# Nonlinear Elliptic Equations With Variable Exponents Involving Singular
Nonlinearity

Hichem Khelifi1 and Youssef El hadfi2
1 Department of Mathematics and Informatics, University of Algiers, Algiers,
Algeria.
2 Street Didouche Mourad Algiers.
Applied Mathematics Laboratory, Badji Mokhtar University-Annaba B.P. 12,
Algeria.
<EMAIL_ADDRESS>
2Laboratory LIPIM, National School of Applied Sciences Khouribga
Sultan Moulay Slimane University, Morocco
<EMAIL_ADDRESS>

###### Abstract

In this paper, we prove the existence and regularity of weak positive
solutions for a class of nonlinear elliptic equations with a singular
nonlinearity, lower order terms and $L^{1}$ datum in the setting of variable
exponent Sobolev spaces. We will prove that the lower order term has some
regularizing effects on the solutions. This work generalizes some results
given in [2].

_K_ eywords Sobolev spaces with variable exponents, Singular Nonlinearity,
Elliptic equation, A given $L^{1}$, Lower order terms.

##  1 Introduction

###  1.1 Introduction of Our Problem

This work is devoted to the study of the nonlinear elliptic problem

$\begin{cases}-\mbox{div}(a(x)|\nabla u|^{p(x)-2}\nabla
u)+b(x)u|u|^{r(x)-1}=\frac{f}{u^{\gamma(x)}},&\mbox{in}\;\;\Omega,\\\
u>0,&\mbox{on}\;\;\Omega,\\\ u=0,&\mbox{on}\;\;\partial\Omega,\end{cases}$
(1.1)

where $\Omega$ is a bounded open subset of $\mathbb{R}^{N}$ ($N\geq 2$) with
Lipschitz boundary $\partial\Omega$, $f$ is a positive (that is $f(x)\geq 0$
and not zero a.e.) function in $L^{1}(\Omega)$, and
$p,r:\overline{\Omega}\rightarrow(0,+\infty)$,
$\gamma:\overline{\Omega}\rightarrow(0,1)$ are continuous functions and
satisfying

$1<p^{-}:=\inf\limits_{x\in\overline{\Omega}}p(x)\leq
p^{+}:=\sup\limits_{x\in\overline{\Omega}}p(x)<N,$ (1.2) $p(x)-1<r(x),$ (1.3)
$0<\gamma^{-}:=\inf_{x\in\overline{\Omega}}\gamma(x)\leq\gamma^{+}:=\sup_{x\in\overline{\Omega}}\gamma(x)<1,\quad\quad\mbox{and}\quad|\nabla\gamma|\in
L^{\infty}(\Omega)$ (1.4)

where $a(x),b(x)$ are measurable functions verifies for some positive numbers
$\alpha,\beta,\mu,\nu$ the following conditions

$0<\alpha\leq a(x)\leq\beta,\quad 0<\mu\leq b(x)\leq\nu.$ (1.5)

Equations with variable exponents appear in various mathematical models. In
some cases, they provide realistic models for the study of natural phenomena
in electro-rheological fluids and important applications are related to image
processing. We refer the reader to [3, 6, 7] and the references therein.

For costant-exponent cases (i.e., $p(x)=p$, $r(x)=r$ and $\gamma(x)=\gamma$),
the existence and regularity of solutions to problem (1.1) are studied in [2,
14, 28, 29]. They proved that the solution was in $W_{0}^{1,q}(\Omega)$ and
$u^{r+\gamma}$ belongs to $L^{1}(\Omega)$, where $q=\frac{pr}{p+1-\gamma}$.
The problem was also considered in [9], when $b(x)=0$ and $\gamma,p$ was a
constants with $0\leq\gamma<1$, $f\in L^{m}(\Omega)$ ($m\geq 1$). The authors
in [9] prove the existence and uniqueness results. If $p(x)=2$ and $\gamma,r$
were constants, the problem (1.1) has been treated in [12].

When the lower-order term does not appear in (1.1) (i.e., $b(x)=0$) and the
exponent $p(x)\equiv p$, the problem (1.1), have been treated in [13], under
the hypothesis $f\in L^{m}(\Omega)$ ($m\geq 1$). If $m=1$ and
$0<\gamma^{-}\leq\gamma(x)\leq\gamma^{+}<1$ the authors proved that the
solution belongs to $W_{0}^{1,q}(\Omega)$, where
$q=\frac{N(p+\gamma^{-}-1)}{N+\gamma^{-}-1}$.

###  1.2  Preliminary Work

For some preliminary results on the Lebesgue and Sobolev spaces with variable
exponent, we only give the definition of $L^{p(.)}(\Omega)$, for more details,
see [15, 16] or monographs [17, 21]. For an open
$\Omega\subset\mathbb{R}^{N}$, let $p:\Omega\rightarrow[1,+\infty)$ be a
measurable function such that

$1<p^{-}=\mathrm{ess\,inf}\,p,\quad p^{+}=\mathrm{ess\,sup}\,p<+\infty.$

We define the Lebesgue space with variable exponent $L^{p(\cdot)}(\Omega)$ to
consist of all measurable functions $u:\Omega\rightarrow\mathbb{R}$ for which
the convex modular

$\rho_{p(\cdot)}(u)=\int\limits_{\Omega}|u|^{p(x)}dx,$

is finite. The expression

$\|u\|_{p(\cdot)}:=\|u\|_{L^{p(\cdot)}(\Omega)}=\inf\left\\{\lambda>0,\
\rho_{p(\cdot)}\left(\frac{u}{\lambda}\right)\leq 1\right\\},$

defines a norm in $L^{p(\cdot)}(\Omega)$, called the Luxemburg norm, and
$\left(L^{p(\cdot)}(\Omega),\ \|u\|_{p(\cdot)}\right)$ is a uniformly convex
Banach space. Its dual space is isomorphic to $L^{p^{\prime}(\cdot)}(\Omega)$,
where $\frac{1}{p(x)}+\frac{1}{p^{\prime}(x)}=1$. For all $u\in
L^{p(\cdot)}(\Omega)$ and $v\in L^{p^{\prime}(\cdot)}(\Omega)$, the Hölder
type inequality

$\left|\int\limits_{\Omega}uvdx\right|\leq\left(\frac{1}{p^{-}}+\frac{1}{p^{\prime-}}\right)\|u\|_{p(\cdot)}\|v\|_{p^{\prime}(\cdot)}\leq
2\|u\|_{p(\cdot)}\|v\|_{p^{\prime}(\cdot)},$

holds true. We define the Sobolev space with variable exponent

$W^{1,p(\cdot)}(\Omega)=\left\\{u\in L^{p(\cdot)}(\Omega)\ \text{and}\ |\nabla
u|\in L^{p(\cdot)}(\Omega)\right\\},$

which is equipped with the following norm

$\|u\|_{1,p(\cdot)}=\|u\|_{W^{1,p(\cdot)}(\Omega)}=\|u\|_{p(\cdot)}+\|\nabla
u\|_{p(\cdot)}.$

The space $\left(W^{1,p(\cdot)}(\Omega),\|u\|_{1,p(\cdot)}\right)$ is a
reflexive Banach space. Next, we define also

$W_{0}^{1,p(\cdot)}(\Omega)=\left\\{u\in W^{1,p(\cdot)}(\Omega),\ u=0\
\text{on}\ \partial\Omega\right\\},$

endowed with the norm $\|.\|_{1,p(\cdot)}$.
The space $W_{0}^{1,p(.)}(\Omega)$ is separable and reflexive provided that
with $1<p^{-}\leq p^{+}<\infty$.

An important role in manipulating the generalized Lebesgue and Sobolev spaces
is played by the modular $\varrho_{p(.)}(\Omega)$ of the space
$L^{p(.)}(\Omega)$. We have the following result

######  Proposition 1 ([17]).

If $(u_{n}),u\in L^{p(\cdot)}(\Omega)$ and $p^{+}<+\infty$, then the following
properties hold true:

  * (i)

$\min\left(\rho_{p(\cdot)}(u)^{\frac{1}{p^{+}}},\rho_{p(\cdot)}(u)^{\frac{1}{p^{-}}}\right)\leq\|u\|_{p(\cdot)}\leq\max\left(\rho_{p(\cdot)}(u)^{\frac{1}{p^{+}}},\rho_{p(\cdot)}(u)^{\frac{1}{p^{-}}}\right),$

  * (ii)

$\min\left(\|u\|_{p(\cdot)}^{p^{-}},\|u\|_{p(\cdot)}^{p^{+}}\right)\leq\rho_{p(\cdot)}(u)\leq\max\left(\|u\|_{p(\cdot)}^{p^{-}},\|u\|_{p(\cdot)}^{p^{+}}\right),$

  * (iii)

$\|u\|_{p(\cdot)}\leq\rho_{p(\cdot)}(u)+1,$

Next, we recall some embedding results regarding variable exponent Lebesgue-
Sobolev spaces. If $p,\theta:\Omega\rightarrow(1,+\infty)$ are Lipschitz
continuous function satisfying (1.2) and $p(x)\leq\theta(x)\leq p^{*}(x)$ for
any $x\in\Omega$, where $p^{*}(x)=\frac{Np(x)}{N-p(x)}$, then there exists a
compact embedding

$W^{1,p(.)}(\Omega)\hookrightarrow\hookrightarrow
L^{\theta(.)}(\Omega)\hookrightarrow L^{\theta^{-}}(\Omega),$ (1.6)

where $\theta^{-}=\inf\limits_{x\in\overline{\Omega}}\theta(x)$.

###  1.3 Statement of main result

We use the following definition of the weak solutions.

######  Definition 1.

Let $f\in L^{1}(\Omega)$. A function $u\in W_{0}^{1,1}(\Omega)$ is a weak
solution to problem (1.1), if

$\displaystyle\forall\omega\subset\subset\Omega,\;\;\exists
c_{\omega}>0\quad\mbox{such that}\quad u\geq c_{\omega}\;\;\mbox{a.e.
in}\;\;\omega,\quad u^{r(x)}\in L^{1}(\Omega),$

and

$\int_{\Omega}a(x)|\nabla u|^{p(x)-2}\nabla u.\nabla\varphi
dx+\int_{\Omega}b(x)u^{r(x)}\varphi
dx=\int_{\Omega}\frac{f\varphi}{u^{\gamma(x)}}dx,$ (1.7)

for every $\varphi\in C^{1}_{0}(\Omega)$.

In this paper we will show the following result.

######  Theorem 1.

Suppose that assumptions (1.2)-(1.4) hold. Let $f\in L^{1}(\Omega)$ be a
positive function ($f(x)\geq 0$ and not zero a.e.). Assume that

$p(x)>1+\frac{1-\gamma(x)}{r(x)}.$ (1.8)

Then, the problem (1.1) has at least one weak solution $u\in
W_{0}^{1,q(.)}(\Omega)$, with

$q(x)=\frac{p(x)}{1+\frac{1-\gamma(x)}{r(x)}}.$ (1.9)

Moreover $u^{r(x)+\gamma(x)}$ belongs to $L^{1}(\Omega)$.

######  Remark 1.

$\bullet$

The assumption (1.4) implies $1<q(.)<p(.).$

$\bullet$

The assumption (1.3) implies $q(.)>p(.)-1.$

In order to prove this result, we will work by approximation, "truncating" the
singular term $\frac{1}{u^{\gamma(x)}}$ so that it becomes not singular at the
origin. We will get some a priori estimates on the solutions $u_{n}$ of the
approximating problems, which will allow us to pass to the limit and find a
solution to problem (1.1).

##  2 Approximating problems

Hereafter, we denote by $T_{k}$ the truncation function at the level $k>0$,
defined by $T_{k}(s)=\max\\{-k,\min\\{s,k\\}\\}$ for every $s\in\mathbb{R}$.

Let $(f_{n})$ ($f_{n}>0$) be a sequence of bounded functions defined in
$\Omega$ which converges to $f>0$ in $L^{1}(\Omega)$, and which verifies the
inequalities $f_{n}\leq n$ and $f_{n}\leq f$ for every $n\geq 1$ (for example
$f_{n}=T_{n}(f)$). Consider the following approximate equation

$\displaystyle\begin{cases}-\mbox{div}(a(x)|\nabla u_{n}|^{p(x)-2}\nabla
u_{n})+b(x)u_{n}|u_{n}|^{r(x)-1}=\frac{f_{n}}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}},&\mbox{in}\;\;\Omega,\\\
u_{n}=0,&\mbox{on}\;\;\partial\Omega.\end{cases}$ (2.1)

######  Theorem 2.

Let $f\in L^{1}(\Omega)$, and let
$r,p:\overline{\Omega}\rightarrow(1,+\infty)$,
$\gamma:\overline{\Omega}\rightarrow(0,1)$ are continuous functions. Assume
that (1.2) and (1.5) holds true. Then the problem (2.1) has a nonnegative
solution $u_{n}\in W_{0}^{1,p(.)}(\Omega)$.

######  Lemma 1.

[26] Suppose that the hypotheses of Theorem 2 are satisfied. Then there exists
at least one solution $u_{n}\in W_{0}^{1,p(.)}(\Omega)\cap L^{\infty}(\Omega)$
to the problem (2.1) in the sense that

$\displaystyle\int_{\Omega}a(x)|\nabla u_{n}|^{p(x)-2}\nabla
u_{n}.\nabla\varphi+\int_{\Omega}b(x)u_{n}|u_{n}|^{r(x)-1}\varphi=\int_{\Omega}\frac{f_{n}}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}}\varphi,$
(2.2)

for every $\varphi\in W_{0}^{1,p(.)}(\Omega)\cap L^{\infty}(\Omega)$.

###### Proof.

This proofs derived from Schauder-Tychonov fixed point Theorem (see, for
example, [24, p. 581], [25, p. 298]). Let $n$ in $\mathbb{N}$ be fixed, let
$v$ be a function in $L^{p(.)}(\Omega),$ we know that the following non
singular problem

$\left\\{\begin{array}[]{cl}-\operatorname{div}\left(a(x)|\nabla
w|^{p(x)-2}\nabla
w\right)+b(x)|w|^{r(x)-1}w=\frac{f_{n}}{\left(|v|+\frac{1}{n}\right)^{\gamma(x)}}&\text{
in }\Omega\\\ w=0&\text{ on }\partial\Omega.\end{array}\right.$ (2.3)

Therefore, the Minty-Browder Theorem (see, e.g. [10]) implies that problem
(2.3) has a unique solution $w\in W_{0}^{1,p(x)}(\Omega).$ Let us define a map

$G:L^{p(.)}(\Omega)\rightarrow L^{p(.)}(\Omega)$

and define $w=G(v)$ to be the unique solution of (2.3). Taking $w$ as test
function, we have

$\alpha\int_{\Omega}|\nabla w|^{p(x)}\leq\int_{\Omega}a(x)|\nabla
w|^{p(x)-2}\nabla w\cdot\nabla
w=\int_{\Omega}\frac{f_{n}w}{\left(|v|+\frac{1}{n}\right)^{\gamma(x)}}\leq
n^{\gamma_{+}+1}\int_{\Omega}|w|.$

By the Sobolev inequality on the left hand side and the Hölder inequality on
the right hand side one has

$\left[\int_{\Omega}|w|^{p^{*}(x)}\right]^{p(x)/p^{*}(x)}\leq
Cn^{\gamma_{+}+1}\left(\int_{\Omega}|w|^{p^{*}(x)}\right)^{\frac{1}{p^{*}(x)}}$

for some constant $C$ independent on $v.$ This implies

$\|w\|_{L^{p^{*}(.)}(\Omega)}\leq(Cn^{\gamma_{+}+1})^{\frac{1}{p(x)-1}}=C_{n}$

so that the ball of $L^{p^{*}(.)}(\Omega)$ of radius $C_{n}$ is invariant for
$G.$ It is easy to prove, using the Sobolev embedding, that $G$ is both
continuous and compact on $L^{p^{*}(.)}(\Omega),$ so that by Schauder’s fixed
point Theorem there exists $u_{n}$ in $W_{0}^{1,p(x)}(\Omega),$ for every
fixed $n$ such that $u_{n}=S\left(u_{n}\right),$ i.e., $u_{n}$ solves

$\left\\{\begin{array}[]{cl}-\operatorname{div}\left(a(x)\left|\nabla
u_{n}\right|^{p(x)-2}\nabla
u_{n}\right)+b(x)\left|u_{n}\right|^{r(x)-1}u_{n}=\frac{f_{n}}{\left(\left|u_{n}\right|+\frac{1}{n}\right)^{\gamma(x)}}&\text{
in }\Omega\\\ u_{n}=0&\text{ on }\partial\Omega\end{array}\right.$ (2.4)

Using as a test function $u_{n}^{-}=\min{\\{u_{n},0\\}}$ , one has $u_{n}\geq
0,$ Since the right hand side of (2.1) belongs to $L^{\infty}(\Omega)$ and we
proceed in the same way as [30] we obtain $u_{n}$ belongs to
$L^{\infty}(\Omega).$ (although its norm in $L^{\infty}(\Omega)$ may depend on
$n$). ∎

######  Lemma 2.

Suppose that the hypotheses of Theorem 2 are satisfied. Then the sequence
$u_{n}$ is increasing with respect to $n$, $u_{n}>0$ in $\Omega$, and for
every $\omega\subset\subset\Omega$ there exists $c_{\omega}>0$ (independent on
$n$) such that

$u_{n}(x)\geq c_{\omega}>0,\quad\forall x\in\Omega,\quad\forall
n\in\mathbb{N}.$ (2.5)

Moreover there exists the pointwise limit $u\geq c_{\omega}$ of the sequence
$u_{n}$.

###### Proof of the Lemma 2.

Due to $0\leq f_{n}\leq f_{n+1}$ and $\gamma(x)>0$, we have that

$\displaystyle-\mbox{div}(a(x)|\nabla u_{n}|^{p(x)-2}\nabla
u_{n})+b(x)u_{n}^{r(x)}=\frac{f_{n}}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}}\quad\leq\frac{f_{n+1}}{\left(u_{n}+\frac{1}{n+1}\right)^{\gamma(x)}}.$

So that

$\displaystyle-\mbox{div}(a(x)|\nabla u_{n}|^{p(x)-2}\nabla
u_{n})+\mbox{div}(a(x)|\nabla u_{n+1}|^{p(x)-2}\nabla
u_{n+1})+b(x)u_{n}^{r(x)}-b(x)u_{n+1}^{r(x)}$ $\displaystyle\qquad\leq
f_{n+1}\Bigg{[}\frac{\left(u_{n+1}+\frac{1}{n+1}\right)^{\gamma(x)}-\left(u_{n}+\frac{1}{n+1}\right)^{\gamma(x)}}{\left(u_{n}+\frac{1}{n+1}\right)^{\gamma(x)}\left(u_{n+1}+\frac{1}{n+1}\right)^{\gamma(x)}}\Bigg{]}.$
(2.6)

We now choose $(u_{n}-u_{n+1})_{+}=\max\\{u_{n}-u_{n+1},0\\}$ as test function
in (2). In the left hand side we use (1.5) and the monotonicity of the
$p(x)-$laplacian operator as well as the monotonicity of the function
$t\rightarrow|t|^{r(x)-1}t$. For the right hand, using the fact that
$\gamma(x)\geq 0$ and $f_{n+1}\geq 0$, we have

$\displaystyle\Bigg{[}\left(u_{n+1}+\frac{1}{n+1}\right)^{\gamma(x)}-\left(u_{n}+\frac{1}{n+1}\right)^{\gamma(x)}\Bigg{]}(u_{n}-u_{n+1})_{+}\leq
0.$ (2.7)

By (2.7), we get

$\alpha\int_{\Omega}|\nabla(u_{n}-u_{n+1})_{+}|^{p(x)}\leq 0,$

which implies that $(u_{n}-u_{n+1})_{+}=0$ a.e. in $\Omega$, that is,
$u_{n}\leq u_{n+1}$ for every $n\in\mathbb{N}$. Since the sequence $(u_{n})$
is increasing with respect to $n$, we only need to prove that (2.5) holds for
$u_{1}$. Using Lemma 1, we know that $u_{1}\in L^{\infty}(\Omega)$, i.e.,
there exists a constant $c_{0}$ (depending only on $\Omega$ and $N$) such that
$\|u_{1}\|_{L^{\infty}(\Omega)}\leq c\|f_{1}\|_{L^{\infty}}(\Omega)\leq
c_{0}$, then

$\displaystyle-\mbox{div}(a(x)|\nabla u_{1}|^{p(x)-2}\nabla
u_{1})+b(x)u_{1}^{r(x)}=\frac{f_{1}}{\left(u_{1}+1\right)^{\gamma(x)}}\geq\frac{f_{1}}{(c_{0}+1)^{\gamma(x)}}\geq
0.$

Since $\frac{f_{1}}{(c_{0}+1)^{\gamma(x)}}$ is not identically zero, the
strong maximum principle implies that $u_{1}>0$ in $\Omega$ (see [31]). Since
$u_{n}\geq u_{1}$ for every $n\in\mathbb{N}$, (2.5) holds for $u_{n}$ (with
the same constant $c_{\omega}$ which is then independent on $n$). ∎

###### Proof of the Theorem 2.

In virtue of the Lemma 1 and Lemma 2, there exists at least one nonnegative
weak solution $u_{n}\in W_{0}^{1,p(.)}(\Omega)\cap L^{\infty}(\Omega)$ of
problem (2.1). ∎

##  3 A priori estimates

In the remainder of this section, we denote by $C_{i}$ $i=1,2,3,...$ various
positive constants depending only on the data of the problem, but not on $n$.

######  Lemma 3.

Let $k>0$ be fixed. The sequence $(T_{k}(u_{n}))$, where $u_{n}$ is a solution
to (2.4), is bounded in $W_{0}^{1,p(.)}(\Omega)$.

###### Proof.

Taking $T_{k}(u_{n})$ as a test function in (2.4), we obtain

$\displaystyle\int_{\Omega}a(x)|\nabla u_{n}|^{p(x)-2}\nabla u_{n}.\nabla
T_{k}(u_{n})+\int_{\Omega}b(x)u_{n}^{r(x)}T_{k}(u_{n})=\int_{\Omega}\frac{f_{n}}{\left(|u_{n}|+\frac{1}{n}\right)^{\gamma(x)}}T_{k}(u_{n}).$

Using (1.5), $f_{n}\leq f$, $T_{k}(u_{n})\neq 0$, and dropping the nonegative
order term, we get

$\int_{\Omega}|\nabla
T_{k}(u_{n})|^{p(x)}dx\leq\frac{k}{\alpha}\|f\|_{L^{1}(\Omega)}.$ (3.1)

As a consequence of Proposition 1 and (3.1), $T_{k}(u_{n})$ is bounded in
$W_{0}^{1,p(.)}(\Omega)$ ∎

######  Lemma 4.

Suppose that the hypotheses of Theorem 1 are satisfied. Then, the sequence
$u_{n}$ is bounded in $W_{0}^{1,q(.)}(\Omega)$, where $q(.)$ is given by
(1.9). Moreover $(u_{n}^{r(x)+\gamma(x)})$ belongs to $L^{1}(\Omega)$.

###### Proof.

Taking $\varphi(x,u)=(u_{n}+1)^{\gamma(x)}-1$, as test function in (2.4), by
(1.4), (1.5), and the fact that for a.e. $x\in\Omega$

$\nabla\varphi(x,u)=\nabla\gamma(x)(u_{n}+1)^{\gamma(x)}\ln(u_{n}+1)+\gamma(x)\frac{\nabla
u_{n}}{(u_{n}+1)^{\gamma(x)}},$

we obtain

$\displaystyle\gamma^{-}\alpha\int_{\Omega}\frac{|\nabla
u_{n}|^{p(x)}}{(1+u_{n})^{1-\gamma(x)}}+\mu\int_{\Omega}u_{n}^{r(x)}\big{[}(u_{n}+1)^{\gamma(x)}-1\big{]}$
$\displaystyle\qquad\leq C_{1}\int_{\Omega}|\nabla
u_{n}|^{p(x)-1}(u_{n}+1)^{\gamma(x)}\ln(u_{n}+1)+\int_{\Omega}f\frac{\big{[}(u_{n}+1)^{\gamma(x)}-1\big{]}}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}}.$

Using the fact that $|u_{n}|^{\theta(x)}\geq
2^{1-\theta^{+}}(1+u_{n})^{\theta(x)}-1$ (here $\theta(x)=r(x)$ and
$\theta(x)=\gamma(x)$), we have

$\displaystyle\gamma^{-}\alpha\int_{\Omega}\frac{|\nabla
u_{n}|^{p(x)}}{(1+u_{n})^{1-\gamma(x)}}+2^{1-r^{+}}\mu\int_{\Omega}(u_{n}+1)^{r(x)+\gamma(x)}$
$\displaystyle\qquad\leq
C_{2}+\frac{1}{2^{1-\gamma^{+}}}\int_{\Omega}f+C_{1}\int_{\Omega}|\nabla
u_{n}|^{p(x)-1}(u_{n}+1)^{\gamma(x)}\ln(u_{n}+1).$ (3.2)

We can estimate the last term in (3) by application of Young’s inequality

$\displaystyle(1+u_{n})^{\gamma(x)}\ln(1+u_{n})|u_{n}|^{p(x)-1}$
$\displaystyle\qquad=(1+u_{n})^{1-\frac{1-\gamma(x)}{p(x)}}\ln(1+u_{n})|u_{n}|^{p(x)-1}(1+u_{n})^{-\frac{(1-\gamma(x))(p(x)-1)}{p(x)}}$
$\displaystyle\qquad\leq
C_{3}(1+u_{n})^{p(x)-(1-\gamma(x))}(\ln(1+u_{n}))^{p(x)}+\varepsilon\frac{|\nabla
u_{n}|^{p(x)}}{(u_{n}+1)^{1-\gamma(x)}}.$ (3.3)

We choose $\varepsilon=\frac{\gamma^{-}\alpha}{2C_{1}}$. By (3) and (3) we
obtain

$\displaystyle\frac{1}{2}\int_{\Omega}\frac{|\nabla
u_{n}|^{p(x)}}{(1+u_{n})^{1-\gamma(x)}}+2^{1-r^{+}}\mu\int_{\Omega}(u_{n}+1)^{r(x)+\gamma(x)}$
$\displaystyle\qquad\leq
C_{4}+C_{5}\int_{\Omega}(u_{n}+1)^{p(x)-(1-\gamma(x))}(\ln(u_{n}+1))^{p(x)}.$
(3.4)

The hypothesis (1.3) implies $(1+t)^{p(x)-1-r(x)-c}(\ln(1+t))^{p(x)}$ is
bounded for all $x\in\overline{\Omega}$ and $t\in\mathbb{R}^{+}$. By another
application of Youngs inequality, we have

$\displaystyle(u_{n}+1)^{p(x)-(1-\gamma(x))}(\ln(u_{n}+1))^{p(x)}$
$\displaystyle\quad=(u_{n}+1)^{r(x)+\gamma(x)+c}(u_{n}+1)^{p(x)-1-r(x)-c}(\ln(u_{n}+1))^{p(x)}$
$\displaystyle\quad\leq\varepsilon(u_{n}+1)^{r(x)+\gamma(x)}+C_{6}.$ (3.5)

Therefore, by (3), (3), we get

$\displaystyle\int_{\Omega}\frac{|\nabla
u_{n}|^{p(x)}}{(1+u_{n})^{1-\gamma(x)}}+\int_{\Omega}(u_{n}+1)^{r(x)+\gamma(x)}$
$\displaystyle\leq C_{7}.$ (3.6)

Since $r(x)\geq 0$ and $\gamma(x)\geq 0$, we have

$\displaystyle\int_{\Omega}u_{n}^{r(x)}\leq\int_{\Omega}(u_{n}+1)^{r(x)}\leq\int_{\Omega}(u_{n}+1)^{r(x)+\gamma(x)}$
$\displaystyle\leq C_{7}.$ (3.7)

The inequality (3.7) implies that $(u_{n}^{r(x)+\gamma(x)})$ is bounded in
$L^{1}(\Omega)$. Let $q(x)<p(x)$, using Young’s inequality and (3.6), we have

$\displaystyle\int_{\Omega}|\nabla u_{n}|^{q(x)}$
$\displaystyle=\int_{\Omega}\frac{|\nabla
u_{n}|^{q(x)}}{(u_{n}+1)^{(1-\gamma(x))\frac{q(x)}{p(x)}}}$ $\displaystyle\leq
C_{8}\int_{\Omega}\frac{|\nabla
u_{n}|^{p(x)}}{(u_{n}+1)^{1-\gamma(x)}}+C_{9}\int_{\Omega}(u_{n}+1)^{(1-\gamma(x))\frac{q(x)}{p(x)-q(x)}}$
$\displaystyle\leq
C_{10}+C_{9}\int_{\Omega}(u_{n}+1)^{(1-\gamma(x))\frac{q(x)}{p(x)-q(x)}}.$
(3.8)

Set

$(1-\gamma(x))\frac{q(x)}{p(x)-q(x)}=r(x).$

Then this equality and (3.7)-(3) yield

$\int_{\Omega}|\nabla u_{n}|^{q(x)}\leq C_{11}.$ (3.9)

∎

######  Lemma 5.

Let $u_{n}$ be a solution to problem (2.4). Then

$\int_{\\{u_{n}>k\\}}u_{n}^{r(x)}\leq\frac{1}{\mu
k^{\gamma^{+}}}\int_{\\{u_{n}>k\\}}f,\quad\forall k>0,$
$\lim\limits_{|E|\rightarrow 0}\int_{E}u_{n}^{r(x)}=0,\quad\mbox{uniformly
with respect to}\;n,$

for every measurable subset $E$ in $\Omega$.

###### Proof.

Let $k>0$ and $\psi_{j}$ be a sequence of increasing, positive, uniformly
bounded $C^{\infty}(\Omega)$ functions, such that
$\psi_{j}(s)\rightarrow\chi_{\\{s>k\\}}$, as $j\rightarrow+\infty$. Choosing
$\psi_{j}(u_{n})$ in (2.4), using (1.5), we get

$\mu\int_{\Omega}u_{n}^{r(x)}\psi_{j}(u_{n})\leq\int_{\Omega}\frac{f_{n}}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}}\psi_{j}(u_{n}).$

Therefore, as $j$ tends to infinity and that
$k^{\gamma^{-}}\leq\left(k+\frac{1}{n}\right)^{\gamma^{-}}\leq\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}$
in the set $\\{u_{n}>k\\}$, we obtain

$\int_{\\{u_{n}>k\\}}u_{n}^{r(x)}\leq\frac{1}{\mu
k^{\gamma^{-}}}\int_{\\{u_{n}>k\\}}f.$ (3.10)

By (3.10), for any measurable subset $E$ in $\Omega$, we have

$\displaystyle\int_{E}u_{n}^{r(x)}$ $\displaystyle=\int_{E\cap\\{u_{n}\leq
k\\}}u_{n}^{r(x)}+\int_{E\cap\\{u_{n}>k\\}}u_{n}^{r(x)}\leq
k^{r^{+}}|E|+\frac{1}{\mu k^{\gamma^{-}}}\int_{\\{u_{n}>k\\}}f.$ (3.11)

Since $f\in L^{1}(\Omega)$, we may choose $k=k_{\varepsilon}$ large enough
such that

$\int_{\\{u_{n}>k\\}}f\leq\varepsilon.$ (3.12)

Therefore, the estimates (3.11)-(3.12) implies that

$\int_{E}u_{n}^{r(x)}\leq k_{\varepsilon}^{r^{+}}|E|+\frac{\varepsilon}{\mu
k_{\varepsilon}^{\gamma^{-}}},$

and the statement of this lemma is thus proved. ∎

######  Lemma 6.

Let $u_{n}$ be a solution to problem (2.4). Then

$\lim\limits_{|E|\rightarrow 0}\int_{E}|\nabla
u_{n}|^{q(x)}=0,\quad\mbox{uniformly with respect to}\;n,$ (3.13)

for every measurable subset $E$ in $\Omega$ and $q(.)$ given by (1.9).

###### Proof.

Let $\varepsilon>$, by Lemma 4, we may choose $k=k_{\varepsilon}$ large enough
such that

$\int_{E\cap\\{u_{n}>k\\}}|\nabla u_{n}|^{q(x)}\leq\varepsilon.$ (3.14)

From the estimate (3.1) and that $q(x)<p(x)$, we deduce

$\int_{E\cap\\{u_{n}\leq k\\}}|\nabla T_{k}(u_{n})|^{q(x)}\leq\varepsilon.$
(3.15)

By (3.14) and (3.15), for any measurable subset $E$ in $\Omega$, we have

$\displaystyle\int_{E}|\nabla u_{n}|^{q(x)}$
$\displaystyle=\int_{E\cap\\{u_{n}\leq k\\}}|\nabla
u_{n}|^{q(x)}+\int_{E\cap\\{u_{n}>k\\}}|\nabla u_{n}|^{q(x)}\leq
2\varepsilon.$

Therefore, we deduce that $|\nabla u_{n}|^{q(x)}$ is equiintegrable in
$L^{1}(\Omega)$. Thus (3.13) is proved. ∎

##  4 Proof of the main theorem

By Lemma 3, the sequence $(u_{n})_{n}$ is bounded in $W_{0}^{1,q(.)}(\Omega)$.
Therefore, there exists a function $u\in W_{0}^{1,q(.)}(\Omega)$ such that (up
to a subsequence)

$\left\\{\begin{array}[]{l}u_{n}\rightharpoonup
u\quad\text{in}\;W_{0}^{1,q(.)}(\Omega),\\\ u_{n}\rightarrow u\quad\text{a.e.
in}\;\Omega.\end{array}\right.$ (4.1)

######  Proposition 2.

If the sequence $T_{k}(u_{n})$ of the truncates of the solutions $u_{n}$ of
(2.4) is bounded in $W_{0}^{1,p(.)}(\Omega)$. Then

$T_{k}(u_{n})\rightarrow T_{k}(u)\quad\mbox{strongly in
}\;W_{0}^{1,p(.)}(\Omega),$ (4.2)

as $n\rightarrow\infty$, for every $k>0$. In particular $\nabla
u_{n}\rightarrow\nabla u$ a.e. in $\Omega$.

###### Proof.

By Lemma 3 $T_{k}(u_{n})$ is bounded in $W_{loc}^{1,p(.)}(\Omega)$, it weakly
converges in this space to its pointwise limit $T_{k}(u)$. Moreover, since
$f_{n}\geq 0$ and $u_{n}\geq 0$ a.e., we have that

$-\mbox{div}(a(x)|\nabla u_{n}|^{p(x)-2}\nabla u_{n})+b(x)u_{n}^{r(x)}\geq 0,$

for all $n\in\mathbb{N}$ and $k>0$.

Now we fix $\phi\in C_{0}^{1}(\Omega)$ such that $0\leq\phi\leq 1$ on $\Omega$
and such that $\phi\equiv 1$ on a fixed subset $\omega$ of $\Omega$. Then,
thanks to the monotonicity of the $p(x)-$laplacian operator, (1.5), and that
$T_{k}(u_{n})\geq T_{k}(u)$ (since $u_{n}\rightarrow u\leq u_{n}$), we can
conclude that the following hold

$\displaystyle 0<\beta\int_{\omega}\left(|\nabla T_{k}(u_{n})|^{p(x)-2}\nabla
T_{k}(u_{n})-|\nabla T_{k}(u)|^{p(x)-2}\nabla
T_{k}(u)\right).\nabla(T_{k}(u_{n})-T_{k}(u))$
$\displaystyle\qquad+\nu\int_{\omega}u_{n}^{r(x)}(T_{k}(u_{n})-T_{k}(u))$
$\displaystyle=\beta\int_{\Omega}\left(|\nabla T_{k}(u_{n})|^{p(x)-2}\nabla
T_{k}(u_{n})-|\nabla T_{k}(u)|^{p(x)-2}\nabla
T_{k}(u)\right).\nabla(T_{k}(u_{n})-T_{k}(u))\phi$
$\displaystyle\qquad+\nu\int_{\Omega}u_{n}^{r(x)}(T_{k}(u_{n})-T_{k}(u))\phi$
$\displaystyle=\beta\int_{\Omega}|\nabla T_{k}(u_{n})\nabla
T_{k}(u_{n})|^{p(x)-2}\nabla[(T_{k}(u_{n})-T_{k}(u))\phi]$
$\displaystyle\qquad-\beta\int_{\Omega}|\nabla T_{k}(u_{n})|^{p(x)-2}\nabla
T_{k}(u_{n}).\nabla\phi[T_{k}(u_{n})-T_{k}(u)]$
$\displaystyle\qquad-\beta\int_{\Omega}|\nabla T_{k}(u)|^{p(x)-2}\nabla
T_{k}(u).\nabla(T_{k}(u_{n})-T_{k}(u))\phi$
$\displaystyle\qquad+\nu\int_{\Omega}u_{n}^{r(x)}(T_{k}(u_{n})-T_{k}(u))\phi$
(4.3)

By Lemma 5, we obtain

$u_{n}^{r(x)}\rightarrow u^{r(x)}\quad\mbox{strongly in }\;L^{1}(\Omega).$

Therefore, since $T_{k}(u_{n})$ strongly converges to $T_{k}(u)$ in
$L^{p(.)}(\Omega)$ (Lemma 3), we have

$\int_{\Omega}u_{n}^{r(x)}(T_{k}(u_{n})-T_{k}(u))\phi\rightarrow
0,\quad\mbox{as}\;\;n\rightarrow\infty.$ (4.4)

We can easily know the fact that $|\nabla T_{k}(u)|^{p(x)-2}\nabla T_{k}(u)\in
L_{loc}^{p^{\prime}(.)}(\Omega)$, and $\nabla(T_{k}(u_{n})-T_{k}(u))\phi$
tends to zero weakly in $L^{p}(\Omega)$, we get

$\int_{\Omega}|\nabla T_{k}(u)|^{p(x)-2}\nabla
T_{k}(u).\nabla(T_{k}(u_{n})-T_{k}(u))\phi\rightarrow
0,\quad\mbox{as}\;\;n\rightarrow\infty.$ (4.5)

$\nabla\phi[T_{k}(u_{n})-T_{k}(u)]$ strongly converges to zero in
$L^{p(.)}(\Omega)$. Thus

$\int_{\Omega}|\nabla T_{k}(u_{n})|^{p(x)-2}\nabla
T_{k}(u_{n}).\nabla\phi[T_{k}(u_{n})-T_{k}(u)]\rightarrow
0,\quad\mbox{as}\;\;n\rightarrow\infty.$ (4.6)

From (4)-(4.6), we have

$\int_{\omega}\left(|\nabla T_{k}(u_{n})|^{p(x)-2}\nabla T_{k}(u_{n})-|\nabla
T_{k}(u)|^{p(x)-2}\nabla
T_{k}(u)\right).\nabla(T_{k}(u_{n})-T_{k}(u))\rightarrow 0,$

then $T_{k}(u_{n})$ strongly converges to $T_{k}(u)$ in
$W_{0}^{1,p(.)}(\omega)$ for all $k>0$, i.e., since $\omega$ is arbitrary,
that $T_{k}(u_{n})$ strongly converges to $T_{k}(u)$
in$W_{loc}^{1,p(.)}(\Omega)$.

Choosing $\phi\equiv 1$ and repeating the same proof, we obtain that
$T_{k}(u_{n})$ strongly converges to $T_{k}(u)$ in $W_{0}^{1,p(.)}(\Omega)$,
then $\nabla u_{n}\rightarrow\nabla u$ a.e. in $\Omega$. ∎

###### Proof of the Theorem 1.

It is easy to pass to the limit in the right hand side of problems (2.4). On
the other hand, using Lemma 2, we have

$0\leq\bigg{|}\frac{f_{n}\varphi}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}}\bigg{|}\leq\frac{\|\varphi\|_{\infty}}{c_{\omega}^{\gamma^{-}}}f,$

for every $\varphi\in C_{0}^{1}(\Omega)$, using Lebesgue Theorem and (4.1), we
conclude that

$\lim\limits_{n\rightarrow\infty}\int_{\Omega}\frac{f_{n}\varphi}{\left(u_{n}+\frac{1}{n}\right)^{\gamma(x)}}=\int_{\Omega}\frac{f\varphi}{u^{\gamma(x)}}.$
(4.7)

By the same argument, we get

$\lim\limits_{n\rightarrow\infty}\int_{\Omega}b(x)u_{n}^{r(x)}\varphi=\int_{\Omega}u^{r(x)}\varphi.$
(4.8)

For the first term, by Proposition 2 we have that

$a(x)|\nabla u_{n}|^{p(x)-2}\nabla u_{n}\rightarrow a(x)|\nabla
u|^{p(x)-2}\nabla u\quad\mbox{a.e. in}\;\Omega,$

furthermore $a(x)|\nabla u_{n}|^{p(x)-2}\nabla u_{n}$ is majorette by
$\beta|\nabla u_{n}|^{p(x)-1}$. Observe that $p(x)-1<q(x)$, by Lemma 6 and
Vitali’s Theorem, we have

$\lim\limits_{n\rightarrow\infty}a(x)|\nabla u_{n}|^{p(x)-2}\nabla
u_{n}.\nabla\varphi=\int_{\Omega}a(x)|\nabla u|^{p(x)-2}\nabla
u.\nabla\varphi.$ (4.9)

Hence from (4.7)-(4.8) we can deduce (1.7). ∎

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