text:
\section{The details of computing $\tilde{\psi}(x,s)$  and the survival probability for the potential $U(x)=\alpha |x|$}
\label{S-I}

In the presence of an absorbing barrier at $x=a$, the quantum potential is given by
\begin{equation}
V(x)=\begin{cases}\displaystyle
\frac{\alpha^2}{4 D} -\alpha \delta(x) &\text{for}~ x< a,\\
\infty &\text{for}~ x\ge a.
\end{cases}
\end{equation}
Therefore, the Schr\"odinger equation (in imaginary time) in the
in the Laplace space, $\tilde{\psi}(x,s)=\int_0^\infty
\psi(x,t)\, e^{-st}\, dt$,  reads
\begin{equation}
D\tilde{\psi}''(x,s) -[\alpha^2/(4 D) -\alpha \delta(x)+s] \tilde{\psi}(x,s)=-\delta(x),
\quad \text{for} ~~ x \le a,
\label{SE-Laplace-SM}
\end{equation}
with boundary conditions $\tilde{\psi}(x\to-\infty,s) =0$ and $\tilde{\psi}(x=a,s)=0$.

We solve \eref{SE-Laplace-SM}
separately for $x < 0$ and $0 < x < a$, and then match the solutions at $x=0$, where the wave function is continuous, but
its first derivatives undergoes a jump due to the $\delta$-function at $x=0$.
In each of these regions
\eref{SE-Laplace-SM} reads $D\tilde{\psi}''(x,s) -[\alpha^2/(4 D) +s] \tilde{\psi}(x,s)=0$.
Therefore, the solutions can be written as
\begin{equation}
\tilde{\psi}(x,s) = \begin{cases}
A_1 e^{px/(2D)} + B_1 e^{-px/(2D)}  &\text{for}~ x <0,\\
A_2 e^{px/(2D)} + B_2 e^{-px/(2D)}  &\text{for}~ 0 < x <a,
\end{cases}
\end{equation}
with $p=\sqrt{\alpha^2+4 Ds}$ and $A_1$, $B_1$, $A_2$ and $B_2$ are four unknown constants to be fixed from
the boundary and the matching conditions.

The boundary condition $\tilde{\psi}(x\to-\infty,s) =0$ implies $B_1=0$.
On the other hand the absorbing boundary condition $\tilde{\psi}(x=a,s)=0$ yields
 \begin{math}
 A_2= - B_2 e^{-p a/D} \, .
 \end{math}
Therefore, the solutions read
\begin{equation}
\tilde{\psi}(x,s) = \begin{cases}
A_1 e^{px/(2D)}  &\text{for}~ x <0,\\
B_2 \left[ 1- e^{- p(a-x)/D}  \right] e^{-px/(2D)}  &\text{for}~ 0 < x <a.
\end{cases}
\end{equation}

The two remaining constants $A_1$ and $B_2$ can be determined by matching the solutions at $x=0$. Integrating \eref{SE-Laplace-SM} in an infinitesimal region across $x=0$ gives the continuity of the solution $\tilde{\psi} (0^+,s) =  \tilde{\psi}(0^-,s) \equiv \tilde{\psi}(0,s)$ and the discontinuity of the derivatives $D \bigl[\tilde{\psi}'(0^+,s)  - \tilde{\psi}'(0^-,s)  \bigr] +\alpha \tilde{\psi}(0,s) = -1$.
The continuity of the solution implies $A_1= B_2\bigl(1-e^{-p a/D}\bigr)$.
Finally, the discontinuity of the solution gives $B_2 = 1/A(p)$ where
\begin{equation}
A(p) = p -\alpha \bigl( 1-e^{-p a/D}\bigr)\, ,
\label{Ap.1}
\end{equation}
with $p=\sqrt{\alpha^2+4 Ds}$.
Therefore,
\begin{equation}
\tilde{\psi}(x,s) = \begin{cases}
\frac{1}{A(p)}\, \bigl[1-e^{-pa/D} \bigr]\, e^{p x/(2D)} &\text{for}~ x\le 0,\\[2mm]
\frac{1}{A(p)}\,\bigl[1 - e^{-p (a-x)/D} \bigr]\, e^{-p x/(2D)} &\text{for}~ 0 \le x \le a,\\
\end{cases}
\label{psi-laplace2-SM}
\end{equation}
as mentioned in the main text.

We can obtain $\psi(x,t)$ by using the Bromwich integral
\begin{equation}
\psi(x,t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \tilde{\psi}(x,s) \, e^{st}\, ds,
\label{eq48.1}
\end{equation}
where $c$ is a real number such that all the singularities of $\tilde{P}(x,s)$ are on the left of the
vertical contour $\mathrm{Re}(s)=c$, in the complex-$s$ plane. The most dominant large $t$ behavior comes
from the singularity closest to the contour, the second dominant contribution comes from the next
singularity and so on.

\begin{figure}
\includegraphics[width=.5\hsize]{p-eqn2.pdf}
\caption{\label{p-eqn2-fig} Graphical representation of the non-trivial (non-zero) solution $p^*$ of the transcedental equation $p=\alpha (1-e^{-p a/D})$, which exists only for $a >a_c=D/\alpha$. For $a\to \infty$, $p^*=\alpha$, whereas for $a\to a_c^+$, we have $p^*\to 0$.}
\end{figure}

Let us first consider the limit $a\to\infty$, i.e., when there is no absorbing barrier. In this case,
we have from \eqref{Ap.1},  $A(p)=p-\alpha$, and hence,
$\tilde{\psi}(x,s)$ has a pole at $s=0$, i.e.,  $p=\alpha$.
There is also a branch-point (and not a pole) at $s=-\alpha^2/(4 D)$, i.e.,  $p=0$.  Therefore, the leading order behavior, that corresponds to the stationary state at $t\to\infty$, comes from the contribution from the pole at $s=0$ and the approach to the stationary state comes from the branch-point at $s=-\alpha^2/(4D)$. Evaluating the residue at $s=0$ gives
\begin{equation}
\psi(x,t\to\infty) = \frac{\alpha}{2D}
\exp\left( -\frac{\alpha |x|}{2D}  \right).
\label{eq48.2}
\end{equation}
Consequently, using $P(x,t) = e^{-\alpha |x|/(2 D)} \psi(x,t)$, we get the stationary distribution
\begin{equation}
P(x,t\to\infty)\equiv p_\mathrm{ss} (x) =\frac{\alpha}{2D} \exp\left( -\frac{\alpha |x|}{D} \right).
\label{eq48.3}
\end{equation}
The contribution from the branch-point gives the approach to the stationary state as
\begin{equation}
\psi(x,t) -\psi(x,t\to\infty) = Q(x,t) \exp\left(-\frac{\alpha^2}{4D} \,t\right)
\label{eq48.4}
\end{equation}
where $Q(x,t)$ is obtained from the integral around the branch-cut from $-\infty$ to $-\alpha^2/(4D)$, and is given by
\begin{equation}
Q(x,t)=\frac{1}{\pi} \int_0^\infty dr\,\frac{e^{-rt} }{\alpha^2+4Dr}
\biggl[
\sqrt{4Dr} \cos\left(\frac{|x|\sqrt{r}}{\sqrt{D}}\right)
-\alpha \sin\left(\frac{|x|\sqrt{r}}{\sqrt{D}}\right)
\biggr].
\label{eq48.6}
\end{equation}
The large $t$  behaviors, for arbitrary $x$,  can be computed by expanding  $[\alpha^2 + 4Dr]^{-1}$ about $r=0$ and carrying out the above integral term by term. The leading order behavior is given by
\begin{equation}
Q(x,t)= \frac{\sqrt{D}\, e^{-x^2/(4Dt)}
}{\sqrt{\pi}\,\alpha^2\,t^{3/2}}
\left(1- \frac{x^2}{2Dt} -\frac{\alpha |x|}{2D} \right)
+O(t^{-5/2}).
\end{equation}

In terms of the energy spectrum of the quantum Hamiltonian $\mathcal{H}$ given in the Schr\"odinger
equation in the main text, there is a single bound state corresponding to energy $E_0=0$, and continuum
spectrum of scattering states with energy $E \ge \alpha^2/(4D)$. Hence, the gap is given by
$\Delta=\alpha^2/(4\,D)-E_0= \alpha^2/(4D)$.  The continuum energy spectrum
from $\alpha^2/(4\,D)$ to $\infty$, manifests as the
branch-cut from $-\alpha^2/(4D)$ to $-\infty$ in the Laplace transform $\tilde{\psi}(x,s)$.

We now turn to finite $a$. As we bring the absorbing wall from $\infty$ to a finite value $a$,
the location of the rightmost pole in \eref{psi-laplace2-SM}
shifts from $s^*(\infty)=0$ to a negative value at $s=s^*(a)$ (with $s^*(a)<0$) that varies
continuously with $a$. In the quantum problem, this corresponds to the fact
that the ground state, while still remains a bound state, its energy $E_0(a)=-s^*(a)$ increases
continuously with decreasing $a$.
On the other hand, the position of the branch-point of \eref{psi-laplace2-SM} remains fixed at
$s=-\alpha^2/(4\,D)$, i.e., at $p=0$, irrespective of $a$.
Therefore, the continuum spectrum for the scattering states is always from
$\alpha^2/(4\,D)$ to $\infty$.
Clearly, as $a$ decreases, at some critical value $a=a_c$, the gap, $\Delta(a)= \alpha^2/(4\,D)-E_0(a)$, between
the scattering band starting at $\alpha^2/(4\, D)$ and the ground state $E_0(a)$
must vanish, triggering a phase transition.

To locate this critical value $a_c$, we look for the rightmost pole
of \eref{psi-laplace2-SM}, i.e., the nonzero solution of the transcendental
equation $A(p)=p -\alpha \bigl( 1-e^{-p a/D}\bigr)=0$. In Fig. (\ref{p-eqn2-fig}) we
plot both $\alpha \bigl( 1-e^{-p a/D}\bigr)$  (solid lines) and $p$ (dashed line) vs $p$. The slope of  $\alpha \bigl( 1-e^{-p a/D}\bigr)$  at $p=0$ is given by $\alpha\, a/D$.
The two curves $p$ and  $\alpha \bigl( 1-e^{-p a/D}\bigr)$  will cross each other at a nonzero value $p*>0$ only
for the slope $\alpha\, a/D>1$, i.e., $a>a_c= D/\alpha$. As $a\to a_c$ from above, the
nonzero solution $p^*\to 0$, triggering the phase transition. In the quantum language,
at $a=a_c$, the gap $\Delta(a)$ vanishes.
The mechanism of this phase transition is thus very similar to the mean-field transition
in ferromagnetic Ising model. For $a<a_c$, $A(p)=0$ has only a trivial solution $p=0$ (which
however is not a pole as it cancels with the numerator of $\tilde{\psi}(x,s)$).
For $a > a_c=D/\alpha$, the pole at $s=s^* <0$
on the real-$s$ line is given by
\begin{equation}
s^*(a) =-\frac{1}{4D} \left[\alpha^2 -p^{*2} (a) \right] \, .
\label{eq60}
\end{equation}
Therefore, for $a >a_c$, the energy of the bound state is given by
\begin{equation}
E_0(a)=-s^*(a) =\frac{1}{4D} \left[\alpha^2 -p^{*2} \right]\, ,
\label{eq61.1}
\end{equation}
and the corresponding gap in the spectrum is given by
\begin{equation}
\Delta(a) \equiv \frac{\alpha^2}{4D} - E_0(a) = \frac{[p^{*}(a)]^2}{4D}.
\label{eq61.2}
\end{equation}
The gap vanishes at $a\to a_c^+$, as $p^*\to 0$. There is no pole, and only the branch-point at
$s= -E_1=-\alpha^2/(4\, D)$,
for $a<a_c$ [see \fref{p-eqn2-fig}]. Therefore, the spectrum remains gapless for $a<a_c$.

For $a>a_c$, $\psi(x,t)$ is given by
\begin{align}
\psi(x,t) = R(x) \, e^{-E_0(a)  t}  + Q(x,t) \, e^{-\alpha^2 t/(4D)} ,
\label{eq61.3}
\end{align}
where $R(x=\lim_{s\to s^*} (s-s^*) \tilde{\psi}(x,s)$ arises from the residue at $s^*$
The prefactor $Q(x,t)$ of the sub-dominant  branch-point contribution is from the contour  integral around the branch-cut from $-\alpha^2/(4D)$ to $-\infty$, and the leading order time dependence is given by $Q(x,t) \sim t^{-3/2}$.

At $a=a_c$,  there is no pole corresponding to $p=0$, and only a branch-point.   In this case
\begin{align}
\psi(x,t) =  Q(x,t) \, e^{-\alpha^2 t/(4D)} ,
\label{eq61.6}
\end{align}
where the leading order time dependence of $Q(x,t)$ is given by $Q(x,t) \sim 1/\sqrt{t}$.

For $a<a_c$, again there is no pole and the contribution comes only from the branch-point. Therefore,
$\psi(x,t)$, is still given by \eref{eq61.6}, however, with a different $Q(x,t)$. In particular, $Q(x,t)
\sim t^{-3/2}$ for large $t$.

\vspace{0.3cm}

{\noindent {\bf {Survival/first-passage probability:}}} The survival probability $S_a(t|0)$ with the starting position $x_0=0$,  can be obtained by integrating  $P(x,t)$ obtained above,  over $x$ fro $0-\infty$ to $a$. Hence the  Laplace transform of the survival probability $\tilde{S}_a(s|0)= \int_0^\infty S_a(t|0)\, e^{-st}\, dt$ is given by
\begin{equation}
\tilde{S}_a(s|0) = \int_{-\infty}^a e^{-\alpha |x|/(2D)}\, \tilde{\psi}(x,s)\, dx,
\end{equation}
where $\tilde{\psi}(x,s)$ is given by \eref{psi-laplace2-SM}.
Performing the integral yields
\begin{equation}
\tilde{S}_a(s|0)=\frac{1}{s} \left[1-\frac{p}{A(p)}\, e^{-(\alpha+p) a/(2D)} \right],
\label{S-Laplace}
\end{equation}
with $p=\sqrt{\alpha^2+4Ds}$ and $A(p) = p -\alpha \bigl( 1-e^{-p a/D}\bigr) $.
The first-passage time distribution is related to the survival probability by
$F_a(t|x_0)=-\partial_t S_a(t|x_0)$. In the Laplace space,
this relation reads $\tilde{F}_a(s|x_0)=1-s\tilde{S}_a(s|x_0)$.
Hence, using \eref{S-Laplace}, we get the Laplace transform of the first-passage distribution as
\begin{equation}
\tilde{F}_a(s|0)=\frac{p}{A(p)}\, e^{-(\alpha+p) a/(2D)}.
\label{F-Laplace}
\end{equation}

By inverting formally the Laplace transform \eqref{S-Laplace},
the survival probability can be expressed as a Bromwich integral
\begin{equation}
S_a(t|0)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \tilde{S}(x_0,s) \, e^{st}\, ds.
\label{eq59}
\end{equation}
Let us first consider the trivial limit $a\to\infty$,
where $\tilde{S}(x_0,s) =1/s$.
The pole at $s=0$ gives $S_\infty(t|0) =1$,
which means $\theta(\infty):=-\lim_{t\to \infty} t^{-1} S_\infty (t|0)=0$.

Now as we bring the absorbing wall from $\infty$ to a finite value $a$, how does the exponent
$\theta(a):=-\lim_{t\to\infty} t^{-1} S_a(t|0)$ change as a function of $a$? It can be checked from
\eref{S-Laplace} that, $s=0$ is no longer a pole for any finite values of $a$. Similarly, although $p=0$
is a trivial solution of $A(p)=0$, it does not correspond to a pole as it cancels with $p$ in the
numerator. Nevertheless, $p=0$ is a branch-point.

For $a > a_c=D/\alpha$, there exists a pole at $s=s^*(a) <0$, on the real-$s$ line, as argued above.
Therefore, for $a>a_c$, the leading contribution at large times, comes from this pole,
and the branch-point gives the subleading correction:
\begin{align}
S_a(t|0) = R  \, \exp\left(-\frac{\alpha^2-p^{*2}}{4D}\, t\right)   + Q(t) \, \exp\left(-\frac{\alpha^2}{4D}\, t\right) ,
\label{eq62}
\end{align}
where $R=\lim_{s\to s^*} (s-s^*) \tilde{S}_a(s|0 )$ arises from the residue at $s^*$ and is given by
\begin{align}
R=\frac{2p^{*2} \exp\left[-\frac{a}{2a_c} \,e^{-p^*a/D}\right]}
{\alpha(\alpha+p^{*}) \left[1-\frac{a}{a_c}\,e^{-p^*a/D}\right]}.
\label{eq63}
\end{align}
The prefactor $Q(t)$ of the branch-point contribution arises from the contour  integral around the branch-cut from $-\infty$ to $-\alpha^2/(4D)$, and is given by
\begin{align}
Q(t)= e^{-\alpha a/(2D)}\int_0^\infty dr \,e^{-rt}\,
\frac{\sqrt{4D r}}
     {\pi \bigl[r+\alpha^2/(4D)\bigr]}\,
     \mathrm{Re} \left[-\frac{e^{-ia\sqrt{r/D}}}{A(i\sqrt{4Dr})}\right],
\label{eq64}
\end{align}
where $\mathrm{Re}[.]$ is the real part of the function inside the square bracket.
In \fref{surv-fig}~(a), we  compare the leading behavior, i.e., the contribution from the pole, given by the first line of \eref{eq62} with numerical simulation results and find very good agreement.

At $a=a_c$, to the leading order,
$A(p)=p^2/(2\alpha) + O(p^3)$, as $p\to 0$. Therefore, as $a\to a_c^+$, the
pole disappears and there is only a branch-point at $s=-\alpha^2/(4D)$,
corresponding to $p=0$. Note that although the location of the branch-point
is independent of $a$, the nature of the singularity  for $a=a_c$ is
different from that for $a\not=a_c$. At, $a=a_c$, following
Eqs.~\eqref{eq64}, the leading order behavior for large  $t$ is given by
\begin{equation}
S_{a_c}(t|0)= Q(t)\,  \exp\left(-\frac{\alpha^2}{4D}\, t\right),
\label{eq68}
\end{equation}
where
\begin{equation}
Q(t)=
e^{-\frac{1}{2}}\,
\frac{4\sqrt{D}}{\alpha\sqrt{\pi\, t}}   +O(t^{-3/2}),
\label{eq68.1}
\end{equation}
\Fref{surv-fig}~(b)  compares this leading behavior with numerical simulation and shows a very good agreement.

For $a<a_c$, there is no pole [see \fref{p-eqn2-fig}], and only a branch-point at $s=-\alpha^2/(4D)$. Therefore,
\begin{equation}
S_a(t|0) =  Q(t) \, \exp\left(-\frac{\alpha^2}{4D}\, t\right),
\label{eq69}
\end{equation}
where $Q(t)$ is given by \eref{eq64}.
For  $ a \ll a_c$, the leading order behavior for large $t$  can be evaluated as
\begin{equation}
Q(t)= \frac{e^{-\alpha a/(2D)}} {(1-a/a_c)^2}
\frac{2a\sqrt{D} }{\alpha^2\sqrt{\pi}\, t^{3/2}}
 +  O(t^{-5/2}),
 \label{eq66}
\end{equation}

Figures~\ref{surv-fig}~(c) and (d)  compare this result with numerical simulations, where $Q(t)$ evaluated exactly by performing numerical integration of  \eref{eq64} in (c), and the leading order behavior of $Q(t)$ given by \eref{eq66} is used in (d). Naturally, the agreement is perfect in \fref{surv-fig}~(c) whereas in \fref{surv-fig}~(d), it becomes better as $a$ moves away from $a_c$.

\begin{figure*}
\includegraphics[width=.45\hsize]{modx1.pdf}~~
\includegraphics[width=.45\hsize]{modx2.pdf}\\[2mm]
\includegraphics[width=.45\hsize]{modx3.pdf}~~
\includegraphics[width=.45\hsize]{modx3a.pdf}
\caption{\label{surv-fig} Survival probability of a Brownian particle in a potential $U(x)=\alpha |x|$,  starting at the position $x_0=0$   and in the presence of an absorbing barrier at $x=a >0$. We set the diffusion coefficient $D=1$ and $\alpha=1$, so that $a_c=D/\alpha=1$. The points are from numerical simulations whereas the solid lines are theoretical results. In (a) $a>a_c$, the theoretical lines plot the dominant behavior given by the first line of \eref{eq62}. In (b), $a=a_c$, the theoretical line is the leading behavior  from \eref{eq68}. In (c) and (d), $a<a_c$, the theoretical lines plot  \eref{eq69} with $Q(t)$ obtained by numerical integration of \eref{eq64} in (c) and  using \eref{eq66} in (d).
}

\end{figure*}

To summarize, the leading asymptotic of $S_a(t|0)$ is given by
\begin{equation}
S_a(t|0) \sim \exp\bigl[-\theta(a) t\bigr],
\label{eq70}
\end{equation}
where
\begin{equation}
\theta(a) = \begin{cases}\displaystyle
\frac{1}{4D}
\bigl[\alpha^2- p^{*2}(a)\bigr] &\quad\text{for}~ a>a_c\\[3mm]
\displaystyle
\frac{\alpha^2}{4D}
 &\quad\text{for}~ a<a_c
\end{cases}
\label{eq71}
\end{equation}
in which $0<p^*(a)<\alpha $ solves the   transcendental
equation  $p =\alpha \bigl( 1-e^{-p a/D}\bigr)$ [see \fref{p-eqn2-fig}]. Note that while $\theta(a)$ varies as a function of $a$ for $a>a_c$, it remains at the constant value $\alpha^2/(4D)$ for $a<a_c$. Near $a=a_c^+$,   we have $p^* = 2D (a-a_c)/a_c^2 + O\bigl[(a-a_c)^2\bigr]$.
Therefore, as $a$ approaches $a_c$ from above,
\begin{equation}
\theta(a) = \frac{\alpha^2}{4D} - \frac{D}{a_c^4}(a-a_c)^2 +O\bigl[ (a-a_c)^3\bigr].
\end{equation}

\vspace{0.3cm}

{\noindent {\bf {Link to the interlace theorem:}}} We now show
how our results are consistent with a powerful
interlace theorem derived in Refs.~\cite{HG2018,HG2019}.
Let us first consider an unconstrained motion
of particle (no absorbing barrier or equivalently $a\to \infty$ limit)
diffusing in a sufficiently confining potential $U(x)$, such that its
quantum counterpart $V(x)=[U'(x)]^2/{4D}- U''(x)/2$ has a discrete spectrum say
$\{E_1<E_2< E_3< \ldots\}$. These energy eigenvalues of the Schr\"odinger
equation in the quantum potential $V(x)$ are identical to
the relaxation eigenvalues of the unconstrained Fokker-Planck
equation, $\partial_t P- D\partial_x^2 P- \partial_x [U'(x)P]=0$ on
the full real line.
Now, imagine that we bring the absorbing barrier from $a=\infty$ to
a finite value $a\ge 0$, i.e., the barrier always stays
to the right of the initial position of the particle which we set to be $x=0$.
If the barrier is to the left of the particle, one can reverse $x$ and
it would be a similar analysis. In this paper, we therefore restrict ourselves to $a\ge 0$.
The presence of the absorbing barrier at $a\ge 0$
changes the spectrum of the Fokker-Planck operator. The new ordered eigenvalues
$\{\lambda_1(a)<\lambda_2(a)<\lambda_3(a) \ldots\}$ obviously
depend on $a$. Note that the lowest eigenvalue $\lambda_1(a)\equiv \theta(a)$
by definition.
Hence as $a$ decreases towards $0$, the whole spectrum $\{\lambda_i(a)\}$
moves upwards. The unconstrained spectrum $\{E_i\}$ of course is independent
of $a$.
According to the interlace theorem, for any given $a$, the $\{\lambda_i\}$'s
are interlaced with $\{E_i\}$'s, i.e., $E_1 \le \lambda_1(a) \le E_2 \le \lambda_2(a) \le E_3 \le \dotsb$,  ---between any pair
of consecutive $E_i$'s, there is  one and only one $\lambda_i$ (see
Fig. (\ref{spectrum_fig})). As $a$
decreases, the whole spectrum $\{\lambda_i(a)\}$ slides upwards, but
always respects the interlacing rule for any $a\ge 0$. In this case, the lowest eigenvalue
$\lambda_1(a)=\theta(a)$ monotonically increases with decreasing $a$ as $a\to 0$.
As mentioned above, this occurs for sufficiently confining potentials, e.g.,
when $U(x)\sim |x|^{\gamma}$ as $x\to -\infty$ with $\gamma>1$.

\begin{figure}
\includegraphics[width=.5  \hsize]{spectrum1.pdf}
\caption{For a sufficiently confining potential in the
unconstrained system (with the absorbing barrier $a\to \infty$),
the discrete energy eigenvalues $\{E_1<E_2<E_3\ldots\}$
of the associated Schr\"odinger equation are shown schematically
on the right (blue). As the barrier location $a\ge 0$ decreases,
the new ordered eigenvalues (first-passage eigenvalues)
$\{\lambda_1(a)<\lambda_2(a)<\lambda_3(a)\ldots\}$ are
shown (schematic) on the left (red). The sets $\{E_i\}$ and
$\{\lambda_i(a)\}$ interlace, i.e., between any two consecutive
$E_i$'s,  there is one and only one $\lambda_i$, as shown schematically
in the figure.}
\label{spectrum_fig}
\end{figure}

In the case $\gamma=1$, e.g., for $U(x)= \alpha |x|$ as $x\to -\infty$,
the Schr\"odinger equation in
the full space (i.e., when the barrier location $a\to \infty$) has
a single discrete eigenvalue (the lowest one) and
a band of continuous eigenvalues, separated from the ground state
by a finite gap $\alpha^2/(4D)$, as shown in Fig. 2 in the main text. As $a$
decreases, the continuous part of the spectrum does not change,
but the lowest discrete eigenvalue increases (see Fig. 2
in the main text). Since the continuous part
of the spectrum is independent of $a$, clearly the discrete eigenvalue can not
increase beyond the gap $\alpha^2/(4D)$. Had it done so,
it will go over the lowest eigenvalue of the continuous sector
and clearly violate the interlacing rule mentioned above. Hence
$\theta(a)$ freezes when it hits the gap value $\alpha^2/(4D)$, at
a certain critical value $a_c=D/\alpha$ as shown in our paper.
Thus this freezing transition and the mechanism behind it is completely
consistent with the interlacing theorem, applied to the discrete
part of the spectrum in the critical case $\gamma=1$ till the
onset of the freezing transition.

\section{The backward Fokker-Planck equation for the survival probability in the Laplace space}

As mentioned in the main text, the backward Fokker-Planck equation satisfied by the survival probability, with the starting position $x_0$, reads
\begin{equation}
\frac{\partial S_a}{\partial t} = D \frac{\partial^2 S_a}{\partial x_0^2} -  U'(x_0)\, \frac{\partial S_a}{\partial x_0}
\end{equation}
with the initial condition $S_a(0|x_0)=1$ and the boundary conditions,
$S_a(t|x_0\to -\infty)=1$ and $S_a(t|x_0=a)=0$.
The Laplace transform  $\tilde{S}_a(s|x_0)= \int_0^\infty S_a(t|x_0)\, e^{-st}\, dt$ then satisfies
\begin{equation}
D \tilde{S}_a''(s|x_0) -U'(x_0) \tilde{S}_a'(s|x_0) -s \tilde{S}_a(s|x_0) = -1,
\label{S-Laplace.1}
\end{equation}
with the boundary conditions
\begin{equation}
\tilde{S}_a(s|x_0\to-\infty)=1/s\quad\text{and}\quad \tilde{S}_a(s|x_0=a)=0.
\end{equation}
Setting
\begin{equation}
\tilde{S}_a(s|x_0) =  \frac{1}{s} + \tilde{q}(x_0),
\end{equation}
$\tilde{q}(x_0)$ satisfies the homogeneous differential equation
\begin{equation}
D \tilde{q}''(x_0) -U'(x_0) \tilde{q}'(x_0) -s \tilde{q}(x_0) = 0,
\label{q2}
\end{equation}
with the boundary conditions $\tilde{q}(x_0\to -\infty)=0$ and $\tilde{q}(x_0=a)=-1/s$.

\section{The details of computing the survival probability for a
potential that  grows faster than $|x|$ as\,  $x\to -\infty$}

The example of the potential considered in the main text is
\begin{equation}
U(x) = \begin{cases}
 \frac{1}{2}\, \mu \, x^2&\text{for}~ x <-b, \\
 \alpha |x| &\text{for}~ -b <x < a,
\end{cases}
\label{potential (i)}
\end{equation}
where $b >0$ and for the sake of continuity of the potential at $x=-b$, we set $\mu =2\alpha/b$.

We have to solve \eref{q2} separately in the three regions, (i)  $x_0<-b$, (ii) $-b < x_0 <0$,  and (iii) $0<x_0 <a$ and then match the solutions for the continuity of the solutions as well as the derivatives at both $x_0=-b$ and $x_0=0$. For $x_0 <-b$, \eref{q2} reads
\begin{equation}
D\tilde{q}''(x_0) -\mu x_0 \tilde{q}'(x_0) -s\tilde{q}(x_0)=0.
\label{qx0}
\end{equation}
Now substituting
\begin{equation}
\tilde{q}(x_0)=e^{\mu x_0^2/(4 D)}\, w(x_0\sqrt{\mu/D}),
\end{equation}
above gives the differential equation
\begin{equation}
w''\bigl(x_0\sqrt{\mu/D}\bigr)  + \left[-\frac{s}{\mu} + \frac{1}{2} - \frac{1}{4} \bigl(x_0\sqrt{\mu/D}\bigr)^2 \right] w\bigl(x_0\sqrt{\mu/D}\bigr) =0,
\end{equation}
whose general solution can be expressed in terms of the two linearly independent parabolic cylinder functions  $D_{-s/\mu}\bigl(x_0\sqrt{\mu/D}\bigr)$ and $D_{-s/\mu} \bigl(-x_0\sqrt{\mu/D}\bigr)$.  Noting that $D_{-s/\mu}\bigl(x_0\sqrt{\mu/D}\bigr)\to \infty$  whereas $D_{-s/\mu}\bigl(-x_0\sqrt{\mu/D}\bigr)\to 0$ as $x_0 \to -\infty$, the solution of \eref{qx0} that tends to zero as $x_0\to -\infty$, is given by
\begin{equation}
\tilde{q}(x_0)= A_1 \, e^{\mu  x_0^2/(4 D)} D_{-\frac{s}{\mu}}\left(-x_0 \sqrt{\frac{\mu}{D}}\right),
\end{equation}
 where  the constant $A_1$ to be determined.

For $-b < x_0 < 0$, \eref{q2} reads
\begin{equation}
D \tilde{q}''(x_0) +\alpha  \tilde{q}'(x_0) -s \tilde{q}(x_0) = 0.
\end{equation}
The general solution is given by
\begin{equation}
\tilde{q}(x_0)= A_2\, e^{(p-\alpha) x_0/(2D)} +  B_2  e^{-(p+\alpha) x_0/(2D)},
\end{equation}
with $p=\sqrt{\alpha^2+4Ds}$ and the constants $A_2$ and $B_2$ to be determined.

Finally, for $0 <x_0 <a$, \eref{q2} reads
\begin{equation}
D \tilde{q}''(x_0) -\alpha  \tilde{q}'(x_0) -s \tilde{q}(x_0) = 0,
\end{equation}
whose general solution is given by
\begin{equation}
\tilde{q}(x_0)= A_3\, e^{(\alpha+p) x_0/(2D)} +  B_3  e^{(\alpha-p) x_0/(2D)}
\end{equation}
with the constant $A_3$ and $B_3$ to be determined.

After determining the five constants by using the four matching conditions at $x_0=-b$ and $x_0=0$,  and the boundary condition at $x_0=a$,  for the starting position $x_0=0$ (taking for simplicity), we get
\begin{equation}
\tilde{S}_a (s|0) = \frac{1}{s} \bigl[1-\tilde{F}_a(s|0)\bigr]
\end{equation}
where the Laplace transform of the first-passage time  distribution is given by
\begin{equation}
\tilde{F}_a(s|0)=\frac{p}{B(s)}\, e^{-(\alpha+p) a/(2D)}\, \chi(s),
\end{equation}
with $p=\sqrt{\alpha^2+4 D s}$ and
\begin{align}
B(s) &= \left[p^2 \left(e^{\frac{p b}{D}}+e^{-\frac{p a}{D}}\right)+3 \alpha ^2 \left(1-e^{-\frac{ p a}{D}}\right) \left(e^{\frac{p b}{D}}-1\right)-\alpha  p \left(4 e^{\frac{p b}{D}}+1-e^{\frac{ p b}{D}} e^{-\frac{p a}{D}}-4 e^{-\frac{p a}{D}}\right)\right] D_{-\frac{s}{\mu }}\left(b \sqrt{\frac{\mu }{D}}\right)\notag\\
&\quad-2 D \left[\alpha  \left(1-e^{-\frac{a p}{D}}\right) \left(e^{\frac{b p}{D}}-1\right)+p \left(e^{-\frac{pa}{D}}-e^{\frac{p b}{D}}\right)\right]\, \sqrt{\frac{\mu }{D}} \,  D_{-\frac{s}{\mu }+1}\left(b \sqrt{\frac{\mu }{D}}\right),
\\
\chi(s)& =\left[p+3 \alpha +(p-3 \alpha ) e^{\frac{b p}{D}}\right] D_{-\frac{s}{\mu }}\left(b \sqrt{\frac{\mu }{D}}\right)+2 D \left(e^{\frac{p b}{D}}-1\right) \, \sqrt{\frac{\mu }{D}} \,\, D_{-\frac{s}{\mu }+1}\left(b \sqrt{\frac{\mu }{D}}\right)
.
\end{align}

After substituting, $s=(p^2-\alpha^2)/(4D)$, it is easy to check that $\tilde{F}_a(s(p)|0)\equiv \tilde{G}_a(p)$ is a symmetric function with respect to $p$, indicating that series expansion of $\tilde{G}_a(p)$ around $p=0$ contains only even powers of $p$.  Therefore, $p=0$, and equivalently $s=-\alpha^2/(4D)$, is not a branch-point. Similarly, it is easy to check that $s=-\alpha^2/(4D)$ is not a pole of $\tilde{F}_a(s|0)$, i.e., $p=0$ is not a pole of $\tilde{G}_a(p)$.

\begin{figure}
\includegraphics[width=.5  \hsize]{bs.pdf}
\caption{$B(s)/(\alpha^2+4Ds)$ as a
function $s$ for $s \le 0$, for certain parameter values $\alpha=D=\mu=1$ and $b=2$.
The red solid line and the blue dashed line plot respectively the real part and the imaginary part
of the function. The zeros of the function given by the points where both real and
imaginary part become zero.  }
\label{bs}
\end{figure}

Although $s=-\alpha^2/(4D)$ is a zero of $B(s)$, it cancels with the numerator as both the numerator and the denominator tend to zero as $\alpha^2+4 Ds$ near $s=-\alpha^2/(4D)$. Therefore, to find the poles of $\tilde{S}_a(s|0)$ one should look at the zeros of $B(s)/(\alpha^2+4Ds)$. This function has infinite number of zeros on the negative $s$ axis [see~\fref{bs}], denoted by $-\infty <\dotsb < s_2^*(a) < s_1^*(a)  < s_0^*(a)  <0$.
Therefore, inverting the Laplace transform, the survival probability is given by
\begin{equation}
S_a(t|0) = \sum_{i=0}^\infty\, R_i (a) \,  e^{s_i^*(a) t},
\end{equation}
where
\begin{equation}
R_i(a) = \lim_{s\to s_i^*} (s-s_i^*)\, \tilde{S}_a(s|0) = \frac{\sqrt{\alpha^2+4 D s_i^*(a)}\,\,  \chi(s_i^*(a))}{\bigl[-s_i^*(a)\bigr]\,\,  B'(s_i^*(a))}\, \exp\left(-\frac{a}{2 D}\left[\alpha + \sqrt{\alpha^2 + 4 D s_i^*(a)}\right]\right).
\end{equation}
Evidently, the leading behavior of the survival probability at large times is given by
\begin{equation}
S_a(t|0) = R_0(a)\,  e^{-\theta(a) \, t} + O(e^{s_1^*(a) t}),
\label{s-quadratic-asym}
\end{equation}
where $\theta(a)=-s_0^*(a)$. We compare this expression with numerical simulation in \fref{s-fig-quadratic} and find very good agreement.

\begin{figure}
\includegraphics[width=.49\hsize]{s1.pdf}~~\includegraphics[width=.49\hsize]{s2.pdf}~
\caption{\label{s-fig-quadratic}Survival probability of a Brownian particle in a potential given by \eref{potential (i)},  starting at the position $x_0=0$  and in the presence of an absorbing barrier at $x=a >0$, for $a=0.5$ and $2.0$ on the left panel and $a=3.0$ om the right panel. We set $D=\alpha=\mu=1$ and $b=2$. The points are from numerical simulations whereas the solid lines plot the leading order theoretical expression given by \eref{s-quadratic-asym}.   }
\end{figure}

\section{ The details of computing the survival probability for a potential that grows slower than $|x|$ as $x\to -\infty$}
\label{logx-potential-left}

The example of the potential considered in the main text is
\begin{equation}
U(x) = \begin{cases}
 c \ln (-x/\lambda) &\text{for}~ x <-b, \\
 \alpha |x| &\text{for}~ -b <x < a,
\end{cases}
\label{potential (iii)}
\end{equation}
where $\lambda>0$ sets a length scale, $b >0$,  and for the sake of continuity of the potential at $x=-b$, we set $c\ln (b/\lambda)=\alpha b$.

We have to solve \eref{q2} separately in the three regions, (i)  $x_0<-b$, (ii) $-b < x_0 <0$,  and (iii) $0<x_0 <a$ and then match the solutions for the continuity of the solutions as well as the derivatives at both $x_0=-b$ and $x_0=0$. In the region $x<-b$, \eref{q2} reads
\begin{equation}
D \tilde{q}''(x_0) -(c/x_0)\,  \tilde{q}'(x_0) -s \tilde{q}(x_0) = 0.
\end{equation}
The solution of this equation, that tends to zero as $x\to -\infty$, is given by
\begin{equation}
\tilde{q}(x_0)=A_1 |x_0|^\nu K_\nu \bigl(|x_0|\sqrt{s/D}\bigr),
\end{equation}
where $A_1$ is a constant to be determined, $\nu=(c/D + 1)/2$,  and $K_\nu(z)$ is the modified Bessel function of the second kind.

In the region $-b <x_0 <0$, \eref{q2} reads
\begin{equation}
D \tilde{q}''(x_0) +\alpha  \tilde{q}'(x_0) -s \tilde{q}(x_0) = 0.
\end{equation}
The general solution is given by
\begin{equation}
\tilde{q}(x_0)= A_2\, e^{(p-\alpha) x_0/(2D)} +  B_2  e^{-(p+\alpha) x_0/(2D)},
\end{equation}
with $p=\sqrt{\alpha^2+4Ds}$ and the constants $A_2$ and $B_2$ to be determined.

Finally, for $0 <x_0 <a$, \eref{q2} reads
\begin{equation}
D \tilde{q}''(x_0) -\alpha  \tilde{q}'(x_0) -s \tilde{q}(x_0) = 0,
\end{equation}
whose general solution is given by
\begin{equation}
\tilde{q}(x_0)= A_3\, e^{(\alpha+p) x_0/(2D)} +  B_3  e^{(\alpha-p) x_0/(2D)}
\end{equation}
with the constant $A_3$ and $B_3$ to be determined.

After determining the five constants by using the four matching conditions at $x_0=-b$ and $x_0=0$,  and the boundary condition at $x_0=a$,  for the starting position $x_0=0$ (taking for simplicity), we get
\begin{equation}
\tilde{S}_a (s|0) = \frac{1}{s} \bigl[1-\tilde{F}_a(s|0)\bigr]
\end{equation}
where the Laplace transform of the first-passage time  distribution is given by
\begin{equation}
\tilde{F}_a(s|0)=\frac{p}{B(s)}\, e^{-(\alpha+p) a/(2D)}\, \chi(s),
\end{equation}
with $p=\sqrt{\alpha^2+4 D s}$ and
\begin{align}
B(s) &=
\left[p^2 e^{p b/D}-\left(\alpha-(\alpha +p) e^{-p a/D}\right) \left(\alpha
\left(e^{p b/D}-1\right)+p\right)\right] K_{\nu }\left(b \sqrt{\frac{s}{D}}\right) \notag\\
&\quad+2 D  \left[p \left(e^{p b/D}-e^{-pa/D}\right)-\alpha  \left(1-e^{-p a/D}\right) \left(e^{p b/D}-1\right)\right] \sqrt{\frac{s}{D}}\,K_{\nu -1}\left(b \sqrt{\frac{s}{D}}\right),
\\
\chi(s)& =
\left[p-\alpha +(\alpha +p) e^{p b/D}\right] K_{\nu }\left(b \sqrt{\frac{s}{D}}\right)
+2 D  \left(e^{p b/D}-1\right) \, \sqrt{\frac{s}{D}}\, K_{\nu -1}\left(b \sqrt{\frac{s}{D}}\right).
\end{align}

Note that  in the limit $a\to\infty$, we get $\tilde{F}_a(s|0)\to 0$,  and consequently, $\tilde{S}_a(s|0) \to 1/s$. This gives, $\lim_{a\to\infty} S(t|0) =1$ at all times,  as expected.
For any finite $a$,
after substituting, $s=(p^2-\alpha^2)/(4D)$, it is easy to check that $\tilde{F}_a(s|0)$ is a symmetric function with respect to $p$, indicating that series expansion of $\tilde{F}_a$ around $p=0$ contains only even powers of $p$.  Therefore, $p=0$, and equivalently $s=-\alpha^2/(4D)$, is not a branch-point.
For several sets of  representative values of the parameters, we have numerically checked that $\tilde{F}_a(s|0)$ does not have any pole, indicating a non-exponential late time behaviour for $F_a(t|0)$ as well as for $S_a(t|0)$.  Anticipating such non-exponential behaviour to show up as branch-point singularities, we analyse $\tilde{F}_a(s|0)$ near $s=0$.  In our case, we choose the potential $U(x)$ to be confining so that there is an equilibrium state (for $a\to \infty$) given by the Boltzmann distribution $P(x, t\to \infty) \propto e^{-U(x)/D}$ that is integrable, i.e.,  $\int e^{-U(x)/D}\, dx$ is finite. This requires $c/D > 1$, and consequently, $\nu = (1+c/D)/2 >1$.

Using the leading behaviour of the modified Bessel function of the second kind near $s=0$, for  a given non-integer $\nu  \in (n,n+1)$ with $n\ge 1$ being an integer,
we get
\begin{align}
K_{\nu }\left(b \sqrt{\frac{s}{D}}\right) &= s^{-\nu/2} \left[ \sum_{m=0}^n \alpha_m \, s^m + \alpha_\nu\, s^\nu + O\bigl(s^{n+1}\bigr)\right],\\
\sqrt{\frac{s}{D}}\, K_{\nu -1}\left(b \sqrt{\frac{s}{D}}\right) &=
s^{-\nu/2} \left[ \sum_{m=1}^n \beta_m \, s^m + \beta_\nu\, s^\nu + O\bigl(s^{n+1}\bigr)\right],
\end{align}
where the coefficients in the above series can be obtained explicitly.
In particular,
\begin{align}
\alpha_0&=2^{\nu -1} b^{-\nu } D^{\nu /2} \Gamma (\nu ),

 \quad
\alpha_\nu = 2^{-\nu -1} b^{\nu } D^{-\frac{\nu }{2}} \Gamma (-\nu ) , \\
\beta_1&= 2^{\nu -2} b^{1-\nu } D^{\frac{\nu }{2}-1} \Gamma (\nu -1), \quad\text{and} \quad
\beta_\nu = 2^{-\nu } b^{\nu -1} D^{-\frac{\nu }{2}} \Gamma (1-\nu ).
\end{align}

The remaining factors in both $B(s)$ and $\chi(s)$ are analytic about $s=0$. Therefore, for a given (finite and fixed)  value of $a$, $\tilde{F}_a(s|0)$ can be expressed as
\begin{equation}
\tilde{F}_a(s|0) = \frac{\bigl[\sum_{j=0}^\infty a_j s^j\bigr]  \left[ \sum_{m=0}^n \alpha_m \, s^m + \alpha_\nu\, s^\nu + O\bigl(s^{n+1}\bigr)\right] + \bigl[ \sum_{j=0}^\infty b_j s^j\bigr]\left[ \sum_{m=1}^n \beta_m \, s^m + \beta_\nu\, s^\nu + O\bigl(s^{n+1}\bigr)\right] }{\bigl[\sum_{j=0}^\infty c_j s^j\bigr]  \left[ \sum_{m=0}^n \alpha_m \, s^m + \alpha_\nu\, s^\nu + O\bigl(s^{n+1}\bigr)\right] + \bigl[ \sum_{j=0}^\infty d_j s^j\bigr]\left[ \sum_{m=1}^n \beta_m \, s^m + \beta_\nu\, s^\nu + O\bigl(s^{n+1}\bigr)\right] },
\label{fp-series1}
\end{equation}
where the coefficients $\{ a_j, b_j, c_j, d_j\}$ can be computed. In particular, the first few coefficients are given by
\begin{align}
a_0&= c_0=2 \alpha ^2 e^{-\frac{\alpha  a}{D}} e^{\frac{\alpha  b}{D}},\quad
a_1= 2 e^{-\frac{a \alpha }{D}} \left[e^{\frac{\alpha  b}{D}} (-a \alpha +2 \alpha  b+3 D)+D\right],
\quad b_0=2 \alpha  D e^{-\frac{a \alpha }{D}} \left(e^{\frac{\alpha  b}{D}}-1\right), \\
c_1&=
e^{-\frac{a \alpha }{D}} \left[2 D \left(2 e^{\frac{\alpha  (a+b)}{D}}-e^{\frac{a \alpha }{D}}+e^{\frac{\alpha  b}{D}}+2\right)-4 \alpha  (a-b) e^{\frac{\alpha  b}{D}}\right], \quad
d_0=2 \alpha  D e^{-\frac{a \alpha }{D}} \left(e^{\frac{a \alpha }{D}}+e^{\frac{\alpha  b}{D}}-2\right).
\end{align}
\Eref{fp-series1} can be expanded in a series as
\begin{equation}
\tilde{F}_a(s|0)=\sum_{m=0}^n \frac{(-1)^m}{m!}\, T_{a,m}(0)\, s^m + \nu\, \Gamma(-\nu)\, a_\nu\,  s^{\nu} +o(s^\nu),
\end{equation}
where $T_{a,m}(0)=\int_0^\infty t^m\, F_a(t|0)\, dt$ is the $m$-th moment of the first-passage time distribution.
In particular, we have $T_{a,0}(0)=a_0/c_0=1$, as expected from normalization of the first-passage time distribution.  The mean first-passage time is given by
\begin{equation}
T_a(0)\equiv T_{a,1}(0)  = \frac{1}{a_0\alpha_0} \left[ \alpha_0 (c_1-a_1) + \beta_1 (d_0-b_0)\right],
\end{equation}
which, after explicit evaluation, agrees with  \eref{Ta-3} obtained from  the general expression
 \eref{mean fp}.

The coefficient of the
most dominant non-analytic term is given by
\begin{equation}
\nu\,\Gamma(-\nu)\, a_\nu = \frac{\beta_\nu (b_0-d_0)}{a_0\alpha_0} ,
\end{equation}
which after evaluation (while using $c\ln (b/\lambda) = \alpha b$) gives
\begin{equation}
a_\nu =
\left(\frac{2D}{\alpha \lambda}\right)\,  \left(e^{\frac{\alpha a}{D}}-1\right)\, \frac{1}{\Gamma(\nu)}\, \left(\frac{\lambda^2}{4D} \right)^\nu\, .
\end{equation}
 Using the relation
\begin{equation}
\int_0^\infty\, dz \left[e^{-z} - \sum_{m=0}^n \frac{(-z)^m}{m!} \right]\, {z^{-(\nu+1)}} = \Gamma(-\nu)
\quad \text{for $n < \nu < n+1$,}
\end{equation}
  we find that the late time behavior of the first-passage time distribution is given by the power-law distribution
\begin{equation}
F_a(t|0)  = \nu\,  a_\nu\, t^{-(\nu+1)} + o( t^{-(\nu+1)}).
\end{equation}
Consequently, the survival probability has the asymptotic power-law decay
\begin{equation}
S_a(t|0) = \int_t^\infty F_a(t'|0)\, dt' =  a_\nu\, t^{-\nu} + o(t^{-\nu}).
\label{S-long-t-log}
\end{equation}
For integer values of $\nu=n$, the Bessel's functions have $\ln s$ singularities, which in turn
gives rise to $\ln s$ singularities in $\tilde{F}_a(s|0)$. Consequently,  the first-passage time distribution and survival probability
have power-low decays accompanied by $\ln t$ corrections.

\section{The mean first-passage time}

The mean first-passage time $T_a(x_0)$ to a position $a$, starting with the position $x_0 <a $
can be computed exactly for arbitrary confining potential $U(x)$~\cite{Risken1984}. Here, for convenience,
we reproduce this proof. The mean first-passage time
is defined by
\begin{equation}
T_a(x_0)= \int_0^\infty\, t \, F_a (t|x_0)\, dt.
\end{equation}
Using $F_a(t|x_0)=-\partial_t S_a(t|x_0)$ in the above integral and then integrating  by parts using the boundary conditions $S_a(t=0|x_0)=1$ and $S_a(t\to\infty|x_0)=0$, we get
\begin{equation}
T_a(x_0)=\int_0^\infty S_a(t|x_0)\, dt \equiv \tilde{S}_a(s=0|x_0).
\end{equation}
Therefore, setting $s=0$ in  \eref{S-Laplace}, we get the differential equation
\begin{equation}
D \frac{d^2 T_a}{d x_0^2} - U'(x_0) \frac{dT_a}{dx_0} =-1.
\end{equation}
Using $W(x_0) = T_a'(x_0)$, we get
\begin{equation}
\frac{dW}{dx_0} - \frac{U'(x_0)}{D} W(x_0) = -\frac{1}{D}.
\end{equation}
Multiplying both sides of  the above equation by $e^{-U(x_0)/D}$ we get
\begin{equation}
\frac{d}{dx_0} \left[ W(x_0)\, e^{-U(x_0)/D}\right] = -\frac{1}{D}\, e^{-U(x_0)/D},
\end{equation}
which can be integrated to
\begin{equation}
W(x_0) e^{-U(x_0)/D} = -\frac{1}{D} \int_{-\infty}^{x_0} e^{-U(z)/D}\, dz,
\end{equation}
where we have used the boundary condition $\lim_{x_0\to -\infty} W(x_0) e^{-U(x_0)/D} = 0$. Therefore,
\begin{equation}
\frac{d T_a}{d x_0} = -\frac{1}{D} e^{U(x_0)/D}\,  \int_{-\infty}^{x_0} e^{-U(z)/D}\, dz.
\end{equation}
Integrating the above equation from $x_0$ to $a$ and using  the boundary condition $T_a(x_0=a)=0$, we get
\begin{equation}
T_a(x_0)  =\frac{1}{D} \int_{x_0}^a\, dy\, e^{U(y)/D}\, \int_{-\infty}^y\, e^{-U(z)/D}\, dz.
\label{mean fp}
\end{equation}

For $U(x)=\alpha |x|$, after performing the integrals (for $x_0=0$) we get,
\begin{equation}
T_a(0)=\frac{2D}{\alpha^2}\left(e^{\alpha a/D}-1\right) -\frac{a}{\alpha}.
\label{Ta-2}
\end{equation}

For the potential given by \eref{potential (i)}, we get
\begin{equation}
T_a(0)= \frac{D}{\alpha^2}\,  \left(e^{\alpha a/D}-1\right) \left[ 2 - e^{-\alpha b/D} +
\sqrt{\frac{\pi}{2}}\, \frac{\alpha}{\sqrt{\mu D}}\, \mathrm{erfc} \left(b \sqrt{\frac{\mu}{2D}}\right)
 \right] -\frac{a}{\alpha}, \quad\text{with}~\mu b= 2\alpha.
\label{Ta-1}
\end{equation}
On the other hand,
for the potential given by \eref{potential (iii)}, we get
\begin{equation}
T_a(0)=\frac{D}{\alpha^2} \, \left(e^{\alpha a/D}-1\right) \left[ 2 - e^{-\alpha b/D} + \frac{\alpha b\, e^{-\alpha b/D}}{2 D (\nu-1)}  \right] -\frac{a}{\alpha}.
\label{Ta-3}
\end{equation}
Note that, by taking the limit $b\to \infty$ in both \eref{Ta-1} and \eref{Ta-3}, one gets back \eref{Ta-2}, as required.

\section{Survival probability for a potential given by \eref{linear-quadratic}}

In the main text, we argued that the freezing transition of the decay rate $\theta(a)$ is robust, i.e, it
occurs for any confining potential $U(x)$ that behaves asymptotically $U(x) \sim |x|$ as $x\to -\infty$.
The actual value of $a_c$ depends on the details of the potential $U(x)$, but the existence of the
freezing transition does not depend on the details of $U(x)$ in the bulk as long as $U(x)\sim |x|$ when
$x\to -\infty$. This argument was based on a general mapping to a quantum problem and using the
properties of the Schr\"odinger equation in one dimension. We showed that the freezing transition
coincides with the vanishing of the gap between the ground state (which is a bound state) and the
continuum of scattering states when $a\to a_c$ from above. We provided an exactly solvable example of
$U(x)$ in Eqs. (7a) and (7b) of the main text. In this appendix, we provide another example of $U(x)$
which differs from the preceding example in the bulk, but still behaves asymptotically $U(x)\sim |x|$
when $x\to -\infty$. We show below, analytically and numerically, that the freezing transition again
occurs, supporting our claim of the robustness of this transition.
We chose the following potential
\begin{equation}
U(x)=\begin{cases}
\alpha |x| & \text{for}~ x<-b \\[2mm]
\frac{1}{2} \mu  x^2 & \text{for}~ x>-b
\end{cases}
\label{linear-quadratic}
\end{equation}
where $b>0$ and $\mu=2\alpha/b$ for continuity.

As before,  the Laplace transform of the survival probability can be found by solving
\eref{S-Laplace} piecewise for $x_0 <-b$ and $x_0 >-b$  and then matching the solutions at $x_0=-b$ as well as using the boundary conditions at $x_0=a$ and $x_0\to -\infty$. Since we have already
solved \eref{S-Laplace} for both linear and quadratic potentials in the examples above, we skip the details here. Moreover,
for simplicity, we set $\alpha=1$, $D=1$, and $b=1$. We find
\begin{equation}
\tilde{S}_a (s|0) = \frac{1}{s} \bigl[1-\tilde{F}_a(s|0)\bigr]
\end{equation}
where the Laplace transform of the first-passage time  distribution is given by
\begin{equation}
\tilde{F}_a(s|0)=\frac{\sqrt{\pi } e^{-\frac{a^2}{2}} 2^{-\frac{s}{4}}}{\Gamma \left(\frac{s+2}{4}\right)}\,
\frac{ \chi(s)}{B(s)},
\end{equation}
where
\begin{align}
\chi(s)  &= 2 \sqrt{2} D_{1-\frac{s}{2}}\left(-\sqrt{2}\right)+2 \sqrt{2} D_{1-\frac{s}{2}}\left(\sqrt{2}\right)+\left(\sqrt{4 s+1}+3\right) \left[D_{-\frac{s}{2}}\left(-\sqrt{2}\right)-D_{-\frac{s}{2}}\left(\sqrt{2}\right)\right],\\
B(s) &= \left[2 \sqrt{2} D_{1-\frac{s}{2}}\left(\sqrt{2}\right)-\left(\sqrt{4 s+1}+3\right) D_{-\frac{s}{2}}\left(\sqrt{2}\right)\right] D_{-\frac{s}{2}}\left(\sqrt{2} a\right) \notag\\
&\quad+2 \sqrt{2} D_{1-\frac{s}{2}}\left(-\sqrt{2}\right) D_{-\frac{s}{2}}\left(-\sqrt{2} a\right)+\left(\sqrt{4 s+1}+3\right) D_{-\frac{s}{2}}\left(-\sqrt{2}\right) D_{-\frac{s}{2}}\left(-\sqrt{2} a\right).
\end{align}

By analyzing $\tilde{F}_a(s|0)$ we find that $s=-1/4$ is a branch-point. Moreover, for $a > a_c \approx
1.06$, $\tilde{F}_a(s|0)$ has a pole at $s^*(a) \in (-1/4,0)$. Therefore, as in \sref{S-I}, here also the
survival probability behaves as $S_a(t|0) \sim e^{-\theta(a) \, t}$ with $\theta(a)=-s^*(a)$, for $a >
a_c$ and $S_a(t|0) \sim t^{-3/2}\, e^{-\theta(a)\, t}$ with $\theta(a) = 1/4$, independent of $a$, for $a
<a_c$.

\begin{figure*}
\includegraphics[width=.45\hsize]{lq1.pdf}\qquad
\includegraphics[width=.45\hsize]{lq2.pdf}\\[5mm]
\includegraphics[width=.5\hsize]{lq3.pdf}
\caption{\label{surv-fig-lq} (a) and (b): Survival probability of a Brownian particle in a
potential given by \eref{linear-quadratic}  starting at the position $x_0=0$
and in the presence of an absorbing barrier at $x=a >0$. We set  $D=1$,  $\alpha=1$, and $b=1$.
The points are from numerical simulations whereas the solid lines are fit to the exponential
function $\propto e^{-\theta(a)\, t}$ where the prefactor to the exponential and the
exponent $\theta(a)$ are chosen to match the simulation points.
(c) The estimated values of $\theta(a)$ as a function of $a$ are shown by points,
together with the theoretically computed $\theta(a)$ by the solid (red) line.
The dashed line shows the inverse of the mean first-passage time calculated exactly
from \eref{mean fp}. The vertical dashed line marks $a_c\approx 1.06$.}
\end{figure*}

We verify the theoretical predication using numerical simulation. In the presence of an absorbing barrier
at $x=a$, using the Langevin dynamics, we compute the survival probability of a Brownian particle
starting from the origin, for various values of $a>0$. The results are shown in \fref{surv-fig-lq}, where
we find that for larger values of $a$, the survival probability behaves as $S_a(t|0) \sim e^{-\theta(a)
\, t}$ with a monotonically decreasing $\theta(a)$ as a function of $a$ [see \fref{surv-fig-lq}~(b)]. On
the other hand, for smaller values of $a$, the survival probability behaves as $S_a(t|0) \sim t^{-3/2}\,
e^{-\theta(a)\, t}$ with $\theta(a) = 1/4$, independent of $a$ [see \fref{surv-fig-lq}~(a)].  The
numerically estimated values of $\theta(a)$ together with the theoretically calculated values are shown
in \fref{surv-fig-lq}~(c), which again shows the freezing transition similar to the one reported in the
main text of the Letter.

\section{Survival probability for a potential given by \eref{sqrt-linear}}

\begin{figure*}
\includegraphics[width=.45\hsize]{Sqrt1.pdf}\qquad
\includegraphics[width=.45\hsize]{Sqrt2.pdf}\\[5mm]
\includegraphics[width=.5\hsize]{Sqrt3.pdf}
\caption{\label{surv-fig-sqrt} Survival probability of a Brownian particle in a
potential given by \eref{sqrt-linear}  starting at the position $x_0=0$
and in the presence of an absorbing barrier at $x=a >0$. We set  $D=1$,  $\alpha=1$,  $b=1$, and $c=1/2$.  The points in (a)-(c) are the same numerical simulation data for the survival probability $S_a(t|0)$ presented in different scales whereas the dashed lines represent different functions.  (a)  The dashed line  indicates that $S_a(t|0)$ has a slower than an exponential $\propto e^{-\theta t}$ decay, where $\theta$ and the proportionality constants are fitting parameters. (b) The dashed line indicates that $S_a(t|0)$ has a faster than a power-law $\propto t^{-3}$ decay. (c) Semi-log plot $S_a(t|0)$ as a function of $\sqrt{t}$. The dashed line demonstrates a stretched exponential decay $S_a(t|0) \sim e^{-r \sqrt{t}}$, where $r$ is a fitting parameter.
 }
\end{figure*}

In the main text, using a mapping to a quantum problem, we argued that if the confining potential $U(x)$ increases slower than $|x|$ as
$x\to -\infty$, then $\theta(a)=0$ for all $a$, indicating a slower than exponential decay with time, of the
first-passage/survival  probability. We have analytically  demonstrated that for $U(x) \sim \ln (-x)$ as $x\to -\infty$, both the first-passage and the survival probability exhibit power law decay with time  [see \sref{logx-potential-left}]. Here we consider another case
\begin{equation}
U(x)=\begin{cases}
2 c \sqrt{-x} & \text{for}~ x<-b \\[2mm]
\alpha |x| & \text{for}~ x>-b
\end{cases}
\label{sqrt-linear}
\end{equation}
 where $b >0$,  and for the sake of continuity of the potential at $x=-b$, we set $c=\alpha \sqrt{b}/2$. Using numerical simulation we show [see \fref{surv-fig-sqrt}] that the survival probability again has a slower than an exponential decay, namely, $S_a(t|0) \sim e^{- r \sqrt{t}}$.

redpajama_set_name: RedPajamaArXiv


Thanks to the Internet, the information marketplace has never been so bustling! So why not take part? With easy-to-use and affordable Web publishing tools, organizations can build consumer trust while strategically positioning their brands. And it starts by crafting compelling content.
In this introductory workshop, we will explore the basics of using a blog as an effective content marketing tool. The primary focus of the workshop is on creating blog content, so some working knowledge of how a blog functions might be helpful.
1) Blogging frequency - How often should you post?
2) Post topics - How do you determine what to write about?
3) Post content - Once you determine what to write, how do you make it compelling?
4) Analysis - How do you use your blog's metrics to shape future content?
Keith is a Vice President at a national marketing/PR agency. He formerly operated his own marketing/PR consultancy, providing services to such clients as Travelzoo, AOL, LexisNexis, FTI Consulting, Guidance Software, DePaul University and others. Prior to that, he was an editor at a national legal magazine and a contributor to such publications as The Huffington Post, The Onion A.V. Club, ChicagoNow and the Chicago Theater Beat.
He also has been active in Chicago's thriving arts community as a performer and writer. He is a creator of the popular reading series Essay Fiesta and Guts & Glory, for which he conducts all marketing and PR activities.
This class was so helpful. Keith is engaging, knowledgeable and a spectacular teacher. I would gladly take a class with him again.
This class was top notch! The teacher is obviously an expert. He was clear, organized, and the class exceeded my expectations. Looking forward to taking more of his classes.
Very knowledgeable and willing to help. Open to customizing lecture for students and has plenty of good advice.
The class was really informative and inspiring. I would take another class with this teacher without hesitation.


A lollipop man has been ordered to remove tinsel from his STOP sign in case it distracts drivers and puts children in danger.
Pupils at Berrywood Primary School in Hedge End, near Southampton, sobbed when the decoration was taken off Kevin Simpson's placard after council officials banned it.
An anonymous passer-by had complained the decoration might obscure the sign.
Stunned parents are now mounting a campaign to allow Mr Simpson to put his tinsel back. They plan to wave banners outside the school and are starting a petition.
The popular father-of-two, who has been doing the job for three years, hopes their campaign will be successful and he will be able to bring festive joy back to his crossing patrol.
Last year Mr Simpson put tinsel round the neck of the lollipop and there were no complaints. This year he decided to extend the decoration further down the pole and round the edge of the circular stop sign.
He said: 'The sign could still be seen very clearly and I also wear distinctive uniform so there is no way I would put the children in danger.
Mother-of-two Maria Ritzema blasted council officers, saying: 'The tinsel was not excessive and I was incensed when I heard it had to be removed.
'We, the parents of Berrywood schoolchildren, believe the world has gone made and Kevin's bosses must be Scrooges.
Sarah Miller, 36, who has two children at the school, said: 'I think it is a shame because the children loved to see the tinsel round the lollipop.
Berrywood School said it was not aware that tinsel had been banned.
Mr Simpson's bosses at Hampshire County Council insisted there were strict guidelines governing dress codes for lollipop men and women.
However council leader, Ken Thornber, said he would look into the case.
He said: 'It's nice to see this lollipop man is trying to spread a little bit of Christmas cheer.
'They do such a vital job and it would be a shame if they could not have some festive fun.
'While I do not know the full details of this matter and I haven't seen the stick, I do know there are guidelines to ensure the safety of children using the crossings as well as the safety of motorists and our school crossing patrol staff.
'Any rules and dress codes are about ensuring the safety of children and they are there for legal and insurance reasons.


text:
December 6, 1958 Issue
By Janice Devine and St. Clair McKelway
The New Yorker, December 6, 1958 P. 45
Anne Ford, who works in Boston for Houghton Mifflin, the publishers, recently came to New York on business. She wanted to get in touch with Mildred Smith, who is editor of Publishers' Weekly. After checking in at the Algonquin Hotel, on the north side of W. 44 St., she put in a call for Miss Smith, but had no luck. She slammed down the phone & glared out of the window. Inside a window across the court she saw a pile of magazines - they were "Publishers' Weekly". Then she saw Miss Smith herself, & remembered that the offices of the magazine were in a building on the south side of W. 45 St. She hoisted her window & began to gesticulate & holler. Miss Smith saw her, recognized her, & came out onto the fire escape. The business matter was settled in a jiffy.
Published in the print edition of the December 6, 1958, issue.
More:CommunicationHotel AlgonquinHotelsTelephones

redpajama_set_name: RedPajamaCommonCrawl


text:
Floating, Drifting, Fleeting
Submitted into Contest #110 in response to: Start your story with the line "We have plenty of time," and/or end it with the line "We were never going to make it."... view prompt
K. Antonio
American Sad Kids
We have plenty of time, Mer-Mer. The fish aren't going anywhere. We'll arrive soon enough, I say.
But what if we get there and it's too late, grandpa? What if Pimento and Sweet Pea are off hiding or end up going to sleep?
I look up to meet Mer-Mer's concerned eyes on the rearview mirror, her bottom lip pouting exaggeratingly and her arms crossed over her lap.
Turtles aren't nocturnal, I say with a chuckle. I'm sure they'll be awake.
I open up the glove compartment, let my hand rummage over loose car documents, scrap papers, and empty boxes of Tic Tacs to find my granddaughter's favorite snack.
Here you go, Mer-Mer, I say while reaching out towards her with a box of crackers, Barnum's Animals, her favorites. The logo contains the famous four: a lion, a polar bear, a gorilla, and an elephant, all with their cubs.
Mer-Mer always asks why the box has those chosen animals. Why not a penguin or a walrus? A sea lion could be cute too.
You're right, I always say. Who knows? Perhaps one day, they'll innovate the logo.
We're about thirty minutes away from the Oregon Coast Aquarium. Soon the mountains will be trailing behind us, ornating our rear with peaks of jagged teeth while we glide with blue blankets, seafoam, and lush greenery to our sides.
I watch Mer-Mer, how she squints to the animals trapped in her podgy fingers, peering to the curves and carved details traced into their golden-brown bodies like a spyglass.
Which animal is your favorite, grandpa? She tilts her head while ensnaring a seal between her teeth.
You know, the one with the... um
The what?
That one with the long neck, I say. You know, that one that has yellow fur and stands tall in the safari.
The giraffe?
Yes, the giraffe, I say. The giraffe's my favorite.
Mer-Mer giggles and shoves her hand into the box. Let's see if I can find one, she says, closing her eyes while sifting her fingers through the sea of crisp crackers like it's a magician's hat.
Got one, she says with an excited voice.
I turn around to the backseat. Mer-Mer's holding the cracker as if it's a coin, valuable and note-worthy like a doubloon.
That's not a... um. It's a camel. But you were pretty close, I say, shining a quick wink before turning my eyes to the road. Now eat up. You know they don't allow any snacks inside the aquarium.
Mer-Mer's seven. She still hasn't noticed my forgetfulness. She simply shrugs off the moments I don't know a word or forget the date from her peripheral like it's a tiny insect, fanned away from her focus by her own bliss.
I keep both hands over the steering wheel, tightly wrapping my fingers around the rim. My body sometimes betrays me, usually when I'm not attentive and allow my mind to wander free.
The doctors call it Lewy Body Dementia, such a funny name for a wicked disease. I once knew a guy named Lewy; he was rather serious, brash even, and had a wide gap between his buckteeth. He wasn't at all funny, maybe, that's the thing about Lewy's, their names make them seem comical and tame, unable to lash out and show their fangs. Lewy was a hunter who skinned deer and sold hide on the market real cheap. He was stern and liked killing; he wasn't known to show mercy. I think this Lewy is the same, a predator that's devouring me slowly.
I'm the animal, being chewed by its teeth.
Mer-Mer loves the aquarium. I've been taking her here ever since she was three. She tells her classmates that I've got an important job here, that sometimes she has to be my sidekick at work, so she enters the place for free.
The other kids and their parents wait in line while we cut through the ribbon of people and walk in using the employee entrance.
Hello, Jan, I say, greeting the guard at her post.
Hey, Derek, she says. I see you brought Meredith to tag along today.
She's been begging all week to come. You know how she is, I say.
Can't get enough of her veggies.
Yuck, I hate vegetables, my granddaughter says. What are you talking about, Jan? I'm here to see Pimento and Sweet Pea!
The sea turtles, right? Jan laughs.
Yeah, the sea turtles, Mer-Mer says while raising herself onto her toes.
Well, let's not keep them waiting any longer. See ya later, Jan.
Bye, Jan.
Bye, Meredith, the guard says.
I love how Mer-Mer sees my job as special or unique; how to her, it's a privilege to come into work with me. I tell her I'm a custodian worker, a fancy word for a janitor, that I have to keep the aquarium clean for all the sea creatures so that they feel more at home.
My boss isn't very fond of children. Dennis thinks it's a risk to have them roaming around the tanks and glass corridors. Little kids and teens are just begging to cause trouble, he generally says.
But Dennis likes Mer-Mer. He's known her from birth and went to school with her mother, Margie. Dennis was there when Margie got pregnant; he helped her by offering a job at the aquarium and getting her into the company's health plan. Dennis was also a big help when my daughter left, swept away by life's currents, by the anonymity and mystical calls of the big city. I asked if I could take a leave from work to help take care of my granddaughter, and he told me to take all the time I needed.
While I mop the floors and wipe the glass that children insist on putting their mouths on to imitate the fish, Mer-Mer walks around the maze of tanks and tunnels, pretending she's a water nymph or sea witch.
She scurries around the blue tunnel and makes wooshing sounds as if she's blasting off in an underwater spaceship.
She's read at least all the plaques in the aquarium tens of times and knows a good chunk of the facts by heart.
What's GPO stand for?
For Giant Pacific Octopus, she says.
And how fast can they...
They can change color in one-tenth of a second. Come on, grandpa! You can do better than that, she says, taunting me.
Alright, alright, I say. You want something a bit tougher, eh? I'll give it to you, Mer-Mer. What are the names of the seven species of sea turtles?
She blows me a raspberry and tells me I need to step it up. The Green Turtle, the Loggerhead, the Hawksbill, Leatherback, Flatback, the Kemp's Riley sea turtle, and my favorite, she says, the Olive Ridley sea turtle!
Haha, great job, kiddo, I say. Follow-up question, when were Pimento and Sweet Pea rescued and brought back to the Oregon Coast Aquarium?
Um...um...I don't know.
In 2021 and 2023, I say. Pimento was brought here a month after you were born. And Sweet Pea was found by a rescue team in Mexico just shy of your second birthday.
Okay, you got me, Mer-Mer says.
Still, you did well, I say while placing the stringy mop head into a bucket of water and suds.
My turn, my turn, it's my turn to ask now.
Hit me, I say.
How many sea creatures does the Oregon Coast Aquarium have?
There are currently 1,600 sea creatures in the exhibit. More than half less than a couple of years ago when we had 3,500.
Mer-Mer looks at me, a mixture of disbelief and sadness.
I know, I say. We've lost quite a lot in these past years.
I think about the numbers, how I remember them so well. How I was here, noticing the fishes dwindle and die out, helping the employees transport species of stingrays and sea leopards to institutions of a higher caliber. How Mer-Mer was only four and cried when one of the whales was put on a truck headed to Costa Rica.
What's the oldest multi-organ animal on Earth?
I know this. Hold on just a sec, I say. It's the um...um...
Mer-Mer sounds off like a buzzer, unwilling to wait for me to recollect the information swimming around in my brain.
It's the jellyfish.
The jelly what?
Haha, don't play dumb, she says.
But I'm not playing. I honestly have no idea what my granddaughter's talking about.
Hey, why don't you go to the dolphin tank and ask Gus if he'll let you help out feeding Gilly and Tails?
Okay, she says, sounding chipper. Can I then ask Gus if I can help out with Pimento and Sweet Pea?
Sure thing, Sweet Pea, I say.
That's not funny, she says with a straight face. Sweet Pea is a turtle's name, you know that.
Mer-Mer turns and skips down the hall, the sound of her flats clapping on the cement ground. She turns back.
Guess what I am? Think, she says while waddling away. She goes on, looking brave and giddy, with her head up high and chest puffed out.
An Emperor Penguin, I say.
She turns the corner and I think to myself. How long?
How long will I get to see my granddaughter in this way?
Her grandma's worried about me working still, but I can't stop; I can't stay at home and wait for the bits and pieces of me to erode. I need to create new memories, new happy snippets to replace the ones I'm bound to lose someday.
Part of me also could never stop working, at least not yet, not while Mer-Mer loves to come here to play. My granddaughter doesn't find my job boring or trivial; it's a task laced with bubbles and underworldly creatures on display, with tentacles and fins, and stories, and colorful words on signs with graphs and blue mirrors and warped reflections and all the fun facts you can drink in.
The doctors say that I could have a good five to eight years in me.
Only five to eight?
Unfortunately, your condition is progressive; there's no cure.
What does that mean? Will I eventually forget everything?
Maybe not everything. Your mind could still retain a few things.
Anything it finds really important.
I wonder if I will remember my granddaughter or if the disease will kill her off, poisoning that branch of my mind and leaving her memory to fall and be stepped on like spoiled fruit.
Don't worry, I say to myself. I grab the mop and bucket and walk to the supply closet, crossing an arch of glass and water, feeling like a creature in a foreign land.
She'll be fine. Eight years, I say. Eight years, and by then, she'll be fifteen, she'll be tall, and strong and well-taught. Mer-Mer won't be small, and she'll know more about the aquarium than anyone I know. She'll still come by and see Pimento and Sweet Pea and help out Gus and talk to Jan by the entrance. Maybe Dennis will help her out with a job, just like he did her mother. Perhaps, she'll stay, or maybe she'll leave to travel the world.
I throw out the dirty water in the small porcelain sink of the cubicle, leave the mop by the dark corner close to the entrance. I then exit the closet, turn off the lights, and lock the place up.
My hand doesn't release the knob. Instead, it tremors as if the handle's delivering an electric shock. My arm's vibrating flimsily like rubber. I tug, I tug, my fingers break loose.
Hey, Derek, are you okay?
I turn around and catch sight of the aquarium's caretaker with my granddaughter by his side.
What's going on? She looks to Gus and then to me. Gus thought we would find you here, she says. Are you okay?
I'm fine, Mer-Mer. Thanks for taking care of her, Gus. I owe you one.
No problem, he says. Meredith is a doll; she reminds me so much of my little girls.
Gus reaches out his hand but stops midway. He murmurs something while seeming unsure of his gesture as if it's offensive. His arm shifts and grabs my shoulder. See you tomorrow, he says.
He turns, pats Mer-Mer on her head, and leaves. My granddaughter is unreactive, standing like a scarecrow, looking at my limb.
I squat down. I think it's about time we go home, I say. Your grandma's probably missing you already.
What's wrong with your arm, grandpa?
I place my hand on her chin, look at the warped silhouette of my body cast in her eyes. Grandpa's just feeling a bit tired. It was a long day, I say.
Mer-Mer nods. Does that mean we can't see Pimento and Sweet Pea one last time before we leave?
Of course not, I say. Let's go see those two together one last time.
We walk down the hall, Mer-Mer's shoulders bump against my waist.
Hey, grandpa?
Did you ever bring mom down to the aquarium? You know, when she was a little girl.
I don't think so, I say.
But being completely honest, I actually don't remember.
03:07 Sep 09, 2021
I've been really fighting with this piece, so any comments or feedback are appreciated. I was partly inspired by Sally Rooney whose work is known for not having quotation marks when referencing dialogue, instead she uses "said/say(s)". So I decided to play a bit with structure and style. I was also in part inspired by the author of "Migrations" and "Once There Were Wolves", Charlotte McConaghy, whose last two books were a mixture of literary and environmental fiction. IF YOU'RE LOOKING FOR A PLOT-DRIVEN STORY. . . THIS IS NOT IT! Than...
Dustin Gillham
K. Antonio, The story is not plot driven, and this makes it more provocative. An educated reader would feel as if he/she were on the outside looking at the subtleties of precious life. The structure added to the ambiance and I am a big fan of authors willing to branch out and play with style. Again, I am grateful to have read your submission. If I might offer one constructive critique, part of me felt like you weren't done editing. I'm not advocating minimalism but you took on a lot with this project and another go with the red pen migh...
I.B. Dunn
K, I am never disappointed when I read your stories. I'm not a technically proficient writer so I only have one test, was the story entertaining, and this was. Very much so. There is so much hidden in plain sight. Fear, loss, love, hope, joy. I know you had to do research for some of the facts but it didn't feel like a class, it felt like learning from the experience of age. Really great work.
Shea West
I stan Sally Rooney😍😍 is that how the kids are saying it these days??? I like that you took that risk and played with format. Seeing as I reside in Oregon and we've got like 1 aquarium I was very easily transported there through your story. Also, strangely enough I was watching a medical drama today where a patient had Lewy Body Disorder!!! This story isn't plot driven, but that didn't bother me much. I enjoy reading about things that are real, a section of the bigger picture if you will. I found it to be endearing!
The reason I ended up choosing Oregon was because they are one of the few places in the USA that rehabilitate Olive Ridley sea turtles. Basically I started with the turtles, and went on from there. 🤣🤣
Well, as a representative of the great PNW I dig it ;)
Priceless Passion
An acquaintance with the beautiful --7698 miles away (from INDIA)-- OREGON COAST AQUARIUM; with the sea world in your beautiful creation-- "FLOATING, DRIFTING, FLEETING" was interesting. If it kept me engaged with Meredith's keenness for sea creatures especially her favourites--- PIMENTO and SWEET PEA, it moved me also by grandpa's mental illness, physical constraints, his sweet calculation of age; of future, of his lovely grand daughter. Also, Thanks for the declaration about the inspired, remodeled style-- WITHOUT QUOTES, in your work...
Fawn Marshall
The lack of quotation marks threw me off for the first couple of paragraphs. Once I realized it was a stylistically choice, though, the story was an easy read. You did a great job of narrative versus speaking voice, and I had no trouble determining which lines were dialogue. "Once There Were Wolves" is next on my reading list, so I was happy to see you mention it in your notes.
Keya Jadav
Wow! You took me back to staring at a cinema play screen, watching an absolute intrigue. It hooked me all the way to the end, reforming like reality in the back of my skull. The choice for the title is appreciable. My heart was torn at the corners seeing what her grandpa was going through, and like any other benignant, hiding it from his granddaughter. Overall, Lovely Job!
Katie Kanning
Hey, I love your story! I'm wondering if I could read it on my podcast, "Unpublished, not Unknown"? It's all about giving voice to indie authors' short stories and spreading their reach a bit further. I'll credit you and link your profile in the show notes. People can listen on Spotify, Apple Podcasts, Youtube, and 5 other locations. It's in its growing stage, so I'd only ask you to share your episode with friends if you like it :) You can check out the format here: https://bio.link/katiek
Hey Katie. Sure! You've already read a few of my author friends' stories from here on Reedsy. I checked on your podcast on Spotify and saw that you credited the authors and all, so it's not a problem for me! Thanks for the kind words and taking an interest on the piece!
Awesome! Thanks for taking a listen. Yes, all the credit to the wonderful authors :) I have a couple questions for you then: 1. What name do you want to be credited to the story? K. Antonio or something else? 2. Is the granddaughters' name pronounced Mer-Mer like it's read or is it Mare-Mare like Meredith is pronounced? I like to at least nail the names down as you imagined them being read. 3. Are there any other links you want me to list in the show notes for you? Like a website or social media.
Yes, Mer-Mer is pronounced like Mare-Mare xD K.Antonio is fine, it's the pen name that I sign all my works with (my first name is unpronounceable to most).😂😂 I'm not a fan of social media so a link to my Reedsy profile is fine!
Okay great! I will let you know here when it's posted. Have a great week
14:09 Oct 04, 2021
Hey, your episode is officially live! Enjoy :) https://bio.link/katiek
22:35 Dec 28, 2021
Hi there! I'm reaching out to all of the authors who have been featured on the UNU podcast. I have a few new opportunities for you! In the quick and easy form linked below, I am offering: 1. A chance to schedule a casual interview with me for the podcast. This will be a basic get to know you, a way for you to advertise your other writing (books if you have them), and a time to laugh and have fun. They would be less than an hour, most likely under 30-min. Whatever works for your schedule. 2. A personal bio page for you on the website I...
Thanks Katie! I'll check it out! Also, wow, you're doing a lot to promote other people's creativity and work, honestly really cool and generous on your part.
It's been so fun to do and so encouraging to me as an aspiring writer :)
a super cool story!! I was sad that he was losing memory and so quickly. great writing with emotional depth!
Cathryn V
Hi there, your story is warm albeit sad, and what I call quiet. The lack of tags didn't bother me at all. The story reads like an essay. The only bump was the line about seeing his reflection in Mer-Mer's eyes. I think I know what you meant but it sounds like he can see himself or maybe I'm just too literal. Anyway, good job on grandpa's interiority.
L Mack
I think the dialogue worked well without quote marks, it added to the vagueness that the grandpa was starting to feel, in my opinion anyway. Great job :)
Jon Casper
We all have an idea of what dementia is, but it's hard to grasp what it's like to suffer from it. I imagine the early stages are a lot like this. Words right on the tip of your tongue. Being reminded of things you know you ought to have remembered. The last glimpse of losing memories of a loved one has to be the most awful. You did a great job portraying the uncertainty and confusion that I'm sure people face in this situation. The relationship between the grandpa and granddaughter was sweet. Great job as always.


Calling all eBook Publishers – Enter Today!
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The second annual eLit Awards are a global awards program committed to illuminating and honoring the very best of English language digital publishing entertainment. The 2011-12 eLit Awards were created as an industry wide, unaffiliated awards program open to all members of the electronic publishing industry. The contest is presented by Jenkins Group Inc., a Michigan-based book publishing and marketing services company that has operated the popular Independent Publisher Book Awards contest since 1996. The eLit Awards celebrate the ever growing market of electronic publishing in the wide variety of reader formats. Hail the revolutionary world of e-books and join the awards program that's highlighting the best in electronic reading entertainment!
Publishers and authors worldwide creating electronic books written in English and created for the global marketplace are eligible for entry in 64 different e-Lit categories. Titles published with a 2011 or 2011 copyright or that were released between January 1, 2011 and January 17, 2012.
Entry fees are designed to save you more money the earlier you enter.
The contest is presented by Jenkins Group, a Michigan-based book publishing and marketing services company that has operated the popular Independent Publisher Book Awards contest since 1996.


Lady Minor is a clothing line dedicated to the woman that is a hopeless romantic. She gets caught up in the occasional daydream, believes men should still send flowers and confess their love in writing. She loves to soak in a tub surrounded by bubbles and candles while reading and sipping champagne. She knows clothes don't make the woman but enhances how she looks and feels.
Lady Minor is for the woman that loves fashion and is up to date with all the latest beauty trends. She's all about catching up with her girls over brunch, is all about her career and her man because why can't we have it all?! She's straightforward, tells it like it is. If you have a problem, she'll drop everything and come over with a bottle of wine or two. She's sweet, seductive and charming with a goofy side.
Lady Minor is the love child of Samantha Jade Minor. That's me. You can call me Sam. I'm the ultimate romantic and champion of lovers. I think women should believe in "The One" and not settle for less. I feel a woman should embrace her curves while showing off her mind. Be strong and confident while maintaining a level of softness and femininity.
I'm a makeup artist, clothing designer and writer. I started designing clothes when I got tired of not finding anything to wear at the mall and wanted a change in fashion. I currently live in Tampa, Fl where I'm a mom above everything. I may be a bit biased, but I pride myself on being a loyal friend and uplifting women everywhere. I love wine (preferably red), lilies, lace (there's just something about a sheer piece of clothing), 90's hip-hop and R&B and to be swept off my feet. I am Lady Minor.


text:
Jin Shin Jyutsu | Auto aplicação em sua casa | Curso Online | Self Help at home | Online Class
Carlos Gutterres & Wayne Hackett
Ready for more
vitality, clarity
and joy?
Learn from two of the most respected and seasoned instructors of Jin Shin Jyutsu® Physio-Philosophy in the world.
Learn the simple practices and philosophical concepts that will help you restore physical, mental and emotional balance with your own hands, literally. This practice helps develop awareness, self confidence and enthusiasm, so we can move with the flow of life and constant transformation processes.
According to Mary Burmeister, the study and practice of this Art enables us to know the essence of human nature, allowing us to perceive, understand and transform our reality without effort, just naturally. This is how we can recognize the wisdom of our bodies and of life, understand their messages and use them to create harmony and health.
The Art of Getting to KNOW (Help) MYSELF
The great interest of Brazilian instuctor Carlos Guterres in the mysteries of nature and human beings has encouraged him, from an early age, to seek answers in philosophy, psychology, numerology, astrology, cabala and other expressions of acient wisdom. He has been dedicated to the study, practise, teaching, research and dissemination of Jin Shin Jyutsu since 2002. He will be ending his activities as an instructor of the Art in 2020.
Wayne Hackett was born and raised in California, having dedicated his early education to dental studies and later specializing in sciences and secondary education. An ever curious mind, eager to find his true path, led him to meet Mary Burmeister in 1977, his teacher and constant source of inspiration. He has since dedicated his life to studying, practicing and teaching Jin Shin Jyutsu all over the world. He has a particular fascination with the esoteric significance of symbols (numbers) and cosmic geometry (planetary relationships).
Who are the instructors?
Carlos Gutterres
Wayne Hackett
What is the purpose of the class?
The focus is on the presentation of the three self-application books written by Mary Burmeister. Teachers teach and encourage each one to take care of themselves, but they also reveal the philosophical aspect that permeates each page of these three books. Suitable for everyone who seeks self-knowledge, the course caters to those who want the first contact with Jin Shin Jyutsu as well as those who already study the Art and would like to access the deeper dimensions of its philosophical aspects. The wealth of this class is also of great value to those dedicated to teaching Jin Shin Jyutsu at the self-help level.
Daily classes from May 24th to 31st.
The course runs from Sunday to Sunday. During the 8 days, live online classes
will happen over 3h00,
with 2 intervals.
São Paulo, BR (GMT -3) 2pm to 5:30pm Lisbon, PT (GMT +1) 6pm to 9:30pm
LA, USA (GMT -7) 10am to 1:30pm
NYC,USA (GMT -1) 1pm to 4:30pm
Hong Kong (GMT +8) 01am to 4:30am (Next Day)
Classes will be recorded and made available for later viewing, in the student area.
The information needed to access the course
and its dynamics will be sent after registration.
Class replays will be up to December 1st
Class Tuition
Registrations on or before May 15
New Students 300 €/ 300 USD
Reviewing students* 180 €/ 180 USD
JSJ Organizers** 150 €/ 150 USD
Registrations as of May 17
#Registration for this class is closed, through this website#
* Reviewing students are those who have attended at least one 5-day self-help class with these 2 instructors.
** JSJ Organizers are JSJ Inc. Organizers.
Questions: +55 41 995 041 042
Class Translator
Maria Rita Guedes
Oficial Class Organizers in Curitiba, Brazil
Maria do Rosário Fonseca da Silva Quitério
Silvia Cristina Patzsch
© 2020 - Foco na Harmonia | Focus on Harmony


text:
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Meet Clare
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WIN A CLARE HAXBY PRINT IN CELEBRATION OF THE SINGAPORE F1 GRAND PRIX
Clare Haxby's painting Singapore Skyline is a night scape of the Marina Bay part of the F1 race circuit. The painting features such landmark buildings as The Fullerton Hotel and The Fullerton Bay Hotel with The Lantern Bar -- the rooftop bar of the hotel that has amazing views across to Marina Bay.
Clare says, "I painted this nightscape to represent Singapore's dramatic skyline of the Central Business District and the waterfront of the Esplanade and Marina Bay. I worked in mixed media using printmaking, acrylic paint and silver leaf and glazes. I had recently completed a nightscape painting for an American client, which inspired me to create this for my own Singapore Landmarks Collection. This section of the CBD is part of the circuits for the Singapore Grand Prix the spectacular Formula one night race."
With a backdrop of its spectacular cityscape, Singapore hosts the only Formula 1 night race on the Formula One calendar this September 18-20, 2015. The timing of the night event means that it can be broadcast live at a convenient time for European TV audiences. The track is also illuminated by a series of projectors, which adapt their output to match the shape of the course.
The atmosphere around the flood-lit Marina Bay Street Circuit is infectious. Utilising public roads around the Marina Bay area, the track includes fast straights as well as plenty of overtaking opportunities. This year the city will come alive with excitement once again!
For the past seven years, Singapore has been setting the stage for this highlight of the Formula One calendar, earning it the title of "the jewel in the Formula One crown."
This September, Multi award-winning, multi-platinum winning bands with famed singers Pharrell Williams and Bon Jovi are set to ignite the Padang stage in their first ever concert in Singapore each night.
City spectators sit in the Stamford Grandstand, The Padang, and Esplanade Waterfront all around the Central Business District as top race car drivers practice and run qualifying sessions on Friday and Saturday, and compete in the Grand Prix race on Sunday night.
This year, to celebrate F1, the #ClareHaxbyStudio are giving you the chance to WIN a Signed Limited Edition Singapore Esplanade print.
To enter simply complete the following -->
1. Follow @ClareHaxby on Instagram
2. Like the "WIN A ESPLANADE PRINT" post on Clare's Instagram ( See post graphic to the right here -->)
3. Tag a friend (or friends!) who loves art in the comment box
Giveaway will end at 5pm on 21/09/2015 * Winner will be announced by 5pm 22/09/2015 Sg time * The prize is not transferable & cannot be exchanged for cash * Winner will be chosen at random * International entries welcome * Winner may be responsible for any shipping charges / customs fees over and above the stated prize value as applicable
DISCLAIMER: As per Instagram guidelines, this giveaway is in no way sponsored, endorsed or administered with Instagram Incorporation. By entering, you confirm you are 18 years of age or older, agree to Instagrams terms of use and release Instagram and hosts of responsibility.
ABOUT SINGAPORE F1 GRAND PRIX
First organised in 1961, the race was initially known as the Orient Year Grand Prix with the race renamed the following year to the Malaysian Grand Prix.
After Singapore attained its independence in 1965, the race was renamed the Singapore Grand Prix. The event was discontinued after 1973 for a variety of reasons, such as the very high danger and unsuitability of the street track for racing, the inconvenience of having to close roads for the event, as well as fatal accidents during the 1972 and 1973 races.
In November 2007 it was announced that the telecommunications company SingTel would sponsor and televise race giving the event the official name Formula 1 SingTel Singapore Grand Prix.
The race was co-funded by the Government of Singapore, footing 60% of the total bill, or S$90 million, out of a total tab of S$150 million.
The 2008 race hosted the famous Amber Lounge after party and in 2010 Singapore became the second location to host the Amber Lounge Fashion show.
For the 2009 race, the circuit was re-profiled slightly, including modifications to turns 1, 2 and 3 to aid overtaking, and also at turn 10 where many accidents occurred in 2008.
In 2012, an agreement for a five-year deal was signed by Singapore GP Pte Ltd, the Singapore Tourism Board and Bernie Ecclestone that the Grand Prix will remain on the Formula One calendar through the year of 2017.
For the 2013 F1 race, the 10th turn of the track, the "Singapore Sling" chicane, was reconfigured so the cars had to navigate a flowing left-hander before accelerating towards the Anderson Bridge.
On 15 April 2014, it was announced that Singapore Airlines would sponsor the Singapore Grand Prix starting from that year.
Entry to the concert is included with any three-day 2015 FORMULA 1 SINGAPORE AIRLINES SINGAPORE GRAND PRIX ticket.
ABOUT GUEST EDITOR MARIE KAZALIA
Find out about Marie at artistmarketingresources.com -and- MarieKazalia.me
Marie Kazalia is an Arts journalist, Fiction author, contributing writer to Xpost Multimedia Magazine, artist agent, Agent and Contributing Writer for ArtScene Guru in Hong Kong. She is represented by Schroder-Romero Gallery in New York City.
"Writing is a creative process for me--even for my blog posts and articles--and especially for my creative short fiction. I am about to e-publish two novellas under one cover Gold Sandals and in the past I've had quite a bit of my poetry published. A poem I wrote in India was requested by the editors for Washing Lines anthology--which sold out on Amazon. The first edition of Washing Lines appeared in 2011 with a reprint edition in 2015.
My interview, with by Editor John Wisniewski of Urban Grafitti, about my writing will appear in the California Journal of Women Writers in a few weeks."
Marie's artwork was recently featured in Professional Artist Magazine
Signed Limited Edition Art Prints
Art Packaging
Starting An Art Collection
Guide To Commissioning Art
Corporate Art Gifts
Art Gift Cards
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text:
Q: How to sort a list in a detail view of a one to many relationship with Core Data and SwiftUI? Meditations has a one to many relationship with stages and I have sorted the Meditations based on an order property, which I change by moving them around when I tap edit and this works. Now I want to do the same with the stages in the detail view of the meditations.
I tried to do a new FetchRequest for the DetailView so that I could use the same method as with meditations, but I couldn't make it specific for the particular meditation (and this might not be the best approach if a new FetchRequest is avoidable I think).
I also tried to sort the list items with .sorted, but then it didn't update right away when the order was changed in edit mode and you had to go back and forth to see the changes.
Is there a way to make one of these approaches work? I am also open to other solutions of course.
This is the view for the meditations:
import SwiftUI
import CoreData

struct ContentView: View {
    @Environment(\.managedObjectContext) private var viewContext

    @FetchRequest(sortDescriptors: [NSSortDescriptor(keyPath: \Meditation.order, ascending: true)])
    private var meditations:FetchedResults<Meditation>

    @State private var add = false

    var body: some View {

        NavigationView {
            List {
                         ForEach(meditations){meditation in
                             NavigationLink(destination: {
                                 DetailView(meditation: meditation)
                             }, label: {
                                 Text(meditation.title ?? "")
                                 Text("\(meditation.order)")
                             })
                         }
                         .onDelete(perform: deleteMeditation)
                         .onMove(perform: moveItem)
                     }
                     .navigationTitle("Meditations")
                     .sheet(isPresented: $add){
                         AddView()
                     }
                     .toolbar {
                         ToolbarItem(placement:.navigationBarTrailing){
                             EditButton()
                         }

                                     ToolbarItem(placement:.navigationBarTrailing){
                                         Button(action: {
                                             add.toggle()
                                         }, label: {
                                             Label("Add Meditation",systemImage: "plus")
                                         })

                                     }
                                 }

        }
    }

    private func deleteMeditation(at offset:IndexSet){
        withAnimation {
            for index in offset{
                let meditationToDelete = meditations[index]
                do{
                    viewContext.delete(meditationToDelete)
                    try viewContext.save()
                }catch{
                    print("Error while deleting Meditation \(error.localizedDescription)")
                }
            }
        }
        }

    private func moveItem(at sets:IndexSet,destination:Int) {
        let itemToMove = sets.first!
        print (itemToMove)
        print (destination)
        if itemToMove < destination {
            var startIndex = itemToMove + 1
            let endIndex = destination - 1
            print (startIndex)
            print (endIndex)
            var startOrder = meditations[itemToMove].order
            while startIndex <= endIndex {
                meditations[startIndex].order = startOrder
                startOrder = startOrder + 1
                startIndex = startIndex + 1
            }
            meditations[itemToMove].order = startOrder
        } else if destination < itemToMove {
            var startIndex = destination
            let endIndex = itemToMove - 1
            var startOrder = meditations[destination].order + 1
            let newOrder = meditations[destination].order
            while startIndex <= endIndex {
                meditations[startIndex].order = startOrder
                startOrder = startOrder + 1
                startIndex = startIndex + 1
            }
            meditations[itemToMove].order = newOrder
        }
        do {
            try viewContext.save()
        }
        catch {
            print(error.localizedDescription)
        }
    }
}

struct ContentView_Previews: PreviewProvider {
    static var previews: some View {
        ContentView()
    }
}

This is the detailview where the stages are shown:

import SwiftUI

struct DetailView: View {
    let meditation : Meditation

    @Environment(\.managedObjectContext) var viewContext
    @Environment(\.dismiss) var dismiss

    @State private var addStage = false
    @State private var showingDeleteAlert = false
    @State private var isPresented = false

    var body: some View {
        VStack {
            Text(meditation.wrappedDetails)
            List{
                ForEach(meditation.meditationStage){meditationStage in
                    MeditationStageDetailCell(meditationStage: meditationStage)
                }
                .onMove(perform: moveStage)
                .onDelete(perform: deleteStage)
            }
            Spacer()
            Button ("Begin"){
                isPresented.toggle()
            }
            .fullScreenCover(isPresented: $isPresented) {
                MeditatingView(meditation: meditation)
            }
            Spacer()
        }
        .navigationTitle(meditation.wrappedTitle)
        .navigationBarTitleDisplayMode(.inline)
        .alert("Delete Meditation?", isPresented: $showingDeleteAlert) {
            Button ("Delete", role: .destructive, action: deleteMeditation)
            Button ("Cancel", role: .cancel) {}
        } message: {
            Text("This action can't be undone")
        }
        .toolbar{
            ToolbarItem(placement:.navigationBarTrailing){
                 Button {
                    showingDeleteAlert = true
                } label: {
                    Label ("Delete this Stage", systemImage: "trash")
                }
            }
            ToolbarItem(placement:.navigationBarTrailing){
                EditButton()
            }
            ToolbarItem(placement:.navigationBarTrailing){
                Button(action: {
                    addStage.toggle()
                }, label: {
                    Label("Add Stage",systemImage: "plus")
                })
            }
        }
        .sheet(isPresented: $addStage){
            AddMeditationStageView(meditation: meditation)
        }
    }

    private func deleteMeditation() {
        viewContext.delete(meditation)
        try? viewContext.save()
        dismiss()
    }

    private func deleteStage(at offset:IndexSet){
            for index in offset{
                let stageToDelete = meditation.meditationStage[index]
                do{
                    viewContext.delete(stageToDelete)
                    meditation.stages -= 1
                    try viewContext.save()
                } catch {
                    print("Error while deleting Meditation \(error.localizedDescription)")
                }
            }
        }

    private func moveStage(at sets:IndexSet,destination:Int) {
        let itemToMove = sets.first!
        print (itemToMove)
        print (destination)
        if itemToMove < destination {
            var startIndex = itemToMove + 1
            let endIndex = destination - 1
            print (startIndex)
            print (endIndex)
            var startOrder = meditation.meditationStage[itemToMove].order
            print(startOrder)
            while startIndex <= endIndex {
                meditation.meditationStage[startIndex].order = startOrder
                startOrder = startOrder + 1
                startIndex = startIndex + 1
            }
            meditation.meditationStage[itemToMove].order = startOrder
        } else if destination < itemToMove {
            var startIndex = destination
            let endIndex = itemToMove - 1
            var startOrder = meditation.meditationStage[destination].order + 1
            let newOrder = meditation.meditationStage[destination].order
            while startIndex <= endIndex {
                meditation.meditationStage[startIndex].order = startOrder
                startOrder = startOrder + 1
                startIndex = startIndex + 1
            }
            meditation.meditationStage[itemToMove].order = newOrder
        }
        do {
            try viewContext.save()
        }
        catch {
            print(error.localizedDescription)
        }
    }
}

A:     @FetchRequest
    private var stages: FetchedResults<MeditationStage>

    init(meditation: Meditation) {
        let sortDescriptors = [SortDescriptor(\MeditationStage.timestamp]
        _stages = FetchRequest(sortDescriptors: sortDescriptors, predicate: NSPredicate(format: "meditation = %@", meditation), animation: .default)
    }

redpajama_set_name: RedPajamaStackExchange


UrbanBound, the first web-based relocation platform that provides companies with a comprehensive relocation experience for their newly hired and transferring employees, has secured $5 million in venture funding. Early-stage investor Grotech Ventures led the Series A round with participation from experienced SaaS investor StarVest Partners. UrbanBound plans to use the funding for aggressive hiring and continued platform innovation. Mike Wilson, a Business Law associate in the firm's Chicago office, advised UrbanBound on the financing.


text:
Twenty-Four Italian Songs and Arias & Voices Toggle Menu
Twenty-Four Italian Songs and Arias & Voices
Submit a Version
Alma del core
Amarilli, mia bella
Caro mio ben
Che fiero costume
Come raggio di sol
Danza, danza, fanciulla gentile
Già il sole dal gange
Lasciatemi morire!
Il mio bel foco
Nel cor più non sento
Non posso disperar
O cessate di piagarmi
O del mio dolce ardor
Per la gloria d'adorarvi
Pietà, Signore!
Pur dicesti, o bocca bella
Se Florindo è fedele
Se tu m'ami, se sospiri
Sebben, crudele
Tu lo sai
Vergin, tutto amor
Le violette
Vittoria, mio core!
Caldara
About the artist(s)
Melvin Claridge
Melvin commenced his operatic training in 1992 at the Ottawa School of Music in Canada. He has sung with the National Youth Choir of The Bahamas in Nassau; the Men and Boys Choir at Christ Church Cathedral and Opera Lyra Ottawa in Ottawa, Canada; and Opera Holland Park and Opera International at the Royal Albert Hall in London, among others. In addition to singing, Melvin works as a solicitor and serves as Vice-Chair of the Board of Trustees at Pegasus Opera Company and is on the Development Committee at OperaUpClose.
Highlights from Melvin's conversation with The Enormity of Now are below.
Who are you and what do you do
I'm Melvin Claridge. I'm a mix of things. I think singing is central to who I am, even though I don't do it as my paid job. I now work as a lawyer but that's not who I am. I don't want to define myself with the job that I do.
My job is a government tax lawyer, and I think I'm an artist, though I don't create much art. I still feel torn between singing and doing a full-time job – I have a strong desire to remain involved in the singing world
How has the journey of your art/career engaged your voice – personally, artistically, politically?
When there have been important things to say, sometimes I've muted my own voice. I want to say something, but then the time has already passed. Politically there's a lot more that I can say, but generally don't. Sometimes, in my capacity of Trustee of Pegasus [Opera Company], my voice and thoughts can be raised.
There are emotional hurdles for me whilst singing, which can lead me to run away. I'll get so far, then run away, backwards and forwards. I feel a lot of regret. It was fear that sent me into finding a full-time job.
What is the voice that you found while finding your voice?
I think I have two voices. One is determined and focussed, a voice that involves hard work. The other, the funny, joking voice, is the escape when things get hard. All I want is a voice that can do the best that it can and it always be honest.
Melvin commenced his operatic training in 1992 at the Ottawa School of Music in Canada. He went on to train with the renowned Italian soprano, Maria Pellegrini, in Ottawa. He has sung with the National Youth Choir of The Bahamas in Nassau; the Men and Boys Choir at Christ Church Cathedral and Opera Lyra Ottawa in Ottawa, Canada; and Opera Holland Park and Opera International at the Royal Albert Hall in London, among others. He currently studies with Neil Baker.
Melvin sang in a previous iteration of 24 Italian Songs and Arias at the Latitude Festival in 2016.
In addition to singing, Melvin works as a solicitor and serves as Vice-Chair of the Board of Trustees at Pegasus Opera Company and is on the Development Committee at OperaUpClose.
Download: Medium-High Acc. (mp3)
Download: Medium-Low Acc. (mp3)
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