The obstacle problem for non-coercive equations with lower order term and $L^{1}$-data

Abstract

The aim of this paper is to study the obstacle problem associated with an elliptic operator having degenerate coercivity, a low order term, and $L^{1}$-data. We prove the existence of an entropy solution to the obstacle problem and show its continuous dependence on the $L^{1}$-data in $W^{1,q}(\varOmega )$ with some $q>1$.

1 Introduction

1.1 Problem setting and main result

Let Ω be a bounded domain in $\mathbb{R}^{N}$ ($N\geq 2$), $1< p<+\infty$, and $\theta \geq 0$. Given functions $g, \psi \in W ^{1,p}(\varOmega )\cap L^{\infty }(\varOmega )$ and data $f\in L^{1}(\varOmega )$, the aim of this paper is to study the obstacle problem for nonlinear non-coercive elliptic equations with lower order term, governed by the operator

$$\begin{aligned} Au=-\operatorname{div} \frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}}+b \vert u \vert ^{r-2}u, \end{aligned}$$

(1)

where $b>0$ is a constant, and $a:\varOmega \times \mathbb{R}^{N} \rightarrow \mathbb{R}^{N}$ is a Carathéodory function, satisfying the following conditions:

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

(2)

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

(3)

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

(4)

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

(5)

for almost every x in Ω and for every ξ, η, ζ in $\mathbb{R}^{N}$ with $\xi \neq \eta$, where $\alpha ,\beta ,\gamma >0$ are constants, and j is a nonnegative function in $L^{p'}( \varOmega )$.

If f has a fine regularity, e.g., $f\in W^{-1,p'}(\varOmega )$, the obstacle problem corresponding to $(f,\psi ,g)$ can be formulated in terms of the inequality

$$\begin{aligned} & \int _{\varOmega }\frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}}\cdot \nabla (u-v)\, \mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}u(u-v)\, \mathrm{d}x \\ & \quad \leq \int _{\varOmega }f(u-v)\,\mathrm{d}x, \quad \forall v\in K_{g,\psi }\cap L^{\infty }(\varOmega ), \end{aligned}$$

(6)

whenever $1\leq r< p$ and the convex subset

$$\begin{aligned} K_{g,\psi }= \bigl\{ v\in W^{1,p}(\varOmega ); v-g \in W^{1,p}_{0}(\varOmega ), v\geq \psi , \text{a.e. in } \varOmega \bigr\} \end{aligned}$$

is nonempty. However, if $f\in L^{1}(\varOmega )$, (6) is not well-defined. Following [1, 3, 5, 19] etc., we are led to the more general definition of a solution to the obstacle problem, using the truncation function

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

Definition 1

An entropy solution of the obstacle problem associated with operator A and functions $(f,\psi ,g)$ with $f\in L^{1}(\varOmega )$ is a measurable function u such that $u\geq \psi$ a.e. in Ω, $\frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}}\in (L^{1}( \varOmega ))^{N}$, $\vert u \vert ^{r-1}\in L^{1}(\varOmega )$, and, for every $s>0$, $T_{s}(u)-T_{s}(g)\in W_{0}^{1,p}(\varOmega )$ and

$$\begin{aligned} & \int _{\varOmega }\frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}}\cdot \nabla \bigl(T_{s}(u-v)\bigr)\,\mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}uT_{s}(u-v) \,\mathrm{d}x \\ & \quad \leq \int _{\varOmega }fT_{s}(u-v)\,\mathrm{d}x, \quad \forall v\in K_{g,\psi }\cap L^{\infty }(\varOmega ). \end{aligned}$$

(7)

Observe that no global integrability condition is required on u nor on its gradient in the definition. As pointed out in [3, 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 $u \in W^{1,1}(\varOmega )$.

Before stating the main result, we make some basic assumptions throughout this paper, i.e., without special statements, we always assume that

$$\begin{aligned} 2-\frac{1}{N}< p< N, \quad\quad 1\leq r< p, \quad\quad 0\leq \theta < \min \biggl\{ \frac{N}{N-1}-\frac{1}{p-1},\frac{p-r}{p-1} \biggr\} ,\quad\quad b>0, \end{aligned}$$

and $\psi ,g\in W^{1,p}(\varOmega )\cap L^{\infty }(\varOmega )$ with $(\psi -g)^{+}\in W^{1,p}_{0}(\varOmega )$ such that $K_{g,\psi }\neq \emptyset$. The following theorem is the main result obtained in this paper.

Theorem 1

Let $f\in L^{1}(\varOmega )$. Then there exists at least one entropy solution u of the obstacle problem associated with $(f,\psi ,g)$. In addition, u depends continuously on f, i.e., 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

$$\begin{aligned} u_{n}\rightarrow u \quad \textit{in } W^{1,q}(\varOmega ),\forall q \in \textstyle\begin{cases} ( \frac{N(r-1)}{N+r-1}, \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} ), & \textit{if }\frac{2N-1}{N-1}\leq r< p, \\ (1,\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} ), & \textit{if } 1\leq r< \min \{\frac{2N-1}{N-1},p\}. \end{cases}\displaystyle \end{aligned}$$

(8)

1.2 Some comments and remarks

Consider the Dirichlet boundary value problem having a form

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

(9)

with $p>1$, $\theta \in (0,1]$, $b\geq 0$, $f\in L^{1}(\varOmega )$. The item $-\operatorname{div}\frac{ \vert \nabla u \vert ^{p-2}\nabla u}{(1+ \vert u \vert )^{\theta (p-1)}}$ may not be coercive when u tends to infinity. Due to this fact, the classical methods used to prove the existence of a solution for elliptic equations, e.g., [14], cannot be applied even if $b=0$ and the data f is regular. Moreover, $\frac{ \vert \nabla u \vert ^{p-2}\nabla u}{(1+ \vert u \vert )^{\theta (p-1)}}$, u and f are only in $L^{1}(\varOmega )$, not in $W^{-1,p'}(\varOmega )$. All these characteristics prevent us from employing the duality argument [17] or nonlinear monotone operator theory [18] directly.

To overcome these difficulties, “cutting” the non-coercivity term and using the technique of approximation, a pseudomonotone and coercive differential operator on $W^{1,p}_{0}(\varOmega )$ can be applied to establish a priori estimates on approximating solutions. As a result, existence of solutions, or entropy solutions, can be obtained by taking limitation for $f\in L^{m}(\varOmega )$, $m\geq 1$, and $b> 0$ due to the almost everywhere convergence of gradients of the approximating solutions, see, e.g., [4, 6, 9,10,11, 15] (see also [1, 2, 7, 12, 13, 16] for $b=0$). However, there is little literature that considers regularities for entropy solutions of obstacle problems governed by (1) and functions $(f,\psi ,g)$ with $f\in L^{1}(\varOmega )$, except [19], in which the authors considered the obstacle problem (7) with $b=0$ and $L^{1}$-data.

Motivated by the study on the non-coercive elliptic equations (9) and the problem considered in [19], in this paper, we consider the obstacle problem governed by (1) and functions $(f,\psi ,g)$ with $f\in L^{1}(\varOmega )$. By the truncation method used in [8] and [19], we prove the existence of an entropy solution and show its continuous dependence on the $L^{1}$-data in $W^{1,q}(\varOmega )$ with some $q\in (1,p)$.

In the following, we give some remarks on our main result and inequalities that will be needed in the proofs. Some notations are provided at the end of this subsection.

Remark 1

• (i) $0\leq \theta < \min \{\frac{N}{N-1}-\frac{1}{p-1}, \frac{p-r}{p-1} \}\Rightarrow r-1<(1-\theta )(p-1)< \frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)}$. Therefore Theorem 1 guarantees $\vert u \vert ^{r-1}\in L^{1}(\varOmega )$, and the second integration in (7) makes sense.

• (ii) We will show that $\frac{a(x, \nabla u)}{(1+ \vert u \vert )^{\theta (p-1)}}\in (L^{1}(\varOmega ))^{N}$ in Proposition 4. Therefore, the first integration in (7) makes sense.

• (iii) $( \frac{N(r-1)}{N+r-1}, \frac{N(p-1)(1-\theta )}{N-1- \theta (p-1)} )\subset (1,\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} )$ if $\frac{2N-1}{N-1}\leq r< p$. Indeed, $\theta < \frac{p-r}{p-1}+\frac{p(r-1)}{N(p-1)}\Leftrightarrow \frac{N(p-1)(1- \theta )}{N-1-\theta (p-1)}>\frac{N(r-1)}{N+r-1}$, while $\frac{2N-1}{N-1}\leq r$ gives $\frac{N(r-1)}{N+r-1}\geq 1$. Thus $u_{n}\rightarrow u$ in $W^{1,q}(\varOmega )$ for all $q\in (1,\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} )$.

• (iv) $r-1<\frac{Nq}{N-q}$. Indeed, by $1\leq r<\frac{2N-1}{N-1}$, there holds $r-1<\frac{N}{N-1}< \frac{Nq}{N-q}$ for any $q>1$, particularly, for $q\in (1, \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} )$. For $r\geq \frac{2N-1}{N-1}$, it suffices to note that $q> \frac{N(r-1)}{N+r-1} \Leftrightarrow r-1< \frac{Nq}{N-q}$.

• (v) $q< p$. Indeed, $0\leq \theta <\frac{N}{N-1}-\frac{1}{p-1}< \frac{N-1}{p-1} \Rightarrow \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)}<p$.

Remark 2

Checking proofs in this paper (e.g., setting $r=1$), one may find that, for $b=0$, (8) holds with

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

(10)

which is the same as [19, Theorem 1]. Thus, Theorem 1 can be seen as an extension of [19, Theorem 1].

Notations

$\Vert u \Vert _{p}:= \Vert u \Vert _{L^{p}(\varOmega )}$ is the norm of u in $L^{p}(\varOmega )$, where $1\leq p\leq \infty$. $\Vert u \Vert _{1,p}:= \Vert u \Vert _{W^{1,p}(\varOmega )}$ is the norm of u in $W^{1,p}(\varOmega )$, where $1< p<\infty$. $p':=\frac{p}{p-1}$ with $1< p<\infty$. $\{u>s\}:=\{x\in \varOmega ;u(x)>s \}$. $\{u\leq s\}:=\varOmega \setminus \{u>s\}$. $\{u< s\}:=\{x\in \varOmega ;u(x)< s\}$. $\{u\geq s\}:=\varOmega \setminus \{u< s\}$. $\{u=s\}:=\{x \in \varOmega ;u(x)=s\}$. $\{t\leq u < s\}:=\{u\geq t\}\cap \{u<s\}$. For a measurable set E in $\mathbb{R}^{N}$, $\vert E \vert :=\mathcal{L}^{N}(E)$, where $\mathcal{L}^{N}$ is the Lebesgue measure of $\mathbb{R}^{N}$. For a real-valued function u, $u^{+}=\max \{u,0\}$, $u^{-}=(-u)^{+}$. Without special statements, positive integers are denoted by n, h, k, $k_{0}$, K. C is a positive constant, which may be different from each other.

2 Lemmas on entropy solutions

It is worthy to note that, for any smooth function $f_{n}$, there exists at least one solution to the obstacle problem (6). Indeed, one can proceed exactly as in [1, 11] to obtain $W^{1,p}$-solutions due to assumptions (2)–(4) on a and $r-1< p$. These solutions, in particular, are also entropy solutions. In this section, using the method of [8] and [19], we establish several auxiliary results on convergence of sequences of entropy solutions when $f_{n}\rightarrow f$ in $L^{1}(\varOmega )$.

Lemma 2

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

$$\begin{aligned} \int _{\{ \vert u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \leq C \bigl(1+t^{r}\bigr), \quad \forall t>0, \end{aligned}$$

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

Proof

Take $v_{0}$ as a test function in (7). For t large enough such that $t- \Vert v_{0} \Vert _{\infty }>0$, we get

$$\begin{aligned} \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla u}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x\leq{}& \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \\ & {} + \int _{\varOmega }\bigl(f-b \vert u \vert ^{r-2}u\bigr) T_{t}(u-v_{0})\,\mathrm{d}x. \end{aligned}$$

(11)

We estimate each integration in the right-hand side of (11). It follows from (3) and Young’s inequality with $\varepsilon >0$ that

$$\begin{aligned}& \int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta ( \vert j \vert + \vert \nabla u \vert ^{p-1})\cdot \vert \nabla v_{0} \vert }{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x}\leq \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta \varepsilon ( \vert j \vert ^{p'}+ \vert \nabla u \vert ^{p})}{(1+ \vert u \vert )^{\theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x}\quad {} + \int _{\{ \vert v_{0}-u \vert < t\}}\frac{\beta C(\varepsilon ) \vert \nabla v_{0} \vert ^{p}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x} \leq \varepsilon \int _{\{ \vert v_{0}-u \vert < t\}}\frac{ \vert \nabla u \vert ^{p}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x \\& \hphantom{\int _{\{ \vert v_{0}-u \vert < t\}}\frac{a(x, \nabla u)\cdot \nabla v_{0}}{(1+ \vert u \vert )^{ \theta (p-1)}}\,\mathrm{d}x}\quad {} +C\bigl( \Vert j \Vert _{p'}^{p'}+ \Vert \nabla v_{0} \Vert _{p}^{p}\bigr), \end{aligned}$$

(12)

$$\begin{aligned}& \begin{aligned}[b] - \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x={}&- \int _{\{ \vert u-v_{0} \vert \leq t\}} b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x \\ & {} - \int _{\{ \vert u-v_{0} \vert > t\}}b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x. \end{aligned} \end{aligned}$$

(13)

Note that on the set $\{ \vert u-v_{0} \vert \leq t\}$,

$$\begin{aligned} \bigl\vert \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \bigr\vert \leq t \bigl\vert t+ \Vert v_{0} \Vert _{\infty } \bigr\vert ^{r-1}\leq C\bigl(1+t ^{r} \bigr), \end{aligned}$$

(14)

where C is a constant depending only on r, $\Vert v_{0} \Vert _{\infty }$.

On the set $\{ \vert u-v_{0} \vert > t\}$, we have $\vert u \vert \geq t- \Vert v_{0} \Vert _{\infty } >0$, thus u and $T_{t}(u-v_{0})$ have the same sign. It follows

$$\begin{aligned} - \int _{\{ \vert u-v_{0} \vert > t\}} b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x\leq 0. \end{aligned}$$

(15)

Combining (13)–(15), we get

$$\begin{aligned}& - \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v_{0}) \,\mathrm{d}x\leq C\bigl(1+t^{r}\bigr), \end{aligned}$$

(16)

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

(17)

Replacing t with $t+ \Vert v_{0} \Vert _{\infty }$ in (17) and noting that $\{ \vert u \vert < t\}\subset \{ \vert v_{0}-u \vert < t+ \Vert v_{0} \Vert _{\infty }\}$, one may obtain the desired result. □

In the rest of this section, let $\{u_{n}\}$ be a sequence of entropy solutions of 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

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 $W^{1,p}(\varOmega )$ for any $k>0$. Thus $T_{k}(u_{n})\rightarrow T _{k}(u)$ strongly in $L^{p}(\varOmega )$ and a.e. in Ω.

Proof

Let s, t, and ε be positive numbers. One may verify that, for every $m,n\geq 1$,

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

(18)

and

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

(19)

Due to $v_{0}=g+(\psi -g)^{+}\in K_{g,\psi }\cap L^{\infty }(\varOmega )$, by Lemma 2, one has

$$\begin{aligned} \int _{\varOmega } \bigl\vert \nabla T_{t}(u_{n}) \bigr\vert ^{p}\,\mathrm{d}x= \int _{\{ \vert u_{n} \vert < t\}} \vert \nabla u_{n} \vert ^{p}\,\mathrm{d}x \leq C(1+t)^{\theta (p-1)}\bigl(1+t^{r} \bigr). \end{aligned}$$

(20)

Note that $T_{t}(u_{n})-T_{t}(g)\in W^{1,p}_{0}(\varOmega )$. By (19), (20), and Poincaré’s inequality, for every $t> \Vert g \Vert _{\infty }$ and for some positive constant C independent of n and t, there holds

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

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

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

(21)

Now we have as in (19)

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

(22)

Using (20) and the fact that $T_{t}(u_{n})-T_{t}(g)\in W^{1,p} _{0}(\varOmega )$ 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 $L^{p}( \varOmega )$. Taking into account (22), there exists $n_{0}=n_{0}(t _{\varepsilon },s)\geq 1$ such that

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

(23)

Combining (18), (21), and (23), 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. The remainder of the lemma is a consequence of the fact that $\{T_{k}(u_{n})\}$ is a bounded sequence in $W^{1,p}(\varOmega )$. □

Proposition 4

There exist a subsequence of $\{u_{n}\}$ and a measurable function u such that, for each q given in (8), we have

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

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

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

To prove Proposition 4, we need two preliminary lemmas.

Lemma 5

There exist a subsequence of $\{u_{n}\}$ and a measurable function u such that, for each q given in (8), we have $u_{n}\rightharpoonup u$ 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

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

(24)

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

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

$$\begin{aligned} & \int _{\varOmega }\frac{a(x,\nabla u_{n})\cdot \nabla (T_{1}(u_{n}-T_{k}(u _{n})))}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2}u _{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \\ & \quad \leq \int _{\varOmega }f_{n}T_{1} \bigl(u_{n}-T_{k}(u_{n})\bigr)\, \mathrm{d}x, \end{aligned}$$

which and (2) give

$$\begin{aligned} \int _{B_{k}} \frac{\alpha \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2} u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \leq \Vert f_{n} \Vert _{1}\leq \Vert f \Vert _{1}+1. \end{aligned}$$

Note that on the set $\{ \vert u_{n} \vert \geq k+1\}$, $u_{n}$ and $T_{1}(u_{n}-T _{k}(u_{n}))$ have the same sign. Then

$$\begin{aligned} \int _{\varOmega } \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x ={}& \int _{D_{k}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \\ & {} + \int _{B_{k}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr)\,\mathrm{d}x \\ & {} + \int _{\{ \vert u_{n} \vert \geq k+1\}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u _{n})\bigr)\,\mathrm{d}x \\ \geq{}& \int _{B_{k}} \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u_{n}) \bigr) \,\mathrm{d}x. \end{aligned}$$

Thus we have

$$\begin{aligned} \int _{B_{k}} \frac{\alpha \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{\theta (p-1)}}\,\mathrm{d}x+ \leq{}& \Vert f \Vert _{1}+1- \int _{B_{k}}b \vert u_{n} \vert ^{r-2}u_{n}T_{1}\bigl(u_{n}-T_{k}(u _{n})\bigr)\,\mathrm{d}x \\ \leq{}& \Vert f \Vert _{1}+1+ \int _{B_{k}}b \vert u_{n} \vert ^{r-1}\,\mathrm{d}x \\ \leq{}& C \biggl(1+ \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{\frac{r-1}{q ^{*}}} \vert B_{k} \vert ^{1-\frac{r-1}{q^{*}}} \biggr) , \end{aligned}$$

(25)

where q is given in (8) and $q^{*}=\frac{Nq}{N-q}$.

Let $s=\frac{q\theta (p-1)}{p}$. Note that $q< p$ and $\frac{ps}{p-q}< q ^{*}$. For $\forall k>0$, we estimate $\int _{B_{k}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x$ as follows:

$$\begin{aligned} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x={}& \int _{B_{k}}\frac{ \vert \nabla u _{n} \vert ^{q}}{(1+ \vert u_{n} \vert )^{s}}\cdot \bigl(1+ \vert u_{n} \vert \bigr)^{s}\,\mathrm{d}x \\ \leq{}& \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{ \theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \int _{B_{k}}\bigl(1+ \vert u _{n} \vert \bigr)^{\frac{ps}{p-q}}\,\mathrm{d}x \biggr)^{\frac{p-q}{p}} \\ \leq{}& C \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{ \theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \vert B_{k} \vert ^{ \frac{p-q}{p}}+ \biggl( \int _{B_{k}} \vert u_{n} \vert ^{\frac{ps}{p-q}}\,\mathrm{d}x \biggr)^{\frac{p-q}{p}} \biggr) \\ \leq{}& C \biggl( \int _{B_{k}}\frac{ \vert \nabla u_{n} \vert ^{p}}{(1+ \vert u_{n} \vert )^{ \theta (p-1)}}\,\mathrm{d}x \biggr)^{\frac{q}{p}} \biggl( \vert B_{k} \vert ^{ \frac{p-q}{p}} + \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{\frac{s}{q ^{*}}} \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{s}{q^{*}}} \biggr) \\ \leq{}& C \biggl( \vert B_{k} \vert ^{\frac{p-q}{p}}+ \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{s}{q ^{*}}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{ \frac{s}{q^{*}}}+ \vert B_{k} \vert ^{1-p_{1}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{p_{1}} \\ & {} + \vert B_{k} \vert ^{1-p_{2}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{p _{2}} \biggr) \quad \text{by (25)} \\ ={}& C \biggl( \vert B_{k} \vert ^{\frac{p-q}{p}}+ \vert B_{k} \vert ^{\frac{p-q}{p}-\frac{s}{q ^{*}}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x \biggr)^{\frac{s}{q^{*}}} \\ & {} + \vert B_{k} \vert ^{1-p_{1}-C_{1}} \vert B_{k} \vert ^{C_{1}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}}\,\mathrm{d}x \biggr)^{p_{1}} \\ & {} + \vert B_{k} \vert ^{1-p_{2}-C_{2}} \vert B_{k} \vert ^{C_{2}} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}}\,\mathrm{d}x \biggr)^{p_{2}} \biggr), \end{aligned}$$

where $p_{1}=\frac{q}{p}\frac{r-1}{q^{*}}$, $p_{2}=\frac{s}{q^{*}}+ \frac{q}{p}\frac{r-1}{q^{*}}$, $C_{1}$ and $C_{2}$ are positive constants to be chosen later.

Note that $\theta <\frac{p-r}{p-1}$, it follows

$$\begin{aligned} \frac{\theta (p-1)}{p}+\frac{r-1}{p}< \frac{p-1}{p}< 1- \frac{1}{N}=1- \frac{1}{q}+\frac{1}{q^{*}}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{\theta q(p-1)}{p}+\frac{q(r-1)}{p}+1< q+\frac{q}{q^{*}} \quad &\Leftrightarrow\quad s+\frac{q(r-1)}{p}+1< q+\frac{q}{q^{*}} \\ & \Leftrightarrow\quad p_{2}+\frac{1-p _{2}}{q^{*}+1}< \frac{q}{q^{*}}. \end{aligned}$$

Note that $p_{1}< p_{2}<1$. Then, for $i=1,2$, we always have

$$\begin{aligned} p_{i}+\frac{1-p_{i}}{q^{*}+1}< \frac{q}{q^{*}}< 1. \end{aligned}$$

From this, we may find positive $C_{i}$ ($i=1,2$) such that

$$\begin{aligned} p_{i}+\frac{1-p_{i}}{q^{*}+1}< p_{i}+C_{i}< \frac{q}{q^{*}}< 1, \quad i=1,2. \end{aligned}$$

(26)

It follows

$$\begin{aligned} \frac{1-p_{i}}{q^{*}+1}< C_{i}\quad \Leftrightarrow \quad 1-p_{i}-C_{i}< C_{i}q^{*}, \quad i=1,2, \end{aligned}$$

which implies

$$\begin{aligned} C_{i}\alpha _{i}q^{*}= \frac{C_{i}q^{*}}{1-p_{i}-C_{i}}>1, \quad i=1,2, \end{aligned}$$

(27)

with $\alpha _{i}=\frac{1}{1-p_{i}-C_{i}}>1$, $i=1,2$. Let $\beta _{i}=\frac{1}{p _{i}+C_{i}}>1$, $i=1,2$. Then we have $\frac{1}{\alpha _{i}}+\frac{1}{ \beta _{i}}=1$ ($i=1,2$).

Since $\vert B_{k} \vert \leq \frac{1}{k^{q^{*}}}\int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x$, $\vert B_{k} \vert ^{1-p_{1}-C_{1}} \leq \vert \varOmega \vert ^{1-p_{1}-C_{1}}$, and $\vert B_{k} \vert ^{1-p_{2}-C_{2}} \leq \vert \varOmega \vert ^{1-p_{2}-C_{2}}$, we have, for $k\geq k_{0}\geq 1$,

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

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

$$\begin{aligned} \sum_{k=k_{0}}^{K} \int _{B_{k}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}& C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{s}{q ^{*}})\frac{p}{q}}} \Biggr)^{\frac{q}{p}}\cdot \Biggl(\sum _{k=k_{0}} ^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{\frac{p-q}{p}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{1}\alpha _{1}}} \Biggr)^{\frac{1}{ \alpha _{1}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}} \,\mathrm{d}x \biggr)^{\beta _{1}(p_{1}+C_{1})} \Biggr)^{\frac{1}{\beta _{1}}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{2}\alpha _{2}}} \Biggr)^{\frac{1}{ \alpha _{2}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \biggl( \int _{B_{k}} \vert u_{n} \vert ^{q ^{*}} \,\mathrm{d}x \biggr)^{\beta _{2}(p_{2}+C_{2})} \Biggr)^{\frac{1}{\beta _{2}}} \\ ={}& C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{s}{q ^{*}})\frac{p}{q}}} \Biggr)^{\frac{q}{p}}\cdot \Biggl(\sum _{k=k_{0}} ^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{\frac{p-q}{p}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{1}\alpha _{1}}} \Biggr)^{\frac{1}{ \alpha _{1}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{p_{1}+C_{1}} \\ & {} +C \Biggl(\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C_{2}\alpha _{2}}} \Biggr)^{\frac{1}{ \alpha _{2}}} \cdot \Biggl(\sum _{k=k_{0}}^{K} \int _{B_{k}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \Biggr)^{p_{2}+C_{2}}. \end{aligned}$$

(28)

Note that

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

(29)

To estimate the first integral in the right-hand side of (29), we compute by using Hölder’s inequality and (24), obtaining

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

(30)

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 $\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}(\frac{p-q}{p}-\frac{s}{q ^{*}})\frac{p}{q}}}$ and $\sum_{k=k_{0}}^{K} \frac{1}{k^{q^{*}C _{i}\alpha _{i}}}$ converge as $K\rightarrow \infty$ due to the fact that $q^{*}(\frac{p-q}{p}-\frac{s}{q^{*}})\frac{p}{q}>1$ and $q^{*}C_{i} \alpha _{i}>1$ by (27), respectively. Combining (28)–(30), we get for $k_{0}$ large enough

$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}& C+C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{ \frac{p-q}{p}} \\ & {} + C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{p _{1}+C_{1}} \\ & {} +C \biggl( \int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}} \,\mathrm{d}x \biggr)^{p _{2}+C_{2}}. \end{aligned}$$

(31)

Since $p>q$, $T_{K}(u_{n})\in W^{1,q}(\varOmega )$, $T_{K}(g)=g\in W^{1,q}( \varOmega )$ 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

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

(32)

Using the fact that

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

(33)

we obtain from (31)–(32)

$$\begin{aligned} \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x\leq{}& C+C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr)^{\frac{q ^{*}}{q}\frac{p-q}{p}} \\ & {} +C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr)^{(p_{1}+C_{1})\frac{q^{*}}{q}} \\ & {} +C \biggl(1+ \int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q} \,\mathrm{d}x \biggr)^{(p_{2}+C_{2})\frac{q^{*}}{q}} . \end{aligned}$$

(34)

Note that $p< N\Leftrightarrow \frac{q^{*}}{q}\frac{p-q}{p}<1$ and $(p_{i}+C_{i})\frac{q^{*}}{q}<1$ by (26). It follows from (34) that, for $k_{0}$ large enough, $\int _{\{ \vert u_{n} \vert \leq K\}} \vert \nabla u_{n} \vert ^{q}\,\mathrm{d}x$ is bounded independently of n and K. Using (32) and (33), we deduce that $\int _{\{ \vert u_{n} \vert \leq K\}} \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x$ 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 $W^{1,q}(\varOmega )$, $u_{n}\rightarrow v$ strongly in $L^{q}(\varOmega )$ and a.e. in Ω. By Lemma 3, $u_{n}\rightarrow u$ in measure in Ω, 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 almost everywhere in Ω to ∇u.

Proof

Define $A(x,u,\xi )=\frac{a(x,\xi )}{(1+ \vert u \vert )^{\theta (p-1)}}$ (for the sake of simplicity, we omit the dependence of $A(x,u,\xi )$ on x). Let $h>0$, $k> \max \{ \Vert g \Vert _{\infty }, \Vert \psi \Vert _{\infty }\}$, and $n\geq h+k$. Take $T_{k}(u)$ as a test function for (7), obtaining

$$\begin{aligned} I_{7}(n,k,h)\leq \int _{\varOmega }f_{n}T_{h} \bigl(u_{n}-T_{k}(u)\bigr)\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2}u_{n}T_{h}\bigl(u_{n}-T_{k}(u) \bigr)\,\mathrm{d}x, \end{aligned}$$

where

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

Note that $r-1< q^{*}$, and $\int _{\varOmega } \vert u_{n} \vert ^{q^{*}}\,\mathrm{d}x$ is uniformly bounded (see the proof of Lemma 5), thus $\vert u_{n} \vert$ converges strongly in $L^{1}(\varOmega )$. Therefore we have

$$\begin{aligned} \lim_{n\rightarrow \infty } \int _{\varOmega } \vert u_{n} \vert ^{r-2}u_{n}T_{h}\bigl(u_{n}-T_{k}(u) \bigr)\,\mathrm{d}x= \int _{\varOmega } \vert u \vert ^{r-2}uT_{h} \bigl(u-T_{k}(u)\bigr)\,\mathrm{d}x. \end{aligned}$$

Then, using the strong convergence of $f_{n}$ in $L^{1}(\varOmega )$, one has

$$\begin{aligned} \lim_{n\rightarrow \infty }I_{7}(n,k,h) \leq \int _{\varOmega }-fT_{h}\bigl(u-T_{k}(u) \bigr)\,\mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}uT_{h} \bigl(u-T_{k}(u)\bigr)\,\mathrm{d}x. \end{aligned}$$

It follows

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

Thanks to Lemma 3 and Lemma 5, we can proceed exactly as [19, Lemma 6] to conclude that, up to subsequence, $\nabla u_{n}\rightarrow \nabla u$ a.e. □

Proof of Proposition 4

We shall prove that $\nabla u_{n}$ converges strongly to ∇u in $L^{q}(\varOmega )$ for each q being given by (8). To do that,we will apply Vitali’s theorem, using the fact that by Lemma 5, $\nabla u_{n}$ is bounded in $L^{q}(\varOmega )$ for each q given by (8). So let $s\in (q,\frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)})$ and $E\subset \varOmega$ be a measurable set. Then we have by Hölder’s inequality

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

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

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

it suffices, thanks to Vitali’s theorem, to show that, for every measurable subset $E\subset \varOmega$, $\int _{E} \vert \frac{a(x, \nabla u_{n})}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \vert \,\mathrm{d}x$ 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 assumptions. For any $q\in (p-1, \frac{N(p-1)(1-\theta )}{N-1-\theta (p-1)} )$, we deduce by Hölder’s inequality

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

 □

Lemma 7

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

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

Proof

See the proof of [19, Lemma 7]. □

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 part, 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 can be proceeded in the same way as in [8] and [19]. 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 (7) associated with $(f_{n},\psi ,g)$, we get

$$\begin{aligned} & \int _{\varOmega }\frac{a(x, \nabla u_{n})}{(1+ \vert u_{n} \vert )^{\theta (p-1)}} \cdot \nabla \bigl(T_{t}(u_{n}-v)\bigr)\,\mathrm{d}x+ \int _{\varOmega }b \vert u_{n} \vert ^{r-2}u _{n}T_{t}(u_{n}-v)\, \mathrm{d}x \\ & \quad \leq \int _{\varOmega }f_{n}T_{t}(u_{n}-v) \,\mathrm{d}x. \end{aligned}$$

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

$$\begin{aligned} \int _{\varOmega }\chi _{n}\nabla _{A} T_{s}(u_{n})\cdot \nabla v\,\mathrm{d}x \geq{}& \int _{\varOmega }-f_{n}T_{t}(u_{n}-v) \,\mathrm{d}x+ \int _{\varOmega }b \vert u _{n} \vert ^{r-2}u_{n}T_{t}(u_{n}-v)\, \mathrm{d}x \\ & {} + \int _{\varOmega }\chi _{n}\nabla _{A} T_{s}(u_{n})\cdot \nabla T_{s}(u _{n})\,\mathrm{d}x, \end{aligned}$$

(35)

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

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

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

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

(36)

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

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

(37)

Moreover, due to the strong convergence of $f_{n}$ to f and $\vert u_{n} \vert ^{r-2}u_{n}$ to $\vert u \vert ^{r-2}u$ (by $r-1< q^{*}$ and the boundedness of $\Vert u_{n} \Vert _{q^{*}}$) in $L^{1}(\varOmega )$, and the weak* convergence of $T_{t}(u_{n}-v)$ to $T_{t}(u-v)$ in $L^{\infty }( \varOmega )$, by passing to the limit in (35) and taking into account (36)–(37), we obtain

$$\begin{aligned} \int _{\varOmega }\chi \nabla _{A} T_{s}(u) \cdot \nabla v\,\mathrm{d}x- \int _{\varOmega }\chi \nabla _{A} T_{s}(u) \cdot \nabla T_{s}(u)\,\mathrm{d}x \geq{}& \int _{\varOmega }-fT_{t}(u-v)\,\mathrm{d}x \\ & {} + \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v) \,\mathrm{d}x, \end{aligned}$$

which can be written as

$$\begin{aligned} \int _{\{ \vert v-u \vert \leq t\}}\chi \nabla _{A} T_{s}(u)\cdot (\nabla v-\nabla u)\,\mathrm{d}x\geq{}& \int _{\varOmega }-fT_{t}(u-v)\,\mathrm{d}x \\ & {} + \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v) \,\mathrm{d}x, \end{aligned}$$

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

$$\begin{aligned} & \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t}( u-v)\,\mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v) \,\mathrm{d}x \\ & \quad \leq \int _{\varOmega }fT_{t}(u-v)\,\mathrm{d}x,\forall t\in (0,\infty ) \setminus \mathcal{O}. \end{aligned}$$

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

$$\begin{aligned} \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t_{k}}( u-v)\,\mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t_{k}}(u-v) \,\mathrm{d}x\leq \int _{\varOmega }fT _{t_{k}}(u-v)\,\mathrm{d}x. \end{aligned}$$

(38)

Since $\nabla (u-v)=0$ a.e. in $\{ \vert u-v \vert =t\}$, the left-hand side of (38) can be written as

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

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

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

(39)

For the right-hand side of (38), we have

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

(40)

Similarly, we have

$$\begin{aligned} \biggl\vert \int _{\varOmega } \vert u \vert ^{r-2}uT_{t_{k}}(u-v) \,\mathrm{d}x- \int _{\varOmega } \vert u \vert ^{r-2}uT _{t}(u-v)\,\mathrm{d}x \biggr\vert &\leq \vert t_{k}-t \vert \cdot \bigl\Vert \vert u \vert ^{r-1} \bigr\Vert _{1} \\ &\rightarrow 0 \quad \text{as } k\rightarrow \infty . \end{aligned}$$

(41)

It follows from (38)–(41) that we have the inequality

$$\begin{aligned} & \int _{\varOmega }\nabla _{A} u\cdot \nabla T_{t}( u-v)\,\mathrm{d}x+ \int _{\varOmega }b \vert u \vert ^{r-2}uT_{t}(u-v) \,\mathrm{d}x \\ & \quad \leq \int _{\varOmega }fT_{t}(u-v)\,\mathrm{d}x,\quad \forall t\in (0,\infty ). \end{aligned}$$

Hence, u is an entropy solution of the obstacle problem associated with $(f,\psi ,g)$. The dependence of the entropy solution on the data $f\in L^{1}(\varOmega )$ is guaranteed by Proposition 4. □