During the holiday season, we've got a lot of parties to attend, from my husband's work party, to the bash that our neighbor's down the street plan to throw. I want to look my best, but I don't have the time to keep my manicure in ship shape, nor do I have the time to change it out to match the outfit I'm wearing. Enter: the holiday nail wrap!
The concept of a nail art wrap is something new to me.
So, what is a nail wrap?
The DIY at-home version of a nail wrap is a thin vinyl sheet with adhesive on one side. Typically, application is as easy as sticking them on your nails, filing them down to your nail length, and applying heat with a hair dryer. That's it! There's no dry time needed, and they last as long as it takes your nails to grow out (I typically wear mine for about 2 1/2 weeks before they start to look like they could use a refresh). If you want to remove them when they're fairly new, you can soak them in acetone to loosen them from your nails, then wipe them off with a cotton ball like you would regular nail polish. But, as they age, they're easy to peel off without any nail polish remover. You can find nail wraps online (Jamberry is a favorite), or at drugstores (Sally Hansen carries a good selection).
Needless to say, I'm a bit obsessed with how easily I can change the color and style of my nails with nail art, especially during this hectic season.
Check out my favorite picks for the holidays season in the slideshow.


For those who don't visit the About page daily, just a couple of small changes.
Posts with longs words are now ignored (the word limit is current 70), most times these are really long URLs or tables which your feed has removed the tags, but not added any spaces.
People who don't post anything for 60 days, will find their name being removed from the BlogBox, as nobody likes an out of date link. At the moment this process is done part manually, part automatically. (The list of when you last posted is created by the computer, so I don't have to trawl through every blog, but the switch to turn you on or off is manual). So don't be alarmed that your blog hasn't been turned back on straight away.
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Case Description Judicial Watch submitted a FOIA request to the FBI for records concerning any interviews conducted by the agency with Barack Obama, Rahm Emanuel, or Valerie Jarrett pertaining to Rod Blagojevich. The FBI acknowledged receipt of the request and denied it under Exemption 7(A) (interference with ongoing investigation or proceeding). Judicial Watch appealed to the Office of Information Policy, which upheld the FBI's denial. Judicial Watch then filed suit.


Brachystelma macropetalum (Schltr.) N.E.Br. in W.H.Harvey & auct. suc. (eds.), Fl. Cap. 5(1): 852 (1908).
Ceropegia macropetala (Schltr.) Bruyns, S. African J. Bot. 112: 432 (2017).
* Brachystelmaria macropetala Schltr., Beibl. Bot. Jahrb. Syst. 51: 51 (1895).
Lasiostelma macropetalum (N.E.Br.) Schltr., J. Bot. 37: 62 (1899).


The Welsh cartoonist Chris Reynolds has been creating Mauretania Comics since 1985. Short detective tales and poetic fragments like the one below thread through a future earth where aliens politely control humanity. On the surface, this world seems much like ours: a place of cool afternoon shadows and gently rolling hills, half-empty trains and sleepy downtown streets. But the closer you look, the weirder it gets. Mysterious figures suddenly appear in childhood photos, family members disappear forever without warning, power outages abound, and certain people gain the power of flight. The loosely plotted comics raise more questions than they answer, leaving behind a lingering sense of existential unease and dread. A new collection of Mauretania Comics, selected and designed by the acclaimed cartoonist Seth, will be published by New York Review Comics on May 1. Below, read "Endless Summer Wells," a strange and melancholy comic that appears in the forthcoming book.
From The New World: Comics from Mauretania, by Chris Reynolds. Excerpt courtesy of New York Review Comics.


text:
ONEFC
Sports Manila
Sterling Brown is the NBA draft sleeper you want your team to take
The SMU senior is flying under the radar in the NBA draft, but he shouldn't be.
A look at recent draft history shows an interesting trend in which players are most often undervalued by NBA teams. In just the past two years, Patrick McCaw, Malcolm Brogdon, Josh Richardson, Norman Powell, and Tyler Johnson have all either been second-round picks or went undrafted. A little farther back Jae Crowder, Danny Green, and Wesley Matthews have been some of the biggest second-round steals of the past 10 years.
The common thread that ties them all together is an ability to play defense that got undervalued by NBA GM's who tend to draft for offense first. All of them but McCaw were seniors when they were drafted.
Who fits the bill this year? SMU's Sterling Brown is the first guy on my list.
Brown is one of those prospects who looks like a poor athlete at first glance due to his physique, but is actually surprisingly explosive. As the brother of former NBA player Shannon Brown, it shouldn't be shocking that he can get up a little.
More importantly than being able to dunk, Brown can move well on defense.
He is attentive keeping track of guys off the ball and usually a step ahead on rotations. On the ball, he has good enough feet and reactions to contain ball handlers. At 6'5 with a 6'9.5 wingspan and a solid 225-pound frame, Brown has the size to comfortably check most shooting guards or small forwards, and even play some power forward in small lineups.
He's not a lockdown one-on-one perimeter defender, but he's smart and instinctive. Playing with consistent effort and paying attention to his team's scheme allows him to be a plus defender. Watch this sequence where he starts in the bottom right of the screen. He trails his man through off-ball screens, closes out on balance, brushes through an on-ball screen, then forces a turnover with his length.
Brown's similarly well-rounded offensive game is what makes him such an appealing prospect.
He has been an efficient player throughout his college career because he knows how to play his role. Career numbers of 45.1 percent from three, 53.4 percent from two, and a 61.6 percent true-shooting mark are rare from a college wing.
His extreme efficiency reveals what there is to like about his profile and shows some of his weaknesses. He is a very good shooter who generally only takes good shots. However, if a team expects him to create for himself at all in the NBA, they will be disappointed.
It's not that he can't handle the ball. He actually plays in pick-and-roll a fair amount and can drive to the basket or pull-up for jumpers off his dribble.
Yet, despite his skill level, Brown's career usage rate at SMU is a measly 16.7 percent. Part of that is due to the talent that has been at SMU and his own selectivity, but lack of elite creativity or athleticism is his primary limitation. The biggest concern with Brown is that because he played a relatively minor role at SMU, he won't be able to translate his game to the NBA level.
But further watching of SMU eases those concerns.
The Mustangs play a motion and pass-heavy style that encourages all five players to get involved. No one dominates the ball. Semi Ojeleye played the role of finisher most often, but Brown was the key to the whole team with his ability to both score efficiently himself and create for others. He averaged over three assists per 40 minutes each of his last three NCAA seasons and even brought the ball up some for SMU.
No, Brown is not going to be a primary or even secondary creator at the NBA level. But his ability to pass the ball and play within the offense will allow him to bring value even in a minor offensive role.
Some of the most valuable role-playing wings in NBA history know how to be selective and not dominate the ball. Guys like Shane Battier, Robert Horry, or even Jared Dudley, have been successful NBA offensive players precisely because they can shoot, make good decisions with the ball, and don't stray from their comfort zone.
It would be optimistic to expect Brown to be as good as any of those guys. Nonetheless, his profile suggests he could play a similar role. His combination of defensive ability and offensive aptitude is exactly what the NBA has missed on recently.
All sorts of teams would love to have Brogdon, McCaw, Richardson, Powell, and Johnson on their teams. Brown is exactly the kind of sleeper that can join that list.

redpajama_set_name: RedPajamaCommonCrawl


1. Cook the rice seperately and keep it aside.
2. Heat oil in a wide vessel and splutter some cumin seeds.
3. Saute the onions, green chillies & ginger.
4. Add finely chopped cilantro & curry leaves.
5. Add chopped potato, mix it, cover and cook.
6. After the potato is well cooked, mix the cooked rice and garam masala.
7. Add salt and lemon juice.
8. Atlast add finely chopped cilantro & green chilly.
- Instead of cooked rice, you could add thick poha (aval) and make aval upma (batata powa) and serve with mint chutney for a quick breakfast, brunch or dinner.
- You could also add some peanuts for some added crunch.


An earnest and accessible anthem of prayer, voiced here for men's choir. The inspired text from Nancy Kraft calls for the healing of our world through compassion, forgiveness, and understanding. Featuring a lovely, uplifting musical setting by Tom Keesecker. The score includes a reproducible page repurposing the anthem's refrain as a congregational prayer response.


Poppy is a verbal baby but she hasn't had any discernible (to people outside her immediate circle) words until this week. Poppy is officially, intentionally, clearly saying "ball" when she sees her ball. It is so cool.
"Wish I Knew You" by The Revivalists. You guys, this song has been on repeat for days.
It's BC Election Day! I didn't manage to vote in the advance polls so I'm heading out today to exercise my right to vote. If you're a BC local, I hope you are too. Not sure who to vote for? This tool can help you decide.
The rain has stopped, the sun is shining, and spring is in the air. It's a great day, even if it did start at 5 this morning because Poppy hates sleep.


text:
Q: How to increase the font size of FCM push notification text How to increase the font size or change the font of FCM push notification body.
 let FCMmessage = {
  to: '/topics/marketing',
  data: data,
  'priority': 'high',
  notification: {
    image: 'url-to-image',
    body:'body'
  }
}

A: Have you tried creating a custom layout with different font sizes and using Notification#setContent(RemoteViews r)?

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}
\label{intro}
In the present paper, we consider instability of radially symmetric
vortex solitons to 2-dimensional nonlinear Schr\"odinger equations
\begin{equation}
  \label{eq:NLS}
\left\{
  \begin{aligned}
& iu_t+\Delta u+f(u)=0 \quad\text{for
$(x,t)\in\mathbb{R}^n\times\mathbb{R}$},
\\
& u(x,0)=u_0(x)\quad\text{for $x\in\mathbb{R}^2$,}
  \end{aligned}\right.
\end{equation}
where $n=2$ and $f(u)=|u|^{p-1}u.$
Let $\omega>0$, $m\in\mathbb{N}\cup\{0\}$, and let
$e^{i(\omega t+m\theta)}\phi_\omega(r)$ be a standing
wave solution of \eqref{eq:NLS} belonging to $H^1(\mathbb{R}^2)$.
Here $r$ and $\theta$ denote polar coordinates in $\mathbb{R}^2$.
Then $\phi_\omega(r)$ is a solution to
\begin{equation}
\label{eq:B}
\left\{
  \begin{aligned}
& \phi''+\frac{1}{r}\phi'-\left(\omega+\frac{m^2}{r^2}\right)\phi+f(\phi)=0
\quad\text{for $r>0$},
\\
& \lim_{r\to0}\frac{\phi(r)}{r^m}=\lim_{r\to0}\frac{\phi'(r)}{mr^{m-1}},\\
& \lim_{r\to\infty}\phi(r)=0.
  \end{aligned}\right.
\end{equation}
We remark that
$e^{im\theta}\phi_\omega(r)$ is a solution to the scalar field equation
\begin{equation}
  \label{eq:sfe}
  \Delta \varphi-\omega \varphi+f(\varphi)=0 \quad\text{for $x\in\mathbb{R}^2$}.
\end{equation}
A standing wave solution of the form $e^{i(\omega t+m\theta)}\phi_\omega(r)$
appears in the study of nonlinear optics (see references in \cite{PWa}).
If $m=0$ and $\phi_\omega(r)$ is positive, then $\phi_\omega$ is
a ground state. Existence and uniqueness of the ground state are
well known (see \cite{BeLi1}, \cite{BLP}, \cite{Kw} and reference therein).

If $m\ne 0$, Iaia and Warchall proved the existence of smooth
solutions to \eqref{eq:B} with any prescribed number of zeroes.
The uniqueness of positive solutions has been proved by
\cite{Mi} by using the
classification theorem of positive solutions due to Yanagida and
Yotsutani \cite{YY}.

Let $c>0$ and let $Q_c$ be a positive solution to
\begin{equation}
  \label{eq:1D}
\left\{
\begin{aligned}
& Q''-cQ+f(Q)=0 \quad\text{for $x\in\mathbb{R}$,}
\\ &  \lim_{x\to\pm\infty}Q(x)=0,
\\ &  Q(0)=\max_{x\in\mathbb{R}}Q(x).
    \end{aligned}
\right.
\end{equation}
Then
\begin{equation}
  \label{eq:Qc}
Q_c(x)=\left(\frac{(p+1)c}{2}\right)^{\frac{1}{p-1}}\sech^{\frac{2}{p-1}}
\left(\frac{(p-1)\sqrt{c}}{2}x\right).
\end{equation}
\par
In \cite{PWa}, Pego and Warchall numerically observe that as spin index
$m$ becomes larger, a solution $\phi_\omega(r)$ to \eqref{eq:B} remains
small initially and then is approximated by $Q_c(r-\bar{r})$ around
$r=\bar{r}$, where $c=\omega+(m^2/\bar{r}^2)$
and $\bar{r}$ is a positive number with
$\bar{r}=O(m)$ as $m\to\infty$ (see also \cite{MoK} and references
 in \cite{PWa}).
One of our goals in the present paper is to explain this phenomena.
Benci and D'Aprile \cite{BdA} studied \eqref{eq:B} in a general
setting and locate the asymptotic peak of solutions (see also \cite{dA}).
Recently, Ambrosetti, Malchiodi and Ni \cite{AMN} have proved
the existence of positive radial solutions concentrating on spheres to
a class of singularly perturbed problem
$$\varepsilon^2\Delta u-Vu+|u|^{p-1}u=0,$$
and obtain their asymptotic profile.
Adopting the argument in \cite{AMN}, we obtain the following.
\begin{theorem}
\label{thm1}
Let $p>1$ and let $\phi_{\omega,m}$ be a positive solution to
\eqref{eq:B}. Then there exists an $m_*\in\mathbb{N}$ such that if $m\ge m_*$,
\begin{gather}
\label{eq:1.5}
\|\phi_{\omega,m}(\cdot)-Q_c(\cdot-\bar{r})\|_{H^2_{r}(\mathbb{R}^2)}
=O(m^{-1/2}),\\
\label{eq:1.6}
\|\phi_{\omega,m}(\cdot)-Q_c(\cdot-\bar{r})\|_{L^\infty(\mathbb{R}^2)}=O(m^{-1}),
\end{gather}
where $\bar{r}=2m/\sqrt{(p-1)\omega}$ and $c=(p+3)\omega/4$.
\end{theorem}
\begin{remark}
 Let $r=ms$, $\varepsilon=1/m$ and $V(r)=\omega+r^{-2}$. Then
\eqref{eq:B} is transformed into
$$
\varepsilon^2\Delta_r \phi-V(r)\phi+f(\phi)=0.$$
Though \cite{AMN} assumes the boundedness of $V(r)$ and cannot be
applied directly to our problem,
a maximum point of $\phi_{\omega,m}(r)$ can be predicted from an
\textit{auxiliary weighted potential} $rV(r)$ introduced by \cite{AMN}.
\end{remark}

Let $\varphi_\omega$ be a ground state to \eqref{eq:sfe}.
As is well known, the standing wave solution $e^{i\omega t}\varphi_\omega$
is stable if $d\|\varphi_\omega\|_{L^2(\mathbb{R}^n)}^2/d\omega>0$ and unstable
if $d\|\varphi_\omega\|_{L^2(\mathbb{R}^n)}^2/d\omega<0$. See e.g. Berestycki-Cazenave
\cite{BeCa}, Cazenave-Lions \cite{CL}, Grillakis-Shatah-Strauss
\cite{GSS1}, \cite{GSS2}, Shatah \cite{Sh},
Shatah-Strauss \cite{ShSt} and  Weinstein \cite{We2}.
Namely, the standing wave solution $e^{i\omega t}\varphi_\omega$
is stable if $1<p<1+4/n$ and unstable if $p\ge1+4/n$.
Grillakis \cite{Gr} proved that every radially symmetric standing wave
solution is linearly unstable if $p>1+4/n$.
However, to the best our knowledge, it remains unknown whether there
exists an unstable standing wave solution with higher energy
in the subcritical case ($1<p<1+4/n$).

Using Theorem \ref{thm1}, we find an unstable direction and
prove $e^{i(\omega t+m\theta)}\phi_\omega(r)$ is unstable in $H^1(\mathbb{R}^2)$
if $p>1$ and $m$ is sufficiently large.
\begin{theorem}
\label{thm2}
Let $p>1$ and $\phi_{\omega,m}$ be as in Theorem \ref{thm1}.
Then there exists an $m_*\in\mathbb{N}$ such that if $m\ge m_*$,
a standing wave solution
$e^{i(\omega t+m\theta)}\phi_\omega$ is linearly unstable.
\end{theorem}

\begin{remark}
By Shatah-Strauss Lemma (see \cite{ShSt2,SW}, see also \cite{Mi2}),
we have orbital instability of the linearly unstable standing wave
solutions.
\end{remark}
\begin{remark}
 If $p<1+4/n$ and $u_0\in H^1(\mathbb{R}^n)$, a solution to \eqref{eq:NLS}
exists globally in time and remains bounded in $H^1(\mathbb{R}^n)$.
Thus the mechanism of instability shown in Theorem \ref{thm2} is quite
different from that of \cite{BeCa} where solutions around a
standing wave solution blow up in finite time.
The instability mechanism we find is close to transversal long-wave
instability of 1-dimensional soliton (see Alexander-Pego-Sachs \cite{APS} for KP equation and Bridges \cite{B1,B2} for nonlinear Schr\"{o}dinger
equation). Theorem \ref{thm1} shows that a profile of vortex soliton
is close to 1D-soliton for large $m$ and thus it becomes possible
to find unstable modes by using perturbation method.
\end{remark}
\begin{remark}
If $\phi_{\omega, m}$ is nonnegative, then
$e^{i(m\theta+\omega t)}\phi_\omega(r)$ is a ground state in the class
$X_m=\{e^{im\theta}v(r)\,|\, v\in H^1_{rad}(\mathbb{R}^2),\;
 v\in L^2_{rad}(\mathbb{R}^2)\}$ and it follows from Grillakis
\textit{et al.} (\cite{GSS1}) that the standing wave solution
$e^{i(m\theta+\omega t)}\phi_\omega(r)$
is stable in the class $X_m$ if $1<p<3$ (\cite{Mi}). Thus the vortex
soliton is stable to the symmetric perturbations in the subcritical case.
\end{remark}
Our plan of the present paper is as follows.
In Section 2, we specify a solution to \eqref{eq:B}  which is expected
to become close to a solution to \eqref{eq:1D} as $m$ tends to infinity.
In Section 3, we investigate some properties of the
linearized operator around an approximate solution constructed is
Section 2. In Section 4, we prove Theorem \ref{thm1} following the
lines of \cite{AMN} and using Liapunov Schmidt method.
Since $\|\phi_\omega(r)\|_{L^2_{r}(\R^2)}$ grows up as $m\to\infty$ whereas
$\|\phi_\omega(r)\|_{L^\infty}$ remains bounded,
we need to estimate both $L^2_{r}(\R^2)$-norm and
$L^\infty_r(\mathbb{R}^2)$-norm of the solution to obtain Theorem \ref{thm1}.
In Section 5, we prove that $e^{im(\omega t+m\theta)}\phi_\omega$ is unstable
to the perturbations of the form $e^{i(m+j)\theta}v(r)$ with $|j|\sim
m^{\min(p-1,1)/6}$ and obtain Theorem \ref{thm2}.
\par
Finally, we introduce several notations.
For Banach spaces $X$ and $Y$, let $B(X,Y)$ be the space of all
bounded linear operators from $X$ to $Y$
and let $\|A\|_{B(X,Y)}$ be the operator norm of an operator
$A\colon X\to Y$. We abbreviate $B(X,X)$ as $B(X)$.
We denote by $D(A)$ and $R(A)$ the domain and the range of the
operator $A$, respectively. We use notations
$\|f\|_{L^2_{r}(\R^2)}=\left(\int_0^\infty|f(r)|^2rdr\right)^{\frac12}$,
$\|f\|_{H^1_r(\mathbb{R}^2)}=\left(
\int_0^\infty\left(|f'(r)|^2+|f(r)|^2\right)rdr\right)^{\frac12}$,
$\Delta_r=\partial_r^2+r^{-1}\partial_r$ and
$\|f\|_{H^2_{r}(\mathbb{R}^2)}=\|(1-\Delta_r)f\|_{L^2_{r}(\R^2)}$.
Various constants will be simply denoted by $C$ and $C_i$
($i\in\mathbb{N}$) in the course of calculations.

\section{An approximation}
\label{sec:2}
In this section, we will construct an approximate solution to
\eqref{eq:B} for large $m$.
Suppose that a positive solution to \eqref{eq:B} is approximated by
$Q_c(r-\bar{r})$ around $r=\bar{r}$ for large $m$.
Let
$\alpha_0=\bar{r}/m$, $\varepsilon=m^{-1}$, $s=r-\bar{r}$ and
$v(s)=\phi_\omega(r)$.
Then \eqref{eq:B} transforms into
\begin{equation}
  \label{eq:B1}
  \left\{
    \begin{aligned}
& v_{ss}+\frac{\varepsilon}{\alpha_0+\varepsilon s}v_s
-\left(\omega+\frac{1}{(\alpha_0+\varepsilon s)^2}\right)v+f(v)=0
\quad\text{for $s\in(-\bar{r},\infty)$,}
\\
& \lim_{s\to-\bar{r}}\frac{v(s)}{(s+\bar{r})^m}
=\lim_{s\to-\bar{r}}\frac{v_s(s)}{m(s+\bar{r})^{m-1}},
\\
& \lim_{s\to\infty}v(s)=0.
    \end{aligned}\right.
\end{equation}
Substituting $v(s)=v_0(s)+\varepsilon v_1(s)+O(\varepsilon^2)$
into \eqref{eq:B} and formally equating the power of $\varepsilon$, we
obtain
\begin{equation}
  \label{eq:approx0}\left\{
  \begin{aligned}
  & v_0''-cv_0+f(v_0)=0,\\
  & \lim_{s\to\pm\infty}v_0(s)=0,
  \end{aligned}\right.
\end{equation}
and
\begin{equation}
  \label{eq:approx1}\left\{
  \begin{aligned}
& v_1''-cv_1+f'(v_0)v_1=-\alpha_0^{-1}v_0'-2\alpha_0^{-3}sv_0,
\\
& \lim_{s\to\pm\infty}v_1(s)=0,
  \end{aligned}\right.
\end{equation}
where $c=\omega+\alpha_0^{-2}$.
Let $v_0(s)=Q_c(s)$, $L_c:=\partial_s^2-c+f'(Q_c)$ and $D(L_c)=H^2(\mathbb{R})$.
Since $\ker(L_c)=\operatorname{span}\{Q_c'\}$, the Fredholm
alternative implies that
\eqref{eq:approx1} has a solution $v_1\in L^2(\mathbb{R})$ if and only if
\begin{equation}
\label{eq:alpha}
\int_\mathbb{R} Q_c'(s)\left(Q_c'(s)+\frac{2s}{\alpha_0^2}Q_c(s)\right)ds
=\int_\mathbb{R} \left(Q_c'(s)^2-\frac{1}{\alpha_0^2}Q_c(s)^2\right)ds=0.
\end{equation}
\begin{lemma}
\label{lem:amplitude}
Let $c=\omega+\alpha_0^{-2}$ and let $Q_c$ be a solution to
\eqref{eq:1D}. If \eqref{eq:alpha} holds,
then $c=(p+3)\omega/4$ and $\alpha_0=2/\sqrt{(p-1)\omega}$.
\end{lemma}
\begin{proof}
By \eqref{eq:1D},
$$
\left(\frac{dQ_c}{dx}\right)^2=cQ_c^2\left(
1-\left(\frac{Q_c}{A}\right)^{p-1}\right),
$$
where $A^{p-1}=(p+1)c/2$.
We compute
\begin{align*}
\int_{-\infty}^\infty Q_c'(x)^2dx=& 2\int_0^A
\left(\frac{dQ_c}{dx}\right)^2\left(-\frac{dx}{dQ_c}(x)\right)dQ_c
\\ =&
2\sqrt{c}\int_0^A u\sqrt{1-\bigl(\frac{u}{A}\bigr)^{p-1}}du
\\ =&
\frac{2}{p-1}\sqrt{c}A^2B\left(\tfrac{2}{p-1},\tfrac{3}{2}\right),
\end{align*}
and
\begin{align*}
\int_{-\infty}^\infty Q_c(x)^2dx =&
2\int_0^AQ_c^2\left(-\frac{dx}{dQ_c}\right)dQ_c
\\=&
\frac{2}{\sqrt{c}}\int_0^A\frac{u}{\sqrt{1-\bigl(\frac{u}{A}\bigr)^{p-1}}}du
\\=&
\frac{2A^2}{(p-1)\sqrt{c}}B\left(\tfrac{2}{p-1},\tfrac12\right).
\end{align*}
Combining the above, we have $c=(p+3)\omega/4$  and
$\alpha_0=2/\sqrt{(p-1)\omega}$.
\end{proof}

Let $\chi(s)$ be smooth nonnegative functions on $\mathbb{R}$ satisfying
$0\le\chi(r)\le1$ and
$$
\chi(r)=\left\{\begin{aligned}
& 1\quad \text{ if $|r|\le 2$,}\\
& 0\quad \text{ if $|r|\ge 3$,}\end{aligned}\right.
$$
and let  $\chi_l(s)=\chi(s/l)$, where
$l=-\frac{2}{\sqrt{c}}\max(1,\frac1{p-1})\log\varepsilon$.
Following \cite{AMN}, we put
\begin{align}
\label{eq:de1}
& \Phi(\varepsilon,\rho)(r)=\chi_l(r-\rho)Q_c(r-\rho),
\quad c=\omega+(\varepsilon\rho)^{-2},
\\
\label{eq:de2}
&\phi_{\omega,m}=\Phi(\varepsilon,\rho)+w,
\end{align}
and search for a positive solution to \eqref{eq:B} for large $m$.
To fix the decomposition \eqref{eq:de2}, we assume
\begin{equation}
  \label{eq:de3}
  \left(w,\partial_\rho\Phi\right)_{L^2_{r}(\R^2)}=0.
\end{equation}
Substituting \eqref{eq:de1} into \eqref{eq:B}, we obtain
\begin{equation}
  \label{eq:2.6}
\mathcal{L}(\varepsilon,\rho)w+R_1(\varepsilon,\rho,w)+R_2(\varepsilon,\rho)=0,
\end{equation}
where $R_2=R_{21}+R_{22}+R_{23}$ and
\begin{align*}
\mathcal{L}(\varepsilon,\rho)=& \Delta_r-\omega-\frac{m^2}{r^2}
+f'(\Phi(\varepsilon,\rho)),
\\
R_1=&f(\Phi(\varepsilon,\rho)+w)-f(\Phi(\varepsilon,\rho))
-f'(\Phi(\varepsilon,\rho))w,
\\
R_{21}=&f(\Phi(\varepsilon,\rho))-\tau_\rho(\chi_lf(Q_c))
\\
R_{22}=&
\left(c-\omega-\frac{m^2}{r^2}\right)\Phi(\varepsilon,\rho)
+\frac{1}{r}\tau_\rho(\chi_lQ_c')
\\
R_{23}=& \tau_\rho\left(\chi_l''Q_c+2\chi_l'Q_c'\right)
+\frac{1}{r}\tau_\rho\left(\chi_l'Q_c\right).
\end{align*}
Here $\tau_h$  denotes the translation, that is, $(\tau_hf)(x)=f(x-h)$.
We will search a solution $(\rho,w)$ to \eqref{eq:de3} and \eqref{eq:2.6}
with $\rho\in(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)$ for large
$m\in\mathbb{N}$.

\section{Spectrum of the linearized operator
$\mathcal{L}(\varepsilon,\rho)$}
In this section, we examine spectral properties of the linearized
operator $\mathcal{L}(\varepsilon,\rho)$.
To begin with, we recall some properties of the operator
$\Delta_r-\omega-m^2/r^2.$
\begin{lemma}
  \label{lem:3.1}
Let $0<\varepsilon<1/2$ and $\mathcal{L}_0(\varepsilon) \colon L^2_{r}(\R^2)\to L^2_{r}(\R^2)$
be a closed operator such that
$$\mathcal{L}_0(\varepsilon)u=\Delta_ru-\omega u-(\varepsilon r)^{-2}u$$
for $u\in C_0^\infty(\mathbb{R}_+).$
Then $\mathcal{L}_0(\varepsilon)$ is a self-adjoint operator with
$$
D(\mathcal{L}_0(\varepsilon))=\{u\in H^2_{r}(\mathbb{R}^2)\,|\,r^{-2}u\in L^2_{r}(\R^2)\}
\quad\text{and}\quad R(\mathcal{L}_0(\varepsilon))=L^2_{r}(\R^2).$$
\end{lemma}
\begin{proof}
Let $X=\{H^2_{r}(\mathbb{R}^2)\,|\, r^{-2}u\in L^2_{r}(\R^2)\}$ be a Hilbert space
equipped with the norm
$\|u\|_{X}=(\|u\|_{H^2_{r}(\mathbb{R}^2)}^2+\|r^{-2}u\|_{L^2_{r}(\R^2)}^2)^{1/2}$.

By Theorem 10.10 and Example 4 in \cite[Appendix to X.1]{RS},
the operator $\mathcal{L}_0(\varepsilon)$ is essentially self-adjoint in
$C_0^\infty(\mathbb{R}_+)$. Thus for any $u\in D(\mathcal{L}_0(\varepsilon))$,
there exist $u_n\in C_0^\infty(\mathbb{R}_+)$ $(n\in\mathbb{N})$ such that
$\mathcal{L}_0(\varepsilon)u_n\to \mathcal{L}_0(\varepsilon)u$ and
 $u_n\to u$ in $L^2_{r}(\R^2)$ as $n\to\infty$.

Integrating by parts, we have
\begin{equation}
\label{eq:3.6}
  \begin{split}
&\|\mathcal{L}_0(\varepsilon)w\|_{L^2_{r}(\R^2)}^2 \\=&
\|(\omega-\Delta_r)w\|_{L^2_{r}(\R^2)}^2+
2\Re\left((\omega-\Delta_r)w,(\varepsilon r)^{-2}w\right)_{L^2_{r}(\R^2)}
+\left\|(\varepsilon r)^{-2}w\right\|_{L^2_{r}(\R^2)}^2
\\ \ge &
\|(\omega-\Delta_r)w\|_{L^2_{r}(\R^2)}^2+
(\varepsilon^{-4}-4\varepsilon^{-2})\left\|r^{-2}w\right\|_{L^2_{r}(\R^2)}^2
\end{split}
\end{equation}
for every $w\in C_0^\infty(\mathbb{R}_+)$.
Eq. \eqref{eq:3.6} yields that $\{u_n\}_{n=1}^\infty$ is a Cauchy
sequence in $X$ and $u_n\to u$ in $X$ as $n\to\infty$.
Thus we have $D(\mathcal{L}_0(\varepsilon))\subset X$.
\par
Next we prove $D(\mathcal{L}_0(\varepsilon))\supset X$.
For every $u\in X$, there exist $u_n\in C_0^\infty(\mathbb{R}_+)$ $(n=1,2,\cdots)$
such that
$\lim_{n\to\infty}\|u_n-u\|_{X}=0$.
Since
$$
\|\mathcal{L}_0(\varepsilon)w\|_{L^2_{r}(\R^2)}\le \max(1,\omega)
\|w\|_{H^2_{r}(\mathbb{R}^2)}+\varepsilon^{-2}\|r^{-2}w\|_{L^2_{r}(\R^2)},$$
we see that $\{\mathcal{L}_0(\varepsilon)u_n\}_{n=1}^\infty$
 and $\{u_n\}_{n=1}^\infty$
are Cauchy sequences in $L^2_{r}(\R^2)$ and that there exists a
$v\in L^2_{r}(\R^2)$ such that $\mathcal{L}_0(\varepsilon)u_n\to v$ as $n\to\infty$.
Since $\mathcal{L}_0(\varepsilon)$ is closed, it follows that
 $v=\mathcal{L}u$ and $u\in D(\mathcal{L}_0(\varepsilon))$. Thus we prove
$D(\mathcal{L}_0(\varepsilon))=X$.

Finally, we will show that $R(\mathcal{L}_0(\varepsilon))=L^2_{r}(\R^2)$.
The self-adjointness of $\mathcal{L}_0(\varepsilon)$ and \eqref{eq:3.6}
implies $$R(\mathcal{L}_0(\varepsilon))^\perp
=\ker(\mathcal{L}_0(\varepsilon))=\{0\}.$$
Hence it follows that $\overline{R(\mathcal{L}_0(\varepsilon))}=L^2_{r}(\R^2)$
and that for every $v\in L^2_{r}(\R^2)$, there exist $u_n\in X$ and $v_n\inL^2_{r}(\R^2)$
$(n\in\mathbb{N})$ such that
$$\mathcal{L}_0(\varepsilon)u_n=v_n\to v\quad\text{in $L^2_{r}(\R^2)$ as $n\to\infty$.}$$
By \eqref{eq:3.6}, there exists $u\in X$ such that $\lim_{n\to\infty}u_n=u$
in $X$. Since $\mathcal{L}_0(\varepsilon)$ is closed we have
$v=\mathcal{L}_0(\varepsilon)u\in R(\mathcal{L}_0(\varepsilon))$.
This completes the proof of Lemma \ref{lem:3.1}.
\end{proof}

Let $\mathcal{P}(\varepsilon,\rho)$ and $\mathcal{Q}(\varepsilon,\rho)$ be
orthogonal projections defined by
\begin{align*}
&\mathcal{P}(\varepsilon,\rho)u=\|\partial_\rho\Phi(\varepsilon,\rho)\|_{L^2_{r}(\R^2)}^{-2}
(u,\partial_\rho\Phi(\varepsilon,\rho))_{L^2_{r}(\R^2)}\partial_\rho\Phi(\varepsilon,\rho),
\\ & \mathcal{Q}(\varepsilon,\rho)=I-\mathcal{P}(\varepsilon,\rho).
\end{align*}
We will show that $\mathcal{L}(\varepsilon,\rho)$ is invertible on
$\mathcal{Q}(\varepsilon,\rho)L^2_{r}(\R^2)$.
\begin{lemma}
  \label{lem:3.2}
Let $w\in H^1_r(\mathbb{R}^2)\cap \{u\,|\, r^{-1}u\in L^2_{r}(\R^2)\}$ and
$$(w,\partial_\rho\Phi(\varepsilon,\rho))_{L^2_{r}(\R^2)}
=(w,\Phi(\varepsilon,\rho)^{\frac{p+1}{2}})_{L^2_{r}(\R^2)}=0.$$
Then there exist positive numbers $\varepsilon_*$ and $c_1$ such that
$$-(\mathcal{L}(\varepsilon,\rho)w,w)_{L^2_{r}(\R^2)} \ge c_1\|w\|_{L^2_{r}(\R^2)}^2$$
for every $\varepsilon\in(0,\varepsilon_*)$ and
$\rho\in(\alpha_0/(2\varepsilon),2\alpha_0/\varepsilon)$.
\end{lemma}
To prove Lemma \ref{lem:3.2}, we need the following.
\begin{lemma}
  \label{lem:l1}
Let $p>1$, $c>0$ and $\lambda_0=(p-1)(p+3)/4$.
Then $$L_c Q_c^{\frac{p+1}{2}}=\lambda_0 cQ_c^{\frac{p+1}{2}}.$$
Furthermore, $\ker(L_c)=\{\beta Q_c'\,|\,\beta\in\mathbb{R}\}$
and there exists a positive number $b$ depending only on of $p$ such that
$\sigma(L_c)\setminus\{0,\lambda_0 c\} \subset (-\infty,-bc].$
\end{lemma}
\begin{proof}
The former part of the lemma can be obtained by a simple computation.
Let $c=1$.
Weyl's essential spectrum theorem tells us that the spectrum of $L_1$
consists of essential spectrum $(-\infty,-1]$ and discrete eigenvalues.
Since $Q_1'$ has exactly one zero and $L_1Q_1'=0$,
it follows from Strum's comparison theorem that $0$ is a second
eigenvalue of $L_1$ and that $\ker(L_1)$ is spanned by $Q_1'$.
Since $L_c(u(c^{1/2}x))=c(L_1u)(c^{1/2}x)$ for every $u\in H^2(\mathbb{R})$, we
have $ \sigma(L_c)=\{c\lambda\,|\, \lambda\in\sigma(L_1)\}.$
Thus we prove Lemma \ref{lem:l1}.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem:3.2}]
Let $\chi_0(s)=1-\chi_l(s)$ and $\chi_1(s)=\chi_l(s)$. By \eqref{eq:Qc}
and the fact that $\mathrm{supp}\chi_0\subset \{r\in\mathbb{R}\,|\, |r|\ge 2l\}$,
\begin{align*}
(\mathcal{L}(\varepsilon,\rho)w,w)_{L^2_{r}(\R^2)}
=&
(\mathcal{L}(\varepsilon,\rho)\chi_1w,\chi_1w)_{L^2_{r}(\R^2)}
+2(\mathcal{L}_0(\varepsilon)\chi_0w,\chi_1w)_{L^2_{r}(\R^2)}
\\&+(\mathcal{L}_0(\varepsilon)\chi_0w,\chi_0w)_{L^2_{r}(\R^2)}
+O(e^{-2(p-1)\sqrt{c}l}\|w\|_{L^2_{r}(\R^2)}^2).
\end{align*}
Integrating by parts and substituting $|\chi_0'(r)|+|\chi_1'(r)|
=O(l^{-1})$ into the resulting equation, we have
\begin{equation}
  \label{eq:L00}
  \begin{split}
-&(\mathcal{L}_0(\varepsilon)\chi_0w,\chi_0w)_{L^2_{r}(\R^2)}\\
=& \int_0^\infty \left((\chi_0w)_r^2+(\omega +(\varepsilon r)^{-2})(\chi_0w)^2
\right)rdr
\\=& \int_0^\infty \chi_0^2\left(w_r^2+\omega w^2+(\varepsilon r)^{-2}w^2
\right)rdr+O(l^{-1}\|w\|_{H^1_r(\mathbb{R}^2)}^2),
  \end{split}
\end{equation}
and
\begin{equation}
  \label{eq:L01}
  \begin{split}
&-(\mathcal{L}_0(\varepsilon)\chi_0w,\chi_1w)_{L^2_{r}(\R^2)}
\\= &
\int_0^\infty \chi_0\chi_1\left(w_r^2+\omega w^2+(\varepsilon r)^{-2} w^2
\right)rdr +O(l^{-1}\|w\|_{H^1_r(\mathbb{R}^2)}^2).
  \end{split}
\end{equation}
Let $U\colon L^2_{r}(\R^2)\to L^2(\mathbb{R}_+)$ be the unitary operator
defined by $U\phi(r)=r^{\frac12}\phi(r)$.
Then
\begin{equation}
  \label{eq:L11}
  \begin{split}
\tau_{-\rho}U\mathcal{L}(\varepsilon)U^{-1}=&\partial_r^2-\omega
-\frac{1-\frac14\varepsilon^2}{(\alpha+\varepsilon r)^2}+f'(\chi_1Q_c)
\\=& L_c+\left(\frac{1}{\alpha^2}-
\frac{1-\frac14\varepsilon^2}{(\alpha+\varepsilon r)^2}\right)
+f'(\chi_1Q_c)-f'(Q_c),
  \end{split}
\end{equation}
where $\alpha=\rho/m$ and $c=\omega+\alpha^{-2}$.
Let $\tilde{\chi}_1$ and $\tilde{\chi}_2$ be smooth nonnegative functions
on $\mathbb{R}$ satisfying
\begin{align*}
& \sup_{r\in\mathbb{R}}|\tilde{\chi}_i'(r)|=O(l^{-1})\quad\text{for $i=0,1$,}\\
& \tilde{\chi}_0(r)=\left\{
  \begin{aligned}
    0 &\quad\text{if $|r|\le l$}\\
    1 &\quad\text{if $|r|\ge 2l$}
  \end{aligned}\right.,\quad
\tilde{\chi}_1(r)=\left\{
  \begin{aligned}
    1 &\quad\text{if $|r|\le 3l$}\\
    0 &\quad\text{if $|r|\ge 4l$}
  \end{aligned}\right..
\end{align*}
Put $\tilde w(r)=(r+\rho)^{1/2}\chi_1(r)w(r+\rho).$
Using $w\perp \partial_\rho \Phi(\varepsilon,\rho)$ and
\begin{equation}
\label{eq:pdP}
\partial_\rho\Phi(\varepsilon,\rho)=-\tau_\rho(\chi_1Q_c)'
-\frac{2\varepsilon}{\alpha^3}\tau_\rho(\chi_1\partial_cQ_c),
\end{equation}
we have
\begin{align*}
  0=&(w,\partial_\rho\Phi(\varepsilon,\rho))_{L^2_{r}(\R^2)}\\
=& -\int_{-\rho}^\infty (\chi_1Q_c)'(r)w(r+\rho)(r+\rho)dr
+O(\varepsilon^{1/2}\|w\|_{L^2_{r}(\R^2)})
\\=&
-\int_\mathbb{R} (\rho+r)^{1/2}\tilde{\chi}_1\tilde{w}Q_c'dr+
O\left((\rho^{1/2}e^{-2\sqrt{c}l}+\varepsilon^{1/2})\|w\|_{L^2_{r}(\R^2)}\right)
\\=&-\rho^{1/2}\int_\mathbb{R}\tilde{w}Q_c'dr+O(\varepsilon^{1/2}l\|w\|_{L^2_{r}(\R^2)}).
\end{align*}
Hence it follows that
\begin{equation}
  \label{eq:L12}
(\tilde{w},Q_c')_{L^2(\mathbb{R})}=O(\varepsilon\log\varepsilon\|w\|_{L^2_{r}(\R^2)}).
\end{equation}
Similarly, we have
\begin{equation}
  \label{eq:L12b}
(\tilde{w},Q_c^{\frac{p+1}{2}})_{L^2_{r}(\R^2)}=O(\varepsilon\log\varepsilon\|w\|_{L^2_{r}(\R^2)}).
\end{equation}
Combining Lemma \ref{lem:l1} with  \eqref{eq:L11}, \eqref{eq:L12} and
\eqref{eq:L12b}, we see that there exist positive constants $C_1$ and
$C_2$ such that
\begin{equation}
  \label{eq:L1}
-(\mathcal{L}(\varepsilon,\rho)\chi_1w,\chi_1w)_{L^2_{r}(\R^2)}
\ge C_1\|\tilde w\|_{H^1(\mathbb{R})}^2 \ge C_2\|\chi_1w\|_{H^1_r(\mathbb{R}^2)}^2.
\end{equation}
Thus by \eqref{eq:L00}, \eqref{eq:L01} and \eqref{eq:L1},
there exist positive numbers $c_1$ and $\varepsilon_*$ such that
$$-(\mathcal{L}(\varepsilon,\rho)w,w)_{L^2_{r}(\R^2)}\ge c_1\|w\|_{L^2_{r}(\R^2)}^2$$
for every $\varepsilon\in(0,\varepsilon_*)$,
$\rho\in(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)$.
\end{proof}
Let $X_1=\mathcal{Q}(\varepsilon,\rho)X$,
$Y_1=\mathcal{Q}(\varepsilon,\rho)L^2_{r}(\R^2)$ and
$\mathcal{A}(\varepsilon,\rho)=
\mathcal{Q}(\varepsilon,\rho)\mathcal{L}(\varepsilon,\rho)
\mathcal{Q}(\varepsilon,\rho)$. Lemma \ref{lem:3.2} yields that
$\mathcal{A}(\varepsilon,\rho)\colon X_1\to Y_1$ is isomorphic.
\begin{corollary}
\label{cor:3.3}
There exist positive numbers  $\varepsilon_*$ and $\nu$ such that
\begin{equation}
  \label{eq:cor3.3}
\|\mathcal{A}(\varepsilon,\rho)^{-1}u\|_{X}\le \nu \|u\|_{L^2_{r}(\R^2)}
\end{equation}
for every $u\in Y_1$, $\varepsilon\in(0,\varepsilon_*)$
and $\rho\in(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)$.
\end{corollary}
\begin{proof}
Let $Q_1$ and $Q_2$ be orthogonal projections such that
\begin{gather*}
Q_1u=\frac{(u,\mathcal{Q}(\varepsilon,\rho)
\Phi(\varepsilon,\rho)^{\frac{p+1}{2}})_{L^2_{r}(\R^2)}}
{\|\mathcal{Q}(\varepsilon,\rho)\Phi(\varepsilon,\rho)^{\frac{p+1}{2}}
\|_{L^2_{r}(\R^2)}^{2}}
\mathcal{Q}(\varepsilon,\rho)\Phi(\varepsilon,\rho)^{\frac{p+1}{2}},
\\
Q_2=\mathcal{Q}(\varepsilon,\rho)-Q_1.
\end{gather*}
Then $\mathcal{A}(\varepsilon,\rho)$ can be written as
$$
\mathcal{A}(\varepsilon,\rho)=
\begin{pmatrix}
Q_1\mathcal{L}(\varepsilon,\rho)Q_1 & Q_1\mathcal{L}(\varepsilon,\rho)Q_2\\
Q_2\mathcal{L}(\varepsilon,\rho)Q_1 & Q_2\mathcal{L}(\varepsilon,\rho)Q_2
\end{pmatrix}.$$
In view of Lemma \ref{lem:l1}, we see that there exists a $c_2>0$ such that
$$
(\mathcal{L}(\varepsilon,\rho)\mathcal{Q}(\varepsilon,\rho)
\Phi(\varepsilon,\rho)^{\frac{p+1}{2}},\mathcal{Q}(\varepsilon,\rho)
\Phi(\varepsilon,\rho)^{\frac{p+1}{2}})_{L^2_{r}(\R^2)}
\ge c_2
\|\mathcal{Q}(\varepsilon,\rho)\Phi(\varepsilon,\rho)^{\frac{p+1}{2}}
\|_{L^2_{r}(\R^2)}^2.$$
Furthermore, we see that
$$
\lim_{\varepsilon\downarrow0}
\left(\|Q_1\mathcal{L}(\varepsilon,\rho)Q_2\|_{B(L^2_{r}(\R^2))}
+\|Q_2\mathcal{L}(\varepsilon,\rho)Q_1\|_{B(L^2_{r}(\R^2))}\right)=0.$$
Combining the above with Lemma \ref{lem:3.2}, we obtain
\begin{equation}
\label{eq:3.7}
\sup_{\varepsilon\in(0,\varepsilon_*)}\sup_{\rho\in
(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)}
\|\mathcal{A}(\varepsilon,\rho)^{-1}\|_{B(L^2_{r}(\R^2))}<\infty.
\end{equation}

Let
\begin{align*}
\mathcal{B}(\varepsilon,\rho)=&
\mathcal{P}(\varepsilon,\rho)\mathcal{L}(\varepsilon,\rho)
+\mathcal{L}(\varepsilon,\rho)\mathcal{P}(\varepsilon,\rho)
-\mathcal{P}(\varepsilon,\rho)\mathcal{L}(\varepsilon,\rho)
\mathcal{P}(\varepsilon,\rho) -f'(\Phi(\varepsilon,\rho)).
\end{align*} Then
\begin{equation}
  \label{eq:3.8}
\mathcal{L}_0(\varepsilon)=\mathcal{A}(\varepsilon,\rho)
+\mathcal{B}(\varepsilon,\rho).
\end{equation}
Using \eqref{eq:3.7}, \eqref{eq:3.8} and the fact that
$$
\sup_{\varepsilon\in(0,\varepsilon_*)}\sup_{\rho\in
(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)}
\left\|\mathcal{B}(\varepsilon,\rho)\right\|_{B(L^2_{r}(\R^2))}<\infty,$$
we have
\begin{equation}
  \label{eq:3.9}
\begin{split}
\|\mathcal{L}_0(\varepsilon)\mathcal{A}(\varepsilon,\rho)^{-1}u\|_{L^2_{r}(\R^2)}
\le C\|u\|_{L^2_{r}(\R^2)}
\end{split}
\end{equation}
for every $u\in Y_1$, $\varepsilon\in(0,\varepsilon_*)$
and $\rho\in(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)$.
Combining \eqref{eq:3.6} and \eqref{eq:3.9}, we obtain \eqref{eq:cor3.3}.
\end{proof}
We will use the lemma below to estimate $L^\infty$-norm of $w$
in the following section.
\begin{corollary}
  \label{cor:3.4}
Let $p>1$. Then there exist positive numbers $\varepsilon_*$ and $C$
such that
\begin{equation}
\label{eq:cor34}
 \|\mathcal{A}(\varepsilon,\rho)^{-1}u\|_{L_r^\infty(\mathbb{R}^2)}
\le C\|u\|_{L_r^\infty(\mathbb{R}^2)}
\end{equation}
for every $u\in L^\infty_r(\mathbb{R}^2)\cap Y_1$,
$\varepsilon\in(0,\varepsilon_*)$ with $\varepsilon^{-1}\in \mathbb{N}$ and $\rho\in
(\alpha_0/(2\varepsilon),2\alpha_0/\varepsilon)$.
\end{corollary}
\begin{proof}
Let $m=\varepsilon^{-1}\in\mathbb{N}$ and
\begin{align*}
& P^{\perp}u=u-\|Q_c'\|_{L^2(\mathbb{R})}^{-2}(u,Q_c')_{L^2(\mathbb{R})}Q_c',\\
& \mathcal{K}(\varepsilon,\rho)
=\mathcal{Q}(\varepsilon,\rho)\left\{
(\tau_\rho\tilde{\chi}_0)\mathcal{L}_0(\varepsilon)^{-1}(\tau_\rho\chi_0)
+U^{-1}\tau_\rho\tilde{\chi}_1P^\perp L_c^{-1}P^\perp\chi_1\tau_{-\rho}
U\right\}\mathcal{Q}(\varepsilon,\rho).
\end{align*}
Noting that $e^{im\theta}\mathcal{L}_0(\varepsilon)u(r)
=(\Delta-\omega)(e^{im\theta}u(r))$, we have
$$\sup_{m\in\mathbb{N}}\|\mathcal{L}_0(\varepsilon)^{-1}\|_{B(L^\infty_r(\mathbb{R}^2))}
<\infty.$$
Furthermore, $L_c\colon P^\perp L^2(\mathbb{R})\to L^2(\mathbb{R})$ has a bounded
 inverse. Hence it follows that
\begin{equation}
  \label{eq:d1}
\sup_{m\ge\varepsilon_*^{-1}}\sup_{\rho\in
(\alpha_0/(2\varepsilon), 2\alpha_0/\varepsilon)}
\|\mathcal{K}(\varepsilon,\rho)\|_{B(L^\infty_r(\mathbb{R}^2))}<\infty.
\end{equation}
We compute
\begin{equation}
  \label{eq:d2}
\begin{split}
& \mathcal{A}(\varepsilon,\rho)\mathcal{K}(\varepsilon,\rho)
\\ =&
\mathcal{A}(\varepsilon,\rho)(\tau_\rho\tilde{\chi}_0)
\mathcal{L}_0(\varepsilon)^{-1}(\tau_\rho\chi_0)
+\mathcal{A}(\varepsilon,\rho) U^{-1}\tau_\rho
\tilde{\chi}_1P^\perp L_c^{-1}P^\perp\chi_1\tau_{-\rho} U.
\\=&I+II.
\end{split}
\end{equation}
Since $\mathcal{A}(\varepsilon,\rho)=\mathcal{L}_0(\varepsilon)-
\mathcal{B}(\varepsilon,\rho)$ and
$\|\mathcal{B}(\varepsilon,\rho)\tau_\rho\tilde{\chi}_0\|_{B(L^\infty)}=
O(e^{-\tilde{p}\sqrt{c}l})$, where $\tilde{p}=\min(1,p-1)$, we have
\begin{equation}
  \label{eq:d3}
\begin{split}
I=& \mathcal{Q}(\varepsilon,\rho)\mathcal{L}_0(\varepsilon)
(\tau_\rho\tilde{\chi}_0)\mathcal{L}_0(\varepsilon)^{-1}(\tau_\rho\chi_0)
-\mathcal{B}(\varepsilon,\rho)
(\tau_\rho\tilde{\chi}_0)\mathcal{L}_0(\varepsilon)^{-1}(\tau_\rho\chi_0)
\\=&
\mathcal{Q}(\varepsilon,\rho)\left\{\tau_\rho(\tilde{\chi}_0\chi_0)
+[\Delta_r,\tau_\rho\tilde{\chi}_0]\mathcal{L}_0(\varepsilon)^{-1}
\tau_\rho\chi_0
-\mathcal{B}(\varepsilon,\rho)(\tau_\rho\tilde{\chi}_0)
\mathcal{L}_0(\varepsilon)^{-1}(\tau_\rho\chi_0)\right\}
\\=&\tau_\rho\chi_0+O(l^{-1})\quad\text{in
 $B(\mathcal{Q}(\varepsilon,\rho)L^\infty_r(\mathbb{R}^2))$.}
\end{split}
\end{equation}
Let
$\mathcal{B}_1=
\mathcal{P}(\varepsilon,\rho)
\mathcal{L}(\varepsilon,\rho)\mathcal{P}(\varepsilon,\rho)
-\mathcal{L}(\varepsilon,\rho)\mathcal{P}(\varepsilon,\rho).$
Then $$\mathcal{A}(\varepsilon,\rho)=
\mathcal{Q}(\varepsilon,\rho)\mathcal{L}(\varepsilon,\rho)
+\mathcal{B}_1(\varepsilon,\rho).$$
In view of the definition of $\mathcal{P}(\varepsilon,\rho)$, \eqref{eq:pdP}
and the fact that $L_cQ_c'=0$, we have
$$\left\|\mathcal{B}_1(\varepsilon,\rho)\right\|_{B(L_r^\infty(\mathbb{R}^2))}
=O(\varepsilon l)$$
for $\varepsilon\in(0,\varepsilon_*)$ and $\rho\in
(\alpha_0/(2\varepsilon),2\alpha_0/\varepsilon)$.
Furthermore \eqref{eq:pdP} implies
$$
\left\|\mathcal{Q}(\varepsilon,\rho)-P^\perp\right\|_{B(L_r^\infty(\mathbb{R}^2))}
=O(\varepsilon).$$
Let $$\mathcal{R}=\alpha^{-2}-(\alpha+\varepsilon r)^{-2}
+f'(\chi_1Q_c)-f'(Q_c).$$
Then we have $\mathcal{L}(\varepsilon,\rho)=U^{-1}(\tau_\rho L_c)U
+\tau_\rho\mathcal{R}$ and
$$
\|(\tau_\rho\mathcal{R}\tilde{\chi}_1)(1-\Delta_r)^{-1}
\|_{B(L^\infty(\mathbb{R}^2))}
=O(\varepsilon l+e^{-2(p-1)\sqrt{c}l}).$$
Combining the above, we have
\begin{equation}
\label{eq:d4}
\begin{split}
II=& \mathcal{Q}(\varepsilon,\rho)U^{-1} \tau_\rho(P^\perp L_c)
\tau_\rho\tilde{\chi}_1 P^\perp L_c^{-1}P^\perp\chi_1\tau_{-\rho}U
+O(\varepsilon l)
\\=&
\mathcal{Q}(\varepsilon,\rho)\tau_\rho(\tilde{\chi}_1\chi_1)+O(l^{-1})
\\=& \tau_\rho\chi_1+O(l^{-1})\quad
\text{in $B(\mathcal{Q}(\varepsilon,\rho)L^\infty_r(\mathbb{R}^2)$.}
\end{split}
\end{equation}
From \eqref{eq:d1}--\eqref{eq:d4}, we deduce
\eqref{eq:cor34}.
\end{proof}
\section{The method of Liapunov-Schmidt}
In this section, we use the method of Liapunov-Schmidt to obtain a
solution to \eqref{eq:de3} and \eqref{eq:2.6}.
Let us translate \eqref{eq:2.6} into a system
\begin{align}
& \label{eq:4.1}
\mathcal{A}(\varepsilon,\rho)w+\mathcal{Q}(\varepsilon,\rho)
R_1(w,\varepsilon,\rho)+\mathcal{Q}(\varepsilon,\rho)R_2(\varepsilon,\rho)=0,
\\ &
\label{eq:4.2}
\mathcal{P}(\varepsilon,\rho)\left(\mathcal{L}(\varepsilon,\rho)w
+R_1(w,\varepsilon,\rho)+R_2(\varepsilon,\rho)\right)=0.
\end{align}
\begin{lemma}
  \label{lem:4.1}
Let $p>1$. Then there exist an $\varepsilon_0>0$ and a $C>0$ such that if
$\varepsilon\in(0,\varepsilon_0]$ and $\rho\in(\alpha_0/(2\varepsilon),
2\alpha_0/\varepsilon)$,
Eq. \eqref{eq:4.1} has a unique solution
$w(\varepsilon,\rho)$ that is continuous in $\varepsilon$ and $\rho$
and satisfies
\begin{gather}
  \label{eq:4.3}
\|w(\varepsilon,\rho)\|_X\le C\varepsilon^{\frac12}\quad\text{as
$\varepsilon\downarrow0$.}
\end{gather}
\end{lemma}
\begin{proof}
Let $T\colon X_1\times(0,\varepsilon_0]\times(\alpha_0/(2\varepsilon),
2\alpha_0/\varepsilon)\to X_1$ be a continuous mapping defined by
$$T(w,\varepsilon,\rho)=-\mathcal{A}(\varepsilon,\rho)^{-1}
\mathcal{Q}(\varepsilon,\rho)
\left\{R_1(w,\varepsilon,\rho)+R_2(\varepsilon,\rho)\right\},
$$ and let
$\widetilde{X}=\{w\in X_1\,|\,\|w\|_{X}\le r_0\},$
where $r_0$ is a positive number to be fixed later.

To begin with, we will show that $T$ maps $\widetilde{X}$ into itself.
We compute
\begin{equation}
\label{eq:4.4}
\begin{split}
\|R_{1}\|_{L^2_{r}(\R^2)} =& \left\|\int_0^1\left\{f'(\Phi(\varepsilon,\rho)+\theta w)
-f'((\Phi(\varepsilon,\rho))\right\}d\theta w\right\|_{L^2_{r}(\R^2)}
\\ \le & \delta(r_0)\|w\|_{L^2_{r}(\R^2)},
\end{split}
\end{equation}
where $\delta(r_0)$ is a positive constant with
$\lim_{r_0\downarrow0}\delta(r_0)=0$.
Eq. \eqref{eq:Qc} and the definition of $\chi_l$ imply
\begin{equation*}
  \begin{split}
\|R_{21}\|_{L^2_{r}(\R^2)} =& \left\|\tau_\rho\left\{(\chi_1^{p-1}-\chi_1)
Q_c^{p-1}\right\}\right\|_{L^2_{r}(\R^2)}
\\ \le & C\rho^{1/2}e^{-2(p-1)\sqrt{c}l},
  \end{split}
\end{equation*}
and
$$ \|R_{23}\|_{L^2_{r}(\R^2)} \le C\rho^{1/2}e^{-2\sqrt{c}l}.$$
Since $\rho^{-1}=O(\varepsilon)$ and
$l=-\frac{2}{\sqrt{c}}\max(1,\frac1{p-1})\log\varepsilon$,
\begin{equation}
  \label{eq:4.5}
\|R_{21}\|_{L^2_{r}(\R^2)}+ \|R_{23}\|_{L^2_{r}(\R^2)}\le C_1\varepsilon^{\frac72}.
\end{equation}
Using \eqref{eq:Qc} and $\alpha^{-2}-(\alpha+\varepsilon
s)^{-2}=\frac{2s}{\alpha^3}\varepsilon+O(\varepsilon^2 s^2),$
we have
\begin{align*}
& \left\|\left(c-\omega-\frac{m^2}{r^2}\right)\tau_\rho(\chi_1Q_c')
\right\|_{L^2_{r}(\R^2)}^2
\\ = & \varepsilon^{-1}\int_{-3l}^\infty(\alpha+\varepsilon s)
\left\{\left(\frac{1}{\alpha^2}-\frac{1}{(\alpha+\varepsilon s)^2}\right)
\chi_1(s)Q_c'(s)\right\}^2ds
\\ \le & C\varepsilon
\end{align*}
for every $\alpha=\varepsilon\rho\in(\alpha_0/2,2\alpha_0)$.
Similarly, we have
$$\left\|\frac{1}{r}\chi_1Q_c'\right\|_{L^2_{r}(\R^2)}\le C\varepsilon^{1/2}.$$
Thus we obtain
\begin{equation}
  \label{eq:4.6}
\|R_{22}\|_{L^2_{r}(\R^2)}\le C_2\varepsilon^{1/2}.
\end{equation}
Combining \eqref{eq:4.4}--\eqref{eq:4.6} with Corollary \ref{cor:3.3}, we have
\begin{equation}
  \label{eq:4.7}
 \|T(w,\varepsilon,\rho)\|_{X}\le
\nu(\delta(r_0)\|w\|_{X}+C_1\varepsilon^{7/2}+C_2\varepsilon^{1/2}).
\end{equation}
Put $r_0=2\nu C_2\varepsilon^{1/2}$. Then $T(\cdot,\varepsilon,\rho)$
maps $\widetilde{X}$ into itself if  $\varepsilon_0$ is  sufficiently
small.
\par
Next, we will show that $T(\cdot,\varepsilon,\rho)$ is a contraction mapping.
For $w_1$, $w_2\in \widetilde{X}$,
\begin{align*}
& \|T(w_1,\varepsilon,\rho)-T(w_2,\varepsilon,\rho)\|_{X}
\\ \le &
\nu\|R_1(w_1,\varepsilon,\rho)-R_1(w_2,\varepsilon,\rho)\|_{L^2_{r}(\R^2)}
\\= &
\nu\left\|\int_0^1\left\{f'(\Phi+\theta w_1+(1-\theta)w_2)-f'(\Phi)\right\}
d\theta(w_1-w_2)\right\|_{L^2_{r}(\R^2)}
\\ \le &
\Lambda\|w_2-w_1\|_{L^2_{r}(\R^2)}
\end{align*}
where
$\Lambda=\nu r_0^{\tilde{p}}
\sup_{\eta\in\widetilde{X}}\|f'(\Phi+\eta)\|_{C^{\tilde{p}}}$
and $\tilde{p}=\min(1,p-1)$.
Taking $\varepsilon_0$ smaller if necessary,
we see that $T(\cdot,\varepsilon,\rho)\colon
\widetilde{X}\to\widetilde{X}$ is a contraction mapping.
Thus we prove that there exists a solution $w(\varepsilon,\rho)$ to
 \eqref{eq:4.1} with $\|w\|_{X}\le 2\nu C_2\varepsilon^{1/2}$
that is continuous in $\varepsilon\in(0,\varepsilon_0)$ and $\rho\in
(\alpha_0/(2\varepsilon),2\alpha_0/\varepsilon)$.
\end{proof}

\begin{corollary}
  \label{cor:4.2}
Let $p>1$. Then there exist an $\varepsilon_0>0$ and a $C>0$ such that
if $\varepsilon\in(0,\varepsilon_0)$, $\varepsilon^{-1}\in\mathbb{N}$ and
$\rho\in (\alpha_0/(2\varepsilon),2\alpha_0/\varepsilon)$,
a solution $w(\varepsilon,\rho)$ to \eqref{eq:4.1} satisfies
\begin{equation}
  \label{eq:4.8}
\|w(\varepsilon,\rho)\|_{L^\infty} \le C\varepsilon.
\end{equation}
\end{corollary}
\begin{proof}
Analogously to \eqref{eq:4.4}--\eqref{eq:4.6}, we have
\begin{align*}
&\|R_{1}\|_{L^\infty}\le \delta(r_0)\|w\|_{L^\infty},
\\
&\|R_2\|_{L^2_{r}(\R^2)}\le \|R_{21}\|_{L^\infty}+\|R_{22}\|_{L^\infty}
+\|R_{23}\|_{L^\infty}=O(\varepsilon),
\end{align*}
where $\delta(r_0)$ is a positive number with
$\lim_{r_0\downarrow0}\delta(r_0)=0$.
Thus by Corollary \ref{cor:3.4},
\begin{align*}
 \|w(\varepsilon,\rho)\|_{L^\infty} \le & C\left(\|R_1\|_{L^\infty}+
\|R_2\|_{L^\infty}\right)
\\ \le &
C\delta(r_0)\|w\|_{L^\infty}+O(\varepsilon).
\end{align*}
Thus we have \eqref{eq:4.8}
\end{proof}
Let
$$
F(\varepsilon,\rho)=\left(\mathcal{L}(\varepsilon,\rho)w(\varepsilon,\rho)
+R_1(w(\varepsilon,\rho),\varepsilon,\rho)+R_2(\varepsilon,\rho),
\partial_\rho\Phi(\varepsilon,\rho)\right)_{L^2_{r}(\R^2)}.$$
By Lemma \ref{lem:4.1}, the system of \eqref{eq:4.1} and \eqref{eq:4.2}
is reduced to an equation
\begin{equation}
   \label{eq:4.10}
F(\varepsilon,\rho)=0.
\end{equation}
\begin{lemma}
  \label{lem:4.2}
Let $p>1$ and let $\varepsilon_0>0$ be a sufficiently small number.
If $\varepsilon\in(0,\varepsilon_0]$, there exists a $\rho=\rho(\varepsilon)\in
(\alpha_0/(2\varepsilon),2\alpha_0/\varepsilon)$ satisfying
\eqref{eq:4.10}.
\end{lemma}
\begin{proof}
Let
$\mathcal{R}_c=
\frac{\varepsilon}{\alpha+\varepsilon r}\partial_r+
\left(\frac{1}{\alpha^2}-\frac{1}{(\alpha+\varepsilon r)^2}\right)
+f'(\chi_1Q_c)-f'(Q_c).$
Using \eqref{eq:Qc}, the definition of $\chi_1$ and the fact that
$L_cQ_c'=0$ and $\rho=O(\varepsilon^{-1})$, we compute
\begin{equation}
  \label{eq:4.11}
  \begin{split}
& \|\mathcal{L}(\varepsilon,\rho)\tau_\rho(\chi_1Q_c')\|_{L^2_{r}(\R^2)}
\\ \le & \|\tau_\rho\chi_1L_cQ_c'\|_{L^2_{r}(\R^2)}
+\|[\partial_r^2,\tau_\rho\chi_1]\tau_\rho Q_c'\|_{L^2_{r}(\R^2)}
+\left\|\tau_\rho(\mathcal{R}_c(\chi_1Q_c'))\right\|_{L^2_{r}(\R^2)}
\\ = & O(\varepsilon^{1/2}).
\end{split}
\end{equation}
Similarly, we have
\begin{equation}
  \label{eq:4.12}
  \begin{split}
& \|\mathcal{L}(\varepsilon,\rho)(\partial_\rho\Phi(\varepsilon,\rho)+
\tau_\rho(\chi_1Q_c'))\|_{L^2_{r}(\R^2)}\\=&
 \left\|\mathcal{L}(\varepsilon,\rho)\tau_\rho\left(\chi_1'Q_c+
\frac{2}{\alpha^3}\varepsilon\chi_1\partial_cQ_c\right)\right\|_{L^2_{r}(\R^2)}
\\ = &O(\varepsilon^{1/2}).
  \end{split}
\end{equation}
By Lemma \ref{lem:4.1}, \eqref{eq:4.11} and \eqref{eq:4.12},
\begin{equation}
  \label{eq:4.13}
  \begin{split}
\left|(\mathcal{L}(\varepsilon,\rho)w,\partial_\rho\Phi(\varepsilon,\rho))_{L^2_{r}(\R^2)}
\right| \le & C\varepsilon^{1/2}\|w\|_{L^2_{r}(\R^2)}=O(\varepsilon).
  \end{split}
\end{equation}
Lemma \ref{lem:4.1} and Corollary \ref{cor:4.2} yield
\begin{equation}
  \label{eq:4.14}
\|R_1\|_{L^2_{r}(\R^2)}\le C\|w\|_{L^\infty}^{\tilde{p}}\|w\|_{L^2_{r}(\R^2)}
=O(\varepsilon^{\tilde{p}+\frac12}),
\end{equation}
where $\tilde p=\min(p-1,1)$.
Combining \eqref{eq:4.5} and \eqref{eq:4.14} with
$$\|\partial_\rho\Phi(\varepsilon,\rho)\|_{L^2_{r}(\R^2)}=O(\varepsilon^{-1/2}),$$ we have
\begin{equation}
  \label{eq:4.15}
\left|(R_1+R_{21}+R_{23},\partial_\rho\Phi)_{L^2_{r}(\R^2)}\right|=O(\varepsilon^{\tilde{p}}).
\end{equation}
In view of \eqref{eq:4.6} and the fact that
$\|\partial_\rho\Phi(\varepsilon,\rho)+\tau_\rho(\chi_1Q_c')\|_{L^2_{r}(\R^2)}
=O(\varepsilon^{1/2}),$
\begin{equation}
  \label{eq:4.16}
 (R_{22},\partial_\rho\Phi(\varepsilon,\rho)+\tau_\rho(\chi_1Q_c'))_{L^2_{r}(\R^2)}
=O(\varepsilon).
\end{equation}
By \eqref{eq:4.13}, \eqref{eq:4.15} and \eqref{eq:4.16},
$$F(\varepsilon,\rho)=-(R_{22},\tau_\rho(\chi_1Q_c'))_{L^2_{r}(\R^2)}
+O(\varepsilon^{\tilde{p}}).$$
Substituting
$$\frac{1}{\alpha^2}-\frac1{(\alpha+\varepsilon s)^2}
=\frac{2\varepsilon}{\alpha^3}s +O(\varepsilon^2s^2)
\quad\text{as $\varepsilon\downarrow0$,}$$
and integrating by parts, we have
\begin{align*}
& (R_{22},\tau_\rho(\chi_1Q_c'))_{L^2_{r}(\R^2)}
\\=&
\frac{1}{\varepsilon}\int_{-\rho}^\infty \left(\frac{1}{\alpha^2}-
\frac{1}{(\alpha+\varepsilon s)^2}\right)\chi_1(s)^2Q_c(s)Q_c'(s)
(\alpha+\varepsilon s)ds\\
& +\int_{-\rho}^\infty \chi_1(s)^2Q_c'(s)^2ds
\\ =&
\int_{-\rho}^\infty\chi_1^2
\left(\frac{2s}{\alpha^2}Q_cQ_c'+Q_c'^2\right)ds+O(\varepsilon)
\\=&
\int_\mathbb{R} \left\{Q_c'^2-\frac1{\alpha^2}Q_c^2\right\}ds+O(\varepsilon).
\end{align*}
Combining the above,  we see that
$$
F(\varepsilon,\rho)=\int_\mathbb{R}
\left(Q_c'(s)^2-(\varepsilon\rho)^{-2}Q_c(s)^2\right)ds
+O(\varepsilon^{\tilde{p}}),
$$
where $c=\omega+(\varepsilon\rho)^{-2}$.
Hence it follows from Lemma \ref{lem:amplitude} and the intermediate
value theorem that \eqref{eq:4.10} has a solution
$\rho=\rho(\varepsilon)$ satisfying
$$\rho=(\alpha_0+o(1))\varepsilon^{-1} \quad\text{as $\varepsilon\downarrow0$.}
$$
Thus we complete the proof of Lemma \ref{lem:4.2}.
\end{proof}
Now, we are in position to prove Theorem \ref{thm1}.
\begin{proof}[Proof of Theorem \ref{thm1}]
Lemmas \ref{lem:4.1} and \ref{lem:4.2} and Corollary \ref{cor:4.2} imply
that there exists a solution $\phi_\omega$ to \eqref{eq:B} satisfying
\eqref{eq:1.5} and \eqref{eq:1.6}.
Suppose that $\phi_\omega$ is a sign-changing solution.
Since $\phi_\omega''\ge0$ and $\phi_\omega'=0$ at the minimum point,
it follows from \eqref{eq:B} that
$$\min_{r>0}\phi_\omega(r)<-\omega^{1/(p-1)}.$$
But this contracts to \eqref{eq:1.6} if $\varepsilon>0$ is
sufficiently small. Thus the solution $\phi_\omega$ to \eqref{eq:B} is
nonnegative. Since a nonnegative solution is unique (see \cite{Mi}),
we obtain Theorem \ref{thm1}.
\end{proof}
\bigskip

\section{Instability of vortex solitons}
In this section, we will prove Theorem \ref{thm2}.
Let $u(x,t)=e^{i\omega t}(e^{im\theta}\phi_\omega(r)+e^{\lambda t}v)$ and
linearize \eqref{eq:NLS} around $v=0$ and $t=0$. Then
\begin{equation}
\label{eq:5.1}
i\lambda v+(\Delta-\omega+\beta_1(r))v
+e^{2im \theta}\beta_2(r)\bar{v}=0,
\end{equation}
where
$$\beta_1(r)=\frac{p+1}2\phi_\omega(r)^{p-1},\quad
\beta_2(r)=\frac{p-1}2\phi_\omega(r)^{p-1}.$$
Put $v=e^{i(j+m)\theta}y_+$, $\bar{v}=e^{i(j-m)\theta}y_-$ and
complexify \eqref{eq:5.1} into a system
\begin{equation}
  \label{eq:5.2}
\left\{
\begin{aligned}
&\left(\Delta_r-\omega-\frac{(m+j)^2}{r^2}+i\lambda+\beta_1(r)
\right)y_+ +\beta_2(r)y_-=0,
\\
& \left(\Delta_r-\omega-\frac{(m-j)^2}{r^2}-i\lambda+\beta_1(r)
\right)y_- +\beta_2(r)y_+=0.
\end{aligned}\right.
\end{equation}
If $\lambda$ is an eigenvalue of the linearized operator,
there exist a $j\in\mathbb{Z}$ and a solution $(y_+,y_-)$ to
\eqref{eq:5.2} that satisfy
$(e^{i(j+m)\theta}y_+(r), e^{i(j-m)\theta}y_-(r)) \in H^1(\mathbb{R}^2,\mathbb{C}^2)$.
We will show the existence of unstable eigenvalues for $j$ with $1\ll j\ll m$.

Let $w_1=y_++y_-$, $w_2=y_+-y_-$, $\varepsilon=m^{-1}$ and
$\delta=j\varepsilon$. Let $s=r-\alpha_0m$.
Then \eqref{eq:5.2} can be rewritten as
\begin{equation}
   \label{eq:5.3}
\mathcal{H}(\varepsilon,\delta)\mathbf{w}=\lambda \mathbf{w},
\end{equation}
where $\mathbf{w}={}^t\!(w_1,w_2)$,
\begin{equation}
\label{eq:5.4}
\mathcal{H}(\varepsilon,\delta)=i
\begin{pmatrix}
 h_{11} & h_{12}\\
h_{21} & h_{22}
\end{pmatrix},
\end{equation}
and
\begin{align*}
h_{11}&= h_{22}=\frac{-2mj}{r^2},
\\
h_{12}&=\Delta_r-\omega-\frac{m^2+j^2}{r^2}+\phi_\omega^{p-1}
\\
h_{21}&=\Delta_r-\omega-\frac{m^2+j^2}{r^2}+p\phi_\omega^{p-1}.
\end{align*}
We remark that
\begin{align*}
\tau_{-\bar{r}}h_{11}=&\tau_{-\bar{r}}h_{22}=
\frac{-2\delta}{(\alpha_0+\varepsilon r)^2}
\\
\tau_{-\bar{r}}h_{12}=&
\partial_r^2+\frac{\varepsilon}{\alpha_0+\varepsilon r}\partial_r
-\omega-\frac{1+\delta^2}{(\alpha_0+\varepsilon r)^2}+\phi_\omega^{p-1}
\\
\tau_{-\bar{r}}h_{21}=&
\partial_r^2+\frac{\varepsilon}{\alpha_0+\varepsilon r}\partial_r
-\omega-\frac{1+\delta^2}{(\alpha_0+\varepsilon r)^2}+p\phi_\omega^{p-1}.
\end{align*}

Before we investigate the spectrum of $\mathcal{H}(\varepsilon,\delta)$,
let us consider the spectrum of a linear operator
$$H(\delta):=i
\begin{pmatrix}
  -2\alpha_0^{-2}\delta & L_--\alpha_0^{-2}\delta^2
\\ L_+-\alpha_0^{-2}\delta^2 & -2\alpha_0^{-2}\delta,
\end{pmatrix}$$
where $L_+=\partial_s^2-c+pQ_c^{p-1}$,  $L_-=\partial_s^2-c+Q_c^{p-1}$,
$D(L_+)=D(L_-)=H^2(\mathbb{R})$ and $c=\omega+\alpha_0^{-2}$.

To begin with, we recall some spectral properties of $H(0)$.
Let
$$
\Phi_1=\begin{pmatrix}0 \\Q_c\end{pmatrix}, \quad
\Phi_2=-i\begin{pmatrix}\partial_cQ_c \\0\end{pmatrix}, \quad
\Phi_3=\begin{pmatrix}Q_c'\\0\end{pmatrix},\quad
\Phi_4=-\frac{i}{2}\begin{pmatrix}0\\sQ_c\end{pmatrix},
$$
and
$$
\Phi_1^*=\theta_1\sigma_2\Phi_2,\quad
\Phi_2^*=\theta_1\sigma_2\Phi_1,\quad
\Phi_3^*=\theta_2\sigma_2\Phi_4,\quad
\Phi_4^*=\theta_2\sigma_2\Phi_3,
$$
where
$$
\sigma_2=\begin{pmatrix}  0 & -i \\ i & 0\end{pmatrix},\quad
\theta_1=2\left(\frac{d}{dc}\|Q_c\|_{L^2(\mathbb{R})}^2\right)^{-1},
\quad \theta_2=4\|Q_c\|_{L^2(\mathbb{R})}^{-2}.
$$
Then we have
\begin{gather}
\label{eq:5.5}
H(0)\Phi_1=0, \quad H(0)\Phi_2=\Phi_1, \quad H(0)\Phi_3=0,
\quad H(0)\Phi_4=\Phi_3,
\\
H(0)^*\Phi_1^*=\Phi_2^*, \quad H(0)^*\Phi_2^*=0,\quad
  H(0)^*\Phi_3^*=\Phi_4^*, \quad H(0)^*\Phi_4^*=0,
\end{gather}
and $\langle \Phi_i, \Phi_j^*\rangle=\delta_{ij}$ for $i$, $j=1,2,3,4$.
Here we denote by $\langle\cdot,\cdot\rangle$ the inner product of
$L^2(\mathbb{R},\mathbb{C}^2)$.
\begin{proposition}[see \cite{We1}]
  \label{prop:5.1}
Let $p>1$ and $p\ne 5$. Then $\lambda=0$ is a discrete eigenvalue of
$H(0)$ with algebraic multiplicity $4$.
\end{proposition}

Using Proposition \ref{prop:5.1}, we investigate the spectrum of $H(\delta)$.
\begin{lemma}
  \label{lem:5.2}
Let $1<p<5$. Then there exist a positive number $\delta_0$ and a neighborhood
$U\subset\mathbb{C}$ of $0$ such that for every $\delta\in(0,\delta_0)$,
$\sigma(H(\delta))\cap U$ consists of algebraically simple eigenvalues
$\lambda_i(\delta)$ $(i=1,2,3,4)$ satisfying
$$
\left|\Re\lambda_1(\delta)-\alpha_0^{-1}\gamma\delta\right|
\le \alpha_0^{-1}\gamma\delta/4,
\quad
\liminf_{\delta\downarrow0}\left(
\delta^{-1}\min_{\substack{1\le i,j\le 4,\\ i\ne j}}
|\lambda_i(\delta)-\lambda_j(\delta)|\right)
>0,$$ where
$$
\gamma=\left(2\frac{\|Q_c\|_{L^2(\mathbb{R})}^2}{{\frac{d}{dc}\|Q_c\|_{L^2(\mathbb{R})}^2}}
\right)^{1/2}.$$
\end{lemma}
\begin{proof}
Let $P_{H}(\delta)$ be a projection defined by
$$
P_H(\delta)=\frac{1}{2\pi i}\oint_{|\lambda|=\rho_0}
\left(\lambda-H(\delta)\right)^{-1}d\lambda,
$$
and let $Q_H(\delta)=I-P_H(\delta)$.
In view of Proposition \ref{prop:5.1}, there exist positive numbers
$\rho_0$ and $\delta_0$ such that $\mathcal{X}_0:=R(P_H(\delta))$ is
4-dimensional for every $\delta\in(0,\delta_0)$.
\par
Let $\mathcal{X}_0$ be a linear subspace whose basis is
$\langle\Phi_1,\Phi_2,\Phi_3,\Phi_4\rangle$.
We decompose $H^2(\mathbb{R};\mathbb{C}^2)$ and $L^2(\mathbb{R};\mathbb{C}^2)$ as
$$
H^2(\mathbb{R};\mathbb{C}^2)=\mathcal{X}_0\oplus Q_H(0)H^2(\mathbb{R};\mathbb{C}^2),\quad
L^2(\mathbb{R};\mathbb{C}^2)=\mathcal{X}_0\oplus Q_H(0)L^2(\mathbb{R};\mathbb{C}^2).
$$
Then  $$H(\delta) =
\begin{pmatrix}
H_{11}(\delta) & H_{12}(\delta) \\
H_{21}(\delta) & H_{22}(\delta)
\end{pmatrix},$$
where
\begin{align*}
&H_{11}(\delta)=P_H(0)H(\delta)P_H(0),\quad
H_{12}(\delta)=P_H(0)H(\delta)Q_H(0)\\
&H_{21}(\delta)=Q_H(0)H(\delta)P_H(0),\quad
H_{22}(\delta)=Q_H(0)H(\delta)Q_H(0).
\end{align*}
By a simple computation, we have
$$H_{11}(\delta)=-2i\alpha_0^{-2}\delta I+
\begin{pmatrix}
0 & 1+b_2\delta^2 & 0 & 0\\
b_1\delta^2 & 0 & 0 & 0\\
0 & 0 & 0 & 1+b_4\delta^2\\
0 & 0 & b_3\delta^2 & 0
\end{pmatrix},
$$
\begin{align*}
H_{12}(\delta)=-i\alpha_0^{-2}\delta^2P_H(0)\sigma_1Q_H(0),
\quad H_{21}(\delta)=-i\alpha_0^{-2}\delta^2Q_H(0)\sigma_1P_H(0),
\end{align*}
where
\begin{align*}
& b_1=\alpha_0^{-2}\theta_1\|Q_c\|_{L^2(\mathbb{R})}^2, \quad
b_2=-\alpha_0^{-2}\theta_1\|\partial_cQ_c\|_{L^2(\mathbb{R})}^2,\\
& b_3=-4\alpha_0^{-4},\quad
b_4=\alpha_0^{-2}\|sQ_c\|_{L^2(\mathbb{R})}^2\|Q_c\|_{L^2(\mathbb{R})}^{-2},
\quad \sigma_1=\begin{pmatrix}  0 & 1\\ 1 & 0\end{pmatrix}.
\end{align*}
\par

First, we investigate the spectrum of $H_{11}(\delta)$.
Suppose $\lambda$ is an eigenvalue of the matrix $H_{11}(\delta)$.
Then
\begin{align*}
& \det(\lambda I-H_{11}(\delta))\\ =&
\left\{(\lambda+2i\alpha_0^{-2}\delta)^2-b_1\delta^2-b_1b_2\delta^4 \right\}
\left\{(\lambda+2i\alpha_0^{-2}\delta)^2-b_3\delta^2-b_3b_4\delta^4\right\}=0.
\end{align*}
Hence there exist eigenvalues $\hat{\lambda}_i$ $(i=1,2,3,4)$ of
$H_{11}(\delta)$ satisfying
\begin{align*}
& \hat{\lambda}_1
=-\delta\left(2i\alpha_0^{-2}-\alpha_0^{-1}\gamma+O(\delta^2)\right),
\quad
\hat{\lambda}_2=
-\delta\left(2i\alpha_0^{-2}+\alpha_0^{-1}\gamma+O(\delta^2)\right),
\\ &
\hat{\lambda}_3=-4i\alpha_0^{-2}\delta\left(1+O(\delta^2)\right),\quad
\hat{\lambda}_4=O(\delta^3).
\end{align*}
\par

Let $R_{ii}(\lambda,\delta)=(\lambda-H_{ii}(\delta))^{-1}$ for $i=1,2$
and let
\begin{gather*}
R_0(\lambda,\delta)=
\begin{pmatrix}
R_{11}(\lambda,\delta) & 0\\
0 & R_{22}(\lambda,\delta)
\end{pmatrix},\\
V_0(\lambda,\delta)=
\begin{pmatrix}
0 & H_{12}(\lambda,\delta)R_{22}(\lambda,\delta)\\
H_{21}(\lambda,\delta)R_{11}(\lambda,\delta) & 0
\end{pmatrix}.
\end{gather*}
We remark that $R_{22}(\lambda,\delta)$ is uniformly bounded for
$\lambda\in U$ and $\delta\in(0,\delta_0)$.
Suppose that $|\lambda-\hat{\lambda_i}|=c_1\delta$, where $c_1\in
(0,\alpha_0^{-1}|\gamma|\delta/4)$ is a constant such that
$|\hat{\lambda}_j-\hat{\lambda}_k|\ge c_1\delta$
for every $j$, $k=1,2,3,4$ with $j\ne k$.
Then in view of the definitions of $H_{12}(\lambda,\delta)$ and
$H_{21}(\lambda,\delta)$,  we have
\begin{equation}
  \label{eq:5.6}
\left\|V_0(\lambda,\delta)\right\|_{B(L^2(\mathbb{R}))}=O(\delta),
\end{equation}
and
\begin{equation}
  \label{eq:5.7}
(\lambda-H(\delta))^{-1}=
R_0(\lambda,\delta)\sum_{i=0}^\infty V_0(\lambda,\delta)^i.
\end{equation}
Now let
\begin{align*}
P_{H,i}(\delta)=& \frac{1}{2\pi
i}\oint_{|\lambda-\hat{\lambda}_i|=c_1\delta}
(\lambda-H(\delta))^{-1}d\lambda,
\\
\widehat{P}_{H,i}(\delta)=&
\frac{1}{2\pi i}\oint_{|\lambda-\hat{\lambda}_i|=c_1\delta}
R_0(\lambda,\delta)d\lambda.
\end{align*}
Combining \eqref{eq:5.6} and \eqref{eq:5.7} with
the fact that
\begin{align*}
\left\|R_0(\lambda,\delta)V_0(\lambda,\delta)\right\|_{B(L^2(\mathbb{R}))}
=&
\left\|
\begin{pmatrix}
0 & R_{11}H_{12}R_{22} \\
R_{22}H_{21}R_{11} & 0
\end{pmatrix}\right\|_{B(L^2(\mathbb{R}))}
=O(\delta),
\end{align*}
we have
$$
\left\| P_{H,i}(\delta)-\widehat{P}_{H,i}(\delta)\right\|=O(\delta)
\quad\text{for every $i= 1,2,3,4$.}
$$
Hence it follows that $R(\widehat{P}_{H,i}(\delta))$ is isomorphic to
$R(P_{H,i}(\delta))$ and that \linebreak
$R(P_{H,i}(\delta))$ is 1-dimensional for $i=1,2,3,4$.
Furthermore, we see that eigenvalues of $H(\delta)$ which lie in $U$
satisfy $|\lambda-\hat{\lambda}_i|< c_1\delta$ for an $i\in\mathbb{N}$ with
$1\le i\le 4$.

 Since $d\|Q_c\|_{L^2(\mathbb{R})}^2/dc>0$ for $p\in(1,5)$, we see that
$\gamma$ is a positive number and that there exist eigenvalues $\lambda_1$ and $\lambda_2$ satisfying
$$
\alpha_0^{-1}\gamma\delta/2<\Re\lambda_1< 3\alpha_0^{-1}\gamma\delta/2,
\quad
-3\alpha_0^{-1}\gamma\delta/2<\Re\lambda_2< -\alpha_0^{-1}\gamma\delta/2.$$
Thus we complete the proof of Lemma \ref{lem:5.2}.
\end{proof}

\begin{proposition}
\label{prop:5.3}
Let $j$, $m\in\mathbb{N}$, $\varepsilon=m^{-1}$ and $\delta=j\varepsilon$.
Let $\beta=\min(p-1,1)/6$. Then there exists an
$m_*\in\mathbb{N}$ such that if $m\ge m_*$, the linearized operator
$\mathcal{H}(\varepsilon,\delta)$ with $j=[m^\beta]$ has an unstable
eigenvalue.
\end{proposition}

\begin{proof}
In order to prove Proposition \ref{prop:5.3}, we will show the
spectrum of $\mathcal{H}(\varepsilon,\delta)$ becomes close to the spectrum
of
$H(\delta)$ as $\varepsilon\downarrow0$.  Let
$$
\mathcal{H}_0=i
\begin{pmatrix}
\frac{-2jm}{r^2} & \Delta_r-\omega-\frac{m^2+j^2}{r^2}\\
\Delta_r-\omega-\frac{m^2+j^2}{r^2} & \frac{-2jm}{r^2}
\end{pmatrix},
$$
and $H_0=U\mathcal{H}_0U^{-1}$.
Let
$$
\mathcal{D}(\lambda)=(\tau_{\bar{r}}\tilde{\chi}_0)(\lambda-H_0)^{-1}
(\tau_{\bar{r}}\chi_0)
+\tau_{\bar{r}}\tilde{\chi}_1(\lambda-H(\delta))^{-1}\chi_1
\tau_{-\bar{r}}.$$
Then we have
$$
\mathcal{D}(\lambda)U(\lambda-\mathcal{H}(\varepsilon,\delta))U^{-1}
=I+R_3+R_4,$$
where
\begin{align*}
& R_3=i(\tau_{\bar{r}})\tilde{\chi}_0(\lambda-H_0)^{-1}\left\{
\begin{pmatrix} 0 & [\partial_r^2,\tau_{\bar{r}}\chi_0]
 \\ [\partial_r^2,\tau_{\bar{r}}\chi_0] & 0\end{pmatrix}
-(\tau_{\bar{r}}\chi_0)\phi_\omega^{p-1} \begin{pmatrix}
0 & 1 \\ p & 0 \end{pmatrix}\right\}
\\ &
R_{4}=i\tau_{\bar{r}}\tilde{\chi}_1(\lambda-H(\delta))^{-1}\left\{
\begin{pmatrix} 0 & [\partial_r^2,\chi_1] \\ [\partial_r^2,\chi_1] & 0\end{pmatrix}
-\chi_1(R_{41}+R_{42})\right\}\tau_{-\bar{r}},
\\ &
R_{41}=
\begin{pmatrix}
\frac{-2\delta}{(\alpha_0+\varepsilon r)^2}+\frac{2\delta}{\alpha_0^2}
& -\frac{1+\delta^2-\frac14\varepsilon^2}{(\alpha_0+\varepsilon r)^2}
+\frac{1+\delta^2}{\alpha_0^2}\\
-\frac{1+\delta^2-\frac14\varepsilon^2}{(\alpha_0+\varepsilon r)^2}
+\frac{1+\delta^2}{\alpha_0^2} &
\frac{-2\delta}{(\alpha_0+\varepsilon r)^2}+\frac{2\delta}{\alpha_0^2}
\end{pmatrix},\\
& R_{42}=\begin{pmatrix}0 & f(\phi_\omega)-f(Q_c) \\
f'(\phi_\omega)-f'(Q_c) & 0\end{pmatrix}.
\end{align*}
We remark that
\begin{gather*}
|[\partial_r^2,\chi_i]\|_{B(L^2(\mathbb{R}),H^{-1}(\mathbb{R}))}=O(l^{-1})
\quad\text{for $i=0,1$,}\\
\|\chi_1R_{41}\|_{B(L^2_{r}(\R^2))}+\|R_{42}\|_{B(L^2_{r}(\R^2))}=O(\varepsilon^{6\beta}l).
\end{gather*}
We have
$$\sup_{\lambda\in\mathbb{C},|\lambda|\le\omega/2}
\|(\lambda-\mathcal{H}_0)^{-1}\|_{B(H^{-2}_r(\mathbb{R}^2),L^2(\mathbb{R}^2))}<\infty,$$
since
$${}^tO\mathcal{H}_0O=i \begin{pmatrix}
\Delta_r-\omega-\frac{(m+j)^2}{r^2} & 0 \\
0 & -\Delta_r+\omega+\frac{(m-j)^2}{r^2} \end{pmatrix},$$
where
$$O=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1\end{pmatrix}.$$
Lemma \ref{lem:5.2} yields that for $\delta\in(0,\delta_0)$,
there exists a $c>0$ such that
$$
\left\|(\lambda-H(\delta))^{-1}\right\|_{B(L^2_{r}(\R^2))}\le C\delta^{-1}$$
for every $\lambda\in U$ with
$\min_{1\le i\le 4}|\lambda-\lambda_i(\delta)|\ge c\delta$ and that
$\Re(\lambda_1(\delta)-c\delta)>0$.
Let $l=\delta^{-3}$. Then it follows from the above that
\begin{align*}
& \|R_3\|_{B(L^2_{r}(\R^2))}=O(\delta^3+e^{-2\sqrt{c}\delta^{-3}}),\\
& \|R_4\|_{B(L^2_{r}(\R^2))}=O(\delta^2+\varepsilon^{6\beta}\delta^{-4}).
\end{align*}
Put
\begin{align*}
\mathcal{P}_{\mathcal{H},1}(\varepsilon,\delta)=& \frac{1}{2\pi i}
\oint_{|\lambda-\lambda_1(\delta)|=c\delta}
(\lambda-\mathcal{H}(\varepsilon,\delta))^{-1}d\lambda,
\\
\mathcal{P}_{H,1}(\varepsilon,\delta)=&
U^{-1}\tau_{\bar{r}}\tilde{\chi}_1P_{H,1}(\delta)\chi_1\tau_{-\bar{r}}U.
\end{align*}
Making use of Cauchy's theorem and noting that
$\delta\sim\varepsilon^{\beta}$, we have
\begin{align*}
& \left\|\mathcal{P}_{\mathcal{H},1}(\varepsilon,\delta)
-\mathcal{P}_{H,1}(\varepsilon,\delta) \right\|_{B(L^2_{r}(\R^2))}
\\ =&
\frac{1}{2\pi}\left\|\oint_{|\lambda|=c\delta}
\left\{(\lambda-\mathcal{H}(\varepsilon,\delta))^{-1}
-U^{-1}\mathcal{D}(\lambda)U\right\}
d\lambda\right\|_{B(L^2_{r}(\R^2))}
\\ \le &
C\delta^{-1}\sup_{|\lambda|=c\delta}(\|R_3\|_{B(L^2(-\bar{r},\infty))}
+\|R_4\|_{B(L^2(-\bar{r},\infty))})
\\ \le & C(\delta+\varepsilon^{6\beta}\delta^{-5})
\\ =& O(\delta).
\end{align*}
From the above, we conclude that the range of
$\mathcal{P}_{\mathcal{H},1}(\varepsilon,\delta)$ is isomorphic to the
range  of $P_{H,1}(\delta)$ and that there exists an eigenvalue $\lambda$ of
$\mathcal{H}(\varepsilon,\delta)$ with $\Re\lambda>0$.
Thus we complete the proof of Proposition \ref{prop:5.3}.
\end{proof}
Now we are in position to prove Theorem \ref{thm2}.
\begin{proof}[Proof of Theorem \ref{thm2}]
Let $\mathfrak{L}$ be the linearized operator of
\eqref{eq:NLS} around $e^{i(\omega t+m\theta)}\phi_\omega$. Then
$$
\mathfrak{L}=i \begin{pmatrix}
\Delta-\omega+\beta_1(r) & e^{2im \theta}\beta_2(r) \\
-e^{-2im \theta}\beta_2(r) & -\Delta+\omega-\beta_1(r)
\end{pmatrix}.$$
Proposition \ref{prop:5.3} tells us that $\mathfrak L$ has unstable
eigenvalues if  $m\in\mathbb{N}$ is large and $p\in(1,5)$. On the other hand,
\cite{Mi3} tells us that $\mathfrak L$ has an unstable eigenvalue if
$p>3$. Hence it follows that $\mathfrak L$ has an unstable eigenvalue
if $p>1$ and $m\in\mathbb{N}$ is sufficiently large.
\end{proof}

\section*{Acknowledgment}
The author would like to express his gratitude to Professor
Jaeyoung Byeon for his useful advice.

redpajama_set_name: RedPajamaArXiv


text:
// January 3, 2022
Sales, 'racist behavior' concerns shut Third Culture's Colorado branches
Two of the Colorado locations of Third Culture, a famous bakery based in Berkeley, close due to the dropping sales and struggle with workers added to "racist behavior" it gets from its clients during the health crisis, Wenter Shyu, the bakery's co-founder, told Denver Post last month.
In January last year, Shyu, her husband, along with chef Sam Butarbutar transferred to Colorado. They established their initial out-of-state branch in Aurora a month after they moved in.
Customers crowded in their location for the opening day, and they opened an additional one last summer in Denver's River North Art District.
But Third Culture's success was brief as they saw the sales drop throughout the pandemic. The bakery also struggled to secure ingredients including rice flour and coconut milk, which they source from California, due to the supply chain shortage.
Last year, Shyu told SFGATE that it had to continue on takeout orders for it to cover operational costs.
"It was really terrifying … we thought we had the possibility of losing the bakery," the co-founder said.
"As Asian Americans, 2021 brought on a lot of racist behavior and confrontational behavior in addition to the pandemic stresses," Shyu furthered in her talk to the Post. "It brought up a whole slew of difficulties. At some point, we were just so exhausted, and it just wasn't quite worth it anymore, to be honest. Which is sad, because our trajectory prior to the pandemic was great."
In November last year, the company shut its location in Denver due to "unsavory" staffing concerns. Before the year ended, its Aurora location also closed.
"In the last six months, we've hired people, in Colorado especially, who start working with us. And then a month later, they make false claims or do things, and then we'd have to terminate them based on the offer letter, and it was just really difficult," she said. "Colorado has some growing up to do."
Meanwhile, the bakery's locations in Berkeley and Walnut Creek remain in service.


text:
Spector share new single 'Funny Way Of Showing It', co-written with The Magic Gang's Jack Kaye
NEW ALBUM NOW OR WHENEVER, OUT OCTOBER 2021 VIA MOTH NOISE
UK HEADLINE TOUR ON SALE NOW
Zeitgeist Agency
Spector have shared new single 'Funny Way Of Showing It', listen below.
'Funny Way Of Showing It' comes in the wake of the band returning with the mammoth 'Catch You On The Way Back In' last month, alongside announcing Now Or Whenever, their first official studio album in six years.
Now Or Whenever is the long awaited third full length release from the cult band. The album follows the critically acclaimed Enjoy It While It Lasts (2012), Moth Boys (2015) and EP compilation Non-Fiction (2020), records which built a culture around the group and accrued them a fiercely loyal fan base, making Spector one of Britain's most-loved alternative outfits.
Where CYOWTBI saw Spector at their most visceral and urgent, 'Funny Way Of Showing It' exudes a slightly more psychedelic slacker feel, rearranging the band's emotionally resonant, indie rock hallmarks into a hazy mid-summer earworm. Co-written with The Magic Gang's Jack Kaye, 'Funny Way Of Showing It' captures the laidback interplay of Jed Cullen's guitar playing and Fred Macpherson's freshly tempered vocals, a theme that runs across the album, with rhythmic instrumentation and blissed out harmonies throughout. "You've been looking for a mint condition first edition / You've been looking for the real deal" sings Fred. "You've been looking for a bit of fervour, look no further / Maybe you just want to feel".
Now Or Whenever is out October 1st on the band's own label Moth Noise, the full tracklist can be found below, alongside the dates for the band's UK headline tour, for which tickets are on sale now.
'Funny Way Of Showing It' is available to stream here.
More info on Now Or Whenever
Following the success of their 2019 tour and singles compilation Non-Fiction, Spector started work on Now or Whenever in the Spring of 2020. With band members locked down in different cities and countries, the core duo of Fred Macpherson and Jed Cullen opted for a more collaborative process than usual, working with friends and contemporaries both online and in the flesh where possible. An improvisatory approach to music and lyrics was encouraged, and in the absence of gigs the songs took on the idea of a theoretical concert of new material, delivered in the spirit of their impassioned live shows. The final songs were recorded in January 2021 at Coastal Sound, Liverpool by Rich Turvey (Blossoms, The Coral) who emphasised performances over production and heavier, stripped down arrangements.
Inspired by the vocal stylings and layered harmonies of British folk rock (Strawbs, Steeleye Span), the drunken riffs of American eighties alternative (The Replacements, Husker Dü) and raucous energy of Antipodean pub rock (Th' Dudes, The Radiators), Now or Whenever is the sound of Spector at their irreverent best. Lyrics explore themes of time, memory, hope, the near future, cars and emotionally repressed adult friendships, but a sense of positivity shines through the fatalistic musings the band have become synonymous with.
It's their first album with live instrumentation throughout, with drums recorded straight to decaying, thirty year old tape by Nicolas Py, the French cornerstone of Spector's touring party for the last four years. Jed Cullen's guitar also bursts into the foreground, experimenting with alternate tunings, drones and equipment left orphaned at the studio by the likes of Clinic and Chris De Burg.
"In some ways this is our first proper guitar album" explains Fred. "And the first album we've made in one studio in one sitting which is crazy. Next year will be the tenth anniversary of Enjoy It While It Lasts which feels like forever, but I think it's taken us that long to work out what's good about us and how to capture it. Which kind of makes Now or Whenever both a sequel and a reboot. Had last year unfolded differently, maybe this would be our big depressing opus, but when positivity's taken away you have to create your own, and somehow that's left us with our most upbeat record yet."
"When there are huge changes happening on a global level and absolutely nothing happening on a local level, people ask you how you've been, what you've been up to etc. and there's not much to say that they don't already know" adds Jed. "So we start talking and writing about the past and the future as if to escape from this paradoxically mundane/extreme present. Then you step back and you realise that the strange conditions of the time are pervading your conversations and it's coming out in the music in ways you didn't expect."
Source Zeitgeist Agency
Zeitgeist Agency: Profile Page
July 14, 2021 4:00am ET by Pressparty
zeitgeist, agency
Shortlink to this content: https://bit.ly/3yV7yZY
All Press Releases By Zeitgeist Agency


text:
package consulo.json.jom.proxy.impl;

import java.lang.reflect.Type;

import javax.annotation.Nonnull;
import javax.annotation.Nullable;
import consulo.annotation.access.RequiredReadAction;
import consulo.json.jom.proxy.JomBadValueExpressionException;
import consulo.json.jom.proxy.JomValueConverter;
import com.intellij.lang.javascript.JSTokenTypes;
import com.intellij.lang.javascript.psi.JSLiteralExpression;
import com.intellij.psi.PsiElement;
import com.intellij.psi.tree.IElementType;
import com.intellij.psi.util.PsiUtilCore;

/**
 * @author VISTALL
 * @since 13.11.2015
 */
public class JomBooleanValue implements JomValueConverter.Converter<Boolean>
{
	private Boolean myDefaultValue;

	public JomBooleanValue(@Nullable Boolean defaultValue)
	{
		myDefaultValue = defaultValue;
	}

	@Override
	public Boolean getDefaultValue()
	{
		return myDefaultValue;
	}

	@RequiredReadAction
	@Override
	public Boolean parseValue(@Nonnull Class type, @Nonnull Type genericType, @Nonnull PsiElement value) throws JomBadValueExpressionException
	{
		if(value instanceof JSLiteralExpression)
		{
			IElementType elementType = PsiUtilCore.getElementType(value.getFirstChild());
			if(elementType == JSTokenTypes.TRUE_KEYWORD)
			{
				return Boolean.TRUE;
			}
			else if(elementType == JSTokenTypes.FALSE_KEYWORD)
			{
				return Boolean.FALSE;
			}
		}
		throw new JomBadValueExpressionException();
	}
}

redpajama_set_name: RedPajamaGithub


Below you will find a list of the churches in Farmersville, CA. These 188 churches are organized by their distance from Farmersville. The closest businesses to Farmersville are Free Will Baptist Church on S. Dode Ave (Farmersville) in the category Churches Baptist Free Will, Church of God of Prophecy on N. Magnolia Ave (Farmersville) in the category Churches, Jehovah's Witnesses on S. Camelia Ave (Farmersville) in the category Churches Jehovahs Witness, Sacred Heart Parish on E. Visalia Rd (Farmersville) in the category Churches Catholic and Pentecostal Church of God on S. Farmersville Blvd (Farmersville) in the category Churches Pentecostal Church Of God.


text:
Early Times Kentucky Whisky 200ml
Straight Whiskey
Early Times Kentucky Whisky
Early Times Kentucky Whisky is a true American. Created in 1860, Early Times grew up during the early days of the country, when most of America was still country. The name Early Times was chosen as a tribute to the old-fashioned method of making whisky, which included mashing grain in small tubs and distilling the whisky in copper stills over open fires. Today, the process of distilling Early Times is much more exacting, but it still delivers the unmistakable mellow taste that is reminiscent of the original American whisky experience.


If you're wondering what the main trends in interior design will be for next year and beyond then the recent exhibition IMM Cologne has shed some light on it, from furniture placement to colour schemes and much more.
What do you do with a shipping container when you no longer need it. Well you turn it into a cafe, hotel, swimming pool, or even a school……obviously. Who'd have thought a corrugated steel box could be so versatile?!
Cold glass sculptor Jack Storms is quite literally a cut above the rest when it comes to his quest for artistic perfection, another level of optical creativity, and the beauty that results from it.
With the 2017 Geneva Motor Show coming to end some 250 cars have been debuted to the public, and the topic on most people's lips has been the venture into the Electric market of luxury car manufacturer Bentley.


We deliver systems built cost effectively using off-the-shelf Cloud based software platforms from service technologies such as Amazon, Google, Dropbox, WordPress and Zoho, whilst meeting the needs for robust scalable secure and sometimes complex IT solutions.
A2Z Cloud made a strategic decision to specialise in the Zoho platform over six years ago.
We therefore fully understand the significant amount of effort, time and commitment it takes a company to understand Zoho as a whole before even beginning to configure a final solution.
We represent a new breed of IT company specifically formed to capitalise on the fundamental shift in online collaborative flexible working, as a result of maturing Cloud based technologies by this we mean web and internet platforms.
Our main objective is to help companies improve efficiency and productivity.
By combining our technical expertise with the Zoho suite of products and a process driven product implementation methodology, A2Z Cloud can design, configure and deploy tailored Cloud solutions to meet the unique needs of individual client organisations.
We understand the importance of streamlining business processes and system automation and integration where possible, and training staff properly for better user adoption and efficiency.
A2Z Cloud is one of only 10 companies around the world who have achieved both Zoho Premium Partner and Zoho Advanced Solutions Provider status.
As a certified deployment and development Zoho partner we have a wealth of working knowledge and experience in bringing this globally recognised suite of products into full operational use with a wide range of organisations.
We practice what we preach as our whole business runs on Zoho. We have amassed a wealth of real world knowledge with the Zoho suite of products and helped businesses, just like yours to achieve the same level of integration and increase in productivity.
By combining our technical expertise with the Zoho suite of products and a process driven product implementation methodology, A2Z Cloud can design, configure and deploy Zoho solutions to meet the unique needs of individual client organisations. This flexible, modular nature of our approach allows A2Z Cloud to jump in at whatever stage is right for the client, and deliver a final high end solution.
Our process driven methodology also helps us identify any functionality that is insufficient, and then in these cases augment a system using rapidly developed "mini Apps" to meet agreed specific requirements. Typically these are built in Zoho Creator, Python, or PHP or equivalent modern platform independent technologies.
Through our close working relationship with the Zoho development team, and the constantly growing platform, we are often already ahead of the game when new features are introduced to the market.
I am really impressed with how the day went, it is a great system, Chris was great at teasing out and challenging our process and Kieron really grasped what we are trying to achieve and offered a plethora of potential ways we could make things easier and our processes more lean, great day and a good start.
I have to say that I have been super impressed with A2Z in their approach and delivery of the project.
Lucy Beck (A2Z Clouds' MD) is a true entrepreneur and Zoholic with tenacity and commitment to succeed. A passionate advocate of Zoho with business and technical acumen in equal measure, she is not afraid of giving a straight-talking opinion respected by customers and the wider Zoho partner network. She has built an exceptionally strong team at A2Z Cloud, fully committed to delivering first class Zoho based solutions, ideal qualities we want in a partner.
Each member of the A2Z Cloud team are unique, and bring their huge amount of experience and professionalism to their work, but more than that, they bring passion and a caring for the solution to be the best it can be within the parameters and deadlines they may be working to.
Working with A2Z Cloud has been an invaluable experience. The team at A2Z Cloud are experts in their field and have a wealth of knowledge about Zoho products. Thank you A2Z Cloud.
The team at A2Z Cloud were incredibly knowledgeable and responsive to my business needs, not just creating a system that works for me in the immediate sense but building something bespoke that can organically move with me through the lifetime of my business.
A2Z Cloud seamlessly took control of delivering web forms to capture data from our website and integrate it with our CRM. It has given us the systems and the confidence that we were looking for to help us manage our business opportunities more quickly and professionally.
Our trainer was so easy to talk to and her teaching methods were very simple and easy to follow. As Zoho 'newbies' this helped us to brainstorm together and learn how to make the best use of Zoho for our business.


Well, today was the last day of classes for the year! It was a great feeling. I kept looking at the clock, counting down the hours, and Bethany and I kept reminding each other of how many hours and minutes were left. Of course, I wasn't as excited as I could have been, just because I know that we still have exams to go. After those, I'll be excited. After all, I'm friggin' graduating! I can't even believe it; it's craziness!
Anyways, I don't feel like writing about today too much, just because it wasn't overly interesting. For Math class, we didn't do math, but watched the French class's videos they made instead. In Accounting was a test, Biology we just reviewed for the exam and played hangman, and in World Religions was another test. I'm just glad, though, that I have almost everything done that I need to get done. Over the past two days, I've done Phys. Ed. questions, a Biology paper on evolution, an English essay on Huckleberry Finn, a book report, and Parenting handouts for Chapter 16. Now all I have left is one more chapter of handouts for Parenting, and then I'm finished! Well, again, except for exams, that is.
Today was a little bit of an up-and-down sort of day. The first half was pretty regular, and then Biology class hit. Mrs. Houtman just read from the textbook for most of the class, and I was honestly pinching myself to try and stay awake. Partway through the class, my allergy medication started to kick into effect, and then I started getting these weird head rushes. Usually that happens if I'm not active at all. Then, either from the medication as well or just because of a lack of sleep, I started getting really tired. My eyes started to close on their own, but I struggled through and kept myself at least half-awake. I tried to keep my mind as active as possible; I started thinking about whether to just stay on the same bus after school and take the extra ten minute walk instead of possibly falling asleep on the bus and missing my stop.


All the GWhere team wishes you a happy new year 2007!!
New graphic contributions in the graphics section!! Thanks to Fredrik.
The Fedora Core 4 RPM are available thanks to Matthias.
New graphic contributions in the graphics section!! Thanks to rstephxp and Olivier.
New graphic contributions in the graphics section!! Thanks to Dario.
Fabien has just published the new Fedora Core 3 RPMs (Athlon 64 bits).
Spanish site translation update by Guillermo.
After a very long waiting, the polls for the GWhere's logo choice are finally open!!
Matthias has just published the new Fedora Core 3 RPMs (menu entries add).
The Fedora Core 3 RPM are available thanks to Matthias.
New graphic contributions in the graphics section!! Thanks to Settra.
New graphic contributions in the graphics section by Wojtas.
New graphic contributions in the graphics section!! Thanks to John and Wojtas.
The first version of the italian site translation by Michele PETRECCA is online.
Michele PETRECCA has just informed me that there is an article about GWhere in the italian Linux Magazine.
The Debian package for GWhere 0.1.3 is available on the site.
The Debian repository is repaired.
A new section was added on the site : the press. This section presents articles about GWhere which were published in press magazines.
All the GWhere team wishes you a happy new year 2004!!
The MySQL data base connexion was optimized thanks a bug report from the OVH team.
The site has undergoes a complete cleaning of the code to be more and more in conformity with the recommendations of the W3C.
Site conforms with XHTML 1.1 and CSS 2.
Site conforms with HTML 4.01 Transitional and CSS 2.
Add greek translation for GWhere 0.1.2 by Dimitris GARANATSIOS.
Add italian translation for GWhere 0.1.2 by Michele PETRECCA.
Guillermo and Victor are working on the spanish site translation. The first Guillermo's translations are online.
GWhere 0.1.2 : FreeBSD port, bugs fix, hungarian translation and new description plugins descript.ion and FILE_ID.DIZ files.
Gentoo installation documentation added by Victor G. Marimon.
Guillermo has just opened a downloads mirror of GWhere.
I've just found the english translation of the Patricia Jung's article about GWhere (the article is becoming old...) : A break for the Disk-Jocket.
A new version of the package for Debian by Luis Parravicini is available on the site.
GWhere 0.1.1 : fix bugs and new description plugin for AVI files (support DivX).
RPMs update to fix plugins detection.
Add spanish translation for GWhere 0.1.0 by Victor G. Marimon.
A new version of the package GWhere 0.1.0 for Debian by Luis Parravicini is available on the site.
All the team of GWhere is happy to announce the arrival of GWhere 0.1.0 to you.
- Because I do not like number 27. And thus I could not publish version 0.0.27.
- The stability of GWhere makes it possible to pass from the alpha state (0.0.x) towards the beta state (0.x.x).
- The win32 port is a major evolution.
- Compatibility with GTK+ 2.x is also a major evolution.
This new version brings the support of a new platform : MS Windows.
Moreover, the project is now compatible with GTK+ 2.x. A significant work was provided to stabilize the application and to fix memories leaks.
Three new translations are available : Czech, Greek, Spanish.
Now GWhere can automatically generate descriptions of your files MP3 and HTML (others plugins will follow).
News options are available in a panel of parameter setting and are recorded (in a configuration file).
The binary name (executable) changed as well as the names of the archives files : they passed to lower case. However the project does not change a name and remains GWhere.
However I make a point particularly of thanking the translators who provide a considerable work and I'm always surprised by their reactivity!!
I also wish to say a big thank you to the testers who work in the shade and, without which, GWhere will not be also stable.
I thank all the people who made me by their suggestions and/or their remarks.
Add spanish translation for GWhere 0.0.26 by Victor G. Marimon.
This patch must be downloaded if you scan media which contains circular links (link to parent folder).
Add greek translation for GWhere 0.0.26 by Dimitris Garanatsios.
Graphics contributions section is opened. A vote will be available soon.
Path of Debian packages in the files Packages.gz and Sources.gz are fixed (thanks Sebastien W.). With your apt-get!!
The last GWhere release goes back to four months, but the project continue to progress. The next release will be a beta version (0.1.0) that's why it make some times to come.
Add czech translation for GWhere 0.0.26 by Thomas Navara.
The Patricia Jung's article about GWhere has been translated in english and is available in Linux Magazine of november month.
GWhere is included in the Gentoo distribution thanks to the Victor G. Marimon's port.
The Debian repository is done on the site. For further information consult the install documentation.
English home page fix (thanks John).
The installation documentation is updated for Debian packages.
The next version of GWhere will be the 0.1.0.
Debian packages move in order to prepare the making of a Debian packages repository (which will run with apt-get).
A new version of the Debian package by Luis Parravicini is available on the download page.
A Debian binary version of GWhere is available thanks Luis Parravicini. It's a first try : don't hesitate to send yours reports.
Protection email address from spam.
Beginning of a campaign to find a new logo, a splash screen and some icons for GWhere.
The campaign is launched on the site of Linux Graphic in the section ArtGNU. The results are availables here.
The german magazine Linux User talk about GWhere in the section "Out of the Box" : the article is available on line.
Fix the language page detection to support Konqueror.
Adding the GWhere API documentation on download.
FAQ update : thanks to Fred for his FAQ proposition.
Adversiting banner : GWhere is a unpaid contribution. But site hosting and domaine reservation aren't free. In this way the banner is to pay back the maintenance costs. Thanks for your comprehension et mail me yours remarks.


A study from the Netherlands shows that sugar in sports drinks slows absorption and increases stomach cramping in running races shorter than 12 miles. Fluids pass through your stomach and are absorbed almost immediately in your intestines. Exercise slows fluid passage from the stomach but does not affect intestinal absorption. Sugar added to drinks can delay stomach emptying to increase risk for cramps.
Another study from the University of Utah, reported in the same journal, shows that taking a salty drink just before exercise increases endurance. Dehydration is the most common cause of fatigue during exercise in fit men and women. This study used salted drinks or placebo (unsalted) drinks with two groups of cyclists, and demonstrated a significant improvement in an endurance time trial as well as better maintenance of blood volume in the group that had the salted drinks.
Anything that increases blood volume should increase endurance. Taking in fluid before exercising increases blood volume, and using salty drinks increases blood volume more than pure water. But a major problem with salty drinks is that they usually taste awful. You can accomplish the same results by drinking water, soda or any other beverage you like and eat a handful of salted peanuts or other salty food before and during your endurance events.
When you're not exercising, don't get in the habit of using sports drinks or any other sugared drinks to quench thirst. They'll add up to a lot of calories with little other nutritional value. Use plain water or a calorie-free beverage instead.


Some days, it seems like the roadways are our own form of the Wild West, with each driver following their own code of law.
One driver lived up to traffic expectations during the morning rush hour earlier this week, blocking the entire HOV lane.
As shown in the video, the dark grey truck is visible between the barriers of the HOV lane, completely perpendicular to the lane's direction.
You can see a car on the shoulder, so it's possible the truck caused or became part of an accident, or the driver potentially planned to offer aid.
It's also possible he is attempting to turn around, possibly planning to drive on the shoulder.
No accidents or injuries are known to be reported around the time of the incident.
What's important is no one seems surprised by the display.
After all, why would they be? It's just another day in Houston traffic.


WHISKY DRAM – the first major festival of single malt whiskey. One day tasting will feature 200 of the best single malt samples from 50 producers in Scotland, Japan, India and Taiwan.
WHY SINGLE MALT? Single malt whiskey – the most honest among drink spirits. This confirms the high demand that the world already exceeds supply. Its aroma and flavor depends on the region (traditions) and the quality of performance, and may be fundamentally different in the same Scotland. From the flower-fruit and a touch of honey in drinks from Speyside to tar, rubber, coal and peat in the Isle and sea salt with iodine in the islands. This drink can be studied for years and each time discovering something new.
Festival WHISKY DRAM – is primarily a site for meeting people with open views. And its main value is a new experience that will get everyone who decides to join. Wine tastings, lectures and acquaintances – all of this is about expanding horizons and development.
Brands represented at the event – the best that can be collected throughout the world.
WHAT ELSE. New independent bottler, which releases always come out a very limited edition, calling on the world market a strong reaction. Whiskey distillery already closed and rare items that can only be found in the cellars of collectors.
Limited Bottling and special filling, including whiskey without mechanical filtration. The lounge area, DJ place and bar with creative cocktails by world-renowned mixology.
The ticket price includes a glass of 300 UAH Spiegelau and 5 pieces for tasting.


It is possible the "Oumuamua" space object that went past the sun in 2017 is an alien object sent toward Earth to observe our civilization.
Scientists said it has high speeds and an unusual trajectory, as it has come from outside of our solar system.
The interstellar visitor appears to be the size of a stadium, reports say. Scientists believe its shape and speed set it apart from the usual comets and asteroids they see.
A pair of Harvard researchers are raising the question of whether this object belongs to aliens, and their comments will appear in a Nov. 12 edition of "The Astrophysical Journal Letters".


text:
Q: How do I contact Xbox Live Account staff via e-Mail? I was looking for a really old account of mine on Xbox live and I found it; it has been about 7 years since it's been used and has become inactive.
Now I cannot remember any info needed to access that account, but I want to email someone who handles things like this at Xbox live headquarters. The only options their xbox.com site is giving is to call them or have them call you, but no email.
I find emailing so much easier and a lot cheaper too.
Does anyone know where i can find an email address to use
or a decent site that will put me into the right direction?
Is there any way of getting through to them without phoning?

A: You can get in contact with Microsoft personnel in the official forum.

But i think this is going to be a lost cause.
If the account has been inactive for 7 years straight, then it is most likely not inactive, but deleted due to a timout.
And seeing as the XBL service as a whole only exists for 10 years now, it actually more than just "most likely" that your account is deleted.

But still, try and you might get lucky.


Paper Seedlings: BEAUTIFULLY SAID: OPTIMIST OR PESSIMIST?
Wonderful quote! And so true! Those like me, of grandmother age, have long agovlearned to avoid those who "make difficulties out of their opportunities!"
I'm an optimist, for sure! Thanks for linking up with us again at #FridayFrivolity!
Another great one Linda thank you. We really enjoy the links you share at Celebrate Your Story. Have a fun weekend!
Great Quote!! My husband and I are the opposite of each other like this...I'm an optimist and he is a Pessimist! :-) Thanks for sharing on My 2 Favorite Things on Thursday! Hope to see you again tomorrow!! Pinned!


To create an image in one's mind of someone or something being inside of something else, such as clothes, an enclosed space, a location, etc. Picture these sofas inside our living room—wouldn't they just be perfect? The old advice for public speaking is to picture the audience in their underwear, but I don't think that really helps. Picturing myself inside of an office building for eight hours a day makes my stomach turn.
1. to form a mental picture of someone wearing something. I can just picture Tony in that baseball uniform. Can you picture yourself in a dress like this?
2. to form a mental picture of someone inside something or some place. I can just picture you in that car! Can you picture yourself in jail?


Fire/rescue units responded to a 5:10 p.m. alarm Wednesday (Jan 16) to Kiholo Bay for a man bitten by a shark.
A 43-year-old man was surfing with a friend at Kiholo Bay when he was bitten on the lower right arm.
Bystanders on the beach were attending to the bitten surfer when medics arrived. The victim was taken by ambulance to North Hawaii Community Hospital.
The beach was closed and shark sighting signs posted after the attack.
Years morning and noticed an unusual number of turtles in the water.
always there. Even seen it at the entrance to the fishponds years ago. People said it went into them at high tide.
Kiholo Bay and Wainanalii Pond on the North end are supposed to be important feeding and sleeping areas for Honu (Green Sea Turtles) that migrate between French Frigate Shoals and the Big Island. I know sometimes outside Honokohau Harbor if there's an injured turtle the Tiger Sharks move on in.


This course provides training in best practice recycled water management. Taking a practical approach the course is targeted to those responsible for aspects of recycled water quality and environmental management in water utilities, irrigators and private water suppliers as well as health and economic regulators.
The trainers delivering the course have been intimately involved in developing and implementing the Australian Guidelines for Water Recycling (AGWR). Dr Dan Deere was a member of the AGWR Risk Management Framework and Integration Working Group, was engaged to develop health aspects of the stormwater recycling guidelines and was part of the Working Group that developed the Requality tool for Framework implementation assessment for the Water Services Association of Australia (WSAA). Dr Daryl Stevens was a member of the AGWR Environmental Risk Working Group, and the Coordinator of the Scientific Services for the Environmental Risk Component. Both Dan and Daryl have been in involved with facilitating and assessing the implementation of the AGWR for recycled water schemes across Australia for the last 20 years. Their experience with derivation of guidelines for water recycling and practical application of these are unmatched in Australia!
Validation of pathogen log reduction in treatment and end use controls.
Management plans and regulatory approvals.
Those developing and implementing recycled water management plans.