text:
\section{Introduction and main results}
 Consider the Ginzburg-Landau equation in ${\mathbb R}^N$
 \begin{equation}
   \label{eq:81}
   -\Delta u=(1-|u|^2)u \,.
 \end{equation}
Much effort has been devoted to classifying the solutions to
\eqref{eq:81} under various assumptions. In the scalar case, the
famous De Giorgi conjecture states that any bounded solution satisfying
$\partial_{x_N}u>0$ must depend on one Euclidean variable only, at
least when $N\leq
8$. This conjecture was proved to be true in dimension $N=2$ by
Ghoussoub and Gui~\cite{gg}, for $N=3$ by Ambrosio and
Cabr\'e~\cite{ac}, and under the additional assumption
$\lim_{x_N\to\pm\infty} u(x',x_N)=\pm 1$, for $4\leq N\leq 8$, by
Savin~\cite{savin}. On the other hand,
the counterexample of del Pino et al.~\cite{dp}  shows that indeed $N=8$
is optimal.

Much less is known in the vector-valued case $u:{\mathbb R}^N\to{\mathbb R}^m$. In that case the
monotony hypothesis no longer makes sense. On the other hand, the
class of locally minimizing solutions in the sense of De Giorgi --- i.e., solutions  that are
minimizing for their own boundary values on every ball --- come up naturally as  blow-up
limits of minimizers of the Ginzburg-Landau energy
\begin{equation*}
  E_\varepsilon(u)=\int \frac{1}{2}|\nabla u|^2+\frac{1}{4\varepsilon^2}(1-|u|^2)^2\,,
\end{equation*}
as $\varepsilon$ goes to zero. In fact, monotone scalar solutions of
\eqref{eq:81} also have a certain local minimality property (see
\cite[Thm~4.4]{aac}).

When $N=m=2$ it follows from the results of Sandier~\cite{s} and Mironescu~\cite{m} that
every local minimizer is either constant or equal to $U(x):=f(|x|)\frac{x}{|x|}$, up to
rotations and translations, where $f$ is the unique solution of the
corresponding ODE. When $N=m=3$ a similar classification was
proved by Millot and Pisante~\cite{mp} under the assumption
\begin{equation*}
  \limsup_{R\to\infty}\frac{E(u;B_R)}{R}<\infty\quad(\text{here }E(u;B_R)=E_1(u;B_R))\,.
\end{equation*}
Under additional assumptions,  Pisante~\cite{p} extended the result to $N=m\geq 4$.
Note that in these cases the existence of a non-constant local
minimizer is a nontrivial fact.

We are interested in the case $N=3$, $m=2$. In this case, it is easy
to deduce from the local minimality property of $U$ that
$V(x,z)=U(x)$ is a local minimizer. We conjecture that it is the only
non-constant
one, up to the obvious symmetries of the problem. As a first step in
this direction we prove:
\begin{theorem}
\label{th:lm}
  If $u:{\mathbb R}^3\to{\mathbb R}^2$ is a local minimizer of the Ginzburg-Landau
  energy such that
\begin{equation*}
  \liminf_{R\to\infty}\frac{E(u;B_R)}{R\ln R}<2\pi\,,
\end{equation*}
 then $u$ is constant.
\end{theorem}
Note that the constant $2\pi$ is optimal since
\begin{equation*}
  \lim_{R\to\infty}\frac{E(V;B_R)}{R\ln R}=2\pi\,.
\end{equation*}
A different assumption was considered by Farina~\cite{f} who proved  that
a local minimizer $u:{\mathbb R}^N\to{\mathbb R}^2$ with $N=3$ or $4$, is constant
provided $\lim_{|x|\to\infty} |u(x)|=1$.

\rth{th:lm} is an easy consequence of the following sharp
$\eta$-ellipticity type result for minimizers in dimension three:
\begin{theorem}
  \label{th:eta}
For every $\gamma\in(0,2\pi)$ and $\lambda>0$  there exists
$\varepsilon_1(\gamma,\lambda)$ such that for every $u_\varepsilon$ which is a minimizer of $E_\varepsilon$ on $B_1$
with $\varepsilon\leq \varepsilon_1(\gamma,\lambda)$ satisfying $E_\varepsilon(u;B_1)\leq \gamma |\ln \varepsilon|$
there holds
\begin{equation}
\label{eq:60}
  \big||u_\varepsilon(0)|-1\big|\leq \lambda\,.
\end{equation}
\end{theorem}
Such a result  first appeared in
the work of Rivi\`ere~\cite{r} for the case
of minimizers in dimension three. There were subsequent
generalizations by  Lin-Rivi\`ere~\cite{lr1,lr2}
and Bethuel-Brezis-Orlandi~\cite{bbo1,bbo}. The result in \cite{bbo}
is the most general, covering the case   of solutions (not necessary
minimizers)
of the Ginzburg-Landau equation in any dimension.

All these results establish the existence of a constant $\gamma>0$
for which the result is true, but no explicit bound is
given. We are able to give the optimal bound, but only for minimizers
in dimension three. Working with minimizers allows us to apply a
construction of an appropriate test function. This is done in
\rprop{prop} below, which plays an important role in the proof of
\rth{th:eta} (see Section~\ref{sec:notation} for notation).
\begin{proposition}
\label{prop}
 Let $u$ be a minimizer for $E$ on $B_R\subset{\mathbb R}^3$, for its
 boundary values on $S_R$.
  If
  \begin{equation}
\label{eq:ineq-prop}
   E^{(T)}(u;S_R)  \leq\gamma\ln R\,,
  \end{equation}
with $\gamma<2\pi$ and
  $R>r_0(\gamma)$ then there exist
  $\alpha=\alpha(\gamma)\in(0,1)$ and a universal $\sigma\in(0,1)$ such that
  \begin{equation}
\label{eq:prop}
    E(u;B_R)\leq \sigma R E^{(T)}(u;S_R)  +C_\gamma R^\alpha\ln R\,.
  \end{equation}
\end{proposition}
\subsubsection*{Acknowledgments} The research was carried out during
several mutual visits at the Technion and Universit\'e
    Paris-Est. The authors acknowledge the hospitality and support of
    these two institutions.

\section{Notation and some preliminary  results}
\label{sec:notation}
We begin by introducing some notation.
By $B_R(a)$ we denote a ball in ${\mathbb R}^N$, $N\geq 2$ (usually for $N=3$) and then $S_R(a)=\partial
B_R(a)$.  Specifically in dimension two,  we denote by $D_R(a)$ a disc
and by
$C_R(a)=\partial D_R(a)$ its boundary. In case $a=0$ we denote for short:
\begin{equation*}
  B_R=B_R(0)\,,~S_R=\partial B_R(0)\,,~D_R=D_R(0)\,,~C_R=\partial D_R(0)\,.
\end{equation*}
 For a set $D$ in ${\mathbb R}^N$, $N\geq 2$, $\varepsilon>0$ and $u\in H^1(D;{\mathbb R}^2)$ we denote:
\begin{equation*}
  E_{\varepsilon}(u;D)=\int_D \frac{1}{2}|\nabla
  u|^2+\frac{1}{4\varepsilon^2}(1-|u|^2)^2\,.
\end{equation*}
For ${\widetilde D}\subseteq S_R(a)\subset{\mathbb R}^N, N\geq 2$, and $u\in H^1({\widetilde D};{\mathbb R}^2)$ we denote
\begin{equation*}
E_{\varepsilon}^{(T)}(u;{\widetilde D})=\int_{{\widetilde D}} \frac{1}{2}|\nabla_T u|^2+\frac{1}{4\varepsilon^2}(1-|u|^2)^2\,.
\end{equation*}
When $\varepsilon=1$ we denote for short $E(u;D)=E_1(u;D)$ and $E^{(T)}(u;{\widetilde D})=E_1^{(T)}(u;{\widetilde D})$.

We denote by ${\widetilde
  D}_\rho^R(a)$ a spherical disc on the sphere $S_R=S_R(0)\subset{\mathbb R}^3$, with center at $a$ and radius
$\rho$ (using the geodesic distance) and by ${\widetilde
  C}_\rho^R(a)=\partial {\widetilde  D}_\rho^R(a)$
its boundary. In case there is no risk of
confusion we shall omit the superscript $R$. We often identify
${\mathbb R}^2$-valued maps with ${\mathbb C}$-valued maps. \\
 We recall the following basic estimates valid for an arbitrary (vector
 valued) solution
 of the Ginzburg-Landau equation \eqref{eq:81}
 in dimension $N\geq2$.
 The first is a $L^\infty$-bound for $u$ and its gradient (\cite[Lemma~
 III.2]{bbo}).
 \begin{lemma}
   \label{lem:L-infty}
Assume $u$ satisfies \eqref{eq:81} in $B_1\subset{\mathbb R}^N$.
Then, there is a  constant $K=K(N)>0$ such that
\begin{align}
  \label{eq:58a}
  \|u\|_{L^\infty(B_{1/2})}\leq K\,,\\
\intertext{and}
  \label{eq:58b}
\|\nabla u\|_{L^\infty(B_{1/2})}\leq K\,.
\end{align}
 \end{lemma}
 \begin{proof}[Sketch of proof]
   For the proof of \eqref{eq:58a} we argue as in
   Brezis~\cite[Remark~3]{brezis} to get, using Kato's inequality,  that $\varphi=(|u|^2-1)^+$
   satisfies
   \begin{equation}
     \label{eq:58}
     \Delta\varphi\geq\varphi^2~\text{ in }B_R.
   \end{equation}
The result then
   follows by the Keller-Osserman theory (see \cite{brezis2} and the
   references therein).

 Once \eqref{eq:58a} is established, \eqref{eq:58b} follows by standard
 elliptic estimates.
 \end{proof}
The second is a version of the monotonicity formula
(\cite[Corollary~II.1]{bbo}):
\begin{lemma}
  \label{lem:mono}
Let $u$ be a solution of \eqref{eq:81} in $B_R\subset{\mathbb R}^N$. Then, the function $r
\mapsto E(u;B_r)/r^{N-2}$ is nondecreasing in $(0,R)$.
\end{lemma}
Another useful result is the following estimate for harmonic functions
in balls.
\begin{lemma}
Let $w\in C^1(\overline{B}_R)$ be a harmonic function in a ball $B_R\subset{\mathbb R}^N$. Then
\begin{equation}
\label{eq:harmonic}
\int_{B_R} |\nabla w|^2\leq \frac{R}{N-1}\int_{\partial B_R} |\nabla_T w|^2\,.
\end{equation}
\end{lemma}
\begin{proof}
First note that since $|\nabla w|^2$ is subharmonic we have
\begin{equation*}
\int_{B_R} |\nabla w|^2\leq\int_0^R \left(\frac{r^{N-1}}{R^{N-1}}\int_{\partial B_R} |\nabla w|^2\right)\,dr=\frac{R}{N}\int_{\partial B_R} |\nabla w|^2\,,
\end{equation*}
i.e.,
\begin{equation}
\label{eq:A1}
N\int_{B_R} |\nabla w|^2\leq R\int_{\partial B_R} \Big(|\nabla_T w|^2+\big|\frac{\partial w}{\partial n}\big|^2\Big)\,.
\end{equation}
On the other hand, Pohozaev identity gives
\begin{equation}
\label{eq:A2}
(N-2)\int_{B_R} |\nabla w|^2=R\int_{\partial B_R} \Big(|\nabla_T w|^2-\big|\frac{\partial w}{\partial n}\big|^2\Big)\,.
\end{equation}
Adding \eqref{eq:A1} with \eqref{eq:A2} yields \eqref{eq:harmonic}.
\end{proof}
\section{Proof of \rprop{prop}}
The next lemma deals with a Ginzburg-Landau minimization problem on a
spherical disc. The proof requires simple modification of the methods
developed for the case of the usual Ginzburg-Landau energy in the
plane with zero degree boundary condition (see \cite{bbh1}).
\begin{lemma}
  \label{lem:ext}
Let $u\in H^1({\widetilde D}_\rho^R(a),{\mathbb R}^2)$ with
\begin{align}
  \label{eq:1}
  \rho/R<1/10\\
\intertext{and}
 \label{eq:tu-energy}
 E^{(T)}(u;{\widetilde C}_\rho^R(a))\leq c_1/\rho\,.
 \end{align}
Then, if $\rho\geq R_0(c_1)$ we have
\begin{equation}
  \label{eq:2}
  \big||u|-1\big|\leq 1/8~\text{ on }{\widetilde C}_\rho^R(a)\,.
\end{equation}
If we further assume that
 \begin{equation}
 \label{eq:tu-degre}
\operatorname{deg}(u,{\widetilde C}_\rho^R(a)) = 0\,,
\end{equation}
then for $\rho\geq R_1(c_1)\geq R_0(c_1) $, any minimizer $v$ of the energy
 $E(\cdot;{\widetilde D}_\rho^R(a))$ for the boundary data
$u$ on ${\widetilde C}_\rho^R(a)$, satisfies:
\begin{equation}
   \big|1-|v|\big|\leq 1/8~\text{ on }
 {\widetilde D}_\rho^R(a)\,.\label{eq:5}
\end{equation}
\end{lemma}
\begin{proof}
The proof of \eqref{eq:2} is easy and standard (see \cite[Lemma
1]{lin}). Indeed, assume that for some point
$x_0\in {\widetilde C}_\rho={\widetilde C}_\rho^R(a)$ we have
\begin{equation}
  \label{eq:3}
  \big||u(x_0)|-1\big|>1/8\,.
\end{equation}
Then, for any $x_1\in {\widetilde C}_\rho$ we have, using H\"older inequality on the arc of
circle  $A(x_0,x_1)\subset {\widetilde C}_\rho$ and \eqref{eq:tu-energy}:
\begin{equation*}
  \big|u(x_0)-u(x_1)\big|\leq \Big(2 E^{(T)}(u;{\widetilde C}_\rho)\Big)^{1/2}|A(x_0,x_1)|^{1/2}\leq\Big(\frac{2c_1}{\rho}\Big)^{1/2}|A(x_0,x_1)|^{1/2}\,.
\end{equation*}
It follows, using \eqref{eq:3},  that there exists $\lambda>0$ such that
\begin{equation}
\label{eq:69}
  \big||u|-1\big|\geq 1/16~\text{ on }\{|x-x_0|<\lambda \rho/ c_1\}\cap {\widetilde C}_\rho\,.
\end{equation}
 Clearly, \eqref{eq:69} implies, for some positive constant $c_2$, that
 \begin{equation*}
   \int_{{\widetilde C}_\rho} (1-|u|^2)^2\geq c_2\rho\,,
 \end{equation*}
which contradicts \eqref{eq:tu-energy} for $\rho$ large
enough.

For the proof of \eqref{eq:5} it is convenient to treat an equivalent problem for a
certain weighted Ginzburg-Landau energy on a planar disc. For that
matter we will perform a change of variables in several steps. Without loss of generality we may assume that $a=(0,\ldots,0,-R)$, the
south pole of $S_R$. Denoting by $\mathcal{S}:S_1\to{\mathbb R}^2$ the standard
stereographic projection
we define a function $\tilde u:D_{\tan(\rho/2R)}\to{\mathbb R}^2$ by $\tilde u(x)=u(R\mathcal{S}^{-1}x)$.
We have
\begin{equation*}
  E(u; {\widetilde D}_\rho^R(a))=
\int_{D_{\tan(\rho/2R)}} \Big\{\frac{1}{2}|\nabla\tilde
u|^2+\frac{R^2(1-|\tilde u|^2)^2}{(1+|x|^2)^2}\Big\}dx\,.
\end{equation*}
 A final rescaling, setting $U(y)=\tilde u(\tan(\rho/2R)y)$, yields
 $U:D_1\to{\mathbb R}^2$ satisfying
\begin{equation*}
  F_\varepsilon(U;D_1)=E(u; {\widetilde D}_\rho^R(a))
\end{equation*}
with
\begin{equation*}
  F_\varepsilon(U;D_1):=\int_{D_1} \Big\{\frac{1}{2}|\nabla U|^2+\frac{p(y)}{4\varepsilon^2}(1-|U|^2)^2\Big\}\,dy\,,
\end{equation*}
where
$$
\varepsilon=1/(2R\tan(\rho/2R))~\text{ and
}~p(y)=1/(1+\tan^2(\rho/2R)|y|^2)^2\,.
$$
Note that by \eqref{eq:1} $p(y)$ is bounded between two positive
constants, and all its derivatives are uniformly bounded as well.
Moreover, by \eqref{eq:tu-energy} we have
\begin{equation*}
  \int_{\partial D_1} \Big\{\frac{1}{2}|\nabla
  U|^2+\frac{p(y)}{4\varepsilon^2}(1-|U|^2)^2\Big\}\leq \tilde c_1\,,
\end{equation*}
where $\tilde c_1$ depends on $c_1$ only.

The proof of \eqref{eq:5} follows by a simple modification of the
arguments in \cite[Thm~2]{bbh1} (in particular the proof of $(95)$
there), see also \cite[Lemma 2]{lin}; the fact that here we deal with
a weighted Ginzburg-Landau energy causes no difficulty since the
weight is smooth, as explained above.

\end{proof}
The next lemma is concerned with an extension problem in a
cylinder. It will be useful for a similar problem on a spherical
cylinder in the course of the proof of \rprop{prop}.
\begin{lemma}
  \label{lem:cyl}
 Let $u,v\in H^1(D_R,{\mathbb R}^2)$ be such that
 \begin{align}
   \label{eq:85}
   u&=v \text{ on }C_R\,,\\
\big|1-|v|\big|&\leq 1/8\text{ and }|u|\leq K~\text{ on }\overline{D}_R\,.\label{eq:86}
 \end{align}
  Then, for any
$H>0$ there
exists $U\in H^1(D_R\times(0,H))$ satisfying
\begin{align}
  U(x,H)&=u(x),\quad x\in D_R\label{eq:7}\\
 U(x,0)&=v(x),\quad x\in D_R\label{eq:8}\\
 U(x,z)&=u(x)=v(x),\quad x\in\partial D_R,\,z\in(0,H)
\intertext{and}
E(U;D_R\times(0,H))&\leq C(H+R^2/H)\,(E(u;D_R)+E(v;D_R))\,.\label{eq:49}
\end{align}
\end{lemma}

\begin{proof}
  We first extend $v$ to the cylinder $\overline{D}_R\times[0,H]$ by letting
  \begin{equation}
    \label{eq:32}
    V(x,z)=v(x)\,,\quad x\in \overline{D}_R,~z\in[0,H]\,.
  \end{equation}
Define the cone $\Gamma$ by
\begin{equation}
  \label{eq:34}
 \Gamma=\{(x,z)\,;\,0<|x|<R\,,\,(H/R)|x|<z<H\}\,.
\end{equation}
Let $w$ be defined in $\overline{D}_R$ by $w=u/v$ and then extend
it to $\overline{D}_R\times[0,H]$ by
\begin{equation}
  \label{eq:33}
W(x,z)=\begin{cases}1 & (x,z)\in \overline{D}_R\times[0,H]\setminus\Gamma\\
w(Hx/z) & (x,z)\in\Gamma
\end{cases}\,.
\end{equation}
Finally we set
\begin{equation}
  \label{eq:35}
U=VW\quad\text{ in }\overline{D}_R\times[0,H]\,.
\end{equation}
Clearly,
\begin{equation}
  \label{eq:36}
 E(W;\overline{D}_R\times[0,H])=E(W;\Gamma)=\int_0^Hdz\,\int_{D_{Rz/H}\times\{z\}}
 \frac{1}{2}|\nabla W|^2+\frac{1}{4}(1-|W|^2)^2\,.
\end{equation}
First note that
\begin{equation}
  \label{eq:37}
  \int_{D_{Rz/H}\times\{z\}}|\nabla_x W|^2=\int_{D_{R}}|\nabla w|^2\,,
\end{equation}
 while
 \begin{equation}
   \label{eq:38}
\begin{aligned}
  \int_{D_{Rz/H}\times\{z\}}\Big|\frac{\partial W}{\partial z}\Big|^2&=
\int_{D_{Rz/H}}\big|\nabla w(Hx/z)\cdot(Hx/z^2)\big|^2\\
&\leq
\frac{1}{H^2} \int_{D_R} |\nabla w(y)|^2|y|^2\,dy\leq
\Big(\frac{R}{H}\Big)^2 \int_{D_R} |\nabla w|^2\,.
\end{aligned}
 \end{equation}
Integrating \eqref{eq:37}--\eqref{eq:38} over $z\in(0,H)$ yields
\begin{equation}
  \label{eq:39}
\int_{D_R\times(0,H)}|\nabla W|^2\leq \big(H+R^2/H\big) \int_{D_R} |\nabla w|^2\,.
\end{equation}
 Since by \eqref{eq:86}
\begin{equation}
\label{eq:uvw}
|\nabla w|\leq C\big(|\nabla u|+|\nabla v|)\,,
\end{equation}
 we deduce from \eqref{eq:39} that
\begin{equation}
 \label{eq:40}
\int_{D_R\times(0,H)}|\nabla W|^2\leq C\big(H+R^2/H\big) \Big(\int_{D_R} |\nabla u|^2+\int_{D_R}|\nabla v|^2\Big)\,.
\end{equation}
Next we turn to estimate the second term in the energy. We have
\begin{multline}
  \label{eq:41}
\int_{D_{Rz/H}}\big(|W(x,z)|^2-1\big)^2\,dx=\int_{D_{Rz/H}}\big(|w(Hx/z)|^2-1\big)^2\,dx\\=(z/H)^2\int_{D_R}(|w(y)|^2-1)^2\,dy\,,
\end{multline}
so integrating \eqref{eq:41} for $z\in(0,H)$ gives
\begin{equation}
  \label{eq:42}
  \int_\Gamma\big(|W|^2-1\big)^2=(H/3) \int_{D_R} (|w(y)|^2-1)^2\,dy\,.
\end{equation}
Noting that
\begin{equation}
  \label{eq:46}
 \big||w|^2-1\big|=\frac{\big||u|^2-|v|^2\big|}{|v|^2}\leq (8/7)^2\Big(\big||u|^2-1\big|+\big||v|^2-1\big|\Big)\,,
\end{equation}
we conclude from \eqref{eq:42} that
\begin{equation}
  \label{eq:47}
 \int_\Gamma\big(|W|^2-1\big)^2\leq CH\Big(\int_{D_R} (|u|^2-1)^2+\int_{D_R} (|v|^2-1)^2\Big)\,.
\end{equation}
We also clearly have
\begin{align}
  \int_{D_R\times(0,H)}|\nabla V|^2&=H\int_{D_R}|\nabla v|^2\,,\label{eq:43}\\
\int_{D_R\times(0,H)}(1-|V|^2)^2&=H\int_{D_R} (1-|v|^2)^2\,.\label{eq:44}
\end{align}
Similarly to \eqref{eq:uvw} we have $|\nabla U|\leq c(|\nabla
V|+|\nabla W|)$. Hence,
from \eqref{eq:40} and \eqref{eq:43} we get that
\begin{equation}
  \label{eq:45}
\int_{D_R\times(0,H)}|\nabla U|^2\leq C\big(H+R^2/H\big) \Big(\int_{D_R} |\nabla u|^2+\int_{D_R}|\nabla v|^2\Big)\,.
\end{equation}
By \eqref{eq:35}, \eqref{eq:86} and \eqref{eq:32}  we have
\begin{equation*}
  \big||U|^2-1\big|\leq
  \big||V|^2|W|^2-|V|^2\big|+\big||V|^2-1\big|\leq (9/8)^2\big||W|^2-1\big|+\big||V|^2-1\big|\,,
\end{equation*}
and applying \eqref{eq:44} and \eqref{eq:47} yields
\begin{equation}
  \label{eq:48}
\int_{D_R\times(0,H)}(1-|U|^2)^2\leq CH\Big(\int_{D_R} (1-|v|^2)^2+\int_{D_R} (1-|u|^2)^2\Big)\,.
\end{equation}
Clearly \eqref{eq:49} follows from \eqref{eq:45}--\eqref{eq:48}.
\end{proof}
\begin{proof}[Proof of \rprop{prop}]
  The proof is divided to several steps.\\[2mm]
\noindent{\underline{Step 1:}} Locating the ``bad discs'' on
$S_R$ and choosing $\alpha$.\\[2mm]

The following result should seem plausible to specialists in the field; we state it as a Lemma and prove it in the appendix.

\begin{lemma}
\label{lem:specialists} Given $\gamma < 2\pi$ and $\Lambda>0$ there exists $R_0>0$ such that if $R>R_0$ and $u:S_R\to{\mathbb R}^2$ satisfies
$$ E^{(T)}(u;S_R)\leq \gamma\ln R,$$
then there exist $\alpha\in(0,1)$ and spherical discs
${\widetilde D}_i={\widetilde D}_{r_i}^R(a_i)$, $i = 1,\dots, k$, such that
\begin{align}
 & |u|>7/8 \text{ on }S_R\setminus \bigcup\limits_{i=1}^k{\widetilde D}_i\,,\label{eq:p1}\\
 &\displaystyle \sum_{i=1}^k r_i \le R^{\alpha}\,,\label{eq:p2}\\
 & E^{(T)}(u,\partial {\widetilde D}_i) \le \displaystyle\frac{2\pi}{r_i}\,,~i=1,\ldots,k\,, \label{eq:p3}\\
 &\operatorname{deg}(u,\partial {\widetilde D}_i) = 0\,,~i=1,\ldots,k\,,\label{eq:p4}\\
 &r_i\geq  \Lambda\,,~i=1,\ldots,k\,. \label{eq:p5}
 \end{align}
\end{lemma}
Applying \rlemma{lem:specialists} with
$\Lambda=R_1(2\pi)$ (see \rlemma{lem:ext}) yields a collection
$\{{\widetilde D}_{ r_i}^R(a_i)\}_{i=1}^k$ satisfying \eqref{eq:p1}--\eqref{eq:p5}.
\vskip 0.2cm

\noindent{\underline{Step 2:}} Extension to a map in $B_R\setminus
B_{R-R^\alpha}$ with a phase on $S_{R-R^\alpha}$.\\[2mm]
\noindent  For each $i=1,\ldots,k$ we apply \rlemma{lem:ext} with
$c_1=2\pi$ to find a map  $v_i$ defined in ${\widetilde D}_i={\widetilde D}_{ r_i}^R(a_i)$
satisfying:
\begin{flalign}
  &v_i=u~\text{ on }\partial {\widetilde D}_i\,,\label{eq:18}\\
 &\big||v_i|-1\big|\leq 1/8~\text{ on }{\widetilde D}_i\,,\label{eq:19}\\
&E^{(T)}(v_i;{\widetilde D}_i)\leq E^{(T)}(u;{\widetilde D}_i)\,.\label{eq:20}
\end{flalign}
Note that for any spherical cylinder of the form
 $$
\mathcal{C}=\Big\{
(r\sin\varphi\cos\theta,r\sin\varphi\sin\theta,\cos\varphi)\,;\,
r\in(R-H,R), \varphi\in[0,\rho/R),\theta\in[0,2\pi)\Big\}
$$
there is a $C^1$ diffeomorphism
$\Psi:D_\rho\times(0,H)\to\mathcal{C}$ given by
\begin{equation*}
  \Psi(y_1,y_2,y_3)=\Big(\frac{y_3+R-H}{R}\Big)\big(y_1,y_2,\sqrt{R^2-y_1^2-y_2^2}\big)\,.
\end{equation*}
A direct computation yields that
\begin{equation}
  \label{eq:6}
  D\Psi=\text{Id}+O(\rho/R)+O(H/R)~\text{ and }D(\Psi^{-1})=\text{Id}+O(\rho/R)+O(H/R)\,.
\end{equation}

Consider each spherical cylinder
$$
\mathcal{C}_i=\Big\{ ry\,;\,y\in {\widetilde D}_i,\,r\in\big(\frac{R-R^\alpha}{R},1\big)\Big\}\,,\quad i=1,\ldots,k\,.
$$
By applying a rotation (sending $a_i$ to $\mathcal{N}$) and then the map $\Psi^{-1}$ we  find a
regular cylinder $D_{r_i}\times(0,R^\alpha)$ on which we perform the
construction of \rlemma{lem:cyl}. Going back to $\mathcal{C}_i$, we get, taking into account \eqref{eq:6},
 a map $U_i\in H^1(\mathcal{C}_i,{\mathbb R}^2)$ satisfying:
\begin{align}
  U_i(x)&=u(x), \quad x\in {\widetilde D}_i\,,\label{eq:21}\\
 U_i(x)&=v_i\big(Rx/(R-R^\alpha)\big), \quad x\in \big((R-R^\alpha)/R\big){\widetilde D}_i\subset S_{R-R^\alpha}\,,\label{eq:22}\\
 U_i(x)&=u\big(Rx/|x|\big)=v_i\big(Rx/|x|\big),
         \quad \frac{Rx}{|x|}\in\partial {\widetilde D}_i\,,\label{eq:23}
\intertext{and}
E(U_i;\mathcal{C}_i)&\leq C(R^\alpha+r_i^2/R^\alpha)\,E^{(T)}(u;{\widetilde D}_i)\leq
            CR^\alpha\,E^{(T)}(u;{\widetilde D}_i)\,.\label{eq:24}
\end{align}
Denoting by $\mathcal{P}_R(x)=Rx/|x|$ the radial projection on $S_R$, we finally
define $U$ in $B_R\setminus B_{R-R^\alpha}$ by
\begin{equation}
  \label{eq:25}
 U(x)=\begin{cases}
 U_i(x) & \text{ if }\mathcal{P}_R(x)\in {\widetilde D}_i \text{ for some }i\\
  u(\mathcal{P}_R(x)) & \text{otherwise}
\end{cases}
\,.
\end{equation}
We clearly have
\begin{equation}
  \label{eq:26}
\begin{aligned}
  U(x)&=u(x) \text{ on }S_R\\
  V(x)&:=U\big|_{S_{R-R^\alpha}}(x) \text{ satisfies
  }7/8\leq |V(x)|\leq K~\text{ on
  }S_{R-R^\alpha}
\end{aligned}
\,.
\end{equation}
Moreover,
\begin{align}
\label{eq:27}
E^{(T)}(V;S_{R-R^\alpha})&\leq E^{(T)}(u;S_{R})\,,\\
\label{eq:28}
E(U;B_R\setminus B_{R-R^\alpha})&\leq CR^\alpha\ln R\,.
\end{align}

Indeed, \eqref{eq:27} follows by summing up the contribution of each
${\widetilde D}_i$ in  \eqref{eq:20} and using \eqref{eq:25}. The inequality
 \eqref{eq:28} follows from \eqref{eq:24}--\eqref{eq:25} and \eqref{eq:ineq-prop}.\\[2mm]
\noindent{\underline{Step 3:}} Extension in $
B_{R-R^\alpha}$.\\[2mm]
 On $S_{R-R^\alpha}$ we may write
 \begin{equation}
   U(x)=\tilde\rho(x)e^{i\tilde\varphi(x)}\,,
 \end{equation}
 with $7/8\leq \tilde\rho\leq K$.
We first extend to $U(x)=\rho(x)e^{i\varphi(x)}$ in
$A(R):=B_{R-R^\alpha}\setminus \overline{B}_{R-2R^\alpha}$ by setting:
\begin{align}
  \rho(x)&=\frac{r-(R-2R^\alpha)}{R^\alpha}\tilde\rho\big((R-R^\alpha)x/|x|\big)+\frac{(R-R^\alpha)-r}{R^\alpha}\label{eq:54}\\
\varphi(x)&=\varphi\big((R-R^\alpha)x/|x|\big)
\,,\label{eq:55}
\end{align}
where
$r=|x|\in(R-2R^\alpha,R-R^\alpha)$. A direct computation gives:
\begin{align}
\label{eq:29}
  \int_{A(R)}\Big|\frac{\partial\rho}{\partial r}\Big|^2&\leq
  \frac{C}{R^\alpha}\ln R\,,\\
 \label{eq:51}\int_{A(R)}\big|\nabla_T\rho\big|^2&\leq CR^\alpha\ln R\,,\\
 \label{eq:52}\int_{A(R)}(1-\rho^2)^2&\leq CR^\alpha\ln R\,,\\
\label{eq:53}\int_{A(R)}|\nabla\varphi|^2&\leq CR^\alpha\ln R\,.
\end{align}
 From \eqref{eq:29}--\eqref{eq:53} it follows that
 \begin{equation}
   \label{eq:30}
 E(U;A(R))\leq CR^\alpha\ln R\,.
 \end{equation}
Moreover, by \eqref{eq:55} and \eqref{eq:25}--\eqref{eq:26} we have
\begin{equation}
  \label{eq:56}
 \int_{S_{R-2R^\alpha}}|\nabla\varphi|^2=
 \int_{S_{R-R^\alpha}}|\nabla\varphi|^2\leq 2(8/7)^2E^{(T)}(u;S_R)\,.
\end{equation}

Finally, denoting by $\Phi$ the harmonic extension of $\varphi$ to
$B_{R-2R^\alpha}$, we set
\begin{equation*}
  U(x)=e^{i\Phi(x)}~\text{ in }B_{R-2R^\alpha}\,.
\end{equation*}
Combining \eqref{eq:harmonic},\eqref{eq:56}, \eqref{eq:28} and
\eqref{eq:30} we obtain
\begin{equation}
  \label{eq:57}E(U;B_R)\leq
 (1/2)(8/7)^2(R-2R^\alpha)E^{(T)}(u;S_R)+CR^\alpha\ln R\,,
\end{equation}
 that clearly implies \eqref{eq:prop} for $R$ large, since $E(u;B_R)\leq E(U;B_R)$.
\end{proof}

\section{Proof of the main results}
We begin with the proof of \rth{th:eta}. It will be
more convenient to prove instead the theorem under the following equivalent
formulation, obtained by rescaling:
\begin{theorem}
  \label{th:main}
For every $\gamma\in(0,2\pi)$ and $\lambda>0$ there exists
$R_1(\gamma,\lambda)$ such that for every $u$ which is a minimizer of $E$ on $B_R$,
with $R\ge R_1(\gamma,\lambda)$, satisfying $E(u;B_R)\leq \gamma R\ln R$
there holds
\begin{equation}
  \label{eq:59}
  \big||u(0)|-1\big|\leq \lambda\,.
\end{equation}
\end{theorem}

\begin{proof}[Proof of \rth{th:main}]  We will use the shorthand
  $E(R)$ for $E(u;B_R)$ and also
  \begin{align*}
    E'(R)&:=\frac{d}{dR}E(R)=\int_{S_R}\frac{1}{2}|\nabla
                            u|^2+\frac{1}{4}(1-|u|^2)^2\,,\\
  e_T(R)&:=E^{(T)}(u;S_R)=\int_{S_R}\frac{1}{2}|\nabla_T
                            u|^2+\frac{1}{4}(1-|u|^2)^2\,.
  \end{align*}
The value of $R_1=R_1(\gamma,\lambda)$ will be determined later. For the moment we
will assume its value is known and in the course of the proof we shall
obtain some constraints that will allow us to detrmine its value
definitively. We first fix $\varepsilon$ satisfying
\begin{equation}
  \label{eq:63}
  \gamma+3\varepsilon<2\pi\,.
\end{equation}
Next we take  any $\beta\in(0,1)$, e.g., $\beta=1/2$. Since
\begin{equation*}
  \int_{R^\beta}^RE'(r)\,dr\leq E(R)<\gamma R\ln
  R=\gamma\int_{R^\beta}^R (\ln r+1)\,dr+\gamma R^\beta\ln(R^\beta)\,,
\end{equation*}
 there exists $\rho_1\in[R^\beta,R]$ such that
 \begin{equation}
   \label{eq:61}
   E'(\rho_1)\leq \gamma\Big(\ln
   \rho_1+1+\frac{R^\beta\ln(R^\beta)}{R-R^\beta}\Big)\leq
   (\gamma+\varepsilon)\ln \rho_1\,,
 \end{equation}
 if $R_1$ is chosen large enough.
Since $e_T(\rho_1)\leq E'(\rho_1)\leq (\gamma+\varepsilon)\ln \rho_1$,
we may apply \rprop{prop} (with $\gamma+\varepsilon$ playing the role
of $\gamma$) to get
\begin{equation}
  \label{eq:62}
  E(\rho_1)\leq \sigma\rho_1e_T(\rho_1)+C\rho_1^\alpha\ln\rho_1\leq \sigma(\gamma+2\varepsilon)\rho_1\ln\rho_1\,,
\end{equation}
provided $R_1$ is large enough. Next we fix $\delta\in(0,1)$ satisfying
\begin{equation}
  \label{eq:64}
  \gamma_0:=\frac{(\gamma+3\varepsilon)}{\delta}<2\pi\,.
\end{equation}
By \eqref{eq:62} we have, for every $r\in[\delta\rho_1,\rho_1]$,
\begin{equation}
  \label{eq:65}
  E(r)\leq
  E(\rho_1)\leq\frac{\sigma(\gamma+2\varepsilon)}{\delta}\,\delta\rho_1\ln\rho_1
<  (\sigma\gamma_0)\delta\rho_1\,\ln(\delta\rho_1)\leq (\sigma\gamma_0)\,r\ln r\,,
\end{equation}
again, if $R_1$ is chosen large enough. Let $M>0$ denote a constant
such that for $R\geq M$ all the inequalities,
\eqref{eq:61},\eqref{eq:62} and \eqref{eq:65} hold true.
Set
\begin{equation}
  \label{eq:66}
  \rho_2:=\inf\{R\in[\widetilde{R}_0,\rho_1]\,:\,E(s)<(\sigma\gamma_0)\,s\ln
  s,\,\forall s\in[R,\rho_1]\}\,,
\end{equation}
 where $\widetilde{R}_0=\widetilde R_0(\gamma_0,\sigma)\geq
 r_0(\gamma_0)$ (see \rprop{prop}) will be
 determined below.
Thanks to \eqref{eq:65} we have $\rho_2\leq\delta\rho_1$.
For $r\in[\rho_2,\rho_1]$ we have, either
\begin{equation}
  \label{eq:67}
  E'(r)\geq \frac{E(r)}{\sigma r}\,,
\end{equation}
or
\begin{equation}
  \label{eq:68}
  E'(r)<\frac{E(r)}{\sigma r}\,.
\end{equation}
 In the latter case, when \eqref{eq:68} holds, we have by \eqref{eq:66},
 \begin{equation*}
   e_T(r)\leq E'(r)<\frac{E(r)}{\sigma r}\leq\gamma_0 \ln r\,.
 \end{equation*}
Applying \rprop{prop} yields
\begin{equation}
  \label{eq:70}
  E'(r) \ge e_T(r)\geq \frac{E(r) - Cr^\alpha\ln r}{\sigma r}.
\end{equation}
The first possibility \eqref{eq:67} clearly implies
\eqref{eq:70}. Hence,   for {\em every} $r\in[\rho_2,\rho_1]$, \eqref{eq:70} holds.
We rewrite \eqref{eq:70} as
\begin{equation}
  \label{eq:71}
  (r^{-1/\sigma}E(r))'\ge -\frac C\sigma r^{\alpha - 1 - 1/\sigma}\ln r\,.
\end{equation}
Integrating \eqref{eq:71} for
 $r\in[\rho_2,\rho_1]$ gives
 \begin{equation}
\label{eq:73}
   \frac{E(\rho_1)}{\rho_1^{1/\sigma}}-\frac{E(\rho_2)}{\rho_2^{1/\sigma}}\geq
-\frac C\sigma\left( \frac{{\rho_1}^{\alpha - 1/\sigma}}{\alpha -
    1/\sigma}\ln \rho_1 -
\frac{{\rho_1}^{\alpha - 1/\sigma}}{(\alpha - 1/\sigma)^2} - \frac{{\rho_2}^{\alpha - 1/\sigma}}{\alpha - 1/\sigma}\ln \rho_2 + \frac{{\rho_2}^{\alpha - 1/\sigma}}{(\alpha - 1/\sigma)^2}\right)\,,
 \end{equation}
whence by \eqref{eq:65},
\begin{equation}
  \label{eq:72}
\begin{aligned}
  E(\rho_2)\leq &\bigl(\frac{\rho_2}{\rho_1}\bigr)^{1/\sigma}
  E(\rho_1)+C(\alpha,\sigma)\rho_2^\alpha\ln\rho_2
    \leq
    \bigl(\frac{\rho_2}{\rho_1}\bigr)^{1/\sigma}\sigma\gamma_0\,(\delta\rho_1)\ln(\delta\rho_1)+C(\alpha,\sigma)\rho_2^\alpha\ln\rho_2\\
&= (\sigma\gamma_0)\delta^{1/\sigma}\rho_2^{1/\sigma}
(\delta\rho_1)^{1-1/\sigma}\ln(\delta\rho_1)
+C(\alpha,\sigma)\rho_2^\alpha\ln\rho_2\,.
\end{aligned}
\end{equation}
Using $\rho_2\leq\delta\rho_1$ and the fact that the function
$t\mapsto (\ln t)t^{1-1/\sigma}$ is decreasing for $t\geq c_0$, we obtain from \eqref{eq:72} that
\begin{equation}
  \label{eq:74}
\begin{aligned}
  E(\rho_2)&\leq
  (\sigma\gamma_0)\delta^{1/\sigma}\rho_2^{1/\sigma}\cdot\rho_2^{1-1/\sigma}\ln \rho_2
+C(\alpha,\sigma)\rho_2^\alpha\ln\rho_2\\
&= (\sigma\gamma_0)\delta^{1/\sigma}\rho_2\ln \rho_2+C(\alpha,\sigma)\rho_2^\alpha\ln\rho_2\,.
\end{aligned}
\end{equation}
Let $r_1$ denote the value of $\rho_2\neq 1$ for which
equality holds between the right hand sides of \eqref{eq:74} and
\eqref{eq:66} (for $s=\rho_2$). That is, $r_1$
satisfies  the equality
\begin{equation}
\label{eq:31}
  (\sigma\gamma_0)\,r_1=
  (\sigma\gamma_0)\delta^{1/\sigma}r_1+C(\alpha,\sigma)r_1^\alpha\,.
\end{equation}
A simple computation gives
\begin{equation}
  \label{eq:4}
  r_1=\left(\frac{C(\alpha,\sigma)}{\sigma\gamma_0(1-\delta^{1/\sigma})}\right)^{\frac{1}{1-\alpha}}\,,
\end{equation}
 and we may indeed assume that $r_1>1$ by replacing,
 if necessary, $C(\alpha,\sigma)$ by a larger constant.
Note that
\begin{equation}
\label{eq:82}
  (\sigma\gamma_0)\,\rho\ln\rho>
 (\sigma\gamma_0)\delta^{1/\sigma}\rho\ln
 \rho+C(\alpha,\sigma)\rho^\alpha\ln\rho\quad\text{ for }\rho>r_1\,.
\end{equation}
 Next we claim that if we take in \eqref{eq:66} $\widetilde{R}_0=\widetilde
R_0(\gamma)$ satisfying
\begin{equation}
  \label{eq:87}
  \widetilde{R}_0\geq \max\{r_0(\gamma_0),r_1,M\}\,,
\end{equation}
 then necessarily
\begin{equation}
  \label{eq:75}
  \rho_2=\widetilde{R}_0\,.
\end{equation}
Indeed, otherwise $\rho_2>\widetilde R_0$
and by \eqref{eq:66} and \eqref{eq:82} we must have
\begin{equation}
  \label{eq:76}
  E(\rho_2)=(\sigma\gamma_0)\rho_2\ln \rho_2>(\sigma\gamma_0)\delta^{1/\sigma}\rho_2\ln
 \rho_2+C(\alpha,\sigma)\rho_2^\alpha\ln\rho_2\,,
\end{equation}
which contradicts the bound in \eqref{eq:74}.

In view of \eqref{eq:75} we know that \eqref{eq:71} holds for all $r\in
[\widetilde{R}_0,\rho_1]$, and integration over this interval yields,
as in \eqref{eq:73}--\eqref{eq:72},
\begin{equation}
  \label{eq:77}
\begin{aligned}
  E(\widetilde{R}_0)&\leq
  \sigma\gamma_0\delta\rho_1^{1-1/\sigma}{\widetilde{R}_0}^{1/\sigma}\ln(\delta\rho_1)+C(\alpha,\sigma){\widetilde{R}_0}^\alpha\ln\widetilde{R}_0\\
&= \sigma\gamma_0\delta{\widetilde{R}_0}^\alpha\Big({\widetilde{R}_0}^{1/\sigma-\alpha}/\rho_1^{1/\sigma-1}\Big) \ln(\delta\rho_1)+C(\alpha,\sigma){\widetilde{R}_0}^\alpha\ln\widetilde{R}_0\,.
\end{aligned}
\end{equation}
Let $\widetilde{R}_1=\widetilde{R}_1(\widetilde{R}_0)$ be determined by the equation
\begin{equation}
  \label{eq:877}
  \widetilde{R}_0^{1/\sigma-\alpha}=\widetilde{R}_1^{\beta(1/\sigma-1)}/\ln (\widetilde{R}_1^\beta)\;(\leq \rho_1^{1/\sigma-1}/\ln\rho_1)\,.
\end{equation}
It follows from  \eqref{eq:77} that for every $R_1\geq \widetilde{R}_1(\widetilde{R}_0)$,
where $\widetilde{R}_0$ satisfies \eqref{eq:87}, we have (under the assumption
of the Theorem, i.e., $E(R)\leq \gamma R\ln R$ for some $R\geq R_1$):
\begin{equation}
  \label{eq:78}
  E(\widetilde{R}_0)\leq
  \sigma\gamma_0\delta{\widetilde{R}_0}^\alpha+C(\alpha,\sigma){\widetilde{R}_0}^\alpha\ln\widetilde{R}_0\leq \widetilde{C}{\widetilde{R}_0}^\alpha\ln\widetilde{R}_0\,,
\end{equation}
for some $\widetilde{C}$ (which actually depends only on $\gamma$).

Finally we turn to the proof of \eqref{eq:59}. It is clearly enough to consider $\lambda<2K$ (where $K$ is given by
\rlemma{lem:L-infty}). Looking for contradiction, assume that
\begin{equation}
\label{eq:89}
  \big||u(0)|-1\big|>\lambda\,.
\end{equation}
Then, by \rlemma{lem:L-infty},
\begin{equation}
\label{eq:79}
  \big||u(x)|-1\big|>\lambda/2\text{
    in }B_{\lambda/2K}\,.
\end{equation}
By \eqref{eq:78} and \eqref{eq:79} and \rlemma{lem:mono} we obtain
\begin{equation}
  \label{eq:80}
\frac{1}{4}\big(\frac{\lambda}{2}\big)^2\big(\frac{\lambda}{2K}\big)^3|B_1|\leq
E(u;B_{\lambda/2K})\leq E(u;B_1)\leq E(\widetilde{R}_0)/\widetilde{R}_0\leq \widetilde{C}{\widetilde{R}_0}^{\alpha-1}\ln\widetilde{R}_0\,.
\end{equation}
Let $T=T(\lambda,\alpha)$ be a large enough value of $\widetilde{R}_0$ for which
\eqref{eq:80} is violated, i.e.,
\begin{equation}
  \label{eq:88}
  \frac{\lambda^5|B_1|}{128K^3}>\widetilde{C} T^{\alpha-1}\ln T.
\end{equation}
Therefore, taking $\widetilde{R}_0=\widetilde{R}_0(\gamma,\lambda)$ satisfying both \eqref{eq:87}
and $\widetilde{R}_0\geq T$, and then setting  $R_1(\gamma,\lambda):=\widetilde{R}_1(\widetilde{R}_0)$ (see
\eqref{eq:877}),  we see that  \eqref{eq:80}, and hence also \eqref{eq:89}, cannot hold for $R\geq
R_1(\gamma,\lambda)$.
 \end{proof}
 \begin{proof}[Proof of \rth{th:lm}]
   Fix any point $x\in{\mathbb R}^3$. For each $\lambda>0$ we may apply
   \rth{th:main} on $B_{R_n}(x)$ for an appropriate sequence $R_n\to\infty$  to conclude
   that $|u(x)|=1$. Hence $|u|\equiv 1$ in ${\mathbb R}^3$ and from
   \eqref{eq:81} we deduce that both components of
   $u=(u_1,u_2)$ are harmonic functions. Assuming without loss of
   generality that $u(0)=(1,0)$, we conclude from the maximum principle
   that $u_1\equiv 1$ and therefore $u\equiv (1,0)$ in ${\mathbb R}^3$.
 \end{proof}

 \section*{Appendix. Proof of \rlemma{lem:specialists}}
 \setcounter{equation}{0}
 \renewcommand{\theequation}{\Alph{section}.\arabic{equation}}
  \setcounter{section}{1}

  \renewcommand{\thelemma}{A}

 \begin{proof}
 We let $\varepsilon = 1/R$ and $v(x) = u(Rx)$. Then $v:S_1\to{\mathbb R}^2$ satisfies
  \begin{equation}\label{bg}E_\varepsilon^{(T)}(v;S_1)\leq \gamma\ln\frac1\varepsilon\,.\end{equation}
  Following \cite{bouquin}, proof of Theorem 5.3, given $\delta>0$
  there exists a collection of disjoint spherical discs that we denote
  $\mathcal D_0$ such that each disc in the family has radius bounded
  below by $\Lambda\varepsilon$, such
  that $|v|> 1 - \delta$ on the complement of $\mathcal D_0$ and such
  that, denoting by $r_0$ the sum of the radii of the discs, we have
  \begin{equation}
    \label{eq:84}
    r_0 \le C\gamma|\ln\varepsilon| \frac{\varepsilon}{\delta^3}\,.
  \end{equation}
  It should be noted that in contrast with the situation considered in
  \cite{bouquin}, the restriction of $v$ to $S_1$ does not necessarily
  satisfy the
  Ginzburg-Landau equation (which would involve the Laplace-Beltrami
  operator on $S_1$). However, \eqref{eq:58b} implies that the restriction of
  $v$ to $S_1$ satisfies an estimate of the form $\|\nabla v\|_{L^\infty}\leq
  C/\varepsilon$, which is what needed to construct $\mathcal D_0$
  using the method of \cite{bouquin}.
  Then we apply the ball-growth procedure as in
  \cite[Thm~4.2]{bouquin}. This yields, for any $t>0$, a family of spherical discs $\mathcal D(t)$ which is increasing and such that the sum of the radii of the discs in $\mathcal D(t)$ is $e^t r_0$. Moreover, denoting
  $$
 \mathcal F(x,r) = E_\varepsilon^{(T)}(v;{\widetilde D}_r(x))
~\text{ and }~\mathcal{F}(\mathcal D(t))=\sum_{{\widetilde D}_r(x)\in\mathcal D(t)}\mathcal F(x,r) \,,
 $$
  we deduce from \cite[Proposition~4.1]{bouquin} and \eqref{bg}
  that, for all $s>0$,
  $$
\gamma|\ln\varepsilon| \ge \mathcal F(\mathcal D(s)) -  \mathcal
F(\mathcal D(0)) \ge \int_0^s \sum_{{\widetilde D}_r(x)\in\mathcal D(t)}
r\frac{\partial\mathcal F}{\partial r} (x,r)\,dt\,.
$$
  Hence, by Fubini's theorem  there exists $t\in(0,s)$ such that $\mathcal D(t) = \{{\widetilde D}_{\rho_i}= {\widetilde D}_{\rho_i}(x_i), 1\le i\le k\}$ satisfies
 $$
\sum_{i=1}^k \rho_i\frac{\partial\mathcal F}{\partial r} (x_i,\rho_i) \le
 \frac{ \gamma|\ln\varepsilon|}s\,.
$$
 Choosing
 $ s = \frac{2\pi +\gamma}{4\pi} |\ln\varepsilon|$,
 we find that,
 \begin{equation}
\label{infpi}
 \sum_{i=1}^k \rho_i\frac{\partial\mathcal F}{\partial r} (x_i,\rho_i) \le
 2\pi\big(\frac{2\gamma}{\gamma+ 2\pi}\big)\,.
\end{equation}
 On the other hand, since $|v|> 1 - \delta$ on the complement of
 $\mathcal D_0$, we have for each $i$
 $$\frac{\partial\mathcal F}{\partial r} (x_i,\rho_i) =
 E_\varepsilon^{(T)}(v;\partial{\widetilde D}_{\rho_i}) \ge \frac{\pi}{\sin \rho_i}
 \operatorname{deg}(v,\partial{\widetilde D}_{\rho_i})^2 (1-\delta)^2
\geq
\frac{\pi}{\rho_i}
 \operatorname{deg}(v,\partial{\widetilde D}_{\rho_i})^2 (1-\delta)^2 \,,
$$
 where we used the fact that the Euclidean radius of the circle
 $\partial{\widetilde D}_{\rho_i}$ is $\sin \rho_i$.
 Substituting  in \eqref{infpi} yields
 \begin{equation}
   \label{eq:83}
   \sum_{i=1}^k \operatorname{deg}(v,\partial{\widetilde D}_{\rho_i})^2 \le \frac2{(1-\delta)^2} \cdot \frac{2\gamma}{\gamma+ 2\pi} <2\,,
 \end{equation}
the last inequality on the R.H.S.~of \eqref{eq:83} holds for
$\delta>0$ small enough,
  $\delta<\delta_0(\gamma)$.
We thus fix any  $\delta\in\big(0,\min(1/8,\delta_0(\gamma))\big)$. Since
$\sum_{i=1}^k \operatorname{deg}(v,\partial{\widetilde D}_{\rho_i}) =0$, it follows from
\eqref{eq:83} that $\operatorname{deg}(v,\partial{\widetilde D}_{\rho_i}) =0$ for all $i$. Indeed,
otherwise we would have at least two nonzero degrees, clearly
violating  \eqref{eq:83}.

 Now the sum of the radii of the discs in $\{{\widetilde D}_{\rho_i}\}_{i=1}^k$ is
 $r = e^t r_0$ hence bounded by $e^s r_0$.  Using \eqref{eq:84} we
 conclude that for some $C>0$ depending on $\gamma$,
 $$ r \le  C|\ln\varepsilon| \varepsilon \times \varepsilon^{-
   \frac{2\pi +\gamma}{4\pi}} =
 C|\ln\varepsilon| \varepsilon^{\frac{2\pi-\gamma}{4\pi}}\leq \varepsilon^{\tilde\alpha}\,,$$
 for every small enough $\varepsilon$, provided  $0<\tilde\alpha<
 \frac{2\pi-\gamma}{4\pi}$. We fix such a value of $\tilde \alpha$.

 The family of spherical discs $\{{\widetilde D}_{\rho_i}\}_{i=1}^k$ is therefore such that $|v|> 1/2$ outside the discs, their total radius is  less than $\varepsilon^{\tilde\alpha}$,  the winding number of $v$ on the boundary of each disc is zero, and such that for each $1\le i\le k$ we have using \eqref{infpi} that
 $$ E_\varepsilon^{(T)}(v;\partial\tilde D_{\rho_i}) \le \frac{2\pi}{\rho_i} \frac{2\gamma}{\gamma+ 2\pi} <  \frac{2\pi}{\rho_i}.$$

 Rescaling back, we see easily that the spherical discs on $S_R$, ${\widetilde D}_{r_i}^R(Rx_i):=R{\widetilde D}_{\rho_i},i=1,\ldots,k$, satisfy
 all the assertions \eqref{eq:p1}--\eqref{eq:p5}  of \rlemma{lem:specialists} with
 $\alpha=1-\tilde\alpha$.
  \end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Use Cmd/Ctrl+F to search Mining Minerals in the Old West.
Mining Minerals in the Old West
Mining Minerals in the Old West looks at the resources that were of interest to Native Americans, the Hispanic and Anglo miners in the Old West. We immediately think of gold and and silver but there were numerous other valuables lying around or buried out in that vast, largely unexplored county. Here, Old West Daily Reader offers a short list of minerals, gemstones, bones (two kinds) and such that might have been of interest to the Indians and the 17th, 18th and 19th century prospectors…
Borax: (1830's to present)
Borax Crystals – Photo: U.S. PD
Chemical Formula: Na2B4O7 • 10H2O
sodium tetraborate decahydrate
Used as a laundry and cleaning product, an insecticide, and a fungicide. Used as a flux in metalwork and a component of enamel glazes. It is the precursor to boric acid. However, the more modern laundry product and the Western TV show it sponsored, is likely why modern forks know anything about borax or the 20 mule teams that originally brought the product from the mines in the California desert to processing for the market.
FYI: There were really only 18 mules, check out the references below… {001}
20 Mule Team in Death Valley, CA
Photo: U.S. PD NPS
Wk. 05, 02/02/1846 – born: "Borax Smith"
Wk. 02, 01/08/1856 – Tuscan Springs discovery
Wk. 35, 08/27/1931 – adios "Borax Smith"! How many mules?
Buffalo Bones: (from the git go… 'til they were used up!)
Well, of course, Native Americans had used buffalo bones for thousands of years. They made tools, implements, weapons and ceremonial objects from them. Buffalo scapula were used as shovels and Indian signboards. They didn't have to mine them, the bones were there to be picked up from the ground on the vast prairie. Fresh bone from hunting success could provide excellent food, the fatty marrow.
When the white man came (1840's to early 1900's), he had different uses for the resource. Here is the story of boneblack and fertilizer.* {001}
*The Originals Index – Cow? What Cow? – Buffalo?
Building stone: dressed stone (1500's to present)
Limestone Block Wall
Photo: U.S. PD? internet
Limestone Block Steps
Here, in the top photos, it's limestone, laid dry as a wall and heavy rectangular blocks of limestone used as steps. The bottom photos are sandstone blocks; one dressed with a smooth face, the other with a rough, irregular face, laid to emphasize the texture and color variety of the stone. Many different types of stone were used throughout the Old West in and near the localities where found. These two are probably the most commonly used, and the easiest to use. {001}
Sandstone Block Wall
(see also: Limestone and Sandstone below)
Clay: (pre-history to present)
Clay, groomed for pottery making
Photo: crop from web pic
Primarily hydrated silicates of alumina (Al₂∙O₃∙2SiO₂∙2H₂O and variations), sometime including other materials such as iron, alkalies or alkaline earth; clay minerals are found worldwide; formed by natural processes eroding and chemically breaking down the earth's surface. Often found in creek beds and wet areas, it is plastic in nature and can easily be formed into shapes. Although there are numerous other uses, its original and primary use has always been for pottery making. Almost always a locally mined product, clay is bulky, heavy and generally of low value, so it isn't usually moved very far. Native American tribes who made pottery certainly knew where the best clays for the purpose in their locality could be found; a river bank or a lake shore, a particular bluff. As the Mexicans and Anglos came on the scene, they certainly noted, and used, the traditional sources, but, no doubt, developed new ones as the needs/requirements for the product expanded. Brick and tile were likely among the first products introduced by these newcomers in addition to their versions of pottery. {001
Coal: (1500's to present)
Coal – Photo: U.S. PD
Lump Coal – Photo: U.S. PD
Relatively common in the west. Mined by every method from a single individual with a pick and shovel to huge steam powered machines. Small local, group and family coal mines were relatively common in the old west. Used where easily available, for domestic and commercial heating. Coal became the primary producer of steam, later in this era (replacing wood), for ships, railroads, factories and a few power plants. As coke, it is essential in the smelting of iron ore.
Bates and Jackson, (1987), defined coal as "a readily combustible rock containing more than 50% by weight and more than 70% by volume of carbonaceous material including inherent moisture, formed from compaction and in duration of variously altered plant remains similar to those in peat. Differences in the kinds of plant materials (type), in degree of metamorphism (rank), and in the range of impurity (grade) are characteristic of coal and are used in classification…". With time, and depending on the depth of burial beneath deposits of overlying geologic strata (which controls the weight of compaction), the plant material undergoes a series of complex chemical reactions that transforms it first into peat, then brown coal or lignite, sub-bituminous, bituminous and finally into anthracite coal. It may be mined and burned at any of these stages. The energy gained from burning increases as the grade of the coal improves. Anthracite coal does not occur in the west. There is no evidence that Indians used coal. {001}
What's in Coal?
Wk. 16, 04/20/1914 – Ludlow Massacre
Photo Gallery Index – Mining Photos – Ludlow Massacre
Photo Gallery Index – Mining Photos – The Mystery of Burning Mountain (video)
Copper: (pre-history to present)
Copper Ore – Photo: U.S. PD
Copper Ore – Photo U.S. PD
(Cu)
Atomic Number: 29
Atomic Weight: 63.65
Copper was likely one of the earliest metals used by man. Found in its native form (pictured below) it can be worked into numerous usable items. Some Native Americans used native copper in this fashion but did not mine or smelt the metal. However, almost anywhere there was copper ore, there was also turquoise and the beauty of this natural stone was appreciated by all. Turquoise was mined and it was definitely an item of trade well before the coming of the Europeans to the west in the 1500's. {001}
Native Copper
Copper Coinage:
Wk. 19, 05/08/1792 – Copper Coinage Act
Wk. 16, 04/22/1864 – Coinage Act of 1864
Diamonds: (Yes and No!)
The Yes part: There are diamonds in North America and even in the Western U.S.
The No part: There has never been a large find of diamond in the West. There have been reports of a few stones; an odd diamond or two had actually turned up during the gold rush, especially near Placerville, California.
There were several excellent hoaxes perpetrated in the West involving diamonds.
"…though it may not pay to hunt for diamonds, yet it always pays to pick them up when you do happen to see them." – A California state geologist of the times (c. 1850's).
The Originals Index – Lost Treasures in the Old West – Hoaxes
Dinosaur Bones: (1850's to present)
Yes! Turns out the West is full of them. They are still out there waiting to be found and today's scientists still mine them and the body of knowledge they represent about times far past. Here is a short look at how it all came to pass in the Old West. Welcome to the "Bone Wars" (1877 – 1892)*. {001}
Allosaurus
Photo: U.S. PD c. 1915
Here is Cope's nearly complete Allosaurus, excavated at Como Bluff in 1879 but not prepared and assembled until 1915; well after his death (1897). The specimen is located at the American Museum of Natural History NY.
*The Originals Index – Range Wars and Feuds – Bone Wars
The Originals Index – Expeditions – Hayden Expedition of 1853
Glass: (7,000 BCE to present)
For the moment we are concerned with obsidian, natural glass. This material flakes well and has been used in the manufacture of tools and weapons points for thousands of years. The obsidian samples shown here are from Glass Buttes, OR. This glass was definitely well traveled in the trade routes, long before the white man or the Hispanics came to the west. Photo: U.S. PD USFS. {001}
The Originals Index – Landmarks and Registers – Glass Buttes
The Originals Index – Trade in the Old West
The Originals Index – Native American Tribes – nice photo of flaked points, top of page.
Gold: (1500's to present)
Gold Nugget – Photo: U.S. PD
Gold on Quartz – Photo U.S. PD
(Ag)
Atomic Weight: 196.97
Gold was recovered in numerous ways
Photo Gallery Index – Mining Photos
Photo Gallery Index – Mining Photos – Lost treasures in the Old West
Photo Gallery Index – Mining Photos – Hoaxes
Photo Gallery Index – Mining Photos – Gold Rushes
Gold Coinage:
Wk. 44, 11/04/1904 – Indian Eagle Coin
Wk. 52, 12/28/1933 – Gold called in
Gravel: (pre-history to present)
In these time, mostly streambed gravel for roads and other construction. As with other bulk loads in the times. you couldn't haul much and not very far. Definitely not something one made (outside of prisons) due to no machinery, no power sources, etc. It generally wasn't worth making back then.
Guano: (1500's to present, earlier in the East)
The primary source of potassium nitrate, KNO3 (salt petre), a natural, solid source of nitrogen. Useful as fertilizer (a source of nitrogen and potassium), in tree stump removal (it supplies nitrogen to the fungi attacking the wood of the stump), sometimes as a food preservative (salted meats). It is an essential component in the manufacture of gunpowder (black powder) and therefore has related uses as a component of rocket propellants and fireworks.
In the times, mining was most often pick and shovel work, mining in caves with large bat populations. There are methods to make the mineral. {001}
Lead: (1500's to present, earlier in the East)
Lead Mining in Missouri c. 1865 – Woodcut
The Loyal West in the Times of the Rebellion
Barber and Howe (1865)
Lead ore
Photo: U.S. PD internet
"Peacock" Lead ore
Just where did they get the lead they slung? Not likely from anywhere in the west. It probably originated in the Southeast Missouri Lead District via the Desloge Family and Desloge Consolidated Lead Company in Desloge, Missouri and Bonne Terre. This outfit was active in lead trading, mining and lead smelting from 1823 to 1929. They were in just the right place to supply the staging areas along the Mississippi River which were supporting the great westward migration.
Finished products such as shot and bullets certainly came west with the traders, but so did ingots for the making of one's own bullets with the tools commonly available or included with the firearm at purchase. There were a number of other common uses such as making water pipes and solders for tin work. Doc Holliday had a lead cylinder among his dental equipment. Along with nitric acid, the good doctor could make nitrous oxide, the "laughing gas", which made him a painless dentist.
There wasn't much incentive to mine lead in the west due to the lack of the necessary facilities to smelt the metal. Then too, there is always the issue of moving heavy or bulky, low value items over long distances, whether ore or finished product. Then, you would still have to compete with established suppliers from the east.
There is usually plenty of lead in the ore with silver, both lead and tin can be by-products of silver mining. Western mined lead was certainly available. {001}
Photo Gallery Index – Weapons Photos Index – Ammunition Then and Now
Photo Gallery Index – Weapons Photos Index – Firearms – Percussion Firearms Accessories
Limestone: (pre-history to present)
Limestone Cliff
Photo: ©? luGher Texture Library
Fossiliferous Limestone
CaCO3 and numerous impurities, fossils, etc.
Used a a construction material as blocks or crushed stone and in the manufacture of glass. A component of cement and thereby, concrete. Used to de-acidify soils and water. {001}
Limestone Blocks 10″
Limestone on Shale
(see also: Building stone: dressed stone above.)
Potash: (1500's ? to present)
(KCl) and numerous variations. Any of various mined and manufactured salts that contain potassium in water-soluble form. Especially potassium carbonate, has been used in bleaching textiles, making glass, and making soap (lye). A component of a variety of fertilizers. {001}
Sandstone: (pre-history to present)
A sandstone arch in Canyonlands<Photo: U.S. PD? internet
Sandstone blocks
Another important, relatively common, construction stone. Almost always locally quarried in the area of use. Numerous colors and textures of sandstone occur. {001}
Silver: (1500's to present)
Silver Nugget – Photo: U.S. PD
Silver ore
A metal certainly known to some Indians in its native form and possibly used as adornment in pre-historic times, but not mined or smelted. {001}
Native Wire Silver
Photo Gallery Index – Mining Photos – Comstock Lode
Photo Gallery Index – Mining Photos – Hoaxes – The Emma Mine
Photo Gallery Index – Mining Photos – Silver Rushes
Silver Coinage:
Wk. 08/02/21/1857 – Coinage Act of 1857
The Originals – Trails – Oregon Trail Memorial Half Dollar
Tin: (1500's to present)
Tin ore
Tin (metal)
Up until the 1700's or so, tin was thought to be a "lighter" lead. The metals are usually found in the same ore, often with silver as well. Used in the manufacture of tinplate, float glass and solder. Tin is an important component in the alloying of bronze and brass. You will see it used as ceiling tile/cover in many old buildings; look carefully at all those old interior shots of saloons. {001}
For related information on this subject in Old West Daily Reader:
Photo Gallery Index – Mining Photos – Mining on the Comstock
Photo Gallery Index – Transportation Photos
Photo Gallery Index – Transportation Photos – Railroads in the West
The Originals Index – Commerce in the Old West
The Originals Index – Expeditions
The Originals Index – Lost Treasures in the Old West
The Originals Index – Mine Names in the Old West
The Originals Index – Western Forts and Trading Posts
The Originals Index – Trade in the Old West – Commerce in the Old West
Wk. 52, 12/28/1933 – Gold Standard
References – Dictionary – Mining Terms…
End: Mining Minerals
{001} C 01/16; E 08/15: F 08/15; P 10/17

redpajama_set_name: RedPajamaCommonCrawl


We have seen some astonishing displays of arrogance recently in politics. First we had the EU referendum in the UK, where almost no politician thought that the vote would be in favour of leaving and consequently failed to put a coherent argument for either remain or leave. This was quickly followed by the election of Donald Trump in the USA. Apart from Trump's own team, who were paid to state the positive to the media, nobody in the establishment thought he stood a hope. Well both things happened due to the arrogance of the establishment refusing to believe that anyone would vote for Brexit or Trump because to do so was to vote against the Establishment itself. After years of having their ego stroked, politicians had mostly developed their egos to the size of the average skyscraper, which in turn had bred in them an arrogance of almost megalithic proportions. They are totally out of touch with the groundswell of opinion and were not prepared for what happens when ordinary men and women make their feelings known. This arrogance is still on display again with those who seek to somehow stop the withdrawal of the UK from the EU because "the public made a mistake, they must realise we are better off in the EU". The same is true of the people marching against the election of Donald Trump in the vain hope that they can somehow swap him in the dead of night for a Hillary. I am not saying that either winning vote was the right outcome for their respective countries, but I am staggered that politicians on both sides of the pond are still trying to overturn the results. I would have thought that elected representatives are charged with doing what is right for their country, not what fits best with their own ideology. This self belief is so strong, they are even prepared to lie to the voting public to get their preferred result. The scare stories about what would happen to both economies is a prime example. Had we believed the harbingers of doom, the Uk's economy would have been on its knees by August 2016 and the USA would be entering a similar economic decline. The opposite has turned out to be true and, whilst there are no doubt some adjustments in terms of currency rates and share prices to be made, both nations are in a good position to prosper. It is high time that politicians ceased their ego stroking and thought about the long term good of the country they represent, instead of short term goals followed by a program of after dinner speaking engagements when they leave office. Note the term leave office. In any other sphere, leaving office after flying high would be failure. But with an ego the size of a skyscraper, the only admission of failure is seen as the failure of the people in not recognizing what was in their best interest.
We have seen some astonishing displays of arrogance recently in politics. First we had the EU referendum in the UK, where almost no politician thought that the vote would be in favour of leaving and consequently failed to put a coherent argument for either remain or leave. This was quickly followed by the election of Donald Trump in the USA. Apart from Trump's own team, who were paid to state the positive to the media, nobody in the establishment thought he stood a hope.
Well both things happened due to the arrogance of the establishment refusing to believe that anyone would vote for Brexit or Trump because to do so was to vote against the Establishment itself. After years of having their ego stroked, politicians had mostly developed their egos to the size of the average skyscraper, which in turn had bred in them an arrogance of almost megalithic proportions. They are totally out of touch with the groundswell of opinion and were not prepared for what happens when ordinary men and women make their feelings known.
This arrogance is still on display again with those who seek to somehow stop the withdrawal of the UK from the EU because "the public made a mistake, they must realise we are better off in the EU". The same is true of the people marching against the election of Donald Trump in the vain hope that they can somehow swap him in the dead of night for a Hillary.
I am not saying that either winning vote was the right outcome for their respective countries, but I am staggered that politicians on both sides of the pond are still trying to overturn the results. I would have thought that elected representatives are charged with doing what is right for their country, not what fits best with their own ideology.
This self belief is so strong, they are even prepared to lie to the voting public to get their preferred result. The scare stories about what would happen to both economies is a prime example. Had we believed the harbingers of doom, the Uk's economy would have been on its knees by August 2016 and the USA would be entering a similar economic decline. The opposite has turned out to be true and, whilst there are no doubt some adjustments in terms of currency rates and share prices to be made, both nations are in a good position to prosper.
It is high time that politicians ceased their ego stroking and thought about the long term good of the country they represent, instead of short term goals followed by a program of after dinner speaking engagements when they leave office.
Note the term leave office. In any other sphere, leaving office after flying high would be failure. But with an ego the size of a skyscraper, the only admission of failure is seen as the failure of the people in not recognizing what was in their best interest.
Bilston › Many politicians have never played fair, but now there's not decency.
Many politicians have never played fair, but now there's not decency.
Sonia › Very well said, SimplyLogical. I agree with every single word!
Very well said, SimplyLogical. I agree with every single word!
The history of the Entroido.


Union Heights School Counseling Program is designed to create a safe and healthy learning environment for all students. Classroom guidance includes Character Counts, bully prevention, anger and conflict management, and careers in addition to many other topics. Mrs. Mountain is also available for individual counseling with students who are having difficulties. She also conducts small group counseling sessions as needs arise.
If you need to see me about individual school counseling for your child, please feel free to contact me. We want to make your child as comfortable in school as we can so he/she is free to learn without too many distractions.


The Law Offices of Allisan Lee Adams, based in Cheshire since 1990, has opened a satellite office in Westbrook, CT. Allisan provides pleasant, prompt and precise legal counsel in the areas of divorce mediation and collaboration, residential real estate, commercial real estate, personal injury, wills, powers of attorney and estate administration.
For over 25 years, Allisan has served individuals and families in Cheshire, Chester , Clinton, Hamden, Killingworth, Meriden, New Haven, North Haven, Old Saybrook, Southington, Wallingford, Waterbury, Westbrook, and other surrounding communities in a caring, compassionate manner in significant matters that impact their lives.
I take a very interactive, team-based approach when working with you to optimize results. I am a meticulous listener, and I will seek your maximum input in order to provide a broad range of creative legal solutions. My goal is to educate my clients on the pros and cons of possible solutions so that we can forge a resolution that best fits their needs.
I provide cost-effective counseling and take pride in providing outstanding value and quality in legal services.I look forward to talking with you about your need for legal services.
Serving CT Families in: Avon, Branford, Berlin, Bethany, Bristol, Chester, Cheshire, Clinton, Danbury, Deep River, Essex, Farmington, Guilford, Hamden, Hartford, Killingworth ,Litchfield, Lyme, Madison, Meriden, Middlebury, Middletown, Milford, Naugatuck, North Haven, New Haven, Old Lyme, Old Saybrook, Oxford, Prospect, Shelton, Southbury, Southington, Seymour, Trumbull, Wallingford, Waterbury, Wolcott, Watertown, Westbrook, West Hartford, Woodbridge, Woodbury, and all surrounding areas.


text:
Home Football Suárez-Real Madrid, solution to their drought and an ill will with a...
Suárez-Real Madrid, solution to their drought and an ill will with a precise origin
The triumph achieved by
before him Villarreal, last Sunday in The ceramic, left a car of good news for the mattress team. To begin with, they won again, after four games without doing so.
Especially important in the face of harassment Real Madrid Y FC Barcelona at this point in the season, especially on the eve of next Sunday's derby against the white box. Then he left a clean sheet after eight games without doing it, that the team needed him.
And what to say about the great goal and the good game of Joao Felix
, recovered something for the cause at a key moment of the course. Among the few things that left a bitter aftertaste was the fact that the team's top scorer, which in this case is Luis Suarez, could not find a door once more.
There are 16 goals that the Uruguayan has at the moment, in 26 games. However, right now he is experiencing his worst situation since he signed for the mattress team. He has not seen the door since the game against him Celta Vigo, on February 8.
475 minutes have passed since then without seeing the door. He failed to score against grenade, the two games against Levante; the appointment of Chelsea in Champions and last weekend's game, against him Villarreal.
What happened? He has stopped having opportunities. Well not really. He is still very close to the positions in which he scored without problems before this streak, but it is true that he shoots less between the three suits. In these last five games without scoring he has added 15 shots, three of them on goal (one at the post, the free-kick he threw against I raised). The average number of shots is three per game despite the fact that in the last two, Chelsea Y VillarrealHe only finished once. That in the aforementioned section, because in the overall season, his average is almost three shots per game (the same as in this drought), but it is true that he has an average in the season of 1.2 shots on goal per match. When in the last five games it has been reduced by half, 0.6 on average. He only shot at the door before grenade, I raised Y Villarreal.
If he Athletic wants to win, a good part of his options will go through the fact that the Uruguayan is reunited with the goal.
A VICTIM WITH WHICH IT IS TAUGHT
Suarez is motivated in an extra way before him Real Madrid. He is especially suspicious of the white team for something that happened many years ago. You have to go back to 2014 when the Uruguayan was the fashion forward in Europe. He came from succeeding in the Liverpool and did a great world. With a big blur, that bite to Chiellini which earned him a severe sanction of the FIFA.
So, the player was in a war between the FC Barcelona and the Real Madrid to sign him. The white club was very interested in him, as well as the Catalan. But after what happened with the Italian defender and the criticism that rained down on the striker, he unchecked himself and did so leaving the player at the foot of the horses, talking about not behaving in accordance with the supposed values that govern the white club and other verbiage .
The South American did not sit well. No longer the fact that the Real
Madrid would pick up the hook, but instead do so by muddying his name because of a mistake he made and for which he paid more than enough. During the following years he would take revenge, making the white team one of his favorite rivals. It has only marked him more, in his career, to Valencia (13), Norwich (12 and Eibar (12).
The Real Madrid he is one of the rivals to whom he has scored 11 goals alongside a Betis, Willem
II Y Seville. No less than 11 goals in 17 games, with a balance of seven wins, four draws and six defeats
Previous articleLaporta: "Five agents have called me"
Next articleTrippier and Carrasco, the ideal plan
La Traviesa: the place of clandestine meetings between Piqué and Clara...
Jordi Martí (Joventut): "It would be a dream to reach the...
Harry, Meghan and the culture of the maximum ego: when the...
Félix Fernández: "Juventut-Baskonia will be a good tie for basketball players"
Doña Letizia and the water polo players: her surprise pose in...
Joana Sanz, sunk after the latest news from Dani Alves: "My...
Juanma Rodríguez (Unicaja): "We go humble and with a lot of...
Dani Alves, transferred from prison to "better guarantee security and coexistence"


Bathroom Vanities And Cabinets Tag: Bathroom Vanity Cabinets or Functional Decoration, Important Facts You Ought to Know about Round Dining Table Sets, DIY Small Computer Desk.
The four poster bed was commonly known as the kings' beds. And it is also associated with fairy tales and romance. Of course, there's a reason behind it. Its quality and design are two main features make this bed exceptionally stunning. And as the year passes by, its design always evolves every year. It makes this furniture ideal to various bedroom styles from traditional to eclectic.


The warm spring months bring with them plenty of sunshine, rain showers and allergies. Most people tend to think of pollution as something that is outside of the home. Pollution takes the form of smog, allergens, haze and ozone outdoors but in fact, the air quality inside the home can be more polluted than the air outside the front door.
Indoor air can be polluted by a number of pesky allergens including dust, formaldehyde, radon and harmful chemicals used in everything from perfume to household cleaners. Luckily, there are several easy steps that can be made in order to improve the indoor air quality in your home or office.
The best place to start with improving the indoor air quality of your home or office is with regular HVAC maintenance. Schedule regular air duct cleaning and system maintenance with your local HVAC technician in the spring and autumn months to help keep things running smoothly all year long.
The introduction of healthy, happy and green houseplants to your indoor space can do wonders for improving air quality while giving your home or office a sophisticated look. Indoor plants work to improve the overall air quality inside by acting as a natural air filter, removing common toxins from the air. NASA recommends homeowners use 15 to 18 good-sized houseplants in the average 1,800 square foot home to help improve the overall air quality.
To keep your HVAC system working at its best, change your air filters regularly. Most HVAC systems require a new filter every 30 days while some recommend a new filter every 90 days. Check with your heating and cooling technician for more details on proper care for your system.
In addition to regularly replacing your HVAC system's air filters, using a vacuum with a certified HEPA filter can easily help to limit the concentration of lead in the home or office. Using a HEPA filter can also help to eliminate other allergens like pollen, dust mites and pet dander.
Stop excess dirt and pollutants from making it through the door by investing in large doormats at every entry to your home or office. Another way to keep those harmful pollutants out of the house is to declare your home a "shoe free" zone by asking that guests take their shoes off at the front door.
Talk to experts like Doc Dancer Heating, Air Conditioning & Generators for more information.


text:
Mon - Fri 9:00 am
/ On the Show
/ Recipes
/ Performances
/ Episodes
/ Be part of the show
Follow The Morning Show on TwitterFollow The Morning Show on FacebookFollow The Morning Show on InstagramEmail The Morning Show
EpisodesExternal Link
Be part of the show
Entertainment - The Morning Show
The story behind the powerful moment between a bride and groom at a Black Lives Matter protest
Sarah Carson and Jenny Ky
Published: Monday, 22 June 2020 12:20 PM GMT+10
The couple who celebrated their wedding during Black Lives Matter protests
The intimate moment quickly circulated around the world.
One of the most striking pictures to come out of the Black Lives Matter protests was a bride and groom, side by side, stepping out of their wedding and into a peaceful protest.
The intimate moment quickly circulated around the world - and it was a symbol of new beginnings, love and hope.
Watch the full interview with the couple above
Watch The Morning Show on Channel 7 and stream it for free on 7plus >>
The couple, Dr Kerry-Anne and Michael Gordon, had actually planned to get married in May - but ended up changing their plans due to COVID-19.
Instead, they tied the knot on June 6th with just 21 friends - and the couple only found out about the Black Lives Matter protests a few nights before their ceremony.
"It was definitely a last-minute addition to our last-minute wedding that we had planned in just over two weeks or so," Michael said.
The couple share an intimate moment surrounded by peaceful protesters. Credit: Supplied
"When the pandemic happened, we realised that we needed to postpone our wedding because it wouldn't have been safe for our families and friends to come together to celebrate our love and our union," Kerry-Anne said.
"We weren't sure if we would be able to do it this year or even next year.
"We were in such a moment of uncertainty that it was best for us to go ahead and tie the knot."
'The most powerful moment'
The couple's first look happened as they were surrounded by peaceful protesters - and Kerry-Anne has described it as "the most powerful moment there was."
"I was feeling the energy and the love and all the excitement from the thousands of people surrounding me," Kerry-Anne said.
"It was just joy. They were so hopeful and peaceful, it was just beautiful.
"As I', feeling all of these emotions within my body, within my soul and my spirit, I felt a release from all our history, all the movement and protests that had happened right before this.
Entertainment - The Morning Show'Not the time to be silent': Why it isn't enough to just be 'not racist'
The Daily EditionHow to talk to your kids about racism and protests
"I really wanted to share this with Mike. I wanted him to really feel the energies of that moment, which was so powerful.
"I called for him, and when I saw him, it was the most magnificent thing you could think of.
"I saw our future, I saw our wedding, I saw the movement and the changes that could happen around the world, and that we're really working together as a nation.
"I felt the most peaceful at that moment. I was so excited."
A national symbol of love
"Honestly, it is difficult to put into words what it is that we felt," Michael said.
"Initially when I came up and grabbed her hands, her hands were shaking.
"She had already been there in the midst of the crowd and all of the positive energy and the well wishes, and I grabbed her hand, looked at her, gave her a kiss, and she had tears running down her eyes.
"That moment, I was really just focused on her, looking right at her and looking at her face.
Dr Kerry-Anne and Michael Gordon changed their wedding plans due to COVID-19. Credit: Supplied
"As we stepped away, I turned around and looked at the crowd and the first thing I remember thinking was, 'Wow, that's a lot of people.'
"It was just all love - the claps, the applause, the comments. And a little bit of clapping from a lot of people makes a lot of noise - and we felt that energy coming from the crowd and through the ground.
"We were all on the same page in that split moment of an instant - and it appears to be a moment that may live on well past what our intentions were for that day."
Entertainment - The Morning ShowLifestyleNorth AmericaRelationshipsFrom The Show
What you need to know about supermarket product limits as Omicron outbreak takes toll
How to lose one kilo every week: Sonia Kruger shares her fun way to get fit
'Vegemite makes me question that': Original 90210 star on romance with Aussie dancer
Her husband died as they chatted in bed. Now Bec is turning her agony into a message of hope
Michael Patrick Sponsored
Money expert reveals which $5 and $50 notes are worth 'a small fortune'
Connect with The Morning Show


We're committed to providing easy,time-saving and money-saving one-stop purchasing service of consumer for Paper Box Carton , Paper Boxes Cartons , Paper Box Set , With us your money in protected your business enterprise in secure . Hope we can be your trustworthy supplier in China . Searching forward for your cooperation .
The company keeps to the operation concept "scientific management, high quality and efficiency primacy, customer supreme for Paper Box Carton , Paper Boxes Cartons , Paper Box Set , Must any of these goods be of interest to you you should let us know. We are going to be satisfied to provide you a quotation on receipt of one's detailed specifications. We've got our private experienced R&D enginners to meet any of one's requriements We appear forward to receiving your enquires soon'and hope to have the opportunity to work together with you in the future. Welcome to check out our company.


I'm an author, editor (both copyeditor and structural 'book doctor'), ghostwriter, and illustrated book project manager, with more than thirty years' experience in the publishing industry. I specialise in non-fiction books in the humanities, the arts (art history, contemporary art, photography), archaeology and biography. Coming from an academic background (I have a PhD in Classics and Archaeology) but with a career in trade publishing, I'm especially practiced in turning scholarly manuscripts into books for a general audience.
Responsible for pitching new titles and commissioning; for content development from initial concept; budget management; management of in-house and freelance editorial and picture research teams; scheduling; detailed text editing and picture approval; design revision.
Responsible for content development, costing, commissioning, text editing, and picture sourcing and approval for books and a quarterly periodical devoted to travel in the eastern Mediterranean.
Responsible for content development from initial concept; budget management; management of in-house and freelance editorial and picture research teams; scheduling; development of database monitoring systems; detailed text editing and picture approval; design revision.
Founding partner of this consulting firm offering specialist publishing, archaeological and architectural services. Publication projects ranged from periodicals to multi-volume archaeological monographs. Publishing clients include Thames & Hudson, English Heritage, University of Bradford, British School of Archaeology at Athens and the Mediterranean Archaeological Trust, and University of Minnesota.
Started as a desk editor, responsible for detailed content and copy-editing of texts for the 34-volume Dictionary of Art. Promoted to Senior Editor six months later, with responsibility for the work of a small team of desk editors, as well as detailed editing of texts in subject areas ranging from Chinese to ancient Cypriot to Jewish art. Promoted to Deputy Editor two years later, with responsibility for the work of desk editor teams; personal negotiation with difficult authors across the Dictionary spectrum; detailed text editing and finalizing all text for non-Western subject areas; writing articles on diverse topics where necessary. Managed a fluctuating team of between five and thirty in-house and freelance editors and support staff. Promoted to Managing Editor following the publication of the Dictionary of Art in 1996. Planned and edited several spin-off volumes, spoke at conferences and trade fairs, wrote for and edited a customer newsletter.
Freelance copy-editor, proofreader and indexer for various publishers, including APA Publications, Ashgate, Blackwell, Brill, Chatto and Windus, Jonathan Cape, Polity Press, Thames and Hudson, University of California Press, Virago Press.
Examines the impact on the British illustrated publishing industry of emigres from Germany and Austria in the first half of the twentieth century, looking in particular at the art publishing houses of Phaidon Press and Thames & Hudson.
Diane is everything I had hoped an editor would be, and more.


text:
Q: What is the question mark's significance in MySQL at "WHERE column = ?"? I am dissecting some code and came across this,
$sql = 'SELECT page.*, author.name AS author, updator.name AS updator '
     . 'FROM '.TABLE_PREFIX.'page AS page '
     . 'LEFT JOIN '.TABLE_PREFIX.'user AS author ON author.id = page.created_by_id '
     . 'LEFT JOIN '.TABLE_PREFIX.'user AS updator ON updator.id = page.updated_by_id '
     . 'WHERE slug = ? AND parent_id = ? AND (status_id='.Page::STATUS_REVIEWED.' OR status_id='.Page::STATUS_PUBLISHED.' OR status_id='.Page::STATUS_HIDDEN.')';

I am wondering what the "?" does in the WHERE statement.  Is it some sort of parameter holder?

A: ? has no special meaning in MySQL WHERE = statements, only in prepared statements
The most common case where we see it is due to special meaning given to ? by several web frameworks like PHP and Rails.
? is just a syntax error at:
CREATE TABLE t (s CHAR(1));
SELECT * FROM t WHERE s = ?;

because it is unquoted, and in:
INSERT INTO t VALUES ('a');
INSERT INTO t VALUES ("?");
SELECT * FROM t WHERE s = '?';

it returns:
s
?

thus apparently without special meaning.
MySQL 5.0 prepared statements
MySQL 5.0 added the prepared statement feature which has similar semantics to the question mark in web frameworks.
Example from the docs:
PREPARE stmt1 FROM 'SELECT SQRT(POW(?,2) + POW(?,2)) AS hypotenuse';
SET @a = 3;
SET @b = 4;
EXECUTE stmt1 USING @a, @b;

Output:
hypotenuse
5

These also escape special characters as expected:
PREPARE stmt1 FROM 'SELECT ? AS s';
SET @a = "'";
EXECUTE stmt1 USING @a;

Output:
s
'

Rails example
In Rails for example, the question mark is replaced by an argument given by a variable of the library's programming language (Ruby), e.g.:
Table.where("column = ?", "value")

and it automatically quotes arguments to avoid bugs and SQL injection, generating a statement like:
SELECT * FROM Table WHERE column = 'value';

The quoting would save us in case of something like:
Table.where("column = ?", "; INJECTION")

A: Prepared statments use the '?' in MySQL to allow for binding params to the statement. Highly regarded as more secure against SQL injections if used properly. This also allows for quicker SQL queries as the request only has to be compiled once and can be reused.

A: The question mark represents a parameter that will later be replaced. Using parameterized queries is more secure than embedding the parameters right into the query.
SQL Server calls this parameterize queries, and Oracle calls it bind variables.
The usage varies with the language that you are executing the query from.
Here is an example of how it is used from PHP.
assuming that $mysqli is a database connection and people is a table with 4 columns.
$stmt = $mysqli->prepare("INSERT INTO People VALUES (?, ?, ?, ?)");

$stmt->bind_param('sssd', $firstName, $lastName, $email, $age);

The 'sssd' is a flag identifying the rest of the parameters, where s represents string and d represents digits.

A: These are prepared statements ,prepared statements offer two major benefits:

The query only needs to be parsed (or prepared) once, but can be
  executed multiple times with the same or different parameters. When
  the query is prepared, the database will analyze, compile and optimize
  its plan for executing the query. For complex queries this process can
  take up enough time that it will noticeably slow down an application
  if there is a need to repeat the same query many times with different
  parameters. By using a prepared statement the application avoids
  repeating the analyze/compile/optimize cycle. This means that prepared
  statements use fewer resources and thus run faster.
The parameters to prepared statements don't need to be quoted; the
  driver automatically handles this. If an application exclusively uses
  prepared statements, the developer can be sure that no SQL injection
  will occur (however, if other portions of the query are being built up
  with unescaped input, SQL injection is still possible).

http://php.net/manual/en/pdo.prepared-statements.php

redpajama_set_name: RedPajamaStackExchange


text:
In Photos: 41st Annual Parade Honors Dr. Martin Luther King Jr.
Civil rights icon U.S. Rep. John Lewis, who marched with King in Alabama more than 50 years ago, participated.
01/21/2019 By Dianna Wray Photography by Daniel Kramer
Watch HPM's Houston In Black and White This MLK Day
Don't miss this thought-provoking program about racial equality.
01/13/2017 By Rafa Farihah
Top Things To Do This Weekend: Jan 16–19 (MLK Weekend Edition)
Mel Chin's Houston takeover begins.
01/14/2015 By Michael Hardy


Contemporary worship - we sing songs to praise God and thank him for his goodness and faithfulness. The word "worship" means "to come close". In our worship to God we are seeking to draw closer to Him.
Informal teaching - during our service we explore the bible together in a relaxed and informal way. We aim to make this appropriate to everyone whether you've been reading the bible for years or have never picked up a bible and are just curious. Our relaxed style means it's ok to interupt and ask questions or make a coment or share something that you feel God is saying.
Prayer ministry - after the service there are people who are happy to pray with you. Whether you would like prayer for healing, a job interview, a family situaiton, or for no particular reason, we are here for you. We believe our God can change situations and we have witnessed this many times so we are confident that He can change yours too.
Communion - we share communion once a month. This is conducted in the usual Baptist manner where the bread and wine are distributed to the congregation. There is no obligation but all who wish to take part in communion are very welcome.


text:
[New Order Explained] - Timeline and Eras feature
Post feature RSS	[New Order Explained] - Timeline and Eras
"New Order covers the period of the Galactic Empire and Imperial Remnant, the timeline spans from the birth of the Galactic Empire in 19 BBY to Second Galactic Civil War in 40 ABY." - said in the summary to our mod. Until then, we have given a fairly general description of the timeline, and we now want to share more precise dates, and also talk about the system of eras changing.
Posted by Du}{ on May 8th, 2017
Era I: 19 BBY - 13 BBY
The Empire is formed, but it hasn't yet acquired it's full power. Conflicts on distant fronts with Separatist remnants are still raging, and the new authority, led by the Emperor Palpatine, are engaged in exterminating the Jedi, as well as everyone helping them. Empire is still using forces, inherited from the Republic, but new technology is making it's way. The Rebellion is an organization consisting of local militias and resistance forces undermining the positions of the Empire and hiding the Jedi from the eyes of the Emperor and Darth Vader.
Bulwark mk I deploys squadrons of the Z-95 headhunters at the early years of rebel movements.
Era II: 12 BBY - 0 BBY
The Dark Times have come. Empire rules the Galaxy through the fear, it's immense forces imposing Emperor's will or crushing those who refuse it. Facing the growing leviathan, numerous resistance movements are organizing into single entity to become the Alliance. "Operation Strike Fear" scenario will take place at the end of this period.
Darktroopers phase 0 are ready for combat at Bespin.
Era III: 0 ABY - 4 ABY
Rebel forces won a series of victories and now openly battle the forces of the Empire, gaining strength.
To compensate, the Empire is developing new weapons while trying to suppress the increasing number of worlds joining the Rebel cause. Striking from the nowhere and gaining resources, troops and materiel from sympathizers, the Alliance is preparing to make it's decisive move against the Empire.
An assortment of rebel vehicles patrols the area on Yavin IV.
Era IV: 5 ABY - 11 ABY
Emperor Palpatine and his right hand, Dart Vader, are dead, the Empire is in disorder and bears great losses. The Alliance, reorganized into the New Republic, captured the Coruscant and forced the Empire and warlords into Deep Core of Outer Rim. While the imperial leadership is demoralized, battles for the throne are fought, and the imperial officers and politicians alike announce their independence and battle between themselves and the remains of imperial authority.
In certain years of this period, the Empire is reorganized to figth the common foe under the leadership of Grand Admiral Thrawn and the reborn Emperor Palpatine.
Imperial Firehawke repulsortank prepares to attack the enemy.
Era V: 12 ABY - 31 ABY
The Imperial Civil War is over: admiral Daala united struggling warlords forces, but is unable to inflict a major defeat to the Republic. Her cause is carried on by next imperial leader, Grand Admiral Pellaeon, who fights several campaigns, but Imperial Remnant still loses it's forces and worlds, which forced him to sign the peace treaty between the Empire and the New Republic. However, the peace in the galaxy does not last long, as new threat looms over the galaxy - the bloodthirsty Yuuzhan Vong, destroying everything in its path. The galaxy is united in every way to repel the uninvited guests.
Viscount Star Defender - a hope to the galaxy.
Era VI: 32 ABY - 40 ABY
The war with the Vongs caused enormous damage to the galaxy. It would seem that there is no reason to fight on the ruins, but over the years new conflicts, perpetrated by Lumiya and Abeloth, will lead the galaxy into a new chaos - the Second Galactic Civil War, followed by emergence of Lost Tribe of Sith.
We would also like to point out that Scenarios are planned for each era, and some of them, to properly portray the events, will be made into full Expansions with it's own features.
Era system
I would also like to say a few words about our new system of era changing in general scenarios.
The vanilla tech levels are now called "eras", and replace each other depending on the time spent in the game. So, the longer you play in the Galactic Conquest, the further your era will move forward. This does not apply to story Scenarios, which take place in specific eras.
In skirmish, you are still forced to purchase the upgrade to a new era, but we are thinking about applying our new system too.
Thank you for reading and following us!
CdtFox May 9 2017
Dark Trooper phase 0, really nice idea! If i understood well, now eras are "random" and you don't have to spend money to improve your tech anymore.
Du}{ May 9 2017
Yes, you're absolutely right :)
Darth_Weasel May 11 2017
Is the tech system supposed to be like ICW's?
Du}{ May 11 2017
No, ICW's tech system depends on deaths of the imperial leaders. Our system depends on time spent in game.
Well that's neat. I don't think I've played a mod with a tech system like that before.
Guest Sep 7 2017
will the mod be placed in the steam workshop
Reply Good karma Bad karma0 votes
Du}{ Oct 15 2017
We're thinking about this :)
Post feature
1,421 (1 today)
Star Wars: New Order Star Wars: Empire at War: Forces of Corruption
Star Wars: Empire at War: Forces of Corruption Real Time Strategy
New Order Dev Team Developer


Description: The Chocolate Moth is a medium sized (21-28mm) moth. The forewings are chocolate to reddish-brown with purplish-gray in the terminal area (outer margins), and the hindwings are brownish-black to brownish-gray. The reniform spot is usually outlined with white. Males have broadly pectinate antennae, but they are simple in females.
Similar species: This is a distinct species and the only one in this genus north of Mexico.
North American Range: The Chocolate Moth is mainly found in coastal areas from Maryland south to Florida and west along the Gulf Coast including Louisiana. It has been documented in several southeastern counties in Virginia.
Flight season and broods: Adults fly from March to September but may fly in all months of the year in the more southern parts of its range.
Habitat and Food Plants: Little is known regarding the habitat and food plant for the Chocolate Moth.
Behavior and Ecology: Little is known regarding the behavior and ecology.
Population trend and potential threats: Population trends and potential threats are unknown. It is considered 'secure' across its entire range by NatureServe (2013).
Management practices: Because little is known regarding the ecology and potential threats, there are no recommended management practices.
References: Covell, C.V. 1984. A Field Guide to the Moths of Eastern North America. Peterson Field Guides, Boston,	Mass. 496pp.
NatureServe. 2013. NatureServe Explorer: An online encyclopedia of life [web application]. Version 7.1. NatureServe, Arlington, Virginia. Available http://www.natureserve.org/explorer. (Accessed: January 23, 2013 ).


Hope you're doing ok, and hope you come back soon.
Did a CYOA chain that included tentacle pet recently and it reminded me of Tentacle Time. Really hope you're doing okay and that you've just been really busy or lost your passwords or something. Hope you come back soon.
Yo major session, you doing all right man? It's been a year since you started your story on this account.
Hoping your doing ok man, your most recent activity was 5 months ago. You have become one of my favorite writers on this site.
Alayne, Ryan Bazana, KuhnKuhn and 3 others like this.
Here's hoping you're doin alright MS.
Ryan Bazana, KuhnKuhn, Ddmkm122 and 2 others like this.
You do it for the lewd. I do it for the hugs. Maybe I can gibs you hugs.
Ddmkm122 and Ryuuto like this.
I'm going to assume that your absence the last month or so has been due to studying, working etc.
Really hope you can find some time to relax and write more glorious consentacles soon.
Ryan Bazana, Ddmkm122, Leonidas333 and 3 others like this.
Also, Happy Birthday for July, just noticed your DOB on the sidebar.
Ryan Bazana and Ddmkm122 like this.
I hope you're okay...if you need some time away, take all the time you need.
Ryan Bazana, Ddmkm122 and Zeushobbit like this.
Hey, how is the next chapter of Tentacle Time coming? It's been a while since the last chapter came out.
Ddmkm122 and Major Session like this.
Going well. Burnt my arm the other day so been hard writing with one hand. But we're going! Should hopefully be done today and then into beta.
Ddmkm122, Musical Blue, Leonidas333 and 2 others like this.
Just thought I would pop over and say that I hope your studies are going well.
Ddmkm122, Scottyzelsnyder and Major Session like this.
Thank you. First assignment down, two to got for this round. Then Practicum. Still, more time for writing while on practicum.
I saw that you had no messages on your profile, so I thought I would say that I love your current story 'Tentacle Time!' and that I wish you luck with your future writings of it and whatever else takes your fancy!
Ddmkm122, DrStrangeLove, Scottyzelsnyder and 1 other person like this.
Thank you! It is much appreciated. Onward to cosentacles!


text:
Bun B's Trill Burgers gets its first brick-and-mortar location in California
Trill Burgers is part of Bored & Hungry, a restaurant based on the Bored Ape NFT.
Joey Guerra March 18, 2022Updated: March 21, 2022, 11:31 am
Bun B and Andy Nguyen of Trill Burgers, which will be available at Bored & Hungry in California.
Bun B's Trill Burgers is finally getting a brick-and-mortar location. But it's not in Houston. Or even Texas.
Trill Burgers is part of the concept behind Andy Nguyen's Bored & Hungry, opening April 9 in Long Beach, Calif. The restaurant will also serve a plant-based Beleaf Burger, fries and milkshakes. It will run as a pop-up for 90 days in California.
H-TOWN HISTORY: Bun B talks about his monumental RodeoHouston performance
Bored & Hungry was born from a Bored Ape, one of the most popular NFTs in the cryptocurrency community. Nguyen, who announced he purchased one earlier this month, is now part of the Bored Ape Yacht Club, a collective of NFT holders that includes Eminem, Paris Hilton and Snoop Dogg.
Coming to real life April 9th, in Long Beach, CA! ?" 🍟 🥤 @BoredNHngry #BAYC #MAYC pic.twitter.com/Em3Ro0mDvG
— AndyTheNguyen.eth (@AndyTheNguyen) March 17, 2022
The Bored & Hungry packaging is based on Nguyen's Bored Ape. He sees the restaurant as "a place to educate those skeptical about this new world. They like to call it a jpeg but we wanted to show them that we're building a new ecosystem."
Trill Burgers launched in August 2021 at a Stick'y Chicken pop-up, a collaboration between Bun B and Nguyen, a California-based culinary entrepreneur; and siblings Patsy and Benson Vivares of Sticky's Chicken. It's currently drawing big crowds daily at RodeoHouston.
"The amazing thing about the NFT space is the opportunity for rapid growth," Bun B says. "With Bored and Hungry, it opens us up to a worldwide community."
"The goal is to give back to the growing Web 3.0 community and open the doors to those who want to learn more about this new Web 3/NFT world," Nguyen said. "Our job is to educate the public about this new future world, and show people that you can create a brand/business out of this IP – taking away the stigma of, 'It's just a jpeg.'"
Joey Guerra
Joey Guerra is the music critic for the Houston Chronicle. He also covers various aspects of pop culture. He has reviewed hundreds of concerts and interviewed hundreds of celebrities, from Justin Bieber to Dolly Parton to Beyonce. He's appeared as a regular correspondent on Fox26 and was head judge and director of the Pride Superstar singing competition for a decade. He has been named journalist of the year multiple times by both OutSmart Magazine and the FACE Awards. He also covers various aspects of pop culture, including the local drag scene and "RuPaul's Drag Race."
More In Special Events
Critics pick things to do: Cirque du Soleil, 'Roe'
Where to celebrate Martin Luther King Jr. Day in Houston, from parades to festivals
By Ana Khan
Critics pick 6 things to do this weekend: 'The Last of Us,' Samantha Fish
By Andrew Dansby,Amber Elliott,Cary Darling
'Pretty Woman,' BowiElvis Fest on the critics' list of 5 things to do this weekend