text:
\section{Introduction}
\setcounter{equation}{0}

This paper is concerned with an inverse boundary value problem for the conductivity equation. We aim to recover the electrical conductivity of a bounded body in ${\mathbb R}^n$, $n\geq2$, as well as the possibly embedded  obstacles  by taking local boundary measurements.
Precisely, let $\Omega$ and $D$ be two bounded domains in ${\mathbb R}^n$ with Lipschitz continuous boundaries  such that $\overline{D}\subseteq \Omega$. In the physical situation, $D$ corresponds to the impenetrable obstacle inside $\Omega$ which consists of possibly several physical components satisfying different boundary conditions. Then the conductivity problem can be
formulated in finding a weak solution $u\in H^1(\Omega\setminus\overline{D})$ such that
\begin{eqnarray}\label{1.1a}
\left\{
 \begin{array}{ll}
{\rm div\,}(\gamma \nabla u) = 0,& \text{in}\;\Omega\setminus\overline{D}\\
\;u = f,                        &\text{on}\;\partial \Omega\\
\;\mathcal{B}u = 0,    & \text{on}\;\partial D
\end{array}
\right.
\end{eqnarray}
for any given voltage potential $f\in H^{1/2}(\partial\Omega)$.  In (\ref{1.1a}), the function
$\gamma\in L^\infty(\Omega\setminus\overline{D})$ denotes the electric conductivity satisfying the ellipticity condition $0<c_0\leq\gamma\leq c_0^{-1}$ a.e. in the domain $\Omega\setminus\overline{D}$ for some positive constant
$c_0\in{\mathbb R}_+$. The operator $\mathcal{B}$ indicates the boundary condition, described by
$\mathcal{B}u: = u$ if $D$ is a sound-soft obstacle and
$\mathcal{B} u: =\partial_\nu u+{\rm i}\lambda u$ with ${\rm Re}(\lambda)\geq 0$ if $D$ is an imperfect obstacle. Here, $\nu$ denotes the exterior unit normal on the boundary $\partial D$. The
impedance boundary condition can be reduced to a Neumann boundary condition if $\lambda=0$.

Under the conditions, (\ref{1.1a}) can be proved to be well-posed in $H^1(\Omega\setminus\overline{D})$. Therefore,
we can define the well-defined Dirichlet-to-Neumann (D-N) map by
\begin{eqnarray}\label{1.2a}
\Lambda_{\gamma,D,\mathcal{B}} f = \gamma\frac{\partial u}{\partial\nu}\Big|_{\partial\Omega} \in H^{-1/2}(\partial\Omega).
\end{eqnarray}
If measurements are taken only on an open subset of $\partial\Omega$, denoted by $\Gamma$, then
the relevant map in (\ref{1.2a}) is called the local D-N map. In this case, we write the map for
$\Lambda^\Gamma_{\gamma,D,\mathcal{B}}$ which indicates the dependence on the measurement boundary
$\Gamma$. Then,
the inverse conductivity problem is to determine all unknown $\gamma, D$ and $\mathcal{B}$ by taking full measurements  $\Lambda_{\gamma,D,\mathcal{B}}$ or partial measurements
$\Lambda^\Gamma_{\gamma,D,\mathcal{B}}$.
This kind of problems can find important applications in diverse fields such as geophysical exploration, nondestructive testing and medical imaging. The first mathematical formulation on the inverse problem can date back to A. P. Calder\'{o}n \cite{Ca80}, where the uniqueness was
considered in recovering $\gamma$ by the knowledge of $\Lambda_{\gamma,D,\mathcal{B}}$ for the case
when no obstacles are embedded
in the domain $\Omega$, i.e., $D=\emptyset$. Substantial progress has been made
in this direction since then; see fundamental papers like \cite{KV84}, \cite{KV85}, \cite{Na96}, \cite{SU87} in both the two-dimensional and three-dimensional cases.
It was shown in \cite{SU87} that the global uniqueness theorem was proved in determining
a sufficiently smooth conductivity or potential for dimension $n\geq3$, by taking measurements over the boundary.
This result was later extended in \cite{B96}, \cite{BT03}, \cite{HT13}, \cite{PPU03} and related references quoted there to the cases of less regularity conductivity or potential. Moreover, in dimension $n=2$, the global
uniqueness theorem was also proved in \cite{Bu08}, \cite{Na96} for a smooth conductivity and \cite{AP06} for the relaxed regularity conductivity, by taking full boundary data.
Compared to the full data case, the inverse conductivity problem by the local D-N map
has been dealt with more recently. We refer to \cite{AV05} \cite{BU02}, \cite{Is07}, \cite{IUY10}, \cite{KSU07} and related references quoted there in both three dimensions
and two dimensions.

The works mentioned above are mainly based on constructing suitable complex geometrical optics solutions associated with the conductivity equation or Schr\"{o}dinger equation. With a different version, a local uniqueness result was proved
\cite{AV05} on the conductivity, by assuming $\gamma$ is a piecewise-constant function
 in a known finite partition of the domain; see also
\cite{D98} for the direct geophysical setting of the problem.

We remark that in all above works except \cite{IUY10}, the inverse conductivity problems were dealt with
only for the case without embedded obstacles in the body. In the current work,
we shall study the inverse problem with impenetrable obstacles embedded in
$\Omega$, which makes the problem become more challenging since unknown embedded obstacles need to be also recovered by full or partial measurements over the boundary. If $D$ is known to be a sound-soft
obstacle in advance, it has been shown in \cite{IUY10} that the conductivity $\gamma$ can be uniquely recovered in
the domain $\Omega\setminus\overline{D}$ by assuming the partial data.
We also mention a related work \cite{YZ17,YZZ13} from Yang {\em et al.} in the field of the inverse scattering, in which the unique recovery of embedded  obstacles and the surrounding inhomogeneous medium described by a piecewise-constant refractive index were considered.
A novel version was proposed in \cite{YZ17,YZZ13} relying on the $L^2$-well-posedness of the interior transmission problem or its modified version for the
Helmholtz equations in a sufficiently small domain, which were recently extended to other cases such as the inverse fluid-solid interaction problem \cite{QYZ17} and
the inverse cavity problem \cite{QY17}. In this paper, motivated by the previous works of the author,
we shall study the inverse conductivity problem by {\em a priori} assuming that the conductivity
is a piecewise-constant function in an unknown finite partition of the domain. Under this assumption,
we are able to prove that the local D-N map defined on an arbitrary open subset of the boundary
can uniquely determine all unknown $\gamma$, $D$ and $\mathcal{B}$.
The approach proposed in this work depends only on the well-posedness of a coupled PDE-system
associated with the conductivity equations in the $H^1$ space, which is
essentially simple and elementary. Moreover, our approach always works whenever in the two dimensional case or a more higher dimensional case.

Next let us outline the underlying main results of this work. We first present some assumptions pertaining to the domain partition. Assume that the bounded domain $\Omega$ consists of a finite number
of disjoint Lipschitz subdomains $\Omega_\ell$, $\ell=1,2,\cdots,N$, such that
$\overline{\Omega}=\bigcup_{1\leq\ell\leq N}\overline{\Omega_\ell}$. And the electrical conductivity $\gamma$ is considered to be
constant in $\Omega_\ell$, $\ell=1,2,\cdots,N$ satisfying a natural condition that $c_{\ell_1}\not=c_{\ell_2}$ if
the interaction $\partial\Omega_{\ell_1}\cap\partial\Omega_{\ell_2}$ contains a non-empty open subset in the ($n-1$)-dimensional manifold of ${\mathbb R}^n$.
Moreover, we also assume that a smooth obstacle
$D$ is embedded inside some subdomain denoted by $\Omega_{\ell_0}$ $(1\leq\ell_0\leq N)$, i.e.,
$\overline{D}\subset\Omega_{\ell_0}$. That is, the material surrounding the obstacle $D$ is assumed to be
homogeneous. Under the above considerations, the conductivity $\gamma$ can be described as the following
function
\begin{eqnarray}\label{1.3a}
\gamma(x) = \sum_{\ell=1}^{N}c_\ell\cdot\chi_{\Omega_\ell}(x), \quad{\rm for\;almost\;every\;}x
\in\Omega\setminus\overline{D},
\end{eqnarray}
where $c_\ell$, $\ell=1,2,\cdots,N$, are unknown real positive constants, and $\chi_{\Omega_\ell}$ denotes the characteristic function in the subdomain $\Omega_\ell$, which is defined by $1$ in $\Omega_\ell$ and $0$ otherwise.
For the non-empty open subset $\Gamma$ of the boundary $\partial\Omega$, we define the following
partial Cauchy data
\begin{eqnarray}\label{1.4a}
\mathcal{C}_{\gamma,D,\mathcal{B}}:
=\{(f|_{\Gamma},(\Lambda_{\gamma,D,\mathcal{B}} f) |_{\Gamma}): f\in H^{\frac{1}{2}}(\partial\Omega)\;
{\rm with}\; f=0\;{\rm on}\;\partial\Omega\setminus\overline{\Gamma}\}
\end{eqnarray}
which will be used to recover internal properties of $\Omega$ in the inverse conductivity problem.
For convenience, we use symbols
$\mathcal{C}_{\gamma_j,D_j,\mathcal{B}_j}$, $j=1,2$, indicating the dependence on two classes of
different conductivities and embedded obstacles. Based on (\ref{1.3a})-(\ref{1.4a}), we now formulate
 our main result in the following theorem.
\begin{theorem}\label{thm1.1}
If $\mathcal{C}_{\gamma_1,D_1,\mathcal{B}_1}=\mathcal{C}_{\gamma_2,D_2,\mathcal{B}_2}$, then $\gamma_1=\gamma_2$, $D_1=D_2$ and $\mathcal{B}_1 = \mathcal{B}_2$.
\end{theorem}

The remaining part of the paper is organized as follows. In Sections \ref{sec2} we introduce an
auxiliary boundary value problem associated with the modified conductivity equation and prove its
well-posedness in a suitable Sobolev space by employing the variational method.
In Section \ref{sec4} we propose
a novel technique for the inverse conductivity problem, also known as Calder\'{o}n problem, of determining the conductivity of a
piecewise-constant function from Cauchy data measured on an arbitrary open subset of the boundary.

\section{A related PDE-model} \label{sec2}
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In this section, we introduce a related PDE-system in $D_0$ described by two modified conductivity equations which are coupled by a pair of Cauchy conditions on the boundary $\partial D_0$:
\begin{eqnarray}\label{2.1}
\left\{
 \begin{array}{ll}
{\rm div\,}(a_1(x) \nabla u_1)-b_1 u_1=\rho_1,& \text{in}\;D_0,\\
{\rm div\,}(a_2(x) \nabla u_2)- b_2 u_2=\rho_2,& \text{in}\; D_0,\\
u_1-u_2=f_1,& \text{on}\;\partial D_0,\\
a_1(x)\partial_\nu u_1-a_2(x)\partial_\nu u_2=f_2,&\text{on}\;\partial D_0,\\
\end{array}
\right.
\end{eqnarray}
where $D_0$ is a simply connected, bounded domain with the Lipschitz boundary $\partial D_0$, and
$\rho_1,\rho_2\in L^2(D_0)$, $f_1\in H^{\frac{1}{2}}(\partial D_0)$ and $f_2\in H^{-\frac{1}{2}}(\partial D_0)$. In (\ref{2.1}), $a_j$, $j=1,2$, are assumed to two bounded measured
functions satisfying $\infty>c_2\geq a_j\geq c_1>0$ for almost every $x\in D_0$, and
$b_j$, $j=1,2$, are also assumed to be two positive constants.

Under some appropriate assumptions on $a_j,b_j$, we shall next prove that (\ref{2.1}) is well-posed
in the classical $H^1(D_0)$ space for the given data $\rho_1,\rho_2,f_1$ and $f_2$.  We remark that the well-posedness of
(\ref{2.1}) will play an important role in the inverse problem of determining a piecewise-constant conductivity $\gamma$. Indeed, as shown in the proof of Theorem \ref{thm1.1} in next section, the regularity of the solutions of (\ref{1.1a}) with a family of special boundary data $f$ having a singularity like $1/|x-z|$ in the three-dimensional case or $1/\ln |x-z|$ in the two-dimensional case
 can be exactly improved uniformly by analyzing the solvability of (\ref{2.1}). This will lead to a contradiction. The detailed discussions can be found in next section.

In order to study the existence of a solution to (\ref{2.1}), we shall adopt the suggestion from \cite{CC06} with
employing a technical variational approach for the coupled Helmholtz equations. To this end, we introdue the Hilbert space $X_{a_2}$ defined by

\begin{eqnarray}\label{2.2}
X_{a_2}:
=\{\bm{v}\in L^2(D_0)^3: {\rm div\,}(a_2(x)\bm{v})\in L^2(D_0),\;{\rm curl\,}\bm{v}=0\;\;\text{in\;}D_0\}
\end{eqnarray}
under the product
$(\bm{v}_1,\bm{v}_2)_{X_{a_2}}:=(\bm{v}_1,\bm{v}_2)_{L^2(D_0)^3}
+({\rm div\,}(a_2\bm{v}_1),{\rm div\,}(a_2\bm{v}_2))_{L^2(D_0)}$ for $\bm{v}_1,\bm{v}_2\in X_{a_2}$. Obviously, it makes sense due to the assumption on $a_2$.
The norm on $X_{a_2}$ is thus given by
\begin{eqnarray*}
\| \bm{v}\|^2_{X_{a_2}}:=\|\bm{v}\|^2_{L^2(D_0)^2}+\|{\rm div\,}(a_2\bm{v})\|^2_{L^2(D_0)}.
\end{eqnarray*}
If $u_2\in H^1(D_0)$ is the solution of the second equation in (\ref{2.1}), we
let $\bm{u}_2:=\nabla u_2$ in $D_0$. Then, one has $\bm{u}_2\in L^2(D_0)^3$ satisfying
\begin{eqnarray}\label{2.3}
{\rm div\,}(a_2\bm{u}_2)=b_2u_2+\rho_2\in L^2(D_0)\quad\text{and}\quad {\rm curl\,}\bm{u}_2=0\;\;\text{in\;}D_0,
\end{eqnarray}
which yields that $\bm{u}_2\in X_{a_2}$. This fact motivates us to find the solution
$(u_1,\bm{u}_2)$ of (\ref{2.1}) with the second equation replaced by (\ref{2.3}) on a new product space $H^1(D_0)\times X_{a_2}(D_0)$. By the Green's theorems, it first follows that
\begin{eqnarray*}
&&\int_{D_0}{\rm div\,}(a_2\bm{u}_2)\cdot{\rm div\,}(a_2\bm{v}){\rm d}x
 =\int_{D_0}(b_2u_2+\rho_2){\rm div\,}(a_2\bm{v}){\rm d}x \\
&&\qquad\qquad\qquad\quad=\int_{\partial D_0}a_2 b_2(\nu\cdot\bm{v})u_2{\rm d}s(x)
          -\int_{D_0}a_2 b_2(\bm{v}\cdot\bm{u}_2){\rm d}x
        +\int_{D_0}\rho_2{\rm div\,}(a_2\bm{v}){\rm d}x\\
&&\qquad\qquad\qquad\quad=\int_{\partial D_0}a_2 b_2(\nu\cdot\bm{v})(u_1-f_1){\rm d}s(x)
      -\int_{D_0}a_2 b_2(\bm{v}\cdot\bm{u}_2){\rm d}x
       +\int_{D_0}\rho_2{\rm div\,}(a_2\bm{v}){\rm d}x
\end{eqnarray*}
and
\begin{eqnarray*}
&&\int_{D_0}({\rm div\,}(a_1\nabla u_1)-b_1u_1)\varphi{\rm d}x
=\int_{D_0}\rho_1\varphi{\rm d}x\\
&&\qquad\qquad\qquad\qquad\quad=\int_{\partial D_0}a_1(\partial_{\nu}u_1)\varphi{\rm d}s(x)
       -\int_{D_0}a_1\nabla u_1\cdot\nabla \varphi{\rm d}x-\int_{D_0}b_1 u_1\varphi{\rm d}x\\
&&\qquad\qquad\qquad\qquad\quad=\int_{\partial D_0}a_2(\nu\cdot\bm{u}_2+f_2)\varphi{\rm d}s(x)
       -\int_{D_0}a_1\nabla u_1\cdot\nabla \varphi{\rm dx}-\int_{D_0}b_1 u_1\varphi{\rm d}x
\end{eqnarray*}
for each $\bm{v}\in X_{a_2}$ and
 $\varphi\in H^1(D_0)$, where the integral on $\partial D_0$ is understood in the dual sense of
$\langle\cdot,\cdot \rangle_{H^{-1/2}\times H^{1/2}}$ due to the term
$\nu\cdot\bm{v}\in H^{-1/2}(\partial D_0)$ for any $\bm{v}\in X_{a_2}$.

Combining above equalities shows that the solution $(u_1,u_2)$ to (\ref{2.1})
satisfies following variation formulation
\begin{eqnarray}\label{2.4}
\mathbb{A}(u_1,\bm{u}_2;\varphi,\bm{v})=\mathbb{F}(\varphi;\bm{v}),
\qquad \text{for}\;(\varphi,\bm{v})\in H^1(D_0)\times X_{a_2},
\end{eqnarray}
where the sesquilinear form $\mathbb{A}(\cdot;\cdot)$
 and the functional $\mathbb{F}(\cdot)$ on $H^1(D_0)\times X_{a_2}(D_0)$ are defined by
\begin{eqnarray*}
\mathbb{A}(\varphi,\bm{v}):
&=&\int_{D}({\rm div\,}(a_2\bm{u}_2)\cdot{\rm div\,}(a_2\overline{\bm{v}})
     +a_2 b_2\bm{u}_2\cdot\overline{\bm{v}}){\rm d}x\\
&&+\int_{D}(a_1\nabla u_1\cdot\nabla \overline{\varphi}+b_1u_1\overline{\varphi}){\rm d}x
      -\int_{\partial D}a_2 b_2(\nu\cdot\overline{\bm{v}})  u_1{\rm d}s
      -\int_{\partial D}a_2(\nu\cdot\bm{u}_2)\overline{\varphi}{\rm d}s
 \end{eqnarray*}
 and
 \begin{eqnarray*}
\mathbb{F}(\varphi,\bm{v}):
&=&\int_{D}\rho_2{\rm div\,}(a_2\overline{\bm{v}}){\rm d}x-\int_{D}\rho_1\overline{\varphi}{\rm d}x
       +\int_{\partial D}a_2f_2\overline{\varphi}{\rm d}s-\int_{\partial D}a_2b_2(\nu\cdot\overline{\bm{v}})f_1{\rm d}s,
\end{eqnarray*}
respectively.

Conversely,
on the other hand, we suppose that $(u_1,\bm{u}_2)$ is one solution to (\ref{2.4}). It follows from the condition ${\rm curl\,}\bm{u}_2=0$ that there
exists some function denoted by $u_2$ such that $u_2\in H^1(D_0)$ with $\bm{u}_2=\nabla(u_2+c)$ for $c\in{\mathbb C}$, since $D_0$ is a simply connected, bounded domain.
Following similar arguments to \cite[Theorem 6.5]{CC06}, we can choose some constant $c_0\in{\mathbb C}$
so that $u_2+c_0$ is the solution to (\ref{2.1}). Thus, the solvability of (\ref{2.1}) is equivalent to that of the variational formulation (\ref{2.4}).

\begin{theorem}\label{thm2.1}
Assume $a_j(x),b_j>c_0>0$ for $j=1,2$. Furthermore, if there exists some positive constant $\varepsilon_0>0$ with  $0<\varepsilon_0<1$ such that
\begin{eqnarray*}
\varepsilon_0 b_1>\left( \frac{1+b_2}{2}\right)^2\qquad{\rm and}\qquad
\varepsilon_0\cdot\left(\inf_{x\in D_0}\frac{a_1(x)}{a_2(x)}\right)>\left( \frac{1+b_2}{2}\right)^2,
\end{eqnarray*}
then the variational formulation $(\ref{2.4})$ has a unique solution in the sense that
\begin{eqnarray}\label{2.5}
\|u_1\|_{H^1(D_0)}+\|\textbf{u}_2\|_{X_{a_2}}
\leq C(\|\rho_1\|_{L^2(D_0)}+\|\rho_2\|_{L^2(D_0)}+\|f_1\|_{H^{1/2}(\partial D_0)}
       +\|f_2\|_{H^{-1/2}(\partial D_0)}).
\end{eqnarray}
\end{theorem}
\begin{proof}
Using Schwarz's inequality with the trace theorem, it is first concluded that
\begin{eqnarray*}
&&|\mathbb{F}(\varphi,\textbf{u}_2)|\\
&\leq&\|\rho_1\|_{L^2(D_0)}\|\varphi\|_{L^2(D_0)}
          +\|\rho_2\|_{L^2(D_0)}\|{\rm div\,}(a_2\textbf{v})\|_{L^2(D_0)} \\
&&\quad  +C\|\varphi\|_{H^1(D_0)}\|f_2\|_{H^{-1/2}(\partial D_0)}
  +C(\|a_2 \textbf{v}\|_{L^2(D_0)}
         +\|{\rm div\,}(a_2\textbf{v})\|_{L^2(D_0)})\|f_1\|_{H^{1/2}(\partial D_0)}\\
&\leq& C(\|\rho_1\|_{L^2(D_0)}+\|\rho_2\|_{L^2(D_0)}
               +\|f_1\|_{H^{1/2}(\partial D_0)}
       +\|f_2\|_{H^{-1/2}(\partial D_0)})(\|\varphi\|_{H^1(D_0)}+\|\textbf{u}_2\|_{X_{a_2}}),
\end{eqnarray*}
which yields that $\mathbb{F}(\cdot,\cdot)$ defines a bounded linear functional on the product space
$H^1(D_0)\times X_{a_2}$.

Let $(\varphi;\bm{v}):=(u_1,\bm{u}_2)$. Then, it holds that
\begin{eqnarray}\nonumber
|\mathbb{A}(u_1,\bm{u}_2;u_1,\bm{u}_2)|
&=&\int_{D_0}|({\rm div\,}(a_2\bm{u}_2)|^2+a_2 b_2|\bm{u}_2|^2){\rm d}x
 +\int_{D_0}(a_1|\nabla u_1|^2+b_1|u_1|^2){\rm d}x \\ \label{2.6}
&& -\int_{\partial D_0}a_2(\nu\cdot\bm{u}_2)\overline{u_1}{\rm d}s(x)
 -\int_{\partial D_0}a_2b_2(\nu\cdot\overline{\bm{u}_2}) u_1{\rm d}s(x).
\end{eqnarray}
In (\ref{2.6}), for the integrals on $\partial D$, we apply the divergence theorem to obtain
\begin{eqnarray}\label{2.7}
\int_{\partial D_0}a_2(\nu\cdot\bm{u}_2)\overline{u_1}{\rm d}s(x)
&=&\int_{D_0}{\rm div\,}(a_2\bm{u}_2)\overline{u_1}{\rm d}x+\int_{D_0}a_2\bm{u}_2\cdot\nabla \overline{u}_1{\rm d}x.
\end{eqnarray}
Taking the equality (\ref{2.7}) into (\ref{2.6}) deduces
\begin{eqnarray}\nonumber
|\mathbb{A}(u_1,\bm{u}_2;u_1,\bm{u}_2)|
&\geq&\int_{D_0}(|{\rm div\,}(a_2\bm{u}_2)|^2+b_1|u_1|^2-(1+b_2)|{\rm div\,}(a_2\bm{u}_2)||u_1|){\rm d}x \\ \label{2.7a}
&& +\int_{D_0}a_2\left(|\bm{u}_2|^2+\left(\frac{a_1}{a_2}\right)|\nabla u_1|^2-(1+b_2)|\bm{u}_2||\nabla u_1|\right){\rm d}x.
\end{eqnarray}
For $0<\varepsilon_0<1$, by the mean value inequality, it holds
\begin{eqnarray}\label{2.7b}
&&(1+b_2)|\bm{u}_2||\nabla u_1|
\leq
\varepsilon_0|\bm{u}_2|^2+\frac{(1+b_2)^2}{4\varepsilon_0}|\nabla u_1|^2,\\ \label{2.7c}
&&(1+b_2)|{\rm div\,}(a_2\bm{u}_2)||u_1|
\leq \varepsilon_0|{\rm div\,}(a_2\bm{u}_2)|^2+\frac{(1+b_2)^2}{4\varepsilon_0}|u_1|^2.
\end{eqnarray}
Define three constants $c_4,c_5$ and $c_6$ by
$c_3:=b_1-(1+b_2)^2/(4\varepsilon_0)$, $c_4:=\inf_{x\in D_0}(a_1/a_2)-(1+b_2)^2/(4\varepsilon_0)$
and $c_5:=\min\{(1-\varepsilon_0), c_0(1-\varepsilon_0), c_3, c_0c_4\}$, respectively.
It follows from the assumptions on $a_j,b_j$ for $j=1,2$, that $c_3, c_4$ and $c_5$ are all positive.
Thus, we conclude from (\ref{2.7a})-(\ref{2.7c}) that
\begin{eqnarray*}
|\mathbb{A}(u_1,\bm{u}_2;u_1,\bm{u}_2)|
&\geq &
c_5\int_{D_0}(|{\rm div\,}(a_2\bm{u}_2)|^2+|u_1|^2+
|\bm{u}_2|^2+|\nabla u_1|^2){\rm d}x,
\end{eqnarray*}
which means that $\mathbb{A}(\cdot;\cdot)$ is coercive on the space
$H^1(D_0)\times X_{a_2}$.
Therefore, the existence
of a solution to (\ref{2.4}) with the stable estimate (\ref{2.5}) follows from Lax-Milgram theorem.
\end{proof}

\begin{corollary}\label{thm2.2}
Let $a_j,b_j$ satisfy the conditions in Theorem $\ref{thm2.1}$. Then, the boundary value problem
$(\ref{2.1})$ has a unique solution $(u_1,u_2)\in H^1(D_0)\times H^1(D_0)$ such that
\begin{eqnarray}\label{2.8}
\|u_1\|_{H^1(D_0)}+\|u_2\|_{H^1(D_0)}
\leq C(\|\rho_1\|_{L^2(D_0)}+\|\rho_2\|_{L^2(D_0)}+\|f_1\|_{H^{\frac{1}{2}}(\partial D_0)}
       +\|f_2\|_{H^{-\frac{1}{2}}(\partial D_0)}).
\end{eqnarray}
\end{corollary}
\begin{proof}
(\ref{2.8}) is a direct consequence of Theorem \ref{thm2.1} by using the equality
$u_2={\rm div\,}(\gamma_2\bm{u}_2)-\rho_2\in L^2(D_0)$ in (\ref{2.1}).
\end{proof}

\section{Proofs of Theorem \ref{thm1.1}}\label{sec4}
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This section is devoted to the proof of Theorem \ref{thm1.1} when the conductivity $\gamma$ is described as a piecewise constant function (\ref{1.3a}). In this case, we can show that
both the embedded obstacle $D$ and coefficients $c_j$, $j=1,2,\cdots,N$, in an unknown finite partition of $\Omega\setminus\overline{D}$ can be uniquely
recovered by local boundary measurements $\mathcal{C}_{\gamma,D,\mathcal{B}}$ on an non-empty open subset $\Gamma$ of $\partial \Omega$. A novel and elementary proof
will be presented which depends only on the well-posedness of the boundary value problem
(\ref{2.1}).\\

{\bf Proofs of Theorem \ref{thm1.1}.}
For $j=1,2$, let $S_j:=\{\Omega_{j,\ell}: \ell=1,2,\cdots,N_j\}$ denote the sets consisting of a finite
number of subdomains of $\Omega$ with respect to two classes of different parameters
$(\gamma_1,D_1,\mathcal{B}_1)$ and $(\gamma_2,D_2,\mathcal{B}_2)$. Without of loss of generality, we may assume that
the boundary of $\Omega_{1,1}$ contains a non-empty open subset of
the measurement boundary $\Gamma$. By the assumption on the obstacle $D_j$, $j=1,2$,
there exists two real number $1\leq\kappa_1\leq N_1$ and $1\leq\kappa_2\leq N_2$ such that
$D_1\subseteq\Omega_{1,\kappa_1}$ and $D_2\subseteq\Omega_{2,\kappa_2}$, respectively.
For convenience, we could still assume $\kappa_1,\kappa_2\geq2$.

In the following, we shall first prove $\gamma_2=\gamma_1$ in $\Omega_{1,1}$ by assuming
$\mathcal{C}_{\gamma_1,D_1,\mathcal{B}_1}=\mathcal{C}_{\gamma_2,D_2,\mathcal{B}_2}$.
Since $\Gamma\cap \partial \Omega_{1,1}\not=\emptyset$, we can choose a point $x_*\in (\Gamma\cap\partial\Omega_{1,1})$, and define the domain
$\mathcal{O}_\varepsilon:=B_{\varepsilon}(x_*)\cap \Omega_{1,1}$ with sufficiently small $\varepsilon>0$ so that
both $\gamma_1$ and $\gamma_2$ are all constants in $\mathcal{O}_\varepsilon$, due to the assumptions on
$\gamma_j$, $j=1,2$. Here, $B_{\varepsilon}(x_*)$ denotes a ball centered at $x_*$ with radii $\varepsilon>0$.
We will first prove that $\gamma_2=\gamma_1$ in $\mathcal{O}_\varepsilon$ by contradiction.
Without loss of generality, assume that $\gamma_1>\gamma_2$ in $\mathcal{O}_\varepsilon$.

For $\ell=1,2$, let $u^{(\ell)}_{j}$, $j\in{\mathbb N}_+$, be the solution to (\ref{1.1a}) with the Dirichlet data
$f(x):=f_j(x)=\chi_\delta(x)\Phi_n(x,x_j)$,
corresponding to the case $(\gamma,D,\mathcal{B}):=(\gamma_\ell,D_\ell,\mathcal{B}_\ell)$.
Here, $\Phi_n(\cdot,x_j)$ denotes the fundamental solution of the Laplace equation
$-\triangle \Phi_n(x,x_j) =\delta_{x_j}(x)$ in ${\mathbb R}^n$, given by
\begin{eqnarray*}
\Phi_n(x,x_j)=
\left\{
 \begin{array}{ll}
\displaystyle \;\;\frac{1}{4\pi}\frac{1}{|x-x_j|}, &\text{for}\;n=3,\\[5mm]
\displaystyle 2\pi\ln|x-x_j|,& \text{for}\;n=2,
\end{array}
\right.
\end{eqnarray*}
$\chi_\delta$ is a smooth cut-off function which is supported in $B_\delta(x_*)$ and satisfies
$|\chi_\delta(x)|\leq 1$ for $x\in{\mathbb R}^n$ and $\chi_\delta(x)=1$ for $x\in B_{\delta/2}(x_*)$,
and the point sequence $x_j$ are defined by
\begin{eqnarray*}
x_j: = x_* + \frac{1}{j}\nu(x_*),\qquad j=J_0,J_0+1,J_0+2,\cdots\cdots,
\end{eqnarray*}
where $J_0>0$ is chosen such that $x_j\in{\mathbb R}^n\setminus\overline{\Omega}$ for all $J_0+{\mathbb N}_+$.
Obviously, one has $f_j = 0$ on $\partial\Omega\setminus\overline{\Gamma}$, if $\delta>0$ is sufficiently small.
This means that
$(f_j|_{\Gamma}, u_j^{(1)}|_{\Gamma}) \in \mathcal{C}_{\gamma_1,D_1,\mathcal{B}_1}$ and
$(f_j|_{\Gamma}, u_j^{(2)}|_{\Gamma})\in \mathcal{C}_{\gamma_2,D_2,\mathcal{B}_2}$ for all large $j\geq J_0$.
Thus, we conclude by $\mathcal{C}_{\gamma_1,D_1,\mathcal{B}_1}=\mathcal{C}_{\gamma_2,D_2,\mathcal{B}_2}$ that $\Lambda_{\gamma_1,D_1,\mathcal{B}_1}f_j = \Lambda_{\gamma_2,D_2,\mathcal{B}_2}f_j$ on $\Gamma$
for all large $j\geq J_0$.. Then, we consider the following boundary value problem
\begin{eqnarray}\label{5.2}
\left\{
 \begin{array}{ll}
{\rm div\,}(\gamma_1 \nabla \Psi_1)-2 \Psi_1=\zeta_1,& \text{in}\;\mathcal{O}_\varepsilon\\
{\rm div\,}(\gamma_2 \nabla \Psi_2)- \Psi_2=\zeta_2,& \text{in}\; \mathcal{O}_\varepsilon\\
\Psi_1-\Psi_2=\varrho_1, & \text{on}\;\partial \mathcal{O}_\varepsilon\\
\gamma_1\partial_\nu \Psi_1- \gamma_2\partial_\nu \Psi_2=\varrho_2,
             &\text{on}\;\partial \mathcal{O}_\varepsilon\\
\end{array}
\right.
\end{eqnarray}
where $\zeta_1:=-2u^{(1)}_j$, $\zeta_2: = -u^{(2)}_j$,
$\varrho_1: = u^{(1)}_j-u^{(2)}_j$ and $\varrho_2:=\gamma_1u^{(1)}_j-\gamma_2u^{(2)}_j$.
We intend to prove the well-posedness of (\ref{5.2}) by utilizing Corollary \ref{thm2.2}, if
$\zeta_1,\zeta_2,\varrho_1$ and $\varrho_2$ can be shown to be bounded uniformly
for $j\geq J_0$ in the corresponding function spaces.

{\bf $\bullet$ The boundedness of $\zeta_1$ and $\zeta_2$.}
Since $\gamma_1$ is constant in $\mathcal{O}_\delta$, we can thus consider
the following Dirichlet problem
\begin{eqnarray}\label{5.3}
\Delta \Theta_j = 0\quad{\rm in\;}\Omega\setminus\overline{D_1},\qquad
\Theta_j=f_j\quad{\rm in\;}\partial\Omega\quad{\rm and}\quad
\mathcal{B}_1\Theta_j=0\quad{\rm on\;}\partial D_1.
\end{eqnarray}
It is noticed that $f_j(\cdot)=\Phi_n(\cdot,x_j)$ on $(B_{\delta/2}(x_*)\cap\partial\Omega)$, which
yields that the difference $\Pi_j(\cdot): = \Theta_j(\cdot)-\Phi_n(\cdot,x_j)$ in
$\Omega\setminus\overline{D_1}$ is bounded uniformly for all $j\geq J_0$ under the classical $H^1$-norm. This is because $\Pi_j(\cdot)$
solves the harmonic equation $\Delta \Pi_j = 0$ in $\Omega\setminus\overline{D_1}$ with the boundary
data $\Pi_j (\cdot)= \Theta_j(\cdot)-\Phi_n(\cdot;x_j)$ vanishing for $x\in (B_{\delta/2}(x_*)\cap\partial\Omega)$. Thus, the variational method can be applied to prove the above
assertion. By the definition of $\Phi_n(\cdot,x_j)$, one has
\begin{eqnarray}\label{5.4}
\|\Theta_j \|_{L^2(\Omega\setminus\overline{D_1})}
+\|\Theta_j \|_{H^1(\Omega\setminus\overline{D_1\cup B_\varepsilon(x_*)})}
\leq C,\quad {\rm for\; sufficiently\;}\varepsilon>0,
\end{eqnarray}
where $C>0$ is independent of the choice of $j\geq J_0$. The estimate (\ref{5.4})
leads in a similar way to that the functions $u_j^{(1)}$ have the same property
\begin{eqnarray}\label{5.5}
\|u_j^{(1)} \|_{L^2(\Omega\setminus\overline{D_1})}
+\|u_j^{(1)} \|_{H^1(\Omega\setminus\overline{D_1\cup B_\varepsilon(x_*)})}
\leq C
\end{eqnarray}
uniformly for all $j\geq J_0$ by considering the difference $u^{(1)}_j(\cdot)-\Theta_j(\cdot)$
which satisfies the equation ${\rm div\,}(\gamma_1\nabla w)=-{\rm div\,}((\gamma_1-\gamma_1|_{\Omega_{1,1}})\nabla\Theta_j)$ in
the domain $\Omega\setminus\overline{D_1}$ in the distribution sense with the homogeneous
boundary data $w=0$ on $\partial \Omega$ and $\mathcal{B}_1w=0$ on $\partial D_1$.
Hence, one has from (\ref{5.5}) that $\zeta_1$ is bounded in $L^2(\mathcal{O}_\delta)$ uniformly all $j\geq J_0$. The same assertion also holds for $\zeta_2$ by a similar discussion.

{\bf $\bullet$ The boundedness of $\varrho_1$ and $\varrho_2$.}
First,  by $\Lambda_{\gamma_1,D_1,\mathcal{B}_1}f_j = \Lambda_{\gamma_2,D_2,\mathcal{B}_2}f_j$ on $\Gamma$,
it is known that $\gamma_1u_j^{(1)} = \gamma_2u_j^{(1)}$ on $\Gamma$. Consequently,
$\varrho_1=0$ and $\varrho_2=0$ on $\Gamma$. We then construct a smooth function
$\eta\in C^\infty({\mathbb R}^n)$ satisfying ${\rm supp}\eta\subseteq B_\delta(x_*)$, $|\eta|\leq1$ in ${\mathbb R}^n$ and $\eta=1$ in $B_{\delta/2}(x_*)$, where $\delta>0$ is sufficiently small such that
$(B_\delta(x_*)\cap\Gamma)\subseteq\Gamma$.
Define
\begin{eqnarray*}
&&\mathbb{H}_1(x): = (1-\eta(x))(u_j^{(1)}(x)-u_j^{(2)}(x)),\quad {\rm in\;} \mathcal{O}_\varepsilon,\\
&&\mathbb{H}_2(x): = (1-\eta(x))(\gamma_1u_j^{(1)}(x)-\gamma_2u_j^{(2)}(x)),\quad {\rm in\;} \mathcal{O}_\varepsilon.
\end{eqnarray*}
It is easy to check that $\varrho_1 = {\rm Tr}_0\mathbb{H}_1$ and
\begin{eqnarray}\label{5.6}
\varrho_2 = {\rm Tr}_1\mathbb{H}_2 - (\partial_\nu(1-\eta))(\gamma_1u_j^{(1)}(x)-\gamma_2u_j^{(2)}(x)),
\end{eqnarray}
where ${\rm Tr}_\ell$, $\ell=1,2$, denote the trace operators defined by
${\rm Tr}_0 v = v|_{\partial\mathcal{O}_\varepsilon}$ and ${\rm Tr}_1 v = (\partial_\nu v)|_{\partial\mathcal{O}_\varepsilon}$ for a smooth function $v$.
Due to $1-\eta=0$ in $B_{\delta/2}(x_*)$, it follows from (\ref{5.5}) that
\begin{eqnarray}\label{5.7}
\|{\rm Tr}_1\mathbb{H}_2-\varrho_2\|_{H^{-1/2}(\partial\mathcal{O}_\varepsilon)}+
\|\mathbb{H}_1\|_{H^1(\mathcal{O}_\varepsilon)} +\|\mathbb{H}_2\|_{H^1(\mathcal{O}_\varepsilon)}
\leq C_\eta
\end{eqnarray}
for all $j\geq J_0$. Furthermore, using the equations
$\Delta u^{(\ell)}_j=0$, one has
$\Delta \mathbb{H}_2=\Delta(1-\eta)(\gamma_1u_j^{(1)}-\gamma_2u_j^{(2)})
+2\nabla(1-\eta)\cdot\nabla(\gamma_1u_j^{(1)}-\gamma_2u_j^{(2)})$
which leads to that
\begin{eqnarray}\label{5.8}
\|\Delta \mathbb{H}_2\|_{L^2(\mathcal{O}_\varepsilon)} \leq C_\eta.
\end{eqnarray}
With the aid of the trace theorems, we arrive from (\ref{5.7})-(\ref{5.8}) at that
\begin{eqnarray}\label{5.9}
\|\varrho_1\|_{H^{1/2}(\partial\mathcal{O}_\varepsilon)}
+\|\varrho_2\|_{H^{-1/2}(\partial\mathcal{O}_\varepsilon)}
\leq C_\eta,
\end{eqnarray}
which finishes the second assertion.

In (\ref{5.2}), let $a_1:=\gamma_1$, $a_2:=\gamma_2$, $b_1:=2$ and $b_2:=1$. Obviously,
$a_1,a_2,b_1,b_2$ satisfy the conditions in Theorem \ref{thm1.1},
due to $\gamma_1>\gamma_2$ in $\mathcal{O}_\varepsilon$. It is then shown by Corollary \ref{thm2.2} that
\begin{eqnarray}\label{5.10}
\sum_{\ell=1}^2\|\Psi_\ell\|_{H^1(\mathcal{O}_\varepsilon)}
\leq C(\|\zeta_1\|_{L^2(\mathcal{O}_\varepsilon)}+\|\zeta_2\|_{L^2(\mathcal{O}_\varepsilon)}
+\|\varrho_1\|_{H^{1/2}(\partial\mathcal{O}_\varepsilon)}
+\|\varrho_2\|_{H^{-1/2}(\partial\mathcal{O}_\varepsilon)})
\leq C
\end{eqnarray}
uniformly for $j\geq J_0$.
Notice that, for boundary data defined in (\ref{5.2}), $(\Psi_1,\Psi_2):=(u_j^{(1)},u_j^{(1)})$
is the unique solution to (\ref{5.2}). Using the trace theorem again, it follows from the above
equality that
\begin{eqnarray*}
+\infty\longleftarrow\|f_j\|_{H^{1/2}(\partial\mathcal{O}_\varepsilon\cap\Gamma)}
\leq\|u_j^{(1)}\|_{H^1(\mathcal{O}_\varepsilon)}\leq C,\quad{\rm as\;}j\to\infty,
\end{eqnarray*}
which clearly leads to a contradiction. Hence, $\gamma_1=\gamma_2$ in $\mathcal{O}_\varepsilon$.

If $\Omega_{1,1}\subseteq \Omega_{2,\ell_0}$ for some $j_0\leq N_2$, then one immediately has
$\gamma_2=\gamma_1$ in $\Omega_{1,1}$. Otherwise, we shall claim that it must hold
\begin{eqnarray}\label{5.1}
\overline{\Omega_{1,1}} &=& \bigcup_{1\leq j\leq M}\overline{\Omega_{2,\ell_j}}\quad {\rm for\;some\;}M\leq N_2,
\end{eqnarray}
which means that $\Omega_{1,1}$ consists of a finite number of elements in $S_2$.
We shall prove (\ref{5.1}) by contradiction.
Assume that it does not hold true for (\ref{5.1}). Then, there exists some subdomain denoted by
$\Omega_{2,j_1}$ in $ S_2$ such that
$(\Omega_{1,1}\cap \Omega_{2,j_1})\not=\emptyset$, $\Omega_{1,1}\nsubseteq \Omega_{2,j_1}$ and
$\Omega_{2,j_1}\nsubseteq \Omega_{1,1}$.
So we can choose two subsets $A_1$ and $A_2$ of ${\mathbb R}^n$ satisfying the following conditions:
${\rm i})$  $A_1\subseteq(\Omega_{1,1}\cap \Omega_{2,j_1})$; ${\rm ii})$ $A_2\subseteq(\Omega_{1,2}\cap \Omega_{2,j_1})$ with $\emptyset\not=\partial A_2\cap(\partial\Omega_{1,1}\cap\partial\Omega_{1,2})$
having a non-empty open subset, denoted by $\Sigma$, in the low-dimensional manifold of ${\mathbb R}^n$. Here, $\Omega_{1,2}$ denotes the element in $S_1$ which shares a non-empty common partial boundary with $\Omega_{1,1}$.
We now take a point $p$ at the boundary $\partial A_1\cap \partial\Omega_{2,j_1}$. Recalling
the assumption on the domain $\Omega\setminus\overline{D}_2$, we can always find a bounded curve $\ell_{x_*,p}\subseteq \Omega_{1,1}$ with two endpoints $x_*$ and $p$ such that $\ell_{x_*,p}\cap A$ ($A\in S_2$) contains at most one point. Therefore, there exists a finite number of small balls $B_{\varepsilon_1}(p_j)$ ($1\leq j\leq M_1<\infty$) in $\Omega_{1,1}$ with $p_j\in\ell_{x_*,p}$, $p_1=x_*$, $p_{M_1}=p$ and sufficiently small
$\varepsilon_1>0$ such that $B_{\varepsilon_1}:=\bigcup_{1\leq j\leq M_1}B_{\varepsilon_1}(p_j)\supset \ell_{x_*,p}$.

Since $\gamma_1$ is constant in $\Omega_{1,1}$, we let $\gamma_1(x):=c_{1,1}\in{\mathbb R}_+$ in $\Omega_{1,1}$.
We next claim that $\gamma_2(x)=c_{1,1}$ in $A'_1: = A_1\cap B_{\varepsilon_1}(p)$. Without loss of generality,
it can be assumed that $\gamma_2(x) =c_{1,1}$ in $B_{\varepsilon_1}\setminus\overline{\Omega_{2,j_1}}$.
For $\ell=1,2$, we now consider the Dirichlet-Green functions $G_\ell(\cdot, y)$, $y\in(\Omega\setminus\overline{D_\ell})$,
which solves the following boundary value problem
 \begin{eqnarray*}
\left\{
 \begin{array}{ll}
{\rm div\,}(\gamma_\ell\nabla G_\ell(\cdot,y)) = -\delta_y(\cdot)& \text{in}\;\Omega\setminus\overline{D_\ell},\\[1.5mm]
 G_\ell(\cdot,y)=0& \text{on}\; \partial \Omega,\\ [1.5mm]
\mathcal{B}_\ell G_\ell(\cdot,y)=0& \text{on}\;\partial D_\ell.
\end{array}
\right.
\end{eqnarray*}
It is easily checked that $G_\ell(\cdot,y)$ has the same singularity with the fundamental solution
$\Phi_n(\cdot,y)$ at $x=y$.
For any $f\in H^{1/2}(\partial\Omega)$ with ${\rm supp}f\subseteq \Gamma_1:=(\Gamma\cap\partial\Omega_{1,1})$,
let $u^{(\ell)}$, $\ell=1,2$, be the solutions to (\ref{1.1a}) with the boundary data $f$, corresponding to
two classes of different parameters $(\gamma_1,D_1,\mathcal{B}_1)$ and $(\gamma_1,D_2,\mathcal{B}_2)$.
Using the representation formulation for $u^{(\ell)}$, $\ell=1,2$, yields
 \begin{eqnarray}\nonumber
 u^{(\ell)}(x)
 &=&\int_{\partial \Omega}\gamma_\ell(y)\left(u^{(\ell)}(y)\frac{\partial G_\ell(x,y)}{\partial\nu(y)}
 -\frac{\partial u^{(\ell)}(y)}{\partial\nu(y)}G_\ell(x,y)\right){\rm d}s(y)\\ \label{5.11}
 &=&\int_{\Gamma_1}\frac{\partial G_\ell(x,y)}{\partial\nu(y)}\gamma_\ell(y) f(y){\rm d}s(y)\qquad\quad {\rm for\;}x\in B_{\varepsilon_1}(p)\setminus\overline{\Omega_{2,j_1}}.
 \end{eqnarray}
Recalling $\Lambda_{\gamma_1,D_1,\mathcal{B}_1}f = \Lambda_{\gamma_2,D_2,\mathcal{B}_2}f$, the unique continuation principle can be thus applied for the Laplace equation to deduce $u^{(1)}(x)=u^{(2)}(x)$ in
$\mathcal{O}_\varepsilon \cap B_{\varepsilon_1}(x_*)$. Then, it holds by turns that $u^{(1)}(x)=u^{(2)}(x)$ in
$B_{\varepsilon_1}\setminus\overline{\Omega_{2,j_1}}$.
Thus, it is shown by (\ref{5.11}) in combination with the fact $\gamma_1(x)=\gamma_2(x)$ in $\mathcal{O}_\varepsilon$ that
\begin{eqnarray}\label{5.12}
\int_{\Gamma_1} \frac{\partial G_1(x,y)}{\partial\nu(y)} f(y) {\rm d}s(y)
=\int_{\Gamma_1} \frac{\partial G_2(x,y)}{\partial\nu(y)}f(y) {\rm ds}(y)
\end{eqnarray}
 for $x\in B_{\varepsilon_1}(p)\setminus\overline{\Omega_{2,j_1}}$ and all $f\in H^{1/2}(\partial\Omega)$ with the support in
 $\Gamma_1$. This means that $\partial_\nu G_1(x,\cdot)=\partial_\nu G_2(x,\cdot)$ on $\Gamma_1$. Moreover,
 one can also concalud from the similar discussion above (\ref{5.12}) that
  \begin{eqnarray}\label{5.13}
 G_1(x,y)=G_2(x,y)\qquad {\rm for\;}x, y\in B_{\varepsilon_1}\setminus\overline{\Omega_{2,j_1}} \;{\rm and \;}\; x\neq y.
 \end{eqnarray}

We are at a position to show that $\gamma_2(x) = c_{1,1}$ in $A'_1$, which will be proved by contradiction.
Assume that $\gamma_2(x)\neq c_{1,1}$ in $A'_1$. Without loss of generality, we may assume
$\gamma_1>\gamma_2$ in $A'_1$. Similar to (\ref{5.2}),
we instead consider the following boundary value problem
 \begin{eqnarray}\label{5.14}
\left\{
 \begin{array}{ll}
{\rm div\,}(\gamma_1 \nabla \Psi_1)-2 \Psi_1=\zeta_1,& \text{in}\;A'_1\\
{\rm div\,}(\gamma_2 \nabla \Psi_2)- \Psi_2=\zeta_2,& \text{in}\; A'_1\\
\Psi_1-\Psi_2=\varrho_1, & \text{on}\;\partial A'_1\\
\gamma_1\partial_\nu \Psi_1- \gamma_2\partial_\nu \Psi_2=\varrho_2,
             &\text{on}\;\partial A'_1\\
\end{array}
\right.
\end{eqnarray}
with $\zeta_1,\zeta_2,\varrho_1$ and $\varrho_2$ defined by
\begin{eqnarray*}
&&\zeta_1:=-2G_1(\cdot,x_j),\quad\zeta'_2:=-G_2(\cdot,x_j),\\
&&\varrho_1:=G_1(\cdot,x_j)-G_2(\cdot,x_j),\quad
\varrho_2:=\gamma_1\partial_\nu G_1(\cdot,x_j)-\gamma_2\partial_\nu G_2(\cdot,x_j),
\end{eqnarray*}
respectively,
where $x_j$ are defined by $x_j:=p+(1/j)\nu(p)\in B_{\varepsilon_1}(p)\setminus\overline{\Omega_{2,j_1}}$ for all large $j>0$.

For $\zeta_j, j=1,2,$ it follows from
the properties of the Dirichlet-Green's function (cf.\cite{Is90}) that there exists a fixed constant $K>0$, depending only on the coefficients of the conductivity equation, such that
$K^{-1}\Phi_n(x,y) \leq G_1(x,y),G_2(x,y)\leq K\Phi_n(x,y)$ for all $x,y\in A'_1$.
This leads to that $\zeta_1$ and $\zeta_2$ are bounded in $L^2(A'_1)$
uniformly for sufficiently large $j>0$.

Furthermore, by the equality (\ref{5.13}) and transmission conditions on $\partial A'_1\cap\partial\Omega_{2,j_1}$, we observe that $\varrho_1=0$ and $\varrho_2=0$ on $\partial A'_1\cap\partial\Omega_{2,j_1}$.
Thus, in order to obtain the
uniform boundedness of $\varrho_1$ and $\varrho_2$ in $H^{1/2}(\partial A'_1)$
and $H^{-1/2}(\partial A'_1)$, respectively,
 it is enough to estimate $G_1(\cdot,x_j)$ and $G_2(\cdot,x_j)$ in the sense of
$H^1(A'_1\setminus\overline{B_{\varepsilon'_1}(p)})$ with $\varepsilon'_1=\varepsilon_1/2$ (see the analysis between (\ref{5.6}) and (\ref{5.9})). Obviously, this holds true due to the properties of the Dirichlet-Green's functions.
Corollary \ref{thm2.2} can be now applied to obtain the estimate
\begin{eqnarray}\label{5.15}
\|\Psi_1\|_{H^1(A'_1)} +\|\Psi_2\|_{H^1(A'_1)} \leq C
\end{eqnarray}
uniformly for sufficiently large $j>0$. Noticing that for each $j>0$,
$(\Psi_1,\Psi_2):=(G_1(\cdot,x_j),G_2(\cdot,x_j))$ is the unique solution to (\ref{5.14}), one has from (\ref{5.15}) that $G_1(\cdot,x_j)$ is bounded in $H^1(A'_1)$ uniformly for all sufficiently $j>0$.
However, this is contradicted with the fact
\begin{eqnarray*}
\|G_1(\cdot,x_j)\|_{H^1(A'_1)}\to\infty\qquad {\rm as \;}j\to\infty.
\end{eqnarray*}
Therefore, one has $\gamma_2(x)=c_{1,1}$ in $A'_1$, which further yields $\gamma_2(x)=c_{1,1}$ in $\Omega_{2,j_1}$,
since $A'_1\subset A_1\subset (\Omega_{1,1}\cap \Omega_{2,j_1})$ and $\gamma_2(x)$ is constant in $\Omega_{2,j_1}$.

Based on the above assertion, we further claim that  $\gamma_1(x)=\gamma_2(x)$ in $A_2$, a previously chosen subdomain in condition ${\rm ii})$ after (\ref{5.1}), which is sill proved by contradiction.
Suppose that $\gamma_1\not=\gamma_2$ in $A_2$. Without loss of generality, we can assume that $\gamma_1>\gamma_2$ in
$A_2$. We now take a new point $p_*\in\Sigma$ (see the definition
in condition ${\rm ii})$), and define the point sequence $y_j: = p_*+(1/j)\nu(p_*)$ for sufficiently large $j>0$ such that $y_j\in(\Omega_{1,1}\cap B_{\varepsilon_2}(p_*))$ for sufficiently small $\varepsilon_2>0$.
Following similar discussions as in deriving (\ref{5.13}), we can first prove that
\begin{eqnarray}\label{5.16}
G_1(x,y_j)=G_2(x,y_j)\qquad {\rm for\;}x_j\not=x\in \Omega_{1,1}\cap B_{\varepsilon_2}(p_*).
\end{eqnarray}
 for each large $j>0$,
This motivates us to consider another new boundary value problem in the subdomain $A'_2:=A_2\cap B_{\varepsilon_2}(p_*)$: \begin{eqnarray}\label{5.17}
\left\{
 \begin{array}{ll}
{\rm div\,}(\gamma_1 \nabla \Psi_1')-2 \Psi_1'=\zeta_1',& \text{in}\;A'_2\\
{\rm div\,}(\gamma_2 \nabla \Psi_2')- \Psi_2'=\zeta_2',& \text{in}\; A'_2\\
\Psi_1'-\Psi_2'=\varrho_1', & \text{on}\;\partial A'_2\\
\gamma_1\partial_\nu \Psi_1'- \gamma_2\partial_\nu \Psi_2'=\varrho_2',
             &\text{on}\;\partial A'_2\\
\end{array}
\right.
\end{eqnarray}
which is similar to (\ref{5.14}) with $A_1'$ and $\zeta_1,\zeta_2,\varrho_1,\varrho_2$ replaced by
$A_2'$ and
\begin{eqnarray*}
&&\zeta'_1:=-2G_1(\cdot,y_j),\quad\zeta'_2:=-G_2(\cdot,y_j),\\
&&\varrho'_1:=G_1(\cdot,y_j)-G_2(\cdot,y_j),\quad
\varrho'_2:=\gamma_1\partial_\nu G_1(\cdot,y_j)-\gamma_2\partial_\nu G_2(\cdot,y_j),
\end{eqnarray*}
respectively.

Using the properties on the Dirichlet-Green's functions again yields that both $\zeta_1'$ and $\zeta_2'$ are
bounded in $L^2(A_2')$ uniformly for all large $j>0$. Moreover, (\ref{5.16}) in combination with the transmission conditions on $\Sigma$ leads to $\varrho_1'=\varrho_2=0$ on $\partial A_2'\cap\Sigma$.
This makes that $\varrho_1'$ and $\varrho_2'$ are bounded in $H^{1/2}(\partial A'_2)$
and $H^{-1/2}(\partial A'_2)$, respectively, uniformly for all large $j>0$; see the analysis between (\ref{5.14})
and (\ref{5.15}). Corollary \ref{thm2.2} can be again applied to obtain the estimate
\begin{eqnarray}\label{5.18}
\|\Psi_1'\|_{H^1(A'_2)} +\|\Psi_2'\|_{H^1(A'_2)} \leq C
\end{eqnarray}
uniformly for sufficiently large $j>0$. However, this is a contradiction since $(\Psi_1',\Psi_2'):=(G_1(\cdot,y_j),G_2(\cdot,y_j))$ is the unique solution to (\ref{5.17}) and
\begin{eqnarray*}
\|G_1(\cdot,y_j)\|_{H^1(A'_2)}\to\infty,\qquad {\rm as \;}j\to\infty.
\end{eqnarray*}
Therefore, we arrive at that $\gamma_1=\gamma_2$ in $A_2$. Notice that this is impossible since
$\gamma_2$ has been proved to be $c_{1,1}$ in $A_2$ and $\gamma_1:=c_{1,2}$, a new constant, is assumed to
be not $c_{1,1}$ (see the conditions between (\ref{1.3a}) and (\ref{1.4a})). Hence, (\ref{5.1}) holds true.

It still needs to be shown that $\gamma_2=c_{1,1}$ in $\Omega_{1,1}$. By (\ref{5.1}), we can easily choose
some subdomain denoted by $\Omega_{2,l_1}$ such that $\Omega_{2,l_1}\subseteq\Omega_{1,1}$ and
$\Gamma\cap\partial\Omega_{2,l_1}\not=\emptyset$. That is, $\partial\Omega_{2,l_1}$ and $\Gamma$ share a non-empty
open subset in a ($n-1$)-dimensional manifold of ${\mathbb R}^n$. By repeating the above process, one can conclude
$\gamma_2=c_{1,1}$ in $\Omega_{2,l_1}$. If $\Omega_{1,1}=\Omega_{2,l_1}$, then the proof is ended. If not, we can
choose another subdomain denoted by $\Omega_{2,l_2}$ such that $\Omega_{2,l_2}\subseteq\Omega_{1,1}$
and $\emptyset\not=\partial\Omega_{2,l_1}\cap\partial\Omega_{2,l_2}$ containing a non-empty open subset
in ($n-1$)-dimensional manifold of ${\mathbb R}^n$. Again repeating the previous analysis leads to
$\gamma_2=c_{1,1}$ in $\Omega_{2,l_2}$. If $\Omega_{1,1}=(\Omega_{2,l_1}\cup\Omega_{2,l_2})$, then the proof is ended.
If not, we continue to choose a new subdomain denoted by $\Omega_{2,l_3}$ such that
$\Omega_{2,l_3}\subseteq\Omega_{1,1}$, and $\emptyset\not=\partial\Omega_{2,l_3}\cap\partial\Omega_{2,l_2}$
or $\emptyset\not=\partial\Omega_{2,l_3}\cap\partial\Omega_{2,l_1}$ contains at least a non-empty open subset
in ($n-1$)-dimensional manifold of ${\mathbb R}^n$; see Figure 2 for a geometry setting. The similar arguments
leads to $\gamma_2=c_{1,1}$ in $\Omega_{2,l_3}$ once again. Thus, we can arrive in turn at
$\gamma_2=c_{1,1}$ in $\Omega_{1,1}$ since $M$ in (\ref{5.1}) is just a bounded positive integer.

Finally, the inverse conductivity problem can be reduced in a similar way to the unique determination of an unknown obstacle
embedded into a known background medium. The standard arguments as in \cite{KP98} implies that
$D_1=D_2$ and $\mathcal{B}_1=\mathcal{B}_2$. The proof is thus complete.

\section*{Acknowledgements}

This work is supported by the NNSF of China under Grant No. 11771349.

redpajama_set_name: RedPajamaArXiv


text:
ARK: Survival Evolved goes full Hunger Games with a F2P spin-off
By Matthew Figueira
Despite ARK: Survival Evolved not even being officially out yet, its creators, Studio Wildcard, have gone ahead and unveiled a spin-off title. It's called Survival of the Fittest, and it looks like it's taken a whole lot of inspiration from The Hunger Games franchise.
Welcome to ARK: Survival of the Fittest, the first ever M.O.S.A. – a Multiplayer Online Survival Arena – that Studio Wildcard designed for the burgeoning wild west of eSports. A spin-off from the most popular open-world Early Access game on Steam ARK: Survival Evolved, ARK: Survival of the Fittest (SotF) pits up to 72 combatants in an action-packed struggle for survival where players are ultimately pushed into an epic final showdown where only one "Tribe" will make it out alive.
How rad? Survival of the Fittest looks chaotic, competitive, and like a whole lot of fun. I VOLUNTEER AS TRIBUTE!
How did this game come about though? Wildcard are still working on ARK: Survival Evolved, aren't they?
ARK: Survival of the Fittest (SotF) originated as a spin-off game mode which came to be with the release of our development kit. The mode was first available as a Total Conversion game system and debuted worldwide with a 70-person slaughter fest of a tournament. The game mode consists of a large-scale battle which takes place on a modified version of the primal ARK, with the objective being to eliminate all other survivors before they can eliminate you.
Following the success of the first tournament, the game mode was made available to the community and had constant iteration right through til the end of 2015. It was clear that a vast number of survivors, not only enjoyed playing the game mode, but they also enjoyed watching it. The ARK: SotF Team wants to work alongside the community, to make this newly refactored game into an epic dinosaur survival arena.
We already have experience working with in an early-access environment, we know what it entails, and with SotF being born from early-access, it only makes sense for it to continue its growth in this program but on a much grander scale, with it's own team, it's own game and it's own community. Learning from experiences in ARK: Survival Evolved, we know through constant iteration and interaction with the community we can truly develop something special, fun and competitive for all survivors out there!"
So yeah, Wildcard essentially just split themselves up. They will have an entirely separate team working on Survival of the Fittest.
Know what's the best part? It's entirely free-to-play, and it's out now! Will you be downloading it?
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Defence of the Ancients? More like Defence of the cabbages! Have you seen my head? I look like a Merino Sheep on pole. NO SHANGE only SHAPPIES! :D
Rush 2019 – A great showing for local esports
The "excitable humans from NAG" as they so lovingly describe themselves are no strangers t…
Destiny 2: Shadowkeep will be around the same size as Rise of Iron
Local weekend box office – Dave Bautista and Chucky get beaten badly by Spider-Man
Weekend box office – Spider-Man swings high above Crawl and Stuber
Sony's next-generation PlayStation VR will be wireless according to a new patent

redpajama_set_name: RedPajamaCommonCrawl


Cartoon Network ; Hanna-Barbera Productions ; Media Asia Films ; Rough Draft Studios.
Burbank, CA : Warner Home Video,  10th anniversary collection.
screenplay, Sadayuki Murai ; producer, Shojiro Abe ; director, Ryutaro Nakamura.
Presented by Terror in Resonance The Animation Committee, Fuji Television Network, Aniplex, Kyoraku Industrial Holdings Co., Dentsu Inc. ; producers, Makoto Kimura, Takamitsu Inoue ; director: Shinichiro Watanabe ; English version by Funimation Entertainment.
Starcom, the U.S. space force : the complete series.
[United States] : distributed by Mill Creek Entertainment, 2015.
Burbank, CA : Distributed by Warner Home Video, [2008?] [Standard format].
Tallahassee, FL : A2ZCDS.COM, c2004.
[United States] : Family Home Entertainment ; Santa Monica, CA : Artisan Entertainment, 2004.
Reboot [videorecording] : the complete series bonus features.
[United States] : Shout! Factory, 2011.
directed by Zondag Entertainment ; executive producers, Steve Barron, Jay Firestone ; produced by Christopher Brough.
Los Angeles : Shout! Factory, c2011.
produced by Bandai Visual, Studio Pierrot, Movic ; English version produced by Pioneer Entertainment (USA) Inc., in association with ZRO Limit Productions and Animaze, Inc. ; original story, Yu Watase ; director, Hajime Kamegaki ; English version producer, Hiroe Tsukamoto.
[United States] : Pioneer Entertainment (USA), ©2002.
The Gumby show. The complete 50s series.
English version by FUNimation Productions, Ltd. ; producers, Carly Hunter, Justin Cook ; producers: Shinichiro Kobayashi ... [et al.] ; director: Sayo Yamamoto.


The man allegedly told police he was dealing to pay bills.
Scott M. Vaughn Jr. Photo by the Dearborn County Sheriff's Office.
(Lawrenceburg, Ind.) - A Sunman man is accused of dealing in methamphetamine.
Scott M. Vaughn Jr., 24, was arrested on Tuesday and charged with Dealing in Methamphetamine at Least Five Grams But Less Than 10 Grams (Level 3 felony), Possession of Methamphetamine (Level 6 felony), Possession of a Syringe (Level 6 felony) and Possession of Paraphernalia (Class C misdemeanor).
According to a probable cause affidavit, Vaughn dealt approximately 6.44 grams of methamphetamine in the Kroger parking lot in Lawrenceburg.
Aurora Police say Vaughn was also in possession of a syringe, a clear plastic baggie containing a clear crystal substance, numerous empty plastic baggies, a piece of straw and $300 cash.
During questioning at the Dearborn County Law Enforcement Center, Vaughn allegedly stated he was only selling methamphetamine to pay bills.


Douglasdale, DouglasGlen and Quarry Park comprise a large area of the southeast sector of Calgary. The community encompasses both sides of Deerfoot from the Bow River to 130th Avenue SE and includes Douglasdale Estates, DouglasGlen, Douglas Ridge, Quarry Park and Douglasdale Meadows as well as the DouglasGlen business park on the east side of the Deerfoot and south of 130th Avenue SE. The community accommodates approximately 12,000 residents, over 4,000 homes and more than 150 businesses!
The DouglasQuarry Community Association ("DQCA") is a non-profit organization run entirely by volunteers.
The DQCA's vision is to be a vibrant community that serves all residents, and is fuelled by the generosity of spirit.
The DQCA's values are the following: volunteerism, open communication, honesty, mutual respect, multiculturalism, social justice, creativity, quality of life, and foremost – community building.


For larger orders we offer laser gravure marking on our olive wood products. Please contact us for an individual offer.
Here you find an article about our products in the German magazine "Mein schönes Landhaus" - Burda Verlag (Release 4/2013).
Please also read the article about the berondi-olive wood honey spoon in the magazine "Living & More" (Release 1/2014).


MCBarb, I just received notice of my payment via Paypal at 9:02 PM 1/15. I like the new report better than before.
You are right. It is there. I agree abt the report being easier. I was waiting to see how we were paid before taking anymore.
I have done three of these shops, and was paid on time. There was a bit of a dust up at the start about getting the forms, reimbursements, and shop specifications nailed down, but that appears to have settled. I am scheduled for two more this month.
It is a violation of the ICA that you sign with each company to disclose what company performs what shop.
Does anybody know if we are also reimbursed if it costs more at an authorized outlet. I just did one that charged $7 (!) over the regular postage amount.
IF you don't know someone in "zone 5", I've created a google sheet where we can list food pantries that have CONFIRMED that they will take usable food items shipped to them. For now I've left the access open -- anyone with this link can add a pantry in. So we can track a problem address, there is a spot to put your Volition ID.
The new company does not have the reimbursement set high enough when shopping at an "approved shipper" where they frequently increase the mailing fee. I escalated my request for additional reimbursement to a manager but never got a response. I will do the CPU locations and the regular Post Office locations but not any more of the "approved shipper" locations because of this. I do appreciate that the fee was increased by $2.
I'm out on the West coast, is there someone on the East Coast who completes these shops who would like to exchange packages? We could just ship the boxes back to each other.
Has anyone asked why, in the instructions, specific symbols are supposed to be found on the marketing materials which make them current but in the locations there are different, newer symbols? In other words, we are supposed to Mark the materials as not current because the symbols do not match the instructions which have not changed in years. I have asked but got no answer. Does anyone know?
Lao, I think they're looking for OLD materials, not newer than. For instance, still having the Christmas stuff up. I'm doing cleanup runs this week. This is one of my favorite companies.
I'm back again with the same question! Does anyone know if these shops are still being done?
What about the shops in PA with the PLCB?
Yes, they are done and are easy. Not sure what PLCB means.
Are they still being done by the same company?
Yep. They are done by the same company as the last few years.
Anyone know if the program changed? I seem to be having a very difficult time connecting with my scheduler, and there are FAR fewer shops up for this quarter.
Weirdly I can't see the shops unless I look at the Map view?


text:
Digital Invoicing Headquarter Address, Email and Contact Number
About Digital Invoicing Software Company
Faqs About Digital Invoicing Software Company
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Q7: What is Digital Invoicing Company Official Email Address?
Digital Invoicing Software Company is the leader in affordable, reliable and customizable invoice software. We have been in business since 2006 and serve small businesses of all sizes as well as medium-sized enterprises.Our flagship product, Invoice2go, is a powerful digital invoicing software designed for entrepreneurs and small businesses who are looking for complete functionality without paying high prices. With a range of benefits that include a mobile-friendly interface, secure connection, and compatibility with Windows 8 (and later), it's the perfect solution for small but growing businesses or professionals on a budget.
Digital Invoicing Software Type: Accounting Software
Software Company Name: Digital Invoicing
Digital Invoicing Headquarter Address: United Kingdom
Digital Invoicing Official Team Contact Number:
Digital Invoicing Website: Official Website: http://www.digitalinvoicing.co.uk
Software Plan Available: Plan
Digital Invoicing Founded In: Year
Digital Invoicing Email Address: Mail @
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Been a while since my last blog, and thought i would make this a more personal update on a couple of car events happening in my world.
Last saturday i was fortunate to be at Rally Drive in Bacchus Marsh for a drive and passenger session in a couple of ex Rally Subaru's . The event was a gift for Fathers day this year from my two sons. One of which came out with me and snapped a couple of shots of his old man trying to make some dust and keep the thing just about on track. The 'drive' car was an RS version which was certainly quick enough for me as a first timer on the dirt, with the passenger lap being in an STI version so completely sideways for most of the lap . A very good morning and one i would highly recommend.
The second part of this Rally week for me is as an official at the ARC/VRC round this weekend based at Lardner Park with action all over the forest sections around Noojee/ Powelltown etc. There really are some excellent spectator sections and fairly close to home for a lot of us and would highly recommend it to anyone who likes to see the best in the country travelling quick and sideways on some awesome forest roads. Rest assured these guys and girls will be a lot quicker than i was.!!
So thats about it for this time , if anyone is thinking of the Rally Drive experience log on to rallydrive.com.au.
Stay safe wherever you drive and will see you at an event soon.


text:
'A player's coach': Hoglund returns to Dell Rapids
by: Grant Sweeter
DELL RAPIDS, S.D. (KELO) — The Dell Rapids boys basketball team began of a three year streak of reaching the state tournament in 2019, but that was snapped in the spring of 2022, when the Quarriers fell in region play. Now, they're looking to get back to state and they'll have a familiar name leading the team.
High School Basketball Preview Show to air Wednesday
Ty Hoglund etched his name as one of the best basketball players in Dell Rapids history, following a standout career as a Quarrier. Hoglund took his talents to Mitchell, to play for Dakota Wesleyan.
The sharp shooting left hander collected more than 2,600 points in his four years in Mitchell, but as he prepared for graduation, the COVID-19 Pandemic began.
"I graduated in 2020, kind of in the middle of COVID and basketball just kind of stopped in my life. I quickly realized, that wasn't how it was supposed to be," Hoglund said.
After a year, Hoglund found himself with an opportunity to become the head coach at his former high school, Dell Rapids.
KELOLAND Sports Update
"When this opportunity came about, I knew it was something I wanted to come back and do. I hopped on the opportunity and I'm very lucky for how it played out," Hoglund said.
Basketball has come full circle for Hoglund as he's now able to give back as a coach. His goal is to empower his players, which is the same thing he got from his college coach, Matt Wilber.
"He was a players coach. He had really great relationships with all of his players and I think that's why guys played so hard for him. I took that and I brought that here, just to be a players coach and want the best for my guys," Hoglund said.
"He kind of gives off the vibe that he is there truly for us. Just to give us a good experience," Dell Rapids senior Brayden Pankonen said.
Now, Hoglund and Quarriers have their eyes on a return trip to the state tournament and they'll look to use the leadership of their eight seniors to do so.
"We just want to put ourselves in a position to win games and if I can help that in any sort of way and build their confidence, so we can get over that hump and take the next step to a state tournament, I think that's kind of our goal for our season," Hoglund said.
"The fact that he cares as much as he does, I can't thank him enough for that, because that's ultimately what I want from my coach. Especially my senior year, he's going to try to get us in the right place and push us in the right direction," Pankonen said.
Dell Rapids opens the season on Friday when they host West Central.
You can hear from the Quarriers and 18 other teams in our High School Basketball Preview Show, which airs on Wednesday at 6:30 p.m.
Tweets by KELOSports
fcast11-17


Below you can view all the homes for sale in the Branch Woods subdivision located in Garner North Carolina. Listings in Branch Woods are updated daily from data we receive from the Raleigh multiple listing service.
To learn more about any of these homes or to receive custom notifications when a new property is listed for sale in Branch Woods, give us a call at 919-759-5792.
Two car garage with utility sink and parking pad. Low HOA dues. Centrally located. MLS#2214026. 2931.


text:
Post Nation
At least 26 dead in South Texas church shooting, officials say
Authorities said several people were killed in a shooting at First Baptist Church in Sutherland Springs in South Texas on Nov. 5. (The Washington Post)
Peter Holley,
Peter Holley
Technology reporter
Eva Ruth Moravec,
Eva Ruth Moravec
Kristine Phillips and
Kristine Phillips
Reporter covering national and breaking news
Wesley Lowery
National correspondent covering law enforcement, justice and their intersection with politics and policy
November 6, 2017 at 10:06 a.m. UTC
SUTHERLAND SPRINGS, Texas — A lone gunman stormed a small community church here on Sunday and fatally shot more than two dozen people before a local resident engaged him in a gun battle as the shooter exited the church. He fled at high speed with the civilian in pursuit before running off the roadway, where he was found dead, authorities said.
Witnesses said a white man in his 20s dressed in all black and wearing a tactical vest started shooting with an assault rifle as he approached First Baptist Church in Sutherland Springs. Police said the gunman killed two people outside before entering the church and spraying bullets at the congregation during morning worship in the country town about 30 miles southeast of San Antonio. Dozens of people were hit with bullets, and the 26 dead ranged in age from 5 to 72, authorities said. One family reported the death of a year-old child. Most were shot in the pews as they worshiped.
After the exchange of gunfire, the armed civilian jumped into the truck of a second resident who had stopped at the scene and the two took off after the shooter. It was "act now, ask questions later," said the truck's driver, Johnnie Langendorff. By the time they caught up with him, however, the fleeing man had crashed his SUV into a ditch. "He might have been unconscious from the crash or something like that, I'm not sure," Langendorff told reporters.
"The local citizen pursued him," said Freeman Martin, a regional director for the Texas Department of Public Safety. "We don't know if it was a self-inflicted gunshot wound or if he was shot by our local resident who engaged him in a gunfight."
An unlikely hero describes gun battle and 95 mph chase with Texas shooting suspect
Two law enforcement officials told The Washington Post that the shooter has been identified as Devin Kelley, 26, a Texas man who lived in a nearby county. Only fragments of Kelley's background emerged in the hours after the shooting, among them that he was court-martialed in 2012 while serving in the U.S. Air Force and sentenced to a year in military prison for assaulting his spouse and child. He was reduced in rank and released with a bad-conduct discharge in 2014.
Nor had authorities speculated publicly about a possible motive as of early Monday morning with the investigation in its very early stages.
President Trump, at a news conference Monday afternoon in Tokyo, said that he thought "mental health" was the problem. "Based on preliminary reports," he said, the shooter was "a very deranged individual, a lot of problems for a long period of time." Trump did not provide the basis for his statement, saying "it's a little bit soon to go into it."
" . . . This isn't a guns situation," he added. "Fortunately someone else had a gun that was shooting in the opposite direction" or it "would have been much worse."
Who is Devin Patrick Kelley, the gunman officials say killed churchgoers in Sutherland Springs?
But residents of this small, out-of-the-way town said they were rattled and dumbfounded at becoming yet another U.S. community victimized by a mass shooting.
The killings left massive and in some instances multigenerational gaps in some families. Joe and Claryce Holcombe lost children, grandchildren and great-grandchildren all at once, a total of eight extended family members, the couple said in a phone interview with The Post.
Their son, Bryan Holcombe, 60, and his wife, Karla Holcombe, 58, were killed. Bryan was associate pastor for the church and walking to preach at the pulpit when he was shot, Joe Holcombe told The Post.
Among the dead was also their granddaughter-in-law, Crystal Holcombe, who was pregnant. She died along with her unborn child and three of her children, Emily, Megan and Greg, according to Joe Holcombe. She had been at church with her husband, John Holcombe, who survived along with two of her other children.
Their grandson, Marc Daniel Holcombe and his infant daughter, who is about a year old, also died, Joe and Claryce Holcombe said.
Neighbors said they became aware of the shooting when they heard loud bangs — and even had bullets spray into their homes — shortly after 11 a.m. on Sunday. Local authorities said the first calls of an active shooter came in about 11:20 a.m., after the gunman opened fire with a Ruger assault-style rifle.
Kevin Jordan, 30, was changing the oil in his Ford Focus ahead of a family road trip when he heard the pops of gunfire. When he stood up and turned his head, he saw a man wearing body armor, a vest and a mask walk down the sidewalk toward the church about 50 yards from his home.
"He was just spraying at the front of the church," Jordan said. "He was shooting outside at first, and then he walked to the door and started shooting inside."
Authorities said that at least 23 of the slain were killed inside the church. They said that two people were shot and killed outside and that one died at a hospital.
Authorities said they found multiple weapons in the gunman's vehicle.
After spotting the shooter, Jordan said, he ran inside his home, scooped up his son, alerted his wife and rushed his family into their bathroom, where they crouched and hid while calling 911. He said the shooter spotted him as he fled and took a shot that went through his front window, nearly hitting his 2-year-old son.
"I looked at the shooter, and he looked right at me," he said. When the shooting stopped, Jordan, who works as a medical assistant, ran to the church, hoping to help.
"I walked inside and just walked out. I couldn't handle it," he said. "It was bad. A lot of blood and bodies. The pews were knocked over. I'm a medical assistant and medical assisting does not prepare you for this."
Tucked a few hundred yards off Highway 87 amid scrubby farmland, the dusty and usually quiet streets of Sutherland Springs, lined with modest one-story family homes and trailers, were swarming with law enforcement Sunday evening. A town with few streetlights that typically goes dark after sundown flashed red and blue with police lights on almost every block.
Federal authorities, including from the FBI and Bureau of Alcohol, Tobacco, Firearms and Explosives were on the scene. The Texas Ranger Division of the state's DPS also is involved in the investigation.
The shooting Sunday came more than a month after a masked gunman stormed into a small community church outside of Nashville and shot seven people, including the pastor, killing one. Authorities said the suspect in that shooting, Emanuel Kidega Samson, might have been motivated by a desire for revenge for a 2015 shooting that targeted black churchgoers in Charleston, S.C. — an attack that left nine people dead.
Trump says Texas shooting is a problem of mental health, not guns
It also came just more than a month after 58 people were killed at a Las Vegas country music festival, in what was the deadliest mass shooting in recent U.S. history; the assailant, Stephen Paddock, killed himself after a lengthy shooting spree from his 32nd-floor Mandalay Bay hotel suite.
Few details were immediately available about Kelley, and attempts to reach his family were unsuccessful Sunday. Kelley had at one point been in the military, enlisting in 2010 and serving as a logistical readiness airman stationed at Holloman Air Force Base in New Mexico, according to Air Force spokeswoman Ann Stefanek, who also confirmed Kelley's 2012 court-martial.
It was unclear Sunday how Kelley obtained the weapon he used or why he targeted the church. Neighbors said the gunshots they heard were thundering.
The scene after a shooting in a Texas church
View PhotosView Photos
Rev. Van Jordan, of Port Arthur, Tex., leads a group prayer while setting up crosses to honor the shooting victims along the edge of a field next to Sutherland Springs First Baptist Church. (Carolyn Van Houten/The Washington Post)
Diana Segura, 69, was in the shower Sunday morning at about 11 a.m. when she was startled by a series of bangs so loud she thought a truck's engine had exploded on the highway behind her home. Minutes later, sirens burst onto her quiet street. She walked outside and saw multiple bodies on the ground outside the First Baptist Church, where she occasionally attends weeknight services.
Standing outside her home down the street from the church, Segura stared at the throng of police cars and emergency vehicles, her head shaking in disbelief.
"This is a small town and nothing never happens here," she said. "We are family here, and that church is always filled with friends."
Frank Pomeroy, the pastor of First Baptist Church, told ABC News that he was not present during the church service but that his teenage daughter, Annabelle Pomeroy, 14, was among the dead.
"She was very quiet, shy, always smiling, and helpful to all," Cynthia Rangel, 50, a resident of nearby Stockdale, said of the teenager. Rangel, a local emergency medical technician, said she knew three individuals who were hospitalized after the shooting and were undergoing surgery. "This just all seems like it's not real."
Dana Fletcher, who owns a business a quarter-mile from the church, said she and her family just moved to Sutherland Springs. She said she was first alerted to the shooting by a call from a reporter.
"My husband and I both are still in shock," she said. "It's a little tiny church that was targeted. It's shocking. It's a bit frightening because it's a little bit close to home."
Carrie Matula, who works at a gas station near the church, told MSNBC she ran out to see what was happening after hearing gunshots.
"I never thought it would happen here," she said. "This is something that happens in a big city. I would have never thought this would have taken place here. It's just too tight a community. It doesn't make sense."
The church is a part of the Southern Baptist Convention, the largest Protestant denomination in the country, with about 15 million members. First Baptist reported an average estimated attendance of 100 in 2015. The church is affiliated with the Southern Baptists of Texas Convention, considered one of Texas's more theologically conservative groups of Southern Baptists.
Kathy Forton, 65, who has lived much of her life in the rural area of Wilson County, said the town has suffered an economic downturn since the county seat was moved to Floresville after a fire many years ago.
"The church helped out in so many ways, " Forton said. "Just the most precious, loving people. The people from the church used their own money to provide turkeys for the community at Thanksgiving and presents for kids at Christmas. The loss of these people is going to devastate that community."
Trump first addressed the shooting while traveling in Asia early Monday, sending his thoughts and prayers to the victims and families.
"This act of evil occurred as the victims and their families were in their place of sacred worship," he said. "We cannot put into words the pain and grief we all feel, and we cannot begin to imagine the suffering of those who lost the ones they so dearly loved. Our hearts are broken, but in dark times — and these are dark times — such as these, Americans do what they do best: We pull together. We join hands. We lock arms, and through the tears and through the sadness we stand strong — oh, so strong."
Authorities reported multiple people are dead and the gunman is "down" after a shooting in a church in Sutherland Springs, Tex., near San Antonio, on Nov. 5.
Phillips and Lowery reported from Washington. Mary Lee Grant in San Antonio and Sarah Pulliam Bailey, Ed O'Keefe, Alex Horton, Samantha Schmidt and Devlin Barrett in Washington contributed to this report.
Tennessee church-shooting suspect had note referencing Dylann Roof attack
Masked gunman rampages through Nashville church; usher uses personal weapon to subdue shooter
Las Vegas shooting motive eludes investigators as new details emerge about gunman Stephen Paddock


text:
Q: How to sort data by column of a table created in Spree? If we create a table in Spree, is there any predefined method to sort its content?
I had found sort_link  @search, does anyone know how to use that?
Also if no predefined methods are available for sorting, what are the other best possible ways to accomplish the task?
Thanks in advance.

A: I got the solution is 'ransack gem', it provides the wonderful way to get the search and sort the results.

A: You should take a look at the order method from ActiveRecord:
http://guides.rubyonrails.org/active_record_querying.html#ordering

redpajama_set_name: RedPajamaStackExchange


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text:
NASCAR adjusts testing schedule
By Dustin LongOct 8, 2015, 10:49 AM EDT
NASCAR announced amendments to the Sprint Cup Rule Book this week.
Among the changes:
Only drivers licensed and approved to compete in the Sprint Cup Series are eligible to participate in NASCAR-approved tire tests.
The tire test scheduled for Auto Club Speedway later this month has been canceled and replaced by a test Oct. 27-29 at Atlanta Motor Speedway. Scheduled to test there will be Richard Childress Racing, Chip Ganassi Racing, Roush Fenway Racing and Joe Gibbs Racing.
A tire test at Homestead-Miami Speedway has been added for Dec. 14-15. Teams scheduled to test will be Stewart-Haas Racing, Hendrick Motorsports, Richard Petty Motorsports and Furniture Row Racing.
NASCAR has a test scheduled Oct. 20 at Michigan International Speedway. Among the items on NASCAR's agenda will be examining a low-downforce package. Denny Hamlin (Joe Gibbs Racing), Chase Elliott (Hendrick Motorsports) and Chris Buescher (Roush Fenway Racing) are scheduled to take part in the test.
Brake caliper and brake rotor part numbers must be declared before opening practice at Charlotte Motor Speedway, Kansas Speedway, Talladega Superspeedway, Texas Motor Speedway and Homestead-Miami Speedway. In 2016, brake caliper and brake rotor part numbers must be declared before opening practice at all events.
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Dong Abay, Coffee Break Island, Lady I, Count Kutu & The Balmers show their support for the Energy [R]evolution.
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* if no node with the given path exists.
-     * thrown if the given version does not match the node's version.
+     * thrown if the given aclVersion does not match the node's aclVersion.
might be handle in another jira..?
better improve Java doc or argument which might not mislead for new users.


text:
Imran Khan's supporters trying hard to defend him
With Imran Khan's popularity graph declining noticeably, his supporters are finding it very difficult to defend him. At first, they claimed that the incriminating audio is fake, then they said it's a personal matter and should not be made public. Imran Khan himself is quoting from the Holy Book about not revealing or publicizing the mistakes and sins of others. He should have thought of all this when he was lambasting his political opponents for much lesser sins. But then, we shouldn't be surprised. Imran Khan has frequently played the religion card to appear very holy to his voters. But now I doubt if they can be fooled. Some of them, the die-hard ones, will of course never think of him as anything but a good man, in fact one of them even tweeted that he will support Imran even if he comes across a video depicting a naked Imran indulging in unmentionable activities.
Such a video clip has been in the news for many months now. There are some who've seen it and say they can't talk about it, it's so shocking. There are reports of other audios and videos in existence. All of them should be publicized repeatedly until Imran Khan himself quits politics. That's the only way the country can be saved.


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				Contents

Asynchronous tests

Combine TestScheduler and async tests

Fulfilling expectations

Triggering some Observable values

Verifying the recorded events with asserts

Conclusion

              Testing your RxSwift code, part 2

                	23/May 2016

            		By Marin Todorov

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In part 1 I looked into writing very basic tests with RxSwift and life was good. But then I wanted to do more and moved on to writing asynchronous tests…

I contributed the code for RxRealm - the Rx extension for RealmSwift owned by the RxSwiftCommunity. For RxRealm I needed some asynchronous tests because Realm’s collections emit notifications (which RxRealm wraps) when the underlaying data changes.

So let’s have a look at some of the unit tests I wrote…

Asynchronous tests

First of all I needed to look into writing asynchronous tests - luckily this is quite easy. The basic structure of such a test is:

func testSomething() {
  let expectation1 = expectationWithDescription("First event")
  let expectation2 = expectationWithDescription("Second event")

  ... whatever code you need to perform the test ...

  someAsyncAPI(didComplete: {
    expectation1.fulfill()
  })

  someOtherAsyncAPI(didComplete: {
    expectation2.fulfill()
  })

  waitForExpectationsWithTimeout(1.0) {error in
     XCTAssertTrue(error == nil)
     ... more asserts ...
  }
}

Let’s look at what this code does:

you need to define all the expectations you’d like this test to wait for before it completes; the string argument is decorative,

whenever the events you are waiting for complete you call fulfill() on your expectations,

finally you need to define a block to be executed when ALL the expectations are fulfilled; waitForExpectationsWithTimeout ’s first parameter is a timeout, which will force the block to execute if the expectations do not fulfill in time.

Combine TestScheduler and async tests

Let’s look at a complete test from the RxRealm test suite.

First I start by defining a single expectation - my expectation will be that a precise number of events were emitted by the observable.

func testResultsTypeChangeset() {
    let expectation1 = expectationWithDescription("Results")

    let realm = realmInMemory(#function)
    clearRealm(realm)

}

Additionally I use two of my helper methods to grab a reference to an in-memory realm and clear all objects that might’ve been stored inside by other tests in the suite.

Now just like in Part 1 I define a dispose bag and a test scheduler:

let bag = DisposeBag()

let scheduler = TestScheduler(initialClock: 0)
let observer = scheduler.createObserver(String)

Note that I’ll be observing a sequence of Strings - you’ll see why in a minute.

Fulfilling expectations

Next I need to create a Results object since I’m going to be testing Results.asObservable():

let messages$ = realm.objects(Message).asObservableChangeset().shareReplay(1)

I’m observing all Message objects stored in my realm.

Now comes the interesting part - first I’ll define when my expectation is fulfilled:

messages$
  .scan(0, accumulator: {acc, _ in return acc+1})
  .filter { $0 == 3 }
  .map {_ in ()}
  .subscribeNext(expectation1.fulfill)
  .addDisposableTo(bag)

I use scan to count the emitted values from my observable, filter helps me to pinpoint in which value I’m interested (the 3rd event), I map the value to Void and subscribe to expectation1.fulfill().

Long story short - when messages$ emits its 3rd event my expectation will fulfill. Neat!

Since I’m not interested in just how many events are emitted but also which events were emitted, I need yet another subscription:

messages$
  .map {results, changes in
    if let changes = changes {
      return "i:\(changes.inserted) d:\(changes.deleted) u:\(changes.updated)"
    } else {
      return "initial"
    }
  }
  .subscribe(observer)
  .addDisposableTo(bag)

If it’s an update event I return a string with the collection indexes that were deleted or inserted, it looks like so:

i:[0] u:[] d:[1,2,4]

This logs all the information about the particular change-set that I’m interested in (and need to verify if the observable emitted the information I expected).

If it’s the initial event for that collection I just return “initial”.

Triggering some Observable values

Now I can start the test scheduler:

scheduler.start()

And, ultimately, perform some operations on my Realm to produce change notifications:

delay(0.1) {
  try! realm.write {
    realm.add(Message("first"))
  }
}
delay(0.2) {
  try! realm.write {
    realm.delete(realm.objects(Message).first!)
  }
}

The code above adds a new object and after a bit of delay deletes it again. This should produce the 3 events I’m expecting: the initial data, the insert notification and the deletion notification.

Verifying the recorded events with asserts

It’s time to (finally) add my asserts:

waitForExpectationsWithTimeout(0.5) {error in
    //do tests here

    XCTAssertTrue(error == nil)
    XCTAssertEqual(observer.events.count, 3)
    XCTAssertEqual(observer.events[0].value.element!, "initial")
    XCTAssertEqual(observer.events[1].value.element!, "i:[0] d:[] u:[]")
    XCTAssertEqual(observer.events[2].value.element!, "i:[] d:[0] u:[]")
}

In the block I perform a number of asserts:

no error happened while fulfilling my expectations

the test observer logged exactly 3 events

the first emitted value is “initial”

the second value is about inserting a value at index 0

the third value is about deleting a value at index 0

Conclusion

It’s amazing how easy writing tests is even when testing two completely de-coupled libraries working together like RealmSwift and RxSwift. As you saw it’s a matter of 10-20 lines of code depending on your setup and I’m sure the code can be simplified even more.

If you want to read through the complete test suite for RxRealm head to the GitHub repo test folder: RxRealm_Tests.

I hope this post has been useful! Do you know a better way to do any of this? Seen a bug? Ping me on Twitter at icanzilb

Hope that post was helpful, and if you want to get in touch you can find me here
Follow @icanzilb

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UK lawmaker seeks police probe into envoy's leaked cables
London [UK], July 9 (ANI): Tom Tugendhat, the House of Commons Foreign Affairs Committee Chairman, has written to the Metropolitan Police Service, asking for a probe into the leak of the secret cables of the British Ambassador to the US.
British envoy Sir Kim Darroch had labelled US President Donald Trump as "inept," "insecure," and "incompetent" in the secret cables which were leaked to a UK newspaper. A UK government official confirmed the leak to CNN while declaring the information in the compromised documents to be true.
"I fear that we are developing a culture of leaks and that would be extremely detrimental to the United Kingdom. They damage our reputation, impact our ability to function effectively and undermine our relations with our allies," Sputnik quoted Tugendhat as saying during a House of Commons meeting on Monday (local time).
"I understand that the Foreign Office has opened an inquiry into this leak. I have today written to the commission of the Metropolitan Police to ask that a criminal investigation also be opened into the leak," he added.
UK Foreign Minister Jeremy Hunt has dismissed Darroch's critique of Trump, adding that the UK will not allow "any interruption" in the "superb" US-UK relations.
Trump, meanwhile, lashed out at the British Ambassador, stating that the US "will no longer deal with" Darroch. He had also added that Darroch had not "served" his country well.
The incident could pose a huge headache for the UK, as it may jeopardise a probable trade deal with Washington after Brexit or UK's impending exit from the European Union (EU). (ANI)
leaked cables
India, Sri Lanka sign contract to upgrade 130 km long railway track
New Delhi [India], July 18 (ANI): India and Sri Lanka on Thursday signed a USD 91.26 million contract to upgrade 130 kilometres long railway track connecting Maho and Omanthai town of the Island Nation.
India closely following peace, reconciliation efforts in Afghanistan: MEA
New Delhi [India] July 18 (ANI): India is closely following the recent developments pertaining to the ongoing peace and reconciliation efforts in Afghanistan and has been in constant touch will all major stakeholders in the matter, the MEA said on Thursday.
Hafiz Saeed being kept in jail superintendent's bungalow
Islamabad [Pakistan], July 18 (ANI): Mumbai terror attack mastermind Hafiz Saeed, who was arrested by Pakistan on charges of terror financing, has been kept in the bungalow of Gujranwala prison superintendent situated next to the jail.
Hindus in Pakistan not free to exercise religious freedom, says...
Washington D.C. [USA], July 18 (ANI): A Sindhi activist has accused Pakistan of implementing discriminatory policies against the minority Hindu community in the country, especially in the Sindh province.
India dubs Saeed's arrest by Pakistan as drama
New Delhi [India], July 18 (ANI): India on Thursday dismissed the recent arrest of 2008 Mumbai terror attack mastermind Hafiz Saeed as a mere "drama".
ICJ verdict is final, binding on Pakistan: MEA
New Delhi [India], July 18 (ANI): India on Thursday said the International Court of Justice (ICJ) verdict on Kulbhushan Jadhav is "final and binding" on Pakistan which is obliged to act and deliver on the directive given to it.
New York [USA], Jul 18 (ANI): Joaquin "El Chapo" Guzman, the Mexican drug kingpin who minted millions as the head of the infamous Sinaloa cartel for several years, was awarded life imprisonment plus 30 years by a US court here on Wednesday (local time).
Former Pak PM Shahid Khaqan Abbasi arrested
Islamabad [Pakistan], July 18 (ANI): Former Pakistan Prime Minister Shahid Khaqan Abbasi was on Thursday arrested by the National Accountability Bureau (NAB) in connection with a corruption case.
New Delhi [India], July 18 (ANI): India on Thursday strongly condemned yesterday's terrorist attack in Iraq's Erbil city, in which three people, including a Turkish diplomat, were killed and expressed its condolences to the families of the deceased.
CPEC stands on shaky legal ground: European researcher
Brussels [Belgium], July 18 (ANI): European researcher and political scientist Siegfried O Wolf has said that the multi-billion dollar China-Pakistan Economic Corridor (CPEC) stands on an extraordinary shaky legal ground.
Kanazawa [Japan], July 18 (ANI): Japan welcomes hundreds of thousands of tourists every year.
Death toll reaches 90 in Nepal floods, landslides
Kathmandu [Nepal], July 18 (ANI): The death toll in floods and landslides triggered by incessant rains since last week in Nepal rose to 90, with about 31 people still missing, the country's home ministry said on Thursday.


Three weeks after he was cut by the Washington Redskins, kicker Shaun Suisham will return to FedEx Field on Sunday with his new team, the Dallas Cowboys.
Suisham, who was cut after he missed a 23-yard field goal attempt that would have clinched an upset over the then-unbeaten New Orleans Saints on Dec. 6, signed with the Cowboys on Monday after Dallas released Nick Folk.
Folk, a Pro Bowl pick as a rookie in 2007, missed a 24-yard try against New Orleans last week. That league-leading 10th miss proved the end for Folk.
Suisham was 18-for-21 with the Redskins this year, but his misses were big. He also missed attempts of 50 and 39 yards in the 7-6 loss to Dallas on Nov. 22.
Redskins coach Jim Zorn said he's glad Suisham found another job so fast, but he wishes it hadn't been with the Cowboys.
Suisham's return to the NFL came as good news to Redskins punter Hunter Smith, his former holder and locker room neighbor. Smith said it was "really hard" seeing what happened to his good friend.
Devin Thomas missed his second straight practice after spraining his right ankle sprain during the second half kickoff of Monday's 45-12 loss to the New York Giants, making it more likely Malcolm Kelly will take his place against the Cowboys.
Kelly caught just seven passes for 73 yards and no touchdowns in five starts before yielding to Thomas, who has 29 catches, 299 yards and three touchdowns in nine games. Kelly had nine catches for 160 yards and no touchdowns in those nine games.
"He's had an excellent week of practice thus far," Zorn said. "He's looked explosive. He's got fresh legs obviously. He's been working to try close that gap.
Cornerback DeAngelo Hall was limited Thursday with a right knee injury that kept him out the three games before the one with the Giants. Zorn expects him to play against Dallas. The coach said the same about defensive tackle Cornelius Griffin, who didn't play against the Giants with a sprained shoulder and was limited the past two days.
Defensive end Andre Carter was limited again because of the torn left biceps he suffered against the Giants but is also expected to start Sunday. Defensive tackle Anthony Montgomery returned to full practice after being limited Wednesday with an ailing hip. Right tackle Stephon Heyer fully participated for a second straight day after missing most of Monday's second half with an ailing left knee.
Defensive tackle Albert Haynesworth was fined $10,000 by the NFL on Thursday for an altercation late in the Redskins' loss to the Giants.
Giants running back Brandon Jacobs was fined $7,500.


text:
Q: Upper bound on $P(n^{-1}\sum_{i=1}^n (X_i - \lambda_i)>t)$ for independent $X_i\sim\operatorname{Poisson}(\lambda_i)$
Let $X_1,\dots,X_n$ be independent random variables, $X_i \sim \operatorname{Poisson}(\lambda_i),$ $i=1,\dots,n.$ Let
  $$S=n^{-1}\sum_{i=1}^n X_i, \quad\quad \lambda=n^{-1}\sum_{i=1}^n \lambda_i.$$
  Find an upper bound for $P(S-\lambda>t)$. What $t$ do we need in order $P(S-\lambda>t)\leq n^{-\tau}$ for some $\tau>0$?

I am not sure how to go about finding an upper bound for $P(S-\lambda > t)$ in this problem. Any help would be appreciated.

A: $\sum_{i=1}^n X_i$ follows Poisson distribution with parameter $\sum_{i=1}^n \lambda_i$. So $S =n^{-1} \sum_{i=1}^n X_i$ has mean $n^{-1}\sum_{i=1}^n \lambda_i = \lambda$ and variance $n^{-2}\sum_{i=1}^n \lambda_i = n^{-1}\lambda$.
According to Cantelli's inequality, $$P(S-\lambda>t) \le \frac {n^{-1}\lambda}{n^{-1}\lambda + t^2}$$
To get $P(S-\lambda>t)\leq n^{-\tau}$, we need $$\frac {n^{-1}\lambda}{n^{-1}\lambda + t^2} = n^{-\tau}$$
So $t=\sqrt{\frac {\lambda(n^{\tau}-1)}n}$


text:
Mike Tyson Admits to Tripping on Toad Venom at Least 53 Times
Boxing legend Mike Tyson says he nearly "died" the first time he tried tripping on toad venom — yet he went on to use the psychedelic drug at least 53 more times.
The former heavyweight champ says using the drug, which is derived from the Sonoran Desert Toad, helped him become more focused, more creative and happier. "In my trips I've seen that death is beautiful," Tyson says. "Life and death both have to be beautiful, but death has a bad rep.
The toad has taught me that I'm not going to be here forever. There's an expiration date." While drug users aren't always the most reliable sources of information, Tyson says he's not the only one who's recognized a positive change as a result of his venom use. "People see the difference," he says. "It speaks for itself.
If you knew me in 1989 you knew a different person. My mind isn't sophisticated enough to fathom what happened, but life has improved.
The toad's whole purpose is to reach your highest potential. I look at the world differently. We're all the same. Everything is love."


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