Existence of solutions for a class of porous medium type equations with lower order terms

Abstract

This paper deals with a class of degenerate quasilinear elliptic equations of the form $-\operatorname{div}(a(x,u,\nabla u))+F(x,u,\nabla u)=f$, where $a(x,u,\nabla u)$ is allowed to degenerate with the unknown u. Under some hypothesis on a, F, and f, we obtain the existence of bounded solutions $u\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$. For the case $f\in L^{1}(\Omega)$, we also prove that there exists at least one renormalized solution.

1 Introduction

This paper concerns the following degenerate problem:

$$(\mathscr{P})\quad \left \{ \textstyle\begin{array}{l@{\quad}l} -\operatorname{div}(a(x,u,\nabla u))+F(x,u,\nabla u)=f &\mbox{in } \Omega, \\ u=0 &\mbox{on } \partial\Omega, \end{array}\displaystyle \right . $$

where Ω is a bounded domain of $\mathbb{R}^{N}$ ($N\geq2$), $f\in L^{q}(\Omega)$ with $q\geq1$ and $a(x,s,\xi)$ is a Carathéodory function. Furthermore, we assume that there exists a continuous function α from $\mathbb{R}^{+}$ into $\mathbb{R}^{+}$ such that $\alpha(0)=0$ and $a(x,s,\xi)\xi\geq\alpha(|s|)|\xi|^{p}$ for any $s\in\mathbb{R}$, $\xi \in \mathbb{R}^{N}$, and almost every x in Ω. Thus problem ($\mathscr{P}$) degenerates for the subset $\{x\in\Omega:u(x)=0\}$.

Problem ($\mathscr{P}$) has important and extensive applications to the fluid dynamics in porous media, in hydrology and in petroleum engineering (see [1, 2]). The simplest model is the stationary case of the porous media equation with zero Dirichlet boundary condition:

$$(\mathrm{P}_{0})\quad \left \{ \textstyle\begin{array}{l@{\quad}l} -\triangle(|u|^{m-1}u)+F(x,u)=f &\mbox{in } \Omega, \\ u=0 &\mbox{on } \partial\Omega, \end{array}\displaystyle \right . $$

which has been widely studied in the literature (see [3–6] and references therein).

For the case $\alpha\equiv\mathrm{constant}>0$, the existence of bounded solutions to problem ($\mathscr{P}$) is proved in [7], when the data f is small in a suitable norm.

Concerning the case that α is a positive function, Porretta and Segura de León investigated the existence results to problem ($\mathscr{P}$); see [8]. We remark that in [8], no sign condition is imposed on F, but the growth of F at infinity need to be controlled. We also point out that a variational inequality related to problem ($\mathscr{P}$) was studied in [9], and similar results can be found in [10] and [11].

In the case $\alpha(0)=0$, $f\in W^{-1,r}(\Omega)\cap L^{1}(\Omega)$ with $r\geq p'$, $r>\frac{N}{p-1}$, Rakotoson proved the existence of a bounded weak solution to problem ($\mathscr{P}$) (see [12]), provided that F satisfies a sign condition. As $F=0$ and $f\in W^{-1,r}(\Omega)$, the existence of solutions to problem ($\mathscr{P}$) has been discussed in [13]. We point out that the parabolic version of [13] has been studied in [14].

As $f\in L^{q}(\Omega)$ with $q\geq\max\{1,\frac{N}{p}\}$, we shall give a direct method to prove the existence of bounded weak solutions to problem ($\mathscr{P}$) in the standard sense, i.e. $u\in W_{0}^{1,p}(\Omega)$. The main difficulty comes from the facts that its modulus of ellipticity vanishes when the solution u vanishes. To overcome this difficulty, we shall firstly establish the $L^{\infty}$ estimate for solution u, by the technique of rearrangement which is differs from the usual Stampacchia $L^{\infty}$ regularity procedure. Then, by constructing suitable approximate problems, and using a priori estimates and a test function method, we shall finish the proof of this existence results.

Furthermore, we will study the case when $f\in L^{1}(\Omega)$. Since no growth conditions are required for ω and β (see (H₂)), it is not obvious that the term $-\operatorname{div}(a(x, u, \nabla u))$ makes sense even as a distribution. To overcome this difficulty, we shall use the concept of renormalized solutions, which is introduced by Diperna and Lions (see [15]). This notion was adapted by many authors to study partial differential equations with measurable data, especially for $L^{1}$ data (see [16–18] for example). We remark that an equivalent notion called entropy solutions, was introduced independently by Bénilan et al. [19].

The main ideas and methods come from [8, 10, 12, 20]. This paper is organized as follows: in Section 2 we give some preliminaries and state the main results; in Section 3, we study the existence of bounded solution to problem ($\mathscr{P}$); in Section 4, we prove the existence of renormalized solution.

2 Some preliminaries and the main results

2.1 Properties of the relative rearrangement

Let Ω be a bounded open subsets of $\mathbb{R}^{N}$, we denote by $|E|$ the Lebesgue measure of a set E. Assume that $u: \Omega \rightarrow\mathbb{R}$ be a measurable function, we define the distribution function $\mu_{u}(t)$ of u as follows:

$$\mu_{u}(t)=\bigl\vert \bigl\{ x\in\Omega: u(x)>t\bigr\} \bigr\vert ,\quad \forall t\in\mathbb{R}. $$

The decreasing rearrangement $u_{\ast}$ of u is defined as the generalized inverse function of $\mu_{u}(t)$, i.e.

$$u_{\ast}(s)=\inf\bigl\{ t\in R: \mu_{u}(t)\leq s\bigr\} , \quad s\in\Omega^{\ast }=\bigl[0,\vert \Omega \vert \bigr]. $$

We recall also that u and $u_{\ast}$ are equi-measurable, i.e.

$$\mu_{u}(t)=\mu_{u_{\ast}}(t),\quad t\in\mathbb{R}, $$

which implies that for any non-negative Borel function ψ we have

$$ \int_{\Omega}\psi\bigl(u(x)\bigr)\, \mathrm{d}x=\int _{0}^{|\Omega|}\psi\bigl(u_{\ast }(s)\bigr)\, \mathrm{d}s, $$

and if $E\subset\Omega$ be a measurable subset, then

$$ \int_{E}u(x)\, dx\leq\int_{0}^{|E|}u_{\ast}(s) \, ds. $$

Using the Fleming-Rishel formula, Hölder’s inequality, and the isoperimetric inequality, we can get the following result (see [7, 9, 12]).

Lemma 2.1

For any non-negative function $u\in W_{0}^{1,1}(\Omega)$, the following chain of inequalities holds:

$$ NC_{N}^{1/N}\mu_{u}(t)^{1-1/N}\leq- \frac{d}{dt}\int_{u>t}|\nabla u|\, \mathrm{d}x\leq \bigl(-\mu_{u}'(t)\bigr)^{1/p'}\biggl(- \frac{d}{dt}\int_{u>t}|\nabla u|^{p}\, \mathrm{d}x\biggr)^{1/p}, $$

where $C_{N}$ denotes the measure of the unit ball in $\mathbb{R}^{N}$.

For more details as regards the theory of rearrangement, we just refer to [21] and the references therein.

2.2 Assumptions and the main results

Let Ω be an open bounded set of $\mathbb{R}^{N}$ ($N\geq2$) and $p>1$, we make the following assumptions.

(H₁):

$a: \Omega\times\mathbb{R}\times\mathbb{R}^{N}\rightarrow \mathbb{R}^{N}$ is a Carathéodory vector function satisfying: there exists a continuous function α from $\mathbb{R}_{+}$ into $\mathbb {R}_{+}$ such that $\alpha(0)=0$ and $\alpha(s)>0$ if $s>0$ and

$$\begin{aligned}& a(x,s,\xi)\xi\geq\alpha\bigl(\vert s\vert \bigr)|\xi|^{p},\quad \forall s\in R, \mbox{a.e. } x\in \Omega, \forall\xi\in\mathbb{R}^{N}, \\& \int_{0}^{+\infty}\alpha^{\frac{1}{p-1}}(s)\, \mathrm{d}s=\int_{0}^{+\infty }\frac{1}{\alpha(s)}\, \mathrm{d}s =+\infty \end{aligned}$$

and

$$ \frac{1}{\alpha}\in L^{1}(0,b) \quad \mbox{for any given }b>0. $$

(H₂):

There exists a Carathéodory vector function ā such that for a.e. $x\in\Omega$, $\forall s\in\mathbb{R}$, $\forall\xi, \xi'\in \mathbb{R}^{N}$ with $\xi\neq\xi'$:

• (i) $a(x,s,\xi)=\alpha(|s|)\bar{a}(x,s,\xi)$.

• (ii) $[\bar{a}(x,s,\xi)-\bar{a}(x,s,\xi')][\xi-\xi']>0$.

• (iii) There exist an increasing function ω from $\mathbb{R}^{+}$ into $\mathbb{R}^{+}$ and a non-negative function $\bar{\omega}\in L^{p'}(\Omega)$ such that

  $$\bigl\vert \bar{a}(x,s,\xi)\bigr\vert \leq\omega\bigl(\vert s\vert \bigr)\bigl[|\xi|^{p-1}+\bar{\omega}(x)\bigr]. $$

• (iv) The function ā is a positively homogeneous of degree $(p-1)$ with respect to the variable ξ, i.e.

  $$\bar{a}(x,s,t\xi)=t^{p-1}\bar{a}(x,s,\xi),\quad \forall t\geq 0. $$

(H₃):

$F:\Omega\times\mathbb{R}\times\mathbb{R}^{N}\rightarrow \mathbb{R}$ is a Carathéodory function, for which there exists an increasing function β from $[0,+\infty)$ into $[0,+\infty)$ vanishing and continuous at zero such that for a.e. $x\in\Omega$, $\forall s\in\mathbb{R}$ and $\forall\xi\in\mathbb {R}^{N}$:

$$\bigl\vert F(x,s,\xi)\bigr\vert \leq\beta\bigl(\vert s\vert \bigr)| \xi|^{p}. $$

(H₄):

$f\in L^{q}(\Omega)$ with $q> \max\{1,\frac{N}{p}\}$.

(H₅):

$\lim_{s\rightarrow \infty}\frac{e^{\gamma(|s|)}}{(1+\phi(|s|))^{p-1}}=0$, where γ and ϕ are defined as follows:

$$ \gamma(s)=\int_{0}^{s} \frac{\beta(|\sigma|)}{\alpha(|\sigma|)}\, \mathrm {d}\sigma;\qquad \phi(s)=\int_{0}^{s} \bigl(\alpha\bigl(|\sigma|\bigr)\bigr)^{\frac{1}{p-1}}e^{\frac{\gamma (|s|)}{p-1}}\, \mathrm{d}\sigma. $$

(2.1)

Remark 2.1

Assumption (H₁) allows us to consider the porous medium operators $\triangle(|u|^{m-1}u)=\operatorname {div}(m|u|^{m-1}\nabla u)$. In this case, it yields $\alpha (|s|)=|s|^{m-1}$, so that the conditions $\alpha(0)=0$ and $\frac {1}{\alpha}\in L^{1}(0,b)$ indicate $1< m<2$. Thus, in this case, the porous medium equation becomes a slow diffusion equation.

We now introduce several auxiliary functions by

$$\begin{aligned}& \tilde{\alpha}(s)=\int_{0}^{s} \alpha^{\frac{1}{p-1}}\bigl(\vert t\vert \bigr)\,\mathrm{d}t, \end{aligned}$$

(2.2)

$$\begin{aligned}& \gamma_{\theta}(s)=\int_{0}^{s} \frac{\beta (|\sigma|)}{\alpha(|\sigma|)+\theta}\,\mathrm{d}\sigma \quad \mbox{for any fixed } \theta>0, \end{aligned}$$

(2.3)

$$\begin{aligned}& \tilde{\gamma}_{\theta}(s)=\int_{0}^{s} \frac {\beta(|g(t)|)}{\alpha(|g(t)|)+\theta}\,\mathrm{d}t \quad \mbox{and}\quad \tilde {\gamma}(s)=\int _{0}^{s}\frac{\beta(|g(t)|)}{\alpha(|g(t)|)}\,\mathrm{d}t. \end{aligned}$$

(2.4)

As usual, the usual truncation function $T_{\theta}$ at level ±θ is defined as $T_{\theta}(s)=\max\{-\theta, \min\{\theta,s\} \}$. Throughout this paper, we use $C(\theta_{1},\theta_{2},\ldots,\theta _{m})$ to denote positive constants depending only on specified quantities $\theta_{1}, \theta_{2},\ldots, \theta_{m}$.

Now we give the definition of weak solutions of problem ($\mathscr{P}$).

Definition 2.1

A measurable function $u\in W_{0}^{1,p}(\Omega)$ is called a weak solution to problem ($\mathscr {P}$), if $a(\cdot, u, \nabla u)\in L^{p'}(\Omega)$ and $F(\cdot, u, \nabla u)\in L^{1}(\Omega)$ such that

$$ \int_{\Omega}a(x,u,\nabla u)\nabla v\,\mathrm{d}x +\int_{\Omega}F(x,u,\nabla u)v\,\mathrm{d}x =\int _{\Omega}f v\,\mathrm{d}x,\quad \forall v \in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega). $$

(2.5)

For the existence of weak solutions, our result is stated as follows.

Theorem 2.1

If assumptions (H₁)-(H₅) hold, then there exists at least one bounded weak solution $u\in L^{\infty}(\Omega)$ to problem ($\mathscr{P}$) in the sense of Definition  2.1.

As we have said before, when dealing with the case $f\in L^{1}(\Omega)$, we shall use the notion of renormalized solution.

Definition 2.2

A measurable function $u: \Omega\rightarrow \mathbb{R}$ is a renormalized solution of problem ($\mathscr{P}$) if

$$\begin{aligned}& T_{k}(u)\in W_{0}^{1,p}(\Omega)\quad \mbox{for any } k\geq0, \end{aligned}$$

(2.6)

$$\begin{aligned}& \lim_{m\rightarrow\infty}\int_{\{m\leq |u|\leq m+1\}}a(x,u, \nabla u)\nabla u\,\mathrm{d}x=0 \end{aligned}$$

(2.7)

and if for any $h\in W^{1,\infty}(\Omega)$ with compact support and $\upsilon\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$, u satisfies

$$ \int_{\Omega}a(x,u,\nabla u)\nabla \bigl(h(u) \upsilon\bigr)\,\mathrm{d}x +\int_{\Omega}F(x,u,\nabla u)h(u) \upsilon\,\mathrm{d}x =\int_{\Omega}fh(u)\upsilon\,\mathrm{d}x. $$

(2.8)

The existence result for $L^{1}$ data is stated as follows.

Theorem 2.2

Assume that (H₁) to (H₃) hold and $\frac{\beta}{\alpha}\in L^{1}(\mathbb{R_{+}})$. If $f\in L^{1}(\Omega)$, then problem ($\mathscr{P}$) admits at least one renormalized solution.

Remark 2.2

In Theorem 2.1, the conditions (H₄) and (H₅) are only needed in proving the $L^{\infty}(\Omega)$ estimate of u. Therefore in Theorem 2.2, we do not need these assumptions. But instead, we need the condition $\frac{\beta}{\alpha}\in L^{1}(\mathbb {R_{+}})$ as in [11]. Moreover, by the result of [22], the solution obtained in Theorem 2.2 belongs to $W_{0}^{1,r}(\Omega)$, provided $2-\frac{1}{N}< p< N$.

3 Existence of weak solution to problem ($\mathscr{P}$)

To prove Theorem 2.1, we first establish the $L^{\infty}$ estimate of solutions to problem ($\mathscr{P}$).

Lemma 3.1

Assume that (H₁) to (H₅) hold. If $u\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$ is a weak solution to problem ($\mathscr{P}$), then u satisfies the following estimate:

$$ \|u\|_{L^{\infty}(\Omega)}\leq M, $$

(3.1)

where M is a constant which depends only on N, p, q, α, β, $\|f\|_{L^{q}(\Omega)}$.

Proof of Lemma 3.1

For $t>0$, $h>0$, let $S_{t, h}$ be a real function defined by

$$ S_{t, h}(\eta)= \textstyle\begin{cases} 1,& \eta>t+h , \\ \frac{\eta-t}{h}, & t\leq\eta\leq t+h , \\ 0, & |\eta|\leq t, \\ \frac{\eta+t}{h}, & -t-h\leq\eta\leq-t , \\ -1,& \eta\leq-t-h. \end{cases} $$

(3.2)

It is easy to see that $S_{t, h}(\phi(u))\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$ and so $S_{t, h}(\phi (u))e^{\gamma_{\theta}(|u|)}\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$, where ϕ and $\gamma _{\theta}$ are defined as in (2.1) and (2.3). Taking $v=e^{\gamma_{\theta}(|u|)}S_{t, h}(\phi(u))$ as a test function in (2.5), we have

$$\begin{aligned} \begin{aligned} &\frac{1}{h}\int_{\{t< |\phi(u)|\leq t+h\}}\phi'(u)e^{\gamma_{\theta}(|u|)}a(x,u, \nabla u)\nabla u\,\mathrm{d}x \\ &\qquad {}+\int_{\{|\phi(u)|> t\}}\bigl\vert S_{t, h}\bigl( \phi(u)\bigr)\bigr\vert \frac{\beta(|u|)}{\alpha(|u|)+\theta}e^{\gamma_{\theta}(|u|)}a(x,u,\nabla u)\nabla u\,\mathrm{d}x \\ &\qquad {} +\int_{\{|\phi(u)|> t\}}F(x,u,\nabla u)e^{\gamma_{\theta}(|u|)}S_{t, h} \bigl(\phi(u)\bigr)\,\mathrm{d}x \\ &\quad =\int_{\{|\phi(u)|> t\}}fe^{\gamma_{\theta}(|u|)}S_{t, h}\bigl( \phi(u)\bigr)\,\mathrm{d}x. \end{aligned} \end{aligned}$$

Then letting $\theta\rightarrow0$, we obtain

$$\begin{aligned}& \frac{1}{h}\int_{\{t< |\phi(u)|\leq t+h\}} \phi'(u)e^{\gamma (|u|)}a(x,u,\nabla u)\nabla u\,\mathrm{d}x \\& \qquad {}+\int_{\{|\phi(u)|> t\}}\bigl\vert S_{t, h}\bigl( \phi(u)\bigr)\bigr\vert \frac{\beta(|u|)}{\alpha(|u|)}e^{\gamma(|u|)}a(x,u,\nabla u)\nabla u\,\mathrm{d}x \\& \qquad {} +\int_{\{|\phi(u)|> t\}}F(x,u,\nabla u)e^{\gamma(|u|)}S_{t, h} \bigl(\phi(u)\bigr)\,\mathrm{d}x \\& \quad =\int_{\{|\phi(u)|> t\}}fe^{\gamma(|u|)}S_{t, h}\bigl( \phi(u)\bigr)\,\mathrm{d}x, \end{aligned}$$

(3.3)

where γ is defined as in (2.1). Notice that $|S_{t, h}(\phi(u))|\leq1$, by (H₁), (H₃), and applying Hölder’s inequality, we deduce from (3.3) that

$$ \frac{1}{h}\int_{\{t< \omega\leq t+h\}}|\nabla\omega|^{p}\, \mathrm{d}x \leq\int_{\{\omega> t\}}|f|e^{\gamma(|u|)}\,\mathrm{d}x\leq \|f\| _{L^{q}(\Omega)}\biggl(\int_{\{\omega> t\}}\bigl\vert e^{\gamma(|u|)}\bigr\vert ^{q'}\,\mathrm {d}x\biggr)^{\frac{1}{q'}}, $$

where $\omega=|\phi(u)|=\phi(|u|)$. Let h tend to zero, we find that

$$ -\frac{d}{dt}\int_{\{\omega>t\}}|\nabla \omega|^{p}\,\mathrm{d}x \leq\int_{\{\omega> t\}}|f|e^{\gamma(|u|)} \,\mathrm{d}x\leq\|f\| _{L^{q}(\Omega)}\biggl(\int_{\{\omega> t\}}\bigl\vert e^{\gamma(|u|)}\bigr\vert ^{q'}\,\mathrm {d}x \biggr)^{\frac{1}{q'}}. $$

(3.4)

Setting

$$z(t)=\sup_{\{|s|>\phi^{-1}(t)\}}\frac{e^{\gamma(|s|)}}{(1+\phi (|s|))^{p-1}}, $$

since ϕ is strictly increasing and $\lim_{s\rightarrow \pm\infty}\phi(s)=0$, we have

$$ \lim_{t\rightarrow +\infty}z(t)=0. $$

(3.5)

Concerning the term $(\int_{\{\omega> t\}}|e^{\gamma(|u|)}|^{q'}\,\mathrm {d}x)^{\frac{1}{q}}$, we have

$$\begin{aligned} \biggl(\int_{\{\omega> t\}}\bigl\vert e^{\gamma(|u|)} \bigr\vert ^{q'}\,\mathrm{d}x\biggr)^{\frac{1}{q}}&= \biggl(\int _{\{\omega> t\}}\biggl(\frac{e^{\gamma(|u|)}}{(1+\omega )^{p-1}}\biggr)^{q'}(1+ \omega)^{q'(p-1)}\,\mathrm{d}x\biggr)^{\frac{1}{q'}} \\ &\leq C(p,q)z(t)\biggl[\biggl(\int_{\{\omega> t\}} \omega^{q'(p-1)}\,\mathrm {d}x\biggr)^{\frac{1}{q'}}+\bigl( \mu_{\omega}(t)\bigr)^{\frac{1}{q'}}\biggr] \\ &\leq C(p,q)z(t)\biggl[\biggl(\int_{0}^{\mu_{\omega}(t)} \omega_{*}^{q'(p-1)}\,\mathrm {d}s\biggr)^{\frac{1}{q'}}+\bigl( \mu_{\omega}(t)\bigr) ^{\frac{1}{q'}}\biggr]. \end{aligned}$$

(3.6)

By (3.4), (3.6), and Lemma 2.1, it follows that

$$\begin{aligned} \begin{aligned}[b] &NC_{N}^{1/N}\mu_{\omega}(t)^{1-1/N} \\ &\quad \leq \bigl(-\mu_{\omega}'(t)\bigr)^{1/p'}\biggl(- \frac{d}{dt}\int_{\{u>t\}}|\nabla \omega|^{p}\, \mathrm{d}x\biggr)^{\frac{1}{p}} \\ &\quad \leq\bigl(-\mu_{\omega}'(t)\bigr)^{1/p'}C(p,q)z^{\frac{1}{p}}(t) \biggl[\biggl(\int_{0}^{\mu _{\omega}(t)}\omega_{*}^{q'(p-1)} \,\mathrm{d}s\biggr)^{\frac{1}{pq'}}+\bigl(\mu_{\omega}(t) \bigr)^{\frac{1}{pq'}}\biggr], \end{aligned} \end{aligned}$$

(3.7)

which indicates that, for $0<\theta<\theta+h<|\Omega|$,

$$\begin{aligned} \frac{\omega_{*}(\theta)-\omega_{*}(\theta+h)}{h} \leq&\frac{C(p,q)}{hNC_{N}^{1/N}}\int_{\omega_{*}(\theta+h)}^{\omega _{*}(\theta)}z^{\frac{1}{p}}(t) \frac{(-\mu_{\omega}'(t))^{1/p'}}{\mu_{\omega}(t)^{1-1/N}} \\ &{}\times\biggl[\biggl(\int_{0}^{\mu _{\omega}(t)} \omega_{*}^{q'(p-1)} \,\mathrm{d}s\biggr)^{\frac{1}{pq'}}+\bigl( \mu_{\omega}(t)\bigr)^{\frac{1}{pq'}}\biggr]\,\mathrm {d}t \\ < &\frac{C(p,q,N)}{h}\sup_{s\in[\omega_{*}(\theta+h),+\infty]}z^{\frac{1}{p}}(s)\int _{\omega_{*}(\theta+h)}^{\omega_{*}(\theta)} \frac{(-\mu_{\omega}'(t))^{1/p'}}{\mu_{\omega}(t)^{1-1/N}} \\ &{}\times\biggl[\biggl(\int _{0}^{\mu _{\omega}(t)}\omega_{*}^{q'(p-1)} \,\mathrm{d}s \biggr)^{\frac{1}{pq'}}+\bigl(\mu_{\omega}(t)\bigr)^{\frac{1}{pq'}}\biggr]\, \mathrm {d}t. \end{aligned}$$

Then we employ (1.15) of [9] to get

$$\begin{aligned} \frac{\omega_{*}(\theta)-\omega_{*}(\theta+h)}{h} < &\frac{C(p,q,N)}{h} \sup_{s\in[\omega_{*}(\theta+h),+\infty]}z^{\frac{1}{p}}(s) \int_{\theta}^{\theta+h} \frac{(-\omega'_{*}(\sigma))^{1/p}}{\sigma^{1-\frac{1}{N}}} \\ &{}\times\biggl[\biggl(\int _{0}^{\sigma}\omega_{*}^{q'(p-1)} \,\mathrm{d}s \biggr)^{\frac{1}{pq'}}+\sigma^{\frac{1}{pq'}}\biggr]\,\mathrm{d}\sigma. \end{aligned}$$

Then letting h tend to zero, we deduce that, for almost $\theta\in [0,|\Omega|]$,

$$ -\omega'_{*}(\theta) < C(p,q,N)\sup_{s\in[\omega_{*}(\theta),+\infty]}z^{\frac {1}{p}}(s) \frac{(-\omega'_{*}(\theta))^{1/p}}{\theta^{1-\frac{1}{N}}}\biggl[\biggl(\int_{0}^{\theta} \omega_{*}^{q'(p-1)} \,\mathrm{d}s\biggr)^{\frac{1}{pq'}}+\theta^{\frac{1}{pq'}} \biggr], $$

which leads, after applying Young’s inequality, to

$$\begin{aligned} -\omega'_{*}(\theta) &< C(p,q,N)\Bigl[\sup _{s\in[\omega_{*}(\theta),+\infty]}z^{\frac {1}{p}}(s)\Bigr]^{p'} \frac{1}{\theta^{(1-\frac{1}{N})p'}}\biggl[\biggl(\int_{0}^{\theta} \omega_{*}^{q'(p-1)} \,\mathrm{d}s\biggr)^{\frac{p'}{pq'}}+\theta^{\frac{p'}{pq'}} \biggr] \\ &\leq C(p,q,N)\sup_{s\in[\omega_{*}(\theta),+\infty]}z^{\frac{p'}{p}}(s) \frac{1}{\theta^{(1-\frac{1}{N})p'}}\bigl[\omega_{*}(0)\theta^{\frac {p'}{pq'}}+\theta^{\frac{p'}{pq'}} \bigr]. \end{aligned}$$

(3.8)

Since $q>\frac{N}{p}$, we have $q_{0}=\frac{p'}{pq'}+\frac {p'}{N}-p'+1>0$. From (3.5), we deduce that there exists $t_{0}>0$ such that

$$C(p,q,N)z^{\frac{p'}{p}}(s)|\Omega|^{q_{0}}\leq\frac{1}{2}\quad \mbox{for all }s\geq t_{0}. $$

Hence, upon integration over $[0,\mu_{\omega}(t_{0})]$, inequality (3.8) gives

$$ \omega_{*}(0) \leq1+2t_{0}, $$

which implies that $\|u\|_{L^{\infty}(\Omega)}\leq\phi^{-1}(1+2t_{0})$. We observe that $t_{0}$ only depends on p, q, N, $|\Omega|$, α, β, thus the proof of Lemma 3.1 is finished. □

To prove Theorem 2.1, we shall consider suitable approximate problems. First of all, we recall the following lemma, proved in [12].

Lemma 3.2

There exists a function $g\in C^{1}(\mathbb{R})$ such that g is odd, strictly increasing, and

$$\begin{aligned}& g'(s)=\alpha\bigl(\bigl\vert g(s)\bigr\vert \bigr)\geq0 \quad \textit{in } \mathbb{R}, \end{aligned}$$

(3.9)

$$\begin{aligned}& g(0)=0,\qquad \lim_{s\rightarrow+\infty}g(s)=+\infty. \end{aligned}$$

(3.10)

For a.e. $x\in\Omega$, $\forall s\in\mathbb{R}$, and $\forall\xi\in \mathbb{R}^{N}$, we define for fixed $\varepsilon>0$:

$$\begin{aligned}& F_{\varepsilon}(x,s,\xi)=\frac{F(x,s,\xi)}{1+\varepsilon|F(x,s,\xi)|}, \\& a_{\varepsilon}(x,s,\xi)=\varepsilon|\xi|^{p-2}\xi+a \bigl(x,g(s),g'(s)\xi\bigr), \\& a_{\varepsilon l}(x,s,\xi)=\varepsilon|\xi|^{p-2}\xi +a\bigl(x,g \bigl(T_{l}(s)\bigr),g'\bigl(T_{l}(s) \bigr)T'_{l}(s)\xi\bigr). \end{aligned}$$

For any fixed $\varepsilon>0$, we introduce the approximate problem

$$(\mathscr{P}_{\varepsilon})\quad \left \{ \textstyle\begin{array}{l@{\quad}l} -\operatorname{div}(a_{\varepsilon}(x,u_{\varepsilon},\nabla u_{\varepsilon }))+F_{\varepsilon}(x,g(u_{\varepsilon}),g'(u_{\varepsilon})\nabla u_{\varepsilon})=f_{\varepsilon}&\textit{in } \Omega, \\ u_{\varepsilon}=0& \textit{on } \partial\Omega, \end{array}\displaystyle \right . $$

where $\{f_{\varepsilon}\}$ satisfy

$$ f_{\varepsilon}\in C_{0}^{\infty}(\Omega)\quad \textit{such that } f_{\varepsilon}\rightarrow f\textit{ strongly in } L^{q}(\Omega) \textit{ as } \varepsilon\rightarrow0. $$

The existence result to problem ($\mathscr{P}_{\varepsilon}$) is stated as follows.

Theorem 3.1

Problem ($\mathscr{P}_{\varepsilon}$) admits at least a solution $u_{\varepsilon}\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega )$ with $\|g(u_{\varepsilon})\|_{L^{\infty}(\Omega)}\leq M_{0}$, where $M_{0}$ is a positive constant depending on M (see Lemma  3.1) and the behavior of function g.

Proof of Theorem 3.1

For any $l>0$, let us consider the following truncated problem:

$$(\mathscr{P}_{\varepsilon l})\quad \left \{ \textstyle\begin{array}{l@{\quad}l} -\operatorname{div}(a_{\varepsilon l}(x,u_{\varepsilon},\nabla u_{\varepsilon }))+F_{\varepsilon}(x,g(T_{l}(u_{\varepsilon})),g'(T_{l}(u_{\varepsilon }))\nabla T_{l}(u_{\varepsilon}))=f_{\varepsilon}& \mbox{in } \Omega, \\ u_{\varepsilon}=0& \mbox{on } \partial\Omega. \end{array}\displaystyle \right . $$

By the classic result (see [23]), problem ($\mathscr {P}_{\varepsilon l}$) admits a solution $u_{\varepsilon}\in W_{0}^{1,p}(\Omega)\in L^{\infty}(\Omega)$. Then using the same argument of Lemma 3.1, we conclude

$$\bigl\Vert g\bigl(T_{l}(u_{\varepsilon})\bigr)\bigr\Vert _{L^{\infty}(\Omega)}\leq M. $$

In view of Lemma 3.2, it is easy to see that $g^{-1}$ is defined well and strictly increasing in $\mathbb{R}$.

Now choosing $l>g^{-1}(M)$, we obtain

$$ \|u_{\varepsilon}\|_{L^{\infty}(\Omega)}\leq g^{-1}(M). $$

(3.11)

Thus we have $T_{l}(u_{\varepsilon})=u_{\varepsilon}$, which implies that $u_{\varepsilon}$ is a weak solution of ($\mathscr{P}_{\varepsilon}$). The proof is finished. □

Proof of Theorem 2.1

Taking $e^{\tilde{\gamma}_{\theta}(|u_{\varepsilon}|)}u_{\varepsilon}$ as a test function in problem ($\mathscr{P}_{\varepsilon}$), we have

$$\begin{aligned}& \int_{\Omega}e^{\tilde{\gamma}_{\theta}(|u_{\varepsilon}|)}a_{\varepsilon}(x,u_{\varepsilon},\nabla u_{\varepsilon})\nabla u_{\varepsilon}\,\mathrm{d}x \\& \qquad {}+\int_{\Omega}|u_{\varepsilon}|\frac{\beta(|g(u_{\varepsilon})|)}{\alpha (|g(u_{\varepsilon})|)+ \theta}e^{\tilde{\gamma}_{\theta}(|u_{\varepsilon}|)}a(x,u_{\varepsilon},\nabla u_{\varepsilon})\nabla u_{\varepsilon}\,\mathrm{d}x \\& \qquad {} +\int_{\Omega}F_{\varepsilon}\bigl(x,g(u_{\varepsilon}),g'(u_{\varepsilon}) \nabla u_{\varepsilon}\bigr)e^{\tilde{\gamma}_{\theta}(|u_{\varepsilon}|)}u_{\varepsilon}\,\mathrm{d}x \\& \quad =\int_{\Omega}f_{\varepsilon}e^{\tilde{\gamma}_{\theta}(|u_{\varepsilon}|)}u_{\varepsilon}\,\mathrm{d}x, \end{aligned}$$

where $\tilde{\gamma}_{\theta}$ is defined as in (2.4), and g is defined as in Lemma 3.2. Then letting θ tend to zero, using assumptions (H₁)-(H₄) and Theorem 3.1 we get

$$ \int_{\Omega}e^{\tilde{\gamma}(|u_{\varepsilon}|)}a_{\varepsilon}(x,u_{\varepsilon},\nabla u_{\varepsilon})\nabla u_{\varepsilon}\,\mathrm{d}x \leq\int_{\Omega}f_{\varepsilon}e^{\tilde{\gamma}(|u_{\varepsilon}|)}u_{\varepsilon}\,\mathrm{d}x, $$

where γ̃ is defined as in (2.4).

In view of Theorem 3.1, (H₁), and (H₂), the above estimate gives

$$ \varepsilon\int_{\Omega}|\nabla u_{\varepsilon}|^{p}+\int_{\Omega}\bigl\vert \nabla g(u_{\varepsilon})\bigr\vert ^{p}\,\mathrm{d}x\leq e^{\tilde{\gamma}(g^{-1}(M))}g^{-1}(M_{0})\|f\|_{L^{1}(\Omega)}. $$

(3.12)

Now denoting $\bar{u}_{\varepsilon}=g(u_{\varepsilon})$, estimates (3.11) and (3.12) imply that $\bar{u}_{\varepsilon}$ is bounded uniformly in $W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$. As a consequence, there exist a subsequence (still denoted by $\{\bar{u}_{\varepsilon}\}$) and a measurable function $\bar{u}\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$ such that

$$\begin{aligned}& \bar{u}_{\varepsilon}\rightharpoonup\bar{u}\quad \mbox{weakly in }W_{0}^{1,p}(\Omega) \mbox{ and weakly}^{\ast}\mbox{ in } L^{\infty }(\Omega), \end{aligned}$$

(3.13)

$$\begin{aligned}& \bar{u}_{\varepsilon}\rightarrow\bar{u} \quad \mbox{a.e. in } \Omega. \end{aligned}$$

(3.14)

In the following, the rest of the proof is divided into several steps.

Step 1: To deal with the difficulty that α vanishes at zero, we define the following truncation function near the origin:

$$ \zeta_{k}(s)=\max\{s,k\}=k+(s-k)_{+},\quad \forall s\in\mathbb{R}, $$

(3.15)

where $k>0$ is a fixed constant. Then we easily get

$$ \zeta_{k}(\bar{u}_{\varepsilon})\rightharpoonup\zeta_{k}( \bar{u})\quad \mbox{weakly in }W_{0}^{1,p}(\Omega) \mbox{ and weakly}^{\ast}\mbox{ in } L^{\infty}(\Omega). $$

(3.16)

Now taking $\rho_{\theta}^{\varepsilon}=e^{\gamma_{\theta}(\bar {u}_{\varepsilon})}[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}(\bar{u})]_{+}$ as a test function in problem ($\mathscr {P}_{\varepsilon}$), by (H₁) we have

$$\begin{aligned}& \int_{\Omega}e^{\gamma_{\theta}(\bar{u}_{\varepsilon})}a(x,\bar {u}_{\varepsilon},\nabla \bar{u}_{\varepsilon})\nabla\bigl[ \zeta_{k}(\bar {u}_{\varepsilon}) - \zeta_{k}(\bar{u})\bigr]_{+}\,\mathrm{d}x \\& \qquad {}+\varepsilon\int_{\Omega}e^{\gamma _{\theta}(\bar{u}_{\varepsilon})}|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}(\bar{u})\bigr]_{+} \,\mathrm{d}x \\& \qquad {} +\int_{\Omega}\frac{\beta(|\bar{u}_{\varepsilon}|)}{\alpha(|\bar {u}_{\varepsilon}|) +\theta}e^{\gamma_{\theta}(\bar{u}_{\varepsilon})} \bigl[ \zeta_{k}(\bar {u}_{\varepsilon}) -\zeta_{k}(\bar{u}) \bigr]_{+}\alpha\bigl( |\bar{u}_{\varepsilon}|\bigr)|\nabla\bar {u}_{\varepsilon}|^{p} \,\mathrm{d}x \\& \qquad {} +\varepsilon\int_{\Omega}\frac{\beta(|\bar{u}_{\varepsilon}|)}{\alpha (|\bar{u}_{\varepsilon}|) +\theta}e^{\gamma_{\theta}(\bar{u}_{\varepsilon})} \bigl[ \zeta_{k}(\bar {u}_{\varepsilon}) -\zeta_{k}(\bar{u}) \bigr]_{+}|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla \bar{u}_{\varepsilon}\,\mathrm{d}x \\& \qquad {} +\int_{\Omega}F_{\varepsilon}(x, \bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon})e^{\gamma_{\theta}(\bar{u}_{\varepsilon})}\bigl[ \zeta_{k}(\bar {u}_{\varepsilon}) -\zeta_{k}(\bar{u})\bigr]_{+} \,\mathrm{d}x \\& \quad \leq\int_{\Omega}f_{\varepsilon}e^{\gamma_{\theta}(\bar{u}_{\varepsilon})}\bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}(\bar{u})\bigr]_{+} \,\mathrm{d}x. \end{aligned}$$

(3.17)

It is easy to see that the fourth term of (3.17) is non-negative. So letting θ tend to zero, the above inequality leads to

$$ I_{1}(\varepsilon)+I_{2}(\varepsilon)\leq I_{3}(\varepsilon), $$

(3.18)

where

$$\begin{aligned}& I_{1}(\varepsilon)=\int_{\Omega}e^{\gamma(\bar{u}_{\varepsilon})}a(x, \bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\nabla\bigl[ \zeta_{k}(\bar {u}_{\varepsilon}) -\zeta_{k}(\bar{u})\bigr]_{+} \,\mathrm{d}x, \\& I_{2}(\varepsilon)=\varepsilon\int_{\Omega}e^{\gamma(\bar {u}_{\varepsilon})}| \nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}(\bar{u})\bigr]_{+} \,\mathrm{d}x, \\& I_{3}(\varepsilon)=\int_{\Omega}f_{\varepsilon}e^{\gamma(\bar {u}_{\varepsilon})}\bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) - \zeta_{k}(\bar{u})\bigr]_{+}\,\mathrm{d}x. \end{aligned}$$

Now we estimate all the terms of (3.18).

Estimate of $I_{2}(\varepsilon)$. Using (3.11), (3.13), and the Hölder inequality, we conclude that

$$\bigl\vert I_{2}(\varepsilon)\bigr\vert \leq\varepsilon e^{\gamma(M_{0})}\biggl(\int_{\Omega} \vert \nabla u_{\varepsilon} \vert ^{p}\,\mathrm{d}x\biggr)^{\frac{p-1}{p}} \biggl[\biggl(\int_{\Omega}\bigl\vert \nabla \zeta_{k}(\bar{u}_{\varepsilon})\bigr\vert ^{p}\,\mathrm {d}x\biggr)^{\frac{1}{p}}+\biggl(\int_{\Omega}\bigl\vert \nabla\zeta_{k}(\bar{u})\bigr\vert ^{p}\,\mathrm {d}x \biggr)^{\frac{1}{p}}\biggr]. $$

Hence, by (3.12) we easily get

$$ \lim_{\varepsilon\rightarrow0}I_{2}(\varepsilon)=0. $$

(3.19)

Estimate of $I_{3}(\varepsilon)$. By (3.11), (3.14), and the Lebesgue dominated convergence theorem, we infer that

$$ \lim_{\varepsilon\rightarrow0}I_{3}(\varepsilon)=0. $$

(3.20)

Estimate of $I_{1}(\varepsilon)$. Since $a(x,s,0)=0$ for a.e. $x\in\Omega$ and every $s\in\mathbb{R}$, we obtain

$$\begin{aligned} I_{1}(\varepsilon) =&\int_{\Omega^{k}_{\varepsilon1}}e^{\gamma(\bar {u}_{\varepsilon})}a(x, \bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \cdot\nabla \bigl[ \bar{u}_{\varepsilon}-\zeta_{k}(\bar{u})\bigr]_{+}\,\mathrm{d}x \\ &{}+\int_{\Omega^{k}_{\varepsilon2}} e^{\gamma(\bar{u}_{\varepsilon})}a(x, \bar{u}_{\varepsilon}, \nabla\bar{u}_{\varepsilon}) \cdot\nabla \bigl[ -k -\zeta_{k}(\bar{u}) \bigr]_{+}\,\mathrm{d}x \\ =&\bar{I}_{11}(\varepsilon)+\bar{I}_{12}(\varepsilon), \end{aligned}$$

(3.21)

where

$$ \Omega^{k}_{\varepsilon1}=\{x\in\Omega:\bar{u}_{\varepsilon}< k\}, \qquad \Omega ^{k}_{\varepsilon2}=\{x\in\Omega:\bar{u}_{\varepsilon} \geq k\}. $$

For the term $\bar{I}_{11}(\varepsilon)$, we can write

$$\begin{aligned} \bar{I}_{11}(\varepsilon) =&\int_{\Omega^{k}_{\varepsilon1}}e^{\gamma(\bar {u}_{\varepsilon})} \bigl[a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla \zeta_{k}(\bar{u}_{\varepsilon})\bigr)- a\bigl(x, \zeta_{k}( \bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u})\bigr)\bigr] \cdot \nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}( \bar{u})\bigr]_{+}\,\mathrm{d}x \\ &{}+\int_{\Omega^{k}_{\varepsilon1}}e^{\gamma(\bar{u}_{\varepsilon})} a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u}) \bigr) \cdot\nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) - \zeta_{k}(\bar{u})\bigr]_{+}\,\mathrm{d}x. \end{aligned}$$

(3.22)

Collecting (3.11), (3.13), (3.14), and (3.16), it is easy to verify that

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega^{k}_{\varepsilon1}}e^{\gamma(\bar{u}_{\varepsilon})} a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u})\bigr) \cdot\nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) - \zeta_{k}(\bar{u})\bigr]_{+}\,\mathrm{d}x=0. $$

(3.23)

Using (3.22), (3.23), (H₁), and (H₂), we find that

$$\begin{aligned} \varlimsup_{\varepsilon\rightarrow0}\bar{I}_{11}(\varepsilon) \geq& \varlimsup_{\varepsilon\rightarrow0}\int_{\Omega^{k}_{\varepsilon1}}\bigl[a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u}_{\varepsilon})\bigr)- a\bigl(x,\zeta_{k}(\bar{u}_{\varepsilon}), \nabla\zeta_{k}(\bar{u})\bigr)\bigr] \\ &{}\cdot\nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}(\bar{u})\bigr]_{+} \,\mathrm{d}x \\ =&\varlimsup_{\varepsilon\rightarrow0}\int_{\Omega}\bigl[a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u}_{\varepsilon})\bigr)- a\bigl(x, \zeta_{k}( \bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u})\bigr)\bigr] \\ &{}\cdot \nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}( \bar{u})\bigr]_{+}\,\mathrm{d}x, \end{aligned}$$

where we have used the fact $a(x,s,0)=0$ for a.e. $x\in\Omega$.

For the term $\bar{I}_{12}(\varepsilon)$, it is easy to get

$$ \lim_{\varepsilon\rightarrow0}\bar{I}_{12}(\varepsilon)=0. $$

The above two convergence results show that

$$\begin{aligned} \varlimsup_{\varepsilon\rightarrow0}I_{1}(\varepsilon) \geq& \varlimsup_{\varepsilon\rightarrow0}\int_{\Omega}\bigl[a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u}_{\varepsilon})\bigr)- a\bigl(x, \zeta_{k}( \bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u})\bigr)\bigr] \\ &{}\cdot \nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}( \bar{u})\bigr]_{+}\,\mathrm{d}x. \end{aligned}$$

(3.24)

Substituting (3.19), (3.20), and (3.24) into (3.18), we conclude

$$\begin{aligned}& \varlimsup_{\varepsilon\rightarrow0}\int_{\Omega}\bigl[a \bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u}_{\varepsilon})\bigr)- a\bigl(x, \zeta_{k}( \bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot \nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}( \bar{u})\bigr]_{+}\,\mathrm{d}x\leq0. \end{aligned}$$

(3.25)

Now choosing $\rho_{\theta}^{\varepsilon}=-e^{\gamma_{\theta}(\bar {u}_{\varepsilon})}[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}(\bar{u})]_{+}$ as a test function in problem ($\mathscr {P}_{\varepsilon}$), by the same arguments as in the proof of (3.25) we arrive at

$$\begin{aligned}& \varlimsup_{\varepsilon\rightarrow0}\int_{\Omega}-\bigl[a \bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u}_{\varepsilon})\bigr)- a\bigl(x, \zeta_{k}( \bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u})\bigr)\bigr] \\& \quad {} \cdot \nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}( \bar{u})\bigr]_{-}\,\mathrm{d}x\leq0. \end{aligned}$$

(3.26)

As a consequence of (3.25) and (3.26), we have

$$ \varlimsup_{\varepsilon\rightarrow0}\int_{\Omega}\bigl[a\bigl(x, \zeta_{k}(\bar{u}_{\varepsilon}),\nabla\zeta_{k}( \bar{u}_{\varepsilon})\bigr)- a\bigl(x, \zeta_{k}( \bar{u}_{\varepsilon}),\nabla\zeta_{k}(\bar{u})\bigr)\bigr] \cdot \nabla \bigl[ \zeta_{k}(\bar{u}_{\varepsilon}) -\zeta_{k}( \bar{u})\bigr]\,\mathrm{d}x\leq0. $$

Then, arguing as in [24], we derive that

$$ \nabla\zeta_{k}(\bar{u}_{\varepsilon})\rightarrow\nabla \zeta_{k}(\bar{u})\quad \mbox{strongly in }\bigl(L^{p}( \Omega)\bigr)^{N} \mbox{ and a.e. in } \Omega. $$

(3.27)

Step 2: For any fixed $k>0$, let us define

$$ \bar{\zeta}_{k}(s)=\min\{s,-k\}=-k+(s+k)_{-},\quad \forall s\in\mathbb {R}. $$

Proceeding as in Step 1, taking $\rho_{\theta}^{\varepsilon}=e^{\gamma _{\theta}(\bar{u}_{\varepsilon})}[ \bar{\zeta}_{k}(\bar{u}_{\varepsilon}) -\bar{\zeta}_{k}(\bar{u})]_{+}$ and $\rho_{\theta}^{\varepsilon}=-e^{-\gamma _{\theta}(\bar{u}_{\varepsilon})}[ \bar{\zeta}_{k}(\bar{u}_{\varepsilon}) -\bar{\zeta}_{k}(\bar{u})]_{-}$ as two test functions in problem ($\mathscr {P}_{\varepsilon}$), we obtain

$$ \nabla\bar{\zeta}_{k}(\bar{u}_{\varepsilon})\rightarrow\nabla\bar{ \zeta }_{k}(\bar{u}) \quad \mbox{strongly in }\bigl(L^{p}( \Omega)\bigr)^{N} \mbox{ and a.e. in } \Omega. $$

(3.28)

By (3.27) and (3.28), it follows that

$$ \chi_{\{|\bar{u}_{\varepsilon}|\geq k\}}\nabla\bar{u}_{\varepsilon}\rightarrow\chi_{\{|\bar{u}|\geq k\}} \nabla\bar{u}\quad \mbox{strongly in }\bigl(L^{p}(\Omega) \bigr)^{N} \mbox{ and a.e. in } \Omega. $$

(3.29)

In the following, we prove that u is a weak solution to problem ($\mathscr{P}$).

Since $u_{\varepsilon}$ is a weak solution to problem ($\mathscr{P}$), it follows that

$$\begin{aligned}& \int_{\Omega}a(x,\bar{u}_{\varepsilon},\nabla \bar{u}_{\varepsilon})\nabla v\,\mathrm{d}x +\varepsilon\int_{\Omega}| \nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla v\, \mathrm{d}x+\int_{\Omega}F_{\varepsilon}(x,\bar{u}_{\varepsilon}, \nabla\bar {u}_{\varepsilon})v\,\mathrm{d}x \\& \quad =\int_{\Omega}f_{\varepsilon}v\,\mathrm{d}x,\quad \forall v\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega). \end{aligned}$$

(3.30)

Concerning the third term on the left-hand side of (3.30), we rewrite it as

$$\begin{aligned}& \int_{\Omega}F(x,\bar{u}_{\varepsilon},\nabla \bar{u}_{\varepsilon}) \upsilon\,\mathrm{d}x \\& \quad = \int_{\{x\in\Omega:|\bar{u}_{\varepsilon}|>k\}} F(x,\bar {u}_{\varepsilon},\nabla \bar{u}_{\varepsilon}) \upsilon\,\mathrm{d}x+ \int_{\{x\in\Omega:|\bar{u}_{\varepsilon}|\leq k\}} F(x, \bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \upsilon\,\mathrm{d}x \\& \quad =I_{1\varepsilon}+I_{2\varepsilon}\quad \mbox{for any fixed } k>0. \end{aligned}$$

(3.31)

To take the limits in $I_{1\varepsilon}$, we next show that

$$ F(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \chi_{\{|\bar {u}_{\varepsilon}|>k\}}\rightarrow F(x,\bar{u},\nabla\bar{u})\chi_{\{ |\bar{u}|>k\}} \quad \mbox{strongly in } L^{1}(\Omega). $$

(3.32)

Indeed, by (3.14) and (3.29), we already know that $F(x,t,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\chi_{\{|\bar {u}_{\varepsilon}|>k\}}\rightarrow F(x,t,\bar{u}, \nabla\bar{u})\chi_{\{ |\bar{u}|>k\}}$ almost everywhere in Ω, it suffices to prove the equi-integrability of this sequence and then apply Vitali’s convergence theorem. Using Theorem 3.1 and (H₃), we get

$$\bigl\vert F(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \chi_{\{|\bar {u}_{\varepsilon}|>k\}}\bigr\vert \leq C_{0}|\nabla\bar{u}_{\varepsilon}|^{p} \chi_{\{ |\bar{u}_{\varepsilon}|>k\}}, $$

where $C_{0}$ is a positive constant independent of ε and k. Then the equi-integrability of $|\nabla\bar{u}_{\varepsilon}|^{p} \chi _{\{|\bar{u}_{\varepsilon}|>k\}}$, which follows from (3.29), indicates that of $F(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\chi_{\{|\bar{u}_{\varepsilon}|>k\}}$. Therefore, (3.32) is proved.

As a conclusion, we have

$$ \lim_{\varepsilon\rightarrow0}I_{1\varepsilon}={\int_{\{ x\in\Omega: |\bar{u}|> k\}}}F(x, \bar{u},\nabla\bar{u})\upsilon\,\mathrm{d}x, $$

so that

$$ \lim_{k\rightarrow0}\lim_{\varepsilon\rightarrow 0}I_{1\varepsilon}= \int_{\Omega}F(x,\bar{u},\nabla\bar{u}) \upsilon\,\mathrm{d}x. $$

(3.33)

Moreover, by assumption (H₃) and (3.12) we get

$$ |I_{2\varepsilon}| \leq\max_{0\leq s\leq k} \beta (s)\int\int _{\{(x,t)\in Q_{\tau}:|\bar{u}_{\varepsilon}(x,t)|\leq k\}} \bigl[|\nabla\bar{u}_{\varepsilon}|^{p}+h(x,t) \bigr]| \upsilon|\,\mathrm{d}x\,\mathrm {d}t \leq C_{1}\max_{0\leq s\leq k} \beta(s), $$

where $C_{1}$ is a positive constant independent of ε and k. Therefore,

$$ \lim_{k\rightarrow0} \lim_{\varepsilon\rightarrow 0}I_{2\varepsilon}=0, $$

(3.34)

since β is a continuous function from $[0,+\infty)$ into $[0,+\infty)$ and $\beta(0)=0$.

It follows from (3.31), (3.33), and (3.34) that

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega}F(x,\bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \upsilon\,\mathrm{d}x=\int _{\Omega}F(x,\bar{u},\nabla\bar{u}) \upsilon\,\mathrm{d}x. $$

(3.35)

Similarly, we have

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega}a(x, \bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon})\nabla v\,\mathrm{d}x=\int _{\Omega}a(x,\bar{u},\nabla\bar {u})\nabla v\,\mathrm{d}x. $$

(3.36)

Furthermore, the same argument as (3.19) shows that

$$ \lim_{\varepsilon\rightarrow0} \varepsilon\int_{\Omega}| \nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla v\, \mathrm{d}x=0. $$

(3.37)

Finally, it is easy to see that

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega}f_{\varepsilon}v\,\mathrm{d}x=\int_{\Omega}f v\,\mathrm{d}x. $$

(3.38)

Now letting ε tend to zero, from (3.36)-(3.38), we deduce that ū satisfies (2.5), with u replaced by ū. Thus, the proof is finished. □

4 Existence of renormalized solution to problem ($\mathscr{P}$)

Proof of Theorem 2.2

By the proof of Theorem 3.1, we deduce that there exists at least one weak solution $u_{\varepsilon}$ satisfying $u_{\varepsilon}\in W_{0}^{1,p}(\Omega)\cap L^{\infty}(\Omega)$ such that

$$\begin{aligned}& \varepsilon\int_{\Omega}|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla v\,\mathrm{d}x+\int _{\Omega}a\bigl(x,g(u_{\varepsilon }),\nabla g(u_{\varepsilon})\bigr)\nabla v\,\mathrm{d}x \\& \quad {} + \int_{\Omega}F_{\varepsilon}\bigl(x,g(u_{\varepsilon}), \nabla g(u_{\varepsilon })\bigr)v\,\mathrm{d}x =\int_{\Omega}f_{\varepsilon} v\,\mathrm{d}x,\quad \forall v \in W_{0}^{1,p}(\Omega), \end{aligned}$$

(4.1)

where $f_{\varepsilon}$ satisfy

$$ f_{\varepsilon}\in C_{0}^{\infty}(\Omega) \quad \mbox{such that } f_{\varepsilon}\rightarrow f\mbox{ strongly in } L^{1}(\Omega) \mbox{ as } \varepsilon\rightarrow0. $$

As before, set $\bar{u}_{\varepsilon}=g(u_{\varepsilon})$. For any given $l>s_{0}$ and $\bar{l}=g^{-1}(l)$, let us take $v=e^{\tilde{\gamma}_{\theta}(|u_{\varepsilon}|)}T_{\bar{l}}(u_{\varepsilon})$ in (4.1), where $s_{0}$ is defined as in the proof of Theorem 3.1. Then sending θ tend to zero, using (H₁)-(H₃) and the fact $\frac{\beta}{\alpha}\in L^{1}(0,+\infty)$, it follows that

$$ \varepsilon\int_{\Omega}\bigl\vert \nabla T_{\bar{l}}(u_{\varepsilon})\bigr\vert ^{p}\,\mathrm {d}x+ \int_{\Omega}\bigl\vert \nabla T_{l}( \bar{u}_{\varepsilon})\bigr\vert ^{p}\,\mathrm {d}x\leq C, $$

(4.2)

where C is a positive constant independent of ε.

Hence, by the Sobolev space embedding theorem, there exist a measurable function ū and a subsequence (still denoted by $\{\bar{u}_{\varepsilon}\}$), such that

$$ \bar{u}_{\varepsilon}\rightarrow\bar{u}\quad \mbox{a.e. in } \Omega $$

(4.3)

and

$$ T_{l}(\bar{u}_{\varepsilon})\rightharpoonup T_{l}(\bar{u}) \quad \mbox{weakly in }W_{0}^{1,p}( \Omega). $$

(4.4)

Step 4.1. In this step, we prove the following result:

$$ \lim_{n\rightarrow\infty} \varlimsup_{\varepsilon\rightarrow 0}{\int _{\{x\in\Omega:n\leq |\bar{u}_{\varepsilon}(x)|\leq n+1\}}}a(x,\bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon})\nabla\bar{u}_{\varepsilon}\,\mathrm{d}x =0. $$

(4.5)

For any integer $n>1$, define $\rho_{n}$ by

$$\rho_{n}(r)=T_{n+1}(r)-T_{n}(r),\quad \forall r\in\mathbb{R}. $$

Obviously, we have

$$ 0< |\rho_{n}|\leq1 \quad \mbox{and} \quad \rho_{n}(r)\rightarrow0\quad \mbox{for any } r \mbox{ as }n \rightarrow\infty. $$

(4.6)

Taking $v=e^{\gamma_{\theta}(|\bar{u}_{\varepsilon}|)}\rho_{n}(\bar {u}_{\varepsilon})$ in (4.1), we get

$$\begin{aligned}& \int_{\Omega} e^{\gamma_{\theta}(|\bar{u}_{\varepsilon}|)}a(x,\bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\nabla\rho_{n}(\bar {u}_{\varepsilon})\,\mathrm{d}x+\int_{\Omega} \rho_{n}(\bar{u}_{\varepsilon})e^{\gamma_{\theta}(|\bar{u}_{\varepsilon}|)}\frac{\beta(|\bar {u}_{\varepsilon}|)}{\alpha(|\bar{u}_{\varepsilon}|)+\theta}a(x, \bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\nabla\bar {u}_{\varepsilon}\,\mathrm{d}x \\& \qquad {} +\int_{\Omega}\varepsilon|\nabla u_{\varepsilon}|^{p-2} \nabla u_{\varepsilon}\nabla\bigl(e^{\gamma_{\theta}(|\bar{u}_{\varepsilon}|)}\rho _{n}( \bar{u}_{\varepsilon})\bigr)\,\mathrm{d}x+\int_{\Omega}F_{\varepsilon}(x, \bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) e^{\gamma_{\theta}(|\bar {u}_{\varepsilon}|)} \rho_{n}(\bar{u}_{\varepsilon})\,\mathrm{d}x \\& \quad = \int_{\Omega} f_{\varepsilon}e^{\gamma_{\theta}(|\bar{u}_{\varepsilon}|)} \rho_{n}(\bar{u}_{\varepsilon})\,\mathrm{d}x. \end{aligned}$$

(4.7)

Passing to the limit as θ tend to zero in (4.7), it follows from (H₁) and (H₃) that

$$ \int_{\{x\in\Omega:n\leq|\bar{u}_{\varepsilon}(x)|\leq n+1\}} e^{\gamma(|\bar{u}_{\varepsilon}|)}a(x, \bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon})\nabla\bar{u}_{\varepsilon}\,\mathrm{d}x\leq \int_{\Omega} f_{\varepsilon}e^{\gamma(|\bar{u}_{\varepsilon}|)} \rho_{n}(\bar {u}_{\varepsilon})\,\mathrm{d}x. $$

(4.8)

Let $\varepsilon\rightarrow0$ and then $n\rightarrow\infty$ in (4.8). Recalling that $\frac{\beta}{\alpha}\in L^{1}(\mathbb{R}_{+})$, using (4.6) we get

$$ \varlimsup_{\varepsilon\rightarrow0}\int_{\{x\in\Omega :n\leq|\bar{u}_{\varepsilon}(x)|\leq n+1\}} a(x, \bar{u}_{\varepsilon },\nabla\bar{u}_{\varepsilon})\nabla\bar{u}_{\varepsilon}\,\mathrm {d}x\leq \int_{\Omega} f e^{\gamma(|\bar{u}|)} \rho_{n}(\bar{u})\,\mathrm{d}x. $$

(4.9)

It is easy to check that $\lim_{n\rightarrow\infty}\int_{\Omega} f e^{\gamma(|\bar{u}|)}\rho_{n}(\bar{u})\,\mathrm{d}x=0$. Thus, passing to the limit as $n\rightarrow\infty$ in (4.9), the desired result (4.5) follows immediately.

Step 4.2. For any fixed $k>0$ and $l>\max\{k,s_{0}\}$, we denote

$$\zeta^{l}_{k}(s)=\max\bigl\{ T_{l}(s),k\bigr\} =k+ \bigl(T_{l}(s)-k\bigr)_{+}, \quad \forall s\in \mathbb{R}. $$

Then we have, in view of (4.3) and (4.4),

$$ \zeta^{l}_{k}(\bar{u}_{\varepsilon}) \rightharpoonup\zeta^{l}_{k}(\bar{u})\quad \mbox{weakly in } W_{0}^{1,p}(\Omega). $$

(4.10)

Let λ be a positive number to be determined, denote

$$\varphi(s)=e^{\lambda s}-1, \quad \forall s\in\mathbb{R} $$

and

$$\rho_{\theta}^{\varepsilon}=e^{\gamma_{\theta}(\bar{u}_{\varepsilon})} \varphi\bigl(\bigl( \zeta^{l}_{k}(\bar{u}_{\varepsilon})-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\bigr) e^{-\gamma_{\theta}(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))}, $$

where $\gamma_{\theta}$ is defined as in (2.3). We now choose a sequence of increasing function $S_{n}\in C^{\infty}(\mathbb {R})$ such that

$$ S_{n}(r)=1 \quad \mbox{for }|r|\leq n;\qquad \operatorname{supp}S_{n}\subset[-n-1,n+1]; \qquad \bigl\Vert S'_{n}\bigr\Vert _{L^{\infty}(\mathbb{R})}\leq1. $$

(4.11)

Taking $v=S_{n}(\bar{u}_{\varepsilon})\rho_{\theta}^{\varepsilon}$ in (4.1), we obtain

$$\begin{aligned}& \hat{I}_{1}(\theta,\varepsilon,n)+\hat{I}_{2}( \theta,\varepsilon,n)+\hat {I}_{3}(\theta,\varepsilon,n) + \hat{I}_{4}(\theta,\varepsilon,n)+\hat{I}_{5}(\theta, \varepsilon,n) \\& \quad \leq\hat{I}_{6}(\theta,\varepsilon,n)+\hat{I}_{7}( \theta,\varepsilon,n) +\hat{I}_{8}(\theta,\varepsilon,n)+ \hat{I}_{9}(\theta,\varepsilon,n), \end{aligned}$$

(4.12)

where

$$\begin{aligned}& \hat{I}_{1}(\theta,\varepsilon,n)=\int_{\Omega} S_{n}(\bar{u}_{\varepsilon })e^{\gamma_{\theta}(\bar{u}_{\varepsilon}) -\gamma_{\theta}(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))} \varphi' \bigl(\bigl(\zeta^{l}_{k}(\bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar {u})\bigr)_{+}\bigr)a(x,\bar{u}_{\varepsilon}, \nabla\bar{u}_{\varepsilon}) \\& \hphantom{\hat{I}_{1}(\theta,\varepsilon,n)={}}{} \cdot\nabla\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar {u})\bigr)_{+}\bigr) \,\mathrm{d}x, \\& \hat{I}_{2}(\theta,\varepsilon,n)=\varepsilon\int _{\Omega} S_{n}(\bar {u}_{\varepsilon})e^{\gamma_{\theta}(\bar{u}_{\varepsilon}) -\gamma_{\theta}(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))} \varphi'\bigl(\bigl(\zeta ^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr)| \nabla u_{\varepsilon }|^{p-2}\nabla u_{\varepsilon} \\& \hphantom{\hat{I}_{2}(\theta,\varepsilon,n)={}}{} \cdot\nabla\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar {u})\bigr)_{+}\bigr) \,\mathrm{d}x, \\& \hat{I}_{3}(\theta,\varepsilon,n)=\int_{\Omega} S_{n}(\bar{u}_{\varepsilon }) \alpha\bigl(\vert \bar{u}_{\varepsilon} \vert \bigr) \frac{\beta(|\bar{u}_{\varepsilon}|)}{\alpha(|\bar{u}_{\varepsilon }|)+\theta}|\nabla \bar{u}_{\varepsilon}|^{p} \rho_{\theta}^{\varepsilon}\, \mathrm{d}x, \\& \hat{I}_{4}(\theta,\varepsilon,n)=\int_{\Omega} S_{n}'(\bar{u}_{\varepsilon })a(x, \bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\nabla\bar {u}_{\varepsilon} \rho_{\theta}^{\varepsilon}\,\mathrm{d}x, \\& \hat{I}_{5}(\theta,\varepsilon,n)=\varepsilon\int _{\Omega} S_{n}'(\bar {u}_{\varepsilon})|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\bar{u}_{\varepsilon} \rho_{\theta}^{\varepsilon}\,\mathrm{d}x, \\& \hat{I}_{6}(\theta,\varepsilon,n)=\int_{\Omega} S_{n}(\bar{u}_{\varepsilon}) \beta\bigl(\vert \bar{u}_{\varepsilon} \vert \bigr)|\nabla\bar{u}_{\varepsilon}|^{p} \rho _{\theta}^{\varepsilon}\,\mathrm{d}x, \\& \hat{I}_{7}(\theta,\varepsilon,n)=\int_{\Omega} S_{n}(\bar{u}_{\varepsilon }) \frac{\beta(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)}{\alpha(|\zeta ^{l}_{k}(\bar{u}_{\varepsilon})|)+\theta} \varphi\bigl(\bigl( \zeta^{l}_{k}(\bar{u}_{\varepsilon})-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\bigr) a(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \nabla\zeta ^{l}_{k}(\bar{u}_{\varepsilon})\,\mathrm{d}x, \\& \hat{I}_{8}(\theta,\varepsilon,n)=\varepsilon\int _{\Omega} S_{n}(\bar {u}_{\varepsilon}) \frac{\beta(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)}{ \alpha(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)+\theta} \varphi\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr)| \nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla \zeta^{l}_{k}(\bar {u}_{\varepsilon})\,\mathrm{d}x, \\& \hat{I}_{9}(\theta,\varepsilon,n)=\int_{\Omega} S_{n}(\bar{u}_{\varepsilon }) |f_{\varepsilon}| \rho_{\theta}^{\varepsilon}\,\mathrm{d}x. \end{aligned}$$

Limit behaviors of $\hat{I}_{2}(\theta,\varepsilon,n)$, $\hat {I}_{5}(\theta,\varepsilon,n)$, and $\hat{I}_{8}(\theta,\varepsilon,n)$. Thanks to (4.11), we have

$$\begin{aligned} \lim_{\theta\rightarrow0}\hat{I}_{2}(\theta,\varepsilon,n) =& \varepsilon\int_{\Omega} S_{n}'( \bar{u}_{\varepsilon})e^{\gamma(T_{n+1}(\bar{u}_{\varepsilon})) -\gamma(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))} \varphi'\bigl(\bigl( \zeta^{l}_{k}(\bar {u}_{\varepsilon})-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\bigr) \\ &{}\times\bigl\vert \nabla T_{n+1}(u_{\varepsilon})\bigr\vert ^{p-2}\nabla T_{n+1}(u_{\varepsilon})\cdot\nabla\bigl(\bigl( \zeta^{l}_{k}(\bar{u}_{\varepsilon })-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\bigr)\,\mathrm{d}x, \end{aligned}$$

and thus

$$\begin{aligned} \Bigl\vert \lim_{\theta\rightarrow0}\hat{I}_{2}(\theta, \varepsilon,n)\Bigr\vert \leq&\varepsilon C_{1}\int _{\Omega } \bigl\vert \nabla T_{n+1}(u_{\varepsilon}) \bigr\vert ^{p-1}\bigl(\bigl\vert \nabla\zeta^{l}_{k}( \bar {u}_{\varepsilon})\bigr\vert +\bigl\vert \nabla\zeta^{l}_{k}( \bar{u})\bigr\vert \bigr)\,\mathrm{d}x \\ \leq&\varepsilon C_{1}\bigl\Vert \nabla T_{n+1}(u_{\varepsilon}) \bigr\Vert ^{p-1}_{L^{p}(\Omega)}\bigl[\bigl\Vert \nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr\Vert _{L^{p}(\Omega)}+\bigl\Vert \nabla\zeta^{l}_{k}(\bar{u}) \bigr\Vert _{L^{p}(\Omega)}\bigr], \end{aligned}$$

where $C_{1}$ is a positive constant independent of ε. Therefore, using (4.2) we get

$$ \lim_{\varepsilon\rightarrow0} \lim_{\theta\rightarrow0} \hat{I}_{2}(\theta,\varepsilon,n)=0. $$

(4.13)

Similarly, we have

$$ \lim_{\varepsilon\rightarrow0} \lim_{\theta\rightarrow0} \hat{I}_{5}(\theta,\varepsilon,n)=0 $$

(4.14)

and

$$ \lim_{\varepsilon\rightarrow0} \lim_{\theta\rightarrow0} \hat{I}_{8}(\theta,\varepsilon,n)=0. $$

(4.15)

Limit behaviors of $\hat{I}_{3}(\theta,\varepsilon,n)$ and $\hat {I}_{6}(\theta,\varepsilon,n)$. Since

$$\begin{aligned} \hat{I}_{3}(\theta,\varepsilon,n) =&\int_{\{x\in\Omega:\bar{u}_{\varepsilon}(x)\neq0\}} S_{n}'(\bar{u}_{\varepsilon}) \alpha\bigl(\bigl\vert T_{n+1}(\bar {u}_{\varepsilon})\bigr\vert \bigr) \frac{\beta(|T_{n+1}(\bar{u}_{\varepsilon})|)}{\alpha(|T_{n+1}(\bar {u}_{\varepsilon}) |)+\theta} \\ &{}\times\bigl\vert \nabla T_{n+1}( \bar{u}_{\varepsilon})\bigr\vert ^{p} \rho_{\theta}^{\varepsilon}\,\mathrm{d}x, \end{aligned}$$

we get

$$ \lim_{\theta\rightarrow0}\hat{I}_{3}(\theta, \varepsilon,n)= \int_{\Omega} S_{n}'( \bar{u}_{\varepsilon})\varphi\bigl(\bigl(\zeta^{l}_{k}(\bar {u}_{\varepsilon}) -\zeta^{l}_{k}(\bar{u})\bigr)_{+} \bigr)e^{\gamma(\bar{u}_{\varepsilon}) -\gamma(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))}\beta\bigl(\vert \bar{u}_{\varepsilon }\vert \bigr)| \nabla \bar{u}_{\varepsilon}|^{p} \,\mathrm{d}x. $$

(4.16)

As far as $\hat{I}_{6}(\theta,\varepsilon,n)$ is concerned, we have

$$ \lim_{\theta\rightarrow0}\hat{I}_{6}(\theta, \varepsilon,n)=\int_{\Omega} S_{n}'( \bar {u}_{\varepsilon})\varphi\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar{u})\bigr)_{+} \bigr)e^{\gamma(\bar{u}_{\varepsilon}) -\gamma(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))}\beta\bigl(\vert \bar{u}_{\varepsilon }\vert \bigr)| \nabla \bar{u}_{\varepsilon}|^{p} \,\mathrm{d}x. $$

(4.17)

Limit behavior of $\hat{I}_{4}(\theta,\varepsilon,n)$. From (4.5) and (4.11), it follows that

$$ \lim_{n\rightarrow\infty} \varlimsup_{\varepsilon\rightarrow 0} \lim _{\theta\rightarrow0}\bigl\vert \hat{I}_{4}(\theta, \varepsilon,n)\bigr\vert =0. $$

(4.18)

Limit behavior of $\hat{I}_{7}(\theta,\varepsilon,n)$. For the term $\hat{I}_{7}(\theta,\varepsilon,n)$, we have

$$\begin{aligned} \lim_{\theta\rightarrow0}\hat{I}_{7}(\theta, \varepsilon,n)&=\int_{\Omega} S_{n}'( \bar {u}_{\varepsilon})\frac{\beta(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)}{ \alpha(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)} \varphi\bigl(\bigl( \zeta^{l}_{k}(\bar{u}_{\varepsilon})-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\bigr) a(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \nabla\zeta ^{l}_{k}(\bar{u}_{\varepsilon})\,\mathrm{d}x \\ &\leq I_{71}(\varepsilon,n)+I_{72}(\varepsilon,n)+I_{73}( \varepsilon,n), \end{aligned}$$

(4.19)

where

$$\begin{aligned}& I_{71}(\varepsilon,n)=\max_{s\in[k,l]}\frac{\beta(|s|)}{ \alpha(|s|)} \int_{\Omega} \bigl[a\bigl(x,\zeta^{l}_{k}( \bar{u}_{\varepsilon}) ,\nabla\zeta^{l}_{k}( \bar{u}_{\varepsilon})\bigr)-a\bigl(x,\zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla\zeta^{l}_{k}(\bar{u})\bigr)\bigr] \\& \hphantom{I_{71}(\varepsilon,n)={}}{}\cdot\nabla\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+} \varphi\bigl(\bigl(\zeta^{l}_{k}(\bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar {u})\bigr)_{+}\bigr)S_{n}'( \bar{u}_{\varepsilon}) \,\mathrm{d}x, \\& I_{72}(\varepsilon,n)= \int_{\Omega} \frac{\beta(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)}{ \alpha(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)} a\bigl(x,\zeta^{l}_{k}( \bar{u}_{\varepsilon}),\nabla\zeta^{l}_{k}(\bar{u})\bigr) \\& \hphantom{I_{72}(\varepsilon,n)={}}{}\cdot\nabla\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+} \varphi\bigl(\bigl(\zeta^{l}_{k}(\bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar {u})\bigr)_{+}\bigr)S_{n}'( \bar{u}_{\varepsilon}) \,\mathrm{d}x \end{aligned}$$

and

$$\begin{aligned} I_{73}(\varepsilon,n) =& \int_{\Omega} \frac{\beta(|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)}{\alpha (|\zeta^{l}_{k}(\bar{u}_{\varepsilon})|)} a\bigl(x,\zeta^{l}_{k}( \bar{u}_{\varepsilon}) ,\nabla\zeta^{l}_{k}( \bar{u}_{\varepsilon})\bigr)\nabla\zeta^{l}_{k}(\bar{u}) \\ &{}\times\varphi\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon}) -\zeta^{l}_{k}(\bar{u})\bigr)_{+} \bigr)S_{n}'(\bar{u}_{\varepsilon})\,\mathrm{d}x. \end{aligned}$$

Combining (4.3) with (4.4), we infer that

$$ \lim_{\varepsilon\rightarrow0}I_{72}(\varepsilon,n)=0 $$

(4.20)

and

$$ \lim_{\varepsilon\rightarrow0}I_{73}(\varepsilon,n)=0. $$

(4.21)

Substituting (4.20) and (4.21) into (4.19), we obtain

$$ \varlimsup_{\varepsilon\rightarrow0} \lim_{\theta\rightarrow0} \hat{I}_{7}(\theta,\varepsilon,n)\leq\varlimsup_{\varepsilon \rightarrow0}I_{71}( \varepsilon,n). $$

(4.22)

Limit behavior of $\hat{I}_{9}(\theta,\varepsilon,n)$. It is straightforward that

$$ \lim_{n\rightarrow\infty} \lim_{\varepsilon\rightarrow0} \lim _{\theta\rightarrow0}\hat{I}_{9}(\theta,\varepsilon,n)=0. $$

(4.23)

Limit behavior of $\hat{I}_{1}(\theta,\varepsilon,n)$. Note that $a(x,s,0)=0$ for a.e. $x\in\Omega$ and every $s\in\mathbb{R}$, and we get

$$\begin{aligned}& \lim_{\theta\rightarrow0}\hat{I}_{1}(\theta, \varepsilon,n) \\& \quad =\int_{\Omega^{k}_{\varepsilon1}}S_{n}'( \bar{u}_{\varepsilon}) \varphi'\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+} \bigr)a(x, \bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \cdot\nabla \bigl(\zeta^{l}_{k}(\bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar {u})\bigr)_{+}\,\mathrm{d}x \\& \qquad {} +\int_{\Omega^{k}_{\varepsilon2}} S_{n}'( \bar{u}_{\varepsilon})e^{\gamma (\bar{u}_{\varepsilon}) -\gamma(l)} \varphi'\bigl(\bigl(l- \zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr)a(x,\bar{u}_{\varepsilon}, \nabla\bar {u}_{\varepsilon})\cdot\nabla\bigl(l-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\,\mathrm{d}x \\& \qquad {} +\int_{\Omega^{k}_{\varepsilon3}} S_{n}'( \bar{u}_{\varepsilon})e^{\gamma (\bar{u}_{\varepsilon}) -\gamma(k)} \varphi'\bigl(\bigl(k- \zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr)a(x,\bar{u}_{\varepsilon}, \nabla\bar {u}_{\varepsilon})\cdot\nabla\bigl(k-\zeta^{l}_{k}( \bar{u})\bigr)_{+}\,\mathrm{d}x \\& \quad =\hat{I}_{21}(\varepsilon)+\hat{I}_{22}( \varepsilon)+\hat {I}_{23}(\varepsilon), \end{aligned}$$

(4.24)

where

$$\begin{aligned}& \Omega^{k}_{\varepsilon1}=\{x\in\Omega:k< \bar{u}_{\varepsilon}< l\}, \\& \Omega^{k}_{\varepsilon2}=\{x\in\Omega:\bar{u}_{\varepsilon}\geq l \}, \\& \Omega^{k}_{\varepsilon3}=\{x\in\Omega:\bar{u}_{\varepsilon}\leq k \}. \end{aligned}$$

Using (4.3), (4.4), and (4.11), it is clear that

$$ \lim_{\varepsilon\rightarrow0}\hat{I}_{22}( \varepsilon)=0 $$

(4.25)

and

$$ \lim_{\varepsilon\rightarrow0}\hat{I}_{23}( \varepsilon)=0. $$

(4.26)

Note that $a(x,s,0)=0$ for a.e. $x\in\Omega$ and every $s\in\mathbb{R}$, the term $\hat {I}_{21}(\varepsilon)$ can be rewritten as follows:

$$\hat{I}_{21}(\varepsilon)=J_{1}(\varepsilon)+J_{2}( \varepsilon), $$

where

$$\begin{aligned}& J_{1}(\varepsilon)=\int_{\Omega} S_{n}'(\bar{u}_{\varepsilon})\bigl[a\bigl(x, \zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a\bigl(x, \zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr)\bigr] \\& \hphantom{J_{1}(\varepsilon)={}}{}\cdot\nabla\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr) \varphi'\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr) \,\mathrm{d}x, \\& J_{2}(\varepsilon)=\int_{\Omega} S_{n}'(\bar{u}_{\varepsilon})a\bigl(x, \zeta^{l}_{k}(\bar{u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr) \\& \hphantom{J_{2}(\varepsilon)={}}{}\cdot\nabla\bigl(\bigl( \zeta^{l}_{k}(\bar{u}_{\varepsilon})-\zeta ^{l}_{k}(\bar{u})\bigr)_{+}\bigr)\varphi'\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr) \,\mathrm{d}x. \end{aligned}$$

By (4.3), (4.4), and (4.10), we find that

$$ \lim_{\varepsilon\rightarrow0}J_{2}(\varepsilon)=0. $$

(4.27)

As a direct consequence of (4.24)-(4.27), we have

$$ \varlimsup_{\varepsilon\rightarrow0} \lim_{\theta\rightarrow0} \hat{I}_{1}(\theta,\varepsilon,n)=\varlimsup_{\varepsilon \rightarrow0}J_{1}( \varepsilon). $$

(4.28)

Choosing $\lambda=2\max_{s\in[k,l]}\frac{\beta(|s|)}{ \alpha(|s|)}$ in the definition of φ, and then combining the limit behaviors of $\hat{I}_{1}(\theta,\varepsilon,n)$-$\hat{I}_{9}(\theta ,\varepsilon,n)$, we get

$$\begin{aligned}& \lim_{n\rightarrow\infty} \varlimsup_{\varepsilon\rightarrow0}\int _{\Omega} S_{n}'(\bar{u}_{\varepsilon}) \bigl[a\bigl(x,\zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a\bigl(x, \zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot\nabla\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr) \varphi'\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u})\bigr)_{+}\bigr) \,\mathrm {d}x\leq0, \end{aligned}$$

which yields

$$\begin{aligned}& \lim_{n\rightarrow\infty} \varlimsup_{\varepsilon\rightarrow0}\int _{\Omega} S_{n}'(\bar{u}_{\varepsilon}) \bigl[a\bigl(x,\zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a\bigl(x, \zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot\nabla\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar {u})\bigr)_{+}\bigr) \,\mathrm{d}x\leq0. \end{aligned}$$

(4.29)

Step 4.3. Choosing $v=-S_{n}(\bar{u}_{\varepsilon })e^{-\gamma_{\theta}(\bar{u}_{\varepsilon}) +\gamma_{\theta}(\zeta^{l}_{k}(\bar{u}_{\varepsilon}))} \varphi((\zeta^{l}_{k}(\bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar{u}))_{-})$ as a test function in (4.1), then arguing as before, we have

$$\begin{aligned}& \lim_{n\rightarrow\infty} \varliminf_{\varepsilon\rightarrow0}\int _{\Omega} S_{n}'(\bar{u}_{\varepsilon}) \bigl[a\bigl(x,\zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a\bigl(x, \zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot\nabla\bigl(\bigl(\zeta^{l}_{k}( \bar{u}_{\varepsilon})-\zeta^{l}_{k}(\bar {u})\bigr)_{-}\bigr) \,\mathrm{d}x\geq0. \end{aligned}$$

(4.30)

It follows from (4.29) and (4.30) that

$$\begin{aligned}& \lim_{n\rightarrow\infty} \varlimsup_{\varepsilon\rightarrow0}\int _{\Omega} S_{n}'(\bar{u}_{\varepsilon}) \bigl[a\bigl(x,\zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a\bigl(x, \zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot\nabla\bigl(\zeta^{l}_{k}(\bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar {u})\bigr)\,\mathrm{d}x\leq0. \end{aligned}$$

(4.31)

Taking into account that $S_{n}'(\bar{u}_{\varepsilon}) a(x,\zeta^{l}_{k}(\bar {u}_{\varepsilon}) ,\nabla\zeta^{l}_{k}(\bar{u}_{\varepsilon}))=a(x,\zeta^{l}_{k}(\bar {u}_{\varepsilon}) ,\nabla\zeta^{l}_{k}(\bar{u}_{\varepsilon}))$ for $n>l$, using (4.31) we get

$$ \varlimsup_{\varepsilon\rightarrow0}\int_{\Omega} a\bigl(x, \zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)\cdot\nabla\bigl( \zeta^{l}_{k}(\bar {u}_{\varepsilon})-\zeta^{l}_{k}( \bar{u})\bigr)\,\mathrm{d}x\leq0, $$

which yields

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega} \bigl[a\bigl(x,\zeta^{l}_{k}(\bar{u}_{\varepsilon}) ,\nabla \zeta^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a\bigl(x, \zeta^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \zeta^{l}_{k}(\bar{u})\bigr)\bigr]\cdot\nabla\bigl(\zeta^{l}_{k}(\bar{u}_{\varepsilon})- \zeta^{l}_{k}(\bar {u})\bigr)\,\mathrm{d}x= 0. $$

(4.32)

Then, arguing as in [24], we derive

$$ \nabla\zeta^{l}_{k}(\bar{u}_{\varepsilon}) \rightarrow\nabla\zeta^{l}_{k}(\bar {u})\quad \mbox{strongly in } \bigl(L^{p}(\Omega)\bigr)^{N} \mbox{ and a.e. in } \Omega. $$

(4.33)

Step 4.4. For any fixed $l>k>0$, we denote

$$ \bar{\zeta}^{l}_{k}(s)=\min\bigl\{ T_{l}(s),-k\bigr\} =-k-\bigl(T_{l}(s)+k\bigr)_{-}, \quad \forall s\in \mathbb{R}. $$

Choosing $v=S_{n}(\bar{u}_{\varepsilon})e^{\gamma_{\theta}(\bar {u}_{\varepsilon}) -\gamma_{\theta}(\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon}))} \varphi((\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon})-\bar{\zeta}^{l}_{k}(\bar {u}))_{+})$ as a test function in (4.1), arguing as before we obtain

$$\begin{aligned}& \lim_{n\rightarrow\infty} \varlimsup_{\varepsilon\rightarrow0}\int _{\Omega} S_{n}'(\bar{u}_{\varepsilon}) \bigl[a\bigl(x,\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon}) , \nabla\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a \bigl(x,\bar{\zeta}^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \bar{\zeta}^{l}_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot\nabla\bigl(\bigl(\bar{\zeta}^{l}_{k}( \bar{u}_{\varepsilon})-\bar{\zeta }^{l}_{k}(\bar{u})\bigr)_{+} \bigr)\,\mathrm{d}x\leq0. \end{aligned}$$

Next choosing $v=-S_{n}(\bar{u}_{\varepsilon})e^{\gamma_{\theta}(\bar{\zeta }^{l}_{k}(\bar{u}_{\varepsilon}))- \gamma_{\theta}(\bar{u}_{\varepsilon})} \varphi((\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon})-\bar{\zeta}^{l}_{k}(\bar {u}))_{-})$ as a test function in (4.1), applying the same argument we get

$$\begin{aligned}& \lim_{n\rightarrow\infty} \varliminf_{\varepsilon\rightarrow0}\int _{\Omega} S_{n}'(\bar{u}_{\varepsilon}) \bigl[a\bigl(x,\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon}) , \nabla\bar{\zeta}^{l}_{k}(\bar{u}_{\varepsilon})\bigr)-a \bigl(x,\bar{\zeta}^{l}_{k}(\bar {u}_{\varepsilon}),\nabla \bar{\zeta}^{l}_{k}(\bar{u})\bigr)\bigr] \\& \quad {}\cdot\nabla\bigl(\bigl(\bar{\zeta}^{l}_{k}( \bar{u}_{\varepsilon})-\bar{\zeta }^{l}_{k}(\bar{u})\bigr)_{-} \bigr) \,\mathrm{d}x\geq0. \end{aligned}$$

Proceeding as in Step 4.3, we infer that

$$ \nabla\bar{\zeta}^{l}_{k}( \bar{u}_{\varepsilon})\rightarrow\nabla\bar {\zeta}^{l}_{k}( \bar{u}) \quad \mbox{strongly in } \bigl(L^{p}(\Omega) \bigr)^{N} \mbox{ and a.e. in } \Omega. $$

(4.34)

As a consequence of (4.33) and (4.34), we have

$$ \chi_{\{|\bar{u}_{\varepsilon}|>k\}}\nabla T_{l}(\bar{u}_{\varepsilon }) \rightarrow\chi_{\{|\bar{u}|>k\}}\nabla T_{l}(\bar{u}) \quad \mbox{strongly in } \bigl(L^{p}(\Omega)\bigr)^{N} \mbox{ and a.e. in } \Omega. $$

(4.35)

Step 4.5. In this step we prove that ū satisfies (2.7), where u is replaced by ū.

For any fixed $m>k$, one has

$$\begin{aligned}& \int_{\{x\in\Omega:m\leq|\bar{u}_{\varepsilon}(x)|\leq m+1\}} a(x,\bar {u}_{\varepsilon},\nabla \bar{u}_{\varepsilon})\nabla\bar {u}_{\varepsilon}\,\mathrm{d}x \\& \quad = \int_{\Omega} a(x,\bar{u}_{\varepsilon},\nabla \bar{u}_{\varepsilon })\bigl[\nabla T_{m+1}(\bar{u}_{\varepsilon})- \nabla T_{m}(\bar{u}_{\varepsilon})\bigr]\,\mathrm{d}x. \end{aligned}$$

(4.36)

Thus, passing to the limit as ε tends to zero in (4.36), we deduce that, for fixed $m>k\geq0$,

$$\begin{aligned}& \lim_{\varepsilon\rightarrow0}\int_{\{x\in\Omega:m\leq|\bar{u}_{\varepsilon}(x)|\leq m+1\}} a(x, \bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\nabla\bar {u}_{\varepsilon}\,\mathrm{d}x \\& \quad = \int_{\Omega} a(x,\bar{u},\nabla\bar{u})\bigl[\nabla T_{m+1}(\bar {u})-\nabla T_{m}(\bar{u})\bigr]\,\mathrm{d}x \\& \quad =\int_{\{x\in\Omega:m\leq|\bar{u}|\leq m+1\}} a(x,\bar{u},\nabla\bar {u})\nabla\bar{u}\, \mathrm{d}x. \end{aligned}$$

(4.37)

Taking the limit as m tends to +∞ in (4.37) and using (4.5), we conclude that ū satisfies (2.7).

In the following, we prove that ū satisfies (2.8). Indeed, by (4.1), we have

$$\begin{aligned}& \int_{\Omega}h(\bar{u}_{\varepsilon})a(x, \bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon})\nabla\upsilon\,\mathrm{d}x + \int_{\Omega}\varepsilon h(\bar{u}_{\varepsilon})|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\upsilon\, \mathrm{d}x \\& \qquad {} +\int_{\Omega}h'(\bar{u}_{\varepsilon})a(x, \bar{u}_{\varepsilon},\nabla \bar{u}_{\varepsilon})\nabla\bar{u}_{\varepsilon} \upsilon\,\mathrm{d}x \\& \qquad {}+\int_{\Omega}\varepsilon h'( \bar{u}_{\varepsilon})|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\bar{u}_{\varepsilon} \upsilon\,\mathrm {d}x \\& \qquad {} +\int_{\Omega}h(\bar{u}_{\varepsilon})F_{\varepsilon}(x, \bar {u}_{\varepsilon},\nabla\bar{u}_{\varepsilon})\upsilon\,\mathrm{d}x \\& \quad = \int _{\Omega}h(\bar{u}_{\varepsilon})f_{\varepsilon}\upsilon\, \mathrm{d}x \end{aligned}$$

(4.38)

for any given $\upsilon\in W^{1,\infty}(\Omega)$ and $h\in W^{1,\infty }(\mathbb{R})$ such that $\operatorname{supp}h\subseteq[-l,l]$ for some $l>0$.

Now we first analyze the fifth term on the left-hand side of (4.38). Recall that $\operatorname{supp}h\subseteq[-l,l]$, we get

$$h(\bar{u}_{\varepsilon})F(x,\bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon})=h(\bar{u}_{\varepsilon})F\bigl(x,T_{l}( \bar{u}_{\varepsilon}),\nabla T_{l}(\bar{u}_{\varepsilon})\bigr). $$

Therefore, for any k satisfying $0< k< l$, one has

$$\begin{aligned}& \int_{\Omega}h(\bar{u}_{\varepsilon})F(x, \bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon}) \upsilon\,\mathrm{d}x \\& \quad = \int_{\{x\in\Omega:|\bar{u}_{\varepsilon}|>k\}} h(\bar {u}_{\varepsilon})F \bigl(x,T_{l}(\bar{u}_{\varepsilon}),\nabla T_{l}(\bar {u}_{\varepsilon})\bigr) \upsilon\,\mathrm{d}x \\& \qquad {} + \int_{\{x\in\Omega:|\bar{u}_{\varepsilon}|\leq k\}} h(\bar {u}_{\varepsilon})F \bigl(x,T_{l}(\bar{u}_{\varepsilon}),\nabla T_{l}(\bar {u}_{\varepsilon})\bigr) \upsilon\,\mathrm{d}x \\& \quad =J_{1\varepsilon}+J_{2\varepsilon}. \end{aligned}$$

(4.39)

Similarly to the proof of (3.33) and (3.34), using (4.3) and (4.35) we obtain

$$\begin{aligned} \lim_{k\rightarrow0} \lim_{\varepsilon\rightarrow 0}J_{1\varepsilon}&= \int_{\Omega}h(\bar{u})F\bigl(x,T_{l}(\bar {u}),\nabla T_{l}(\bar{u})\bigr) \upsilon\,\mathrm{d}x \\ &=\int_{\Omega}h(\bar{u})F(x,\bar{u},\nabla\bar{u}) \upsilon\, \mathrm{d}x \end{aligned}$$

(4.40)

and

$$ \lim_{k\rightarrow0} \lim_{\varepsilon\rightarrow 0}J_{2\varepsilon}=0, $$

(4.41)

which imply that

$$ \lim_{\varepsilon\rightarrow 0}\int_{\Omega}h(\bar {u}_{\varepsilon})F(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \upsilon\,\mathrm{d}x=\int_{\Omega}h(\bar{u})F(x,\bar{u},\nabla \bar{u}) \upsilon\,\mathrm{d}x. $$

(4.42)

Similarly, we have

$$ \lim_{\varepsilon\rightarrow 0}\int_{\Omega}h'( \bar {u}_{\varepsilon})a(x,\bar{u}_{\varepsilon},\nabla\bar{u}_{\varepsilon}) \nabla\bar{u}_{\varepsilon}\upsilon\,\mathrm{d}x=\int_{\Omega}h'(\bar {u})a(x,\bar{u},\nabla\bar{u})\nabla\bar{u}\upsilon\, \mathrm{d}x $$

(4.43)

and

$$ \lim_{\varepsilon\rightarrow 0}\int_{\Omega}h(\bar {u}_{\varepsilon})a_{\varepsilon}(x,\bar{u}_{\varepsilon},\nabla\bar {u}_{\varepsilon}) \nabla\upsilon\,\mathrm{d}x=\int_{\Omega}h( \bar {u})a(x,\bar{u},\nabla\bar{u}) \nabla\upsilon\,\mathrm{d}x. $$

(4.44)

As far as the second term of the left-hand side of (4.38) is concerned, by (4.1) we easily get

$$\begin{aligned}& \biggl\vert \int_{\Omega}\varepsilon h(\bar{u}_{\varepsilon})| \nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\upsilon\, \mathrm{d}x\biggr\vert \\& \quad =\biggl\vert \int_{\Omega}\varepsilon h( \bar{u}_{\varepsilon})\bigl|\nabla T_{\tilde{l}}(u_{\varepsilon})\bigr|^{p-2}\nabla T_{\tilde{l}}(u_{\varepsilon})\nabla\upsilon\, \mathrm {d}x\biggr\vert \\& \quad \leq\varepsilon\sup_{\sigma\in[-l,l]}\bigl\vert h(\sigma)\bigr\vert \bigl\Vert \nabla T_{\tilde{l}}(u_{\varepsilon})\bigr\Vert ^{p-1}_{L^{p}(\Omega)}\|\nabla \upsilon\|_{L^{p}(\Omega)},\quad \mbox{where } \tilde{l}=g^{-1}(l), \end{aligned}$$

thus

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega}\varepsilon h( \bar{u}_{\varepsilon})|\nabla u_{\varepsilon}|^{p-2}\nabla u_{\varepsilon}\nabla\upsilon\,\mathrm{d}x=0. $$

(4.45)

Reasoning as in (4.45), one has

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega}\varepsilon h'(\bar{u}_{\varepsilon})|\nabla u_{\varepsilon}|^{p-2} \nabla u_{\varepsilon}\nabla\bar{u}_{\varepsilon} \upsilon\,\mathrm{d}x=0. $$

(4.46)

Finally, it is clear that

$$ \lim_{\varepsilon\rightarrow0}\int_{\Omega}h(\bar {u}_{\varepsilon})f_{\varepsilon}\upsilon\,\mathrm{d}x=\int _{\Omega}h(\bar {u})f\upsilon\,\mathrm{d}x. $$

(4.47)

Then, letting ε tend to zero in (4.38), we conclude from (4.42)-(4.47) that ū satisfies (2.8). Hence, ū is a renormalized solution to problem ($\mathscr{P}$). □