The obstacle problem for nonlinear noncoercive elliptic equations with $L^{1}$-data

Abstract

In this paper, we study the obstacle problem governed by nonlinear noncoercive elliptic equations with $L^{1}$-data. We prove the existence of an entropy solution and show its continuous dependence on the $L^{1}$-data in $W^{1,q}(\varOmega )$ with $q>1$.

1 Introduction

Let $\varOmega \subset \mathbb{R}^{N}$ ($N\geq 2$) be a bounded domain, $p>1$, $\theta \geq 0$, $f\in L^{1}(\varOmega )$, and $g, \psi \in W^{1,p}(\varOmega )\cap L^{ \infty }(\varOmega )$. We consider the obstacle problem governed by the noncoercive operator

$$\begin{aligned} Au= \operatorname{div} \frac{a(x, \nabla u)}{(1+b(x) \vert u \vert )^{\theta (p-1)}} \end{aligned}$$

(1)

associated with $(f,\psi ,g)$, where $a:\varOmega \times \mathbb{R}^{N} \rightarrow \mathbb{R}^{N}$ is a Carathéodory function satisfying:

$$\begin{aligned} &a(x,\xi )\cdot \xi \geq \alpha \vert \xi \vert ^{p}, \end{aligned}$$

(2)

$$\begin{aligned} & \bigl\vert a(x,\xi ) \bigr\vert \leq \beta \bigl(j(x)+ \vert \xi \vert ^{p-1}\bigr), \end{aligned}$$

(3)

$$\begin{aligned} &\bigl(a(x, \zeta )-a(x, \eta )\bigr) (\zeta -\eta )>0, \end{aligned}$$

(4)

$$\begin{aligned} & \bigl\vert a(x, \xi )-a(x, \eta ) \bigr\vert \leq \gamma \textstyle\begin{cases} \vert \xi -\eta \vert ^{p-1} & \text{if }1< p< 2, \\ (1+ \vert \xi \vert + \vert \eta \vert )^{p-2} \vert \xi -\eta \vert & \text{if }p\geq 2, \end{cases}\displaystyle \end{aligned}$$

(5)

for almost every $x\in \varOmega$ and every $\xi ,\zeta ,\eta \in \mathbb{R}^{N}$ with $\zeta \neq \eta$, where α, β, γ are positive constants, j is a nonnegative function in $L^{ \frac{p}{p-1}}(\varOmega )$, and b is an $L^{\infty }$-function satisfying, with some $B\geq 0$,

$$\begin{aligned} 0\leq b(x)\leq B \end{aligned}$$

(6)

for almost every $x\in \varOmega$.

If $f\in W^{-1,p'}(\varOmega )$, then the obstacle problem associated with $(f,\psi ,g)$ is formulated in terms of the inequality

$$\begin{aligned} \int _{\varOmega }\frac{a(x, \nabla u)}{(1+b(x) \vert u \vert )^{\theta (p-1)}}\cdot \nabla (v-u)\,dx+ \int _{\varOmega }f(v-u)\,dx\geq 0 \quad \forall v\in K_{g,\psi }\cap L^{\infty }(\varOmega ) \end{aligned}$$

(7)

whenever $K_{g,\psi }= \{v\in W^{1,p}(\varOmega ); v-g \in W^{1,p} _{0}(\varOmega ), v\geq \psi \text{ a.e. in }\varOmega \} \neq \emptyset$. However, the second integration in (7) is not well defined for $f\in L^{1}(\varOmega )$. Following [1, 3, 5], and so on, we are led to a more general definition of a solution to the obstacle problem with data $f\in L^{1}(\varOmega )$, using the truncation function

$$\begin{aligned} T_{s}(r)=\max \bigl\{ -s,\min \{s,r\}\bigr\} , \quad s,r\in \mathbb{R}. \end{aligned}$$

Definition 1

An entropy solution to the obstacle problem associated with $(f,\psi ,g)$ is a measurable function u such that $u\geq \psi$ a.e. in Ω, $T_{s}(u)-T_{s}(g)\in W_{0}^{1,p}(\varOmega )$ for every $s>0$, and

$$\begin{aligned} \int _{\varOmega }\frac{a(x, \nabla u)}{(1+b(x) \vert u \vert )^{\theta (p-1)}} \cdot \nabla T_{s}(v-u) \,dx+ \int _{\varOmega }fT_{s}(v-u)\,dx \geq 0 \quad \forall v \in K_{g,\psi }\cap L^{\infty }(\varOmega ). \end{aligned}$$

(8)

Observe that in the definition a global integrability condition is required neither on u nor on its gradient. As pointed out in [8], if $T_{s} (u) \in W^{1,p}(\varOmega )$ for all $s > 0$, then there exists a unique measurable vector field $U:\varOmega \rightarrow \mathbb{R}^{N}$ such that $\nabla (T_{s} (u)) = \chi _{\{\vert u\vert < s\}}U$ a.e. in Ω, $s > 0$, which, in fact, coincides with the standard distributional gradient of ∇u whenever $u \in W^{1,1}(\varOmega )$.

The motivation of this paper comes from the study on the Dirichlet boundary value problem

$$\begin{aligned} \textstyle\begin{cases} \operatorname{div}\frac{ \vert \nabla u \vert ^{p-2}\nabla u}{(1+ \vert u \vert )^{\theta (p-1)}}=f & \text{in } \varOmega , \\ u=0 & \text{on } \partial \varOmega . \end{cases}\displaystyle \end{aligned}$$

(9)

Indeed, for the p-Laplacian equation, that is, $\theta =0$ in (9), the existence and regularity of solutions when f has a fine regularity have been well studied. However, under weaker summability assumptions on f, for example, $f\in L^{1}( \varOmega )$, the gradient of u (and even u itself) may not be in $L^{1}(\varOmega )$. In this case, it is possible to give a meaning to solutions of problem (9) by using the concept of entropy solutions. The works on the theory of entropy solutions for p-Laplacian equations have been applied to unilateral problems in [5, 7, 17], and so on and have been extended in [8, 20] to the obstacle problems with $L^{1}$-data in Sobolev spaces with variable exponents and Orlicz–Sobolev spaces, respectively. We remark that the classical obstacle problem for elliptic operators with nonlinear variational energies was considered in [12] and linear elliptic systems involving Radon measures were considered in [19]. Parabolic problems with irregular obstacles and nonstandard $p(x,t)$-growth were considered in [10] and references therein.

If $0<\theta \leq 1$, then due to the lack of coercivity, the standard Leray–Lions surjectivity theorem cannot be used for the establishment of existence of solutions even in the case $f\in W^{-1,p'}(\varOmega )$. To overcome this difficulty, “cutting” the nonlinearity and using the technique of approximation, a pseudomonotone coercive differential operator on $W^{1,p}_{0}(\varOmega )$ can be applied to establish a priori estimates on approximating solutions. Then by the almost everywhere convergence for the gradients of the approximating solutions, the existence and regularity of solutions (or entropy solutions) to problems of the form (9) can be obtained by taking limitation (see, e.g., [1]). For different summability of the data f, Alvino, Boccardo, Ferone, Orsina, Trombetti, et al. have done a lot of work on the existence and regularity of solutions (or entropy solutions) to problems of the form (9) (see [1, 2, 6, 13, 18] and references therein). Particularly, a whole range of existence results have been proven in [6] for $p=2$ and f regular enough, showing that solutions are in some Sobolev space $W^{1,q}_{0}(\varOmega )$ ($1< q\leq 2$) (see also [13, 14, 18]). Nevertheless, little literature has considered the obstacle problem for noncoercive elliptic equations, particularly, for noncoercive elliptic equations with $L^{1}$-data.

Motivated by this, we study the obstacle problem governed by (1) and $(f,\psi ,g)$ with $L^{1}$-data. The main ideas in this paper originate from [1, 8], which can be also applied to the study on a large class of elliptic/parabolic equations [9, 15, 21], potential theory [22], and Schrödinger equations [11, 16]. Throughout this paper, without special statements, we always assume that

$$\begin{aligned} 2-\frac{1}{N}< p< N, \qquad 0\leq \theta (p-1)< \frac{N(p-1)}{N-1}-1, \end{aligned}$$

(10)

$f\in L^{1}(\varOmega )$, $\psi ,g\in W^{1,p}(\varOmega )\cap L^{\infty }( \varOmega )$ satisfy $(\psi -g)^{+}\in W^{1,p}_{0}(\varOmega )$, and $K_{g,\psi }\neq \emptyset$.

Note that (10) implies that

$$\begin{aligned} 0\leq \theta < \frac{N}{N-1}-\frac{1}{p-1}, \quad \text{and} \quad \frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)}>1. \end{aligned}$$

The main result in this paper is the following:

Theorem 1

There exists at least one entropy solution u to the obstacle problem associated with $(f,\psi ,g)$. In addition, u depends continuously on f, that is, if $f_{n}\rightarrow f$ in $L^{1}(\varOmega )$ and $u_{n}$ is a solution to the obstacle problem associated with $(f_{n},\psi ,g)$, then

$$\begin{aligned} u_{n}\rightarrow u \quad \textit{in } W^{1,q}(\varOmega ) \textit{ for all } q\in \biggl(1, \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} \biggr). \end{aligned}$$

Notations

• $\Vert u\Vert _{p}=\Vert u\Vert _{L^{p}(\varOmega )}$ is the norm of $L^{p}(\varOmega )$, where $1\leq p\leq \infty$.

• $\Vert u\Vert _{1,p}=\Vert u\Vert _{W^{1,p}(\varOmega )}$ is the norm of $W^{1,p}(\varOmega )$, where $1< p<\infty$.

• $p'=\frac{p}{p-1}$ with $1< p<\infty$.

• $u^{+}=\max \{u,0\}$, $u^{-}=(-u)^{+}$ for a real-valued function u.

• C is a constant, which may be different from line to line.

• $\{u>s\}=\{x\in \varOmega ;u(x)>s\}$.

• $\{u\leq s\}=\varOmega \setminus \{u>s\}$.

• $\{u< s\}=\{x\in \varOmega ;u(x)< s\}$.

• $\{u\geq s\}=\varOmega \setminus \{u< s\}$.

• $\{u=s\}=\{x\in \varOmega ;u(x)=s\}$.

• $\{t\leq u < s\}=\{u\geq t\}\cap \{u<s\}$.

• $\mathcal{L}^{N}$ is the Lebesgue measure in $\mathbb{R}^{N}$.

• $\vert E\vert =\mathcal{L}^{N}(E)$ for a measurable set E.

2 Preliminaries on entropy solutions

It is worth noting that, for any function $f_{n}$ smooth enough, there exists at least one solution to the obstacle problem (7). Indeed, we can proceed exactly as in Theorem 1.1 of [1] to obtain $W^{1,p}$-solutions due to assumptions (2)–(6), which, particularly, are also entropy solutions. In this section, we establish several auxiliary results on convergence of sequences of entropy solutions as $f_{n}\rightarrow f$ in $L^{1}(\varOmega )$. The main techniques used in this section come from [1, 8]. We start with a priori estimate.

Lemma 2

Let $v_{0}\in K_{g,\psi }\cap L^{\infty }(\varOmega )$, and let u be an entropy solution to the obstacle problem associated with $(f,\psi ,g)$. Then, we have

$$\begin{aligned} \int _{\{ \vert u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+b(x) \vert u \vert )^{\theta (p-1)}} \,dx\leq C \bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p}^{p}+ \Vert f \Vert _{1}\bigl(t+ \Vert v_{0} \Vert _{\infty }\bigr) \bigr) \quad \forall t>0, \end{aligned}$$

where C is a positive constant depending only on α, β, and p.

Proof

For $t>0$, taking $v_{0}$ as a test function in (8), we compute

$$\begin{aligned} \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla u}{(1+b(x) \vert u \vert )^{ \theta (p-1)}}\,dx &\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+b(x) \vert u \vert )^{\theta (p-1)}}\,dx \\ &\quad {} + \int _{\varOmega }f\bigl(T_{t}(v_{0}-u)\bigr)\,dx. \end{aligned}$$

It follows from (2), (3), and Young’s inequality with $\varepsilon >0$ that

$$\begin{aligned} \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\alpha \vert \nabla u \vert ^{p}}{(1+b(x) \vert u \vert )^{ \theta (p-1)}}\,dx &\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta ( \vert j \vert + \vert \nabla u \vert ^{p-1})\cdot \vert \nabla v_{0} \vert }{(1+b(x) \vert u \vert )^{\theta (p-1)}} \,dx+t \Vert f \Vert _{1} \\ &\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta \varepsilon ( \vert j \vert ^{p'}+ \vert \nabla u \vert ^{p})}{(1+b(x) \vert u \vert )^{\theta (p-1)}}\,dx \\ &\quad {} + \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta C(\varepsilon ) \vert \nabla v _{0} \vert ^{p}}{(1+b(x) \vert u \vert )^{\theta (p-1)}}\,dx+ t \Vert f \Vert _{1} \\ &\leq \varepsilon \int _{\{ \vert v_{0}-u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+b(x) \vert u \vert )^{ \theta (p-1)}}\,dx \\ &\quad {} +C\bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p}^{p}\bigr)+t \Vert f \Vert _{1}. \end{aligned}$$

Thus we have

$$\begin{aligned} & \int _{\{ \vert v_{0}-u \vert < t\}} \frac{ \vert \nabla u \vert ^{p}}{(1+b(x) \vert u \vert )^{\theta (p-1)}}\,dx\leq C\bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p}^{p}+t \Vert f \Vert _{1}\bigr). \end{aligned}$$

(11)

Replacing t with $t+\Vert v_{0}\Vert _{\infty }$ in (11) and noting that $\{\vert u\vert < t\}\subset \{\vert v_{0}-u\vert < t+\Vert v_{0}\Vert _{\infty }\}$, we obtain the desired result. □

In the rest of this section, let $\{u_{n}\}$ be a sequence of entropy solutions to the obstacle problem associated with $(f_{n},\psi ,g)$ and assume that

$$\begin{aligned} f_{n}\rightarrow f \quad \text{in } L^{1}(\varOmega ) \quad \text{and} \quad \Vert f_{n} \Vert _{1}\leq \Vert f \Vert _{1}+1. \end{aligned}$$

Lemma 3

For $k>0$ large enough, there exists a measurable function u such that $u_{n}\rightarrow u$ in measure and $T_{k}(u_{n})\rightharpoonup T_{k}(u)$ weakly in $W^{1,p}(\varOmega )$. Thus $T_{k}(u_{n})\rightarrow T_{k}(u)$ strongly in $L^{p}(\varOmega )$, and up to a subsequence, $T_{k}(u_{n})\rightarrow T_{k}(u)$ a.e. in Ω.

Proof

Let $s,t,\varepsilon >0$. We can verify that, for all $m,n\geq 1$,

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n}-u_{m} \vert >s\bigr\} \bigr) &\leq \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t \bigr\} \bigr)+ \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{m} \vert >t\bigr\} \bigr) \\ &\quad {}+{\mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert T_{t}(u_{n})-T_{t}(u_{m}) \bigr\vert >s\bigr\} \bigr)} \end{aligned}$$

(12)

and

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t\bigr\} \bigr)=\frac{1}{t^{p}} \int _{\{ \vert u_{n} \vert >t\}}t ^{p}\,dx\leq \frac{1}{t^{p}} \int _{\varOmega } \bigl\vert T_{t}(u_{n}) \bigr\vert ^{p} \,dx. \end{aligned}$$

(13)

Since $v_{0}=g+(\psi -g)^{+}\in K_{g,\psi }\cap L^{\infty }(\varOmega )$, by Lemma 2 we have

$$\begin{aligned} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n}) \bigr\vert ^{p}\,dx &= \int _{\{ \vert u_{n} \vert < t\}} \vert \nabla u_{n} \vert ^{p}\,dx \\ &\leq C(1+Bt)^{\theta (p-1)}\bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p} ^{p}+ \Vert f \Vert _{1}\bigl(t+ \Vert v_{0} \Vert _{\infty } \bigr)\bigr). \end{aligned}$$

(14)

Note that $T_{t}(u_{n})-T_{t}(g)\in W^{1,p}_{0}(\varOmega )$. By (13), (14), and Poincaré’s inequality, for every $t>\Vert g\Vert _{\infty }$ and for some positive constant C independent of n and t, we have

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t\bigr\} \bigr) &\leq \frac{1}{t^{p}} \int _{\varOmega } \bigl\vert T _{t}(u_{n}) \bigr\vert ^{p}\,dx \\ &\leq \frac{2^{p-1}}{t^{p}} \int _{\varOmega } \bigl\vert T_{t}(u_{n})-T_{t}(g) \bigr\vert ^{p} \,dx+\frac{2^{p-1}}{t^{p}} \Vert g \Vert _{p}^{p} \\ &\leq \frac{C}{t^{p}} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n})- \nabla T_{t}(g) \bigr\vert ^{p} \,dx+\frac{2^{p-1}}{t^{p}} \Vert g \Vert _{p}^{p} \\ &\leq \frac{C}{t^{p}} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n}) \bigr\vert ^{p}\,dx+\frac{C}{t ^{p}} \Vert g \Vert _{1,p}^{p} \\ &\leq \frac{C(1+t)^{1+\theta (p-1)}}{t^{p}}. \end{aligned}$$

Since $0\leq \theta <1$, there exists $t_{\varepsilon }>0$ such that

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n} \vert >t\bigr\} \bigr)< \frac{\varepsilon }{3} \quad \forall t\geq t_{\varepsilon }, \forall n\geq 1. \end{aligned}$$

(15)

Now, as in (13), we have

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert T_{t_{\varepsilon }}(u_{n})-T_{t_{\varepsilon }}(u _{m}) \bigr\vert >s\bigr\} \bigr) &=\frac{1}{s^{p}} \int _{\{ \vert T_{t_{\varepsilon }} (u_{n})-T_{t_{\varepsilon }}(u_{m}) \vert >s \}} s^{p}\,dx \\ &\leq \frac{1}{s^{p}} \int _{\varOmega } \bigl\vert T_{t_{\varepsilon }}(u_{n})-T _{t_{\varepsilon }}(u_{m}) \bigr\vert ^{p}\,dx. \end{aligned}$$

(16)

Using (14) and the fact that $T_{t}(u_{n})-T_{t}(g)\in W^{1,p} _{0}(\varOmega )\cap L^{\infty }(\varOmega )$ again, we see that $\{T_{t _{\varepsilon }}(u_{n})\}$ is a bounded sequence in $W^{1,p}(\varOmega )$. Thus, up to a subsequence, $\{T_{t_{\varepsilon }}(u_{n})\}$ converges strongly in $L^{p}(\varOmega )$. By (16) there exists $n_{0}=n _{0}(t_{\varepsilon },s)\geq 1$ such that

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert T_{t_{\varepsilon }}(u_{n})-T_{t_{\varepsilon }}(u _{m}) \bigr\vert >s\bigr\} \bigr)< \frac{\varepsilon }{3} \quad \forall n,m\geq n_{0}. \end{aligned}$$

(17)

Combining (12), (15), and (17), we obtain

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \vert u_{n}-u_{m} \vert >s\bigr\} \bigr)< \varepsilon , \quad \forall n,m\geq n_{0}. \end{aligned}$$

Hence $\{u_{n}\}$ is a Cauchy sequence in measure, and therefore there exists a measurable function u such that $u_{n}\rightarrow u$ in measure. Note that $T_{k}(u_{n})-T_{k}(g)\in W^{1,p}_{0}(\varOmega )$. By (14) and Poincaré’s inequality we conclude that, for fixed k, $\{T_{k}(u_{n})\}$ is a bounded sequence in $W^{1,p}(\varOmega )$. Therefore, $T_{k}(u_{n})\rightarrow T_{k}(u)$ strongly in $L^{p}(\varOmega )$, and, up to a subsequence, $T_{k}(u_{n})\rightarrow T_{k}(u)$ a.e. in Ω. □

Proposition 4

There exist a subsequence of $\{u_{n}\}$ and a measurable function u such that, for each $q\in (1,\frac{N(p-1)(1-\theta )}{N-1- \theta (p-1)} )$, we have

$$\begin{aligned} u_{n}\rightarrow u \quad \textit{strongly in } W^{1,q}( \varOmega ). \end{aligned}$$

Furthermore, if $0\leq \theta < \min \{\frac{1}{N-p+1},\frac{N}{N-1}- \frac{1}{p-1}\}$, then

$$\begin{aligned} \frac{a(x, \nabla u_{n})}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla u)}{(1+b(x) \vert u \vert )^{\theta (p-1)}} \quad \textit{strongly in } \bigl(L^{1}(\varOmega )\bigr)^{N}. \end{aligned}$$

To prove Proposition 4, we need two preliminary lemmas.

Lemma 5

There exists a subsequence of $\{u_{n}\}$ and a measurable function u such that for each $q\in (1,\frac{N(p-1)(1-\theta )}{N-1- \theta (p-1)} )$, we have that $u_{n}\rightharpoonup u$ weakly in $W^{1,q}(\varOmega )$ and $u_{n}\rightarrow u$ strongly in $L^{q}(\varOmega )$.

Proof

Let $k>0$ and $n\geq 1$. Define $D_{k}=\{\vert u_{n}\vert \leq k\}$ and $B_{k}=\{k\leq \vert u_{n}\vert < k+1\}$. Using Lemma 2 with $v_{0}=g+(\psi -g)^{+}$, we get

$$\begin{aligned} \int _{D_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}} \,dx\leq C(1+k), \end{aligned}$$

(18)

where C is a positive constant depending only on α, β, p, $\Vert j\Vert _{p'}$, $\Vert f\Vert _{1}$, $\Vert \nabla v_{0}\Vert _{p}$, and $\Vert v_{0} \Vert _{\infty }$.

Taking the function $T_{k}(u_{n})$ with $k>\{ \Vert g\Vert _{\infty },\Vert \psi \Vert _{\infty }\}$ as a test function for the problem associated with $(f_{n},\psi ,g)$, we obtain

$$\begin{aligned} \int _{\varOmega }\frac{a(x,\nabla u_{n})\cdot \nabla T_{1}(u_{n}-T_{k}(u _{n}))}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}}\,dx\leq \int _{\varOmega }-f _{n}T_{1} \bigl(u_{n}-T_{k}(u_{n})\bigr)\,dx, \end{aligned}$$

which, together with (2), gives

$$\begin{aligned} \int _{B_{k}}\frac{\alpha \vert \nabla u_{n} \vert ^{p}}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}}\,dx\leq \Vert f_{n} \Vert _{1}\leq \Vert f \Vert _{1}+1. \end{aligned}$$

(19)

Let $q\in (1,\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} )$ and $r=\frac{q\theta (p-1)}{p}$. Noting that $\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)}< p$, it follows $q< p$. Let $\varTheta =\theta (p-1)$. Since $\frac{B-1}{A-1}\leq \frac{B}{A}$ for all $A\geq B>1$, we have $\frac{p-\varTheta -1}{N-\varTheta -1}\leq \frac{p-\varTheta }{N-\varTheta }$, which implies $q<\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)}=\frac{N(p-1)-N\varTheta }{N-1-\varTheta }<\frac{N(p-\varTheta )}{N-\varTheta }$. So we get $\frac{\varTheta }{p-q}<\frac{N}{N-q}$. It follows that $\frac{pr}{p-q}=\frac{q\theta (p-1)}{p-q} =\frac{q\varTheta }{p-q} <\frac{Nq}{ N-q}:=q^{*}$. For all $k>0$, we estimate $\int _{B_{k}}\vert \nabla u_{n}\vert ^{q}\,dx$:

$$\begin{aligned} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,dx &= \int _{B_{k}}\frac{ \vert \nabla u _{n} \vert ^{q}}{(1+b(x) \vert u_{n} \vert )^{r}}\cdot \bigl(1+b(x) \vert u_{n} \vert \bigr)^{r}\,dx \\ &\leq \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+b(x) \vert u_{n} \vert )^{ \theta (p-1)}}\,dx \biggr)^{\frac{q}{p}} \biggl( \int _{B_{k}}\bigl(1+b(x) \vert u _{n} \vert \bigr)^{\frac{pr}{p-q}}\,dx \biggr)^{\frac{p-q}{p}} \\ &\leq C \vert B_{k} \vert ^{\frac{p-q}{p}}+C \biggl( \int _{B_{k}} \vert u_{n} \vert ^{ \frac{pr}{p-q}}\,dx \biggr)^{\frac{p-q}{p}}\quad \text{by (19) and (6)} \\ &\leq C \vert B_{k} \vert ^{\frac{p-q}{p}} +C \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,dx \biggr)^{\frac{r}{q^{*}}} \cdot \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{r}{q ^{*}}}. \end{aligned}$$

Since $\vert B_{k}\vert \leq \frac{1}{k^{q^{*}}}\int _{B_{k}}\vert u_{n}\vert ^{q^{*}} \,dx$, for $k\geq k_{0}\geq 1$, we have

$$\begin{aligned} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,dx &\leq C \biggl( \frac{1}{k^{q^{*}}} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,dx \biggr)^{ \frac{p-q}{p}} +C\frac{1}{k^{q^{*} (\frac{p-q}{p}-\frac{r}{q^{*}} )}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,dx \biggr)^{ \frac{p-q}{p}} \\ &\leq \frac{2C}{k^{q^{*} (\frac{p-q}{p}-\frac{r}{q^{*}} )}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,dx \biggr)^{\frac{p-q}{p}}. \end{aligned}$$

Summing up from $k=k_{0}$ to $k=K$ and using Hölder’s inequality, we have

$$\begin{aligned} \sum_{k=k_{0}}^{K} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,dx\leq C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{r}{q ^{*}})\frac{p}{q}}} \Biggr)^{\frac{q}{p}}\cdot \Biggl(\sum _{k=k_{0}} ^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,dx \Biggr)^{\frac{p-q}{p}}. \end{aligned}$$

(20)

Note that

$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,dx= \int _{D_{k_{0}}} \vert \nabla u_{n} \vert ^{q}\,dx+\sum_{k=k_{0}}^{K} \int _{B_{k}} \vert \nabla u _{n} \vert ^{q} \,dx. \end{aligned}$$

(21)

To estimate the first integral in the right-hand side of (21), using Hölder’s inequality, (18), and (6), we obtain

$$\begin{aligned} \int _{D_{k_{0}}} \vert \nabla u_{n} \vert ^{q}\,dx &\leq \biggl( \int _{D_{k _{0}}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}} \,dx \biggr)^{\frac{q}{p}} \biggl( \int _{D_{k_{0}}}\bigl(1+b(x) \vert u_{n} \vert \bigr)^{ \frac{pr}{p-q}} \,dx \biggr)^{\frac{p-q}{p}} \\ &\leq C, \end{aligned}$$

(22)

where C depends only on α, β, B, p, θ, $\Vert j\Vert _{p'}$, $\Vert f\Vert _{1}$, $\Vert \nabla v_{0}\Vert _{p}$, $\Vert v_{0}\Vert _{\infty }$, and $k_{0}$.

Note that $\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{r}{q ^{*}})\frac{p}{q}}}$ converges since $q^{*} (\frac{p-q}{p}-\frac{r}{q ^{*}} )\frac{p}{q}>1$. Combining (20)–(22), we get, for $k_{0}$ large enough,

$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,dx\leq C+C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,dx \biggr)^{ \frac{p-q}{p}}. \end{aligned}$$

(23)

Since $p>q$, $T_{K}(u_{n})\in W^{1,q}(\varOmega )$ and $T_{K}(g)=g\in W ^{1,q}(\varOmega )$ for $K>\Vert g\Vert _{\infty }$. Hence $T_{K}(u_{n})-g\in W ^{1,q}_{0}(\varOmega )$. Using the Sobolev embedding $W^{1,q}_{0}(\varOmega )\subset L^{q^{*}}(\varOmega )$ and Poincaré’s inequality, we obtain

$$\begin{aligned} \bigl\Vert T_{K}(u_{n}) \bigr\Vert ^{q}_{q^{*}} &\leq 2^{q-1}\bigl( \bigl\Vert T_{K}(u_{n})-g \bigr\Vert ^{q}_{q ^{*}}+ \Vert g \Vert ^{q}_{q^{*}}\bigr) \\ &\leq C\bigl( \bigl\Vert \nabla \bigl(T_{K}(u_{n})-g \bigr) \bigr\Vert ^{q}_{q}+ \Vert g \Vert ^{q}_{q^{*}}\bigr) \\ &\leq C \bigl( \bigl\Vert \nabla T_{K}(u_{n}) \bigr\Vert ^{q}_{q}+ \Vert \nabla g \Vert ^{q}_{q}+ \Vert g \Vert ^{q}_{q^{*}} \bigr) \\ &\leq C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,dx \biggr). \end{aligned}$$

(24)

Using the fact that

$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,dx\leq \int _{\{ \vert u_{n} \vert \leq K\}} \bigl\vert T_{K}( u_{n}) \bigr\vert ^{q^{*}} \,dx\leq \bigl\Vert T _{K}( u_{n}) \bigr\Vert ^{q^{*}}_{q^{*}}, \end{aligned}$$

(25)

from (23)–(24) we obtain

$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,dx\leq C+C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,dx \biggr)^{\frac{q ^{*}}{q}\frac{p-q}{p}}. \end{aligned}$$

(26)

Note that $p< N\Leftrightarrow \frac{q^{*}}{q}\frac{p-q}{p}<1$. It follows from (26) that, for $k_{0}$ large enough, $\int _{\{\vert u_{n}\vert \leq K\}}\vert \nabla u_{n}\vert ^{q}\,dx$ is bounded independently of n and K. Using (24) and (25), we deduce that $\int _{\{\vert u_{n}\vert \leq K\}}\vert u_{n}\vert ^{q^{*}}\,dx$ is also bounded independently of n and K. Letting $K\rightarrow \infty$, we deduce that $\Vert \nabla u_{n}\Vert _{q}$ and $\Vert u_{n}\Vert _{q^{*}}$ are uniformly bounded independently of n. Particularly, $u_{n}$ is bounded in $W^{1,q}(\varOmega )$. Therefore, there exist a subsequence of $\{u_{n}\}$ and a function $v\in W^{1,q}(\varOmega )$ such that $u_{n}\rightharpoonup v$ weakly in $W^{1,q}(\varOmega )$ and $u_{n}\rightarrow v$ strongly in $L^{q}(\varOmega )$ and a.e. in Ω. By Lemma 3, $u_{n}\rightarrow u$ in measure in Ω, and we conclude that $u=v$ and $u\in W^{1,q}(\varOmega )$. □

Lemma 6

There exist a subsequence of $\{u_{n}\}$ and a measurable function u such that $\nabla u_{n}$ converges to ∇u almost everywhere in Ω.

Proof

The proof is similar to that of [1, Thm. 4.1] and can be also found in [4]. Here we sketch only the main steps due to slight modifications. For $r_{2}>1$, let $\lambda =\frac{q}{pr_{2}}<1$, where q is the same as in Lemma 5. Define $A(x,u, \xi )=\frac{a(x,\xi )}{(1+b(x)\vert u\vert )^{\theta (p-1)}}$ (for simplicity, we omit the dependence of $A(x,u,\xi )$ on x) and

$$\begin{aligned} I(n)= \int _{\varOmega } \bigl(\bigl(A(u_{n},\nabla u_{n})-A(u_{n},\nabla u)\bigr) \cdot \nabla (u_{n}-u) \bigr)^{\lambda }\,dx. \end{aligned}$$

We fix $k>0$ and split the integral in $I(n)$ on the sets $\{\vert u\vert >k\}$ and $\{\vert u\vert \leq k\}$, obtaining

$$\begin{aligned} I_{1}(n,k)= \int _{\{\vert u\vert >k\}} \bigl(\bigl(A(u_{n},\nabla u_{n})-A(u_{n},\nabla u)\bigr)\cdot \nabla (u_{n}-u) \bigr)^{\lambda }\,dx \end{aligned}$$

and

$$\begin{aligned} I_{2}(n,k)= \int _{\{\vert u\vert \leq k\}} \bigl(\bigl(A(u_{n},\nabla u_{n})-A(u_{n}, \nabla u)\bigr)\cdot \nabla (u_{n}-u) \bigr)^{\lambda }\,dx. \end{aligned}$$

For $I_{2}(n,k)$, we have

$$\begin{aligned} I_{2}(n,k)\leq I_{3}(n,k)= \int _{\varOmega } \bigl(\bigl(A_{n}(u_{n},\nabla u _{n})-A_{n}\bigl(u_{n},\nabla T_{k}(u)\bigr)\bigr)\cdot \nabla \bigl(u_{n}-T_{k}(u) \bigr) \bigr)^{\lambda }{d}x. \end{aligned}$$

Fix $h>0$ and split $I_{3}(n,k)$ on the sets $\{\vert u_{n}-T_{k}(u)\vert > h\}$ and ${\{\vert u_{n}-T_{k}(u)\vert \leq h\}}$, obtaining

$$\begin{aligned} I_{4}(n,k,h)= \int _{\{\vert u_{n}-T_{k}(u)\vert >h\}} \bigl(\bigl(A_{n}(u_{n},\nabla u _{n})-A_{n}\bigl(u_{n},\nabla T_{k}(u)\bigr)\bigr)\cdot \nabla \bigl(u_{n}-T_{k}(u) \bigr) \bigr)^{\lambda }\,dx \end{aligned}$$

and

$$\begin{aligned} I_{5}(n,k,h) &= \int _{\{ \vert u_{n}-T_{k}(u) \vert \leq h\}} \bigl(\bigl(A_{n}(u_{n}, \nabla u_{n})-A_{n}\bigl(u_{n},\nabla T_{k}(u)\bigr)\bigr)\cdot \nabla \bigl(u_{n}-T_{k}(u) \bigr) \bigr)^{\lambda }\,dx \\ &= \int _{\varOmega } \bigl(\bigl(A_{n}(u_{n},\nabla u_{n})-A_{n}\bigl(u_{n},\nabla T _{k}(u)\bigr)\bigr)\cdot \nabla T_{h}\bigl(u_{n}-T_{k}(u) \bigr) \bigr)^{\lambda }\,dx \\ &\leq \vert \varOmega \vert ^{1-\lambda } \biggl( \int _{\varOmega }\bigl(A_{n}(u_{n},\nabla u _{n})-A_{n}\bigl(u_{n},\nabla T_{k}(u)\bigr)\bigr)\cdot \nabla T_{h}\bigl(u_{n}-T_{k}(u) \bigr) \,dx \biggr)^{\lambda } \\ &= \vert \varOmega \vert ^{1-\lambda } \bigl(I_{6}(n,k,h) \bigr)^{\lambda }. \end{aligned}$$

For $I_{6}(n,k,h)$, we can split it as the difference $I_{7}(n,k,h)-I _{8}(n,k,h)$, where

$$\begin{aligned} I_{7}(n,k,h)= \int _{\varOmega }A(u_{n},\nabla u_{n})\cdot \nabla T_{h}\bigl(u _{n}-T_{k}(u)\bigr)\,dx \end{aligned}$$

and

$$\begin{aligned} I_{8}(n,k,h)= \int _{\varOmega }A\bigl(u_{n},\nabla T_{k}(u) \bigr)\cdot \nabla T_{h}\bigl(u _{n}-T_{k}(u) \bigr)\,dx. \end{aligned}$$

Note that $\vert \nabla u_{n}\vert$ is bounded in $L^{q}(\varOmega )$ and $\lambda p r_{2}=q$. Due to Lemmas 3 and 5, in the same way as Theorem 4.1 in [1], we can get that

$$\begin{aligned}& \lim_{k\rightarrow \infty }\limsup_{n\rightarrow \infty }I_{1}(n,k)=0,\quad\quad \lim_{h\rightarrow \infty }\limsup_{k\rightarrow \infty } \limsup _{n\rightarrow \infty }I_{4}(n,k,h)=0, \\& \lim_{n\rightarrow \infty }I_{8}(n,k,h)=0. \end{aligned}$$

For $I_{7}(n,k,h)$, let $k> \max \{\Vert g\Vert _{\infty },\Vert \psi \Vert _{\infty }\}$ and $n\geq h+k$. Take $T_{k}(u)$ as a test function for (8), obtaining

$$\begin{aligned} I_{7}(n,k,h)\leq \int _{\varOmega }-f_{n}T_{h} \bigl(u_{n}-T_{k}(u)\bigr)\,dx. \end{aligned}$$

Using the strong convergence of $f_{n}$ in $L^{1}(\varOmega )$, we have

$$\begin{aligned} \lim_{n\rightarrow \infty }I_{7}(n,k,h) \leq \int _{\varOmega }-fT_{h}\bigl(u-T _{k}(u)\bigr) \,dx. \end{aligned}$$

Note that $h>0$ and $\lim_{k\rightarrow \infty }T_{h}(u-T_{k}(u))=0$. It follows that

$$\begin{aligned} \lim_{k\rightarrow \infty }\lim_{n\rightarrow \infty }I_{7}(n,k,h) \leq 0. \end{aligned}$$

Putting together all the limitations and noting that $I(n)\geq 0$, we have

$$\begin{aligned} \lim_{n\rightarrow \infty }I(n)=0. \end{aligned}$$

The same arguments as in [1] give that, up to a subsequence, $\nabla u_{n}(x)\rightarrow \nabla u(x)$ a.e. □

Proof of Proposition 4

We shall prove that $\nabla u_{n}$ converges strongly to $\nabla u_{n}$ in $L^{q}(\varOmega )$ for each $q\in (1,\frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)} )$. To this end, we will apply Vitalli’s theorem, using the fact that by Lemma 5, $\nabla u_{n}$ is bounded in $L^{q}(\varOmega )$ for each $q\in (1,\frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)} )$. Let $r\in (q,\frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)} )$, and let $E\subset \varOmega$ be a measurable set. We have, by Hölder’s inequality,

$$\begin{aligned} \int _{E} \vert \nabla u_{n} \vert ^{q}\,dx\leq \biggl( \int _{E} \vert \nabla u_{n} \vert ^{r} \,dx \biggr)^{\frac{q}{r}}\cdot \vert E \vert ^{\frac{r-q}{r}}\leq C \vert E \vert ^{ \frac{r-q}{r}}\rightarrow 0 \end{aligned}$$

uniformly in n as $\vert E\vert \rightarrow 0$. From this and from Lemma 6 we deduce that $\nabla u_{n}$ converges strongly to ∇u in $L^{q}(\varOmega )$.

Now assume that $0\leq \theta < \min \{\frac{1}{N-p+1}, \frac{N}{N-1}-\frac{1}{p-1} \}$. Note that since $\nabla u_{n}$ converges to ∇u a.e. in Ω, to prove the convergence

$$\begin{aligned} \frac{a(x, \nabla u_{n})}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla u)}{(1+b(x) \vert u \vert )^{\theta (p-1)}} \quad \text{strongly in } \bigl(L^{1}(\varOmega )\bigr)^{N}, \end{aligned}$$

it suffices, thanks to Vitalli’s theorem, to show that, for every measurable subset $E\subset \varOmega$, $\int _{E} \vert \frac{a(x, \nabla u_{n})}{(1+b(x)\vert u_{n}\vert )^{\theta (p-1)}} \vert \,dx$ converges to 0 uniformly in n as $\vert E\vert \rightarrow 0$. Note that $p-1<\frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)}$ by the assumptions. For any $q\in (p-1, \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)})$, we deduce by Hölder’s inequality that

$$\begin{aligned} \int _{E} \biggl\vert \frac{a(x, \nabla u_{n})}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}} \biggr\vert \,dx &\leq \beta \int _{E}\bigl(j+ \vert \nabla u_{n} \vert ^{p-1}\bigr)\,dx \\ &\leq \beta \Vert j \Vert _{p'} \vert E \vert ^{\frac{1}{p}}+ \beta \biggl( \int _{E} \vert \nabla u_{n} \vert ^{q}\,dx \biggr)^{\frac{p-1}{q}} \vert E \vert ^{\frac{q-p+1}{q}} \\ &\rightarrow 0 \quad \text{uniformly in }n\text{ as } \vert E \vert \rightarrow 0. \end{aligned}$$

 □

Lemma 7

There exists a subsequence of $\{u_{n}\}$ such that, for all $k>0$,

$$\begin{aligned} \frac{a(x, \nabla T_{k}(u_{n}))}{(1+b(x) \vert T_{k}(u_{n}) \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla T_{k}(u))}{(1+b(x) \vert T_{k}(u) \vert )^{\theta (p-1)}} \quad \textit{strongly in } \bigl(L^{1}(\varOmega )\bigr)^{N}. \end{aligned}$$

Proof

Let k be a positive number. It is well known that if a sequence of measurable functions $\{u_{n}\}$ with uniform boundedness in $L^{p}(\varOmega )$ ($p>1$) converges in measure to u, then $u_{n}$ converges strongly to u in $L^{1}(\varOmega )$. First note that the sequence $\{\frac{a(x, \nabla T_{k}(u_{n}))}{(1+b(x)\vert T_{k}(u _{n})\vert )^{\theta (p-1)}} \}$ is bounded in $L^{p'}(\varOmega )$. Indeed, we have by (3) and Lemma 2,

$$\begin{aligned} \int _{\varOmega } \biggl\vert \frac{a(x, \nabla T_{k}(u_{n}))}{(1+b(x) \vert T_{k}(u _{n}) \vert )^{\theta (p-1)}} \biggr\vert ^{p'}\,dx &\leq \beta \Vert j \Vert ^{p'} _{p'}+\beta \int _{\varOmega }\frac{ \vert \nabla T_{k}(u_{n}) \vert ^{p}}{(1+b(x) \vert T _{k}(u_{n}) \vert )^{\theta p}}\,dx \\ &\leq \beta \Vert j \Vert ^{p'}_{p'}+\beta \int _{\varOmega }\frac{ \vert \nabla T_{k}(u _{n}) \vert ^{p}}{(1+b(x) \vert T_{k}(u_{n}) \vert )^{\theta (p-1)}}\,dx \\ &\leq C. \end{aligned}$$

Next, it suffices to show that there exists a subsequence of $\{u_{n}\}$ such that

$$\begin{aligned} \frac{a(x, \nabla T_{k}(u_{n}))}{(1+b(x) \vert T_{k}(u_{n}) \vert )^{\theta (p-1)}} \rightarrow \frac{a(x, \nabla T_{k}(u))}{(1+b(x) \vert T_{k}(u) \vert )^{\theta (p-1)}} \quad \text{in measure}. \end{aligned}$$

Note that $u_{n},u\in W^{1,q}(\varOmega )$, where q is the same as in Proposition 4. The a.e. convergence of $u_{n}$ to u and the fact that $\nabla u_{n}\rightarrow \nabla u$ in measure imply that

$$\begin{aligned} \nabla T_{k}(u_{n})\rightarrow \nabla T_{k}(u) \quad \text{in measure}. \end{aligned}$$

Let s, t be positive numbers and write $\nabla _{A}u=\frac{a(x, \nabla u)}{(1+b(x)\vert u\vert )^{\theta (p-1)}}$. Define

$$\begin{aligned} &E_{n}=\bigl\{ \bigl\vert \nabla _{A} T_{k}(u_{n})-\nabla _{A} T_{k}(u) \bigr\vert >s\bigr\} , \\ &E_{n}^{1}=\bigl\{ \bigl\vert \nabla T_{k}(u_{n}) \bigr\vert >t\bigr\} , \\ &E_{n}^{2}=\bigl\{ \bigl\vert \nabla T_{k}(u) \bigr\vert >t\bigr\} , \\ &E_{n}^{3}=E_{n}\cap \bigl\{ \bigl\vert \nabla T_{k}(u_{n}) \bigr\vert \leq t\bigr\} \cap \bigl\{ \bigl\vert \nabla T _{k}(u) \bigr\vert \leq t\bigr\} . \end{aligned}$$

Note that $E_{n}\subset E_{n}^{1}\cup E_{n}^{2}\cup E_{n}^{3}$ for each $n\geq 1$. Since by Lemma 5 the sequence $\{u_{n}\}$ and the function u are uniformly bounded in $W^{1,q}(\varOmega )$, we obtain

$$\begin{aligned} &\mathcal{L}^{N}\bigl(E_{n}^{1}\bigr)\leq \frac{1}{t^{q}} \int _{\varOmega } \bigl\vert \nabla T_{k}(u_{n}) \bigr\vert ^{q}\,dx\leq \frac{1}{t^{q}} \int _{\varOmega } \vert \nabla u _{n} \vert ^{q}\,dx \leq \frac{C}{t^{q}}, \\ &\mathcal{L}^{N}\bigl(E_{n}^{2}\bigr)\leq \frac{1}{t^{q}} \int _{\varOmega } \bigl\vert \nabla T_{k}(u) \bigr\vert ^{q}\,dx\leq \frac{1}{t^{q}} \int _{\varOmega } \vert \nabla u \vert ^{q} \,dx \leq \frac{C}{t^{q}}. \end{aligned}$$

We deduce that, for any $\varepsilon >0$, there exists $t_{\varepsilon }>0$ such that

$$\begin{aligned} \mathcal{L}^{N}\bigl(E_{n}^{1}\bigr)+ \mathcal{L}^{N}\bigl(E_{n}^{2}\bigr)< \frac{\varepsilon }{3} \quad \forall t\geq t_{\varepsilon }, \forall n\geq 1. \end{aligned}$$

(27)

Note that, for $a\geq b\geq 0$ and $\tau \geq 0$, we have the inequality

$$\begin{aligned} \biggl\vert \frac{1}{(1+a)^{\tau }}-\frac{1}{(1+b)^{\tau }} \biggr\vert = \biggl\vert \frac{ \tau (b-a)}{(1+c)^{1+\tau }} \biggr\vert \leq \tau \vert b-a \vert \quad \text{for some } c\in [b,a]. \end{aligned}$$

From (3), (5), and (6) we deduce that, in $E_{n}^{3}$,

$$\begin{aligned} s&< \bigl\vert \nabla _{A} T_{k}(u_{n})- \nabla _{A} T_{k}(u) \bigr\vert \\ &= \biggl\vert \frac{a(x,\nabla T_{k}(u_{n}))-a(x,\nabla T_{k}(u))}{(1+b(x) \vert T _{k}(u_{n}) \vert )^{\theta (p-1)}} \\ &\quad {} + \biggl(\frac{1}{(1+b(x) \vert T_{k}(u_{n}) \vert )^{\theta (p-1)}}-\frac{1}{(1+b(x) \vert T _{k}(u) \vert )^{\theta (p-1)}} \biggr)a\bigl(x,\nabla T_{k}(u)\bigr) \biggr\vert \\ &\leq \theta (p-1)B \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert \cdot \beta \bigl(j+ \bigl\vert \nabla T _{k}(u) \bigr\vert ^{p-1}\bigr) \\ &\quad {} + \gamma \textstyle\begin{cases} \vert \nabla T_{k}(u_{n})-\nabla T_{k}(u) \vert ^{p-1} & \text{if }1< p< 2, \\ \vert \nabla T_{k}(u_{n})-\nabla T_{k}(u) \vert (1+ \vert \nabla T_{k}(u_{n}) \vert + \vert \nabla T_{k}(u) \vert )^{p-2} & \text{if }p\geq 2 \end{cases}\displaystyle \\ &\leq C_{0}j \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert \\ &\quad {} +C_{0}\bigl(1+t^{p-1}+t^{p-2}\bigr) \bigl( \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert + \bigl\vert \nabla T _{k}(u_{n})-\nabla T_{k}(u) \bigr\vert \bigr), \end{aligned}$$

which leads to $E^{3}_{n}\subset F_{1} \cup F_{2}$ with

$$\begin{aligned} &F_{1}=\biggl\{ j \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert >\frac{s}{2C_{0}}\biggr\} , \\ &F_{2}= \biggl\{ \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert + \bigl\vert \nabla T_{k}(u_{n})- \nabla T_{k}(u) \bigr\vert >\frac{s}{2C_{0}(1+t^{p-1}+t^{p-2})} \biggr\} . \end{aligned}$$

In $F_{1}$, we have

$$\begin{aligned} \mathcal{L}^{N}(F_{1})=\frac{2C_{0}}{s} \int _{F_{1}}\frac{s}{2C_{0}} \,dx< \frac{2C_{0}}{s} \int _{F_{1}}j \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert \,dx. \end{aligned}$$

By Lemma 3 we deduce that there exists $n_{0}=n_{0}(S,C _{0},\varepsilon )$ such that

$$\begin{aligned} \mathcal{L}^{N}(F_{1})\leq \frac{\varepsilon }{3} \quad \forall n\geq n_{0}. \end{aligned}$$

(28)

Note that $F_{2}\subset F_{3} \cup F_{4}$ with

$$\begin{aligned} &F_{3}= \biggl\{ \bigl\vert T_{k}(u_{n})- T_{k}(u) \bigr\vert >\frac{s}{4C_{0}(1+t^{p-1}+t ^{p-2})} \biggr\} , \\ &F_{4}= \biggl\{ \bigl\vert \nabla T_{k}(u_{n})- \nabla T_{k}(u) \bigr\vert >\frac{s}{4C_{0}(1+t ^{p-1}+t^{p-2})} \biggr\} . \end{aligned}$$

Using the convergence in measure of $\nabla T_{k}(u_{n})$ to $\nabla T_{k}(u)$ and of $T_{k}(u_{n})$ to $T_{k}(u)$, for $t=t_{ \varepsilon }$, we obtain the existence of $n_{1}=n_{1}(s,\varepsilon )\geq 1$ such that

$$\begin{aligned} \mathcal{L}^{N}(F_{2})\leq \mathcal{L}^{N}(F_{3}) + \mathcal{L}^{N}(F _{4}) < \frac{\varepsilon }{3} \quad \forall n\geq n_{1}. \end{aligned}$$

(29)

Combining (27), (28), and (29), we obtain

$$\begin{aligned} \mathcal{L}^{N}\bigl(\bigl\{ \bigl\vert \nabla _{A}T_{k}(u_{n})- \nabla _{A}T_{k}(u) \bigr\vert >s\bigr\} \bigr)< \varepsilon \quad \forall n\geq \max \{n_{0},n_{1}\}. \end{aligned}$$

Hence the sequence $\{\nabla _{A}T_{k}(u_{n})\}$ converges in measure to $\nabla _{A}T_{k}(u)$, and the lemma follows. □

3 Proof of the main result

Now we have gathered all the lemmas needed to prove the existence of an entropy solution to the obstacle problem associated with $(f,\psi ,g)$. In this section, let $f_{n}$ be a sequence of smooth functions converging strongly to f in $L^{1}(\varOmega )$ with $\Vert f_{n}\Vert _{1} \leq \Vert f\Vert _{1}+1$. We consider the sequence of approximated obstacle problems associated with $(f_{n},\psi ,g)$. The proof originates from [8]. We provide details for readers’ convenience.

Proof of Theorem 1

Let $v\in K_{g,\psi }\cap L^{\infty }(\varOmega )$. Taking v as a test function in (8) associated with $(f_{n},\psi ,g)$, we get

$$\begin{aligned} \int _{\varOmega } \frac{a(x, \nabla u_{n})}{(1+b(x) \vert u_{n} \vert )^{\theta (p-1)}}\cdot \nabla T_{t}(v-u_{n}) \,dx\geq \int _{\varOmega }-f_{n}T_{t}(v-u_{n}) \,dx. \end{aligned}$$

Since $\{\vert v-u_{n}\vert < t\}\subset \{\vert u_{n}\vert < s\}$ with $s=t+\Vert v\Vert _{\infty }$, the previous inequality can be written as

$$\begin{aligned} \int _{\varOmega }\chi _{n}\nabla _{A} T_{s}(u_{n})\cdot \nabla v\,dx \geq \int _{\varOmega }-f_{n}T_{t}(v-u_{n}) \,dx+ \int _{\varOmega }\chi _{n}\nabla _{A} T_{s}(u_{n})\cdot \nabla T_{s}(u_{n}) \,dx, \end{aligned}$$

(30)

where $\chi _{n}=\chi _{\{\vert v-u_{n}\vert < t\}}$ and $\nabla _{A}u=\frac{a(x, \nabla u)}{(1+b(x)\vert u\vert )^{\theta (p-1)}}$. It is clear that $\chi _{n} \rightharpoonup \chi$ weakly* in $L^{\infty }(\varOmega )$. Moreover, $\chi _{n}$ converges a.e. to $\chi _{\{\vert v-u\vert < t\}}$ in $\varOmega \setminus \{\vert v-u\vert =t\}$. It follows that

$$\begin{aligned} \chi = \textstyle\begin{cases} 1 & \text{in } \{ \vert v-u \vert < t\}, \\ 0 & \text{in } \{ \vert v-u \vert >t\}. \end{cases}\displaystyle \end{aligned}$$

Note that we have $\mathcal{L}^{N}(\{\vert v-u\vert =t\})=0$ for a.e. $t\in (0,\infty )$. So there exists a measurable set $\mathcal{O} \subset (0,\infty )$ such that $\mathcal{L}^{N}(\{\vert v-u\vert =t\})=0$ for all $t\in (0,\infty )\setminus \mathcal{O}$. Assume that $t\in (0, \infty )\setminus \mathcal{O}$. Then $\chi _{n}$ converges weakly* in $L^{\infty }(\varOmega )$ and a.e. in Ω to $\chi =\chi _{\{\vert v-u\vert < t \}}$. Since $\nabla T_{s}(u_{n})$ converges a.e. to $\nabla T_{s}(u)$ in Ω (Proposition 4), by Fatou’s lemma we obtain

$$\begin{aligned} \liminf_{n\rightarrow \infty } \int _{\varOmega }\chi _{n}\nabla _{A} T_{s}(u _{n})\cdot \nabla T_{s}(u_{n}) \,dx\geq \int _{\varOmega }\chi \nabla _{A} T_{s}(u)\cdot \nabla T_{s}(u)\,dx. \end{aligned}$$

(31)

Using the strong convergence of $\nabla _{A} T_{s}(u_{n})$ to $\nabla _{A} T_{s}(u)$ in $L^{1}(\varOmega )$ (Lemma 7) and the weak* convergence of $\chi _{n}$ to χ in $L^{\infty }(\varOmega )$, we obtain

$$\begin{aligned} \lim_{n\rightarrow \infty } \int _{\varOmega }\chi _{n}\nabla _{A} T_{s}(u _{n})\cdot \nabla v\,dx= \int _{\varOmega }\chi \nabla _{A} T_{s}(u) \cdot \nabla v\,dx. \end{aligned}$$

(32)

Moreover, since $f_{n}$ converges to f in $L^{1}(\varOmega )$ and $T_{t}(v-u_{n})$ converges to $T_{t}(v-u)$ weakly* in $L^{\infty }( \varOmega )$, by passing to the limit in (30) and taking into account (31)–(32) we obtain

$$\begin{aligned} \int _{\varOmega }\chi \nabla _{A} T_{s}(u)\cdot \nabla v\,dx- \int _{\varOmega }\chi \nabla _{A} T_{s}(u)\cdot \nabla T_{s}(u)\,dx \geq \int _{\varOmega }-fT_{t}(v-u)\,dx, \end{aligned}$$

which can be written as

$$\begin{aligned} \int _{\{\vert v-u\vert \leq t\}}\chi \nabla _{A} T_{s}(u)\cdot (\nabla v-\nabla u)\,dx\geq \int _{\varOmega }-fT_{t}(v-u)\,dx \end{aligned}$$

or, since $\chi =\chi _{\{\vert v-u\vert < t\}}$ and $\nabla (T_{t}(v-u))= \chi _{\{\vert v-u\vert < t\}}\nabla (v-u)$,

$$\begin{aligned} \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t}( v-u)\,dx\geq \int _{\varOmega }-fT_{t}(v-u)\,dx, \quad \forall t\in (0, \infty )\setminus \mathcal{O}. \end{aligned}$$

For $t\in \mathcal{O}$, we know that there exists a sequence $\{t_{k}\}$ in $(0,\infty )\setminus \mathcal{O}$ such that $t_{k}\rightarrow t$ due to $\vert \mathcal{O}\vert =0$. Therefore we have

$$\begin{aligned} \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t_{k}}( v-u)\,dx\geq \int _{\varOmega }-fT_{t_{k}}(v-u)\,dx \quad \forall k\geq 1. \end{aligned}$$

(33)

Since $\nabla (v-u)=0$ a.e. in $\{\vert v-u\vert =t\}$, the left-hand side of (33) can be written as

$$\begin{aligned} \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t_{k}}( v-u)\,dx= \int _{\varOmega \setminus \{\vert v-u\vert =t\}}\chi _{\{\vert v-u\vert < t_{k}\}}\nabla _{A} u \cdot \nabla ( v-u)\,dx. \end{aligned}$$

The sequence $\chi _{\{\vert v-u\vert < t_{k}\}}$ converges to $\chi _{\{\vert v-u\vert < t \}}$ a.e. in $\varOmega \setminus \{\vert v-u\vert =t\}$ and therefore converges weakly* in $L^{\infty }(\varOmega \setminus \{\vert v-u\vert =t\})$. We obtain

$$\begin{aligned} \lim_{k\rightarrow \infty } \int _{\varOmega }\nabla _{A} u\cdot \nabla T _{t_{k}}( v-u)\,dx &= \int _{\varOmega \setminus \{\vert v-u\vert =t\}} \chi _{\{\vert v-u\vert < t\}}\nabla _{A} u\cdot \nabla ( v-u)\,dx \\ &= \int _{\varOmega }\chi _{\{\vert v-u\vert < t\}}\nabla _{A} u\cdot \nabla ( v-u) \,dx \\ &= \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t}( v-u)\,dx. \end{aligned}$$

(34)

For the right-hand side of (33), we have

$$\begin{aligned} \biggl\vert \int _{\varOmega }-fT_{t_{k}}(v-u)\,dx- \int _{\varOmega }-fT_{t}(v-u) \,dx \biggr\vert \leq \vert t_{k}-t \vert \Vert f \Vert _{1}\rightarrow 0 \quad \text{as }k\rightarrow \infty . \end{aligned}$$

(35)

It follows from (33)–(35) that we have the inequality

$$\begin{aligned} \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t}( v-u)\,dx\geq \int _{\varOmega }-fT_{t}(v-u)\,dx \quad \forall t\in (0, \infty ). \end{aligned}$$

Hence, u is an entropy solution of the obstacle problem associated with $(f,\psi ,g)$. The regularity of the entropy solution u is guaranteed by Proposition 4. □