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%macros de Louis
\def\bom{\bar\Omega}
\def\dom{\partial\Omega}
\def\intbas#1{\int_{#1}^\infty}
\def\N{\mathbb{N}}
\def\om{\Omega}
\def\R{\mathbb{R}}
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\author[S. Dumont, L. Dupaigne, O. Goubet, V. R\u{a}dulescu]{S. Dumont, L. Dupaigne, O. Goubet, V. R\u{a}dulescu}

\title[Back to the Keller-Osserman condition for boundary blow-up solutions.]{Back to the Keller-Osserman condition for boundary blow-up solutions.}


\address[Serge Dumont, Louis Dupaigne, Olivier Goubet]{LAMFA CNRS UMR 6140 \\
Universit\'e de Picardie Jules Verne \\
33 rue Saint-Leu 80039 Amiens cedex.} \address[Vicentiu
R\u{a}dulescu]{Department of Mathematics, University of Craoiva\\
200585 Craoiva, Romania}



\begin{document}
\email{serge.dumont@u-picardie.fr}
\email{louis.dupaigne@u-picardie.fr}
\email{olivier.goubet@u-picardie.fr} \email{vicradulescu@yahoo.com}

\date{27 April 2006}

\begin{abstract}
This article is concerned with the existence, uniqueness and
numerical approximation of boundary blow up solutions for elliptic
PDE's as $\Delta u=f(u)$ where $f$ satisfies the so-called
Keller-Osserman condition. We {\bleu characterize} existence of such
solutions {\bleu for non-monotone $f$} . As an example, we construct
an infinite family of boundary blow up solutions for the equation
${\bleu \Delta u=u^2(1+\cos u)}$ on a ball. We {\bleu prove}
uniqueness {\bleu (on balls) when} $f$ is increasing and convex
{\bleu in a} neighborhood of infinity and we discuss and perform
some numerical computations to approximate such boundary blow-up
solutions.
\end{abstract}

\maketitle

\noindent
{\it \footnotesize 2000 AMS Mathematics Subject Classification}. {\scriptsize 35J60}.\\
{\it \footnotesize Key words}. {\scriptsize Elliptic equations,
boundary blow�-up, Keller-Osserman condition}

\

\centerline{ In memory of Thomas Lachand-Robert}


\section{Introduction}

\subsection{Setting the problem}

{\rm .} Let $f$ be a nonnegative function defined on $[0,+\infty[$
such that $f(0)=0$. We assume, for the sake of simplicity that $f$
is a $C^1$ function. Considering $\Omega$ a smooth bounded domain of
$\mathbb{R}^D$, $D\ge1$, we seek $u>0$ a smooth function such that
\begin{equation}\label{problem}
\left\{\begin{split}
\Delta u=f(u) \; in \; \Omega, \\
u=+\infty \; on \; \partial\Omega,  \\
\end{split}\right.
\end{equation}
where the boundary condition is to be understood as :
$$
\lim_{x\to x_0} u(x) = +\infty\quad \forall x_0\in\partial\Omega.
$$
\noindent To prove existence of such a boundary blow-up solution,
{\bleu it is classically assumed that} $f$ is a nondecreasing
function with suitable growth rate at infinity, as demonstrated
independently by Keller \cite{K} and Osserman \cite{O}.

In this article, we study existence, asymptotic behaviour,
uniqueness and numerical approximation of solutions of
\eqref{problem}, when $f$ may exhibit \it non monotone \rm
behaviour.

\subsubsection{Existence results}

Existence of solutions of \eqref{problem} is closely related to the
following growth conditions : for $s\in[0,+\infty)$, let
$F(s)=\int_0^s f(t)\,dt$ and define $\Phi
:(0,+\infty)\to(0,+\infty]$ by
$$
\Phi(\alpha)=\frac1{\sqrt2}\int_{\alpha}^\infty
\frac{ds}{\sqrt{F(s)-F(\alpha)}}\;,
$$
where we let by convention $\Phi(\alpha)=+\infty$, whenever the
integral is divergent or $F(s)=F(\alpha)$ on a set of positive
measure.
\begin{defi}
{\rm We say that $f$ satisfies the Keller-Osserman condition
whenever
\begin{equation}\label{KO}
\exists \alpha>0\quad \Phi(\alpha)<\infty.
\end{equation}
We say that $f$ satisfies the Sharpened Keller-Osserman condition
whenever
\begin{equation}\label{SKO}
\liminf_{\alpha\to\infty} \Phi(\alpha) = 0.
\end{equation}
}\end{defi}
\label{remsko} Clearly, the Sharpened Keller-Osserman
condition implies the classical one. Under fairly weak assumptions
on the nonlinearity $f$, it turns out that both conditions are
equivalent :
\begin{pro}\label{KOegalSKO}
Assume there exist $\eta,T>0$ such that $t\to\frac{F(t)}{t^\eta}$ is
nondecreasing for $t>T$. Assume \eqref{KO} holds for some
$\alpha>0$. Then,
\begin{equation*}
\lim_{\alpha\to\infty} \Phi(\alpha) = 0.
\end{equation*}
\end{pro}
\noindent See the Appendix for a proof. In the special case
$f(u)=u^2(1+\cos u)$,
$\limsup_{\alpha\to\infty}\Phi(\alpha)=+\infty$. However,
\eqref{SKO} still holds (see the Appendix). This raises the
following question :
\begin{op}\label{OP1}
Does there exist a nonnegative $C^1$ function $f$ such that
\eqref{KO} holds for some $\alpha>0$ but $ \liminf_{\alpha\to\infty}
\Phi(\alpha) \in (0,+\infty]$ ?
\end{op}
%, or \cite{AR} for a proof in the case where $f$ is a nondecreasing function (which includes the model nonlinearity $f(u)=u^p$ with $p>1$).\end{rmk}
With these definitions in mind, our main results concerning
existence read as follows :
\begin{thm}\label{th1}
$f$ satisfies the Keller-Osserman condition if and only if
\eqref{problem} admits (at least) a positive boundary blow-up
solution on {\bf some\rm} ball.
\end{thm}

\begin{thm}\label{th2}
$f$ satisfies the Sharpened Keller-Osserman condition if and only if
\eqref{problem} admits (at least) a positive boundary blow-up
solution on {\bf any\rm} (smooth bounded) domain $\Omega$.
\end{thm}
In particular, Theorem \ref{th2} implies existence of boundary
blow-up solutions for functions such as $f(u)=u^2(1+\cos u)$.

\subsubsection{Asymptotic behaviour}
The blow-up rate of solutions of \eqref{problem} is determined
implicitely by the following theorem :
\begin{thm}\label{blow-up-rate}
Assume $\Omega$ satisfies an interior and an exterior sphere
condition at every point of its boundary. Assume \eqref{SKO} holds
and let $u$ denote any positive solution of \eqref{problem}. Then,
\begin{equation*}
    \lim_{x\to x_0}\frac{\displaystyle{\int_{u(x)}^\infty}
    \frac{dt}{\sqrt{2F(t)}}}{\delta(x)}=1,
\end{equation*}
where $\delta(x)={\rm dist}(x,\partial\Omega)$.
\end{thm}
\subsubsection{Uniqueness results}
{\bleu In view of the maximum principle, it seems natural to only
assume that $f$ is nondecreasing in order to obtain uniqueness. To
the best of our knowledge, no proof (or counter-example) of such a
statement has been given yet. Extra requirements such as the
convexity of $f$ or the monotony of $f(u)/u$ are needed in the
proofs found in the literature (see e.g. \cite{MV}). When the domain
is a ball, we relax such assumptions as follows:}
\begin{thm}\label{th-uniqueness}
Assume that {\bleu $f$ is nondecreasing and} that $f$ is convex in a
neighborhood of $+\infty$. Assume also that $f$ satisfies the
Keller-Osserman condition \eqref{KO}. Then on any ball $B(0,R)$
there exists a unique boundary blow up solution {\bleu of
\eqref{problem}}.
\end{thm}

\begin{rmk}
{\rm The same result also holds if $f$ is nondecreasing, $f(u)/u$ is
nondecreasing in a neighbourhood of $+\infty$ and \eqref{KO} holds.}
\end{rmk}

\noindent In dimension $D=1$, a necessary and sufficient condition
for uniqueness can be derived. Namely, we have the following
\begin{prop}Assume that $f$ satisfies the Sharpened Keller-Osserman
condition. Then \eqref{problem} admits a unique solution on
$\Omega=]-R,R[$ if and only if the equation
$$
\Phi(\alpha)=R
$$
admits exactly one solution.
\end{prop}
As a straightforward consequence, we obtain
\begin{cor}
Assume that $f$ satisfies the Sharpened Keller-Osserman condition.
\eqref{problem} admits a unique solution on {\bf any\rm} domain
$\Omega=]-R,R[$ if and only if $\Phi:(0,\infty)\to(0,\infty)$ is
one-to-one.
\end{cor}

\begin{rmk}
{\rm In particular, if $f$ is nondecreasing, one can easily show
that $\Phi$ is one-to-one.}
%If $F(u)/u^2$ is nondecreasing, I don't have a direct proof that $\Phi$ is one-to-one. Any idea appreciated.
\end{rmk}
\section{Existence of boundary blow-up solutions}

\subsection{Minimality Principle}\label{subsection2.1}

We restate the well-known sub and supersolution method (see
\cite{CS2} and \cite{CS}) and derive elementary but important
corollaries.

\begin{pro}\label{mis}
Consider $\Omega$ a bounded domain of $\mathbb{R}^D$ such that all
boundary points are regular, $f\in C(\R)$ and $g\in
C(\partial\Omega)$. Assume there exist two functions $\sub u$,
$\super u\in C(\bom)$ such that $\sub u\le \super u$ and
\begin{equation}\label{solution}
\left\{
\begin{split}
\Delta \sub u\ge f(\sub u) \; in \; \mathcal D'(\Omega),& \qquad\text{(resp. $\Delta \super u\le f(\super u) \; in \; \mathcal D'(\Omega)$)}\\
\sub u\le g \; on \; \partial\Omega,&  \qquad\text{(resp. $\super u\ge g \; on \; \dom$),} \\
\end{split}
\right.
\end{equation}
Then problem
\begin{equation}\label{dirichlet}
\left\{
\begin{split}
\Delta  u = f(u) \; in \; \mathcal D'(\Omega), \\
u= g \; on \; \partial\Omega.\\
\end{split}
\right.
\end{equation}
possesses at least one solution $u\in C(\bom,\R)$ such that $\sub
u\le u\le \super u$.
\end{pro}

\begin{cor}\label{mp}(Minimality Principle)
Make the same assumptions as in Proposition \ref{mis}. Then there
exists a unique solution $u\in C(\bom)$ of \eqref{dirichlet} such
that $\sub u\le u$ and $u|_\omega \le \super v$ for any open subset
$\omega$ of $\om$ and any function $\super v\in C(\bar\omega)$
satisfying
\begin{equation}\label{localsupersolution}
\left\{
\begin{split}
\Delta  \super v \le f(\super v) \; in \; \mathcal D'(\omega), \\
\super v\ge \sub u \; in \; \omega,\\
\super v\ge u \; on \; \partial\omega.\\
\end{split}
\right.
\end{equation}
We call $u$ the minimal solution of \eqref{dirichlet} relative to
$\sub u$.
\end{cor}
A complete proof of Corollary \ref{mp} is given in the Appendix. We
present here a short proof in the case where $f$ is a locally
Lipschitz function.

\noindent Proof:

\it Uniqueness \rm : Let $u_1$, $u_2$ be two such solutions.
Choosing $\omega=\Omega$ and $\super v=u_2$ in the statement of
Corollary \ref{mp}, we conclude that $u_1\le u_2$. Reversing the
roles of $u_1$ and $u_2$, we conclude that $u_1=u_2$.

\it Existence : \rm

\noindent Let $\Lambda=\underset{[\min\sub u,\max\super
u]}{\sup}|f'|$, $u_0=\sub u$ and for $k\ge1$, define $u_k\in
C(\bom)$ inductively by
\begin{equation*}
\left\{
\begin{split}
\Delta  u_k-\Lambda u_k = f(u_{k-1})-\Lambda u_{k-1} \; in \; \mathcal D'(\Omega), \\
u_k= g \; on \; \partial\Omega.\\
\end{split}
\right.
\end{equation*}
Then it is known that the sequence $(u_k)$ is nondecreasing and
converges to a solution $u\in C^2(\om)\cap C(\bom)$ of
\eqref{dirichlet}, which satisfies in addition $\sub u\le u \le
\super u$.

Let $\super v\in C(\bar\omega)$ verify \eqref{localsupersolution}
and assume by contradiction that the set $\omega_1:=\{x\in\omega\;
:\; \super v(x)<u(x)\}$ is non empty. Clearly $\omega_1$ is open.
Working if necessary with a connected component of $\omega_1$, we
assume that $\omega_1$ is connected. We prove by induction that
$\super v\ge u_k$ in $\omega_1$ for all $k\in\N$. Passing to the
limit as $k\to\infty$, we then obtain a contradiction with the
definition of $\omega_1$.

By assumption, $\super v\ge \sub u=u_{0}$ in $\omega_1$. Given
$k\ge1$, assume that $\super v\ge u_{k-1}$ in $\omega_1$. In
particular, we have that $\super v(x)\in[\min \sub u,\max\super u]$
for $x\in \omega_1$.

Observe that if $x\in\partial\omega_1$ then either $\super
v(x)=u(x)$, either $x\in\partial\omega$, whence $\super v(x)\ge
u(x)$. Since $u\ge u_k$, we conclude that $\super v\ge u\ge u_k$ on
$\partial\omega_1$. Hence,
\begin{equation*}
\left\{
\begin{split}
\Delta (\super v- u_k)-\Lambda (\super v - u_k) \le f(\super v) - f(u_{k-1})-\Lambda (\super v - u_{k-1})\le0 \; in \; \mathcal D'(\omega_1) \\
\super v- u_k\ge 0 \; on \; \partial\omega_1.\\
\end{split}
\right.
\end{equation*}
By the (weak) Maximum Principle, $\super v\ge u_{k}$ in $\omega_1$.

\begin{rmk}\label{remmax}{\rm Applying the Minimality Principle to \eqref{dirichlet} with nonlinearity $-f$,
we also obtain the existence and uniqueness of a maximal solution
relative to $\super u$, defined in the straightforward way.}
\end{rmk}

\begin{rmk}\label{remsim}
{\rm Assume $\Omega=B_R$ is a ball centered at the origin and $g$ is
a positive constant. If $\sub u$ is radial , one easily sees that
the minimal solution $u$ relative to $\sub u$ is radial : just apply
the Minimality Principle \ref{mp} with $\super v(x)=u(O(x))$, where
$O\in\mathcal{O}_D$ is an arbitrary rotation of the Euclidean space.
A well-known result of Gidas-Ni-Nirenberg \cite{GNN} states that
{\bf any\rm} solution $u> g$ is radially symmetric, provided $f$ is
e.g. locally Lipschitz.

Finally, letting $\phi(r)=u(x)$ for $r=|x|$, it follows from
standard ODE theory that $\phi'(0)=0$ and $\phi'(r)>0$ in $(0,R)$.}
\end{rmk}


\begin{cor}\label{mpb}(Minimality Principle for blow-up solutions)
Let $\om$ be a bounded domain of $\mathbb{R}^D$ such that all
boundary points are regular, $f\in C(\R)$. Assume there exist a
function $\sub u\in C(\bom)$ such that $\Delta\sub u\ge f(\sub u)$
in $\mathcal D'(\om)$ and a function $v\in C(\om)$ such that $\Delta
v\le f(v)$ in $\mathcal D'(\om)$, $\lim_{x\to x_0}v(x)=+\infty$ for
all $x_0\in\partial\Omega$ and $v\ge\sub u$.

Then there exists a unique solution $u\in C(\om)$ of \eqref{problem}
such that $\sub u\le u$ and $u|_{\omega} \le \super v$ for any open
subset $\omega\subset\Omega$ and any $\super v\in C(\omega)$
satisfying
\begin{equation}\label{localblowupsupersolution}
\left\{
\begin{aligned}
\Delta  \super v &\le f(\super v)& \; in \; \mathcal D'(\omega), \\
\super v&\ge \sub u &\; in \; \omega,\\
\lim_{x\to x_0}\super v(x)&=+\infty &\; for \; all \;
x_0\in\partial\omega.
\end{aligned}
\right.
\end{equation}
We call $u$ the minimal solution of \eqref{problem} relative to
$\sub u$.
\end{cor}

For a proof, see the Appendix.

\begin{rmk}{\rm
In contrast with Remark \ref{remmax}, there \it does not \rm exist
in general a maximal boundary blow-up solution of \eqref{problem}.
See {\bleu Section \ref{6.3}} for enlightening counter-examples.}
\end{rmk}

\subsection{Existence of solutions on some
ball}\label{subsection2.2}
%We first prove that \eqref{KO} is sufficient.
In this section, we prove that \eqref{KO} implies the existence of a
boundary blow-up solution on some ball. First, we state and prove a
useful technical lemma

\begin{lem}\label{lemmetechnique}
Let $\phi\in C^2(0,R)$ be a nondecreasing function solving
\begin{equation}\label{uradial0}
\phi''+\frac{D-1}{r}\phi'=f(\phi)\quad\text{in $(0,R)$}
\end{equation}
Then, given $0<r_1<r_2<R$,
$$
\frac{1}{\sqrt
2}\int_{\phi(r_1)}^{\phi(r_2)}\frac{1}{\sqrt{F(s)-F(\phi(r_1))}}\,ds\ge
\frac{1}{D-2}{r_1}\left(1-\left(\frac{r_1}{r_2}\right)^{D-2}\right),
$$
if $D\neq2$ and
$$
\frac{1}{\sqrt
2}\int_{\phi(r_1)}^{\phi(r_2)}\frac{1}{\sqrt{F(s)-F(\phi(r_1))}}\,ds\ge
{r_1}\ln{\frac{r_2}{r_1}},
$$
if $D=2$.
\end{lem}

\noindent Proof: for $r\in (r_1,r_2)$, \eqref{uradial0} is
equivalent to
\begin{equation}\label{uradial2}
\frac{d}{dr}\left(r^{D-1}\phi'\right)=r^{D-1}f(\phi).
\end{equation}
Multiplying the above equation by $r^{D-1}\phi'$ and integrating
between $r_1$ and $r$, we obtain
\begin{equation*}
\begin{split}
\frac12r^{2D-2}\left(\phi'(r)\right)^2&\ge
\frac12\left(r^{2D-2}\left(\phi'(r)\right)^2-r_1^{2D-2}\left(\phi'(r_1)\right)^2\right)\\
&= \int_{r_1}^r t^{2D-2}f(\phi(t))\phi'(t)\,dt\\ &\ge
r_1^{2D-2}\left( F(\phi(r))-F(\phi(r_1)) \right).\\
\end{split}
\end{equation*}
So,
$$
\frac1{\sqrt
2}\frac{\phi'(r)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}\ge\left(\frac{r_1}{r}\right)^{D-1}.
$$
Integrating the above equation between $r_1$ and $r_2$, we obtain
the desired result.\qed

Assume now \eqref{KO} holds for some $\alpha>0$. If $D\neq2$, assume
temporarily that $\Phi(\alpha)<\frac1{|D-2|}$. Applying Proposition
\ref{mis} with $\sub u=0$ and $\super u =\alpha$, let $u$ be the
minimal solution relative to $\sub u$ of
\begin{equation*}
\left\{
\begin{split}
\Delta  u = f(u) \; in \; B_{1}, \\
u= \alpha \; on \; \partial B_{1}.\\
\end{split}
\right.
\end{equation*}
Using Remark \ref{remsim} and letting $\tilde\alpha=u(0)$,
$\phi(r):=u(x)$ for $r=|x|$ solves \eqref{uradial0}, subject to the
initial conditions $\phi(0)=\tilde{\alpha}$ and $\phi'(0)=0$. $\phi$
can thus be extended on some maximal interval $(0,R)$.
{\bleu Assume temporarily that  $R<\infty$. Then $u$ is a boundary blow-up solution on $B_R$.
Indeed, by definition of $R$, we must have either $\phi(R)=+\infty$ or $\phi'(R)=+\infty$. In the latter case, multiply
\eqref{uradial0} by $\phi'$ and integrate between $0$ and $r$ to obtain that
$
\frac12 (\phi')^2\le F(\phi)
$. Hence $F(\phi(R))=+\infty$, $\phi(R)=+\infty$ and $u$ is a boundary blow-up solution. It remains to prove that $R<\infty$. }

Assume by contradiction that $R=\infty$. Apply Lemma
\ref{lemmetechnique} between $r_1=1$ and $r_2>1$ :
$$
\Phi(\alpha)\ge\frac{1}{D-2}\left(1-\left(\frac{1}{r_2}\right)^{D-2}\right),
$$
if $D\ge3$ and
$$
\Phi(\alpha)\ge \ln r_2,
$$
if $D=2$. Letting $r_2$ converge to $\infty$, we obtain a
contradiction if {\bleu either $D=2$ or} $\Phi(\alpha)<\frac{1}{|D-2|}$.

If $D\neq2$ and $\Phi(\alpha)\ge\frac{1}{|D-2|}$, choose $K>0$ so
large that $\frac{1}{K}\Phi(\alpha)<\frac1{|D-2|}$. The above proof
provides a boundary blow-up solution $u$ of \eqref{problem} on some
ball $B_R$, when $f$ is replaced by ${K^2}f$. $\tilde u(x):=u(x/K)$
is then a boundary blow-up solution of \eqref{problem} with
nonlinearity $f$ on $B_{RK}$.\qed
\begin{rmk}\label{remarksomeball}{\rm
Let $B$ be a ball of radius $R$ and assume $\sub u\in C(\bar B)$ is
such that $\Delta \sub u\ge f(\sub u)$ in $B$. Assume $\eqref{KO}$
holds for some $\alpha\ge\sup_B \sub u$ and let $\super u=\alpha$.
Using Proposition \ref{mis}, let $u$ be the minimal solution
relative to $\sub u$ of
\begin{equation*}
\left\{
\begin{split}
\Delta  u = f(u) \; in \; B, \\
u= \alpha \; on \; \partial B.\\
\end{split}
\right.
\end{equation*}
Repeating the above proof, we conclude that $u$ can be extended to a
radially symmetric boundary blow-up solution on some ball $\tilde B$
of radius $\tilde R>R$, satisfying $u\ge\sub u$ in $B$.}
\end{rmk}
%\
%
%Conversely, we prove that \eqref{KO} is sufficient. Take a boundary
%blow-up solution $v$, solving \eqref{problem} on some ball $B_R$.
%Using Proposition \ref{mis} with $\sub u=0$, $\super u=v$,
%$\Omega=B_r$, $r<R$ and $g=\sup_{\partial B_r}v$, one can easily
%construct a radial boundary blow-up solution. Applying Lemma
%\ref{lemmetechnique2}, we then have
%$$
%\Phi(\alpha)\le R<\infty.
%$$
%
%\qed
\subsection{Existence of solutions on small balls}\label{subsection2.3} Assume
\eqref{SKO} holds. By Theorem \ref{th1}, \eqref{problem} has a
solution on some ball and we may define
$$
R_0:=\inf\{R>0\; :\; \text{\eqref{problem} has a solution in
$B_R$}\}
$$
We assume by contradiction that $R_0>0$. Let $(\beta_n)$ be a
sequence of real numbers increasing to infinity and satisfying
$$
\lim_{\beta_n\to\infty}\Phi(\beta_n)=0.
$$
Applying Proposition \ref{mis} with $\sub u=0$ and $\super u
=\beta_n$, let $u_n$ be the minimal solution relative to $\sub u$ of
\begin{equation*}
\left\{
\begin{split}
\Delta  u_n = f(u_n) \; in \; B_{R_0/2}, \\
u_n= \beta_n \; on \; \partial B_{R_0/2}.\\
\end{split}
\right.
\end{equation*}
By Remark \ref{remsim}, letting $\alpha_n=u_n(0)$,
$\phi_n(r):=u_n(x)$ for $r=|x|$ solves \eqref{uradial0} with initial
conditions $\phi_n(0)=\alpha_n$ and $\phi_n'(0)=0$. By definition of
$R_0$, $\phi_n$ can be extended so that $\phi_n$ remains a solution
of \eqref{uradial0} in $(0,R_0)$. Now apply Lemma
\ref{lemmetechnique} with $r_1=R_0/2$ and $r_2=R_0$ :
$$
\Phi(\beta_n)\ge
\frac{1}{D-2}\frac{R_0}{2}\left(1-\left(\frac{1}{2}\right)^{D-2}\right),
$$
if $D\ge3$ and
$$
\Phi(\beta_n)\ge \frac{R_0}{2}\ln(2),
$$
if $D=2$. Passing to the limit as $n\to\infty$, we obtain a
contradiction in both cases. We have just proved that
\begin{equation}\label{infR}
\inf\{R>0\; :\; \text{\eqref{problem} has a solution in $B_R$}\}=0.
\end{equation}


\begin{rmk}\label{remarksmallballs}{\rm
Let $B$ be a ball of radius $R$ and assume $\sub u\in C(\bar B)$ is
such that $\Delta \sub u\ge f(\sub u)$ in $B$. Using Remark
\ref{remarksomeball} and working as above, one can show that {\bleu
$\inf\{\tilde R>R\; :\; \text{\eqref{problem} has a solution $u$ in
$B_{\tilde R}$ such that $u\ge\sub u$ in $B$}\}=R$}.}
\end{rmk}

\subsection{Existence of solutions on smooth domains}\label{subsection2.4} We assume
here that (\ref{SKO}) is valid. Applying Proposition \ref{mis} with
$\sub u=0$ and $\super u=N$, $N\in\N$, let $u_N$ be the minimal
solution relative to $\sub u$ of \eqref{dirichlet} with $g\equiv N$. For $x\in\Omega$,
choose a ball $B(x,r)\subset\om$ such that there exists a boundary
blow-up solution $u_r$ on $B(x,r)$. This is always possible since
\eqref{infR} holds. Applying the Minimality Principle \ref{mp} with
$\bar v=u_r$, we conclude that $0\le u_N\le u_r$ in $B(x,r)$. In
particular, the sequence $(u_N)$ is uniformly bounded in $B(x,r/2)$.

Let $K$ denote an arbitrary compact subset of $\om$. Covering $K$ by
finitely many balls $B(x_i,r_i/2)$, we conclude that $(u_N)$ is
uniformly bounded on $K$ by a constant depending only on $K$ and
$f$. Applying the Minimality Principle \ref{mp} with $\bar
v=u_{N+1}$, we can also infer that $(u_N)$ is a nondecreasing
sequence. Using these two facts and elliptic regularity, we conclude
that $(u_N)$ converges to a function $u$ solving $\Delta u = f(u)$
in $\Omega$.

Fix a point $x_0\in\dom$ and an arbitrary sequence $(x_k)$ in $\om$
converging to $x_0$. Then, since $u\ge u_N$,
$$
\liminf_{k\to\infty} u(x_k)\ge \liminf_{k\to\infty} u_N(x_k)=N.
$$
Letting $N$ converge to infinity, we conclude that $u$ is a boundary
blow-up solution of \eqref{problem} in $\om$.

%\subsubsection{Existence of solutions on arbitrary domains}
%Let $(\om_n)$ denote a sequence of smooth domains such that
%$\om_n\subset\om$ and $\cup\om_n=\om$. By the previous section,
%there exists a blow-up solution $u^n$ on each $\om_n$. We take the
%minimal such solution associated to $\sub u=0$. By the Minimality
%Principle \ref{mp}, $u^{n+1}\le u^n$ in $\om_{n}$. In particular,
%$(u^n)$ is uniformly bounded in any compact subset of $\om$ and must
%therefore converge to some function $u\in C(\om)$ solving $\Delta u
%= f(u)$ in $\om$.
%
%Fix $N>0$. Applying Proposition \ref{mis} with $\sub u=0$ and
%$\super u=N$, let $u_N$ be the minimal solution relative to $\sub u$
%of \eqref{equn}. Working on any subdomain $\omega$ strongly included
%in $\Omega_n$ and such that $u^n\ge N$ on $\partial\omega$, the
%Minimality Principle \ref{mp} implies that $u^n\ge u_N$ in
%$\Omega_n$. Passing to the limit as $n\to\infty$, we conclude that
%$u\ge u_N$ in $\Omega$. Hence, given $x_0\in\dom$,
%$$
%\liminf_{x\to x_0} u(x) \ge \liminf_{x\to x_0} u_N(x)=N.
%$$
%Since this holds true for all $N>0$, we conclude that $u$ is a
%boundary blow-up solution on $\Omega$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%Fix $x_0\in\dom$ and an arbitrary sequence $(x_k)$ in $\om$
%converging to $x_0$. Let $M>0$. Since $u_n=\infty$ on
%$\partial\om_n$, there exist $k_0,n_0\in\N$ such that $u_n(x_k)\ge
%M$ for $n\ge n_0$ and $k\ge k_0$.

%Then
%\begin{pro}\label{away}
%Assume that the Sharpened Keller-Osserman condition \eqref{SKO}
%holds true. Consider any radially symmetric solution $u(x)=\phi(r)$
%to $\Delta u=f(u)$ such that $r\rightarrow \phi(r)$ is
%nondecreasing. Then for any $K$ compact subset, there exists $M$
%that depends only on $K$ and $f$ such that $\sup_{x\in K}|u(x)|\leq
%M$.
%\end{pro}
%\noindent Proof: we argue by contradiction. Assume that there exists
%$\varepsilon>0$ in $(0,1)$ and a sequence of solutions $\phi_n$ and
%$r_n$ in $[0,1]$ such that $\phi_n(r_n)\rightarrow +\infty$ and
%$r_n\leq 1-\varepsilon$. Choosing $r_n$ closer to 1 if necessary, we
%may also assume that $\phi_n(r_n)=\beta_n$, where $\beta_n$ is such
%that $\lim_{n\to\infty}\Phi(\beta_n)=0$.
%Since $\phi_n$ is
%nondecreasing we may assume without loss of generality that $r_n\geq
%\frac{1}{2}$.
%For any radially symmetric solution the following ODE is valid
%\begin{equation}\label{ODE}
%\frac{\partial}{\partial r}(r^{D-1}\dot{\phi})=r^{D-1}f(\phi).
%   \end{equation}
%\noindent Then for $r_n\leq r\leq 1$,
%\begin{equation}\label{C1}
%r^{2D-2}|\dot{\phi}_n(r)|^2=r_n^{2D-2}|\dot{\phi}_n(r_n)|^2
%+2\int_{r_n}^r s^{2D-2}f(\phi_n(s))\dot{\phi}_n(s)ds.
%\end{equation}
%\noindent We now use that $f(\phi_n(s))\dot{\phi}_n(s)\geq 0$ to
%obtain
%\begin{equation}\label{C2}
%r^{2D-2}|\dot{\phi}_n(r)|^2\geq 2r_n^{2D-2}\int_{r_n}^r
%f(\phi_n(s))\dot{\phi}_n(s)ds=
%2r_n^{2D-2}(F(\phi_n(r))-F(\phi_n(r_n)).
%\end{equation}
%\noindent Therefore
%\begin{equation}\label{C3}
%\frac{\dot{\phi}_n(r)}{\sqrt{F(\phi_n(r))-F(\phi_n(r_n)}}\geq
%\sqrt{2}r_n^{D-1},
%\end{equation}
%\noindent that we integrate to obtain
%\begin{equation}\label{C4}
%\int_{r_n}^r\frac{\dot{\phi}_n(s)}{\sqrt{F(\phi_n(s))-F(\phi_n(r_n))}}ds\geq
%\sqrt{2}r_n^{2D-2}(r-r_n),
%\end{equation}
%\noindent and then perform the change of variable $t=\phi_n(s)$
%\begin{equation}\label{C5}
%\int_{\phi_n(r_n)}^{\phi_n(r)}\frac{dt}{\sqrt{F(t)-F(\phi_n(r_n))}}\geq
%\sqrt{2}r_n^{D-1}(r-r_n).
%\end{equation}
%\noindent Let now majorize $\phi_n(r)<+\infty$ and then let
%$r\rightarrow 1$ to obtain
%\begin{equation}\label{C6}
%\int_{\phi_n(r_n)}^{+\infty}\frac{dt}{\sqrt{F(t)-F(\phi_n(r_n))}}\geq
%\sqrt{2}r_n^{D-1}\varepsilon.
%\end{equation}
%\noindent Let $n$ goes to $+\infty$ contradicts (\ref{MKOstrong})
%since $r_n\geq \frac{1}{2}$. $\square$

\subsection{Proof of Theorem \ref{th1} completed.}

By Section \ref{subsection2.2}, we know that if \eqref{KO} holds,
there exists a blow-up solution on some ball. Conversely, assume
that $u>0$ solves \eqref{problem} on some ball $B$ of radius $R$ {\bleu centered at the origin}.
By Corollary
\ref{mpb}, we may always assume that $u$ is the minimal solution
relative to $\sub u=0$ of \eqref{problem}. In particular $u$ is
radial and we define $\phi(r)=u(x)$ for $r=|x|$, so that $\phi$
solves \eqref{uradial2} in $(0,R)$. Multiplying \eqref{uradial2} by
$r^{D-1}\phi'$ and integrating between $0$ and $r$, we obtain
$$
\frac12 r^{2D-2}\phi'(r)^2 = \int_0^r
t^{2D-2}f(\phi(t))\phi'(t)\;dt\le
r^{2D-2}\left[F(\phi(r))-F(\phi(0))\right].
$$
Integrating once more between $0$ and $R$,
\begin{equation}\label{estimation-Phi}
0\le\int_0^R\frac{\phi'(r)}{\sqrt{2\left[F(\phi(r))-F(\phi(0))\right]}}\;dr\le
R,
\end{equation}
which implies \eqref{KO} with $\alpha=\phi(0)$.

\subsection{Proof of Theorem \ref{th2} completed.}

By Section \ref{subsection2.4}, we know that if \eqref{SKO} holds,
there exists a blow-up solution on any domain. Conversely, given
$n\in\N$, assume that $u_n>0$ solves \eqref{problem} on the ball $B$
of radius $1/n$ centered at the origin. By Corollary \ref{mpb}, we
may always assume that $u_n$ is the minimal solution relative to
$\sub u=0$. In particular $u_n$ is radial. Let now $\beta_n=u_n(0)$.
We claim that $(\beta_n)$ is unbounded. Taking a subsequence if
necessary, we {\bleu then have} that $\lim_n \beta_n=\infty$ and
\eqref{SKO} follows from \eqref{estimation-Phi} applied with
$R=1/n$.

It remains to prove that $(\beta_n)$ is unbounded. If not, up to a
subsequence, $(\beta_n)$ converges to some $\beta\ge0$. By
\eqref{estimation-Phi} applied with $R=1/n$, we have
\begin{equation*}
0\le\int_{\beta_n}^\infty\frac{dt}{\sqrt{2\left[F(t)-F(\beta_n)\right]}}\;dr\le
1/n.
\end{equation*}
By Fatou's lemma, we conclude that
\begin{equation*}
\int_\beta^\infty\frac{dt}{\sqrt{2\left[F(t)-F(\beta)\right]}}\;dr=0,
\end{equation*}
which is not possible.

\section{Asymptotic behaviour of solutions}
\subsection{Blow-up rate of radially symmetric solutions}
\begin{pro}\label{existenceball}
Assume that $f$ satisfies the Keller-Osserman condition \eqref{KO}.
Assume $\phi$ is a radially symmetric and monotone boundary blow-up
solution on the unit ball. Then, for $r\sim 1$,
\begin{equation}\label{asymptotic}
\int_{\phi(r)}^{+\infty}\frac{dt}{\sqrt{F(t)}}\sim \sqrt{2}(1-r)
\end{equation}
\end{pro}

Proof: %We assume for simplicity that $D\ge3$, the cases $D=1,2$ being straightforward adaptations.
Multiplying \eqref{uradial2} by $r^{D-1}\phi'$ and
integrating by parts, we easily obtain that given $r\in(0,1)$,
\begin{equation}\label{eq-Gphi}
    \frac{\left(\phi'\right)^2(r)}{2} = F(\phi(r)) - G_\phi(r),
\end{equation}
where
\begin{equation*}
    G_\phi(r)=\frac{2D-2}{r}\int_0^r\left(\frac{s}{r}\right)^{2D-1}F(\phi(s))ds.
\end{equation*}
We claim that
\begin{equation}\label{asymptotics-of-Gphi}
    G_\phi(r)= o (F(\phi(r))),\qquad\text{as $r\to1$.}
\end{equation}
Let indeed $\epsilon>0$. Then, since $F$ is nondecreasing,
\begin{align*}
\frac{G_\phi(r)}{F(\phi(r))}
&=\frac{2D-2}{r}\int_0^{1-\epsilon}\left(\frac{s}{r}\right)^{2D-1}\frac{F(\phi(s))}{F(\phi(r))}ds
\\
&\qquad\qquad\qquad\qquad\qquad +\frac{2D-2}{r}\int_{1-\epsilon}^r\left(\frac{s}{r}\right)^{2D-1}\frac{F(\phi(s))}{F(\phi(r))}ds.\\
&\le C \frac{F(\phi(1-\epsilon))}{F(\phi(r))}  \,\,\,\,\qquad
+C\epsilon.
\end{align*}
Letting $r\to1$ and then $\epsilon\to0$, we obtain the desired
result. Returning to \eqref{eq-Gphi}, we obtain
\begin{equation*}
    1 - \frac{\phi'}{\sqrt{2F(\phi)}} = 1 -
    \left[1-\frac{G_\phi}{F(\phi)}\right]^{1/2}.
\end{equation*}
Combining this with \eqref{asymptotics-of-Gphi}, it follows that for
$r\sim1$,
\begin{equation*}
    1 - \frac{\phi'}{\sqrt{2F(\phi)}} \sim \frac{G_\phi}{2F(\phi)}
\end{equation*}
and, integrating between $r$ and $1$,
\begin{equation*}
    (1-r) - \int_{\phi(r)}^\infty \frac{dt}{\sqrt{2F(t)}} \sim
    \int_r^1 \frac{G_\phi(s)}{2F(\phi(s))}ds{\bleu = o(1),}
\end{equation*}
which implies \eqref{asymptotic}.

%Consider any blow up solution $\phi$ to the equation that is
%radially symmetric and monotone.
%\begin{equation}\label{uradial}
%\phi''+\frac{D-1}{r}\phi'=f(\phi)\quad\text{in $(0,R)$}
%\end{equation}
% for $r\in (r_1,1)$, \eqref{uradial} is equivalent
%to
%$$
%\frac{d}{dr}\left(r^{D-1}\phi'\right)=r^{D-1}f(\phi).
%$$
%\noindent that we multiply by $r^{D-1}\phi'$. We then have,
%integrating between $r_1$ and $r$ (we use the fact that
%$f(\phi)\phi'\ge 0$)
%\begin{equation}\label{C7}\begin{split}
%2r_1^{2D-2}(F(\phi(r))-F(\phi(r_1)))\leq
%r^{2D-2}|\dot{\phi}(r)|^2-r_1^{2D-2}|\dot{\phi}(r_1)|^2\leq \\
%2r^{2D-2}\int_{r_1}^{r} f(\phi(z))\dot{\phi}(z)dz=
%2r^{2D-2}(F(\phi(r))-F(\phi(r_1))). \end{split}\end{equation}
%
%\noindent Therefore, skipping some unnecessary terms
%\begin{equation}\label{C8}\begin{split}
%\sqrt{2}r_1^{D-1}\leq
%\frac{\dot{\phi}(r)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}\leq \left(
%\frac{\dot{\phi}^2(r_1)}{F(\phi(r))-F(\phi(r_1))}+2\right)^{1/2}\leq
%\\ \frac{\dot{\phi}(r_1)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}+\sqrt{2}.
% \end{split}\end{equation}
%
%\noindent We first proceed to the lower bound for the blow up. From
%(\ref{C8}), we also have
%
%\begin{equation}
%\frac{\dot{\phi}(r)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}\leq
%\frac{\dot{\phi}(r_1)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}+\sqrt{2}.
%\end{equation}
%
%\noindent We integrate in $r$ between $r$ and $1$ (still with
%$r>r_1$) to obtain
%
%\begin{equation}\begin{split}
%\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)}}\leq
%\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)-F(\phi(r_1))}}\leq \\
%\int_r^1\frac{\dot{\phi}(r_1)ds}{\sqrt{F(\phi(s))-F(\phi(r_1))}}+\sqrt{2}(1-r)\leq
%\\
%\sqrt{2}(1-r)\left(\frac{\dot{\phi}(r_1)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}+1\right)
%\end{split}\end{equation}
%\noindent We let $r\rightarrow 1$ to get
%\begin{equation}
%\limsup\left(
%\frac{1}{1-r}\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)}}\right)\leq
%\sqrt{2}.
%\end{equation}
%\noindent We now proceed to the upper bound. We integrate in $r$
%between $r$ and $1$ (still with $r>r_1$) the inequality
%\begin{equation}
%\sqrt{2}r_1^{D-1}\leq
%\frac{\dot{\phi}(r)}{\sqrt{F(\phi(r))-F(\phi(r_1))}}.
% \end{equation}
%\noindent to obtain
%\begin{equation}
%\sqrt{2}r_1^{D-1}(1-r)\leq
%\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)-F(\phi(r_1))}}.
% \end{equation}
% \noindent Fix $\varepsilon$ a small positive number and chose
% $r_1$ such that $r_1^{D-1}=1-\varepsilon$. On the one hand, for $r$
% close enough to $1$ we have
%\begin{equation}
%\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)-F(\phi(r_1))}}\leq
%(1+\varepsilon)\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)}} .
% \end{equation}
% \noindent Therefore
% \begin{equation}
%\sqrt{2}(1-\varepsilon)\leq (1+\varepsilon) \liminf\left(
%\frac{1}{1-r}\int_{\phi(r)}^{+\infty}\frac{dz}{\sqrt{F(z)}}\right).
%\end{equation}
%\noindent $\varepsilon \rightarrow 0$ concludes the proof. $\square$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{The general case}
%
%In this section $\Omega$ is a bounded open connected domain of
%$\mathbb{R}^D$ whose boundary is smooth. We now state
%
%\begin{thm}
%Assume that the Modified Keller-Osserman condition is valid. Then
%there exists a boundary blow up solution to (\ref{problem}).
%\end{thm}
%
%\noindent Proof: we consider the increasing sequence $u_N$ provided
%by Proposition \ref{mis}. The main issue is to bound this sequence
%away from the boundary. We begin with a local argument. Consider a
%ball $B(x_0,\alpha)$ whose closure is included in $\Omega$.
%Introduce $\phi_\alpha$ the radially symmetric solution on that ball
%which blows up at $\partial B(x_0,\alpha)$, that was constructed in
%the previous section. $\phi_\alpha$ is a supersolution for the
%problem
%\begin{equation}\label{localproblem}
%\begin{split}
%\Delta u_N=f(u_N) \; in \; B(x_0,\alpha), \\
%u_N\leq N \; on \; \partial B(x_0,\alpha).  \\
%\end{split}
%\end{equation}
%\noindent If we go back to the proof of the monotone iteration
%scheme, one can prove that $u_N\leq \phi_\alpha$ on the ball. Since
%the Modified Keller-Osserman condition is valid, we now have a bound
%for $\phi_\alpha$ and then to $u_N$ in a smaller ball, say
%$B(x_0,\alpha')$, for $\alpha'<\alpha$.
%\smallskip
%We now extend this local argument. Consider a compact subset $K$ of
%$\Omega$. We can find a finite covering of $K$ by $\cup
%B(x_j,\alpha'_j)$ whose closures are in $\Omega$. For each $j$, we
%introduce $\alpha_j>\alpha'_j$ and we apply the local argument. The
%bound on $K$ follows promptly.
%\smallskip
%We then pass to the limit when $N$ converges to $+\infty$. Setting
%$u(x)=\lim u_N(x)=\sup_Nu(x)$, due to standard elliptic estimates we
%are able to pass to the limit on any compact subset of $\Omega$ in
%the equation $\Delta u=f(u)$. We just have to prove that $u$ blows
%up at the boundary. For that purpose, let us write
%$N=\lim_{x\rightarrow \partial\Omega}u(x)\leq \liminf_{x\rightarrow
%\partial\Omega}u(x)$ and then let $N$ converges to $+\infty$. $\square$
%
%
%\begin{rmk}
%Actually $u$ constructed by this argument is the smaller boundary
%blow-up solution to that problem. In fact any other boundary blow-up
%solution is a supersolution for the problem (\ref{Nsolution}); then we pass
%to the limite when $N$ goes to the infinity.
%\end{rmk}
%
%\begin{rmk}
%Assume that we can approximate any point $x_0$ of $\partial\Omega$
%by a sequence of balls $B_n$ tangent to the boundary whose radii
%converge to $0$. Let $\phi_n$ be the radially symmetric solution
%constructed on $B_n$ that blows up on its boundary. Can we compare
%$\phi_n$ and $u$ on $B_n$ to get a lower bound for the blow up ? not
%so sure. What about the upper bound ?
%\end{rmk}

\subsection{Blow-up rate of solutions on smooth domains} Let $u$ be
a blow-up solution on a domain $\Omega$, which satisfies an interior
and an exterior sphere condition at any boundary point. Fix
$x_0\in\dom$ and let $B_R\subset\om$ denote a small ball which is
tangent to $\dom$ at $x_0$. Fix $\eta\in(0,1)$. Let $\sub
u:=u|_{B_{\eta R}}$. By Remark \ref{remarksomeball}, there exists a
radial boundary blow-up solution $v$ defined on some ball $\tilde
B\supset B_{\eta R}$, such that $v\ge \sub u$ in $B_{\eta R}$. Let
$K>0$ such that $K\tilde B=B_{R}$ and let $v_K(x):=v((x-x_1)/K+x_1)$, where $x_1$ is the center of $B_{R}$.
Then $v_K$ solves
\begin{equation*}
\begin{split}
\Delta v_K=\frac1{K^2}f(v_K) \; in \; B_{R}, \\
v_K=+\infty \; on \; \partial B_{R}.  \\
\end{split}
\end{equation*}
%Given any $N\in\N$ such that $u|_{B_{\eta R}}\le N$, apply
%Proposition \ref{mis} with $\sub u=u|_{B_{\eta R}}$ and $\super
%u=N$. We obtain the minimal solution $u_N$ relative to $\sub u$ of
%\begin{equation*}
%\left\{
%\begin{split}
%\Delta  u_N = f(u_N) \; in \; B_{\eta R}, \\
%u_N= N \; on \; \partial B_{\eta R}.\\
%\end{split}
%\right.
%\end{equation*}
%Let $(R_n)$ denote a minimizing sequence for \eqref{infR}. Choosing
%$R>0$ so that $\eta R=R_n$, we conclude that there exists a boundary
%blow-up solution $w$ on $B_{\eta R}$. By the Minimality Principle,
%$u_N\le w$  in $B_{\eta R}$, hence $(u_N)$ is uniformly bounded on
%compact subsets of $B_{\eta R}$.

%Then $(u_N)$ converges to a radial boundary blow-up solution $v$ on
%$B_{\eta R}$. Remembering that $v\ge \sub u=u|_{B_{\eta R}}$ and
Since $u(x_1)\le v(x_1)= v_K(x_1)$, Proposition \ref{existenceball}
implies that
$$
K\int_{u(x_1)}^{+\infty}\frac{dt}{\sqrt{F(t)}}\ge
K\int_{v_K(x_1)}^{+\infty}\frac{dt}{\sqrt{F(t)}}\sim \sqrt{2} R.
$$
Letting $R\to0$, we then have
$$
K\liminf_{x\to
x_0}\frac{\int_{u(x)}^{+\infty}\frac{dt}{\sqrt{F(t)}}}{\delta(x)}\ge
\sqrt 2.
$$
By Remark \ref{remarksmallballs}, we may take $K$ arbitrarily close
to $1/\eta$. Also, $0<\eta<1$ was chosen arbitrarily, so letting
$K,\eta\to1$, we finally obtain
$$
\liminf_{x\to
x_0}\frac{\int_{u(x)}^{+\infty}\frac{dt}{\sqrt{F(t)}}}{\delta(x)}\ge
\sqrt 2.
$$
Choose another ball $B_{R'}\subset\R^N\setminus\bom$ which is
tangent to $\dom$ at $x_0$ and a concentric ball $B_{R''}$ with
$R''>R'$ so large that $\Omega\subset B_{R''}$. Finally, let
$A=B_{R''}\setminus B_{R'}$. Let $v$ denote the minimal boundary
blow-up solution (relative to $\sub u=0$) on $A$. By the Minimality
Principle \ref{mpb}, we deduce that $u\ge v$ in $\om$. Applying
Proposition \ref{existenceball} (which still holds on an annulus)
with $v$, we conclude that
$$
\limsup_{x\to
x_0}\frac{\int_{u(x)}^{+\infty}\frac{dt}{\sqrt{F(t)}}}{\delta(x)}\le
\limsup_{x\to
x_0}\frac{\int_{v(x)}^{+\infty}\frac{dt}{\sqrt{F(t)}}}{\delta(x)}
\le\sqrt 2.
$$
{\bleu This finishes the proof of Theorem \ref{blow-up-rate}. {\bleu
More can be said about the asymptotic behaviour of solutions
provided $F$ satisfies some extra growth assumption :
\begin{lem}\label{lemsaussol}
Let $u,v$ denote two radially symmetric boundary blow-up solutions defined on the unit ball $B$.
Assume there exist $\beta>0$ and $M>0$ such that $\frac{F(v)}{v^2}\ge \beta^2\frac{F(u)}{u^2}$ whenever $M\le u\le v$.
Then $u(r)\sim v(r)$ on $\partial B.$
\end{lem}

\noindent Proof: We recall from the proof of
of Proposition \ref{existenceball} that for $\phi$ a radially
symmetric boundary blow up solution and for $r\sim1$,
\begin{equation*}
    (1-r) - \int_{\phi(r)}^\infty \frac{dt}{\sqrt{2F(t)}} \leq C
    \int_r^1 \frac{G_\phi(s)}{F(\phi(s))}ds \leq  C\int_r^1
    \frac{\phi(s)}{\sqrt{F(\phi(s))}}ds.
\end{equation*}
\noindent Using that $\frac{\dot{\phi(s)}}{\sqrt{2F(\phi(s))}}\sim
1$, we then obtain that, introducing $K(r)$ such that
$1-r=\int_{K(r)}^\infty \frac{dt}{\sqrt{2F(t)}}$,

\begin{equation}\label{saussol}
\int_{K(r)}^{\phi(r)} \frac{dt}{\sqrt{2F(t)}} \leq
C\int_{\phi(r)}^\infty \frac{t}{F(t)}dt.
\end{equation}

\noindent Since $F$ is increasing, we thus obtain

\begin{equation}\label{saussol2}
\left(1-\frac{K(r)}{\phi(r)} \right)\leq
C\frac{\sqrt{F(\phi(r))}}{\phi(r)}\int_{\phi(r)}^\infty
\frac{t}{F(t)}dt.
\end{equation}

\noindent Since $\frac{F(v)}{v^2}\geq \beta^2 \frac{F(u)}{u^2}$  for $u\leq v$ large
enough, (\ref{saussol2}) implies that

\begin{equation*}
\left(1-\frac{K(r)}{\phi(r)} \right)\leq
\frac{C}{\beta}\int_{\phi(r)}^\infty \frac{dt}{\sqrt{F(t)}}.
\end{equation*}

\noindent The classical Keller-Osserman \eqref{KO} condition gives the result.
$\square$

\begin{cor}
Assume either that $f$ is convex on some interval $[a,+\infty)$ or that $f(t)/t$ is nondecreasing on $[a,+\infty)$.
Then the result of the previous lemma holds.
\end{cor}

\noindent Proof : Assume $f$ is convex in $[a,+\infty)$ and let
$G(t)=F(t+a)-F(a)-f(a)t$. Then $G(0)=G'(0)=0$ and $G'$ is convex in
$\R^+$. So $G(t)/t^2$ is nondecreasing, i.e. $t\to \frac{F(t+a)-
L(t)}{t^2}$ is nondecreasing, where $L(t)=F(a)+f(a)t$ is affine.
Observe that $\lim_{t\to\infty} F(t)/t^2=+\infty$ since the
Keller-Osserman condition \eqref{KO} implies
$$
\frac{u}{2\sqrt{F(u)}}\le\int_{u/2}^{u}\frac{dt}{\sqrt{F(t)}}=
o(1)\qquad\text{as $u\to\infty$}.
$$
It follows that there exists $\beta>0$ such that
$\frac{F(v)}{v^2}\ge \beta^2\frac{F(u)}{u^2}$ for $u\le v$ large
enough and we may apply Lemma \ref{lemsaussol}. The case where
$f(t)/t$ is nondecreasing on $[a,+\infty)$ is similar, so we skip
it. }
\section{The uniqueness result}

\noindent
\begin{lem}
Assume that $f$ is {\bleu nonde}creasing on $[0,+\infty)$ and convex
in a neighborhood of $+\infty$ (say $[a,+\infty)$). Consider two
radially symmetric boundary blow-up solutions such that $u(r)\leq
v(r)$ on $ B(0,R)$ then in fact $u=v$ everywhere.
\end{lem}

\noindent Proof: Set $R=1$ for the sake of simplicity. Since $f$ is
{\bleu nonde}creasing, either $u(0)=v(0)$ (and then $u=v$
everywhere) or $u(r)<v(r)$ everywhere since
$\partial(r^{D-1}(\dot{v}-\dot{u}))=r^{D-1}(f(v)-f(u))$ {\bleu so}
the map $r\rightarrow v(r)-u(r)$ is {\bleu nonde}creasing. Assume
then $u(0)<v(0)$.

Let $\varepsilon>0$. Consider the set $\omega_\varepsilon=\{r \in
[0,1); \forall s<r, \; (1+\varepsilon)u(s)<v(s) \}$. If
$\varepsilon$ is small enough, $0\in\omega_\varepsilon$. Due to
lemma {\bleu \ref{lemsaussol}}, $R=1\not\in\omega_\varepsilon$ since
$u\sim v$ close to the boundary. Then introduce
$r_\varepsilon^0=\sup\omega_\varepsilon$, {\bleu which} satisfies
$0<r_\varepsilon^0<1$. We now have

\begin{equation}\label{coucou}
v(0)-u(0)\leq v(r_\varepsilon^0)-u(r_\varepsilon^0)=\varepsilon
u(r_\varepsilon^0).
\end{equation}

\noindent Then either $ r_\varepsilon^0$ converges towards $R=1$
when $\varepsilon \rightarrow 0$ or, letting $\varepsilon
\rightarrow 0$ $u(0)=v(0)$ and the proof is over.

\noindent Introduce now $a$ such that $f$ is convex on
$[a,+\infty)$. Introduce $R_0$ such that $u(r)\geq a$ for $r \geq
R_0$. Then for $\varepsilon$ small enough $r_\varepsilon^0>R_0$. Set
$w(r)=(1+\varepsilon)u(r)$. Then, on the annulus
$R_0<s<r_\varepsilon^0$, using the convexity

\begin{equation}\begin{aligned}\label{coucou2}
\Delta(v-w)&=f(v)-(1+\varepsilon)f(u)\\
&\geq
f(v)-(1+\varepsilon)(f(w)-f(a)\frac{u-a}{w-a}-(1+\varepsilon)f(a)
\\ &\geq f(v)-f(w)+\frac{\varepsilon}{w-a}(af(w)-wf(a)).
\end{aligned}\end{equation}

\noindent Observe now that the map $X\rightarrow
\frac{Xf(a)-af(X)}{X-a}$ is majorized by some constant $C$ for
$X\geq a$. Then introducing $\chi$ that satisfies $-\Delta \chi=1$
with homogeneous Dirichlet condition at $R=1$ ($\chi(1)=0$)

\begin{equation}\label{prout}
\Delta(v-w-C\varepsilon\chi)\geq 0
\end{equation}

\noindent and by the maximum principle, for any $r$ in
$R_0<r<r_\varepsilon^0$

{\bleu
\begin{equation}\label{glou}
v(r)-w(r)-C\varepsilon\chi(r)\leq
\max(-C\varepsilon\chi(r_\varepsilon^0) ,
v(R_0)-w(R_0)-C\varepsilon\chi(R_0)).
\end{equation}
}
\noindent Then letting $\varepsilon \rightarrow 0$ we obtain that
for any fixed $r$ such that $R_0<r<1$,

\begin{equation}\label{vlan}
v(r)-u(r)\leq v(R_0)-u(R_0).
\end{equation}

\noindent Since the map $r\rightarrow v(r)-u(r)$ is {\bleu
nonde}creasing, this implies that $v(r)-u(r)$ is {\it constant} on
$[R_0,1)$. By standard ODE theory, this implies that $v$ and $u$
coincide everywhere on $[0,1]$. $\square$

\begin{rmk}
{\rm Since $f$ is nondecreasing, there exists $\sub U,\super U$ the
minimal and the maximal boundary blow-up solutions of the problem
(the latter can be obtained e.g. as the monotone limit of $u(R)$ as
$R\to 1^-$, where $u(R)$ denotes the minimal boundary blow-up
solution on $B_R$). Clearly both $\sub U$ and $\super U$ are radial
and they coincide by the previous lemma. Since any solution $u$ of
the problem must stay between $\sub U$ and $\super U$, Theorem
\ref{th-uniqueness} follows. Alternatively, according to a result of
Poretta-Veron \cite{PV}, any boundary blow-up solution is radially
symmetric if $f$ is convex in a neighborhood of $+\infty$, whence
again Theorem \ref{th-uniqueness} follows from the previous lemma.}
\end{rmk}

\begin{rmk}{\rm
The previous lemma is still valid if we substitute the assumption
$\frac{f(u)}{u}$ increasing in a neighborhood of  infinity to the
convexity assumption. Since the proofs are easier they are left as
an exercice to the reader.}
\end{rmk}}


%\subsection{The case of $\Omega=B$ the unit ball of $\mathbb{R}^D$}
%
%To begin with, we state
%\begin{thm}\label{uniquenessontheball}
%Consider the problem (\ref{problem}), where $f$ satisfies the
%classical Keller-Osserman condition and if
%$\frac{F(u)}{u^{2-\varepsilon}}$ is nondecreasing for large $u$s
%there exist a unique radially symmetric boundary blow-up solution on
%$\Omega=B$ the unit ball of $\mathbb{R}^D$.
%\end{thm}
%
%\begin{rmk}
%In the one dimensional case, we may consider the case
%$\varepsilon=0$. We introduce such small number for technical
%computations.
%\end{rmk}
%
%
%\subsubsection{Existence of the minimal boundary blow-up solution}
%
%We just have to prove that the Modified Keller-Osserman condition is
%valid. For that purpose we state and prove
%
%\begin{lem}
%If $f$ satisfies the classical Keller-Osserman condition and if
%$\frac{F(u)}{u^2}$ is nondecreasing for large $u$s, then the
%modified Keller-Osserman condition is valid in its strong form.
%\end{lem}
%
%\noindent Proof: for any $C\geq 1$,
%\begin{equation}\label{C10}
%\int_\lambda^{+\infty}\frac{dt}{\sqrt{F(t)-F(\lambda)}}\leq
%\int_\lambda^{C\lambda}\frac{dt}{\sqrt{F(t)-F(\lambda)}}+
%\int_{C\lambda}^{+\infty}\frac{dt}{\sqrt{F(t)-F(\lambda)}}.
%\end{equation}
%\noindent One the one hand, since $\frac{F(u)}{u^2}$ is
%nondecreasing for $\lambda$ large enough, then $F(t)-F(\lambda) \geq
%(\frac{t^2}{\lambda^2}-1)F(\lambda)$ and then
%\begin{equation}\label{C11}
%\int_\lambda^{C\lambda}\frac{dt}{\sqrt{F(t)-F(\lambda)}}\leq
%\frac{1}{\sqrt{F(\lambda)}}\int_\lambda^{C\lambda}\frac{dt}{\sqrt{\frac{t^2}{\lambda^2}-1}}\leq
%\frac{\lambda}{\sqrt{F(\lambda)}}\int_1^{C}\frac{dt}{\sqrt{t^2-1}}.
%\end{equation}
%On the other hand, since $\frac{F(u)}{u^2}$ is nondecreasing for
%$\lambda$ large enough, then for $t$ larger than $C\lambda$,
%$F(t)-F(\lambda) \geq (1-\frac{1}{C})F(t)$ and then
%\begin{equation}\label{C12}
%\int_{C\lambda}^{+\infty}\frac{dt}{\sqrt{F(t)-F(\lambda)}}\leq
%\sqrt{\frac{C}{C-1}}\int_{C\lambda}^{+\infty}\frac{dt}{\sqrt{F(t)}}.
%\end{equation}
%\noindent We now let $\lambda$ converge to $+\infty$ in (\ref{C11})
%and (\ref{C12}). $\square$

%\subsubsection{Uniqueness of radially symmetric boundary blow-up solution}

%We now prove Theorem \ref{uniquenessontheball}. Consider $u$ the
%minimal boundary blow-up solution constructed in the Section 2. Set
%$u(x)=u(r)$ for the sake of simplicity. Consider $v(r)$ another
%radially symmetric boundary blow up solution. Introduce $H(r)=\ln
%(\frac{v}{u})$. Since $u\leq v$ then $H$ is nonnegative. We know
%that $u$ and $v$ have the same behavior near the boundary (see
%Theorem \ref{existenceball}). Then $H(1)=0$. We now compute
%\begin{equation}\label{C13}
%\dot{H}(r)= \frac{\dot{v}}{v}-\frac{\dot{u}}{u}.
%\end{equation}


%\noindent On the other hand we have (see (\ref{C1}) pretending that
%$r_n=0$)
%\begin{equation}\label{C14}
% g(r)=r^{2D-2}\frac{\dot{v}^2}{v^2}(r)= \frac{1}{v^2(r)}\int_0^r
% s^{2D-2}f(v(s)\dot{v}(s)ds.
%\end{equation}
%\noindent Due to the implicit function theorem $r$ is also a
%function of $v$ and  we then have (the dot represents the derivative
%with respect to $r$)
%\begin{equation}\label{C15}\begin{split}
% \frac{\partial}{\partial v} g= -\frac{2}{v^3}\int_0^r
% s^{2D-2}f(v(s)\dot{v}(s)ds+ \frac{1}{v^2}\frac{\partial}{\partial
% r}(\int_0^r  s^{2D-2}f(v(s)\dot{v}(s)ds)\frac{\partial r}{\partial
% v}=\\
%-\frac{2}{v^3}\int_0^r
% s^{2D-2}f(v(s)\dot{v}(s)ds+\frac{1}{v^2}r^{2D-2}f(v(r))=\\
%-\frac{2r^{2D-2}}{v^3}F(v(r))+
%(2D-2)\int_0^rs^{2D-3}F(v(s))ds+\frac{r^{2D-2}}{v^3}v(r)f(v(r)).
%\end{split}\end{equation}
%\noindent At that stage, we observe that
%$\frac{1}{v^{3-\varepsilon}}\left((2-\varepsilon)F(v)-vf(v)
%\right)=-\frac{\partial}{\partial
%v}\frac{F(v)}{v^{2-\varepsilon}}\geq 0$. Therefore
%\begin{equation}\label{C16}
%\frac{v^3}{r^{2D-2}}\frac{\partial}{\partial v} g\leq -\varepsilon
%F(v)+\frac{(2D-2)}{r^{2D-2}}\int_0^rs^{2D-3}F(v(s))ds.
%\end{equation}
%\noindent Let us pretend tat we are able to prove that
%\begin{equation}\label{C17}
%\lim_{r\rightarrow 1} \frac{1}{F(v(r))}\int_0^rs^{2D-3}F(v(s))ds=0
%\end{equation}
%Then $\dot{H}\leq 0$ for $r$ close to $1$ (since in that range $u$
%and $v$ are large). Since $H(1)=0$, we just have proved that $u=v$
%in a neighborhood of $1$. Then, by Cauchy-Lipschitz Theorem, $u=v$
%everywhere. To complete the proof of the proposition, we check
%(\ref{C17}). Consider $\alpha$ a small positive number. Then, for
%$r\geq 1-\alpha$,
%\begin{equation}\label{C18}\begin{split}
%\int_0^rs^{2D-3}F(v(s))ds\leq \int_0^\alpha
%s^{2D-3}F(v(s))ds+\int_\alpha^r s^{2D-3}F(v(s))ds\leq
%\\F(v(\alpha))\int_0^\alpha s^{2D-3}ds+ \alpha F(v(r)).
%\end{split}\end{equation}
%\noindent $\alpha$ being fixed, we now divide by $F(v(r))$ and let
%$r\rightarrow 1$. $\square$

\section{Discrete equations}

We are concerned with finite difference approximations of
(\ref{problem}) {\bleu when $D=1$ or $D=2$} on a cube or a ball.
{\bleu After introducing some notation, we observe that both the
maximum principle and the minimality principle extend to the case of
finite difference operators.}  We conclude this section with some
theoretical error estimates, assuming that $f$ is a nondecreasing
function.

\subsection{Finite differences}

To begin with, consider the interval $[-1,1]$ or the unit square
$[-1,1]^2$. Consider a uniform grid {\bleu$\Omega_h$} {\bleu which}
mesh size is $h=\frac{1}{L}$ for some integer $L$. The nodes on the
grid are respectively $jh$ {\bleu if D=1}, $-L\leq j\leq L$, or
$(ih,jh)$ {\bleu if $D=2$}, with $-L\leq i,j\leq L$.

The discrete Laplace operator is then defined on each point/node of
the grid respectively by
\begin{equation}\label{discretelaplace1D}
(\Delta_h U)_j= \frac{1}{h^2}(-2U_{j}+U_{j+1}+U_{j-1}),
\end{equation}
{\bleu if $D=1$ and}
\begin{equation}\label{discretelaplace2D}
(\Delta_h U)_{i,j}=
\frac{1}{h^2}(-4U_{i,j}+U_{i,j+1}+U_{i,j-1}+U_{i+1,j}+U_{i-1,j}),
\end{equation}
if $D=2$. {\bleu In the above, } $U$ is a vector in
$\mathbb{R}^{2L+1}$ (or $\mathbb{R}^{(2L+1)2}$ in 2D) which
components are the $U_j\simeq u(jh)$ (or $U_{i,j}\simeq u(ih,jh)$).
{\bleu If $D=2$}, we then solve $\Delta_h U=f(U)$, i.e. $(\Delta_h
U)_{i,j}=f(U_{i,j})$ {\bleu for all interior nodes $(i,j)$ and set
$U_{\pm L,j}=U_{i,\pm L}=N$ at all boundary nodes, where $N$ is a
fixed large constant. We work accordingly when $D=1$.}

It is standard to prove that the matrix $\Delta_h$ has positive
inverse, i.e. the entries of the inverse matrix are positive.
Therefore, the maximum principle is valid (see \cite{Ciarlet}).
Actually, if $U$ satisfies $\Delta_hU\leq 0$ on the interior nodes
of the grid and $U\geq 0$ on the boundary, then $U\geq 0$
everywhere. Here and throughout this section we write $U\geq 0$ iff
$U_{i,j}\geq 0$ for all $(i,j)$ nodes of the grid. We shall use the
same notation for $B$ a matrix: $B\geq 0$ iff the entries of $B$ are
all non negative.

{\bleu When working on the unit ball, we use a slightly modified
scheme.} Focussing on radially symmetric functions, we approximate
the equation
\begin{equation}
\frac{1}{r} \frac{\partial}{\partial r}(r\frac{\partial u}{\partial
r}) =f\qquad\text{for $r\in (0,1)$}.
\end{equation}
Discretize  $[0,1]$ by setting $Lh=1$, $0\leq j\leq L$. At $j=L$ set
$U_L=N$ (boundary condition). For {\bleu $0<j<L$} solve
\begin{equation}\frac{1}{r}\partial
(r\partial u)\simeq \frac{1}{jh} D^+ (jh D^- U)=
\frac{-2U_j+U_{j+1}+U_{j-1}}{h^2}+\frac{U_{j+1}-U_j}{jh^2}=F_j,
\end{equation}
\noindent where $(D^+U)_j=\frac{U_{j+1}-U_j}{h}$,
$(D^-U)_j=\frac{U_{j}-U_{j-1}}{h}$. It remains to define the
equation at $j=0$. For that purpose, we use the symmetry property
$u(h)=u(-h)$ and the approximation
$\frac{\dot{u}(0)}{0}=\ddot{u}(0)$ to set
$$ \frac{4}{h^2}(-U_0+U_1)=F_0.$$
This approximation of the Laplace operator satisfies the maximum
principle. {\bleu Indeed, it can be easily checked that} if $F_j\geq
0$ then $j\rightarrow U_j$ is increasing. The {\bleu maximum
principle} follows promptly.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Computing an approximation}

We aim to solve the following problem
\begin{equation}
\label{problemdicretfini}
\begin{split}
\Delta_h u_h=f(u_h) \; in \; \Omega, \\
u_h=N \; on \; \partial\Omega,  \\
\end{split}
\end{equation}
for $N$ large enough.

We expect that $u_h$ {\bleu is} an approximation for $u$, the
minimal boundary blow up solution corresponding to
$\underline{u}=0$. {\bleu As in Proposition \ref{mis}, $u_h$ is
obtained by monotone iteration,} starting from the discrete
subsolution 0. We claim that Proposition \ref{mis} and Corollary
\ref{mp} are valid for the finite difference approximation. The
proof follows the guidelines of the continuous case and is left as
an exercice to the reader.

An approximation of the solution to problem
(\ref{problemdicretfini}) is recursively obtained by the following
discrete iterative scheme:

Consider {\bleu $u^k\in \mathbb{R}^{(2L+1)D}$}, where
$L=\frac{1}{h}$, recursively defined by
\begin{equation}\label{algo1}
%\begin{split}{l}
\begin{array}{l}
u^0=0\qquad\text{and for\ $k\geq 0$,}\\
\\
\left\{\begin{array}{l}
(\Delta_h-\Lambda_N Id)u^{k+1}=f(u^k)-\Lambda_Nu^k \quad in\ \Omega_h\\
u^{k+1}=N\qquad on\ \partial\Omega_h\\
\end{array}\right.
\end{array}%\end{split}
\end{equation}
where $\Lambda_N=\sup_{[0,N]}f'$.

Therefore the error between $u$, the minimal boundary blow up
solution, and $u^k$ the $k^{th}$ iterate of (\ref{algo1}) can be
{\bleu split} as follows:

\begin{equation}
I_h(u)-u^k=I_h(u-u_N)+(I_h(u_N)-u_h)+(u_h-u^k)
\end{equation}

where $I_h$ is the interpolation operator defined by
$I_h(u)_i=u(x_i)$ when $D=1$ (respectively by $I_h(u)_a=u(a)$ when
$D=2$ for a node $a=(ih,jh)$ on the grid), and $u_N$ is the solution
of
\begin{equation}\label{problemfini}
\left\{\begin{aligned}
\Delta u=f(u) \; in \; \Omega, \\
u=N \; on \; \partial\Omega.  \\
\end{aligned}\right.
\end{equation}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Error estimate}

Throughout this section, we assume that $f$ is a convex increasing
function and that $\Omega=[-1,1]^D$. We first bound from above the
rate of convergence of the algorithm (\ref{algo1}).
\begin{lem}\label{lem52}
Let $u^k\in\mathbb{R}^{(2L+1)D}$ be given by \eqref{algo1} and
$u_h=\lim_{k\to\infty}u^k$. Then, there exist constants $C=C(D)$,
$\mu=\mu(D)>0$ and $\lambda=\lambda(f)<\Lambda_N$ such that
 \begin{equation}
 \|u^k-u_h\|_{\ell^{\infty}} \leq C\left(1-\frac{\lambda}{\Lambda_N}\right)^{k}
 \min\left(1,\frac1{h^{D/2}}\left[1-\frac{\mu}{\Lambda_N}+\frac{\mu^2}{\Lambda_N^2}\right]^{k}\right)
  \|u^0-u_h\|_{\ell^{\infty}}
 \end{equation}
 \end{lem}

\noindent Proof : In the sequel let us denote by $a$ a node of the
grid (that is $a=ih$ in 1D or $a=(ih,jh)$ in 2D). $(u^k)$ is a
nondecreasing sequence in ${\mathbb R}^{(2L+1)D}$ (i.e. $u^k_a\leq
u^{k+1}_a$ for each node $a$). By the mean value theorem, there
exists $\theta_a\in (u^k_a,(u_h)_a)$ such that
\begin{equation}
f(u^k_a)-f((u_h)_a)=f'(\theta_a)(u^k_a-(u_h)_a).
\end{equation}

\noindent Therefore

\begin{equation}\label{cpe}
\left|\left((u_h)_a-\frac{f((u_h)_a)}{\Lambda_N}\right)-\left((u^k)_a-\frac{f((u^k)_a)}{\Lambda_N}\right)\right|
\leq (1-\frac{\lambda}{\Lambda_N})((u_h)_a-u^k_a),
\end{equation}
\noindent where $\lambda=\inf f'$.

\noindent On the other hand
\begin{equation}\label{employabilite}
u_h-u^k=\left(Id-\frac{\Delta_h}{\Lambda_N}\right)^{-1}\left(\left(u_h-\frac{f(u_h)}
{\Lambda_N}\right)-\left(u^k-\frac{f(u^k)}{\Lambda_N}\right)\right),
\end{equation}
\noindent where $f(u)$ denotes the vector which components are
$f(u)_a=f(u_a)$. The key argument is to observe that the matrix
$Id-\frac{\Delta_h}{\Lambda_N}$ satisfies the maximum principle.
Therefore,
\begin{equation}\label{flexibilite}
0\leq u_h-u^{k+1}\leq
\left(1-\frac{\lambda}{\Lambda_N}\right)\left(Id-\frac{\Delta_h}{\Lambda_N}\right)^{-1}(u_h-u^k),
\end{equation}
\noindent where inequalities hold component by component. We thus
obtain that

\begin{equation}\label{separabilite}
\|u_h-u^{k}\|_{\ell^{\infty}} \leq
\left(1-\frac{\lambda}{\Lambda_N}\right)^k\left\|\left(Id-\frac{\Delta_h}{\Lambda_N}\right)^{-k}\right\|_{\mathcal
L(\ell^{\infty})}\|u_h-u^0\|_{\ell^{\infty}}.
\end{equation}

\noindent On the one hand, the maximum principle implies that
\begin{equation}\label{precarite}
\left\|\left(Id-\frac{\Delta_h}{\Lambda_N}\right)^{-k}\right\|_{\mathcal
L(\ell^{\infty})}\le C,
\end{equation}
for some constant $C$ depending only on the dimension $D$.

On the other hand, since the spectrum of $-\Delta_h$ lies in a
segment $[\mu, \frac{\mu}{h^2}]$ (see \cite{Ciarlet}) and
$\frac1{\sqrt{(2/h+1)D}}\|\cdot\|_{\ell^{2}}\le
\|\cdot\|_{\ell^{\infty}}\le \|\cdot\|_{\ell^{2}}$ in
$\mathbb{R}^{(2L+1)D}$, we get

\begin{equation}\label{abrogation}
||(Id-\frac{\Delta_h}{\Lambda_N})^{-k}||_{\mathcal
L(\ell^{\infty})}\leq C
h^{-D/2}||(Id-\frac{\Delta_h}{\Lambda_N})^{-k}||_{\mathcal
L(\ell^{2})}=C
h^{-D/2}\left(\frac{1}{1+\frac{\mu}{\Lambda_N}}\right)^{k}.
\end{equation}
\noindent Using that $\frac{1}{1+\frac{\mu}{\Lambda_N}}\leq
1-\frac{\mu}{\Lambda_N}+\frac{\mu^2}{\Lambda_N^2}$ and collecting
\eqref{separabilite}, \eqref{precarite} and \eqref{abrogation}, the
proof is over. $\square$

\bigskip

We now provide an upper bound for $u_h-I_h(u_N)$.
\begin{lem}\label{lem53}
Assume $f$ is convex. Let $u_h$ be the solution of
(\ref{problemdicretfini}) and $u_N$ be the solution of
(\ref{problemfini}). Then,
\begin{equation}
\|u_h-I_h(u_N)\|_{\ell^\infty}\leq Ch^2\alpha(N,f)
\end{equation}
where {\bleu $\alpha(N,f)=\|u^{(4)}_N\|_{L^\infty}$} is a constant
depending only on $N$ and $f$.
\end{lem}

\noindent Proof:

For the sake of simplicity we will denote $I_h(u_N)$ by ${u_N}$,
this introduces no confusion.

We write down the proof for the 2D problem, leaving the 1D case as
an exercise to the reader. Let $a=(ih,jh)$ be a node on the grid. By
the mean value theorem, for each node $a$, there exist $\xi,\eta$ in
$\mathbb R^2$ such that
\begin{equation}
|\xi - a|<h\ {\rm and}\ |\eta-a|<h,
\end{equation}
and
\begin{equation}
\Delta_h(u_N)_a-(\Delta u_N)_a=ch^2(\frac{\partial^4 u}{\partial
x^4}(\xi) + \frac{\partial^4 u}{\partial y^4}(\eta)  ).
\end{equation}
Therefore
\begin{equation}
\left\{\begin{array}{l} \Delta_h(u_h-u_N)_a=f(u_h)_a-f(u_N)_a -ch^2
(\frac{\partial^4 u}{\partial x^4}(\xi)
+ \frac{\partial^4 u}{\partial y^4}(\eta)  )\ in\ \Omega_h\\
(u_h-u_N)_a=0 \ on\ \partial \Omega_h\\
\end{array}\right.
\end{equation}
Consider $\omega=u_h-u_N$, then working as in the previous lemma we
obtain
\begin{equation}
w_a=\left[(\Delta_h+\Lambda_N)^{-1}
\left(Id-D\left(\frac{f'(\theta)}{\Lambda_N}\right)\right)\Lambda_N\right]
ch^2 \left(\frac{\partial^4 u}{\partial x^4}(\xi) + \frac{\partial^4
u}{\partial y^4}(\eta)  \right),
\end{equation}
where $\theta_a\in(u_N(a),(u_h)_a)$.

We therefore obtain
\begin{equation}
\|w_a\|_{\ell^{\infty}} \leq
\frac{1}{\Lambda_N}\frac{4}{1+\frac{4}{\Lambda_N}}
\|\Delta_h^{-1}\|_{\mathcal L(\ell^{\infty})} Ch^2
\|u^{(4)}_N\|_{L^{\infty}},
\end{equation}
where {\bleu $\|u^{(4)}_N\|_{L^{\infty}}=\max_{|\alpha|\leq
4}(\|\partial^{\alpha}u_N\|_{L^\infty})$}.\qed

\noindent
\begin{rmk}
{\rm When $D=1$ and $f(u)=u^p$ with $p\geq 2$, the constant
$\alpha(N,f)$ is given by
$$
\alpha(N,f)=\|u_N^{(4)}\|_{L^\infty}=\frac{p(3p-1)}{p+1} N^{2p-1}.
$$}
\end{rmk}
\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Error estimate for$\|u-u_N\|$}

Assume here that the Sharpened Keller-Osserman condition (\ref{SKO})
is valid. Consider then a sequence {\bleu$(\alpha_N)$} such that
$\Phi(\alpha_N)$ converges towards $0$. Consider the minimal
solution $u_N$ of

\begin{equation}\label{a}\begin{split}
\Delta u_N&=f(u_N) \; \quad{\rm in} \; \Omega=B(0,1), \\
u_N&=\alpha_N \; \quad{\rm on} \; \partial\Omega.
\end{split}\end{equation}

\noindent Then one may wonder how $u_N$ approximates the minimal
boundary blow up solution $u$ defined on the unit ball. We first
state a qualitative result

\begin{pro}
There exists $R_N$ such that $u_N$ is a boundary blow up solution on
$B(0,R_N)$. Moreover
\begin{equation}\label{b}
R_N-1 \sim
\frac{1}{\sqrt{2}}\int_{\alpha_N}^{+\infty}\frac{dt}{\sqrt{F(t)}}.
\end{equation}
\end{pro}

\begin{rmk}{\rm
This proposition shows that when we plot {\bleu the} approximation
$u_N$, we plot in fact a boundary blow-up solution on a ball that is
close to the unit ball.}
\end{rmk}

\noindent Let us proceed to the proof in the case where $D\geq 3$.
The cases $D=1,2$ are very similar and then omitted. Assume that
$u_N$ extends to $\mathbb{R}^D$. Then by Lemma \ref{lemmetechnique}

\begin{equation}\label{c}
\frac{1}{\sqrt{2}}\int_{u_N(1)}^{u_N(r)}\frac{dt}{\sqrt{F(t)-F(u_N(1))}}\geq
\frac{1}{D-2}(1-(\frac{1}{r})^{D-2}).
\end{equation}

\noindent Here we have used that $u_N$ is radially symmetric.
Therefore since $u_N(1)=\alpha_N$, $\Phi(\alpha_N)\geq
\frac{1}{D-2}(1-(\frac{1}{r})^{D-2})$. $N \rightarrow +\infty$ leads
to a contradiction. The estimate (\ref{b}) comes from Proposition
\ref{existenceball}. $\square$

\begin{rmk}{\rm
Observe that if $f(u)=u^p$, then
$u_N(r)=R_N^{-\frac{2}{p-1}}u(rR_N)$. In that case, $0\leq
u(r)-u_N(r)\leq C(R_N-1)(\dot{u}(r)+u(r)).${\bleu The inequality} is
sharp for some numerical constant $C$. To prove the estimate in a
more general context, we need extra hypotheses.}
\end{rmk}

\begin{defi}\label{d}
{\rm Consider $g:[0,+\infty)\rightarrow [0,+\infty)$ a function. We
say that $g$ is {\it strongly increasing} if the function
$$\rho(\lambda)=\inf_{u\geq 0}\frac{g(\lambda u)}{g(u)}$$
\noindent is a $C^1$ increasing function on $[1,+\infty)$ that
satisfies $\dot{\rho}(1)\neq 0$.}
\end{defi}

\noindent A strongly increasing function is increasing in the usual
sense. $g(u)=u^p$, $p>0$ is strongly increasing. $g(u)=\ln(u+1)$ is
not.

\noindent We now state and prove

\begin{pro}\label{e}
Assume that $\frac{f(u)}{u}$ is strongly increasing. Then
$$0\leq u(r)-u_N(r)\leq C(R_N-1)(\dot{u}(r)+u(r)). $$
\end{pro}

\noindent Proof:  $v(r)=\lambda u_N(rR_N)$ is a blow up function on
the unit ball. We have

\begin{equation}\label{f}
\Delta v=\lambda R_N^2 f\left(\frac{v}{\lambda}\right) \leq
\frac{R_N^2}{\rho(\lambda)}f(v).
\end{equation}

\noindent For $N$ large enough, we choose $\lambda_N$ close to $1$
such that $\rho(\lambda_N)=R_N^2$. Then $v$ is a blow-up
supersolution to (\ref{problem}). Since $u$ is the minimal blow up
solution, then $u(r)\leq v(r)$. Therefore

\begin{equation}\label{g}\begin{split}
0\leq u(r)-u_N(r)\leq u(r)-\frac{1}{\lambda_N}{\bleu
v}(\frac{r}{R_N})\leq
\\ (R_N-1)\dot{u}(r)+(1-\frac{1}{\lambda_N})u(rR_N^{-1})\leq
C(R_N-1)(\dot{u}(r)+u(r)),
\end{split}\end{equation}
\noindent since $(1-\frac{1}{\lambda_N})\leq
2\frac{R_N-1}{\dot{\rho}(1)}$. $\square$



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Numerical computations}

In this section, we present some numerical results obtained with our
method of approximation.
\begin{rmk}{\rm
At this stage, we would like to point out that our method is
self-contained, and does not use the knowledge of the boundary blow
up behavior of the solution. In fact, as in \cite{Ki}, one can
introduce another approximate problem such as taking
$\Omega_{\varepsilon}\subset\Omega$ where
$dist(\Omega,\Omega_{\varepsilon})\leq \varepsilon$, and solve the
problem
\begin{equation}
\left\{\begin{aligned}
\Delta u_\varepsilon=f(u_\varepsilon) \; in \; \Omega_\varepsilon, \\
u_\varepsilon(x)=K(x) \; on \; \partial\Omega_\varepsilon,  \\
\end{aligned}\right.
\end{equation}
where
$$
\frac{1}{\sqrt{2}}\int_{K(x)}^{+\infty} \frac{dt}{\sqrt{F(t)}} =
dist(x,\partial \Omega).
$$}
\end{rmk}
We discuss our numerical results successively on three examples:
\begin{itemize}
\item $f(u)=u^2$ (i.e. $\frac{f(u)}{u}$ is increasing),
\item $f(u)=u^2(2+\cos u)$,
\item $f(u)=u^2(1+\cos u)$.
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{$f(u)=u^2$}

Since $\frac{f(u)}{u}$ is increasing, on any domain we have a unique
boundary blow up solution (see e.g. \cite{AR} and references
therein).

 \begin{figure}[th]
\centerline{\epsfig{file=phi2.eps,width=7cm,angle=-90}}
%\vspace*{8pt}
%\centering\includegraphics{phi2.eps,width=5cm,angle=-90}
\caption{$\Phi$ when $f(u)=u^2$}\label{phi2}
\end{figure}

We see in figure (\ref{phi2}) that $\Phi$ is a strictly decreasing
function. On figure (\ref{ics}) below some boundary blow up
solutions obtained respectively on $[-1,1]$, $[-1,1]^2$,
$B(0,1)\subset \mathbb{R}^2$ (we plot $i\mapsto U(ih,0)$ in 2D).

\begin{figure}[th]
\centerline{\epsfig{file=coupicd.eps,width=7cm,angle=-00}}
\caption{Solutions on the interval, the square and the
disk}\label{ics}
\end{figure}

For the sake of completness, we plot in 1D the values of $u_h(0)$
for several values of $h$ and we compare to theorical results (see
Proposition \ref{e}).

\begin{figure}[th]
\centerline{\epsfig{file=u0-2.eps,width=7cm,angle=0}}
\caption{Convergence of $u_h(0)$ to $u(0)$ versus $h$}\label{conv}
\end{figure}
\bigskip

\begin{rmk}
{\rm The value of $u(0)$ is obtained by solving} $$\displaystyle
\frac{1}{\sqrt{2}} \int_{u(0)}^{+\infty}
\frac{dt}{\sqrt{F(t)-F(u(0))}}=1.$$
\end{rmk}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{$f(u)=u^2(2+\cos u)$}

This function $f$ satisfies the Sharpened Keller-Osserman condition
\eqref{SKO} {\bleu(see the Appendix for a proof in a similar case)}.

Figure (\ref{phi3}) below shows that $\Phi(\alpha)$ tends to 0 when
$\alpha$ tends to $+\infty$ (then the Sharpened Keller-Osserman
condition is valid).

\begin{figure}[th]
\centerline{\epsfig{file=phi3.eps,width=7cm,angle=-90}}
\caption{$\Phi$ when $f(u)=u^2(2+\cos(u))$}\label{phi3}
\end{figure}

But $\Phi$ is not a decreasing function ; for instance for {\bleu
$c\approx 0.49$} there exist $\alpha\neq\beta$ such that
$\Phi(\alpha)=\Phi(\beta)=c$. Therefore, at least in 1D, uniqueness
does not hold.
\bigskip

\noindent Remark : The uniqueness result for $f(u)=u^2(2+\cos u)$ in
$B(0,1)\subset \mathbb{R}^D$ is still an open question for $D\geq
2$.
\bigskip

We now plot an approximation of the minimal boundary blow-up radial
solution on $B(0,1)\subset \mathbb{R}^2$.

\begin{figure}[th]
\centerline{\epsfig{file=sr2.eps,width=7cm,angle=-90}}
\caption{Solution on the disk when $f(u)=u^2(2+\cos u)$}\label{sr2}
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{$f(u)=u^2(1+\cos u)$}\label{6.3}

We first state

\begin{pro}\label{skocos}
The function {\bleu $f(u)=u^2(1+\cos u)$}  satisfies  the Sharpened
Keller-Osserman condition \eqref{SKO}. Moreover,
$$
\lim_{\alpha \rightarrow (2k+1)\pi} \Phi(\alpha) = +\infty.
$$
\end{pro}

\noindent For a proof, see the Appendix (Proposition
\ref{fonction-cos}).
\bigskip

Figure (\ref{phi1}) shows that for any domain, there exist an
infinite sequence of boundary blow up solutions.

\begin{figure}[th]
\centerline{\epsfig{file=phi1.eps,width=5cm,angle=-90}}
\caption{$\Phi$ when $f(u)=u^2(1+\cos(u))$}\label{phi1}
\end{figure}

{\bleu When $D=1$, this follows from the fact that $\Phi(\alpha)=R$
admits an infinite number of solutions $\alpha$. When $D\ge2$, fix
an integer $m$ and observe that $\alpha_m=(2m+1)\pi$ is a
subsolution. Let $u_m$ denote the minimal boundary blow-up solution
relative to $\sub u=\alpha_m$. Infinitely many $u_m$'s must be
distinct. Indeed, choose $m_1$ such that $\alpha_{m_1}>u_0(0)$. Then
$u_{m_1}(0)\ge \alpha_{m_1}>u_0(0)$. Repeating this process
inductively yields infinitely many distinct solutions $u_{m_k}$.}

We plot in figure \ref{sol1} approximations of different boundary
blow-up solutions on the interval $[-1,1]$.

\begin{figure}[th]
\centerline{\epsfig{file=sol1.eps,width=7cm,angle=-90}}
\caption{Solutions when $f(u)=u^2(1+\cos u)$}\label{sol1}
\end{figure}

Figure \ref{sr3} shows two radial approximations of different
boundary blow up solutions on the unit ball in $\mathbb{R}^2$.
\begin{figure}[th]
\centerline{\epsfig{file=sr3.eps,width=5cm,angle=-90}}
\caption{Solutions on the disk when $f(u)=u^2(1+\cos u)$}\label{sr3}
\end{figure}




\section{Appendix}

\subsection{Regarding the two forms of the Keller-Osserman condition}
We present the proof of Proposition \ref{KOegalSKO} and consider the
special case $f(u)=u^2(1+\cos u)$.

\noindent Proof of Proposition \ref{KOegalSKO}. Given
$\epsilon\in(0,1)$, we recall that
$
(1-\epsilon)^{-1/2} = \sum_{n=0}^\infty a_n\epsilon^n,
$
where
\begin{equation}\label{asymptotics-an}
a_n=\frac{1}{n!}\frac{1\cdot3\cdots(2n-1)}{2^n}=\frac{(2n)!}{4^n(n!)^2}\sim\frac{C}{\sqrt
n}.
\end{equation}
By Tonelli's theorem we deduce that given $\alpha>T$
\begin{equation}\label{Phi-en-serie}
\sqrt2\Phi(\alpha) =
\int_\alpha^\infty\frac{dt}{(F(t)-F(\alpha))^{1/2}} =
\sum_{n=0}^\infty a_n F(\alpha)^n\int_\alpha^\infty
\frac{dt}{F(t)^{n+1/2}}.
\end{equation}
Choose $N\in\N$ so large that $(N+3/2)\eta>1$. Since $F$ is
nondecreasing, we have on the one hand
\begin{equation}\label{N-premiers-termes}
    \sum_{n=0}^N a_n F(\alpha)^n\int_\alpha^\infty
\frac{dt}{F(t)^{n+1/2}}\le \left(\sum_{n=0}^N a_n\right)
\int_\alpha^\infty \frac{dt}{F(t)^{1/2}}.
\end{equation}
On the other hand, since $F(t)/t^\eta$ is nondecreasing for
$t>\alpha$,
\begin{equation}\label{termes-suivants}
\begin{aligned} \sum_{n=N+1}^\infty a_n
F(\alpha)^n\int_\alpha^\infty \frac{dt}{F(t)^{n+1/2}}&\le&
\sum_{n=N+1}^\infty a_n
\frac{\alpha^{\eta(n+1/2)}}{F(\alpha)^{n+1/2}}F(\alpha)^n
\int_\alpha^\infty\frac{dt}{t^{\eta(n+1/2)}}\\
&\le&\left(\sum_{n=N+1}^\infty
a_n\frac1{\eta(n+1/2)-1}\right)\frac\alpha{\sqrt{F(\alpha)}}\\
&\le& C\frac\alpha{\sqrt{F(\alpha)}},
\end{aligned}
\end{equation}
where we used \eqref{asymptotics-an} in the last inequality. Now,
$$
\frac\alpha{\sqrt{F(\alpha)}}\le
2\int_{\alpha/2}^{\alpha}\frac{dt}{\sqrt{F(t)}} = o(1).
$$
Collecting the above inequality, \eqref{Phi-en-serie},
\eqref{N-premiers-termes} and \eqref{termes-suivants}, we obtain the
desired result by letting $\alpha\to\infty$.\qed

\

\begin{pro}\label{fonction-cos}
Let $f(u)=u^2(1+\cos u)$. Then,
\begin{align*}
\limsup_{\alpha\to\infty} \Phi(\alpha) &= \infty,\\
\liminf_{\alpha\to\infty} \Phi(\alpha) &= 0.
\end{align*}
\end{pro}

\noindent Proof. \noindent Set $\alpha=(2k+1)\pi$. For $t$ close to
$\alpha$,  $F(t)-F(\alpha)\sim \alpha^2(t-\alpha)^2$. Therefore
$\Phi(\alpha)=+\infty$. In particular, $\limsup_{\alpha\to\infty}
\Phi(\alpha) = \infty$. To compute the $\liminf$, introduce
$\varepsilon>0$. Set $\alpha=2k\pi$ and assume that $k$ is large
enough with respect to $\varepsilon$ to ensure that
$\alpha\varepsilon \geq \frac{\pi}{2}$ and that for all $u\geq
\alpha$
\begin{equation}\label{}
F(u)\geq \frac{1}{1+\varepsilon}(\frac{u^3}{3}+2u).
\end{equation}

\noindent On the one hand

\begin{equation}\label{}
\int_\alpha^{\alpha+\frac{\pi}{2}}\frac{dt}{(F(t)-F(\alpha))^{1/2}}
\leq
\int_\alpha^{\alpha+\frac{\pi}{2}}\frac{dt}{\alpha(t-\alpha)^{1/2}}
\leq C/\alpha.
\end{equation}


\noindent On the other hand


\begin{equation}\label{}
\int_{\alpha+\frac{\pi}{2}}^{\alpha(1+\varepsilon)}
\frac{dt}{(F(t)-F(\alpha))^{1/2}} \leq
\int_{\alpha}^{\alpha(1+\varepsilon)}
\frac{dt}{(F(\alpha+\frac{\pi}{2})-F(\alpha))^{1/2}} \leq
C\varepsilon.
\end{equation}

\noindent Finally, introducing $G(u)=\frac{u^3}{3}+2u$,

\begin{align}
\int_{\alpha(1+\varepsilon)}^{+\infty}
\frac{dt}{(F(t)-F(\alpha))^{1/2}} &\leq
\int_{\alpha(1+\varepsilon)}^{+\infty}
\frac{dt}{(\frac{1}{1+\varepsilon}G(t)-G(\alpha))^{1/2}}\\ &\leq
(1+\varepsilon)\int_{\alpha}^{+\infty}
\frac{ds}{(G(s)-G(\alpha))^{1/2}}.
\end{align}

\noindent This last integral converges to $0$ when $\alpha$
converges to $+\infty$, by applying Proposition \ref{KOegalSKO} to
$G$. Therefore, using Chasles theorem

\begin{equation}\label{}
\limsup_{\alpha\rightarrow +\infty}
 \Phi(\alpha) \leq 0+ \sqrt{2}C\varepsilon+0.
\end{equation}

\noindent $\varepsilon\rightarrow 0$ completes the proof of the
proposition.
\subsection{Minimality Principle}

We present here the proof of Corollary \ref{mp} and Corollary
\ref{mpb}.

\

\noindent Proof of Corollary \ref{mp}. Let $\sub u, \super u$ be the
sub and supersolution given in the statement of Proposition
\ref{mis}. Let $(I,>)$ denote the set of all finite families
containing $\super u$ of supersolutions of \eqref{dirichlet} which
stay above $\sub u$, ordered by inclusion : $i\in I$ if there exist
$n\in \N$ and supersolutions $\super v_k\in C(\bom)$, $1\le k\le n$
(i.e. \eqref{solution} holds when $\super u$ is replaced by $\super
v_k$) with $\super v_k\ge \sub u$, such that $i=\{\super u, \super
v_1,\dots,\super v_n\}$.

$I$ is non-empty since $\{\super u\}\in I$. $I$ is filtrating
increasing, i.e. if $i_1,i_2\in I$ there exists $i_3\in I$ such that
$i_3>i_1,i_2$ (take e.g. $i_3=i_1\cup i_2$). We prove that given
$i=\{\super u,\super v_1,\dots,\super v_n\}\in I$ there exists a
solution $u_i\in C(\bom)$ of \eqref{dirichlet} such that
$u_i\le\super v$ for all $\super v\in i$. Let indeed $u^0$ denote
the solution given by Proposition \ref{mis}. Following \cite{CS},
since $\sub u\le u^0\le \super u$, $u^0$ is also a solution of
\eqref{dirichlet}, when $f$ is replaced by the truncation $f^0\in
C(\bom\times\R)$ defined by
\begin{equation*}
f^0(x,u)=\left\{
\begin{split}
f(\sub u(x))& \quad\text{if $u<\sub u(x)$,}\\
f(u)& \quad\text{if $\sub u(x)\le u\le \super u(x)$,}\\
f(\super u(x))& \quad\text{if $u>\super u(x)$.}\\
\end{split}
\right.
\end{equation*}
In \cite{CS}, the authors prove that in fact \it any \rm solution
$u$ of \eqref{dirichlet} with nonlinearity $f^0$ satisfies $\sub
u\le u\le \super u$ (and solves the problem with the original
nonlinearity $f$). For convenience, we reproduce here the argument
of Cl\'ement and Sweers : take any solution $u$ of \eqref{dirichlet}
with nonlinearity $f^0$. Assume by contradiction that
$\Omega_+:=\{x\in\Omega \; :\; u(x)>\super u(x)\}$ is non-empty.
Working if necessary on a connected component of $\Omega_+$, we may
also assume that $\Omega_+$ is connected. For $x\in\dom_+$, either
$u(x)=\super u(x)$ or $x\in\dom$, so that $u(x)=g(x)\le\super u(x)$.
Hence,
\begin{equation*}
\left\{
\begin{split}
\Delta  (\super u-u)\le f(\super u)-f^0(x,u)=0 \; in \; \mathcal D'(\Omega_+), \\
\super u-u\ge 0 \; on \; \partial\Omega_+.\\
\end{split}
\right.
\end{equation*}
By the (weak) Maximum Principle, $\super u\ge u$ in $\Omega_+$,
which is a contradiction. Hence, $u\le\super u$ and we can prove
similarly that $u\ge\sub u$.

Define now the truncation $f^1\in C(\bom\times\R)$ of $f^0$
associated to $\super v_1$ by :
\begin{equation*}
f^1(x,u)=\left\{
\begin{split}
f^0(x,\sub u(x))& \quad\text{if $u<\sub u(x)$,}\\
f^0(x,u)& \quad\text{if $\sub u(x)\le u\le \super v_1(x)$,}\\
f^0(x,\super v_1(x))& \quad\text{if $u>\super v_1(x)$.}\\
\end{split}
\right.
\end{equation*}
Clearly, $\sub u$ and $\super v_1$ are a sub and a super solution of
\eqref{dirichlet} with nonlinearity $f^1$. Applying Proposition
\ref{mis} (which still holds for non-autonomous nonlinearities, see
\cite{CS}), we can thus construct a solution $u^1$ of \eqref{dirichlet} with nonlinearity $f^1$, satisfying
$\sub u\le u^1\le \super v_1$. Clearly, $u^1$ is a solution of the
problem with nonlinearity $f^0$ and, as we mentioned earlier, we
must have $u^1\le \super u$. Repeating the process inductively, we
obtain a solution $u_i:=u^n$ such that $\sub u\le u_i\le \super u,
\super v_1, \dots, \super v_n$.

Note that $u_i$ may not be unique. Nevertheless, using the Axiom of
Choice on the set of all such solutions, we can construct a
well-defined generalized sequence $(u_i)_{i\in I}$, contained in the
set $K$ of all solutions $u$ satisfying $\sub u\le u\le \super u$.

By standard elliptic estimates, $K$ is a compact subset of $C(\bom)$
so there exists a generalized subsequence $(u_{\phi(j)})_{j\in J}$
converging to a solution $u$ of \eqref{dirichlet}.

Choose now an arbitrary supersolution $\super v\ge \sub u$ and let
$i_1:=\{\super v,\super u\}\in I$. Given $\epsilon>0$, let $j_0\in
J$ such that $j>j_0\Longrightarrow
\|u_{\phi(j)}-u\|_\infty<\epsilon$. Also choose $j_1\in J$ such that
$j>j_1\Longrightarrow \phi(j)>i_1$. Finally pick $j_3>j_1,j_2$.
Then, for $j>j_3$,
$$
u\le \|u_{\phi(j)}-u\|_\infty + u_{\phi(j)}\le \epsilon + \super v.
$$
Letting $\epsilon\to0$, we conclude that $u\le\super v$ for any
supersolution $\super v\ge\sub u$. Clearly, $u$ is the unique such
solution.

It remains to prove that given any subdomain $\omega$ and any
function $\super v\in C(\bar\omega)$ satisfying
\eqref{localsupersolution}, $u\le\super v$. Fix such a function $\super v$ and define $h^k\in C(\bom\times\R)$, $k=0,1$, by
\begin{equation*}
h^0(x,t)=\left\{
\begin{split}
f(\sub u(x))& \quad\text{if $t<\sub u(x)$,}\\
f(t)& \quad\text{if $\sub u(x)\le t\le  u(x)$,}\\
f(u(x))& \quad\text{if $t>u(x)$.}\\
\end{split}
\right.
\end{equation*}
and
\begin{equation*}
h^1(x,t)=\left\{
\begin{split}
h^{1}(x,\sub u(x))& \quad\text{if $t<\sub u(x)$,}\\
h^{1}(x,t)& \quad\text{if $x\in\Omega\setminus\omega$ or if $x\in\omega$ and $\sub u(x)\le t\le \super v(x)$,}\\
h^{1}(x,\super v(x))& \quad\text{if $x\in\omega$ and $t>\super v(x)$.}\\
\end{split}
\right.
\end{equation*}
Working as before, we may solve \eqref{dirichlet} with nonlinearity $h^1$ and obtain a solution $\tilde u$ of \eqref{dirichlet} with nonlinearity $f$ such that
$\sub u\le \tilde u\le u$ and $\tilde u|_{\omega}\le \super v$. Since $\tilde u$ is a (super)solution of \eqref{dirichlet}, we also have $u\le\tilde u$.
Hence $u|_{\omega}=\tilde u|_\omega\le\super v$.


%(we call any such $\super v$ a local
%supersolution), $u\le\super v$. Let $(I_2,>)$ be the set of all
%finite families containing $u$ of local supersolutions of
%\eqref{dirichlet}, ordered by inclusion : $i\in I_2$ if there exist
%$n\in \N$ and local supersolutions $\super v_k\in C(\bar\omega_k)$,
%$1\le k\le n$, (i.e. $\omega_k$ is an open subset of $\om$ and
%\eqref{localsupersolution} holds when $\super v$ is replaced by
%$\super v_k$), such that $i=\{u, \super v_1,\dots,\super v_n\}$.
%Working as above, we can construct a generalized sequence
%$(u_{i})_{i\in I_2}$ of solutions belonging to $K$ such that
%$u_i\le\super v$ for all $\super v\in i$ : one just needs to work
%with truncations of the form $h^k\in C(\bom\times\R)$ defined
%inductively by
%\begin{equation*}
%h^0(x,t)=\left\{
%\begin{split}
%f(\sub u(x))& \quad\text{if $t<\sub u(x)$,}\\
%f(t)& \quad\text{if $\sub u(x)\le t\le  u(x)$,}\\
%f(u(x))& \quad\text{if $t>u(x)$.}\\
%\end{split}
%\right.
%\end{equation*}
%and for $k=1,\dots,n$,
%\begin{equation*}
%h^k(x,t)=\left\{
%\begin{split}
%h^{k-1}(x,\sub u(x))& \quad\text{if $t<\sub u(x)$,}\\
%h^{k-1}(x,t)& \quad\text{if $x\in\Omega\setminus\omega$ or if $x\in\omega$ and $\sub u(x)\le t\le \super v_k(x)$,}\\
%h^{k-1}(x,\super v_k(x))& \quad\text{if $x\in\omega$ and $t>\super v_k(x)$.}\\
%\end{split}
%\right.
%\end{equation*}
%Passing to a generalized subsequence, we obtain a solution $\tilde
%u$ satisfying $\sub u\le \tilde u\le \super v$ for all local
%supersolutions $\super v$. In particular $\tilde u\le\super v$ for
%all (global) supersolutions $\super v\ge\sub u$ and $\tilde u$ must
%therefore coincide with $u$. \qed

\

\noindent Proof of Corollary \ref{mpb}. Clearly, there exists a most
one such solution. Let now $N$ denote any integer larger than $\|\sub u\|_{L^\infty(\Omega)}$ and let $u_N$ denote the minimal solution relative to $\sub u$ of
\begin{equation*}
\left\{
\begin{split}
\Delta  u_N = f(u_N) \; in \; \mathcal D'(\Omega), \\
u_N= N \; on \; \partial\Omega.
\end{split}
\right.
\end{equation*}
Using the Minimality Principle \ref{mp}, one can easily show that
the sequence $(u_N)$ is nondecreasing and that $u_N\le N$. Take any
(smooth) open set $\omega\subset\subset\Omega$ such that $v\ge N$ on
$\partial\omega$. By the Minimality Principle \ref{mp} again, we
conclude that $v|_{\omega}\ge u_N|_\omega$. Since this holds for any
such $\omega$, we conclude that $v\ge u_N$ in $\Omega$. In
particular the sequence $(u_N)$ is bounded on compact subsets of
$\Omega$ and using elliptic regularity, we conclude that $(u_N)$
converges to a blow-up solution $u$ of \eqref{problem} such that
$\sub u\le u\le v$.

Now take $\omega\subset\Omega$ open and $\super v\in C(\omega)$
satisfying \eqref{localblowupsupersolution}. Take $\tilde
\omega\subset\subset\omega$ such that $\super v\ge N$ on
$\partial\tilde\omega$. Applying the Minimality Principle \ref{mp},
we have that $u_N|_{\tilde\omega}\le \super v|_{\tilde\omega}$.
Again, since $\tilde\omega\subset\subset\omega$ is arbitrary, we
conclude that $u_N|_{\omega}\le \super v$. Letting $N\to\infty$
yields the desired inequality. \qed




%\subsection{Finite differences case}
%
%The same method applies straightforwardly for the discrete Laplace
%operator on the unit interval (discretized with a mesh of size $h$)
%or in $2D$ square that reads respectively
%\begin{equation}\label{discretelaplace1D}
%(\Delta_h U)^j= \frac{1}{h^2}(-2U^{j}+U^{j+1}+U^{j-1}),
%\end{equation}
%\begin{equation}\label{discretelaplace2D}
%(\Delta_h U)^{i,j}=
%\frac{1}{h^2}(-4U^{i,j}+U^{i,j+1}+U^{i,j-1}+U^{i+1,j}+U^{i-1,j}),
%\end{equation}
%
%\noindent In fact, these operators satisfy the maximum principle
%(\cite{Ciarlet}). So $-\Delta_h+\Lambda_N Id$ do as well. For the
%case of the unit ball of $\mathbb{R}^2$, we use the following finite
%difference scheme to compute the radially symmetric solutions to
%\begin{equation}\label{radialdiscrete}
%\Delta_h U= F,
%\end{equation}
%\noindent where $F$ is radially symmetric. Discretize $r$ in $[0,1]$
%by setting $Lh=1$, $0\leq j\leq L$. At $j=L$ set $U^L=N$ (boundary
%condition). For $0<j<N$ solve
%\begin{equation}\frac{1}{r}\partial
%(r\partial u)\simeq \frac{1}{jh} D^+ (jh D^- U)=
%\frac{-2U^j+U^{j+1}+U^{j-1}}{h^2}+\frac{U^{j+1}-U^j}{jh^2}=F_j,
%\end{equation}
%\noindent where $(D^+U)_j=\frac{U^{j+1}-U^j}{h}$,
%$(D^-U)_j=\frac{U^{j}-U^{j-1}}{h}$. It remains to define the
%equation at $j=0$. For that purpose, we use the symmetry property
%$u(h)=u(-h)$ and the approximation
%$\frac{\dot{u}(0)}{0}=\ddot{u}(0)$ to set
%$$ \frac{4}{h^2}(-U^0+U^1)=F^0.$$
%This approximation of the Laplace operator does satisfies the
%maximum principle: it is an exercice to prove that if $F^j\geq 0$
%then $j\rightarrow U^j$ is increasing. The result follows promptly.

\vskip 1cm

 \centerline{ACKNOWLEDGEMENT} This article is dedicated to
the memory of T. Lachand-Robert. This work was performed with the
financial support of PAI Egide Brancusi. The authors would like to
thank H. Brezis for pointing out reference \cite{PV}. Thanks also to
H. Brezis and A. Farina for stimulating discussions and for their
invaluable mathematical culture.

\vskip 1cm

\begin{thebibliography}{99}
\bibitem{AR} A. Aftalion and W. Reichel, {\it
Existence of two boundary blow-up solutions for semilinear elliptic
equations}, J. Diff Eq., 141, (1997), 400-421.

\bibitem{BM} C. Bandle and M. Marcus, {\sl Large solutions of
semilinear elliptic equations: existence, uniqueness and asymptotic
behavior}, J. Math. Anal., 58, (1992), 9-24.

\bibitem{BM2} C. Bandle and M. Marcus, {\sl Asymptotic behaviour of solutions
and their derivatives, for semilinear elliptic problems with blow-up
on the boundary}, Ann. Inst. Henri Poincar\'e, 12 (1995), 155-171.

\bibitem{Ciarlet} P. Ciarlet, {\sl Introduction \`a l'analyse num\'erique
matricielle et \`a l'optimisation}, Masson, 1988.

\bibitem{CS2} Ph. Cl\'ement and G. Sweers, {\it Existence and
multiplicity results for a semilinear elliptic eigenvalue problem},
Ann. della Scola Norm. Sup. di Pisa, (4), 14, n 1, (1987), 97-121.

\bibitem{CS} Ph. Cl\'ement and G. Sweers, {\it Getting a solution
between sub and supersolutions without monotone iteration}, Rend.
Istit. Mat. Univ. Trieste, 19, n. 2, (1987), 189-194.

\bibitem{DG2} Y. Du and Z. Guo, {\it Liouville type results and eventual flatness of positive solutions
for $p$-laplacian equations }, Adv. in Diff. Eq., vol. 7, n 12,
(2002), 1479-1512.

\bibitem{DG} Y. Du and Z. Guo, {\it Uniqueness and layer analysis
for boundary blow-up solutions}, J. Maths. Pures et Appl. 83 (2004),
739-763.

\bibitem{GNN} B. Gidas, W-M. Ni and L. Nirenberg, {\it Symmetry
and related properties via the maximum principle}, Comm. Math. Phy.
 68, (1979), 209-243.

\bibitem{K} J. Keller, {\it On solutions to $\Delta u=f(u)$},
Comm. Pure and Applied Math. 10, (1957), 503-510.

\bibitem{Ki} S. Kishenassamy, {\it Recent progress in boundary blow
up}, Elliptic and parabolic problems, Prog. in nonlinear diff.
equations appl. 63, Birkhauser, Basel, (2005), 329-341.

\bibitem{LM} A. Lazer and P. McKenna, {\it Asymptotic behaviour of
solutions to boundary blow-up problems}, Diff. and Int. Equations,
vol 7, n 4, (1994), 1001-1019.

\bibitem{MV} M. Marcus and L. V\'eron, {\it Existence and uniqueness results for large solutions of general nonlinear elliptic
equations}, J. Evolution Equ., 3, (2003), 637-652.

\bibitem{O} R. Osserman, {\it On the inequality $\Delta u\geq
f(u)$}, Pacific J. Math. 7, (1957), 1641-1647.

\bibitem{PV} A. Poretta and L. V\'eron, {\it Symmetry of large
solutions of nonlinear elliptic equations in a ball}, to appear.


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