Let \(\Omega\subset\mathbb{R}^{N}\) be a bounded domain with the segment property, \(N\geq2\), M be an N-function, M̄ 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 [2]).
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., [30] or [20]). 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 [1]. 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., [31]) 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 [32], 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 [33].
The proof of the following proposition is similar to the proof of Lemma 2.2 in [34].
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)
\(\exp(G(T_{k}(u_{n})))\varphi\)
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)
\(\exp(-G(T_{k}(u_{n})))\varphi\)
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)
\(u_{n}\rightarrow u\)
a.e. in Ω;
-
(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)
\(T_{k}(u_{n})\rightarrow T_{k}(u)\)
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 [2], 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)
\(\{a(x,T_{k}(u_{n}), DT_{k}(u_{n}))\}_{n}\)
is bounded in
\(L_{\bar{M}}(\Omega)^{N}\);
-
(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 [2], 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 [2], 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 [35], 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). □