text:
\section{Introduction}

Let $\Omega \subset \mathbb{R}^2$  be a bounded domain with smooth boundary $ \partial  \Omega$. For each $q\in L^\infty(\Omega)$, consider the so called Dirichlet to Neumann map (DtN) given by
\begin{equation} \label{eq:Lambda}
\begin{array}{rcl}
\Lambda_q : H^{1/2}(\partial \Omega) &\to& H^{-1/2}(\partial \Omega)\\
f& \to &\frac{\partial u}{\partial n}|_{\partial \Omega}.
\end{array}
\end{equation}
where $u$ is the solution of the following problem
\begin{equation}\label{problemi}
\left\{  \begin{array}{ll}
\Delta u + q(x) u=0, \hspace{0.5cm} x \in \Omega,
\\ u= f, \hspace{2.2cm} \partial \Omega,
\end{array}     \right.
\end{equation}
and $\frac{\partial u}{\partial n}|_{\partial \Omega}$ denotes the normal derivative of $u$ on the boundary $\partial \Omega$.

Note that the uniqueness of $u$ as solution of (\ref{problemi}) requires that $0$ is not a Dirichlet eigenvalue of $\Delta + q$. A sufficient condition to guarantee that $\Lambda_q$ is well-defined is to assume $q(x)<\lambda_1$, the first Dirichlet eigenvalue of the Laplace operator in $\Omega$, since in this case, the solution in (\ref{problemi}) is unique. We assume that this condition holds and let us define the space
$$
L^\infty_{< \lambda_1}(\Omega) = \{q\in L^\infty (\Omega), \mbox{ s. t. } q(x) < \lambda_1, \mbox{ a. e. } \} .
$$

In this work we are interested in the following  map
\begin{equation} \label{eq_mapl}
\begin{array}{rcl}
\Lambda: L^\infty_{<\lambda_1} (\Omega) & \to & \mathcal{L}(H^{1/2}(\partial \Omega);H^{-1/2}(\partial \Omega))\\
q & \to & \Lambda_q.
\end{array}
\end{equation}
This has an important role in inverse problems where the aim is to recover the potential $q$ from boundary measurements. In practice, these boundary measurements correspond to the associated DtN map and therefore, the mathematical statement of the classical inverse problem consists in the inversion of $\Lambda$.

It is known that $\Lambda$ is one to one as long as $q\in L^p$ with $p>2$ (see \cite{BIY}, [Bl\.{a}sten, Imanuvilov and Yamamoto 2015]). Therefore, the inverse map $\Lambda^{-1}$ can be defined in the range of $\Lambda$. There are, however, two related important and difficult questions that are not well understood: a characterization of the range of $\Lambda$ and its stability i.e., a quantification of the difference of two potentials, in the $L^\infty$ topology in terms of the distance of their associated DtN maps. Obviously, this stability will affect to the efficiency of any inversion or reconstruction algorithm to recover the potential from the DtN map (see \cite{T1}, [Tejero 2016] and \cite{T2}, [Tejero 2018]).

The first question, i.e. the characterization of the range of $\Lambda$ is widely open. To our knowledge, the further result is due to Ingerman in \cite{I} [Ingerman 2000], where a difficult characterization is obtained for the adherence with respect to a certain topology.
Concerning the stability, there are some results when we assume that the potential $q$ has some smoothness. In particular, if $q\in H^{s}(\Omega)$ with $s>0$, the following
$\log-$stability condition is known (see \cite{BIY}, [Bl\.{a}sten, Imanuvilov and Yamamoto 2015]),
\begin{equation} \label{eq_cal}
\| q_1-q_2\|_{L^\infty} \leq V( \| \Lambda_{q_1}-\Lambda_{q_2} \|_{\mathcal{L}(H^{1/2};H^{-1/2})}) ,
\end{equation}
where $V(t)=C\log(1/t)^{-\alpha}$ for some constants $C,\alpha>0$.
Stronger stability conditions are known in some particular cases. For example, in \cite{BHQ}, [Beretta, De Hoop and Qiu 2013] it is shown that when $q$ is piecewise constant and all the components where it takes a constant value touch the boundary, the stability is Lipschitz, i.e. there exists a constant $C>0$ such that
$$
\| q_1-q_2\|_{L^\infty} \leq
C \| \Lambda_{q_1}-\Lambda_{q_2} \|_{\mathcal{L}(H^{1/2};H^{-1/2})} .
$$

In this work we try to understand better the situation by considering the simplest case of a disk with one-step radial potentials $q$.  More precisely, we give some results on the range of $\Lambda$ and its stability when we restrict to the particular case $\Omega=B(0,1)=\left\{ x \in \mathbb{R}^2: \; r=|x| <1 \right\} $ and
$
q\in F\subset L^\infty(\Omega)$
given by
\begin{equation} \label{def:F}
F=\{ q\in L^\infty(\Omega) \; : \; q(r)=\gamma \chi_{(0,b)}(r), \; r=|x|, \; b\in(0,1), \; \gamma \in [0,1] \},
\end{equation}
where $\chi_{(0,b)}(r)$ is the characteristic function of the interval $(0,b)$. Note that $F$ is a two-parametric family depending on $\gamma$ and $b$.

It is worth mentioning that, as we show below, the solution of (\ref{problemi}) is unique for all $b\in(0,1)$ and $\gamma \geq 0$, and therefore the DtN map is well defined for all these one step potentials. However, we restrict ourselves to the bounded set $F$ to simplify.

Even in this simple case, a complete analytic answer to the previous questions is unknown. Therefore, we have considered a numerical approach based on a discrete sampling of the set $F$. Given an integer $N>0$ we define $h=1/N$ and
\begin{eqnarray} \nonumber
F_h&=&\{ q\in L^\infty(\Omega) \; : \; q(r)=\gamma \chi_{(0,b)}(r), \; b=hi, \; \gamma= h_j , \\ && \qquad  i=1,...,N-1, \; j=0,...,N \} . \label{def:Fh}
\end{eqnarray}
Note that $F_h$ has $N(N-1)+1$ functions from $F$. As $h\to 0$ we obtain a better description of $F$ and, in particular, we should recover the stability properties for $q\in F$.

Concerning the stability of $\Lambda$, we show that it fails in the sense that inequality (\ref{eq_cal}) does not hold for any continuous function $V(t)$ with $V(0)=0$. The proof relies on the ideas in \cite{A} [Alessandrini 1988] where the analogous result is obtained for the conductivity problem.

We also obtain some partial stability results when $b$ and $\gamma$ are fixed. To state them we define the subsets $F_b\subset F$, for $b\in(0,1)$, by
\begin{equation} \label{def_Fb}
F_b=\{ q\in L^\infty(\Omega) \; : \; q(r)=\gamma \chi_{(0,b)}(r), \; \gamma \in[0,1] \},
\end{equation}
and $G_\gamma\subset F$, for $\gamma\in [0,1]$, by
\begin{equation} \label{def_Gg}
G_\gamma=\{ q\in L^\infty(\Omega) \; : \; q(r)=\gamma \chi_{(0,b)}(r), \; b \in(0,1) \}.
\end{equation}
We prove that, for fixed $b_0>0$, Lipschitz stability holds, if we restrict ourselves to $F_{b_0}$. Therefore, the lack of stability is due to the change of $b_0$ rather than to changes in $\gamma$. Concerning $G_\gamma$, we prove that if $b\geq b_0>0$
there is stability of the DtN map with respect to $b$ for potentials in $G_\gamma$. This suggests a possible stability with respect to the $L^1$-norm of the potentials which is sensitive to the position of the discontinuities, when consider discontinuous functions.  In fact, we show numerical evidences of such stability when considering potentials in $F$.

For the range of $\Lambda$ we give a characterization in terms of the first two eigenvalues of the DtN map. We also analyze the region where the stability constant is larger and, therefore, the potentials for which any recovering algorithm for $q$ from the DtN map will have more difficulties.

It is worth mentioning that the results in this paper cannot be easily translated into the closely related, and more classical, conductivity problem where (\ref{problemi}) is replaced by
\begin{equation}\label{ecuacion_divergencia}
\left\{  \begin{array}{ll}
-\mbox{ div } a(x) \nabla v =0, \hspace{0.5cm} x \in \Omega,
\\ v= f, \hspace{2.2cm} \partial \Omega,
\end{array}     \right.
\end{equation}
and the Dirichlet to Neumann map, or voltage to current map, is given by
\begin{equation} \label{eq:Lambda_1}
\begin{array}{rcl}
\Lambda_a : H^{1/2}(\partial \Omega) &\to& H^{-1/2}(\partial \Omega)\\
f& \to &a\frac{\partial v}{\partial n}|_{\partial \Omega}.
\end{array}
\end{equation}
We refer to the review paper \cite{Ul} [Uhlmann 2014] and the references therein for theoretical results on the DtN map in this case.

The rest of this paper is divided as follows: In section 2 below we characterize the DtN map in terms of its eigenvalues  using polar coordinates, in section 3 and 4 we analyze the stability and range results respectively. In section 5 we briefly describe the main conclusions and finally section 5 contains the proofs of the theorems stated in the previous sections.

\section{The Dirichlet to Neumann map}\label{sec1}

In this section we characterize the Dirichlet to Neumann map in the case of a disk. System (\ref{problemi}) in polar coordinates reads
\begin{equation}\label{problema1}
\left\{   \begin{array}{lll}
r^2 \frac{\partial ^2 v}{\partial r^2} +r \frac{\partial v}{\partial r}+ \frac{\partial^2 v}{\partial \theta^2}+r^2q(r)v=0, \hspace{0.3cm}  (r, \theta) \in (0,1) \times [0, 2 \pi), \\  \lim_{r \longrightarrow 0, r>0}v(r, \theta) < \infty, \\ v(1, \theta)=g(\theta), \hspace{0.5cm} \theta \in [0,2 \pi),
\end{array}  \right.
\end{equation}
where $v(r, \theta )= u(r \cos \theta , r \sin \theta) \; \textrm{ and } g(\theta)=f(\cos \theta, \sin \theta)$ is a periodic function.

An orthonormal basis in $L^2(0,2\pi)$ is given by $\{e^{in\theta}\}_{n\in \mathbb{Z}}$. Here we use this complex basis to simplify the notation but in the analysis below we only consider the subspace of real valued functions.  Therefore, any function $g\in L^2(0,2\pi)$ can be written as
\begin{equation} \label{eq:gt}
g(\theta)=\sum_{n \in \mathbb{Z}} g_n  e^{in\theta}, \hspace{0.5cm} g_n=\frac{1}{2\pi}\int_0^{2 \pi} g(t)e^{-int}dt,
\end{equation}
and $\| g \|_{L^2(0,2\pi)}^2 = \sum_{n\in \mathbb{Z}} |g_n|^2$. Associated to this basis we define the usual Hilbert spaces $H^\alpha_{\#}$, for $\alpha >0$, as
$$
H^\alpha_{\#}=\{ g \; : \; \| g\|_\alpha^2 = \sum_{n\in\mathbb{Z}}(1+n^2)^{\alpha} |g_n|^2 <\infty \}.
$$

The Dirichlet to Neumann map in this case is defined as
\begin{equation} \label{eq:Lambda_r}
\begin{array}{rcl}
\Lambda_q : H^{1/2}_{\#}(0,2\pi) &\to& H^{-1/2}_{\#}(2\pi)\\
g& \to &\frac{\partial v}{\partial r}(1,\cdot),
\end{array}
\end{equation}
where $v$ is the unique solution of  (\ref{problema1}).

In the above basis the Dirichlet to Neumann map turns out to be diagonal. In fact, we have the following result:

\begin{theorem} \label{th1}
Let $\Omega$ be the unit disk and $q\in F$. Then,
\begin{eqnarray} \label{eq:Dtn}
\Lambda_q\left(e^{in\theta}\right)&=&c_n e^{in\theta}, \quad n\in \mathbb{Z},
\end{eqnarray}
where
\begin{eqnarray}
\label{c0}
c_0&=&\frac{- b\sqrt{\gamma}J_1(\sqrt{\gamma}b)}{b\log b\sqrt{\gamma}J_1(\sqrt{\gamma}b)+J_0(\sqrt{\gamma}b)}, \\ \label{c0_b}
c_{n}&=&c_{-n}=n\frac{ J_{n-1}(\sqrt{\gamma}b)-b^{2n}J_{n+1}(\sqrt{\gamma}b) }{J_{n-1}(\sqrt{\gamma}b)+ b^{2n}J_{n+1}(\sqrt{\gamma}b)   }, \quad n\in \mathbb{N},
\end{eqnarray}
and $J_n(r)$ are the Bessel functions of first kind.
\end{theorem}

Note that the range of $\Lambda$, when restricted to $F$, is characterized by the set of sequences $\{ c_n \}_{n\geq 0}$ of the form (\ref{c0})-(\ref{c0_b}) for all possible $b,\gamma$. In particular, when $q=0$ we have
\begin{equation}\label{potencial_nulo}
c_{n}=n, \hspace{0.4cm} n=0,1,2, \cdot \cdot \cdot,
\end{equation}
 and this sequence must be in the range of $\Lambda$.

The norm of $\Lambda_q$, when restricted to $F$, is given by
\begin{equation} \label{eq:norm_l}
\|\Lambda_q \|_{\mathcal{L}(H^{1/2}_{\#}; H^{-1/2}_{\#})}=\sup_{n\geq 0} \frac{|c_n|}{1+n}.
\end{equation}

\begin{proof} (of Theorem \ref{th1})
We first compute $c_0$ in (\ref{eq:Dtn}). As the boundary data at $r=1$  in (\ref{problema1}) is the constant $g(\theta)=1$, we assume that $v(r,\theta)$ is radial, i.e. $v(r,\theta)=a_0(r)$. Then, $a_0$ should satisfy
 \begin{equation}\label{problema_0}
\left\{   \begin{array}{ll}
r^2a_0''+ra_0'+r^2q(r)a_0=0, \hspace{0.5cm} 0<r<1, \\ a_0(1)=1, \hspace{0.5cm} \lim_{r \longrightarrow 0, r>0}a_0(r)< \infty.
\end{array}     \right.
\end{equation}
For $r\in (0,b)$ we solve the ODE with the boundary data at $r=0$, while for $r\in(b,1)$ we use the boundary data at $r=1$.  In the first case the ODE is the Bessel ODE or orden 0 and therefore we have
$$
a_0(r)= \left\{
\begin{array}{ll}
A_0J_0(\sqrt{\gamma} r), & r\in (0,b), \\
1+C_0 \log r, & r\in (b,1),
\end{array}
\right.
$$
where $J_0$ is the Bessel function of the first kind and $A_0$, $C_0$ are constants.  These are computed by imposing continuity of $a_0$ and $a_0'$ at $r=b$. In this way, we obtain
$$
\left\{
\begin{array}{ll}
A_0J_0(\sqrt{\gamma} b) = 1+C_0 \log b \\
A_0\sqrt{\gamma} J_0'(\sqrt{\gamma} b) = C_0 \frac1{b}.
\end{array}
\right.
$$
Solving the system for $A_0$ and $C_0$ and taking into account that $\Lambda_q(1)=\frac{\partial v}{\partial r}(1,\theta)=a_0'(1)=C_0$ we easily obtain (\ref{eq:Dtn}).

Similarly, to compute $c_n$ in (\ref{eq:Dtn})  we have to consider $g(\theta)=e^{in \theta}$ in (\ref{problema1}) and therefore we assume that the solution $v(r,\theta)$  can be written in separate variables, i.e. $v(r,\theta)=a_n(r)e^{in\theta}$. Then, $a_n$ must satisfy
\begin{equation}\label{problema_coseno}
\left\{   \begin{array}{ll}
r^2a_n''+ra_n'+\left(r^2q(r)-n^2 \right)a_n=0, \hspace{0.5cm} 0<r<1, \\ a_n(1)=1, \hspace{0.5cm} \lim_{r \longrightarrow 0, r>0}a_n(r)< \infty , \quad n\geq 1.
\end{array}     \right.
\end{equation}
 As in the case of $c_0$, for $r\in (0,b)$ we solve the ODE with the boundary data at $r=0$, while for $r\in(b,1)$ we use the boundary data at $r=1$. We have
$$
a_n(r)= \left\{
\begin{array}{ll}
A_nJ_n(\sqrt{\gamma} r), & r\in (0,b), \\
C_n (r^n-r^{-n})+r^n, & r\in (b,1),
\end{array}
\right.
$$
where $A_n$, $C_n$ are constants.  These are computed by imposing continuity of $a_n$ and $a_n'$ at $r=b$. In this way, we obtain
$$
\left\{
\begin{array}{ll}
A_nJ_n(\sqrt{\gamma} b) = C_n (b^n-b^{-n}) +b^n \\
A_n\sqrt{\gamma} J_n'(\sqrt{\gamma} b) = n C_n (b^{n-1}+b^{-n-1})+nb^{n-1}.
\end{array}
\right.
$$
Solving the system for $A_n$ and $C_n$ we obtain in particular
$$
C_n=\frac{-b^nJ'_n(\sqrt{\gamma}b) +n\frac{b^{n-1}}{\sqrt{\gamma}}J_n(\sqrt{\gamma}b) }{-(b^{-n-1}+b^{n-1}) \frac{n}{\sqrt{\gamma}}J_n(\sqrt{\gamma}b) -(b^{-n}-b^n)J'_n(\sqrt{\gamma}b)} .
$$
We simplify this expression using the well known identity
$$
2J'_n(r)=J_{n-1}(r)-J_{n+1}(r),
$$
and we obtain,
$$
C_n=\frac{- J_{n+1}(\sqrt{\gamma}b) }{b^{-2n}J_{n-1}(\sqrt{\gamma}b) +J_{n+1}(\sqrt{\gamma}b)}.
$$
Now, taking into account that $\Lambda_q(e^{in\theta})=\frac{\partial v}{\partial r}(1,\theta)=a_n'(1)e^{in \theta}=(2nC_n+n)e^{in\theta}$ we easily obtain (\ref{eq:Dtn}).
\end{proof}
\begin{remark}
In this proof of Theorem \ref{th1} we do not use the restriction $\gamma\leq 1$ that satisfy the potentials in $F$. In fact, the statement of the theorem still holds for any step potential, as in $F$, but with any arbitrary large $\gamma \geq 0$.
\end{remark}

\section{Stability}

In this section we focus on the stability results for the map $\Lambda$. Some results are analytical and they are stated as theorems. The proofs are given in the appendix below.  We divide this section in three subsections where we consider the negative stability result for $q\in F$ norm  and some partial results when we consider the subsets $F_b$ and $G_\gamma$ defined in (\ref{def_Fb}) and (\ref{def_Gg}).

\subsection{Stability for $q\in F$}

The first result in this section is the lack of any stability property when $q\in F$. In particular, we prove that inequality  (\ref{eq_cal}) fails, for any continuous function $V(t)$ with $V(0)=0$.

\begin{theorem} \label{th_st1}
Given $q_0\in F$, there exists a sequence $\{q_k\}_{k\geq 1}\subset F$ such that $\|q_0-q_k\|_{L^{\infty}}=\gamma >0$ for all $k\geq 1$, while
\begin{equation}\label{C2b}
\| \Lambda_{q_0}-\Lambda_{q_k}\|_{\mathcal{L}(H^{1/2}_{\#}; H^{-1/2}_{\#})} \to 0 , \quad \mbox{ as $k\to \infty$}.
\end{equation}
\end{theorem}

This result contradicts any possible stability result of the DtN map at $q_0\in F$. Roughly speaking the idea is that the eigenvalues of $\Lambda$, given in Theorem \ref{th1} above, depend continuously on $b$, unlike the $L^\infty$ norm of the potentials. A detailed proof of Theorem \ref{th_st1} is given in the appendix below.

\subsection{Partial stability}

We give now two partial stability results when we fix $b$ and $\gamma$, respectively.
\begin{theorem} \label{th 0.1}
Given $b\in(0,1)$ and $q_1,q_2\in F_b$, we have
\begin{equation}\label{estimacion_estabilidad}
 \|q_1-q_2 \|_{L^\infty} \leq \frac{(4,8765)^2}{b^4}     \| \Lambda_{q_1}-\Lambda_{q_2} \|_{\mathcal{L}(H^{1/2}_{\#}; H^{-1/2}_{\#})} .
\end{equation}
On the other hand, given $\gamma \in (0,1]$ and $q_1,q_2\in G_\gamma$, we have
\begin{equation}\label{estimacion_estabilidad_2}
\left| b_1-b_2  \right| \leq \frac{15}{2 \gamma b^3} \| \Lambda_{q_1}-\Lambda_{q_2} \|_{\mathcal{L}(H^{1/2}_{\#}; H^{-1/2}_{\#})},
\end{equation}
where $b=\min \{ b_1,b_2\}$.
\end{theorem}

The proof of this theorem is in the appendix below.

Inequality (\ref{estimacion_estabilidad}) provides a Lipschitz stability result for $\Lambda$ when $b$ is fixed. This shows that the lack of Lipschitz stability is related to variations in the position of the discontinuity, which is the main idea in the negative result given in Theorem \ref{th_st1}.

A numerical quantification of this Lipschitz stability for $b$ fixed is easily obtained. We fix $b=b_0$ and consider
$$
F_{h,b_0}=\{ q\in L^\infty(\Omega) \; : \; q(r)=\gamma \chi_{(0,b)}(r), \; b=b_0, \; \gamma=0+hj, \; j=1,...,1/h-1 \} .
$$
and, for $q_0\in F_{h,b_0}$,
\begin{equation} \label{eq_s2}
C_2(h,q_0,b_0)= \max_{q\in F_{h,b_0}} \frac{\| q_0-q \|_{L^\infty}}{\| \Lambda_{q_0} - \Lambda_{q} \|_{\mathcal{L}(H^{1/2}_{\#}; H^{-1/2}_{\#})}},
\end{equation}
then $C_2(h,q_0,b_0)$ remains bounded as $h\to 0$ for all $q_0\in F_h$. In Figure \ref{fig:estabilidad_b_iguales} we show the behavior of $C_2(h,q_0,b_0)$ when $h=10^{-4}$ for different values of $b_0$. To illustrate the behavior with respect to $b\to 0$ we plot in the left hand side of Figure \ref{fig:estabilidad_b_iguales} the graphs of the functions
\begin{equation} \label{eq_s1}
C_{2,\min}(b)=\min_{q\in F_{h,b}}C_2(10^{-4},q,b), \mbox{ and } C_{2,\max}(b)=\max_{q\in F_b}C_2(10^{-4},q,b).
\end{equation}
We see that both constants become larger for small values of $b$. We also see that both graphs are close in this logarithmic scale. However, the range of the interval $[C_{2,\max}(b),C_{2,\min}(b)]$ is not small, as showed in the right hand side of Figure \ref{fig:estabilidad_b_iguales}.

\begin{figure}[htbp]
  \centering
 \subfigure[]{  \includegraphics[width=0.45\textwidth]{estabilidadbiguales.png}}
  \subfigure[]{  \includegraphics[width=0.45\textwidth]{diferenciaC2biguales.png}}
  \caption{Numerical estimate of the stability constant $C_2$ in (\ref{eq_s2}) for $h=10^{-4}$. To illustrate the behavior on $b$ we plot the maximum and minimum value when $q\in F_{h,b}$ with respect to $b$  in logarithmic scale (left), and its range in normal scale (right). }\label{fig:estabilidad_b_iguales}
\end{figure}

Concerning inequality (\ref{estimacion_estabilidad_2}) in Theorem \ref{th 0.1}, it provides a stability result of  $\Lambda$ with respect to the position of the discontinuity. In particular, this means that we can expect some Lipschitz stability if we consider a norm for the potentials that is sensitive to the position of the discontinuity. This is not the case for the $L^\infty$ norm but it is true for the $L^p$-norm for some $1\leq p <\infty$. In particular,
\begin{equation} \label{eq_s2b}
C_2(h,q_0,b_0)= \max_{q\in F_{h,b_0}} \frac{\| q_0-q \|_{L^1}}{\| \Lambda_{q_0} - \Lambda_{q} \|_{\mathcal{L}(H^{1/2}_{\#}; H^{-1/2}_{\#})}},
\end{equation}
is bounded as $h\to 0$ and $b\geq b_0 >0$. In Figure \ref{fig:estabilidad_norma1} we show the values when $h=10^{-4}$. We observe that the constant blows up as $b\to 0$.

\begin{figure}[htbp]
  \centering
\includegraphics[width=0.50\textwidth]{estabilidadnorma1.png}
 \caption{ $C_2(h,q)$ for $b>b_0$ when $h=10^{-4}$.}
 \label{fig:estabilidad_norma1}
 \end {figure}

\section{Range of the DtN map}

In this section we are interested in the range of $\Lambda$ when $q\in F$, i.e. the set of sequences $\{ c_n \}_{n\geq 0}$ of the form (\ref{c0})-(\ref{c0_b}) for all possible $b,\gamma \in[0,1]\times[0,1]$.

As $F$ is a bi-parametric family of potentials, it is natural to check if we can characterize the family $\{ c_n \}_{n\geq 0}$ with only the first two coefficients $c_0$ and $c_1$. In this section we give numerical evidences of the following facts:
\begin{enumerate}
\item  The first two coefficients $c_0$ and $c_1$ in (\ref{c0})-(\ref{c0_b}) are the most sensitive with respect to $(b,\gamma)$ and, therefore, the more relevant ones to identify $b$ and $\gamma$ from the DtN map.
\item The function
\begin{eqnarray} \label{DtNh}
\Lambda^h:F_h &\to& \mathbb{R}^2 \\ \nonumber
q &\to & (c_0,c_1),
\end{eqnarray}
is injective. This means, in particular, that the DtN map  can be characterized by the coefficients $c_0$ and $c_1$, when restricted to functions in $F_h$.  We also illustrate the set of possible coefficients $c_0,c_1$.
\item The lack of stability for $\Lambda$ is associated to  higher density of points in the range of $\Lambda^h$. This occurs when either $b$ or $\gamma$ are close to zero.
\end{enumerate}

\subsection{Sensitivity of $c_n$}

To analyze the relevance and sensitivity of the coefficients $c_n=c_n(b,\gamma)$ to identify the parameters $(b,\gamma)$ we have computed their range when $(b,\gamma)\in [0,1]\times [0,1]$, and the norm of their gradients. As we see in (\ref{table_1}) the range decreases for large  $n$. This means that, for larger values of $n$, the variability of $c_n$ is smaller and they are likely to be less relevant to identify $q$.

However, even if the range of $c_n$ becomes smaller for large $n$ they could be more sensitive to small perturbations in $(b,\gamma)$ and this would  make them useful to distinguish different potentials. But this is not the case.  In Figure \ref{fig:gradientegammaconstante} we show that for the given values of $\gamma=0.1,0.34,0.67,0.99$ and $b\in (0,1]$ the gradients of the first two coefficients, with respect to $(b,\gamma)$, are larger than the others. Therefore, we conclude that the two first coefficients $c_0$ and $c_1$ are the most sensitive, and therefore relevant, to identify the potential $q$.

We also see in Figure \ref{fig:gradientegammaconstante} that these gradients are very small for $b<<1$. This means, in particular, that identifying potentials with small $b$ from the DtN map should be more difficult.
\begin{table} \label{table_1}
$$
\begin{array}{|l|l|}
\mbox{Coefficient} & \mbox{Range} \\
\hline
c_0  & 0.5523 \\
\hline
c_1  & 0.2486 \\
\hline
c_2  & 0.1588 \\
\hline
c_3  & 0.1157 \\
\hline
c_4  & 0.0904 \\
\hline
c_5  & 0.0736 \\
\end{array}
$$
\caption{Range of the coefficients, i.e. for each $c_n$ the range is defined as $\max_{q\in F_h} c_n - \min_{q\in F_h} c_n$.}
\end{table}

\begin{figure}[htbp]
  \centering
\includegraphics[width=0.99\textwidth]{gradientegammaconstante.png}

 \caption{Norm of the gradient of the coefficients $c_n(\gamma,b)$ in terms of $b\in(0,1)$ for different values of $\gamma$. We see that the gradients of higher coefficients $n\geq 2$ are smaller than those of the first two ones. We also observe that these gradients become small for small values of $b$.   } \label{fig:gradientegammaconstante}
 \end {figure}

\subsection{Range of the DtN in terms of $c_0,c_1$}

Now we focus on the range of the DtN in terms of the relevant coefficients $(c_0,c_1)$, i.e. the range of the map $\Lambda^h$ in
(\ref{DtNh}): $R(\Lambda^h)$. In Figure \ref{fig:graficoc0c1} we show this range.

\begin{figure}[htbp]
  \centering
\includegraphics[width=0.50\textwidth]{graficoc0c1.png}
 \caption{Range of the discrete DtN map in \ref{DtNh}} \label{fig:graficoc0c1}
 \end {figure}
Coordinates lines for fixed $\gamma$ and $b$ are given in Figure \ref{fig:curvasbgammaconstante}. We observe that $R(\Lambda^h)$ is a convex set between the curves
\begin{eqnarray*}
&& r_{low}: \{ (c_0(\gamma,1), c_1(\gamma,1)),  \mbox{ with } \gamma\in[0,1]\},  \\
&& r_{up}: \{ (c_0(1,b), c_1(1,b)),  \mbox{ with } b\in[0,1]\}.
\end{eqnarray*}

Note also that, in the $c_0,c_1$ plane, the length of the coordinates lines associated to  $b$ constant are segments that become smaller as $b\to 0$. Analogously, the length of those associated to constant $\gamma$ become smaller as $\gamma\to 0$. Thus, the region where either $b$ or $\gamma$ are small produces the higher density of points in the range of $\Lambda^h$. This corresponds to the upper left part of its range (see Figure \ref{fig:graficoc0c1}). On the other hand, this Figure provides a numerical evidence of the injectivity of $\Lambda^h$ too. In fact, any point inside  $R(\Lambda^h)$ is the intersection of two coordinates lines associated to some unique $b_0$ and $\gamma_0$.

\begin{figure}[htbp]
  \centering
 \subfigure[]{  \includegraphics[width=0.4\textwidth]{previobconstante.png}}
  \subfigure[]{  \includegraphics[width=0.4\textwidth]{curvasbconstante.png}}
  \subfigure[]{  \includegraphics[width=0.4\textwidth]{previogammaconstante.png}}
  \subfigure[]{  \includegraphics[width=0.4\textwidth]{curvasgammaconstante.png}}
  \caption{Coordinates lines of the map $\Lambda^h$ defined in (\ref{DtNh}). The upper figure contains the coordinates lines associated to  $b$ constant, while the lower one those corresponding to $\gamma$ constant. }\label{fig:curvasbgammaconstante}
\end{figure}

The higher density of points in the upper left hand side of the range of $\Lambda^h$ should correspond to potentials $q$ with large stability constant $C_2(h,q)$. In Figure \ref{fig:curvanivelC2_1} we show the level sets of $C_2(h,q)$ for $h=10^{-4}$ and different $q\in F_h$. The region with larger constant corresponds to small values of $b$ (upper right  figure) and larger values of $c_1$ (upper left and lower figures). On the other hand, the region with lower stability constant is for $b$ close to $b=1$, which corresponds to the lower part of the range of $\Lambda^h$ when $c_0$ is small.

\begin{figure}[htbp]
  \begin{center}
	\begin{tabular}{cc}
\includegraphics[width=0.50\textwidth]{curvanivelC2_1.png} &
\includegraphics[width=0.50\textwidth]{curvanivelC2_2.png}
\end{tabular}

\includegraphics[width=0.50\textwidth]{curvanivelC2_3.png}
\end{center}
 \caption{Level sets of the $C_2(b,\gamma)$ for $q\in F_h$ and $h=10^{-4}$ in terms of $(b,\gamma)$ (upper left) and in terms of $ (c_0,c_1)$ (upper right). A zoom of the upper left region in this last figure is in the lower figure. Regions separated by the level sets are indicated: region I corresponds to the potentials with stability constant larger that $10^7$, region II corresponds to those with stability constant lower that $10^7$ but larger than $10^6$, and so on. } \label{fig:curvanivelC2_1}
 \end {figure}

It is interesting to analyze the set of potentials with the same coefficient $c_0$ or $c_1$. We give in Figure \ref{fig:curvasc0c1constante} the coordinates lines of the inverse map $(\Lambda^h)^{-1}$. When increasing the value of either $c_0$ (light lines) or $c_1$ (dark lines) we obtain lines closer to the left part of the $(b,\gamma)$ region. We see that the angle between coordinate lines  becomes  very small for $b$ small. In this region, close points could be the intersection of coordinates lines associated to not so close parameters $(b,\gamma)$. This agrees with the region where the stability constant is larger.

\begin{figure}[htbp]
  \centering
 \subfigure[]{  \includegraphics[width=0.45\textwidth]{c0c1constanteenc0c1.png}}
  \subfigure[]{  \includegraphics[width=0.45\textwidth]{c0c1constanteenbgamma.png}}

  \caption{Coordinates lines of the map $(\Lambda^h)^{-1}$ defined in (\ref{DtNh})}\label{fig:curvasc0c1constante}
\end{figure}

\section{ Conclusions}

We have considered the relation between the potential in the Schr\"odinger equation and the associated DtN map in one of the simplest situations,
i.e. for the subset of radial one step potentials in dimension 2. In particular we have focused on two difficult problems: the stability of the map $\Lambda$ (defined in \ref{eq_mapl}) and its range.
In this case, the map $\Lambda$ is easily characterized in terms of the Bessel functions and this allows us to give some analytical and numerical results on these problems.
We have proved the lack of any possible stability result, by adapting the argument in \cite {A} [Alessandrini 1988] for the conductivity problem. We have also obtained some partial Lipschitz stability when the position of the discontinuity is fixed in the potential and numerical evidences of the stability with respect to the $L^1$ norm. Finally, we have characterized numerically the range of $\Lambda$ in terms of the first two eigenvalues of the DtN map and given some insight in the regions where stability of $\Lambda$ is worse.

\section*{Acknowledgements}
The first three authors have been partially supported  by project MTM2017-85934-C3-3-P from the MICINN (Spain).
The fourth author has been partially supported by project MTM2016-80474-P of MINECO, Spain.

redpajama_set_name: RedPajamaArXiv


text:
Quinto Sol, Pachamama @ Cal State L.A. 5/4/07
Written by Matthew M
Quinto Sol, Pachamama, Olmeca, Cihualtl Tonali @ Cal State L.A. 5/4/07
Featured Artists: Cihualtl Tonali; Olmeca; Pachamama (Estela); Quinto Sol;
Genre: Hip Hop; Reggae
Brought together by Cal State L.A.'s Cross Cultural Center, the night's performance was dedicated towards celebrating Cinco de Mayo one day early. My girlfriend and I arrived close to 8 o' clock and walked into the Golden Eagle ballroom where the event was taking place. We walked in right as a group of Mexican mariachi style performers began playing. I'm not sure if they had a group name but they consisted of two violins, a bass guitar, and two acoustic guitars. The violins had an awesome sound and the guitar solos were very catchy as well. An appreciation for musical instruments allowed me to really enjoy their music despite the fact that I had somewhat taken their genre of music for granted in the past. After their performance, Pachamama took the stage.
Pachamama had an excellent stage presence and during their set, reminded the audience to keep in mind the violence that had occurred on "May Day" in Downtown Los Angeles. Pachamama has a ska/reggae/funk style that is easy to get into. Vocalist, Guillermo has a strong voice that beautifully accents the instrumental sounds. The bass guitar, drums, and vocals are awesome and the saxophone completes the ska/reggae feel. Pachamama conveys a bold East Los style that I definitely recommend checking out. Overall, Pachamama had an awesome set despite minor problems with the sound board. Afterwards, Cihualtl Tonali took the stage.
Cihualtl Tonali is composed of two young females with an underground hip-hop style. Their lyrics were rhythmic, fun, and thought-provoking. They are talented young females. They were joined with Guillermo from Pachamama on backup conga drums. After their set, Olmeca took the stage.
Olmeca has a free-style underground rap style. Olmeca is an alumnus from Cal State L.A. and had one or two songs about student frustration that I'm sure most young individuals in the room could find relation. Not all of Olmeca's beats were necessarily original; in fact one beat was called out by an audience member to be borrowed from 3-6 Mafia. However, Olmeca added his own flavor of lyrics, style, and poetry to the previously known beats. Olmeca's sound system was his iPod connected to the rooms speakers. At one point during his set, Olmeca brought up some freestylers from the audience to help him brainstorm and play around with a new beat. The freestylers were somewhat talented for being called out on the spot. Quinto Sol took the stage afterwards.
I had never seen Quinto Sol perform before but I was excited to see them since I had previously heard of their popularity. Quinto Sol had a cool laid-back reggae sound. They started their set off with a narration backed by instrumentals. They played for possibly a good half-hour to an hour. There were three saxophones on stage which I thought was awesome since it allowed for some great saxophone solos. The guitar maneuvered in and out of a funk sounding distortion similar to Pachamama's style. Quinto Sol had a really really awesome sound that absolutely entranced me in their music. I plan to see them again as soon as I get the opportunity. Each performance of the night was amazing.

redpajama_set_name: RedPajamaCommonCrawl


text:
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Indoor and Outdoor Recreation Facilities!
This oak-shaded playground at Dublin Park has slides, swings, and outdoor toys as well as benches and picnic tables.
Dublin offers a facility with basketball courts, meeting rooms, walking track, indoor swimming pool, soccer fields, tennis courts and an outdoor walking trail.
Baseball, softball, soccer and football fields mark this sprawling park as well as batting cages, pavilions and playground.
The Hogan Family YMCA features a wellness center with exercise equipment, TV and computer screens, an aquatic center, walking track, Ping-Pong and foosball area, multi-purpose rooms for martial arts and dance lessons as well as basketball courts and an indoor soccer facility.
Madison's premier gala event of the year features the Mayor's presentation of the year in review and future of the city. Held annually at the US Space & Rocket Center, a portion of the proceeds are donated to the Mayor's Scholarship Fund by the event host, the Madison Chamber of Commerce.
Join the City of Madison for inflatable activities, games, snacks and egg hunt at Dublin Park. Special needs Eggstravaganza takes place the same day inside the Dublin Park Main Building.
This annual festival showcases the best foods, businesses and music that Madison has to offer. No matter what your tastes, there is something for everyone to enjoy at this wonderfully diverse event.
These 5k and 10k courses are perfect for any level runner. Fun and challenging with awards given in many categories.
On Hughes Road, south of Discovery Middle School, the Farmer's Market is open from 8am to Noon each Saturday from April – October. Enjoy fresh local produce, cheese, herbs, baked goods and the work of local artisans.
Come to Dublin Park and enjoy inflatables, games, live entertainment, face painting, food trucks and an amazing fireworks display!
Local businesses show off their wares to the community in a fun, family-friendly atmosphere. Lots of activities for the kids including face paining, inflatables, SARTEC rescue dogs, and more!
On the third Saturday of September, Madison comes alive with the sound of motorcycles participating in the Trail of Tears ride. The ride commemorates the forced march of the Cherokee from their tribal lands in Tennessee and North Carolina to Oklahoma.
Downtown Madison's Main and Front Streets are transformed the first Saturday in October with festival booths, food, live performances and kids activities. The day beings with a parade at 8:00am and ends at 4:00pm with the Miss Madison Street Festival Pageant.
Held the second Saturday in December, the parade starts at 5:30pm on Plaza Drive. It proceeds to Hughes Road, then heads into Historic Downtown Madison. Bring the kids and catch the candy thrown from the floats!


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Print | Volkswagen regards itself as the one and only brand. The Golf is considered to be the de facto car. There's a reason for this: Volkswagen sells more cars than any other brand. Also, the Golf is the best selling model and therefore the standard. Now Volkswagen introduces the Golf Mk 7 and just because of its heritage there's no doubt it will be another success. The question is: will the new Golf only be the standard because of its sales figures or also because of its qualities?


Maitland K, et al. "Mortality after fluid bolus in African children with severe infection". The New England Journal of Medicine. 2011. 364(26):2483-2495.
In Sub-Saharan African pediatric patients presenting in shock, does early resuscitation with either saline or albumin compare to a no-bolus strategy in terms of all-cause mortality at 48 hours?
Among Sub-Saharan African children meeting a particular definition of shock, saline and albumin resuscitation appears to increase mortality when compared to a no-bolus strategy.
Despite a lack of clinical evidence supporting its efficacy, fluid resuscitation was incorporated into practice guidelines for the treatment of children with septic shock in 2002. Most patients receive crystalloids such as normal saline or lactated Ringer's solution (LR), although there are theoretical benefits to colloids such as albumin. While trials such as SAFE and ALBIOS attempt to address this debate, the question of whether fluid resuscitation is necessary at all has resurfaced. Resource-limited areas like sub-Saharan Africa often do not have adequate intensive care capabilities and often do not provide fluid resuscitation as their standard of care. This provided an ideal locale to compare the local standard of care (no fluids) with fluid resuscitation according to conventional Western guidelines.
Published in 2011, the Fluid Expansion as Supportive Therapy (FEAST) study enrolled 3,141 Sub-Saharan children with severe febrile illness and impaired organ perfusion, and randomized them to receive either albumin, saline, or no volume resuscitation. At 48 hours, mortality was higher with albumin (10.6%) and saline (10.5%) as compared to no volume resuscitation (7.3%). This was independent of malaria status, coma, severe anemia, base deficit, or lactate level.
The outcomes of this trial have been controversial. The trial design was limited by the lack of blood pressure as inclusion, and only one of three WHO criteria for shock was required. As such, not all children may have been in shock and may have received unnecessary fluid boluses. Additionally, half of the study population had malaria parasitemia; the study's results may therefore help address fluid resuscitation strategies in patients infected with malaria, but may not adequately address the fluid management in undifferentiated hypovolemic shock. Given these controversies, the WHO guidelines did not include the outcomes of FEAST in developing their guidelines on fluid resuscitation in children.
Before extrapolating the FEAST findings to the developed world, future research would need to explore similar fluid resuscitation strategies in the context of other resuscitative measures including readily available intensive care.
Children received IV maintenance fluids (2.5-4 ml/kg/h); antibiotics; antimalarial, antipyretic, and anticonvulsant drugs; treatment for hypoglycemia and transfusion with 20 ml/kg of whole blood if Hgb <5 g/dl according to national guidelines.
A structured clinical case-report was completed at admission, 1, 4, 8, 24, and 48 hours. Assessment of neurologic sequelae occurred at 4 weeks by an independent clinician unaware of study assignment.
Comparisons are albumin vs. saline vs. no bolus. RR is relative risk, CI is confidence interval.
In January 2011, the trial was stopped for futility, as the combined intervention arms had significantly higher mortality (RR 1.46, 95% CI 1.13-1.90, P=0.004) than the control arm.
The risk of death was similar at 1 hour with a persistent trend for increased mortality up to day 2 in the two fluid bolus groups. Most (87%) of deaths occurred within the first 24 hours and very few occurred after 48 hours.
In Stratum B (hypotensive patients) there was no difference between saline and albumin (RR 1.23; 95% CI 0.70-2.16; P=0.45).
Investigators identified no subgroup (coma, malaria, severe anemia, age, gender, severe acidosis) for which fluid resuscitation was beneficial.
As blood pressure was not a component of the inclusion criteria, it's possible that children who were not in shock were given volume resuscitation when it wasn't necessary.
While the authors excluded children with severe malnutrition by arm circumference, it is unclear how the authors further determined nutritional status.
The original manuscript did not analyze mortality by cause of death; however, the authors provided details assessing all-cause mortality in a subsequent analysis.
The external validity of these findings in the developed world is uncertain, given access to ventilator support and intensive care in much of the developed world. Differences in access to resources, immunization and nutrition status of African children, as well as epidemiological differences in disease processes and prevalence (eg, malaria) further limits generalizability.
↑ Carcillo JA, et al. "Clinical practice parameters for hemodynamic support of pediatric and neonatal patients in septic shock." Critical Care Medicine. 2002;30(6):1365-1378.
↑ 2.0 2.1 2.2 2.3 2.4 2.5 Multiple authors. "Correspondence: Mortality after fluid bolus in African children with sepsis." The New England Journal of Medicine. 2011;365:1348-1353.
↑ 4.0 4.1 WHO writers. "Pocket book of hospital care for children: Second edition." WHO Int. Published 2013. Accessed 2014-11-07.
↑ 5.0 5.1 Kiguli S, et al. "WHO guidelines on fluid resuscitation in children: Missing the FEAST data." BMJ. 2014;348:f7003.
↑ Duke T and Mason E. "WHO guidelines on fluid resuscitation in children with shock." The Lancet. 2014;383(9915):411-414.


text:
By afaqs! Published: 14 Sep 2021, 11:17 IST
HUL's Surf Excel moves to 50% recycled plastic bottles
HUL partnered with Banyan Nation to innovate and create bottles that are less dependent on virgin plastic thereby reducing the net plastic generated.
Hindustan Unilever's Surf Excel, today, announced its transition to recyclable bottles made from 50% post-consumer recycled (PCR) plastic and use of 100% biodegradable actives in its formulation. Surf Excel Matic Liquid was launched in 2016 to give a better laundry experience to Washing Machine users. Over time the business grew and so did its plastic footprint. To live by the clean future principles of the organisation, HUL partnered with Banyan Nation to innovate and create bottles that are less dependent on virgin plastic thereby reducing the net plastic generated. By the end of this year, the brand will cross an annual consumption of 900 tonnes of recycled plastic. HUL is also working on no plastic innovations like the refill machine setup at a store in Mumbai.
Talking about this transition, Prabha Narasimhan, executive director and vice president, Home Care, Hindustan Unilever (HUL) said, "Every decision we make is evaluated, not just on its benefit to the consumers but also on its impact on the environment, now and in the long term, to ensure what is best for the consumer is also good for the larger world we inhabit. Why should the consumer be forced to make a choice between the two?"
"All innovations have a scorecard on various parameters and its positive impact on our clean future agenda is a key criterion to meet for it to move ahead in the organisation's innovation funnel" says Mr. Sanjiv Mehta, the Chairman and MD of Hindustan Unilever. "If we don't set higher standards for ourselves, we will not be able to meet the needs of the future generations. We take great pride in innovating to meet the consumers' needs, some of these aren't critical for the present but are key for the future. This innovation is yet another step in that direction."
CEO and Co-founder of Banyan Nation, Mani Kishore Vajipeyajula said, "Banyan Nation's mission is to ensure that the use of safe and high-quality recycled plastic becomes the norm in mainstream consumer products and packaging. HUL's alignment with our mission and unwavering commitment to sustainability has resulted in over 100 million bottles being made from Banyan Nation's recycled granules and made available to consumers across India."
(We got this information in a press release).
Surf ExcelHindustan Unilever Limited (HUL)


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@Pioneer1 every human on Earth came from Africa, including everyone in Asia, or the Asian peninsula.
@Mel Hopkins from my quote about "marrying up" you take that to mean that I believe people are commodities? The statement was made partially in jest. But surely, you don't believe a Meghan would have any interest in marrying someone like me (assuming we were the same age) over someone in the British royal family.
"marrying up" you take that to mean that I believe people are commodities?
@Troy Yes! But not just you; our entire western civilization society! "Marrying up" or "She can do better" "He married beneath him" Those sayings are predicated on value. Something useful or valuable is a commodity .
He married beneath him" Those sayings are predicated on value. Something useful or valuable is a commodity .
Brother, Troy, does it not depend on the beneficiary? I think royalty there has much to gain having a Black woman in the family, I see Dutchess of Essex as an international value asset. I think the Prince married-up! 'What I think/ is true for me as well as true of a fact. Our Black women are humans this world is unworthy.
Come on @Kalexander2 Megahn is a white-looking multimillionaire celebrity she is not Nay-Nay from the projects. Let's not pretend that all people are the same or viewed the same way.
@Mel Hopkins, people are not commodities for he reasons described above, a barrel is crude oil is, a bustle of corn is, but people are not -- we are unique, so we can not be "commodities." No two people or groups of people are equal.
We, necessarily, made assessments of the "quality" of people every day. Whether you are hiring a welder, surgeon, or basketball player some people are better than others depending upon how they are being measured.
If people were truly commodities it would not matter if Mehgan married me, Pioneer, or Prince Harry. But the reality is that we are not commodities; a wealthy Prince is worth more than struggling bookseller. Hence "marrying up" is an appropriate description.
This is not a western construct; we, in the west, are more likely to marry across class, race, religion, than many other cultures.
Brother – does a Black person need to be from Selma, Georgia, the projects, or Africa to be of value to the Black community? Or, do they need to be wealthy, famous, of profound intellect, or dark-skinned to be respected as a Black person? What's that you say? NO. Then it really doesn't matter whether he/she loves or hate their own community to instill a sense of pride and motivation. It's not her politics nor her origin, she's Black; whether she likes, accepts it or not. Can your daughters learn something, anything from the Black Duchess of Essex?
Can your daughters learn something, anything from the Black Duchess of Essex?
No, nothing I can think of that would take precedence over the other things that they need to know to survive here in America. But maybe I'm missing something. Tell me what Meghan can teach my daughter besides how to nab a wealthy white man?
@Troy: If nothing else, an embedded, verifiable sense of choice to follow, avoid, or go her own way. As a Black man, I've chosen the latter; in spite of the fact I hate uncle Tom's Ben Carson, Herman Cain, and sell-out Barack Obama the pride they have confirms, so can I achieve, follow, lead, or go my way.
Maxine Waters of the U.S. Congress, like Mehgan she is like any other Black women; with a maternal instinct even with all the scars, cracks, and crevices of any human being. Let's not get caught up in what she recognized, do or don't do for her female Black community. This isn't "what have you done for me lately" kind of value.
Sure Mel commodities have value, but the value must be the same for each unit for it to be a commodity. If every ounce of gold had a different price it could not be traded reliably and would not be a commodity. The way you are using the word is confusing for this reason.
If your point is to say that people have value; I say of course they do.
Or are you saying that we all have the same value, and this is why "marrying up" does not make sense?
K2 by your description Mehgan is like any woman. Again this is a notion that I reject because not all women (or men) are the same, some have more "value" than others when it comes to marriage. The reality is marriage, at least a successful one, is more like a great business relationship with "benefits."
Sure Mel commodities have value, but the value must be the same for each unit for it to be a commodity.
@Troy as I'm sitting here watching bloomberg - and looking at the board -the value changes depending on demand and that's how we treat people. We do it in the work place - and it seems to carry over to relationships. Again, I'm not saying "YOU" - I'm saying our society does it when it comes to people and I dream a world when we simply look at our community based on our goals. You wrote Harry wouldn't marry Gabby in billion years... I get the sense that's statement was made on some arbitrary value assignment. Is that correct?
Added commentary - edit people to relationships.
Again this is a notion that I reject because not all women (or men) are the same, some have more "value" than others when it comes to marriage. The reality is marriage, at least a successful one, is more like a great business relationship with "benefits."
Granted, the business of marriage we're all involved with, in one way or another, for some benefit. However, what's the differences that cause you to feel that way? What is the crucial meaning of the difference?
The reality is marriage, at least a successful one, is more like a great business relationship with "benefits."
That's what I say!!! GOAL MATES!!!
I was in Morocco some time ago and was shocked to learn it was the single female royalty capital of the world, to me that is, Black women, mostly from Africa with wealth, prestige, and profound careers. Many left their home country to avoid jail, exile or other repercussions for cheating their people or ripping off their wealthy husbands. Even more shocked that this information is never shared on the media.
Now, I wouldn't want my daughter to do the same, I admire the spirit, independence, and boldness of those women which is what Mehgan does for Black women worldwide. Pride is a powerful motivator, something females, especially Black utilize better than us males.
@Kalexander2 what wouldn't you want your daughter to do?
avoid jail, exile or other repercussions?
And do you believe these women were guilty without due process?
what wouldn't you want your daughter to do?
Because doing anything worthy of jail, exile, or repercussions is self-defeating, following rather than leading or going on one's own way. OH YEAH, with or without due process, they're guilty!! Amneh, one sister from Mozumbic Africa was royalty in her village, she split with over 2 billion euros of the village and her husband's family's cash. But it's not the point I'm trying to make here, nor does it matter to anyone except the women and their victims. It's the independence, spirit, and bold determination to do what they wanted, for selfish gains or not. They survive!!! I recall being offered, more than once to wed for immigration to the U.S., I'm not going to tell you that if not for the notoriety I would have jumped without being asked twice. Besides, I have a wife already and was too afraid of getting busted for abusing my citizenship for money.
Yeah, sister Cynique, the fairy tale princess crap impressional young women gleam is just that; a fairy tale. The scandalous royal family is about to come down to earth now that a real earthling has taken-up residence in Buckingham Palace. They're about to find out that Black folks don't leave when asked nicely, you have to throw us out. Smile.
OH YEAH, with or without due process, they're guilty!!
@Kalexander2 Here in America, we are still have due process. We are innocent until PROVEN guilty, And if anyone is following the rules without question that makes them a follower who is obedient to authority, That is the antithesis of being a leader. I asked because your comment was sending mixed messages and I needed clarification.
It's the independence, spirit, and bold determination to do what they wanted, for selfish gains or not.  They survive!!!
YES!!! Now that's someone with the spirit of a leader.
HEY, so am I a follower because I didn't abuse my citizenship by marrying for money, by following the rules?
@Kalexander2 how would you abuse your citizenship by marrying for money? Are you a U.S. national? Marrying for money isn't illegal in U.S.
how would you abuse your citizenship by marrying for money? Are you a U.S. national? Marrying for money isn't illegal in U.S.
I'm natural born and reared U.S. citizen, but regardless, though I'm not familiar with the exact statute/ordinance consider marriage to a foreigner for money and then apply for a visa, residency, or citizenship is forbidden. No one had to tell me this and I may wrong, I doubt it. You're right most anything one does for money isn't illegal, except if it to do something that impacts the integrity of laws and privileges. Such as applying for immigration status for a foreigner married to a U.S. citizen as part of a payment. Now, does that make a follower, fool, or avoid legal complications for a monetary benefit?
You're right most anything one does for money isn't illegal, except if it to do something that impacts the integrity of laws and privileges.
@Kalexander2 Ok, now I understand your position. In your comment above you left out the part about marrying for money to gain citizenship for yourself or someone else.
Besides, following rules without question is not the same as honoring one's integrity.
Based on the information you've provided, I wouldn't now if that makes you a follower or even obedient to authority. This could be one area that you feel it's wrong to take money for something you'd do for free. There's no way I could know your threshold .
Yeah, well, why else would a wealthy foreign woman want to marry a Black American man, certainly not for love which is the other reason I refused, I knew it would just for the money. And no person should face possible abuse by a male gold-digger. Truth is, one side of me says I'm a fool for passing up an opportunity for, whatever; and another says I'm a rebel because I went against the norm, putting principle before profit. I grow and please myself by doing or not doing what is considered normal. Most men would have jumped instantly at such an offer.
I simply refuse to believe in the "out of Africa" theory.
Do I have proof or a ton of solid evidence to support my disbelief?
Outside of my observations of the phenotypical patterns of people around the globe....no.
But given the ever changing history that anthropologists, paleontologists, and others in Western science offers the world every few decades.....I'm sure before this century is out they'll have a NEW theory of where humans supposedly came from.
Megahn is a white-looking multimillionaire celebrity she is not Nay-Nay from the projects. Let's not pretend that all people are the same or viewed the same way.
I don't even think they see her as actually "Black".
AfroAmericans see her as Black because we've been raised with that erroneous "one drop" rule.
People are NOT commodities. We are members of a networked community. So, marriage is not a step-up but rather a strategic partnership based on goals.
Since marriage is also government-sanctioned some people marry to expand their territories.
Married people may not be commodities but MARRIAGE itself is usually a commodity and a financial agreement in THIS society as well as European, African, and Asian societies.
Historically speaking, people RARELY got married for "love"....but out of financial, social, and political benefit and it was usually for the benefit of the men in the relationship and not the women.
Even though a poor woman may have married a wealthier man, it wasn't the woman who benefited from the arrangement so much as it was her father and brothers who got some control of the money and land involved.
Personally, I would have NO PROBLEM marrying a woman for my economic well being or hers AS LONG AS IT'S UNDERSTOOD between us that that's what we're marrying for.
In my opinion, LYING and BREAKING OATHS TO THE SUPREME BEING is wrong, not simply chasing after wealth or status.
So if I lied for the reasons I married her or took an oath to marry her for a particular reason and then broke it.....that would be "wrong".
But simply marrying for money or financial gain in and of itself isn't wrong.
Notwithstanding how any of us feel, Meghan's marriage to the prince has or will have an effect on the Black community worldwide. Perhaps not an economic or political effect but of attitudes of sorts. Some may see it as a Black woman making it good in a bad, dark time for Black folk and benefit from some positive attitude; others may see it as a realization of false pretense of moving up in the world, enlighten by a negative attitude.
No matter what the attitude, Meghan owns nothing to the Black community; she's not the cause of issues facing Black people and she cannot be a solution, even she was the queen. Though it means nothing to me or my life, "I'm happy for her," and pray she doesn't fall victim as did Princess Dianna. Never bought that 'car accident' excuse for her murder!
Never bought that 'car accident' excuse for her murder!
@Kalexander2Nope... Me neither! LOL! I still think Queen Elizabeth II is straight up gansta. I know I wouldn't cross her. I remember shopping in London - and the taxes on my stuff equaled the cost of goods. The queen don't play. And one would have to wonder why her dynasty hasn't been overthrown like Ethiopia Monarchy.
MARRIAGE itself is usually a commodity and a financial agreement in THIS society as well as European, African, and Asian societies.
@Pioneer1 It is amazing that so many people don't question why they have to get a state sanctioned marriage license. Marriage is a state-sanctioned business. Children are (fruit of the labor) of marriage. But so many folks are under the fairy tale impression it's because " I love him/her" - NOPE! In fact it has nothing to do with religion either - - as some familiar with marriage laws - say the state allows G-d to be party to the marriage but here in the U.S. ; God is optional.
@Cynique Yes, I saw that too. According to the New York Times article, (October 13, 2017) Joel Makonnen had dated Ariana Austin for 12 years before he admitted that he was the great-grandson of Haile Selassie, Ethiopia's emperor who was ousted (1974 ) After an civil war the monarchy was overthrown in 1991. Makonnen is a prince without a country. Still, the story of their wedding brought Ethiopia back into mainstream news.
After your post, I started thinking about how stories of folks in the African diaspora don't get coverage like European diaspora in america . For example. if Ariana Austin met Prince Yoel in 2005; you mean to tell me, no other media knew he was living here in the U.S. Where are the stories on him prior to his wedding? He could only keep his identity a secret because he knew American Media wouldn't cover him.
@Troy I agree we have to do better with getting media coverage for our community. I know I've been responsible for getting my fair share of folks on record - but I'm just one journalist/media consultant. I knew about Meghan before the world started talking about her - because I watched the first season of suits and I'm huge hallmark channel watcher... I thought it was funny how they would cast her in racially ambiguous roles... and then she got all this hate mail when Suits cast Wendell Pierce as her overprotecting father. Folks started complaining that she hid that she was black - and she responded that she never said she wasn't black to her critics.
But anyway; this thread reminded me of when I pitched a story to an Essence editor in 2007 that black men were getting "snipped" (vasectomies) in record numbers - they were interested at first - then shelved it. Ten years later, I saw a similar story in the NY Post where white men said they are getting snipped to avoid gold-diggers and child support. Maybe if I'd pitch baby mama drama Essence would have ran with it.
s my understanding that the Arabs in Africa were descended from the Arabs of Asian nations like Iraq, Syria, Saudi, ect...who invaded and occupied Africa centuries ago.
THE MENES, the ARMENIANS, the MEDES, MEDO-PERSIA, the ACHAEMENIDS, ... and some of these people were concentrated in THE WEST initially. Nevertheless, they are still defined as ORIENTAL, but not necessarily ASIAN. Some the early people, from my research shows that they were, as you said, BLACK. They interacted with the ancestors of the Khurds and Greater Babylon and etc.
You wrote Harry wouldn't marry Gabby in billion years... I get the sense that's statement was made on some arbitrary value assignment.
Yes arbitrary, subjective, and often unjust -- particularly as they relate to Black people. Mel do you think Harry would marry Gabby? Does anyone here think he would (for all the reasons one might image) or am I just making stuff up?
Ebony is far from the magazine it once was. I've written about that here a couple of times here, but Ebony is not unique. We have lost so much media there is not one Black owned daily newspaper in all of America --despite all of our education, wealth, and talent.
I've heard the argument that the Middle East is actually part of northeast Africa but the West tries to seperate it.
To be honest, I could see it that way....but barely.
As far as I'm concerned it's just as much if not more a part of Asia than Africa.
I believe since most Black people in the United States are either Muslim or Christian they have an interest in seeing what their religions call the "holy land" as part of Africa.....similar to how a lot of our people want to make Jesus and the other biblical characters Black.
I'm beyond that at this point.
Before the Caucasian Medes and Persians moved into Iran the original peoples there were a Black people known as Elamites.
Before for the Caucasian Syrians moved into the Palestine coastal regions, the original people's in that area were Black and built a city called "Salem" that Caucasian later came in, continued to build upon, and now call "Jeru-SALEM"
They were a tall and strong Black people and when the Caucasians saw them they called them "Giants".
Infact, the original inhabitants of Italy and Greece were Black people known as Etruscans.
So clearly Black isn't limited to Africa.
Japeth is also related to the Greco-Roman deity JUPITER.
According to the myth, when Japeth and his family moved down from the Caucasus mountains into Greece...him and many of his offspring were deified as Greek gods and heros.
@Pioneer1 the problem is that white people put Egypt which is firmly on the African Continent in a nonsensically named place they made up called the "middle East." Clearly Egypt is in Africa -- full stop, end of story.
the problem is that white people put Egypt which is firmly on the African Continent in a nonsensically named place they made up called the "middle East." Clearly Egypt is in Africa -- full stop, end of story.
Is a dowry or a prenuptial the price of the marriage. While that doesn't make a commodity is does make a price.
I know Egypt is part of Africa as well as Lybia and Morocco but they are still considered part of the Middle East as well.
Middle East is more of a CULTURAL term than a geographical one.
Juar like the term Latin America.
Latin American countries are found in both North AND South America....two continents.
Likewise, the "Middle East" also spans two continents....Africa and Asia.
So you are saying Syrians are culturally the same as Egyptians, and Israelis the same as iranians?!
Latin America and the middle east are not analogous at all.
Yeah, Latin America is a cultural term; that's ridiculous. Latin America is comprised mostly of Latin peoples within the Americas, it's geography. Middle Eastern Peoples are Africans, Arabs, And Jews. In fact, the term "Middle East" was probably first originated in the 1850's by Britons but became popular when Americans designated it as the area between Arabia and India. Hence, Middle East.
Right again, there's no data anywhere that would support (even) remotely Latin America and the Middle East are comparable.
Any elementary school student knows Morocco is Africa, never considered part of the Middle East.
I KNOW where those nations are located, but IT DOESN'T MATTER whether or not the country is in Africa or not the term "Middle East" has more to do with culture than continent.
Egypt, Lybia, and Morocco may be in Africa but it's CONTROLLED by Arabs and it's culture is an Arabic culture.
And yes Israel and Iran is part of Middle Eastern culture because of their history.
Hebrew is a Semetic language like Arabic.
Israel has a large population of Arabs still within it's borders.
Persian or Farsi is written in Arabic and Iran is a predominately Muslim nation.
I understand YOUR point that these nations are in the continent of Africa, but you're looking at maps and I'm talking about how many of the people from those nations see THEMSELVES.
...especially those from Egypt and Morocco that I've talked to.
They share far more in common with other Arabs from places like Syria and Bahrain, than they do Black folks from Zimbabwe or Angola.
Neither you nor I invented the term "Middle East".
I'm not sure how productive it would be for us to wrangle over it's exact meaning of a tern neither of us invented or have control over.
If the media and even the Eyptian government wants to call Egypt the Middle East and align it with Syria and Iraq....what can me, you, or any of the Black controlled nations in Africa do about it?
@Pioneer1, I'm not confused. But it is pointless to argue over geography because maps clearly show Morocco and Egypt are in Africa.
As far as the Middle East is concerned it was something created by white folks, like the national borders of present day Africa, created without regard to the cultures or input from the people who lived there.
If your definition of being in the middle East is, "CONTROLLED by Arabs and it's culture," how do you explain Israel?
Besides you, who else considers Morocco to be in the Middle East?
No need to bring in issues of politics, religion, or the white man into this. Just admit you made a mistake, learn from the experience, and move on -- what is the big deal?
Pioneer, I'm not confused. But it is pointless to argue over geography because maps clearly show Morocco and Egypt are in Africa.
Besides you, who else considers Morocco to be in the Middle East.
You are right, brother Troy; in fact, I'm about 8-hours from the Egyptian border, visited there several times, driving from Israel; yet they call Israel part of the Middle East, even area's in Palestine they call Israel. It's all Africa with African Christian, Muslim, and Jewish religion cultures and people. One cannot say they are of Arab cultures only; not Egypt, Libya, nor Morocco. They are part of the African continent where Black African, Arabs, and Jews live. Simple as that, period.
Even the United Nations, though they recognize something called the 'State of Isreal," they actually consider what is called Israel as Palestine. Isreal is just a government State without a country, until the 1947 war where even Jews called a "Beautiful Bride Married to Another" as Palestine. Colorful words to that successfully hide their true intention from this stupid world.
You must realize brother, it's easy for Europeans to win-over foolish Black folk because they understand we'll never question what they say.
I guess the establishment of an Embassy in Israel is icing on the cake.
Are any of the Black folks in your your neck of the woods interested in the royal wedding?
Remember, you don't have to copy what I wrote in your reply.
Interestingly, even to my surprise Arabs don't seem to care about the Embassy in Jerusalem. Many never really had access when it was Tel Aviv. I was there several times to cancel a resident visa for my wife, we decided it best she does not come to the U.S. Something about fear of me being harmed should she face Arab hating Islamaphobic white folk. She's right too, I'll make them kill me! unfortunately, my compassion is not passive.
The royal wedding, many of the women folk in my wife's family, and others have positive comments about it, but also see it as a Black woman being submissive to a Whiteman. I constantly hear commentary that American Blacks are becoming worse than whites. That's probably because they expected Barack Obama to stand-up to Netanyahu more forcefully. At any rate, I can easily get away from conversations I disdain by denying I speak Arabic or Hebrew unless I'm at one the universities here or in Palestine.
My wife won't allow me to go to bars, but she did surprise me on New Years with a bottle of 50-year-old Glenfiddich Single Malt Scotch Whiskey. And no, she's stingy to pay $23,000, she got it from an auction house in Dubai for $2,000., more than 2/3rd remain. You'll never gtuess how many Russian females are here; more Russian Jews then Middle Eastern Jews and most, if all are deeply involved in criminal activity.