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# Existence results for some nonlinear elliptic equations with measure data in Orlicz-Sobolev spaces

An Erratum to this article was published on 10 July 2017

This article has been updated

## Abstract

We prove the existence results in the setting of Orlicz spaces for the following nonlinear elliptic equation:

$$A(u)+g(x,u,Du)=\mu,$$

where A is a Leray-Lions operator defined on $$D(A)\subset W_{0}^{1}L_{M}(\Omega)$$, while g is a nonlinear term having a growth condition with respect to Du, but does not satisfy any sign condition. The right-hand side μ is a bounded Radon measure data.

## Introduction

Let Ω be a bounded domain in $$\mathbb{R}^{N}$$. In the classical Sobolev space $$W^{1,p}_{0}(\Omega)$$, Porretta  studied the solution of the following problem:

$$-\operatorname{div} a(x,u,D u)=H(x,u,D u)+\mu,$$
(1)

where a is supposed to satisfy a polynomial growth condition with respect to u and Du, H has natural growth with respect to Du without any sign condition (i.e., $$H(x,s,\xi )s\geq0$$), that is, a and H satisfy

1. (a1)

$$|a(x,s,\xi)|\leq\beta(k(x)+|s|^{p-1}+|\xi|^{p-1})$$, $$k(x)\in L^{p'}(\Omega)$$, $$\beta>0$$, $$p>1$$, $$\frac{1}{p}+\frac{1}{p'}=1$$,

2. (H)

$$|H(x,s,\xi)|\leq\gamma(x)+g(s)|\xi|^{p}$$, $$\gamma(x)\in L^{1}(\Omega)$$, and $$g:\mathbb{R}\rightarrow\mathbb{R}^{+}$$ is continuous, $$g\geq0$$, $$g\in L^{1}(\mathbb{R})$$,

for almost every $$x\in\Omega$$, for all $$s\in\mathbb{R}$$, $$\xi\in \mathbb{R}^{N}$$. The right-hand side μ is a nonnegative bounded Radon measure on Ω. The model example is the equation

$$-\Delta_{p}(u)+g(u)|D u|^{p}=\mu$$

in Ω coupled with a Dirichlet boundary condition.

Aharouch et al.  proved the existence results in the setting of Orlicz spaces for the unilateral problem associated to the following equation:

$$A(u)+g(x,u,D u)=f,$$
(2)

where $$A(u)=-\operatorname{div} a(x,u,D u)$$ is a Leray-Lions operator defined on $$D(A)\subset W_{0}^{1}L_{M}(\Omega)$$, a and g satisfy the following growth conditions:

1. (a2)

$$|a(x,s,\xi)|\leq c(x)+k_{1}\bar{P}^{-1}(M(k_{2}|s|))+k_{3}\bar {M}^{-1}(M(k_{4}|\xi|))$$, $$k_{1},k_{2},k_{3},k_{4} \geq0$$, $$c(x)\in E_{\bar{M}}(\Omega)$$,

2. (g)

$$|g(x,s,\xi)|\leq\gamma(x)+\rho(s)M(|\xi|)$$, $$\gamma(x)\in L^{1}(\Omega)$$, and $$\rho:\mathbb{R}\rightarrow\mathbb{R}^{+}$$ is continuous, $$\rho\geq0$$, $$\rho\in L^{1}(\mathbb{R})$$,

for almost every $$x\in\Omega$$, for all $$s\in\mathbb{R}$$, $$\xi\in \mathbb{R}^{N}$$, where M and P are N-functions such that $$P\ll M$$. The right-hand side f belongs to $$L^{1}(\Omega)$$. The obstacle is a measurable function.

Youssfi et al.  proved the existence of bounded solutions of problem (2) whose principal part has a degenerate coercivity, where g does not satisfy the sign condition and f is an appropriate integrable source term.

Some elliptic equations in Orlicz spaces with variational structure of the form

$$\int_{\Omega}M\bigl(|Du|\bigr)\,dx$$

have been studied, where $$u:\Omega\rightarrow\mathbb{R}^{N}$$, $$\Omega \subset\mathbb{R}^{n}$$ is a bounded open set (see, e.g., ). The associated Euler-Lagrange system is

$$-\operatorname{div} \biggl(M'\bigl(|Du|\bigr)\frac{Du}{|Du|} \biggr)=0\quad (\mbox{see}, \textit{e.g.}, \mbox{}).$$

In this case methods from the calculus of variations can be used and regularity of solutions can be shown. However, the assumptions are strong. For example, it is needed that M satisfies $$\Delta_{2}$$ condition in  and .

The purpose of this paper is to study the existence of a solution for the following nonlinear Dirichlet problem:

$$A(u)+g(x,u,D u)=\mu,$$
(3)

where $$A(u)=-\operatorname{div} a(x,u,D u)$$ is a Leray-Lions operator defined on $$D(A)\subset W_{0}^{1}L_{M}(\Omega)$$ having the following growth condition:

$$\bigl|a(x,s,\xi)\bigr|\leq\beta \bigl[c(x)+\bar{M}^{-1} \bigl(M\bigl(|s|\bigr) \bigr)+ \bar{M}^{-1} \bigl(M\bigl(|\xi |\bigr) \bigr) \bigr],\quad \beta>0, c(x)\in E_{\bar{M}}(\Omega)$$

for almost every $$x\in\Omega$$, for all $$s\in\mathbb{R}$$, $$\xi\in \mathbb{R}^{N}$$, g is a nonlinear term having the growth condition (g) without any sign condition, and μ is a nonnegative bounded Radon measure on Ω. When trying to relax the restriction on a and H in Eq. (1), we are led to replace Sobolev spaces by Orlicz-Sobolev spaces without assuming any restriction on M (i.e., without the $$\Delta _{2}$$ condition). The choice $$M(t)=t^{p}$$, $$p>1$$, $$t>0$$ leads to . A nonstandard example is $$M(t)=t\ln(1+t)$$, $$t>0$$ (see, e.g., [8, 9]). Taking $$M(t)=e^{t}-1$$, $$t>0$$, M does not satisfy $$\Delta_{2}$$-condition. Moreover, the elimination of the term g in Eq. (3) can lead to . A specific example to which our result applies includes the following:

$$-\operatorname{div} \biggl(a(u)\frac{M(|Du|)Du}{|Du|^{2}} \biggr)+ a'(u)\int _{0}^{|D u|}\frac{M(t)}{t}\,dt=\delta,$$

where $$a(s)$$ is a smooth function, and δ is a Dirac measure.

This paper is organized as follows. In Section 2, we recall some preliminaries and some technical lemmas which will be needed in Section 3. In Section 3, we first prove that there exist solutions in $$W_{0}^{1}E_{M}(\Omega)$$ for approximate equations by using a linear functional analysis method; next, following [13, 10], we prove the existence results for problem (9)-(10) and show that solutions belong to Orlicz-Sobolev spaces $$W_{0}^{1}L_{B}(\Omega)$$ for any $$B\in\mathcal{P}_{M}$$, where $$\mathcal{P}_{M}$$ is a special class of N-functions (see Theorem 3.1 below).

For some classical results on equations, we refer to .

## Preliminaries

### N-function

Let $$M:\mathbb{R}^{+}\to\mathbb{R}^{+}$$ be an N-function; i.e., M is continuous, convex with $$M(u)>0$$ for $$u>0$$, $$M(u)/u\to 0$$ as $$u\to0$$, and $$M(u)/u\to\infty$$ as $$u\to\infty$$. Equivalently, M admits the representation $$M(u)=\int_{0}^{u}\phi(t)\,dt$$, where $$\phi:\mathbb{R}^{+}\to\mathbb{R}^{+}$$ is a nondecreasing, right-continuous function with $$\phi(0)=0$$, $$\phi(t)>0$$ for $$t>0$$, and $$\phi(t)\to\infty$$ as $$t\to\infty$$.

The conjugated N-function of M is defined by $$\bar{M}(v)=\int_{0}^{v}\psi(s)\,ds$$, where $$\psi:\mathbb{R}^{+}\to\mathbb{R}^{+}$$ is given by $$\psi(s)=\sup\{t:\phi(t)\leq s\}$$.

The N-function M is said to satisfy the $$\Delta_{2}$$ condition if, for some $$k>0$$,

$$M(2u)\leq kM(u),\quad \forall u\geq0.$$

The N-function M is said to satisfy the $$\Delta_{2}$$ condition near infinity if, for some $$k>0$$ and $$u_{0}>0$$, $$M(2u)\leq kM(u)$$, $$\forall u\geq u_{0}$$ (see [19, 20]).

Moreover, one has the following Young inequality:

$$\forall u, v\geq0,\quad uv\leq M(u)+\bar{M}(v).$$

We will extend these N-functions into even functions on all $$\mathbb{R}$$.

Let P, Q be two N-functions, $$P\ll Q$$ means that P grows essentially less rapidly than Q; i.e., for each $$\varepsilon>0$$, $$P(t)/Q(\varepsilon t)\to0$$ as $$t\to\infty$$. This is the case if and only if $$\lim_{t\to\infty} Q^{-1}(t)/P^{-1}(t)=0$$ (see [19, 21]).

### Orlicz spaces

Let Ω be an open subset of $$\mathbb{R}^{N}$$ and M be an N-function. The Orlicz class $$K_{M}(\Omega)$$ (resp. the Orlicz space $$L_{M}(\Omega)$$) is defined as the set of (equivalence classes of) real valued measurable functions u on Ω such that

$$\int_{\Omega}M \bigl(u(x) \bigr)\,dx< +\infty \quad \biggl( \mbox{resp. } \int_{\Omega}M \biggl(\frac{u(x)}{\lambda} \biggr) \,dx< + \infty \mbox{ for some }\lambda>0 \biggr).$$

$$L_{M}(\Omega)$$ is a Banach space under the Luxemburg norm

$$\|u\|_{(M)}=\inf \biggl\{ \lambda>0: \int_{\Omega}M \biggl(\frac{u(x)}{\lambda} \biggr)\,dx\leq1 \biggr\} ,$$

and $$K_{M}(\Omega)$$ is a convex subset of $$L_{M}(\Omega)$$ but not necessarily a linear space.

The closure in $$L_{M}(\Omega)$$ of the set of bounded measurable functions with compact support in Ω̄ is denoted by $$E_{M}(\Omega)$$. The equality $$E_{M}(\Omega)=L_{M}(\Omega)$$ holds if and only if M satisfies the $$\Delta_{2}$$ condition for all t or for t large according to whether Ω has infinite measure or not. The dual space of $$E_{M}(\Omega)$$ can be identified with $$L_{\bar {M}}(\Omega)$$ by means of the pairing $$\int_{\Omega}u(x)v(x)\,dx$$, and the dual norm of $$L_{\bar{M}}(\Omega)$$ is equivalent to $$\|\cdot\| _{(\bar{M})}$$.

### Orlicz-Sobolev spaces

We now turn to the Orlicz-Sobolev spaces. The class $$W^{1}L_{M}(\Omega)$$ (resp., $$W^{1}E_{M}(\Omega)$$) consists of all functions u such that u and its distributional derivatives up to order 1 lie in $$L_{M}(\Omega)$$ (resp., $$E_{M}(\Omega)$$). The classes $$W^{1}L_{M}(\Omega)$$ and $$W^{1}E_{M}(\Omega)$$ of such functions may be given the norm

$$\|u\|_{\Omega,M}=\sum_{|\alpha|\leq1}\bigl\| D^{\alpha}u \bigr\| _{(M)}.$$

These classes will be Banach spaces under this norm. We refer to spaces of the forms $$W^{1}L_{M}(\Omega)$$ and $$W^{1}E_{M}(\Omega)$$ as Orlicz-Sobolev spaces. Thus $$W^{1}L_{M}(\Omega)$$ and $$W^{1}E_{M}(\Omega)$$ can be identified with subspaces of the product of $$N+1$$ copies of $$L_{M}(\Omega)$$. Denoting this product by $$\Pi L_{M}$$, we will use the weak topologies $$\sigma(\Pi L_{M},\Pi E_{\bar{M}})$$ and $$\sigma(\Pi L_{M},\Pi L_{\bar{M}})$$. If M satisfies $$\Delta_{2}$$ condition (near infinity only when Ω has finite measure), then $$W^{1}L_{M}(\Omega)=W^{1}E_{M}(\Omega)$$.

The space $$W_{0}^{1}E_{M}(\Omega)$$ is defined as the (norm) closure of the Schwartz space $$\mathcal{D}(\Omega)$$ in $$W^{1}E_{M}(\Omega)$$ and the space $$W^{1}_{0}L_{M}(\Omega)$$ as the $$\sigma(\Pi L_{M},\Pi E_{\bar{M}})$$ closure of $$\mathcal{D}(\Omega)$$ in $$W^{1}L_{M}(\Omega)$$.

We recall that a sequence $$u_{n}$$ converges to u for the modular convergence in $$W^{1}L_{M}(\Omega)$$ if there exists $$\lambda>0$$ such that

$$\int_{\Omega}M \biggl(\frac{|D^{\alpha}u_{n}-D^{\alpha}u|}{\lambda} \biggr)\,dx\rightarrow0 \quad\mbox{as } n\rightarrow\infty \mbox{ for all } |\alpha|\leq1.$$

Let $$W^{-1}L_{\bar{M}}(\Omega)$$ (resp. $$W^{-1}E_{\bar{M}}(\Omega)$$) denote the space of distributions on Ω which can be written as sums of derivatives of order ≤1 of functions in $$L_{\bar{M}}(\Omega)$$ (resp. $$E_{\bar{M}}(\Omega)$$). It is a Banach space under the usual quotient norm.

If the open set Ω has the segment property, then the space $$\mathcal{D}(\Omega)$$ is dense in $$W^{1}_{0}L_{M}(\Omega)$$ for the modular convergence and thus for the topology $$\sigma(\Pi L_{M},\Pi L_{\bar{M}})$$. Consequently, the action of a distribution in $$W^{-1}L_{\bar{M}}(\Omega)$$ on an element of $$W^{1}_{0}L_{M}(\Omega)$$ is well defined. The dual space of $$W^{1}_{0}E_{M}(\Omega)$$ is $$W^{-1}L_{\bar{M}}(\Omega )$$ and the dual space of $$W^{-1}E_{\bar{M}}(\Omega)$$ is $$W^{1}_{0}L_{M}(\Omega )$$ (see [21, 22]).

For the above results, the readers can also be referred to [8, 2325].

We recall some lemmas which will be used later.

### Lemma 2.1

(see )

For all $$u\in W_{0}^{1}L_{M}(\Omega)$$, one has

$$\int_{\Omega}M\bigl(|u|/\operatorname{diam}\Omega\bigr)\,dx\leq\int _{\Omega} M\bigl(|Du|\bigr)\,dx,$$

where diamΩ is the diameter of Ω.

### Lemma 2.2

(see )

If the open set Ω has the segment property, $$u\in W_{0}^{1}L_{M}(\Omega)$$, then there exists $$\lambda>0$$ and a sequence $$u_{k}\in\mathcal {D}(\Omega)$$ such that for any $$|\alpha|\leq1$$, $$\rho_{M}(|D^{\alpha}u_{k}-D^{\alpha}u|/\lambda)\rightarrow0$$, $$k\rightarrow\infty$$.

### Definition 2.1

(see )

Let $$V_{m}=\operatorname{span}\{\omega_{1},\ldots,\omega_{m}\}$$; then $$u_{m}\in V_{m}$$ is called a Galerkin solution of $$A(u)=f$$ in $$V_{m}$$ if and only if

$$\bigl(A(u_{m}),v \bigr)=(f,v)\quad \forall v\in V_{m}.$$

The proof of the following lemma can be found in Lemma 5.12.1 in .

### Lemma 2.3

Let $$f:\mathbb{R}^{m}\rightarrow\mathbb{R}^{m}$$ be a continuous mapping with

$$\lim_{|x|\rightarrow\infty}\frac{\langle x,f(x)\rangle}{|x|}=a,$$
(4)

where a is a constant with $$-\infty\leq a<0$$ or $$0< a\leq+\infty$$, $$|\cdot|$$ is a norm in $$\mathbb{R}^{m}$$, $$\langle\cdot,\cdot\rangle$$ is an inner product defined as $$\langle x,f(x)\rangle=\sum_{i=1}^{m}x_{i}f_{i}(x)$$ with $$x=(x_{1},x_{2},\ldots,x_{m})$$ and $$f(x)=(f_{1}(x),f_{2}(x),\ldots,f_{m}(x))$$. Then the range of f is the whole of $$\mathbb{R}^{m}$$.

### Proof

Let $$u_{0}\in\mathbb{R}^{m}$$ and define $$f^{*}(x)=f(x)-u_{0}$$. Then $$f^{*}$$ satisfies (4). Consequently, it is sufficient to prove that the range of any map satisfying (4) contains the origin.

If $$0< a\leq+\infty$$, using (4) we see that we may choose r large enough so that

$$\frac{\langle x,f(x)\rangle}{|x|}>0 \quad\mbox{for }|x|=r.$$
(5)

But from (5), it follows that the mapping

$$w(\xi)=-\frac{rf(\xi)}{|f(\xi)|},\quad |\xi|\leq r.$$

Then $$w:B(0,r)\rightarrow B(0,r)$$ is continuous where $$B(0,r)=\{x\in \mathbb{R}^{m}, |x|\leq r\}$$. By the Brouwer fixed point theorem, f is continuous from $$B(0,r)\subset\mathbb{R}^{m}$$ into $$B(0,r)$$, and f has a fixed point, i.e., there exists $$x\in B(0,r)$$ such that $$x=w(x)$$. Then

$$|x|=\bigl|w(x)\bigr|= \biggl\vert -\frac{rf(x)}{|f(x)|} \biggr\vert =r,$$

which implies that

$$\frac{\langle x,f(x)\rangle}{|x|}=\frac{\langle x,-\frac {x}{r}|f(x)|\rangle}{|x|} =-\frac{|f(x)|\langle x,x\rangle}{r|x|}=-\frac{|f(x)|\cdot |x|^{2}}{r|x|} =-\frac{|f(x)|\cdot|x|}{r}< 0.$$

It is a contradiction with (5). Therefore, f is surjective.

If $$-\infty\leq a<0$$, then let $$g=-f$$. Thanks to (4), we have

$$\lim_{|x|\rightarrow\infty}\frac{\langle x,g(x)\rangle}{|x|}=-a.$$

From this we deduce that g is surjective. Therefore −g is surjective, too. Immediately, f is surjective, i.e., the range of f is the whole of $$\mathbb{R}^{m}$$. □

### Remark 2.1

Let V be a vector space of finite dimension and $$A:V\rightarrow V^{*}$$ be a continuous mapping with

$$\lim_{\|u\|_{V}\rightarrow\infty}\frac{(A(u),u)}{|x|}=a,$$
(6)

where a is the constant in Lemma 2.3 and $$V^{*}$$ is the dual space of V, then A is surjective.

Clearly, condition (4) is weaker than the one of Lemma 5.12.1 in .

### Remark 2.2

If condition (4) is replaced by

$$\lim_{|x|\rightarrow\infty}\frac{|\langle x,f(x)\rangle|}{|x|}=a,$$

then f is not surjective. For example, let $$f(x)=|x|$$, then $$f:\mathbb {R}\rightarrow\mathbb{R}$$ is continuous and

$$\frac{|\langle x,f(x)\rangle|}{|x|}=\frac{| x\cdot|x||}{|x|} =|x|\rightarrow+\infty \quad\mbox{as }|x| \rightarrow+\infty.$$

However, the range of f is $$[0,+\infty)$$. Therefore, Lemma 1 in Landes  should be without absolute.

### Lemma 2.4

(see  and )

If a sequence $$u_{n}\in L_{M}(\Omega)$$ converges a.e. to u and if $$u_{n}$$ remains bounded in $$L_{M}(\Omega)$$, then $$u\in L_{M}(\Omega)$$ and $$u_{n}\rightharpoonup u$$ for $$\sigma(L_{M}, E_{\bar{M}})$$.

### Lemma 2.5

(see )

Let $$u_{k}, u\in L_{M}(\Omega)$$. If $$u_{k}\rightarrow u$$ with respect to the modular convergence, then $$u_{k}\rightarrow u$$ for $$\sigma(L_{M}, L_{\bar{M}})$$.

For N-function M, $$\mathcal{T}_{0}^{1,M}(\Omega)$$ is defined as the set of measurable functions $$u:\Omega\rightarrow\mathbb{R}$$ such that for all $$k>0$$ the truncated functions $$T_{k}(u)\in W_{0}^{1}L_{M}(\Omega)$$ with $$T_{k}(s)=\max(-k,\min(k,s))$$.

The following lemmas will be applied to the truncation operators.

### Lemma 2.6

(see [2, 23] and )

Let $$F:\mathbb{R}\rightarrow\mathbb{R}$$ be uniformly Lipschitzian with $$F(0) = 0$$. Let M be an N-function, and let $$u\in W^{1}L_{M}(\Omega)$$ (resp. $$W^{1}E_{M}(\Omega)$$). Then $$F(u)\in W^{1}L_{M}(\Omega)$$ (resp. $$W^{1}E_{M}(\Omega)$$). Moreover, we have $$\frac{\partial}{\partial x_{i}}F(u)=F^{\prime}(u)\frac{\partial }{\partial x_{i}}u$$, a.e. in $$\{x\in\Omega|u(x)\notin D\}$$, and $$\frac {\partial}{\partial x_{i}}F(u)=0$$, a.e. in $$\{x\in\Omega|u(x)\in D\}$$, where D is the set of discontinuity points of $$F^{\prime}$$.

### Lemma 2.7

(see )

If $$u\in W^{1}L_{M}(\Omega)$$, then $$u^{+},u^{-}\in W^{1}L_{M}(\Omega)$$ and

$$Du^{+}= \textstyle\begin{cases} Du, &\textit{if }u>0, \\ 0, & \textit{if }u\leq0, \end{cases}\displaystyle \quad\textit{and} \quad Du^{-}= \textstyle\begin{cases} 0, & \textit{if }u\geq0, \\ -Du,& \textit{if }u< 0. \end{cases}$$
(7)

### Lemma 2.8

(see )

For every $$u\in\mathcal {T}_{0}^{1,M}(\Omega)$$, there exists a unique measurable function $$v:\Omega\rightarrow\mathbb{R}$$ such that $$DT_{k}(u)=v\chi_{\{|u|< k\}}$$ almost everywhere in Ω for every $$k>0$$. Define the gradient of u as the function v, and denote it by $$v=Du$$.

## Existence theorem

Let $$\Omega\subset\mathbb{R}^{N}$$ be a bounded domain with the segment property, $$N\geq2$$, M be an N-function, be a complementary function of M. Assume that M is twice continuously differentiable. Denote by $$\mathcal{P}_{M}$$ the following subset of N-functions defined as:

\begin{aligned} \mathcal{P}_{M} =& \biggl\{ B:\mathbb{R}^{+}\rightarrow \mathbb{R}^{+}: N\mbox{-function}: B \mbox{ is twice continuously differentiable}, \\ &{} B^{\prime\prime}/B^{\prime}\leq M^{\prime\prime}/M^{\prime}; \int_{0}^{1}B\circ H^{-1} \bigl(1/t^{1-1/N} \bigr)\,dt< \infty \biggr\} , \end{aligned}

where $$H(r)=M(r)/r$$. Assume that there exists $$Q\in\mathcal{P}_{M}$$ such that

$$Q\circ H^{-1} \mbox{ is an }N\mbox{-function}.$$
(8)

Let μ be a bounded nonnegative Radon measure on Ω. We consider the following Dirichlet problem:

\begin{aligned}& A(u)+g(x,u,D u)=\mu \quad\mbox{in } \Omega, \end{aligned}
(9)
\begin{aligned}& u=0, \quad\mbox{on } \partial\Omega, \end{aligned}
(10)

where $$A:D(A)\subset W_{0}^{1}L_{M}(\Omega)\to W^{-1}L_{\bar{M}}(\Omega)$$ is a mapping given by $$A(u)=-\mathop{\operatorname{div}} a(x,u,D u)$$. $$a:\Omega\times\mathbb{R}\times\mathbb{R}^{N}\to\mathbb{R}^{N}$$ is a Carathéodory function satisfying for a.e. $$x\in\Omega$$ and all $$s\in\mathbb{R}$$, $$\xi,\eta\in\mathbb{R}^{N}$$ with $$\xi\neq\eta$$:

\begin{aligned}& \bigl|a(x,s,\xi)\bigr|\leq\beta \bigl[c(x)+\bar{M}^{-1} \bigl(M\bigl(|s|\bigr) \bigr)+ \bar{M}^{-1} \bigl(M\bigl(|\xi|\bigr) \bigr) \bigr], \end{aligned}
(11)
\begin{aligned}& { \bigl[}a(x,s,\xi)-a(x,s,\eta) \bigr] [\xi-\eta]>0, \end{aligned}
(12)
\begin{aligned}& a(x,s,\xi)\xi\geq\alpha M\bigl(|\xi|\bigr), \end{aligned}
(13)

where $$\alpha,\beta>0$$, $$k_{1},k_{2}\geq0$$, $$c(x)\in E_{\bar{M}}(\Omega)$$.

$$g:\Omega\times\mathbb{R}\times\mathbb{R}^{N}\to\mathbb{R}$$ is a Carathéodory function satisfying for a.e. $$x\in\Omega$$ and all $$s\in\mathbb{R}$$, $$\xi\in\mathbb{R}^{N}$$:

$$\bigl|g(x,s,\xi)\bigr|\leq\gamma(x)+\rho(s)M\bigl(|\xi|\bigr),$$
(14)

where $$\rho:\mathbb{R}\rightarrow\mathbb{R}^{+}$$ is a continuous positive function which belongs to $$L^{1}(\mathbb{R})$$ and $$\gamma(x)$$ belongs to $$L^{1}(\Omega)$$. For example, $$g(x,u,D u)=\gamma(x)+|\sin u|e^{-u}M(|D u|)$$ (see ).

We have the following theorem.

### Theorem 3.1

Assume that (8)-(14) hold. Then there exists at least one solution of the following problem:

$$\textstyle\begin{cases} u\in\mathcal{T}_{0}^{1,M}(\Omega)\cap W_{0}^{1}L_{B}(\Omega),& \forall B\in\mathcal{P}_{M}, \\ \langle A(u),\phi\rangle+\int_{\Omega}g(x,u,D u)\phi \,dx =\langle\mu,\phi\rangle,&\forall\phi\in\mathcal{D}(\Omega). \end{cases}$$
(15)

### Remark 3.1

It is well known that there exists a sequence $$\mu_{n}\in\mathcal {D}(\Omega)$$ such that $$\mu_{n}$$ converges to μ in the distributional sense with $$\|\mu_{n}\|_{L^{1}(\Omega)}\leq\|\mu\|_{\mathcal{M}_{b}(\Omega)}$$ and $$\mu_{n}$$ is nonnegative if μ is nonnegative.

### Remark 3.2

(1) Benkirane and Bennouna [30, Remark 2.2] give some examples of N-functions M for which the set $$\mathcal{P}_{M}$$ is not empty. For example, assume that the N-function M is defined only at infinity, and let $$M(t)=t^{2}\log t$$ and $$B(t)=t\log t$$, then $$H(t)=t \log t$$ and $$H^{-1}(t)=t(\log t)^{-1}$$ at infinity (see, e.g.,  or ). Hence, the N-function B belongs to $$\mathcal{P}_{M}$$.

(2) Let $$M(t)=|t|^{p}$$ and $$B(t)=|t|^{q}$$, then $$B\in\mathcal{P}_{M}\Leftrightarrow1< q<\tilde{p}=\frac{N(p-1)}{N-1}$$ and $$p>2-\frac{1}{N}$$. So that we find the same result given in . Our theorem gives a refinement of the regularity result. For example, take $$B_{1}(t)=\frac{t^{\tilde{p}}}{\log^{\alpha}(e+t)}$$ with $$\alpha>1$$.

We have the following proposition.

### Proposition 3.1

Assume that (8)-(14) hold. Then, for any $$n\in \mathbb{N}$$, there exists at least one solution $$u_{n}\in W_{0}^{1}E_{M}(\Omega)$$ of the following approximate equation:

$$\int_{\Omega} \bigl[a(x,u,Du)Dv+g_{n}(x,u,Du)v \bigr]\,dx=\int_{\Omega}\mu_{n}v\,dx, \quad\forall v \in W_{0}^{1}L_{M}(\Omega),$$
(16)

where $$g_{n}(x,s,\xi)=\frac{g(x,s,\xi)}{1+\frac{1}{n}|g(x,s,\xi)|}$$.

### Proof

Denote $$V=W_{0}^{1}E_{M}(\Omega)$$. Define $$A_{n}:V\rightarrow V^{\ast}$$,

$$(A_{n}u,w):=\int_{\Omega} \bigl[a(x,u,Du)Dw(x) +g_{n}(x,u,Du)w(x) \bigr]\,dx,\quad \forall w\in V.$$

Then $$A_{n}$$ is well defined. Indeed, from (11) we have

$$\int_{\Omega}\bar{M} \biggl(\frac{1}{3\beta}\bigl|a(x,u,Du)\bigr| \biggr)\,dx \leq\int_{\Omega}\frac{1}{3} \bigl[\bar{M} \bigl(c(x) \bigr)+M\bigl(|u|\bigr) +M\bigl(|Du|\bigr) \bigr]\,dx< \infty.$$

Therefore, $$a(x,u,Du)\in(L_{\bar{M}}(\Omega))^{N}$$. On the other hand, for every fixed n, $$\int_{\Omega}\bar{M}(|g_{n}(x,u, Du)|)\,dx\leq\bar{M}(n) \operatorname{meas}(\Omega)<\infty$$. Thus $$g_{n}(x,u, Du)\in L_{\bar{M}}(\Omega)$$.

There exists a sequence $$\{w_{j}\}_{n=1}^{\infty}\subset\mathcal{D}(\Omega)$$ such that $$\{w_{j}\}_{n=1}^{\infty}$$ dense in V. Let $$V_{m}=\operatorname{span}\{w_{1},\ldots, w_{m}\}$$ and consider $$A_{n}|_{V_{m}}$$. $$\int_{\Omega}|Du|\,dx$$ and $$\|Du\|_{(M)}$$ to be two norms of $$V_{m}$$ equivalent to the usual norm of finite dimensional vector spaces.

Claim: the mapping $$u\rightarrow A_{n}|_{V_{m}}u:V_{m}\rightarrow V_{m}^{*}$$ is continuous. Indeed, if $$u_{j}\rightarrow u$$ in $$V_{m}$$ and there exists $$\varepsilon_{0}>0$$ such that

$$\|A_{n}|_{V_{m}}u_{j}-A_{n}|_{V_{m}}u \|_{V_{m}^{\ast}}\geq\varepsilon_{0},$$
(17)

and since $$u_{j}\rightarrow u$$ strongly in $$V_{m}$$,

$$\int_{\Omega}M\bigl(2|u_{j}-u|\bigr)\,dx\rightarrow0 \quad \mbox{and}\quad \int_{\Omega}M\bigl(2|Du_{j}-Du|\bigr)\,dx \rightarrow0,$$

then there exists a subsequence of $$\{u_{j}\}$$ still denoted by $$\{u_{j}\}$$ and $$f_{1}, f_{2}\in L^{1}(\Omega)$$ such that $$M(2|u_{j}-u|)\leq f_{1}$$ and $$M(2|Du_{j}-Du|)\leq f_{2}$$. By the convexity of M, we deduce that

$$M\bigl(|u_{j}|\bigr)\leq \frac{1}{2}M\bigl(2|u_{j}-u|\bigr)+ \frac{1}{2}M\bigl(2|u|\bigr)\leq\frac{1}{2}f_{1} + \frac{1}{2}M\bigl(2|u|\bigr).$$
(18)

Similarly,

$$M\bigl(|Du_{j}|\bigr)\leq\frac{1}{2}f_{2} + \frac{1}{2}M\bigl(2|Du|\bigr).$$
(19)

For $$\forall w\in V_{m}$$, by (11), (18), (19) and Young inequality, one has

\begin{aligned} & \bigl|a(x,u_{j},Du_{j})Dw(x) +g_{n}(x,u_{j},Du_{j})w(x)\bigr| \\ &\quad\leq \beta \bigl[c(x)+ \bar{M}^{-1} \bigl(M\bigl(|u_{j}|\bigr) \bigr)+ \bar{M}^{-1}M\bigl(|Du_{j}|\bigr) \bigr]|Dw| +n|w| \\ &\quad\leq \beta \bigl[\bar{M} \bigl(c(x) \bigr)+3M\bigl(|Dw|\bigr) + M\bigl(|u_{j}|\bigr)+M\bigl(|Du_{j}|\bigr) \bigr] + \bigl[\bar{M}(n)+M\bigl(|w|\bigr) \bigr] \\ &\quad\leq\beta \biggl[\bar{M} \bigl(c(x) \bigr)+3M\bigl(|Dw|\bigr)+ \frac{1}{2}f_{1} +\frac{1}{2}M\bigl(2|u|\bigr)+\frac{1}{2}f_{2} + \frac{1}{2}M\bigl(2|Du|\bigr) \biggr] \\ &\qquad{} +\bar{M}(n)+M\bigl(|w|\bigr). \end{aligned}
(20)

Hence $$(A_{n}|_{V_{m}}u_{j},w)<\infty$$ for all $$w\in V_{m}$$. By the Banach-Steinhaus theorem $$\{\|A_{n}|_{V_{m}}u_{j}\|_{V_{m}^{\ast}}\}_{j}$$ is bounded. Hence $$\{A_{n}|_{V_{m}}u_{j}\}_{j}$$ is relatively sequently compact in $$V_{m}^{\ast}$$. Passing to a subsequence if necessary, there exists $$\eta_{n}\in V_{m}^{\ast}$$ such that

$$\|A_{n}|_{V_{m}}u_{j}-\eta_{n} \|_{V_{m}^{\ast}}\rightarrow 0.$$

On the other hand, passing to a subsequence if necessary,

$$u_{j}(x)\rightarrow u(x) \quad\mbox{a.e. in } \Omega \quad\mbox{and} \quad Du_{j}(x)\rightarrow Du(x) \quad\mbox{a.e. in } \Omega.$$

By the Lebesgue theorem, we know that for each $$w\in V_{m}$$,

$$\lim_{j\rightarrow\infty}(A_{n}|_{V_{m}}u_{j},w) =(A_{n}|_{V_{m}}u,w).$$

Hence $$A_{n}|_{V_{m}}u=\eta_{n}$$, it is a contradiction with (17).

Thanks to (13) and Lemma 2.1, for all $$u\in V_{m}$$,

\begin{aligned} (A_{n}u,u) =& \int_{\Omega} \bigl[a(x,u,Du)Du +g_{n}(x,u,Du)u \bigr]\,dx \\ \geq& \int_{\Omega} \bigl[\alpha M\bigl(|Du|\bigr)-n|u| \bigr]\,dx \\ \geq& \alpha\int_{\Omega} M\bigl(|Du|\bigr)\,dx-\int_{\Omega} \biggl[\bar{M} \biggl(\frac {1}{\alpha_{0}}(n\operatorname{diam}\Omega) \biggr) +M \biggl(\alpha_{0}\frac{|u|}{\operatorname{diam}\Omega} \biggr) \biggr]\,dx \\ \geq& \alpha\int_{\Omega} M\bigl(|Du|\bigr)\,dx- \bar{M} \biggl( \frac{1}{\alpha _{0}}(n\operatorname{diam}\Omega) \biggr)\operatorname{meas}\Omega- \int_{\Omega} \alpha_{0}M\bigl(|Du|\bigr)\,dx \\ =&(\alpha-\alpha_{0})\int_{\Omega} M\bigl(|Du|\bigr)\,dx- \bar{M} \biggl(\frac{1}{\alpha _{0}}(n\operatorname{diam}\Omega) \biggr) \operatorname{meas}\Omega, \end{aligned}
(21)

where $$\alpha_{0}=\min\{\frac{\alpha}{2},1\}$$. By Lemma 2.1, one has $$\|u\|_{(M)}\leq \operatorname{diam}\Omega\|Du\| _{(M)}$$. It follows that $$\|u\|_{\Omega, M}\leq(1+\operatorname{diam}\Omega)\| Du\|_{(M)}$$. We have

$$\frac{\int_{\Omega} M(|Du|)\,dx}{\|u\|_{\Omega, M}} \geq\frac{1}{ 1+\operatorname{diam}\Omega} \frac{\int_{\Omega} M(|Du|)\,dx}{\|Du\| _{(M)}} \geq \frac{1}{1+\operatorname{diam}\Omega}$$
(22)

since $$\int_{\Omega}M(u)\,dx>\|u\|_{(M)}$$ whenever $$\|u\|_{(M)}>1$$. Combining (21) and (22), one has

$$\frac{(A_{n}u,u)}{\|u\|_{\Omega, M}}\geq\frac{1}{1+\operatorname{diam}\Omega}.$$
(23)

By Remark 2.1, $$A_{n}$$ is surjective, i.e., there exists a Galerkin solution $$u_{m}\in V_{m}$$ for every m such that

$$(A_{n}u_{m},v)=(\mu_{n},v),\quad \forall v\in V_{m}.$$
(24)

We will show that the sequence $$\{u_{m}\}$$ is bounded in V.

In fact, for every $$u_{m}\in V$$, if $$\|u_{m}\|_{\Omega, M}\rightarrow \infty$$, then by (23), $$(A_{n}u_{m},u_{m})\rightarrow\infty$$. It is a contradiction with (24). Therefore $$\{u_{m}\}$$ is bounded in V.

It follows from (20) that we can deduce $$\{\| A_{n}|_{V_{m}}u_{m}\|_{V^{*}}\}_{m}$$ is bounded. So we can extract a subsequence $$\{u_{k}\}_{k=1}^{\infty}$$ of $$\{ u_{m}\}_{m=1}^{\infty}$$ such that

$$u_{k}\rightharpoonup u_{0} \quad\mbox{in } V \mbox{ for } \sigma(\Pi L_{M}, \Pi E_{\bar{M}}),\qquad A_{n}u_{k}\rightharpoonup\xi_{n} \quad\mbox{in } V^{*} \mbox{ for } \sigma (\Pi L_{\bar{M}}, \Pi E_{M}),$$
(25)

as $$k\rightarrow\infty$$ and $$(\xi_{n},w)=(\mu_{n},w)$$ for all $$w\in\bigcup _{m=1}^{\infty}\{w_{m}\}$$. By the density of $$\{w_{m}\}$$, we get

$$(\xi_{n},w)=(\mu_{n},w),\quad \forall w\in V.$$

By the imbedding theorem (see, e.g., ) we have

$$u_{k}\rightarrow u_{0} \quad\mbox{strongly in } L_{M}(\Omega) \mbox{ as }k\rightarrow\infty.$$
(26)

Hence, passing to a subsequence if necessary

$$u_{k}(x)\rightarrow u_{0}(x) \quad \mbox{a.e. }x \in\Omega \mbox{ as }k\rightarrow\infty.$$
(27)

On the other hand, thanks to (26), we have

$$\int_{\Omega}g_{n}(x,u_{k},Du_{k}) (u_{k}-u_{0})\,dx\rightarrow0 \quad\mbox{and}\quad\int _{\Omega}\mu_{n}(u_{k}-u_{0})\,dx \rightarrow0$$

as $$k\rightarrow\infty$$. Thus we obtain that

\begin{aligned} & \int_{\Omega}a(x,u_{k},Du_{k}) (Du_{k}-Du_{0})\,dx \\ &\quad=\int_{\Omega}\mu_{n}(u_{k}-u_{0}) \,dx -\int_{\Omega}g_{n}(x,u_{k},Du_{k}) (u_{k}-u_{0})\,dx \rightarrow0. \end{aligned}
(28)

Fix a positive real number r and define $$\Omega_{r}=\{x\in\Omega :|Du_{0}(x)|\leq r\}$$ and denote by $$\chi_{r}$$ the characteristic function of $$\Omega_{r}$$.

Taking $$s\geq r$$, one has

\begin{aligned} 0 \leq&\int_{\Omega_{r}} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0}) \bigr] (Du_{k}-Du_{0})\,dx \\ \leq&\int_{\Omega_{s}} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0}) \bigr] (Du_{k}-Du_{0})\,dx \\ =&\int_{\Omega_{s}} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0} \chi_{s}) \bigr] (Du_{k}-Du_{0} \chi_{s})\,dx \\ \leq&\int_{\Omega} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0} \chi_{s}) \bigr] (Du_{k}-Du_{0} \chi_{s})\,dx. \end{aligned}

On the other hand,

\begin{aligned} &\int_{\Omega}a(x,u_{k},Du_{k}) (Du_{k}-Du_{0})\,dx \\ &\quad=\int_{\Omega} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0} \chi_{s}) \bigr] (Du_{k}-Du_{0} \chi_{s})\,dx \\ &\qquad{} -\int_{\Omega}a(x,u_{k},Du_{k})Du_{0} \chi_{\Omega\backslash\Omega_{s}}\,dx +\int_{\Omega}a(x,u_{k},Du_{0} \chi_{s}) (Du_{k}-Du_{0}\chi_{s}) \,dx. \end{aligned}

Therefore

\begin{aligned} & \int_{\Omega} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0} \chi_{s}) \bigr] (Du_{k}-Du_{0} \chi_{s})\,dx \\ &\quad=\int_{\Omega}a(x,u_{k},Du_{k}) (Du_{k}-Du_{0})\,dx \\ & \qquad{} +\int_{\Omega}a(x,u_{k},Du_{k})Du_{0} \chi_{\Omega\backslash\Omega_{s}}\,dx -\int_{\Omega}a(x,u_{k},Du_{0} \chi_{s}) (Du_{k}-Du_{0}\chi _{s}) \,dx. \end{aligned}
(29)

In view of (28) the first term of the right-hand side of (29) tends to 0 as $$k\rightarrow\infty$$.

$$\{a(x,u_{k},Du_{k})\}_{k}$$ is bounded in $$(L_{\bar{M}}(\Omega))^{N}$$. Indeed, for every $$w\in(E_{M}(\Omega))^{N}$$,

\begin{aligned} & \int_{\Omega}a(x,u_{k},Du_{k})w\,dx \\ &\quad=\int_{\Omega}\mu_{n}w\,dx -\int _{\Omega}g_{n}(x,u_{k},Du_{k})w \,dx \\ &\quad\leq\|\mu_{n}\|_{\bar{M}}\cdot\|w\|_{(M)}+\|n \|_{\bar{M}}\cdot\|w\|_{(M)} =\bigl(\|\mu_{n} \|_{\bar{M}}+\|n\|_{\bar{M}}\bigr)\|w\|_{(M)}< +\infty. \end{aligned}

By the Banach-Steinhaus theorem, $$\{\|a(x,u_{k},Du_{k})\|_{\bar{M}}\} _{k}$$ is bounded.

Thus, there exists $$h\in(L_{\bar{M}}(\Omega))^{N}$$ such that (for a subsequence still denoted by $$\{u_{k}\}$$)

$$a(x,u_{k},Du_{k})\rightharpoonup h \quad\mbox{in } \bigl(L_{\bar{M}}(\Omega ) \bigr)^{N} \mbox{ for } \sigma(\Pi L_{\bar{M}},\Pi E_{M}).$$

It follows that the second term of the right-hand side of (29) tends to $$\int_{\Omega\backslash\Omega_{s}}hDu_{0}\,dx$$ as $$k\rightarrow\infty$$.

Since $$a(x,u_{k},Du_{0}\chi_{s})\rightharpoonup a(x,u_{0},Du_{0}\chi_{s})$$ strongly in $$(E_{\bar{M}}(\Omega))^{N}$$, while by (25) $$Du_{k}-Du_{0}\chi_{s}\rightharpoonup Du_{0}-Du_{0}\chi_{s}$$ tends weakly in $$(E_{M}(\Omega))^{N}$$ for $$\sigma((L_{M}(\Omega))^{N},(E_{\bar {M}}(\Omega))^{N})$$, the third term of the right-hand side of (29) tends to $$-\int_{\Omega}a(x,u_{0},Du_{0}\chi_{s})(Du_{0}-Du_{0}\chi_{s})\,dx =-\int_{\Omega\backslash\Omega_{s}}a(x,u_{0},0)Du_{0}\,dx$$.

Therefore,

\begin{aligned} &\int_{\Omega} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0} \chi_{s}) \bigr] (Du_{k}-Du_{0} \chi_{s})\,dx \\ &\quad=\int_{\Omega\backslash\Omega_{s}} \bigl[h-a(x,u_{0},0) \bigr]Du_{0}\,dx +\varepsilon(k). \end{aligned}

We have then proved that

\begin{aligned} 0 \leq&\limsup_{k\rightarrow\infty} \int_{\Omega_{r}} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0}) \bigr] (Du_{k}-Du_{0})\,dx \\ =&\int_{\Omega\backslash\Omega_{s}} \bigl[h-a(x,u_{0},0) \bigr]Du_{0}\,dx. \end{aligned}

Using the fact that $$[h-a(x,u_{0},0)]Du_{0}\in L^{1}(\Omega)$$ and letting $$s\rightarrow\infty$$, we get, since $$\operatorname{meas}(\Omega\backslash \Omega_{s})\rightarrow0$$,

$$\int_{\Omega_{r}} \bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0}) \bigr] (Du_{k}-Du_{0})\,dx\rightarrow0 \quad\mbox{as }k \rightarrow \infty,$$

which gives

$$\bigl[a(x,u_{k},Du_{k})-a(x,u_{k},Du_{0}) \bigr] (Du_{k}-Du_{0})\,dx \rightarrow0 \quad\mbox{a.e. in } \Omega_{r}$$
(30)

(for a subsequence still denoted by $$\{u_{k}\}$$), say, for each $$x\in \Omega_{r}\backslash Z$$ with $$\operatorname{meas}(Z)=0$$. As the proof of Eq. (3.23) in , we can construct a subsequence such that

$$Du_{k}(x)\rightarrow Du_{0}(x) \quad\mbox{a.e. in }\Omega.$$
(31)

Consequently, we get

$$a(x,u_{k},Du_{k})\rightarrow a(x,u_{0},Du_{0}) \quad\mbox{a.e. in }\Omega,$$

and

$$g_{n}(x,u_{k},Du_{k})\rightarrow g_{n}(x,u_{0},Du_{0}) \quad\mbox{a.e. in } \Omega.$$

By Lemma 2.4, we get

$$a(x,u_{k},Du_{k})\rightharpoonup a(x,u_{0},Du_{0}) \quad \mbox{in } \bigl(L_{\bar{M}}(\Omega) \bigr)^{N} \mbox{ for } \sigma \bigl( \bigl(L_{\bar{M}}(\Omega ) \bigr)^{N}, \bigl(E_{M}( \Omega) \bigr)^{N} \bigr),$$

and

$$g_{n}(x,u_{k},Du_{k})\rightharpoonup g_{n}(x,u_{0},Du_{0}) \quad\mbox{in } \bigl(L_{\bar{M}}(\Omega) \bigr)^{N} \mbox{ for }\sigma \bigl( \bigl(L_{\bar {M}}(\Omega) \bigr)^{N}, \bigl(E_{M}( \Omega) \bigr)^{N} \bigr).$$

Then

$$\int_{\Omega} \bigl[a(x,u_{k},Du_{k})Dw+g_{n}(x,u_{k},Du_{k})w \bigr]\,dx \rightarrow\! \int_{\Omega} \bigl[a(x,u_{0},Du_{0})Dw+g_{n}(x,u_{0},Du_{0})w \bigr]\,dx$$

for every $$w\in V$$. Thus, we get $$(A_{n}u_{k},w)\rightarrow(A_{n}u_{0},w)$$ for every $$w\in V$$. It follows that $$A_{n}u_{0}=\xi_{n}$$. Therefore,

$$(A_{n}u_{0},w)=(\mu_{n},w), \quad\forall w\in W_{0}^{1}E_{M}(\Omega).$$

Furthermore, by Lemmas 2.2 and 2.5, we have

$$(A_{n}u_{0},v)=(\mu_{n},v),\quad \forall v\in W_{0}^{1}L_{M}(\Omega).$$

Hence, for every n, there exists at least one solution $$u_{n}$$ of (16) with $$u_{n}\in W_{0}^{1}E_{M}(\Omega)$$. □

### Remark 3.3

From Proposition 3.1, we have the following approximate equations:

$$\int_{\Omega} \bigl[a(x,u_{n},Du_{n})Dv+g_{n}(x,u_{n},Du_{n})v \bigr]\,dx=\int_{\Omega}\mu_{n}v\,dx, \quad\forall v \in W_{0}^{1}L_{M}(\Omega),$$
(32)

where $$u_{n}\in W_{0}^{1}E_{M}(\Omega)$$.

### Remark 3.4

Clearly, condition (11) is weaker than

$$\bigl|a(x,s,\xi)\bigr|\leq\beta \bigl[c(x)+\bar{P}^{-1} \bigl(M\bigl(|s|\bigr) \bigr)+\bar{M}^{-1} \bigl(M\bigl(|\xi|\bigr) \bigr) \bigr],$$
(33)

whenever $$P\ll M$$. If condition (11) is replaced by (33) in Proposition 3.1, the approximate equations (16) has at least one solution $$u_{n}\in W_{0}^{1}L_{M}(\Omega)$$ by the classical result of .

The proof of the following proposition is similar to the proof of Lemma 2.2 in .

### Proposition 3.2

Assume that (8)-(14) hold true, and let $$\{u_{n}\} _{n}$$ be a solution of the approximate problem (16). Let $$\varphi\in W_{0}^{1}L_{M}(\Omega)\cap L^{\infty}(\Omega)$$ with $$\varphi\geq0$$. Then

1. (1)

$$\exp(G(T_{k}(u_{n})))\varphi$$ can be taken as a test function in (32) and

\begin{aligned} &\int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(G(u_{n}) \bigr)D\varphi \,dx \\ &\quad\leq\int_{\Omega}\mu_{n}\exp \bigl(G(u_{n}) \bigr)\varphi \,dx +\int_{\Omega} \gamma(x) \exp \bigl(G(u_{n}) \bigr)\varphi \,dx; \end{aligned}
(34)
2. (2)

$$\exp(-G(T_{k}(u_{n})))\varphi$$ can be taken as a test function in (32) and

\begin{aligned} &\int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(-G(u_{n}) \bigr)D\varphi \,dx +\int_{\Omega} \gamma(x)\exp \bigl(-G(u_{n}) \bigr)\varphi \,dx \\ &\quad\geq\int_{\Omega}\mu_{n}\exp \bigl(-G(u_{n}) \bigr)\varphi \,dx. \end{aligned}
(35)

### Proof

(1) Choosing $$\exp(G(T_{k}(u_{n})))\varphi$$ as a test function in (32), we have

\begin{aligned} &\int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr) \frac{\rho (T_{k}(u_{n}))}{\alpha} DT_{k}(u_{n})\varphi \,dx \\ &\qquad{} +\int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)D\varphi \,dx \\ &\qquad{} +\int_{\Omega}g_{n}(x,u_{n},Du_{n}) \exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\varphi \,dx \\ &\quad=\int_{\Omega}\mu_{n}\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\varphi \,dx, \end{aligned}
(36)

which implies by (13)

\begin{aligned} &\int_{\Omega}\alpha M \bigl(\bigl|DT_{k}(u_{n})\bigr| \bigr)\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr) \frac{\rho (T_{k}(u_{n}))}{\alpha} \varphi \,dx \\ &\quad\leq\int_{\Omega}a \bigl(x,T_{k}(u_{n}),DT_{k}(u_{n}) \bigr)\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr) \frac{\rho(T_{k}(u_{n}))}{\alpha} DT_{k}(u_{n})\varphi \,dx \\ &\quad= \int_{\Omega}a(x,u_{n},Du_{n})\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\frac{\rho (T_{k}(u_{n}))}{\alpha} DT_{k}(u_{n})\varphi \,dx. \end{aligned}

Since $$T_{k}(u_{n})\rightarrow u_{n}$$ and $$DT_{k}(u_{n})\rightarrow Du_{n}$$ a.e. in Ω as $$k\rightarrow\infty$$, by the Fatou lemma, we get

\begin{aligned} &\int_{\Omega}\alpha M\bigl(|Du_{n}|\bigr)\exp \bigl(G(u_{n}) \bigr) \frac{\rho(u_{n})}{\alpha } \varphi \,dx \\ &\quad\leq\liminf_{k\rightarrow\infty}\int_{\Omega}\alpha M \bigl(\bigl|DT_{k}(u_{n})\bigr| \bigr)\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr) \frac{\rho(T_{k}(u_{n}))}{\alpha } \varphi \,dx \\ &\quad\leq\liminf_{k\rightarrow\infty} \int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr) \frac{\rho (T_{k}(u_{n}))}{\alpha} DT_{k}(u_{n})\varphi \,dx. \end{aligned}

On the other hand, the functions $$a(x,u_{n},Du_{n})D\varphi$$, $$g_{n}(x,u_{n},Du_{n})\varphi$$, and $$\mu_{n}\varphi$$ are summable, and the functions $$\exp(G(T_{k}(u_{n})))$$ are bounded in $$L^{\infty}(\Omega )$$; so Lebesgue’s dominated convergence theorem may be applied in the remaining integrals. Indeed, thanks to (11) and Young inequality, one has

\begin{aligned} &\bigl|a(x,u_{n},Du_{n})\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)D\varphi\bigr| \\ &\quad\leq e^{\frac{\|\rho\|_{L^{1}(\mathbb{R})}}{\alpha}}\beta \bigl[c(x)+\bar {M}^{-1} \bigl(M\bigl(|u_{n}|\bigr) \bigr) +\bar{M}^{-1} \bigl(M\bigl(|Du_{n}|\bigr) \bigr) \bigr]|D\varphi| \\ &\quad\leq e^{\frac{\|\rho\|_{L^{1}(\mathbb{R})}}{\alpha}}\beta \bigl[\bar {M} \bigl(c(x) \bigr) +M\bigl(|u_{n}|\bigr) +M\bigl(|Du_{n}|\bigr)+3M\bigl(|D\varphi|\bigr) \bigr]. \end{aligned}

Since $$a(x,u_{n},Du_{n})\exp(G(T_{k}(u_{n})))D\varphi\rightarrow a(x,u_{n},Du_{n})\exp(G(u_{n}))D\varphi$$ a.e. in Ω as $$k\rightarrow\infty$$, and by Lebesgue’s dominated convergence theorem, we deduce that

$$\int_{\Omega}a(x,u_{n},Du_{n})\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)D\varphi \,dx \rightarrow \int_{\Omega}a(x,u_{n},Du_{n})\exp \bigl(G(u_{n}) \bigr)D\varphi \,dx$$

as $$k\rightarrow\infty$$. Since $$g_{n}(x,u_{n},Du_{n})\exp(G(T_{k}(u_{n})))\varphi\rightarrow g_{n}(x,u_{n},Du_{n})\exp(G(u_{n}))\varphi$$ a.e. in Ω as $$k\rightarrow\infty$$, and

$$\bigl|g_{n}(x,u_{n},Du_{n})\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\varphi\bigr| \leq n e^{\frac{\|\rho\|_{L^{1}(\mathbb{R})}}{\alpha}}\|\varphi\| _{\infty},$$

by Lebesgue’s dominated convergence theorem one has

$$\int_{\Omega}g_{n}(x,u_{n},Du_{n}) \exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\varphi \,dx \rightarrow\int_{\Omega}g_{n}(x,u_{n},Du_{n}) \exp \bigl(G(u_{n}) \bigr)\varphi \,dx$$

as $$k\rightarrow\infty$$. Since $$\mu_{n}\exp(G(T_{k}(u_{n})))\varphi \rightarrow\mu_{n}\exp(G(u_{n}))\varphi$$ a.e. in Ω as $$k\rightarrow\infty$$, and

$$\bigl|\mu_{n}\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\varphi\bigr| \leq e^{\frac{\|\rho\|_{L^{1}(\mathbb{R})}}{\alpha}}\mu_{n}\|\varphi\| _{\infty},$$

we have

$$\int_{\Omega}\mu_{n}\exp \bigl(G \bigl(T_{k}(u_{n}) \bigr) \bigr)\varphi \,dx \rightarrow\int _{\Omega}\mu_{n} \exp \bigl(G(u_{n}) \bigr) \varphi \,dx$$

as $$k\rightarrow\infty$$.

Thus, letting k tend to ∞ in (36), we obtain

\begin{aligned} &\int_{\Omega}M\bigl(|Du_{n}|\bigr)\exp \bigl(G(u_{n}) \bigr)\rho(u_{n}) Du_{n}\varphi \,dx \\ &\qquad{} +\int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(G(u_{n}) \bigr)D\varphi \,dx +\int_{\Omega}g_{n}(x,u_{n},Du_{n}) \exp \bigl(G(u_{n}) \bigr)\varphi \,dx \\ &\quad\leq\int_{\Omega}\mu_{n}\exp \bigl(G(u_{n}) \bigr)\varphi \,dx. \end{aligned}
(37)

By (14), (37) is reduced to (34).

(2) Similarly, taking $$\exp(-G(T_{k}(u_{n})))\varphi$$ as a test function in (32), we obtain (35). □

### Proposition 3.3

Assume that (8)-(14) hold true, and let $$\{u_{n}\} _{n}$$ be a solution of the approximate problem (16). Then, for all $$k>0$$, there exists a constant C (which does not depend on the n and k) such that

$$\int_{\Omega}M \bigl(\bigl|DT_{k}(u_{n})\bigr| \bigr)\,dx\leq Ck.$$
(38)

### Proof

Let $$\varphi=T_{k}(u_{n})^{+}$$ in (34). Also let $$G(\pm\infty )=\frac{1}{\alpha}\int_{0}^{\pm\infty}\rho(s)\,ds$$ which are well defined since $$\rho\in L^{1}(\mathbb{R})$$, then $$G(-\infty)\leq G(s)\leq G(+\infty)$$ and $$|G(\pm\infty)|\leq\|\rho\|_{L^{1}(\mathbb{R})}/\alpha$$. We have

$$\int_{\Omega}a(x,u_{n},Du_{n})\exp \bigl(G(u_{n}) \bigr)DT_{k}(u_{n})^{+} \,dx \leq e^{\frac{\|\rho\|_{L^{1}(\mathbb{R})}}{\alpha}}k \bigl[\|\mu\|_{\mathcal {M}_{b}(\Omega)} +\bigl\| \gamma(x) \bigr\| _{L^{1}(\Omega)} \bigr].$$

Immediately, by (13) we get

$$\int_{\Omega}M \bigl(\bigl|DT_{k}(u_{n})^{+}\bigr| \bigr)\,dx \leq Ck .$$
(39)

Similarly, let $$\varphi=T_{k}(u_{n})^{-}$$ in (35). We obtain

$$\int_{\Omega}M \bigl(\bigl|DT_{k}(u_{n})^{-}\bigr| \bigr)\,dx \leq Ck .$$
(40)

Combing (39) and (40), we deduce (38). □

### Proposition 3.4

Assume that (8)-(14) hold true, and let $$\{u_{n}\} _{n}$$ be a solution of the approximate problem (16). Then there exists a measurable function u such that for all $$k>0$$ we have (for a subsequence still denoted by $$\{u_{n}\}_{n}$$),

1. (1)

$$u_{n}\rightarrow u$$ a.e. in Ω;

2. (2)

$$T_{k}(u_{n})\rightharpoonup T_{k}(u)$$ weakly in $$W_{0}^{1}E_{M}(\Omega)$$ for $$\sigma(\Pi L_{M}, \Pi E_{\bar{M}})$$;

3. (3)

$$T_{k}(u_{n})\rightarrow T_{k}(u)$$ strongly in $$E_{M}(\Omega )$$ and a.e. in Ω.

### Proof

Since

\begin{aligned} & M \biggl(\frac{k}{\operatorname{diam}\Omega} \biggr)\operatorname{meas} \bigl\{ \bigl|T_{k}(u_{n})\bigr|=k \bigr\} \\ &\quad= \int_{\{|T_{k}(u_{n})|=k\}}M \biggl(\frac{|T_{k}(u_{n})|}{\operatorname{diam}\Omega} \biggr)\,dx \leq \int_{\Omega}M \bigl(\bigl|DT_{k}(u_{n})\bigr| \bigr)\,dx \leq Ck \end{aligned}

and $$\{|u_{n}|>k\}\subset\{|T_{k}(u_{n})|=k\}$$, we get

$$\operatorname{meas}\bigl\{ |u_{n}|>k\bigr\} \leq\operatorname{meas} \bigl\{ \bigl|T_{k}(u_{n})\bigr|=k \bigr\} \leq\frac{Ck}{M(\frac{k}{\operatorname{diam}\Omega})}$$

for all n and for all k. Similar to the proof of Proposition 4.3 in , assertions (1)-(3) hold. □

### Proposition 3.5

Assume that (8)-(14) hold true, and let $$\{u_{n}\} _{n}$$ be a solution of the approximate problem (16). Then, for all $$k>0$$,

1. (1)

$$\{a(x,T_{k}(u_{n}), DT_{k}(u_{n}))\}_{n}$$ is bounded in $$L_{\bar{M}}(\Omega)^{N}$$;

2. (2)

$$Du_{n}\rightarrow Du$$ a.e. in Ω (for a subsequence) as $$n\rightarrow\infty$$.

### Proof

(1) Let $$w\in(E_{M}(\Omega))^{N}$$ be arbitrary. By condition (11) and Young inequality, we have

\begin{aligned} &\int_{\Omega}a \bigl(x,T_{k}(u_{n}), DT_{k}(u_{n}) \bigr)w\,dx \\ &\quad\leq\beta\int_{\Omega} \bigl[\bar {M} \bigl(c(x) \bigr)+M(k)+M \bigl(\bigl|DT_{k}(u_{n})\bigr| \bigr)+3M\bigl(|w|\bigr) \bigr] \,dx \\ &\quad\leq\beta \biggl[\int_{\Omega}\bar{M} \bigl(c(x) \bigr) \,dx+M(k) \operatorname{meas}\Omega+\int_{\Omega}M \bigl(\bigl|DT_{k}(u_{n})\bigr| \bigr)\,dx+3\int_{\Omega}M\bigl(|w|\bigr) \,dx \biggr] \\ &\quad\leq\beta \biggl[\int_{\Omega}\bar{M} \bigl(c(x) \bigr) \,dx+M(k) \operatorname{meas}\Omega+Ck+3\int_{\Omega}M\bigl(|w|\bigr)\,dx \biggr]=C(k)< +\infty, \end{aligned}

where $$C(k)$$ is a constant independent of n.

By the Banach-Steinhaus theorem $$\{\|a(x,T_{k}(u_{n}), DT_{k}(u_{n}))\| _{\bar{M}}\}_{n}$$ is bounded; this completes the proof of assertion (1).

(2) Let $$\Omega_{s}=\{x\in\Omega||DT_{k}(u_{n})|< s\}$$ and denote by $$\chi_{s}$$ the characteristic function of $$\Omega_{s}$$. Clearly, $$\Omega _{s}\subset\Omega_{s+1}$$ and $$\operatorname{meas}(\Omega\backslash\Omega _{s})\rightarrow0$$ as $$s\rightarrow\infty$$.

Step (i). We shall show the following assertion:

$$\lim_{j\rightarrow\infty}\limsup_{n\rightarrow\infty}\int _{\{-(j+1)\leq u_{n}\leq-j\}}a(x,u_{n}, Du_{n})Du_{n} \,dx=0.$$
(41)

Indeed, the term in (35) with $$\mu_{n}$$ can be neglected since it is nonnegative. Hence

$$-\int_{\Omega}a(x,u_{n},Du_{n}) \exp \bigl(-G(u_{n}) \bigr)D\varphi \,dx \leq\int_{\Omega} \gamma(x)\exp \bigl(-G(u_{n}) \bigr)\varphi \,dx.$$
(42)

Taking $$\varphi=T_{1}(u_{n}-T_{j}(u_{n}))^{-}$$ in (42), we obtain

\begin{aligned} &\int_{\{-(j+1)\leq u_{n}\leq-j\}}a(x,u_{n},Du_{n})Du_{n} \exp \bigl(-G(u_{n}) \bigr)\,dx \\ & \quad \leq\int_{\Omega}\gamma(x)\exp \bigl(-G(u_{n}) \bigr)T_{1} \bigl(u_{n}-T_{j}(u_{n}) \bigr)^{-}\,dx. \end{aligned}

Since $$|\gamma(x)\exp(-G(u_{n}))T_{1}(u_{n}-T_{j}(u_{n}))^{-}| \leq e^{\frac{\|\rho\|_{L^{1}(\mathbb{R})}}{\alpha}}|\gamma(x)|$$, we deduce

$$\lim_{j\rightarrow\infty}\lim_{n\rightarrow\infty} \int _{\Omega}\gamma (x)\exp \bigl(-G(u_{n}) \bigr)T_{1} \bigl(u_{n}-T_{j}(u_{n}) \bigr)^{-}\,dx=0,$$

by Lebesgue’s dominate convergence theorem, which implies (41).

Step (ii). Taking $$\varphi =(T_{k}(u_{n})-T_{k}(v_{i}))^{-}[1-|T_{1}(u_{n}-T_{j}(u_{n}))|]$$ and $$\varphi =(T_{k}(v_{i})-T_{k}(u_{n}))^{-}[1-|T_{1}(u_{n}-T_{j}(u_{n}))|]$$ in (42) with $$j>k$$, as in , we can deduce that, by passing to a subsequence if necessary,

$$DT_{k}(u_{n})\rightarrow DT_{k}(u) \quad \mbox{a.e. in } \Omega,$$
(43)

and

$$Du_{n}\rightarrow Du \quad\mbox{a.e. in } \Omega.$$
(44)

□

### Proof of Theorem 3.1

(1) We are going to show that as $$n\rightarrow\infty$$,

$$g_{n}(x,u_{n}, Du_{n}) \rightarrow g(x,u, Du)\quad \mbox{in } L^{1}(\Omega).$$
(45)

Indeed, taking $$v=\exp(-G(T_{k}(u_{n})))\int_{T_{k}(u_{n})}^{0}\rho (s)\chi_{\{s<-h\}}\,ds$$ as a test function in (32), we have

\begin{aligned} & \int_{\Omega}a(x,u_{n},D u_{n})DT_{k}(u_{n}) \frac{\rho(T_{k}(u_{n}))}{\alpha}\exp \bigl(-G \bigl(T_{k}(u_{n}) \bigr) \bigr) \int_{T_{k}(u_{n})}^{0}\rho(s)\chi_{\{s< -h\}}\,ds \,dx \\ &\qquad{} + \int_{\Omega}a(x,u_{n},Du_{n})D T_{k}(u_{n}) \exp \bigl(-G \bigl(T_{k}(u_{n}) \bigr) \bigr)\rho \bigl(T_{k}(u_{n}) \bigr) \chi_{\{T_{k}(u_{n})< -h\}}\,dx \\ &\quad=\int_{\Omega}g_{n}(x,u_{n},Du_{n}) \exp \bigl(-G \bigl(T_{k}(u_{n}) \bigr) \bigr) \int _{T_{k}(u_{n})}^{0}\rho(s)\chi_{\{s< -h\}}\,ds\,dx \\ &\qquad{} - \int_{\Omega}\mu_{n}\exp \bigl(-G \bigl(T_{k}(u_{n}) \bigr) \bigr)\int_{T_{k}(u_{n})}^{0} \rho(s)\chi_{\{s< -h\}}\,ds\,dx. \end{aligned}

Using (13) and by Fatou’s lemma and Lebesgue’s theorem, we can deduce that

\begin{aligned} & \int_{\Omega}\alpha M\bigl(|D u_{n}|\bigr)\exp \bigl(-G(u_{n}) \bigr) \frac{\rho(u_{n})}{\alpha} \int_{u_{n}}^{0} \rho(s)\chi_{\{s< -h\}}\,ds\,dx \\ &\qquad{} + \int_{\Omega}\alpha M\bigl(|D u_{n}|\bigr)\exp \bigl(-G(u_{n}) \bigr) \rho(u_{n})\chi_{\{u_{n}< -h\}}\,dx \\ &\quad\leq\int_{\Omega}g_{n}(x,u_{n},Du_{n}) \exp \bigl(-G(u_{n}) \bigr) \int_{u_{n}}^{0} \rho(s)\chi_{\{s< -h\}}\,ds\,dx \\ &\qquad{} - \int_{\Omega}\mu_{n}\exp \bigl(-G(u_{n}) \bigr)\int_{u_{n}}^{0} \rho(s)\chi _{\{s< -h\}}\,ds\,dx, \end{aligned}

which implies that

\begin{aligned} &\int_{\Omega}\alpha M\bigl(|D u_{n}|\bigr)\exp \bigl(-G(u_{n}) \bigr) \rho(u_{n})\chi_{\{u_{n}< -h\}}\,dx \\ &\quad\leq \int_{\Omega}\gamma(x) \exp \bigl(-G(u_{n}) \bigr) \int_{u_{n}}^{0}\rho(s)\chi_{\{s< -h\}}\,ds \,dx \\ & \qquad{}- \int_{\Omega}\mu_{n}\exp \bigl(-G(u_{n}) \bigr)\int_{u_{n}}^{0} \rho(s)\chi_{\{ s< -h\}}\,ds\,dx. \end{aligned}

Since $$\rho\geq0$$, we get

$$\int_{u_{n}}^{0}\rho(s)\chi_{\{s< -h\}}\,ds \leq \int_{-\infty}^{-h}\rho(s)\,ds.$$

Hence we have

\begin{aligned} & \int_{\Omega}M\bigl(|D u_{n}|\bigr)\exp \bigl(-G(u_{n}) \bigr)\rho(u_{n})\chi_{\{u_{n}< -h\}}\,dx \\ & \quad\leq\frac{1}{\alpha}e^{\frac{\|\rho\|_{L^{1}(\mathbb {R})}}{\alpha}} \int_{-\infty}^{-h} \rho(s)\,ds \bigl(\|\gamma\|_{L^{1}(\Omega)}+\|\mu\|_{\mathcal{M}_{b}(\Omega)}\bigr) =C\int _{-\infty}^{-h}\rho(s)\,ds. \end{aligned}

Consequently, one has

$$\int_{\Omega}M\bigl(|D u_{n}|\bigr)\rho(u_{n}) \chi_{\{u_{n}< -h\}}\,dx \leq C\int_{-\infty}^{-h}\rho(s) \,ds.$$

Letting $$h\rightarrow+\infty$$, one has

$$\int_{-\infty}^{-h}\rho(s)\,ds\rightarrow0.$$

Therefore,

$$\lim_{h\rightarrow+\infty}\sup_{n\in N} \int _{\{u_{n}< -h\}}M\bigl(|D u_{n}|\bigr)\rho(u_{n})\,dx=0.$$

Taking $$v=\exp(G(T_{k}(u_{n})))\int_{0}^{T_{k}(u_{n})}\rho(s)\chi_{\{ s>h\}}\,ds$$ as a test function in (32), similarly we obtain that

$$\lim_{h\rightarrow+\infty}\sup_{n\in N} \int _{\{u_{n}>h\}}M\bigl(|D u_{n}|\bigr)\rho(u_{n})\,dx=0.$$

Hence,

$$\lim_{h\rightarrow+\infty}\sup_{n\in N} \int _{\{|u_{n}|>h\}}M\bigl(|D u_{n}|\bigr)\rho(u_{n})\,dx=0.$$
(46)

Following the proof of step 1 in Theorem 3.1 of , we can deduce (45).

(2) We will prove that $$a(x,u_{n},D u_{n})\rightharpoonup a(x,u,Du)$$ weakly for $$\sigma(\Pi L_{Q\circ H^{-1}}, \Pi E_{\overline{Q\circ H^{-1}}})$$.

By (44), we have

$$a(x,u_{n},D u_{n})\to a(x,u,D u) \quad\mbox{a.e. in } \Omega.$$
(47)

By $$Q\in\mathcal{P}_{M}$$ and (8), one has $$Q^{\prime\prime}/Q^{\prime}\leq M^{\prime\prime}/M^{\prime}$$. Then

$$\int\frac{Q^{\prime\prime}(t)}{Q^{\prime}(t)}\,dt \leq\int\frac{M^{\prime\prime}(t)}{M^{\prime}(t)}\,dt.$$

Thus, there exists a constant C such that $$\ln|Q^{\prime}(t)|\leq\ln|M^{\prime}(t)|+C$$. Therefore,

$$Q^{\prime}(t)\leq CM^{\prime}(t).$$

It implies that

$$Q(r)=\int_{0}^{r}Q^{\prime}(t)\,dt\leq C \int_{0}^{r}M^{\prime }(t)\,dt=CM(r).$$
(48)

Let $$s=H(r)$$, then $$s=\frac{M\circ H^{-1}(s)}{H^{-1}(s)}$$. By Young inequality we have

$$M\circ H^{-1} \biggl(\frac{s}{2} \biggr)=\frac{s}{2} \cdot H^{-1} \biggl(\frac{s}{2} \biggr) \leq\frac{1}{2} \bar{M}(s)+\frac{1}{2}M\circ H^{-1} \biggl(\frac{s}{2} \biggr).$$

Hence

$$M\circ H^{-1} \biggl(\frac{s}{2} \biggr)\leq\bar{M}(s).$$
(49)

In view of (48) and (49), we get

$$\int_{\Omega}Q\circ H^{-1} \biggl( \frac{1}{2}c(x) \biggr)\,dx \leq C\int_{\Omega}M\circ H^{-1} \biggl( \frac{1}{2}c(x) \biggr)\,dx \leq C\int _{\Omega}\bar{M} \bigl(c(x) \bigr)\,dx< \infty.$$
(50)

Since $$\bar{M}^{-1}(M(|D u_{n}|))\leq2\frac{M(|D u_{n}|)}{|D u_{n}|}$$, we have

$$\frac{1}{2}\bar{M}^{-1} \bigl(M\bigl(|D u_{n}|\bigr) \bigr) \leq \frac{M(|D u_{n}|)}{|D u_{n}|}.$$

Hence

$$\int_{\Omega}Q\circ H^{-1} \biggl(\frac{1}{2} \bar{M}^{-1} \bigl(M\bigl(|D u_{n}|\bigr) \bigr) \biggr)\,dx \leq\int _{\Omega}Q\circ H^{-1} \biggl(\frac{M(|D u_{n}|)}{|D u_{n}|} \biggr) \,dx =\int_{\Omega}Q\bigl(|D u_{n}|\bigr)\,dx,$$

and

$$\int_{\Omega}Q\circ H^{-1} \biggl(\frac{1}{2} \bar{M}^{-1} \bigl(M\bigl(|u_{n}|\bigr) \bigr) \biggr)\,dx \leq\int _{\Omega} Q\bigl(|u_{n}|\bigr)\,dx.$$

For $$t>0$$, by taking $$T_{h}(u_{n}-T_{t}(u_{n}))$$ as a test function in (32), from (14) and (46), we can deduce that

$$\int_{\{t< |u_{n}|\leq t+h\}}a(x,u_{n},D u_{n})D u_{n}\,dx \leq Ch,$$

where C is a constant independent of n, h, t, which gives

$$\frac{1}{h}\int_{\{t< |u_{n}|\leq t+h\}}M\bigl(|D u_{n}|\bigr)\,dx\leq C,$$

and by letting $$h\to0$$,

$$-\frac{d}{dt}\int_{\{|u_{n}|>t\}}M\bigl(|D u_{n}|\bigr)\,dx\leq C.$$

Let now $$B\in\mathcal{P}_{M}$$. Following the lines of , it is easy to deduce that

$$\int_{\Omega}B\bigl(|D u_{n}|\bigr)\,dx\leq C,\quad \forall n.$$

This implies that $$\{u_{n}\}$$ is bounded in $$W_{0}^{1}L_{Q}(\Omega)$$ and converges to u strongly in $$L_{Q}(\Omega)$$. Consequently, using the convexity of $$Q\circ H^{-1}$$ and by (50), we have

\begin{aligned} &\int_{\Omega}Q\circ H^{-1} \biggl(\frac{|a(x,u_{n},D u_{n})|}{6\beta} \biggr)\,dx \\ &\quad\leq\frac{1}{3}\int_{\Omega}Q\circ H^{-1} \biggl(\frac{1}{2}c(x) \biggr)\,dx +\frac{1}{3}\int _{\Omega}Q\circ H^{-1} \biggl(\frac{1}{2}\bar {M}^{-1} \bigl(M\bigl(|u_{n}|\bigr) \bigr) \biggr)\,dx \\ &\qquad{} +\frac{1}{3}\int_{\Omega}Q\circ H^{-1} \biggl( \frac{1}{2}\bar {M}^{-1} \bigl(M\bigl(|D u_{n}|\bigr) \bigr) \biggr)\,dx \\ &\quad\leq\frac{1}{3} \biggl[C\int_{\Omega}\bar{M} \bigl(c(x) \bigr)\,dx+\int_{\Omega }Q\bigl(|u_{n}|\bigr)\,dx+\int _{\Omega}Q\bigl(|D u_{n}|\bigr)\,dx \biggr]\leq C, \end{aligned}

where C is independent of n. Thus we get

$$a(x,u_{n},D u_{n})\rightharpoonup a(x,u,D u) \quad\mbox{weakly for } \sigma(\Pi L_{Q\circ H^{-1}} \Pi E_{\overline{Q\circ H^{-1}}}).$$
(51)

Thanks to (45) and (51) we can pass to the limit in (32) and we obtain that u is a solution of (15). □

## Change history

• ### 10 July 2017

An erratum to this article has been published.

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## Acknowledgements

The authors are highly grateful for the referees’ careful reading and comments on this paper. The first author was supported by ’Chen Guang’ Project (supported by Shanghai Municipal Education Commission and Shanghai Education Development Foundation) (10CGB25), and Shanghai Universities for Outstanding Young Teachers’ Scientific Research Selection and Training Special Fund (sjq08011). The second author was supported by the National Natural Science Foundation of China (11371279).

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Correspondence to Xiaochun Fang.

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The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

The original version of this article was revised: In the publication of this article (1), there was an error in the References. The error in reference 3: ‘Youssfi, A, Benkirane, A, Moumni, A: Bounded solutions of unilateral problems for strongly nonlinear equations in Orlicz spaces. Electron. J. Qual. Theory Differ. Equ. 2013, 21 (2013)’ Should instead read: ‘Youssfi, A, Benkirane, A, El Moumni, M: A: Bounded solutions of unilateral problems for strongly nonlinear equations in Orlicz spaces. Electron. J. Qual. Theory Differ. Equ. 2013, 21 (2013)’.

An erratum to this article is available at https://doi.org/10.1186/s13661-017-0839-0.

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