\documentclass[reqno]{amsart}
\usepackage{amssymb}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 20, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/20\hfil An incorrectly posed problem]
{An incorrectly posed problem for nonlinear elliptic equations}

\author[S. G. Georgiev \hfil EJDE-2009/20\hfilneg]
{Svetlin G. Georgiev}

\address{Svetlin Georgiev Georgiev \newline
Department of Differential Equations
University of Sofia, Sofia, Bulgaria}
\email{sgg2000bg@yahoo.com}

\thanks{Submitted January 6, 2008. Published January 23, 2009.}
\subjclass[2000]{35J60, 35J65, 35B05}
\keywords{Nonlinear elliptic equation; incorrectly posed problems}

\begin{abstract}
 We study properties of solutions to non-linear
 elliptic problems involving the Laplace operator on the
 unit sphere. In particular, we show that solutions do not
 depend continuously on the initial data.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}


\section{Introduction}

In this paper we study  properties of  solutions to the
initial-value problem
\begin{gather} \label{e1.1}
u_{rr}+{{n-1}\over r}u_r+{1\over {r^2}}\Delta_S u=f(r, u),
\quad r\geq r_0,\\
  u|_{r= r_0}=u_0\in X,\quad {u_r}|_{r= r_0}=u_1\in Y, \label{e1.2}
\end{gather}
where $n\geq 2$, $r_0\geq 1$ is suitable chosen and fixed number,
$X$ and $Y$ are Banach spaces,
$f\in \mathcal{C}([r_0, \infty))\times \mathcal{C}^1(\mathbb{R}^1)$,
$f(r, 0)=0$ for every $r\geq r_0$,
$a|u|\leq f_u'(r, u)\leq b|u|$ for every $r\geq r_0$,
$u\in \mathbb{R}^1$, $a$ and $b$ are  positive constants,
$\Delta_S$ is the
Laplace operator on the unit sphere $S^{n-1}$. More precisely we
prove that the initial-value problem \eqref{e1.1}-\eqref{e1.2} is
incorrectly posed in the following sense.

When we say that \eqref{e1.1}-\eqref{e1.2} is
incorrectly posed when the following happens:
\eqref{e1.1}-\eqref{e1.2} has exactly one solution $u(r)\in X$
for each $u_0\in X$, $u_1\in Y$;
there exists $\epsilon>0$ such that for every  $\delta>0$,
we have: $\|u_0-u_0'\|_X<\delta$, $\|u_1-u_1'\|_Y<\delta$
and $\|u-u'\|_{X}\geq \epsilon$,
where $u$ is a solution with initial data $u_0, u_1$,
and  $u'$ is a solution with initial data $u_0', u_1'$.

In this article, we obtain the following results using the same
approach as in \cite{g1,g2,g3,g4},

\begin{theorem} \label{thm1.1}
Let $n\geq 2$, $r_0\geq 1$,
$f\in \mathcal{C}([r_0, \infty))\times \mathcal{C}^1(\mathbb{R}^1)$
$f(r, 0)=0$ for every $r\geq r_0$, and $X=Y=L^2(S^{n-1})$.
Assume that there are positive constants, $a\leq b$, such that
$a|u|\leq f_u'(r, u)\leq b|u|$ for every $r\geq r_0$ and
every $u\in \mathbb{R}$.   Then
\eqref{e1.1}-\eqref{e1.2} is incorrectly posed.
\end{theorem}

\begin{theorem} \label{thm1.2}
Let $n\geq 2$, $r_0\geq 1$,
$f\in \mathcal{C}([r_0, \infty))\times \mathcal{C}^1(\mathbb{R}^1)$,
$f(r, 0)=0$ for every $r\geq r_0$,  $X=\mathcal{C}^2(S^{n-1})$
and $Y=\mathcal{C}^1(S^{n-1})$.
Assume that there are positive constants, $a\leq b$, such that
$a|u|\leq f_u'(r, u)\leq b|u|$ for every $r\geq r_0$ and
every $u\in \mathbb{R}$. Then  \eqref{e1.1}-\eqref{e1.2} is
incorrectly posed.
\end{theorem}

This paper is organized as follows. In section 2 we
prove our main results.
In the appendix we prove  results needed for the proof of
Theorems \ref{thm1.1} and \ref{thm1.2}.

\section{Proof of Main Results}

Here and bellow we will assume that $r_0\geq 1$ and $n\geq 2$.
First we will consider the initial-value problem
\begin{gather}
\label{e2.1}  u_{rr}+{{n-1}\over r}u_r+{1\over {r^2}}\Delta_S u=f(u),
  \quad r\geq r_0,\\
\label{e2.2}  u(r)|_{r= r_0}=u_0\in L^2(S^{n-1}),
u_r(r)|_{r= r_0}=u_1\in {L}^{2}(S^{n-1}),
\end{gather}
where $\Delta_S$ is the Laplace operator on the unit sphere $S^{n-1}$,
$f\in \mathcal{C}^1(\mathbb{R}^1)$, $f(0)=0$, $a|u|\leq f'(u)\leq b|u|$
for every $u\in \mathbb{R}^1$, $a\leq b$ are fixed positive constants.

For fixed positive constants $n\geq 2$, $r_0\geq 1$, $a$, $b$, $a\leq b$,
we suppose that the constants  $A$, $B$, $c_1$, $d_1$ satisfy
the following conditions
\begin{equation}
\begin{gathered}
r_0\leq c_1\leq d_1,\\
A\geq B>0,\\
{a\over {2A}}{{d_1^n}\over {(d_1+1)^n}}\geq 1.
\end{gathered} \label{ei1}
\end{equation}

\noindent\textbf{Example.}
Let $n\geq 1$, $r_0\gg 1$, $A=2$, $B=1$,
$a=r_0^{10n}$,$b=2r_0^{10n}$, $c_1=r_0+1$, $d_1=r_0+2$.

Let $N$ be the set
\begin{align*}
N=\Bigl\{&u(r): u(r)\in \mathcal{C}^{2}([r_0, \infty)),\;
u(\infty)=u_r(\infty)=0, \\
& r^{\alpha}|\partial_r^{\beta}u(r)|\leq 1
\; \forall r\geq r_0,\; \forall \alpha
\in \mathbb{N}\cup \{0\},\beta=0, 1,\\
&u(r)\geq 0\; \forall r\geq r_0, u(r)\leq {1\over B}
\; \forall  r\geq r_0,\\
&u(r)\geq {1\over A}\; \forall r\in [c_1, d_1],\;
u(r)\in L^2([r_0, \infty))\; \Bigr\}.
\end{align*}
For  $n\geq 1$, $f(u)\in \mathcal{C}^1(\mathbb{R}^1)$,
$a|u|\leq f'(u)\leq b|u|$, where $a\geq b$ are  positive constants,
and $u\in N$ we define the operator and the initial values
\begin{gather*}
P(u)=\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f(u)d\tau ds,\\
u_0=\int_{r_0}^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f(u)d\tau ds,\quad
u_1=-{1\over {r_0^n}}\int_{r_0}^{\infty}\tau^n f(u)d\tau.
\end{gather*}

\begin{theorem} \label{thm2.1}
Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}^1(\mathbb{R}^1)$,
and $f(0)=0$. Assume that there exist positive constants
 $a\leq b$ such that $a|u|\leq f'(u)\leq b|u|$.
Then \eqref{e2.1}-\eqref{e2.2} has exactly one solution $u\in N$.
\end{theorem}

\begin{proof}
 First we  prove that $P:N\to N$.
Let $u\in N$ be fixed. Then
\\
\textbf{(1)} Since $f\in \mathcal{C}^1([r_0, \infty))$,
$u\in \mathcal{C}^2([r_0, \infty))$
we have that $P(u)\in \mathcal{C}^2([r_0, \infty))$. Also we have
\begin{gather*}
P(u)_{|_{r=\infty}}=0,\\
{{\partial P(u)}\over {\partial r}}=-{1\over {r^n}}\int_r^{\infty}
\tau^n f(u)d\tau,\\
{{\partial P(u)}\over {\partial r}}_{|_{r=\infty}}=0.
\end{gather*}

\noindent\textbf{(2)} Let $\alpha\in \mathbb{N}\cup\{0\}$. We choose
$k\in \mathbb{N}$ such that
$k\geq \alpha+3$ and $ {b\over {2B(k-1)}}<1$.
Then
$$
r^{\alpha}P(u)=r^{\alpha}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
\tau^n f(u)d\tau ds\,.
$$
Now we use that for $u\in N$, we have  $u\geq 0$ for every $r\geq r_0$,
$f(0)=0$, $f'(u)\leq bu$, from here $f(u)\leq {b\over 2}u^2$, since
$u\leq {1\over B}$ for every $r\geq r_0$ we get $f(u)\leq {b\over {2B}}u$.
Then
\begin{align*}
r^{\alpha}P(u)
&\leq {b\over {2B}}r^{\alpha}\int_r^{\infty}{1\over {s^n}}
 \int_s^{\infty}\tau^n ud\tau ds \\
&=r^{\alpha}{b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
 \tau^{n+k}{1\over {\tau^k}} ud\tau ds \quad
\text{(use that $\tau^{n+k}u\leq 1$)}\\
&\leq {b\over {2B}}r^{\alpha}\int_r^{\infty}{1\over {s^n}}
 \int_s^{\infty}{1\over {\tau^k}} d\tau ds \\
& \leq {b\over {2B}}{1\over {(k-1)(n+k-2)}}{1\over {r_0^{n+k-\alpha-2}}}
 \leq 1.
\end{align*}
In the above inequality we use our choice of the constant $k$. Also,
\begin{align*}
\big|r^{\alpha}{{\partial P(u)}\over {\partial r}}\big|
&\leq {b\over {2B}}r^{\alpha}{1\over {r^n}}\int_r^{\infty}\tau^n ud\tau \\
&=r^{\alpha}{b\over {2B}}{1\over {r^n}}\int_r^{\infty}\tau^{n+k}
 {1\over {\tau^k}} ud\tau \quad
\text{(use $\tau^{n+k}u\leq 1$)}\\
& \leq r^{\alpha}{b\over {2B}}\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}{1\over {\tau^k}} d\tau ds \\
& \leq {b\over {2B}}{1\over {(k-1)}}{1\over {r_0^{n+k-\alpha-1}}}\leq 1.
\end{align*}
In the above inequality we use our choice of the constant $k$.

\noindent\textbf{(3)} First we note that for $u\in N$ we have
$f(u)\geq a u^2/2$.
Therefore for every $r\geq r_0$ we have
$$
P(u)\geq {a\over 2}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n u^2 d\tau ds\geq 0.
$$

\noindent\textbf{(4)} Let $r\in [c_1, d_1]$. Then
$$
P'(u)=\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f'(u)d\tau ds\geq
a\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n ud\tau ds\geq 0.
$$
Therefore, for $u\in N$ the function $P(u)$ is increase function of $u$.
Since for every $r\in [c_1, d_1]$ we have that $u\geq 1/A$ we get
\begin{align*}
P(u)&\geq P({1\over A})
=\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n f
\bigl({1\over A}\bigr)d\tau ds \\
&\geq {a\over {2A^2}}\int_{d_1}^{d_1+1}{1\over {s^n}}
  \int_{d_1}^{d_1+1}\tau^n d\tau ds \\
& \geq {a\over {2A^2}}{{d_1^n}\over {(d_1+1)^n}}\geq {1\over A},
\end{align*}
in the above inequality we use \eqref{ei1}.

\noindent\textbf{(5)} Choose  $k\in \mathbb{N}$  such that
$$
k>3,\quad {b\over {2(k-1)(n+k-2)}}<1.
$$
Then
\begin{align*}
P(u)&\leq {b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
\tau^n ud\tau ds \\
&\leq {b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
\tau^{k+n}{1\over {\tau ^k}} ud\tau ds \\
&\leq {b\over {2B}}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
{1\over {\tau ^k}}d\tau ds \\
&={b\over {2B(k-1)(n+k-2)r_0^{n+k-2}}}\leq {1\over B}.
\end{align*}

\noindent\textbf{(6)} Now we prove that $P(u)\in L^2([r_0, \infty))$.
Indeed,
\begin{align*}
\|P(u)\|_{L^2([r_0, \infty))}^2
&=\int_{r_0}^\infty \Bigl(\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}\tau^n f(u)d\tau ds\Bigr)^2 dr \\
& \leq {{b^2}\over 4} \int_{r_0}^\infty
\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n u^2d\tau ds
 \Bigr)^2 dr \\
& \leq {{b^2}\over 4} \int_{r_0}^\infty
\Bigl(\int_r^{\infty}{1\over {s^n}}
\int_s^{\infty}\tau^{k+n}u {u\over {\tau^k}}d\tau ds\Bigr)^2 dr\quad
\text{(use  that $\tau^{k+n}u\leq 1$)}\\
&\leq {{b^2}\over 4} \int_{r_0}^\infty
 \Bigl(\int_r^{\infty}{1\over {s^n}}
 \int_s^{\infty} {u\over {\tau^k}}d\tau ds\Bigr)^2 dr\leq
\quad \text{(use H\"older's inequality)}\\
& \leq {{b^2}\over 4} \int_{r_0}^\infty
\Bigl(\int_r^{\infty}{1\over {s^n}}
\Bigl(\int_s^{\infty}{1\over {\tau^{2k}}}d\tau\Bigr)^{1/2}
\Bigl(\int_s^{\infty}u^2d\tau\Bigr)^{1/2} ds\Bigr)^2 dr
\\
& \leq {{b^2}\over {4(2k-1)(n+k-{3\over 2})^2(2n+2k-4)r_0^{2n+2k-4}}}
\|u\|_{L^2([r_0, \infty))}^2<\infty,
\end{align*}
because $u\in L^2([r_0, \infty))$.
From (1)--(6) we conclude that $P:N\to N$.

Now we prove that the operator $P$ has exactly one fixed point in $N$.
Let $u_1, u_2\in N$ are fixed and $\alpha=\|u_1-u_2\|_{L^2([r_0, \infty))}$.
We choose the constant $k\in \mathbb{N}$  large so that
$Q_1/\alpha<1$,
where
$$
Q_1={{2b^2}\over
{B^2({4\over 3}k-1)^{3\over 2}(n+k-{7\over 4})^2(2n+2k-{9\over 2})
r_0^{2n+2k-{9\over 2}}}}.
$$
Then
\begin{align*}
&\|P(u_1)-P(u_2)\|_{L^2([r_0, \infty))}^2 \\
&=\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n
 (f(u_1)-f(u_2))d\tau ds\Bigr)^2 dr\quad
 \text{(mean value theorem)}\\
&=\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}\tau^n
f'(\xi)(u_1-u_2)d\tau ds\Bigr)^2 dr \\
& \leq \int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}\tau^n
  |f'(\xi)\|u_1-u_2|d\tau ds\Bigr)^2 dr \\
&\quad \text{(use that $|f'(\xi)|\leq b|\xi|\leq {b\over B}$,
$|\xi|\leq \max\{|u_1|, |u_2|\}$)}\\
&\leq {{b^2}\over {B^2}}
\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\tau^n
|u_1-u_2|d\tau ds\Bigr)^2 dr \\
&={{b^2}\over {B^2}}
\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
\sqrt{\tau^{2k+2n}
|u_1-u_2|}{1\over {\tau^k}}\sqrt{|u_1-u_2|}d\tau ds\Bigr)^2 dr \\
&\quad \text{(use that $\sqrt{\tau^{2k+2n}|u_1-u_2|}\leq \sqrt{2}$)}\\
&\leq {{2b^2}\over {B^2}}
\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
\sqrt{|u_1-u_2|}{1\over {\tau^k}}d\tau ds\Bigr)^2 dr \quad
\text{(H\"older's inequality)}\\
&\leq {{2b^2}\over {B^2}}
\int_{r_0}^{\infty}\Bigl(\int_r^{\infty}{1\over {s^n}}\Bigl(\int_s^{\infty}
{1\over {\tau^{{4k}\over 3}}}d\tau\Bigr)^{3/4}
\Bigl({|u_1-u_2|}^2d\tau\Bigr)^{1/4} ds\Bigr)^2 dr \\
& \leq Q_1\|u_1-u_2\|_{L^2([r_0, \infty))};
\end{align*}
i.e.,
$$
\|P(u_1)-P(u_2)\|_{L^2([r_0, \infty))}^2\leq
Q_1\|u_1-u_2\|_{L^2([r_0, \infty))}.
$$
From this,
$$
\|P(u_1)-P(u_2)\|_{L^2([r_0, \infty))}^2\leq
{{Q_1}\over {\alpha}}\alpha\|u_1-u_2\|_{L^2([r_0, \infty))}\leq
{{Q_1}\over {\alpha}}\|u_1-u_2\|_{L^2([r_0, \infty))}^2.
$$
For our next step we need the theorem in \cite[page 294]{k1}:
\begin{quote}
 Let $B$ be the complete metric space for which $AB\subset B$
 and for the operator $A$ satisfies the condition
$$
\rho(Ax, Ay)\leq L(\alpha, \beta)\rho(x, y),\quad
x, y\in B,\alpha\leq \rho(x, y)\leq \beta,
$$
where $L(\alpha, \beta)<1$ for $0<\alpha\leq\beta<\infty$.
 Then the operator $A$ has exactly one fixed point in  $B$.
\end{quote}

From the above result and our choice of $k$ we conclude that the
operator $P$ has exactly one fixed point $u\in N$.  Consequently
$u$ is a solution to the problem \eqref{e2.1}-\eqref{e2.2}.
In the appendix we will prove that the set $N$ is closed subset of
the space $L^2([r_0, \infty))$. We have that $u_0\in
L^2(S^{n-1})$, $u_1\in L^2(S^{n-1})$.
\end{proof}

\begin{theorem} \label{thm2.2}
Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}^1(\mathbb{R}^1)$,
and $f(0)=0$.
Assume that there exists positive constants, $a\leq b$, such that
$a|u|\leq f'(u)\leq b|u|$. Then \eqref{e2.1}-\eqref{e2.2}
is incorrectly posed.
\end{theorem}

\begin{proof} On the contrary, suppose that \eqref{e2.1}-\eqref{e2.2}
is correctly posed. Let $u$ is the solution from
Theorem \ref{thm2.1}. We choose $\epsilon $ such that
$0<\epsilon<1/Q_2$,
where
$$
Q_2={{b^2}\over {4(4k-1)^{1/2}(n+k-{5\over 4})^2
(2n+2k-{7\over 2})r_0^{2n+2k-{7\over 2}}}}.
$$
Then there exists $\delta=\delta(\epsilon)>0$ such that
$$
\|u_0\|_{L^2(S^{n-1})}<\delta, \quad \|u_1\|_{L^2(S^{n-1})}<\delta
$$
imply
$$
\|u\|_{L^2([r_0, \infty))}<\epsilon.
$$
From the definition of $u$, we have
\begin{align*}
\|u\|_{L^2([r_0, \infty))}^2
&=\int_{r_0}^{\infty} \Bigl(\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}\tau^n f(u)d\tau ds\Bigr)^2 dr \\
&\leq {{b^2}\over 4} \int_{r_0}^{\infty}
\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
 \tau^n u^2d\tau ds\Bigr)^2 dr \\
&= {{b^2}\over 4} \int_{r_0}^{\infty}
\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}\sqrt{\tau^{2k+2n}u}
u^{3\over 2}{1\over {\tau^k}}d\tau ds\Bigr)^2 dr \\
&\quad \text{(use that $\sqrt{\tau^{2k+2n}u} \leq 1$)}\\
&\leq {{b^2}\over 4} \int_{r_0}^{\infty}
\Bigl(\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
u^{3\over 2}{1\over {\tau^k}}d\tau ds\Bigr)^2 dr \quad
\text{(H\"older's inequality)}\\
&\leq {{b^2}\over 4} \int_{r_0}^{\infty}
 \Bigl(\int_r^{\infty}{1\over {s^n}}
 \Bigl(\int_s^{\infty} u^{2}d\tau\Bigr)^{3/4}
 \Bigl({1\over {\tau^{4k}}}d\tau\Bigr)^{1/4} ds\Bigr)^2 dr \\
&\leq Q_2\|u\|_{L^2([r_0, \infty))}^3;
\end{align*}
i.e.,
$$
\|u\|_{L^2([r_0, \infty))}^2\leq {{Q_2}}\|u\|_{L^2([r_0, \infty))}^3.
$$
From this,,
$$
\|u\|_{L^2([r_0, \infty))}\geq {1\over {Q_2}}>\epsilon
$$
which is a contradiction. Consequently the
problem \eqref{e2.1}-\eqref{e2.2} is incorrectly posed.
\end{proof}

\begin{theorem} \label{thm2.3}
Let $n\geq 2$, $r_0\geq 1$, $f\in \mathcal{C}^1(\mathbb{R}^1)$,
and $f(0)=0$.
Assume that there are positive constants
$a\leq b$ such that $a|u|\leq f'(u)\leq b|u|$.
Then the problem
\begin{gather}
 u_{rr}+{{n-1}\over r}u_r+{1\over {r^2}}\Delta_S u=f(u),
   \quad r\geq r_0, \label{e2.1b} \\
\label{e2.3}  u(r)_{|_{r= r_0}}=u_0\in \mathcal{C}^2(S^{n-1}),\quad
u_r(r)_{|_{r=r_0}}=u_1\in \mathcal{C}^{1}(S^{n-1}),
\end{gather}
is incorrectly posed.
\end{theorem}

\begin{proof}
Let us suppose that \eqref{e2.1b}-\eqref{e2.3} is
correctly posed, and let
$$
Q_3={b\over {2(k-1)(n+k-2)r_0^{n+k-2}}}.
$$
Then for
$0<\epsilon<1/Q_3^2$,
there exists $\delta=\delta(\epsilon)>0$ such that
$$
\|u_0\|_{\mathcal{C}^2(S^{n-1})}<\delta,\quad
\|u_1\|_{\mathcal{C}^1(S^{n-1})}<\delta,
$$
imply
$$
\max_{r\in [r_0, \infty)}|u|<\epsilon,\quad
\max_{r\in [r_0, \infty)}|u_r|<\epsilon,\quad
\max_{r\in [r_0, \infty)}|u_{rr}|<\epsilon,
$$
where $u$ is the solution from the Theorem \ref{thm2.1}.
From the definition of $u$, and $k\in \mathbb{N}$, we have
\begin{align*}
u(r)&\leq {b\over 2}\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}\tau^n u^2 d\tau ds\\
&={b\over 2}\int_r^{\infty}{1\over {s^n}}
  \int_s^{\infty}\sqrt {\tau^{2k+2n}u}
u^{3\over 2}{1\over {\tau ^k}} d\tau ds \\
& \leq {b\over 2}\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
u^{3\over 2}{1\over {\tau^k}} d\tau ds\\
&\leq {b\over 2}(\max_{r\in [r_0, \infty)}u)^{3\over 2}
\int_r^{\infty}{1\over {s^n}}\int_s^{\infty}
{1\over {\tau^k}} d\tau ds\\
&\leq Q_3(\max_{r\in [r_0, \infty)}u)^{3\over 2}.
\end{align*}
From this it follows that
$$
Q_3(\max_{r\in [r_0, \infty)}u)^{1/2}\geq 1,\quad
\text{or}\quad
\max_{r\in [r_0, \infty)}u>{1\over {Q_3^2}}>\epsilon,
$$
which is a contradiction with our assumption. Consequently
\eqref{e2.1b}-\eqref{e2.3} is incorrectly posed.
\end{proof}

The proofs of Theorems \ref{thm1.1} and \ref{thm1.2}
follow from the method used in the proof of Theorems \ref{thm2.2} and
\ref{thm2.3}.

\section{Appendix}

\begin{lemma} \label{lem3.1}
The set $N$ is a closed subset of $L^2([r_0, \infty))$.
\end{lemma}

\begin{proof}
Let $\{u_n\}$ is a sequence of elements in $N$ for which
$$
\lim_{n\to\infty}\|u_n-\tilde{u}\|_{L^2([r_0, \infty))}=0,
$$
where $\tilde{u}\in L^2([r_0, \infty))$. Since $P(u)$ is a
 continuous differentiable  function of $u$,
for $r\in [r_0, c_1]$ and $u\in N$ we have
\begin{align*}
P'(u)&=\int_r^{\infty}{1\over {s^{n}}}\int_s^{\infty}\tau^{n}
f'(u)d\tau ds \\
& \geq a\int_{c_1}^{d_1}{1\over {s^{n}}}\int_{c_1}^{d_1}\tau^{n}
u d\tau ds \\
& \geq {a\over A}{{c_1^n}\over {d_1^n}}(d_1-c_1)^2.
\end{align*}
From this, it follows that for every $u\in N$ there exists
$$
L=\min_{r\in [r_0, c_1]}|P'(u)(r)|>0.
$$
Let
$$
M_1=\max_{r\in [r_0, c_1]}\Bigl|{{\partial}\over {\partial r}}P'(u)(r)\Bigr|.
$$
Now we  prove that for every $\epsilon >0$ there exists
$\delta=\delta(\epsilon)>0$
such that from $|x-y|<\delta$ we have
$$
|u_m(x)-u_m(y)|<\epsilon\quad  \forall m\in \mathbb{N}.
$$
We suppose that there exists ${\tilde \epsilon}>0$ such that for
every $\delta>0$ there exist natural number $m$ and
 $x, y\in [r_0, \infty)$, $|x-y|<\delta$ for which
$|u_m(x)-u_m(y)|\geq {\tilde\epsilon}$.
We choose ${\tilde{\tilde \epsilon}}$ such that
$0<{\tilde {\tilde \epsilon}}<L{\tilde\epsilon}$. We note that $P(u_m)(x)$
is uniformly continuous for $x\in [r_0, \infty)$.
For $u\in N$ $P(u)(r)$ is uniformly
continuous function for $r\in [r_0, \infty)$
because $P(u)(r)\in \mathcal{C}([r_0, \infty))$
and as in the proof of the Theorem \ref{thm2.1} we have that there exists
positive constant $C$ such that
$\bigl|{{\partial}\over {\partial r}}P(u)(r)\bigr|\leq C$.
Then there exists $\delta_1=\delta_1({\tilde {\tilde \epsilon}})>0$
such that for every natural $m$ we have
$$
|P(u_m)(x)-P(u_m)(y)|<{\tilde {\tilde \epsilon}},\quad \forall
 x, y\in [r_0, \infty):|x-y|<\delta_1.
$$
Consequently we can choose
$$
0<\delta<\min\bigl\{c_1-r_0, \delta_1, {{(L{\tilde\epsilon}-{\tilde {\tilde\epsilon}})B}
\over {M_1}}\bigr\}
$$
such that there exist natural number $m$ and $x_1, x_2\in [r_0, \infty)$
for which
$$
|x_1-x_2|<\delta,\quad |u_m(x_1-x_2+r_0)-u_m(r_0)|\geq {\tilde\epsilon}.
$$
In particular,
\begin{equation}
|P(u_m)(x_1-x_2+r_0)-P(u_m)(r_0)|<{\tilde {\tilde \epsilon}}.
\label{e3.1}
\end{equation}
Let us suppose for convenience that $x_1-x_2>0$. Then $x_1-x_2<c_1-r_0$
and for every $u\in N$ we have $P'(u)(x_1-x_2+r_0)\geq L$.
Then from the middle point theorem
we have
$P(0)=0$, $P(u_m)(x_1-x_2+r_0)=P'(\xi)(x_1-x_2+r_0)u_m(x_1-x_2+r_0)$,
$P(u_m)(r_0) = P'(\xi)(r_0) u_m(r_0)$,
\begin{align*}
&|P(u_m)(x_1-x_2+r_0)-P(u_m)(r_0)| \\
&=|P'(\xi)(x_1-x_2+r_0)u_m(x_1-x_2+r_0)-P'(\xi)(r_0)u_m(r_0)| \\
&=|P'(\xi)(x_1-x_2+r_0)u_m(x_1-x_2+r_0)-
P'(\xi)(x_1-x_2+r_0)u_m(r_0) \\
&\quad +P'(\xi)(x_1-x_2+r_0)u_m(r_0)-P'(\xi)(r_0)u_m(r_0)|\\
&\geq |P'(\xi)(x_1-x_2+r_0)u_m(x_1-x_2+r_0)-
P'(\xi)(x_1-x_2+r_0)u_m(r_0)| \\
&\quad -|P'(\xi)(x_1-x_2+r_0)u_m(r_0)-P'(\xi)(r_0)u_m(r_0)| \\
&=|P'(\xi)(x_1-x_2+r_0)u_m(x_1-x_2+r_0)-
P'(\xi)(x_1-x_2+r_0)u_m(r_0)|\\
&\quad -
\Bigl|{{\partial}\over {\partial r}}P'(\xi)\Bigr\|x_1-x_2\|u_m(r_0)| \\
&\geq
L{\tilde \epsilon}-M_1\delta{1\over B}\geq {\tilde {\tilde\epsilon}},
\end{align*}
which is a contradiction with \eqref{e3.1}. Therefore, for every
$\epsilon>0$ there exists $\delta=\delta(\epsilon)>0$ such that
from $|x-y|<\delta$ follows
\begin{equation}
|u_m(x)-u_m(y)|<\epsilon\quad \forall m\in \mathbb{N}.
\label{e3.2}
\end{equation}
On the other hand from the definition of the set $N$ we have that
for every natural number $m$
\begin{equation}
u_m(r)\leq {1\over B}\quad \forall r\geq r_0. \label{e3.3}
\end{equation}
 From this inequality and \eqref{e3.2} it follows that the set
$\{u_m\}$ is a compact subset of the space
$\mathcal{C}([r_0, \infty))$. Therefore there is a subsequence
$\{u_{n_k}\}$ and function $u\in \mathcal{C}([r_0, \infty))$ for which
$$
|u_{n_k}(x)-u(x)|<\epsilon\quad \forall x\in [r_0, \infty).
$$

Now we suppose that  that $u\neq \tilde{u}$ a.e. in $[r_0, \infty)$.
Then there exist $\epsilon_1>0$ and subinterval
$\Delta\subset [r_0, \infty)$ such that
$\mu(\Delta)>0$ and
$$
|u-\tilde{u}|>\epsilon_1\quad \text{for } r\in \Delta.
$$
Let $\epsilon>0$ is chosen such that
\begin{equation}
\epsilon<{{\epsilon_1(\mu(\Delta))^{1/2}}\over
{\mu(\Delta)^{1/2}+1}}. \label{e3.4}
\end{equation}
Then, for every $n_k\in \mathbb{N}$ sufficiently large,
we have $\|u_{n_k}-\tilde{u}\|_{L^2([r_0, \infty))}<\epsilon$,
\begin{align*}
\epsilon\mu(\Delta)
&=\epsilon\int_{\Delta} dx\\
&>\int_{\Delta}|u_{n_k}-{u}|dx
 =\int_{\Delta}|u_{n_k}-\tilde{u}+\tilde{u}-u|dx\\
&\geq \int_{\Delta}|\tilde{u}-u|dx-\int_{\Delta}|u_{n_k}-\tilde{u}|dx \\
&\geq \epsilon_1\mu(\Delta)-\Bigl(\int_{\Delta}|u_{n_k}-\tilde{u}|^2dx\Bigr)
 ^{1/2} \bigl(\mu(\Delta)\bigr)^{1/2} \\
&\geq\epsilon_1\mu(\Delta)-\|u_{n_k}-\tilde{u}\|_{L^2([r_0, \infty))}
  \bigl(\mu(\Delta)\bigr)^{1/2} \\
&>\epsilon_1\mu(\Delta)-\epsilon\bigl(\mu(\Delta)\bigr)^{1/2},
\end{align*}
which is a contradiction  with \eqref{e3.4}. From this,
$u=\tilde{u}$ a.e. in $[r_0, \infty)$,
$|u_n-u|^2=|\tilde{u}-u_n|^2$ a.e. in $[r_0, \infty)$,
$\|u_n-u\|_{L^2([r_0,\infty))}=\|u_n-\tilde{u}\|_{L^2([r_0, \infty))}$.
Consequently, for every sequence $\{u_n\}$ from elements of the set $N$,
which is convergent in $L^2([r_0, \infty))$, there exists a function
$u\in \mathcal{C}([r_0, \infty))$, $u\in L^2([r_0, \infty))$ for which
$$
\lim_{n\to \infty}\|u_n-u\|_{L^2([r_0, \infty))}=0.
$$

Bellow we will suppose that
$\{u_n\}$ is a sequence from elements of the set $N$,
which is convergent in $L^2([r_0, \infty))$. Then there exists a function
$u\in \mathcal{C}([r_0, \infty))$, $u\in L^2([r_0, \infty))$ for which
$$
\lim_{n\to \infty}\|u_n-u\|_{L^2([r_0, \infty))}=0.
$$

Now we suppose that $u(\infty)\ne 0$. Then there exist sufficiently large
$Q>0$, a large natural number $m$ and  $\epsilon_2>0$ for which
$$
u_m(r)=0,\quad u(r)>\epsilon_2,\quad \forall r\geq Q.
$$
We choose
\begin{equation}
0<\epsilon_3<\epsilon_2. \label{e3.5}
\end{equation}
Then, for every  $n\in \mathbb{N}$ sufficiently large,
we have $|u_n(r)-u( r)|<\epsilon_3$
and
\begin{align*}
\epsilon_3&>\int_Q^{Q+1}|u_n(r)-u(r)|dr \\
&\geq \int_Q^{Q+1}(|u(r)|-|u_n( r)|)dr \\
&=\int_Q^{Q+1}|u(r)|dr>\epsilon_2,
\end{align*}
which is a contradiction with \eqref{e3.5}. Therefore,
$u(\infty)=0$.

Now we  prove that ${{\partial}\over {\partial r}}u(r)$ exists for every
$r\geq r_0$. Let us suppose that there exists $r_1\in [r_0, \infty)$
such that
${{\partial}\over {\partial r}}u(r_1)$ does not exists. Then for every
$h>0$, which is enough small, exists $\epsilon_4>0$ such that
$$
\Bigl|{{u(r_1+h)-u(r_1)}\over h}\Bigr|>\epsilon_4,
$$
and
\begin{equation}
0<\epsilon_5<{h\over 2}\epsilon_4, \label{e3.6}
\end{equation}
 such that
$|{{u_n(r_1+h)-u(r_1)}}|<\epsilon_5$.
From this,
\begin{align*}
\epsilon_5&>|u_n(r_1+h)-u(r_1+h)| \\
&=|u_n(r_1+h)-u(r_1)+u(r_1)-u(r_1+h)| \\
&\geq |u(r_1)-u(r_1+h)|{1\over h} h-|u_n(r_1+h)-u(r_1)|\\
&\geq \epsilon_4 h-\epsilon_5,
\end{align*}
which is a contradiction of our choice of $\epsilon_5$.
Therefore ${{\partial}\over {\partial r}}u(r)$ exists for every
$r\in [r_0, \infty)$. As in above we can see that
$u(r)\in \mathcal{C}^2([r_0, \infty))$
$u_r(\infty)=0$.

Now we suppose that there exists interval $\Delta_2\subset [r_0, \infty)$
such that
$$
u(r)\geq {1\over B}+\epsilon_7\quad \text{for }r\in \Delta_2.
$$
Let $n\in \mathbb{N}$ be large and $\epsilon_8>0$  chosen such that
\begin{equation}
|u_n(r)-u(r)|<\epsilon_8\quad \text{for } r\in \Delta_2,
0<\epsilon_8<\epsilon_7. \label{e3.7}
\end{equation}
From this, for $r\in \Delta_2$, we have
$$
\epsilon_8>|u_n(r)-u(r)|\geq |u(r)|-|u_n(r)|\geq
{1\over B}+\epsilon_7-{1\over B}=\epsilon_7,
$$
which is a contradiction with \eqref{e3.7}. Therefore,
$u(r)\leq {1\over B}$ for every $r\geq r_0$.

Now we suppose that there exists interval $\Delta_3\subset [c_1, d_1]$
for which $u(r)< {1\over A}$ for every $r\in \Delta_3$.
From this, there exists $\epsilon_9>0$ such that
$u(r)\leq {1\over A}-\epsilon_9$ for $r\in \Delta_3$. Also, let
\begin{equation}
0<\epsilon_{10}<\epsilon_9 \label{e3.8}
\end{equation}
and $ n\in \mathbb{N}$ is enough large such that
$\epsilon_{10}>|u_n(r)-u(r)|$
for $r\in \Delta_3$. Then for $r\in \Delta_3$ we have
$$
\epsilon_{10}>|u_n(r)-u(r)|\geq
|u_n(r)|-|u(r)|\geq {1\over A}-{1\over A}+\epsilon_9,
$$
which is a contradiction with \eqref{e3.8}. Consequently, for
every $r\in [c_1, d_1]$ we have $u(r)\geq {1\over A}$.

Now we suppose that there exist $\alpha\in \mathbb{N}\cup\{0\}$,
interval $\Delta_4\subset [r_0, \infty)$ and $\epsilon_{11}>0$
such that
$$
|r^{\alpha}u( r)|>1+\epsilon_{11}\quad\text{for } r\in \Delta_4.
$$
Let $\epsilon_{12}>0$ and $n\in \mathbb{N}$ be chosen such that
\begin{equation}
|r^{\alpha}(u_n(r)-u(r))|<\epsilon_{12}\quad\text{for }
r\in\Delta_4,\; 0<\epsilon_{12}<\epsilon_{11}.
\label{e3.9}
\end{equation}
From this,
$$
\epsilon_{12}>|r^{\alpha}(u_n(r)-u(r))|\geq |r^{\alpha}u(r)|-
r^{\alpha}|u_n(r)|\geq\epsilon_{11},
$$
which is a contradiction with \eqref{e3.9}. Therefore for every
$\alpha\in \mathbb{N}\cup\{0\}$ and for every $r\in [r_0, \infty)$
we have $r^{\alpha}u(r)\leq 1$. After we use the same arguments we
can see that for every $\alpha\in \mathbb{N}\cup\{0\}$ and for every
$r\in [r_0, \infty)$  we have $r^{\alpha}|u_r(r)|\leq 1$.

Now we suppose that there exist interval $\Delta_5\subset [r_0, \infty)$ and
$\epsilon_{13}>0$ such that for $r\in \Delta_5$ we have
$u(r)<-\epsilon_{13}$.
Let $n\in \mathbb{N}$ is enough large and $\epsilon_{14}>0$ are
fixed for which
\begin{equation}
|u_n(r)-u(r)|<\epsilon_{14}\quad \text{for } r\in \Delta_5,\quad
0<\epsilon_{14}<\epsilon_{13}. \label{e3.10}
\end{equation}
Then for $r\in \Delta_5$ we have
$$
\epsilon_{14}>u_n(r)-u(r)>\epsilon_{13}
$$
which is a contradiction with \eqref{e3.10}.
\end{proof}

\begin{thebibliography}{00}

\bibitem{b1} Bandle, C., M. Essen.
 On the solutions of quasilinear elliptic problems with boundary
 blow - up, PDE of Elliptic Type, (Cortona, 1992),
 Sympos. Math., Vol. 35, Cambridge University Press, 1994, pp. 93-111.

\bibitem{c1} Cirstea, F., V. Radulescu;
 Uniqueness of the blow - up boundary solution of logistic equations
with absorption, C. R. Acad. Sci. Paris, Ser. I 335(2002), 447-452.

\bibitem{g1} Georgiev, S.
 Blow up of solutions for Klein-Gordon equations in the
Reissner-Nordstr\"om metric, Electron. J. Diff. Eqns., Vol 2005(2005),
No 67, pp. 1-22.

\bibitem{g2}  Georgiev, S.;
 Blow-up of the solutions of nonlinear wave equation in
Reissner-Nordstr\"om metric, Dynamics of PDE, Vol. 3, No. 4, 295-329, 2006.

\bibitem{g3}  Georgiev, S.;
Uniform continuity of the solution map for nonlinear wave
equation, Electon. J. Qualitive Theory Diff. Eqns., Vol 2007(2007),
 No 12, pp. 1-14.

\bibitem{g4} Georgiev, S.;
 Blow - up of the solutions of nonlinear wave equation, BVP, 2007, Q2.

\bibitem{g5} Ghergu, M., V. Radulescu.;
 Singular Elliptic Problems. Bifurcation and Asymptotic Analysis,
Oxford Lecture Series in Mathematics and its Applications, Vol. 37,
Oxford University Press, 2008.

\bibitem{k1}  Krasnosel'skii, M., P.Zabrejko;
Geometrical Methods of Nonlinear Analysis,
Nauka, Moscow, 1975. In Russian

\bibitem{m1}  Marcus, M., L. Veron.;
 Existence and uniqueness results for large solutions of general
nonlinear elliptic equations, J. Evol. Equations 3(2003), 637- 652.

\bibitem{r1}  Radulescu, V.;
 Singular phenomena in nonlinear elliptic problems.
From blow - up boundary solutions to equations with singular
nonlinearities, in Handbook of Differential Equations:
Stationary Partial Differential Equations, Vol. 4
(Michel Chipot Editor) North Holland Elsevier Science,
Amsterdam, 2007, pp. 483 - 591.

\end{thebibliography}

\end{document}