text:
\section{Introduction}

The question of existence of initial data for isolated multiple black hole systems  can be best analyzed according to whether any of the black holes is extremal or not. This is related
to the different topologies in the initial manifold. For non-extremal black holes, the initial manifold contains one asymptotically flat (AF) end for each black hole, but for
 extremal black holes, the associated ends are asymptotically cilindrical (AC) \cite{dain10d}. This change in topology translates into different behaviours of the initial data
 near the ends, and results in that
 for extremal black holes, the familiar theorems for elliptic
equations that use standard weighted Sobolev spaces on AF manifolds, need to be adapted.

The treatment of the Lichnerowicz equation (arising from the conformal method) in the non-extremal case, \textit{i.e.} initial manifold only
with AF ends, is  similar to the case of
compact manifolds, see for instance \cite{Choquet99}, \cite{Chrusciel02a}, \cite{Maxwell:2004yb}, \cite{Maxwell:2005bd} (cf. \cite{Chrusciel:2010my}).
We focus here on the extremal case, that is, when there is one AF end representing the region at
spatial infinity, and at least two AC ends, representing the extremal black holes.

In the literature, the many ends are usually treated individually, with cutoff functions that single out one asymptotic end at a time.
In this line, Chrusciel \textit{et al}, \cite{Chrusciel:2012np}, prove that if the scalar curvature of the seed metric is
positive, a solution to the Lichnerowicz equation on a manifold with a finite number of AF and AC ends, exists.
This is particularly relevant after the result of Leach \cite{Leach:2016jra} on manifolds with one AF and one AC end that states that if the
Riemannian manifold is Yamabe positive, then there exists a conformal transformation to a metric with positive scalar curvature
(see Proposition 3.5 in \cite{Leach:2016jra}). Also in the same article, an existence result for the existence of far from Constant Mean Curvature (CMC) initial data
describing an isolated black hole is given.

Sobolev spaces arise naturally when studying the elliptic equations coming from the Einstein constraints. Concerning manifolds with one AF and one AC end, Bartnik \cite{Bartnik86} introduced weighted Sobolev spaces $W'^{k,p}_\delta$, that not only
describe asymptotically flat solutions, but that also single out a point in the initial
manifold in such a way that by an appropriate choice of $\delta$, the origin in $\mathbb R^3$ may be made to describe another asymptoric end. These weighted spaces
were crucially used in \cite{Dain:2010uh,gabach09} for studying CMC, single, extremal black hole initial data, and further results concerning
these spaces were proven.

In this article we present weighted Sobolev spaces with weights that reflect the existence of more than one black hole. We also extend the
results in \cite{Bartnik86, Dain:2010uh,gabach09}. The aim of these results is to provide an adequate framework  to study initial
data for many black holes. In particular we use this framework to study existence of initial data for Einstein equations that
describe many extremal black holes.

We will consider initial data with one AF end and several AC ends. The initial data is the set $(M,g_{ij},K_{ij}, E^i,B^i)$ subjected to the constraint equations
\begin{equation}
R+K^2-K_{ij}K^{ij}=2(E_iE^i+B_iB^i),
\end{equation}
\begin{equation}\label{mom}
D_jK^j\,_i-D_iK=0,
\end{equation}
\begin{equation}\label{maxcon}
D_iE^i=0,\qquad D_iB^i=0,
\end{equation}
where $K=K_{ij}g^{ij}$, $D_i$ and $R$ are respectively the covariant derivative and
the curvature scalar associated to the metric $g_{ij}$.

In the Conformal Method with constant mean curvature, we consider the rescaling
\begin{equation}\label{rescaling}
g_{ij}=\Phi^4\tilde g_{ij},\quad K_{ij}=\Phi^{-2}\tilde K_{ij},\quad E^i=\Phi^{-6}\tilde E^i,\quad B^i=\Phi^{-6}\tilde B^i.
\end{equation}
The constraint equations in terms of the conformal quantities are
\begin{eqnarray}
\label{constEq}   \t{\Delta}\Phi = \frac{1}{8}\t{R}\Phi+\frac{\t{K}^2-\t{K}_{ij}\t{K}^{ij}}{8\Phi^7}-\frac{\t{E}_i\t{E}^i+\t{B}_i\t{B}^i}{4\Phi^3}, \\
  \t{D}_j\t{K}^j\,_i-\t{D}_i\t{K}+4\Phi^{-1}\t{K}\t{D}_i\Phi = 0, \\
  \t{D}_i\t{E}^i = 0, \\
  \t{D}_i\t{B}^i = 0.
\end{eqnarray}

As a concrete example, one of the most relevant multi-black hole initial data was found by Majumdar and Papapetrou \cite{Papapetrou45,Majumdar47}
and consists of $N$ charged black holes with equal mass and charge parameters for each black hole. It is interpreted as a set of extremal Reissner-Nordstr\"om punctures,
held in equilibrium
by the balance between the gravitational atraction and the electrostatic repulsion.
The space metric and electromagnetic potential are given by
\begin{equation}
ds^2=\Phi^{4}ds^2_{flat},
\end{equation}

\begin{equation}
A=\pm \Phi^{-2}dt,
\end{equation}
where $ds^2_{flat}$ is the 3-metric of Euclidean space, and
\begin{equation}
\Phi=\sqrt{1+\sum_i^N\frac{m_i}{|x-x_i|}},
\end{equation}
where $m_i, \; i=1,\ldots,N$, is the positive electric charge of the $i-$th black hole located at $x=x_i$.
This solution contains $N+1$ ends, the end $|x|\to\infty$ is AF and the $N$ ends
$x\to x_i$ are AC. The latter can be seen by taking the limit of the metric and ckecking that it goes to the standard metric on the cylinder
(see
\cite{Acena:2015rxa} for details).

The article is organized as follows. In section \ref{Sobolev} we define the functional spaces needed for dealing with many AC ends. We prove estimates, study the behaviour of functions and properties of operators on these spaces.
In section \ref{initial} we apply these results to study perturbations
of axially symmetric initial data for many extremal black holes.

\section{Weighted Sobolev spaces}\label{Sobolev}

The functional spaces that we need in order to deal with many asymptotic ends are extensions of the standard weighted Lebesgue and Sobolev spaces. For comparison we state the norms of those spaces as
presented by Bartnik \cite{Bartnik86}. Given locally $p$-measurable functions in $\mathbb R^n$ and $\mathbb R^n\setminus\{0\}$ respectively, define
\begin{equation}\label{LLnorm00}
 L^p_\delta:\qquad ||u||_{p,\delta} := \left\{\begin{array}{cc}
                                  \left(\int_{\mathbb{R}^n}|u|^p\, \sigma^{-\delta p -n} dx\right)^\frac{1}{p},&p<\infty\\
                                  \mbox{ess sup}_{\mathbb R^n}(\sigma^{-\delta}|u|),& p=\infty
                                           \end{array}\right.
\end{equation}
\begin{equation}\label{Lnorm00}
  W^{k,p}_\delta:\qquad ||u||_{k,p,\delta} := \sum_{j=0}^k ||D^ju||_{p,\delta-j},
\end{equation}

\begin{equation}\label{LLnorm0}
 L'^p_\delta:\qquad ||u||'_{p,\delta} := \left\{\begin{array}{cc}
                                  \left(\int_{\mathbb{R}^n\backslash\{0\}}|u|^p\, r^{-\delta p -n} dx\right)^\frac{1}{p},&p<\infty\\
                                  \mbox{ess sup}_{\mathbb R^n\setminus\{0\}}(r^{-\delta}|u|),& p=\infty
                                           \end{array}\right.
\end{equation}
\begin{equation}\label{Lnorm0}
  W'^{k,p}_\delta:\qquad ||u||'_{k,p,\delta} := \sum_{j=0}^k ||D^ju||'_{p,\delta-j},
\end{equation}
where $\sigma:=(1+r^2)^{1/2}$. Both $\sigma$ and $r$ are in $L^1_{loc}(\mathbb R^3)$ and $L^1_{loc}(\mathbb R^3\setminus\{0\})$ respectively. The unprimed set \eqref{LLnorm00}, \eqref{Lnorm00}
is best suited to problems involving AF initial data, with only one asymptotic end. For single black hole initial data, with two asymptotic ends,
one AF and one at the origin, the spaces \eqref{LLnorm0}, \eqref{Lnorm0} are more appropriate, as one locates the black hole puncture precisely at $r=0$.
In order to extend these spaces to account for many punctures we introduce the following definitions.

\begin{definition}\label{def1}
Let $\Pi:=\{ x_i\in \mathbb{R}^n,\;i=1,\ldots,N\}$,  and the function $w\in L^1_{loc}(\mathbb{R}^n\setminus \Pi)$ be
\begin{equation}
 w:= \left(\sum_{i=1}^N\frac{1}{r_i}\right)^{-1},
\end{equation}
where $r_i:=|x-x_i|$ is the Euclidean distance from the point $ x_i$.
Then we define the norms and associated functional spaces
\begin{equation}\label{Lnorm}
L'^p_{w,\delta}:\qquad ||u||'_{w,p,\delta} := \left\{\begin{array}{cc}
                                  \left(\int_{\mathbb{R}^n\setminus\Pi}|u|^p\, w^{-\delta p -n} dx\right)^\frac{1}{p},&p<\infty\\
                                  \textnormal{ess sup}_{\mathbb R^n\setminus\Pi}(w^{-\delta}|u|),& p=\infty
                                           \end{array}\right.
\end{equation}

\begin{equation}\label{Wnorm}
  W'^{k,p}_{w,\delta}:\qquad ||u||'_{w,k,p,\delta} := \sum_{j=0}^k ||D^ju||'_{w,p,\delta-j}.
\end{equation}
\end{definition}
The particular choice of $w$ responds to its behaviour near the $N$ punctures and the extra asymptotic end.
If we approach one of the punctures, say $l$, we have $r_l \rightarrow 0$, and then
\begin{equation}
 w \rightarrow r_l.
\end{equation}
Also, if $r_l \rightarrow \infty$, then all $r_i \rightarrow\infty$, and
\begin{equation}
 w \rightarrow \frac{r}{N}.
\end{equation}
It is clear that $w$ reduces to $r$  and the norms \eqref{Lnorm}-\eqref{Wnorm} reduce to \eqref{LLnorm0}-\eqref{Lnorm0} when there is only one puncture at the origin ($N =1,\,r_1=r$).
Moreover, Definition \ref{def1} treats each puncture in the same manner, \textit{i.e}. there is one $\delta$ for all the punctures. This is particularly relevant for the case where
all the black holes are extremal. If some of the ends are AC and some are AF (apart from the region at infinity), then a different weight should be used, with specific behaviour near
the two types of punctures.

The importance of the behaviour of the function $w$ lies in making the standard norms and the new norms equivalent when we restrict the domain to a neighborhood of only one end. For this, we need to recenter the standard norms with the origin in the corresponding puncture. To show that they are indeed equivalent we need to isolate each end, including the $r\rightarrow\infty$ end, and therefore we define the following quantities
\begin{equation}
d_{ij} := | x_i- x_j|,
\end{equation}
\begin{equation}
R_i := \frac{1}{3}\min_{j\neq i}\{d_{ij}\},
\end{equation}
\begin{equation}
R := 3\max_{i}\{| x_i|\}.
\end{equation}
With these quantities we define the sets
\begin{equation}
B_i := \{r_i \leq R_i\},
\end{equation}
\begin{equation}
B := \{r \geq R\}.
\end{equation}
It is straightforward to show that
\begin{equation}
B_i \cap B = \emptyset, \qquad B_i \cap B_j = \emptyset,\qquad i\neq j.
\end{equation}
Therefore each set $B_i$, $B$ contains only one end, and we show that the norms restricted to each of these sets are equivalent, which is a direct corollary to the following proposition.
\begin{proposition}\label{prop1}
If $r_i\leq R_i$, then there is a constant $C_i$, $0<C_i<1$, such that
 \begin{equation}\label{ineqw1}
 C_i r_i \leq w < r_i.
\end{equation}
If $r \geq R$, then
\begin{equation}\label{ineqw2}
\frac{1}{2N}r < w < \frac{2}{N}r.
\end{equation}
\end{proposition}
\begin{proof}
We first prove \eqref{ineqw1}. The right hand side inequality follows from
the definition of $w$
\begin{equation}\label{desIzq}
\frac{1}{w} = \sum_j \frac{1}{r_j} > \frac{1}{r_i},\qquad \forall\, i=1,\ldots,N.
\end{equation}
For the left hand side inequality, as $r_i \leq R_i$, then for $j\neq i$
\begin{equation}
r_j \geq d_{ij} - R_i > 0
\end{equation}
and
\begin{eqnarray}
\frac{1}{w} & = & \frac{1}{r_i} + \sum_{j\neq i} \frac{1}{r_j} \leq  \frac{1}{r_i} + \sum_{j\neq i} \frac{1}{d_{ij} - R_i} \\
& \leq & \frac{1}{r_i} + \left(\sum_{j\neq i} \frac{1}{d_{ij} - R_i}\right)\frac{R_i}{r_i}
 =  \frac{1}{r_i}\left(1 + R_i \sum_{j\neq i} \frac{1}{d_{ij} - R_i}\right).
\end{eqnarray}
If we define
\begin{equation}
C_i := \left(1 + R_i \sum_{j\neq i} \frac{1}{d_{ij} - R_i}\right)^{-1}
\end{equation}
the proof of \eqref{ineqw1} is complete.

Now we prove \eqref{ineqw2}. If $r\geq R$,
\begin{equation}
\frac{r}{2}\leq r-\frac{R}{2} < r_i < r+\frac{R}{2} <2r,
\end{equation}
therefore
\begin{equation}
\frac{1}{2r} < \frac{1}{r_i} < \frac{2}{r},
\end{equation}
and
\begin{equation}
\frac{N}{2r} < \frac{1}{w} < \frac{2N}{r},
\end{equation}
which gives \eqref{ineqw2}.
\end{proof}
For the corollary we denote by $||\cdot||'_{w,k,p,\delta;B_i}$ the norm with the domain of integration restricted to $B_i$ (resp. for the domain $B$). In the case of the standard norms, $||\cdot||'_{k,p,\delta;B_i}$ also means that in the integration the replacement $r\rightarrow r_i$ has been made, centering the norm at $x_i$.
\begin{corollary}\label{equivNorms}
 The norms $||\cdot||'_{k,p,\delta;B_i}$ and $||\cdot||'_{w,k,p,\delta;B_i}$ (resp. $||\cdot||'_{k,p,\delta;B}$ and $||\cdot||'_{w,k,p,\delta;B}$) are equivalent.
\end{corollary}

Next we prove important estimates on functions $u\in W'^ {k,p}_{w,\delta}$ near the asymptotic ends. This is analogous to and follows Lemma A.1 in \cite{Dain:2010uh}.
\begin{lemma}\label{decay}
Assume $u\in W'^{k,p}_{w,\delta}$ with  $n-kp<0$, then we have the following estimate
 \begin{equation}
   \label{eq:41}
   w^{-\delta}|u|\leq C \left\lVert u  \right\rVert'_{w,k,p,\delta}.
 \end{equation}
Moreover, we have
 \begin{equation}
\label{eq:44}
 \lim_{r_i\to 0}w^{-\delta}  |u|= \lim_{r\to \infty}w^{-\delta}  |u|= 0.
\end{equation}
\end{lemma}

\begin{proof}
Restricting the domain of $u$ to $B_i$ there are constants $C_k$ such that
\begin{equation}
w^{-\delta}|u| \leq C_1 r_i^{-\delta}|u|\leq C_2 ||u||'_{k,p,\delta;B_i} \leq C_3 ||u||'_{w,k,p,\delta;B_i} \leq C_4 ||u||'_{w,k,p,\delta},
\end{equation}
where we have used first Proposition \ref{prop1}, then Lemma A.1 in \cite{Dain:2010uh}, noticing that in that lemma the inequality and the
limits can also be proven separately for a ball around the origin and for the asymptotic region $r>R$, therefore we can use separately the
bounds in each of $B_i$ and in $B$. For the third inequality Corollary \ref{equivNorms} was used and the last inequality comes
from the fact that the norm of $u$ on the whole domain is bigger than its norm restricted to $B_i$. The same sequence of inequalities
hold for the domain $B$. Also, the set
\begin{equation}
A = \{r_i \geq R_i, r\leq R\},
\end{equation}
is compact and hence the function
\begin{equation}
w^{-\delta}|u|
\end{equation}
has a maximum there, this together with the previous inequalities means that there exists a constant $C$ such that \eqref{eq:41}
is satisfied in $\mathbb{R}^n\backslash \Pi$.
Also, using the same argument as in Lemma A.1 in \cite{Dain:2010uh}, we have \eqref{eq:44}.
\end{proof}

Another important result that will be useful later is the following.

\begin{lemma}\label{lemab1}
 If $u\in W'^{1,p}_{w,-1/2}$, then $w^{\frac{1}{2}-\frac{n}{p}}u\in L^p$ and $w^{\frac{3}{2}-\frac{n}{p}}\partial u \in L^p$.
\end{lemma}
\begin{proof}
\begin{equation}
 \|u\|_{W'^{1,p}_{w,-1/2}} = \|u\|_{L'^p_{w,-1/2}} + \|\partial u\|_{L'^p_{w,-3/2}} = \|w^{\frac{1}{2}-\frac{n}{p}}u\|_{L^p} + \|w^{\frac{3}{2}-\frac{n}{p}}\partial u\|_{L^p},
 \end{equation}
 and therefore if $\|u\|_{W'^{1,p}_{w,-1/2}}$ is bounded, then $\|w^{\frac{1}{2}-\frac{n}{p}}u\|_{L^p}$ and $\|w^{\frac{3}{2}-\frac{n}{p}}\partial u\|_{L^p}$ are also bounded.
\end{proof}

The main result we prove in this section is the fact that the Laplace operator is an isomorphism $\Delta: W'^{k+2,p}_{w,\delta}\rightarrow W'^{k,p}_{w,\delta-2}$. We  follow
Bartnik's arguments in the proof of Theorem 1.7 in \cite{Bartnik86}. We need to exclude some exceptional values of $\delta$. The exceptional values are $\{ m'\in\mathbb{Z}, m'\neq -1,-2,\ldots,3-n\}$, $\delta$ is said to be nonexceptional if it is not one of those values, and it is convenient to define
\begin{equation}
 m = \max \,\{ m' \mbox{ exceptional}, m'<\delta \}.
\end{equation}

\begin{theorem}\label{lemaIsom}
 Let $\delta$ be nonexceptional, $1<p<\infty$, and $k$ a non-negative integer. Then the Laplace operator
 \begin{equation}
 \Delta: W'^{k+2,p}_{w,\delta}\rightarrow W'^{k,p}_{w,\delta-2}
 \end{equation}
 is an isomorphism and there exists a constant $C=C(n,p,\delta,k)$ such that
 \begin{equation} \label{bound1}
 \|u\|'_{w,k+2,p,\delta}\leq C\|\Delta u\|'_{w,k,p,\delta-2}.
 \end{equation}
\end{theorem}
\begin{proof}
As noted in \cite{Bartnik86} we only need to prove the $k=0$ case. We start considering the convolution kernel $K(x,y)$,
 \begin{equation}\label{kernelLap}
 K(x,y) = \left\{ \begin{array}{ll}
                   |x-y|^{2-n}, & \mbox{if }2-n<\delta<0,\\
                   |x-y|^{2-n}-|y|^{2-n}\sum_{j=0}^m P_j^\lambda\left(\frac{x\bullet y}{|x||y|}\right)\left(\frac{|x|}{|y|}\right)^j, & \mbox{if } m\geq 0,\\
                   |x-y|^{2-n}-|x|^{2-n}\sum_{j=0}^m P_j^\lambda\left(\frac{x\bullet y}{|x||y|}\right)\left(\frac{|y|}{|x|}\right)^j, & \mbox{if } m<2-n,
                  \end{array}\right.
 \end{equation}
 where $x\bullet y$ is the Euclidean scalar product of $x$ and $y$, and $P_j^\lambda$ are the ultraspherical functions. Associated with this kernel we have the operator $K$,
 \begin{equation}
 Ku(x) = \int_{\mathbb{R}^n\backslash\Pi} K(x,y)u(y)\,dy.
 \end{equation}
 We present only in detail the first case in \eqref{kernelLap}, as the three cases are similar and our main interest concerns $n=3$, $\delta=-1/2$.
Let us define the associated kernel
 \begin{equation}
 K'(x,y) = w(x)^{-\delta-n/p}\, K(x,y)\, w(y)^{\delta-2+n/p}.
 \end{equation}
Then, the Nirenberg-Walker Lemma 2.1 in \cite{Nirenberg73} implies that $K'$ is a bounded $L^p\to L^p$ operator, and the equality
 \begin{equation}
 \|Ku\|'_{w,p,\delta} = \|K'(w^{-\delta+2-n/p}u)\|_{L^p},
 \end{equation}
means that
 \begin{equation}
 \|Ku\|'_{w,p,\delta} = \|K'(w^{-\delta+2-n/p}u)\|_{L^p} \leq C\|w^{-\delta+2-n/p}u\|_{L_p} = C \|u\|'_{w,p,\delta-2},
 \end{equation}
 where the last equality comes directly from the definition of the norms. This shows that $K$ is a bounded $L'^p_{w,\delta-2}\rightarrow L'^p_{w,\delta}$ operator.
The distributional identities
\begin{equation}
\Delta_xK(x,y)=\Delta_yK(x,y)=\delta(x-y) \qquad \mbox{ in }\mathbb{R}^n\backslash\Pi
\end{equation}
imply that $K(\Delta u)=u$ for all $u\in C_c^\infty(\mathbb{R}^n\backslash\Pi)$,
and therefore
\begin{equation}
\|u\|_{L'^p_{w,\delta}}=\|K(\Delta u)\|_{L'^p_{w,\delta}}\leq C \|\Delta u\|_{L'^p_{w,\delta-2}}\qquad \forall u\in W'^{2,p}_{w,\delta}.
\end{equation}
This and the elliptic estimate
\begin{equation}\label{elli}
\|u\|_{W'^{2,p}_{w,\delta}}\leq c(\|\Delta u\|_{L'^p_{w,\delta-2}}+\|u\|_{L'^p_{w,\delta}})
\end{equation}
give the desired estimates \eqref{bound1}. Note that the proof of \eqref{elli} uses the same arguments as in the case of standard, not weighted Sobolev spaces.
The rest of the proof follows the same lines as Bartnik's.
\end{proof}

\section{Axially symmetric multi-black hole initial data}\label{initial}

The main application of the Sobolev spaces presented in section \ref{Sobolev} that we want to discuss here is on the deformation of extremal black hole initial data.
We refer the reader to \cite{Acena:2015rxa} where the one black hole case is treated. The general argument and several calculations can readily be extended from one to
several black holes, and we present here mainly the path needed to understand the theorem and its proof and the arguments that need special care, some particular details are left to the appendix.

Axial symmetry is not required for the validity of the main arguments presented below, but it simplifies the treatment of certain bounds and asymptotic behaviours. Also it is
necessary when dealing with quasilocal angular momentum. The case without axial symmetry will be given elsewhere.

From now on we assume axial symmetry, namely, there is a spacelike Killing vector field, $\eta$, on $M$, with complete closed orbits, such that
\begin{equation}\label{axSym}
\c{L}_\eta g_{ij}=0,\quad\c{L}_\eta K_{ij}=0,\quad\c{L}_\eta E^i=0,\quad\c{L}_\eta B^i=0,
\end{equation}
and as we want the axial symmetry to be reflected in the conformal metric,
\begin{equation}
 \c{L}_\eta\Phi = 0.
\end{equation}
Physically this condition implies that the black holes are all aligned, situated on the symmetry axis $\Gamma$. The perturbed initial data will also have this property.

We also take the initial data to be time-rotation symmetric, which implies that the data is maximal
\begin{equation}
 K = 0 \Rightarrow \t{K}=0,
\end{equation}
and that $(M,g_{ij})$ is in the positive Yamabe class. We use cylindrical coordinates $(z,\rho,\phi)$ adapted to the axial symmetry
\begin{equation}
\eta = \partial_\phi.
\end{equation}
Using the symmetry it can be shown that there are scalar potentials $\omega$, $\psi$, $\chi$ that contain the information about the extrinsic curvature and the electromagnetic fields,
and that they do not depend on $\phi$. Also, all the constraint equations but \eqref{constEq} are automatically satisfied.

We define the conformal metric as
\begin{equation}
 \t{g}_{ij} = e^{2q}(dz^2+d\rho^2)+\rho^2 d\phi^2,
\end{equation}
where $q=q(z,\rho)$. Then,
\eqref{constEq} takes the form
\begin{eqnarray}\label{LicEq}
 \Delta\Phi = -\frac{1}{4}\Delta_2q\Phi - \frac{(\partial\omega)^2}{16\rho^4\Phi^7}-\frac{(\partial\psi)^2+(\partial\chi)^2}{4\rho^2\Phi^3},
\end{eqnarray}
where
\begin{equation}
 \Delta_2 := \partial_z^2 + \partial_\rho^2,\quad
 \Delta := \partial_z^2 + \partial_\rho^2 + \rho^{-1}\partial_\rho,
\end{equation}
and consider that $\Phi = \Phi(z,\rho)$.

Given a set of functions $(\Phi_0,q_0,\omega_0,\psi_0,\chi_0)$ that satisfy \eqref{LicEq}, we take arbitrary smooth axially symmetric functions with compact support outside the
symmetry axis, $q,\,\omega,\,\psi,\,\chi$, and a parameter $\lambda$, perturb the given functions as follows
\begin{equation}
 q \rightarrow q_0+\lambda q, \quad \omega \rightarrow \omega_0+\lambda \omega, \quad \psi \rightarrow \psi_0+\lambda \psi, \quad \chi \rightarrow \chi_0+\lambda \chi,
\end{equation}
and write the perturbed solution as
\begin{equation}
 \Phi = \Phi_0 + u.
\end{equation}
Then equation \eqref{LicEq} can be written as an equation for $u$, namely
\begin{eqnarray}\label{LicEqPert}
 \Delta u & = & -\frac{1}{4}\Delta_2q_0 u -\frac{1}{4}\lambda \Delta_2 q(\Phi_0+u) - \frac{(\partial\omega_0+\lambda\partial\omega)^2}{16\rho^4(\Phi_0+u)^7}+\frac{(\partial\omega_0)^2}{16\rho^4\Phi_0^7} \\
&& - \frac{(\partial\psi_0+\lambda\partial\psi)^2}{4\rho^2(\Phi_0+u)^3}+\frac{(\partial\psi_0)^2}{4\rho^2\Phi_0^3} - \frac{(\partial\chi_0+\lambda\partial\chi)^2}{4\rho^2(\Phi_0+u)^3}+\frac{(\partial\chi_0)^2}{4\rho^2\Phi_0^3}.
\end{eqnarray}
In order to consider existence of solutions we associate with this equation the map $G$,
\begin{eqnarray}
 G(\lambda,u) & := & \Delta u + \frac{1}{4}\Delta_2q_0 u + \frac{1}{4}\lambda \Delta_2 q(\Phi_0+u) + \frac{(\partial\omega_0+\lambda\partial\omega)^2}{16\rho^4(\Phi_0+u)^7} - \frac{(\partial\omega_0)^2}{16\rho^4\Phi_0^7} \\
&& + \frac{(\partial\psi_0+\lambda\partial\psi)^2}{4\rho^2(\Phi_0+u)^3} - \frac{(\partial\psi_0)^2}{4\rho^2\Phi_0^3} + \frac{(\partial\chi_0+\lambda\partial\chi)^2}{4\rho^2(\Phi_0+u)^3} - \frac{(\partial\chi_0)^2}{4\rho^2\Phi_0^3},
\end{eqnarray}
where solutions of \eqref{LicEqPert} are given by
\begin{equation}\label{eqG}
 G(\lambda,u) = 0.
\end{equation}
In particular, the background solution is
\begin{equation}
 G(0,0)=0.
\end{equation}

The main result of this section is the following theorem, an extension of theorem 2.1 of \cite{Acena:2015rxa}. We use
the standard notation $H'^k_{w,\delta} = W'^{k,2}_{w,\delta}$.

\begin{theorem}\label{thm}
Let $q, \omega, \psi, \chi\in C^\infty_0(\mathbb R^3\setminus\Gamma)$ be arbitrary smooth axially symmetric functions. Then,
there is $\lambda_0>0$ such that for all $\lambda\in (-\lambda_0, \lambda_0)$  there
exists a solution $u(\lambda)\in H'^{2}_{w,-1/2}$ of equation
\eqref{eqG}. The solution $u(\lambda)$ is continuously differentiable in
$\lambda$ and satisfies $\Phi_0+u(\lambda)>0$. Moreover, for small $\lambda$
and small $u$ (in the norm $ H^{'2}_{w,-1/2}$) the solution $u(\lambda)$ is the
unique solution of equation \eqref{eqG}.
\end{theorem}

\begin{proof}
The proof of this theorem follows the lines of \cite{Acena:2015rxa} and makes strong use of the Inverse Function theorem. We will ommit some arguments that are carried over from one cylindrical end to many such ends.
The proof can be thought of as showing that $u$ gives rise to a good conformal factor and collecting the necessary conditions for the Inverse Function theorem to hold.

\begin{itemize}
\item \textit{$\Phi_0+u$ is positive.}

We are considering the map $G:\mathbb{R}\times H'^2_{w,-1/2}\rightarrow L'^2_{w,-5/2}$, but for a general $u\in H'^2_{w,-1/2}$
the function $\Phi=\Phi_0+u$ does not have a definite sign. In order for $\Phi$ to be a
conformal factor we need it to be positive. As we assume $\Phi_0$ to be a  conformal factor, then we can
conjecture that if $u$ is small enough, then $\Phi$ is also going to be a  conformal factor. This can be achieved by restricting $u$ to be in a ball around the origin in $H'^2_{w,-1/2}$. We therefore define the subset $V$ of $H'^2_{w,-1/2}$ as
\begin{equation}
 V = \{v\in H'^2_{w,-1/2} : ||v||_{H'^2_{w,-1/2}}<\xi\},
\end{equation}
where $\xi>0$ is a constant that can always be chosen and whose particular value depends on $\Phi_0$. We restrict the map,
$G:\mathbb{R}\times V\rightarrow L'^2_{w,-5/2}$, having that if $u\in V$ then
\begin{equation}
 \Phi_0+u > 0.
\end{equation}

\item\textit{The map $G$ is well defined}

To prove that the map $G:\mathbb{R}\times V\rightarrow L'^2_{w,-5/2}$ is well defined we need to show that $||G(\lambda,u)||_{L'^2_{w,-5/2}}$ is bounded whenever $\lambda \in \mathbb{R}$ and $u\in V$. This is accomplished using the definition of the $H'^2_{w,-1/2}$ norm, the triangle inequality, the asymptotic conditions on the background functions and the compact support of $q$, $\omega$, $\psi$ and $\chi$,  together with the inequalities presented in Appendix \ref{useIneq}.

\item \textit{Partial Fr\'echet derivatives of $G$}

To obtain the partial Fr\'echet derivatives of $G$ and to show that $G$ is $C^1$ we start by calculating the directional derivatives of $G$, namely
\begin{equation}
 \left.\frac{d}{dt}G(\lambda+t\gamma,u)\right|_{t=0},\qquad \frac{d}{dt}G(\lambda,u+tv)\Bigg|_{t=0},
\end{equation}
and propose them as the partial Fr\'echet derivatives of $G$,
\begin{eqnarray}
 D_1G(\lambda,u)[\gamma] & = &\Bigg[\frac{1}{4}\Delta_2q(\Phi_0+u)+\frac{\partial\omega(\partial\omega_0+\lambda\partial\omega)}{8\rho^4(\Phi_0+u)^7} \\
 && +\frac{\partial\psi(\partial\psi_0+\lambda\partial\psi)}{2\rho^2(\Phi_0+u)^3}+\frac{\partial\chi(\partial\chi_0+\lambda\partial\chi)}{2\rho^2(\Phi_0+u)^3}\Bigg]\gamma,\\
 D_2G(\lambda,u)[v] & = & \Delta v + \Bigg[\frac{1}{4}\Delta_2 q_0+\frac{1}{4}\lambda\Delta_2q - \frac{7(\partial\omega_0+\lambda\partial\omega)^2}{16\rho^4(\Phi_0+u)^8} \\
 && -\frac{3(\partial\psi_0+\lambda\partial\psi)^2}{4\rho^2(\Phi_0+u)^4}-\frac{3(\partial\chi_0+\lambda\partial\chi)^2}{4\rho^2(\Phi_0+u)^4}\Bigg]v.
\end{eqnarray}
Now it can be shown that these operators are bounded using the properties mentioned in the point above, showing that they are linear operators between the following spaces
\begin{equation}
D_1G(\lambda,u):\mathbb{R}\rightarrow L'^2_{-5/2}, \qquad D_2G(\lambda,u):H'^2_{-1/2}\rightarrow L'^2_{-5/2}.
\end{equation}
A lengthy but straightforward calculation shows that
\begin{equation}
 \lim_{\gamma\ra0}\frac{||G(\lambda+\gamma,u)-G(\lambda,u)-D_1G(\lambda,u)[\gamma]||_{L'^2_{-5/2}}}{|\gamma|}=0,
\end{equation}
\begin{equation}
 \lim_{v\ra0}\frac{||G(\lambda,u+v)-G(\lambda,u)-D_2G(\lambda,u)[v]||_{L'^2_{-5/2}}}{||v||_{H'^2_{-1/2}}}=0
\end{equation}
and therefore $D_1G$ and $D_2G$ are indeed the partial Fr\'echet derivatives of $G$. It can also be checked that
the derivatives are continuous, and then $G$ is $C^1$.

\item \textit{The map $D_2G(0,0):H'^2_{w,-1/2}\rightarrow L'^2_{w,-1/2}$ is an isomorphism}

Associated with the map $D_2G(0,0)$ we define the operator $\mathcal{L}$ through
\begin{equation}
\mathcal{L}v := -\Delta v+\alpha v,
\end{equation}
where
\begin{equation}\label{alpha0}
 \alpha=-\frac{\Delta_2q_0}{4}+7\frac{(\partial\omega_0)^2}{16\rho^4\Phi_0^8}+3\frac{(\partial\psi_0)^2+(\partial\chi_0)^2}{4\rho^2\Phi_0^4}.
\end{equation}
We want to show that the equation
\begin{equation}\label{ecL}
\mathcal Lu:=-\Delta u+\alpha u=f \qquad \mbox{in}\; \mathbb R^{3}\setminus \Pi
\end{equation}
has a unique solution $u\in H'^2_{w,-1/2}$ for each $f\in L'^2_{w,-1/2}$. The steps for this are the same as in \cite{Acena:2015rxa},
see also the references therehin. The steps of the proof are as follows. First the Yamabe condition is used to show that for all $f\in C_0^\infty$, $f\neq0$,
\begin{equation}\label{yamflat}
 \int_M|\partial f|^2+\alpha f^2 d\mu>0,
 \end{equation}
where $\alpha$ is given by \eqref{alpha0} and  the norm and volume
element in \eqref{yamflat} are computed with respect to the flat metric. This is used to show that the bilinear form
\begin{equation}\label{Bil}
B[u,v]:=\int_{\mathbb{R}^3\setminus\Pi}\partial u\cdot \partial v+\alpha uv\,d\mu,
\end{equation}
with $u,v\in H^{'1}_{w,-1/2}$, which is well defined in virtue of Lemma \ref{lemab1} and corresponds to
the linear operator $\mathcal L$, is indeed an inner product. Using equation \eqref{alpha0} and  the standard Holder inequality for $L^2$ spaces (note that by Lemma \ref{lemab1}, $\partial u,\;\partial v,\; uw^{-1},\;v^{-1}\in L^2$) it is shown that the linear functional $\ell(\cdot):=B[\cdot,v]$ is
bounded for all $v\in H^{'1}_{w,-1/2}$. With these conditions fulfilled, the Riesz Representation Theorem states that there exists a unique weak solution, $u\in H^{'1}_{w,-1/2}$, of $\mathcal Lu=f$, for each $f\in L^{'2}_{w,-5/2}$. The last step is to show that the weak solution is indeed in $H^ {'2}_{w,-1/2}$. This follows the same argument as in \cite{gabach09}, the only new ingredient is the use of Theorem \ref{lemaIsom}.
\end{itemize}
This completes the proof, as we have shown that the conditions for the Inverse Function theorem to hold are true in the case at hand.
\end{proof}

redpajama_set_name: RedPajamaArXiv


The Statue of Unity, a 182-metre giant structure built in honour of Sardar Vallabhbhai Patel, will be dedicated to the nation on Wednesday by Prime Minister Narendra Modi. The imposing monument, touted as the world's tallest statue, is twice the height of Statue of Liberty and is built on an islet, Sadhu Bet, near the Sardar Sarovar Dam in Narmada district. After Modi unveils the statue, three IAF planes would fly past and create the tricolour in the sky by emitting saffron, white and green colours. PM Modi would also inaugurate the 'Wall of Unity' (a monument defining India's unity) near the statue. At that time, three Jaguar fighter jets would fly past the structure at a low altitude.
1. There will be dance and musical performances by artistes drawn from 29 states and two Union Territories. Some of other major attractions include inauguration of a 17-km-long Valley of Flowers, a tent city for tourists near the statue and a museum dedicated to the life and times of Patel.
2. A viewing gallery at a height of 135 metres has been created inside the statue to enable tourists to have a view of the dam and nearby mountain ranges.
3. The statue has been built using over 70,000 tonnes of cement, 18,500 tonnes of reinforcement steel, 6,000 tonnes of structural steel and 1,700 metric tonnes of bronze, which was used for the outer cladding of the structure.
4. The statue's construction had received opposition from some tribal groups in Narmada district. Meanwhile, local tribal leaders have announced a boycott of Wednesday's event claiming "mass destruction" of natural resources due to the project.
5. In an open letter addressed to Modi, sarpanches of 22 villages situated near the Sardar Sarovar dam on the Narmada river have said villagers will not welcome the prime minister when he arrives for the function.
6. The Congress said the "real tribute" to Sardar Vallabhbhai Patel would be that his order banning the RSS should be placed at the foot of the 'Statue of Unity' in Gujarat. Senior Congress leader Anand Sharma, at a press conference, alleged that the Bharatiya Janata Party does not have any idols of its own and highlighted that Patel was from the Congress and also a former president of the party.
7. The CPI flayed the Narendra Modi government for "splurging" nearly Rs 3,000 crore on building Sardar Patel's gigantic statue at a time when debt-ridden farmers are "being driven to suicide" and millions of people are defecating in the open for want of sanitation facilities.


One of the Lowcountry's leading cake designers, Linda has the reputation of her cakes tasting as good as they look!
Please take a few minutes to browse through our website for a portfolio of our work, testimonials and information.
Designing Appointments should be made early to assure availability.
One word - WOW!!! your cake was everything I thought it would be - a moist, sweet, delicious HIT! It took everything I had to actually freeze, and not eat, the top tier. As you can imagine, the cake (and some sand was devoured! I will send pictures when we get them developed.
Thank you so much for the Beautiful cake. It was just how I pictured it. We got lots of compliments. Also, the picture of the mustang on the groom's cake was a huge success, They both tasted great! I knew from our first meeting that I had made a great choice with you.


text:
Buy from $6.71
See More by guyver
Kingdom Hearts - Sora and Kairi by NorthWester1
Kpop by OnyxHotelPrince
Power Rangers by kiananuva12
In honor of Harold Ramis May you haunt Bill Murray for the rest of his life.
© 2014 - 2022 guyver
jackcrowder
Ramis did so many projects beyond Ghostbusters.
Not to mention Groundhog Day which was all Murray.
I can't think of any reason he'd haunt Murray.
In fact, Ramis didn't have an improv wizard like Murray for YEAR ZERO and it flopped.
Because Bill held up the plans to move forward with Ghostbusters 3 for years.
The scripts weren't any good.
The script submitted by Ramis last summer was not only good but headed to a studio. When Asked from day one about the possibility of being involved with a third film Bill was against it. Going as far to say he would only do it if they killed him off in the first five minutes. Sounds like he had a problem with the whole idea.
Just don't take the comedian as a grudge type.
In life or death.
www.youtube.com/watch?v=6ShyYH…
This says it all.
www.telegraph.co.uk/culture/fi…
A cameo isn't something I think such a gentle good man as Ramis would have worried about.
Maybe if it had been his script.
Maybe if it had been his directing.
But a cameo?
Pshaw, man.
Dang!! He passed away today?!
Yes. Harold Ramis as Egon was my childhood hero. I have never felt this sad for the passing of any actor EVER.
Damn...I've read up what got him too...shucks...69 is too dang young in this day and age. Well, I dunno how religious he was, but its my prayer he's now helping God laugh in spades and enjoying helping all of Heaven now laugh raucously too.
I was shocked to hear it myself. His comedy was as unique as his look.

redpajama_set_name: RedPajamaCommonCrawl


text:
rally Ultimatecarpage.com > Cars by brand > Great Britain > Lotus
Racing cars > Other Rally Cars
Exige R-GT
Lotus Exige R-GT
Country of origin: Great Britain
What better way to demonstrate Lotus' superior ride and handling than by re-entering the adrenaline filled Rally Championships? Showcasing the new Exige S' phenomenal performance, the Exige R-GT will participate in the newly formed FIA GT category of the FIA Rally Championship on asphalt in Monte Carlo, Tour de Corse and San Remo.
Powered by the same V6 3.5 litre engine as the new road Exige, significant changes have been made to the gearbox and the restrictors on the engine in order to control the power. Lotus will initially partner with several customer teams following the tried and tested method in which Lotus entered IndyCar and returned to Le Mans.
For Group Lotus' Director of Motorsport Claudio Berro the return to rally has an extra special meaning: "My motorsport career started in rally, in fact, I even took part in the Talbot Sunbeam Lotus' Italian Rally championship winning a race in San Marino back in 1981 so this is a very nostalgic moment for me. As we unveiled the car in Frankfurt, alongside the original championship winning car, I had the chance to get back inside and the smell was exactly the same - it brought back many happy memories for me."
"With the new R-GT rules in rally, it's a very interesting time for us to return to the sport. Naturally our approach this time around will be different to when we won the championship three decades ago, but I think our philosophy is definitely the same, we want to compete and ultimately we want to win. Our new rally car, the Exige R-GT is a dream come true for those who are passionate about GT racing - this will be their car."
The beauty of the Exige R-GT is that although it's a professionally set up race car, you don't need to be a pro to drive it - the race car has been developed alongside the road car allowing the technical team to create the perfect race machine, high performance but without losing the sense of fun to drive and ease of use. Developing the race version of the Exige in parallel with the road car has allowed the Lotus technicians to make rapid advances in terms of safety and reliability.
Add your comments on the Lotus Exige R-GT


text:
Campus NewsFaculty & Staff
Professor publishes new textbook on implant prosthodontics
Stacey Hudson2 min readOctober 13, 2021
Dr. Todd Schoenbaum
Todd Schoenbaum, professor in the Department of Restorative Sciences in the Dental College of Georgia, has published a new textbook titled Implant Prosthodontics: Protocols and Techniques for Fixed Implant Restorations with Quintessence International Publishing Group.
"I created this book to enhance the overall skill and expertise of our profession in the treatment of implant patients," Schoenbaum said. "The motivation for it for me is ensuring that our patients get better quality care, and the best way to do that is to help ensure that our colleagues are performing at the highest possible level."
The book was designed as a concise clinical guide for restoring implants with fixed restorations — such as crowns and bridges — in patients who are experiencing the loss of some adult teeth. It provides the essentials of implant prosthodontic workflows and protocols based on the best available current evidence.
This is Schoenbaum's second textbook on the topic, but this book is designed from an educator's perspective and dives into the fundamentals. In order to make it as efficient and practical a tool as possible, the text is easily searchable by procedure and includes a detailed guide on how to properly execute each procedure.
Each chapter provides guidance on selecting appropriate protocol or components, explains the rationale for the given procedure, provides a step-by-step protocol and discusses special considerations and potential complications.
"It's a very visual learning experience with nearly 700 diagrams and photographs of procedures and case studies, more than 25 protocols, and it has a multimedia component with the QR codes that link to instructional videos," Schoenbaum said.
To make the book maximally applicable and to remove any erroneous extrapolation from one system to another, the author has included directions and photographic demonstrations for three of the major implant manufacturers: Nobel Biocare, Straumann and BioHorizons.
Implant dentistry is still relatively new to the profession, according to Schoenbaum. While it has grown in popularity over the last 30 years, he hopes clinicians will benefit from the additional training and knowledge in the book.
"I would say interest in implant dentistry continues to increase. Traditionally, we have not been able to dedicate enough time to implants in dental education, so I created this book as a way to enhance the overall quality of care our profession provides to patients. I hope in some small way it achieves this goal," Schoenbaum said.
The textbook is available for pre-order now, with full release no later than January 2022.
Read about Schoenbaum's other recent project, an academic journal.
Tagged asDental College of GeorgiaRestorative SciencesTodd Schoenbaum
Stacey Hudson
Stacey Hudson is communications coordinator for the Dental College of Georgia.
Student organization celebrates women in dentistry with annual Friendsgiving
Faculty Club awards scholarships to health sciences students
Values Week 2021: Dentistry student lives with integrity
Raysean Ricks
Written by	Stacey Hudson
Chelsea Paulding crowned Miss Augusta University 2022
Listen: Miss Augusta University begins her reign on 'In the Wild' podcast


text:
Tag Archives: Hocus Pocus
Halloween Movies to Watch With Your Kids
Like almost everything else about 2020, Halloween is going to look a little different. The CDC has released its guidelines for Halloween, Día de los Muertos, and Thanksgiving to help protect individuals, families, friends, and communities from COVID-19. Based on these recommendations, many families and communities are looking for Trick or Treating alternatives this year.
One timeless tradition is to get your scare on through a Halloween-themed TV show or spooky movie. Here are a few age-based suggestions to help you make Halloween 2020 memorable for all the right reasons.
Super Monsters Save Halloween (2018) – Netflix
The Super Monsters are preschoolers whose parents just happen to be the world's most famous monsters. In this 30 minute Netflix special, they show preschoolers how the holiday's scares are mostly make-believe by explaining the tricks behind haunted houses and spooky decorations. Super Monsters: Vida's First Halloween is another fun option.
Room on the Broom (2013) – Netflix
Based on a book by Julia Donaldson, Room on the Broom features a friendly witch who's quick with a smile and kind to the animals who ask her for a ride on her broom.
Kiki's Delivery Service (1989) – Amazon Prime Video/HBO Max
Kiki is a good witch who uses her broom and her flying ability to start a delivery service. It's from Studio Ghibli, so you know it's beautifully animated.
The Addams Family (2019) – Amazon Prime Video
The adorably kooky family moves to the suburbs where Wednesday's friendship with the daughter of a local reality show host causes problems. There are plenty of jokes for both children and adults to keep everyone entertained throughout the film.
The Worst Witch (2017) – Netflix
Mildred Hubble lives an ordinary life until the day that Maud Spellbody crashes her broomstick into their balcony. Maud introduces Mildred to Cackle's Academy – a school for young witches where under the watchful eye of friendly headmistress Miss Cackle and scary deputy Miss Hardbroom, Mildred begins her training. But no matter how hard she tries, her spells have a habit of going badly wrong, causing chaos. Will Mildred always be The Worst Witch?
Hotel Transylvania (2012) – Amazon Prime Video/Hulu/Netflix
Dracula, who operates a high-end resort away from the human world, goes into overprotective dad mode when a boy discovers the resort and falls for the count's teenaged daughter.
The movie is from the POV of the monsters, and they're the good guys. Plus, if your kids love this one, two sequels follow.
Is it a Halloween movie or a Christmas movie? The world may never agree, so feel free to start watching Jack Skellington, king of Halloween Town, attempt to bring Christmas to his friends on repeat from now through December 5.
Corpse Bride (2005) – Amazon Prime Video
When a shy groom practices his wedding vows, he accidentally marries a ghost. Written and directed by Tim Burton, the movie is exactly the style you'd expect from a Tim Burton film. Sweet, macabre, and fun for kids and adults alike.
Halloweentown (2004) – Amazon Prime Video/Disney Now
After learning she is a witch, a girl helps save a town full of other supernatural creatures. It's cute and all-around wholesome entertainment, and a Disney Channel Original Movies classic, as any 90's kid will tell you.
Hocus Pocus (1993) – Amazon Prime Video/Disney+/Select theatres – check local listings
Max and Thora's family moves them to Salem, where they struggle to fit in. Max accidentally awakens a trio of diabolical and slightly goofy witches who were executed in the 17th century. There's a reason this Disney classic is at the top of so many must-watch lists.
The Witches (1990) – Amazon Prime Video
Netflix is promoting a remake with Anne Hathaway as the Grand High Witch, but she's got big pointy shoes to fill as Angelica Houston is magnificent in the 1990 original. Luke and his grandmother Helga stumble onto a witch convention and must stop them, even after he has been turned into a mouse.
Ghostbusters (1988) – Amazon Prime Video/Disney+/Select theatres – check local listings
IMDb describes Ghostbusters as "Three former parapsychology professors set up shop as a unique ghost removal service," which may be the lamest description of a classic movie I've ever heard. Bill Murray, Dan Aykroyd, Harold Ramis, Ernie Hudson, Rick Moranis, and Sigourney Weaver helm one of Hollywood's biggest hits.
The Worst WitchHuluHotel TransylvaniaThanksgivingCorpse BrideThe Nightmare Before ChristmasHalloweentownAmazon Prime VideoHocus PocusCDC has released its guidelines for HalloweenThe WitchesDía de los MuertosGhostbustersSuper Monsters Save HalloweenRoom on the BroomKiki's Delivery ServiceThe Addams FamilyNetflix
slot deposit dana on Friday, November 9: Let's Get Social!
https://totalslot88.ornop.org on What to Consider Before Buying Tech for Your Child


A little snowflake slowly drifted on the air descending from one of the winter clouds in the sky. He was one of the last snowflakes that had left the cloud and as he slowly fell, the sun started breaking through the cloud cover: a beautiful afternoon sun that made the little snowflake sparkle and glitter as it turned in the air. He rode on the wind and enjoyed the feeling of complete freedom, thinking, "I'm the prettiest of all little snowflakes. One in a million and nothing would ever compare to my beauty." He was perfect, as all little snowflakes are.
Obsessed with his own beauty, he noticed nothing else around him, and was surprised when he gently landed on top of a cushion of snow on a window sill. A gust of wind swept the snowflake towards the window, right up against the windowpane in which he admired his own reflection, glittering, smiling and very content.
It was one of those short winter days where it seemed to get dark far too early. In fact, it was late afternoon on the 24th of December, the day before Christmas, and already the daylight was fading. As the sun slowly went down behind the hills far away, the reflection in the window became fainter and fainter and the little snowflake started getting frustrated that he almost couldn't see himself in the window. Straining the eyes to see clearer, he suddenly noticed movement. What was this? Was there more to the world than his own quiet reflection? Surely, that couldn't be! As darkness descended, the movements behind the window became clearer. Yes, there was movement. The little snowflake watched as if he was in a trance. It was a family, it seemed. They had just arrived home with the last minute shopping before all the shops closed. They carried the bags in. Some full of groceries and others full of strange decorations; men in red suits wearing long hats, snowmen, Christmas hearts and other amazing stuff.
The man of the family slowly walked over to the fireplace to light a fire. It was a cold day and a fire would be so nice, warm and cosy. The father laid the logs in a shape like a pyramid and lit the fire. The snowflake watched in wonder as the flame caught the small wood and spread to the larger logs; the warm orange light seemed to pull him closer and closer and the feeling of warmth radiated through the window. The mother and the two girls came into the living room carrying boxes which were carefully put on the coffee table and opened. You could see the excitement in the eyes of the girls as they looked inside the various boxes. The older girl pulled out four big Christmas stockings and gave them to her mother for her to hang the stockings up at the front of the fireplace–one Christmas stocking for each of them. The snowflake thought it very strange to hang stockings by the fire. Maybe it was to warm their feet later?
The Christmas tree, standing proud, heavy with decorations and sweets, looked a bit empty at the top. "Strange," thought the snowflake, "how can that be?" He was soon to find out, as the mother sat with a box in her lap and asked the smallest of the girls to come over and open it.
With pride and a great big smile the smallest girl carefully took out the most perfect thing the snowflake had ever seen. An angel! She was made of delicate white paper. She had great big, almost see-through wings and silver streaks in her hair. Beautiful and fragile. The father grabbed the girl around her waist and lifted her up. She was holding the angel out in front of herself, looking at the angel. The father held the little girl in outstretched arms and she carefully placed the angel at the very top of the tree. The angel completed the Christmas tree beautifully and it took the breath away from the little snowflake when the angel turned her head and looked at him. The look was so quick that it was almost impossible to notice, but he saw her looking, just before the lights went out in the living room.
Outside it was dark as night and cold, too. The little snowflake wasn't that cold, he was made out of frozen water after all, but he felt chilly never the less. Maybe it was because he had looked at the fire in the fireplace? Now there were only the dark orange embers left.
The whole house and the world, too, seemed still as if all were asleep and slowly the snowflake drifted off to sleep too. He dreamt of angels and Christmas trees and laughter and children and in the background there was a faint ringing of bells. Strangely enough the ringing seemed to be getting louder and louder. The ringing woke up the little snowflake and he realized that the ringing was not from the dream at all. There, up high in the night sky was a sleigh, a red sleigh with reindeers in front and sacks of presents in the back. Driving the sleigh was a man dressed all in red, with a big white beard, a smile on his face and pointy hat on his head. He looked like the nicest man anyone could ever imagine.
"This must be the Santa Claus," the little snowflake thought as the sleigh landed on the roof of the house.
The man in red got out of the sleigh, grabbed one of the sacks from the back of his sleigh and climbed down through the chimney landing inside the fireplace. Santa let the sack slide off his shoulders and gently put it onto the floor, before taking out a list from the inside pocket of his red coat. He looked at it for a while, smiled and mumbled, "OK. Ho ho ho!" After rolling up the list and putting it back in the inside pocket he started taking presents out of the big sack. Some went into the socks hanging on the fireplace and others he placed underneath the Christmas tree. Santa looked around with a satisfied smile on his face, went to the table where he picked up a glass of sherry and drank it with pleasure. Then he picked up a biscuit and the carrots laid out for the reindeers and climbed back up the chimney. The snowflake could hear Santa on the roof, getting into the sleigh and setting off with a "Merry Christmas, Ho ho ho," as he drifted back into sleep.
When the snowflake woke again it was still very early. The sun was just rising; the sky such a bright pink that was reflected in the glittering frosty morning snow. The birds were waking up too and singing their happy morning songs.
It wasn't long before the snowflake heard footsteps inside the house, running eagerly down the stairs. The footsteps belonged to the two girls, who both ran into the living room, still wearing their nightgowns and fluffy slippers. Oh, how the little snowflake enjoyed watching the girls looking around the living room in amazement at all the presents in the stockings and under the Christmas tree.
The larger of the girls went over to the fireplace and took two of the stockings down. The snowflake now noticed there were names on the stockings. The larger girl handed the stocking with the name Matilde on it to the small girl. The stocking she kept for herself had the name Karoline on it. "Oh," thought the snowflake, "that must be their names, then." Karoline and Matilde started opening their presents, Karoline helping her younger sister whenever she had problems with the band around the presents. What wonderful presents they got: warm winter mitts, a sleigh to share, pink dresses for Barbie dolls and even a fairytale castle for the dolls to live in.
The snowflake looked on from the outside and was so envious of their happiness and presents. "If only I was in-there," he thought. "If only I could play with them and all their toys." Not long after, the mum and dad entered the room and gave each of the girls a hug and a big kiss. "Merry Christmas! What wonderful presents you have got," they said. The dad went over to the fireplace and took the last two stockings down from the fireplace and they opened their presents while they looked at the children playing.
When they had opened all the presents, the mum told the girls to go up and get dressed and then they could go out to play with the new sleigh. Matilde and Karoline got up and ran away, up to their rooms to change. In no time at all they were back, all excited about going outside to play with their new sleigh. The mum came to them at the door with the new mitts in her hand and helped them put on coats, mitts and scarfs and get the sleigh outside. The snowflake heard the door slam shut and shortly after the two girls walked past on the way to the hill at the end of one of the fields. A short while later he could see them playing, both of them riding downhill on the one sleigh, and after, Karoline kindly dragging the sleigh with Matilde sitting on it, back up the hill.
Down they went again, giggling and screaming with joy. For hours they played on the little hill until the mother called them in to have their lunch. The girls reluctantly started back. Again Karoline dragged her little sister on the sleigh. They both had red cheeks and happy sparkles in their eyes. The older girl suddenly let go of the rope of the sleigh, let her sister sit there and walked towards the window, the window where the snowflake was sitting. "She is coming to play with me! I am the luckiest snowflake ever." He was getting all excited thinking of all the games they would play together.
But of course the girl heard nothing. She kept pushing the snow together and made a nice round snowball. She turned around and looked at her sister with a cheeky grin on her face and got ready to throw the snowball toward Matilde who was still sitting on the sleigh. Matilde realized what her sister was going to do, she screamed and jumped off the sleigh and started running towards the house laughing and screaming in joy. Karoline took aim and threw the ball, just a bit ahead of where her sister was running but, oops, suddenly their cat Gokke jumped out of the snow and got hit by the snowball instead of Matilde. The snowball broke to pieces and scattered in the coat of the cat. "Phew," thought the snowflake, "I can breathe again. But what happened and where am I?" He was stuck in the fur of Gokke, the cat, who was now running as fast as he could towards the house. The snowflake looked around. "We-hey! This is fun. I am riding a cat," he laughed. "What an adventure." The cat jumped through the cat-flap and ran into the living room where he hid behind the big chair next to the fireplace.
Earlier in the day, the dad had lit the fire again and it was now burning lively and hot. The little snowflake could feel the heat from the fire, how wonderful it felt against his frozen body. There was a loud giggle and thumps against the door. The girls were back! Matilde opened the door and both girls stumbled into the hallway. As the door was opened a gust of wind came in and went through the house. A nice fresh wind from the outside.
The draft lifted the angel off the top of the Christmas tree. The snowflake watched as the angel slowly got carried on the wind and flew towards him, down from great heights and right to him. His heart beat faster. "Is this what love feels like?" he wondered.
Behind him the angel burst into flames and at the same time the little snowflake that was now melted to water evaporated into the air and got carried out with the smoke, up through the chimney, into the sky, back where he had come from.
This entry was posted in Regina Holm, Short Story on December 21, 2011 by lauralibricz.
Bamberg is a city in Upper Franconia with approximately 70,000 inhabitants. The city was built on seven hills like Rome: Cathedral Hill, Michaelsberg, Kaulberg/Obere Pfarre, Stefansberg, Jakobsberg, Altenburger Hill and Abtsberg. Bamberg was spared during the bombings in the Second World War, and carries the title of 'UNESCO World Cultural Heritage' because of the well-maintained medieval and baroque architecture. Not only the beautiful old town and 'Little Venice' on the banks of the river Regnitz attract many tourists; the city is known for its diverse and independent beer tradition. And, after 69 years of US Army presence in Bamberg, USAG Bamberg closed in September 2014.
In 1007, the German King Henry II created a new Catholic diocese that would aid in the final conquest of paganism in the area around Bamberg. The Catholic hold on this area remained securely intact with every new territory they acquired. For a short time Bamberg was even the center of the Holy Roman Empire. And they enjoyed many a century of power, until the reformers challenged their position. The first was Jan Hus in the early 15th Century and then, one hundred years later, the more well-known Protestant Martin Luther.
Then in the 17th Century, Bamberg suffered greatly during the Thirty Years War. The plague returned to Germany along with regiments of mercenaries. The country was experiencing what they call the 'Little Ice Age,' the cold-crisis that no expert could explain.
But what was it that turned organized religions, Catholics and Protestants alike, to seek blame for bad weather, failed crops and plagues of vermin among the city's inhabitants? Women, children and any sorts who fell out of the ranks were accused of witchcraft and brutally tortured and executed. Were they really to blame for the atrocities of the Thirty Years War?
The guidelines for the proper handling of potential witches were printed in 1488 by two German Dominican monks with the authorization from the Pope. This guide was called the Malleus Maleficarum, or Witches Hammer. This book offered the church fathers an explanation for the climactic problems, the plague or any of the other epidemics in general.
Was it just a book or was it some kind of collective insanity? Bamberg was the center ring of the witch executions of 17th Century Germany, along with the city of Würzburg. All in all, during this 20 year period between 1616 and 1636, 1000 women, children and the occasional man (even the city's mayor and the city council) were executed in Bamberg in the most brutal ways. Some sources say that the victims of this senseless wielding of power number 40,000 – 100,000 throughout Europe. Some go as far to say it was closer to a million.
This entry was posted in Bamberg on December 9, 2011 by lauralibricz.
It depends on the cookie, really. What would you do for a vanilla wafer, a sugar cookie, or a store-bought ginger snap? Or a chocolate covered oreo, a warm chocolate chip cookie with vanilla ice cream melting on the top, or a chocolate-chocolate chip macadamia that was taken out of the oven after just eight minutes, so it's still a bit raw in the middle?
I never did a lot of cooking. I only really started doing more than noodles or potatoes or an occasional boiled chicken ten years or so ago. But I've been baking cookies for as long as I can remember.
I like to experiment with the recipes and, like anything else in the world, there are just a few ground rules. It's like a science project, right Cathy? You need some dry stuff, some wet stuff, some fatty stuff and something to glue it all together. Sounds appetizing.
Adding an extra egg makes the cookies cakey-er, but that isn't always the desired result. American recipes call for salt and baking soda, but I find they get too salty, especially if I'm leaving some of the sugar out, or using this wonderful dark-dark brown sugar. I substitute baking powder and get good results. And I never let them bake all the way. I take them out just before they are really done–with chocolate cookies even more so, because mine are so dark, you can't see if they're brown or burnt. Maybe I should turn the light on.
10. Chocolate-Ghiradelli chocolate chip with cashews or macadamias, warm, ice cream, whipped cream and choco syrup.
This entry was posted in Food and Drink on December 5, 2011 by lauralibricz.


If Google has its way, we'll be driven BY our cars in the future. But when Armageddon comes, our robocars might not prove too useful. In a world of empty plains and sawed-off shotguns, you may need more than just your right foot to get around. Good thing for you that on your next trip to the Thunderdome, you'll be able to drive yourself.
Additionally, he has nearly 25-years experience in racing. Beginning at age eight racing three-wheeled ATVs in the AMA, Emile transitioned to car racing after attending the Jim Russell Racing Drivers School. It was there that he won his first scholarship to race, winning the 1990 Graduate Run-Offs. His driving career progressed through the ranks culminating in factory-backed efforts in production-based sedan racing in the U.S. Emile has also worked for car manufacturers such as Ferrari, BMW, and Audi as a product trainer, demonstration driver, and driving instructor in the U.S. and in Latin America.
A self proclaimed "Car Freak," Emile somehow finds time between testing cars, writing about cars, and instructing aspiring drivers, to paint and illustrate his favorite automotive subjects. Clearly, his passion for the automobile comes through in everything he does; "As shallow as it may sound to non-car people, cars are my life. I live, breathe, and dream about them, and feel very fortunate that I've been able to make a living by surrounding myself with cars."


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Senior Momentum
Asia's Aging Population Has a Need for Projects in the Healthcare Sector
Article Change Management, Healthcare 1 March 2018
Senior Momentum: Asia's Aging Population Has a Need for Projects in the Healthcare Sector (2018). PM Network, 32(3), 11.
As Asia's population grows older, there's a need for new projects in the healthcare sector.
Asia's population is aging faster than anywhere else in the world.
60% of the world's older-than-65 population will live in Asia by 2030.
BOOMING INVESTMENT
Healthcare projects are being launched all over the continent to cater to the demographic shift.
US$20 trillion Amount needed to fund elderly healthcare in Asia Pacific by 2030—a nearly fivefold increase from 2015
Healthcare expenditure per elderly person between 2015 and 2030 (in US$)
COUNTRY 2015 2030
Singapore 8,200 37,400
China 850 3,900
Vietnam 140 1,300
600,000 square meters (6,458,346 square feet)
Size of the redesigned Health City Novena after an expansion program slated to finish by 2030. The integrated health complex in Singapore is growing to deal with the country's aging population crisis. One project slated to close in 2022 will deliver an elderly-focused 500-bed rehabilitation facility.
Sources: Voice of Asia, Deloitte Insights, 2017; Advancing Into the Golden Years, Marsh & McLennan Companies, 2016; Health City Novena
Change Management 1 November 2020
Drive-By Innovation
By Ali, Ambreen | Hendershot, Steve | Hermans, Amanda | Thomas, Jen | Wilkinson, Amy To lure hungry patrons who are wary of indoor dining in the age of the coronavirus, restaurants are launching a feast of new projects to deliver better drive-thru service.
Innovation, Change Management 1 November 2020
Fashion, Forward
By Ali, Ambreen | Hendershot, Steve | Hermans, Amanda | Thomas, Jen | Wilkinson, Amy COVID-19 didn't kill the fashion show—it's simply accelerating the industry's shift to digital events. Ever since Shanghai Fashion Week canceled its in-person event and unveiled plans for its first…
Adjusting for Uncertainty
Forget business as usual. The global pandemic and ensuing economic uncertainty have forever reshaped the way the world works—from paused projects to newly virtual teams.
Innovation, Change Management, Organizational Culture August 2020
The Innovation Imperative
Organizations must invest in building a culture—and project teams— that can turn cutting-edge ideas into reality, according to new PMI research.
Article Change Management, Teams 1 March 2020
Organizations and their teams will have to adapt—and quickly—if they want to maintain their competitive advantage, according to PMI's latest Pulse of the Profession®.


From the moment you buy a Kubota you can be assured of first-class service.
We provide the highest standards of professionalism as well as great product knowledge and understanding of your needs.
To ensure you receive first class customer and aftersales service time after time we have made substantial investments in our parts and customer service resources.
Our Technicians are trained by Kubota and work in our fully equipped workshops.
We are proud to have been awarded the following service excellence awards by Kubota as part of our service training.


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Latest Arrival : Y-3 Kaiwa Trainers | Aphrodite Menswear Blog
Latest Arrival : Y-3 Kaiwa Trainers
Y-3 Kaiwa trainer brings sport-style to the street.
The Y-3 collection first hit the streets in 2002, with the idea of redefining streetwear. Blurring the line between sports and streetwear by fusing both together. Y-3's signature style combines the excellence of German-engineered garments with the creativity of Japanese designer Yohji Yamamoto who is well known for his avant-garde tailoring featuring Japanese design aesthetics. Resulting in modern pieces of Japanese and German streetwear which have inspired people around the globe, gaining a large following in their almost 20-year existence.
The Y-3 Kaiwa trainer is part of the exclusive Y-3 collection available at Aphrodite, which takes inspiration from the past. Taking designs that were released in the 70s, 80s and the 90s producing a contemporary twist on classic designs. The unique idea of redesigning old silhouettes by taking inspiration from today's styles and trends has led to a thread of classic designs which all have a uniqueness shaped by the decade when they were crafted.
Y-3 kaiwa trainers have a minimalist design of a running shoe silhouette. The Kaiwa is the first-ever Y-3 shoe to feature pieces directly from the archives of both collaborators. The silhouette was originally designed by Yamamoto and Adidas and now has been updated with a contemporary touch, infused with modern technology. A woven upper offers enhanced comfort with a lightweight EVA midsole that gives extra cushioning to support your feet throughout the day.
A lace closure helps give a secure fit to the trainer, allowing the shoe to be worn as both a sports trainer made for exercise or a fashion trainer crafted to make a bold statement on the street. Stylish detailing of geometric shapes in leather creates a modern and striking exterior. The Y-3 KAIWA shoe is finalised with a suede Y3 logo tab to the heel of the shoe completing the symmetrical look of the updated trainer.
The Y-3 Kaiwa trainer is available at Aphrodite.
Tagsy-3, Y-3 kaiwa
Fresh Drop of the Iconic Stone Island Overshirt
Aphrodite Clothing Black Friday Deals : Footwear


"I was a nuke ET and an instructor in BS (90-99), now father to a nuke EM in the sub fleet. You're blog about the significance of a SO that is your best friend and a partner that compliments your skill set making you a great team is essential. More importantly I've found that truly the most important thing is 'open and honest communication' at all times, especially where stressful jobs and distance come into place.
I moved insanely fast (as many nukes do) when I met my wife. It was Feb. 93. Within a couple weeks we were exclusive, within a month or so, we found an apartment together, engaged a month after that and married in July 93. I left San Diego in Sept. of 93 on my first 6 month run, like no other, because I was never coming back, changing home ports to Newport News, VA at the end to decommission.
In Dec. of 93, I pulled into Florida for a week and I flew her out to meet me. We had a blast, put her back on a plane, and about a month later…Western Union telegram (no email back then), she was pregnant…go figure his birthdate is 9-25-94. We really got lucky because honestly we hadn't spent enough dedicated time together to even know if we really had the whole package, besides love…because many times, it's not enough. Marriage takes a ton of love, patience, compassion, and communication to get you through the roller coaster of life's obstacles, especially those that the Navy conveniently offers.
We had many bad years, but continued to fight through as we raised our boys. We've now been married for 23 years, are empty nesters with both of our sons serving in the Navy, and have a stronger relationship today than we've ever had, but none of it was easy.
Honestly I think the only reason we didn't self destruct in the early years was because my wife is 5 years older than me and she had also served in the Navy as a conventional ET, so she fully understood what she was buying into. We met in a country bar…she was recently discharged and was out looking for her cowboy, instead she met a Sailor playing cowboy and after 6 years of active duty, she did another 6 years as a dependent.


Deborah Pack is the Human Resources Director for the South Carolina State Library, since February 2016. Deborah has over 13 years of professional human resources experience in the public and private sectors. She is responsible for implementing human resource management strategies that enable the South Carolina State Library to recruit, train, and retain a high performing and motivated workforce. Deborah has experience in benefits, employee relations, talent acquisition, performance management, and compensation. Deborah received a Bachelor of Arts in Behavioral Science and a Master of Education from Columbia College. She holds a SHRM Senior Certified Professional (SHRM-SCP) certification, and is a member of the Society of Human Resource Management.


Thought it might be useful to give some things that have helped me with the TextBlade so when others get it, the experience I've had may be useful.
As everyone probably knows by now, the reduced reach will be a problem at first, especially for touch typists probably. The ergonomic design will be as well if you aren't used to that angled design. Most commonly, it is the pinky keys that are missed as lot. For me, once I got over my initial adjustment, it was hitting too low.
When typing, it helped me to stop and think before typing that my fingers could actually feel that they were in the well of the 8 keys. Too often I'd start typing without doing that and would be all over the place. For some reason it was the "A" key that I was always too low on.
After I had a lot of days practice, I can still have cases where I just can't seem to type well. One thing that has helped in the last day was that I put a marker on the lap desk I use (all my typing is on my lap) for how the space blade should be placed. In my case, it seems my best typing has the right side lower than the left so I place the homemade lap desk squared up evenly on my lap with a line on it for how I want the space blade angled. I may adjust this over time, as I notice where most errors occur, but in the meantime, it guarantees I'll be consistent at least.
While it seems that almost everything works bettter on this technology, there are some issues. Some are fairly direct trade-offs. Such as that, while you have to hold the space bar to type a number (unless you activate the special number pad lock), this extra step is offset by the fact that the numbers are so much easier to reach. Even I can do it. But another issue is how the combinations work out depending on what you are typing.
If typing words and then a number, such as "in 1492", the problem is that after you type the space, you must release the space bar and then put it back down to get the number 1. At least for me, that is NOT a natural pattern! As a result, I'll get a typo often for the first number after a space. Might not be an issue if you don't type fast. And I've never sat down to just practice such things. Maybe once I do I won't find it to be an issue.
Hopefully others can jump in and give more hints. For example, orand had a video that showed what may be the most efficient way to pack up and open up the textblade. And I think he may have also said that, if only using one nano stand as a case, to put the long side against the stainless steel of the textblade because it will slide a little easier.
As I was reading about the space number combination, I thought of a possible mental solution. What if you designated specific spots on the SpaceBlade for space and green modifier? That way your brain would be thinking you were hitting two separate keys and the second one would stop feeling like a space.
Obviously I don't have TextBlade to practice with, but I've always typed where the space key was kind of like the "enter" or "confirm" after the word, so a word will roll off with a space as the last character.
Does this mean that pressing the space key again for the green layer likely won't be an issue? Do you guys do this too and still have the problem or do you do something else?
The multi-use nature of the WT keyboard assumes you are typing correctly. And not doing nonsense like I do where I only use my left pinky for all shift commands, or other such things.
What if you designated specific spots on the SpaceBlade for space and green modifier?
Maybe, but have to be careful about too many areas. We can do hot corners which have cmd and alt at each end. Maybe an option could be made to let them do something different or in addition. Something I just thought about, but haven't tried, is to teach myself to alternate which thumb I use. So if I want space then number, I might hit space with the left them then use the right thumb to get the number. It would give the feeling of changing keys even though you really wouldnt have.
Update: I misread your post and thought you were talking about officially programming sections of the space bar, but apparently you were thinking pretty much the same thing I mentioned.
Does this mean that pressing the space key again for the green layer likely won't be an issue?
I'm just not clear on your situation. It certainly isn't an issue to press space again. Its just that the pattern of a plain space and then having to hit space again, as a separate stroke, for a number just isnt natural. The tendency is to want to keep the thumb down but if you do that, you won't get the space - it will be waiting fo something from the green layer.
In our phone conversation, Mark mentioned that the world's fastest typist doesnat use the shift key - just caps lock. I assume that means just using the left side. This makes it much like how TB works. You tap shift and it waits for the next key before turning off caps. Or tap it twice to get a true caps lock. I'm training myself to do this.


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Hernando deputies: Neighbor admitted to shooting at kids riding ATV
Huggies puts fathers on diaper packages for the first time
Our Aim Foundation: Saving lives and creating opportunity
Piece of Bayshore Boulevard history up for sale
Hillsborough Co. could reduce trash pickup days
Transportation tax funds to be re-established after legal battles
Columbia Restaurant's nearly 100-year-old grand piano restored to former glory
By Lloyd Sowers, FOX 13 News
TAMPA, Fla. (FOX 13) - Music brought Cesar and Adela Gonzmart together. At 21, he was a violin soloist with the Havana Symphony; she was a Julliard graduate and concert pianist.
Even when the couple took over the Columbia restaurant from Adela's father, the music came right along with them. The 1921 Steinway piano she first played remained at the restaurant for decades.
"It's got a lot of memories," said Richard Gonzmart, the couple's son. "It's played 'Happy Birthday' to thousands of people."
But after Cesar and then Adela passed away, her old piano appeared to be at the end of its life. The keys were damaged, the inner workings tarnished, and the wood heavily damaged by termites.
Richard looked into getting it restored; even Steinway thought the piano was beyond repair.
"They said, 'Listen, it doesn't make any sense. You could buy a new piano. This would cost you more.' But a new piano doesn't have the memories and wasn't played by my mom," he said.
Richard was determined to bring his mother's piano back to life. His daughter, Andrea Gonzmart Williams, knew her dad would not take no for an answer.
"With my father, he never gives up, so there was not a doubt in my mind for him to do whatever it took to bring it back to what it used to be," she said. "There was not a doubt."
The answer came months later. It was no small task to turn back time, but all of the parts and pieces of the nearly 100-year-old piano were put back together -- good as new.
"I had tears, it's a dream come true that I didn't think would happen," Richard said. "Gave it the original finish, the original colors and new strings, new ivories, etc."
That special piano is now back at the Columbia.
"I know that my mom and my dad will be looking down knowing that they're remembered and loved," he said.
Rattlesnake bite leaves man in hospital on due date of his second child
By Dan Matics, FOX 13 News
A Sarasota man is recovering in the hospital after being bitten by a rattlesnake.
He should have been in the hospital Wednesday for a different reason: The bird of his second child.
The baby hasn't come yet, so Earl Hjertstedt has to worry about his wife – and his recovery – from a hospital bed.
By Josh Cascio, FOX 13 News
Hillsborough County's garbage pickup contracts are set to expire next year.
While county officials eye new deals, they're asking the public to weigh in on some potential options.
One of the more controversial choices includes cutting trash pickup from twice a week to just once.
Stolen car suspects nabbed on Howard Frankland Bridge
It was smooth sailing for Tampa-bound traffic on the Howard Frankland Bridge Wednesday until deputies tracked down and pulled over a stolen car.
The traffic caused a mess for anyone needing to get somewhere on time. Pinellas County Deputies stopped the stolen vehicle near the middle span of the bridge.
It started in St. Petersburg, where stolen cars have been a topic of discussion. On Monday morning, the driver of a stolen Infinity hit and killed a 26-year-old woman.


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Q: How to build Java webhook for new Google actions builder I created the java webhook in past for dialogflow where 1-1 mapping of intent used to be done. Now I am working with actions builder & not getting how to define and build the handlers using Java.
Before @ForIntent was used to map the intents with java methods. Now how should I map the methods with any scene or handlers.

A: At the moment there isn't a client library or a SDK to use in order to implement a webhook in Java for the Actions Builder.
You have to implement this manually, exposing and endpoint able to receive POST requests with a HandlerRequest object as input param, as specified in this json-schema
Once created your webhook, to select the handler to use you have to check the value under handler.name field of the request and invoke the right handler. To avoid hardcoded if/switch you could use a strategy pattern and reflection/bean loading to invoke the specific handler, using the value achieved by the field handler.name.

redpajama_set_name: RedPajamaStackExchange


Every kid's dream...the snow day! However, for high school principals, it's a nightmare. Not only do you have to make up that day, but you're responsible for getting the word out to all the parents.
Nevertheless, when you live in Nebraska, a snow day is inevitable. So you might as well make the most out of it. And for Conestoga High School Principal, David Friedli, you get creative. There's nothing like a parody song set to Billy Joel's "We Didn't Start The Fire".
Ladies and gents, the moment you've all been waiting for..."We Didn't Need A Snow Day".


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Home World Chris Hopkins is the new prime minister of New Zealand
Chris Hopkins is the new prime minister of New Zealand
Labor MP Chris Hopkins has been named as the new Prime Minister of New Zealand after the resignation of the female Prime Minister of New Zealand, Jacinda Ardern.
According to a foreign news agency, pressure was seen in New Zealand to elect a prime minister from the ancient indigenous Maori race, after which the names of Chris Hopkins, Justice Minister Kerry Allen and Immigration Minister Michael Wood were considered for the country's prime minister. were
However, Labor MP Chris Hopkins has been nominated for the new Prime Minister of New Zealand.
Jacinda will formally tender her resignation to Governor-General King Charles III on February 7, after which the Governor-General will appoint Chris Hopkins as Prime Minister if he receives the support of the House of Representatives.
It should be noted that on January 19, New Zealand Prime Minister Jacinda Ardern announced her resignation and said that she has no intention of contesting the next election.
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The key lime compensates for its more petite size with a strong aroma, higher acidity and thinner rind. The combination makes for a superior lime flavor and a better e-juice. This refreshing citrus will provide a real pick-me-up with a just squeezed burst of flavor.
Other Blends that include "Key Lime Delight"


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The Package | 19 Stories
Madras High Court rejects Vedanta's plea to reopen Sterlite copper plant
In their judgement running to 815 pages, the judges agreed that the prohibitory orders had been rightly passed for violation of laws relating to pollution control
August 18, 2020 11:41 am | Updated 01:21 pm IST - CHENNAI
Mohamed Imranullah S.
CPI (M) cadres distributing sweets near Rajaji Park in Thoothukudi on Tuesday welcoming the High Court's verdict
The Madras High Court on Tuesday rejected a plea by natural resources company Vedanta seeking permission to reopen its Sterlite copper smelting plant in Thoothukudi . The plant was sealed by the Tamil Nadu government on May 28, 2018 due to environmental concerns.
Justices T.S. Sivagnanam and V. Bhavani Subbaroyan dismissed all 10 writ petitions filed by Vedanta challenging various orders passed by the State government and the Tamil Nadu Pollution Control Board (TNPCB) refusing consent to operate, to stop production, to disconnect electricity and to seal the premises.
The Hindu explains: Sterlite protests
In their judgment running to 815 pages, the judges agreed with Advocate General Vijay Narayan for the State and senior counsel C.S. Vaidyanathan, representing TNPCB, that the prohibitory orders had been rightly passed for violation of the laws relating to pollution control.
After the delivery of the verdict, senior counsel C. Aryama Sundaram, representing Vedanta, urged the court to order status quo for three, or at least two weeks, so that nothing adverse takes place until the judgement could be taken on appeal to the Supreme Court.
Sterlite Copper expresses shock over Madras High Court verdict
However, the judges refused to pass any such order. The senior judge in the Bench said it may not be appropriate to add anything after they had delivered their verdict. He said, the verdict would have been delivered in March itself but for the pandemic.
He told the lawyers that the judgement was dictated to three different personal assistants over the phone before everything could be collated and corrected. It was an enormous effort, given that the lawyers argued the case for 42 days and submitted voluminous documents running to thousands of pages.
Sterlite verdict: A tense Thoothukudi celebrates
Posthumous Padma Vibhushan for Mulayam Singh Yadav, ORS pioneer Dilip Mahalanabis
Jaishankar slams Pompeo for 'disrespectful' comments on Sushma Swaraj
Youth outfits defy 'ban' to hold public screenings of BBC documentary on PM Modi
Hockey World Cup 2023, HIGHLIGHTS Quarterfinals: Netherlands defeats Korea 5-1; Germany thumps England in shootout 4-3


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Two more healthcare workers arrested in hospital poisoning probe
By Pat Hurst 12th December 2018 23rd December 2018 News EnglandHospitals, Lancashire, News England, NHS, North West
Detectives investigating allegations of mistreatment and neglect of stroke victims at a hospital have arrested two more people.
A man and a woman, described by Lancashire Police as healthcare professionals, were arrested on suspicion of administering any poison or noxious thing with the intent to injure and ill treatment or wilful neglect, at Blackpool Victoria Hospital.
They are currently in custody.
Police were initially contacted by Blackpool Teaching Hospitals NHS Foundation Trust on November 8 after concerns were raised about the care provided to some patients.
The claims related to the alleged mistreatment and neglect of patients on the stroke unit that month.
A third healthcare professional was arrested last month on suspicion of administering any poison or noxious thing with the intent to injure and ill treatment or wilful neglect and bailed until February 10, 2019
All three have been suspended.
Lancashire Police said officers remain in the early stages of their investigation and are working closely with Blackpool Teaching Hospitals NHS Foundation Trust and HM Coroner for Blackpool and Fylde.
A number of post-mortem examinations have been conducted as part of police inquiries.
Det Chief Insp Jill Johnston, of Lancashire Police said: "We have made two more arrests as part of an on-going investigation into allegations of mistreatment of patients at Blackpool Victoria Hospital.
"This is a complicated and sensitive inquiry which remains at its early stages.
"I understand further arrests might be concerning but I would like to reassure people we have a dedicated team of detectives working on this investigation.
"We are offering support to those families who have had loved ones identified as potentially being involved. They are being kept fully updated and supported throughout the process by specially trained officers.
"Some post-mortem examinations have been conducted to assist our inquiries.
"Our priority and the priority of Blackpool Teaching Hospitals NHS Foundation Trust is to ensure the safety of patients. We are working closely with the trust as well as HM Coroner for Blackpool and Fylde as part of the investigation."
Anyone with information can contact police on 101 quoting log reference 0553 of November 14. Information can also be given to police via: https://mipp.police.uk
Copyright (c) Press Association Ltd. 2018, All Rights Reserved. Picture (c) Google Maps.
Man arrested over suspected murder of elderly relative at Essex care home
Scottish budget 'misses opportunity to tackle poverty', according to charities
Recovery Worker
Berkshire, Newbury
Hays London Ebury Gate


text:
Q: Pagination with spatie always shows paginateroute::paginateroute in url Why is my url like this
example.test/blog/paginateroute::paginateroute.page/2 using spatie/laravel-paginateroute
my route :
Route::paginate('blog', [ 'as' => 'blog', 'uses' => 'BlogController@index' ] );

on views
@if(PaginateRoute::hasPreviousPage())
     «
@endif

@if(PaginateRoute::hasNextPage($posts))
     »
@endif

what is solution for remove paginateroute::paginateroute. from url ?


- Stream of consciousness. - Personal? Political?
It was stream of consciousness!
but we can initiate the consciousness stream.
Do you believe, Mr. Martins, in the stream of consciousness?
No, yeah, no, it reads as stream of consciousness.
Will you tell one-liners, stream of consciousness?
We must all arouse the political consciousness of the people.
- ♪ Two plus three equals 23 ♪ - This is just stream of consciousness.
From a political point of view or personal?
a front-row seat to the sclerotic political gridlock.
...do you mean myself as like a unified person as the source of consciousness?


text:
Q: Interacting with libscap from Python using ctypes I have been working on a wrapper around sysdig libscap library using ctypes. It is all working good except one thing that is starting to frustrate me. Libscap uses the uthash library for implementing hash table like data structures in C language. I have this declarations for structures used by uthash.
if sizeof(c_void_p) == 4:
     ptrdiff_t = c_int32
elif sizeof(c_void_p) == 8:
     ptrdiff_t = c_int64

# uthash.h
class UT_hash_table(Structure):
    pass
class UT_hash_handle(Structure):
    pass
UT_hash_handle._fields_ = [ ("tbl", POINTER(UT_hash_table)),
                 ("prev", c_void_p),
                 ("next", c_void_p),
                 ("hh_prev", POINTER(UT_hash_handle)),
                 ("hh_next", POINTER(UT_hash_handle)),
                 ("key", c_void_p),
                 ("keylen", c_uint),
                 ("hashv", c_uint)]

class UT_hash_bucket(Structure):
    _fields_ = [ ("hh_head", POINTER(UT_hash_handle)),
                 ("count", c_uint),
                 ("expand_mult", c_uint)]

UT_hash_table._fields_ = [
                 ("buckets", POINTER(UT_hash_bucket)),
                 ("num_buckets", c_uint),
                 ("log2_num_buckets", c_uint),
                 ("num_items", c_uint),
                 ("tail", POINTER(UT_hash_handle)),
                 ("hho", ptrdiff_t),
                 ("ideal_chain_maxlen", c_uint),
                 ("nonideal_items", c_uint),
                 ("ineff_expands", c_uint),
                 ("noexpand", c_uint),
                 ("signature", c_uint32),
                 ("bloom_sig", c_uint32),
                 ("bloom_bv", POINTER(c_uint8)),
                 ("bloom_nbits", c_char)]

There is also another structure scap_threadinfo, which acts as hash table, and holds the list of processes parsed from /proc filesystem.
class scap_threadinfo(Structure):
        _fields_ = [ ("tid", c_uint64),
                     ("pid", c_uint64),
                     ("ptid", c_uint64),
                     ("comm", c_char * SCAP_MAX_PATH_SIZE),
                     ("exe", c_char * SCAP_MAX_PATH_SIZE),
                     ("args", c_char * SCAP_MAX_PATH_SIZE),
                     ("args_len", c_uint16),
                     ("env", c_char * SCAP_MAX_ENV_SIZE),
                     ("env_len", c_uint16),
                     ("cwd", c_char * SCAP_MAX_PATH_SIZE),
                     ("fdlimit", c_uint64),
                     ("flags", c_uint32),
                     ("uid", c_uint32),
                     ("gid", c_uint32),
                     ("vmsize_kb", c_uint32),
                     ("vmrss_kb", c_uint32),
                     ("vmswap_kb", c_uint32),
                     ("pfmajor", c_uint64),
                     ("pfminor", c_uint64),
                     ("scap_fdinfo", POINTER(scap_fdinfo)),
                     ("hh", UT_hash_handle)]

The last member of this structure marks it as hasheable. The problem comes when I call scap_get_proc_table function, next and prev pointers of UT_hash_handle are always evaluated as None. Other members like exe or comm have correct value for the first process on my system.Any help is appreciated.


text:
24 Hour Home Care > Blog > 24 Hour Home Care > 24 Hour Home Care Named to Inc. Magazine's List of America's Fastest-Growing Private Companies for the Sixth Consecutive Year
24 Hour Home Care Named to Inc. Magazine's List of America's Fastest-Growing Private Companies for the Sixth Consecutive Year
September 5, 2018 24 Hour Home Care
Los Angeles, CA- 24 Hour Home Care is elated to announce their 4,108 placement on Inc. Magazine's List of America's Fastest-Growing Private Companies for the sixth year in a row.
Introduced in 1982, this prestigious list ranks the nation's most profitable companies that has experienced significant growth in the past year.
2017 marked a year of growth, expansion and innovation for 24 Hour Home Care. In the past twelve months alone, 24 Hour Home Care opened the highest number of new offices in one year since its inception in 2008. These offices include Southern California offices of Carlsbad, Pasadena and Fullerton. 24 Hour Home Care's Disability Services Division also expanded into Riverside, where over 30,000 English and Spanish-speaking families are given the care they need. Over the past year, new partnerships with industry leading companies were established, such as an innovative pilot program with Cedars-Sinai called "Care Extenders," that provides support in the home for at-risk patients.
Giving back to the community has always been a priority for 24 Hour Home Care. Despite having a busy year, in 2017 alone 24 Hour Home Care was able to donate close to $48,000 through their Proceeds to Charity Program to help contribute to research on illnesses and diseases. Since the inception of the program, 24 Hour Home Care has donated over $230,000 to various community partners. For 24 Hour Home Care, growth is not solely defined by numbers of revenue, but rather the impact made in the community as well.
"We are incredibly proud to be named on Inc. Magazine's list for the sixth year in a row," CEO and Co-Founder David Allerby says, "24 Hour Home Care started as a dream Ryan Iwamoto, Tyner Brenneman-Slay and myself shared. When we opened our doors in 2008, our goal was to provide quality care and customer service to our clients. 10 years later we employ over 7,000 professional caregivers and internal employees, serving over 10,000 clients and providing over 15 million hours of care. Our growth is attributed to the hard work and dedication of our staff and the passion we have to help others. We are so proud of this company and look forward to delivering quality care and service to our clients and the communities in which we serve."
About 24 Hour Home Care
24 Hour Home Care is a Los Angeles-based company that provides high-quality, customized, professional caregiving services to seniors and individuals with developmental disabilities, allowing them to continue full, active and healthy lifestyles. Founded by David Allerby, Tyner Brenneman-Slay, and Ryan Iwamoto in 2008, 24 Hour Home Care has expanded to 16 locations throughout California, Arizona and Dallas, hiring over 3,000 employees. Throughout the past several years, 24 Hour Home Care has received numerous awards, including the owners being awarded with the Ernst & Young Entrepreneur of the Year Award in 2017 and the company being named to Inc. Magazine's List of Fastest Growing Private Companies, the Inc. 5000, for the sixth consecutive year in 2018. 24 Hour Home Care has received additional accolades including being listed by Forbes Magazine as the #24 Most Promising Company in America. To learn more, visit us here


Carnival Against Vivisection - Other Targets in Herefordshire!
The police are sure to be out in force in Herefordshire for the Carnival Against Vivisection but Sequani cannot run without certain key suppliers in the Herefordshire area.
I'm sure the police can spare a couple of extra hundred officers to protect all of these other animal abusers in the area.
Here are a few of the most important Sequani (and HLS suppliers) in the area.
Develop cage systems for lab animals used by Sequani, HLS and many other vivisection labs in the UK and overseas. No CCTV. No Fences.
Auditors for Sequani Limited, these people are integral to the running of Sequani as they have to be audited for government legislation or risk losing animal licences.
Builders and civil engineers for Sequani, they have received many demos and have still not dropped the animal abusers!


text:
<!DOCTYPE html>

	<meta charset="utf-8">
	<title>Backchart Module: canvasjs/collection</title>

	<link type="text/css" rel="stylesheet" href="styles/sunlight.default.css">

	<link type="text/css" rel="stylesheet" href="styles/site.cosmo.css">

			Backchart

					Modules

							amcharts/collection

							amcharts/model

							amcharts/view

							base/collection

							base/model

							base/view

							canvasjs/collection

							canvasjs/model

							canvasjs/view

							echarts/collection

							echarts/model

							echarts/view

							util/parser

					Events

							beforeRender

							beforeValidateData

							removed

							reseted

							seted

							rendered

							beforeRender

Module: canvasjs/collection

    canvasjs/collection

Backbone chart CanvasJS collection

Source:

        backchart.canvasjs/collection.js, line 34

Requires

module:jquery

module:base/backbone

					Open source project and code licensed under MIT License(https://github.com/xch89820/backchart/blob/master/LICENSE-MIT)

		Documentation generated by JSDoc 3.3.0-beta3
		on Fri Apr 24 2015 10:15:25 GMT+0800 (CST) using the DocStrap template.

redpajama_set_name: RedPajamaGithub