Differential equations of divergence form in separable Musielak-Orlicz-Sobolev spaces

Abstract

In this paper, we study the existence of weak solutions for differential equations of divergence form

$$-\operatorname {div}\bigl(a_{1}(x,Du)\bigr)+a_{0}(x,u)=f(x,u,Du), $$

in Ω coupled with a Dirichlet or Neumann boundary condition in separable Musielak-Orlicz-Sobolev spaces where $a_{1}$ satisfies the growth condition, the coercive condition, and the monotone condition, and $a_{0}$ satisfies the growth condition without any coercive condition or monotone condition. The right-hand side $f:\Omega\times\mathbb{R}\times\mathbb{R}^{N}\rightarrow\mathbb {R}$ is a Carathéodory function satisfying a growth condition dependent on the solution u and its gradient Du. We prove the existence of weak solutions by using a linear functional analysis method. Some sufficient conditions guarantee the existence enclosure of weak solutions between sub- and supersolutions. Our method does not require any reflexivity of the Musielak-Orlicz-Sobolev spaces.

1 Introduction

Let $\Omega\subset\mathbb{R}^{N}$ be a bounded domain with Lipschitz boundary. Le [1] established a sub-supersolution method for variational inequalities with Leray-Lions operators in Sobolev spaces with variable exponents. Following [1], Fan [2] established a sub-supersolution method for the differential equations of divergence form

$$ -\operatorname {div}\bigl(a_{1}(x,Du)\bigr)+a_{0}(x,u)=f(x,u), $$

(1.1)

in Ω coupled with Neumann or Dirichlet boundary condition in reflexive Musielak-Orlicz-Sobolev spaces $W_{0}^{1}L_{\Phi}(\Omega)$. Here $a_{1}$ and $a_{0}$ are supposed to satisfy growth conditions, coercive conditions, and monotone conditions, that is,

$$\begin{aligned}& \bigl\vert a_{1}(x,\xi)\bigr\vert \leq b_{1}\varphi \bigl(x, \vert \xi \vert \bigr)+g(x), \end{aligned}$$

(1.2)

$$\begin{aligned}& a_{1}(x,\xi)\xi\geq b_{2}\Phi\bigl(x, \vert \xi \vert \bigr)-h(x), \end{aligned}$$

(1.3)

$$\begin{aligned}& \bigl[a_{1}(x,\xi)-a_{1}(x,\eta)\bigr](\xi-\eta)\geq0, \end{aligned}$$

(1.4)

and

$$\begin{aligned}& \bigl\vert a_{0}(x,t)\bigr\vert \leq b_{1}\varphi \bigl(x, \vert t\vert \bigr)+g(x), \end{aligned}$$

(1.5)

$$\begin{aligned}& a_{0}(x,t)t\geq b_{2}\Phi\bigl(x, \vert t\vert \bigr)-h(x), \end{aligned}$$

(1.6)

$$\begin{aligned}& \bigl[a_{0}(x,s)-a_{0}(x,t)\bigr](s-t)\geq0, \end{aligned}$$

(1.7)

for $x\in\Omega$, $s, t\in\mathbb{R}$ and $\xi,\eta\in\mathbb{R}^{N}$, where $b_{1}, b_{2} >0$, $g\in E_{\overline{\Phi}}(\Omega)$, $g\geq0$, $h \in L^{1}(\Omega)$, and $h\geq0$. The right-hand side $f:\Omega \times\mathbb{R}\rightarrow\mathbb{R}$ is a Carathéodory function.

Liu et al. [3] proved the existence of weak solutions for (1.1) with $a_{0}=0$ in reflexive Musielak-Orlicz-Sobolev spaces.

However, there exist some nonreflexive Musielak-Orlicz-Sobolev spaces. For example, let $\Phi(x,t)=(1+\frac{t}{p(x)})\ln(1+\frac {t}{p(x)})-\frac{t}{p(x)}$, for $x\in\Omega$ and $t>0$, where $p:\Omega \rightarrow\mathbb{R}$ is a measurable function such that $1< p_{-}:=\inf_{x\in\Omega}p(x)\leq p(x)\leq p_{+}:=\sup_{x\in\Omega}p(x)< +\infty$. Then the Musielak-Orlicz-Sobolev space $W^{1}L_{\Phi}(\Omega)$ is separable and nonreflexive.

The purpose of this paper is to weaken the restriction of reflexivity of the Musielak-Orlicz spaces in [2] and study the existence of solutions for the following nonlinear problem:

$$ -\operatorname {div}\bigl(a_{1}(x,Du)\bigr)+a_{0}(x,u)=f(x,u,Du), $$

(1.8)

in Ω coupled with Dirichlet or Neumann boundary condition, where $a_{1}$ satisfies the growth condition, the coercive condition, and the monotone condition, and $a_{0}$ satisfies the growth condition without any coercive condition or monotone condition. The right-hand side $f:\Omega\times\mathbb{R}\times\mathbb{R}^{N}\rightarrow\mathbb{R}$ is a Carathéodory function satisfying a growth condition dependent on the solution u and its gradient Du.

One needs the following coercive condition of Φ in [2]:

$$ \Phi(x,\alpha u) \geq\alpha G(\alpha)\Phi(x,u),\quad \mbox{for } x\in \Omega , t\in\mathbb{R} \mbox{ and } \alpha>0, $$

(1.9)

where $G:(0,+\infty)\rightarrow\mathbb{R}$ is a function such that $G(\alpha)\rightarrow+\infty$ as $\alpha\rightarrow+\infty$. We will point out that the condition (1.9) can be omitted.

This paper is organized as follows: Section 2 contains some preliminaries and some technical lemmas which will be needed. We establish some basic properties for Musielak-Orlicz functions and some necessary and sufficient conditions for Musielak-Orlicz functions satisfying the $\Delta_{2}$ condition. In Section 3, we establish a linear functional analysis method for differential equations of divergence form to prove the existence of weak solutions for (1.8) with Dirichlet boundary or Neumann boundary condition in separable Musielak-Orlicz-Sobolev spaces. We give the enclosure of weak solutions between sub- and supersolutions by using a sub-supersolution method. Our method does not require any monotonicity or coercivity of $a_{0}$. We point out that the coercive condition (1.9) of Φ can be omitted because of the reflexivity of the Musielak-Orlicz-Sobolev spaces in [2].

We refer to some results of sub-supersolution methods for variational inequalities and the existence of solutions for differential equations studied in variable exponent Sobolev or Orlicz-Sobolev spaces (see, e.g., [4–11]). For some results we also refer to [12–14].

In this paper, we always assume that $\Omega\subset\mathbb{R}^{N}$ is a bounded domain with Lipschitz boundary and denote by $L^{0}(\Omega)$ the set of all real measurable functions defined on Ω.

2 Preliminaries

Now we list briefly some definitions and facts about Musielak-Orlicz-Sobolev spaces; for more details see [2, 15, 16], and [17].

A real function Φ defined on $\Omega\times\mathbb{R}_{+}$, where $\mathbb{R}_{+}=[0,+\infty)$, will be said a generalized N-function (i.e. a Musielak-Orlicz function), denoted by $\Phi\in N(\Omega )$, if it satisfies the following conditions:

• (i) $\Phi(x,u)$ is an N-function of the variable $u\geq0$ for every $x\in\Omega$, i.e. is a convex, nondecreasing, continuous function of u such that $\Phi(x, 0)=0$, $\Phi(x, u)>0$ for $u > 0$, and we have the conditions

  $$\lim_{u\rightarrow0^{+}}\sup_{x\in\Omega}\frac{\Phi(x,u)}{u}=0,\qquad \lim_{u\rightarrow+\infty}\inf_{x\in\Omega}\frac{\Phi(x,u)}{u}=+\infty. $$

• (ii) $\Phi(x,u)$ is a measurable function of x for all $u \geq0$.

Equivalently, Φ admits the representation

$$ \Phi(x,u)= \int_{0}^{u}\varphi(x,\tau)\,d\tau, $$

(2.1)

where $\varphi(x,u)$ is the right-hand derivative of $\Phi(x,\cdot)$ at u, for a fixed $x\in\Omega$ and all $u\geq0$. Then for every $x\in \Omega$, $\varphi(x,\tau)$ is a right-continuous and nondecreasing function of $\tau\geq0$, $\varphi(x,0)=0$, $\varphi(x,\tau)> 0$ for $\tau> 0$, and $\lim_{u\rightarrow+\infty}\inf_{x\in\Omega}\varphi (x,\tau)=+\infty$.

Let $\Phi\in N(\Omega)$, then $\Phi(x,u)\leq u\varphi(x,u)\leq\Phi (x,2u)$, for $x\in\Omega$, $u\geq0$.

The complementary function Φ̅ to a Musielak-Orlicz function Φ is defined as follows:

$$\overline{\Phi}(x,v)=\sup_{u\geq0}\bigl\{ uv-\Phi(x,u)\bigr\} , \quad \mbox{for all } v\geq0, x\in\Omega. $$

Then Φ̅ is a Musielak-Orlicz function and Φ is also the complementary function to Φ̅. Equivalently, Φ̅ admits the representation

$$ \overline{\Phi}(x,v)= \int_{0}^{v}\phi(x,\sigma)\,d\sigma, $$

(2.2)

where ϕ is given by

$$ \phi(x,\sigma)=\sup\bigl\{ \tau: \varphi(x,\tau)\leq\sigma\bigr\} , \quad \mbox{for all } x\in\Omega. $$

(2.3)

Similar to the proof in [18], we can deduce that

$$ \phi\bigl(x,\varphi(x,u)\bigr)\geq u, \qquad \varphi\bigl(x,\phi(x,v) \bigr)\geq v, \quad \mbox{for } u\geq0, v\geq0 \mbox{ and } x\in\Omega, $$

(2.4)

and

$$\begin{aligned}& \phi\bigl(x,\varphi(x,u)-\varepsilon\bigr)\leq u, \quad \mbox{for } u\geq0, 0< \varepsilon\leq\varphi(x,u) \mbox{ and } x\in\Omega, \\& \varphi\bigl(x,\phi(x,v)-\varepsilon\bigr)\leq v, \quad \mbox{for } v\geq0, 0< \varepsilon\leq\phi(x,v) \mbox{ and } x\in\Omega. \end{aligned}$$

For $\Phi\in N(\Omega)$, the following inequality is called the Young inequality:

$$ uv \leq\Phi(x,u)+\overline{\Phi}(x,v),\quad \mbox{for all } u, v \geq0, x\in\Omega, $$

(2.5)

and the equality holds if and only if $u=\phi(x,v)$ or $v=\varphi (x,u)$, i.e.

$$ u\varphi(x,u)= \Phi(x,u)+\overline{\Phi}\bigl(x,\varphi(x,u) \bigr), \qquad \phi(x,v)v= \Phi\bigl(x,\phi(x,v)\bigr)+\overline{\Phi}(x,v). $$

(2.6)

Let $\Phi\in N(\Omega)$. Φ is said to satisfy the $\Delta_{2}$ condition ($\Phi\in\Delta_{2}$, for short), if there exist a positive constant $K>1$ and a nonnegative function $h\in L^{1}(\Omega)$ such that

$$ \Phi(x,2u)\leq K \Phi(x,u)+h(x), \quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

(2.7)

Clearly, by the proof of Proposition 1.3(6) in [2], if $\Phi\in\Delta_{2}$, then there exist $K>1$ and a nonnegative function $h\in L^{1}(\Omega)$ such that

$$ \overline{\Phi}\bigl(x,\varphi(x,u)\bigr)\leq(K-1)\Phi(x,u)+h(x),\quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

(2.8)

For each $x\in\Omega$, the inverse function of $\Phi(x,\cdot)$ is denoted by $\Phi^{-1}(x,\cdot)$, i.e.

$$\Phi^{-1}\bigl(x,\Phi(x,u)\bigr)= \Phi\bigl(x,\Phi^{-1}(x,u) \bigr)=u, \quad \mbox{for } u\geq0. $$

Let $\Psi,\Upsilon\in N(\Omega)$. $\Psi\preceq\Upsilon$ means that Ψ is weaker than ϒ, i.e., there exist positive constants $K_{1}$, $K_{2}$ and a nonnegative function $h_{1} \in L^{1}(\Omega)$ such that

$$ \Psi(x,u)\leq K_{1}\Upsilon(x, K_{2}u)+h_{1}(x), \quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

(2.9)

Φ is called locally integrable, if $\int_{\Omega}\Phi(x,u)\,dx<\infty$ for every $u>0$.

The following assumptions will be used.

($\Phi_{1}$):

$\inf_{x\in\Omega}\Phi(x,1)=c_{1}>0$.

($\Phi_{2}$):

For every $t_{0}>0$ there exists $c=c(t_{0})>0$ such that

$$ \inf_{x\in\Omega}\frac{\Phi(x,t)}{t}\geq c $$

(2.10)

and

$$ \inf_{x\in\Omega}\frac{\overline{\Phi}(x,t)}{t}\geq c, $$

(2.11)

for all $t\geq t_{0}$.

Obviously, (2.10) implies ($\Phi_{1}$).

Let $\Phi\in N(\Omega)$. The Musielak-Orlicz space (i.e. the generalized Orlicz space) $L_{\Phi}(\Omega)$ is defined by

$$L_{\Phi}(\Omega)= \biggl\{ u\in L^{0}(\Omega): \int_{\Omega}\Phi \biggl(x,\frac{\vert u(x)\vert }{\lambda} \biggr)\,dx< \infty, \mbox{ for some } \lambda >0 \biggr\} , $$

with the (Luxemburg) norm

$$\Vert u\Vert _{\Phi}=\inf \biggl\{ \lambda>0: \int_{\Omega}\Phi \biggl(x,\frac {\vert u(x)\vert }{\lambda} \biggr)\,dx\leq1 \biggr\} . $$

Moreover, the set

$$K_{\Phi}(\Omega)= \biggl\{ u\in L^{0}(\Omega): \int_{\Omega}\Phi \bigl(x,\bigl\vert u(x)\bigr\vert \bigr)\,dx< \infty \biggr\} , $$

will be called the Musielak-Orlicz class (i.e. the generalized Orlicz class). A function $u\in L^{0}(\Omega)$ will be called a finite element of $L_{\Phi}(\Omega)$, if $\lambda u\in K_{\Phi}(\Omega)$ for every $\lambda> 0$. The space of all finite elements of $L^{0}(\Omega )$ will be denoted by $E_{\Phi}(\Omega)$. Then $K_{\Phi}(\Omega)$ is a convex subset of $L_{\Phi}(\Omega)$, $L_{\Phi}(\Omega)$ is the smallest vector subspace of $L^{0}(\Omega)$ containing $K_{\Phi}(\Omega )$, and $E_{\Phi}(\Omega)$ is the largest vector subspace of $L^{0}(\Omega)$ contained in $K_{\Phi}(\Omega)$.

If Φ is locally integrable, then $E_{\Phi}(\Omega)$ is a separable space, and $E_{\Phi}(\Omega)=K_{\Phi}(\Omega)=L_{\Phi}(\Omega)$ if and only if $\Phi\in\Delta_{2}$.

If Φ is locally integrable and satisfy (2.10), then $(E_{\Phi}(\Omega))^{\ast}=L_{\overline{\Phi}}(\Omega)$. Moreover, if Φ̅ is locally integrable satisfying (2.11), and $\Phi, \overline{\Phi} \in\Delta_{2}$, then $L_{\Phi }(\Omega)$ is reflexive.

The Musielak-Orlicz-Sobolev space $W^{1}L_{\Phi}(\Omega)$ is defined by

$$W^{1}L_{\Phi}(\Omega)=\bigl\{ u\in L_{\Phi}(\Omega): \forall \vert \alpha \vert \leq1, D^{\alpha}u\in L_{\Phi}(\Omega) \bigr\} , $$

where $\alpha=(\alpha_{1},\ldots, \alpha_{N})$ with nonnegative integers $\alpha_{i}$, $i=1,\ldots, N$, $\vert \alpha \vert = \vert \alpha_{1}\vert + \vert \alpha_{2}\vert + \cdots+ \vert \alpha_{N}\vert$ and $D^{\alpha}u$ denote the distributional derivatives.

Let

$$\varrho_{\Phi}(u)=\sum_{\vert \alpha \vert \leq1} \int_{\Omega}\Phi\bigl(x,\bigl\vert D^{\alpha }u(x)\bigr\vert \bigr)\,dx \quad \mbox{and}\quad \Vert u\Vert _{\Phi,\Omega}=\inf \biggl\{ \lambda>0: \varrho _{\Phi} \biggl(\frac{u}{\lambda} \biggr)\leq1 \biggr\} , $$

for $u \in W^{1}L_{\Phi}(\Omega)$. $\varrho_{\Phi}(u)$ is a convex modular and $\Vert u\Vert _{\Phi,\Omega}$ is a norm on $W^{1}L_{\Phi}(\Omega)$, respectively. The pair $(W^{1}L_{\Phi}(\Omega), \Vert u\Vert _{\Phi,\Omega})$ is a Banach space if Φ is locally integrable and satisfies ($\Phi_{1}$).

Taking $\Phi(x,u)=\Phi(u)$, $W^{1}L_{\Phi}(\Omega)$ is the Orlicz-Sobolev space. Taking $\Phi(x,\vert u\vert )=\vert u\vert ^{p(x)}$, $W^{1}L_{\Phi }(\Omega)$ is the variable exponent Sobolev space $W^{1,p(\cdot)}(\Omega)$.

It is easy to see that

$$W^{1}L_{\Phi}(\Omega)=\bigl\{ u\in L_{\Phi}(\Omega): \vert Du\vert \in L_{\Phi}(\Omega )\bigr\} . $$

Denote $\Vert Du\Vert _{\Phi}=\Vert \vert Du\vert \Vert _{\Phi}$ and $\Vert u\Vert _{1,\Phi}=\Vert u\Vert _{\Phi }+\Vert Du\Vert _{\Phi}$. Then $\Vert u\Vert _{1,\Phi}$ and $\Vert u\Vert _{\Phi,\Omega}$ are two equivalent norms.

The space $W^{1}L_{\Phi}(\Omega)$ will always be identified to a subspace of the product $\prod_{\vert \alpha \vert \leq1}L_{\Phi}(\Omega)=\prod L_{\Phi}$; this subspace is $\sigma(\prod L_{\Phi}, \prod E_{\overline {\Phi}})$ closed. Let $W_{0}^{1}L_{\Phi}(\Omega)$ be the $\sigma(\prod L_{\Phi}, \prod E_{\overline{\Phi}})$ closure of the Schwartz space $\mathcal{D}(\Omega)$ in $W^{1}L_{\Phi}(\Omega)$.

Let $W^{1}E_{\Phi}(\Omega)=\{u\in E_{\Phi}(\Omega): \forall \vert \alpha \vert \leq1, D^{\alpha}u\in E_{\Phi}(\Omega)\}$, and $W_{0}^{1}E_{\Phi}(\Omega)$ is the (norm) closure of $\mathcal {D}(\Omega)$ in $W^{1}L_{\Phi}(\Omega)$.

The proof of the following lemma is similar to [19].

Lemma 2.1

Let measΩ be bounded, $\Phi\in N(\Omega)$, and φ is the right-hand derivative of Φ. Then

$$ \frac{\int_{\Omega}\varphi(x,\vert Du\vert )\vert Du\vert \,dx}{\int_{\Omega }\vert Du\vert \,dx}\rightarrow+\infty, \quad \textit{if } \int_{\Omega} \vert Du\vert \,dx\rightarrow+\infty. $$

(2.12)

Proof

Let us assume that there is a sequence $\{u_{n}\}$ with $\int_{\Omega} \vert Du_{n}(x)\vert \,dx\rightarrow+\infty$ and $K_{0}<\infty$ such that

$$\frac{\int_{\Omega}\varphi(x,\vert Du_{n}(x)\vert )\vert Du_{n}(x)\vert \,dx}{\int_{\Omega }\vert Du_{n}(x)\vert \,dx}\leq K_{0}. $$

Since $\Phi\in N(\Omega)$, there exists $R>0$ such that

$$\inf_{x\in\Omega}\varphi(x,R)\geq\inf_{x\in\Omega} \frac{\Phi(x,R)}{R}>2K_{0}. $$

We define $\widetilde{\Omega}(R,n):=\{x\in\Omega| \vert D u_{n}(x)\vert \geq R\}$ and take for all n with $\int_{\Omega} \vert Du_{n}(x)\vert \,dx \geq2R \operatorname {meas}\Omega$, then

$$\begin{aligned} &\frac{\int_{\Omega}\varphi(x,\vert Du_{n}(x)\vert )\vert Du_{n}(x)\vert \,dx}{\int_{\Omega }\vert Du_{n}(x)\vert \,dx} \\ &\quad \geq\inf_{x\in\Omega}\varphi(x,R) \frac{\int_{\widetilde{\Omega }(R,n)}\vert Du_{n}(x)\vert \,dx}{\int_{\widetilde{\Omega }(R,n)}\vert Du_{n}(x)\vert \,dx+R\cdot \operatorname {meas}(\Omega)} \\ &\quad \geq\frac{1}{2} \inf_{x\in\Omega}\varphi(x,R) >K_{0}. \end{aligned}$$

This is a contradiction, thus (2.12) holds. □

Lemma 2.2

(see [20], Remark 2.1)

Let V be a vector space of finite dimension and $A:V\rightarrow V'$ be a continuous mapping with

$$\lim_{\Vert u\Vert _{V}\rightarrow+\infty}\frac{(A(u),u)}{\Vert u\Vert _{V}}=+\infty, $$

where $V'$ is the dual space of V, then A is surjective.

Lemma 2.3

(see [21], Lemma 2.1)

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

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

Here $u^{+}=\max\{u,0\}$, $u^{-}=-\min\{u,0\}$. This lemma holds in $W_{0}^{1}L_{\Phi}(\Omega)$ as well.

Lemma 2.4

(see [17])

If a sequence $g_{n}\in L_{\overline{\Phi }}(\Omega)$ converges in measure to a measurable function g and if $g_{n}$ remains bounded in $L_{\overline{\Phi}}(\Omega)$, then $g\in L_{\overline{\Phi}}(\Omega)$ and $g_{n}\rightarrow g$ for $\sigma (L_{\overline{\Phi}}(\Omega),E_{\Phi}(\Omega))$.

The following propositions refer to Theorems 1.6-1.8 in [16], Theorem 4.2 in [22], and Theorem 2.1 in [18].

Proposition 2.1

Let $\Phi\in N(\Omega)$ and

$$ \Phi_{1}(x,u)=a\Phi(x,bu) \quad (a, b>0), \textit{ for all } u \geq0, x\in\Omega. $$

(2.13)

Then $\Phi_{1}\in N(\Omega)$ and the complementary function $\overline {\Phi_{1}}$ to $\Phi_{1}$ is given by

$$ \overline{\Phi_{1}}(x,v)=a\overline{\Phi}\biggl(x, \frac{v}{ab}\biggr),\quad \textit{for all } v\geq0, x\in\Omega, $$

(2.14)

where Φ̅ is the complementary function to Φ.

Proof

It is easy to see that $\Phi_{1}\in N(\Omega)$. We only need to show (2.14). By (2.1) and (2.13), we can deduce that

$$\varphi_{1}(x,\tau)=ab\varphi(x,b\tau), \quad \mbox{for all } \tau\geq0, x \in \Omega, $$

where φ and $\varphi_{1}$ are the right-hand derivatives of Φ and $\Phi_{1}$, respectively.

From (2.3), $\phi_{1}(x,\sigma) =\frac{1}{b}\sup\{b\tau: \varphi(x,b\tau)\leq\frac{\sigma}{ab}\}=\frac{1}{b}\phi(x,\frac{\sigma }{ab})$, $\forall\sigma\geq0$ and $x\in\Omega$.

For $\forall v\geq0$, by (2.2), $\overline{\Phi}_{1}(x,v) =a\int_{0}^{v}\phi(x,\frac{\sigma}{ab})\,d\frac {\sigma}{ab}$, $\forall v\geq0$ and $x\in\Omega$. Define $s =\frac{\sigma}{ab}$. Then $\overline{\Phi}_{1}(x,v)=a\int_{0}^{\frac{v}{ab}}\phi(x,s)\,ds= a\overline{\Phi}(x,\frac{v}{ab})$, $\forall v\geq0$ and $x\in\Omega$. □

Proposition 2.2

Let $\Phi_{1}, \Phi_{2}\in N(\Omega)$ and

$$ \Phi_{1}(x,u)\leq\Phi_{2}(x,u)+h(x), \quad \textit{for some } h\in L^{1}(\Omega), \textit{ all } u\geq0 \textit{ and }x \in\Omega. $$

(2.15)

Then

$$\overline{\Phi_{2}}(x,v)\leq\overline{\Phi_{1}}(x,v)+h(x),\quad \textit{for all } v\geq0 \textit{ and }x\in\Omega, $$

where $\overline{\Phi_{1}}$ and $\overline{\Phi_{2}}$ are the complementary functions to $\Phi_{1}$ and $\Phi_{2}$, respectively.

Proof

By (2.5) and (2.6), one has $\Phi_{2}(x,\phi_{2}(x,v))+\overline{\Phi_{2}}(x,v)=\phi_{2}(x,v)\cdot v \leq\Phi_{1}(x,\phi_{2}(x,v))+\overline{\Phi_{1}}(x,v)$, $\forall v\geq0$ and $x\in\Omega$.

In view of (2.15), $\Phi_{2}(x,\phi_{2}(x,v)) +h(x) \geq\Phi _{1}(x,\phi_{2}(x,v))$, $\forall v\geq0$ and $x\in\Omega$. Therefore, $\overline{\Phi_{2}}(x,v) \leq\overline{\Phi _{1}}(x,v)+h(x)$, $\forall v\geq0$ and $x\in\Omega$. □

Proposition 2.3

Let $\Phi\in N(\Omega)$ and its complementary function is Φ̅. φ and ϕ are given by (2.1) and (2.2), respectively. Then the following assertions are equivalent.

• (1) $\Phi\in\Delta_{2}$.

• (2) $\forall l_{1}>1$, there exist $K'>1$ and a nonnegative function $\tilde{h}_{1}\in L^{1}(\Omega)$ such that

  $$\Phi(x,l_{1}u)\leq K' \Phi(x,u)+\tilde{h}_{1}(x), \quad \textit{for all } u\geq 0 \textit{ and a.e. } x\in\Omega. $$

• (3) $\forall l_{2}>1$, there exist $\varepsilon\in(0,1)$ and a nonnegative function $\tilde{h}_{2}\in L^{1}(\Omega)$ such that

  $$\Phi\bigl(x,(1+\varepsilon)u\bigr) \leq l_{2} \Phi(x,u)+ \tilde{h}_{2}(x), \quad \textit {for all } u\geq0 \textit{ and a.e. } x\in\Omega. $$

• (4) $\forall l_{3}>1$, there exist $\delta>0$ and a nonnegative function $\tilde{h}_{3}\in L^{1}(\Omega)$ such that

  $$(l_{3}+\delta)\overline{\Phi}(x,v)\leq\overline{\Phi}(x,l_{3}v)+ \tilde {h}_{3}(x),\quad \textit{for all } v\geq0 \textit{ and a.e. } x\in\Omega. $$

• (5) $\forall l_{4}>1$, there exist $l_{0}>1$ and a nonnegative function $\tilde{h}_{4}\in L^{1}(\Omega)$ such that

  $$\overline{\Phi}(x,v)\leq\frac{1}{l_{0}l_{4}}\overline{\Phi }(x,l_{4}v)+ \tilde{h}_{4}(x), \quad \textit{for all } v\geq0 \textit{ and a.e. } x\in\Omega. $$

• (6) There exist $l_{5}>1$ and a nonnegative function $\tilde {h}_{5}\in L^{1}(\Omega)$ such that

  $$\overline{\Phi}(x,v)\leq\frac{1}{2l_{5}}\overline{\Phi }(x,l_{5}v)+ \tilde{h}_{5}(x),\quad \textit{for all } v\geq0 \textit{ and a.e. } x\in\Omega. $$

• (7) There exist $l_{6}>0$ and a nonnegative function $\tilde {h}_{6}\in L^{1}(\Omega)$ such that

  $$u\varphi(x,2u)\leq l_{6}u\varphi(x,u)+\tilde{h}_{6}(x), \quad \textit{for all } u\geq0 \textit{ and a.e. } x\in\Omega. $$

• (8) $\forall m_{1}>1$, there exist $l_{7}>0$ and a nonnegative function $\tilde{h}_{7}\in L^{1}(\Omega)$ such that

  $$u\varphi(x,m_{1}u)\leq l_{7}u\varphi(x,u)+ \tilde{h}_{7}(x), \quad \textit{for all } u\geq0 \textit{ and a.e. } x\in\Omega. $$

Proof

(1)⇒(2). Since $\Phi\in\Delta_{2}$, by (2.7), there exist $K>1$ and a nonnegative function $h\in L^{1}(\Omega)$ such that $\Phi(x,2u) \leq K \Phi(x,u)+ h(x)$, $\forall u\geq0$ and a.e. $x\in\Omega$. For every $l_{1}>1$, there exists $n\in\mathbb{N}$ such that $2^{n}\geq l_{1}$. Then

$$\begin{aligned} \Phi(x,l_{1}u) & \leq\Phi\bigl(x,2^{n}u\bigr) \leq K \Phi \bigl(x,2^{n-1}u\bigr)+ h(x) \\ &\leq K^{2} \Phi\bigl(x,2^{n-2}u\bigr) + (K+1)h(x) \\ & \leq\cdots\leq K^{n} \Phi(x,u) + \bigl(K^{n-1}+\cdots+K+1 \bigr)h(x) \\ & = K^{n} \Phi(x,u) + \frac{K^{n}-1}{K-1}h(x), \end{aligned}$$

$\forall u\geq0$ and a.e. $x\in\Omega$. Taking $K'=K^{n}$ and $\tilde {h}_{1}=\frac{K^{n}-1}{K-1}h(x)$, we can deduce the assertion (2).

(2)⇒(3). For every $l_{2}>1$, by the assertion (2), there exist $K'>l_{2}$ and a nonnegative function $\tilde{h}_{1}\in L^{1}(\Omega)$ such that

$$\Phi(x,2u)\leq K' \Phi(x,u)+\tilde{h}_{1}(x),\quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

Take $\varepsilon=\frac{l_{2}-1}{K'-1}$, then $\varepsilon\in(0,1)$. Hence,

$$\begin{aligned} \Phi\bigl(x,(1+\varepsilon)u\bigr) &=\Phi\bigl(x,(1-\varepsilon)u+2\varepsilon u \bigr) \leq(1-\varepsilon)\Phi(x,u) + \varepsilon\Phi(x,2u) \\ & \leq(1-\varepsilon)\Phi(x,u) + K'\varepsilon\Phi(x,u)+ \varepsilon \tilde{h}_{1}(x) =l_{2} \Phi(x,u)+\varepsilon \tilde{h}_{1}(x), \end{aligned}$$

for all $u\geq0$ and a.e. $x\in\Omega$. Taking $\tilde{h}_{2}=\varepsilon\tilde{h}_{1}$, we complete the assertion (3).

(3)⇒(4). By the assertion (3), $\forall l_{3}>1$, there exist $\varepsilon\in(0,1)$ and a nonnegative function $\tilde {h}_{2}\in L^{1}(\Omega)$ such that

$$\Phi\bigl(x,(1+\varepsilon)u\bigr) \leq l_{3} \Phi(x,u)+ \tilde{h}_{2}(x),\quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

It implies that $\frac{1}{l_{3}}\Phi(x,(1+\varepsilon)u)\leq \Phi (x,u)+ \frac{1}{l_{3}}\tilde{h}_{2}(x)$. Denote $\Phi_{1}(x,u)=\frac {1}{l_{3}}\Phi(x,(1+\varepsilon)u)$. By Proposition 2.1, $\overline{\Phi_{1}}(x,v)=\frac{1}{l_{3}}\overline{\Phi}(x,\frac {l_{3}}{1+\varepsilon}v)$, $\forall v\geq0$ and a.e. $x\in\Omega$. By Proposition 2.2, we get

$$\overline{\Phi_{1}}(x,v) \leq\frac{1}{l_{3}}\overline{\Phi} \biggl(x,\frac {l_{3}}{1+\varepsilon}v\biggr)+ \frac{1}{l_{3}}\tilde{h}_{2}(x) \leq\frac {1}{l_{3}(1+\varepsilon)}\overline{\Phi}(x,l_{3}v)+ \frac {1}{l_{3}} \tilde{h}_{2}(x), $$

$\forall v\geq0$, and a.e. $x\in\Omega$. Thus, we have $l_{3}(1+\varepsilon)\overline{\Phi_{1}}(x,v) \leq\overline{\Phi }(x,l_{3}v)+(1+\varepsilon)\tilde{h}_{2}(x)$. Taking $\delta=l_{3}\varepsilon$ and $\tilde{h}_{3}=(1+\varepsilon )\tilde{h}_{2}$, we complete the assertion (4).

(4)⇒(5). By the assertion (4), $\forall l_{4}>1$, there exist $\delta>0$ and a nonnegative function $\tilde{h}_{3}\in L^{1}(\Omega)$ such that

$$(l_{4}+\delta)\overline{\Phi}(x,v)\leq\overline{\Phi}(x,l_{4}v)+ \tilde {h}_{3}(x), \quad \forall v\geq0 \mbox{ and a.e. } x\in\Omega. $$

Hence, $\overline{\Phi}(x,v)\leq\frac{1}{l_{4}(1+\frac{\delta }{l_{4}})}\overline{\Phi}(x,l_{4}v)+\frac{1}{l_{4}(1+\frac{\delta }{l_{4}})}\tilde{h}_{3}(x)$. Taking $l_{0}=1+\frac{\delta}{l_{4}}$ and $\tilde{h}_{4}=\frac {1}{l_{4}(1+\frac{\delta}{l_{4}})}\tilde{h}_{3}$, we complete the assertion (5).

(5)⇒(1). By the assertion (5), $\forall l_{4}>1$, there exist $l_{0}>1$ and a nonnegative function $\tilde{h}_{4}\in L^{1}(\Omega)$ such that

$$\overline{\Phi}(x,v)\leq\frac{1}{l_{0}l_{4}}\overline{\Phi }(x,l_{4}v)+ \tilde{h}_{4}(x), \quad \forall v\geq0 \mbox{ and a.e. } x\in \Omega. $$

By Proposition 2.1 and Proposition 2.2, we obtain $\Phi(x,l_{0}u)\leq l_{0}l_{4}\Phi(x,u)+l_{0}l_{4}\tilde {h}_{4}(x)$, $\forall u\geq0$ and a.e. $x\in\Omega$. Take $n_{0}\in\mathbb{N}$ such that $l_{0}^{n_{0}}\geq2$. Then $\Phi(x,2u)\leq\Phi(x,l_{0}^{n_{0}}u)\leq l_{0}^{n_{0}}l_{4}^{n_{0}}\Phi(x,u)+\frac {l_{0}^{n_{0}}l_{4}^{n_{0}}-1}{l_{0}l_{4}-1}\tilde{h}_{4}(x)$. Denote $l_{0}^{n_{0}}l_{4}^{n_{0}}=K$ and $\frac {l_{0}^{n_{0}}l_{4}^{n_{0}}-1}{l_{0}l_{4}-1}\tilde{h}_{4}=h$. We deduce (2.7), i.e. $\Phi\in\Delta_{2}$.

(6)⇒(1). Define $\Psi_{1}(x,v)=\frac{1}{2l_{5}}\overline {\Phi}(x,l_{5}v)$. By Proposition 2.1, $\overline{\Psi_{1}}(x,u)=\frac {1}{2l_{5}}\Phi(x,2u)$, $\forall u\geq0$ and a.e. $x\in\Omega$. By Proposition 2.2, $\Phi(x,2u)\leq2l_{5}\Phi(x,u)+2l_{5}\tilde{h}_{5}(x)$, $\forall u\geq 0$ and a.e. $x\in\Omega$. Therefore, $\Phi\in\Delta_{2}$.

Similarly, (1) implies (6).

(1)⇒(7). By (2), there exist $K'>0$ and $\tilde{h}_{1}\in L^{1}(\Omega)$ such that

$$\Phi(x,4u)\leq K' \Phi(x,u)+\tilde{h}_{1}(x), \quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

On the other hand, we have $2u\varphi(x,2u)\leq\Phi(x,4u)$ and $\Phi (x,u)\leq u\varphi(x,u)$, for $x\in\Omega$, $u\geq0$. Hence,

$$u\varphi(x,2u)\leq\frac{K'}{2} u\varphi(x,u)+\frac{1}{2}\tilde {h}_{1}(x), \quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

Consequently, the assertion (7) holds by taking $l_{6}=\frac{K'}{2}$ and $\tilde{h}_{6}=\frac{1}{2}\tilde{h}_{1}$.

(7)⇒(8). For every $m_{1}>1$, there is $n_{0}\in\mathbb {N}^{+}$ such that $2^{n_{0}}\geq m_{1}$. Then $u\varphi(x,m_{1}u)\leq u\varphi(x,2^{n_{0}}u)\leq l_{6}^{n_{0}}u\varphi(x,u)+\frac {l_{6}^{n_{0}}-1}{l_{6}-1}\tilde{h}_{6}(x)$, $\forall u\geq0$ and a.e. $x\in\Omega$. Taking $l_{7}=l_{6}^{n_{0}}$ and $\tilde{h}_{7}=\frac {l_{6}^{n_{0}}-1}{l_{6}-1}\tilde{h}_{6}$, we complete (8).

(8)⇒(1). For every $l_{1}>1$, we have $\Phi(x,l_{1}u)\leq l_{1}u\varphi(x,l_{1}u)$. By (8), there exist $l_{7}>0$ and $\tilde{h}_{7}\in L^{1}(\Omega)$ such that

$$u\varphi(x,l_{1}u)\leq l_{7}u\varphi\biggl(x, \frac{u}{2}\biggr)+\tilde{h}_{7}(x),\quad \mbox{for all } u\geq0 \mbox{ and a.e. } x\in\Omega. $$

It follows that $\Phi(x,l_{1}u)\leq l_{1}l_{7}u\varphi(x,\frac{u}{2})+ l_{1}\tilde {h}_{7}(x)\leq 2l_{1}l_{7}\Phi(x,u)+ l_{1}\tilde{h}_{7}(x)$, for all $u\geq0$ and a.e. $x\in\Omega$. Taking $K'=2l_{1}l_{7}$ and $\tilde{h}_{1}=l_{1}\tilde{h}_{7}$, we deduce (2). Immediately, (1) holds. □

Example 2.1

Let $\Phi(x,\vert t\vert )=(1+\frac{\vert t\vert }{p(x)})\ln(1+\frac{\vert t\vert }{p(x)})-\frac {\vert t\vert }{p(x)}$, for $x\in\Omega$ and $t\in\mathbb{R}$, where $p:\Omega \rightarrow\mathbb{R}$ is a measurable function such that $1< p_{-}\leq p(x)\leq p_{+}< +\infty$. Then $\varphi(x,\vert t\vert )=\frac{1}{p(x)}\ln(1+\frac{\vert t\vert }{p(x)})$, $\phi (x,\vert s\vert )=p(x)(\exp(p(x)\vert s\vert )-1)$ and $\overline{\Phi}(x,\vert s\vert )=\exp(p(x)\vert s\vert )-p(x)\vert s\vert -1$. It follows that $\Phi\in N(\Omega)$ and $\Phi\in\Delta_{2}$. But $\overline{\Phi}\notin\Delta_{2}$. Moreover, both Φ and Φ̅ are locally integrable. Therefore, $L_{\Phi} (\Omega)$ is separable, but $L_{\Phi} (\Omega)$ is not reflexive.

Remark 2.1

Let $\Phi(x,\vert t\vert )=\exp(p(x)\vert t\vert )-1$, for $x\in\Omega$ and $t\in\mathbb {R}$, where $p:\Omega\rightarrow\mathbb{R}$ is a measurable function such that $1< p_{-}\leq p(x)\leq p_{+}<+\infty$. It is worth noting that Φ does not satisfy the condition $\lim_{u\rightarrow0^{+}}\sup_{x\in\Omega}\frac{\Phi(x,u)}{u}=0$. Therefore, $\Phi\notin N(\Omega)$.

Clearly, by (2.9), Proposition 2.1 and Proposition 2.2, we can deduce the following proposition.

Proposition 2.4

If $\Phi\preceq\Psi$, then $\overline{\Psi}\preceq\overline{\Phi}$.

3 Existence theorems

Let $\Phi\in N(\Omega)$, and satisfy the condition

(Φ):

$\Phi\in\Delta_{2}$, Φ̅ is a complementary function to Φ, both Φ and Φ̅ are locally integrable and satisfy $(\Phi_{2})$.

We assume that there exists $\Psi\in N(\Omega)$ satisfying the condition

(Ψ):

$\Psi\in\Delta_{2}$, Ψ̅ is a complementary function to Ψ, both Ψ and Ψ̅ are locally integrable and satisfy $(\Phi_{2})$, $\Phi\preceq\Psi$, and the embedding $W^{1}L_{\Phi}(\Omega )\hookrightarrow L_{\Psi}(\Omega)$ is compact.

Note that, in this case, the spaces $L_{\Phi}(\Omega)$, $L_{\Psi }(\Omega)$, $W^{1}L_{\Phi}(\Omega)$, $W_{0}^{1}L_{\Phi}(\Omega)$ are separable Banach spaces.

For $u, v\in L^{0}(\Omega)$, we denote $u\wedge v=\min\{u,v\}$, $u\vee v=\max\{u,v\}$, $u^{+}:=u\vee0$, $u^{-}:=-u\wedge0$, $u\leq v\Leftrightarrow u(x)\leq v(x)$ for a.e. $x\in\Omega$.

Let $a_{1}:\Omega\times\mathbb{R}^{N}\rightarrow\mathbb{R}^{N}$ be a Carathéodory function satisfying the following conditions:

($A_{1}$):

For a.e. $x\in\Omega$ and all $\xi, \eta\in\mathbb{R}^{N}$,

$$\begin{aligned}& \bigl\vert a_{1}(x,\xi)\bigr\vert \leq b_{1}\overline{ \Phi}^{-1}\bigl(x,\Phi\bigl(x, \vert \xi \vert \bigr) \bigr)+g_{1}(x), \end{aligned}$$

(3.1)

$$\begin{aligned}& a_{1}(x,\xi)\xi\geq b_{2}\Phi\bigl(x, \vert \xi \vert \bigr)-g_{2}(x), \end{aligned}$$

(3.2)

$$\begin{aligned}& \bigl[a_{1}(x,\xi)-a_{1}(x,\eta)\bigr](\xi-\eta)> 0,\quad \xi\neq \eta, \end{aligned}$$

(3.3)

where $b_{1},b_{2}>0$, $g_{1}\in E_{\overline{\Phi}}(\Omega)$, $g_{1}\geq0$, $g_{2}\in L^{1}(\Omega)$, and $g_{2}\geq0$.

Let $a_{0}:\Omega\times\mathbb{R}\rightarrow\mathbb{R}$ be a Carathéodory function satisfying the following conditions:

($A_{0}$):

For a.e. $x\in\Omega$ and all $t \in\mathbb{R}$,

$$ \bigl\vert a_{0}(x,t)\bigr\vert \leq b_{1} \overline{\Phi}^{-1}\bigl(x,\Phi\bigl(x, \vert t\vert \bigr) \bigr)+g_{1}(x), $$

(3.4)

where $b_{1}>0$, $g_{1}\in E_{\overline{\Phi}}(\Omega)$, and $g_{1}\geq0$.

Example 3.1

• (1) Let $\Phi(x,\vert t\vert )=\frac{1}{p(x)}\vert t\vert ^{p(x)}$, $a_{1}(x,\xi)=\vert \xi \vert ^{p(x)-2}\xi$, for $x\in\Omega$ and $t\in\mathbb{R}$, where $p:\Omega\rightarrow\mathbb{R}$ is a measurable function such that $2\leq p_{-}\leq p(x)\leq p_{+}< +\infty$. Then Φ satisfies (Φ) and we get the $p(x)$-Laplace operator $\operatorname {div}(\vert Du\vert ^{p(x)-2}Du)$.

• (2) Let $\Phi(x,\vert t\vert )=\frac{1}{p(x)}[(1+\vert t\vert ^{2})^{p(x)/2}-1]$, $a_{1}(x,\xi )=(1+\vert \xi \vert ^{2})^{(p(x)-2)/2}\xi$, for $x\in\Omega$ and $t\in\mathbb{R}$, where $p:\Omega\rightarrow\mathbb{R}$ is a measurable function such that $2\leq p_{-}\leq p(x)\leq p_{+}<+\infty$. Then Φ satisfies (Φ) and we obtain the generalized mean curvature operator $\operatorname {div}((1+\vert Du\vert ^{2})^{(p(x)-2)/2}Du)$. Moreover, by Proposition 2.3(6), $\overline{\Phi}\in\Delta_{2}$.

• (3) Let $\Phi(x,\vert t\vert )=(1+\frac{\vert t\vert }{p(x)})\ln(1+\frac{\vert t\vert }{p(x)})-\frac {\vert t\vert }{p(x)}$, for $x\in\Omega$ and $t\in\mathbb{R}$, where $p:\Omega \rightarrow\mathbb{R}$ is a measurable function such that $1< p_{-}\leq p(x)\leq p_{+}< +\infty$. Clearly, it can be verified that Φ satisfies (Φ). Put $a_{1}(x,\xi)=\varphi(x,\vert \xi \vert )\frac{\xi}{ \vert \xi \vert }$, and $a_{0}(x,t)=\varphi(x,\vert t\vert )$, for $x\in\Omega$, $t\in\mathbb{R}$ and $\xi\in\mathbb{R}^{N}$, where $\varphi(x,\vert t\vert )=\frac{1}{p(x)}\ln(1+\frac {\vert t\vert }{p(x)})$. Then $a_{1}$ and $a_{0}$ satisfy ($A_{1}$) and ($A_{0}$), respectively.

Remark 3.1

Clearly, the condition (1.2) (resp. (1.5)) implies (3.1) (resp. (3.4)).

Consider the following Dirichlet boundary value problem:

$$ \textstyle\begin{cases} -\operatorname {div}(a_{1}(x,Du))+a_{0}(x,u)=f(x,u,Du), \quad \mbox{in } \Omega, \\ u=0, \quad \mbox{on } \partial\Omega, \end{cases} $$

(3.5)

where $f:\Omega\times\mathbb{R}\times\mathbb{R}^{N}\rightarrow\mathbb {R}$ is a Carathéodory function. Denote by F the Nemytskii operator associated to f, that is,

$$F(u) (x)=f\bigl(x,u(x),Du(x)\bigr),\quad \mbox{for } x\in\Omega. $$

A function u is called a (weak) solution of (3.5) if $u\in W_{0}^{1}L_{\Phi}(\Omega)$, $F(u)\in L_{\overline{\Psi}}(\Omega)$ and u satisfies the equation

$$ \int_{\Omega}a_{1}(x,Du)Dv\,dx+ \int_{\Omega}a_{0}(x,u)v\,dx= \int_{\Omega }f(x,u,Du)v\,dx, \quad \mbox{for all } v\in W_{0}^{1}L_{\Phi}(\Omega). $$

(3.6)

A function u is called a subsolution (resp. supersolution) of (3.5) if $u\in W_{0}^{1}L_{\Phi}(\Omega)$, $F(u)\in L_{\overline{\Psi}}(\Omega)$ and (3.6) holds with ‘=’ replaced by ‘≤’ (resp. ‘≥’) for every nonnegative functions v in $W_{0}^{1}L_{\Phi }(\Omega)$ (see [2]).

Theorem 3.1

Suppose that $\underline{u}_{1}, \ldots, \underline{u}_{k}$ and $\overline{u}_{1}, \ldots, \overline{u}_{m}$ are subsolutions and supersolutions of (3.5), respectively, that satisfy

$$\underline{u}:=\underline{u}_{1}\vee\underline{u}_{2}\vee \cdots\vee \underline{u}_{k} \leq\overline{u}_{1}\wedge \overline{u}_{2}\wedge \cdots\wedge\overline{u}_{m}:= \overline{u}. $$

Let (Φ), (Ψ), ($A_{1}$), ($A_{0}$) hold. Assume the nonlinear term g satisfies the following local growth condition:

$$ \bigl\vert f(x,t, \xi)\bigr\vert \leq q(x)+b_{3} \overline{\Phi}^{-1}\bigl(x,\Phi \bigl(x,\vert t\vert \bigr) \bigr)+b_{4}\overline{\Psi}^{-1}\bigl(x,\Phi\bigl(x,\vert \xi \vert \bigr)\bigr), $$

(3.7)

for a.e. $x\in\Omega$ and $\forall t\in[\underline{u}(x),\overline {u}(x)]$, with $q\in E_{\overline{\Psi}}(\Omega)$, $b_{3}, b_{4}>0$. Then there exists a solution u of (3.5) such that $\underline{u} \leq u \leq\overline{u}$.

Proof

Denote $V=W_{0}^{1}L_{\Phi}(\Omega)$. For $x\in\Omega$, we put

$$Tu(x)=\textstyle\begin{cases} \overline{u}(x), & \mbox{if }u(x)>\overline{u}(x), \\ u(x), & \mbox{if } \underline{u}(x)\leq u(x)\leq\overline{u}(x), \\ \underline{u}(x), & \mbox{if }u(x)< \underline{u}(x), \end{cases}\displaystyle \quad \mbox{for } u\in V. $$

Then $Tu =u\vee\underline{u}+u\wedge\overline{u}-u$. By Remark 3.1 in [2], $T: V\rightarrow V$ is continuous. It is easy to see that T is bounded. From Proposition 2.4, we obtain $F(Tu)\in L_{\overline {\Psi}}(\Omega)$, $\forall u\in V$.

We define the cutoff function $l:\Omega\times\mathbb{R}\rightarrow \mathbb{R}$ given by

$$l(x,s)=\textstyle\begin{cases} \overline{\Phi}^{-1}(x,\Phi(x,s-\overline{u}(x))), & \mbox{if } s>\overline{u}(x), \\ 0, & \mbox{if } \underline{u}(x)\leq s\leq\overline{u}(x),\\ -\overline{\Phi}^{-1}(x,\Phi(x,\underline{u}(x)-s)), & \mbox{if } s< \underline{u}(x), \end{cases} $$

for $x\in\Omega$, $s\in\mathbb{R}$. Then l satisfies the following condition:

$$ \bigl\vert l(x,s)\bigr\vert \leq\overline{\Phi}^{-1} \bigl(x,\Phi\bigl(x,2\vert s\vert \bigr)\bigr) +\overline{\Phi }^{-1} \bigl(x,\Phi\bigl(x,2\bigl\vert \overline{u}(x)\bigr\vert \bigr)\bigr) + \overline{\Phi}^{-1}\bigl(x,\Phi \bigl(x,2\bigl\vert \underline{u}(x) \bigr\vert \bigr)\bigr), $$

(3.8)

for $x\in\Omega$ and all $s\in\mathbb{R}$.

For all $u\in V$, since $\Phi\in\Delta_{2}$, there exists $K_{1}>1$ such that

$$\begin{aligned} & \int_{\Omega} l(x,u)u\,dx \\ &\quad = \int_{\{u>\overline{u}\}}\overline{\Phi}^{-1}\bigl(x, \Phi(x,u- \overline {u})\bigr) (u-\overline{u})\,dx \\ &\qquad {}+ \int_{\{u>\overline{u}\}}\overline{\Phi}^{-1}\bigl(x, \Phi(x,u- \overline {u})\bigr)\overline{u}\,dx \\ &\qquad {}+ \int_{\{u< \underline{u}\}}\overline{\Phi}^{-1}\bigl(x, \Phi(x, \underline {u}-u)\bigr) (\underline{u}-u)\,dx \\ &\qquad {}- \int_{\{u< \underline{u}\}}\overline{\Phi}^{-1}\bigl(x, \Phi(x, \underline {u}-u)\bigr)\underline{u}\,dx \\ &\quad \geq \int_{\{u>\overline{u}\}}\Phi(x,u-\overline{u})\,dx - \int_{\{u>\overline{u}\}}\biggl[\frac{1}{2}\Phi(x,u-\overline{u})+\Phi \bigl(x,2\vert \overline{u}\vert \bigr)\biggr]\,dx \\ &\qquad {}+ \int_{\{u< \underline{u}\}}\Phi(x,\underline{u}-u)\,dx - \int_{\{u< \underline{u}\}}\biggl[\frac{1}{2}\Phi(x,\underline{u}-u)+\Phi \bigl(x,2\vert \underline{u}\vert \bigr)\biggr]\,dx \\ &\quad =\frac{1}{2} \int_{\{u>\overline{u}\}}\Phi(x,u-\overline{u})\,dx - \int_{\{u>\overline{u}\}}\Phi\bigl(x,2\vert \overline{u}\vert \bigr)\,dx \\ &\qquad {}+\frac{1}{2} \int_{\{u< \underline{u}\}}\Phi(x,\underline{u}-u)\,dx - \int_{\{u< \underline{u}\}}\Phi\bigl(x,2\vert \underline{u}\vert \bigr)\,dx \\ &\quad \geq \frac{1}{2} \int_{\{u>\overline{u}\}}\biggl[2\Phi\biggl(x,\frac{\vert u\vert }{2}\biggr)-\Phi \bigl(x,\vert \overline{u}\vert \bigr)\biggr]\,dx - \int_{\Omega}\Phi\bigl(x,2\vert \overline{u}\vert \bigr)\,dx \\ &\qquad {}+\frac{1}{2} \int_{\{u< \underline{u}\}}\biggl[2\Phi\biggl(x,\frac{\vert u\vert }{2}\biggr)-\Phi \bigl(x,\vert \underline{u}\vert \bigr)\biggr]\,dx \\ &\qquad {}- \int_{\Omega}\Phi\bigl(x,2\vert \underline{u}\vert \bigr)\,dx \\ &\quad \geq \int_{\{u>\overline{u}\}\cup\{u< \underline{u}\}}\Phi\biggl(x,\frac{\vert u\vert }{2}\biggr)\,dx -C \\ &\qquad {}+ \int_{\{\overline{u}\leq u \leq\underline{u}\}}\biggl[\Phi\biggl(x,\frac{\vert u\vert }{2}\biggr) -\Phi \biggl(x,\frac{\vert \overline{u}\vert \vee \vert \underline{u}\vert }{2}\biggr)\biggr]\,dx \\ &\quad = \int_{\Omega}\Phi\biggl(x,\frac{\vert u\vert }{2}\biggr)\,dx-C \\ &\quad \geq \frac{1}{K_{1}} \int_{\Omega}\Phi\bigl(x,\vert u\vert \bigr)\,dx-C, \end{aligned}$$

(3.9)

for some constant $C>0$ independent of u, where $\{u<\underline{u}\} =\{x\in\Omega:u(x)<\underline{u}(x)\}$, $\{u>\overline{u}\}=\{x\in\Omega :u(x)>\overline{u}(x)\}$, and $\{\underline{u}\leq u \leq\overline{u}\}=\{x\in\Omega:\underline {u}(x)\leq u(x) \leq\overline{u}(x)\}$.

Let us consider the auxiliary equation of finding $u\in V$ such that

$$\begin{aligned} &\int_{\Omega}a_{1}(x,Du)Dv\,dx+ \int_{\Omega}a_{0}(x,Tu)v\,dx+ \lambda \int _{\Omega}l(x,u)v\,dx \\ &\quad = \int_{\Omega}F(Tu)v\,dx, \quad \forall v\in V, \end{aligned}$$

(3.10)

where $\lambda>0$ is a parameter to be specified later.

Define $\Gamma_{T}: V\rightarrow V^{*}$,

$$(\Gamma_{T} u,v) := \int_{\Omega}a_{1}(x,Du)Dv\,dx+ \int_{\Omega }a_{0}(x,Tu)v\,dx+ \lambda \int_{\Omega}l(x,u)v\,dx- \int_{\Omega}F(Tu)v\,dx, $$

$\forall v\in V$. Then $\Gamma_{T}$ is well defined.

Since $\Phi\in\Delta_{2}$, there exists a sequence $\{w_{n}\}\subset V$ such that $\{w_{n}\}$ is dense in V. Let $V_{m}=\operatorname {span}\{ w_{1},\ldots,w_{m}\}$ and consider $\Gamma_{T}|_{V_{m}}$. For every $u\in V_{m}$, $\Vert Du\Vert _{\Phi}$ and $\int_{\Omega} \vert Du\vert \,dx$ are two norms of $V_{m}$ equivalent to the usual norm of finite dimensional vector spaces.

Similar to the proof of Proposition 3.1 in [20], we can deduce that the mapping $u\rightarrow\Gamma _{T}|_{V_{m}}u:V_{m}\rightarrow V_{m}^{*}$ is continuous.

In view of (3.7), one has

$$\begin{aligned} &\biggl\vert \int_{\Omega}F(Tu)u\,dx\biggr\vert \\ &\quad \leq C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} +2b_{3} \int_{\Omega }\Phi\bigl(x,\vert u\vert \bigr)\,dx +b_{3} \int_{\Omega}\Phi\bigl(x,\vert \overline{u}\vert \bigr) \,dx+b_{3} \int_{\Omega}\Phi \bigl(x,\vert \underline{u}\vert \bigr)\,dx \\ & \qquad {}+b_{4}\varepsilon_{1} \int_{\Omega}\Phi\bigl(x,\vert Du\vert \bigr)\,dx +b_{4} \int_{\Omega }\varepsilon_{1}\Psi\biggl(x, \frac{1}{\varepsilon_{1}}\vert \overline{u}\vert \vee \vert \underline{u}\vert \biggr)\,dx +b_{4} \int_{\Omega}\Psi\bigl(x,\vert \overline{u}\vert \bigr)\,dx \\ &\qquad {}+b_{4} \int_{\Omega}\Psi\bigl(x,\vert \underline{u}\vert \bigr)\,dx +b_{4} \int_{\Omega}\Phi \bigl(x,\vert D\overline{u}\vert \bigr)\,dx +b_{4} \int_{\Omega}\Phi\bigl(x,\vert D\underline{u}\vert \bigr)\,dx, \end{aligned}$$

(3.11)

for all $u\in V$, where $\varepsilon_{1}=\frac{b_{2}}{2b_{4}}$ and the constant $C^{*}>0$.

Thanks to (3.4) and (2.8), there exist $K_{2}>1$ and a nonnegative function $h\in L^{1}(\Omega)$ such that

$$\begin{aligned} &\biggl\vert \int_{\Omega}a_{0}(x,Tu)u\,dx\biggr\vert \\ & \quad \leq b_{1}(K_{2}-1) \int_{\Omega} \bigl[\Phi\bigl(x, \vert u\vert \bigr) +\Phi \bigl(x, \vert \underline {u}\vert \bigr)+\Phi\bigl(x, \vert \overline{u} \vert \bigr)\bigr]\,dx +b_{1} \int_{\Omega} h(x)\,dx \\ &\qquad {} +(b_{1}+1) \int_{\Omega}\Phi\bigl(x, \vert u\vert \bigr)\,dx+ \int_{\Omega}\overline{\Phi }\bigl(x, \bigl\vert g_{1}(x)\bigr\vert \bigr)\,dx \\ & \quad = (b_{1}K_{2}+1) \int_{\Omega} \Phi\bigl(x, \vert u\vert \bigr)\,dx +C, \end{aligned}$$

(3.12)

for all $u\in V$, where the constant $C>0$ is independent of u.

Let $\lambda> K_{1}(b_{1}K_{2}+1+2b_{3})$. Combining (3.2), (3.9), (3.11), and (3.12), we obtain

$$\begin{aligned} (\Gamma_{T}u,u) \geq{}& \frac{b_{2}}{2} \int_{\Omega}\Phi\bigl(x, \vert Du\vert \bigr)\,dx + \biggl( \frac{\lambda}{K_{1}}- b_{1}K_{2}-1-2b_{3} \biggr) \int_{\Omega }\Phi\bigl(x,\vert u\vert \bigr)\,dx \\ &{} -C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} \\ \geq{}& \frac{b_{2}}{2} \int_{\Omega}\Phi\bigl(x, \vert Du\vert \bigr)\,dx-C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi}, \end{aligned}$$

(3.13)

for all $u\in V$ and some $C>0$ independent of u. By Proposition 1.9 in [2], there exists $C_{1}>0$ such that $\Vert u\Vert _{\Phi}\leq C_{1}\Vert Du\Vert _{\Phi}$. In view of (3.13), for all $u\in V_{m}$, we have

$$\begin{aligned} \frac{(\Gamma_{T}|_{V_{m}} u,u)}{\Vert u\Vert _{1,\Phi}} &\geq \frac{b_{2} \int_{\Omega}\Phi(x, \vert Du\vert )\,dx}{2(1+C_{1})\Vert Du\Vert _{\Phi }} -\frac{C}{\Vert u\Vert _{1,\Phi}}- C^{*}\Vert q\Vert _{\overline{\Psi}} \\ &\geq\frac{b_{2} \int_{\Omega}\Phi(x, \vert Du\vert )\,dx}{2C_{2}(1+C_{1})\int _{\Omega} \vert Du\vert \,dx}-\frac{C}{\Vert u\Vert _{1,\Phi}}- C^{*}\Vert q\Vert _{\overline{\Psi}}, \end{aligned}$$

for some constant $C_{2}>0$. By Lemma 2.1, we get

$$ \frac{(\Gamma_{T}|_{V_{m}} u,u)}{\Vert u\Vert _{1,\Phi}} \rightarrow+\infty, \quad \mbox{as } \Vert u\Vert _{1,\Phi}\rightarrow+\infty. $$

(3.14)

By Lemma 2.2, there exists a Galerkin solution $u_{m}\in V_{m}$ for every $m\in\mathbb{N}$ such that $(\Gamma_{T} u_{m},v)=0$, $v\in V_{m}$. Using the density of $\{w_{m}\}$, we deduce that

$$ (\Gamma_{T} u_{m},v)=0,\quad \forall v\in V. $$

(3.15)

For $u\in V$, define $\rho(u)=\int_{\Omega}(\Phi(x, \vert Du\vert )+\Phi(x,\vert u\vert ))\,dx$ and $\Vert u\Vert _{\rho} =\inf\{\lambda>0: \rho (\frac{u}{\lambda} )\leq1\}$. Then $\Vert u\Vert _{\rho}$ is a norm of V equivalent to $\Vert u\Vert _{1,\Phi}$ (see [2]).

Taking $\alpha_{0}=\min \{\frac{b_{2}}{2},\frac{\lambda}{K_{1}}- b_{1}K_{2}-1-2b_{3} \}$, we have

$$\begin{aligned} (\Gamma_{T}u,u) \geq{}& \alpha_{0} \biggl[ \int_{\Omega}\Phi\bigl(x, \vert Du\vert \bigr)\,dx+ \int _{\Omega}\Phi\bigl(x,\vert u\vert \bigr)\,dx\biggr]-C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} \\ \geq{}& \alpha_{0}\bigl(\Vert u\Vert _{\rho}-\varepsilon \bigr) \biggl[ \int_{\Omega}\Phi\biggl(x, \frac {\vert Du\vert }{\Vert u\Vert _{\rho}-\varepsilon}\biggr)\,dx+ \int_{\Omega}\Phi\biggl(x,\frac{\vert u\vert }{\Vert u\Vert _{\rho}-\varepsilon}\biggr)\,dx\biggr] \\ & {}-C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} \\ \geq{}& \alpha_{0}\bigl(\Vert u\Vert _{\rho}-\varepsilon \bigr) -C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi}, \end{aligned}$$

for all $u\in V$, as $\Vert u\Vert _{1,\Phi}$ is large enough. Therefore, by (3.15), we get a sequence $\{u_{m}\}$ that is bounded in V. Hence, there exist $u_{0}\in V$ and a subsequence $\{ u_{k}\}$ of $\{u_{m}\}$, such that

$$\begin{aligned}& u_{k}\rightharpoonup u_{0}\quad \mbox{weakly in } V \mbox{ for } \sigma\Bigl(\prod L_{\Phi}, \prod E_{\overline{\Phi}} \Bigr), \end{aligned}$$

(3.16)

$$\begin{aligned}& u_{k}\rightarrow u_{0} \quad \mbox{strongly in } L_{\Psi}(\Omega), \end{aligned}$$

(3.17)

$$\begin{aligned}& u_{k}\rightarrow u_{0}\quad \mbox{a.e. in } \Omega, \end{aligned}$$

(3.18)

as $k\rightarrow\infty$.

By (3.4) and (3.8), $\{a_{0}(x, Tu_{k})\}$ and $\{ l(x,u_{k})\}$ are bounded in $L_{\overline{\Phi}}(\Omega)$. By Lemma 2.4,

$$a_{0}(x, Tu_{k})\rightharpoonup a_{0}(x, Tu_{0}) \quad \mbox{weakly in } L_{\overline{\Phi}}(\Omega) \mbox{ for } \sigma(L_{\overline{\Phi}}, E_{\Phi}) $$

and

$$l(x, u_{k})\rightharpoonup l(x, u_{0}) \quad \mbox{weakly in } L_{\overline {\Phi}}(\Omega) \mbox{ for } \sigma(L_{\overline{\Phi}}, E_{\Phi}), $$

as $k\rightarrow\infty$.

On the other hand, by the Lebesgue theorem, we deduce that

$$\int_{\Omega}a_{0}(x, Tu_{k}) (u_{k}-u_{0})\,dx\rightarrow0,\qquad \int_{\Omega}l(x, u_{k}) (u_{k}-u_{0}) \,dx\rightarrow0, \quad \mbox{as } k\rightarrow\infty. $$

Thanks to (3.7), $\{F(Tu_{k})\}$ is bounded in $L_{\overline {\Psi}}(\Omega)$. Hence,

$$\int_{\Omega}F(Tu_{k}) (u_{k}-u_{0}) \,dx\rightarrow0, \quad \mbox{as } k\rightarrow\infty. $$

Thus we obtain

$$ \int_{\Omega}a_{1}(x, Du_{k}) (Du_{k}-Du_{0})\,dx\rightarrow0,\quad \mbox{as } k\rightarrow \infty. $$

(3.19)

Similar to the proof of Proposition 3.1 in [20], we can construct a subsequence still denoted by $\{u_{k}\}$ such that

$$ Du_{k}\rightarrow Du_{0} \quad \mbox{a.e. in } \Omega, \mbox{ as } k\rightarrow \infty. $$

(3.20)

Hence

$$ a_{1}(x, Du_{k})\rightarrow a_{1}(x, Du_{0}) \quad \mbox{a.e. in } \Omega, \mbox{ as } k \rightarrow\infty. $$

(3.21)

In view of (3.1), $\{a_{1}(x, Du_{k})\}$ is bounded in $(L_{\overline{\Phi}}(\Omega))^{N}$, then by Lemma 2.4, we have

$$ a_{1}(x, Du_{k})\rightarrow a_{1}(x, Du_{0}) \quad \mbox{weakly in } \bigl(L_{\overline{\Phi}}( \Omega)\bigr)^{N} \mbox{ for } \sigma\bigl(\bigl(L_{\overline{\Phi }}( \Omega)\bigr)^{N}, \bigl(E_{\Phi}(\Omega)\bigr)^{N} \bigr), $$

(3.22)

as $k\rightarrow\infty$. Similarly,

$$F(Tu_{k})\rightharpoonup F(Tu_{0})\quad \mbox{weakly in } L_{\overline{\Psi }}(\Omega) \mbox{ for } \sigma(L_{\overline{\Psi}}, E_{\Psi}), \mbox{as } k\rightarrow\infty. $$

Hence, $(\Gamma_{T} u_{k},v)=(\Gamma_{T} u_{0},v)$, $\forall v\in V$. By (3.15), $(\Gamma_{T} u_{0},v)=0$, $\forall v\in V$, i.e., $u_{0}$ is a solution of (3.10).

For every $m\in\mathbb{N}$, taking $v=(u_{m}-\overline{u})^{+}\in V$ in (3.15) as a test function, we get

$$\begin{aligned} & \int_{\Omega}\bigl[a_{1}(x,Du_{m})-a_{1}(x,D \overline{u})\bigr] D(u_{m}-\overline{u})^{+}\,dx \\ &\qquad {}+ \int_{\Omega}\bigl[a_{0}(x,Tu_{m}) -a_{0}(x,\overline{u})\bigr](u_{m}-\overline {u})^{+}\,dx +\lambda \int_{\Omega}l(x,u_{m}) (u_{m}- \overline{u})^{+}\,dx \\ &\quad \leq \int_{\Omega}\bigl[F(Tu_{m})-F(\overline{u}) \bigr](u_{m}-\overline{u})^{+}\,dx. \end{aligned}$$

(3.23)

By (3.3), we have

$$\begin{aligned} & \int_{\Omega}\bigl[a_{1}(x,Du_{m})-a_{1}(x,D \overline{u})\bigr] D(u_{m}-\overline{u})^{+}\,dx \\ &\quad = \int_{\{u_{m}>\overline{u}\}}\bigl[a_{1}(x,Du_{m})-a_{1}(x,D \overline {u})\bigr] D(u_{m}-\overline{u})\,dx\geq0. \end{aligned}$$

Since

$$\int_{\Omega}\bigl[a_{0}(x,Tu_{m})-a_{0}(x, \overline{u})\bigr] (u_{m}-\overline {u})^{+}\,dx=0 $$

and

$$\int_{\Omega}\bigl[F(Tu_{m})-F(\overline{u}) \bigr](u_{m}-\overline{u})^{+}\,dx=0, $$

we get

$$0\geq \int_{\Omega}l(x,u_{m}) (u_{m}- \overline{u})^{+}\,dx \geq \int_{\{u_{m}>\overline{u}\}} \Phi(x,u_{m}-\overline{u})\,dx\geq0. $$

It follows that $u_{m}\leq\overline{u}$. Using arguments similar to those above we can prove that $u_{m}\geq \underline{u}$.

Thanks to (3.18), one has $\underline{u} \leq u_{0}\leq \overline{u}$. From the definitions of $l(\cdot,u_{0}(\cdot))$ and $Tu_{0}$, we have

$$l\bigl(x,u_{0}(x)\bigr)=0, \qquad a_{0}\bigl(x,Tu_{0}(x) \bigr)=a_{0}\bigl(x,u_{0}(x)\bigr) $$

and

$$f\bigl(x,Tu_{0}(x),DTu_{0}(x)\bigr)=f\bigl(x,u_{0}(x),Du_{0}(x) \bigr), $$

for a.e. $x\in\Omega$. We note that then (3.10) reduces to (3.6), which completes the proof. □

Remark 3.2

Our proof does not need the conditions $\overline{\Phi}\in\Delta_{2}$ and $(\Phi_{3})$ in [2].

Remark 3.3

Our method needs the strict monotonicity (3.3) of $a_{1}$, but does not require monotonicity (1.7) or coercivity (1.6) of $a_{0}$. However, if $\overline{\Phi}\in\Delta_{2}$, then we can deduce (3.22) by following the lines of Theorem 4.1 in [23] when (3.3) is replaced by (1.4).

Remark 3.4

Assume that (1.7) holds and the assumptions of Theorem 3.1 hold. If $f(x,u,Du)=f(x)\in L_{\overline{\Psi}}(\Omega)$, then it is easy to see that (3.5) has a unique solution.

Remark 3.5

Now we consider the following Neumann boundary value problem:

$$ \textstyle\begin{cases} -\operatorname {div}(a_{1}(x,Du))+a_{0}(x,u)=f(x,u,Du), \quad \mbox{in } \Omega, \\ a_{1}(x,Du)\cdot\gamma=0, \quad \mbox{on } \partial\Omega, \end{cases} $$

(3.24)

where γ is the outward unit normal to ∂Ω.

We also assume that there is a function $G:[k,+\infty)\rightarrow\mathbb {R}$ for some $k>0$ such that $G(s)\rightarrow+\infty$ as $s\rightarrow +\infty$ and

$$ \Phi(x,s u)\geq G(s)s\Phi(x,u)-s h(x),\quad \mbox{for all } s>0, u \geq0, \mbox{ a.e. }x\in\Omega, $$

(3.25)

and some $h\in L^{1}(\Omega)$, $h\geq0$.

Assume that (3.25) holds and the assumptions of Theorem 3.1 hold. Replacing V by $W^{1}L_{\Phi}(\Omega)$ in the proof of Theorem 3.1, and (3.13)-(3.14) by the following lines, we can deduce a similar theorem to Theorem 3.1 for the Neumann boundary value problem (3.24).

$$\begin{aligned} (\Gamma_{T}u,u) \geq{}&\frac{b_{2}}{2} \int_{\Omega}\Phi\bigl(x, \vert Du\vert \bigr)\,dx + \biggl( \frac{\lambda}{K_{1}}- b_{1}K_{2}-1-2b_{3} \biggr) \int_{\Omega }\Phi\bigl(x,\vert u\vert \bigr)\,dx \\ &{}-C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} \\ \geq{}& \alpha_{0} \biggl[ \int_{\Omega}\Phi\bigl(x, \vert Du\vert \bigr)\,dx+ \int_{\Omega}\Phi \bigl(x,\vert u\vert \bigr)\,dx\biggr]-C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi}, \end{aligned}$$

(3.26)

for all $u\in V$ and some $C>0$ independent of u, where $\alpha_{0}=\min\{\frac{b_{2}}{2},\frac{\lambda}{K_{1}}- b_{1}K_{2} -1-2b_{3}\}$.

Combining (3.25) and (3.26), we can deduce that, for any $\varepsilon>0$,

$$\begin{aligned} &(\Gamma_{T}u,u) \\ &\quad \geq \alpha_{0} \biggl[ \int_{\Omega}\Phi\biggl(x, \bigl(\Vert u\Vert _{\rho}- \varepsilon\bigr) \frac{\vert Du\vert }{\Vert u\Vert _{\rho}-\varepsilon}\biggr)\,dx+ \int_{\Omega}\Phi\biggl(x,\bigl(\Vert u\Vert _{\rho}- \varepsilon\bigr)\frac{\vert u\vert }{\Vert u\Vert _{\rho}-\varepsilon}\biggr)\,dx\biggr] \\ &\qquad {}-C- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} \\ &\quad \geq \alpha_{0}\bigl(\Vert u\Vert _{\rho}-\varepsilon \bigr)G\bigl(\bigl(\Vert u\Vert _{\rho}-\varepsilon \bigr)\bigr) \biggl[ \int_{\Omega}\Phi\biggl(x, \frac{\vert Du\vert }{\Vert u\Vert _{\rho}-\varepsilon}\biggr)\,dx+ \int _{\Omega}\Phi\biggl(x,\frac{\vert u\vert }{\Vert u\Vert _{\rho}-\varepsilon}\biggr)\,dx\biggr] \\ &\qquad {} -\alpha_{0}\bigl(\Vert u\Vert _{\rho}-\varepsilon\bigr) \int_{\Omega}\bigl\vert h(x)\bigr\vert \,dx-C- C^{*} \Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi} \\ &\quad \geq \alpha_{0}\bigl(\Vert u\Vert _{\rho}-\varepsilon \bigr)G\bigl(\bigl(\Vert u\Vert _{\rho}-\varepsilon \bigr)\bigr) - \alpha_{0}\bigl(\Vert u\Vert _{\rho}-\varepsilon\bigr) \int_{\Omega}\bigl\vert h(x)\bigr\vert \,dx -C \\ &\qquad {}- C^{*}\Vert q\Vert _{\overline{\Psi}} \Vert u\Vert _{1,\Phi}, \end{aligned}$$

$\forall u\in V$, as $\Vert u\Vert _{1,\Phi}$ is large enough. Since ε is arbitrary, we get

$$(\Gamma_{T}u,u)\geq\alpha_{0}\Vert u\Vert _{\rho}G\bigl(\Vert u\Vert _{\rho}\bigr) - \alpha_{0}\Vert u\Vert _{\rho} \int_{\Omega}\bigl\vert h(x)\bigr\vert \,dx -C- C^{*}\Vert q\Vert _{\overline {\Psi}} \Vert u\Vert _{1,\Phi}, $$

$\forall u\in V$, as $\Vert u\Vert _{1,\Phi}$ is large enough. Therefore, we obtain

$$\frac{(\Gamma_{T}|_{V_{m}} u,u)}{\Vert u\Vert _{1,\Phi}} \rightarrow+\infty,\quad \mbox{as } \Vert u\Vert _{1,\Phi}\rightarrow+\infty. $$

Proposition 3.1

If $\overline{\Phi}\in\Delta_{2}$, then there are functions $h\in L^{1}(\Omega)$, $h\geq0$, and $G:[k,+\infty)\rightarrow\mathbb{R}$ for some $k>2$ such that $G(s)\rightarrow+\infty$ as $s\rightarrow+\infty$ and (3.25) holds.

Proof

The proof of (3.25) is similar to the proof of Lemma 3.14 of [24].

Since $\overline{\Phi}\in\Delta_{2}$, there exist a positive constant $k>1$ and a nonnegative function $h\in L^{1}(\Omega)$ such that $\overline{\Phi}(x,2v)\leq k \overline{\Phi}(x,v)+h(x)$, for all $v\geq 0$ and a.e. $x\in\Omega$. Necessarily, $k>2$. Defining a function $F:[1,+\infty)\rightarrow[k,+\infty)$ by

$$F(r)=r\bigl((1-\lambda)k^{n}+\lambda k^{n+1}\bigr)\quad \mbox{if } r\in\bigl[2^{n},2^{n+1}\bigr] \mbox{ and } r=(1- \lambda)2^{n}+\lambda2^{n+1}, $$

we obtain

$$\begin{aligned} \overline{\Phi}(x,rv)&\leq\bigl[(1-\lambda)k^{n}+\lambda k^{n+1}\bigr] \overline {\Phi}(x,v)+\biggl[(1-\lambda) \frac{k^{n}-1}{k-1}+\lambda\frac {k^{n+1}-1}{k-1}\biggr]h(x) \\ &\leq\bigl[(1-\lambda)k^{n}+\lambda k^{n+1}\bigr] \overline{ \Phi }(x,v)+\bigl[(1-\lambda)k^{n}+\lambda k^{n+1}\bigr]h(x) \\ &\leq F(r) \overline{\Phi}(x,v)+ \frac{F(r)}{r} h(x). \end{aligned}$$

Hence $\frac{1}{F(r)}\overline{\Phi}(x,rv)\leq \overline{\Phi}(x,v)+ \frac{1}{r}h(x)$. Taking $\Psi_{1}(x,v)=\frac{1}{F(r)}\overline{\Phi}(x,rv)$, by Proposition 2.1 and Proposition 2.2, we have $\Phi(x,u)\leq\frac{1}{F(r)}\Phi(x,\frac{F(r)}{r}u) + \frac {1}{r}h(x)$, for all $u\geq0$ and a.e. $x\in\Omega$. It follows that $F(r)\Phi(x,u)\leq\Phi(x,\frac{F(r)}{r}u) +\frac {F(r)}{r} h(x)$, for all $u\geq0$ and a.e. $x\in\Omega$. Since $\frac{F(r)}{r}$ strictly increases from k to +∞ as $r\in[1,+\infty)$, its reciprocal function $G(s)$ is well defined and strictly increases from 1 to +∞ as $s\in[k,+\infty)$, and we have $sG(s)\Phi(x,u)\leq\Phi(x,su) + s h(x)$, i.e.

$$\Phi(x,su) \geq sG(s)\Phi(x,u) - s h(x),\quad \mbox{for } s\geq k, u\geq0 \mbox{ and a.e. } x\in\Omega. $$

 □

Remark 3.6

Clearly, (1.9) can be replaced by (3.25) in the proof of Theorem 2.1 in [2]. Therefore, by Proposition 3.1, the condition (1.9) can be omitted since $\overline{\Phi}\in\Delta_{2}$ in [2].

Denote $\mathcal{S}=\{u\in W_{0}^{1}L_{\Phi}(\Omega): u \mbox{ is a solution of (3.5) and } \underline{u}\leq u\leq \overline{u}\}$. Under the assumptions of Theorem 3.1, the solution set $\mathcal{S}$ is nonempty and we can deduce the following corollary.

Corollary 3.1

Under the assumptions of Theorem  3.1, the following assertions about $\mathcal{S}$ are true.

• (a) The set $\mathcal{S}$ is compact in $W_{0}^{1}L_{\Phi}(\Omega)$.

• (b) $\mathcal{S}$ is a direct set in both directions, that is, if $u_{1}, u_{2}\in\mathcal{S}$ then there exist $u, v \in\mathcal {S}$ such that $u\geq u_{1}\vee u_{2}$ and $v\leq u_{1}\wedge u_{2}$.

• (c) $\mathcal{S}$ has least and greatest elements with respect to the ordering ‘≤’, that is, there are $u_{*}, u^{*}\in\mathcal {S}$ such that $u_{*}\leq u\leq u^{*}$, for all $u\in\mathcal{S}$.