text:
\section{Introduction}

In this paper we will consider the following problem:
\begin{equation} \label{eq1}
  \begin{cases}
    -\Delta u=\Psi_{\alpha}u^{p-1} &\text{in $\Omega$},\\ u>0
    &\text{in $\Omega$},\\ u=0 &\text{on $\partial \Omega$},
  \end{cases}
\end{equation}
where $\Omega=\{x\in {\mathbb R^N}|\;1<|x|<3\}$ is an annulus in
$\mathbb R^N$, $N\geq 3$, $\alpha>0$, $p>2$ and $\Psi_{\alpha}$ is
the radial function
\[
\Psi_{\alpha}(x)=\left| |x|-2 \right|^{\alpha}.
\]
This equation can be seen as a natural extension to the annular
domain $\Omega$ of the celebrated H\'{e}non equation with
Dirichlet boundary conditions (see \cite{henon,ni})
\begin{equation} \label{eq:henon}
\begin{cases}
-\Delta u = |x|^\alpha |u|^{p-1} &\hbox{for $|x| < 1$} \\
u = 0 &\hbox{if $|x|=1$}.
\end{cases}
\end{equation}
Actually, the weight function $\Psi_\alpha$ reproduces on $\Omega$
a similar qualitative behavior of $|\cdot|^\alpha$ on the unit
ball $B$ of $\mathbb{R}^N$.

A standard compactness argument shows that the infimum
\begin{equation} \label{eq:infimum}
\inf_{\substack{u \in H_0^1(B) \\ u \neq 0}} \frac{\int_B |\nabla
u|^2\ dx}{\left( \int_B |x|^\alpha |u|^p \ dx\right)^{2/p}}
\end{equation}
is achieved for any~$2<p<2^*$ and any~$\alpha>0$. In 1982, Ni
proved in \cite{ni} that the infimum
\begin{equation} \label{eq:infimum-rad}
\inf_{\substack{u \in H_{0,\mathrm{rad}}^1(B) \\ u \neq 0}}
\frac{\int_B |\nabla u|^2\ dx}{\left( \int_B |x|^\alpha |u|^p\ dx
\right)^{2/p}}
\end{equation}
is achieved for any $p \in (2,2^* + \frac{2\alpha}{N-2})$ by a
function in $H_{0,\mathrm{rad}}^1(B)$, the space of radial
$H_0^1(B)$ functions. Thus, radial solutions of \eqref{eq:henon}
exist also for (Sobolev) supercritical exponents $p$. Actually,
radial $H_0^1$ elements show a power--like decay away from the
origin (as a consequence of the Strauss Lemma, see
\cite{AM,strauss}) that combines with the weight $|x|^\alpha$ and
provides the compactness of the embedding $H_{0,\mathrm{rad}}^1(B)
\subset L^p(B)$ for any $2<p<2^*+\frac{2\alpha}{N-2}$.

A natural question is whether any minimizer of \eqref{eq:infimum}
must be radially symmetric in the range $2<p<2^*$ and $\alpha>0$. Since
the weight $| \cdot|^\alpha$ is an increasing function,
neither rearrangement arguments nor the moving plane techniques of
\cite{gnn} can be applied.

Reverting to the case $\alpha >0$, Smets \textit{et al.} proved in
\cite{ssw}
some symmetry--breaking results for \eqref{eq:henon}. They proved
that minimizers of \eqref{eq:infimum} (the so-called
\emph{ground-state
  solutions}, or least energy solutions) cannot be radial, at least
for $\alpha$ large enough. As a consequence, \eqref{eq:henon} has
at least two solutions when $\alpha$ is large (see also
\cite{sw}).

Later on, Serra proved in \cite{serra03} the existence of at least
one non--radial solution to \eqref{eq:henon} in the critical case
$p=2^*$, and in \cite{badialeserra} the authors proved the
existence of more than one solution to the same equation also for
some supercritical values of $p$. These solutions are non-radial
and they are obtained by minimization under suitable symmetry
constraints.

Quite recently, Cao and Peng proved in \cite{caopeng} that, for
$p$ sufficiently close to $2^*$, the ground-state solutions of
\eqref{eq:henon} possess a unique maximum point whose distance
from $\partial B$ tends to zero as $p \to 2^*$.

This kind of result was improved in \cite{peng}, where multibump
solutions for the H\'{e}non equation with almost critical Sobolev
exponent $p$ are found, by means of a finite--dimensional
reduction. These solution are not radial, though they are
invariant under the action of suitable subgroups of $O(N)$, and
they concentrate at boundary points of the unit ball of $\mathbb{R}^N$ as
$p \to 2^*$. The r\^{o}le of $\alpha$ is however a static one.
(For more results for $p\approx 2^*$ see also \cite{PiSe05}).

\medskip

In this paper we will prove that similar phenomena take place for
problem (\ref{eq1}) on the annulus $\Omega$. In
Section~\ref{paralpha}, we present some estimates for the least
energy radial solutions of \eqref{eq1} when $p<2^*$ is kept fixed
but $\alpha \to +\infty$. These will lead us to a first
symmetry--breaking result, stating that for $\alpha$ sufficiently
large there exist at least two solutions of \eqref{eq1}: a global
minimizer of the associated Rayleigh quotient, and a global
minimizer \emph{among radial functions}.

In Section 3, another symmetry--breaking is proved, with $\alpha$
fixed and $p \to 2^*$. To show this phenomenon, we will use a
decomposition lemma in the spirit of P.L.~Lions' concentration and
compactness theory (\cite{PLL}), and inspired by \cite{caopeng}.
It will turn out that global minimizers of the same Rayleigh
quotient concentrate as $p \to 2^*$ at precisely one point of the
boundary $\partial \Omega$, which has two connected components. A
second nonradial solution can then be found in a tricky but
natural way, by minimization over functions that are ``heavier''
on the opposite connected component of $\partial\Omega$.

In Section 4, a third nonradial solution is singled out, by means
of a linking argument. Roughly speaking, the previous nonradial
solutions can be used to build a mountain pass level. In
particular,
 this third solution will not be a local minimizer of
the Rayleigh quotient.

 Section 5 describes the behavior of
ground-state solutions of \eqref{eq1} as $\alpha \to +\infty$ and
$p<2^*$ is kept fixed. Although the conclusion is not as precise
as in the case $p \to 2^*$, we can nevertheless show that a sort
of concentration near the boundary $\partial \Omega$ still
appears.

 We would like to stress that the existence
of non-radial solutions in the annulus in the almost critical case
$p\approx 2^*$ is not by now a surprise. When the weight
disappears, i.e. $\alpha=0$, Brezis and Nirenberg proved in
\cite{BN} that the ground state solution of $-\Delta u = u^p$ in
$H_0^1$ is not a radial function. Indeed, the authors proved that
both a radial and a non-radial (positive) solution arise as $p
\approx 2^*$. Their simple continuation argument can be adapted to
cover our weighted equation. Subgroups of $O(N)$ are used in
\cite{byeon01} for the equation $-\Delta u + u = f(u)$, and some
refined properties of symmetric solutions are proved. We refer to
the bibliography of that paper for more references.

 For more results about asymptotic estimates for
solutions of the H\'{e}non equation with $\alpha$ large, see
\cite{byeonwangI,byeonwang}.

\section{Symmetry breaking for $\alpha$ large }\label{paralpha}

Let $H_{0,\mathrm{rad}}^1(\Omega)$ be the space of radially
symmetric functions of $H_0^1(\Omega)$. With a slight but common
abuse of notation, we will systematically write $u(x)=u(|x|)$ for
$u \in H_{0,\mathrm{rad}}^1(\Omega)$.

Consider the minimization problem
\begin{equation} \label{eq:6} S_{\alpha,p}^{\rm rad} = \inf_{u \in
    H_{0,{\rm rad}}^1 (\Omega) \setminus \{0\}} R_{\alpha,p} (u),
\end{equation}
where
\begin{equation}\label{eq2}
  R_{\alpha,p}(u)= \frac{\int_{\Omega}|\nabla
    u|^2 \, dx}{\left(\int_{\Omega}{\Psi_{\alpha}} |u|^p \, dx\right)^\frac
    2p},\qquad u\in H^1_0(\Omega)\setminus \{0\}
\end{equation}
is the \emph{Rayleigh quotient} associated to \eqref{eq1}. It is known
that any minimizers of \eqref{eq:6} can be scaled so as to become
solutions of \eqref{eq1}. Therefore, we will use freely this fact in
the sequel.
\begin{remark}
  Unlike the result of~\cite{ni}, the fact that the annulus $\Omega$
  does not contain the origin implies the existence of a radial
  solution of \eqref{eq1} for any $p>2$. Indeed, the embedding
  $H_{0,\mathrm{rad}}^1(\Omega)\subset L^q(\Omega)$ is compact for
  every $q \geq 1$, and therefore the infimum \eqref{eq:6}
  is achieved by a (radial) function.
\end{remark}

In the next Proposition, we provide an estimate of the
\textit{energy} $S^{\mathrm{rad}}_{\alpha,p} $ as $\alpha \to
\infty$.

\begin{proposition} \label{prop:2}
Let $p>2$. As $\alpha \to \infty$, there exist two constants $C_1$
and $C_2$ depending on $p$ such that
\begin{equation} \label{eq:11}
0< C_1 \leq \frac{S_{\alpha,p}^{\rm rad}}{ \alpha^{1+2/p}} \leq
C_2 < + \infty.
\end{equation}
Moreover, for any $M>2$ it is possible to choose the constants
$C_1$ and $C_2$ independent of $p\in (2,M]$.
\end{proposition}

\begin{proof}
The upper bound $C_2$ can be obtain exactly as in \cite{ssw}: we
fix a positive, radial function $\psi \in C_0^\infty(\Omega)$, and
set $\psi_\alpha (x) =\psi_\alpha (|x|) = \psi (\alpha
(|x|-3+3/\alpha))$. Then
\begin{eqnarray*}
\int_\Omega |\nabla \psi_\alpha|^2 \, dx &=& \omega_{N-1}
\int_{3-\frac{2}{\alpha}}^3
                          \left( \psi'_\alpha(r) \right)^2 r^{N-1} \, dr\\
&=&  \omega_{N-1} \int_{1}^3
                          \alpha^2\psi'(s)^2 \left( \frac{s}{\alpha}
                          +3-\frac{3}{\alpha} \right)^{N-1}
                          \alpha^{-1} \, ds \\
&=& \alpha\ \omega_{N-1} \int_{1}^3 \psi'(s)^2 \left(
\frac{s+3\alpha-3}{\alpha s}
                          \right)^{N-1} \, s^{N-1}\, ds \\
&\leq& 3^{N-1} \alpha \int_{\Omega} | \nabla \psi |^2 \, dx, \quad
\hbox{(since $1 \leq \frac{s+3\alpha-3}{\alpha s}\leq 3$)}
\end{eqnarray*}
and
\begin{equation*}
\int_\Omega \Psi_\alpha  \psi_\alpha^p \, dx \geq \left(1 -
\frac{2}{\alpha} \right)^{\alpha} \alpha^{-1} \int_\Omega \psi^p
\, dx.
\end{equation*}
This proves that $S_{\alpha,p}^{\mathrm{rad}} \leq C(\alpha, p)
\alpha^{1+\frac{2}{p}}$, where
\[
C(\alpha, p) =3^{N-1}\frac{\int_\Omega|\nabla \psi|^2 dx}{(1-\frac
2\alpha)^\frac{2\alpha}{p}(\int_\Omega \psi^p(x)\, dx)^\frac 2p}
\leq C_2 \quad\hbox{for any $p>2$ and $\a>1$}.
\]
To find the lower bound $C_1$, we will perform some scaling. Let
us define the functions $\psi_1 \colon [1,2] \to [1,2]$ and
$\psi_2 \colon [2,3] \to [2,3]$ as follows:
\begin{equation}
\psi_1 (r) = 2- (2- r)^\beta, \quad \psi_2 (r) = 2 +  (r-2)^\beta,
\end{equation}
where $\beta\in (0,1)$ will be chosen hereafter. It is clear that
we can obtain a piecewise $C^1$ homeomorphism  $\psi \colon [1,3]
\to [1,3]$ by gluing $\psi_1$ and $\psi_2$. Now, for any radial
function $u \in H_0^1(\Omega)$, we set $v(\rho) = u(\psi(\rho))$
and compute:
\begin{eqnarray}\label{eq:8}
\int_\Omega |\nabla u|^2 \, dx
&=&\omega_{N-1} \int_{1}^{3} |u'(r)|^2 r^{N-1} \, dr \notag \\
&\geq&\omega_{N-1} \int_{1}^{3} |u'(r)|^2  \, dr \notag \\
&=& \omega_{N-1} \left(\int_{1}^{2} |v'(\rho)|^2
\frac{1}{\psi_1'(\rho)} \, d\rho +
 \int_{2}^3 |v'(\rho)|^2 \frac{1}{\psi_2'(\rho)} \, d\rho\right) \notag \\
&=& \omega_{N-1} \, \frac{1}{\beta}\left( \int_{1}^{2}
|v'(\rho)|^2 (2-\rho)^{1-\beta} \, d\rho \right. \notag \\
&\qquad& \left. {} +\int_{2}^{3}
|v'(\rho)|^2 (\rho-2)^{1-\beta} \, d\rho \right)\notag \\
&=& \omega_{N-1} \, \frac{1}{\beta} \int_{1}^{3}
|v'(\rho)|^2 |\rho-2|^{1-\beta} \, d\rho \notag\\
&\geq&  \omega_{N-1} \, \frac{1}{\beta} \int_{1}^{3} |v'(\rho)|^2
|\rho-2| \, d\rho.
\end{eqnarray}
Choosing $\beta = 1/(\alpha+1)$,
\begin{eqnarray}
\int_\Omega \Psi_\alpha (x) |u(|x|)|^p \, dx
&=& \omega_{N-1} \int_{1}^{3}\Psi_\alpha (r)|u(r)|^p r^{N-1}\; dr\notag \\
&\leq&3^{N-1} \omega_{N-1} \int_{1}^{3}\Psi_\alpha (r)|u(r)|^p\; dr\notag \\
&=&3^{N-1}\omega_{N-1}\left( \int_{1}^{2} \Psi_\alpha
(\psi_1(\rho)) |v(\rho)|^p
\psi_1'(\rho) \, d\rho\right.  \notag \\
&& \qquad + \left. \int_{2}^3 \Psi_\alpha
(\psi_2(\rho))|v(\rho)|^p \psi_2'(\rho) \,
 d\rho\right) \notag \\
&=& 3^{N-1}\omega_{N-1} \beta \int_{1}^3 |v(\rho)|^p \, d \rho.
\label{eq:14}
\end{eqnarray}
Since we are integrating over $\Omega$ and $0 \notin
\Omega$, the integral $\int_{1/2}^1 |v'(\rho)|^2 \rho^{N-1} \, d\rho$
is finite if and only if $\int_{1/2}^1 |v'(\rho)|^2 \, d\rho$ is
finite.
  Therefore,
\begin{equation} \label{eq:7}
R_{\a,p}(u) \geq {C}{\alpha^{1+\frac{2}{p}}} \inf_{\substack{ v
\in H^1_0 (\Omega) \\ v \neq 0}} \frac{\int_{1}^3 |v'(\rho)|^2 |4
\rho -3| \, d\rho}{\left(\int_{1}^3 |v(\rho)|^p \, d\rho
\right)^{2/p}}.
\end{equation}
where $C$ depends only on $N$. To complete the proof, we will show
that the right-hand side of \eqref{eq:7} is greater than zero.
This follows from some general Hardy--type inequality (see
\cite{opic}, Theorem 1.14), but we present here an elementary
proof for the sake of completeness. Indeed, given $v \in
H^1_{0,\mathrm{rad}} (\Omega)$, we can write for $\rho\in [1,2]$
\begin{eqnarray*}
|v(\rho)| &=& |v(\rho)-v(1)| \leq \int_{1}^\rho |v'(t)|
|2-t|^{1/2} \frac{dt}
{|2-t|^{1/2}} \\
&\leq& \left( \int_{1}^\rho |v'(t)|^2 |2-t| \, dt \right)^{1/2}
\left( \int_{1}^\rho \frac{dt}{|2-t|} \right)^{1/2} \\
&\leq& \left( \int_{1}^3 |v'(t)|^2 |2-t| \, dt \right)^{1/2}
\left( - \log |2-\rho| \right)^{1/2}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
  \int_{1}^{2} |v(\rho)|^p \, d\rho &\leq&  \left( \int_{1}^3 |v'(\rho)|^2 |2-\rho| \, d\rho \right)^{p/2}
   \int_{1}^{2} \left( -\log (2-\rho) \right)^{p/2} \, d\rho \\
  &=&  \left( \int_{1}^3 |v'(\rho)|^2 |2-\rho| \, d\rho \right)^{p/2}
  \int_0^\infty t^{p/2} e^{-t} \, dt \\
  &=&  \Gamma \left( \frac{p+2}{2} \right)
  \left( \int_{1}^3 |v'(\rho)|^2 |2-\rho| \, d\rho \right)^{p/2},
\end{eqnarray*}
and in a similar way
\[
\int_{2}^{3} |v(\rho)|^p\, d\rho \leq  \Gamma \left( \frac{p+2}{2}
\right) \left( \int_{1}^3 |v'(\rho)|^2 |2-\rho| \, d\rho
\right)^{p/2}.
\]
Therefore
\[
{\int_{1}^{3} |v'(\rho)|^2 |2-\rho| \, d\rho} \geq \frac{1}{2^{
2/p}\Gamma \left( \frac{p+2}{2}
\right)^{2/p}}\left(\int_{1}^3|v(\rho)|^p\;d\rho\right)^{\frac
2p}.
\]
This implies that  the infimum in \eqref{eq:7} is strictly
positive and for any $M>2$ there exists a constant $C_1>0$ such
that $2^{-2/p}\geq C_1\, \Gamma \left( \frac{p+2}{2}
\right)^{2/p}$ for any $p\in (2,M]$, since the Gamma function is
positive, $C^\infty$ and  $\Gamma \left( \frac{p+2}{2} \right)\sim
(p/2)^{p/2}{\rm e}^{-p/2}\sqrt{\pi p}$ for $p\to +\infty$. We
finally collect \eqref{eq:8} and \eqref{eq:14} to get the desired
conclusion
\[
S_{\alpha,p}^{\mathrm{rad}} \geq C_1 \, \alpha^{1+\frac{2}{p}}.
\]
\end{proof}

Set now
\begin{equation}\label{eq3}
S_{\alpha,p}=\inf_{\substack{u \in H_0^1(\Omega) \\ u \neq
0}}R_{\alpha,p}(u).
\end{equation}
It is easily proved that for $p$ subcritical ~$S_{\alpha,p}$ is
attained by a function $u_{\alpha,p}$ that  satisfies (up to a
scaling) equation \eqref{eq1}.

In order to prove that { any} solution  $u_{\alpha,p}$ is not
radial (at least for $\a$ large) we need
 an estimate from above of the level
$S_{\a,p}$.
\begin{lemma}
Let $p\in (2,2^*)$. There exists $\bar \alpha$ such that for
$\a\geq\bar\alpha$
\begin{equation} \label{eq:S-asym}
S_{\a,p}\le C\a^{2-N+\frac{2N}{p}}.
\end{equation}
\end{lemma}
\begin{proof} The proof essentially follows the same techniques    of
\cite{ssw}.

Let $\psi$ be a positive smooth function with support in the unit
ball
  $B$. Let us consider
  $\psi_\a(x)=\psi(\a(x-x_\a))$, where $x_{\a}=(3-\frac
  1\a,0,\ldots,0)$. Since $\psi_\a$ has support in the ball
  $B(x_\a,\frac 1\a)$, by the change of variable $y=\a(x-x_\a)$ we
  obtain:
\begin{equation*}
\int_{\O} \Psi_\alpha(x) \psi^p_{\a}(x)\, dx =
\int_{B(x_\alpha,\frac 1\alpha)}||x|-2|^\alpha \psi_\a^p(x)\, dx
\geq\left(1-\frac 2\alpha\right)^\alpha \alpha^{-N}\int_{B}
\psi^p(y)\, dy
\end{equation*}
Moreover
\begin{equation*}
\int_{\O} |\nabla \psi_{\a}|^2\, dx = \a^2\int_{\O} |\nabla
\psi(\a(x-x_\a))|^2\, dx =\a^{2-N}\int_{B}|\nabla \psi|^2\, dx,
\end{equation*}
so that
\begin{equation*}
S_{\a,p}\le R_{\a,p}(\psi_\a)\le  C{\alpha^{2-N+\frac{2N}{p}}}
\end{equation*}
for all $\alpha$ sufficiently large. This proves the Lemma.
\end{proof}

By comparing \eqref{eq:S-asym} and \eqref{eq:11}, we deduce a
first symmetry--breaking result.

\begin{theorem}
  Let $p \in (2,2^*)$. For $\alpha$  sufficiently large, { any} ground
  state $u_{\alpha,p}$ is a non-radial function.
\end{theorem}
\begin{proof}
  From \eqref{eq:S-asym} and \eqref{eq:11} it follows that
  $S_{\alpha,p} < S_{\alpha,p}^{\mathrm{rad}}$ when $\alpha$ is large.
\end{proof}

\vspace{.2cm}
\section{Symmetry breaking as $p\to 2^*$}

In this section we consider  $\alpha$ fixed, $p$ close to $2^*$
and we establish the following
\begin{theorem}\label{3sol}
Let $\a>0$. For $p $ close to $2^*$ the quotient $R_{\alpha,p}$
has at least two non radial local minima.
\end{theorem}

We briefly explain how the proof proceeds. Of course, we already
know that any global minimizer of $R_{\alpha,p}$ produces a first
solution $u_{\alpha,p}$.  As the Theorem \ref{conc} states, this
solution concentrates at precisely one point of the boundary
$\partial\Omega$. Since this boundary has two connected
components, we will minimize $R_{\alpha,p}$ over the set $\Lambda$
of $H_0^1$ functions which ``concentrate'' at the opposite
component of the boundary. A careful estimate is proved in order
to show that minimizers fall inside the interior of $\Lambda$.

Consider now any minimizer~$u_{\alpha,p}$. As in \cite{caopeng} we have a
description of the profile of $u_{\alpha,p}$ as $p \to 2^*$.

\begin{theorem}\label{conc}
Let $p\in (2,2^*)$ and $\alpha >0$.  { Any} minimum $u_{\alpha,p}$
of $R_{\alpha, p}(u)$ in $H_0^1\setminus\{ 0\}$ satisfies (passing
to a subsequence) for some $x_0\in \partial \Omega$
\begin{itemize}
  \item[i)] $|\nabla u_{\alpha,p}|^2 \rightarrow \mu \delta_{x_0}$
   weakly in the sense of measure as $p\rightarrow 2^*$,
\item[ii)] $ |u_{\alpha,p}|^{2^*}\rightarrow \nu \delta_{x_0}$
weakly in the sense of measure as $p\rightarrow 2^*$,
\end{itemize}
where $\mu>0$ and $\nu >0$ are such that $\mu \geq
S_{0,2^*}\nu^{2/2^*}$ and $\delta_x$ is the Dirac mass at $x$.
\end{theorem}

\begin{proof}
This result can be proven by using, with suitable modifications,
the same arguments of \cite{caopeng}.
\end{proof}
 To get a second local minimizer, we will assume without loss
of generality that { any} $u_{\alpha,p}$ concentrates at some
point on the sphere $|x|=3$ (a similar argument holds if
$u_{\alpha,p}$ concentrates at some point $x$ with $|x|=1$). After
a rotation, we can even assume that { any} $u_{\alpha,p}$
concentrates at the point $(3,0,\dots,0)$.

Let
\[
\Omega^-=\left\{x\in\mathbb{R}^N \mid 1<|x|<2 \right\} ,\quad
  \Omega^+=\left\{x\in\mathbb{R}^N \mid 2<|x|<3 \right\}
\]
and
\[
\Sigma =\left\{u\in H_0^1 \setminus\{0\} \mid  \int_{\Omega
^+}|\nabla u|^2\ dx=
 \int_{\Omega ^-}|\nabla u|^2\ dx\right\}.
\]
Let us denote
\[
T_{\alpha,p}=\inf_{u \in \Sigma} R_{\alpha,p}(u).
\]
We have the following uniform estimate for $T_{\alpha,p}$.
\begin{proposition}\label{equipart}
  Let $\a>0$. There exists $\delta>0$ such that
\[
\liminf_{p\to 2^*}T_{\alpha,p}>S_{0,2^*}+\delta.
\]
\end{proposition}
\begin{proof}
We first prove that  $T_{\alpha,p} $ is achieved by a function
$v_{\a,p}\in \Sigma$. Consider a minimizing sequence $\{u_n\}$ for
$T_{\alpha,p} $. We can exploit the homogeneity of $R_{\alpha,p}$
and assume that $\int_{\Omega}|\nabla u_n|^2 \, dx=1 $. Up to a
subsequence, $u_n$ converges to
$v=v_{\a,p}$  weakly in $H^1_0(\Omega)$ and strongly in
$L^q(\Omega)$, for all $q \in (2,2^*)$. All we have to check is
that $v\in \Sigma$ (proving \textit{a posteriori} that the
convergence of $u_n$ to $v$ is strong). From the strong convergence in
$L^q(\Omega)$ we have that
\begin{equation} \label{sigma}
R_{\alpha,p}(v)\le\frac{1}{\left(\int_{\Omega}\Psi_\alpha(x)|v|^p
dx\right)^{2/p}}=T_{\alpha,p}
\end{equation}
and in particular $v\neq 0$. It is not restrictive to suppose that
$v \geq 0$ in $\Omega$. For the sake of contraddiction, assume that
$$
\int_{\Omega^+}|\nabla v|^2<\frac12.
$$
Fix a nonnegative smooth function $\varphi_1\in
C_0^{\infty}({\Omega^+})$,  $\varphi_1\neq 0$ and $\delta\geq 0$.
Setting $u=v+\delta\varphi_1$ from the positivity of $v$ and
$\varphi_1$, we have, for $\delta>0$,

\begin{equation} \label{sigma2}
\int_{\Omega}\Psi_\alpha(x)|v|^p
dx<\int_{\Omega}\Psi_\alpha(x)|u|^p dx.
\end{equation}

Now,
$$
\int_{\Omega^+}|\nabla u|^2 dx=\int_{\Omega^+}|\nabla
v|^2dx+2\delta\int_{\Omega^+}\nabla v\cdot\nabla\varphi_1 dx
+\delta^2\int_{\Omega^+}|\nabla \varphi_1|^2dx
$$
If we define $g_1:[0,+\infty)\to \mathbb R$ by
$$
g_1(\delta)=\int_{\Omega^+}|\nabla
v|^2dx+2\delta\int_{\Omega^+}\nabla v\cdot\nabla\varphi_1 dx
+\delta^2\int_{\Omega^+}|\nabla \varphi_1|^2dx
$$
we see that $g_1$ is continuous and $ g_1(0)<\frac
12,\quad\displaystyle \lim_{\delta\to+\infty}g_1(\delta)=+\infty.
$ Hence there exists $\delta_1>0$ with $g_1(\delta_1)=\frac12$. We
 can reason in an analogous way if $\int_{\Omega^-}|\nabla
v|^2dx< 1/2$ in order to find $\delta_2\geq 0$ and $ \varphi_2\geq
0$  such that $\int_{\Omega^-}|\nabla
(v+\delta_2\varphi_2)|^2dx=\frac 12$ .

 From  \eqref{sigma2},  this
shows that there exists
$w=v+\delta_1\varphi_1+\delta_2\varphi_2\in \Sigma$ such that
$$
R_{\alpha,p}(w)<T_{\alpha,p}
$$
which gives a contraddiction. Finally we must have that
$v_{\alpha,p}\in\Sigma,$  is a minimum point.

\vspace{.2cm}

 Moreover for any $\alpha>0,$ and $ 2<p<2^*$
we have
 $T_{\alpha,p}\geq S_{\alpha,p}$. We want to prove that  the inequality is
 strict  at least for $p\to 2^*$. Indeed assume on the contrary
 that
 \[
 \liminf_{p\to 2^*}T_{\alpha,p}=\liminf_{p\to
 2^*}R_{\alpha,p}(v_{\alpha,p})=S_{0,2^*}.
 \]
From the definition of $S_{0,2^*}$ and H$\ddot{\rm o}$lder
inequality we get, for a subsequence $p=p_k\to 2^*$
\begin{multline} \label{eq:comebrezis}
S_{0,2^*}\leq \frac{\int_{\Omega}|\nabla v_{\a,p}|^2\
dx}{\left(\int_{\Omega}| v_{\a,p}|^{2^*}\
dx\right)^{2/{2^*}}}\leq{
|\Omega|^{\frac{(2^*-p)2}{2^*p}}}\frac{\int_{\Omega}|\nabla
v_{\a,p}|^2\ dx}{(\int_{\Omega}| v_{\a,p}|^{p}\
dx)^{2/{p}}} \\
\leq{ |\Omega|^{\frac{(2^*-p)2}{2^*p}}}
 \frac{\int |\nabla
v_{\a,p}|^2\ dx}{(\int_{\Omega}\Psi_\alpha(x)| v_{\a,p}|^{p}\
dx)^{2/{p}}}=S_{0,2^*}+o(1)
\end{multline}
since the weight satisfies  $\Psi_\alpha(x)\leq 1$. In particular
\[
\frac{\int_{\Omega}|\nabla v_{\a,p}|^2\ dx}{\left(\int_{\Omega}|
v_{\a,p}|^{2^*}\ dx\right)^{2/{2^*}}}\to S_{0,2^*},
\]
and $v_{\a,p}$ is a minimizing sequence of $S_{0,2^*}$.

In the same way as Cao and Peng did in {\cite{caopeng}}, Theorem
1.1,  we can prove that $v_{\alpha,p}$ concentrates at precisely
one point one of the boundary $\partial \Omega$. This contradicts
the fact that $\int_{\Omega ^+}|\nabla v_{\alpha,p}|^2\ dx=
 \int_{\Omega ^-}|\nabla v_{\alpha,p}|^2\
dx$.
\end{proof}

Consider now the points
\[
x_{0,\varepsilon} = x_0= \left(3-\frac{1}{|\log
\varepsilon|},0,\ldots,0\right), \quad x_{1,\varepsilon} = x_1=
\left( 1+\frac{1}{|\log \varepsilon|},0,\ldots,0 \right)
\]
and
\[
U(x) = \frac{1}{\left(1 +|x| \right)^{(N-2)/2}}.
\]
We recall that $S_{0,2^*}$ is not achieved on any proper subset of
$\mathbb{R}^N$, and that it is independent of $\Omega$. However, it is known
that $S_{0,2^*}(\mathbb{R}^N)$ is achieved, and all the minimizers can be
written in the form
\[
\mathcal{U}_{\theta,y}(x)=\frac{1}{\left(\theta^2+|x-y|^2
  \right)^\frac{N-2}{2}}, \quad \hbox{$\theta>0$, $y \in \mathbb{R}^N$}.
\]

We set
\[
U^i_\varepsilon(x)=\varepsilon^{-\frac{N-2}{2}} \
U\left(\frac{x-x_i}{\sqrt\varepsilon}\right)=\frac{1}{(\varepsilon+|x-x_i|^2)^{\frac{N-2}{2}}},
\]
and denote by $\varphi_i$ ($i=0,1$) two \textit{cut-off} functions
such that $0\le\varphi_i\le 1$, $|\nabla\varphi_i|\le C|\log
\varepsilon|$ for some constant $C>0$, and
\[
\varphi_i (x) =
\begin{cases}
  1, &\text{if $|x-x_i| < \frac{1}{2|\log \varepsilon|}$}  \\[10pt]
  0, &\text{if $|x-x_i| \geq \frac{1}{|\log \varepsilon|}$}.
\end{cases}
\]
The following Lemma shows that the truncated functions
\begin{equation} \label{eq:instanton}
  u^i_\varepsilon(x)=\varphi_i(x)U^i_\varepsilon(x), \quad i=0,1,
\end{equation}
are almost minimizers for $S_{0,2^*}$. We omit the proof of this fact,
since it is an easy modification of the argument of Cao and Peng in
\cite{caopeng}.

\begin {lemma}\label{zero}
  Let $\a >0$. There results
  \[
  \lim_{p\to
    2^*}R_{\alpha,p}(u^i_\varepsilon)=S_{0,2^*}+K(\varepsilon)
  \]
  with $\displaystyle\lim_{\varepsilon \to 0}K(\varepsilon)=0.$
\end{lemma}

\begin{remark}
A direct consequence of Lemma \ref{zero} is that
    $S_{0,2^*}=S_{\alpha, 2^*}$. Indeed $S_{0,2^*}\leq S_{\alpha,2^*}$
    since $\Psi_\alpha (|x|)\leq 1$ . On the other hand by Lemma
    \ref{zero}
\[
 R_{\alpha,2^*}(u^i_\varepsilon)= \lim_{p\to
 2^*}R_{\alpha,p}(u^i_\varepsilon)=S_{0,2^*}+K(\varepsilon).
\]
Therefore $S_{0,2^*}+K(\varepsilon)\geq S_{\alpha,2^*}$ for every
$\varepsilon >0$. Letting $\varepsilon \to 0$ we conclude
$S_{0,2^*}\geq S_{\alpha,2^*}$.
\end{remark}

\bigskip

\noindent We are now ready to prove the Theorem \ref{3sol}.
\begin{proof}[Proof of Theorem \ref{3sol}]
 Let $u_{\alpha,p}$ be a ground state solution. Let
us suppose that it concentrates on the outer boundary.  Consider
the open subset
\[
\Lambda=\left\{u\in H^1_0(\Omega):\;\int_{\Omega^-}|\nabla u|^2\
dx> \int_{\Omega^+}|\nabla u|^2\ dx\right\}.
\]
The infimum of $R_{\alpha,p}$ on $\overline\Lambda$ is achieved.
However it cannot be achieved on the boundary $\partial
\Lambda=\Sigma$. Indeed, by Proposition \ref{equipart},
$$\displaystyle \inf_\Sigma
R_{\alpha,p}>S_{0,2^*}+\delta \;\;{\rm as }\;\;p\to 2^*
$$
 and $$\displaystyle\inf_{\Lambda}R_{\alpha,p}(u)\le
R_{\alpha,p}(u^1_{\varepsilon})\to
S_{0,2^*}+K_1(\varepsilon)\;\;{\rm as }\;\;p\to 2^*$$
 since
$u^1_\varepsilon\in \Lambda $ for $\varepsilon $ small enough.
Then the infimum is achieved in a interior point of $\Lambda$ and
is therefore a critical point of $R_{\alpha,p}$.

\end{proof}

\begin{remark}
  Theorem \ref{conc} shows that { any} ground state solution
 ``concentrates" in a single point at the boundary as $p\to 2 ^*$ and
  consequently this solution is not radial. This symmetry breaking can
  be also proved by using a continuation argument as in
  \cite{BN}. Indeed, \eqref{eq:comebrezis} shows that $\lim_{p \to
    2^*} S_{\alpha,p} = S_{0,2^*}$, and since $S_{0,2^*}<S^{\rm
    rad}_{0,2^*}$ we conclude as in \cite{BN} that ground states of
  $S_{\alpha,p}$ cannot be radially symmetric as $p \to 2^*$.
\end{remark}

\section{existence of a third non-radial solution}

In the previous section we proved the existence of two solutions
of \eqref{eq1} which are  local minima of the Rayleigh quotient
for $p$ near $2^*$. One would expect another critical point of
$R_{\alpha,p}$ located in some sense between these minimum points.
This is precisely the idea we are going to pursue further in the
current section.

For ${\varepsilon}$ small enough let $u^i_{\varepsilon}=\varphi_i
U^i_{\varepsilon}$, $i\in \{0,1\}$, be defined as in
\eqref{eq:instanton}.  We will prove that $R_{\alpha,p}$ has the
Mountain Pass geometry.

\bigskip

Let us introduce the \textit{mountain--pass level}
\[
\beta=\beta(\alpha,p)=\inf_{\gamma\in\Gamma}\max_{t\in
[0,1]}R_{\alpha,p}(\gamma(t)),
\]
where $\Gamma =\{\gamma\in C([0,1],H^1_0(\Omega)) \mid
\gamma(0)=u^0_{\varepsilon},\; \gamma(1)=u^1_{\varepsilon}\}$ is
the set of continuous paths joining $u_\ge^0$ with $u_\ge^1$.
We claim that $\beta$ is a critical value for $R_{\alpha,p}$.

We begin to prove that $\beta$ is larger, uniformly with respect
to~$\ge$, than the values of the functional $R_{\alpha,p}$ at the
points $u_\ge^0$ and $u_\ge^1$.

\begin{lemma}\label{lemma:cont} Let
  $M_{\varepsilon}=\max\{R_{\alpha,p}(u^0_{\varepsilon}),R_{\alpha,p}(u^1_{\varepsilon})\}$.
  There exists $\sigma>0$ such that $\beta\geq
  M_{\varepsilon}+\sigma$ uniformly with respect
to~$\ge$.
\end{lemma}

\begin{proof}
  We prove that there exists $\sigma$ such that for all $\gamma \in
  \Gamma$
\[
\max R_{\alpha,p}(\gamma(t))\geq M_{\varepsilon}+\sigma.
\]
A simple continuity argument shows that for every $\gamma \in
\Gamma$ there exists $t_{\gamma}$ such that $\gamma(t_{\gamma})\in
\Sigma$, where
\[
\Sigma =\left\{u\in H_0^1 \setminus\{0\} \mid  \int_{\Omega
^+}|\nabla u|^2\ dx=
 \int_{\Omega ^-}|\nabla u|^2\
dx\right\}.
\]
Indeed the map $t \in [0,1] \mapsto \int_{\Omega^{+}} |\nabla
\gamma (t)|^2 \, dx - \int_{\Omega^{-}} |\nabla \gamma (t)|^2 \,
dx$ is continuous and it  takes a negative value at $t=0$ and a
positive value at $t=1$. Now Proposition \ref{equipart} implies,
for $p$ near $2^*$ the existence of $\delta>0$ with
\[
\max_{t\in [0,1]} R_{\alpha,p}(\gamma(t))\geq
R_{\alpha,p}(\gamma(t_{\gamma}))\geq \inf_{u\in \Sigma
}R_{\alpha,p}(u)\geq S_{0,2^*}+\delta.
\]
On the other hand, for $\varepsilon$ sufficiently small,
\[
M_{\varepsilon}<S_{0,2^*}+\frac{\delta}{2}.
\]
This concludes the proof.
\end{proof}

The previous estimate allows us to show that $\beta$ is a critical
level for $R_{\alpha,p}$. Therefore a further nonradial solution
to \eqref{eq1} arises.

\begin{proposition}\label{prop:MP}
There exist  $\bar{\alpha} > 0$ and $2<\bar p<2^*$ such that   for
all $\a\geq\bar{\a}$ and $\bar p\le p< 2^*$ it results that
$\beta$ is a critical value for $R_{\a,p}$ and it is attained by a
non-radial function.
\end{proposition}
\begin{proof}
  From the previous result we can apply a deformation argument (see
  \cite{AM,st}) to prove that $\beta$ is a critical level and it is
  attained (since the PS condition is satisfied) by a
  function $w$. From the asymptotic estimate \eqref{eq:11} for the
  radial level $S_{\alpha,p}^{\mathrm{rad}}$, one has that there exists a
  constant $C$ independent from $p$ such that
  \[
  S_{\alpha,p}^{\mathrm{rad}}\geq C \alpha^{1+2/p}.
  \]
  In particular
  \[
  S_{\alpha,p}^{\mathrm{rad}}\to +\infty \quad\text{as $\alpha \to
    +\infty$}.
  \]
  This allows us to choose $\alpha_0$ such that
  $S_{\alpha,p}^{\mathrm{rad}}\geq 3 S_{0,2^*}$ for all
  $\alpha\geq\alpha_0$.

  Define $\zeta\in\Gamma$ by $\zeta(t)=t
  u^1_{\varepsilon}+(1-t)u^0_{\varepsilon}$ for all~$t \in [0,1]$, and
  let $\tau \in [0,1]$ be such that $R_{\alpha,p}(\zeta(
  \tau))=\max_{t\in [0,1]} R_{\alpha,p}(\zeta(t))$.

\vspace{.1cm}

Since $u^1_{\varepsilon}$ and $u^0_{\varepsilon}$ have disjoint
supports one has, for $\varepsilon $ sufficiently small,
\begin{eqnarray*}
  R_{\alpha,p}(w) &=& \beta\le R_{\alpha,p}(\zeta(\tau))=
  \frac{\int_{\Omega}|\nabla (\tau
    u^1_{\varepsilon}+(1-\tau)u^0_{\varepsilon})|^2\ dx}
  {\left(\int_{\Omega}\Psi_{\alpha}|\tau
      u^1_{\varepsilon}+(1-\tau)u^0_{\varepsilon}|^p \ dx\right)^{2/p}}
  \\[10pt]
  &=&
  \frac{\int_{\Omega}\tau^2|\nabla
    u^1_{\varepsilon}|^2\ dx+ \int_{\Omega}(1-\tau)^2|\nabla
    u^0_{\varepsilon}|^2\ dx}{{\left(\tau^p \int_{\Omega}\Psi_{\alpha}|
        u^1_{\varepsilon}|^p\
        dx+(1-\tau)^p\int_{\Omega}\Psi_{\alpha}|u^0_{\varepsilon}|^p \ dx\right)^{2/p}}}
  \\[10pt]
  &\leq& \frac{\tau^2\int_{\Omega}|\nabla u^1_{\varepsilon}|^2\
    dx}{{\left(\tau^p \int_{\Omega}\Psi_{\alpha}|
        u^1_{\varepsilon}|^p\ dx\right)^{2/p}}}+\frac{(1-\tau)^2\int_{\Omega}|\nabla
    u^0_{\varepsilon}|^2\ dx}{{\left((1-\tau)^p
        \int_{\Omega}\Psi_{\alpha}|
        u^0_{\varepsilon}|^p\ dx\right)^{2/p}}} \\[10pt]
  &=&R_{\alpha,p}(u^0_{\varepsilon})+R_{\alpha,p}(u^1_{\varepsilon})\leq
  2 M_{\varepsilon}<3 S_{0,2^*}\le S_{\alpha,p}^{\mathrm{rad}}.
\end{eqnarray*}
This concludes the proof.
\end{proof}
\vspace{.2cm}

\section{Behaviour of the ground-state solutions for $\alpha$
large}

This section is devoted to the analysis of  a ground state
solution as $\alpha \to +\infty$. Even in this case this solution
tends to ``concentrate" at the boundary~$\partial\Omega$. However,
this concentration is much weaker than the concentration as $p \to
2^*$.

\bigskip

We use the notation~$C(r_1,r_2)=\{x\in \mathbb R^N \mid
r_1<|x|<r_2\}$. Let $\delta$ be sufficiently small (say
$\delta<\frac12$) and $\phi$ be a smooth cut-off function such
that $0\le\phi\le 1$ with
\begin{equation} \label{eq:phi}
\phi(x) =
\begin{cases}
1, &x\in C(1,1+\delta) \cup C(3-\delta,3) \\
0, &x\in C(2-\delta,2+\delta)
\end{cases}
\end{equation}
From now on, since $p\in(2,2^*)$ is fixed we denote a ground state
solution of problem (\ref{eq1}) $u_{\a,p}$ with $u_{\a}$.
\begin{proposition}\label{prop:1}
  Let $u_{\alpha}$ be such that $R_{\alpha,p}(u_\a)=S_{\alpha,p}$. If $\phi$ is
  as in \eqref{eq:phi}, then
\begin{equation}
R_{\alpha, p}(\phi u_\a) = S_{\alpha,p} + o(S_{\alpha,p})
\quad\hbox{as $\alpha \to +\infty$}.
\end{equation}
\end{proposition}
\begin{proof}
It is not restrictive, by the homogeneity of $R_{\alpha,p}$, to
assume $\int_{\Omega}|\nabla u_\a|^2\ dx=1$. We split the proof
into two steps.

\noindent\textbf{Step 1.} We claim that
\begin{equation} \label{eq:1}
\int_{\Omega}{\Psi_{\alpha}} (u_\a \phi)^p\ dx= \int_{\Omega}{\Psi_{\alpha}} u_\a^p\ dx\
+ o \left( \int_{\Omega}{\Psi_{\alpha}} u_\a^p\ dx \right)
\end{equation}
Indeed, suppose on the contrary that
\[
\limsup_{\alpha\to\infty}\frac{\int_{\Omega}{\Psi_{\alpha}} u_\a^p(1-\phi^p)\
dx}{\int_{\Omega}{\Psi_{\alpha}} u_\a^p\ dx}=\beta>0
\]
This implies that, up to some subsequence,
\[
\frac{\int_{\Omega}{\Psi_{\alpha}} u_\a^p(1-\phi^p)\ dx}{\int_{\Omega}{\Psi_{\alpha}}
u_\a^p\ dx}>\beta/2>0
\]
Since $1-\phi^p\equiv 0$ on $C(1,1 +\delta) \cup C(3-\delta,3)$ we
have
\begin{eqnarray*}
\int_{\Omega}{\Psi_{\alpha}} u_\a^p(1-\phi^p)\ dx &=&
\int_{C{(1+\delta,3-\delta)}}{\Psi_{\alpha}} u_\a^p(1-\phi^p)\
dx \\
&\leq& (1-\delta)^{\alpha}\int_{\Omega}u_{\a}^p(1-\phi^p)\
dx\le(1-\delta)^{\alpha}\int_{\Omega}u_{\a}^p\ dx.
\end{eqnarray*}
Therefore
\[
\int_{\Omega}u_{\a}^p\ dx\geq
(1-\delta)^{-\alpha}\int_{\Omega}{\Psi_{\alpha}} u_\a^p(1-\phi^p)\ dx
\]
Now
\[
\frac{\int_{\Omega}u_{\a}^p\ dx}{\int_{\Omega}{\Psi_{\alpha}} u_{\a}^p\
dx}\geq (1-\delta)^{-\alpha}\, \frac{\int_{\Omega}{\Psi_{\alpha}}
u_\a^p(1-\phi^p)\ dx}{{\int_{\Omega}{\Psi_{\alpha}} u_{\a}^p}\ dx}\geq
(1-\delta)^{-\alpha}\, {\frac{\beta}{2}}.
\]
Since $S_{\alpha,p}^{p/2}=\left( \int_{\Omega}{\Psi_{\alpha}} u_\a^p\, dx
\right)^{-1}$ the last inequality can be written as
\[
S_{\alpha,p}^{p/2}\geq \frac{\beta}{2} \,
\frac{(1-\delta)^{-\alpha}}{{\int_{\Omega}u_{\a}^p}\ dx} \geq
{{\frac{\beta}{2}}} (1-\delta)^{-\alpha}S^{p/2}_{0 ,p},
\]
where
\[
S_{0,p}=\inf_{u\neq 0}\frac{\int_{\Omega}|\nabla u|^2\,
dx}{(\int_{\Omega} u^p\, dx)^{2/p}}
\]
On the other hand  from \eqref{eq:S-asym} one has the estimate
\[
S_{\alpha,p}^{p/2}\le C \alpha ^{p-\frac{N}{2}p+ N},
\]
which gives a contradiction for $\alpha$ large. This proves
\eqref{eq:1}.

\noindent \textbf{Step 2.} Now we prove that
\begin{equation} \label{eq:2}
\int_{\Omega} |\nabla {u_\a\phi}|^2\ dx= \int_{\Omega} |\nabla
{u_\a}|^2\ dx\; +\ o( 1) =1+o(1).
\end{equation}
It is not difficult to prove that $u_\a$ satisfies the problem
\begin{equation}\label{eq5}
\begin{cases}
 -\Delta u_\a=S_{\a,p}^{p/2}\Psi_{\alpha}u_\a^{p-1} &\text{in $\Omega$},\\
 u_\a>0 &\text{in $\Omega$},\\
u_\a =0 &\text{on $\partial \Omega$},
\end{cases}
\end{equation}
Since $\| \nabla u_\a \|_2=1$, up to subsequences, we have that,
as $\alpha \to \infty$,
\[
u_\alpha \to u \quad\hbox{weakly in $H_0^1(\Omega)$, strongly in
  $L^q(\Omega)$, and a.e.}
\]
We now prove that $u=0$. Indeed, multiplying equation \eqref{eq5}
by a smooth function $\psi$ with $\supp \psi  \subset \subset \O$
and integrate, we obtain
\[
\int_{\Omega} \nabla u_\a \nabla\psi\, dx=\int_{\Omega}
S_{\a,p}^{p/2} \Psi_{\alpha}u_\a^{p-1}\psi\, dx \to 0 , \quad
\alpha\to+\infty
\]
since, by \eqref{eq:S-asym},  $S_{\alpha,p}^{p/2}{\Psi_{\alpha}}\to 0$
uniformly on $\supp \psi$ and $u_\a$ is uniformly bounded in $L^q$
for $1\le q< 2^*$. Hence $\int_{\Omega} \nabla u \cdot \nabla
\varphi \, dx = 0$ for all $\varphi \in C_0^\infty(\Omega)$. Since
$u \in H_0^1(\Omega)$, this implies that $u=0$.

Now we estimate the difference
\begin{multline} \label{eq:20}
\left| \int_{\Omega} |\nabla {u_\a}|^2\;dx-\int_{\Omega} |\nabla
{(u_\a\phi)}|^2\;dx\right|\le
\\
\le \int_{\Omega} |\nabla {u_\a}|^2(1-\phi^2)\;dx+\int_{\Omega}
|\nabla \phi|^2u_\a^2\;dx+2\left|\int_{\Omega} \nabla {u_\a}\nabla
\phi {u_\a}\phi\;dx\right|
\end{multline}
The last terms tend to zero thanks to the strong convergence in
$L^q$ for all $q\in[1,2^*)$.  In order to estimate the term
$\int_{\Omega} |\nabla {u_\a}|^2(1-\phi^2)\ dx$, we multiply
\eqref{eq5} by $u_\a(1-\phi^2)=u_\a\eta$ and integrate.  Since
$\supp\eta=\supp (1-\phi^2)\subset\subset\Omega$ we have
\[
\int_{\Omega} \nabla {u_\a}\nabla(\eta u_\a )\,
dx=\int_{\Omega}S_{\a,p}^{p/2}\Psi_{\alpha}u_\a^{p}\eta\, dx ,
\]
so that
\begin{eqnarray*}
\left|\int_{\Omega}| \nabla {u_\a}|^2\eta\, dx\right| &\leq&
\left|\int_{\Omega}u_\a\nabla\eta\nabla u_\a\, dx\right|+
\left|\int_{\Omega}S_{\a,p}^{p/2}\Psi_{\alpha}u_\a^{p}\eta\, dx
\right|
\\
&\leq& ||\nabla\eta||_{\infty}\int_{\supp \eta}|\nabla u_\a u_\a\,
| \, dx+\left|\int_{\supp
\eta}S_{\a,p}^{p/2}\Psi_{\alpha}u_\a^{p}\eta
 \, dx \right|\to 0.
\end{eqnarray*}
\end{proof}

In proposition \ref{prop:1} we proved that the infimum of the
Rayleigh quotient $R_{\alpha,p}$ is essentially attained by the
function $\phi u_\alpha$. Thanks to the definition of $\phi$, we
can decompose $\phi u_\alpha=u_{\alpha,
  1}+u_{\alpha, 2}$, where $u_{\alpha, 1}$ vanishes in $C(2-\delta,
3)$ and $u_{\alpha, 2}$ vanishes in $C( 1,2 +\delta )$. The
following proposition is the main  step in order to prove that the
function $\phi u_\alpha$ concentrates at  the boundary.

\begin{proposition} \label{prop:3}
Let $\phi u_\alpha=u_{\alpha, 1}+u_{\alpha, 2}$, where $\operatorname
 {supp} u_{\alpha, 1} \subset C(1,2 -\delta)$ and
 $\operatorname{supp}u_{\alpha, 2} \subset C(2 +\delta,3)$, and
 $\lambda_\alpha = {\int_\Omega \Psi_\alpha u_{\alpha,1}^p \, dx}
 \big/ {\int_\Omega \Psi_\alpha u_{\alpha,2}^p \, dx}$. If
 $\lim_{n\to \infty}\lambda_{\alpha_n}=L$ for a sequence
 $\alpha_n\to \infty$ then either $L=0$ or $L=+\infty$.
\end{proposition}

\begin{remark}
For the quantity $\lambda_\alpha = {\int_\Omega \Psi_\alpha
u_{\alpha,1}^p \, dx} \big/ {\int_\Omega \Psi_\alpha
u_{\alpha,2}^p \, dx}$, we cannot exclude the case
$\limsup_{\alpha \to +\infty} \lambda_\alpha=+\infty$ and
$\liminf_{\alpha \to +\infty} \lambda_\alpha=0$. If  a
\emph{uniqueness} result for the minimizer $u_\alpha$ were known,
then it would be easy to conclude that $\alpha \mapsto u_\alpha$
is continuous. Therefore $\lambda_\alpha$ would be continuous,
too, and we could replace both the lower and the upper limit by
a unique limit. In general, one does not expect such a uniqueness
property for any $p$ and any $\alpha$.
 However, when $p \approx 2^*$ we suspect that the uniqueness argument of \cite{PiSe05} may be applied to our
 setting.
\end{remark}

\begin{proof}
By the definition of $u_{\alpha, 1}$ and $u_{\alpha, 2}$ we have
\begin{equation}
R_{\a,p}(\phi u_\alpha)=\frac{\int_{\Omega}|\nabla
u_{\alpha,1}|^2\, dx+\int_{\Omega}|\nabla u_{\alpha,2}|^2\,
dx}{\left(\int_{\Omega}{\Psi_{\alpha}} u_{\alpha,1}^p dx+\int_{\Omega}{\Psi_{\alpha}}
u_{\alpha,2}^pdx\right)^{\frac 2p}}.
\end{equation}
 Since
$u_\alpha$ is a positive solution, we can say that $\lambda_\alpha > 0$. We obtain the
following identity:
\begin{eqnarray}\label{eq7}
R_{\a,p}(\phi u_\alpha) &=&\frac{\int_{\Omega}|\nabla
u_{\alpha,1}|^2\, dx+\int_{\Omega}|\nabla u_{\alpha,2}|^2\,
dx}{\left(\lambda_\alpha\int_{\Omega}{\Psi_{\alpha}}
u_{\alpha,2}^p\, dx+\int_{\Omega}{\Psi_{\alpha}} u_{\alpha,2}^p\, dx\right)^{2/p}}\nonumber \\
&=& \frac{\int_{\Omega}|\nabla u_{\alpha,1}|^2\,
dx}{\left(\lambda_\alpha+1\right)^{2/p} \left(\int_{\Omega}{\Psi_{\alpha}}
u_{\alpha,2}^p \, dx\right)^{2/p}}+\frac{\int_{\Omega}|\nabla
u_{\alpha,2}|^2 \, dx}{\left(\lambda_\alpha+1\right)^{2/p}
\left(\int_{\Omega}{\Psi_{\alpha}} u_{\alpha,2}^p \, dx\right)^{2/p}} \nonumber \\
&=& \frac{\lambda_\alpha^{\frac 2p}\int_{\Omega}|\nabla
u_{\alpha,1}|^2\, dx}{\left(\lambda_\alpha+1\right)^\frac
2p\left(\int_{\Omega}{\Psi_{\alpha}} u_{\alpha,1}^p \,
dx\right)^{2/p}}+\frac{\int_{\Omega}|\nabla u_{\alpha,2}|^2\,
 dx}{\left(\lambda_\alpha+1\right)^\frac 2p\left(\int_{\Omega}{\Psi_{\alpha}}
u_{\alpha,2}^p \, dx\right)^{2/p}}.
\end{eqnarray}
By the definition of $S_{\alpha,p}$ each quotient
$R_{\a,p}(u_{\alpha,1})$ and $R_{\a,p}(u_{\alpha,2})$ in the last
term is greater than or equal to $S_{\alpha,p}$. Therefore  by
Proposition \ref{prop:1} and equation \eqref{eq7} one obtains
\begin{equation}\label{eq6}
S_{\alpha,p}+o(S_{\alpha,p})\geq \frac {1+\lambda_\alpha^\frac
2p}{(\lambda_\alpha+1)^\frac 2p}S_{\alpha,p}.
\end{equation}
We notice that the function $f(x)=\frac {1+x^{2/p}}{(x+1)^{2/p}}$
is strictly greater than 1 for every $x>0$, $f(0)=1$ and $f(x) \to
1$ as $x\to
  +\infty$. Moreover it is increasing in $[0,1]$ and
decreasing in $[1,+\infty)$ and $\max_{x >0} f(x) =
f(1)=2^{1-2/p}$. Let  $L\in \Lambda$ and  $\{\alpha_n\}$  a
sequence such that $\lambda_{\alpha_n} \to L$ as $n \to +\infty$.
Passing to the limit
 in \eqref{eq6}, we obtain that $1\geq\frac
{1+L^{2/p}}{(L+1)^{2/p}}$ and so either $L=+\infty$, or $L=0$.
\end{proof}

\begin{corollary} \label{coroll5}
With the notation of Proposition \ref{prop:3}, for any sequence $\{\alpha_n\}$
such that  $\lambda_{\alpha_n}\rightarrow 0$ one has
\begin{equation}
\lim_{n \to +\infty} \frac{\int_\Omega
|\nabla u_{\alpha_n,1} |^2\, dx}{\int_\Omega |\nabla
u_{\alpha_n,2}|^2 \, dx} =0.
\end{equation}
\end{corollary}
\begin{proof}
Let
\[
\xi_{\alpha}=\frac{\int_\Omega |\nabla u_{\alpha,1} |^2\,
dx}{\int_\Omega |\nabla u_{\alpha,2}|^2 \, dx}
\]
and  suppose that
$\limsup_{n\to\infty}\xi_{\a_n}> 0$. Up to subsequences,
$\xi_{\a_n}>\xi>0$ for some $\xi$. Therefore we have
\begin{eqnarray*}
S_{\alpha_n,p}+o(S_{\alpha_n,p}) &=& \frac{\int_{\Omega}|\nabla
u_{\alpha_n,1}|^2\ dx+\int_{\Omega}|\nabla u_{\alpha_n,2}|^2\
dx}{\left(\int_{\Omega} {\Psi_{\alpha_n}}u_{\alpha_n,1}^p
dx+\int_{\Omega}{\Psi_{\alpha_n}} u_{\alpha_n,2}^pdx\right)^{\frac
2p}}  \\
&=& \frac{(1+\xi_{\a_n})\int_{\Omega}|\nabla u_{\alpha_n,2}|^2\
dx}{\left(\int_{\Omega}{\Psi_{\alpha_n}}
u_{\alpha_n,2}^p dx\right)^{\frac 2p}(1+\lambda_{\a_n})} \\
&\geq& R_{\a_n,p}(u_{\a_n,2})\frac{1+\xi}{1+o(1)}\geq
(1+\xi)S_{\alpha_n,p}+o(S_{\alpha_n,p}),
\end{eqnarray*}
which is a contradiction. Hence
\[
\xi_{\alpha_n}=\frac{\int_\Omega |\nabla u_{\alpha_n,1} |^2\,
dx}{\int_\Omega |\nabla u_{\alpha_n,2}|^2 \, dx}\to 0.
\]
\end{proof}

\begin{remark}
An immediate consequence of the previous results is that in
particular for any $\alpha_n$ such that $\lambda_{\alpha_n}\to 0$
\begin{equation} \label{eq:strong}
\lim_{n \to +\infty} \int_\Omega |\nabla u_{\alpha_n,1}|^2 \, dx
=0.
\end{equation}
\end{remark}
\vspace{.2cm}

\section{Acknowledgements}
We would like to thank E.~Serra for suggesting the problem and for
his constant support.

redpajama_set_name: RedPajamaArXiv


text:
Winsor, G McLeod
Entry updated 29 November 2021. Tagged: Author.
(1856-1939) UK author in whose first sf novel, Station X (1919), a psychic Invasion from Mars is repelled by an Earth-Venus alliance, despite the Martians' use of an Antigravity device and advanced Weapons to power a hijacked warship; in 1975 the book was reprinted with an introduction by Richard Gid Powers which mystifyingly claims it to be an important work. In The Mysterious Disappearances (1926; vt Vanishing Men 1927) a Mad Scientist discovers a new Element which allows him to levitate his victims, committing the sort of "impossible crime" so popular in the detective fiction of the early decades of the twentieth century. [JC/PN]
George McLeod Winsor
born Gateshead. County Durham [now Tyne and Wear]: 1856
died Isleworth, Middlesex: 27 July 1939
Station X (London: Herbert Jenkins, 1919) [hb/uncredited]
The Mysterious Disappearances (London: Faber and Gwyer, 1926) [hb/J Abbey]
Vanishing Men (New York: William Morrow, 1927) [vt of the above: hb/]

redpajama_set_name: RedPajamaCommonCrawl


text:
Javad Hardani (born 22 March 1984) is a Paralympic athlete from Iran. He competes in throwing events, and is classified F38, a class for athletes with cerebral palsy.

Javad competed in the 2008 Summer Paralympics where he won the F37/38 discus and finished third in F37/38 javelin. In 2012, at the London Paralympic Games he succeeded in winning a gold medal in the F37/38 discus, breaking the F38 world record in the process.

References

External links

1984 births
Living people
People from Ahvaz
Paralympic athletes of Iran
Paralympic gold medalists for Iran
Paralympic silver medalists for Iran
Paralympic bronze medalists for Iran
Athletes (track and field) at the 2004 Summer Paralympics
Athletes (track and field) at the 2008 Summer Paralympics
Athletes (track and field) at the 2012 Summer Paralympics
Athletes (track and field) at the 2016 Summer Paralympics
Medalists at the 2004 Summer Paralympics
Medalists at the 2008 Summer Paralympics
Medalists at the 2012 Summer Paralympics
Medalists at the 2016 Summer Paralympics
World record holders in Paralympic athletics
Paralympic medalists in athletics (track and field)
Iranian male discus throwers
Iranian male javelin throwers
Iranian male shot putters
21st-century Iranian people
Medalists at the 2010 Asian Para Games
Medalists at the 2018 Asian Para Games
World Para Athletics Championships winners
Medalists at the World Para Athletics Championships

redpajama_set_name: RedPajamaWikipedia


Dr. Sepehr Egrari fuses the experience of a skilled surgeon with the guiding philosophy that "beauty can buoy the spirit and enhance lives." He leads the exquisite Bellevue, Washington accredited facility as its sole physician. Dr. Egrari maintains an impressive roster of credentials. He is certified and recently re-certified by the American Board of Plastic Surgery and the American Board of Surgery, both of which are under theAmerican Board of Medical Specialties. This dual certification is truly the ultimate that a plastic surgeon can achieve in training. Dr. Egrari is involved with a rigorous program of Continual Medical Education (CME).
Dr. Egrari commenced his medical training in Chicago, Illinois. After graduating with top honors from medical school, he completed a rigorous residency in general surgery in Los Angeles. Following this he began a coveted fellowship in plastic and reconstructive surgery at one of country's most respected programs: The University of California at Davis Medical Center in Sacramento, California. Throughout his academic career Dr. Egrari has written extensively about a variety of diverse surgical issues and has presented his work internationally. He is currently involved in multiple exciting and innovative clinical trials. Also, over the years he has established a trusted practice through his vast experience in aesthetic surgery.
Dr. Egrari is on active staff at Overlake Hospital Medical Center. He is a senior member of the PAC Committee at Overlake Hospital and is the immediate past Chair of the Development Committee of the American Society of Plastic Surgeons. He has also served on the Board of Directors of the American Society of Plastic Surgeons. He is a member of the Nominating Committee of The American Society for Aesthetic Plastic Surgery. He also serves on the Board of Advisors of Mona Foundation. Dr. Egrari is a Maliniac Fellow of the American Society of Plastic Surgeons, and a member of the President's Circle of the Aesthetic Surgery Education and Research Foundation, honors reserved for a handful of individuals who have made significant contributions to these societies and foundations.


Getting glammed up for a winter wedding? Whether you're the blushing bride-to-be or a winter wedding guest, these winter lashes are all a flutter.
We have exclusively hand-picked a selection of our Eylure false lashes that we think are perfect for a winter wedding wonderland. This edited collection is suited to both the bride-to-be and the wedding guests. With a range of different styles to suit different tastes, every one of these lash styles will add a flourish to your winter wedding style.
With different styles to suit your individual needs, we love how these wedding fake lashes ooze romance and glamour. From full and fluttery to wispy and demure, no winter wedding outfit will be complete without these luscious lashes.


text:
<?php
namespace Scalr\Service\Aws\Rds\DataType;

use Scalr\Service\Aws\Rds\AbstractRdsDataType;

/**
 * VpcSecurityGroupMembershipData
 *
 * @author   Vitaliy Demidov   <vitaliy@scalr.com>
 * @since    19.03.2013
 */
class VpcSecurityGroupMembershipData extends AbstractRdsDataType
{

    /**
     * The name of the VPC security group
     *
     * @var string
     */
    public $vpcSecurityGroupId;

    /**
     * The status of the VPC Security Group.
     *
     * @var string
     */
    public $status;

    /**
     * Constructor
     *
     * @param   string     $vpcSecurityGroupId optional The name of the VPC security group
     * @param   string     $status             optional The status of the VPC Security Group
     */
    public function __construct($vpcSecurityGroupId = null, $status = null)
    {
        parent::__construct();
        $this->vpcSecurityGroupId = $vpcSecurityGroupId;
        $this->status = $status;
    }
}

redpajama_set_name: RedPajamaGithub


Borealis Carbondale 27.5+ rims. Double walled carbon construction. This is a great option for those looking for a summer rim that will hold up to whatever you throw at it, while still allowing for the use of up to a 4-inch tire. Double walled construction and wide rim profile gives you a stiff, durable rim and a nice tire profile. Rims are tubeless-ready and include rim tape and valves.


text:
The Wood Stove Decathlon PopMech People's Choice Award
This weekend on the National Mall in Washington, D.C., the best and brightest ideas for cleaner-burning wood stoves are on display.
By David Agrell
WASHINGTON — Both the old guard and a surprising upstart turned heads this weekend at the Wood Stove Decathlon at the National Mall.
Travis Industries earned the Popular Mechanics People's Choice award for its Lopi Cape Cod stove, earning 5021 votes. More than 15,000 ballots were cast online since the award was announced in March, and in person through paper balloting over the weekend. An established stove manufacturer with more than 500 employees and 35 years of industry experience, Travis took advantage of its large customer base and established online presence. Entering an innovative and affordable stove into the competition didn't hurt, either.
Woodstock Soapstone and Unforgettable Fire also put up impressive numbers, and the top three vote getters accounted for 75 percent of the total balloting. PopMech editor-in-chief Jim Meigs presented the award, which is not part of the official decathlon scoring process. Meigs also gave a brief talk on the importance of energizing public interest in emerging technologies, especially when you want your product to sell—a concept not lost on Travis.
"It's no good developing technology if no one wants it," Travis's president, Kurt Rumens, said earlier in the day.
The University of Maryland's Mulciber team collected the most paper ballots over the weekend, thanks to plenty of local support. More significant, though, was that the decathlon's lone university team ranked first in emissions and fourth in efficiency after one round of testing. (There will be at least two rounds of emissions and efficiency tests.) Team Mulciber mostly comprises graduate and undergraduate students from the Clark School's Fire Protection Engineering Department and is arguably the competition underdog.
"We're fire protection majors; we're used to putting fires out," Mulciber team captain Taylor Myers said. "As of about a year ago, the only experience we had with stoves was sitting in front of them to keep warm."
Myers and his team designed the stove from scratch, perhaps unencumbered by centuries of bad stove-building habits. The Mulciber features include an automation system that controls combustion conditions, particulate scrubbers, and a coaxial chimney stack that acts as a heat recovery system, forcing hot air back into the firebox, where it can continue the combustion process.
Still, Mulciber has a way to go. Decathlon judges will rank each team in five categories. In addition to earning objective scores for emissions and efficiency, each stove will be ranked for affordability, market appeal, and level of innovation. The competition will continue through Tuesday, with the overall winner announced at 4 pm that afternoon. The grand prize is $25,000.
The decathlon is open to the public and includes panels by industry and energy policy experts. Among the upcoming topics are the Future of Wood Stove Technology (10 am on Monday), which will be moderated by Meigs; Sustainability of Biomass Resources (11 am on Monday); and Forest Harvesting Practices (11 am on Tuesday).
David Agrell is PopMech's associate home editor. For competition updates, follow him on on Twitter: @dagrell.
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text:
ETs UFOsRecent Articles
Jim Marrs: Secret Government Program Used Telepathy to Contact Aliens
Alexander LightNovember 28, 2015
by Greg White
Military-trained psychic spies were able to view non-humans on Earth and elsewhere through remote viewing, argued Jim Marrs at this year's Secret Space Program conference in Bastrop, Texas.
Jim Marrs is a former newspaper journalist who has written for a myriad of news outlets, including the Fort Worth Star-Telegram.
He is perhaps best known for his 1989 book Crossfire: The Plot That Killed Kennedy, which reached the New York Times paperback non-fiction best seller list and later served as the basis for Oliver Stone's film JFK.
SRI International developed a method of spying called "remote viewing" in the 1970s, which caught the attention of the CIA and was later passed on to the U.S. Army. Remote viewing is a form of telepathy that perceives places, persons and actions through extrasensory perception (ESP).
According to Marrs, everyone has the ability to do remote viewing. It is not limited to time and place. You can use it to look into the past and future.
However, the further you look into the future, the fuzzier it gets, which suggests the future is not determined, explained Marrs.
US Military launches StarGate program
A US military program, known as StarGate, sought to locate and map Soviet Union submarines using remote viewing. The process involved a group of individuals who attempted to visualize the coordinates on a world map, given the magnitude and longitude of the submarines.
The program had apparently worked effectively for more than twenty years.
Army intelligence officer Joe McMoneagle is said to be the first man to be taught remote viewing during the StarGate project. The original group consisted of six remote viewers. During the course of the program, there was an interest in UFOs. One of the best remote viewers, according to Marrs, went by the name of Pat Price.
According to Price, he found four alien bases on Earth. They were located on Mount Perdido in Pyrenees, Mount Inyangani in Zimbabwe, Mount Ziel in Australia and Mount Haye's in Alaska. The organization originally filed the report away.
In 1998, Skip Atwater, executive officer of the grill flame unit, was responsible for giving members of the group odd tasks. He recalled Price's four alien bases and gave them the coordinates for each. All the members came back and reported that all the bases were on Earth and run by aliens.
Marrs believes that aliens have visited Earth many times throughout human history. Nevertheless, he believes the frequency of visits spiked with the rise of nuclear weapons.
The Aztec Incident
Marrs also referenced the 1948 Aztec story as evidence that remote viewing has been used to communicate with aliens. According to various reports, a UFO crashed in the New Mexico desert in 1948. Alien corpses were taken from the site by the U.S. Air Force, which many believe were dissected.
There has been a lot of controversy about the Aztec story. The reason not many people know about the Aztec story, according to Marrs, is because Aztec is near the Colorado border.
Military officials had been tracking the object prior to the crash. When the UFO landed, military units in Colorado immediately rushed to the site and disposed the evidence before anyone had a chance to see it.
In 2009, Randy Barnes of the Aztec Library Association contracted with Problems > Solutions > Innovations, headed by Lyn Buchanan, who was once in charge of the army's remote viewing unit. Eleven remote viewers were summoned to take a look at Aztec to see if the incident really happened.
The remote viewers found that a large, silver object designed for transportation flew to the ground. Many people had observed the object. They found "biologicals" in the craft. Some witnesses tried to help the creatures but couldn't do so due to language barriers.
Eventually, the army showed up. There was some sort of negotiation among the parties involved. They took the biologicals inside the craft to an undisclosed location.
According the report, the aliens could use remote viewing to manipulate their captures. Government officials couldn't let the aliens loose. As a result, an official decision was made to execute the aliens.
Marrs's claims are mind-bending and provocative. He draws upon multiple lines of evidence to support how remote viewing has substantiated the existence of UFOs.
For more information, check out Marrs's book, Alien Agenda.
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text:
Local communities' solidarity in face of unprecedented floods in British Columbia
November 30, 2021 Clare Hudson, British Columbia, CWI-Canada Canada
Meritt, BC, flooding, 17 November 2021 (Photo: Stepan/CC)
After a year like no other, which has seen record-breaking heatwaves and wildfires that ravaged towns and led to thousands of hectares of land being destroyed, British Columbia (BC), Canada, has faced yet another historical disaster in the form of unprecedented floods. Undoubtedly linked to global climate change, the resulting loss of life, homes, and livelihoods, has impacted all who live in BC which is now under its third state of emergency this year.
On Friday 11 November, meteorologists warned of extreme rainfall coming to Southern BC. A month's worth of rain fell in just 48 hours and the devastating impact that ensued is hard to fathom. News agencies struggled to keep up with the fast pace of events that were unfolding, and by the evening of 12 November, all roads and railroads going into and out of the lower mainland were cut off from the rest of Canada.
Mudslides and washed-out roads trapped thousands of motorists and severed the supply arteries into and out of the lower mainland, which includes the city of Vancouver. Aerial footage showed railroads, which are vital to transport goods to and from the lower mainland, severed apart with train carriages hanging off the edge of cliffs created by mudslides that tumbled down mountainsides.
The town of Merritt, with a population of 7,000, was evacuated, which was closely followed by thousands of residents being evacuated from Abbotsford, which lies approximately 70kms southeast of Vancouver. Many more communities have been left isolated and only reachable by air. The southern part of Vancouver Island was temporarily cut off from the rest of the Island due to the Malahat stretch of highway being washed out. The resulting loss of life and damage including that of thousands of animals in agricultural centres, such as Abbotsford, is on an epic scale.
A state of emergency was declared a few days later which sets limits on how much gas and food people can buy. But panic hit fast with the knowledge supply chains were at best delayed, and many gas stations and groceries stores were empty before the state of emergency was enforced. Supply routes are being used through the US border and limited access is now open on a few routes to only essential vehicles. A full evaluation of the extent of damage to infrastructure is in the early stages as the flood water continues to recede.
The cost of repairing the damage will run into billions (Canadian $). This price tag will be expected to be paid by the working class of BC and Canada, who will suffer the most in the aftermath of this disaster.
Merritt now joins Lytton, the town that was destroyed by wildfires in the summer, as another town in the province destructed by the consequence of global climate change. Global climate change, which, under capitalism, is fuelled by the drive for profit over people, is another problem of this system that the working-class and youth of BC are faced with.
British Columbia's many problems
BC has lost 7,000 people to the opioid overdose epidemic since 2016. Housing in BC is unaffordable for the majority. Many workers and youth are at the mercy of an overinflated rental market and low pay, which forces many to continue to live in poverty. The ongoing injustice and discrimination faced by First Nation and Indigenous communities, such as a lack of access to clean water, destruction of land, and access to safe housing, and services such as health care and education continue to disproportionally impact their communities. This is a fight for all workers and youth – and can see working-class people gaining knowledge of the plight of the First Nation and Indigenous communities, and both building class solidarity.
These issues, combined with living with the fallout from the covid-19 pandemic and a crippled health care system, have taken their toll on the mental health and wellbeing of many.
However, despite the devastation of the past few weeks, community solidarity and human ingenuity have helped to pull people together. Reports of people handing out food and blankets to stranded motorists, providing homes to strangers to stay out of the cold, in addition to footage of hundreds of people forming a human chain to lay sandbags at a dam to prevent further flooding, to people with boats and dinghies pulling cattle and livestock out of rising floodwaters, has shown the true power workers have when organising together. The line workers, construction workers, city workers and more have been working tirelessly to regain power, dig and rebuild roads, and lift those left stranded to safety.
These acts of solidarity can form the basis for a revaluation as to how we can fight together, organise, and rebuild our communities. Global climate change is here, and what has taken place in BC in 2021 is a warning as to the type of events that will likely continue to unfold internationally. The need for a socialist society, based on democratic planning for the needs of the majority, has never been more urgent. Capitalism as a system has proved it is unworkable – the task ahead is to change it.
We demand:
People before profit: For the creation of democratic community organisations to decide and plan rebuilding efforts. For community-led emergency planning and mitigation.
Inclusion of scientific evidence to ensure knowledge regarding both local and international environmental factors and needs, geography, technology, and future planning is utilised for the benefit of all, not big business interests.
For the right to self-determination of First Nation and Indigenous communities: For an end to discrimination, racism, poverty, unsafe water supply, and for equitable access to quality housing and services.
For the planning and provision of shelter, food, and services during environmental crises.
No to big business profiteering out of environmental disasters; working-class people and youth should not pay the price of climate change. For a cap on the cost of fuel, food, and services.
End poverty: For the immediate building of affordable housing and an end to low pay. Access to free education and training for workers and youth.
For the building of a mass party of the working class, youth, and oppressed minorities, with a socialist programme.
For a democratically planned, environmentally friendly economy, as part of a socialist society, based on need and not profit.
Chilean elections – defeat the far right presidential candidate, José Antonio Kast!
Covid-19: New variant emerges but same old capitalist disease


The first thing you must do whenever you buy deodorant or any product that goes on your skin or in your body is to read the ingredients. They're usually in very tiny print on the back of the label. If you have trouble reading small print, carry a magnifying glass around with you so you can read before you buy. This is critical because what you put on your skin can be absorbed directly into your bloodstream. And harmful ingredients in your blood can have potentially deadly effects on your body.
A word of caution: Plant extracts listed on labels can already contain parabens and propylene glycol. These do not have to be listed on the label. Very few herbs supplied to manufacturers are free of parabens or other synthetic preservatives unless they are extracted in vegetable glycerine. To make sure that the products you buy are healthy and free of these harmful chemicals, call the deodorant manufacturer and ask them for detailed information.
by Dr. Christine H. Farlow, D.C. is "The Ingredients Investigator." She has been researching ingredient safety since 1991. She is the author of three books, including the new, second edition of Dying To Look Good.
Right now I'm using Tom's brand. I'm still researching the right brand of deodorant to use. Anything that you use daily is worth investigating. Feel free to leave a comment of healthy products that you use.
by Dr. Bill Douglass Explains Why He Does NOT Recommend Fluoride.
•Binds to calcium ions and affects utilization.
•Affects a number of enzymes responsible for energy production, transmission of nerve impulses and blood coagulation.
•Increases levels of potassium in the blood.
Fluoride Exposure and ToxicityThe controversy over fluoride poisoning or fluorosis revolves around high fluoride levels in drinking water. The accepted range of fluoride in water in 0.7 to 1.2 ppm. Drinking water with levels of fluoride exceeding 2ppm should not be consumed, especially by children. Levels moderately higher than this may cause mild signs and symptoms of poisoning but levels exceeding 10 ppm can cause serious symptoms and even result in death.
Other methods of excessive fluoride exposure include accidental or intentional consumption of insecticides or exposure to fluoride dust in industrial settings.
Signs and Symptoms of Fluoride Poisoning•White specks or streak to yellow-brown discoloration of the teeth.
•Damage of tooth enamel – teeth are more prone to cavities.
•Nausea with or without vomiting.
•Low levels of calcium (hypocalcemia), magnesium (hypomagnesemia) and glucose (hypoglycemia) in the blood.
•High levels of potassium in the blood (hyperkalemia).
In severe cases of fluoride poisoning, there may be signs of shock and cardiac arrest is possible.
Potential problems: One ingredient, triethanolamine, is a possible human carcinogen, and three other ingredients contain harmful additives linked to cancer or other health problems: sodium lauroyl sarcosinate, TEA-stearate and tocopherol. Further, sodium lauroyl sarcosinate is a penetration enhancer that could increase your exposure to the harmful ingredients. The soap also contains six ingredients that have not been assessed for safety by the CIR and one ingredient (fragrance) that could be an allergen.
Note: Each formula contains slightly different ingredients. Those listed are for the Oily Skin formula. For specifics on the other formulas, please visit the EWG's Skin Deep Report.
What's lurking in the soap YOU use? Visit EWG's Skin Deep Report to find out.
Potential problems: One ingredient, lecithin, may contain harmful impurities linked to cancer or other health problems, while four ingredients, lecithin, propylene glycol, tetrasodium EDTA and trisodium HEDTA are penetration enhancers. Propylene glycol poses risks of other health concerns, and 14 ingredients haven't been assessed for safety.
Potential problems: One ingredient, triethanolamine, is a possible human carcinogen and may be linked to other health problems as well. Two ingredients, propylene glycol and sodium laureth sulfate, are penetration enhancers, 13 ingredients have not been assessed for safety, one ingredient (fragrance) could be an allergen, and two ingredients, propylene glycol and sodium stearate, may pose other health concerns.
Potential problems: Two ingredients, PEG-6 and sodium PCA, may contain harmful impurities linked to cancer or other health problems. PEG-6, specifically, may be linked to breast cancer. One ingredient, tetrasodium EDTA, is a penetration enhancer and 15 ingredients have not been safety tested.
Potential problems: One ingredient, butylated hydroxytoluene, is a possible human carcinogen, and another ingredient, tocopheryl acetate, may be linked to cancer or other health problems. Two ingredients, propylene glycol and tetrasodium EDTA, are penetration enhancers, 13 ingredients have not been safety tested, fragrance may be an allergen and propylene glycol may also present other health concerns.
Regular use helps prevent itching and flaking, and may prevent hair loss.
Provides a balanced, healthy scalp.
May help relieve psoriasis (or eczema) on problem skin.
Infuses hair with essential oils of jojoba and green tea to nourish and leave hair glossy.
With regular use, scalp should become balanced and well nourished providing an excellent foundation for healthy hair.
NO talc, mineral oil, FD& C dyes, animal byproducts, formaldehyde-releasing preservatives, and any harsh or unhealthy ingredients.
Find out more about Ecco Bella Holistic Remedies Hair & Scalp Intensive Therapy's Cost-Saving 2-pack Now!
Potential problems: One ingredient, coal tar, is a known or probable human carcinogen and two ingredients, cocamide DEA and PEG-55 propylene glycol oleate, may be linked to cancer or other health problems (PEG-55 propylene glycol oleate may be linked to breast cancer). Four ingredients, menthol, propylene glycol, sodium laureth sulfate and tetrasodium EDTA, are penetration enhancers, and five ingredients have not been safety tested. The fragrance in this product may be an allergen, and four ingredients, ethylparaben, methylparaben, propylene glycol and propylparaben, may pose additional health concerns.
Potential problems: Three ingredients, butylated hydroxytoluene, hydrochloric acid and polyacrylic acid, are possible human carcinogens, and one ingredient, ketoconazole, is a pregnancy concern that is a known, suspected or possible reproductive toxin. Four ingredients, cocamide MEA, polyquaternium-7, quaternium-15 and sodium cocoyl sarcosinate, may be linked to cancer or other health problems, and three ingredients, sodium cocoyl sarcosinate, sodium laureth sulfate and tetrasodium EDTA, are penetration enhancers. Seven ingredients have not been safety tested, the fragrance may be an allergen and cocamide MEA may pose additional health concerns.
Potential problems: Six ingredients, acetamide MEA, cocamide MEA, cocamidopropyl betaine, diazolidinyl urea, PEG-150 distearate and tocopherol, may be linked to cancer or other health problems (PEG-150 distearate may be linked to breast cancer). Three ingredients, propylene glycol, sodium laureth sulfate and tetrasodium EDTA, are penetration enhancers and 44 ingredients haven't been safety tested. Further, five ingredients, cocamide MEA, methylparaben, PEG-150 distearate, propylene glycol and propylparaben, may pose other potential health concerns.
Potential problems: Four ingredients, cocamide DEA, cocamidopropyl betaine, polyquaternium-10 and polyquaternium-7, may contain harmful impurities linked to cancer or other health problems. Five ingredients, disodium EDTA, menthol, propylene glycol, sodium laureth sulfate and tetrasodium EDTA, are penetration enhancers. Fifteen ingredients have not been safety tested, the fragrance may be an allergen, and four ingredients, ethylparaben, methylparaben, propylene glycol and propylparaben, pose other health concerns.
Potential problems: Eight ingredients, cocamide DEA, cocamidopropyl betaine, diazolidinyl urea, DMDM hydantoin, lauramide DEA, octoxynol-1, polyquaternium-10 and polyquaternium-7, may contain harmful impurities linked to cancer or other health problems (octoxynol-1 may be linked to breast cancer). Two ingredients, disodium EDTA and propylene glycol, are penetration enhancers, 18 ingredients haven't been safety tested, the fragrance may be an allergen, and four ingredients, methylparaben, octoxynol-1, propylene glycol and propylparaben, may pose additional health concerns.
Because the personal care products we use daily expose us to a chemical cocktail with completely unknown health effects (some of us use as many as 25 different products containing more than 200 different chemical compounds daily!), and the products' manufacturers are not likely to start cleaning up their acts anytime soon, your best bet to protect your health is to seek out pure and safe personal care products. Sixwise.com has done the research for you and is happy to offer the following safe alternatives for your own personal care.
1. Ecco Bella Holistic Remedies Hair & Scalp Intensive Therapy, an exclusive combination of essential oils including cedarwood, juniper, neem, organic herbs and green tea, all blended in a non-greasy Jojoba oil base. Perfect for itchy, flaky scalps, this treatment contains no talc, mineral oil, FD& C dyes, animal byproducts, formaldehyde-releasing preservatives, or any harsh or unhealthy ingredients.
For more of the purest and truly natural soaps and other personal cleansing products on the market, see the new line of Personal Care Products we have found! And keep watching for more as we are busy analyzing more products in this area that make "natural" claims to recommend and offer ONLY those that really live up to those claims!
Feel free to do your own reasearch and share with the group of readers on my blog using the comment section below. I'm a firm believer that if we intake toxics into our bodies on a daily basis, it will cause long term health issues in the future. Why should you sit around and ignore what you're using on your body every day? Do you really believe that everything you're using is safe?
I'm not the type of person that just sits around and waits for something to happen to me in the future. I just hope that your not either.
This entry was posted on Wednesday, November 10th, 2010 at 11:17 pm	and is filed under Nutrition. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
OMG. I did pretty good on the test but not perfect. My toothpaste, Colgate Total, is the one recommended by my DDS but it contains triclosan, propylene glycol and, of course, a small amount of fluoride. I don't swallow it but perhaps I should look for an alternative. NYC water contains 1 ppm fluoride so that is safe and within the recommended range. Fluoride in small amounts helps to prevent cavities.
The aluminum additives in deodorants are the antiperspirants. Stay cool, don't perspire. Don't work out?
I passed all the other tests but what about cosmetics especially face creams and lotions? Bet there's a lot of bad stuff in those. A biologist friend once said that the best moisturizer to put on one's face is avocado.


The administrative law judges presiding over a disability hearing have access to the applicant's complete file. This includes the medical records and other documentation that the applicants and their disability attorneys submitted to the hearing office in advance as part of their preparation. As a result, some administrative law judges may have already decided to approve a claim for Social Security Disability benefits on the basis of the relevant documents. The judge may have few questions for the claimant or the disability lawyer in this case.
However, some hearings take a longer time. This can include cases where the judge requests testimony from an expert witness such as a medical or vocational expert. These witnesses may be asked to determine hypothetical questions about the applicant, such as his or her level of functionality and ability to perform certain types of work. Some experts may be asked about the availability of certain types of jobs and whether it is even feasible for the applicant to obtain them. The judge and the disability lawyers may ask additional questions to surface key areas of expert knowledge related to the applicant's condition.
People who have faced early denials in their applications for disability benefits can still find hope in a disability hearing. By working with a disability lawyer, people may be able to improve their likelihood of approval at the hearing stage.


UNIQUE WATERFRONT DEVELOPEMENT OR LARGE ESTATE PROPERTY. Boat, fish, Kayak, or canoe from this beautiful parcel! This is one of the few larger developable parcels left on Bayou Grande and is perfect for a large estate like the one next door to it or a waterfront residential development. With the addition of 4 Coast Guard Cutters at NAS and many high level personnel coming to Corry Station for Homeland Security, that part of town will need more housing and soon! About 3/4 of the parcel is not in the official flood zone and with the restricted MDR zoning this 6+ acres could be developed into about 18 homes that would NOT be in flood zone or wetlands. More Navy & retirees are coming into this area and another waterfront residential community is now needed on Bayou Grande, a great fishing and boating Bayou that's close to the Gulf of Mexico. This part of Bayou Grande is very beautiful and fishing is great! There is County right-of-way access but the road was never built. The Gerald Road County right-of-way continues to the east along the north boundary of the parcel to access the property. Property is wooded and could be transformed into a great waterfront community with community dock. The western property line can be seen from the end of Wade Ave on the other side of the cleared property and home that lies there. The distance from the corner of Wade & Gerald to the northwest corner is about 310 feet. Information presented is believed to be true but Buyer should perform own due diligence and check info presented for accuracy.
Information is deemed reliable but not guaranteed. Based on the information from the Pensacola Association of Realtors (PAMLS). IDX information is provided exclusively for consumers' personal, non-commercial use, and may not be used for any purpose other than to identify prospective properties consumers may be interested in purchasing. Information is based on data available to the associate, including county records. The information has not been verified by the associate and should be verified by the buyer. Some of the active listings appearing on this site may be listed by other REALTORS®. If you are interested in those active listings, our company may represent you as the buyer's agent. If the active listing you are interested in is our company's active listing, you may speak to one of our agents regarding your options for representation. Listing courtesy of REAL ESTATE COUNSELORS. Last updated on 2019-04-22.


text:
East End Angel
Kay Brellend
Two sisters living in London's tough East End … will Jennifer and Kathy be tough enough to survive?
Katherine and Jennifer Finch are like chalk and cheese: Katherine is a midwife, dedicating her life to helping others – and Jennifer is a prostitute.
Leaving home age 15, Jennifer must learn to live on her wits in London's notorious East End. Soon, however, she is battling drink and drugs, and is pregnant with a child. Her labour is nearly fatal, but her sister steps in to save the day.
Kathy continues to tend to the poor of the East End and help her sister – and she of all people knows that the society that Jennifer keeps is damaging her. Despite her best intentions, however, she begins to fall for an East End wide boy. She knows that he is no good, but her heart will not listen to her head – and what's more, he has promised to change for her …
Praise for Kay Brellend from five-star reviews on
Amazon.co.uk:
Brilliant read! Really takes you through a journey of hardship love hate etc. Was gripped from beginning to end!!!
By Woo
This book was excellent, I didn't want it to end. Reading it made me feel like I was living in the era the book was set in.
By Amanda Bicknell
What a wonderful read…having been born near this area l could see the places and people as described, could not wait to read a few more chapters each night but was so sad when l finished it.
By LondonSue
loved this book the sort you pickup and can't put back down, the setting and characters really come to life, it's as if you can see it and breath it. Wonderful.
By K
Kay Brellend, the third of six children, was born in North London but now lives in a Victorian farmhouse in Suffolk. Under a pseudonym she has written sixteen historical novels published in England and North America. This is her first novel set in the twentieth century and was inspired by her grandmother's reminiscences about her early life in Campbell Road, Islington.
More Books by Kay Brellend
The Windmill Girls
Women on the Home Front
Kay Brellend 3-Book Collection
The Campbell Road Girls
Rosie's War
Connie's Courage
Hettie of Hope Street
The Grafton Girls
London Belles


text:
Green County Inmate Search
State: All States Alabama Alaska Arizona Arkansas California Colorado Connecticut Delaware District of Columbia Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine Maryland Massachusetts Michigan Minnesota Mississippi Missouri Montana Nebraska Nevada New Hampshire New Jersey New Mexico New York North Carolina North Dakota Ohio Oklahoma Oregon Pennsylvania Rhode Island South Carolina South Dakota Tennessee Texas Utah Vermont Virginia Washington West Virginia Wisconsin Wyoming
Search Green County, WI Inmate Records
Green County Quick Facts
County Crimes by Type
Crimes % in 2016 - TOTAL ALL AGENCIES
The Timeline of the County Crime Trends
Green County Jail Demographics
A Timeline of Yearly Data Pop Total from 2005-2015
A Timeline of Yearly Female/Male Incarceration for the Years 2005-2015
A Timeline of Green County Incarceration by Gender
Green County Prison Records by Race & Ethnicity
County Jail Statistics
Local Jail, Pretrial and Admit Rates
GreenCounty Incarceration Rates
Similar Counties to Green County
Pretrial vs. Admit Rates Timeline
The Average Number of Incarcerations per Day
Green County, WI Jails
Green County, WI jails hold prisoners after an arrest or people who have been transferred to the county from a detention center. Green County holds 2 jails with a total of 23,732 inmates. These correctional facilities have private cells for extremely violent criminals or controversial suspects. Green County jails are subject to health and medical restrictions and have to provide medical care to all its inmates.
Green County Population Total
11,286 / 27 / 3
Female Population, Rate & Count
11,640 / 395 / 46
Male Population, Rate & Count
Local Jail Rate
Local Jail Count (the average # of people in jail for one day)
Population Total Confined
Pretrial Rate
Admit Rate
Admit Count
Yearly Prisoner Count
Green County is well known for its Property Crime Totals, it has a share of 47.20% of all the crimes committed in the county - 76 Property Crime Total cases. Murder And Nonnegligent Manslaughter is the least trendy crime of the area with only 0.00% out of all the crimes in Green County.
Murder and Nonnegligent Manslaughter
Larceny Theft
Violent Crime Total 8 9 10 9 11 11 7 6 9 10 5
Murder And Nonnegligent Manslaughter 1 1 2 0 0 0 0 0 0 0 0
Revised Rape 0 0 0 0 0 0 0 0 0 0 0
Legacy Rape 2 4 3 1 5 2 0 1 2 1 1
Robbery 0 1 0 0 0 0 0 0 0 0 1
Aggravated Assault 5 3 5 8 6 9 7 5 7 9 3
Property Crime Total 182 161 167 162 169 223 169 197 125 107 117
Burglary 52 33 46 26 53 65 46 53 34 29 55
Larceny Theft 124 116 110 125 109 144 120 121 85 72 59
Motor Vehicle Theft 6 12 11 11 7 14 3 23 6 6 3
Arson 0 0 1 1 0 0 0 0 0 0 0
The county has 37,153 people confined with a jail population density of 41 prisoners per jail. When breaking down the population by gender, females are a minority compared to male prisoners, being 11,769 women and 11,963 men.
Male Population
Female Population
2005 11,640 11,286
Male Rate
Female Rate
The county has an arrest rate of 4,076 and a local jail rate of 218 inmates per year and a pretrial rate of 172 pre trials per 100,000 residents aged from 15 to 64. The jail admit rate for the area equals 2,926, which is lower compared to the statistics of 2005 - 4,321 incarcerations a year.
The average jail count for Green is 52 incarcerations in a day. The numbers have decreased during the last 10 years, when there were on average 52 imprisonments a day.
Green County Incarceration Rates
Jail Rate
2005 4,321 31
Jail Count
There is not enough data to compare these counties.
Compare based on:
Incarcertion Rate
Jail Population
Green County WI Jail


text:
Early Texas Newspapers - 9 Matching Results
Sort: By Creator
The Standard (Clarksville, Tex.), Vol. 7, No. 47, Ed. 1 Friday, October 1, 1886
Description: Weekly newspaper from Clarksville, Texas that includes local, state and national news along with some advertising.
Creator: DeMorse, Charles
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 1, Ed. 1 Friday, January 1, 1886
Description: Monthly newspaper from Chappell Hill, Texas that includes local news and commentary related to the Chappell Hill Female College.
Creator: Felder, Kate & Tarrant, Mamie
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 2, Ed. 1 Monday, February 1, 1886
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 3, Ed. 1 Monday, March 1, 1886
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 5-6, Ed. 1 Saturday, May 1, 1886
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 7, Ed. 1 Wednesday, September 1, 1886
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 8, Ed. 1 Friday, October 1, 1886
Creator: Felder, Kate; Tarrant, Mamie; Thompson, Lulu & Burkhart, Grace
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 9, Ed. 1 Monday, November 1, 1886
The Philomathean (Chappell Hill, Tex.), Vol. 1, No. 10, Ed. 1 Wednesday, December 1, 1886
The Northern Standard, from Clarksville, TX 1 1
The Philomathean 8 8 The Standard 1 1
Red River County, TX 1 1 Washington County, TX 8 8
January 1 1 February 1 1 March 1 1 May 1 1 September 1 1 October 2 2 November 1 1 December 1 1


Mahou Tsukai ni Taisetsu na Koto ~Natsu no Sora~… The perfect summer anime. The best one to air this summer, all genres included, at least according to yours truly. What could more fitting than a slice-of-life-with-a-touch-of-magic show depicting the life of a young girl, Suzuki Sora, going from Hokkaido to Tokyo to learn magic for a month ? Especially with such gorgeous visuals ? Between the most photorealistic backgrounds you'll see in a while and characters with that distinct "Beck" flavor to them, the life of Sora is a real feast for your eyes. Add to that a beautiful soundtrack composed of celtic-inspired melodies and acoustic songs sung by a girl near a train station which Sora passes by everyday, and you've got something exceptionally pleasing for the senses.
And if that wasn't good enough, the story and characters themselves are very muck likeable, each one in their own way. From the innocent Sora, to the grumpy Gouta, and including their classmates, and even their teachers, you're bound to find at least some of them attaching. Thus, even though most of the series is slow-paced, it never feels boring. There's a plenty of things that make you want to come back to Shimokitazawa (the Tokyo neighborhood Sora is living in) to see how events unfold : what will Sora's next client be like ? Will Gouta be able to use magic properly ? And, as becomes evident pretty soon : when are Sora and Gouta going to go out together ?
I'll have to admit, they make one of the, if not the, cutest couple I've ever seen in an anime. And the way their relationship evolves feels completely natural, including the way they "confess", without really saying much. I couldn't help but to go "awww" when they held hands together the first time. For once, this is a show where there's no love triangle, no one ends up in tears because they were left out. Natsu no Sora is like condensed happiness for everybody involved, including the viewer.
That is, until episode 10.
In one of the most dramatic plot twisting I know of, your whole perspective of the show changes. I won't spoil it here, but let it be known that this will have almost anyone feel a huge emotional shock. It's at the same time both brilliant and cruel. It does turn a fairly inconsequential slice-of-life anime to something with a much greater impact, but you can't help but to feel that it is incredibly unfair. You half-expect something like Sora having to separate from Gouta, her first love, at the end of summer, and you're prepared to feel pretty sorry for both of them, but you aren't prepared for what's happening.
If Natsu no Sora was a TV drama, the "shocking truth" would have been revealed right at the start, and the overall tone of the show would have been greatly modified. But here, because of all the happy moments you've shared with Sora, you feel robbed of something. Or rather, you feel that she's the one being robbed of the happiness that should have been hers. The last episode even goes as far as having Gouta and Sora's mother meeting, but without knowing who the other is. It's the last coincidence that makes you wish things would have turned differently.
But in a way only japanese anime do, you're left with a strange mix of emotions at the end of the twelfth episode. You want to cry, to smile, to scream it's unfair, to be happy for how Sora's summer went, in the end. So you go through a little bit of everything, and it's a feeling like no other. And that's why the jaw-dropping revelations of Mahou Tsukai ni Taisetsu na Koto ~Natsu no Sora~ episode 10 get this second spot in my list of 2008's anime moments.


text:
Q: Summing a aggregate function I am trying to sum a total that i have from an aggregate function by using a nested select statement and then i have a group by at the end but I am getting an error that the keyword by is not expected.
Here is the code:
SELECT SUM(TOTAL) as CURTOTAL, SUM(LYTOTAL) as CURLYTOTAL
from
(select
pf.*, ps.*, st.*, RANK() OVER (ORDER BY JULsaless desc) as TOTRANK,
 JANSALESS + FEBSALESS + MARSALESS + APRSALESS + MAYSALESS + JUNSALESS + JULSALES as TOTAL,
JANLYSAL + FEBLYSAL + MARLYSAL + APRLYSAL + MAYLYSAL + JUNLYSAL + JULLYSAL as LYTOTAL`
from
payssfile pf left joinpayssspec ps on pf.str#` = ps.str# and pf.item# = ps.item#  join storefile
st on
pf.str# = st.str# where(year = 2015 and totaltype = '' and pf.str# =38))` group by pf.str#;`

then when I go to run that i get this message:

The keyword BY was not expected here.  A syntax error was detected at
  keyword BY.  The partial list of valid tokens is FOR USE SKIP WAIT
  WITH FETCH ORDER UNION EXCEPT OPTIMIZE.

A: Whatever your DBMS is, you are missing an alias.
You should always assign an alias for the derived table you're using.
Example of a derived table
select...
from (
-- this part of the query is a derived table:
select ... from ...
) as <ALIAS_HERE> -- must give derived table an alias

This is why you have an error near GROUP BY
In your particular example just replace
)) group by pf.str#;

with
)) foo group by pf.str#;
-- here you will name your derived table foo, to be able to call it's columns by foo.column_name

Here's a reference to a similar question on SO.

A: it looks like you haven't given your sub-query the alias pf, pf is the name of the table within the sub-query, but not sub-query itself. As such, the group by function is broken.

redpajama_set_name: RedPajamaStackExchange


Peter Ukstins is the Director of Integrated Construction for James G. Davis Construction. As Director, Peter works directly with operations staff to help develop, educate and promote the use of advanced construction techniques, methods, schedules, tools and strategic planning. Peter has a special focus on Quality Management, Lean Construction and Sustainability. Drawing from his award-winning operations experience, he champions DAVIS' approach to craftsmanship at every level.
With two decades of experience in construction, Peter is fearless, forward-thinking and innovative; his credo is as simple as it is successful - build it right the first time. Every time. From small interior build-outs to high-rise residential condos, Peter does sweat the small stuff without ever losing sight of the bigger picture – guaranteeing the highest level of quality, craftsmanship and accuracy for every project.
Peter is known for his exacting standards and unwavering commitment to quality. Most recently, Peter has been focused on developing and implementing the DAVIS quality management program from the ground up. He established the use of an inspection management system company-wide. He maintains the corporate database to track quality issues and provides reporting to operations on inspection statistics and the use of best practices.
Peter graduated from the University of New Hampshire in1996. His construction experience includes both base buildings and interior projects, multi-family high rise residential, single family residential, higher education, healthcare, historical renovations and government projects.
Peter became involved with the Construction Quality Executives Council during 2013. He is a member of the Lean Construction Institute and serves as Chair for the DC Metro Community of Practice. He has served as Co-chair for the 2015 Green Building Tours Committee and Board of Directors for his local volunteer Fire Department.


Firstly I put "expert" in inverted comas because my understanding is due to personal experience and research, I don't have a psychologist or psychiatrist qualification. However I do know a lot about narcissistic behaviours and felt I should have known better.
Well over on Twitter I was tagged in a post which in a nutshell stated that Parental Alienation was made up by men to further abuse women. I must admit that triggered me. From experience, women are just as capable as men of alienating the other parent and I know how devastating that can be so I replied to the post saying that PA is a very real issue and both men and women can experience it. I said I felt it was unwise to make it about gender.
They retaliated with lots of claims about how women are abused by everyone when they make claims of domestic or child abuse. Again I reiterated that it wasn't a gender issue.
This carried on for a few days and I began to see that they were no interested in a healthy debate, they simply wanted to win (which I pointed out to them). Now at this point my advice to clients is always ignore and go grey rock, don't feed them.
I kept responding to their more and more outrageous claims – at one point they actually used the phrase "men are physically stronger" to support their argument that all men are abusers and all women are victims. I attempted to reason with them that gender stereotyping is dangerous but they kept coming back with two tweets at a time.
And I kept taking the bait. Until they accused me of gaslighting. That was my jolt that I needed to step away and cut contact. Which is exactly what I did, asking them not to contact me again to which they responded with some more insults and now they have blocked me (they still follow me though).
I share this not to name and shame but just to highlight how easily I was drawn into their supply. Their comments triggered me and before I knew it I was down the rabbit hole! I recognised their behaviours but still engaged, fuelled by my passion to try and help them see how dangerous their statements were. If they had presented their cause in a less alienating, controlling and aggressive way, I would have enjoyed listening to their perspectives. But instead I was left feeling like I had let myself down by engaging with them.
So if you are struggling with enforcing those boundaries, I understand. It happens so easily that you don't even know what is going on until it is too late. So don't beat yourself up. Remember that they are very clever at knowing which buttons to press and how to engage you. And if you do find yourself embroiled, get out as soon as you recognise what is going on. It's never too late to go grey rock.
If you'd like any help in dealing with narcissistic abuse or parental alienation, please do get in touch at enquiries@thenurturingcoach.co.uk or join our Facebook Group.


Well, I guess I just need some new images, to fill my needs.
I know I can do it myself, but maybe you have some good ideas?
can i ask u for what website it is ?
I can do that, but I can wait for him to reply with more info on how he want it to look.
If you just share something of yours as well.