\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2015 (2015), No. 305, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2015 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2015/305\hfil Nonlinear elliptic equations] {Nonlinear elliptic equations with general growth in the gradient related to Gauss measure} \author[Y. Tian, C. Ma, F. Li \hfil EJDE-2015/305\hfilneg] {Yujuan Tian, Chao Ma, Fengquan Li} \address{Yujuan Tian (corresponding author) \newline School of Mathematical Sciences, Shandong Normal University, Jinan 250014, China} \email{tianyujuan0302@126.com} \address{Chao Ma \newline School of Mathematical Sciences, University of Jinan, Jinan 250022, China} \email{chaos\_ma@163.com} \address{Fengquan Li \newline School of Mathematical Sciences, Dalian University of Technology, Dalian 116024, China} \email{fqli@dlut.edu.cn} \thanks{Submitted July 22, 2015. Published December 16, 2015.} \subjclass[2010]{35J60, 35J70, 35J15} \keywords{Comparison results; symmetrization; gauss measure; \hfill\break\indent nonlinear elliptic equation} \begin{abstract} In this article, we establish a comparison result through symmetrization for solutions to some problems with general growth in the gradient. This allows to get sharp estimates for the solutions, obtained by comparing them with solutions of simpler problems whose data depend only on the first variable. Furthermore, we use such result to prove the existence of bounded solutions. All the above results are based on the study of a class of nonlinear integral operator of Volterra type. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} Let $\Omega$ be an open subset of $\mathbb{R}^n$. We consider the Dirichlet problem whose prototype is \begin{equation} \begin{gathered} -\operatorname{div}(\varphi(x)|\nabla u|^{p-2}\nabla u)=H(x,u,\nabla u) \quad \text{in }\Omega,\\ u=0\quad \text{on }\partial \Omega, \end{gathered}\label{P1} \end{equation} where $|H(x,s,\xi)|\leq \varphi(x)\left(f(x)+\theta|\xi|^q\right)$ with $p-10$ and $\varphi(x)=(2\pi)^{-n/2}\exp\big(-\frac{|x|^{2}}{2}\big)$ is the density of Gauss measure. Problem \eqref{P1} is related to the generator of Ornstein-Uhlenbeck semigroup. As $\Omega$ is bounded, the operator in \eqref{P1} is uniformly elliptic. In this case, it is well known that one can use Schwarz symmetrization to estimate the solutions of elliptic equations in terms of the solutions of radially symmetric problems. This kind of issue has been faced in \cite{Diaz2,Talenti1,Mossino1,Alvino1} for linear equations. As regards nonlinear equations, for the case of $p-1$ growth in the gradient, comparison results are obtained in \cite{Betta1,Messano1}. Optimal summability of solutions are discussed in \cite{Cianchi}. For the case of $p$ growth in the gradient, using Schwarz symmetrization, the existence of bounded solutions are obtained in \cite{Diaz1,Grenon1,Grenon2,Messano1}. For the case of $q (p-1\tau\}$, i.e. $$ \Phi(\tau)=\gamma_{n}(\{x\in\mathbb{R}^n:x_{1}>\tau\}) =\frac{1}{\sqrt{2\pi}}\int_{\tau}^{+\infty}\exp\big(-\frac{t^2}{2}\big)dt, \tau\in \mathbb{R}. $$ Observe that (see \cite{Tian2}) \begin{gather}\label{2.1} \exp\Big(-\frac{\Phi^{-1}(t)^{2}}{2}\Big)\leq \alpha t(1-\log t)^{1/2} , \quad t\in(0,\gamma_n(\Omega)), \\ \label{2.2}\exp\Big(-\frac{\Phi^{-1}(t)^{2}}{2}\Big) \geq \beta t(1-\log t)^{1/2}, \quad t\in(0,\gamma_n(\Omega)), \end{gather} where $\alpha$ and $\beta$ are positive constants depending on $\gamma_n(\Omega)$. Now we give the notion of rearrangement. \begin{definition} \label{def2.1}\rm If $u$ is a measurable function in $\Omega$ and $\mu(t)=\gamma_{n}(\{x\in\Omega:|u|>t\})$ is the distribution function of $u$, then we define the decreasing rearrangement of $u$ with respect to Gauss measure as \[ u^{\star}(s)=\inf\{t\geq0:\mu(t)\leq s\}, \quad s\in[0,\gamma_n(\Omega)]. \] If $\Omega^{\sharp}=\{x=(x_1,x_2,\cdots,x_n)\in\mathbb{R}^n:x_1>\lambda\}$ is the half-space such that $\gamma_n(\Omega)=\gamma_n(\Omega^{\sharp})$, then \[ u^\sharp(x)=u^\star(\Phi(x_1)), \quad x\in\Omega^{\sharp} \] denote the increasing Gauss symmetrization of $u$ (or Gauss symmetrization of $u$). \end{definition} Similarly, the decreasing Gauss symmetrization of $u$ is $$ u_{\sharp}(x)=u_{\star}(\Phi(x_1)), \quad x\in\Omega^{\sharp}, $$ with $$ u_{\star}(s)=u^{\star}(\gamma_n(\Omega)-s),\quad s\in(0,\gamma_n(\Omega)). $$ The properties of rearrangement with respect to Gauss measure or a positive measure have been widely considered, see \cite{Chong,Carlen,Rakotoson1,Rakotoson2,Talenti2} for instance. Here we just recall that (a) (Hardy-Little inequality) \begin{align*} \int_0^{\gamma_n(\Omega)}u_\star(s)v^\star(s)ds &=\int_{\Omega^{\sharp}}u_\sharp(x)v^\sharp(x)d\gamma_n \leq\int_{\Omega}|u(x)v(x)|d\gamma_{n}\\ &\leq\int_{\Omega^{\sharp}}u^\sharp(x)v^\sharp(x)d\gamma_n= \int_0^{\gamma_n(\Omega)}u^\star(s)v^\star(s)ds, \end{align*} where $u$ and $v$ are measurable functions. (b) (Polya-Sz\"{e}go principle) Let $u\in W_0^{1,p}(\varphi,\Omega)$ with $10, k(x)\geq0$ and $k\in L^{p'}(\varphi,\Omega)$; \item[(A3)] $|H(x,s,\xi)|\leq \varphi(x)\left(f(x)+\theta|\xi|^q\right)$, with $\theta>0$, $p-1\frac{1}{\gamma'} (\gamma M_0)^{\frac{1}{1-\gamma}}$. Thus, the above theorem proposes an example to show that \eqref{P2} has no symmetric solution without the assumption \eqref{1.2}. \end{remark} Now, the comparison results can be stated by the following theorem. \begin{theorem} \label{thm3.3} Assume that {\rm (A1)--(A3)} hold. Let $u$ be a solution to \eqref{P3} and $v$ be a solution to \eqref{P2} such that $v(x)=v^\sharp(x)$. Then \begin{gather}\label{1.6} u^{\sharp}(x)\leq v(x), \quad x\in\Omega^{\sharp}, \\ \label{1.7}\int_{\Omega}\eta(|\nabla u|^p)\varphi\,dx \leq \int_{\Omega^\sharp}\eta(|\nabla v|^p)\varphi\,dx, \end{gather} where $\eta$ is a concave and nondecreasing function on $[0,+\infty)$. Moreover, if $f(x)\not\equiv 0$, the equality in \eqref{1.6} holds if and only if \begin{equation}\label{1.8} \begin{gathered} \Omega=\Omega^\sharp,\\ u(x)=\delta u^\sharp(x), \quad\text{a.e. } x\in\Omega^{\sharp},\\ a_1(x,u,\nabla u)=\delta\varphi|D_1 u^{\sharp}|^{p-2}D_1 u^{\sharp}, \quad \text{a.e. }x\in\Omega^\sharp,\\ \sum_{i=2}^n D_ia_i(x,u,\nabla u)=0 \quad \text{in } \mathcal{D}'(\Omega^{\sharp}),\\ H(x,u,\nabla u)=\delta\varphi(f^{\sharp}(x)+\theta|D_1 u^{\sharp}|^{q}), \quad \text{a.e. }x\in\Omega^\sharp \end{gathered} \end{equation} modulo a rotation with $\delta=\pm 1$. \end{theorem} \begin{remark} \label{rmk3.3} \rm Equality \eqref{1.8} implies that the comparison result \eqref{1.6} is sharp in the sense that as the equality holds, problem \eqref{P3} is equivalent to its ``symmetrized" problem \eqref{P2} modulo a rotation. \end{remark} The estimates we have found can be applied to prove the existence result by using the well known approximation techniques \cite{Boccardo}. \begin{theorem} \label{thm3.4} Let {\rm (A1)--(A3)} and \eqref{1.2} hold. Assume that $$ [a(x,s,\xi_1)-a(x,s,\xi_2)]\cdot(\xi_1-\xi_2)>0, \quad\text{for } \xi_1\neq\xi_2 . $$ Then there exists at least one solution to \eqref{P3}. \end{theorem} \section{Results for Volterra integral operators} This section is devoted to study a class of Volterra integral operator. We prove three results, i.e. the comparison principle, the existence of fixed point and the nonexistence of fixed point for the Volterra integral operator, which will be useful in proving the main results of this paper. Assume $h(s,\xi) :[0,T]\times\mathbb{R}\to\mathbb{R}$ is a Carath\'{e}odory function. Consider the following Volterra integral operator (see \cite{pasic1,Korkut}) \begin{equation*} K: D(K)\subseteq C([0,T])\to C([0,T]), \quad K\psi(t)=\int_0^th(\tau,\psi(\tau))d\tau,\quad \forall\psi\in D(K). \end{equation*} \begin{definition} \label{def4.1}\rm We say that the operator $K$ has property (m) if for all $\psi_1, \psi_2\in D(K)$ and $a\in[0,T)$, there exist constants $b\in (a,T]$ and $m(a,b,\psi_1, \psi_2)\in [0,1)$ such that for any $t\in(a,b]$, \begin{equation}\label{4.1} \|h(\cdot,\psi_1(\cdot))-h(\cdot,\psi_2(\cdot))\|_{L^1(a,t)} \leq t m(a,b,\psi_1, \psi_2)\big\|\frac{\psi_1 -\psi_2}{s}\big\|_{L^\infty(a,t)}. \end{equation} \end{definition} \begin{lemma} \label{lem4.1} Let the operator $K$ satisfy property (m) and $h(t,\cdot)$ be nondecreasing for a.e. $t\in[0,T]$. If $ u, v \in D(K)$ are such that $u\leq Ku$, $v\geq Kv$, then we have \begin{equation}\label{3.4} u\leq v. \end{equation} In particular, the equation $w=Kw$ possesses at most one solution in $D(K)$. \end{lemma} \begin{proof} We argue by contradiction. If \eqref{3.4} does not hold, then there must exist $a\in[0,T)$ and $b_1\in(a,T]$ such that $u(t)\leq v(t)$ for $t\in[0,a]$ and $u(t)>v(t)$ for $t\in(a,b_1]$. Set $b_2=\min\{b_1,b\}$ with $b$ is the constant of property (m). Then $u(t)>v(t)$ in $(a,b_2]$. Since $u\leq Ku$, $v\geq Kv$ and $h(t,\cdot)$ is nondecreasing for a.e. $t\in[0,T]$, it follows that for $\forall t\in(a,b_2]$, \begin{equation}\label{4.3} \begin{split} |u(t)-v(t)|&=u(t)-v(t)\leq \int_0^t\left(h(\tau,u(\tau))-h(\tau,v(\tau))\right)d\tau \\ &=\int_0^a\left(h(\tau,u(\tau))-h(\tau,v(\tau))\right)d\tau+ \int_a^t\left(h(\tau,u(\tau))-h(\tau,v(\tau))\right)d\tau \\ &\leq\int_a^t\left(h(\tau,u(\tau))-h(\tau,v(\tau))\right)d\tau \\ &\leq\|h\left(\cdot,u(\cdot)\right) -h\left(\cdot,v(\cdot)\right)\|_{L^1(a,t)}. \end{split} \end{equation} Using property (m), we obtain \begin{equation}\label{4.31} |u(t)-v(t)|\leq t m(a,b,u, v)\|\frac{u-v}{s}\|_{L^\infty(a,t)}, \quad \forall t\in(a,b_2]. \end{equation} Taking the maximum over $t\in(a,b_2]$ and noting $m(a,b,u, v)\in [0,1)$, we have $$ \|\frac{u-v}{s}\|_{L^\infty(a,b_2)}\leq m(a,b,u,v) \|\frac{u -v}{s}\|_{L^\infty(a,b_2)} <\|\frac{u -v}{s}\|_{L^\infty(a,b_2)}, $$ which is a contradiction. Thus $$ u\leq v, \forall u, v \in D(K), $$ and the uniqueness claim easily follows. Then the lemma is proved. \end{proof} Let \begin{equation*}%\label{3.6} K\psi(s)=\int_0^sf^\star(\tau)+\theta\Big(\sqrt{2\pi} \exp\big(\frac{\Phi^{-1}(\tau)^{2}}{2}\big)\Big)^\gamma\psi^\gamma(\tau)d\tau, \quad s\in[0,\gamma_n(\Omega)]. \end{equation*} We shall deal with two types of domains \begin{gather*}%\label{3.7} D_1(K)=\{\psi\in C([0,\gamma_n(\Omega)]): M\geq0, 0\leq\psi(s)\leq Ms\},\\ %\label{3.8} D_2(K)=\{\psi\in C([0,\gamma_n(\Omega)]): M_{\psi}\geq0, 0\leq\psi(s)\leq M_{\psi}s\}, \end{gather*} and let $R_i(K)$ be the range of $K$ on $D_i(K)$, $i=1,2$. \begin{lemma} \label{lem4.2} Let $M=(\gamma M_0)^{\frac{1}{1-\gamma}}$ in $ D_1(K)$. If \eqref{1.2} holds, then the equation $w=Kw$ has a unique solution in $ D_1(K)$. Furthermore, the solution is also unique in $D_2(K)$. \end{lemma} \begin{proof} First, we prove that $R_1(K)\subseteq D_1(K)$. For $\forall\psi\in D_1(K)$ and $s\in[0,\gamma_n(\Omega)]$, we have by \eqref{2.2} that \begin{equation}\label{3.9} \begin{split} K\psi(s)&=\int_0^sf^\star(\tau)+\theta\Big(\sqrt{2\pi} \exp\big(\frac{\Phi^{-1}(\tau)^{2}}{2}\big)\Big)^\gamma\psi^\gamma(\tau)d\tau \\ &\leq s\|f\|_{L^{\infty}(\Omega)}+\theta\big(\sqrt{2\pi}\beta^{-1}\big)^\gamma \int_0^s\Big(\frac{1}{\tau(1-\log \tau)^{1/2}}\Big)^\gamma(M\tau)^\gamma d\tau\\ &\leq \left(\|f\|_{L^{\infty}(\Omega)}+\theta\left(\sqrt{2\pi}\beta^{-1}M(1-\log \gamma_n(\Omega))^{-1/2}\right)^\gamma\right) s\\& = \left(\|f\|_{L^{\infty}(\Omega)}+M_0M^{\gamma}\right) s. \end{split} \end{equation} Under the assumption \eqref{1.2}, noting that $M=(\gamma M_0)^{\frac{1}{1-\gamma}}$, we obtain $$ \|f\|_{L^{\infty}(\Omega)}+M_0M^{\gamma}\leq M. $$ Thus, $$ 0\leq K\psi(s)\leq Ms, \quad s\in[0,\gamma_n(\Omega)]. $$ This implies that $R_1(K)\subseteq D_1(K)$. Next, we verify that the operator $K$ is compact with respect to the uniform topology of $C([0,\gamma_n(\Omega)])$. To prove this, let us first show the equicontinuity of $R_1(K)$ in $C([0,\gamma_n(\Omega)])$. Take any $K\psi\in R_1(K)$. For any $0\leq aa$ such that $0\leq m(a,b_0,\psi_1,\psi_2)<1$. Thus \eqref{4.8} implies that $K$ satisfies property (m) in $D_2(K)$. Since $ h(t,\psi)$ is nondecreasing for $t\in[0,\gamma_n(\Omega)]$, we have by Lemma \ref{lem4.1} that the solution of $w=Kw$ is unique in $D_2(K)$. Thus the proof is complete. \end{proof} For $f(x)\equiv f_0$, we have the following nonexistence result. \begin{lemma} \label{lem4.3} Let \eqref{1.4} hold. Then the equation $z=Kz$ has no solution in $D_1(K)$. \end{lemma} \begin{proof} We argue by contradiction. Assume that there exists a solution $z\in D_1(K)$ of $z=Kz$. Let $z_0(s)=f_0s$ and \begin{equation}\label{3.32} z_{m}(s)=\theta\int_0^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^\gamma z_{m-1}^\gamma(\tau)d\tau, \quad s\in[0,\gamma_n(\Omega)]. \end{equation} First, we claim that \begin{equation}\label{3.34} z_m(s)\geq sA(s)\Big(\frac{f_0}{A(s)}\Big)^{\gamma^m}\gamma^{\frac{m}{\gamma-1}}, \quad m\geq1 \end{equation} where \[ A(s)=\Big[\frac{\gamma^{\gamma'}}{\theta(\gamma-1)} \left(\frac{\alpha}{\sqrt{2\pi}}\right)^{\gamma}\frac{s}{F_1(s)} \Big]^{\frac{1}{\gamma-1}},\quad F_1(s)=\int_0^s(1-\ln\tau)^{-{\frac{\gamma}{2}}}d \tau. \] To prove \eqref{3.34}, set \begin{equation*}%\label{3.36} F_{m+1}(s)=\int_0^s \left( 1-\ln \tau\right)^{-\gamma/2} F^{\gamma}_m(\tau)d\tau. \end{equation*} Thus, \begin{equation}\label{3.35} z_m(s)\geq\theta^{\frac{\gamma^m-1}{\gamma-1}} \Big(\frac{\sqrt{2\pi}}{\alpha}\Big)^{\frac{\gamma^{m+1}-\gamma}{\gamma-1}} f_0^{\gamma^m} s^{-\frac{\gamma^m-\gamma}{\gamma-1}}F_m(s). \end{equation} Indeed, by \eqref{2.1} and \eqref{3.32}, we have \begin{align*} z_{1}(s)&=\theta \int_0^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^\gamma z_0^\gamma(\tau)d\tau\\ & \geq\theta \Big(\frac{\sqrt{2\pi}}{\alpha}\Big)^{\gamma}f^{\gamma}_0 \int_0^s\Big(\frac{1}{\tau(1-\ln\tau)^{1/2}}\Big)^\gamma {\tau}^\gamma d\tau\\ &=\theta \Big(\frac{\sqrt{2\pi}}{\alpha}\Big)^{\gamma}f^{\gamma}_0F_1(s), \end{align*} which proves \eqref{3.35} for $m=1$. Assume it holds for $m-1$. Then \eqref{3.35} can be established by induction on $m$. Next, treat the term $F_m(s)$ in \eqref{3.35}. Actually, \begin{equation}\label{3.37} F_m(s)\geq \prod_{\delta=1}^{m} \Big(\frac{\gamma-1}{\gamma^\delta-1}\Big)^{\gamma^{m-\delta}} F_1^{\frac{\gamma^m-1}{\gamma-1}}(s), m\geq1. \end{equation} The case $m=1$ is obvious. Suppose \eqref{3.37} holds for some $m$. Then \begin{align*} F_{m+1}(s)&=\int_0^s \left( 1-\ln \tau\right)^{-\gamma/2} F^{\gamma}_m(\tau)d\tau\\ & \geq\prod_{\delta=1}^{m} \Big(\frac{\gamma-1}{\gamma^\delta-1}\Big)^{\gamma^{m+1-\delta}} \int_0^s \left( 1-\ln \tau\right)^{-\gamma/2} F^{\frac{\gamma^{m+1}-\gamma}{\gamma-1}}_1(\tau)d\tau\\ &= \prod_{\delta=1}^{m}\Big(\frac{\gamma-1}{\gamma^\delta-1}\Big)^{\gamma^{m+1-\delta}} \int_0^s F^{\frac{\gamma^{m+1}-\gamma}{\gamma-1}}_1(\tau)dF_1(\tau)\\ &= \prod_{\delta=1}^{m+1}\Big(\frac{\gamma-1}{\gamma^\delta-1} \Big)^{\gamma^{m+1-\delta}} F^{\frac{\gamma^{m+1}-1}{\gamma-1}}_1(s), \end{align*} which proves \eqref{3.37}. Moreover, \begin{equation} \label{3.39} \begin{split} \prod_{\delta=1}^{m}\Big(\frac{\gamma-1}{\gamma^\delta-1}\Big)^{\gamma^{m-\delta}} &\geq \prod_{\delta=1}^{m} \Big(\frac{\gamma-1}{\gamma^\delta}\Big)^{\gamma^{m-\delta}} =\left(\gamma-1\right)^{\sum_{\delta=1}^{m}\gamma^{m-\delta}} \gamma^{-\sum_{\delta=1}^{m}\delta\gamma^{m-\delta}}\\ &=\left(\gamma-1\right)^{\frac{\gamma^m-1}{\gamma-1}} \gamma^{\frac{m}{\gamma-1}-\frac{\gamma(\gamma^m-1)}{(\gamma-1)^2}}. \end{split} \end{equation} Combining \eqref{3.35}, \eqref{3.37} and \eqref{3.39}, we know that \eqref{3.34} holds. On the other hand, $A(s)$ is a decreasing continuous function on $(0,\gamma_n(\Omega)]$ and $\lim_{s\to 0^+}A(s)=+\infty$. Then the range of $A$ is $[A(\gamma_n(\Omega)), +\infty)$. Recalling that $f_0\geq A(\gamma_n(\Omega)))$, thus there must exist a constant $s^\ast\in(0,\gamma_n(\Omega)]$ such that $f_0=A(s^\ast)$. As $s\geq s^\ast$, it follows that $\frac{f_0}{A(s)}\geq1$ and \begin{equation*} z_m(s)\geq sA(s)\gamma^{\frac{m}{\gamma-1}}. \end{equation*} Note that $\lim_{m\to\infty}\gamma^{\frac{m}{\gamma-1}}=+\infty$. We conclude that $\sum_{m=0}^{\infty}z_m(s)=+\infty$ for $s\in [s^\ast,\gamma_n(\Omega)]$. On the other hand, by inducing on $k$, it is easy to prove that \begin{equation}\label{3.33} z(s)\geq\sum_{m=0}^kz_m(s), \quad \forall k\in \mathbb{N}. \end{equation} Indeed, since $z(s)=Kz(s)\geq f_0s=z_0(s)$, we obtain \[ z(s)=f_0s+\theta\int_0^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^\gamma z^\gamma(\tau)d\tau \geq z_0(s)+z_1(s), \] which proves the claim for $k=1$. Now assume \eqref{3.33} holds for some $k\in N$. Then \begin{align*} z(s)&\geq z_0(s)+\theta\int_0^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^\gamma \Big(\sum_{m=0}^kz_m(\tau)\Big)^\gamma d\tau\\ & \geq z_0(s)+\theta\sum_{m=0}^k\int_0^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^\gamma z_m^\gamma(\tau)d\tau\\ &=z_0(s)+\sum_{m=0}^kz_{m+1}(s) =\sum_{m=0}^{k+1}z_m(s). \end{align*} Thus for $k\in \mathbb{N}$, \eqref{3.33} holds. However, recalling that $z\in D_1(K)$, we obtain $$ Ms\geq z(s)\geq \sum_{m=0}^{\infty}z_m(s)=+\infty, \quad s\in [s^\ast,\gamma_n(\Omega)], $$ which is a contradiction. Thus the lemma is proved. \end{proof} \section{Proofs of the main results} First, let us enunciate the following lemma, the proof of which is not supplied here since it follows the same lines as in \cite{Ferone1,Messano1}. \begin{lemma} \label{lem5.1} Let $u$ be a solution of \eqref{P1} and $v$ be a solution of \eqref{P2} such that $v(x)=v^\sharp(x)$. Then \begin{equation} \label{3.12} \begin{split} &\big(-u^{\star'}(s)\big)^{{p-1}} \Big(\frac{1}{\sqrt{2\pi}} \exp\Big(-\frac{\Phi^{-1}(s)^{2}}{2}\Big)\Big)^p\\ &\leq\int_0^sf^\star(\tau)\exp \Big[\theta\int_\tau^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\sigma)^{2}}{2}\Big)\Big)^{p-q} \left(-u^{\star'}(\sigma)\right)^{q-p+1}d\sigma\Big] d\tau \end{split} \end{equation} and \begin{equation} \label{3.13} \begin{split} &\big(-v^{\star'}(s)\big)^{{p-1}} \Big(\frac{1}{\sqrt{2\pi}} \exp\Big(-\frac{\Phi^{-1}(s)^{2}}{2}\Big)\Big)^p\\ &=\int_0^sf^\star(\tau)\exp \Big[\theta\int_\tau^s\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\sigma)^{2}}{2}\Big)\Big)^{p-q} \big(-v^{\star'}(\sigma)\big)^{q-p+1}d\sigma\Big] d\tau, \end{split} \end{equation} a.e. $s\in(0,\gamma_n(\Omega))$. \end{lemma} \begin{proof}[Proof of Theorem \ref{thm3.1}] Let \begin{equation}\label{3.14} v(x)=V(\Phi(x_1)) =\int_{\Phi(x_1)}^{\gamma_n(\Omega)}\Big(\sqrt{2\pi} \exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^{p'}w^{\frac{1}{p-1}}(\tau)d \tau, x\in\Omega^\sharp, \end{equation} where $w$ is the unique solution of $w=Kw$ obtained in Lemma \ref{lem4.2}. Clearly, $v(x)=v^{\sharp}(x)$, $x\in\Omega^{\sharp}$. By \eqref{3.14} and \eqref{2.2}, \begin{align*} |\nabla v(x)| &=D_1v(x)=-\frac{1}{\sqrt{2\pi}}V'(\Phi(x_1)) \exp\big(-\frac{x_1^{2}}{2}\big)\\ &= \Big(\sqrt{2\pi}\exp\big(\frac{x_1^{2}}{2}\big)\Big)^{\frac{1}{p-1}} w^{\frac{1}{p-1}}(\Phi(x_1))\\ &\leq\big(\sqrt{2\pi}\big)^\frac{1}{p-1}\left(\beta\Phi(x_1)(1-\log \Phi(x_1))^{1/2}\right)^{-\frac{1}{p-1}}(M\Phi(x_1))^{\frac{1}{p-1}}\\ & \leq \Big(\sqrt{2\pi}\beta^{-1} M(1-\log \gamma_n(\Omega))^{-1/2}\Big)^{\frac{1}{p-1}}. \end{align*} Moreover, since $10$, we take \[ \psi(x)=\begin{cases} \operatorname{sign} u(x) &\text{if } |u(x)|>t+h,\\ \frac{(|u(x)|-t)\operatorname{sign} u(x)}{h} &\text{if } t<|u(x)|\leq t+h,\\ 0 &\text{otherwise} \end{cases} \] as the test function in \eqref{e3.1} and let $h\to 0$. Since $\delta u=|u|=u^{\sharp}=v^{\sharp}=v$, we have by Hardy-Littlewood inequality that \begin{equation} \label{3.51} \begin{aligned} &-\frac{d}{dt}\int_{u^{\sharp}>t}|\nabla u^{\sharp}|^p\varphi \,dx \\ &=-\frac{d}{dt}\int_{|u|>t}|\nabla u|^p\varphi \,dx\leq-\frac{d}{dt}\int_{|u|>t}a(x,u,\nabla u)\nabla u\,dx\\ &=\int_{|u|>t}H(x,u,\nabla u)udx \leq\int_{|u|>t}fu\varphi\,dx +\theta\int_{|u|>t}|\nabla u|^qu\varphi\,dx\\ &\leq \int_{u^{\sharp}>t}f^{\sharp}u^{\sharp}\varphi\,dx +\theta\int_{u^{\sharp}>t}|\nabla u^{\sharp}|^qu^{\sharp}\varphi\,dx\\ &=\int_{v>t}f^{\sharp}v\varphi\,dx+\theta\int_{v>t}|\nabla v|^qv\varphi\,dx\\ & =-\frac{d}{dt}\int_{v>t}|\nabla v|^p\varphi\,dx =-\frac{d}{dt}\int_{u^{\sharp}>t}|\nabla u^{\sharp}|^p\varphi\,dx. \end{aligned} \end{equation} Thus, the above equalities hold. In particular, \begin{gather}\label{3.54} -\frac{d}{dt}\int_{|u|>t}a(x,u,\nabla u)\nabla u\,dx =-\frac{d}{dt}\int_{u^{\sharp}>t}|\nabla u^{\sharp}|^p\varphi\,dx, \\ \label{3.53} \int_{|u|>t}H(x,u,\nabla u)udx =\int_{u^{\sharp}>t}f^{\sharp}u^{\sharp}\varphi \,dx +\theta\int_{u^{\sharp}>t}|\nabla u^{\sharp}|^qu^{\sharp}\varphi\,dx,\\ \label{3.52} \int_{u^{\sharp}>t}fu^{\sharp}\varphi \,dx =\int_{u^{\sharp}>t}f^{\sharp}u^{\sharp}\varphi \,dx. \end{gather} By \cite[Lemma 4.3]{Tian3}, from \eqref{3.52} on has $f=f^{\sharp}$ a.e. $x\in\Omega^{\sharp}$. Then combine \eqref{3.53} and assumption (iii) to discover that \[ H(x,u,\nabla u)=\delta(f^{\sharp}\varphi+\theta|D_1 u^{\sharp}|^q\varphi), \quad \text{a.e. }x\in\Omega^{\sharp} \] Similarly, from \eqref{3.54} and assumption (i) we have \[ \delta a(x,u,\nabla u)\nabla u^{\sharp}=|\nabla u^{\sharp}|^p\varphi. \] Recalling \eqref{3.27}, we have \begin{gather*} D_1 u^{\sharp}(x)=D_1 v^{\sharp}(x) =\Big(\sqrt{2\pi} \exp\big(\frac{x_1^2}{2}\big)\Big)^{\frac{1}{p-1}} w^{\frac{1}{p-1}}(\Phi(x_1))>0, \quad \text{a.e. } x\in\Omega^{\sharp},\\ D_iu^\sharp(x)=0, \quad i=2,3,\dots,n. \end{gather*} Hence, \[ a_1(x,u,\nabla u)=\delta|D_1 u^{\sharp}|^{p-2}D_1 u^{\sharp}\varphi, \quad \text{a.e. } x\in\Omega^{\sharp}. \] From the definition of solution it follows that for all $\psi\in W_0^{1,p}(\varphi,\Omega^\sharp)\cap L^{\infty}(\Omega^\sharp)$, \begin{align*} &\int_{\Omega^\sharp}\delta|D_1 u^{\sharp}|^{p-2}D_1 u^{\sharp} D_1\psi \varphi\,dx +\int_{\Omega^\sharp}\sum_{i=2}^n a_i(x,u,\nabla u)D_i\psi\,dx\\ &=\int_{\Omega}a(x,u,\nabla u)\nabla \psi \,dx =\int_{\Omega}H(x,u,\nabla u)\psi \,dx\\ &=\int_{\Omega^\sharp}\delta(f^{\sharp}\varphi+\theta|D_1 u^{\sharp}|^q\varphi)\psi \,dx =\int_{\Omega^\sharp}\delta(f^{\sharp}\varphi +\theta|D_1 v|^q\varphi)\psi\,dx\\ &=\int_{\Omega^\sharp}\delta|D_1 v|^{p-2}D_1 vD_1\psi \varphi\,dx =\int_{\Omega^\sharp}\delta|D_1 u^{\sharp}|^{p-2} D_1 u^{\sharp}D_1\psi \varphi \,dx. \end{align*} Then \[ \int_{\Omega^\sharp}\sum_{i=2}^n a_i(x,u,\nabla u)D_i\phi \,dx=0, \quad \forall \psi\in W_0^{1,p}(\varphi,\Omega^\sharp)\cap L^{\infty}(\Omega^\sharp), \] which completes the proof. \end{proof} \section{Appendix} In this section give an example to Remark \ref{rmk3.1}. Assume that $v$ is a solution of the problem \begin{equation} \label{P4} \begin{gathered} -D_1(\varphi|D_1v|^{p-2}D_1v)=\varphi+\theta|D_1v|^q\varphi \quad \text{in } \Omega^\sharp,\\ v\in W^{1,p}_0(\varphi,\Omega^\sharp)\cap L^{\infty}(\Omega^\sharp) \end{gathered} \end{equation} and $Y$ is the solution of the problem \begin{equation} \label{P5} \begin{gathered} -D_1(\varphi|D_1Y|^{p-2}D_1Y)=\varphi \quad \text{in } \Omega^\sharp,\\ Y=0 \quad\text{on } \partial\Omega^\sharp. \end{gathered} \end{equation} For any given $k>0$, let $$ T_k(s)=\begin{cases} k & \text{if } s>k,\\ s & \text{if } |s|\leq k,\\ -k & \text{if } s<-k. \end{cases} $$ Take $T_k((v-Y)^-)$ as the test function in \eqref{P4} and \eqref{P5}, and subtract the two results. We obtain \[ 0\geq\int_{\{0<(v-Y)^-\leq k\}}\varphi(|D_1 v|^{p-2}D_1 v-|D_1 Y|^{p-2}D_1 Y)(Y-v)dx\geq 0. \] Thus, $\gamma_n(\{0<(v-Y)^-\leq k\})=0$. Let $k\to +\infty$. Then $\gamma_n(\{(v-Y)^->0\})=0$, which implies $v\geq Y$ a.e. in $\Omega^{\sharp}$. However, if $p\geq 2$, \begin{align*} \|v\|_{L^\infty(\Omega^{\sharp})} &\geq\|Y\|_{L^\infty(\Omega^{\sharp})}=Y^\star(0)\\ &=\int_0^{\gamma_n(\Omega)} \Big(\sqrt{2\pi}\exp\Big(\frac{\Phi^{-1}(\tau)^{2}}{2}\Big)\Big)^{p'} \tau^{\frac{1}{p-1}} d\tau\\ &\geq C \int_0^{\gamma_n(\Omega)} \Big(\frac{1}{\tau(1-\log \tau)^{1/2}}\Big)^{p'} \tau^{\frac{1}{p-1}} d\tau\\ &= C \int_0^{\gamma_n(\Omega)} (1-\log \tau)^{-\frac{p'}{2}} \tau^{-1} d\tau=+\infty. \end{align*} This contradicts with $v\in L^{\infty}(\Omega^\sharp)$. As $p\geq 2$, problem \eqref{P4} has no solution. \subsection*{Acknowledgements} This research was supported by the National Natural Science Foundation of China (Grant No. 11501333), by the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11426146), by the Promotive Research Fund for Excellent Young and Middle-Aged Scientists of Shandong Province (Grant No. BS2012SF026), by the Natural Science Foundation of Shandong Province (Grant No. ZR2015AL015) and by the Doctoral Fund of University of Jinan (Grant No. XBS1337). \begin{thebibliography}{99} \bibitem{Cianchi} A. Alberico, A. Cianchi; \emph{Optimal summability of solutions to nonlinear elliptic problems}, Nonlinear Analysis, 67 (2007), 1775-1790. \bibitem{Alvino1} A. Alvino, P. L. Lions, G. Trombetti; \emph{Comparison results for elliptic and parabolic equations via Schwarcz symmetrization}, Ann. Inst. H. Pincar\'{e} Anal. Non Lin\'{e}aire., 7 (1990), 37-65. \bibitem{Betta1} M. F. 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