\documentclass[twoside]{article} \usepackage {amsfonts} \pagestyle{myheadings} \markboth{ Generalized boundary value problems } { Laurent V\'eron } \begin{document} \setcounter{page}{313} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize USA-Chile Workshop on Nonlinear Analysis, \newline Electron. J. Diff. Eqns., Conf. 06, 2001, pp. 313--342. \newline http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Generalized boundary value problems for nonlinear elliptic equations % \thanks{ {\em Mathematics Subject Classifications:} 35J60, 35J67, 35D05. \hfil\break\indent {\em Key words:} Dirichlet Problem, Radon and Borel measures. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Published January 8, 2001. } } \date{} \author{ Laurent V\'eron } \maketitle \begin{abstract} We give here an overview of some recent developments in the study of the description of all the positive solutions of \begin{equation} -\Delta u+|u|^{q-1}u=0 \label{NLE} \end{equation} in a domain $\Omega$ where $q>1$. \end{abstract} \newtheorem{theorem}{Theorem} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} \newcommand{\abs}[1]{{\left | #1 \right| }} \newcommand{\norm}[1]{{\| #1 \|}} \section{Introduction} Given a partial differential equation in a domain $\Omega$ of ${\mathbb R}^{N}$, a natural question is to find a way to describe all its solutions by means of their possible boundary value. For example, if $\Omega$ is smooth, any nonnegative harmonic function $u$ in $\Omega$ admits a boundary trace which is a Radon measure $\mu$ on $\partial\Omega$ and the Riesz-Herglotz representation formula holds, \begin{equation} u(x)=\int_{\partial\Omega}P(x,y)d\mu(y) \end{equation} for any $x\in \Omega$, where $P(x,y)$ is the Poisson kernel in $\Omega$. If $\Omega$ is not smooth, the Poisson kernel is replaced by the Martin kernel and a representation formula holds. This article is concerned with is the description of the positive solutions of (\ref {NLE}) and this study is known as the nonlinear trace theory. A description as in the linear case is still far out of reach, but thank to the works of Le Gall \cite {LG1,LG2,LG3}, Marcus and V\'eron \cite {MV1,MV2,MV3,MV4,MV5}, Dynkin and Kuzneztsov \cite {DK1,DK2,DK3,DK4,DK5} this program is now well advanced. In this survey I want to present the historical progression which led to the nonlinear trace problem, and in particular present the preliminary works of Gmira and V\'eron (1989-1991) \cite {GmV} dealing with the measure boundary data and the isolated boundary singularities, the question of existence and uniqueness of the large solutions in particular in non-smooth domains and the role of the Borel measures and the boundary capacities framework for representing all the positive solutions. We shall also give applications to conformal deformations of hyperbolic space. The associated boundary value problem (or generalised Dirichlet problem) is the following, \begin{equation} \begin{array}{c} -\Delta u+|u|^{q-1}u=0,\mbox { in } \Omega , \cr u=g\mbox { on } \partial \Omega , \end{array} \label{NLDP}\end{equation} where $g$ may be a function (regular or not, a Radon measure, a generalized Borel measure). \section{The regular non-linear Dirichlet problem in regular domains} By using convex analysis, $L^{p}$ and Schauder's regularity theory of elliptic equations, it is classical that for any $g$ in $C^{2,\alpha}(\partial\Omega)$ there exists a unique $u$ in $C^{2,\alpha}(\bar\Omega)$ solution of (\ref {NLDP}). The extension to merely integrable boundary data is due to Brezis \cite {Br} \begin{theorem}Assume $g\in L^{1}(\partial \Omega)$, then there exists a unique function $u\in L^{1}(\Omega)\cap L^{q}(\Omega,\rho_{\Omega}dx)$ , where $\rho_{\Omega}(x)=\mathop{\rm dist}(x,\partial \Omega$ such that \begin{equation} \int_{\Omega}\left(-u\Delta \zeta+|u|^{q-1}u\zeta\right )dx =-\int_{\partial \Omega}\frac{\partial \zeta}{\partial \nu}gdH_{N-1} \label{integral formulation} \end{equation} $\forall \zeta\in C_{0}^{1,1}(\bar\Omega)$. Moreover the mapping $g\mapsto u=P_{\Omega}^q(g)$ is increasing and \begin{equation} {\norm {u_{1}-u_{2}}}_{L^{1}(\Omega)} + {\norm {\rho_{\Omega}\left (h(u_{1})-h(u_{2})\right )}}_{L^{1}(\Omega)} \leq C{\norm {g_{1}-g_{2}}}_{L^{1}(\partial \Omega)} \label{stability 1} \end{equation} where $u_{j}=P_{\Omega}^q(g_{j})$, $j=1,\,2$ and $h(r)=|r|^{q-1}r$. \end{theorem} In this statement, $dH_{N-1}$ denotes the $(N-1)$-dimensional Hausdorff measure. Notice also that $\zeta\in C_{0}^{1,1}(\bar\Omega)\Rightarrow \abs{ \zeta (x)}\leq C\rho_{\Omega}(x)$, which gives its meaning to the condition $u\in L^{q}(\Omega,\rho_{\Omega}dx)$. The key point in Brezis proof is the following result \begin{lemma}For any $(f,g)\in L^{1}(\Omega,\rho_{\Omega}dx)\times L^{1}(\partial \Omega)$ there exists a unique $v\in L^{1}( \Omega)$ such that \begin{equation} -\int_{\Omega}v\Delta \zeta dx =\int_{\Omega}f\zeta dx-\int_{\partial \Omega}\frac{\partial \zeta}{\partial \nu}gdH_{N-1} \label{linear integral formulation } \end{equation} $\forall \zeta\in C_{0}^{1,1}(\bar\Omega)$. Moreover, if $\zeta\geq 0$, \begin{equation} -\int_{\Omega}|v|\Delta \zeta dx +\int_{\partial \Omega}\frac{\partial \zeta}{\partial \nu}{\abs g} dH_{N-1} \leq \int_{\Omega}f\zeta sign (v)dx \label{linear integral formulation 1} \end{equation} and the same estimate holds if one replaces $\abs v$ by $v_{+}$. \end{lemma} \section{The a priori estimate of Keller and Osserman} One of the most striking aspects of the equation (\ref {NLE}) is the fact that all the solutions are locally uniformly bounded. More precisely, by using suitable local radial supersolutions of (\ref {NLE}), Keller \cite {Ke} and Osserman \cite {Os} proved independently that there always holds \begin{equation} \abs {u(x)}\leq C(N,q)\rho_{\Omega}(x)^{2/(1-q)}. \label{KO estimate} \end{equation} One of the important consequence of this inequality consists in the construction of positive solution of (\ref {NLDP}) with a boundary value $g$ belonging to $C(\partial \Omega;[0,\infty])$. %%THEOREM%% \begin{theorem} Let $\Omega$ be a Lipschitz bounded domain. Then for each $g$ in \\ $C(\partial \Omega;[0,\infty])$, there exists at least one solution of (\ref {NLDP}). \end{theorem} \paragraph{Proof.} For any positive integer $k$ let $u_{k}$ be the solution of \[ \begin{array}{c} -\Delta u_{k}+{\abs u_{k}}^{q-1}u_{k}=0,\mbox { in } \Omega , \cr u_{k}=g_{k}=\min (k,g)\mbox { on } \partial \Omega , \end{array} \] By the maximum principle, the sequence $\{u_{k}\}$ is positive and increasing. Moreover it is locally bounded thanks to (\ref {KO estimate}). Therefore it converges to some solution $u$ of (\ref {NLE}). If $x_{0}\in \partial\Omega $ is such that $g(x_{0})<\infty$, then the continuity of $g$ and the elliptic equations boundary regularity theory imply that the set of functions $\{u_{k}\}$ remains equicontinuous in some neighborhood of $x_{0}$. Consequently the limit function $u$ achieves the value of $g$ in a relative neighborhood of $x_{0}$ on $\partial\Omega $. If $g(x_{0})=\infty$, then $\{u_{k}(x_{0})\}$ is not bounded from above. The Lipschitz regularity assumption can be relaxed and replaced by the Wiener regularity criterion. The problem of uniqueness remains open up to 1993 when Kondratiev and Nikishkin \cite {KN} discovered that there may exist infinitely many solutions in the framework of boundary data in the class of continuous functions with possibly infinite value. This will be completely understood in the framework of the boundary trace theory. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Measure boundary data} In this section $\Omega$ is a smooth bounded domain. The Poisson formula which expresses the Poisson potential of a boundary Radon measure is the following \begin{equation} P_{\mu}(x)=\int_{\Omega}P(x,y)d\mu(y) \label{Poisson formula} \end{equation} The first extension of Theorem 1 to measure boundary data, is due to Gmira and V\'eron \cite {GmV} for the existence part (1989-1991) and Marcus and V\'eron \cite {MV4} for the stability part (1996). \begin{theorem}Suppose that $10$ such that $$C^{-1}\rho(x){\abs {x-y}}^{-N}\leq P(x,y)\leq C\rho(x){\abs {x-y}}^{-N}$$ for all $(x,y)\in\Omega\times\partial\Omega$. Consequently $$\displaylines{ {\norm {P(.,y)}}_{L^p(\Omega)}\leq K_{p,\Omega}\quad \forall 12$). This treatment involves the notion of removable sets which are associated to Bessel capacities. It will be presented in Section 9. \paragraph{Remark.}A more elaborated analytic tool (weighted Marcinkiewicz spaces) allowed Gmira and V\'eron \cite{GmV} to prove an existence and uniqueness result for the general problem \begin{equation} \begin{array}{c} -\Delta u+g(u)=0,\mbox { in } \Omega , \cr u=\mu \mbox { on } \partial \Omega , \end{array} \label{NLDP 2}\end{equation} where $g$ is continuous and non-decreasing, $\mu\in {\mathcal M}(\partial\Omega)$, and $$\int_{\Omega}{\abs {g(P_{\abs \mu})}}\rho (x)dx<\infty.$$ By a solution we mean an integrable function $u$ defined in $\Omega$ such that $g(u)\in L^{1}(\Omega;\rho dx)$ which satisfies, for every $\zeta\in C^{1,1}_{0}(\bar \Omega)$, \begin{equation} \int_{\Omega}\left(-u\Delta \zeta+g(u)\zeta\right )dx =-\int_{\partial \Omega}\frac{\partial \zeta}{\partial \nu}\zeta d\mu. \label{integral formulation 3} \end{equation} The above integrability condition appears to be more general, however in the particular case where $g(r)=r|r|^{q-1}$, it may not be satisfied even with $\mu=hdx\in L^{1}(\partial\Omega)$ when $q\geq (N+1)/(N-1)$. It is then natural to isolate the singular part of the measure by writing the Lebesgue decomposition $$\mu=\mu_{S}+\mu_{R}$$ where $\mu_{R}$ is the regular part (with respect to the (N-1)-Hausdorff measure), and $\mu_{S}$, the singular one. If $g$ satisfies a $\Delta_{2}$-condition, $${\abs g(r+r')}\leq \theta ({\abs g(r)}+{\abs g(r')})$$ for some $\theta>0$, whenever $rr'\geq 0$, Marcus and V\'eron \cite {Ve3}(1996) proved the existence of a solution (always unique) under the mere condition $$\int_{\Omega}{\abs g(P_{\abs \mu_{S}})}\rho (x)dx<\infty.$$ The case of the exponential is different. For example if $g(r)=e^{2r}$ the existence of a solution of (\ref{NLDP 2}) is insured if, for some $p\in (1,\infty]$, there holds \begin{equation} \begin{array}{c} \exp(2P_{\mu_{S}})\in L^{p/(p-1)}(\partial \Omega),\\ \exp(2\mu_{R})\in L^p(\partial \Omega).\end{array} \end{equation} This was obtained by Grillot and V\'eron in 1997 \cite {GrV}. \section{Isolated singularities} As we have seen it above, if $10$ the function $u_{k,a}=P^q_{\Omega}(k\delta_{a}$ is a solution of (\ref {NLE}) which vanishes on $\partial\Omega\setminus \{a\}$. Moreover, when $k$ increases, it is the same with $u_{k,a}$. From the Osserman-Keller estimate, this sequence is locally uniformly bounded in $\Omega$ , therefore it converges to some positive solution $u_{{\infty,a}}$ of (\ref {NLE}). By using some local estimate on the boundary it can be checked that $\{u_{k,a}\}$ is equicontinuous on any compact subset of $\Omega$. Therefore $u_{{\infty,a}}$ vanishes on $\partial\Omega\setminus \{a\}$. This function $u_{{\infty,a}}$ is a solution of (\ref{NLE}) has the maximal blow-up rate at $a$ among the solutions vanishing on $\partial\Omega\setminus \{a\}$. As for the behaviour of $u_{{k,a}}$ near $a$ it can be obtained from perturbation theory, since the blow-up estimate on Poisson's kernel yields to $$0\leq P^q_{\Omega}(k\delta_{a})\leq P_{k\delta_{a}}= kP_{\delta_{a}}=kP(x,a),$$ that is $$ 0\leq u_{{k,a}}\leq Ck {\abs {x-a}}^{-N}\rho_{\Omega}$$ Finally it is possible to prove that the non-linear term is negligible near $a$ in some sense and \begin{equation} \lim_{x\to a}\frac{u_{{k,a}}}{P(x,a)}=k. \label{weak singularity} \end{equation} Always in the range $10$ such that $u=P^q_{\Omega}(h+k\delta_{a})$ and $u\approx u_{k,a}$ in the sense of (\ref {weak singularity}); (iii) or $u=P^q_{\Omega}(h+\infty\delta_{a}) =\lim_{k\to\infty}u=P^q_{\Omega}(h+k\delta_{a})$ and $u\approx u_{\infty,a}$ in the sense of (\ref {strong singularity}). \end{theorem} \paragraph{Remark.}It is always possible to assume that $u$ vanishes on the boundary except the point $a$. Actually, if it is not the case, it follows by the maximum principle that there exists a solution $\tilde u$ of (\ref {NLE}) vanishing on $\partial \Omega \setminus \{a\}$ and such that $$\abs {u(x)-\tilde u (x)}\leq \sup_{y\in\partial\Omega}\abs {h(y)},\qquad \forall x\in\Omega.$$ \paragraph{Remark.}The above classification can be extended to changing sign solutions when $(N+2)/N\leq q<(N+1)/(N-1)$: in case (ii) there is no sign restriction on $k$, and in case (iii), we have either $u\approx u_{\infty,a}$ or $u\approx -u_{\infty,a}$. \paragraph{Sketch of the proof of Theorem 4.} The full proof is highly technical, therefore we shall restrict ourselves to the case where $\partial\Omega$ is flat near $a$. By scaling and translating it can be supposed that $a=0$ and $\partial\Omega \supset T_{0}\partial\Omega\cap \{x:\,\abs x\leq 1\}$ We shall not impose the positivity of $u$ in order to see at what level this condition versus $(N+2)/N\leq q<(N+1)/(N-1)$ appears. Let $(r,\sigma)\in {\mathbb R}_{+}\times S^{N-1}$ be the spherical coordinates, then $u$ solution of (\ref{NLE}) satisfies \begin{equation} \partial _{rr}u+\frac{N-1}{r}\partial _{r}u+\frac{1}{r^{2}}\Delta_{S^{N-1}}u=|u|^{q-1}u \label{NLE 2} \end{equation} in $(0,1]\times S^{N-1}_{+}$, and vanishes on $(0,1]\times \partial S^{N-1}_{+}$. Setting $t=\ln (1/r)$ and $w=r^{2/(q-1)}u$, then $w$ is bounded and it satisfies \begin{equation} \partial _{tt}w+\beta_{q,N}\partial _{t}w+\Delta_{S^{N-1}}w+\alpha_{q,N}w={\abs w}^{q-1}w \label{NLE 3} \end{equation} in $[0,\infty)\times S^{N-1}_{+}$, where $$\beta_{q,N}=2\frac{q+1}{q-1}-N \mbox { and } \alpha_{q,N}=\frac {2}{q-1}\left (\frac {2q}{q-1}-N\right). $$ Since $w$ is bounded in $[0,\infty)\times S^{N-1}_{+}$ and vanishes on $[0,\infty)\times \partial S^{N-1}_{+}$, it follows from the regularity theory of linear elliptic equations that $\nabla_{S} ^{\alpha}\partial ^{\beta}_{t}w$ remains uniformly bounded for any $\abs {\alpha}+\abs {\beta}\leq 3$ (here $\nabla_{S}$ is the covariant derivative symbol). Multiplying (\ref {NLE 3}) by $\partial _{t}w$ gives \begin{equation} \beta_{q,N}\int_{S^{N-1}_{+}}(\partial _{t}w)^{2}d\sigma = \frac{dE}{dt} \label{energy variation} \end{equation} with $$E=\int_{S^{N-1}_{+}}\left(\frac{1}{2} {\abs {\nabla_{S}w}}^{2} +\frac{1}{q+1}{\abs {w}}^{q+1}- \frac{\alpha_{q,N}}{2}{\abs {w}}^{2}-\frac{1}{2}(\partial _{t}w)^{2}\right)d\sigma $$ But $\beta_{q,N}\neq 0$ since $q\neq (N+2)/(N-2)$. Since $E$ is bounded, \begin{equation} \int_{0}^{\infty}\int_{S^{N-1}_{+}}(\partial _{t}w)^{2}d\sigma dt<\infty. \label{energy estimate} \end{equation} By differentiating (\ref {NLE 3}) with respect to $t$ and multiplying by $\partial _{tt}w$, there also holds \begin{equation} \int_{0}^{\infty}\int_{S^{N-1}_{+}}(\partial _{tt}w)^{2}d\sigma dt<\infty, \label{energy estimate 2} \end{equation} and by using the previous regularity estimates and (\ref {energy estimate}), (\ref {energy estimate 2}), it follows \begin{equation}\lim_{t\to\infty}\partial _{t}w= \lim_{t\to\infty}\partial _{tt}w=0, \label{limits} \end{equation} in $L^{2}(S^{N-1}_{+})$ and actually uniformly on $\bar S^{N-1}_{+}$. Consequently, the $\omega$-limit set of the trajectory ${\mathcal T}=\bigcup _{t\geq 0}\{w(t,.)\}$ is included into a non-empty, compact and connected component of the set $$ {\mathcal S}=\left\{\varphi\in C^{2}(S^{N-1}_{+}):\; \Delta_{S^{N-1}}\varphi+\alpha_{q,N}\varphi= {\abs \varphi}^{q-1}\varphi\mbox { in }S^{N-1}_{+},\; \varphi =0 \mbox { on }\partial S^{N-1}_{+}\right\} $$ {\bf 1-} If $\alpha_{q,N}\leq N-1=\lambda_{1}(S^{N-1}_{+})$, which is equivalent to $q\geq (N+1)/(N-1)$, ${\mathcal S}$ is reduced to the zero function. This is easily done by multiplying by $\varphi$, and integrating over $S^{N-1}_{+}$. \noindent {\bf 2-} If $2N=\lambda_{2}(S^{N-1}_{+})\leq \alpha_{q,N}<\lambda_{1}$, which is equivalent to $(N+2)/N\leq q<(N+1)/(N-1)$, the set ${\mathcal S}$, besides the zero function, contains two nonzero elements, $\omega$ and $-\omega$, which keep constant sign (($\omega >0$ for example), and are rotationally invariant. \noindent {\bf 3-} If $10$ such that $$\abs {w(t,\sigma)}\leq Ce^{-\varepsilon t}$$ for $t\geq 0$. This type of estimate is not easy to obtain. It is obtained by contradiction and the fact that either $\alpha_{q,N}$ is not an eigenvalue of $-\Delta_{S}$ in $W^{1,2}_{0}(S^{N-1}_{+})$, or, if it is, the equivariance properties of $w$ are not compatible with the ones of the element of $\ker(-\Delta_{S}-\alpha_{q,N}I)$. From this follows an improved estimate of the following type $$\abs {w(t,\sigma)}\leq C \min (e^{-\theta \varepsilon t},e^{(N-1-2/(q-1))t})$$ where $\theta=\theta(N,q)>1$. This estimate is of linear type and easy to obtained by a representation formula. Since $\theta$, a final estimate of the type $$\abs {w(t,\sigma)}\leq C e^{(N-1-2/(q-1))t}$$ is derived in a finite number of iterations. The projection of $w$ onto $\ker(-\Delta_{S}-(N-1)I)$ yields to the existence of some $\varphi\in \ker(-\Delta_{S}-(N-1)I)$ such that $$\lim_{t\to\infty}e^{(2/(q-1)-N+1)t}w(t,.)=\varphi (.)$$ in the $C^{2}(S^{N-1}_{+}$-topology. Finally, if $\varphi =0$, comparison of $u$ with $\pm \varepsilon P(.,0)$ implies that $u$ is actually the zero-function. \paragraph{Remark.} If the assumptions on $q$ and the sign of $u$ are relaxed, the only thing which can be proved is that the limit set of the trajectory $\mathcal T$ is included into a connected component of the set $\mathcal S$. However, in such cases exept for the constant-sign solutions, and the zero function, the connected components of $\mathcal S$ are continuum. Therefore it is not possible to assert that $w$ converges precisely to a particular element. \section{Large solutions} Let $\Omega$ be a domain with a compact boundary. By a large solution of (\ref{NLE}) we mean a positive solution $u$ such that \begin{equation} \lim_{\rho(x)\to 0}u(x)=\infty. \label{boundary blow-up} \end{equation} Since it is classical to approximate $\Omega$ by an increasing sequence $\Omega_{n}$ of smooth subdomains, on each of them we can construct a positive solution $u_{n}$ of (\ref{NLE}) in $\Omega_{n}$ with infinite value on the boundary , we obtain a decreasing sequence of solutions of (\ref{NLE}), each of them dominating in $\Omega_{n}$ any solution $u$ of (\ref{NLE}). Therefore the sequence $\{u_{n}\}$ converges to the maximal solution $\bar u_{\Omega}$ of (\ref{NLE}) in $\Omega$ (this is essentially the results of Keller and Osserman). The questions which are raised are therefore \smallskip \noindent {\bf 1}- Existence of large solution or, equivalently, is the maximal solution a large solution? This question is linked to the regularity of $\partial\Omega$. \smallskip \noindent {\bf 2}- Uniqueness of the large solution.\\ The first question which is a problem linked to the regularity of $\partial\Omega $, and it has been solved in the case $q=2$ by Dhersin and Le Gall \cite {DLG} , by probabilistic methods. If $a\in\partial\Omega$, they proved that a Wiener-type criterion at the point $a$, associated to some Bessel capacity is a necessary and sufficient condition for the existence of a solution $u$ of (\ref{NLE}) such that $$\lim_{x\to a}u(x)=\infty.$$ Their methods heavily relies on probability theory and the question remains open to cover the full range of exponent $q>1$. The following easy to prove result is essentially due to Marcus and V\'eron \cite {MV1} (although not explicitely written in this reference, it clearly corresponds to their cone condition with dimension $0$). \begin{theorem}Suppose $10$ such that, for any $a\in\partial\Omega$, there exists a segment $I_{\varepsilon,a}\subset \Omega^c$ with length $\varepsilon$ and $a$ as one of its end-points. It is also proved in \cite {MV1} that if $q\geq N/(N-2)$, the blow-up rate of the maximal solution may be much weaker than the one given by (\ref{explicit solution}). \smallskip The first result of uniqueness of the large solution are due to Iscoe \cite{Is}, Bandle and Marcus \cite {BM1, BM2} and V\'eron \cite {Ve4}. The technique used by Bandle and Marcus or V\'eron needs a $C^{2}$-regularity of $\partial\Omega$ and relies on a precise blow-up estimate of any large solution $u$ at the boundary, namely \begin{equation} \lim _{\rho(x)\to 0}\rho^{2/(q-1)}(x)u(x)=C(q)>0 \label{precise blow-up} \end{equation} From this estimate and the maximum principle it follows $$u\leq (1+\varepsilon)\tilde u \quad \mbox { in } \Omega ,$$ if $u$ and $\tilde u$ are two maximal solutions, and this holds for any $\varepsilon>0$. Letting $\varepsilon\to 0$ and exchanging the role of $u$ and $\tilde u$ implies $u=\tilde u$. The result of Iscoe, only stated in the case $q=2$ gives uniqueness without the possible existence in the case where $\Omega$ is star-shapped with respect to some point, say 0. It use the equi-variant properties of equation (\ref{NLE})under the transformation $$u\mapsto u_{k}, \quad \mbox { where } u_{k} (x)=k^{2/(q-1)}u(kx), \quad \forall k>0.$$ If $u$ and $\tilde u$ are two large solutions in $\Omega$, then $u_{k}$ is a large solution in $\Omega_{k}=k^{-1}\Omega$. If $k>1$, $\Omega_{k}\subset\Omega$ and therefore $u_{k}\geq \tilde u$. Letting $k\to 1$ and exchanging again the role of $u$ and $\tilde u$ implies uniqueness. \smallskip In 1995 Marcus and V\'eron \cite {MV1} proved a uniqueness result extending Iscoe's one to a wide class of domains \begin{theorem}Suppose $\partial\Omega$ is locally the graph of a continuous function, then there exists at most one large solution of (\ref{NLE}) in $\Omega$. \end{theorem} The proof of this result is not easy. The key point is that the assumption on $\partial\Omega$ implies that any two large solutions $u$ and $\tilde u$ satisfy \begin{equation}\lim_{\rho(x) \to 0}\frac {u(x)}{\tilde u(x)}=1. \label{similar blow-up} \end{equation} Finally, in 1997 Marcus and V\'eron \cite{MV2} introduced a new technique for proving uniqueness of a wide class of solutions of (\ref{NLE}) with positively homogeneous boundary data. In the case of uniqueness of large solutions, their results is the following \begin{theorem}Suppose $11$ such that \begin{equation} K^{-1}\leq \frac{u(x)}{\bar u_{\Omega}(x)}\leq K,\quad \forall x\in \Omega. \label{quotient estimate} \end{equation} Since $\partial\Omega=\partial\bar\Omega^c$; for any $a\in\partial\Omega$, there exists a sequence $\{a_{n}\}\subset\bar\Omega^c$ such that $a_{n}\to a$ when n goes to infinity. Because $u$ blows-up on $\partial\Omega$ and the explicit solution $U_{a_{n}}$ (defined in (\ref {explicit solution})), is finite, $U_{a_{n}}\leq u$, and letting n go to infinity, $U_{a}\leq u$ in $\Omega$. If $x$ is any point in $\Omega$, let $a\in \partial\Omega$ such that $\rho_\Omega(x)=\abs {x-a}$. Then $$u(x)\geq U_{a}(x)=\alpha_{q,N}^{1/(q-1)}{\abs {x-a}}^{2/(1-q)}= \alpha_{q,N}\rho_\Omega^{2/(1-q)}(x).$$ But, by the Keller-Osserman estimate (\ref{KO estimate}), $$\bar u_{\Omega}(x)\leq C(q,N)\rho_\Omega^{2/(1-q)}(x).$$ Consequently (\ref {quotient estimate}) holds with $K(N,q)=C(q,N)/\alpha_{q,N}^{1/(q-1)})$. \smallskip \noindent{\bf Step 2} End of the proof. Set $$w=u-\frac {1}{2K}(\bar u-u).$$ Because of Step 1, $$\left (\frac {1}{2}+\frac {1}{2K}\right )u0$ near $a_{j}$ (we assume $N>2$, the cases $N=2,1$ needing only very minor modifications). However, it has been noticed by Chasseigne and Vazquez \cite {CV} that a large solution $u$ is unique if one imposes the strongest blow-up on $\partial\Omega$, that is $$\lim_{\rho (x)\to 0}\rho (x)^{N-2}u(x)=\infty.$$ \section{ The boundary trace} For simplicity of exposition, in the next sections, we shall restrict ourself to the case where $\Omega=B=\{x\in{\mathbb R}^{N}:\,\abs x<1\}$, although all the results we shall propose extend to a $C^{2}$ bounded domain. We recall that $(r,\sigma)$ are the spherical coordinates in $R^{N}$. It is classical, and easy to check, that every positive harmonic function $u$ in $B$ possesses a boundary trace given by a positive Radon measure $\mu\in{\mathcal M}(\partial B)$ which is attained in the sense of weak convergence of measures: \begin{equation} \lim_{r\to 1}\int_{S^{N-1}}u(r,\sigma)\zeta (\sigma)d\sigma = \int_{S^{N-1}}\zeta (\sigma)d\mu \label{linear trace} \end{equation} for every $\zeta\in C(S^{N-1})$ and $u=P_{{\mu}}$. In this writing, we identify $\partial B$ and $S^{N-1}$. The trace result is still valid if harmonic is replaced by super-harmonic \cite {Do} and a representation formula holds. Moreover, the positivity assumption on $u$ can be replaced by an integrability condition such as $\Delta u\in L^{1}(B;(1-r)dx)$ and in that case the boundary trace is a general Radon measure on $\partial B$. \smallskip The existence of a boundary trace for positive solutions of (\ref {NLE}) is due to Le Gall in 1993 \cite{LG1, LG2} when $q=2$. Actually, in this pioneering work, Le Gall gives a probabilistic representation of any positive solution of (\ref {NLE}) in this case. The notion of trace that we present is due to Marcus and V\'eron \cite {MV2}(1995); it is a purely analytic presentation and is extendible to a wider class of nonlinearities. \begin{theorem}Suppose $q>1$ and $u$ is a positive solution of (\ref {NLE}) in $B$. Then for any $\omega\in\partial B$ we have the following alternative. \smallskip \noindent (i) Either for every relatively open neighbourhood $U\subset \partial B$ of $\omega$ \begin{equation} \lim_{r\to 1}\int_{U}u(r,\sigma)d\sigma =\infty, \label{strong trace} \end{equation} \smallskip \noindent (ii) or there exists a relatively open neighbourhood $U\subset \partial B$ of $\omega$ such that for every $\zeta\in C^\infty (U)$ \begin{equation} \lim_{r\to 1}\int_{U}u(r,\sigma)\zeta (\sigma)d\sigma = \ell (\zeta), \label{weak trace} \end{equation} where $\ell$ is a positive linear functional on $C^\infty (U)$. \end{theorem} If $V$ is an open domain of $S^{N-1}$, we denote by $\varphi_{V}$ the first eigenfunction of $-\Delta $ in $W^{1,2}_{0}(V)$ with the normalization condition $$\max_{V}\varphi_{V}=1.$$ The corresponding eigenvalue is quoted $\lambda_{V}$ , and if the relative boundary of $V$ is $C^{2}$, the Hopf lemma applies with $\nu_{S}$ the normal outward unit vector to $V$ (tangent to $S^{N-1}$), $$\frac{\partial \varphi_{V}}{\partial \nu_{S}}<0 \quad \mbox { on } \partial V.$$ \begin{lemma}Let $V$ be an open domain of $S^{N-1}$ with a $C^{2}$ boundary, $u$ a positive solution of (\ref{NLE}) in $B$ and $\alpha>(N+1)/(N-1)$ . Then we have the following dichotomy. \\ (i) Either $$\int_{0}^{1}\int_{V}u^q\varphi_{V}^{\alpha}(1-r)r^{N}drd\sigma =\infty,$$ and in that case $$\lim_{r\to 1}\int_{V}(u\varphi_{V}^{\alpha})(r,\sigma)d\sigma =\infty,$$ (ii) or $$\int_{0}^{1}\int_{V}u^q\varphi_{V}^{\alpha}(1-r)r^{N}drd\sigma <\infty,$$ and in that case, for any $C^{2}$ function on $V$ which satisfies \begin{equation} \abs\zeta\leq k\varphi_{V}^{\alpha}\quad \mbox{ and } \abs{\Delta \zeta}\leq k\varphi_{V}^{\alpha -2}\label{inequalities for varphi} \end{equation} for some $k>0$, the following limit exists $$ \lim_{r\to 1}\int_{V}u(r,\sigma)\zeta d\sigma=\ell (\zeta),$$ and the mapping $\zeta\mapsto \ell (\zeta)$ is a positive linear functional defined on the subset of $C^{2}$ functions which satisfy (\ref {inequalities for varphi}). \end{lemma} \paragraph{Proof.} {\bf Step 1}. There holds $$I_{\alpha}=\int_{V}{\abs {\Delta_ S^{N-1}\varphi_{V}^{\alpha}}}^{q/(q-1)} \varphi_{V}^{-q\alpha/(q-1)}d\sigma<\infty. $$ From Hopf boundary Lemma $$\varphi_{V}(\sigma)\geq C_{1} \mathop{\rm dist_{_{S^{N-1}}}}(\sigma,\partial V)$$ for any $\sigma\in V$, where $\mathop{\rm dist_{_{S^{N-1}}}}$ is the geodesic distance on $S^{N-1}$ and $C_{1}>0$. Since $$\Delta_ S^{N-1}\varphi_{V}^{\alpha}= -\alpha\lambda_{V}\varphi_{V}^{\alpha}+\alpha (\alpha -1)\varphi_{V}^{\alpha -2}{\abs{\nabla \varphi_{V}}}^{2},$$ then $${\abs {\Delta_ S^{N-1}\varphi_{V}^{\alpha}}}^{q/(q-1)} \leq C_{2}\left(\mathop{\rm dist_{_{S^{N-1}}}}(.,\partial V)\right)^{q(\alpha -2)/(q-1)},$$ and finally $${\abs {\Delta_ S^{N-1}\varphi_{V}^{\alpha}}}^{q/(q-1)} \varphi_{V}^{-q\alpha/(q-1)}\leq C_{3}\left (\mathop{\rm dist_{_{S^{N-1}}}} (. ,\partial V) \right )^{q((\alpha -2)-\alpha)/(q-1)}.$$ But $\alpha>(N+1)/(N-1)$, and the claim follows follows. \smallskip \noindent {\bf Step 2} Reduction of the equation. We shall assume $N>2$ , the case $N=2$ requiring only minor technical modifications. We transform the equation (\ref {NLE 2}) satisfied by $u$ in polar coordinates by setting $$s=\frac{r^{N-2}}{N-2} \quad \mbox { and } u(r,\sigma)=r^{2-N}v(s,\sigma),$$ and we get \begin{equation}s^{2}\frac {\partial ^{2}v}{\partial s^{2}}+\frac {1}{(N-2)^{2}}\Delta_{S^{N-1}}v-Ks^{N/(N-2)-q}v^q=0, \label{NLE 4} \end{equation} in $(0,a)\times S^{N-1}$, where $K=K(N)>0$ and $a=(N-2)^{-1}$ , from which follows $$ s^{2}\frac {d ^{2}}{d s^{2}}\int_{V}v\varphi_{V }^\alpha d\sigma+\frac {1}{(N-2)^{2}} \int_{V}v\Delta_{S^{N-1}}\varphi_{V }^\alpha d\sigma -Ks^{N/(N-2)-q}\int_{V}v^q\varphi_{V }^\alpha d\sigma =0. $$ We set $$X(s)= \int_{V}v\varphi_{V }^\alpha d\sigma\quad \mbox { and } \quad Y(s)=\left (\int_{V}v^q\varphi_{V }^\alpha d\sigma\right ). $$ Since, by H\"older's inequality and Step 1, $$Y(s)\leq I_{\alpha}^{1-1/q}Y(s),$$ it infers \begin{equation}\begin{array}{l} (i)\quad s^{2}X''+JY-Ks^{N/(N-2)-q}Y^q\geq 0\\ (ii)\quad s^{2}X''-JY-Ks^{N/(N-2)-q}Y^q\leq 0 \end{array} \label{inequalities} \end{equation} where $J=I^{1-1/q}(N-2)^{-2}$. \smallskip \noindent {\bf Step 3} End of the proof. {\it Case 1}: suppose $\int_{0}^{1}\int_{V}u^q\varphi_{V}^{\alpha}(1-r)r^{N}drd\sigma =\infty$. Since (\ref {inequalities}-i) implies $$X''\geq AY^q-B$$ on $[a/2,a)$, where $A>0$ and $B>0$ do not depend on $s$, a double integration gives $$X(s)\geq X(a/2)+(s-a/2)X'(a/2)+A\int_{a/2}^s(s-\tau)Y^q(\tau)d\tau -B(s-a/2)^{2}/2,$$ and $\displaystyle {\lim_{s\to a}X(s)=\infty}$, which is assertion (i). \noindent{\it Case 2}: suppose $\int_{0}^{1}\int_{V}u^q\varphi_{V}^{\alpha}(1-r) r^{N}drd\sigma <\infty$. Then $\int_{a/2}^{a}(a-s)Y^qds<\infty$. Moreover $X''\leq AY^q+B$, and $$\frac{d^{2}}{ds^{2}}\left (X(s)+A\int_{a/2}^s(s-\tau)Y^q(\tau)d\tau -\frac{B}{2}(s-a/2)^{2}\right ) \leq 0, $$ which infers that $s\mapsto X(s)$ admits a finite limit at infinity. If $\zeta$ is a $C^{2}$-function satisfying (\ref {inequalities for varphi}), we set $X_{\zeta}(s)=\int_{V}v\zeta d\sigma$ and derive \begin{equation} s^{2}\frac {d ^{2}X_{\zeta}} {ds^{2}}+\frac {1}{(N-2)^{2}} \int_{V}v\Delta_{S^{N-1}}\zeta d\sigma -Ks^{N/(N-2)-q}\int_{V}v^q\zeta d\sigma =0\label{NLDI} \end{equation} from (\ref{NLE 4}). But $$\abs {\int_{V}v\Delta_{S^{N-1}}\zeta d\sigma}\leq k\left (\int_{V} v^q \varphi_{V}^{\alpha}d\sigma \right)^{1/q} \left (\int_{V}\varphi_{V}^{\alpha -2q/(q-1)}d\sigma \right)^{1-1/q} $$ and $$\abs {\int_{V}v^q\zeta d\sigma}\leq k\int_{V}v^q\varphi_{V}^{\alpha} d\sigma. $$Consequently $$\int_{a/2}^{a}\abs {\int_{V}v\Delta_{S^{N-1}}\zeta d\sigma}(a-s)ds<\infty \,,\quad \int_{a/2}^{a}\abs {\int_{V}v^q\zeta d\sigma}(a-s)ds<\infty.$$ Integrating twice the equality (\ref {NLDI}) implies that $s\mapsto X_{\zeta}(s)$ admits a finite limit at infinity and this limit depends linearly of $\zeta$. moreover it is nonnegative if such is the case of $\zeta$. We call this limit $\ell (\zeta)$. \paragraph{\bf Proof of Theorem 9.} If $\omega\in S^{N-1}$ and there exists an open neighbourhood $V$ such that Lemma 1-i holds, there existence of a nonegative Radon measure $\mu_{V} $ such that \begin{equation} \lim_{r\to 1}\int_{V}u(r,\sigma)\zeta (\sigma)d\sigma = \int_{V}\zeta d\mu \quad \forall \zeta\in C_{c}(V) \label{regular} \end{equation} If such a neighbourhood does not exist, we are in case (i). \smallskip Let ${\mathcal R}$ be the set of the $\omega\in S^{N-1}$ such that (ii) holds; ${\mathcal R}$ is relatively open and there exists a unique (non-negative) Radon measure $ \mu$ such that $\mu_{\vline V}=\mu_{V}$. The set ${\mathcal S}=S^{N-1}\setminus {\mathcal R}$ is closed. \paragraph{Definition.} The couple $({\mathcal S},\mu)$ is called the boundary trace of $u$. The set ${\mathcal S}$ is the singular part of this trace and the measure $\mu$ on ${\mathcal R}$, the regular part of the trace. We denote $$\mathop{\rm tr_{_{\partial B}}}(u)=\mathop{\rm tr}(u)=({\mathcal S},\mu).$$ For convenience it is often useful to introduce the Borel measure framework. Actually there is a one to one correspondence between the family $CM$ of couples $({\mathcal S},\mu)$ where ${\mathcal S}$ is a compact subset of $\S^{N-1}$ and $\mu$ a positive Radon measures on ${\mathcal R}=S^{N-1}\setminus {\mathcal S}$ and the set ${\mathcal B}^{+}_{reg}(S^{N-1})$ of outer regular, positive Borel measures on $S^{N-1}$. If $\beta\in {\mathcal B}^{+}_{reg}$, the singular set ${\mathcal R}_{\beta}$ and the singular part ${\mathcal S}_{\beta}$ are defined as follows: $${\mathcal R}_{\beta}=\{\omega\in S^{N-1}:\;\exists U \mbox { rel. open neighborhood of }\omega\mbox { s.t. }\beta (U)<\infty\}, $$ $${\mathcal S}_{\beta}=\{\omega\in S^{N-1}:\;\forall U \mbox { rel. open neighborhood of }\omega\mbox { s.t. }\beta (U)=\infty\}.$$ The correspondence $CM\leftrightarrow {\mathcal B}^{+}_{reg}(S^{N-1})$ is given by \begin{equation} {\mathcal M}(({\mathcal S},\mu))=\bar\mu \mbox { where }\bar\mu (A)= \left\{\begin{array}{l}\mu (A) \mbox { if } A\subseteq {\mathcal R},\\ \infty \mbox { if } A\cap {\mathcal S}\neq \emptyset, \end{array}\right. \label{correspondence} \end{equation} for any Borel subset $A$ of $S^{N-1}$, and \begin{equation} {\mathcal M}^{-1}(\beta)=({\mathcal S}_{\beta},\beta_{\vline {\mathcal R}_{\beta}}).\label{correspondence 2} \end{equation} With this correspondence, we denote $$\mathop{\rm Tr}(u)={\mathcal M}(\mathop{\rm tr}(u))\in {\mathcal B}^{+}_{reg}(S^{N-1}),$$ and a more general formulation for (\ref {NLDP}) is therefore what we call the generalized Dirichlet problem, namely \begin{equation} \begin{array}{c} -\Delta u+u^q=0,\mbox { in } B , \cr \mathop{\rm Tr}(u)=\bar\mu \in {\mathcal B}^{+}_{reg}(S^{N-1}). \end{array} \label{GDP}\end{equation} The problem is completely different according $10$ $$\lim_{r\to 1}\int_{D_{\eta}(\omega)}u(r,\sigma)d\sigma =\infty,$$ where $D_{\eta}(\omega)=\{\sigma\in S^{N-1}:\mathop{\rm dist_{_{S^{N-1}}}}(\sigma,\omega)< \eta\}$ We recall that $u_{k}(x)=k^{2/(q-1)}u(kx)$, and $u_{k}$ is a soluton of (\ref {NLE}) in $B_{1/k}=k^{-1}B$. For $0<\varepsilon<1$ we denote $$M_{\varepsilon,\eta}=\int_{D_{\eta}(\omega)}u(1-\varepsilon,\sigma)d\sigma.$$ For $m>0$ large enough and $\eta$ small enough there exists $\varepsilon=\varepsilon (\eta,m)>0$ such that $m=M_{\varepsilon,\eta}$. Let $\chi_{D_{\eta}(\omega)}$ be the characteristic function of the set $D_{\eta}(\omega)$ and $w_{\eta}$ be the solution of (\ref{NLE}) in $B$ with boundary value $u_{1-\varepsilon}(1,.)\chi_{D_{\eta}(\omega)}$. Clearly $w_{\eta}\leq u_{1-\varepsilon}$. Since $\lim_{\eta\to 0}\varepsilon=0$, then $\lim_{\eta\to 0}w_{\eta}=P^q_{B}(m\delta_{\omega})=u_{m,\omega}$. Moreover $\lim_{\varepsilon\to 0}u_{1-\varepsilon}=u$. Therefore $u_{m,\omega}\leq u$, for any $m>0$, which implies (\ref{blow-up}). \paragraph{Remark.} The proof below cannot work in the supercritical case, and actually the result does not hold as we shall see it below. \smallskip It is proved by Le Gall \cite {LG1, LG2} in the case $N=2=q$, and by Marcus and V\'eron \cite {MV2} in the general case, that the boundary data problem (\ref {GDP}) is well posed. The method of Le gall is mainly a probabilistic one while Marcus and V\'eron use only analytic tools. \begin{theorem} Suppose $10$ set $$\Phi (r,s)=\left\{\begin{array}{l} \frac{r^q-s^q}{r-s}\quad \mbox { if } r\neq s,\\ qs^{q-1}\quad \mbox { if } r=s.\end{array}\right.$$ By convexity, $$\left\{\begin{array}{c}r_{0}\geq s_{0},\, r_{1}\geq s_{1}\\ r_{1}\geq r_{0},\, s_{1}\geq s_{0},\end{array}\right\} \Longrightarrow \Phi (r_{1},s_{1})\geq\Phi (r_{0},s_{0}). $$ We set (with the notations of Lemmas 3-4) $${\underline u}_{n}={\underline u}_{n,\mu} \quad \mbox {and} \quad {\overline u}_{n}={\overline u}_{n,\mu},$$ and $$Z_{n,\mu}={\overline u}_{n,\mu}-{\underline u}_{n,\mu} \quad \mbox {and} \quad \lambda_{n,\mu}= \Phi ({\overline u}_{n,\mu},{\underline u}_{n,\mu}).$$ Then $$\Delta Z_{n,\mu}= \lambda_{n,\mu}Z_{n,\mu}$$ and $$\Delta (Z_{n,\mu}-Z_{n,0})- \lambda_{n,\mu}(Z_{n,\mu}-Z_{n,0}) =(\lambda_{n,\mu}-\lambda_{n,0})Z_{n,0}.$$ Since $${\overline u}_{n,\mu}\geq {\underline u}_{n,\mu}, \;{\overline u}_{n,\mu}\geq {\overline u}_{n,0}\quad \mbox {and}\quad {\overline u}_{n,0}\geq {\underline u}_{n,0}, \;{\underline u}_{n,\mu}\geq{\underline u}_{n,0},$$ $\lambda_{n,\mu}\geq\lambda_{n,0}$. Therefore $$\Delta (Z_{n,\mu}-Z_{n,0})- \lambda_{n,\mu}(Z_{n,\mu}-Z_{n,0}) \geq 0.$$ Moreover $Z_{n,\mu}=Z_{n,0}$ on $\partial B$. Therefore $$Z_{n,\mu}\leq Z_{n,0}\quad \mbox {in}\quad B.$$ Actually an approximation argument is needed to take into account the fact that the solutions have infinite values in ${\overline \mathcal S}_{1/n}$. Conclusion follows by letting $n\to\infty$. \begin{lemma} There exits $L=L(N,q)>1$ such that $${\overline u}_{{\mathcal S},0}\leq L{\underline u}_{{\mathcal S},0}.$$ \end{lemma} \paragraph{Sketch of the proof.} {\it Case 1} Suppose $\omega\in {\mathcal S}$. Since $$U_{\omega}(r,\omega)\leq{\underline u}_{{\mathcal S},0}(r,\omega) \leq {\overline u}_{{\mathcal S},0}(r,\omega)\leq \bar u_{B}(r,\omega)$$ where $\bar u_{B}$ is the maximal solution of (\ref{NLE}) in $B$, and $$U_{\omega}(r,\omega)\geq K(N,q)\bar u_{B}(r,\omega),$$ the estimate holds. \smallskip \noindent {\it Case 2} Suppose $\omega\in {\mathcal R}$. The way for obtaining this estimate is a bit more technical, however, if we assume that $B$ is replaced ${\mathbb R}_{+}^{N}$ and $\partial B$ by the hyperplane $H={\mathbb R}^{N-1}$, the argument goes as follows. By a change of coordinates, it can be assumed that $0=\omega$ is the origin. Let $D_{\rho}\subset H$ be the largest open (N-1)-ball with center $0$ such that $D_{\rho}\subseteq {\mathcal R}$ ($\rho$ is finite since ${\mathcal S}\neq\emptyset$), and let $a\in\partial D_{\rho}\cap {\mathcal S}$. If we denote $x=(x_{1},x')$ the current point in ${\mathbb R}_{+}\times{\mathbb R}^{N-1}$, then $$u_{\infty ,\omega}(x_{1},0)\leq {\underline u}_{{\mathcal S},0}(x_{1},0) \leq {\overline u}_{{\mathcal S},0}(x_{1},0) \leq {\overline u}_{D^c_{\rho},0}(x_{1},0).$$ In order to estimate the bounds of the quotient $$ Q_{\rho}(x_{1})= \frac{u_{\infty ,\omega}(x_{1},0)} {{\overline u}_{D^c_{\rho},0}(x_{1},0)},$$ when $x_{1}$ runs from $0$ to $\infty$, a scaling argument shows that it can be assumed that $\rho=1$. For small values of $x_{1}$ the lower bounds is positive and given by Hopf's lemma, while for large values of $x_{1}$, the lower bound follows from the fact that $$\lim_{x_{1}\to\infty} x_{1}^{2/(q-1)}u_{\infty ,\omega}(x_{1},0)= \omega (x_{1}/\abs {x_{1}}),$$ (in this formula $\omega (.)$ is the spherical function defined in (\ref {strong singularity})) while $$\lim_{x_{1}\to\infty} x_{1}^{2/(q-1)}{\overline u}_{D^c_{\rho},0}(x_{1},0)=\left (\frac {2(q+1)}{(q-1)^{2}}\right )^{1/(q-1)}.$$ Consequently there exists $L_{1}>0$ such that $${\overline u}_{{\mathcal S},0}(x_{1},0)\leq L {\underline u}_{{\mathcal S},0}(x_{1},0).$$ In the real situation where we are dealing with $B$ and not with ${\mathbb R}_{+}^{N}$, we proceed by contradiction and scaling, reducing the situation to the half space case. \paragraph{Proof of Theorem 10.} It follows the proof of Theorem 8 (in a simplified way). By Lemma 5 it is sufficient to prove uniqueness in the case where $\mathop{\rm tr}(u)=({\mathcal S},0)$. Assuming that $${\underline u}_{{\mathcal S},0}\neq{\overline u}_{{\mathcal S},0} \Longrightarrow {\underline u}_{{\mathcal S},0} <{\overline u}_{{\mathcal S},0}, $$ then $$w={\underline u}_{{\mathcal S},0}-\frac {1}{2L} ({\overline u}_{{\mathcal S},0}-{\underline u}_{{\mathcal S},0})$$ is a supersolution of (\ref {NLE}) dominated by ${\underline u}_{{\mathcal S},0}$, but dominating the subsolution $(1/2+1/(2L)){\underline u}_{{\mathcal S},0}$. Therefore there exists a solution $\tilde u$ of (\ref{NLE}) such that $$(\frac {1}{2}+\frac {1}{2L}){\underline u}_{{\mathcal S},0} \leq \tilde u\leq w<{\underline u}_{{\mathcal S},0}.$$ It follows that $tr (\tilde u)=({\mathcal S},0)$, which contradicts the minimality of ${\underline u}_{{\mathcal S},0}$. \section{Generalized Dirichlet problem: the supercritical case} As we have seen it in Sections 3-4, neither any Radon measure, nor any closed subset of $\partial B$ are eligible for being the regular or the singular part of the boundary trace $({\mathcal S},\mu)$ of a positive solution of (\ref {NLE}) in$B$. Roughly speaking, a Radon measure is admissible if it does not charge too thin sets, while a closed subset is admissible for being a singular set if it is not too ramified. Moreover there is a compatibility requirement between ${\mathcal S}$ and $\mu$. Those different notions will be made rigourous with the help of boundary Bessel capacities. The results that we present are due to Le Gall \cite {LG2} in the case $q=$2, Dynkin and Kuznetsov in the case $12$. Because of the technical difficulties of the various aspects of the theory of supercritical boundary trace we shall essentially restrict ourself to present the main results in the case $q>2$, with some ideas of how to obtain them. {\it Up to now a unified proof covering all the cases $q\geq (N+1)/(N-1)$ is still missing}. \subsection* {Removable sets} \paragraph{Defintion.} (i) A subset $E\subset \partial B$ is said {\bf q-removable} if any non-negativefunction $u\in C^{2}(B)\cap C(\bar B\setminus E)$ which satisfies (\ref{NLE}) in $B$ and vanishes on $\partial B\setminus E$ is identically zero. \noindent (ii) A subset $E\subset \partial B$ is said {\bf conditionally q-removable} if any non-negative function $u\in C^{2}(B)\cap C(\bar B\setminus E)$ which satisfies (\ref{NLE}) in $B$ and vanishes on $\partial B\setminus E$ is such that $u\in L^q(B;(1-r)dx)$. \smallskip The main result concerning the removability of set is the following \begin{theorem} Assume $q\geq (N+1)/(N-1)$ and $E\subset \partial B$ is a Borel set. Then the following assertions are equivalent. \smallskip \noindent(i) $E$ is q-removable \noindent(ii) $E$ is conditionally q-removable \noindent (iii) $C_{2/q,q'}(E)=0$. \smallskip Moreover an arbitrary set $A\subset \partial B$ is q-removable if and only if every closed subset of $A$ is q-removable. \end{theorem} In the statement of the Theorem, $C_{2/q,q'}(E)$ denote the Bessel capacities of order $2/q$ and exponent $q$. In the space ${\mathbb R}^d$ the Bessel capacities $C_{\alpha,p}(E)$ ($\alpha >0$, $11$ the spaces are defined by induction. From this spaces (which coincide with the classical Sobolev spaces when $\alpha$ is not an integer and $p=q$), we define the capacity of a compact subset $E\in{\mathbb R}^d$ by $C_{B_{\alpha}^{p,q}}(E)$ by $$C_{B_{\alpha}^{p,q}}(E)=\inf\left\{{\norm f}_{B_{\alpha}^{p,q}}:\;f\in {\mathcal S}({\mathbb R}^d),\;f\geq 1 \mbox { on }E \right \}.$$ The $C_{B_{\alpha}^{p,q}}$-capacity of an open set is defined by the supremum of the capacities of its compact subsets, and the capacity of a general set by the infimum of the capacities of the open sets in which it is contained. \smallskip When $\alpha>0$, $10$ such that for any subset of ${\mathbb R}^d$, $$K^{-1}C_{\alpha,p}(E)\leq C_{B_{\alpha}^{p,p}}(E)\leq K C_{\alpha,p}(E).$$ On a submanifold of ${\mathbb R}^{N}$, the capacity is defined by using local charts. \smallskip In the forthcoming lemmas, we give the strategy for proving Theorem 12. \begin{lemma}Suppose that $q\geq (N+1)/(N-1)$ and let $E\subset \partial B$ be compact and such that $C_{2/q,q'}(K)=0$. Then $K$ is conditionally q-removable. \end{lemma} \paragraph{Sketch of the proof.} In the case $q>2$ Marcus and V\'eron \cite {MV5} proceed as follows: If $\eta\in C^{2}(\partial B)$ is such that $0\leq \eta\leq 1$ is identically $1$ is a neighborhood of $K$ set $\zeta=(1-P_{\eta})^{2q'}\varphi_{1}$, where $\varphi_{1}$ is the positive first eigenfunction of $-\Delta$ in $W^{1,2}_{0}(B)$. Then it is proved by approximation that \begin{equation} \int_{B}(-u\Delta \zeta+u^q\zeta)dx=0. \label{estimate for zeta} \end{equation} From (\ref {estimate for zeta}) and H\"older's inequality, $$\int_{B}u^q\zeta dx\leq \left(\int_{B}u^q\zeta dx\right)^{1/q} \left(\int_{B}\zeta^{-q'/q}{\abs {\Delta \zeta}}^{q'}dx\right)^{1/q'}. $$ Using sharp regularizing estimates of the Poisson potential, $$\int_{B}\zeta^{-q'/q}{\abs {\Delta \zeta}}^{q'}dx\leq C_{1}{\norm \eta}_{W^{2/q,q'}}+C_{2}. $$ From the assumption on $K$, there exists a sequence of functions $\eta_{k}\in C^{2}(\partial B)$ such that $0\leq \eta_{k}\leq 1$ and ${\norm {\eta_k}}_{W^{2/q,q'}}\to 0$ as $k\to\infty$. Since this implies in particular that $1-P_{\eta_k}\to 1$, it follows $$\int_{B}u^q\varphi_{1} dx\leq C_{2},$$ which implies the claim. \begin{lemma} Any conditionally q-removable closed subset $E\subset\partial B$ is q-removable. \end{lemma} \paragraph{Proof.} If $E$ is not q-removable, there exists a nonnegative nonzero solution $u$ of (\ref {NLE}) vanishing on $\partial B\setminus E$. Therefore $Tr (u)$ is a Borel measure which is zero outside $E$. Since $E$ is conditionally q-removable, $u^q\in L^{1}(B;(1-r)dx)$. Therefore the singular part of the boundary trace of $u$ is empty and $$Tr (u)=\mu\in {\mathcal M_{+}}(\partial B).$$ For $n\geq 1$, $\tilde u_{n}=nu$ is a supersolution of (\ref {NLE}) with boundary trace $n\mu$. Therefore there exists a nonnegative solution $u_{n}$ of (\ref {NLE}) such that $Tr (u_{n})=n\mu$, and in particular $$\int_{B}(\lambda_{1}u_{n}+u_{n}^q)\varphi_{1}dx=-n\int_{\partial B}\frac{\partial\varphi_{1}}{\partial\nu}d\mu$$ Letting $n\to\infty$ implies that the increasing sequence $\{u_{n}\}$ converges to some solution $u_{\infty}$ in $B$ with $Tr(u_{\infty})=0$ outside $E$. Moreover, $$\int_{B}(\lambda_{1}u_{\infty}+u_{\infty}^q)\varphi_{1}dx=\infty\Longrightarrow \int_{B}u_{\infty}^q\varphi_{1}dx=\infty.$$ This contradicts the fact that $u_{\infty}^q\in L^{1}(B;(1-r)dx)$ because $E$ is conditionally q-removable. \paragraph{Remark.} It is clear that if a subset $E\in \partial B$ is q-removable, it is conditionnaly q-removable. \smallskip By sharp linear estimates on the Poisson potential, the following result holds \begin{lemma} Suppose $q\geq 2$, then $\mu\mapsto P_{B}(\mu)$ maps ${\mathcal M_{+}}(\partial B)\cap W^{-2/q,q}(\partial B)$ into $L^q(B;(1-r)dx)$, and there holds $${\norm {P_\mu}}_{L^q(B;(1-r)dx)} \leq C (q) {\norm {\mu}}_{W^{-2/q,q}(\partial B)}.$$ \end{lemma} As an important consequence we have \begin{cor}Suppose $q\geq 2$ and $\mu\in{\mathcal M_{+}}(\partial B)\cap W^{-2/q,q}(\partial B)$. Then there exists $u=P_{B}^q(\mu)$. \end{cor} \paragraph{Proof.} By Lemma 9, $P_{\mu}\in L^q(B;(1-r)dx)$. For $k>0$ we denote $\{u_{k}\}$ the solution of $$ \begin{array}{c} -\Delta u_{k}+(\min u_{k},k)^q=0,\mbox { in } B , \cr u_{k}=\mu\mbox { on } \partial B , \end{array} $$ Clearly $u_{k}\leq P_{\mu}$ which implies that the sequence $\{(\min u_{k},k)^q\}$ is uniformly integrable in $L^{1}(B;(1-r)dx)$. When $k$ increases, $u_{k}$ decreases and converges to some $u$ which belongs to $L^{1}(B)\cap L^{q}(B;(1-r)dx)$. Letting $k\to\infty$ in the weak formulation (which is valid for every $\zeta\in C_{0}^{1,1}(\bar B)$) $$\int_{B}(-u_{k}\Delta \zeta +\zeta(\min u_{k},k)^q)dx=-\int_{B}\frac {\partial \zeta}{\partial \nu}d\mu,$$ infers that $u=P_{B}^q(\mu)$. From a dual definition of the $C_{2/q,q'}$-capacity of a closed subset $E\subseteq\partial B$, follows the implication \begin{equation} C_{2/q,q'}(E)>0\Longrightarrow \exists \mu\in {\mathcal M_{+}}\cap W^{-2/q,q}(\partial B) \; s.t.\; \mu (E)=\mu (\partial B)>0. \label{dual capacity}\end{equation} Consequently there holds \begin{lemma}Suppose $q\geq \max (2,(q+1)/(q-1)$. If $E\subseteq\partial B$ is closed and such that $C_{2/q,q'}(E)>0$, then $E$ is not conditionally q-removable. \end{lemma} \paragraph{Proof.} By Lemma 9 there exists $\mu\in {\mathcal M_{+}}\cap W^{-2/q,q}(\partial B)$ such that $\mu (E)=\mu (\partial B)>0$. By Corollary 1, $u=P_{B}^q(\mu)$ exists and $\mathop{\rm tr}(u)=(\emptyset,\mu )$, which implies in particular that $u$ vanishes on $\partial B\setminus E$. The proof of Theorem 12 is completed by the next result whose proof is too technical to be presented here. \begin{lemma} A set $A\subset \partial B$ is q-removable if and only if every closed subset of $A$ is q-removable. \end{lemma} \subsection* {q-traces} \paragraph{Definition.} A Radon measure $\mu$ on $\partial B$ is called a q-trace if there exists a solution $u$ of (\ref {NLE}) such that $u=P^q_{B}(\mu)$. The set of q-traces is denoted by ${\mathcal M}^q(\partial B)$. The main result concerning q-traces is \begin{theorem} Assume $q\geq (N+1)/(N-1)$. A measure $\mu$ on $\partial B$ is a q-trace if and only if for any Borel subset $A\subseteq\partial B$ \begin{equation}\mu (A)=0\quad \mbox {whenever }\quad C_{2/q,q'}(A)=0. \label{condition}\end{equation} \end{theorem} Besides what has been proved in the preceding subsection, the two next lemmas are needed, which both are mere adaptations of results of Baras and Pierre \cite {BP}, and Meyers \cite {Me}. \begin{lemma} Let $\mu\in {\mathcal M}\cap W^{-2/q,q}(\partial B)$. Then $\mu$ does not charge the sets with $C_{2/q,q'}$-capacity zero. \end{lemma} \begin{lemma}Suppose $\mu\in {\mathcal M}^{+}(\partial B)$ does not charge the sets with $C_{2/q,q'}$-capacity zero. Then there exists a sequence $\{\mu_{n}\}\subset W^{2/q,q'}(\partial B) \cap {\mathcal M}^{+}(\partial B)$, such that $\{\mu_{n}\}$ converges in increasing to $\mu$. \end{lemma} \paragraph{Sketch of the proof of Theorem 13.} We shall restrict ourselves to the cases $q\geq 2$ and the measures are nonnegative, the general case needing some approximation argument. Let $\mu\in {\mathcal M}_{+}^q(\partial B)$ and $E\subset \partial B$ be a Borel subset such that $C_{1/q,q'}(E)=0$. Set $\mu_{E}=\chi_{E}\mu$. Since $0\leq \mu_{E}\leq\mu$, the construction given in Corollary 1 yields $\mu_{E}\in {\mathcal M}_{+}^q(\partial B)$ and $P_{B}^q(\mu_{E})\leq P_{B}^q(\mu)$. But $C_{1/q,q'}(E)=0$ and Theorem 12 implies that $E$ is q-removable. therefore $P_{B}^q(\mu_{E})=0$ and consequently $\mu_{E}=0$. \smallskip Conversely, let $\mu\in {\mathcal M}^{+}(\partial B)$ which does not charge the sets with $C_{2/q,q'}$-capacity zero. Then there exists an increasing sequence of positive measures $\{\rho_{n}\}$ belonging to $W^{-2/q,q}(\partial B)$ and converging to $\mu$. Then $\rho_{n}\in {\mathcal M}_{+}^q(\partial B)$, and the sequence $\{u_{n}\}=\{ P_{B}^q(\rho_{n})\}$ is increasing and converges to some $u$. Moreover \begin{equation} \int_{B}(\lambda_{1}u_{n}+u_{n}^q)\varphi_{1}dx=-\int_{\partial B}\frac {\partial \varphi_{1} }{\partial \nu}d\rho_{n}.\label{estimate 10} \end{equation} Because the right-hand side of (\ref {estimate 10}) is convergent, the same holds for the left-hand side (by the Beppo-Levi theorem). Therefore $u\in L^q(B,(1-r)dx)\cap L^{1}(B) $, which is sufficient to derive that $u=P_{B}^q(\mu)$ from the weak formulation of the fact that $u_{n}=P_{B}^q(\rho_{n})$. \subsection* {The generalized Dirichlet problem} Given a Borel measure $\bar \mu\in {\mathcal B}^{+}_{reg}(\partial B)$ with ${\mathcal M}^{-1}(\bar \mu)=({\mathcal S},\mu )\in CM$, we recall that $${\mathcal S}_{\varepsilon}=\{\sigma \in S^{N-1}:\mathop{\rm dist_{_{S^{N-1}}}}(S^{N-1},{\mathcal S})\leq\varepsilon\}.$$ and $$\mu_{\varepsilon}=\chi_{{\mathcal S}_{\varepsilon}^c}\mu.$$ Let ${\overline u}_{{\mathcal S}_{\varepsilon}}$ be the maximal solution of (\ref {NLE}) in $B$ with $\mathop{\rm tr}({\overline u}_{{\mathcal S}_{\varepsilon}})=({\mathcal S}_{\varepsilon},0)$. Because of the construction it is easy to see that $$0<\varepsilon<\delta \Longrightarrow {\overline u}_{{\mathcal S}_{\varepsilon}} \leq {\overline u}_{{\mathcal S}_{\delta}}.$$ When $\varepsilon\to 0$, ${\overline u}_{{\mathcal S}_{\varepsilon}}$ converges locally uniformly in $B$ to a nonnegative solution $u^{*}$ of (\ref {NLE}). Let ${\mathcal S}_{q}^*$ be the singular part of the boundary trace of $u^{*}$. Clearly ${\mathcal S}_{q}^*\subseteq {\mathcal S}$. \smallskip If ${\mathcal R}=\partial B\setminus {\mathcal R}$ and $\mu\in {\mathcal M}_{+}({\mathcal R})$ is such that $\mu_{K}=\chi_{K}\mu$ is a q-trace for any compact subset $K\subset {\mathcal R}$, we denote $u_{K}=P_{B}^q(\mu_{K})$. If $K_{n}$ is an increasing sequence of compact subset of ${\mathcal R}$ such that $\bigcup K_{n}={\mathcal R}$, the sequence $\{u_{K_{n}}\}$ is increasing and converges to a solution $\check u$ of (\ref {NLE}). Let $\partial _{\nu}{\mathcal S}$ be the singular part of the boundary trace of $\check u$. Again it is easy to verify that $\partial _{\nu}{\mathcal S}\subseteq {\mathcal S}$. The following result is the main result concerning the solvability of (\ref {GDP}) in the supercritical case. \begin{theorem}Assume $q\geq (N+1)/(N-1)$ and let $\bar\mu\in {\mathcal B}_{reg}^{+}(\partial B)$, with regular set ${\mathcal R}={\mathcal R}_{\bar\mu}$, regular part $\mu=\bar\mu_{\vline {\mathcal R}_{\bar\mu}}$ and singular part ${\mathcal S}={\mathcal S}_{\bar\mu}=\partial B\setminus {\mathcal R}_{\bar\mu}$. Then problem (\ref {GDP}) possesses a maximal solution $\bar u_{\bar\mu}$ if and only if the following two conditions are satisfied: \noindent (i) For every Borel subset $A\subset {\mathcal R}$, $C_{2/q,q'}(A)=0\Longrightarrow \mu (A)=0$. \noindent (ii) ${\mathcal S}={\mathcal S}_{q}^*\cup \partial _{\nu}{\mathcal S}$. \end{theorem} We shall not give the proof, although it is interesting to note that a key inequality which give the condition (ii) for solving (\ref {GDP}) is the following \begin{equation}\max(\check u,u^{*})\leq \bar u_{\bar\mu}\leq \check u+u^{*}, \label{key inequality}\end{equation} in which formula $u^{*}$ and $\check u$ have been defined above. \smallskip A striking difference between the subcritical case and the supercritical case is the loo of uniqueness. It was first proved by Le Gall \cite {LG3} in the case $q=2$ and then by Marcus and V\'eron \cite {MV5} in the general case $q\geq (N+1)/(N-1)$ that there may exists infinitely many solutions of (\ref {GDP}) with a given singular trace. In particular, for every $\gamma>0$, there exists a solution $u_{\gamma}$ with $\mathop{\rm tr}(u_{\gamma})=(\partial B,0)$ and such that $u_{\gamma}(0)<\gamma$. Moreover, by sharpening the construction of the $u_{\gamma}$ it is proved in \cite {MV5} that for any $\varepsilon>0$ there exists a Borel subset $A\subset \partial B$ with $\mathop{\rm meas} (\partial B\setminus A)<\varepsilon$ and a solution $u$ of (\ref {GDP}) such that $Tr(u)=(\partial B,0)$ and $\displaystyle {\lim_{r\to 1}}u(r,\sigma)=0$ for every $\sigma \in A$. This clearly indicates that the formulation of boundary traces in terms of the usual topology on $\partial B$ is not sufficient to describe the variety of phenomena which may occur at the boundary and that a sharper notion is needed. A first and important step towards this direction has been made by Dynkin and Kuznetsov \cite {DK5}\cite {Ku1, Ku2} in a series of recent papers. In these works they introduce a topology thiner than the usual one in ${\mathbb R}^d$ and they introduce a new class of problems in which some uniqueness is proved. {\it However the problem of finding The notion of boundary trace which gives rise to a one to one and onto correspondence between the set of all positive solutions of (\ref {GDP}) and their boundary trace is still open}. \subsection* {Conformal deformations of hyperbolic space} The problem of conformal deformation of Riemannian metrics is one of the most interesting field of applications of semilinear elliptic equations. We perform the identification of the hyperbolic N-space ${\mathbb H} ^{N}$ with $(B,g_{H})$, where $$g_{H}(x)=\left(\frac {2}{1-{\abs x}^{2}}\right)^{2}\eta (x)$$ with $\eta_{ij}=\delta_{ij}$. Given $K\in C^\infty (B)$ with $N > 2$, three classical problems arising from conformal geometry are the following \noindent I- Does there exists a positive function $v$ on such that the metric $g_{v}=v^{4/(N-2)}$ has scalar curvature $K$ ? \noindent II- Is $g_{v}$ a complete Riemannian metric ? \noindent III- If so is $v$ unique in the class of complete Riemannian metrics conformal to $g_{H}$ with scalar curvature $K$ ? \smallskip It is classical that the function $v$ satisfies \begin{equation} C_N\Delta v+K(x)v^{(N+2)/(N-2)}=0 \quad \mbox {in}\quad B \end{equation} with $C_N=4(N-1)/(N-2)$. This problem has been thoroughly studied by Loewner and Nirenberg \cite {LN}(1974), Ni (1982), Aviles and McOwen (1985-1988) and more recently (1993-94) by Ratto, Rigoli and V\'eron \cite {RRV}. An interesting case which is associated to supercritical trace problems occurs when $K$ is nonpositive (at least near $\partial B$ ). The completeness assumption means that the geodesic distance from inside up to the boundary is always infinite, which in this case is equivalent to \begin{equation} \int_{0}^{1}v^{(N-2)/2}(\gamma (t))dt=\infty \label{completeness}\end{equation} for any $\gamma\in C^{0,1}([0,1];\bar B)$, with $\gamma([0,1))\subset B$, and $\gamma (1)\in\partial B$. This means that $v$ has some kind of blow-up near $\partial \Omega$. Many existence results concerning this equation have been proven. 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Math. 59, 231-250 (1992). \end{thebibliography} \noindent{\sc Laurent V\'eron }\\ Laboratoire de Math\'ematiques et Physique Th\'eorique \\ CNRS UMR 6083 \\ Facult\'e des Sciences et Techniques \\ Parc de Grandmont \\ F 37200 Tours France\\ e-mail: veronl@univ-tours.fr \end{document}