Proof of Theorem 2.6
We consider
$$ \textstyle\begin{cases} -\Delta _{\Phi}u= H(u, x, h_{1}(\zeta (u,x)), h_{2}(\zeta (u,x))), \quad x\in \Omega , \\ u=0, \quad x\in \partial \Omega , \end{cases} $$
(3.1)
where
$$ H(u, x, s, t)= J_{1}(u)s+ J_{2}(u)t-\gamma (u,x). $$
We have the following claims:
Claim 1. Problem (3.1) has a solution in \(W^{1,\Phi}_{0}(\Omega )\cap L^{\infty}(\Omega )\).
Define \(B:W_{0}^{1, \Phi}(\Omega ): \to W^{-1, \Phi}(\Omega )\) as
$$ \begin{aligned} \bigl(B(u), w\bigr)&= \int _{\Omega}-\Delta _{\Phi}uw- \int _{\Omega}\bigl[h_{1}\bigl(\zeta (u,x) \bigr)J_{1}(u)+h_{2}\bigl(\zeta (u,x)\bigr)J_{2}(u) - \gamma (u,x)\bigr]w \\ &= \int _{\Omega}\rho \bigl( \vert \nabla u \vert \bigr) (\nabla u \cdot \nabla w)- \int _{\Omega}\bigl[h_{1}\bigl(\zeta (u,x) \bigr)J_{1}(u)+h_{2}\bigl(\zeta (u,x)\bigr)J_{2}(u) \\ &\quad {}-\gamma (u,x)\bigr]w,\quad \forall u, w\in W_{0}^{1, \Phi}( \Omega ), \end{aligned} $$
where ρ satisfies \((\rho _{1})\), \((\rho _{2})\), and \((\rho _{3})\).
First, we want to show that B is continuous, bounded, and coercive.
It is easy to see that the conditions on ρ and the continuity of \(h_{1}\) and \(h_{2}\) guarantees that B is bounded and continuous.
According to \((\rho _{3})'\), there exist \(\kappa , s\in (1,N)\) such that
$$ \kappa \le \frac{\rho (t)t^{2}}{\Phi (t)}\le s,\quad \forall t>0, $$
which implies that
$$ \begin{aligned} \frac{(B(u), u)}{ \Vert u \Vert _{1, \Phi}}&= \frac{\int _{\Omega}\rho ( \vert \nabla u \vert ) \vert \nabla u \vert ^{2}-\int _{\Omega}[h_{1}(\zeta (u,x))J_{1}(u)+h_{2}(\zeta (u,x))J_{2}(u)-\gamma (u,x)]u}{ \Vert u \Vert _{1, \Phi}} \\ &\ge \frac{\kappa \int _{\Omega}\Phi ( \vert \nabla u \vert )-\int _{\Omega}[h_{1}(\zeta (u,x))J_{1}(u)+h_{2}(\zeta (u,x))J_{2}(u)-\gamma (u,x)]u}{ \Vert u \Vert _{1, \Phi}}. \end{aligned} $$
From the Lemma 2.3 and Lemma 2.1 in [12], we have
$$ \min \bigl\{ \Vert \nabla u \Vert _{L^{\Phi}}^{\kappa}, \Vert \nabla u \Vert _{L^{\Phi}}^{s} \bigr\} =\xi _{0}\bigl( \Vert \nabla u \Vert _{L^{\Phi}}\bigr)\le \int _{\Omega}\Phi \bigl( \vert \nabla u \vert \bigr) $$
and
$$ \int _{\Omega}\Phi \bigl( \vert \nabla u \vert \bigr)\ge \int _{\Omega}\Phi \biggl(\frac{ \vert u \vert }{d}\biggr), $$
then we deduce
$$ \begin{aligned} \frac{(B(u), u)}{ \Vert u \Vert _{1, \Phi}}&\ge \frac{\frac{\kappa}{2}\min \{ \Vert \nabla u \Vert _{L^{\Phi}}^{\kappa}, \Vert \nabla u \Vert _{L^{\Phi}}^{s}\}+\frac{\kappa}{2}\min \{ \Vert \frac{u}{d} \Vert _{L^{\Phi}}^{\kappa}, \Vert \frac{u}{d} \Vert _{L^{\Phi}}^{s}\}}{ \Vert u \Vert _{1, \Phi}} \\ &\quad {}- \frac{\int _{\Omega}[h_{1}(\zeta (u,x))J_{1}(u)+h_{2}(\zeta (u,x))J_{2}(u)-\gamma (u,x)]u}{ \Vert u \Vert _{1, \Phi}}. \end{aligned} $$
It follows that
$$ \begin{aligned} \frac{\kappa \int _{\Omega}\Phi ( \vert \nabla u \vert )}{ \Vert u \Vert _{1, \Phi}} &= \frac{\kappa \int _{\Omega}\Phi ( \vert \nabla u \vert )}{ \vert \nabla u \vert _{L^{\Phi}}+ \vert u \vert _{L^{\Phi}}} \\ &\ge \frac{\frac{\kappa}{2}\min \{ \Vert \nabla u \Vert _{L^{\Phi}}^{\kappa}, \Vert \nabla u \Vert _{L^{\Phi}}^{s}\}+\frac{\kappa}{2}\min \{ \Vert \frac{u}{d} \Vert _{L^{\Phi}}^{\kappa}, \Vert \frac{u}{d} \Vert _{L^{\Phi}}^{s}\}}{ \vert \nabla u \vert _{L^{\Phi}}+ \vert u \vert _{L^{\Phi}}} \\ &=\frac{\kappa}{2} \frac{\min \{ \Vert \nabla u \Vert _{L^{\Phi}}^{\kappa}, \Vert \nabla u \Vert _{L^{\Phi}}^{s}\}+\min \{ \Vert \frac{u}{d} \Vert _{L^{\Phi}}^{\kappa}, \Vert \frac{u}{d} \Vert _{L^{\Phi}}^{s}\}}{ \vert \nabla u \vert _{L^{\Phi}}+ \vert u \vert _{L^{\Phi}}} \to \infty \end{aligned} $$
if \(\|u\|_{1, \Phi}\to \infty \). Then we have
$$ \begin{aligned} \frac{(B(u), u)}{ \Vert u \Vert _{1, \Phi}}\to \infty \quad \bigl( \Vert u \Vert _{1, \Phi}\to \infty \bigr). \end{aligned} $$
Hence we can conclude that the operator B is coercive.
In the end, we will prove that operator B is pseudomonotone, i.e., if
$$ u_{n}\rightharpoonup u \quad \text{in } W_{0}^{1,\Phi}( \Omega )\cap L^{ \infty}(\Omega )$$
and
$$ \lim_{n\to \infty}\sup \bigl(B(u_{n}), (u_{n}-u) \bigr)\le 0,$$
then
$$ \lim_{n\to \infty}\inf \bigl(B(u_{n}), (u_{n}-w)\bigr)\ge \bigl(B(u), (u-w)\bigr), \quad \forall w\text{ in } W_{0}^{1,\Phi}(\Omega )\cap L^{\infty}( \Omega ). $$
(3.2)
From
$$ \int _{\Omega}\bigl[h_{1}\bigl(\zeta (u_{n},x)\bigr)J_{1}(u_{n})+g\bigl(\zeta (u_{n},x)\bigr)J_{2}(u_{n})- \gamma (u_{n},x)\bigr](u_{n}-u)\to 0 $$
and
$$ \limsup_{n\to \infty}\bigl(B(u_{n}), (u_{n}-u) \bigr)\le 0, $$
we obtain
$$ \limsup_{n\to \infty} \int _{\Omega}\rho \bigl( \vert \nabla u_{n} \vert \bigr) \bigl( \nabla u_{n}\cdot \nabla (u_{n}-u)\bigr)\le 0. $$
(3.3)
From Lemma 3.1 in [12], we infer
$$ \begin{aligned} \bigl\Vert \nabla (u_{n}-u) \bigr\Vert _{L^{\Phi}}\le \int _{\Omega}\Phi \bigl( \bigl\vert \nabla (u_{n}-u) \bigr\vert \bigr). \end{aligned} $$
(3.4)
From Lemma 2.5, we can obtain a \(k_{0}>0\) such that
$$ \begin{aligned} &\Phi \bigl( \bigl\vert \nabla (u_{n}-u) \bigr\vert \bigr) \\ &\quad\le \frac{[\Phi ( \vert \nabla u_{n} \vert )+\Phi ( \vert \nabla u \vert )]^{\frac{1}{\kappa +1}}}{k_{0}^{\frac{\kappa}{\kappa +1}}} \\ &\quad\quad {}\times \bigl[\rho \bigl( \vert \nabla u_{n} \vert \bigr) \bigl(\nabla u_{n}\cdot \nabla (u_{n}-u)\bigr)- \rho \bigl( \vert \nabla u \vert \bigr) \bigl(\nabla u\cdot \nabla (u_{n}-u) \bigr)\bigr]^{ \frac{\kappa}{\kappa +1}}, \end{aligned} $$
(3.5)
that is,
$$ \begin{aligned} & \int _{\Omega}\Phi \bigl( \bigl\vert \nabla (u_{n}-u) \bigr\vert \bigr) \\ &\quad\le \int _{\Omega} \biggl\{ \frac{[\Phi ( \vert \nabla u_{n} \vert )+\Phi ( \vert \nabla u \vert )]^{\frac{1}{\kappa +1}}}{k_{0}^{\frac{\kappa}{\kappa +1}}} \\ &\quad \quad {}\times \bigl[\rho \bigl( \vert \nabla u_{n} \vert \bigr) \bigl( \nabla u_{n}\cdot \nabla (u_{n}-u)\bigr)- \rho \bigl( \vert \nabla u \vert \bigr) \bigl(\nabla u\cdot \nabla (u_{n}-u) \bigr)\bigr]^{ \frac{\kappa}{\kappa +1}} \biggr\} \\ &\quad\le \biggl\{ \int _{\Omega} \biggl[ \frac{[\Phi ( \vert \nabla u_{n} \vert )+\Phi ( \vert \nabla u \vert )]^{\frac{1}{\kappa +1}}}{k_{0}^{\frac{\kappa}{\kappa +1}}} \biggr]^{\kappa +1} \biggr\} ^{\frac{1}{\kappa +1}} \\ &\quad\quad {}\times \biggl\{ \int _{\Omega}\bigl[\rho \bigl( \vert \nabla u_{n} \vert \bigr) \bigl(\nabla u_{n} \cdot \nabla (u_{n}-u) \bigr)-\rho \bigl( \vert \nabla u \vert \bigr) \bigl(\nabla u\cdot \nabla (u_{n}-u)\bigr)\bigr] \biggr\} ^{\frac{\kappa}{\kappa +1}}. \end{aligned} $$
(3.6)
Since \(u_{n}\rightharpoonup u\), we have
$$ \int _{\Omega}\rho \bigl( \vert \nabla u \vert \bigr) \bigl( \nabla u\cdot \nabla (u_{n}-u)\bigr) \to 0, $$
which, together with (3.3), guarantees that
$$ \int _{\Omega}\rho \bigl( \vert \nabla u_{n} \vert \bigr) \bigl(\nabla u_{n}\cdot \nabla (u_{n}-u)\bigr)- \rho \bigl( \vert \nabla u \vert \bigr) \bigl(\nabla u\cdot \nabla (u_{n}-u)\bigr)\to 0\quad \text{as } n \to +\infty . $$
(3.7)
From (3.5), (3.6), and (3.7), we have
$$ \int _{\Omega}\Phi \bigl( \bigl\vert \nabla (u_{n}-u) \bigr\vert \bigr)\to 0, $$
that is,
$$ \bigl\Vert \nabla (u_{n}-u) \bigr\Vert _{L^{\Phi}}\to 0. $$
Therefore,
$$ \begin{aligned} \Vert u_{n}-u \Vert _{1, \Phi}= \Vert u_{n}-u \Vert _{L^{\Phi}}+ \bigl\Vert \nabla (u_{n}-u) \bigr\Vert _{L^{ \Phi}}\to 0, \end{aligned} $$
which implies that (3.2) is true.
According to Lemma 2.2.2 in [21], there is a \(u\in W_{0}^{1,\Phi}(\Omega )\cap L^{\infty}(\Omega )\) such that for \(\forall w\in W_{0}^{1,\Phi}(\Omega )\),
$$ \bigl(B(u), w\bigr)=0. $$
Therefore, we know that u is a (weak) solution of problem (3.1).
Claim 2. We show that the solution u of problem (3.1) obtained above is a solution of (1.1).
We shall prove that
$$ \begin{aligned} \underline{w}_{*}\le u\le \overline{w}^{*} \quad \text{in } \Omega . \end{aligned} $$
(3.8)
Choosing \(w=(u-\overline{w}^{*})_{+}\) as a test function, we have
$$ \begin{aligned} \int _{\Omega}-\Delta _{\Phi}u\bigl(u-\overline{w}^{*} \bigr)_{+} &= \int _{\Omega}\bigl[H\bigl(x, u, h_{1}\bigl(\zeta (u,x) \bigr), h_{2}\bigl(\zeta (u,x)\bigr)\bigr)- \gamma (u,x)\bigr]\bigl(u- \overline{w}^{*}\bigr)_{+} \\ &= \int _{\Omega}\bigl[h_{1}\bigl(\zeta (u,x) \bigr)J_{1}(u)+ h_{2}\bigl(\zeta (u,x)\bigr)J_{2}(u)- \gamma (u,x)\bigr]\bigl(u-\overline{w}^{*}\bigr)_{+}. \end{aligned} $$
(3.9)
Define
$$ \Omega _{1}:=\bigl\{ x\in \Omega \mid u>\overline{w}^{*} \bigr\} . $$
Then
$$ \begin{aligned} & \int _{\Omega}\bigl[h_{1}\bigl(\zeta (u,x) \bigr)J_{1}(u)+ h_{2}\bigl(\zeta (u,x)\bigr)J_{2}(u)- \gamma (u,x)\bigr]\bigl(u-\overline{w}^{*}\bigr)_{+} \\ &\quad= \int _{\Omega _{1}}+ \int _{\Omega -\Omega _{1}}\bigl[h_{1}\bigl(\zeta (u,x) \bigr)J_{1}(u)+ h_{2}\bigl(\zeta (u,x)\bigr)J_{2}(u)- \gamma (u,x)\bigr]\bigl(u-\overline{w}^{*}\bigr)_{+} \\ &\quad= \int _{\Omega _{1}}\bigl[h_{1}\bigl(\zeta (u,x) \bigr)J_{1}(u)+ h_{2}\bigl(\zeta (u,x)\bigr)J_{2}(u)- \gamma (u,x)\bigr]\bigl(u-\overline{w}^{*}\bigr)_{+}+0 \\ &\quad= \int _{\Omega _{1}}\bigl[h_{1}\bigl(\overline{w}^{*} \bigr)J_{1}(u)+ h_{2}\bigl( \overline{w}^{*} \bigr)J_{2}(u)-\bigl(u-\overline{w}^{*}\bigr)_{+}^{\nu} \bigr]\bigl(u- \overline{w}^{*}\bigr)_{+}. \end{aligned} $$
(3.10)
Since Ψ and Λ are increasing, from Lemma 2.1 and \(|\zeta (u,x)|\leq \overline{w}^{*}\), we have
$$ \biggl\{ \varsigma >0\Bigm| \int _{\Omega} \Psi \biggl( \frac{ \vert \zeta (u,x) \vert }{\varsigma} \biggr) \le 1 \biggr\} \supseteq \biggl\{ \varsigma >0\Bigm| \int _{\Omega} \Psi \biggl( \frac{\overline{w}^{*}}{\varsigma} \biggr) \le 1 \biggr\} $$
and
$$ \biggl\{ \varsigma >0\Bigm| \int _{\Omega} \Lambda \biggl( \frac{ \vert \zeta (u,x) \vert }{\varsigma} \biggr) \le 1 \biggr\} \supseteq \biggl\{ \varsigma >0\Bigm| \int _{\Omega} \Lambda \biggl( \frac{\overline{w}^{*}}{\varsigma} \biggr) \le 1 \biggr\} , $$
which implies that
$$ J_{1}\bigl(\zeta (u,x)\bigr)\leq J_{1} \bigl(\overline{w}^{*}\bigr),\qquad J_{2}\bigl(\zeta (u,x)\bigr) \leq J_{2}\bigl(\overline{w}^{*}\bigr). $$
(3.11)
From (3.9), (3.10), and (3.11), we have
$$ \int _{\Omega}-\Delta _{\Phi}u\bigl(u-\overline{w}^{*} \bigr)_{+}\le \int _{ \Omega}\bigl[h_{1}\bigl(\overline{w}^{*} \bigr)J_{1}\bigl(\overline{w}^{*}\bigr)+h_{2} \bigl( \overline{w}^{*}\bigr)J_{2}\bigl(\overline{w}^{*} \bigr)-\bigl(u-\overline{w}^{*}\bigr)_{+}^{ \nu} \bigr]\bigl(u-\overline{w}^{*}\bigr)_{+}. $$
By Definition 2.2, we have
$$ \int _{\Omega}-\Delta _{\Phi}u\bigl(u-\overline{w}^{*} \bigr)_{+}\le \int _{ \Omega}\bigl[-\Delta _{\Phi}\overline{w}^{*}- \bigl(u-\overline{w}^{*}\bigr)_{+}^{\nu}\bigr] \bigl(u- \overline{w}^{*}\bigr)_{+}. $$
Hence
$$ \int _{\Omega}-\Delta _{\Phi}u\bigl(u-\overline{w}^{*} \bigr)_{+}+ \int _{\Omega} \Delta _{\Phi}\overline{w}^{*} \bigl(u-\overline{w}^{*}\bigr)_{+}\le \int _{ \Omega}\bigl[-\bigl(u-\overline{w}^{*} \bigr)_{+}^{\nu +1}\bigr]\le 0, $$
i.e.,
$$ \int _{\Omega}\bigl(\rho \bigl( \vert \nabla u \vert \bigr) \nabla u-\rho \bigl( \bigl\vert \nabla \overline{w}^{*} \bigr\vert \bigr) \nabla \overline{w}^{*}\bigr)\cdot \nabla \bigl(u- \overline{w}^{*}\bigr)_{+}\le \int _{\Omega}\bigl[-\bigl(u-\overline{w}^{*} \bigr)_{+}^{\nu +1}\bigr]\le 0. $$
(3.12)
From Lemma 2.5, there exists a \(k_{0}>0\) such that
$$ \begin{aligned} & \int _{\Omega}\bigl(\rho \bigl( \vert \nabla u \vert \bigr) \nabla u-\rho \bigl( \bigl\vert \nabla \overline{w}^{*} \bigr\vert \bigr)\nabla \overline{w}^{*}\bigr)\cdot \nabla \bigl(u- \overline{w}^{*}\bigr)_{+} \\ &\quad \ge \int _{\Omega}k_{0} \frac{\Phi ( \vert \nabla u-\nabla \overline{w}^{*} \vert )^{\frac{\kappa +1}{\kappa}}}{(\Phi ( \vert \nabla u \vert )+\Phi ( \vert \nabla \overline{w}^{*} \vert ))^{\frac{1}{\kappa}}} \frac{\nabla (u-\overline{w}^{*})_{+}}{\nabla (u-\overline{w}^{*})}. \end{aligned} $$
(3.13)
Since
$$ \int _{\Omega}k_{0} \frac{\Phi ( \vert \nabla u-\nabla \overline{w}^{*} \vert )^{\frac{\kappa +1}{\kappa}}}{(\Phi ( \vert \nabla u \vert )+\Phi ( \vert \nabla \overline{w}^{*} \vert ))^{\frac{1}{\kappa}}} \frac{\nabla (u-\overline{w}^{*})_{+}}{\nabla (u-\overline{w}^{*})}= \int _{\Omega _{1}}k_{0} \frac{\Phi ( \vert \nabla u-\nabla \overline{w}^{*} \vert )^{\frac{\kappa +1}{\kappa}}}{(\Phi ( \vert \nabla u \vert )+\Phi ( \vert \nabla \overline{w}^{*} \vert ))^{\frac{1}{\kappa}}} $$
and Φ is continuous, we obtain that there is an \(M_{1}>0\) such that
$$ \int _{\Omega _{1}}k_{0} \frac{\Phi ( \vert \nabla u-\nabla \overline{w}^{*} \vert )^{\frac{\kappa +1}{\kappa}}}{(\Phi ( \vert \nabla u \vert )+\Phi ( \vert \nabla \overline{w}^{*} \vert ))^{\frac{1}{\kappa}}}= \frac{k_{0}}{M_{1}} \int _{\{u>\overline{w}^{*}\}}\Phi \bigl( \bigl\vert \nabla u- \nabla \overline{w}^{*} \bigr\vert \bigr)^{\frac{\kappa +1}{\kappa}}. $$
(3.14)
From (3.12), (3.13), and (3.14), we have
$$ \int _{\{u>\overline{w}^{*}\}}\Phi \bigl( \bigl\vert \nabla u-\nabla \overline{w}^{*} \bigr\vert \bigr)^{ \frac{\kappa +1}{\kappa}}\le 0. $$
From Lemma 2.2 in [11] and [14], we obtain
$$ \int _{\{u>\overline{w}^{*}\}}\Phi \biggl(\frac{ \vert u-\overline{w}^{*} \vert }{d} \biggr)^{\frac{\kappa +1}{\kappa}} \le \int _{\{u>\overline{w}^{*}\}} \Phi \bigl( \bigl\vert \nabla u-\nabla \overline{w}^{*} \bigr\vert \bigr)^{\frac{\kappa +1}{\kappa}} \le 0, $$
where \(d=\mathrm{diam}(\Omega )\). Therefore, we can conclude that
$$ \bigl\vert \bigl\{ u>\overline{w}^{*}\bigr\} \bigr\vert =0, $$
and then \(u\le \overline{w}^{*}\).
A similar argument shows that \(u\geq \underline{w}_{*}\).
Therefore, (3.8) is true and thus u is a solution of problem (1.1).
The proof is completed. □
Proof of Theorem 2.7
In order to get positive solutions of problem (1.3), we study the following problem:
$$ \textstyle\begin{cases} -\Delta _{\Phi}u=(u+\frac{1}{n})^{\beta} \Vert u \Vert _{L^{\Psi}}^{\alpha}, \quad x\in \Omega , \\ u=0, \quad x\in \partial \Omega , \end{cases} $$
(3.15)
for \(n\geq 1\). We will use Theorem 2.6 to discuss problem (3.15).
First, we will construct a supersolution u̅ of problem (3.15).
From Lemma 2.4, problem (2.1) has a unique positive \(z_{\lambda}\in W_{0}^{1, \Psi}(\Omega )\) which satisfies
$$ \begin{aligned} 0< z_{\lambda}(x)\le K\lambda ^{\frac{1}{\kappa -1}}, \quad x\in \Omega \end{aligned} $$
(3.16)
for \(\lambda >0\) big enough, where K is independent of λ.
Let \(M=K\lambda ^{\frac{1}{\kappa -1}}\). Then
$$ \begin{aligned} K\lambda ^{\frac{1}{\kappa -1}}< z_{\lambda}(x)+M\le 2K \lambda ^{ \frac{1}{\kappa -1}}, \quad x\in \Omega . \end{aligned} $$
The condition \(0<\alpha <\kappa -1\) implies that there is a \(\lambda >1\) big enough such that
$$ \lambda ^{\frac{\alpha}{\kappa -1}} \Vert 2K \Vert _{L^{\Psi}}^{\alpha}\le \lambda ,\qquad M=K\lambda ^{\frac{1}{\kappa -1}}>1$$
and (3.16) holds. Hence
$$ \biggl(z_{\lambda}+M+\frac{1}{n}\biggr)^{\beta} \Vert z_{\lambda}+M \Vert _{L^{\Psi}}^{ \alpha}\le \Vert z_{\lambda}+M \Vert _{L^{\Psi}}^{\alpha}\le \lambda ^{ \frac{\alpha}{\kappa -1}} \Vert 2K \Vert _{L^{\Psi}}^{\alpha}\le \lambda $$
and
$$ -\Delta _{\Phi}(z_{\lambda}+M)= -\Delta _{\Phi}z_{\lambda}= \lambda \geq \biggl(z_{\lambda}+M+\frac{1}{n} \biggr)^{\beta} \Vert z_{\lambda}+M \Vert _{L^{ \Psi}}^{\alpha}. $$
Therefore, \(z_{\lambda}+M\) is a supersolution of (3.15).
Second, we will construct a positive subsolution \(\underline{u}_{*}\) of problem (3.15).
Define \(d(x):=\mathrm{dist}(x,\partial \Omega )\), then by a direct calculation one can deduce that \(|\nabla d(x)|=1\). Because ∂Ω is \(C^{2}\), we can get a constant \(\tau >0\) such that \(d\in C^{2}(\overline{\Omega _{3\tau}})\) with \(\overline{\Omega _{3\tau}}:=\{x\in \overline{\Omega}:d(x)\le 3\tau \} \) (see [9, 10]). Let \(\varpi \in (0, \tau )\). Define
$$ \eta (x):= \textstyle\begin{cases} e^{\vartheta d(x)}-1, &\text{for } d(x)< \varpi , \\ e^{\vartheta \varpi}-1+\int _{\varpi}^{d(x)}\vartheta e^{\vartheta d(x)}( \frac{2\tau -t}{2\tau -\varpi})^{\frac{s}{\kappa -1}}\,dt,&\text{for } \varpi \le d(x)\le 2\tau , \\ e^{\vartheta \varpi}-1+\int _{\varpi}^{2\tau}ke^{\vartheta d(x)}( \frac{2\tau -t}{2\tau -\varpi})^{\frac{s}{\kappa -1}}\,dt,&\text{for } 2\tau < d(x), \end{cases} $$
where \(\vartheta >0\) is an arbitrary number. Direct computations imply that
$$ -\Delta _{\Phi}(\mu \eta ) = \textstyle\begin{cases} -\vartheta \Theta (x)\frac{d}{dt}(\rho (t)t)|_{t=\Theta (x)} - \rho (\Theta (x))\Theta (x)\Delta d ,&\text{for } d(x)< \varpi , \\ \frac{\Theta _{0}(\frac{s}{\kappa -1})\chi (x)^{\frac{s}{\kappa -1}-1}}{2\tau -\varpi} \frac{d}{dt}(\rho (t)t)|_{t=\Theta _{0}\chi (x)^{ \frac{s}{\kappa -1}}} \\ \quad {}-\rho (\Theta _{0}\chi (x)^{\frac{s}{\kappa -1}})\Theta _{0}\chi (x)^{ \frac{s}{\kappa -1}}\Delta d,&\text{for } \varpi \le d(x) \le 2\tau , \\ 0, &\text{for } 2\tau < d(x), \end{cases} $$
with \(\Theta (x)=\mu \vartheta e^{\vartheta d(x)}\), \(\Theta _{0}=\mu \vartheta e^{\vartheta \varpi}\), and \(\chi (x)=\frac{2\tau -d(x)}{2\tau -\varpi}\) for all \(\mu >0\).
There are three cases: (1) \(d(x)<\varpi \); (2) \(\varpi < d(x)<2\tau \); and (3) \(d(x)>2\tau \).
(1) We consider the case \(d(x)<\varpi \).
Since Δd is a bounded function near ∂Ω and \(\kappa >1\), there is a ϑ large enough such that
$$ \begin{aligned} -\Delta _{\Phi}(\mu \eta )&= - \mu \vartheta ^{2}e^{\vartheta d(x)} \frac{d}{dt}\bigl(\rho (t)t \bigr)\bigg|_{t=\mu \vartheta e^{\vartheta d(x)}}- \rho \bigl(\mu \vartheta e^{\vartheta d(x)}\bigr)\mu \vartheta e^{\vartheta d(x)} \Delta d \\ &\le -\vartheta ^{2}\mu e^{\vartheta d(x)}(\kappa -1)\rho \bigl(\mu \vartheta e^{\mu \vartheta e^{\vartheta d(x)}}\bigr)-\rho \bigl(\mu \vartheta e^{ \vartheta d(x)}\bigr) \mu \vartheta e^{\vartheta d(x)}\Delta d \\ &=\mu \vartheta e^{\vartheta d(x)}\rho \bigl(\mu \vartheta e^{\vartheta d(x)}\bigr) \bigl(- \vartheta (\kappa -1)-\Delta d\bigr) \\ &\le 0, \end{aligned} $$
which implies that
$$ \begin{aligned} -\Delta _{\Phi}(\mu \eta )\le 0\le (\mu \eta )^{\beta} \vert \mu \eta \vert _{L^{ \Psi}}^{\alpha}, \end{aligned} $$
when \(d(x)<\varpi \) and ϑ is large enough.
(2) We consider the case \(\varpi < d(x)<2\delta \).
From the condition \((\rho _{3})\) and Lemma 2.3, we have
$$ \begin{aligned} &\mu \vartheta e^{\vartheta \varpi} \biggl(\frac{s}{\kappa -1} \biggr) \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\frac{s}{\kappa -1}-1} \biggl( \frac{1}{2\tau -\varpi} \biggr)\frac{d}{dt}\bigl(\rho (t)t \bigr)\bigg|_{t=\mu \vartheta e^{\vartheta \varpi} (\frac{2\tau -d(x)}{2\tau -\varpi} )^{\frac{s}{\kappa -1}}} \\ &\quad\le \mu \vartheta e^{\vartheta \varpi} \biggl(\frac{s}{\kappa -1} \biggr) \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\frac{s}{\kappa -1}-1} \biggl(\frac{s-1}{2\tau -\varpi} \biggr)\rho \biggl(\mu \vartheta e^{ \vartheta \varpi} \biggl(\frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{ \frac{s}{\kappa -1}} \biggr) \\ &\quad\le \biggl(\frac{s}{\kappa -1} \biggr) \biggl(\frac{s-1}{2\tau -\varpi} \biggr) \frac{s\Phi (\mu \vartheta e^{\vartheta \varpi} (\frac{2\tau -d(x)}{2\tau -\varpi} )^{\frac{s}{\kappa -1}} )}{\mu \vartheta e^{\vartheta \varpi} (\frac{2\delta -d(x)}{2\tau -\varpi} )^{\frac{s}{\kappa -1}}} \frac{1}{\frac{2\tau -d(x)}{2\tau -\varpi}} \\ &\quad\le \biggl(\frac{s^{2}}{\kappa -1} \biggr) \biggl(\frac{s-1}{2\tau -\varpi} \biggr)\max \biggl\{ \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{s-1} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{s(\frac{s}{\kappa -1})-( \frac{s}{\kappa -1}+1)}, \\ &\quad\quad \bigl(\mu \vartheta e^{ \vartheta \varpi}\bigr)^{\kappa -1} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\kappa (\frac{s}{\kappa -1})-(\frac{s}{\kappa -1}+1)} \biggr\} \Phi (1). \end{aligned} $$
(3.17)
Now \(s, \kappa >1\) implies \(\kappa (\frac{s}{\kappa -1} )-s (\frac{s}{\kappa -1}+1 ), s (\frac{s}{\kappa -1} )-s (\frac{s}{\kappa -1}+1 )>0\), which, together with \(0\le \frac{2\tau -d(x)}{2\tau -\varpi}\le 1\) and (3.17), guarantees that
$$\begin{aligned} &\mu \vartheta e^{\vartheta \varpi} \biggl(\frac{s}{\kappa -1} \biggr) \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\frac{s}{\kappa -1}-1} \biggl( \frac{1}{2\tau -\varpi} \biggr)\frac{d}{dt}\bigl(\rho (t)t \bigr)\bigg|_{t=\mu \vartheta e^{\vartheta \varpi} ( \frac{2\delta -d(x)}{2\tau -\varpi} )^{\frac{s}{\kappa -1}}} \\ &\quad \le \biggl(\frac{s^{2}}{\kappa -1} \biggr) \biggl(\frac{s-1}{2\tau -\varpi} \biggr)\Phi (1) \max \bigl\{ \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{s-1}, \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{\kappa -1}\bigr\} \\ &\quad =C_{1} \biggl(\frac{1}{2\tau -\varpi} \biggr)\max \bigl\{ \bigl(\mu \vartheta e^{ \vartheta \varpi}\bigr)^{s-1}, \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{ \kappa -1}\bigr\} , \end{aligned}$$
(3.18)
where \(C_{1}=\frac{s^{2}(s-1)\Phi (1)}{\kappa -1}\) is a constant independent of μ and ϑ. Similarly, one has
$$ \begin{aligned} & \biggl\vert \rho \biggl(\mu \vartheta e^{\vartheta \varpi} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\frac{s}{\kappa -1}} \biggr)\mu \vartheta e^{\vartheta \varpi} \frac{(2\tau -d(x))^{\frac{s}{\kappa -1}}}{(2\tau -\varpi )^{\frac{s}{\kappa -1}}} \Delta d \biggr\vert \\ &\quad\le \rho \biggl(\mu \vartheta e^{\vartheta \varpi} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\frac{s}{r-1}}\biggr)\mu \vartheta e^{ \vartheta \varpi} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{ \frac{s}{\kappa -1}}\sup_{\overline{\Omega _{3\tau}}} \vert \Delta d \vert \\ &\quad\le C \frac{\Phi (\mu \vartheta e^{\vartheta \varpi} (\frac{2\tau -d(x)}{2\tau -\varpi} )^{\frac{s}{\kappa -1}} )}{\mu \vartheta e^{\vartheta \varpi} (\frac{2\tau -d(x)}{2\tau -\varpi} )^{\frac{s}{\kappa -1}}} \\ &\quad\le C\max \biggl\{ \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{s-1} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{s(\frac{s}{\kappa -1})-( \frac{s}{\kappa -1}+1)}, \\ &\quad\quad \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{\kappa -1} \biggl( \frac{2\tau -d(x)}{2\tau -\varpi} \biggr)^{\kappa ( \frac{s}{\kappa -1})-(\frac{s}{\kappa -1}+1)} \biggr\} \\ &\quad\le C_{2}\max \bigl\{ \bigl(\mu \vartheta e^{\vartheta \varpi} \bigr)^{s-1}, \bigl( \mu \vartheta e^{\vartheta \varpi}\bigr)^{\kappa -1} \bigr\} , \end{aligned} $$
(3.19)
where \(C_{2}\) is a constant independent of ϖ, ϑ, and μ. Thus from (3.18) and (3.19) we have
$$ \begin{aligned} -\Delta _{\Phi}u\le \max \biggl\{ \frac{C_{1}}{2\tau -\varpi}, C_{2} \biggr\} \max \bigl\{ \bigl(\mu \vartheta e^{\vartheta \varpi}\bigr)^{s-1}, \bigl(\mu \vartheta e^{\vartheta \varpi} \bigr)^{\kappa -1} \bigr\} , \end{aligned} $$
when \(\varpi < d(x)<2\tau \).
Let \(\varpi =\frac{ln2}{\vartheta }\) and \(\mu =e^{-\vartheta }\), then \(e^{\vartheta \varpi}=2\). Since
$$ \begin{aligned} \eta (x)&=e^{\vartheta \varpi}-1+ \int _{\varpi}^{d(x)}\vartheta e^{ \vartheta d(x)} \biggl( \frac{2\tau -t}{2\tau -\varpi} \biggr)^{ \frac{s}{\kappa -1}}\,dt \\ &>2-1+2\vartheta \int _{\varpi}^{d(x)} \biggl( \frac{2\tau -t}{2\tau -\varpi} \biggr)^{\frac{s}{\kappa -1}}\,dt \\ &=1+\vartheta C_{3} \\ &\ge 1, \end{aligned} $$
where \(C_{3}>0\) is a constant, we have that when μ is small enough and n is large enough,
$$\begin{aligned} \biggl(\mu \eta +\frac{1}{n} \biggr)^{\beta} \vert \mu \eta \vert _{L^{\Psi}}^{\alpha}& \ge \vert \mu \eta \vert _{L^{\Psi}}^{\alpha} \\ &={\inf}^{\alpha} \biggl\{ \varsigma >0: \int _{\Omega}\Psi \biggl( \frac{ \vert \mu \eta \vert }{\varsigma} \biggr)< 1 \biggr\} \\ &={\inf}^{\alpha} \biggl\{ \tau \mu >0: \int _{\Omega}\Psi \biggl( \frac{ \vert \mu \eta \vert }{\tau \mu} \biggr)< 1 \biggr\} \\ &=\mu ^{\alpha}{\inf}^{\alpha} \biggl\{ \tau >0: \int _{\Omega}\Psi \biggl( \frac{ \vert \eta \vert }{\tau} \biggr)< 1 \biggr\} \\ &\ge \mu ^{\alpha}C_{4}, \end{aligned}$$
where \(C_{4}>0\) is a constant independent of \(\vartheta >0\).
Since \(0<\alpha <\kappa -1\), we have the result
$$ \begin{aligned} \lim_{\vartheta \to +\infty} \frac{\vartheta ^{\kappa -1}}{e^{\vartheta (\kappa -1-\alpha )}}=0. \end{aligned} $$
In view of
$$ \begin{aligned} -\Delta _{\Phi}(\mu \eta )\le \max \biggl\{ \frac{C_{1}}{2\tau -\varpi}, C_{2} \biggr\} \max \bigl\{ 2^{s-1}, 2^{\kappa -1}\bigr\} \biggl( \frac{\vartheta }{e^{\vartheta }} \biggr)^{\kappa -1}, \end{aligned} $$
choose a \(\vartheta _{0}>0\) large enough such that
$$ \begin{aligned} C_{4}\ge \max \biggl\{ \frac{C_{1}}{2\tau -\frac{\ln 2}{\vartheta }}, C_{2} \biggr\} \max \bigl\{ 2^{s-1}, 2^{\kappa -1}\bigr\} \biggl( \frac{\vartheta ^{\kappa -1}}{e^{\vartheta (\kappa -1-\alpha )}} \biggr) \end{aligned} $$
for all \(\vartheta \ge \vartheta _{0}\).
Thus,
$$ \begin{aligned} -\Delta _{\Phi}(\mu \eta )\le \biggl(\mu \eta +\frac{1}{n} \biggr)^{\beta} \vert \mu \eta \vert _{L^{\Psi}}^{\alpha} \end{aligned} $$
in the case \(\varpi < d(x)<2\tau \) for \(\vartheta >0\) large enough.
(3) We consider the case \(d(x)>2\tau \).
Obviously,
$$ \begin{aligned} -\Delta _{\Phi}(\mu \eta )=0\le \biggl(\mu \eta +\frac{1}{n} \biggr)^{ \beta} \vert \mu \eta \vert _{L^{\Psi}}^{\alpha}. \end{aligned} $$
It is obvious that \(\underline{w}_{*}\leq \overline{w}^{*}\) if M is large enough and μ is small enough. And \((\underline{w}_{*},\overline{w}^{*})\) is a sub-supersolution pair of problem (3.15). Now Theorem 2.6 guarantees that problem (3.15) has a solution \(u_{n}\) which satisfies \(0<\mu \eta \le u_{n}\le z_{\lambda}+M\).
Now we consider the set \(\{u_{n}\}\).
From Lemma 2.2 in [12], one has that \(\|u\|_{1, \Phi}\) and \(|\!|\!|\nabla u|\!|\!|_{L^{\Phi}}\) defined on \(W_{0}^{1, \Phi}\) are equivalent. And from the proof of the coercivity of the operator B, we know that if \(|\!|\!|\nabla u|\!|\!|_{L^{\Phi}}>1\), then
$$ \int _{\Omega}\Phi \bigl( \vert \nabla u \vert \bigr)\ge |\!|\!|\nabla u|\!|\!|_{L^{\Phi}}, $$
that is,
$$ \int _{\Omega}\Phi \bigl( \vert \nabla u \vert \bigr)\ge \Vert u \Vert _{1, \Phi}, $$
when \(\|u\|_{1, \Phi}>1\).
If \(\|u_{n}\|_{1, \Phi}\le 1\), then \({u_{n}}\) is bounded in \(W_{0}^{1, \Phi}(\Omega )\) naturally.
If \(\|u_{n}\|_{1, \Phi}>1\), then
$$ \Vert u_{n} \Vert _{1, \Phi}\le \int _{\Omega}\Phi \bigl( \vert \nabla u_{n} \vert \bigr). $$
By the condition \((\rho _{3})'\) and due to
$$ \int _{\Omega}-\Delta _{\Phi}u_{n}u_{n}= \int _{\Omega}u_{n} \biggl(u_{n}+ \frac{1}{n} \biggr)^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{\alpha}, $$
we have
$$ \kappa \int _{\Omega}\Phi \bigl( \vert \nabla u_{n} \vert \bigr)\le \int _{\Omega}\phi \bigl( \vert \nabla u_{n} \vert \bigr) \vert \nabla u_{n} \vert ^{2}= \int _{\Omega}u_{n} \biggl(u_{n}+ \frac{1}{n} \biggr)^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{\alpha}, $$
which, together with \(\alpha \ge 0\), \(-1<\beta <0 \), gives
$$ \int _{\Omega}\Phi \bigl( \vert \nabla u_{n} \vert \bigr)\le \frac{1}{\kappa} \int _{\Omega} \overline{w}^{* \beta +1} \bigl\Vert \overline{w}^{*} \bigr\Vert _{L^{\Psi}}^{\alpha}, $$
that is,
$$ \Vert u_{n} \Vert _{1, \Phi}\le \frac{1}{\kappa} \int _{\Omega}\overline{w}^{* \beta +1} \bigl\Vert \overline{w}^{*} \bigr\Vert _{L^{\Psi}}^{\alpha}. $$
Therefore, \(\{u_{n}\}\) is bounded in \(W_{0}^{1, \Phi}(\Omega )\).
Since \(W_{0}^{1, \Phi}(\Omega )\) is reflexive, \(\{u_{n}\}\) has weakly convergent subsequences in \(W_{0}^{1,\Phi}(\Omega )\cap L^{\infty}(\Omega )\), and we still use \(u_{n}\) to denote its subsequence. From the analysis in [3], we have
$$ \begin{aligned} u_{n}\rightharpoonup u\quad \text{in } W_{0}^{1,\Phi}(\Omega )\cap L^{ \infty}(\Omega ) \end{aligned} $$
and
$$ \begin{aligned} u_{n}(x)\stackrel{\text{a.e.}}{\to} u(x), \quad x\in \Omega . \end{aligned} $$
Since
$$ \underline{w}_{*}\le u_{n}\le \overline{w}^{*}, \quad x\in \Omega ,$$
Lebesgue theorem implies
$$ u_{n} \to u \quad \text{in } L^{q}(\Omega ) \ \forall q\in [1, + \infty ).$$
(3.20)
Since \(u_{n}\) is a (weak) solution of (3.15) for all , we have
$$ \begin{aligned} \int _{\Omega}-\Delta _{\Phi}u_{n}w= \int _{\Omega} \biggl(u_{n}+ \frac{1}{n} \biggr)^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{\alpha}w, \end{aligned} $$
for all \(w\in W_{0}^{1,\Phi}(\Omega )\).
Denoting \(w=u_{n}-u\), we have
$$ \begin{aligned} \int _{\Omega}-\Delta _{\Phi}u_{n}(u_{n}-u)= \int _{\Omega} \biggl(u_{n}+ \frac{1}{n} \biggr)^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{\alpha}(u_{n}-u). \end{aligned} $$
Since
$$ \biggl(u_{n}+\frac{1}{n} \biggr)^{\beta}\le \underline{w}_{*}^{\beta}, \quad x\in \Omega , $$
one has
$$ \begin{aligned} \int _{\Omega} \biggl(u_{n}+\frac{1}{n} \biggr)^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{ \alpha} \vert u_{n}-u \vert &\le \int _{\Omega}\underline{w}_{*}^{\beta} \vert u_{n}-u \vert \Vert u_{n} \Vert _{L^{ \Psi}}^{\alpha} \\ &\le \biggl[ \int _{\Omega} \bigl(\underline{w}_{*}^{\beta} \Vert u_{n} \Vert _{L^{ \Psi}}^{\alpha} \bigr)^{p} \biggr]^{\frac{1}{p}} \biggl[ \int _{\Omega} \vert u_{n}-u \vert ^{q} \biggr]^{\frac{1}{q}}, \end{aligned} $$
where \(p, q>1\), \(\frac{1}{p}+\frac{1}{q}=1\), and \(\beta \in (-1,0)\). From (3.20), we have
$$ \biggl[ \int _{\Omega} \bigl(\underline{w}_{*}^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{ \alpha} \bigr)^{p} \biggr]^{\frac{1}{p}} \biggl[ \int _{\Omega} \vert u_{n}-u \vert ^{q} \biggr]^{\frac{1}{q}}\to 0,$$
and so
$$ \begin{aligned} \int _{\Omega} \biggl(u_{n}+\frac{1}{n} \biggr)^{\beta} \Vert u_{n} \Vert _{L^{\Psi}}^{ \alpha} \vert u_{n}-u \vert \to 0\quad \text{as }n\to +\infty , \end{aligned} $$
which implies
$$ \begin{aligned} \int _{\Omega}-\Delta _{\Phi}u_{n}(u_{n}-u) \to 0. \end{aligned} $$
Obviously,
$$ \begin{aligned} \int _{\Omega}-\Delta _{\Phi}u (u_{n}-u)\to 0. \end{aligned} $$
(3.21)
Similar to the previous proof, from (3.4), (3.6), and (3.21), we have
$$ u_{n}\to u\quad \text{in } W_{0}^{1,\Phi}(\Omega ) \cap L^{\infty}( \Omega ),$$
and so
$$ \Vert u_{n} \Vert _{L^{\Psi}}^{\alpha}\to \Vert u \Vert _{L^{\Psi}}^{\alpha}.$$
Therefore, taking the limit as \(n\to \infty \) in (3.15), we have
$$ -\Delta _{\Phi}u=u^{\beta} \Vert u \Vert _{L^{\Psi}}^{\alpha}.$$
The limit value u is just the solution which we are looking for, and it satisfies \(\underline{w}_{*}\le u\le \overline{w}^{*}\), obviously. Therefore, the proof is finished. □