Diffusive convective elliptic problem in variable exponent space and measure data

Safimba Soma; Ibrahime Konaté; Adama Kaboré

doi:10.4153/S0008414X24001196

Diffusive convective elliptic problem in variable exponent space and measure data

Part of: Elliptic equations and systems

Published online by Cambridge University Press: 23 January 2025

Safimba Soma

Ibrahime Konaté and

Adama Kaboré

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Safimba Soma*: Affiliation:
Laboratoire de Mathématiques et d’Informatique (LA.M.I), UFR, Sciences Exactes et Appliquées, Université Joseph KI-ZERBO, 03 BP 7021 03 Ouagadougou, Burkina Faso
Ibrahime Konaté: Affiliation:
Laboratoire de Science et technologie (LaST), UFR, Sciences et Technologies, Université Thomas SANKARA, 12 BP 417 12 Ouagadougou, Burkina Faso e-mail: ibrakonat@yahoo.fr
Adama Kaboré: Affiliation:
Laboratoire de Mathématiques et d’Informatique (LA.M.I), Institut des Sciences et de Technologie, Ecole Normale Superieure, 01 BP 1757 01 Ouagadougou, Burkina Faso e-mail: kaboreadama59@yahoo.fr
*: e-mail: safimba.soma@ujkz.bf

Article contents

Abstract
Introduction
Preliminaires
Assumptions and main results
Existence of solution for a regular right hand side data
Proof of Theorem
Uniqueness of solution
References

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Abstract

In this article, we study a class of convective diffusive elliptic problem with Dirichlet boundary condition and measure data in variable exponent spaces. We begin by introducing an approximate problem via a truncation approach and Yosida’s regularization. Then, we apply the technique of maximal monotone operators in Banach spaces to obtain a sequence of approximate solutions. Finally, we pass to the limit and prove that this sequence of solutions converges to at least one weak or entropy solution of the original problem. Furthermore, under some additional assumptions on the convective diffusive term, we prove the uniqueness of the entropy solution.

Keywords

Sobolev spaces variable exponent entropy solution maximal monotone graph Radon measure

MSC classification

Primary: 35J65: Nonlinear boundary value problems for linear elliptic equations 35J20: Variational methods for second-order elliptic equations 35J25: Boundary value problems for second-order elliptic equations

Information

Type: Article
Information: Canadian Journal of Mathematics , Volume 78 , Issue 5 , October 2026 , pp. 1505 - 1528

DOI: https://doi.org/10.4153/S0008414X24001196 [Opens in a new window]
Copyright: © The Author(s), 2025. Published by Cambridge University Press on behalf of Canadian Mathematical Society

1 Introduction

Solving partial differential equations and variational problems combined with assumptions of $p(x)$ -growth has undergone significant evolution, both through the theoretical development of mathematics in Sobolev spaces with variable exponent and through their accuracy applications in modeling various real-word phenomena. Indeed, fluids that change their chemical properties when subjected to an electric field can be efficiently modeled in Sobolev spaces with variable exponents [Reference Antontsev and Rodrigues1, Reference Diening10, Reference Pfeiffer, Mavroidis, Bar-Cohen and Dolgin24]. A Leray–Lions type operator with $p(.)$ -growth also appears in biology, as it was discovered that blood exhibits electrorheological fluid properties. In [Reference Chen, Levine and Rao8], Chen et al. demonstrated the importance of such equations in image processing. For example, such operator can be used to search for a perfect image from a noisy one.

The aim of this article is to study the existence and uniqueness of solution for the following nonlinear elliptic problem.

(1.1) $$ \begin{align} (\mathcal{P})\left \lbrace \begin{array}{l} -\text{div}\ a(x,\nabla u) + \beta (u)+ \text{div} \phi (u) \ \ni \ \mu \ \ \text{in} \ \ \Omega \\ u = 0 \ \ \text{on}\ \ \partial \Omega , \end{array} \right. \end{align} $$ left parenthesis script upper P right parenthesis StartLayout Enlarged left brace 1st Row minus div a left parenthesis x comma nabla u right parenthesis plus beta left parenthesis u right parenthesis plus div phi left parenthesis u right parenthesis contains as member mu in normal upper Omega 2nd Row u equals 0 on partial differential normal upper Omega comma EndLayout

where $\Omega $ is an open bounded domain of $\mathbb {R}^N\ (N\geq 2)$ , $\beta $ is a maximal monotone graph with bounded domain on $\mathbb {R}$ (i.e., $\overline {\text {dom}(\beta )}=[m,M]\subset \mathbb {R}$ ) such that $0\in \beta (0)$ and $\mu $ is a Radon diffuse measure.

In the literature, there are numerous works related to the problem $(\mathcal {P})$ , but it’s important to emphasize that none of these studies have addressed the problem $(\mathcal {P})$ under measure data and with $\text {div}\phi \neq 0$ simultaneously. Going into detail, when $\beta $ is assumed to be a continuous and nondecreasing function with $div \phi =0$ , the authors in [Reference Bonzi and Ouaro6] proved the existence and uniqueness of entropy solution to the problem $(\mathcal {P})$ when the right-hand side datum belongs to $L^1$ . For other works in the same direction, we refer to [Reference Benboubker, Chrayteh, Moumni and Hjaji2, Reference Boccardo and Gallouet4, Reference Boccardo, Gallouet and Orsina5, Reference Dal Maso, Murat, Orsina and Prignet9, Reference Dolzmann, Hungerbehler and Muller11]. In the context of classical Sobolev space with constant exponent, Soma et al. [Reference Igbida, Ouaro and Soma16] analyzed the existence and uniqueness of solution of problem $(\mathcal {P})$ when the convective diffusive term $\phi $ is null (see also [Reference Bénilan, Boccardo, Gallouet, Gariepy, Pierre and Vázquez3]). Furthermore, they also obtained in [Reference Nyanquini, Ouaro and Soma22] the existence and uniqueness of the entropy solution in the framework of variable exponent spaces and measure data. In the case of the right-hand side being in $L^1$ , Wittbold and Zimmermann [Reference Wittbold and Zimmermann26] used the bi-monotone technique and the comparison principle to prove the existence and uniqueness of the renormalized solution to the problem $(\mathcal {P})$ .

The aim of this article is to extend the main results of [Reference Wittbold and Zimmermann26] to the framework of measure data on the right-hand side. However, due to the lack of regularity in the measure data, we cannot use the same method, therefore, we must proceed differently. To achieve our goal, we first construct an approximate problem $(\mathcal {P}_\epsilon )$ through approximation by truncation and Yosida regularization. Then, using the technique of maximal monotone operators in Banach spaces, we ensure the existence of a sequence of solutions to the problem $(\mathcal {P}_\epsilon )$ . We conclude by proving that this sequence of solutions converges to the solution of the problem $(\mathcal {P})$ .

The remaining part of this article is organized as follows: In Section 2, we introduce some preliminary results that can be useful throughout the article. In Section 3, we present the necessary assumptions on the data of the problem and also we provide the main results. In Section 4, we prove the existence of at least one weak and/or entropy solution. In Section 5, we explore the question of uniqueness of the solution.

2 Preliminaires

Let $\Omega $ be a bounded open domain in $\mathbb {R}^N \ (N\geq 3)$ with smooth boundary $\partial \Omega $ . In this entire article, $p(.):\overline {\Omega }\longrightarrow \mathbb {R}^+$ is a continuous function satisfying

(2.1) $$ \begin{align} 1<p^-:=\min _{x\in \overline{\Omega}}p(x)\leq p^+ : = \max _{x\in \overline{\Omega}}p(x)< \infty. \end{align} $$

We define the set

$$\begin{align*}C_+(\overline{\Omega})=\bigg \{ p\in C(\overline{\Omega}): \min _{x\in \overline{\Omega}}p(x)>1 \bigg \rbrace .\end{align*}$$ StartLayout 1st Row upper C Subscript plus Baseline left parenthesis normal upper Omega overbar right parenthesis equals StartSet p element of upper C left parenthesis normal upper Omega overbar right parenthesis colon min Underscript x element of normal upper Omega overbar Endscripts p left parenthesis x right parenthesis greater than 1 EndSet period EndLayout

For any $p\in C_+(\overline {\Omega })$ , the variable exponent Lebesgue space is defined by

$$\begin{align*}L^{p(.)}(\Omega):=\bigg \lbrace u:\Omega \rightarrow \mathbb{R}\ \text{measurable} : \int _\Omega |u|^{p(x)}dx < \infty \bigg \rbrace.\end{align*}$$

If the exponent is bounded, i.e., $p^+<\infty $ , then the expression

$$\begin{align*}\|u\|_{p(.)}:= \inf \bigg \{ \lambda>0: \int _\Omega \bigg |\frac{u(x)}{\lambda } \bigg |^{p(x)}dx \leq 1 \bigg \} \end{align*}$$

defines a norm in $L^{p(.)}(\Omega )$ called the Luxemburg norm. Then $(L^{p(.)}(\Omega ),\|u\|_{p(.)})$ is a separable Banach space. Moreover, if $1<p^-\leq p^+<\infty $ , then $L^{p(.)}(\Omega )$ is uniformly convex, hence reflexive, and its dual space is isomorphic to $L^{p'(.)}(\Omega )$ , where $\frac {1}{p(x)}+\frac {1}{p'(x)}=1$ in $\Omega $ .

The $p(.)$ -modular of the $L^{p(.)}(\Omega )$ space is the mapping $\rho _{p(.)}: L^{p(.)}(\Omega )\longrightarrow \mathbb {R}$ defined by

$$\begin{align*}\rho_{p(.)}(u):=\int _\Omega |u|^{p(x)}dx.\end{align*}$$

For any $u\in L^{p(.)}(\Omega )$ , the following inequality (see [Reference Fan12, Reference Fan and Zhao13]) will be used later.

(2.2) $$ \begin{align} \min \left \lbrace \|u\|^ {p^-}_{p(.)}; \ \|u\|^ {p^+}_{p(.)} \right \rbrace \leq \rho _{p(.)}(u)\leq \max \left \lbrace \ \|u\|^ {p^-}_{p(.)}; \ \|u\|^ {p^+}_{p(.)} \right \rbrace. \end{align} $$

For any $u\in L^{p(.)}(\Omega )$ and $v\in L^{p'(.)}(\Omega )$ , we have the Hölder type inequality (see [Reference Kovacik and Rakosnik19]).

(2.3) $$ \begin{align} \bigg | \int _\Omega uvdx \bigg |\leq \bigg ( \frac{1}{p^-}+\frac{1}{q^-} \bigg ) \ \|u\|_ {p(.)}\|v\|_ {q(.)}. \end{align} $$

If $\Omega $ is bounded and $p,\ q\in C_+(\overline {\Omega })$ such that $p(x)\leq q(x)$ for any $x\in \Omega $ , then the embedding $L^{q(.)}(\Omega ) \hookrightarrow L^{p(.)}(\Omega )$ is continuous (see [Reference Kovacik and Rakosnik19, Theorem 2.8]).

Proposition 2.1 ([Reference Kovacik and Rakosnik19])

For $u_n, u \in L^{p(x)} (\Omega )$ and $p_+<\infty $ , the following assertions hold true.

(i) $\|u\|_{p(.)}<1$ (resp, $=1, \>1$ ) if and only if $\rho _{p(.)}(u)<1$ (resp, $=1, \>1$ );
(ii) $\|u\|_{p(.)}>1$ imply $\|u\|_{p(.)}^{p_-}\leq \rho _{p(.)} (u) \leq \|u\|_{p(.)}^{p_+}$ , and $\|u\|_{p(.)}<1$ imply $\|u\|_{p(.)}^{p_+}\leq \rho _{p(.)} (u) \leq \|u\|_{p(.)}^{p_-}$ ;
(iii) $\|u_n\|_{p(.)} \rightarrow 0$ if and only if $\rho _{p(.)} (u_n)\rightarrow 0$ , and $\|u_n\|_{p(.)}\rightarrow \infty $ if and only $\rho _{p(.)} (u_n)\rightarrow \infty $ .

Now, we define the variable exponent Sobolev space as follows

$$\begin{align*}W^{1,p(.)}(\Omega):=\bigg \{ u\in L^{p(.)}(\Omega) : \ |\nabla u|\in L^{p(.)}(\Omega) \bigg \},\end{align*}$$

with the norm

$$\begin{align*}||u||_{1,p(.)}= \|u\|_{p(.)} + \|\nabla u\|_{p(.)}.\end{align*}$$

For a measurable function $u:\Omega \longrightarrow \mathbb {R}$ , we introduce the following notation

$$\begin{align*}\rho_{1,p(.)}(u):=\int _\Omega |u|^{p(x)}dx+\int _\Omega |\nabla u|^{p(x)}dx.\end{align*}$$

We denote by $W_0^{1,p(.)}(\Omega )$ the closure of $C^\infty _0(\Omega )$ in $W^{1,p(.)}(\Omega )$ .

The Sobolev exponent is defined as $p^*(x)=\frac {Np(x)}{N-p(x)}$ if $p(x)<N$ and $p^*(x)=\infty $ if $p(x)\geq N$ .

Proposition 2.2 (see [Reference Wang, Fan and Ge25, Reference Yao27])

For $u\in W^{1,p(.)}(\Omega )$ , the following properties hold

(i) $\|u\|_{1,p(.)}>1 \Rightarrow \|u\|^{p^-}_{1,p(.)}<\rho _{1,p(.)}(u)<\|u\|^{p^+}_{1,p(.)}$ ;
(ii) $\|u\|_{1,p(.)}<1 \Rightarrow \|u\|^{p^+}_{1,p(.)}<\rho _{1,p(.)}(u)<\|u\|^{p^-}_{1,p(.)}$ ;
(iii) $\|u\|_{1,p(.)}<1$ (respectively, $=1, \>1$ ) $\Longleftrightarrow \rho _{1,p(.)} (u)<1$ (respectively, $=1, \>1$ ).

Theorem 2.3 ([Reference Fan and Zhao13, Reference Harjulehto and Hästö14])

(i) Assuming $1<p_-\leq p^+ <\infty $ , the space $W^{1,p(.)}(\Omega )$ is a separable and reflexive Banach space.
(ii) If $q \in C_+(\overline {\Omega })$ and $q(x)<p^*(x)$ for any $x\in \Omega $ , then the embedding $W_0^{1,p(.)}(\Omega )\hookrightarrow \hookrightarrow L^{q(.)}(\Omega )$ is continuous and compact.
(iii) Poincaré inequality: there exists a constant $C>0$ , such that
$$\begin{align*}\|u\|_{p(.)}\leq C\|\nabla u\|_{p(.)}, \ \forall u\in W_0^{1,p(.)}(\Omega).\end{align*}$$
(iv) Sobolev–Poincaré inequality: there exists a constant $C>0$ , such that
$$\begin{align*}\|u\|_{p^*(.)}\leq C\|\nabla u\|_{p(.)}, \ \forall u\in W_0^{1,p(.)}(\Omega).\end{align*}$$

Remark 2.4 By (iii) of Theorem 2.3, we deduce that $\|\nabla u\|_{p(.)}$ and $\| u\|_{1,p(.)}$ are equivalent norms in $W_0^{1,p(.)}(\Omega )$ .

We denote by $\mathcal {L}^N$ the N-dimensional Lebesgue measure of $\mathbb {R}^N$ and by $\mathcal {M} _b(\Omega )$ the space of bounded Radon measures in $\Omega $ , equipped with its standard norm $||.||_{\mathcal {M} _b(\Omega )}$ . Note that, if $\mu $ belongs to $\mathcal {M} _b(\Omega )$ , then $|\mu |(\Omega )$ (the total variation of $\mu $ ) is a bounded positive measure on $\Omega $ .

Given $\mu \in \mathcal {M} _b(\Omega )$ , we say that $\mu $ is diffuse with respect to the capacity $W_0^{1,p(.)}(\Omega )$ ( $p(.)$ -capacity for short) if $\mu (A)=0$ , for every set A such that $Cap_{p(.)}(A,\Omega )=0$ .

For every $A \subset \Omega $ , we denote

$$\begin{align*}S_{p(.)}(A) = \{ u\in W^{1,p(.)}_0(\Omega)\cap C_0(\Omega) : u = 1 \ \text{on} \ A, u\geq 0 \ \text{on} \ \Omega\}.\end{align*}$$

The $p(.)$ -capacity of every subset A with respect to $\Omega $ is defined by

$$\begin{align*}Cap_ {p(.)} (A, \Omega)= \displaystyle \inf _{u\in S_{p(.)}(A)} \bigg \{ \int _\Omega |\triangledown u|^{p(x)}dx \bigg \}.\end{align*}$$

In the case $ S_{p(.)}(A) = \emptyset $ , we set $Cap_ {p(.)} (A, \Omega ) = +\infty $ .

The set of bounded Radon diffuse measure in the variable exponent setting is denoted by $\mathcal {M}_b^{p(.)}(\Omega )$ . We will use the following decomposition result of bounded Radon diffuse measure due to Nyanquini et al. (see [Reference Nyanquini, Ouaro and Soma22]).

Theorem 2.5 Let $p(.): \overline {\Omega } \longrightarrow (1,+\infty )$ be a continuous function and $\mu \in \mathcal {M}_b(\Omega )$ . Then $\mu \in \mathcal {M}^{p(.)}_b(\Omega )$ if and only if $\mu \in L^1(\Omega )+W^{-1,p'(.)}(\Omega )$ .

Lemma 2.6 Let $\Omega $ be a bounded open subset of $\mathbb {R}^N$ ( $N \geq 1$ ). If $u\in W^{1,p(x)}_0(\Omega )$ , then

$$\begin{align*}\int _\Omega div (u)dx=0.\end{align*}$$

If $\gamma $ is a maximal monotone operator defined on $\mathbb {R}$ , by $\gamma _0$ we denote the main section of $\gamma $ ; i.e.,

$$\begin{align*}\gamma_0(s) = \left \lbrace \begin{array}{l} \text{minimal absolute value of } \ \gamma (s) \ \ \text{if} \ \gamma (s)\ \neq \emptyset \\ +\infty \ \text{if} \ [s, +\infty)\cap D(\gamma) = \emptyset \\ -\infty \text{if} \ (-\infty , s]\cap D(\gamma) = \emptyset. \end{array} \right. \end{align*}$$

StartLayout 1st Row gamma 0 left parenthesis s right parenthesis equals StartLayout Enlarged left brace 1st Row minimal absolute value of gamma left parenthesis s right parenthesis if gamma left parenthesis s right parenthesis not equals normal empty set 2nd Row plus infinity if left bracket s comma plus infinity right parenthesis intersection upper D left parenthesis gamma right parenthesis equals normal empty set 3rd Row negative infinity if left parenthesis negative infinity comma s right bracket intersection upper D left parenthesis gamma right parenthesis equals normal empty set period EndLayout EndLayout

We also recall an important result on convergence (see [Reference Nyanquini, Ouaro and Soma22]).

Lemma 2.7 Let $(\beta _n)_{n\geq 1}$ be a sequence of maximal monotone graphs such that $\beta _n \rightarrow \beta $ in the sense of the graph (for $(x, y) \in \beta $ , there exists $(x_n, y_n)\in \beta _n$ such that $x_n\rightarrow x$ and $y_n \rightarrow y$ ). We consider two sequences $(z_n)_{n\geq 1}\subset L^1(\Omega )$ and $(w_n)_{n\geq 1}\subset L^1(\Omega )$ .

We suppose that: $\forall n\geq 1, w_n \in \beta _n(z_n)$ , $(w_n)_{n\geq 1}$ is bounded in $L^1(\Omega )$ and $z_n \rightarrow z$ in $L^1(\Omega )$ . Then,

$$\begin{align*}z\in dom (\beta).\end{align*}$$

Throughout the article, we use the truncation function $T_k$ , $(k>0)$ defined by

(2.4) $$ \begin{align} T_k(s) =\max \{-k,\min \{k;s\} \}. \end{align} $$

It is obvious that $\displaystyle \lim _{k\rightarrow \infty } T_k(s)=s$ and $|T_k(s)|=\min \{|s|;k\}$ .

We define $\mathcal {T}_0^{1,p(.)}(\Omega )$ as the set of the measurable function $u:\Omega \longrightarrow \mathbb {R}$ such that $T_k(u)\in W^{1,p(.)}_0(\Omega )$ .

We denote by

$$\begin{align*}H_\epsilon = \min \bigg ( \frac{s^+}{\epsilon}; 1 \bigg ) \ \text{and} \ sign_0^+ (s) = \left \lbrace \begin{array}{l} 1 \ \text{if} \ s> 0, \\ 0 \ \text{if} \ s \leq 0, \end{array} \right. \end{align*}$$ upper H Subscript epsilon Baseline equals min left parenthesis StartFraction s Superscript plus Baseline Over epsilon EndFraction semicolon 1 right parenthesis and s i g n 0 Superscript plus Baseline left parenthesis s right parenthesis equals StartLayout Enlarged left brace 1st Row 1 if s greater than 0 comma 2nd Row 0 if s less than or equals 0 comma EndLayout

Remark that as $\epsilon $ goes to 0, $H_\epsilon (s)$ goes to $sign_0^+(s)$ .

To outline our definition of solution and the principal results, we set

$$\begin{align*}\text{ int(dom} (\beta)) = (m,M) \ \text{with} \ -\infty < m \leq 0 \leq M < +\infty. \end{align*}$$

For any $r\in \mathbb {R}$ and any measurable function u on $\Omega $ , $[u = r]$ , $[u \leq r]$ and $[u \geq r]$ denote the set $\{ x\in \Omega : u(x) = r\}$ , $\{ x\in \Omega : u(x) \leq r\}$ , $\{ x\in \Omega : u(x) \geq r \}$ , respectively.

3 Assumptions and main results

3.1 Assumptions

We study the problem $(\mathcal {P})$ under the following assumptions on the data.

Let $\Omega $ be a bounded open domain in $\mathbb {R}^N \ (N\geq 2)$ with smooth boundary domain $\partial \Omega $ .

We assume that $p(.)$ verifies (2.1) and $a:\Omega \times \mathbb {R}^N\longrightarrow \mathbb {R}^N$ denotes a Carathéodory function satisfying the following conditions.

$(H_1)$ there exists a positive constant $C_1$ such that

(3.1) $$ \begin{align} |a(x,\xi)|\leq C_1\bigg (j(x)+|\xi|^{p(x)-1}\bigg ), \end{align} $$ StartAbsoluteValue a left parenthesis x comma xi right parenthesis EndAbsoluteValue less than or equals upper C 1 left parenthesis j left parenthesis x right parenthesis plus StartAbsoluteValue xi EndAbsoluteValue Superscript p left parenthesis x right parenthesis minus 1 Baseline right parenthesis comma

for almost every $x\in \Omega $ and for every $\xi \in \mathbb {R}^N$ , where j is a non-negative function in $L^{p'(.)}(\Omega )$ , with $\frac {1}{p(x)}+\frac {1}{p'(x)}=1$ ;

$(H_2)$ for all $\xi ,\eta \in \mathbb {R}^N$ with $\xi \neq \eta $ and for every $x\in \Omega $ ,

(3.2) $$ \begin{align} (a(x,\xi)-a(x,\eta )).(\xi -\eta)>0, \end{align} $$

$(H_3)$ there exists a positive constant $C_2$ such that

(3.3) $$ \begin{align} a(x,\xi).\xi \geq C_2|\xi|^{p(x)}, \end{align} $$

for $\xi \in \mathbb {R}^N$ and almost every $x \in \Omega $ .

$(H_4)$ $\overline {\text {dom}(\beta )}=[m,M]\subset \mathbb {R}$ where $-\infty <m\leq 0\leq M<+\infty $ .

$(H_5)$ $\phi : \mathbb {R}\longrightarrow \mathbb {R}^N$ is a continuous function with $\phi (0)=0$ and there exists a constant $C_3>0$ such that

(3.4) $$ \begin{align} \forall s\in \mathbb{R}, \ |\phi (s)|\leq C_3 |s|^{p(x)-1}. \end{align} $$

3.2 Notions of solutions and main results

Definition 3.1 Let $\mu \in \mathcal {M}_b^{p(.)}(\Omega )$ . We say that a couple $(u, b)\in W_0^{1, p(.)}(\Omega )\times L^1(\Omega )$ is a weak solution of problem $(\mathcal {P})$ if there exists $\nu \in \mathcal {M}^{p(.)}_b(\Omega )$ satisfying $\nu \ \bot \ \mathcal {L}^N$ and

(3.5) $$ \begin{align} \left \lbrace \begin{array}{l} u \in \beta (u)\mathcal{L}^N- a.e. \ \text{in} \ \Omega,\ b \in \beta (u)\mathcal{L}^N- a.e. \ \text{in} \ \Omega,\\ \nu ^+ \ \text{is concentrated on }\ [u = M], \\ \nu ^- \ \text{is concentrated on }\ [u = m], \end{array} \right. \end{align} $$

StartLayout 1st Row StartLayout Enlarged left brace 1st Row u element of beta left parenthesis u right parenthesis script upper L Superscript upper N Baseline minus a period e period in normal upper Omega comma b element of beta left parenthesis u right parenthesis script upper L Superscript upper N Baseline minus a period e period in normal upper Omega comma 2nd Row nu Superscript plus Baseline is concentrated on left bracket u equals upper M right bracket comma 3rd Row nu Superscript minus Baseline is concentrated on left bracket u equals m right bracket comma EndLayout EndLayout

such that

(3.6) $$ \begin{align} \displaystyle \int _\Omega a(x,\nabla u).\nabla \varphi dx + \int _\Omega b\varphi dx + \int _\Omega \varphi d\nu -\int _\Omega \phi (u).\nabla \varphi dx =\int _\Omega \varphi d\mu, \end{align} $$

for any $\varphi \in W_0^{1, p(.)}(\Omega )\cap L^\infty (\Omega )$ .

Moreover,

(3.7) $$ \begin{align} \lim _{n\to +\infty}\int _ {\{n<|u|<n+1\}} |\nabla u| ^{p(x)}dx=0. \end{align} $$

Definition 3.2 Let $\mu \in \mathcal {M}_b^{p(.)}(\Omega )$ . An entropy solution of problem $(\mathcal {P})$ is a couple $(u, b)\in W_0^{1, p(.)}(\Omega )\times L^1(\Omega )$ such that (3.5) holds and

(3.8) $$ \begin{align} \int _\Omega a(x,\nabla u).\nabla T_k(u - \varphi )dx &+ \int _\Omega bT_k(u - \varphi )dx -\int _\Omega \phi (u).\nabla T_k(u-\varphi )dx \notag \\ &\leq \int _\Omega T_k(u - \varphi)d\mu, \end{align} $$

StartLayout 1st Row 1st Column integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla upper T Subscript k Baseline left parenthesis u minus phi right parenthesis d x 2nd Column plus integral Underscript normal upper Omega Endscripts b upper T Subscript k Baseline left parenthesis u minus phi right parenthesis d x minus integral Underscript normal upper Omega Endscripts phi left parenthesis u right parenthesis period nabla upper T Subscript k Baseline left parenthesis u minus phi right parenthesis d x 2nd Row 1st Column Blank 2nd Column less than or equals integral Underscript normal upper Omega Endscripts upper T Subscript k Baseline left parenthesis u minus phi right parenthesis d mu comma EndLayout

where $k> 0$ and $\varphi \in W_0^{1, p(.)}(\Omega )\cap L^\infty (\Omega )$ such that $\varphi \in dom\beta $ .

Theorem 3.3 Assuming $(H_1)-(H_5)$ and $\mu \in \mathcal {M}_b^{p(.)}(\Omega )$ . Then, the problem $(\mathcal {P})$ admits at least one renormalized solution in the sense of Definition 3.1.

The connection between our notion of weak solution and the entropy solution is formulated as follows.

Theorem 3.4 A solution of problem $(\mathcal {P})$ in the sense of Definition 3.1 is also an entropy solution.

Proof Let $(u, b)$ be a weak solution of $(\mathcal {P})$ and $\varphi \in W_0^{1,p(.)}(\Omega ) \cap L^\infty (\Omega )$ such that $\varphi \in dom (\beta )$ .

For any $k> 0$ , taking $T_k(u - \varphi )$ as a test function in (3.6) one obtains

(3.9) $$ \begin{align} \int _\Omega a(x,\nabla u).\nabla T_k(u - \varphi)dx &-\int _\Omega \phi (u).\nabla T_k(u - \varphi) dx + \int _\Omega bT_k(u - \varphi)dx \nonumber\\ & + \int _\Omega T_k(u - \varphi)d\nu= \int _\Omega T_k(u - \varphi)d\mu. \end{align} $$

Neglecting the positive term $\displaystyle \int _\Omega T_k(u - \varphi )d\nu $ (see [Reference Konaté and Ouaro18]), we obtain (3.8).

4 Existence of solution for a regular right hand side data

In this section we study the following problem

(4.1) $$ \begin{align} (P^\phi_{g,\gamma})\left \lbrace \begin{array}{l} -\text{div}\ a(x,\nabla u) + g(u)+ \text{div} \phi (u) \ = \gamma \ \ \text{in} \ \ \ \Omega \\ u = 0 \text{on}\ \ \partial \Omega , \end{array} \right. \end{align} $$

where g is a continuous and nondecreasing function such that $g(0)=0$ and $\gamma \in L^\infty (\Omega )$ .

Theorem 4.1 Under assumptions $(H_1)-(H_3)$ , the problem $ (P^\phi _{g,\gamma })$ admits at least one weak solution in the following sense:

$u\in W_0^{1,p(.)}\cap L^\infty (\Omega ), \ g(u)\in L^\infty (\Omega )$ and

(4.2) $$ \begin{align} \displaystyle \int _\Omega a(x,\nabla u).\nabla \varphi dx + \int _\Omega g(u)\varphi dx -\int _\Omega \phi (u).\nabla \varphi dx =\int _\Omega \varphi \gamma dx. \end{align} $$

Proof For any $k>0$ , let us consider the following problem

$$ \begin{align*}(P^\phi_{T_k(g),\gamma})\left \lbrace \begin{array}{l} -\text{div}\ a(x,\nabla u_k) + T_k(g(u_k))+ \text{div} \phi (u_k) \ = \gamma \ \ \text{in} \ \ \ \Omega \\ u_k = 0 \text{on}\ \ \partial \Omega , \end{array} \right. \end{align*} $$

Theorem 4.2 Under assumptions $(H_1)-(H_3)$ , the problem $ (P^\phi _{T_k(g),\gamma })$ admits at least one weak solution in the following sense:

$u\in W_0^{1,p(.)}(\Omega )$ and

(4.3) $$ \begin{align} \displaystyle \int _\Omega a(x,\nabla u_k).\nabla \varphi dx + \int _\Omega T_k(g(u_k))\varphi dx -\int _\Omega \phi (u_k).\nabla \varphi dx =\int _\Omega \varphi \gamma dx. \end{align} $$

for any $\varphi \in W_0^{1,p(.)}(\Omega )\cap L^\infty (\Omega )$ .

Moreover

(4.4) $$ \begin{align} \forall k>\|\gamma\|_\infty, \ |g(u_k)|\leq \|\gamma\|_\infty \ \text{a.e. in} \ \Omega. \end{align} $$

Proof We define the operators $A_1$ , $A_2$ and $A:= A_1+A_2$ , acting from $W_0^{1, p(.)}(\Omega )$ into its dual $W^{-1, p'(.)}(\Omega )$ as follows

$$\begin{align*}\langle A_1 u, \varphi \rangle \ = \int _\Omega \displaystyle \bigg( a(x,\nabla u)-\phi (u)\bigg). \nabla \varphi dx, \ \forall \ u, \varphi \ \in W_0^{1,p(.)}(\Omega) \end{align*}$$

and

$$\begin{align*}\langle A_2 u, \varphi \rangle \ = \int _\Omega T_k(g(u))\varphi dx, \ \forall \ u, \varphi \ \in W_0^{1,p(.)}(\Omega). \end{align*}$$

We have

$$ \begin{align*} \bigg |\displaystyle \int _\Omega T_k(g(u)) \varphi dx \bigg |& \leq \displaystyle\int _\Omega |T_k(g(u))| |\varphi| dx\\ & \leq k\displaystyle\int _\Omega |\varphi| dx\\ & \leq k\bigg(\frac{1}{p_-}+\frac{1}{p_-'}\bigg) (\text{meas}(\Omega)+1)^{\frac{1}{p'_-}}\|\varphi\|_{p(.)}\\ &\leq C_4\|\varphi\|_{1,p(.)}. \end{align*} $$

StartLayout 1st Row 1st Column vertical bar integral Underscript normal upper Omega Endscripts upper T Subscript k Baseline left parenthesis g left parenthesis u right parenthesis right parenthesis phi d x vertical bar 2nd Column less than or equals integral Underscript normal upper Omega Endscripts StartAbsoluteValue upper T Subscript k Baseline left parenthesis g left parenthesis u right parenthesis right parenthesis EndAbsoluteValue StartAbsoluteValue phi EndAbsoluteValue d x 2nd Row 1st Column Blank 2nd Column less than or equals k integral Underscript normal upper Omega Endscripts StartAbsoluteValue phi EndAbsoluteValue d x 3rd Row 1st Column Blank 2nd Column less than or equals k left parenthesis StartFraction 1 Over p Subscript minus Baseline EndFraction plus StartFraction 1 Over p prime Subscript minus EndFraction right parenthesis left parenthesis meas left parenthesis normal upper Omega right parenthesis plus 1 right parenthesis Superscript StartFraction 1 Over p prime Super Subscript minus EndFraction Baseline StartMetric phi EndMetric Subscript p left parenthesis period right parenthesis Baseline 4th Row 1st Column Blank 2nd Column less than or equals upper C 4 StartMetric phi EndMetric Subscript 1 comma p left parenthesis period right parenthesis Baseline period EndLayout

According to $(H_1)$ , one has

$$ \begin{align*} \bigg|\displaystyle \int _\Omega a(x,\nabla u).\nabla \varphi dx\bigg| & \leq \int _\Omega |a(x,\nabla u)| |\nabla \varphi| dx\\ & \leq C_1\int _\Omega j(x)|\nabla \varphi |dx + C_1\int _\Omega |\nabla u|^{p(x)-1}|\nabla \varphi |dx \\ & \leq C_1\bigg(\frac{1}{p^-}+\frac{1}{(p')^-}\bigg)\bigg(\|j\|_{p'(.)} +\||\nabla u|\|^{p(x)-1}_{p(.)}\bigg)\|\nabla \varphi\|_{p(.)}\\ & \leq C_5\|\varphi\|_{1,p(.)}. \end{align*} $$

StartLayout 1st Row 1st Column vertical bar integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla phi d x vertical bar 2nd Column less than or equals integral Underscript normal upper Omega Endscripts StartAbsoluteValue a left parenthesis x comma nabla u right parenthesis EndAbsoluteValue StartAbsoluteValue nabla phi EndAbsoluteValue d x 2nd Row 1st Column Blank 2nd Column less than or equals upper C 1 integral Underscript normal upper Omega Endscripts j left parenthesis x right parenthesis StartAbsoluteValue nabla phi EndAbsoluteValue d x plus upper C 1 integral Underscript normal upper Omega Endscripts StartAbsoluteValue nabla u EndAbsoluteValue Superscript p left parenthesis x right parenthesis minus 1 Baseline StartAbsoluteValue nabla phi EndAbsoluteValue d x 3rd Row 1st Column Blank 2nd Column less than or equals upper C 1 left parenthesis StartFraction 1 Over p Superscript minus Baseline EndFraction plus StartFraction 1 Over left parenthesis p Superscript prime Baseline right parenthesis Superscript minus Baseline EndFraction right parenthesis left parenthesis StartMetric j EndMetric Subscript p prime left parenthesis period right parenthesis Baseline plus StartMetric StartAbsoluteValue nabla u EndAbsoluteValue EndMetric Subscript p left parenthesis period right parenthesis Superscript p left parenthesis x right parenthesis minus 1 Baseline right parenthesis StartMetric nabla phi EndMetric Subscript p left parenthesis period right parenthesis Baseline 4th Row 1st Column Blank 2nd Column less than or equals upper C 5 StartMetric phi EndMetric Subscript 1 comma p left parenthesis period right parenthesis Baseline period EndLayout

By using the growth condition $(H_5)$ on $\phi $ , one obtains

$$ \begin{align*} \bigg|\displaystyle \int _\Omega \phi (u).\nabla \varphi dx\bigg| & \leq \int _\Omega |\phi (u)||\nabla \varphi | dx\\ & \leq C_3\int _\Omega |u|^{p(x)-1}|\nabla \varphi |dx\\ & \leq C_3\bigg(\frac{1}{p^-}+\frac{1}{(p')^-}\bigg)\||u|^{p(x)-1}\|_{p'(.)} \|\nabla \varphi\|_{p(.)}\\ & \leq C_6\|\varphi\|_{1,p(.)}. \end{align*} $$

StartLayout 1st Row 1st Column vertical bar integral Underscript normal upper Omega Endscripts phi left parenthesis u right parenthesis period nabla phi d x vertical bar 2nd Column less than or equals integral Underscript normal upper Omega Endscripts StartAbsoluteValue phi left parenthesis u right parenthesis EndAbsoluteValue StartAbsoluteValue nabla phi EndAbsoluteValue d x 2nd Row 1st Column Blank 2nd Column less than or equals upper C 3 integral Underscript normal upper Omega Endscripts StartAbsoluteValue u EndAbsoluteValue Superscript p left parenthesis x right parenthesis minus 1 Baseline StartAbsoluteValue nabla phi EndAbsoluteValue d x 3rd Row 1st Column Blank 2nd Column less than or equals upper C 3 left parenthesis StartFraction 1 Over p Superscript minus Baseline EndFraction plus StartFraction 1 Over left parenthesis p Superscript prime Baseline right parenthesis Superscript minus Baseline EndFraction right parenthesis StartMetric StartAbsoluteValue u EndAbsoluteValue Superscript p left parenthesis x right parenthesis minus 1 Baseline EndMetric Subscript p prime left parenthesis period right parenthesis Baseline StartMetric nabla phi EndMetric Subscript p left parenthesis period right parenthesis Baseline 4th Row 1st Column Blank 2nd Column less than or equals upper C 6 StartMetric phi EndMetric Subscript 1 comma p left parenthesis period right parenthesis Baseline period EndLayout

Claim 1: the operator A is bounded.

Indeed, for any $u,\varphi \in W_0^{1,p(.)}(\Omega )$ , one has

$$ \begin{align*} \bigg | \langle A u, \varphi \rangle \bigg| &\leq C_7\|\varphi\|_{1,p(.)}. \end{align*} $$ StartLayout 1st Row 1st Column StartAbsoluteValue left angle bracket upper A u comma phi right angle bracket EndAbsoluteValue 2nd Column less than or equals upper C 7 StartMetric phi EndMetric Subscript 1 comma p left parenthesis period right parenthesis Baseline period EndLayout

Therefore, A is bounded.

Claim 2: A is coercive.

Indeed, since the divergence theorem implies $\displaystyle \int _\Omega \phi (u).\nabla u dx=0$ , and $\displaystyle \int _\Omega T_k(g(u))u dx\geq ~0$ , thanks to Proposition 2.1 and the Poincaré-type inequality, one obtains

$$ \begin{align*} \langle A u, u \rangle & = \displaystyle \int _\Omega a(x,\nabla u).\nabla u dx + \int _\Omega T_k(g(u))u dx -\int _\Omega \phi (u).\nabla u dx\\ &\geq \displaystyle \int _\Omega a(x,\nabla u).\nabla u dx\\ &\geq C_1 \int _\Omega |\nabla u|^{p(x)}dx\\ &\geq C_1\|\nabla u\|^\alpha _{p(.)}\\ & \geq C_8\| u\|^\alpha _{1,p(.)} \end{align*} $$

StartLayout 1st Row 1st Column left angle bracket upper A u comma u right angle bracket 2nd Column equals integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla u d x plus integral Underscript normal upper Omega Endscripts upper T Subscript k Baseline left parenthesis g left parenthesis u right parenthesis right parenthesis u d x minus integral Underscript normal upper Omega Endscripts phi left parenthesis u right parenthesis period nabla u d x 2nd Row 1st Column Blank 2nd Column greater than or equals integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla u d x 3rd Row 1st Column Blank 2nd Column greater than or equals upper C 1 integral Underscript normal upper Omega Endscripts StartAbsoluteValue nabla u EndAbsoluteValue Superscript p left parenthesis x right parenthesis Baseline d x 4th Row 1st Column Blank 2nd Column greater than or equals upper C 1 StartMetric nabla u EndMetric Subscript p left parenthesis period right parenthesis Superscript alpha Baseline 5th Row 1st Column Blank 2nd Column greater than or equals upper C 8 StartMetric u EndMetric Subscript 1 comma p left parenthesis period right parenthesis Superscript alpha EndLayout

where

$$\begin{align*}\alpha =\left \lbrace \begin{array}{l} p_+ \ \text{if} \ \|\nabla u\|_{p(.)} \leq 1, \\ p_-\ \text{if} \ \|\nabla u\|_{p(.)}> 1. \end{array} \right. \end{align*}$$ StartLayout 1st Row alpha equals StartLayout Enlarged left brace 1st Row p Subscript plus Baseline if StartMetric nabla u EndMetric Subscript p left parenthesis period right parenthesis Baseline less than or equals 1 comma 2nd Row p Subscript minus Baseline if StartMetric nabla u EndMetric Subscript p left parenthesis period right parenthesis Baseline greater than 1 period EndLayout EndLayout

Thus, we obtain

$$\begin{align*}\frac{\langle Au, u\rangle}{\|u\|_{1,p(x)}}\longrightarrow \infty \ \ \text{as}\ \ \|u\|_{1,p(.)}\longrightarrow \infty.\end{align*}$$ StartLayout 1st Row StartFraction left angle bracket upper A u comma u right angle bracket Over StartMetric u EndMetric Subscript 1 comma p left parenthesis x right parenthesis Baseline EndFraction long right arrow infinity as StartMetric u EndMetric Subscript 1 comma p left parenthesis period right parenthesis Baseline long right arrow infinity period EndLayout

Claim 3: $A_1$ is of type $(M)$ .

Indeed, let $(u_n)_{n\in \mathbb {N}}$ be a sequence in $W_0^{1,p(.)}(\Omega )$ such that

(4.5) $$ \begin{align} \left \lbrace \begin{array}{l} u_n\rightharpoonup u \ \text{in}\ W_0^{1,p(.)}(\Omega)\ \text{as}\ n\to \infty, \\ A_1 u_n\rightharpoonup \chi \ \text{in} \ W^{-1, p'(.)}(\Omega) \ \text{as}\ n\to \infty, \\ \displaystyle \lim _{n\to \infty}\sup \langle A_1 u_n, u_n \rangle \leq \langle \chi , u\rangle. \end{array} \right. \end{align} $$

StartLayout 1st Row StartLayout Enlarged left brace 1st Row u Subscript n Baseline right harpoon with barb up u in upper W 0 Superscript 1 comma p left parenthesis period right parenthesis Baseline left parenthesis normal upper Omega right parenthesis as n right arrow infinity comma 2nd Row upper A 1 u Subscript n Baseline right harpoon with barb up chi in upper W Superscript negative 1 comma p prime left parenthesis period right parenthesis Baseline left parenthesis normal upper Omega right parenthesis as n right arrow infinity comma 3rd Row limit Underscript n right arrow infinity Endscripts sup left angle bracket upper A 1 u Subscript n Baseline comma u Subscript n Baseline right angle bracket less than or equals left angle bracket chi comma u right angle bracket period EndLayout EndLayout

Let us set $h_n(x)= b(x,u_n,\nabla u_n)$ where $b(x,s,\varphi )=a(x,\varphi )-\phi (s)$ , $\forall (x,s,\varphi )\in \Omega \times \mathbb {R}\times \mathbb {R}^N$ .

Then, one has

(4.6) $$ \begin{align} |h_n(x)|=|a(x,\nabla u_n)-\phi (u_n)|\leq C_9(j(x)+|u_n|^{p(x)-1}+|\nabla u_n|^{p(x)-1}), \end{align} $$

where $C_9=\max \{C_1,C_3\}$ .

We aim to show that

$$\begin{align*}\langle A_1u_n, u_n \rangle \longrightarrow \langle \chi , u\rangle \ \text{as}\ n\longrightarrow \infty, \ \text{where}\ \chi =A_1 u.\end{align*}$$

Due to the compact embedding $W_0^{1,p(.)}(\Omega )\hookrightarrow \hookrightarrow L^{p(.)}(\Omega )$ , one has $u_n\rightarrow u$ in $L^{p(.)}(\Omega )$ as $n\to \infty $ (up to a subsequence still denoted $(u_n) _{n\in \mathbb {N}}$ ). Since $(u_n) _{n\in \mathbb {N}}$ is a bounded sequence in $W_0^{1,p(.)}(\Omega )$ , one can deduce from (4.6) that $(h_n)_{n\in \mathbb {N}}$ is bounded in $(L^{p'(.)}(\Omega ))^N$ . Therefore, there exists a function $h\in (L^{p'(.)}(\Omega ))^N$ such that

(4.7) $$ \begin{align} h_n\rightharpoonup h\ \text{in}\ (L^{p'(.)}(\Omega))^N \ \text{as}\ n\rightarrow \infty. \end{align} $$

For all $\varphi \in W_0^{1,p(.)}(\Omega )$ , one has

(4.8) $$ \begin{align} \langle\chi , \varphi\rangle =\lim _{n\to \infty}\langle A_1 u_n,\varphi\rangle=\lim _{n\to \infty} \int _\Omega h_n(x).\nabla \varphi dx = \int _\Omega h(x).\nabla \varphi dx=\langle A_1u , \varphi\rangle. \end{align} $$

This implies that $\chi =A_1u$ .

Applying (4.5), one obtains

(4.9) $$ \begin{align} \displaystyle \limsup_{n\to +\infty} \langle A_1 u_n, u_n \rangle \leq \int _\Omega h(x).\nabla u dx. \end{align} $$

Using $(H_2)$ , for any $\varphi \in (L^{p(.)}(\Omega ))^N$ , one has

$$\begin{align*} & \int_\Omega (b(x,u_n,\nabla u_n)-b(x,u_n,\varphi)).(\nabla u_n-\varphi)dx \\ & \quad = \int_\Omega (a(x,\nabla u_n)-a(x,\varphi)).(\nabla u_n-\varphi)dx\geq 0. \end{align*}$$

StartLayout 1st Row 1st Column Blank 2nd Column integral Underscript normal upper Omega Endscripts left parenthesis b left parenthesis x comma u Subscript n Baseline comma nabla u Subscript n Baseline right parenthesis minus b left parenthesis x comma u Subscript n Baseline comma phi right parenthesis right parenthesis period left parenthesis nabla u Subscript n Baseline minus phi right parenthesis d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts left parenthesis a left parenthesis x comma nabla u Subscript n Baseline right parenthesis minus a left parenthesis x comma phi right parenthesis right parenthesis period left parenthesis nabla u Subscript n Baseline minus phi right parenthesis d x greater than or equals 0 period EndLayout

This is equivalent to

(4.10) $$ \begin{align} \int_\Omega h_n.\nabla u_n dx- \int_\Omega h_n.\varphi dx - \int_\Omega b(x,u_n,\varphi).(\nabla u_n-\varphi) dx \geq 0. \end{align} $$

Since $u_n\rightharpoonup u$ in $W_0^{1,p(.)}(\Omega )$ , then up to a subsequence still denoted $(u_n)_{n\in \mathbb {N}}$ one has $u_n\rightarrow u$ in $L^{p(.)}(\Omega )$ , $u_n\rightarrow u$ a.e in $\Omega $ as $n\to \infty $ , and $|u_n|\leq v \in L^{p(.)}(\Omega )$ .

Since the function $b(x,s,\varphi )$ is continue with respect to s, on has

$$\begin{align*}b(x,u_n,\varphi)\longrightarrow b(x,u,\varphi)\ \text{a.e in}\ \Omega.\end{align*}$$

On the other hand, one has

$$\begin{align*}|b(x,u_n,\varphi)|\leq C_9 \bigg(j(x)+|v|^{p(x)-1}+|\varphi|^{p(x)-1}\bigg)\in (L^{p'(.)}(\Omega))^N.\end{align*}$$

Then, using Lebesgue dominated convergence theorem, one obtains

$$\begin{align*}b(x,u_n,\varphi)\longrightarrow b(x,u,\varphi)\ \text{a.e in}\ (L^{p'(.)}(\Omega))^N.\end{align*}$$

Therefore, we have

$$\begin{align*}\lim _{n\to \infty}\int_\Omega b(x,u_n,\varphi).(\nabla u_n-\varphi) dx=\int_\Omega b(x,u,\varphi).(\nabla u-\varphi) dx\end{align*}$$

and

$$\begin{align*}\lim _{n\to 0}\int_\Omega h_n.\varphi dx =\int_\Omega h.\varphi dx.\end{align*}$$

Passing to the limit as $n\to \infty $ in (4.10) and using (4.9), we obtain

(4.11) $$ \begin{align} \int_\Omega \bigg (h-b(x,u,\varphi)\bigg).(\nabla u-\varphi)dx\geq 0. \end{align} $$ integral Underscript normal upper Omega Endscripts left parenthesis h minus b left parenthesis x comma u comma phi right parenthesis right parenthesis period left parenthesis nabla u minus phi right parenthesis d x greater than or equals 0 period

By considering $\tilde {\varphi }\in (\mathcal {D}(\Omega ))^N$ and replacing in (4.11) $\varphi $ by $\nabla u +t\tilde {\varphi }$ , $t\in \mathbb {R}$ , one obtains

(4.12) $$ \begin{align} (-t)\int_\Omega \bigg (h-b(x,u,\nabla u +t\tilde{\varphi})\bigg).\tilde{\varphi} dx\geq 0. \end{align} $$ left parenthesis negative t right parenthesis integral Underscript normal upper Omega Endscripts left parenthesis h minus b left parenthesis x comma u comma nabla u plus t phi overtilde right parenthesis right parenthesis period phi overtilde d x greater than or equals 0 period

Dividing the above inequality by $t>0$ and by $t<0$ , then letting t go to 0, one can deduce from Lebesgue’s dominated convergence theorem that

$$\begin{align*}\int_\Omega \bigg (h-b(x,u,\nabla u)\bigg).\tilde{\varphi} dx= 0,\end{align*}$$ integral Underscript normal upper Omega Endscripts left parenthesis h minus b left parenthesis x comma u comma nabla u right parenthesis right parenthesis period phi overtilde d x equals 0 comma

this implies that $h=b(x,u,\nabla u)$ . Hence $A_1u=\chi $ .

Claim 4: $A_2$ is monotone and weakly continue.

Since for any $k>0$ , the function $T_k(g)$ is non-decreasing and satisfies $T_k(g(0))=0$ , one has

$$\begin{align*}\langle A_2 u -A_2\varphi,u-\varphi \rangle \ = \int _\Omega\bigg ( T_k(g(u))- T_k(g(\varphi))\bigg)(u-\varphi) dx \geq 0.\end{align*}$$

Hence, $A_2$ is monotone.

Let $(u_n)_{n\in \mathbb {N}}$ be a sequence in $W_0^{1,p(.)}(\Omega )$ such that $u_n\rightharpoonup u$ in $W_0^{1,p(.)}(\Omega )$ as $n\to \infty $ . Then, for all $\varphi \in W_0^{1,p(.)}(\Omega )$ , one has

$$\begin{align*}\langle A_2 u_n -A_2u,\varphi \rangle \ = \int _\Omega\bigg ( T_k(g(u_n))- T_k(g(u))\bigg)\varphi dx.\end{align*}$$

Since $u_n\rightharpoonup u$ in $W_0^{1,p(.)}(\Omega )$ , up to a subsequence still denoted $(u_n)_{n\in \mathbb {N}}$ , one has $u_n\rightarrow u$ in $L^{p(.)}(\Omega )$ , $u_n\rightarrow u$ a.e in $\Omega $ as $n\to \infty $ , and $|u_n|\leq v \in L^{p(.)}(\Omega )$ .

By the continuity of the function $T_k(g)$ , it follows that

$$\begin{align*}\bigg ( T_k(g(u_n))- T_k(g(u))\bigg)\varphi \rightarrow 0 \ \text{a.e in} \ \Omega \ \text{as}\ n\to \infty.\end{align*}$$ StartLayout 1st Row left parenthesis upper T Subscript k Baseline left parenthesis g left parenthesis u Subscript n Baseline right parenthesis right parenthesis minus upper T Subscript k Baseline left parenthesis g left parenthesis u right parenthesis right parenthesis right parenthesis phi right arrow 0 a period e in normal upper Omega as n right arrow infinity period EndLayout

Moreover,

$$\begin{align*}\bigg|\bigg ( T_k(g(u_n))- T_k(g(u))\bigg)\varphi \bigg|\leq 2k|\varphi| \in L^1(\Omega).\end{align*}$$

Leveraging the Lebesgue dominated convergence theorem, one arrives at

$$\begin{align*}\int _\Omega \bigg( T_k(g(u_n))- T_k(g(u))\bigg)\varphi dx =0.\end{align*}$$ StartLayout 1st Row integral Underscript normal upper Omega Endscripts left parenthesis upper T Subscript k Baseline left parenthesis g left parenthesis u Subscript n Baseline right parenthesis right parenthesis minus upper T Subscript k Baseline left parenthesis g left parenthesis u right parenthesis right parenthesis right parenthesis phi d x equals 0 period EndLayout

Therefore, $A_2u_n \rightharpoonup A_2u$ as $n\to \infty $ .

Since A is the sum of an operator of type $(M)$ and a monotone, weakly continuous operator, A is of type $(M)$ . Adding the fact that A is bounded and coercive, we conclude that A is surjective.

Therefore, for any $L\in W^{-1,p'(.)}(\Omega )$ , there exists at least one solution $u\in W_0^{1,p(.)}(\Omega )$ such that $A(u_k)=L$ .

Setting $L(\varphi )=\displaystyle \int _\Omega \gamma \varphi dx$ , we conclude that the problem $(P^\phi _{T_k(g),\gamma })$ admits at least one solution.

To complete the proof of Theorem 4.2, it remains to prove (4.4). To this end, we take $H_\epsilon (u_k-R)$ as a test function in (4.2), where $\epsilon>0$ and $R>0$ is a real to be specified later. One obtains

(4.13) $$ \begin{align} \displaystyle &\int _\Omega a(x,\nabla u_k).\nabla H_\epsilon (u_k-R) dx \notag\\ &\qquad + \int _\Omega T_k(g(u))H_\epsilon (u_k-R) dx -\int _\Omega \phi (u_k).\nabla H_\epsilon (u_k-R) dx \notag\\ &\quad =\int _\Omega \gamma H_\epsilon (u_k-R) dx. \end{align} $$

StartLayout 1st Row 1st Column Blank 2nd Column integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u Subscript k Baseline right parenthesis period nabla upper H Subscript epsilon Baseline left parenthesis u Subscript k Baseline minus upper R right parenthesis d x 2nd Row 1st Column Blank 2nd Column plus integral Underscript normal upper Omega Endscripts upper T Subscript k Baseline left parenthesis g left parenthesis u right parenthesis right parenthesis upper H Subscript epsilon Baseline left parenthesis u Subscript k Baseline minus upper R right parenthesis d x minus integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript k Baseline right parenthesis period nabla upper H Subscript epsilon Baseline left parenthesis u Subscript k Baseline minus upper R right parenthesis d x 3rd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts gamma upper H Subscript epsilon Baseline left parenthesis u Subscript k Baseline minus upper R right parenthesis d x period EndLayout

For the first term of (4.13), one has

$$\begin{align*}\int _\Omega a(x,\nabla u_k).\nabla H_\epsilon (u_k-R) dx = \frac{1}{\epsilon}\int _{\{|u_k-R|<\epsilon\}}a(x,\nabla u_k).\nabla u_k \geq 0.\end{align*}$$

By setting $\psi _\epsilon (u_k)=\displaystyle \int _0^{u_k}\phi (s)\chi _{\{0\leq |u_k-R|<\epsilon \}}(s) ds$ , one obtains

$$ \begin{align*} \int _\Omega \phi (u_k).\nabla H_\epsilon (u_k-R) dx & = \int _\Omega \frac{1}{\epsilon}\phi (u_k).\nabla u_k \chi _{\{0\leq |u_k-R|<\epsilon\}}dx\\ & = \int _\Omega \nabla .\bigg (\int _0^{u_k}\phi (s)\chi _{\{0\leq |u_k-R|<\epsilon\}}(s) ds \bigg)dx\\ &= \int _{\partial \Omega} \psi _\epsilon (u_k).\nu d\sigma =0 \ \ \ \ (\text{as} \ u_k=0 \ \ \text{on}\ \partial \Omega). \end{align*} $$

StartLayout 1st Row 1st Column integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript k Baseline right parenthesis period nabla upper H Subscript epsilon Baseline left parenthesis u Subscript k Baseline minus upper R right parenthesis d x 2nd Column equals integral Underscript normal upper Omega Endscripts StartFraction 1 Over epsilon EndFraction phi left parenthesis u Subscript k Baseline right parenthesis period nabla u Subscript k Baseline chi Subscript left brace 0 less than or equals StartAbsoluteValue u Sub Subscript k Subscript minus upper R EndAbsoluteValue less than epsilon right brace Baseline d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts nabla period left parenthesis integral Subscript 0 Superscript u Subscript k Baseline Baseline phi left parenthesis s right parenthesis chi Subscript left brace 0 less than or equals StartAbsoluteValue u Sub Subscript k Subscript minus upper R EndAbsoluteValue less than epsilon right brace Baseline left parenthesis s right parenthesis d s right parenthesis d x 3rd Row 1st Column Blank 2nd Column equals integral Underscript partial differential normal upper Omega Endscripts psi Subscript epsilon Baseline left parenthesis u Subscript k Baseline right parenthesis period nu d sigma equals 0 left parenthesis as u Subscript k Baseline equals 0 on partial differential normal upper Omega right parenthesis period EndLayout

Consequently, (4.13) becomes

(4.14) $$ \begin{align} \int _\Omega T_k(g(u))H_\epsilon (u_k-R) dx\leq \int _\Omega\gamma H_\epsilon (u_k-R) dx. \end{align} $$

Using the inequality above, one can deduce (4.4) (see [Reference Ibrango and Ouaro15, Reference Konaté and Ouaro17, Reference Nyanquini, Ouaro and Soma22] for the details).

Setting $k=k_0=\|\gamma \|_\infty +1$ , Theorem 4.1 is a consequence of Theorem 4.2.

5 Proof of Theorem 3.3

This section is devoted to the proof of Theorem 3.3.

5.1 Approximate problem

For every $\epsilon> 0 $ , we consider the Yosida regularization $\beta _\epsilon :\mathbb {R}\rightarrow \mathbb {R} $ of $\beta $ (see [Reference Brezis7]), given by

$$\begin{align*}\beta _\epsilon = \dfrac{1}{\epsilon}(I - (I + \epsilon \beta)^{-1}).\end{align*}$$

We emphasize that the function $\beta _\epsilon $ is both non-decreasing and Lipschitz-continuous

Since $\mu $ belongs to $\mathcal {M}_b^{p(.)}(\Omega )$ , so, by Theorem 2.5, it can be decomposed as $\mu = f - \text {div}(F)$ , where $f\in L^1(\Omega )$ and $F\in ( L^{p'(.)}(\Omega ))^N$ .

By introducing the function $f_\epsilon (x) = T_{\frac {1}{\epsilon }}(f(x))$ for $a.e. x\in \Omega $ , the regularized form of the measure $\mu $ is given by

$$\begin{align*}\mu _\epsilon = f_\epsilon - \nabla. \ F \ \text{for any} \ \epsilon> 0.\end{align*}$$

Therefore, one has $\mu _{\epsilon } \in \mathcal {M}_b^{p(.)}(\Omega ), \; \mu _{\epsilon } \rightharpoonup \mu $ and $\mu _{\epsilon } \in L^{\infty }(\Omega )$ .

Then, we consider the following approximating scheme problem.

(5.1) $$ \begin{align} Pb(\beta _\epsilon,\phi)(\mu_\epsilon)\left \lbrace \begin{array}{l} \beta _\epsilon( u_\epsilon) - \text{div} a (x,\nabla u_\epsilon) + \text{div} \phi (u_\epsilon) = \mu_\epsilon \ \text{in} \ \ \ \Omega \\ u_\epsilon = 0 \ \ \text{on}\ \ \partial \Omega. \end{array} \right. \end{align} $$

Theorem 5.1 Let $(H_1)-(H_3)$ hold true. Then, the problem $Pb(\beta _\epsilon ,\phi )(\mu _\epsilon )$ admits at least one weak solution $u_\epsilon $ in the sense that $u_\epsilon \in W_0^{1, p(.)}(\Omega )$ , $\beta _\epsilon (u_\epsilon )\in L^1(\Omega )$ and $\forall \varphi \ \in \ W_0^{1, p(.)}(\Omega )\cap L^\infty (\Omega )$ ,

(5.2) $$ \begin{align} \int _\Omega a(x,\nabla u_\epsilon). \nabla \varphi dx + \int _\Omega \beta _\epsilon (u_\epsilon)\varphi dx -\int _\Omega \phi (u_\epsilon).\nabla \varphi dx = \int _\Omega \varphi d \mu_\epsilon. \end{align} $$

Proof We just need to set $g=\beta _\epsilon $ and $\gamma =\mu _\epsilon $ in Theorem 4.1.

5.2 A priori estimates

Now, we derive a priori estimates for the sequence of solutions $(u_\epsilon )_{\epsilon>0}$ which will enable us to obtain the necessary convergence results.

Proposition 5.2 Let $k>0$ and $u_\epsilon $ be a solution to the problem $Pb(\beta _\epsilon ,\phi )(\mu _\epsilon )$ . Then,

(i) there exist a constant $C_{10}>0$ such that
(5.3) $$ \begin{align} \int _{\{|u_\epsilon|\leq k\}} |\nabla u_\epsilon|^{p(x)}dx \leq C_{10}, \end{align} $$
(ii) the sequence $(\beta _\epsilon (u_\epsilon ))_{\epsilon> 0}$ is uniformly bounded in $L^1(\Omega )$ ,
(iii) the sequence $(\beta _\epsilon (T_k(u_\epsilon )))_{\epsilon> 0}$ is uniformly bounded in $L^1(\Omega )$ .

Proof Taking $\varphi = T_k(u_\epsilon )$ as a test function in (5.2) we obtain

(5.4) $$ \begin{align} \int _\Omega a(x,\nabla u_\epsilon).\nabla T_k(u_\epsilon) dx & + \int _\Omega \beta _\epsilon(u_\epsilon) T_k(u_\epsilon)dx -\int _\Omega \phi (u_\epsilon).\nabla T_k(u_\epsilon)dx \nonumber\\ &\quad = \int _\Omega f_\epsilon T_k(u_\epsilon)dx +\int _\Omega F .\nabla T_k(u_\epsilon)dx. \end{align} $$

StartLayout 1st Row 1st Column integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u Subscript epsilon Baseline right parenthesis period nabla upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 2nd Column plus integral Underscript normal upper Omega Endscripts beta Subscript epsilon Baseline left parenthesis u Subscript epsilon Baseline right parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x minus integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript epsilon Baseline right parenthesis period nabla upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts f Subscript epsilon Baseline upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x plus integral Underscript normal upper Omega Endscripts upper F period nabla upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x period EndLayout

The third term of (5.4) is zero. Indeed, we have

(5.5) $$ \begin{align} \int _\Omega \phi (u_\epsilon).\nabla T_k(u_\epsilon)dx & = \int _\Omega \phi (T_k(u_\epsilon)).\nabla T_k(u_\epsilon) dx \nonumber\\ & = \int _\Omega \nabla \bigg (\int _0^{T_k(u_\epsilon)} \phi (s)ds\bigg ) dx =0, \end{align} $$

StartLayout 1st Row 1st Column integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript epsilon Baseline right parenthesis period nabla upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis period nabla upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts nabla left parenthesis integral Subscript 0 Superscript upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis Baseline phi left parenthesis s right parenthesis d s right parenthesis d x equals 0 comma EndLayout

The remainder of the proof follows in the same manner as [Reference Ouaro, Ouédraogo and Soma23] (see also [Reference Konaté and Ouaro18]).

Proposition 5.3 ([Reference Ouaro, Ouédraogo and Soma23])

Let $u_\epsilon $ be a weak solution of $Pb(\beta _\epsilon ,\phi )(\mu _\epsilon )$ and let $k> 0$ large enough. Then, we have

(5.6) $$ \begin{align} meas\{ |u_\epsilon|> k \}\leq \dfrac{C(\mu, \Omega)}{\min \{\beta _\epsilon (k), |\beta _\epsilon (-k)| \}} \end{align} $$

and

(5.7) $$ \begin{align} meas \bigg \{ | \nabla u_\epsilon |> k \bigg \} \leq \dfrac{C_{11}(k + 1)}{k^{p^-}} + \dfrac{C(\mu, \Omega)}{\min \{\beta _\epsilon (k), |\beta _\epsilon (-k)|}, \end{align} $$

where $C_{11}$ is a positive constant.

5.3 Convergence results

Proposition 5.4 ([Reference Ouaro, Ouédraogo and Soma23])

Let $u_\epsilon $ be a weak solution of $Pb(\beta _\epsilon ,\phi )(\mu _\epsilon )$ . Then, there exists $u\in W_0^{1, p(.)}(\Omega )\subset \mathcal {T}_0^{1,p(.)}(\Omega )$ such that $u\in dom (\beta )$ a.e. in $\Omega $ and

(5.8) $$ \begin{align} u_\epsilon \longrightarrow u\ \text{in measure and a.e. in } \ \Omega \ \text{as}\ \epsilon \longrightarrow 0. \end{align} $$

Lemma 5.5 For every function $h \in W^{1,+\infty }(\mathbb {R}), h \geq 0$ with $\text {supp}(h)$ compact,

(5.9) $$ \begin{align} \displaystyle \limsup_{\epsilon \rightarrow 0} \int_{\Omega} \Big[a(x, \nabla u_{\epsilon})-\phi (u_\epsilon) \Big].\nabla\Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx \leq 0, \end{align} $$

(5.10) $$ \begin{align} \displaystyle \limsup _{\delta\to 0} \limsup _{\epsilon\to 0} \int_{\{\delta <|u_\epsilon| < \delta +1\}} a(x, \nabla u_{\epsilon}).\nabla u_{\epsilon}dx \leq 0 \end{align} $$

and

(5.11) $$ \begin{align} \displaystyle \limsup _{\epsilon\to 0} \int_\Omega a(x,\nabla u_{\epsilon}).[\nabla T_k(u_{\epsilon})-\nabla T_k(u)]dx \leq 0. \end{align} $$

Proof By choosing $\displaystyle h(u_{\epsilon })(T_{k}(u_{\epsilon })-T_{k}(u))$ as a test function in (5.2), one obtains

(5.12) $$ \begin{align} \left \lbrace \begin{array}{l} \displaystyle \int_{\Omega} \Big[a(x, \nabla u_{\epsilon})-\phi (u_\epsilon) \Big].\nabla \Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx \\ \displaystyle \quad + \int_{\Omega} \beta _\epsilon (u_\epsilon) h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx\\ \displaystyle = \int_{\Omega} f_{\epsilon} h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx + \int_{\Omega} F_\epsilon.\nabla\Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx. \end{array} \right. \end{align} $$

StartLayout 1st Row StartLayout Enlarged left brace 1st Row integral Underscript normal upper Omega Endscripts left bracket a left parenthesis x comma nabla u Subscript epsilon Baseline right parenthesis minus phi left parenthesis u Subscript epsilon Baseline right parenthesis right bracket period nabla left bracket h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis right bracket d x 2nd Row plus integral Underscript normal upper Omega Endscripts beta Subscript epsilon Baseline left parenthesis u Subscript epsilon Baseline right parenthesis h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis d x 3rd Row equals integral Underscript normal upper Omega Endscripts f Subscript epsilon Baseline h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis d x plus integral Underscript normal upper Omega Endscripts upper F Subscript epsilon Baseline period nabla left bracket h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis right bracket d x period EndLayout EndLayout

$\bullet $ Let us start by proving (5.9). The following inequality holds

(5.13) $$ \begin{align} \displaystyle \limsup_{\epsilon \rightarrow 0} \int_{\Omega} \beta _\epsilon (u_\epsilon) h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx \geq 0. \end{align} $$

Indeed, for any $r>0$ sufficiently small we set

$$ \begin{align*}u_r = \left(u\wedge (M-r)\right )\vee (m+r).\end{align*} $$

According to [Reference Ouaro, Ouédraogo and Soma23], for any $k>0$ , $\displaystyle T_k(u_r)\in W^{1,p(.)}_0(\Omega )$ , one has

$$ \begin{align*}\displaystyle \int_{\Omega} h(u_{\epsilon})(\beta_{\epsilon}(u_{\epsilon}) - \beta_{\epsilon}(u_r))(T_{k}(u_{\epsilon})-T_{k}(u_r))dx \geq 0,\end{align*} $$

and

$$ \begin{align*} \displaystyle\int_{\Omega} \beta _\epsilon ( u_\epsilon) h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx &\geq \int_{\Omega} h(u_{\epsilon})\beta _\epsilon ( u_r)(T_{k}(u_{\epsilon})-T_{k}(u_r))dx \\ & \quad + \int_{\Omega} h(u_{\epsilon})\beta _\epsilon ( u_\epsilon)(T_{k}(u_r)-T_{k}(u))dx\\ & =:I_{\epsilon,r}+J_{\epsilon,r}. \end{align*} $$

StartLayout 1st Row 1st Column integral Underscript normal upper Omega Endscripts beta Subscript epsilon Baseline left parenthesis u Subscript epsilon Baseline right parenthesis h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis d x 2nd Column greater than or equals integral Underscript normal upper Omega Endscripts h left parenthesis u Subscript epsilon Baseline right parenthesis beta Subscript epsilon Baseline left parenthesis u Subscript r Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript epsilon Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis right parenthesis d x 2nd Row 1st Column Blank 2nd Column plus integral Underscript normal upper Omega Endscripts h left parenthesis u Subscript epsilon Baseline right parenthesis beta Subscript epsilon Baseline left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis d x 3rd Row 1st Column Blank 2nd Column equals colon upper I Subscript epsilon comma r Baseline plus upper J Subscript epsilon comma r Baseline period EndLayout

Having in mind that $ m+r\leq u_r \leq M-r$ , one has (see [Reference Ouaro, Ouédraogo and Soma23])

$$\begin{align*}\displaystyle\limsup_{\epsilon \rightarrow 0} I_{\epsilon,r} = \int_{\Omega} h(u)\beta_{0}(u_r)(T_{k}(u)-T_{k}(u_r))dx \geq 0.\end{align*}$$

We treat the term $J_{\epsilon ,r}$ as follows

$$ \begin{align*}\displaystyle J_{\epsilon,r}: = \int_{\Omega} h(u_{\epsilon})\beta _\epsilon ( u_\epsilon)(T_{k}(u_r)-T_{k}(u))dx:= A_{\epsilon,r}+B_{\epsilon,r}+C_{\epsilon,r}+D_{\epsilon,r},\end{align*} $$

where

$$\begin{align*}A_{\epsilon,r}&:=\int_{\Omega} h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u))d\mu _\epsilon, \qquad\qquad\qquad\\B_{\epsilon,r}&:=- \int_{\Omega} h(u_{\epsilon}) a(x, \nabla u_{\epsilon}).\nabla (T_{k}(u_r)-T_{k}(u))dx, \qquad \\C_{\epsilon,r}&:=-\int_{\Omega} h'(u_\epsilon)(T_{k}(u_r)-T_{k}(u)) a(x, \nabla u_{\epsilon}).\nabla u_{\epsilon} dx,\\D_{\epsilon,r}&:=-\int_{\Omega} \phi (u_\epsilon). \nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx. \qquad\qquad\end{align*}$$

StartLayout 1st Row 1st Column upper A Subscript epsilon comma r 2nd Column colon equals integral Underscript normal upper Omega Endscripts h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis d mu Subscript epsilon Baseline comma 2nd Row 1st Column upper B Subscript epsilon comma r 2nd Column colon equals minus integral Underscript normal upper Omega Endscripts h left parenthesis u Subscript epsilon Baseline right parenthesis a left parenthesis x comma nabla u Subscript epsilon Baseline right parenthesis period nabla left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis d x comma 3rd Row 1st Column upper C Subscript epsilon comma r 2nd Column colon equals minus integral Underscript normal upper Omega Endscripts h prime left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis a left parenthesis x comma nabla u Subscript epsilon Baseline right parenthesis period nabla u Subscript epsilon Baseline d x comma 4th Row 1st Column upper D Subscript epsilon comma r 2nd Column colon equals minus integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript epsilon Baseline right parenthesis period nabla left bracket h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis right bracket d x period EndLayout

According to [Reference Ouaro, Ouédraogo and Soma23], one has $\displaystyle \lim _{r\to 0}A_{\epsilon ,r}=\lim _{r\to 0}B_{\epsilon ,r}=\lim _{r\to 0}C_{\epsilon ,r}=0$ .

$$ \begin{align*} D_{\epsilon,r}&=\displaystyle\int_{\Omega} \phi (u_\epsilon). \nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx\\ &=\int_{\Omega} \phi (T_l(u_\epsilon)).\nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx, \end{align*} $$

StartLayout 1st Row 1st Column upper D Subscript epsilon comma r 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript epsilon Baseline right parenthesis period nabla left bracket h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis right bracket d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript l Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis period nabla left bracket h left parenthesis u Subscript epsilon Baseline right parenthesis left parenthesis upper T Subscript k Baseline left parenthesis u Subscript r Baseline right parenthesis minus upper T Subscript k Baseline left parenthesis u right parenthesis right parenthesis right bracket d x comma EndLayout

where $l>0$ is such that $\text {supp}h\subset ]-l,l[$ .

Thanks to $(H_5)$ , one has

$$\begin{align*}|\phi (T_l(u_\epsilon))|\leq |T_l(u_\epsilon)|^{p(x)-1}\leq (l+1)^{p(x)-1}\leq (l+1)^{p^+-1}.\end{align*}$$

It follows that $(\phi (T_l(u_\epsilon )))_\epsilon $ is uniformly bounded. Adding the fact that $\nabla \Big [ h(u_{\epsilon })(T_{k}(u_r)-T_{k}(u)) \Big ] \rightharpoonup 0$ in $(L^{p(.)}(\Omega ))^N$ (see [Reference Ouaro, Ouédraogo and Soma23]) as $r\to 0$ , one obtains

$$\begin{align*}\lim _{r\to 0}D_{\epsilon,r} =\int_{\Omega} \phi (T_l(u_\epsilon)). \nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx=0.\end{align*}$$

From above results, one deduces that $\displaystyle \lim _{r\to 0} J_{\epsilon ,r}=0$ and (5.13).

Therefore, passing to the limit as $\epsilon \to 0$ in (5.12), one obtains (5.9).

$\bullet $ Taking $\varphi _\delta (u_\epsilon )=T_1(u_\epsilon -T_\delta (u_\epsilon ))$ as test function in (5.2), one obtains

(5.14) $$ \begin{align} &\displaystyle \int _\Omega a(x,\nabla u_\epsilon ). \nabla T_1(u_\epsilon -T_\delta (u_\epsilon )) dx + \int _\Omega \beta _\epsilon (u_\epsilon )T_1(u_\epsilon -T_\delta (u_\epsilon )) dx \notag \\ &\qquad -\int _\Omega \phi (u_\epsilon).\nabla T_1(u_\epsilon -T_\delta (u_\epsilon)) dx \notag \\ & \quad = \int _\Omega f_\epsilon T_1(u_\epsilon -T_\delta (u_\epsilon)) dx +\int _\Omega F.\nabla T_1(u_\epsilon -T_\delta (u_\epsilon)) dx. \end{align} $$

On the other hand, one has

$$\begin{align*}\displaystyle \int _\Omega \phi (u_\epsilon).\nabla \varphi_\delta (u_\epsilon) dx =\int _\Omega \nabla \bigg (\int _0^{\varphi_\delta (u_\epsilon)}\phi ((\varphi_\delta)^{-1}\circ (s))ds \bigg)dx=0.\end{align*}$$

The rest of the proof of (5.10) and (5.11) follow the same lines as in [Reference Ouaro, Ouédraogo and Soma23].

The following results are necessary for the sequel.

Lemma 5.6 Let $u_\epsilon $ be a weak solution of $Pb(\beta _\epsilon ,\phi )(\mu _\epsilon )$ and $k>0$ . Then

(5.15) $$ \begin{align} \phi (T_k(u_\epsilon))\longrightarrow \phi (T_k(u)) \text{ in} \ L^{p'(.)}(\Omega) \ \text{as}\ \epsilon\to 0, \end{align} $$

(5.16) $$ \begin{align} \displaystyle\lim _{\epsilon\to 0}\int _\Omega \phi (u_\epsilon).\nabla \Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx=0, \end{align} $$

(5.17) $$ \begin{align} \lim _{\epsilon\to 0} \int_{\Omega} f_{\epsilon} h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx=0 \end{align} $$

and

(5.18) $$ \begin{align} \lim _{\epsilon\to 0} \int_{\Omega} F_\epsilon.\nabla\Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx=0. \end{align} $$

Proof $\bullet $ Since $\phi (T_k(u_\epsilon ))\longrightarrow \phi (T_k(u))$ a.e. in $\Omega $ , the growth condition $(H_5)$ implies that

$$\begin{align*}|\phi (T_k(u_\epsilon))|\leq C_3 |T_k(u_\epsilon)|^{p(x)-1}\in L^{p'(.) }(\Omega).\end{align*}$$

On the other hand, the sequence $(|T_k(u_\epsilon )|^{p(x)-1})_{\epsilon>0}$ is bounded in $L^{p'(.) }(\Omega )$ and $|T_k(u_\epsilon )|^{p(.)-1} \longrightarrow |T_k(u)|^{p(.)-1}$ in $L^{p'(.)}(\Omega )$ as $\epsilon \to 0$ . Thanks to the generalized Lebesgue convergence theorem, we obtain (5.15).

$\bullet $ For $l>0$ such that $\text {supp}h\subset ]-l,l[$ , one has

$\displaystyle \int _{\Omega } \phi (u_\epsilon ) .\nabla \Big [ h(u_{\epsilon })(T_{k}(u_\epsilon )-T_{k}(u)) \Big ]dx=\int _{\Omega } \phi (T_l(u_\epsilon )) .\nabla \Big [ h(u_{\epsilon })(T_{k}(u_\epsilon )-T_{k}(u)) \Big ]dx$ .

Using the convergence (5.15), one deduces (5.16).

For the proofs of (5.17) and (5.18), see [Reference Ouaro, Ouédraogo and Soma23].

Proposition 5.7 [Reference Ouaro, Ouédraogo and Soma23]

Let $u_\epsilon $ be a weak solution of $Pb(\beta _\epsilon ,\phi )(\mu _\epsilon )$ with $k>0$ . Then, as $\epsilon \to 0$ , we have

(i) $a(x,\nabla T_k(u_\epsilon ))\rightharpoonup a(x,\nabla T_k(u))$ in $(L^{p'(.)}(\Omega ))^N$ ,
(ii) $\nabla T_k(u_\epsilon )\rightarrow \nabla T_k(u)$ a.e. in $\Omega $ ,
(iii) $a(x,\nabla T_k(u_\epsilon ))\cdot \nabla T_k(u_\epsilon )\longrightarrow a(x,\nabla T_k(u))\cdot \nabla T_k(u)$ a.e. in $\Omega $ and strongly in $L^1(\Omega )$ ,
(iv) $\nabla T_k(u_\epsilon )\rightarrow \nabla T_k(u)$ in $(L^{p(.)}(\Omega ))^N$ .

Remark 5.8 Since $T_k$ is continuous, for $k> 0$ , it follows that $T_k(u_\epsilon ) \rightarrow T_k(u)$ a.e. in $\Omega $ . Finally, applying Lemma 2.7, we deduce that for all $k> 0$ , $T_k(u)\in dom (\beta )$ a.e. in $\Omega $ . Therefore, since $T_k(u) \in dom (\beta )$ , we conclude that $u\in dom (\beta ) \ a.e.$ in $\Omega $ , and sine $dom (\beta )$ is bounded, we have $u\in W_0^{1, p(.)}(\Omega )$ .

Lemma 5.9 [Reference Ouaro, Ouédraogo and Soma23]

For any $ h \in C^{1}_{c}(\mathbb {R}) \text { and } \xi \in W^{1,p(.)}_{0}(\Omega )\cap L^{\infty }(\Omega )$ ,

$$\begin{align*}\nabla[h(u_{\epsilon})\varphi] \longrightarrow \nabla[h(u)\varphi] \text{ strongly in } L^{p(.)}(\Omega) \text{ as } \epsilon \rightarrow 0. \end{align*}$$

5.4 Existence of solution

Let us introduce, for any $l_0>0$ , the function $h_0$ defined by

(i) $h_0\in C_c^1(\mathbb {R})$ , $h_0(r)\geq 0$ , for all $r\in \mathbb {R}$ ,
(ii) $h_0(r)=1$ if $|r|\leq l_0$ and $h_0(r)=0$ if $|r|\geq l_0+1$ .

To demonstrate the Theorem 3.3, one chooses $h_0(u_\epsilon )\varphi $ as a test function in (5.2) to obtain

(5.19) $$ \begin{align} \int _ \Omega a (x,\nabla u_\epsilon).\nabla (h_0(u_\epsilon )\varphi dx &-\int _\Omega \phi(u_\epsilon).\nabla [h_0(u_\epsilon)\varphi] dx + \int _\Omega \beta _\epsilon(u _\epsilon )h_0(u_\epsilon)\varphi dx \notag \\ & \quad = \int _\Omega f _\epsilon h_0(u_\epsilon)\varphi dx + \int _\Omega F.\nabla [h_0(u_\epsilon)\varphi]dx, \end{align} $$

where $\varphi \in W^{1, p(.)}_0(\Omega )\cap L^\infty (\Omega )$ .

By applying the same arguments as [Reference Ouaro, Ouédraogo and Soma23], one can express

(5.20) $$ \begin{align} \lim _{\epsilon \to 0} \int _ \Omega a (x,\nabla u_\epsilon).\nabla (h_0(u_\epsilon )\varphi dx & = \int _ \Omega a (x,\nabla u).\nabla (h_0(u)\varphi dx \notag \\ &= \int _ \Omega a(x,\nabla u).\nabla \varphi dx, \end{align} $$

StartLayout 1st Row 1st Column limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u Subscript epsilon Baseline right parenthesis period nabla left parenthesis h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi d x 2nd Column equals integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla left parenthesis h 0 left parenthesis u right parenthesis phi d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla phi d x comma EndLayout

(5.21) $$ \begin{align} \lim _{\epsilon \to 0} \int _\Omega f _\epsilon h_0(u_\epsilon)\varphi dx=\int _\Omega f h_0(u)\varphi dx =\int _\Omega f\varphi dx, \end{align} $$

and

(5.22) $$ \begin{align} \lim _{\epsilon \to 0} \int _\Omega F.\nabla [h_0(u_\epsilon)\varphi]dx=\int _\Omega F.\nabla [h_0(u)\varphi]dx=\int _\Omega F.\nabla \varphi dx. \end{align} $$

According to Lemma 5.9 and the convergence (5.15), one has

$$ \begin{align*} \lim _{\epsilon \to 0}\int _\Omega \phi(u_\epsilon).\nabla [h_0(u_\epsilon)\varphi] dx & = \int _\Omega \phi(T_{l_0+1}(u_\epsilon)).\nabla [h_0(u_\epsilon)\varphi] dx\\ &=\int _\Omega \phi(T_{l_0+1}(u)).\nabla [h_0(u)\varphi] dx\\ &=\int _\Omega \phi(u).\nabla [h_0(u)\varphi] dx \end{align*} $$

StartLayout 1st Row 1st Column limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts phi left parenthesis u Subscript epsilon Baseline right parenthesis period nabla left bracket h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi right bracket d x 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript l 0 plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis period nabla left bracket h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi right bracket d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript l 0 plus 1 Baseline left parenthesis u right parenthesis right parenthesis period nabla left bracket h 0 left parenthesis u right parenthesis phi right bracket d x 3rd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis u right parenthesis period nabla left bracket h 0 left parenthesis u right parenthesis phi right bracket d x EndLayout

Therefore,

(5.23) $$ \begin{align} \lim _{\epsilon \to 0}\int _\Omega \phi(u_\epsilon).\nabla [h_0(u_\epsilon)\varphi] dx=\int _\Omega \phi(u).\nabla \varphi dx. \end{align} $$

In order to pass to the limit in the sequence $(\beta _\epsilon ( u_\epsilon ))_{\epsilon>0}$ as $\epsilon $ goes to 0, we need the following lemmas.

Lemma 5.10 [Reference Nassouri, Ouaro and Urbain20]

Let j be a lower semi-continuous function on $\mathbb {R}$ with $\overline {dom(j)} =[m,M]\subset \mathbb {R}$ , and let $j_\epsilon $ be a sequence of lower semi-continuous functions such that

$$\begin{align*}j_\epsilon (t)\geq 0, \forall t\in [m,M], \ \text{and}\ j_\epsilon \uparrow j \ \text{as}\ \epsilon \downarrow 0.\end{align*}$$

Consider two sequences $(v_\epsilon )_{\epsilon>0}$ and $(z_\epsilon )_{\epsilon>0}$ of measurable functions on $\Omega $ satisfying

$$ \begin{align*}\left \lbrace \begin{array}{l} v_\epsilon \longrightarrow v\mathcal{L}^N \ \text{a.e in} \ \Omega, \ v\in dom (j)\mathcal{L}^N a.e. \ \text{in} \ \Omega, \\ \forall \epsilon>0, \ z_\epsilon \in \partial j_\epsilon (v_\epsilon)\mathcal{L}^N \ \text{a.e in} \ \Omega. \end{array} \right. \end{align*} $$

Assume that there exists $z\in \mathcal {M}^{p(.)}_b(\Omega ) \cap [(W^{1,p(.)}(\Omega ))^*+L^1]$ such that for all $\varphi \in C_C^1(\Omega )$ , $\varphi \geq 0$ ,

(5.24) $$ \begin{align} \liminf _{\epsilon\to 0} \int _\Omega (t-v_\epsilon)\varphi h_0(v_\epsilon)z_\epsilon dx \geq \int _\Omega (t-v)\varphi dz, \ \forall t\in \mathbb{R}. \end{align} $$

Then,

$$ \begin{align*}\left \lbrace \begin{array}{l} z = w\mathcal{L}^N + z_s \ \text{with}\ \nu \ \bot \ \mathcal{L}^N, \ w \in \partial j(v)\mathcal{L}^N \ \text{a.e in} \ \Omega, \ w\in L^1(\Omega),\\ z ^+_s \ \text{is concentrated on }\ [u = M], \ z ^-_s\ \text{is concentrated on }\ [u = m]. \end{array} \right. \end{align*} $$

Lemma 5.11 Let $l_0>0$ such that $\mathcal {D}(\beta )=[m,M]\subset [-l_0,l_0]$ . Then, there exists $\sigma \in \mathcal M_{b}^{p(.)}(\Omega )$ such that $ h_{0}(u_{\epsilon })\beta _{\epsilon }(u_{\epsilon }) \stackrel {*}{\rightharpoonup } \sigma , \ \text {as} \ \epsilon \rightarrow 0. $

Proof $\bullet $ According to Proposition 5.2-(ii), for any $k>0,$ the sequence $\displaystyle (h_{0}(u_{\epsilon })\beta _{\epsilon }(u_{\epsilon }))_{\epsilon> 0}$ is bounded in $\displaystyle L^{1}(\Omega )$ . Then, there exists $\displaystyle \sigma \in \mathcal M_{b}(\Omega )$ such that $h_{0}(u_{\epsilon })\beta _{\epsilon }(u_{\epsilon }) \stackrel {*}{\rightharpoonup } \sigma \text { in } \mathcal {M}_{b}(\Omega ) \text { as } \epsilon \rightarrow 0.$

$\bullet $ One can write $\sigma \in \mathcal {M}_b^{p(.)}(\Omega )\cap (W^{-1,p'(.)}+L^1(\Omega ))$ .

Indeed, for any $\varphi \in \mathcal {D}(\Omega )$ , one has

$$ \begin{align*} \int _\Omega \varphi d\sigma &= \int _\Omega h_0(u)\varphi d\sigma = \lim _{\epsilon \to 0} \int _\Omega h_{0}(u_{\epsilon})\varphi \beta_{\epsilon}(u_{\epsilon})dx\\ &= -\lim _{\epsilon \to 0} \int _\Omega \bigg [ a(x,\nabla T_{l_0+1}(u_\epsilon))- \phi (T_{l_0+1}(u_\epsilon))\bigg ].\nabla [h_0(u_\epsilon)\varphi]dx\\ &\quad + \lim _{\epsilon \to 0}\int _\Omega f_\epsilon h_0(u_\epsilon)\varphi dx +\lim _{\epsilon \to 0}\int _\Omega F.\nabla [h_0(u_\epsilon)\varphi] dx \\ &= -\int _\Omega a(x, \nabla T_{l_0+1}(u)).\nabla [h_0(u)\varphi ]dx +\int _\Omega \phi (T_{l_0+1}(u).\nabla [h_0(u)\varphi] dx\\ &\quad + \displaystyle\int _\Omega fh_0(u)\varphi dx+\int _\Omega F.\nabla [h_0(u)\varphi] dx\\ & = -\displaystyle\int _\Omega a(x, \nabla u).\nabla \varphi dx+\int _\Omega \phi (u).\nabla \varphi dx + \int _\Omega f\varphi dx + \int _\Omega F.\nabla \varphi dx. \end{align*} $$

StartLayout 1st Row 1st Column integral Underscript normal upper Omega Endscripts phi d sigma 2nd Column equals integral Underscript normal upper Omega Endscripts h 0 left parenthesis u right parenthesis phi d sigma equals limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi beta Subscript epsilon Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 2nd Row 1st Column Blank 2nd Column equals minus limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts left bracket a left parenthesis x comma nabla upper T Subscript l 0 plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis minus phi left parenthesis upper T Subscript l 0 plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis right bracket period nabla left bracket h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi right bracket d x 3rd Row 1st Column Blank 2nd Column plus limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts f Subscript epsilon Baseline h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi d x plus limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts upper F period nabla left bracket h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi right bracket d x 4th Row 1st Column Blank 2nd Column equals minus integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla upper T Subscript l 0 plus 1 Baseline left parenthesis u right parenthesis right parenthesis period nabla left bracket h 0 left parenthesis u right parenthesis phi right bracket d x plus integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript l 0 plus 1 Baseline left parenthesis u right parenthesis period nabla left bracket h 0 left parenthesis u right parenthesis phi right bracket d x 5th Row 1st Column Blank 2nd Column plus integral Underscript normal upper Omega Endscripts f h 0 left parenthesis u right parenthesis phi d x plus integral Underscript normal upper Omega Endscripts upper F period nabla left bracket h 0 left parenthesis u right parenthesis phi right bracket d x 6th Row 1st Column Blank 2nd Column equals minus integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla phi d x plus integral Underscript normal upper Omega Endscripts phi left parenthesis u right parenthesis period nabla phi d x plus integral Underscript normal upper Omega Endscripts f phi d x plus integral Underscript normal upper Omega Endscripts upper F period nabla phi d x period EndLayout

Therefore, $\sigma =\text {div} a(x,\nabla u)-\text {div} \phi (u)+\mu $ in $\mathcal {D}'(\Omega )$ and $\sigma \in \mathcal {M}_b^{p(.)}(\Omega )\cap (W^{-1,p'(.)}(\Omega )+~L^1(\Omega ))$ .

Remark 5.12 The measure $\sigma $ can be written as $\sigma = b\mathcal {L}^N + \nu \ \ \text {with}\ \nu \ \bot \ \mathcal {L}^N$ such that all the properties of (3.5) hold.

Indeed, for any $\varphi \in C_c^1(\Omega ), \ t\in \mathbb {R}$ , one has

$$ \begin{align*} &\lim _{\epsilon \to 0} \int _\Omega (t-u_\epsilon)h_{0}(u_{\epsilon})\varphi \beta_{\epsilon}(u_{\epsilon})dx\\ &\quad = -\lim _{\epsilon \to 0} \int _\Omega \bigg [ a(x,\nabla T_{l_0+1}(u_\epsilon))- \phi (T_{l_0+1}(u_\epsilon))\bigg ].\nabla [(t-u_\epsilon)h_0(u_\epsilon)\varphi]dx \\ &\qquad + \lim _{\epsilon \to 0}\int _\Omega f_\epsilon (t-u_\epsilon)h_0(u_\epsilon)\varphi dx +\lim _{\epsilon \to 0}\int _\Omega F.\nabla [(t-u_\epsilon)h_0(u_\epsilon)\varphi] dx \\ &\quad = -\int _\Omega a(x, \nabla T_{l_0+1}(u)).\nabla [(t-u)h_0(u)\varphi ]dx \\ &\qquad +\int _\Omega \phi (T_{l_0+1}(u)).\nabla [(t-u)h_0(u)\varphi] dx \\ & \qquad +\int _\Omega (t-u)fh_0(u)\varphi dx+\int _\Omega F.\nabla [(t-u)h_0(u)\varphi] dx \\ &\quad = -\int _\Omega a(x, \nabla u).\nabla [(t-u)h_0(u)\varphi ]dx +\int _\Omega \phi (u).\nabla [(t-u)h_0(u)\varphi] dx \\ &\qquad +\int _\Omega (t-u)fh_0(u)\varphi dx+\int _\Omega F.\nabla [(t-u)h_0(u)\varphi] dx \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts left parenthesis t minus u Subscript epsilon Baseline right parenthesis h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi beta Subscript epsilon Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 2nd Row 1st Column Blank 2nd Column equals minus limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts left bracket a left parenthesis x comma nabla upper T Subscript l 0 plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis minus phi left parenthesis upper T Subscript l 0 plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis right bracket period nabla left bracket left parenthesis t minus u Subscript epsilon Baseline right parenthesis h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi right bracket d x 3rd Row 1st Column Blank 2nd Column plus limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts f Subscript epsilon Baseline left parenthesis t minus u Subscript epsilon Baseline right parenthesis h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi d x plus limit Underscript epsilon right arrow 0 Endscripts integral Underscript normal upper Omega Endscripts upper F period nabla left bracket left parenthesis t minus u Subscript epsilon Baseline right parenthesis h 0 left parenthesis u Subscript epsilon Baseline right parenthesis phi right bracket d x 4th Row 1st Column Blank 2nd Column equals minus integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla upper T Subscript l 0 plus 1 Baseline left parenthesis u right parenthesis right parenthesis period nabla left bracket left parenthesis t minus u right parenthesis h 0 left parenthesis u right parenthesis phi right bracket d x 5th Row 1st Column Blank 2nd Column plus integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript l 0 plus 1 Baseline left parenthesis u right parenthesis right parenthesis period nabla left bracket left parenthesis t minus u right parenthesis h 0 left parenthesis u right parenthesis phi right bracket d x 6th Row 1st Column Blank 2nd Column plus integral Underscript normal upper Omega Endscripts left parenthesis t minus u right parenthesis f h 0 left parenthesis u right parenthesis phi d x plus integral Underscript normal upper Omega Endscripts upper F period nabla left bracket left parenthesis t minus u right parenthesis h 0 left parenthesis u right parenthesis phi right bracket d x 7th Row 1st Column Blank 2nd Column equals minus integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla u right parenthesis period nabla left bracket left parenthesis t minus u right parenthesis h 0 left parenthesis u right parenthesis phi right bracket d x plus integral Underscript normal upper Omega Endscripts phi left parenthesis u right parenthesis period nabla left bracket left parenthesis t minus u right parenthesis h 0 left parenthesis u right parenthesis phi right bracket d x 8th Row 1st Column Blank 2nd Column plus integral Underscript normal upper Omega Endscripts left parenthesis t minus u right parenthesis f h 0 left parenthesis u right parenthesis phi d x plus integral Underscript normal upper Omega Endscripts upper F period nabla left bracket left parenthesis t minus u right parenthesis h 0 left parenthesis u right parenthesis phi right bracket d x EndLayout

Setting $v_\epsilon = u_\epsilon $ and $z_\epsilon =\beta _\epsilon $ in Lemma 5.10, one can deduce (3.5). Since $\nu = (f-b)-div (a(x,\nabla u)-F)$ in $\mathcal {D}'(\Omega )$ , one has also $\nu \in \mathcal {M}_b^{p(.)}(\Omega ))$ .

Using the results above, by letting $\epsilon \rightarrow 0$ , one obtains

(5.25) $$ \begin{align} \int _ \Omega a (x,\nabla u).\nabla \varphi dx-\int _\Omega \phi (u).\nabla \varphi dx + \int _\Omega b\varphi dx + \int _\Omega \varphi d\nu = \int _\Omega \varphi d\mu. \end{align} $$

Now, we focus on the proof of (3.7) to end the demonstration.

For that, one chooses $T_1(u_\epsilon -T_n(u_\epsilon ))$ as test function in (5.2) to obtain

(5.26) $$ \begin{align} \int _ \Omega a(x,\nabla u_\epsilon) &.\nabla T_1(u_\epsilon-T_n(u_\epsilon))dx + \int _\Omega \beta _\epsilon (u_\epsilon)T_1(u_\epsilon-T_n(u_\epsilon))dx \nonumber \\ &\quad +\int _\Omega \phi (u_\epsilon).\nabla T_1(u_\epsilon-T_n(u_\epsilon)) dx = \int _\Omega T_1(u_\epsilon-T_n(u_\epsilon))d \mu_\epsilon. \end{align} $$

Observing that $\displaystyle \int _\Omega \beta _\epsilon (u_\epsilon )T_1(u_\epsilon -T_n(u_\epsilon ))dx\geq 0$ and $\nabla T_1(u_\epsilon -T_k(u_\epsilon ))=\nabla u_\epsilon \chi _{\{n<|u_\epsilon |<n+1\}}$ , (5.26) becomes

$$\begin{align*}\int _ {\{n<|u_\epsilon|<n+1\}} a(x,\nabla u_\epsilon) .\nabla u_\epsilon dx + \int _{\{n<|u_\epsilon|<n+1\}} \phi (u_\epsilon).\nabla u_\epsilon dx\end{align*}$$

$$\begin{align*}\leq \int _\Omega f_\epsilon T_1(u_\epsilon-T_n(u_\epsilon))dx + \int _{\{n<|u_\epsilon|<n+1\}} F.\nabla u_\epsilon dx. \end{align*}$$

Using $(H_3)$ , we deduce that

$$\begin{align*}C_3\int _ {\{n<|u_\epsilon|<n+1\}} |\nabla u_\epsilon| ^{p(x)}dx + \int _{\{n<|u_\epsilon|<n+1\}} \phi (u_\epsilon).\nabla u_\epsilon dx \end{align*}$$

(5.27) $$ \begin{align} \leq \int _\Omega f_\epsilon T_1(u_\epsilon-T_n(u_\epsilon))dx + \int _{\{n<|u_\epsilon|<n+1\}} F.\nabla u_\epsilon dx. \end{align} $$

Let us consider $\displaystyle \Phi (t)=\int _0^t\phi (\tau )d\tau $ . Then $\Phi (T_n(u_\epsilon ))\in (W_0^{1,p(x)}(\Omega ))^N$ ,

By using Lemma 2.6, one gets

(5.28) $$ \begin{align} &\int _{\{n<|u_\epsilon|<n+1\}} \phi (u_\epsilon).\nabla u_\epsilon dx \notag \\ &\quad =\int _\Omega \phi (T_{n+1}(u_\epsilon)).\nabla T_{n+1}(u_\epsilon )dx-\int _\Omega \phi (T_n(u_\epsilon)).\nabla T_n(u_\epsilon) dx\notag \\ &\quad =\int _\Omega \text{div} \Phi ((T_{n+1}(u_\epsilon)))dx -\int _\Omega \text{div} \Phi (T_n(u_\epsilon))dx=0. \end{align} $$

StartLayout 1st Row 1st Column Blank 2nd Column integral Underscript left brace n less than StartAbsoluteValue u Subscript epsilon Baseline EndAbsoluteValue less than n plus 1 right brace Endscripts phi left parenthesis u Subscript epsilon Baseline right parenthesis period nabla u Subscript epsilon Baseline d x 2nd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript n plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis period nabla upper T Subscript n plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x minus integral Underscript normal upper Omega Endscripts phi left parenthesis upper T Subscript n Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis period nabla upper T Subscript n Baseline left parenthesis u Subscript epsilon Baseline right parenthesis d x 3rd Row 1st Column Blank 2nd Column equals integral Underscript normal upper Omega Endscripts div normal upper Phi left parenthesis left parenthesis upper T Subscript n plus 1 Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis right parenthesis d x minus integral Underscript normal upper Omega Endscripts div normal upper Phi left parenthesis upper T Subscript n Baseline left parenthesis u Subscript epsilon Baseline right parenthesis right parenthesis d x equals 0 period EndLayout

Consequently, (5.27) becomes

$$\begin{align*}C_3\int _ {\{n<|u_\epsilon|<n+1\}} |\nabla u_\epsilon| ^{p(x)}dx \leq \int _\Omega f_\epsilon T_1(u_\epsilon-T_n(u_\epsilon))dx + \int _{\{n<|u_\epsilon|<n+1\}} F.\nabla u_\epsilon dx.\end{align*}$$

Arguing similarly as in [Reference Ouaro, Ouédraogo and Soma23], one obtains the rest of the proof of the condition (3.7).

Lemma 5.13 Suppose that $\phi $ is a Lipschitz function. let $s \in W^{1, p(.)}_0(\Omega ), \ \sigma \ in \ \mathcal {M}^{p(.)}_b(\Omega )$ and $\lambda \in \mathbb {R}$ such that

(5.29) $$ \begin{align} \left \lbrace \begin{array}{l} s \leq \lambda \ a.e. \ \text{in} \ \Omega \ (resp. \ s \geq \lambda) \\ \sigma = -\text{div} \ a(x,\nabla s) + \text{div}\ \phi (s) \ in \ \mathcal{D}'(\Omega). \end{array} \right. \end{align} $$

Then,

(5.30) $$ \begin{align} \int _{[s = \lambda]}\varphi d\sigma \geq 0, \end{align} $$

(resp.)

(5.31) $$ \begin{align} \int _{[s =\lambda]}\varphi d\sigma \leq 0, \end{align} $$

for any $\varphi \in C_c^1(\Omega )$ , $\varphi \geq 0$ .

Proof For $n\geq 1$ , we consider the function $\theta _n$ defined by

$$\begin{align*}\theta _n(r) = \inf \{1, (nr-n\lambda+1)^+\}, \ \forall r\in \mathbb{R}.\end{align*}$$

Note that $\theta _n(r)$ converges to $\chi _{[\lambda , \infty )}(r)$ for every $r\in \mathbb {R}$ , so $\theta _n(s (x))$ converges to $\chi _{[\lambda , \infty )}(s (x))$ at every x where $s (x)$ is defined.

Since s is defined quasi everywhere and $\chi _{[\lambda , \infty )}\circ s = \chi _{\{x\in \Omega : s (x) = \lambda \}}$ , then the convergence of $\theta _n(s)$ to $\chi _{[\lambda , \infty )}(s )$ is quasi everywhere.

Therefore, since $\sigma $ is diffuse, then $\theta _n(s)$ converges to $\chi _{\{ x\in \Omega : s (x) = \lambda \}}$ , $\sigma $ -a.e. in $\Omega $ .

$\forall \varphi \in C_c^1(\Omega )$ such that $\varphi \geq 0$ , one has

$$ \begin{align*} \int _{[s = \lambda ]}\varphi d\sigma &= \lim _{n\to +\infty}\int _\Omega \varphi \theta _n(s)d\sigma\\ & = \lim _{n\to +\infty}\int _\Omega a(x,\nabla s) .\nabla [\varphi \theta _n(s)] dx+\lim _{n\to +\infty}\int _\Omega \text {div}\ \phi (s)(\varphi \theta _n(s))dx\\ & \geq \int _\Omega \theta _n(s) a(x,\nabla s) .\nabla \varphi dx+\lim _{n\to +\infty}\int _\Omega \text {div}\ \phi (s)(\varphi \theta _n(s))dx. \end{align*} $$

StartLayout 1st Row 1st Column integral Underscript left bracket s equals lamda right bracket Endscripts phi d sigma 2nd Column equals limit Underscript n right arrow plus infinity Endscripts integral Underscript normal upper Omega Endscripts phi theta Subscript n Baseline left parenthesis s right parenthesis d sigma 2nd Row 1st Column Blank 2nd Column equals limit Underscript n right arrow plus infinity Endscripts integral Underscript normal upper Omega Endscripts a left parenthesis x comma nabla s right parenthesis period nabla left bracket phi theta Subscript n Baseline left parenthesis s right parenthesis right bracket d x plus limit Underscript n right arrow plus infinity Endscripts integral Underscript normal upper Omega Endscripts div phi left parenthesis s right parenthesis left parenthesis phi theta Subscript n Baseline left parenthesis s right parenthesis right parenthesis d x 3rd Row 1st Column Blank 2nd Column greater than or equals integral Underscript normal upper Omega Endscripts theta Subscript n Baseline left parenthesis s right parenthesis a left parenthesis x comma nabla s right parenthesis period nabla phi d x plus limit Underscript n right arrow plus infinity Endscripts integral Underscript normal upper Omega Endscripts div phi left parenthesis s right parenthesis left parenthesis phi theta Subscript n Baseline left parenthesis s right parenthesis right parenthesis d x period EndLayout

Since $\phi $ is a Lipschitz function, one has

$$\begin{align*}\int _\Omega \text{div}\ \phi (s) (\theta _n(s))dx=\int _\Omega (\theta _n(s)) \phi ' (s)\cdot \nabla s dx.\end{align*}$$

It follows that

$$ \begin{align*} \bigg |\int _\Omega \text{div}\ \phi (s) (\theta _n(s))dx \bigg |&=\bigg |\int _\Omega (\theta _n(s)) \phi ' (s)\cdot \nabla s dx \bigg |\\ &\leq \|\varphi\|_\infty \int _{\{\lambda-\frac{1}{n}\leq s \leq \lambda\}} |\phi ' (s)| \ |\nabla s|dx\\ & \longrightarrow 0 \ \text{as}\ n\to +\infty. \end{align*} $$

StartLayout 1st Row 1st Column StartAbsoluteValue integral Underscript normal upper Omega Endscripts div phi left parenthesis s right parenthesis left parenthesis theta Subscript n Baseline left parenthesis s right parenthesis right parenthesis d x EndAbsoluteValue 2nd Column equals StartAbsoluteValue integral Underscript normal upper Omega Endscripts left parenthesis theta Subscript n Baseline left parenthesis s right parenthesis right parenthesis phi prime left parenthesis s right parenthesis dot nabla s d x EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column less than or equals StartMetric phi EndMetric Subscript infinity Baseline integral Underscript left brace lamda minus StartFraction 1 Over n EndFraction less than or equals s less than or equals lamda right brace Endscripts StartAbsoluteValue phi prime left parenthesis s right parenthesis EndAbsoluteValue StartAbsoluteValue nabla s EndAbsoluteValue d x 3rd Row 1st Column Blank 2nd Column long right arrow 0 as n right arrow plus infinity period EndLayout

On the other hand, we have

$$ \begin{align*} \bigg | \int _\Omega \theta _n(s) a(x,\nabla s) .\nabla \varphi dx \bigg |& \leq \|\nabla \varphi\|_\infty \int _{\{\lambda-\frac{1}{n}\leq s \leq \lambda\}} |a(x,\nabla s)|dx\\ & \longrightarrow 0 \ \text{as}\ n\to +\infty. \end{align*} $$

StartLayout 1st Row 1st Column StartAbsoluteValue integral Underscript normal upper Omega Endscripts theta Subscript n Baseline left parenthesis s right parenthesis a left parenthesis x comma nabla s right parenthesis period nabla phi d x EndAbsoluteValue 2nd Column less than or equals StartMetric nabla phi EndMetric Subscript infinity Baseline integral Underscript left brace lamda minus StartFraction 1 Over n EndFraction less than or equals s less than or equals lamda right brace Endscripts StartAbsoluteValue a left parenthesis x comma nabla s right parenthesis EndAbsoluteValue d x 2nd Row 1st Column Blank 2nd Column long right arrow 0 as n right arrow plus infinity period EndLayout

Hence, the relation (5.30) holds.

In the case where $s \geq \lambda $ , one reasons similarly as above after setting $\tilde {s} = -s$ , $\tilde {\lambda } = -\lambda $ and $\tilde {a}(x,z )=a(x,-z)$ to obtain (5.31).

Remark 5.14 Moreover, if $\phi $ is a Lipschitz function, then a weak solution u of problem $(\mathcal {P})$ satisfies

(5.32) $$ \begin{align} \nu ^+ \leq \mu _s\lfloor \ [u = M], \end{align} $$

(5.33) $$ \begin{align} \nu ^- \leq - \mu _s\lfloor \ [u = m]. \end{align} $$

Indeed, since

$$\begin{align*}\nu = \text{div}\ a(x,\nabla u)-\text{div} \ \phi (u) - b\mathcal{L}^N + \mu ,\end{align*}$$

one has

$$\begin{align*}\mu - \nu - b\mathcal{L}^N = - \text{div}\ a(x,\nabla u)+\text{div} \ \phi (u).\end{align*}$$

According to Lemma 5.13, the proof follows the same approach as in [Reference Igbida, Ouaro and Soma21, Theorem 1.3]).

Remark 5.15 In the case where the right-hand side data is a regular function (for example, an $L^1$ -function), one has $\mu _s=0$ , so that $\nu ^+=\nu ^-=0$ and the notion of weak solution in this article coincides with the usual one.

6 Uniqueness of solution

The study of the uniqueness of the solution depends on additional conditions on the convection term

Theorem 6.1 Let $\phi $ be a Lipschitz function. If $(u_1, b_1)$ and $(u_2, b_2)$ are two solutions of (4.1), then

(6.1) $$ \begin{align} \int _\Omega (b_1-b_2)\text{sign} _0(u_1-u_2)dx=0. \end{align} $$

Proof By choosing $\varphi = u_2$ and $\varphi = u_1$ as tests functions in (3.8) for $(u_1,b_1)$ and $(u_2,b_2)$ , respectively, we obtain

$$\begin{align*}\int _\Omega a (x,\nabla u_1).\nabla T_k(u_1 - u_2)dx + \int _\Omega b_1T_k(u_1 - u_2)dx-\int _\Omega \phi (u_1).\nabla T_k(u_1 - u_2) dx \end{align*}$$

(6.2) $$ \begin{align} \leq \int _\Omega T_k(u_1 - u_2)d\mu \end{align} $$

and

$$\begin{align*}\int _\Omega a (x,\nabla u_2).\nabla T_k(u_2 - u_1)dx + \int _\Omega b_2T_k(u_2 - u_1)dx-\int _\Omega \phi (u_2).\nabla T_k(u_1 - u_2) dx \end{align*}$$

(6.3) $$ \begin{align} \leq \int _\Omega T_k(u_2 - u_1)d\mu. \end{align} $$

By adding (6.2) and (6.3), we obtain

$$\begin{align*}\int _\Omega \left(a (x,\nabla u_1)- a (x,\nabla u_2)\right).\nabla T_k(u_1 - u_2)dx+ \int _\Omega (b_1 - b_2)T_k(u_1 - u_2)dx \end{align*}$$

(6.4) $$ \begin{align} -\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx\leq 0. \end{align} $$ StartLayout 1st Row minus integral Underscript normal upper Omega Endscripts left parenthesis phi left parenthesis u 1 right parenthesis minus phi left parenthesis u 2 right parenthesis right parenthesis period nabla upper T Subscript k Baseline left parenthesis u 2 minus u 1 right parenthesis d x less than or equals 0 period EndLayout

Since $a(x,.)$ is monotone, the first term of (6.4) is non-negative, and we deduce from (6.4) that

$$\begin{align*}\int _\Omega (b_1 - b_2)T_k(u_1 - u_2)dx-\int _\Omega \phi (u).\nabla T_k(u_2 - u_1) dx\leq 0.\end{align*}$$

Dividing the above inequality by $k>0$ , we get

(6.5) $$ \begin{align} \frac{1}{k}\int _\Omega (b_1 - b_2)T_k(u_1 - u_2)dx-\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx\leq 0. \end{align} $$

Setting $A_k:=\{0\leq |u_1-u_2|\leq k\}$ , the second term of (6.5) gives

$$ \begin{align*} &\bigg |-\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx \bigg |\\ &=\bigg |\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla (u_2 - u_1)\chi _{A_k} dx \bigg |\\ & \leq \frac{1}{k}\int _\Omega \bigg |\bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla (u_2 - u_1)\chi _{A_k}\bigg | dx\\ &\leq \frac{C}{k}\int _\Omega |u_2 - u_1| \ |\nabla (u_2 - u_1)|\chi _{A_k}dx\\ & \leq C\int _\Omega |\nabla (u_2 - u_1)|\chi _{A_k}dx. \end{align*} $$

StartLayout 1st Row 1st Column Blank 2nd Column StartAbsoluteValue minus StartFraction 1 Over k EndFraction integral Underscript normal upper Omega Endscripts left parenthesis phi left parenthesis u 1 right parenthesis minus phi left parenthesis u 2 right parenthesis right parenthesis period nabla upper T Subscript k Baseline left parenthesis u 2 minus u 1 right parenthesis d x EndAbsoluteValue 2nd Row 1st Column Blank 2nd Column equals StartAbsoluteValue StartFraction 1 Over k EndFraction integral Underscript normal upper Omega Endscripts left parenthesis phi left parenthesis u 1 right parenthesis minus phi left parenthesis u 2 right parenthesis right parenthesis period nabla left parenthesis u 2 minus u 1 right parenthesis chi Subscript upper A Sub Subscript k Subscript Baseline d x EndAbsoluteValue 3rd Row 1st Column Blank 2nd Column less than or equals StartFraction 1 Over k EndFraction integral Underscript normal upper Omega Endscripts StartAbsoluteValue left parenthesis phi left parenthesis u 1 right parenthesis minus phi left parenthesis u 2 right parenthesis right parenthesis period nabla left parenthesis u 2 minus u 1 right parenthesis chi Subscript upper A Sub Subscript k Subscript Baseline EndAbsoluteValue d x 4th Row 1st Column Blank 2nd Column less than or equals StartFraction upper C Over k EndFraction integral Underscript normal upper Omega Endscripts StartAbsoluteValue u 2 minus u 1 EndAbsoluteValue StartAbsoluteValue nabla left parenthesis u 2 minus u 1 right parenthesis EndAbsoluteValue chi Subscript upper A Sub Subscript k Subscript Baseline d x 5th Row 1st Column Blank 2nd Column less than or equals upper C integral Underscript normal upper Omega Endscripts StartAbsoluteValue nabla left parenthesis u 2 minus u 1 right parenthesis EndAbsoluteValue chi Subscript upper A Sub Subscript k Subscript Baseline d x period EndLayout

Since

$$\begin{align*}|\nabla (u_2 - u_1)|\chi _{A_k} \longrightarrow 0 \ \text{a.e. in} \ \Omega \ \text{as}\ k\rightarrow 0\end{align*}$$

and

$$\begin{align*}\bigg ||\nabla (u_2 - u_1)|\chi _{A_k} \bigg |\leq |\nabla (u_2 - u_1)|\in L^1(\Omega).\end{align*}$$ StartAbsoluteValue EndAbsoluteValue nabla left parenthesis u 2 minus u 1 right parenthesis StartAbsoluteValue chi Subscript upper A Sub Subscript k Subscript Baseline EndAbsoluteValue less than or equals StartAbsoluteValue nabla left parenthesis u 2 minus u 1 right parenthesis EndAbsoluteValue element of upper L Superscript 1 Baseline left parenthesis normal upper Omega right parenthesis period

By Lebesgue’s dominated convergence theorem, one obtains

$$\begin{align*}\lim _{k\to 0}\int _\Omega |\nabla (u_2 - u_1)|\chi _{\{0\leq |u_1-u_2|\leq k\}}dx =0.\end{align*}$$

Therefore,

$$\begin{align*}\lim _{k\to 0}-\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx =0.\end{align*}$$

For the first term of (6.5), we have

$$\begin{align*}\frac{1}{k}(b_1 - b_2)T_k(u_1 - u_2)\rightarrow (b_1 - b_2)\text{sign}_0(u_1 - u_2) \ a.e. \ \text{in}\ \Omega, \ \text{as}\ k\rightarrow 0\end{align*}$$

and

$$\begin{align*}\bigg |\frac{1}{k}(b_1 - b_2)T_k(u_1 - u_2) \bigg |\leq (b_1 - b_2)\in L^1(\Omega).\end{align*}$$ StartAbsoluteValue StartFraction 1 Over k EndFraction left parenthesis b 1 minus b 2 right parenthesis upper T Subscript k Baseline left parenthesis u 1 minus u 2 right parenthesis EndAbsoluteValue less than or equals left parenthesis b 1 minus b 2 right parenthesis element of upper L Superscript 1 Baseline left parenthesis normal upper Omega right parenthesis period

Hence,

$$\begin{align*}\lim _{k\to 0}\frac{1}{k}\int _\Omega (b_1 - b_2)T_k(u_1 - u_2)dx=0.\end{align*}$$

By taking the limit as $k\to 0$ in (6.5), we arrive at (6.1).

Corollary 6.2 Let $\phi $ be a Lipschitz function and let $\beta $ be a continuous, increasing function on $\mathbb {R}$ . Then $b_1=b_2$ a.e. in $\Omega $ .

Proof Let $\beta $ be a continuous and increasing function on $\mathbb {R}$ . One can deduce that

$$\begin{align*}(b_1-b_2)\text{sign}_0(u_1-u_2)=|b_1-b_2|.\end{align*}$$

Then, using Theorem 6.1, it follows that

(6.6) $$ \begin{align} \|b_1-b_2\|_{L^1(\Omega)}=0. \end{align} $$

Hence, $b_1=b_2$ a.e. in $\Omega $ .

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Article contents

Diffusive convective elliptic problem in variable exponent space and measure data

Abstract

Keywords

MSC classification

Information

1 Introduction

2 Preliminaires

Proposition 2.1 ([Reference Kovacik and Rakosnik19])

Proposition 2.2 (see [Reference Wang, Fan and Ge25, Reference Yao27])

Theorem 2.3 ([Reference Fan and Zhao13, Reference Harjulehto and Hästö14])

3 Assumptions and main results

3.1 Assumptions

3.2 Notions of solutions and main results

4 Existence of solution for a regular right hand side data

5 Proof of Theorem 3.3

5.1 Approximate problem

5.2 A priori estimates

Proposition 5.3 ([Reference Ouaro, Ouédraogo and Soma23])

5.3 Convergence results

Proposition 5.4 ([Reference Ouaro, Ouédraogo and Soma23])

Proposition 5.7 [Reference Ouaro, Ouédraogo and Soma23]

Lemma 5.9 [Reference Ouaro, Ouédraogo and Soma23]

5.4 Existence of solution

Lemma 5.10 [Reference Nassouri, Ouaro and Urbain20]

6 Uniqueness of solution

References

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