\documentclass[reqno]{amsart} \usepackage{amsfonts} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 58, pp. 1--19.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2003 Southwest Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/58\hfil Positive solutions for polyharmonic problems] {Existence of positive solutions for some polyharmonic nonlinear boundary-value problems} \author[H. M\^{a}agli, F. Toumi, \& M. Zribi\hfil EJDE--2003/58\hfilneg] {Habib M\^{a}agli, Faten Toumi, \& Malek Zribi} \address{Habib M\^{a}agli \hfill\break D\'{e}partement de Math\'{e}matiques, Facult\'{e} des Sciences de Tunis, Campus universitaire, 1060 Tunis, Tunisia} \email{habib.maagli@fst.rnu.tn} \address{Faten Toumi \hfill\break D\'{e}partement de Math\'{e}matiques, Facult\'{e} des Sciences de Tunis, Campus universitaire, 1060 Tunis, Tunisia} \email{toumifeten@yahoo.fr} \address{Malek Zribi \hfill\break D\'{e}partement de Math\'{e}matiques, Facult\'{e} des Sciences de Tunis, Campus universitaire, 1060 Tunis, Tunisia} \email{malek.Zribi@insat.rnu.tn} \date{} \thanks{Submitted April 2, 2003. Published May 20, 2003.} \subjclass[2000]{34B27, 35J40} \keywords{Green function, positive solution, Schauder fixed point theorem, \hfill\break% \indent singular polyharmonic elliptic equation} \begin{abstract} We present existence results for the polyharmonic nonlinear elliptic boundary-value problem \begin{gather*} (-\Delta )^m u=f(\cdot,u) \quad \hbox{in }B \\ (\frac{\partial }{\partial \nu })^j u=0\quad \hbox{on }\partial B, \quad 0\leq j\leq m-1. \end{gather*} (in the sense of distributions), where $B$ is the unit ball in $\mathbb{R}^n$ and $n\geq 2$. The nonlinearity $f(x,t)$ satisfies appropriate conditions related to a Kato class of functions $K_{m,n}$. Our approach is based on estimates for the polyharmonic Green function with zero Dirichlet boundary conditions and on the Schauder fixed point theorem. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Boggio \cite{b2} gave an explicit expression for the Green function $G_{m,n}$ of $(-\Delta )^m $ on the unit ball $B$ of $\mathbb{R}^n $ $(n\geq 2) $, with Dirichlet boundary conditions $(\frac{\partial }{\partial \nu }% )^{j}u=0 $, $0\leq j\leq m-1$. In fact, he proved that for each $x,y$ in $B$, \begin{equation} \label{e1.1} G_{m,n}(x,y)=k_{m,n}{| x-y| }^{2m-n}\int_{1}^{\frac{[x,y] }{| x-y| }} \frac{(v^{2}-1)^{m-1}}{v^{n-1}}\,dv \end{equation} where $\frac{\partial }{\partial \nu }$ is the outward normal derivative, $m$ is a positive integer, $k_{m,n}$ is a positive constant and $[x,y]^{2}=| {x-y% }|^{2}+(1-| x|^{2})(1-| y|^{2})$, for $x,y$ in $B$. Hence, from its expression, it is clear that $G_{m,n}$ is positive in $B^{2}$, which does not hold for the Green function of the biharmonic or $m$-polyharmonic operator in an arbitrary bounded domain (see for example \cite{g1}). Only for the case $m=1$, we have not this restriction. Grunau and Sweers \cite{g2} derived from Boggio's formula some interesting estimates on the Green function $G_{m,n}$ in $B$, including a 3G-Theorem, which holds in the case $m=1$ for the Green function $G_{\Omega }$ of an arbitrary bounded $C^{1,1}$-domain $\Omega $ (see \cite{c1} and \cite{z3}). When $m=1$, the 3G-Theorem has been exploited to introduce the classical Kato class of functions $K_{n}(\Omega )$, which was used in the study of some nonlinear differential equations (see \cite{m4,z2}). Definition and properties of the class $K_{n}(\Omega )$ can be found in \cite{a1,c1}. Recently, Bachar et al \cite{b1} improved the inequalities of Grunau and Sweers \cite{g2} satisfied by $G_{m,n}$ in $B$. For instance, they gave a new form of the 3G-Theorem (see inequality \eqref{e1.2} below and its proof in the Appendix). \begin{theorem}[3G-theorem] \label{thm3G} There exists $C_{m,n}>0$ such that for each $x,y,z\in B$, we have \begin{equation} \label{e1.2} \frac{G_{m,n}(x,z)G_{m,n}(z,y)}{G_{m,n}(x,y)} \leq C_{m,n}\big[ \big( \frac{\delta (z)}{\delta (x)}\big)^m G_{m,n}(x,z) +\big( \frac{\delta (z)}{\delta (y)}\big)^m G_{m,n}(y,z)\big], \end{equation} where $\delta (x)=1-| x|$. \end{theorem} When $m=1$, this new form of the 3G-Theorem has been proved for the Green function $G_{\Omega }$ in an arbitrary bounded $C^{1,1}$-domain $\Omega $, by Kalton and Verbritsky \cite{k1} for $n\geq 3$ and by Selmi \cite{s1} for $% n=2$. In \cite{b1}, the authors used this 3G-Theorem to define and study a new Kato class of functions on $B$ denoted by $K_{m,n}:=K_{m,n}(B) $ (see Definition \ref{def1} below). In the case $m=1$, this class was introduced for a bounded $C^{1,1}$-domain $\Omega $ in $\mathbb{R}^n $, in \cite{m5} for $n\geq 3$ and in \cite{m2} and \cite{z1} for $n=2$. Moreover, it has been shown that $K_{1,n}( \Omega )$ contains properly the classical Kato class $K_{n}( \Omega )$. \begin{definition} \label{def1} \rm A Borel measurable function $\varphi $ defined on $B$ belongs to the class $K_{m,n}$ if $\varphi $ satisfies the condition \begin{equation} \label{e1.3} \lim_{\alpha \to 0}\Big( \sup_{x\in B}\int_{B\cap B(x,\alpha )} \big( \frac{\delta (y)}{\delta (x)}\big)^m G_{m,n}(x,y)|\varphi (y)|dy\Big) =0. \end{equation} \end{definition} The properties of the class $K_{m,n}$ were used in \cite{b1}, to study a singular nonlinear differential polyharmonic equation \begin{equation*} ( -\Delta )^m u+\varphi ( .,u) =0,\quad \text{in }B\backslash \{ 0\} , \end{equation*} with boundary conditions $(\frac{\partial }{\partial \nu })^{j}u=0$ on $% \partial B$, $0\leq j\leq m-1$. The function $\varphi $ satisfies $| \varphi ( x,t) | \leq tq( x,t)$, where $q$ is a nonnegative Borel measurable function in $B\times (0,\infty ) $ which is required to satisfy some other hypotheses related to the class $K_{m,n}$. The plan for this paper is as follows: In Section 2, we recall some estimates on the Green function $G_{m,n}$ and some properties of functions belonging to the Kato class $K_{m,n}( B)$. In section 3, we study the polyharmonic boundary-value problem \begin{equation} \label{P} \begin{gathered} ( -\Delta )^m u=f( \cdot,u)\quad \text{ in $B$ (in the sense of distributions)} \\ ( \frac{\partial }{\partial \nu })^{j}u=0\quad \text{on }\partial B \quad 0\leq j\leq m-1. \end{gathered} \end{equation} The function $f$ satisfies the following hypotheses: \begin{itemize} \item[(H1)] The function $f$ is a nonnegative Borel measurable function on $% B\times ( 0,\infty )$, which is continuous and non-increasing with respect to the second variable. \item[(H2)] For each $c>0$, the function $x\to \frac{f( x,c( \delta ( x) )^m ) }{( \delta( x) )^{m-1}}$ is in $K_{m,n}$. \item[(H3)] For each $c>0$, $f( .,c) $ is positive on a set of positive measure. \end{itemize} To study problem (P), we assume $m\geq n\geq 2$. So we show that for $% G_{m,n} $ there exists $C>0$ such that for each $x,y\in B$, \begin{equation*} \frac{1}{C}( \delta ( x) )^m G_{m,n}(0,y)\leq G_{m,n}(x,y)\leq CG_{m,n}(0,y), \end{equation*} which is a fundamental inequality. Then by similar techniques to those used by Masmoudi and Zribi \cite{m6}, we prove that \eqref{P} has a positive continuous solution $u$ satisfying $a( \delta ( x) )^m \leq u(x)\leq b(\delta ( x) )^{m-1}$, where $a,b$ are positive constants. Note that for $m=1$, using the complete maximum principle argument, which does not hold for $m\geq 2$, M\^{a}agli and Zribi \cite{m4} established an existence and an uniqueness result for the problem \eqref{P} in a bounded $% C^{1,1}$ domain $\Omega $ of $\mathbb{R}^n$ ($n\geq 3$), where the function $% f$ is required to satisfy the hypotheses (H1), (H3), and \begin{itemize} \item[(H0)] For each $c>0$, $f( .,c)$ is in $K_{n}(\Omega )$. \end{itemize} In section 4, we shall study the following nonlinear polyharmonic problem in $B$, where $m\geq 1,n\geq 2$, \begin{equation} \label{Q} \begin{gathered} ( -\Delta )^m u=g( .,u) \quad \text{in $B$ (in the sense of distributions)} \\ (\frac{\partial }{\partial \nu })^{j} u=0,\quad \text{ on} \partial B,\quad 0\leq j\leq m-1\,. \end{gathered} \end{equation} We Assume that $g$ verifies the following hypotheses: \begin{itemize} \item[(H4)] The function $g$ is nonnegative Borel measurable function on $% B\times ( 0,\infty )$, and is continuous with respect to the second variable. \item[(H5)] There exist $p,q:B\to ( 0,\infty )$ nontrivial Borel measurable functions and $h,k:( 0,\infty )\to [ 0,\infty )$ nontrivial and nondecreasing Borel measurable functions satisfying \begin{equation*} p( x) h( t) \leq g( x,t) \leq q(x) k( t), \end{equation*} for $( x,t) \in B\times (0,\infty )$, such that \item[(A1)] $p\in L_{\mathrm{loc}}^{1}( B)$. \item[(A2)] The function $\theta ( x) :=q(x)/( \delta ( x) )^{m-1}$ is in $% K_{m,n}$. \item[(A3)] $\lim_{t\to 0^{+}} h( t)/t=+\infty$. \item[(A4)] $\lim_{t\to +\infty } k(t)/t=0$. \end{itemize} Under these hypotheses, we will prove that \eqref{Q} has a positive continuous solution $u$ satisfying $\ a( \delta ( x) ) ^m \leq u(x)\leq b( \delta ( x) )^{m-1}$, where $a,b$ are positive constants. This result is a follow up to the one of Dalmasso \cite{d1}, who studied the problem \eqref{Q} with more restrictive conditions on the function $g$. Indeed, he assumed that $g$ is nondecreasing with respect to the second variable and satisfies \begin{equation*} \lim_{t\to 0^{+}}\min_{x\in \overline{B}}\frac{g( x,t) }{t} =+\infty \quad \text{and}\quad \lim_{t\to +\infty }\max_{x\in \overline{B}} \frac{g(x,t) }{t% }=0. \end{equation*} He proved the existence of positive solution and he gave also an uniqueness result for positive radial solution when $g( x,t) =g(|x|,t)$. On the other hand, we note that when $m=1$, Brezis and Kamin \cite{b3} proved the existence and the uniqueness of a positive solution for the problem \begin{gather*} -\Delta u=\rho (x)u^{\alpha }\quad \text{in }\mathbb{R}^n , \\ \liminf_{| x| \to \infty }u(x)=0, \end{gather*} with $0<\alpha <1$ and $\rho $ is a nonnegative measurable function satisfying some appropriate conditions. To simplify our statements, we define the following convenient notations: \begin{itemize} \item $B=\{x\in \mathbb{R}^n : | x| <1\}$ with $n\geq 2$. \item $s\wedge t=\min (s,t)$ and $s\vee t=\max (s,t)$, for $s,t\in \mathbb{R}$. \item $C_{0}(B)=\{ w\in C( B):\lim_{| x|\to 1}w(x)=0\}$ \item For $x,y\in B$, we define: $[ x,y]^{2}=| x-y|^{2}+(1-| x|^{2})(1-|y|^{2})$, $\delta (x)=1-| x|$, and $\theta (x,y)=[x,y]^{2}-| x-y|^{2}=(1-| x|^{2})(1-| y|^{2})$. \newline Note that $[x,y]^{2}\geq 1+| x|^{2}| y|^{2}-2| x| | y| =(1-| x| | y|)^{2}$. So that \begin{equation} \label{e1.4} \delta (x)\vee \delta (y)\leq \lbrack x,y]. \end{equation} \item Let $f$ and $g$ be two positive functions on a set $S$. We call $% f\sim g$, if there is $c>0$ such that \begin{equation*} \frac{1}{c}g( x) \leq f( x) \leq cg( x), \quad \text{for all }x\in S. \end{equation*} We call $f\preceq g$, if there is $c>0$ such that \begin{equation*} f( x) \leq cg( x), \quad \text{for all }x\in S. \end{equation*} \end{itemize} The following properties will be used several times.\newline For $s,t\geq 0$, we have \begin{gather} s\wedge t\sim \frac{st}{s+t}, \label{e1.5} \\ (s+t)^{p}\sim s^{p}+t^{p},\quad p\in \mathbb{R}^{+}. \label{e1.6} \end{gather} Let $\lambda ,\mu >0$ and $0<\gamma \leq 1$, then we have, \begin{gather} 1-t^{\lambda }\sim 1-t^{\mu },\quad \text{for }t\in [ 0,1], \label{e1.7} \\ \log( 1+t) \preceq t^{\gamma },\quad \text{for }t\geq 0, \label{e1.8} \\ \log( 1+\lambda t) \sim \log( 1+\mu t) ,\quad \text{for }t\geq 0, \label{e1.9} \\ \log( 1+t^{\lambda }) \sim t^{\lambda }\log( 2+t),\quad\text{for }t\in [0,1]. \label{e1.10} \end{gather} On $B^{2}$ (that is $( x,y) \in B^{2}$), we have \begin{gather} \theta (x,y)\sim \delta (x)\delta (y), \label{e1.11} \\ [ x,y]^{2}\sim | x-y|^{2}+\delta (x)\delta (y)\,. \label{e1.12} \end{gather} \section{Properties of the Green function and Kato class} For this paper to be self contained, we shall recall some results concerning the Green function $G_{m,n}(x,y)$ and the class $K_{m,n}$. The next result is due to Grunau and Sweers in \cite{g2}. \begin{proposition} \label{prop1} On $B^{2}$, we have the following statements: \begin{enumerate} \item For $2mn$, \[ G_{m,n}(x,y)\sim (\delta (x)\delta (y))^{m-\frac{n}{2}} \Big(1\wedge \frac{(\delta (x)\delta (y))^{n/2}}{{| x-y| }^n }\Big). \] \end{enumerate} \end{proposition} \begin{corollary} \label{coro1} On $B^{2}$, we have \begin{enumerate} \item If $2mn$, \[ G_{m,n}(x,y)\sim \frac{(\delta (x)\delta (y))^m }{( {| x-y| }% ^{2}+(\delta (x)\delta (y)))^{n/2}} \sim \frac{(\delta (x)\delta (y))^m }{[x,y]^n }. \] \end{enumerate} \end{corollary} The proof of this corollary follows immediately from Proposition \ref{prop1} and the statements \eqref{e1.5}--\eqref{e1.7} and \eqref{e1.9}--\eqref{e1.12}. \begin{corollary} \label{coro2} For each $x,y\in B$ such that $| x-y| \geq r$, we have \begin{equation} \label{e2.1} G_{m,n}(x,y)\preceq \frac{(\delta (x)\delta (y))^m }{r^n }. \end{equation} \noindent Moreover, on $B^{2}$ we have \begin{gather} (\delta (x)\delta (y))^m \preceq G_{m,n}(x,y), \label{e2.2}\\ (\delta (x))^m \wedge (\delta (y))^m ,\text{ if }m\geq n. \label{e2.3} \end{gather} \end{corollary} The assertions of this corollary are obviously obtained by using the estimates in Corollary \ref{coro1} and the inequalities \eqref{e1.4} and $| x-y| \leq [ x,y] \preceq 1$. Now we recall some properties of functions belonging to the class $K_{m,n}$. \begin{lemma} \label{lm1} Let $\varphi$ be a function in $K_{m,n}$. Then the function $x\to ( \delta ( x) )^{2m}\varphi (x)$ is in $L^{1}( B)$. \end{lemma} \begin{proof} Let $\varphi \in K_{m,n}$, then by \eqref{e1.3} there exists $\alpha >0$ such that for each $x\in B$, \[ \int_{B( x,\alpha ) \cap B}\big( \frac{\delta (y)}{\delta (x)} \big)^m G_{m,n}(x,y)| \varphi (y)| dy\leq 1. \] Let $x_{1},\dots ,x_{p}$ in $B$ such that $B\subset \cup_{1\leq i\leq p} B( x_{i},\alpha ) $. Then by \eqref{e2.2}, there exists $C>0$ such that for all $\ i\in \{ 1,\dots p\} $ and $y\in B(x_{i},\alpha ) \cap B$, we have \[ ( \delta (y))^{2m}\leq C( \frac{\delta (y)}{\delta (x_{i})})^m G_{m,n}(x_{i},y). \] Hence, we have \begin{align*} \int_{B}( \delta (y))^{2m}| \varphi (y)| dy &\leq C \sum_{1\leq i\leq p} \int_{B( x_{i},\alpha ) \cap B}\big( \frac{\delta (y)}{\delta (x_{i})}\big)^m G_{m,n}(x_{i},y)| \varphi (y)| dy \\ &\leq Cp<\infty . \end{align*} This completes the proof. \end{proof} Throughout the paper, we will use the notation \begin{equation*} \| \varphi \|_{B}:=\sup_{x\in B}\int_{B}( \frac{\delta (y)}{\delta (x)})^m G_{m,n}(x,y)| \varphi (y)| dy, \end{equation*} for a measurable function $\varphi $ on $B$. \begin{proposition} \label{prop2} Let $\varphi $ be a function in $K_{m,n}$, then $\|\varphi \|_{B}<\infty $. \end{proposition} \begin{proof} Let $\varphi \in K_{m,n}$ and $\alpha >0$. Then we have \begin{align*} \int_{B}\big( \frac{\delta (y)}{\delta (x)}\big)^m G_{m,n}(x,y)|\varphi (y)| dy &\leq \int_{B\cap B( x,\alpha ) }\big( \frac{\delta (y)}{\delta (x)}\big)^m G_{m,n}(x,y)| \varphi (y)| dy \\ &\quad +\int_{B\cap B^{c}( x,\alpha ) }\big( \frac{\delta (y)}{\delta (x)}\big)^m G_{m,n}(x,y)| \varphi (y)| dy. \end{align*} Now, by \eqref{e2.1}, we have \[ \int_{B\cap B^{c}( x,\alpha ) }( \frac{\delta (y)}{\delta (x)% })^m G_{m,n}(x,y)| \varphi (y)| dy\preceq \frac{1}{\alpha ^n }\int_{B}( \delta (y))^{2m}| \varphi (y)| dy, \] then the result follows from \eqref{e1.3} and Lemma \ref{lm1}. \end{proof} The next result is due to Bachar et al \cite{b1}. Since reference \cite{b1} is not available, we have chosen to reproduce it here. \begin{proposition} \label{prop3} There exists a constant $C>0$ such that for all $\varphi \in K_{m,n}$ and $h$ a nonnegative harmonic function in $B$, we have \begin{equation} \label{e2.4} \int_{B}G_{m,n}(x,y)( \delta (y))^{m-1}h(y)| \varphi (y)| dy\leq C\| \varphi \|_{B}( \delta (x))^{m-1}h(x), \end{equation} for all $x$ in $B$. \end{proposition} \begin{proof} Let $h$ be a nonnegative harmonic function in $B$. So by Herglotz representation theorem \cite[p, 29]{h1}, there exists a nonnegative measure $\mu $ on $\partial B$ such that \[ h(y)=\int_{\partial B}P(y,\xi )\mu ( d\xi ) , \] where $P(y,\xi )=\frac{1-| y|^{2}}{| y-\xi |^n }, $ for $y\in B$ and $\xi \in \partial B$. So we need only to verify (2.4) for $h(y)=P(y,\xi )$ uniformly in $\xi \in \partial B$. From expression \eqref{e1.1} of $G_{m,n}$, it is clear that for each $x,y\in B$, we have \[ G_{m,n}(x,y)\sim \frac{(\theta (x,y))^m }{[{x,y]}^n }( 1+o(1-| y|^{2}) ) . \] Hence for $x,y,z$ in $B$, \[ \frac{G_{m,n}(y,z)}{G_{m,n}(x,z)}=\frac{( 1-| y|^{2}) ^m [{x,z]}^n }{( 1-| x|^{2})^m [y,z{]}^n }( 1+o( 1-| z|^{2}) ) , \] which implies \begin{equation} \label{e2.5} \underset{z\to \xi }{\lim }\frac{G_{m,n}(y,z)}{G_{m,n}(x,z)} =\frac{( 1-| y|^{2})^m }{( 1-| x|^{2})^m }\frac{| x-\xi |^n }{| y-\xi |^n }% \sim \big( \frac{\delta (y)}{\delta (x)}\big)^{m-1} \frac{P(y,\xi )}{P(x,\xi )}. \end{equation} Thus by Fatou's lemma and \eqref{e1.2}, we deduce that \begin{align*} &\int_{B}G_{m,n}(x,y)( \frac{\delta (y)}{\delta (x)})^{m-1} \frac{P(y,\xi )}{P(x,\xi )}| \varphi (y)| dy \\ &\preceq \liminf_{z\to \xi } \int_{B}G_{m,n}(x,y)\frac{G_{m,n}(y,z)}{G_{m,n}(x,z)}| \varphi (y)| dy \\ &\preceq \sup_{x\in B}\int_{B}( \frac{\delta (y)}{\delta (x)})^m G_{m,n}(x,y)| \varphi (y)| dy=\| \varphi\|_{B}. \end{align*} Which completes the proof. \end{proof} For a nonnegative measurable function $\varphi $ on $B$ and $x\in B$, we define \begin{equation*} V\varphi ( x) =\int_{B}( \delta (y))^{m-1}G_{m,n}(x,y)\varphi (y)dy. \end{equation*} \begin{corollary} \label{coro3} Let $\varphi \in K_{m,n}$. Then we have \begin{equation} \label{e2.6} \| V\varphi \|_{\infty }<\infty . \end{equation} Moreover, the function $x \mapsto ( \delta ( x) )^{2m-1}\varphi ( x) $\ is in $L^{1}( B)$. \end{corollary} \begin{proof} Put $h\equiv 1$ in (2.4) and using Proposition \ref{prop2}, we get \eqref{e2.6}. On the other hand, by \eqref{e2.2}, it follows that \[ \int_{B}( \delta (y))^{2m-1}| \varphi (y)| dy\preceq \int_{B}G_{m,n}(0,y)( \delta (y))^{m-1}| \varphi (y)| dy. \] Hence the result follows from \eqref{e2.6}. \end{proof} \begin{example} \label{ex1} \rm If $n\geq 2m$, for $p>\frac{n}{2m}$ we have $L^{p}(B)\subset K_{m,n}$. Furthermore, if $n<2m$ then for $p>1$ we have \[ \frac{1}{( \delta (.) )^{2m-n}}L^{p}(B)\subset K_{m,n}. \] Indeed, these inclusions are obtained by using the estimates on Corollary \ref{coro1}, \eqref{e1.4} and the H\"{o}lder inequality. \end{example} \begin{example} \label{ex2} \rm Let $\rho $ be the function defined in $B$ by $\rho ( x) =\frac{1}{\delta ( x)^{\lambda }}$. Then shown in \cite{b1}, $\rho \in K_{m,n}$ if and only if $\lambda <2m$ and we have the following estimates for $V\rho $ in $B$, \begin{enumerate} \item $\delta ( x)^m \preceq V\rho ( x) \preceq \delta ( x)^{3m-\lambda -1}$, if $2m-1<\lambda<2m $. \item $\delta ( x)^m \preceq V\rho ( x) \preceq \delta ( x)^m \log( \frac{2}{\delta ( x)})$, if $\lambda =2m-1$. \item $V\rho ( x) \sim \delta ( x)^m $, if $\lambda <2m-1$. \end{enumerate} \end{example} The properties in Propositions \ref{prop4} and \ref{prop5} below are useful for our existence results. However, to establish them we need the next key Lemma. \begin{lemma} \label{lm2} Let $x_{0}\in \overline{B}$, then for each $\varphi \in K_{m,n}$, \begin{equation} \lim_{\alpha \to 0}\big( \sup_{x\in B}\int_{B\cap B(x_{0},\alpha )}\big( \frac{\delta (y)}{\delta (x)}\big) ^m G_{m,n}(x,y)|\varphi (y)|dy\big) =0\,. \end{equation} Also for a positive harmonic function $h$ in $B$, we have \begin{equation} \lim_{\alpha \to 0}\big( \sup_{x\in B}\int_{B\cap B(x_{0},\alpha )}\big(\frac{\delta (y)}{\delta (x)}\big)^{m-1}\frac{h(y) }{h(x)}G_{m,n}(x,y)|\varphi (y)|dy\big) =0. \end{equation} \end{lemma} \begin{proof} Let $\varepsilon >0$, then by \eqref{e1.3}, there exists $r>0$ such that \[ \sup_{z\in B}\int_{B\cap B(z,r)}\big( \frac{\delta ( y) }{\delta ( z) }\big)^m G_{m,n}(z,y)|\varphi (y)|dy\leq \varepsilon \] Let $x_{0}\in \overline{B}$ and $\alpha >0$. Then by \eqref{e2.1} we have for each $x\in B$, \begin{align*} &\int_{B\cap B(x_{0},\alpha )}( \frac{\delta ( y) }{\delta ( x) })^m G_{m,n}(x,y)|\varphi (y)|dy \\ &\leq \int_{B\cap B(x,r)}\big( \frac{\delta ( y) }{\delta ( x) }\big)^m G_{m,n}(x,y)|\varphi (y)|dy\\ &\quad +\int_{B\cap B(x_{0},\alpha )\cap B^{c}(x,r)}\big( \frac{\delta ( y) }{\delta ( x) }\big)^m G_{m,n}(x,y)|\varphi (y)|dy \\ &\preceq \varepsilon +\int_{B\cap B(x_{0},\alpha )}( \delta ( y) )^{2m}|\varphi (y)|dy. \end{align*} Hence, using Lemma \ref{lm1} and letting $\alpha \to 0$, claim (2.7) follows. Now to prove (2.8), using again Herglotz representation theorem, we need only to verify the assertion for $h(y)=P(y,\xi )$ uniformly in $\xi \in \partial B$, where $P(y,\xi )=\frac{1-| y|^{2}}{|y-\xi |^n }$, for $y\in B$ and $\xi \in \partial B$. Let $x\in B$, then by Fatou's Lemma and (2.5), we deduce that \begin{align*} &\int_{B\cap B(x_{0},\alpha )}\big( \frac{\delta (y)}{\delta (x)} \big)^{m-1}\frac{P(y,\xi )}{P(x,\xi )}G_{m,n}(x,y)|\varphi (y)|dy \\ &\preceq \liminf_{z\to \xi } \int_{B\cap B(x_{0},\alpha )}G_{m,n}(x,y) \frac{G_{m,n}(y,z)}{G_{m,n}(x,z)}|\varphi (y)|dy \\ &\preceq \sup_{x\in B}\int_{B\cap B(x_{0},\alpha )} \big( \frac{\delta (y)}{\delta (x)}\big)^m G_{m,n}(x,y)|\varphi (y)|dy, \end{align*} Then by (2.7), we get (2.8) when $\alpha \to 0$. \end{proof} \begin{proposition} \label{prop4} Let $\varphi \in K_{m,n}$. Then the following function is in $C_{0}( B)$, \[ v(x):=\frac{1}{( \delta (x))^{m-1}}V\varphi (x)\,. \] \end{proposition} \begin{proof} Let $x_{0}\in B$ and $\alpha >0$. Let $x,z\in B\cap B(x_{0},\alpha )$, then \begin{align*} | v(x)-v(z)| &\leq \int_{B}\big| \frac{G_{m,n}(x,y)}{( \delta (x))^{m-1}} -\frac{G_{m,n}(z,y)}{( \delta(z))^{m-1}}\big| ( \delta (y))^{m-1}|\varphi (y)|dy \\ &\leq 2\sup_{\xi \in B}\int_{B\cap B(x_{0},2\alpha )} \big( \frac{\delta (y)}{\delta (\xi )}\big)^{m-1}G_{m,n}(\xi ,y)|\varphi (y)|dy \\ &\quad +\int_{B\cap B^{c}(x_{0},2\alpha )}\big| \frac{G_{m,n}(x,y)}{% ( \delta (x))^{m-1}}-\frac{G_{m,n}(z,y)}{( \delta (z))^{m-1}}\big| ( \delta (y))^{m-1}|\varphi (y)|dy. \end{align*} If $| x_{0}-y| \geq 2\alpha $ then $| x-y| \geq \alpha $ and $| z-y| \geq \alpha $. Moreover, by \eqref{e2.1} for all $x\in B\cap B(x_{0},\alpha )$ and $y\in \Omega :=B\cap B^{c}(x_{0},2\alpha )$, we have \[ \big( \frac{\delta (y)}{\delta (x)}\big)^{m-1}G_{m,n}(x,y)\preceq (\delta ( y) )^{2m-1}. \] Since when $y\in \Omega $, the function $x\to \frac{G_{m,n}(x,y)}{( \delta ( x) )^{m-1}}$ is continuous in $B\cap B(x_{0},\alpha )$, then by (2.8), Corollary \ref{coro3} and the dominated convergence theorem, we obtain that \[ \int_{B}| \frac{G_{m,n}(x,y)}{( \delta (x))^{m-1}} -\frac{G_{m,n}(z,y)}{( \delta (z))^{m-1}}| ( \delta(y))^{m-1}|\varphi (y)|dy\to 0 \] as $| x-z| \to 0$. Hence, we deduce that $v$ is continuous in $B$. Next, we show that $v(x)\to 0$ as $\delta (x) \to 0$. Let $x_{0}\in \partial B$, $\alpha >0$ and $x\in B(x_{0},\alpha )$, then \begin{align*} | v(x)| &\leq \int_{B\cap B(x_{0},2\alpha )} \big(\frac{\delta (y)}{\delta (x)}\big)^{m-1}G_{m,n}(x,y)|\varphi (y)|dy \\ &\quad+\int_{B\cap B^{c}(x_{0},2\alpha )} \big( \frac{\delta (y)}{\delta (x)}\big)^{m-1}G_{m,n}(x,y)|\varphi (y)|dy. \end{align*} Since $\lim_{\delta ( x) \to 0}\frac{G_{m,n}(x,y)}{( \delta (x))^{m-1}}=0$, as in the above argument, we get $\lim_{x\to x_{0}}v(x)=0$. Hence $v\in C_{0}( B)$. \end{proof} For a nonnegative function $\rho $ in $K_{m,n}$, we define \begin{equation*} M_{\rho }:=\{ \varphi \in K_{m,n} :| \varphi | \preceq \rho \} . \end{equation*} By similar arguments as in the proof of the above Proposition, we can prove the following statement. \begin{proposition} \label{prop5} For any nonnegative function $\rho \in K_{m,n}$, the family of functions $\{ V\varphi :\varphi \in M_{\rho }\}$ is relatively compact in $C_{0}( B)$. \end{proposition} \section{First existence result} In this section, we consider the case $m\geq n\geq 2$ to study problem % \eqref{P}. The main result that we shall prove is the following. \begin{theorem} \label{thm1} Assume (H1)--(H3). Then the problem \eqref{P} has a positive continuous solution $u$. Moreover, there exist two positive constants $a$ and $b$ such that for each $x\in B$, \[ a( \delta ( x) )^m \leq u(x)\leq b( \delta ( x) )^{m-1}\,. \] \end{theorem} To prove this theorem, we state an existence result for the following boundary-value problem (in the sense of distributions) \begin{equation} \label{Pl} \begin{gathered} ( -\Delta )^m u=f( .,u) \quad \text{in }B \\ u=\lambda \quad \text{ on }\partial B, \\ \big( \frac{\partial }{\partial \nu }\big)^{j}u=0,\quad \text{on }\partial B,\quad 1\leq j\leq m-1. \end{gathered} \end{equation} where $\lambda >0$. For the next theorem we need the hypothesis \begin{itemize} \item[(H2')] For each $c>0$, the function $x\to \frac{f( x,c)}{( \delta ( x) )^{m-1}}$ is in $K_{m,n}$. \end{itemize} Note that hypothesis (H2) implies (H2'). \begin{proposition} \label{prop6} Suppose that $f$ satisfies (H1), (H3), and (H2'). Then for each $\lambda >0$, problem \eqref{Pl} has a positive solution $u_{\lambda }\in C( \overline{B}) $, such that for each $x\in B$, \[ u_{\lambda }(x)=\lambda +\int_{B}G_{m,n}(x,y)f(y,u_{\lambda }(y))dy. \] \end{proposition} \begin{proof} Let $\lambda >0$. Then by (H2'), the function $\rho (y):=\frac{f( y,\lambda ) }{( \delta ( y))^{m-1}}\in K_{m,n}$ and so by Corollary \ref{coro3}, we have $\beta :=\lambda +\| V\rho \|_{\infty }<\infty$. Let $Y$ be the convex set given by \[ Y=\left\{ u\in C( \overline{B}) :\lambda \leq u\leq \beta \right\} . \] We consider the integral operator $T$ on $Y$, defined by \[ Tu(x)=\lambda +\int_{B}G_{m,n}(x,y)f(y,u(y))dy. \] We shall prove that $T$ has a fixed point in $Y$. Since for $u\in Y$ and $y\in B$, by (H1) we have \[ \frac{f(y,u(y))}{( \delta ( y) )^{m-1}}\leq \frac{% f(y,\lambda )}{( \delta ( y) )^{m-1}}=\rho (y), \] then using (H2'), we deduce that the function $y\to \frac{f(y,u(y))}{( \delta ( y) )^{m-1}}$ is in $M_{\rho }$. So from Proposition \ref{prop5}, we deduce that $TY$ is relatively compact in $C( \overline{B}) $. In particular, for all $u\in Y$, $Tu\in C( \overline{B}) $ and so it is clear that $TY\subset Y$. Now, we aim to prove the continuity of $T$ in $Y$. Let $(u_{k})_{k}$ be a sequence in $Y$ which converges uniformly to $u\in Y$. Then since $f$ is continuous with respect to the second variable, we deduce by the dominated convergence theorem that \[ \forall x\in B,\text{ }Tu_{k}(x)\to Tu(x)\quad \text{as }k\to \infty . \] As $TY$ is relatively compact in $C( \overline{B}) $, then \[ \| Tu_{k}-Tu\|_{\infty }\to 0\quad \text{as }k\to \infty . \] Thus we have proved that $T$ is a compact mapping from $Y$ to itself. Hence, by Schauder fixed point theorem, there exists a function $u_{\lambda }\in Y$ such that \[ u_{\lambda }( x) =\lambda +\int_{B}G_{m,n}(x,y)f(y,u_{\lambda}(y))dy. \] Finally, we need to verify that $u_{\lambda }$ is a solution for problem \eqref{Pl}. Since by (H1) we have for each $y\in B$, $f(y,u_{\lambda }(y))\leq f(y,\lambda )=( \delta ( y) )^{m-1}\rho ( y) $, then we deduce from Corollary \ref{coro3} that the function $y\to f(y,u_{\lambda }(y))$ is in $L_{\rm loc}^{1}(B) $. So it is clear that $u_{\lambda }$ satisfies (in the sense of distributions) the elliptic differential equation \[ (-\Delta )^m u_{\lambda }=f(.,u_{\lambda })\quad\text{in }B. \] Furthermore, by (H2'), we have \[ 0\leq \frac{u_{\lambda }( x) -\lambda }{( \delta ( x) )^{m-1}}\leq \frac{1}{( \delta ( x) ) ^{m-1}}V\rho (x). \] This implies from Proposition \ref{prop4} that $\lim_{\delta ( x) \to 0}\frac{u_{\lambda }( x) -\lambda }{( \delta( x) )^{m-1}}=0$. Namely, $u_{\lambda }$ satisfies the boundary conditions $u_{\lambda }=\lambda $ and $( \frac{\partial }{\partial \nu })^{j}u_{\lambda }=0$, on $\partial B$ for $1\leq j\leq m-1$. This ends the proof. \end{proof} In the sequel, we consider a sequence $( \lambda_{k})_{k}$ of positive real numbers, decreasing to zero. We denote by $u_{k}$ the solution of the problem $(P_{\lambda_{k}})$ given by Proposition \ref{prop6} and satisfying for each $x\in B$, \begin{equation} \label{e3.1} u_{k}( x) =\lambda_{k}+\int_{B}G_{m,n}(x,y)f(y,u_{k}(y))dy. \end{equation} \begin{lemma} \label{lm3} There exists a positive constant $a$ such that for all $k\in \mathbb{N}$, and $x\in B$, $u_{k}( x) \geq a( \delta (x))^m .$ \end{lemma} \begin{proof} By \eqref{e2.2} and (2.3), we remark that on $B$, \[ G_{m,n}(0,y)\sim ( \delta ( y) )^m . \] Then\ by \eqref{e2.2} and (2.3) again, we deduce that there exists a constant $c>1$ such that we have for each $x,y\in B$ \[ \frac{1}{c}( \delta ( x) )^m G_{m,n}(0,y)\leq G_{m,n}(x,y)\leq cG_{m,n}(0,y). \] This implies by (3.1) that \begin{equation} u_{k}( x) \leq c\big( \lambda _{k}+\int_{B}G_{m,n}(0,y)f(y,u_{k}(y))dy\big) =cu_{k}( 0) . \end{equation} and \begin{align*} u_{k}( x) &\geq \frac{1}{c}\big( \delta ( x) ) ^m ( \lambda_{k}+\int_{B}G_{m,n}(0,y)f(y,u_{k}(y))dy\big) \\ &\geq \frac{1}{c}\big( \delta ( x) )^m ( \inf_{k\in \mathbb{N}}u_{k}( 0) \big) . \end{align*} We claim that $a=\frac{1}{c}( \inf_{k\in \mathbb{N}}u_{k}(0) ) >0$. Assume on the contrary that there exists a subsequence $( u_{k_{p}}( 0) )_{p}$ which converges to zero. In particular, for $p$ large enough, we have $u_{k_{p}}( 0) \leq 1$, which implies with (3.3) and (H1) that \[ u_{k_{p}}( 0) =\lambda _{k_{p}}+\int_{B}G_{m,n}(0,y)f(y,u_{k_{p}}(y))dy \geq \lambda_{k_{p}}+\int_{B}G_{m,n}(0,y)f(y,c)dy. \] Thus, by letting $p$ to $\infty $, we reach a contradiction from hypothesis (H3). This completes the proof. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1}] Let $a$ be the constant given in Lemma \ref{lm3}, then by hypothesis (H2), we deduce that the function \[ \rho ( y) :=\frac{f(y,a( \delta ( y) ) ^m )}{( \delta ( y) )^{m-1}}\in K_{m,n}. \] Since for each $k\in \mathbb{N}$ and $y\in B$, by (H1) we have \[ \frac{f(y,u_{k}( y) )}{( \delta ( y) ) ^{m-1}}\leq \frac{f(y,a( \delta ( y) )^m )}{( \delta ( y) )^{m-1}}=\rho ( y) \,. \] Then the function $y\to \frac{f(y,u_{k}( y) )}{(\delta ( y) )^{m-1}}$ is in $M_{\rho }$. So using Proposition \ref{prop5}, we deduce from (3.2) that the family $( u_{k})_{k}$ is relatively compact in $C(\overline{B})$. Then it follows that there exists a subsequence $( u_{k_{p}})_{p}$ which converges uniformly to a function $u\in C(\overline{B})$. Moreover, by Lemma \ref{lm3}, we have $u(x)\geq a( \delta ( x) )^m $, for each $x\in B$. Hence, using the continuity of $f$ with respect to the second variable, we apply the dominated convergence theorem in (3.2) to obtain that \[ u(x)=\int_{B}G_{m,n}(x,y)f(y,u(y))dy. \] Finally, by Lemma \ref{lm3} and hypothesis (H1), for each $y\in B$, we have \[ f(y,u(y))\leq f(y,a( \delta ( y) )^m )=( \delta ( y) )^{m-1}\rho ( y) . \] Then we deduce from Corollary \ref{coro3} that the function $y\to f(y,u(y))$ is in $L_{\rm loc}^{1}( B) $. So $u$ satisfies (in the sense of distributions) the elliptic differential equation \[ (-\Delta )^m u=f(.,u)\quad\text{in }B. \] Furthermore, we have for $x\in B$, \[ a\delta ( x) \leq \frac{u(x)}{( \delta ( x) )^{m-1}}\leq \frac{1}{( \delta (x))^{m-1}}V\rho ( x) , \] which together with Proposition \ref{prop4} imply that $u$ satisfies the boundary conditions $( \frac{\partial }{\partial \nu })^{j}u=0$, on $\partial B$, for $0\leq j\leq m-1$ and that there exists a positive constant $b$ such that \[ a( \delta ( x) )^m \leq u(x)\leq b( \delta ( x) )^{m-1}. \] This completes the proof. \end{proof} \begin{corollary} \label{coro4} Let $\varphi \in C(\partial B)$ and $\psi \in C^{1}(\partial B)$ be nonnegative functions on $\partial B$ and $f$ satisfies (H1)--(H3), then the polyharmonic boundary-value problem \begin{equation} \label{P0} \begin{gathered} ( -\Delta )^m u=f( .,u) \quad \text{ in $B$ (in the sense of distributions)}, \\ ( -\frac{\partial }{\partial \nu })^{m-1}u=\psi ,\quad (-\frac{\partial }{\partial \nu })^{m-2}u=\varphi ,\quad (\frac{\partial }{\partial \nu })^{j}u=0\quad \text{on }\partial B\quad \text{for }0\leq j\leq m-3, \end{gathered} \end{equation} has a positive continuous solution $u$. Moreover there exists a positive constant $a$ such that \[ u(x)\geq a( \delta ( x) )^m . \] \end{corollary} \begin{proof} Let $h$ be the solution of the Dirichlet problem \begin{gather*} ( -\Delta )^m h=0\quad \text{in }B \\ ( -\frac{\partial }{\partial \nu })^{m-1}h=\psi ,\quad (-\frac{\partial }{\partial \nu })^{m-2}h=\varphi ,\quad ( \frac{\partial }{\partial \nu })^{j}h=0,\quad \text{on }\partial B, \text{ for }0\leq j\leq m-3. \end{gather*} Then as in \cite{g3}, for $x\in B$ we have \[ h(x)=\int_{\partial B}K_{m,n}(x,y)\varphi ( y) d\omega (y)+\int_{\partial B}L_{m,n}(x,y)\psi ( y) d\omega (y), \] where \begin{gather*} L_{m,n}(x,y)=\frac{1}{2^m ( m-2)! \omega_{n}} \frac{(1-| x|^{2})^m }{| x-y|^{n+2}}[n(1-| x|^{2})+( m+2-n) | x-y|^{2}],\\ K_{m,n}(x,y)=\frac{1}{2^{m-1}( m-1) !\omega_{n}}\frac{(1-| x|^{2})^m }{| x-y|^n } \end{gather*} for $x,y\in B$, and $\omega_{n}$ denotes the $( n-1) $ dimensional surface area of the unit ball. For $m\geq n\geq 2$, we have evidently $L_{m,n}>0$ and so $h$ is nonnegative on $B$. Using this fact, we can easily see that the function $f_{0}$\ defined on $B\times ( 0,\infty ) $ by \[ f_{0}( x,t) =f(x,t+h(x)) \] satisfies (H1)--(H3). Hence by Theorem \ref{thm1}, the problem \begin{gather*} ( -\Delta )^m v=f_{0}( .,v) \quad \text{in $B$ (in the sense of distributions)} \\ ( \frac{\partial }{\partial \nu })^{j}v=0,\quad \text{on }\partial B,\quad \text{for }0\leq j\leq m-1\,. \end{gather*} has a positive solution $v\in C_{0}( B)$ satisfying $v(x)\geq a( \delta ( x) )^m$, where $a$ is a positive constant. Let $u=v+h$. Then $u$ is the desired solution for the problem \eqref{P0}. This completes the proof. \end{proof} \begin{remark} \label{rmk2} \rm Let $f$ satisfy (H1), (H3), and \begin{itemize} \item[(H2'')] For each $c>0$, the function $x\to \frac{f( x,c( \delta ( x) ) ^m ) }{( \delta ( x) )^{m+n-1}}$ is in $K_{m,n}$. \end{itemize} Then problem \eqref{P} has a positive solution $u$ satisfying $u(x)\sim ( \delta ( x) )^m $. Indeed, we note that (H2'') implies (H2), so by Theorem \ref{thm1}, problem \eqref{P} has a positive solution satisfying that for each $x\in B$ \[ u(x)=\int_{B}G_{m,n}(x,y)f(y,u(y))dy \] and $u(x)\geq a( \delta ( x) )^m $. Now, if $m\geq n$, we have by Corollary \ref{coro1} that $G_{m,n}(x,y)\sim \frac{(\delta (x)\delta (y))^m }{[ x,y]^n }$, which by \eqref{e1.4} implies that \[ G_{m,n}(x,y)\preceq ( \delta ( x) )^m ( \delta ( y) )^{m-n}. \] Hence for each $x\in B$, we have \begin{equation} a( \delta ( x) )^m \leq u(x)\preceq ( \delta ( x) )^m \int_{B}( \delta ( y) ) ^{m-n}f(y,a( \delta ( y) )^m )dy. \end{equation} Since $f$ satisfies (H2''), we deduce by Corollary \ref{coro3}, that $u(x)\sim ( \delta ( x) )^m$. \end{remark} \begin{remark} \label{rmk3} \rm Let $\psi ( r,.) =\max_{| x| =r}f(x,.)$, for $r\in [ 0,1] $ and suppose that for all $c>0$, \begin{equation} \int_{0}^{1}r^{n-1}( 1-r)^{m-1}\psi ( r,c( 1-r)^m ) dr<\infty . \end{equation} Then the solution $u$ of \eqref{P} satisfies $u(x)\sim (\delta ( x) )^m$. Indeed, by Theorem \ref{thm1} and (H1), we have \begin{equation} a( \delta ( x) )^m \leq u(x)\leq \int_{B}G_{m,n}(x,y)f(y,a( \delta ( y) )^m )dy. \end{equation} On the other hand using \eqref{e1.1}, we have \[ G_{m,n}(x,y)\preceq | x-y|^{2m-n}\big( \frac{[ x,y] ^{2}}{| x-y|^{2}}-1\big)^{m-1}\int_{1}^{\frac{[ x,y% ] }{| x-y| }}\frac{dv}{v^{n-1}}. \] Now since $\frac{[ x,y]^{2}}{| x-y|^{2}}-1\sim \frac{\delta ( x) \delta ( y) }{| x-y|^{2}}$, we deduce that \[ G_{m,n}(x,y)\preceq ( \delta ( x) \delta ( y) )^{m-1}G_{1,n}(x,y). \] Hence it follows from (3.6) that \[ u(x)\preceq ( \delta ( x) )^{m-1}\int_{B}( \delta ( y) )^{m-1}G_{1,n}(x,y)\psi (| y|,a( \delta ( y) )^m )dy. \] By similar calculus as in \cite[p.538]{m4}, we have by (3.6) that for $x\in B$, \[ \int_{B}( \delta ( y) )^{m-1}G_{1,n}(x,y)\psi (|y| ,a( \delta ( y) )^m )dy \preceq \delta ( x) . \] This implies that $u(x)\sim ( \delta ( x) )^m $. \end{remark} \begin{example} \label{ex3} \rm Let $\alpha >0$ and $\lambda 0$ such that $Mk(b)\leq b$. On the other hand, by (A1) the function $p$ is a nontrivial nonnegative function in $L_{\rm loc}^{1}( B) $, then there exists $r\in (0,1)$ such that \[ 0<\int_{B( 0,r) }p(y)dy<\infty . \] Furthermore, from \eqref{e2.2} there exists $c>0$ such that for each $x,y\in B$ \[ G_{m,n}(x,y)\geq c( \delta ( x) )^m ( \delta ( y) )^m . \] Hence, since by (A3) we have $\lim_{t\to 0}\frac{h(t)}{t}=+\infty $, then there exists $a>0$ such that \[ c( 1-r)^m h( a( 1-r)^m ) \int_{B(0,r) }p(y)dy\geq a. \] Let $\Lambda $ be the convex set \[ \Lambda =\{ u\in C_{0}(B):a( \delta ( x) ) ^m \leq u(x)\leq b( \delta ( x) )^{m-1}\} \] and $T$ be the operator defined on $\Lambda $ by \[ Tu(x)=\int_{B}G_{m,n}(x,y)g(y,u(y))dy. \] We shall prove that $T$ has a fixed point. We first note that for $u\in \Lambda $ and $y\in B$, we have by (H5) \[ \frac{g(y,u(y))}{( \delta ( y) )^{m-1}}\leq \frac{q(y)k(u(y))}{( \delta ( y) )^{m-1}} \leq k(b)\frac{q(y)}{( \delta ( y))^{m-1}}:=k(b)\theta (y). \] Then we deduce that the function $y\to \frac{g(y,u(y))}{( \delta ( y) )^{m-1}}\in M_{\theta }$. Thus by Proposition \ref{prop5}, we obtain that the family $T\Lambda $ is relatively compact in $C_{0}(B)$ We need now to verify that for $u\in \Lambda $, we have \[ a( \delta ( x) )^m \leq Tu(x)\leq b( \delta ( x) )^{m-1}. \] Let $u\in \Lambda $ and $x\in B$, then by (H5), we have \begin{align*} Tu(x) &\leq \int_{B}G_{m,n}(x,y)q(y)k(u(y)) \\ &\leq ( \delta ( x) )^{m-1}\Big[ k(b)\int_{B}\big( \frac{\delta (y)}{\delta (x)}\big)^{m-1}G_{m,n}(x,y)\theta ( y) dy\Big] \\ &\leq Mk(b)( \delta ( x) )^{m-1} \\ &\leq b( \delta ( x) )^{m-1}. \end{align*} On the other hand from (H5) and \eqref{e2.2}, we have \begin{align*} Tu(x) &\geq c( \delta ( x) )^m \int_{B}( \delta ( y) )^m p(y)h(u(y))dy \\ &\geq ( \delta ( x) )^m \Big[ c( 1-r) ^m h( a( 1-r)^m ) \int_{B(0,r)}p(y)dy\Big] \\ &\geq a( \delta ( x) )^m . \end{align*} Thus we have proved that $T\Lambda \subset \Lambda$. Now we aim to prove the continuity of $T$ in $\Lambda $. We consider a sequence $( u_{k})_{k}$ in $\Lambda $ which converges uniformly to $u$ in $\Lambda $. Then since $g$ is continuous with respect to the second variable, we deduce by the dominated convergence theorem that for all $x\in B$, \[ Tu_{k}(x)\to Tu(x)\quad \text{as }k\to \infty . \] Since $T\Lambda $ is relatively compact in $C_{0}(B)$, we have the uniform convergence. Hence $T$ is a compact mapping from $\Lambda $ to itself. Then by the Schauder fixed point theorem, we deduce that there exists a function $u\in \Lambda $ such that \[ u(x)=\int_{B}G_{m,n}(x,y)g(y,u(y))dy. \] So $u$ satisfies (in the sense of distributions) the elliptic differential equation \[ (-\Delta )^m u=g(.,u)\text{ }in\text{ }B. \] Moreover, since $u$ satisfies \[ a( \delta ( x) ) \leq \frac{u(x)}{( \delta ( x) )^{m-1}}\preceq \frac{1}{( \delta ( x) )^{m-1}}V\theta ( x) , \] we deduce by Proposition \ref{prop4} that $\lim_{\delta ( x)\to 0}\frac{u(x)}{( \delta ( x) )^{m-1}}=0$ and so $u$ satisfies the boundary conditions $( \frac{\partial }{\partial \nu })^{j}u=0$, on $\partial B$ for $0\leq j\leq m-1$. This completes the proof. \end{proof} \begin{example} \label{ex4} \rm Let $\lambda n$, from Proposition \ref{prop1}, we have \[ A(x,y)\sim (| {x-y}|^{2}\vee (\delta (x)\delta (y)))^{1/2}. \] Then the result holds by similar arguments as in case 1. 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