neighborhood of $0$. But, if $\| u\|_{(A)}<1$, then $\rho (u)\leq \| u\| _{(A)}<1$. \end{proof} \begin{proposition}\label{P4} Let $A$ be an $N$-function. Then, one has: \begin{itemize} \item[(i)] If $\rho (u)=\int_{\Omega }A(u(x))dx=1$, then $\| u\| _{(A)}=1$; \item[(ii)] if, in addition, $A$ satisfies a $\Delta _{2}$-condition, then $\rho (u)=\int_{\Omega }A(u(x))dx=1$ if and only if $\| u\|_{(A)}=1$. \end{itemize} \end{proposition} \begin{proof} (i) Indeed, we have \begin{equation*} 1=\rho (u)=\int_{\Omega }A(\frac{u(x)}{1})dx\geq \| u\| _{(A)}, \end{equation*} the last inequality being justified by the definition of the $\| \cdot \| _{(A)}$-norm. If $\| u\| _{(A)}<1$, then (see Theorem \ref{T2}), we have \begin{equation*} \int_{\Omega }A(u(x))dx\leq \| u\| _{(A)}<1, \end{equation*} which is a contradiction. (ii) Taking into account the result given by (i), the ``only if'' implication has to be proved. Now, since $\| u\| _{(A)}=1$, we can write \begin{equation*} \rho (u)=\int_{\Omega }A(\frac{u(x)}{1})dx=\int_{\Omega }A( \frac{u(x)}{\| u\| _{(A)}})dx\leq 1. \end{equation*} The strict inequality cannot hold. Indeed, if for some $u$ with $\| u\| _{(A)}=1$, we have $\int_{\Omega }A(u(x))dx<1$, then there exists $\varepsilon >0$ such that $\int_{\Omega }A(u(x))dx+\varepsilon <1$. From Proposition \ref{P3}, $\lim_{\lambda \to 1_{+}}\rho (\lambda u)=\rho (u)$, therefore, there exists $\delta >0$, such that for each $\lambda $ with $| \lambda -1| <\delta $, we have \begin{equation*} \big| \int_{\Omega }A(\lambda u(x))dx-\int_{\Omega }A(u(x))dx\big| <\varepsilon . \end{equation*} It follows that, for $1<\lambda <1+\delta $, $\int_{\Omega }A(\lambda u(x))dx<\int_{\Omega }A(u(x))dx+\varepsilon <1$. Since $\int_{\Omega }A(\lambda u(x))dx<1$, we infer that $\| u\|_{(A)}\leq \frac{1}{\lambda }<1$, which is a contradiction. \end{proof} \begin{proposition}\label{P5} Let $A$ be an $N$-function which satisfies the $\Delta _{2}$-condition. If $\| u\| _{(A)}>\varepsilon $, then there exists $\eta >0$ such that $\int_{\Omega }A(u(x))dx>\eta $. \end{proposition} \begin{proof} Let $u$ be such that $\| u\| _{(A)}>\varepsilon $. Assume that the assertion in the proposition is not true, therefore for each $\eta$ we have $\int_{\Omega }A(u(x))dx\leq \eta $. This means that $\rho (u)=\int_{\Omega }A(u(x))dx=0$. Then, the $\Delta_{2}$-condition implies that, $\rho ( 2^{p}u) \leq k^{p}\rho (u)=0$, therefore $\rho (2^p u) =0$. Consequently $\| 2^{p}u\| _{A}\leq \rho (2^p u) +1=1$, therefore $\| u\| _{(A)}\leq \|u\| _{A}\leq \frac{1}{2^{p}}<\varepsilon $ for $p$ large enough, which is a contradiction. \end{proof} \begin{definition}\label{D3} \rm The space $( X,\| \cdot \| _{X}) $ is called uniformly convex if for each $\varepsilon \in (0,2]$ there exists $\delta (\varepsilon )\in (0,1]$ such that for $u,v\in X$ with $\|u\| _{X}=\| v\| _{X}=1$ and $\|u-v\| _{X}\geq \varepsilon $, one has $\| \frac{u+v}{2}\| _{X}\leq 1-\delta (\varepsilon )$. \end{definition} \begin{theorem}\label{T10} Let $A(u)=\int_{0}^{| u| }p(t)\,dt$ be an $N$-function. Suppose that the function $p(t)/t$ is nondecreasing on $(0,\infty )$. If the $N$-function $A$ satisfies the $\Delta _{2}$-condition, then $W_0^m E_A(\Omega)$ endowed with the norm \begin{equation*} \| u\| _{m,A}=\| \sqrt{T[u,u]}\| _{(A)} \end{equation*} is uniformly convex. \end{theorem} \begin{proof} We start with the following technical remark: if the $N$-function $A$ satisfies a $\Delta _{2}$-condition and $\int_{\Omega }A(u(x))dx<1-\eta $ for some $0<\eta <1$, there is $\delta >0$ such that $\| u\| _{(A)}<1-\delta $. In the contrary case, there is $u$ satisfying $\int_{\Omega }A(u(x))dx<1-\eta $ for which $\| u\|_{(A)}\geq 1-\delta $ for any $\delta>0$. In particular inequality $\int_{\Omega }A(u(x))dx<1-\eta $ may be satisfied for some $u$ with $\| u\| _{(A)}>1/2$. On the other hand, every $u$ satisfying $\int_{\Omega }A(u(x))dx<1-\eta $ has to satisfy $\| u\| _{(A)}<1$ (see Theorem \ref{T2} and Proposition \ref{P4}). Put $a=1/\| u\|_{(A)}$. Clearly $12$, it follows that $[(a-1)k+2-a]\cdot (1-\eta )<1$, which is a contradiction. Now, let $\varepsilon >0$ be and $u,v\in W_{0}^{m}E_{A}(\Omega )$ such that $\| u\| _{m,A}=\| \sqrt{T[u,u]}\| _{(A)}=1$ , $\| v\| _{m,A}=\| \sqrt{T[v,v]}\| _{(A)}=1$ and $\| u-v\| _{m,A}=\| \sqrt{T[u-v,u-v]} \| _{(A)}>\varepsilon $. Then $\| \frac{u-v}{2}\| _{m,A}=\| \sqrt{T[\frac{u-v}{2},\frac{u-v}{2}]}\| _{(A)}> \frac{\varepsilon }{2}$. From Proposition \ref{P5} it follows that there exists $\eta >0$ such that $\int_{\Omega }A\big( \sqrt{T[\frac{u-v }{2},\frac{u-v}{2}]}\big)\,dx>\eta $. On the other hand, from Proposition \ref{P4}, we have $\int_{\Omega }A\big( \sqrt{T[u,u]}\big)\,dx= \int_{\Omega }A( \sqrt{T[v,v]})\,dx=1$. Taking into account Proposition \ref{P2}, we obtain that $\int_{\Omega}A\left( \sqrt{T[\frac{u+v}{2},\frac{u+v}{2}]}\right)\,dx<1-\eta $. From the above remark, we conclude that there is a $\delta >0$ depending on $\varepsilon $ such that $\| \frac{u+v}{2}\| _{m,A}=\| \sqrt{T[ \frac{u+v}{2},\frac{u+v}{2}]}\| _{(A)}<1-\delta $. \end{proof} \section{Duality mapping on $\big( W_{0}^{m}E_{A}(\Omega ),\| \cdot \| _{m,A}\big) $} Let $X$ be a real Banach space and let $\varphi :\mathbb{R}_{+}\to \mathbb{R}_{+}$ be a gauge function, i.e. $\varphi $ is continuous, strictly increasing, $\varphi (0)=0$ and $\varphi (t)\to \infty $ as $t\to \infty $. By duality mapping corresponding to the gauge function $\varphi $ we understand the multivalued mapping $J_{\varphi }:X\to \mathcal{P}(X^{\ast })$, defined as follows: \begin{equation} \begin{gathered} J_{\varphi }0=\{0\}, \\ J_{\varphi }x=\varphi (\| x\|) \{u^{\ast }\in X^{\ast };\| u^{\ast }\| =1,\langle u^{\ast },x\rangle =\| x\| \},\quad \text{if }x\neq 0. \end{gathered} \label{301} \end{equation} According to the Hahn-Banach theorem it is easy to see that the domain of $J_{\varphi }$ is the whole space: \begin{equation*} D(J_{\varphi })=\left\{ x\in X;J_{\varphi }x\neq \emptyset \right\} =X. \end{equation*} Due to Asplund's result \cite{[As]}, \begin{equation} J_{\varphi }=\partial \psi , \psi (x)=\int_{0}^{\| x\| }\varphi (t)dt, \label{302} \end{equation} for any $x\in X$ and $\partial \psi $ stands for the subdifferential of $\psi $ in the sense of convex analysis. By the preceding definition, it follows that $J_{\varphi }$ is single valued if and only if $X$ is smooth, i.e. for any $x\neq 0$ there is a unique element $u^{\ast }(x)\in X^{\ast }$ having the metric properties \begin{equation} \langle u^{\ast }(x),x\rangle =\| x\| , \quad \| u^{\ast }(x)\| =1 \label{303} \end{equation} But it is well known (see, for example, Diestel \cite{[Di]}) that a real Banach space $X$ is smooth if and only if its norm is differentiable in the G\^{a}teaux sense, i.e. at any point $x\in X$, $x\neq 0$ there is a unique element $\| \cdot \| '(x)\in X^{\ast }$ such that, for any $h\in X$, the following equality \begin{equation*} \lim_{t\to 0}\frac{\| x+th\| -\| x\| }{t}=\langle \| \cdot \| '(x),h\rangle \end{equation*} holds. Since, at any $x\neq 0$, the gradient of the norm satisfies \begin{equation} \| \| \cdot \| '(x)\| =1, \quad \langle \| \cdot \| '(x),x\rangle =\| x\| \label{304} \end{equation} and it is the unique element in the dual space having these properties, we immediately get that: if $X$ is a smooth real Banach space, then the duality mapping corresponding to a gauge function $\varphi $ is the single valued mapping $J_{\varphi }:X\to X^{\ast }$, defined as follows: \begin{equation} \begin{gathered} J_{\varphi }0=0, \\ J_{\varphi }x=\varphi (\| x\|) \| \cdot \| '(x),\quad \text{if }x\neq 0. \end{gathered} \label{305} \end{equation} \begin{remark}\label{R4} \rm By coupling (\ref{305}) with the Asplund's result quoted above, we get: if $X$ is smooth, then \begin{equation} J_{\varphi }x=\psi '(x)= \begin{cases} 0 &\text{if }x=0\\ \varphi ( \| x\|) \| \cdot \| '(x) &\text{if }x\neq 0, \end{cases} \label{306} \end{equation} where $\psi $ is given by \eqref{302}. \end{remark} From (\ref{304}) and (\ref{305}), it follows that \begin{equation} \begin{gathered} \| J_{\varphi }x\| =\varphi (\| x\|) , \\ \langle J_{\varphi }x,x\rangle =\varphi (\| x\|) \| x\| ,\quad\text{for all }x\in X. \end{gathered} \label{307} \end{equation} The following surjectivity result will play an important role in what follows: \begin{theorem} \label{T11} If $X$ is a real reflexive and smooth Banach space, then any duality mapping $J_{\varphi }:X\to X^{\ast }$ is surjective. Moreover, if $X$ is also strictly convex, then $J_{\varphi }$ is a bijection of $X$ onto $X^{\ast }$. \end{theorem} In proving the surjectivity of $J_{\varphi }$, the main ideas are as follows: (for more details, see Browder \cite{[Br]}, Lions \cite{[Li]}, Deimling \cite{[De]}) (i) $J_{\varphi }$ is monotone: \begin{equation*} \langle J_{\varphi }x-J_{\varphi }y,x-y\rangle \geq \left( \varphi (\| x\|) -\varphi \left( \| y\| \right) \right) \left( \| x\| -\| y\| \right) \geq 0, \forall x,y\in X. \end{equation*} The first inequality is a direct consequence of (\ref{307}) while the second one follows from $\varphi $ being increasing. (ii) Any duality mapping on a real smooth and reflexive Banach space is demicontinuous: \begin{equation*} x_{n}\to x\Rightarrow J_{\varphi }x_{n}\rightharpoonup J_{\varphi }x. \end{equation*} Indeed, since $\left( x_{n}\right) _{n}$ is bounded and $\| J_{\varphi }x_{n}\| =\varphi \left( \|x_{n}\| \right) $, it follows that $\left( J_{\varphi}x_{n}\right) _{n}$ is bounded in $X^{\ast }$. Since $X^{\ast }$ is reflexive, in order to prove $J_{\varphi }x_{n}\rightharpoonup J_{\varphi }x$ it is enough to prove that $J_{\varphi }x$ is the unique point in the weak closure of $\left( J_{\varphi}x_{n}\right) _{n}$. (iii) $J_{\varphi }$ is coercive, in the sense that \begin{equation*} \frac{\langle J_{\varphi }x,x\rangle }{\| x\| }=\varphi (\| x\|) \to \infty \text{ as }\| x\| \to \infty . \end{equation*} According to a well-known surjectivity result due to Browder (see, for example, Browder \cite{[Br]}, Lions \cite{[Li]}, Zeidler \cite{[Ze]}, Deimling \cite{[De]}), if $X$ is a reflexive real Banach space, then any monotone, demicontinuous and coercive operator $T:X\to X^{\ast }$ is surjective. Consequently, from (i), (ii), (iii) and the Browder's surjectivity result above mentioned it follows that, under the hypotheses of Theorem \ref{T11}, $J_{\varphi }$ is surjective. It can be shown that if $X$ is a strictly convex real Banach space, then any duality mapping $J_{\varphi }:X\to \mathcal{P}(X^{\ast })$ is strictly monotone, in the following sense: if $x,y\in X$ and $x\neq y$, then, for any $x^{\ast }\in J_{\varphi }x$ and $y^{\ast }\in J_{\varphi }y$ one has $\langle x^{\ast }-y^{\ast },x-y\rangle >0$. Clearly, the strict monotonicity implies the injectivity: if $x,y\in X$ and $x\neq y$ then $J_{\varphi }x\cap J_{\varphi }y=\emptyset $. In particular, if the strictly convex real Banach space $X$ is also a smooth one, then any duality mapping $J_{\varphi }:X\to X^{\ast }$ is strictly monotone: \begin{equation*} \langle J_{\varphi }x-J_{\varphi }y,x-y\rangle >0 , \quad \forall x,y\in X, x\neq y, \end{equation*} and, consequently, injective. \begin{corollary}\label{C3} If $X$ is a reflexive and smooth real Banach space having the Kade\v{c}-Klee property, then any duality mapping $J_{\varphi}:X\to X^{\ast }$ is bijective and has a continuous inverse. Moreover, \begin{equation} J_{\varphi }^{-1}=\chi ^{-1}J_{\varphi ^{-1} }^{\ast }, \label{308} \end{equation} where $J_{\varphi ^{-1} }^{\ast }:X^{\ast }\to X^{\ast \ast }$ is the duality mapping on $X^{\ast }$ corresponding to the gauge function $\varphi ^{-1}$ and $\chi :X\to X^{\ast \ast }$ is the canonical isomorphism defined by $\langle \chi (x),x^{\ast }\rangle =\langle x^{\ast },x\rangle $, for all $x\in X$, for all $x^{\ast }\in X^{\ast }$. \end{corollary} \begin{proof} The existence of $J_{\varphi }^{-1}$ follows from Theorem \ref{T11}. As far as formula (\ref{308}) is concerned, first we shall prove that, under the hypotheses of Corollary \ref{C3}, any duality mapping on $X^{\ast }$ (in particular, that corresponding to the gauge function $\varphi ^{-1}$) is single valued. This is equivalent with proving that $X^{\ast }$ is smooth. The smoothness of $X^{\ast }$ will be proved by using the (partial) duality between strict convexity and smoothness given by the following theorem due to Klee (see Diestel \cite[Chapter 2, \S 2, Theorem 2]{[Di]}): \begin{equation*} X^{\ast }\text{ smooth (strictly convex) }\Rightarrow X \text{ strictly convex (smooth).} \end{equation*} Clearly, if $X$ is reflexive, then \begin{equation*} X^{\ast }\text{ smooth (strictly convex) }\Leftrightarrow X \text{ strictly convex (smooth).} \end{equation*} Now, by the hypotheses of Corollary \ref{C3}, $X$ is reflexive and smooth. Also, by the same hypotheses, $X$ possesses the Kade\v{c}-Klee property, that means: $X$ is strictly convex and \begin{equation} \left[ x_{n}\rightharpoonup x\text{ and }\| x_{n}\| \to \| x\| \right] \Rightarrow x_{n}\to x. \label{309} \end{equation} Consequently, $X$ being reflexive, smooth and strictly convex so is $X^{\ast}$. Let us prove that equality (\ref{308}) holds or, equivalently, \begin{equation} \chi J_{\varphi }^{-1}x^{\ast }=J_{\varphi ^{-1} }^{\ast }x^{\ast }, \forall x^{\ast }\in X^{\ast }. \label{310} \end{equation} From the definition of duality mappings, $J_{\varphi ^{-1} }^{\ast }x^{\ast }$ is the unique element in $X^{\ast \ast }$ having the metric properties \begin{equation} \begin{gathered} \langle J_{\varphi ^{-1} }^{\ast }x^{\ast },x^{\ast }\rangle =\varphi ^{-1}\left( \| x^{\ast }\| \right) \| x^{\ast }\| , \\ \| J_{\varphi ^{-1} }^{\ast }x^{\ast }\| =\varphi ^{-1}\left( \| x^{\ast }\| \right) . \end{gathered}\label{311} \end{equation} We shall show that $\chi J_{\varphi }^{-1}x^{\ast }$ possesses the same metric properties and then the result follows by unicity. Putting $x^{\ast }=J_{\varphi }x$ it follows (by definition of $J_{\varphi }$) that \begin{gather*} x^{\ast }=\varphi (\| x\|) , \\ \langle x^{\ast },x\rangle =\varphi (\| x\|) \| x\| =\varphi ^{-1}\left( \| x^{\ast }\| \right) \|x^{\ast }\| \end{gather*} and, consequently, we deduce that \begin{equation} \begin{gathered} \langle \chi J_{\varphi }^{-1}x^{\ast },x^{\ast }\rangle =\langle \chi (x),x^{\ast }\rangle =\langle x^{\ast },x\rangle =\varphi ^{-1}\left( \| x^{\ast }\| \right) \| x^{\ast }\| , \\ \| \chi J_{\varphi }^{-1}x^{\ast }\| =\| \chi (x)\| =\| x\| =\varphi ^{-1}(\| x\|) \end{gathered} \label{312} \end{equation} Equality (\ref{310}) follows by comparing (\ref{311}) and (\ref{312}) and using the uniqueness result evoked above. Formula (\ref{308}) is basic in proving the continuity of $J_{\varphi }^{-1}$. Indeed, let $x_{n}^{\ast }\to x^{\ast }$ in $X^{\ast }$. As any duality mapping on a reflexive Banach space, $J_{\varphi ^{-1} }^{\ast }$ is demicontinuous, $J_{\varphi ^{-1} }^{\ast }x_{n}^{\ast }\rightharpoonup J_{\varphi ^{-1} }^{\ast }x^{\ast }$. Consequently, we deduce that \begin{equation} J_{\varphi }^{-1}x_{n}^{\ast }=\chi ^{-1}J_{\varphi ^{-1} }^{\ast }x_{n}^{\ast }\rightharpoonup \chi ^{-1}J_{\varphi ^{-1} }^{\ast }x^{\ast }=J_{\varphi }^{-1}x^{\ast }. \label{313} \end{equation} On the other hand, \begin{equation} \| J_{\varphi }^{-1}x_{n}^{\ast }\| =\| \chi ^{-1}J_{\varphi ^{-1} }^{\ast }x_{n}^{\ast }\| =\| J_{\varphi ^{-1} }^{\ast }x_{n}^{\ast }\| =\varphi ^{-1}\left( \| x_{n}^{\ast }\| \right) \to \varphi ^{-1}(\| x\|) =\| J_{\varphi }^{-1}x^{\ast }\| . \label{314} \end{equation} From (\ref{313}), (\ref{314}) and the Kade\v{c}-Klee property of $X$, we infer that $J_{\varphi }^{-1}x_{n}^{\ast }\to J_{\varphi }^{-1}x^{\ast }$. \end{proof} \begin{corollary}\label{C4} If $X$ is a weakly locally uniformly convex, reflexive and smooth real Banach space, then any duality mapping $J_{\varphi }:X\to X^{\ast }$ is bijective and has a continuous inverse given by (\ref{308}). \end{corollary} \begin{proof} Since any weakly locally uniformly convex Banach space has the Kade\v{c}-Klee property (see Diestel[Chapter 2, \S 2, Theorems 3 and 4(iii)]\cite{[Di]}) the result follows by Corollary \ref{C3}. \end{proof} \begin{theorem} \label{T12} Let $\varphi $ be a gauge function. The duality mapping on $(W_{0}^{m}E_{A}(\Omega )$, $\| u\| _{m,A})$ is the single valued operator $J_{\varphi }:W_{0}^{m}E_{A}(\Omega )\to \big(W_{0}^{m}E_{A}(\Omega )\big) ^{\ast }$ defined by \[ J_{\varphi }u=\psi '(u)= \begin{cases} 0 &\text{if }u=0\\ \varphi ( \| u\| _{m,A}) \| \cdot \| _{m,A}'(u) &\text{if }u\neq 0, \end{cases} \] where \begin{equation*} \psi (u)=\int_{0}^{\| u\| _{m,A}}\varphi (t)dt=\Phi \left( \| u\| _{m,A}\right) , \quad \forall u\in W_{0}^{m}E_{A}(\Omega ), \end{equation*} where $\Phi $ is the $N$-function generated by $\varphi $ and, for any $u\neq 0$, $\| \cdot \| _{m,A}'(u)$ being given by (\ref{5}). \end{theorem} This result immediately follows by Theorem \ref{T9} and Remark \ref{R4}. \section{Nemytskij operator on $L_{A}(\Omega )$} We recall that $f:\Omega \times \mathbb{R}\to \mathbb{R}$ is a \textit{Carath\'{e}odory function} if it satisfies: \begin{itemize} \item[(i)] for each $s\in \mathbb{R}$, the function $x\to f(x,s)$ is Lebesgue measurable in $\Omega $; \item[(ii)] for a.e. $x\in \Omega $, the function $s\to f(x,s)$ is continuous in $\mathbb{R}$. \end{itemize} We make convention that in the case of a Carath\'{e}odory function, the assertion $x\in \Omega $ to be understood in the sense a.e. $x\in \Omega $. \begin{proposition}[{\cite[Theorem 17.1]{[KR]}}] Suppose that $f:\Omega \times \mathbb{R}\to \mathbb{R}$ is a Carath\'{e}o\-dory function. Then, for each measurable function $u$, the function $N_{f}u:\Omega \to R$, given by \begin{equation} (N_{f}u)(x)=f(x,u(x)),\quad \text{for each }x\in \Omega \label{ec5.1} \end{equation} is measurable in $\Omega $. \end{proposition} \begin{definition}\label{D4} \rm Let $\mathcal{M}$ be the set of all measurable functions $u:\Omega \to \mathbb{R}$, $f:\Omega \times \mathbb{R}\to \mathbb{R}$ be a Carath\'{e}odory function. The operator $N_{f}:\mathcal{M}\to \mathcal{M}$ given by \eqref{ec5.1} is called \textit{Nemytskij operator} defined by Carath\'{e}odory function $f$. \end{definition} Theorem here below states sufficient conditions when Nemytskij operator maps a Orlicz class $K_{A}(\Omega )$ into another Orlicz class $K_{B}(\Omega )$, being at the same time continuous and bounded. The following result is useful. \begin{theorem}\label{T13} Let $A$ and $B$ be two $N$-functions and $f:\Omega \times \mathbb{R}\to \mathbb{R}$ be a Carath\'{e}o\-dory function which satisfies the growth condition \begin{equation} | f(x,u)| \leq c(x)+bB^{-1}\left( A(u)\right) , x\in \Omega , u\in \mathbb{R}, \label{ec5.2} \end{equation} where $c\in K_{B}(\Omega )$ and $b\geq 0$ is a constant. Then the following statements are true: \begin{itemize} \item[(i)] If $B$ satisfies the $\Delta _{2}$-condition, then $N_{f}$ is well-defined and mean bounded from $K_{A}(\Omega )$ into $K_{B}(\Omega )=E_{B}(\Omega )$. Moreover, $N_{f}:\big( E_{A}(\Omega ),\| \cdot \| _{(A)}\big) \to \big( E_{B}(\Omega ),\| \cdot \| _{(B)}\big) $ is continuous; \item[(ii)] If both $A$ and $B$ satisfy the $\Delta _{2}$-condition, then $N_{f}:\big( E_{A}(\Omega ),\| \cdot \| _{(A)}\big) \to \big(E_{B}(\Omega ),\| \cdot \| _{(B)}\big) $ is norm bounded. \end{itemize} \end{theorem} \begin{proof} Let us first remark that the well-definedness of $N_{f}$ as well as the continuity and the boundedness on every ball $B(0,r)\subset L_{A}(\Omega )$, with $r<1$, may be obtained as consequences Theorem 17.6 in Krasnosel'skij and Rutickij (\cite{[KR]}). The proof of this theorem is quite complicated; that is why a direct proof of Theorem \ref{T13}, including the supplementary result given by (ii), will be given below. (i) Let $u,v\in \mathbb{R}$. Since $B$ is convex and satisfies the $\Delta_{2}$-condition, one has \begin{equation} B(u+v)=B\big( 2\cdot \frac{1}{2}(u+v)\big) \leq \frac{k}{2}\left( B(u)+B(v)\right) . \label{ec5.3} \end{equation} Let $p$ be such that $2^{p}\geq b$. Since $B$ satisfies the $\Delta _{2}$-condition, one has \begin{equation} B(bu)\leq B(2^{p}u)\leq k^{p}B(u). \label{ec5.4} \end{equation} Now, let $u\in K_{A}(\Omega )$. By using (\ref{ec5.2}), (\ref{ec5.3}), ( \ref{ec5.4}) and integrating on $\Omega $, we have \begin{equation} \begin{aligned} \int_{\Omega }B\left[ N_{f}(u)(x)\right] \,dx &=\int_{\Omega }B\left( | f(x,u(x))| \right) \,dx\\ &\leq \frac{k}{2}\int_{\Omega }B\left( c(x)\right) \,dx +\frac{k^{p+1}}{2}\int_{\Omega }A( u(x)) \,dx<\infty , \end{aligned} \label{ec5.5} \end{equation} saying then $N_{f}( K_{A}(\Omega )) \subset L_{B}(\Omega)=E_{B}(\Omega )$. From (\ref{ec5.5}) it follows that, if $u\in K_{A}(\Omega )$ and $\int_{\Omega }A( u(x)) \,dx\leq $const., then \begin{equation*} \int_{\Omega }B\left[ N_{f}(u)(x)\right]\,dx\leq \frac{k}{2} \int_{\Omega }B\left( c(x)\right)\,dx+{\rm const.}; \end{equation*} therefore $N_{f}$ transforms mean bounded sets in $K_{A}(\Omega )$ into mean bounded sets in $E_{B}(\Omega )$. Now, let us consider $u\in E_{A}(\Omega) $. For the continuity of $N_{f}$, it suffices to show that every sequence $( u_{n})_{n}\subset E_{A}(\Omega) $ such that \begin{equation*} \lim_{n\to \infty }\| u_{n}-u\| _{A}=0 \end{equation*} has a subsequence $( u_{n_{k}}) _{k}$ such that $N_{f}(u_{n_{k}}) \to N_{f}(u)$ as $k\to \infty $, in $L_{B}(\Omega )=E_{B}(\Omega )$. Indeed, let $( u_{n}) _{n}$ be a sequence as above. By using Lemma \ref{L2}, it follows that there exists a subsequence $(u_{n_{k}}) _{k}\subset (u_n) _{n}$ and $h\in K_{A}(\Omega) $ such that \begin{equation} \lim_{k\to \infty }u_{n_{k}}(x)=u(x), \quad\text{a.e. }x\in \Omega \label{ec5.6} \end{equation} and \begin{equation} | u_{n_{k}}(x)| \leq | h(x)| , \quad\text{a.e. } x\in \Omega , k\in \mathbb{N}. \label{ec5.7} \end{equation} The function $f$ being a Carath\'{e}odory function, it is clear that \begin{equation*} \lim_{k\to \infty }N_{f}( u_{n_k}) (x)=N_{f}(u)(x), \quad\text{a.e. }x\in \Omega , \end{equation*} therefore, \begin{equation} \lim_{k\to \infty }B\left( N_{f}( u_{n_k}) (x)-N_{f}(u)(x)\right) =0, \quad\text{a.e. }x\in \Omega . \label{ec5.8} \end{equation} On the other hand, from (\ref{ec5.2}) it follows that \begin{equation*} | N_{f}( u_{n_{k}}) (x)| =| f(x,u_{n_{k}}(x))| \leq c(x)+bB^{-1}\left( A(h(x))\right) , \quad\text{a.e. }x\in \Omega , k\in \mathbb{N}. \end{equation*} Consequently, by using a similar argument to that in (\ref{ec5.5}) and taking into account (\ref{ec5.7}), one obtains \begin{equation*} B\left( N_{f}( u_{n_k}) (x)\right) \leq \frac{k}{2}B\left( c(x)\right) +\frac{k^{p+1}}{2}A\left( h(x)\right) , \end{equation*} and by using a similar argument to that in (\ref{ec5.5}), one has \begin{equation*} B\left( N_{f}(u) (x)\right) \leq \frac{k}{2}B\left( c(x)\right) + \frac{k^{p+1}}{2}A\left( h(x)\right) ; \end{equation*} therefore, by (\ref{ec5.3}) and the preceding two inequalities, one obtains \begin{align*} B\left( N_{f}( u_{n_k}) (x)-N_{f}(u)(x)\right) &\leq \frac{k}{2}B\left( N_{f}( u_{n_k}) (x)\right) +\frac{k}{2}B\left( N_{f}(u)(x)\right) \\ &\leq \frac{k^{2}}{2}B( c(x)) +\frac{k^{n+2}}{4}[A\left( h(x)\right) +A( u(x)) ]. \end{align*} Since the right term of this inequality is in $L^{1}(\Omega) $ and (\ref{ec5.8}) holds, by applying Lebesgue's dominated convergence theorem, it follows that \begin{equation*} \lim_{k\to \infty }\int_{\Omega }B\left( N_{f}( u_{n_k}) (x)-N_{f}(u)(x)\right) \,dx=0, \end{equation*} that is the subsequence $\left( N_{f}(u_{n_{k}})\right) _{k}$ converges in mean to $N_{f}(u)$. The $N$-function $B$ satisfying the $\Delta _{2}$ -condition, it follows that the subsequence $\left( N_{f}(u_{n_{k}})\right) _{k}$ converges in norm to $N_{f}(u)$, therefore the operator $N_{f}$ is continuous. (ii) Now, let us suppose that $A$ satisfies the $\Delta _{2}$-condition. If the set $\mathcal{M}\subset E_{A}(\Omega) $ is norm bounded, then, from $\Delta _{2}$-condition, it follows that $\mathcal{M}$ is mean bounded, therefore $N_{f}(\mathcal{M})$ is also mean bounded. But any mean bounded set is norm bounded too. \end{proof} Now, let us consider the functional $\mathcal{G}:E_{A}(\Omega )\to \mathbb{R}$ given by \begin{equation} \mathcal{G}(u)=\int_{\Omega }G\left( x,u(x)\right) \,dx, \label{ec5.9} \end{equation} where \begin{equation} G\left( x,s\right) =\int_{0}^{s}f(x,\tau )\,d\tau . \label{ec5.10} \end{equation} We recall the following result concerning the differentiability of the functional $F$. \begin{theorem}[{\cite[Theorem 18.1]{[KR]}}] \label{T14} Let $f:\Omega \times \mathbb{R} \to \mathbb{R}$ be a Carath\'{e}odory function. Assume that there exists an $N$-function $M$ such that \begin{equation} | f(x,u)| \leq c(x)+b\overline{M}^{-1}\left( M(u)\right) , x\in \Omega , u\in \mathbb{R}, \label{ec5.12} \end{equation} where $\overline{M}$ is the complementary $N$-function to $M$, $c\in K_{ \overline{M}}(\Omega )$, $b\geq 0$ is a constant and $\overline{M}$ satisfies the $\Delta _{2}$-condition. Then, the functional $\mathcal{G}$, given by \eqref{ec5.9}, is of class $\mathcal{C}^{1}$ on $E_{M}(\Omega )$, with Fr\'{e}chet derivative given by \begin{equation} \langle \mathcal{G}'(u),h\rangle =\int_{\Omega }N_{f}(u)(x)h(x)\,dx=\int_{\Omega }f\left( x,u(x)\right) \cdot h(x)\,dx, \quad u,h\in E_{M}(\Omega ). \label{ec5.13} \end{equation} \end{theorem} \section{An existence result for \eqref{ec1.2}, \eqref{ec1.3}, via a Leray-Schauder technique} Since any $u\in W_{0}^{m}E_{A}(\Omega )$ satisfies the boundary conditions \eqref{ec1.3} (see Theorem \ref{T3}), the idea is to prove the existence of an element $u\in W_{0}^{m}E_{A}(\Omega )$ which satisfies also (\ref{ec1.2}) in a sense that will be clarified. First, we shall prove the following result. \begin{proposition}\label{P7} Let $A(u)=\int_{0}^{| u| }a(t)\,dt$ be an $N$-function which satisfies the $\Delta _{2}$-condition. Suppose that the function $\frac{a(t)}{t}$ is nondecreasing on $(0,\infty )$. Then \begin{equation*} J_{a}:W_{0}^{m}E_{A}(\Omega )\to \big( W_{0}^{m}E_{A}(\Omega )\big) ^{\ast } \end{equation*} is a bijection, with monotone, bounded and continuous inverse. \end{proposition} \begin{proof} According to Theorem \ref{T10}, $W_{0}^{m}E_{A}(\Omega )$ is uniformly convex (in particular, reflexive) and smooth (Theorem \ref{T9}) The result follows by Corollary \ref{C4}. \end{proof} In what follows, we give a meaning of the right member in (\ref{ec1.2}) as operator acting from $W_{0}^{m}E_{A}(\Omega )$ into $\left(W_{0}^{m}E_{A}(\Omega )\right) ^{\ast }$. To do it, let us first remark that if $M_{\alpha }$, $| \alpha | 0}\frac{tM_{\alpha}'(t)}{M_{\alpha }(t)}, \quad | \alpha | 0}\frac{ta(t)}{A(t)}, \end{equation*} then the operator $P=J_{a}^{-1}\circ i^{\ast }\circ N\circ i$ has a fixed point in $W_{0}^{m}E_{A}(\Omega )$ or equivalently, problem (\ref{ec1.2}), (\ref{ec1.3}) has a solution. Moreover, the solution set of problem (\ref{ec1.2}), (\ref{ec1.3}) is compact in $W_{0}^{m}E_{A}(\Omega )$. \end{theorem} \begin{proof} In order to prove that the compact operator $P$ has a fixed point, it suffices to prove that the set \begin{equation*} \mathcal{S}=\{u\in W_{0}^{m}E_{A}(\Omega )\mid \exists t\in [ 0,1] \text{ such that }u=tPu\} \end{equation*} is bounded in $W_{0}^{m}E_{A}(\Omega )$. To do this, a technical lemma is needed. \end{proof} \begin{lemma}\label{L5} Let $A(u)=\int_{0}^{| u| }a(t)\,dt$ be an $N$-function. \begin{itemize} \item[(a)] If $p_{0}=\inf_{t>0}\frac{ta(t)}{A(t)}$, then for any $t>1$, one has $A(t)\geq A(1)t^{p_{0}}$; \item[(b)] If $A$ satisfies the $\Delta _{2}$-condition and \begin{equation*} p^{\ast }=\sup_{t>0}\frac{ta(t)}{A(t)}, \end{equation*} then $\infty >p^{\ast }>1$ and for any $u\in L_{A}(\Omega )$ with $\| u\| _{(A)}>1$, one has \begin{equation} \int_{\Omega }A(u(x))dx\leq \| u\| _{(A)}^{p^{\ast }}. \label{602} \end{equation} \end{itemize} \end{lemma} \begin{proof} (a) First, we remark that, from Young's equality, we have \begin{equation} \frac{ta(t)}{A(t)}>1, \quad \textit{for any }t>0, \label{600} \end{equation} therefore $p_{0}\geq 1$. Integrating the inequality \begin{equation*} \frac{a(\tau )}{A(\tau )}\geq \frac{p_{0}}{\tau }, \quad \tau >0. \end{equation*} over the interval $[1,t]$, we obtain \begin{equation} A(t)\geq A(1)t^{p_{0}},\quad\text{for }t>1. \label{601} \end{equation} (b) According to (\ref{600}), $p^{\ast }>1$. Since $A$ satisfies the $\Delta_{2}$-condition, $kA(t)\geq A(2t)>ta(t)$; therefore \begin{equation*} \frac{ta(t)}{A(t)}2. \end{equation*} Thus, $p^{\ast }$ is finite and \begin{equation*} \frac{a(\tau )}{A(\tau )}\leq \frac{p^{\ast }}{\tau }, \quad \tau >0. \end{equation*} Now, let $u$ be such that $\| u\| _{(A)}>1$. Integrating over the interval $[\frac{| u(x)| }{\| u\| _{(A)}},| u(x)| ]$, we obtain \begin{equation} A(u(x))\leq A(\frac{| u(x)| }{\| u\| _{(A)}})\| u\| _{(A)}^{p^{\ast }}. \label{604} \end{equation} Integrating over $\Omega $ and taking into account that \begin{equation*} \int_{\Omega }A(\frac{| u(x)| }{\| u\| _{(A)}})dx=1, \end{equation*} it follows (\ref{610}). Now, let $u\in \mathcal{S}$, $u=tJ_{a}^{-1}(i^{\ast }Ni)u$, $t\in (0,1]$. Then $J_{a}(\frac{u}{t})=(i^{\ast }Ni)u$, therefore (see (\ref{610})), we have \begin{equation} \langle J_{a}(\frac{u}{t}),\frac{u}{t}\rangle =\frac{1}{t} \langle (i^{\ast }Ni)u,u\rangle =\frac{1}{t} \sum_{\vert \alpha | 1$, then from (\ref{602}), \begin{equation*} \int_{\Omega }M_{\alpha }(D^{\alpha }(u(x)))dx\leq \| D^{\alpha }(u)\| _{(M_{\alpha })}^{\gamma _{\alpha }}\leq c^{\gamma }\| u\| _{m,A}^{\gamma }, \end{equation*} with $\gamma \geq 1$. Consequently, if $u\in W_{0}^{m}E_{A}(\Omega )$, we have \begin{equation} \int_{\Omega }M_{\alpha }(D^{\alpha }u(x))dx\leq c^{\gamma }\| u\| _{m,A}^{\gamma }+c\| u\| _{m,A}, | \alpha | 0$ and $r>0$ such that $I(u)\geq r$ for $\| u\| =\rho $; \item[(G2)] There exists $e\in X$ with $\| e\| >\rho $ such that $I(e)\leq 0$. \end{itemize} Let \begin{equation*} \Gamma =\{\gamma \in C([0,1];X);\gamma (0)=0,\gamma (1)=e\} \end{equation*} and set \begin{equation*} c=\inf_{\gamma \in \Gamma }\max_{0\leq t\leq 1}I(\gamma (t)) \text{ {} {}}(c\geq r). \end{equation*} Then, there is a sequence $(u_n) _{n}$ in $X$ such that \begin{equation*} I(u_{n})\to c\quad \text{and}\quad I'(u_{n})\to 0\text{ as } n\to \infty . \end{equation*} \end{theorem} \begin{remark} \label{R5} \rm Let $A$ be the $N$-function given by (\ref{ec1.1}). Suppose that $A$ satisfies the $\Delta _{2}$-condition. Then, according to Lemma \ref{L5}(b), \begin{equation*} \infty >p^{\ast }=\sup_{u>0}\frac{ua(u)}{A(u)}>1. \end{equation*} Therefore, \begin{equation*} \frac{a(u)}{A(u)}\leq \frac{p^{\ast }}{u}, u>0. \end{equation*} Integrating over the interval $[ t_{1},t_{2}] $, $t_{2}>t_{1}>0$, we obtain \begin{equation} \frac{A(t_{2})}{A(t_{1})}\leq \big( \frac{t_{2}}{t_{1}}\big) ^{p^{\ast }}. \label{ec6.99} \end{equation} In particular, for $00}\frac{ta(t)}{A(t)}, \quad p^{\ast }=\sup_{t>0}\frac{ta(t)}{A(t)}, \quad p_{\ast }=\liminf_{t\to \infty }\frac{tA_{\ast }'(t)}{A_{\ast }(t)}, \label{ec6.0} \end{equation} $A_{\ast }$ being the Sobolev conjugate of $A$. Suppose that the following conditions hold: \begin{itemize} \item[(H1)] there exists a positive constant $C>0$ such that \begin{equation} A(t)\geq C\cdot t^{p_{0}}, \forall t\in (0,1)\text{;} \label{ec6.56} \end{equation} \item[(H2)] there exist the $N$-functions $M_{\alpha }$, $| \alpha| 0$ and $\theta _{\alpha}>p^{\ast }$ such that \begin{equation} 0<\theta _{\alpha }G_{\alpha }(x,s)\leq sg_{\alpha }(x,s),\quad\text{for a.e. }x\in \Omega \label{ec6.55} \end{equation} and all $s$ with $| s| \geq s_{\alpha }$ and $p^{\ast }$ is given by (\ref{ec6.0}); \item[(H5)] $p_{0}0$ such that \begin{equation} G_{\alpha }(x,s)<\mu _{\alpha }A(s),\quad\text{for }x\in \Omega ,\; 0<| s| 0$ such that \begin{equation*} \frac{g_{\alpha }(x,s)}{a(s) }<\mu _{\alpha } ,\quad \text{for }x\in \Omega , 0<| s| -\mu _{\alpha }a(| s| ) ,\quad\text{for }x\in \Omega , s\in (-s_{\alpha },0). \label{ec6.7} \end{equation} Thus \begin{equation} | g_{\alpha }(x,s)| <\mu _{\alpha}a(| s| ),\quad\text{for }x\in \Omega , | s| 0, \end{equation*} in particular \begin{equation*} \lim_{s\to \infty }\frac{M_{\alpha }(s)}{A_{\ast }(s)}=0. \end{equation*} Consequently, there exist $s_{\alpha }'>s_{\alpha }$ such that \begin{equation} M_{\alpha }(s)\leq C_{\alpha }'A_{\ast }(s),\quad \forall s\geq s_{\alpha }' \label{ec6.21} \end{equation} The definition of $p_{\ast }$ implies that there exists $\mu \in (0,p_{\ast }-p_{0})$ and $s_{\alpha }''>s_{\alpha}'$ such that \begin{equation} \frac{A_{\ast }'(s)}{A_{\ast }(s)}\geq \frac{p_{\ast }-\mu }{s} ,\quad\text{for }s\geq s_{\alpha }''. \label{ec6.22} \end{equation} Let us denote by $S_{\alpha }=\max \left( s_{\alpha}',s_{\alpha }''\right) $, $k_{\alpha }=\frac{S_{\alpha }}{s_{\alpha }}>1$, $| \alpha | 0. \label{ec6.29} \end{equation} Since $\| u\| _{m,A}<1$, one can consider the interval $\big[ \sqrt{T[u,u](x)}, \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\big] $. By integrating in (\ref{ec6.29}) over this interval, we obtain \begin{equation*} A\left( \sqrt{T[u,u](x)}\right) \leq \| u\| _{m,A}^{p_{0}}\cdot A\Big( \frac{\sqrt{T[u,u](x)}}{\| u\| _{m,A}}\Big) . \end{equation*} By integrating again this inequality on $\Omega $, we find that \begin{equation} \int_{\Omega }A\left( \sqrt{T[u,u](x)}\right) \,dx\leq \| u\| _{m,A}^{p_{0}}. \label{ec6.30} \end{equation} Consequently, taking into account (\ref{ec6.28}) and (\ref{ec6.30}) and summing by $\alpha $, we have \begin{equation} \sum_{| \alpha | C\| u\| _{m,A}^{p_{0}}-\frac{C}{2}\| u\| _{m,A}^{p_{0}}-D\cdot \| u\| _{m,A}^{p_{\ast }-\mu } =\| u\| _{m,A}^{p_{0}}\big[ \frac{C}{2}-D\| u\| _{m,A}^{p_{\ast }-\mu -p_{0}}\big] . \] So, for \[ \| u\| _{m,A}=\rho \leq \min \Big( 1,\frac{1}{ ( \max_{| \alpha | \frac{C}{6}\rho ^{p}>0$. Now, we shall verify the hypothesis (G2) of the mountain pass theorem. Let $\theta _{\alpha }$ and $s_{\alpha }$ be as in (H4). We shall deduce that for any $\alpha $ with $| \alpha |0,\quad\text{for a.e. }x\in \Omega \text{ and } | s| \geq s_{\alpha }. \label{73} \end{equation} Then, for a.e. $x\in \Omega $ and $\tau \geq s_{\alpha }$, from (\ref{ec6.55}), we have \begin{equation*} \frac{\theta _{\alpha }}{\tau }\leq \frac{g_{\alpha }(x,\tau )}{G_{\alpha }(x,\tau )}. \end{equation*} Integrating from $s_{\alpha }$ to $s\geq s_{\alpha }$, it follows that \begin{equation*} \frac{s^{\theta _{\alpha }}}{s_{\alpha }^{\theta _{\alpha }}}\leq \frac{G_{\alpha }(x,s)}{G_{\alpha }(x,s_{\alpha })}, \end{equation*} which implies \begin{equation} G_{\alpha }(x,s)\geq G_{\alpha }(x,s_{\alpha })\cdot \frac{s^{\theta _{\alpha }}}{s_{\alpha }^{\theta _{\alpha }}},\quad\text{for a.e. }x\in \Omega \text{ and }s\geq s_{\alpha }, \label{74} \end{equation} for any $\alpha $ with $| \alpha | 0$, for a.e. $x\in \Omega $, and $\mathop{\rm vol}\big( \Omega _{1}^{\alpha }(u)\big) >0$, $| \alpha | 0. \] On the other hand, if $x\in \Omega \setminus \Omega _{\lambda }^{\alpha }(u)$, then $\lambda D^{\alpha }u(x)1$, we have \begin{equation*} A\left( \lambda \| u\| _{m,A}\right) \leq A(1)\lambda ^{p^{\ast }}\| u\| _{m,A}^{p^{\ast }}-\sum_{| \alpha | p^{\ast }$ for any $\alpha $ with $| \alpha | 1$ and the second hypothesis of the mountain pass theorem satisfied. \end{proof} \begin{lemma} \label{L8} Under the hypotheses of Theorem \ref{T17}, the functional $F$ given by (\ref{ec6.3}) has the following property: any sequence $(u_n) _{n}\subset W_{0}^{m}E_{A}(\Omega )$ for which $\left(F(u_{n})\right) _{n}$ is bounded and $F'(u_{n})\to 0$ as $n\to \infty $, is bounded. \end{lemma} \begin{proof} Let $(u_{n})_{n}\subset W_{0}^{m}E_{A}(\Omega )$ be a sequence such that $\left( F(u_{n})\right) _{n}$ is bounded and $F'(u_{n})\to 0$ as $n\to \infty $. We shall show that the sequence $(u_{n})_{n}$ is bounded in $W_{0}^{m}E_{A}(\Omega )$. Indeed, let us put \begin{equation*} \theta =\min_{| \alpha | s_{\alpha }\}$, and $\Omega _{\alpha ,n}^{'}=\Omega \backslash \Omega _{\alpha ,n}$. Clearly \begin{equation} \label{904} \begin{aligned} &\sum_{| \alpha | 0$, the above inequality gives \begin{equation*} \langle \Psi '(u)-\Phi '(u-th),h\rangle \geq 0, \quad \forall h\in W_{0}^{m}E_{A}(\Omega ). \end{equation*} By letting $t\to 0$, one has \begin{equation*} \langle \Psi '(u)-\Phi '(u),h\rangle \geq 0, \quad \forall h\in W_{0}^{m}E_{A}(\Omega ). \end{equation*} Thus $F'(u)=\Phi '(u)-\Psi '(u)=0$. \end{proof} \section{Examples} In this section, some examples illustrating the above existence results are given. To do this, some prerequisites are needed. \begin{lemma}\label{L9} Let $A:\mathbb{R}\to\mathbb{R}_{+}$, $A(t)=\int_{0}^{| t| }a(s)\,ds$, be an $N$-function and $\overline{A}$, the complementary $N$-function of $A$. \begin{itemize} \item[(i)] Suppose that \begin{equation*} p^{\ast }=\sup_{t>0}\frac{ta(t)}{A(t)}<\infty \quad\text{and}\quad p_{0}=\inf_{t>0}\frac{ta(t)}{A(t)}>1. \end{equation*} Then both $A$ and $\overline{A}$ satisfy the $\Delta _{2}$-condition. \item[(ii)] Suppose, in addition, that $p^{\ast }0$ such that \begin{equation} A(t)\geq Ct^{\gamma }, \quad \forall t\in (0,A^{-1}(\delta )). \label{870} \end{equation} Then $A_{\ast }$, the Sobolev conjugate of $A$, can be defined. \end{itemize} \end{lemma} \begin{proof} (i) Since $p^{\ast }<\infty $, $A$ satisfies the $\Delta _{2}$-condition. (see \cite[Theorem 4.1]{[KR]}). Since $p_{0}>1$, $\overline{A}$ satisfies the $\Delta _{2}$-condition (see \cite[Theorem 4.3]{[KR]}). (ii) It is sufficient to prove that conditions \eqref{ec2.4} and (\ref{ec2.5}) are satisfied (see Theorem \ref{T5}). Indeed, it follows from (\ref{870}) that \begin{equation*} A^{-1}(\tau )\leq c_{1}\cdot \tau ^{1/\gamma }, \tau \in (0,\delta ), \end{equation*} with $C_{1}=C^{-1/\gamma }$. Consequently, \begin{equation*} \int_{t}^{\delta }\frac{A^{-1}(\tau )}{\tau ^{(N+1)/N}}\,d\tau \leq c_{1}\cdot \frac{N\gamma }{N-\gamma } \Big( \delta ^{\frac{N-\gamma }{ N\gamma }}-t^{\frac{N-\gamma }{N\gamma }}\Big) . \end{equation*} Without loss of generality, we may assume that $0<\delta <1$ and then \begin{align*} \lim_{t\to 0}\int_{t}^{1}\frac{A^{-1}(\tau )}{\tau ^{(N+1)/N}}\,d\tau &=\lim_{t\to 0}\Big( \int_{t}^{\delta }\frac{A^{-1}(\tau )}{\tau ^{(N+1)/N}}\,d\tau +\int_{\delta }^{1}\frac{A^{-1}(\tau )}{\tau ^{(N+1)/N}}\,d\tau \Big)\\ &\leq c_{1}\cdot \frac{N\gamma }{N-\gamma }\delta ^{\frac{N-\gamma }{N\gamma }}+\int_{\delta }^{1}\frac{A^{-1}(\tau )}{\tau ^{(N+1)/N}}\,d \tau <\infty . \end{align*} Thus \eqref{ec2.4} is satisfied. To prove that (\ref{ec2.5}) is also satisfied, we first remark that, from (\ref{ec6.99}), one has $A(t)\leq \frac{t^{p^{\ast }}}{\left(A^{-1}(1)\right) ^{p^{\ast }}}$, for any $t>A^{-1}(1)$. It follows that $A^{-1}(\tau )\geq c'\cdot \tau ^{1/p^{\ast }}$, for any $\tau >1$, with $c'=A^{-1}(1)$. Consequently, for any $t>1$, \begin{equation*} \int_{1}^{t}\frac{A^{-1}(\tau )}{\tau ^{(N+1)/N}}\,d\tau \geq c'\cdot \frac{Np^{\ast }}{N-p^{\ast }}\Big( t^{\frac{N-p^{\ast }}{ Np^{\ast }}}-1\Big) \end{equation*} and, since $p^{\ast }0$, $1\leq i\leq n$, $p_{i+1}>p_{i}>1$, $1\leq i\leq n-1$, $p_{n}0$ and $\theta _{\alpha }>p_{n}$ such that \begin{equation} 0<\theta _{\alpha }G_{\alpha }(x,s)\leq sg_{\alpha}(x,s), \quad\text{for a.e. }x\in \Omega \label{852} \end{equation} and all $s$ with $| s| \geq s_{\alpha }$. \end{itemize} Under these conditions, the problem (\ref{ec1.2}), (\ref{ec1.3}) has a nontrivial weak solution. \end{example} \begin{proof} Before giving the proof, we underline that the function $a$ given by hypothesis (i) appears in \cite{[GMU]}, in the following context (see \cite[example 3.1]{[GMU]}: if $a$ is given by (i) and \begin{equation*} f(s)=\sum_{j=1}^{m}\beta _{j}| s| ^{\delta _{j}-1}s, \end{equation*} with $\beta _{j}>0$, for $j=1,\dots ,m$, and $\delta _{j+1}>\delta _{j}>1$, for $j=1,\dots ,m-1$, satisfying $N>p_{n}$ and \begin{equation*} p_{n}-1<\delta _{m}<\frac{N( p_{n}-1) +p_{n}}{N-p_{n}}, \end{equation*} then, the problem \begin{gather*} -\left( r^{N-1}a(u')\right) '=r^{N-1}f(u)\quad \text{in }( 0,R) \\ u'(0)=0=u(R) \end{gather*} has a positive solution and therefore the problem \begin{gather*} -\mathop{\rm div}\big( \frac{a(| Du| )}{| Du| } Du\big) =f(u)\quad \text{in }\Omega =B_{\mathbb{R}^{N}}\left( 0,R\right) \\ u=0\quad \text{on }\partial \Omega \end{gather*} has a positive radial solution of class $\mathcal{C}^{1}$. The idea of the proof is as follows: The preceding assumptions entail that the hypotheses of Theorem \ref{T17} are fulfilled. To do this, first we compute the numerical characteristics $p_{0}$, $p^{\ast }$ and $p_{\ast }$, given by (\ref{ec6.0}). Let $h:(0,\infty )\to\mathbb{R}$, defined by $h(t)=\frac{ta(t)}{A(t)}$, with \begin{equation} A(t)=\sum_{i=1}^{n}\frac{a_{i}}{p_{i}}| t| ^{p_{i}}. \label{853} \end{equation} By direct calculus, we obtain $\lim_{t\to 0}h(t)=p_{1}$ and $p_{1}0$. Thus $p_{0}=p_{1}>1$. Analogously, $\lim_{t\to \infty }h(t)=p_{n}$ and $h(t)0$. Thus $p^{\ast }=p_{n}0. \label{855} \end{equation} Since $p^{\ast }=p_{n}1$ and $p^{\ast }1$. To conclude that (H2) in Theorem \ref{T17} is also satisfied, we have to prove that inequalities (\ref{ec6.4}) hold. Indeed, since $\overline{M}_{\alpha }(s)=\frac{| s| ^{q_{\alpha }'}}{q_{\alpha}'}$, $\frac{1}{q_{\alpha }}+\frac{1}{q_{\alpha}'}=1$ (see Remark \ref{R3}), then \begin{equation*} | s| ^{q_{\alpha }-1}=\left( q_{\alpha }-1\right) ^{\frac{1}{q_{\alpha }'}}\overline{M}_{\alpha }^{-1}\left( M_{\alpha }(s)\right) . \end{equation*} Consequently, (\ref{851}) rewrites as \begin{equation} | g_{\alpha }(x,s)| \leq c_{\alpha }+d_{\alpha }\left( q_{\alpha }-1\right) ^{\frac{1}{ q_{\alpha }'}}\overline{M}_{\alpha }^{-1}\left( M_{\alpha }(s)\right) , x\in \Omega , s\in \mathbb{R} , | \alpha | 0$, $1\leq i\leq n$, $p_{i+1}>p_{i}>1$, $1\leq i\leq n-1$, $p_{1}\geq 2$, $p_{n}2$, $2\leq i\leq n$, it easily follows that $\frac{a(t)}{t}$ is strictly increasing for $t>0$. As $M_{\alpha }$ functions, which increase essentially more slowly than $A_{\ast }$ near infinity and satisfy the $\Delta _{2}$-condition as well as the growth conditions (\ref{607}), we shall take $M_{\alpha }(s)=\frac{ | s| ^{q_{\alpha }}}{q_{\alpha }}$, $| \alpha | 0}\frac{tM_{\alpha }'(t)}{M_{\alpha }(t)}=q_{\alpha }, | \alpha | 0$ and $\theta _{\alpha}>p+1$ such that the conditions (\ref{852}) hold. \end{itemize} Under these conditions, the problem (\ref{ec1.2}), (\ref{ec1.3}) has a nontrivial weak solution. \end{example} \begin{proof} The idea of the proof is that used for Example \ref{Ex1}, namely, we shall show that the preceding assumptions entail the fulfillment of those of Theorem \ref{T17}. To do this, first we compute the numerical characteristics $p_{0}$, $p^{\ast }$ and $p_{\ast }$, given by (\ref{ec6.0}). Let $h:(0,\infty )\to\mathbb{R}$, defined by $h(t)=\frac{ta(t)}{A(t)}$, with \begin{equation} A(t)=\frac{t^{p}}{p}\sqrt{t^{2}+1}-\frac{1}{p}\int_{0}^{t}\frac{\tau ^{p}}{\sqrt{\tau ^{2}+1}}\,d\tau , t>0. \label{863} \end{equation} First, by direct calculus, we obtain $\lim_{t\to 0}h(t)=p$ and, since \begin{equation} h(t)=p+\frac{pI(t)}{t^{p}\sqrt{t^{2}+1}-I(t)}, I(t)=\int _{0}^{t}\frac{\tau ^{p}}{\sqrt{\tau ^{2}+1}}\,d\tau , \label{865} \end{equation} one has $p0$. Consequently $p_{0}=p>1$. Secondly, one has \begin{equation} \frac{pI(t)}{t^{p}\sqrt{t^{2}+1}-I(t)}<1, \quad \forall t>0. \label{864} \end{equation} Indeed, let $f(t)=( p+1) I(t)-t^{p}\sqrt{t^{2}+1}$. Since $f(0)=0$ and $f'(t)<0$, for all $t>0$, inequality (\ref{864}) follows. From (\ref{865}) and (\ref{864}), we infer that $h(t)0$ and, since $\lim_{t\to \infty }h(t)=p+1$, we conclude that $p^{\ast }=p+11$ and $p^{\ast }0}\frac{tM_{\alpha}'(t)}{M_{\alpha }(t)} =q_{\alpha }, \quad | \alpha | N-1$; \item[(ii)] there exist $q_{\alpha }$, $10. \label{972} \end{equation} By direct calculus, we obtain $\lim_{t\to 0}h(t)=p+1$ and, since \begin{equation} h(t)=p+\frac{pI(t)}{t^{p}\ln (1+t) -I(t)}, I(t)=\int_{0}^{t}\frac{\tau ^{p}}{1+\tau }\,d\tau , \label{973} \end{equation} one has $p0$. Consequently $p_{0}=p>1$. To compute $p^{\ast }$, we shall show that \begin{equation} \frac{pI(t)}{t^{p}\ln (1+t) -I(t)}<1, \quad \forall t>0. \label{974} \end{equation} Indeed, let $f(t)=f(t)=(p+) I(t)-t^{p}\ln (1+t) $. Since $f(0)=0$ and $f'(t)<0$, for all $t>0$, inequality (\ref{974}) follows. From (\ref{973}) and (\ref{974}), we infer that $h(t)0$ and, since $\lim_{t\to 0}h(t)=p+1$, we conclude that $p^{\ast }=p+1\frac{2}{p+1}t^{p+1}, \quad \forall t\in (0,\underline{\delta } =A^{-1}\left( \delta \right) ). \label{975} \end{equation} Since $p^{\ast }=p+11$, therefore \begin{equation*} \lim_{s\to \infty }\frac{s}{\left( A(s)\right) ^{\frac{1}{ q_{\alpha }}+\frac{1}{N}}}\leq \lim_{s\to \infty } \frac{s}{\left( A(1)\right) ^{\frac{1}{q_{\alpha }}+\frac{1}{N} }s^{\big( \frac{1}{q_{\alpha }}+\frac{1}{N}\big) p}}=0. \end{equation*} Consequently (see Example \ref{Ex3}), the arguments continue. Finally, the last condition in Theorem \ref{T15} is satisfied since \begin{equation*} \gamma _{\alpha }=\sup_{t>0}\frac{tM_{\alpha }'(t)}{M_{\alpha }(t)}=q_{\alpha }, | \alpha | 0$; \item[(i)] The Carath\'{e}odory functions $g_{\alpha }:\Omega \times\mathbb{R}\to\mathbb{R}$, $| \alpha | 0$ and $\theta _{\alpha}\geq p+1$ such that conditions (\ref{852}) hold. \end{itemize} Under these conditions, the problem (\ref{ec1.2}), (\ref{ec1.3}) has a nontrivial weak solution. \end{example} \begin{proof} The idea of the proof is as follows: the preceding assumptions entail that the hypotheses of Theorem \ref{T17} are fulfilled. To do this, first we compute the numerical characteristics $p_{0}$, $p^{\ast }$ and $p_{\ast }$, given by (\ref{ec6.0}). Let $h:(0,\infty )\to\mathbb{R}$, defined by $h(t)=\frac{ta(t)}{A(t)}$, with \begin{equation} A(t)=\frac{t^{p}}{p}\ln \left( 1+\alpha +t\right) -\frac{1}{p} \int_{0}^{t}\frac{\tau ^{p}}{1+\alpha +\tau }\,d\tau , t>0. \label{883} \end{equation} By direct calculus, we obtain $\lim_{t\to 0}h(t)=p$ and, since \begin{equation} h(t)=p+\frac{pI(t)}{t^{p}\ln \left( 1+\alpha +t\right) -I(t)}, I(t)=\int_{0}^{t}\frac{\tau ^{p}}{1+\alpha +\tau }\,d\tau \text{, } \label{884} \end{equation} one has $p0$. Consequently, $p_{0}=p>1$. To estimate $p^{\ast }$, we shall prove the existence of a constant $00}h(t)\leq p+C_{0}. \end{equation*} Indeed, the system \begin{equation} \begin{gathered} t-C(1+\alpha +t)\ln (1+\alpha +t)=0 \\ 1-C-C\ln (1+\alpha +t)=0 \end{gathered} \label{888} \end{equation} admits a unique solution $(t_{0},C_{0})$. Clearly \begin{equation*} 00$. This is true, since \begin{equation} \frac{pI(t)}{t^{p}\ln \left( 1+\alpha +t\right) -I(t)}\leq C_{0}, \forall t>0. \label{885} \end{equation} Indeed, let $f(t)=\left( p+C_{0}\right) I(t)-t^{p}\ln \left( 1+\alpha +t\right) $, for all $t\geq 0$. One has $f'(t)=\frac{pt^{p-1}}{1+\alpha +t}g(t)$ with $g(t)=t-C_{0}(1+\alpha +t)\ln (1+\alpha +t)$, $t\geq 0 $. Since $(t_{0},C_{0})$ is the unique solution of (\ref{888}), it may easily show that $t_{0}>0$, $g'(t)=0$ and $g(t_{0})=0=\max_{t\geq 0}g(t)$. Consequently, $g(t)\leq 0$, $t\geq 0$, which implies $f'(t)\leq 0$, for all $t\geq 0$. Since $f(0)=0$, it follows that $f(t)\leq 0$, for all $t\geq 0$, which is equivalent with (\ref{885}). This calculus explicit the claim concerning inequality (6.19) in \cite{[CGMS]}. Clearly, since $a(t)\geq \ln ( 1+\alpha) \cdot t^{p-1}$, for all $t\geq 0$, it follows that \begin{equation} A(t)\geq \frac{\ln (1+\alpha) }{p}t^{p}, \quad \forall t\geq 0. \label{886} \end{equation} Since $p^{\ast }1$ and $p^{\ast }