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.
$$ \begin{align} (\mathcal{P})\left \lbrace \begin{array}{l} -\text{div}\ a(x,\nabla u) + \beta (u)+ \text{div} \phi (u) \ \ni \ \mu \ \ \text{in} \ \ \Omega \\ u = 0 \ \ \text{on}\ \ \partial \Omega , \end{array} \right. \end{align} $$
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
$$ \begin{align} 1<p^-:=\min _{x\in \overline{\Omega}}p(x)\leq p^+ : = \max _{x\in \overline{\Omega}}p(x)< \infty. \end{align} $$
We define the set
$$\begin{align*}C_+(\overline{\Omega})=\bigg \{ p\in C(\overline{\Omega}): \min _{x\in \overline{\Omega}}p(x)>1 \bigg \rbrace .\end{align*}$$
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.
$$ \begin{align} \min \left \lbrace \|u\|^ {p^-}_{p(.)}; \ \|u\|^ {p^+}_{p(.)} \right \rbrace \leq \rho _{p(.)}(u)\leq \max \left \lbrace \ \|u\|^ {p^-}_{p(.)}; \ \|u\|^ {p^+}_{p(.)} \right \rbrace. \end{align} $$
For any
$u\in L^{p(.)}(\Omega )$
and
$v\in L^{p'(.)}(\Omega )$
, we have the Hölder type inequality (see [Reference Kovacik and Rakosnik19]).
$$ \begin{align} \bigg | \int _\Omega uvdx \bigg |\leq \bigg ( \frac{1}{p^-}+\frac{1}{q^-} \bigg ) \ \|u\|_ {p(.)}\|v\|_ {q(.)}. \end{align} $$
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.,
$$\begin{align*}\gamma_0(s) = \left \lbrace \begin{array}{l} \text{minimal absolute value of } \ \gamma (s) \ \ \text{if} \ \gamma (s)\ \neq \emptyset \\ +\infty \ \text{if} \ [s, +\infty)\cap D(\gamma) = \emptyset \\ -\infty \text{if} \ (-\infty , s]\cap D(\gamma) = \emptyset. \end{array} \right. \end{align*}$$
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
$$ \begin{align} T_k(s) =\max \{-k,\min \{k;s\} \}. \end{align} $$
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
$$\begin{align*}H_\epsilon = \min \bigg ( \frac{s^+}{\epsilon}; 1 \bigg ) \ \text{and} \ sign_0^+ (s) = \left \lbrace \begin{array}{l} 1 \ \text{if} \ s> 0, \\ 0 \ \text{if} \ s \leq 0, \end{array} \right. \end{align*}$$
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
$$ \begin{align} |a(x,\xi)|\leq C_1\bigg (j(x)+|\xi|^{p(x)-1}\bigg ), \end{align} $$
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 $
,
$$ \begin{align} (a(x,\xi)-a(x,\eta )).(\xi -\eta)>0, \end{align} $$
$(H_3)$
there exists a positive constant
$C_2$
such that
$$ \begin{align} a(x,\xi).\xi \geq C_2|\xi|^{p(x)}, \end{align} $$
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
$$ \begin{align} \forall s\in \mathbb{R}, \ |\phi (s)|\leq C_3 |s|^{p(x)-1}. \end{align} $$
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
$$ \begin{align} \left \lbrace \begin{array}{l} u \in \beta (u)\mathcal{L}^N- a.e. \ \text{in} \ \Omega,\ b \in \beta (u)\mathcal{L}^N- a.e. \ \text{in} \ \Omega,\\ \nu ^+ \ \text{is concentrated on }\ [u = M], \\ \nu ^- \ \text{is concentrated on }\ [u = m], \end{array} \right. \end{align} $$
such that
$$ \begin{align} \displaystyle \int _\Omega a(x,\nabla u).\nabla \varphi dx + \int _\Omega b\varphi dx + \int _\Omega \varphi d\nu -\int _\Omega \phi (u).\nabla \varphi dx =\int _\Omega \varphi d\mu, \end{align} $$
for any
$\varphi \in W_0^{1, p(.)}(\Omega )\cap L^\infty (\Omega )$
.
Moreover,
$$ \begin{align} \lim _{n\to +\infty}\int _ {\{n<|u|<n+1\}} |\nabla u| ^{p(x)}dx=0. \end{align} $$
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
$$ \begin{align} \int _\Omega a(x,\nabla u).\nabla T_k(u - \varphi )dx &+ \int _\Omega bT_k(u - \varphi )dx -\int _\Omega \phi (u).\nabla T_k(u-\varphi )dx \notag \\ &\leq \int _\Omega T_k(u - \varphi)d\mu, \end{align} $$
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
$$ \begin{align} \int _\Omega a(x,\nabla u).\nabla T_k(u - \varphi)dx &-\int _\Omega \phi (u).\nabla T_k(u - \varphi) dx + \int _\Omega bT_k(u - \varphi)dx \nonumber\\ & + \int _\Omega T_k(u - \varphi)d\nu= \int _\Omega T_k(u - \varphi)d\mu. \end{align} $$
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
$$ \begin{align} (P^\phi_{g,\gamma})\left \lbrace \begin{array}{l} -\text{div}\ a(x,\nabla u) + g(u)+ \text{div} \phi (u) \ = \gamma \ \ \text{in} \ \ \ \Omega \\ u = 0 \text{on}\ \ \partial \Omega , \end{array} \right. \end{align} $$
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
$$ \begin{align} \displaystyle \int _\Omega a(x,\nabla u).\nabla \varphi dx + \int _\Omega g(u)\varphi dx -\int _\Omega \phi (u).\nabla \varphi dx =\int _\Omega \varphi \gamma dx. \end{align} $$
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
$$ \begin{align} \displaystyle \int _\Omega a(x,\nabla u_k).\nabla \varphi dx + \int _\Omega T_k(g(u_k))\varphi dx -\int _\Omega \phi (u_k).\nabla \varphi dx =\int _\Omega \varphi \gamma dx. \end{align} $$
for any
$\varphi \in W_0^{1,p(.)}(\Omega )\cap L^\infty (\Omega )$
.
Moreover
$$ \begin{align} \forall k>\|\gamma\|_\infty, \ |g(u_k)|\leq \|\gamma\|_\infty \ \text{a.e. in} \ \Omega. \end{align} $$
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
$$ \begin{align*} \bigg |\displaystyle \int _\Omega T_k(g(u)) \varphi dx \bigg |& \leq \displaystyle\int _\Omega |T_k(g(u))| |\varphi| dx\\ & \leq k\displaystyle\int _\Omega |\varphi| dx\\ & \leq k\bigg(\frac{1}{p_-}+\frac{1}{p_-'}\bigg) (\text{meas}(\Omega)+1)^{\frac{1}{p'_-}}\|\varphi\|_{p(.)}\\ &\leq C_4\|\varphi\|_{1,p(.)}. \end{align*} $$
According to
$(H_1)$
, one has
$$ \begin{align*} \bigg|\displaystyle \int _\Omega a(x,\nabla u).\nabla \varphi dx\bigg| & \leq \int _\Omega |a(x,\nabla u)| |\nabla \varphi| dx\\ & \leq C_1\int _\Omega j(x)|\nabla \varphi |dx + C_1\int _\Omega |\nabla u|^{p(x)-1}|\nabla \varphi |dx \\ & \leq C_1\bigg(\frac{1}{p^-}+\frac{1}{(p')^-}\bigg)\bigg(\|j\|_{p'(.)} +\||\nabla u|\|^{p(x)-1}_{p(.)}\bigg)\|\nabla \varphi\|_{p(.)}\\ & \leq C_5\|\varphi\|_{1,p(.)}. \end{align*} $$
By using the growth condition
$(H_5)$
on
$\phi $
, one obtains
$$ \begin{align*} \bigg|\displaystyle \int _\Omega \phi (u).\nabla \varphi dx\bigg| & \leq \int _\Omega |\phi (u)||\nabla \varphi | dx\\ & \leq C_3\int _\Omega |u|^{p(x)-1}|\nabla \varphi |dx\\ & \leq C_3\bigg(\frac{1}{p^-}+\frac{1}{(p')^-}\bigg)\||u|^{p(x)-1}\|_{p'(.)} \|\nabla \varphi\|_{p(.)}\\ & \leq C_6\|\varphi\|_{1,p(.)}. \end{align*} $$
Claim 1: the operator A is bounded.
Indeed, for any
$u,\varphi \in W_0^{1,p(.)}(\Omega )$
, one has
$$ \begin{align*} \bigg | \langle A u, \varphi \rangle \bigg| &\leq C_7\|\varphi\|_{1,p(.)}. \end{align*} $$
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
$$ \begin{align*} \langle A u, u \rangle & = \displaystyle \int _\Omega a(x,\nabla u).\nabla u dx + \int _\Omega T_k(g(u))u dx -\int _\Omega \phi (u).\nabla u dx\\ &\geq \displaystyle \int _\Omega a(x,\nabla u).\nabla u dx\\ &\geq C_1 \int _\Omega |\nabla u|^{p(x)}dx\\ &\geq C_1\|\nabla u\|^\alpha _{p(.)}\\ & \geq C_8\| u\|^\alpha _{1,p(.)} \end{align*} $$
where
$$\begin{align*}\alpha =\left \lbrace \begin{array}{l} p_+ \ \text{if} \ \|\nabla u\|_{p(.)} \leq 1, \\ p_-\ \text{if} \ \|\nabla u\|_{p(.)}> 1. \end{array} \right. \end{align*}$$
Thus, we obtain
$$\begin{align*}\frac{\langle Au, u\rangle}{\|u\|_{1,p(x)}}\longrightarrow \infty \ \ \text{as}\ \ \|u\|_{1,p(.)}\longrightarrow \infty.\end{align*}$$
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
$$ \begin{align} \left \lbrace \begin{array}{l} u_n\rightharpoonup u \ \text{in}\ W_0^{1,p(.)}(\Omega)\ \text{as}\ n\to \infty, \\ A_1 u_n\rightharpoonup \chi \ \text{in} \ W^{-1, p'(.)}(\Omega) \ \text{as}\ n\to \infty, \\ \displaystyle \lim _{n\to \infty}\sup \langle A_1 u_n, u_n \rangle \leq \langle \chi , u\rangle. \end{array} \right. \end{align} $$
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
$$ \begin{align} |h_n(x)|=|a(x,\nabla u_n)-\phi (u_n)|\leq C_9(j(x)+|u_n|^{p(x)-1}+|\nabla u_n|^{p(x)-1}), \end{align} $$
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
$$ \begin{align} h_n\rightharpoonup h\ \text{in}\ (L^{p'(.)}(\Omega))^N \ \text{as}\ n\rightarrow \infty. \end{align} $$
For all
$\varphi \in W_0^{1,p(.)}(\Omega )$
, one has
$$ \begin{align} \langle\chi , \varphi\rangle =\lim _{n\to \infty}\langle A_1 u_n,\varphi\rangle=\lim _{n\to \infty} \int _\Omega h_n(x).\nabla \varphi dx = \int _\Omega h(x).\nabla \varphi dx=\langle A_1u , \varphi\rangle. \end{align} $$
This implies that
$\chi =A_1u$
.
Applying (4.5), one obtains
$$ \begin{align} \displaystyle \limsup_{n\to +\infty} \langle A_1 u_n, u_n \rangle \leq \int _\Omega h(x).\nabla u dx. \end{align} $$
Using
$(H_2)$
, for any
$\varphi \in (L^{p(.)}(\Omega ))^N$
, one has
$$\begin{align*} & \int_\Omega (b(x,u_n,\nabla u_n)-b(x,u_n,\varphi)).(\nabla u_n-\varphi)dx \\ & \quad = \int_\Omega (a(x,\nabla u_n)-a(x,\varphi)).(\nabla u_n-\varphi)dx\geq 0. \end{align*}$$
This is equivalent to
$$ \begin{align} \int_\Omega h_n.\nabla u_n dx- \int_\Omega h_n.\varphi dx - \int_\Omega b(x,u_n,\varphi).(\nabla u_n-\varphi) dx \geq 0. \end{align} $$
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
$$ \begin{align} \int_\Omega \bigg (h-b(x,u,\varphi)\bigg).(\nabla u-\varphi)dx\geq 0. \end{align} $$
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
$$ \begin{align} (-t)\int_\Omega \bigg (h-b(x,u,\nabla u +t\tilde{\varphi})\bigg).\tilde{\varphi} dx\geq 0. \end{align} $$
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
$$\begin{align*}\int_\Omega \bigg (h-b(x,u,\nabla u)\bigg).\tilde{\varphi} dx= 0,\end{align*}$$
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
$$\begin{align*}\bigg ( T_k(g(u_n))- T_k(g(u))\bigg)\varphi \rightarrow 0 \ \text{a.e in} \ \Omega \ \text{as}\ n\to \infty.\end{align*}$$
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
$$\begin{align*}\int _\Omega \bigg( T_k(g(u_n))- T_k(g(u))\bigg)\varphi dx =0.\end{align*}$$
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
$$ \begin{align} \displaystyle &\int _\Omega a(x,\nabla u_k).\nabla H_\epsilon (u_k-R) dx \notag\\ &\qquad + \int _\Omega T_k(g(u))H_\epsilon (u_k-R) dx -\int _\Omega \phi (u_k).\nabla H_\epsilon (u_k-R) dx \notag\\ &\quad =\int _\Omega \gamma H_\epsilon (u_k-R) dx. \end{align} $$
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
$$ \begin{align*} \int _\Omega \phi (u_k).\nabla H_\epsilon (u_k-R) dx & = \int _\Omega \frac{1}{\epsilon}\phi (u_k).\nabla u_k \chi _{\{0\leq |u_k-R|<\epsilon\}}dx\\ & = \int _\Omega \nabla .\bigg (\int _0^{u_k}\phi (s)\chi _{\{0\leq |u_k-R|<\epsilon\}}(s) ds \bigg)dx\\ &= \int _{\partial \Omega} \psi _\epsilon (u_k).\nu d\sigma =0 \ \ \ \ (\text{as} \ u_k=0 \ \ \text{on}\ \partial \Omega). \end{align*} $$
Consequently, (4.13) becomes
$$ \begin{align} \int _\Omega T_k(g(u))H_\epsilon (u_k-R) dx\leq \int _\Omega\gamma H_\epsilon (u_k-R) dx. \end{align} $$
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.
$$ \begin{align} Pb(\beta _\epsilon,\phi)(\mu_\epsilon)\left \lbrace \begin{array}{l} \beta _\epsilon( u_\epsilon) - \text{div} a (x,\nabla u_\epsilon) + \text{div} \phi (u_\epsilon) = \mu_\epsilon \ \text{in} \ \ \ \Omega \\ u_\epsilon = 0 \ \ \text{on}\ \ \partial \Omega. \end{array} \right. \end{align} $$
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
$\forall \varphi \ \in \ W_0^{1, p(.)}(\Omega )\cap L^\infty (\Omega )$
,
$$ \begin{align} \int _\Omega a(x,\nabla u_\epsilon). \nabla \varphi dx + \int _\Omega \beta _\epsilon (u_\epsilon)\varphi dx -\int _\Omega \phi (u_\epsilon).\nabla \varphi dx = \int _\Omega \varphi d \mu_\epsilon. \end{align} $$
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
$$ \begin{align} \int _\Omega a(x,\nabla u_\epsilon).\nabla T_k(u_\epsilon) dx & + \int _\Omega \beta _\epsilon(u_\epsilon) T_k(u_\epsilon)dx -\int _\Omega \phi (u_\epsilon).\nabla T_k(u_\epsilon)dx \nonumber\\ &\quad = \int _\Omega f_\epsilon T_k(u_\epsilon)dx +\int _\Omega F .\nabla T_k(u_\epsilon)dx. \end{align} $$
The third term of (5.4) is zero. Indeed, we have
$$ \begin{align} \int _\Omega \phi (u_\epsilon).\nabla T_k(u_\epsilon)dx & = \int _\Omega \phi (T_k(u_\epsilon)).\nabla T_k(u_\epsilon) dx \nonumber\\ & = \int _\Omega \nabla \bigg (\int _0^{T_k(u_\epsilon)} \phi (s)ds\bigg ) dx =0, \end{align} $$
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
$$ \begin{align} meas\{ |u_\epsilon|> k \}\leq \dfrac{C(\mu, \Omega)}{\min \{\beta _\epsilon (k), |\beta _\epsilon (-k)| \}} \end{align} $$
and
$$ \begin{align} meas \bigg \{ | \nabla u_\epsilon |> k \bigg \} \leq \dfrac{C_{11}(k + 1)}{k^{p^-}} + \dfrac{C(\mu, \Omega)}{\min \{\beta _\epsilon (k), |\beta _\epsilon (-k)|}, \end{align} $$
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
$$ \begin{align} u_\epsilon \longrightarrow u\ \text{in measure and a.e. in } \ \Omega \ \text{as}\ \epsilon \longrightarrow 0. \end{align} $$
Lemma 5.5 For every function
$h \in W^{1,+\infty }(\mathbb {R}), h \geq 0$
with
$\text {supp}(h)$
compact,
$$ \begin{align} \displaystyle \limsup_{\epsilon \rightarrow 0} \int_{\Omega} \Big[a(x, \nabla u_{\epsilon})-\phi (u_\epsilon) \Big].\nabla\Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx \leq 0, \end{align} $$
$$ \begin{align} \displaystyle \limsup _{\delta\to 0} \limsup _{\epsilon\to 0} \int_{\{\delta <|u_\epsilon| < \delta +1\}} a(x, \nabla u_{\epsilon}).\nabla u_{\epsilon}dx \leq 0 \end{align} $$
and
$$ \begin{align} \displaystyle \limsup _{\epsilon\to 0} \int_\Omega a(x,\nabla u_{\epsilon}).[\nabla T_k(u_{\epsilon})-\nabla T_k(u)]dx \leq 0. \end{align} $$
Proof By choosing
$\displaystyle h(u_{\epsilon })(T_{k}(u_{\epsilon })-T_{k}(u))$
as a test function in (5.2), one obtains
$$ \begin{align} \left \lbrace \begin{array}{l} \displaystyle \int_{\Omega} \Big[a(x, \nabla u_{\epsilon})-\phi (u_\epsilon) \Big].\nabla \Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx \\ \displaystyle \quad + \int_{\Omega} \beta _\epsilon (u_\epsilon) h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx\\ \displaystyle = \int_{\Omega} f_{\epsilon} h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx + \int_{\Omega} F_\epsilon.\nabla\Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx. \end{array} \right. \end{align} $$
$\bullet $
Let us start by proving (5.9). The following inequality holds
$$ \begin{align} \displaystyle \limsup_{\epsilon \rightarrow 0} \int_{\Omega} \beta _\epsilon (u_\epsilon) h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx \geq 0. \end{align} $$
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
$$ \begin{align*} \displaystyle\int_{\Omega} \beta _\epsilon ( u_\epsilon) h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx &\geq \int_{\Omega} h(u_{\epsilon})\beta _\epsilon ( u_r)(T_{k}(u_{\epsilon})-T_{k}(u_r))dx \\ & \quad + \int_{\Omega} h(u_{\epsilon})\beta _\epsilon ( u_\epsilon)(T_{k}(u_r)-T_{k}(u))dx\\ & =:I_{\epsilon,r}+J_{\epsilon,r}. \end{align*} $$
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
$$\begin{align*}A_{\epsilon,r}&:=\int_{\Omega} h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u))d\mu _\epsilon, \qquad\qquad\qquad\\B_{\epsilon,r}&:=- \int_{\Omega} h(u_{\epsilon}) a(x, \nabla u_{\epsilon}).\nabla (T_{k}(u_r)-T_{k}(u))dx, \qquad \\C_{\epsilon,r}&:=-\int_{\Omega} h'(u_\epsilon)(T_{k}(u_r)-T_{k}(u)) a(x, \nabla u_{\epsilon}).\nabla u_{\epsilon} dx,\\D_{\epsilon,r}&:=-\int_{\Omega} \phi (u_\epsilon). \nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx. \qquad\qquad\end{align*}$$
According to [Reference Ouaro, Ouédraogo and Soma23], one has
$\displaystyle \lim _{r\to 0}A_{\epsilon ,r}=\lim _{r\to 0}B_{\epsilon ,r}=\lim _{r\to 0}C_{\epsilon ,r}=0$
.
$$ \begin{align*} D_{\epsilon,r}&=\displaystyle\int_{\Omega} \phi (u_\epsilon). \nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx\\ &=\int_{\Omega} \phi (T_l(u_\epsilon)).\nabla\Big[ h(u_{\epsilon})(T_{k}(u_r)-T_{k}(u)) \Big]dx, \end{align*} $$
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
$$ \begin{align} &\displaystyle \int _\Omega a(x,\nabla u_\epsilon ). \nabla T_1(u_\epsilon -T_\delta (u_\epsilon )) dx + \int _\Omega \beta _\epsilon (u_\epsilon )T_1(u_\epsilon -T_\delta (u_\epsilon )) dx \notag \\ &\qquad -\int _\Omega \phi (u_\epsilon).\nabla T_1(u_\epsilon -T_\delta (u_\epsilon)) dx \notag \\ & \quad = \int _\Omega f_\epsilon T_1(u_\epsilon -T_\delta (u_\epsilon)) dx +\int _\Omega F.\nabla T_1(u_\epsilon -T_\delta (u_\epsilon)) dx. \end{align} $$
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
$$ \begin{align} \phi (T_k(u_\epsilon))\longrightarrow \phi (T_k(u)) \text{ in} \ L^{p'(.)}(\Omega) \ \text{as}\ \epsilon\to 0, \end{align} $$
$$ \begin{align} \displaystyle\lim _{\epsilon\to 0}\int _\Omega \phi (u_\epsilon).\nabla \Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx=0, \end{align} $$
$$ \begin{align} \lim _{\epsilon\to 0} \int_{\Omega} f_{\epsilon} h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))dx=0 \end{align} $$
and
$$ \begin{align} \lim _{\epsilon\to 0} \int_{\Omega} F_\epsilon.\nabla\Big[ h(u_{\epsilon})(T_{k}(u_{\epsilon})-T_{k}(u))\Big]dx=0. \end{align} $$
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
$$ \begin{align} \int _ \Omega a (x,\nabla u_\epsilon).\nabla (h_0(u_\epsilon )\varphi dx &-\int _\Omega \phi(u_\epsilon).\nabla [h_0(u_\epsilon)\varphi] dx + \int _\Omega \beta _\epsilon(u _\epsilon )h_0(u_\epsilon)\varphi dx \notag \\ & \quad = \int _\Omega f _\epsilon h_0(u_\epsilon)\varphi dx + \int _\Omega F.\nabla [h_0(u_\epsilon)\varphi]dx, \end{align} $$
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
$$ \begin{align} \lim _{\epsilon \to 0} \int _ \Omega a (x,\nabla u_\epsilon).\nabla (h_0(u_\epsilon )\varphi dx & = \int _ \Omega a (x,\nabla u).\nabla (h_0(u)\varphi dx \notag \\ &= \int _ \Omega a(x,\nabla u).\nabla \varphi dx, \end{align} $$
$$ \begin{align} \lim _{\epsilon \to 0} \int _\Omega f _\epsilon h_0(u_\epsilon)\varphi dx=\int _\Omega f h_0(u)\varphi dx =\int _\Omega f\varphi dx, \end{align} $$
and
$$ \begin{align} \lim _{\epsilon \to 0} \int _\Omega F.\nabla [h_0(u_\epsilon)\varphi]dx=\int _\Omega F.\nabla [h_0(u)\varphi]dx=\int _\Omega F.\nabla \varphi dx. \end{align} $$
According to Lemma 5.9 and the convergence (5.15), one has
$$ \begin{align*} \lim _{\epsilon \to 0}\int _\Omega \phi(u_\epsilon).\nabla [h_0(u_\epsilon)\varphi] dx & = \int _\Omega \phi(T_{l_0+1}(u_\epsilon)).\nabla [h_0(u_\epsilon)\varphi] dx\\ &=\int _\Omega \phi(T_{l_0+1}(u)).\nabla [h_0(u)\varphi] dx\\ &=\int _\Omega \phi(u).\nabla [h_0(u)\varphi] dx \end{align*} $$
Therefore,
$$ \begin{align} \lim _{\epsilon \to 0}\int _\Omega \phi(u_\epsilon).\nabla [h_0(u_\epsilon)\varphi] dx=\int _\Omega \phi(u).\nabla \varphi dx. \end{align} $$
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
$\varphi \in C_C^1(\Omega )$
,
$\varphi \geq 0$
,
$$ \begin{align} \liminf _{\epsilon\to 0} \int _\Omega (t-v_\epsilon)\varphi h_0(v_\epsilon)z_\epsilon dx \geq \int _\Omega (t-v)\varphi dz, \ \forall t\in \mathbb{R}. \end{align} $$
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
$$ \begin{align*} \int _\Omega \varphi d\sigma &= \int _\Omega h_0(u)\varphi d\sigma = \lim _{\epsilon \to 0} \int _\Omega h_{0}(u_{\epsilon})\varphi \beta_{\epsilon}(u_{\epsilon})dx\\ &= -\lim _{\epsilon \to 0} \int _\Omega \bigg [ a(x,\nabla T_{l_0+1}(u_\epsilon))- \phi (T_{l_0+1}(u_\epsilon))\bigg ].\nabla [h_0(u_\epsilon)\varphi]dx\\ &\quad + \lim _{\epsilon \to 0}\int _\Omega f_\epsilon h_0(u_\epsilon)\varphi dx +\lim _{\epsilon \to 0}\int _\Omega F.\nabla [h_0(u_\epsilon)\varphi] dx \\ &= -\int _\Omega a(x, \nabla T_{l_0+1}(u)).\nabla [h_0(u)\varphi ]dx +\int _\Omega \phi (T_{l_0+1}(u).\nabla [h_0(u)\varphi] dx\\ &\quad + \displaystyle\int _\Omega fh_0(u)\varphi dx+\int _\Omega F.\nabla [h_0(u)\varphi] dx\\ & = -\displaystyle\int _\Omega a(x, \nabla u).\nabla \varphi dx+\int _\Omega \phi (u).\nabla \varphi dx + \int _\Omega f\varphi dx + \int _\Omega F.\nabla \varphi dx. \end{align*} $$
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
$$ \begin{align*} &\lim _{\epsilon \to 0} \int _\Omega (t-u_\epsilon)h_{0}(u_{\epsilon})\varphi \beta_{\epsilon}(u_{\epsilon})dx\\ &\quad = -\lim _{\epsilon \to 0} \int _\Omega \bigg [ a(x,\nabla T_{l_0+1}(u_\epsilon))- \phi (T_{l_0+1}(u_\epsilon))\bigg ].\nabla [(t-u_\epsilon)h_0(u_\epsilon)\varphi]dx \\ &\qquad + \lim _{\epsilon \to 0}\int _\Omega f_\epsilon (t-u_\epsilon)h_0(u_\epsilon)\varphi dx +\lim _{\epsilon \to 0}\int _\Omega F.\nabla [(t-u_\epsilon)h_0(u_\epsilon)\varphi] dx \\ &\quad = -\int _\Omega a(x, \nabla T_{l_0+1}(u)).\nabla [(t-u)h_0(u)\varphi ]dx \\ &\qquad +\int _\Omega \phi (T_{l_0+1}(u)).\nabla [(t-u)h_0(u)\varphi] dx \\ & \qquad +\int _\Omega (t-u)fh_0(u)\varphi dx+\int _\Omega F.\nabla [(t-u)h_0(u)\varphi] dx \\ &\quad = -\int _\Omega a(x, \nabla u).\nabla [(t-u)h_0(u)\varphi ]dx +\int _\Omega \phi (u).\nabla [(t-u)h_0(u)\varphi] dx \\ &\qquad +\int _\Omega (t-u)fh_0(u)\varphi dx+\int _\Omega F.\nabla [(t-u)h_0(u)\varphi] dx \end{align*} $$
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
$$ \begin{align} \int _ \Omega a (x,\nabla u).\nabla \varphi dx-\int _\Omega \phi (u).\nabla \varphi dx + \int _\Omega b\varphi dx + \int _\Omega \varphi d\nu = \int _\Omega \varphi d\mu. \end{align} $$
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
$$ \begin{align} \int _ \Omega a(x,\nabla u_\epsilon) &.\nabla T_1(u_\epsilon-T_n(u_\epsilon))dx + \int _\Omega \beta _\epsilon (u_\epsilon)T_1(u_\epsilon-T_n(u_\epsilon))dx \nonumber \\ &\quad +\int _\Omega \phi (u_\epsilon).\nabla T_1(u_\epsilon-T_n(u_\epsilon)) dx = \int _\Omega T_1(u_\epsilon-T_n(u_\epsilon))d \mu_\epsilon. \end{align} $$
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
$$\begin{align*}C_3\int _ {\{n<|u_\epsilon|<n+1\}} |\nabla u_\epsilon| ^{p(x)}dx + \int _{\{n<|u_\epsilon|<n+1\}} \phi (u_\epsilon).\nabla u_\epsilon dx \end{align*}$$
$$ \begin{align} \leq \int _\Omega f_\epsilon T_1(u_\epsilon-T_n(u_\epsilon))dx + \int _{\{n<|u_\epsilon|<n+1\}} F.\nabla u_\epsilon dx. \end{align} $$
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
$$ \begin{align} &\int _{\{n<|u_\epsilon|<n+1\}} \phi (u_\epsilon).\nabla u_\epsilon dx \notag \\ &\quad =\int _\Omega \phi (T_{n+1}(u_\epsilon)).\nabla T_{n+1}(u_\epsilon )dx-\int _\Omega \phi (T_n(u_\epsilon)).\nabla T_n(u_\epsilon) dx\notag \\ &\quad =\int _\Omega \text{div} \Phi ((T_{n+1}(u_\epsilon)))dx -\int _\Omega \text{div} \Phi (T_n(u_\epsilon))dx=0. \end{align} $$
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
$$ \begin{align} \left \lbrace \begin{array}{l} s \leq \lambda \ a.e. \ \text{in} \ \Omega \ (resp. \ s \geq \lambda) \\ \sigma = -\text{div} \ a(x,\nabla s) + \text{div}\ \phi (s) \ in \ \mathcal{D}'(\Omega). \end{array} \right. \end{align} $$
Then,
$$ \begin{align} \int _{[s = \lambda]}\varphi d\sigma \geq 0, \end{align} $$
(resp.)
$$ \begin{align} \int _{[s =\lambda]}\varphi d\sigma \leq 0, \end{align} $$
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
$$ \begin{align*} \int _{[s = \lambda ]}\varphi d\sigma &= \lim _{n\to +\infty}\int _\Omega \varphi \theta _n(s)d\sigma\\ & = \lim _{n\to +\infty}\int _\Omega a(x,\nabla s) .\nabla [\varphi \theta _n(s)] dx+\lim _{n\to +\infty}\int _\Omega \text {div}\ \phi (s)(\varphi \theta _n(s))dx\\ & \geq \int _\Omega \theta _n(s) a(x,\nabla s) .\nabla \varphi dx+\lim _{n\to +\infty}\int _\Omega \text {div}\ \phi (s)(\varphi \theta _n(s))dx. \end{align*} $$
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
$$ \begin{align*} \bigg |\int _\Omega \text{div}\ \phi (s) (\theta _n(s))dx \bigg |&=\bigg |\int _\Omega (\theta _n(s)) \phi ' (s)\cdot \nabla s dx \bigg |\\ &\leq \|\varphi\|_\infty \int _{\{\lambda-\frac{1}{n}\leq s \leq \lambda\}} |\phi ' (s)| \ |\nabla s|dx\\ & \longrightarrow 0 \ \text{as}\ n\to +\infty. \end{align*} $$
On the other hand, we have
$$ \begin{align*} \bigg | \int _\Omega \theta _n(s) a(x,\nabla s) .\nabla \varphi dx \bigg |& \leq \|\nabla \varphi\|_\infty \int _{\{\lambda-\frac{1}{n}\leq s \leq \lambda\}} |a(x,\nabla s)|dx\\ & \longrightarrow 0 \ \text{as}\ n\to +\infty. \end{align*} $$
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
$$ \begin{align} \nu ^+ \leq \mu _s\lfloor \ [u = M], \end{align} $$
$$ \begin{align} \nu ^- \leq - \mu _s\lfloor \ [u = m]. \end{align} $$
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
$$ \begin{align} \int _\Omega (b_1-b_2)\text{sign} _0(u_1-u_2)dx=0. \end{align} $$
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
$$\begin{align*}\int _\Omega a (x,\nabla u_1).\nabla T_k(u_1 - u_2)dx + \int _\Omega b_1T_k(u_1 - u_2)dx-\int _\Omega \phi (u_1).\nabla T_k(u_1 - u_2) dx \end{align*}$$
$$ \begin{align} \leq \int _\Omega T_k(u_1 - u_2)d\mu \end{align} $$
and
$$\begin{align*}\int _\Omega a (x,\nabla u_2).\nabla T_k(u_2 - u_1)dx + \int _\Omega b_2T_k(u_2 - u_1)dx-\int _\Omega \phi (u_2).\nabla T_k(u_1 - u_2) dx \end{align*}$$
$$ \begin{align} \leq \int _\Omega T_k(u_2 - u_1)d\mu. \end{align} $$
By adding (6.2) and (6.3), we obtain
$$\begin{align*}\int _\Omega \left(a (x,\nabla u_1)- a (x,\nabla u_2)\right).\nabla T_k(u_1 - u_2)dx+ \int _\Omega (b_1 - b_2)T_k(u_1 - u_2)dx \end{align*}$$
$$ \begin{align} -\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx\leq 0. \end{align} $$
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
$$ \begin{align} \frac{1}{k}\int _\Omega (b_1 - b_2)T_k(u_1 - u_2)dx-\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx\leq 0. \end{align} $$
Setting
$A_k:=\{0\leq |u_1-u_2|\leq k\}$
, the second term of (6.5) gives
$$ \begin{align*} &\bigg |-\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla T_k(u_2 - u_1) dx \bigg |\\ &=\bigg |\frac{1}{k}\int _\Omega \bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla (u_2 - u_1)\chi _{A_k} dx \bigg |\\ & \leq \frac{1}{k}\int _\Omega \bigg |\bigg (\phi (u_1)-\phi (u_2)\bigg ).\nabla (u_2 - u_1)\chi _{A_k}\bigg | dx\\ &\leq \frac{C}{k}\int _\Omega |u_2 - u_1| \ |\nabla (u_2 - u_1)|\chi _{A_k}dx\\ & \leq C\int _\Omega |\nabla (u_2 - u_1)|\chi _{A_k}dx. \end{align*} $$
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
$$\begin{align*}\bigg ||\nabla (u_2 - u_1)|\chi _{A_k} \bigg |\leq |\nabla (u_2 - u_1)|\in L^1(\Omega).\end{align*}$$
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
$$\begin{align*}\bigg |\frac{1}{k}(b_1 - b_2)T_k(u_1 - u_2) \bigg |\leq (b_1 - b_2)\in L^1(\Omega).\end{align*}$$
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
$$ \begin{align} \|b_1-b_2\|_{L^1(\Omega)}=0. \end{align} $$
Hence,
$b_1=b_2$
a.e. in
$\Omega $
.
