text:
\section{Introduction}
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In the last twenty years, sampling methods for shape reconstruction in inverse scattering problems have attracted a lot of interest.
Typical examples include the Linear Sampling Method by Colton and Kirsch \cite{ColtonKirsch}, the Singular Sources Method by Potthast \cite{Potthast2000}
and the Factorization Method by Kirsch \cite{Kirsch98}.
The basic idea is to design an indicator which is big inside the underlying scatterer and relatively small outside.
We refer to the monographs of Cakoni and Colton \cite{CC2014}, Colton and Kress \cite{CK} and Kirsch and Grinberg \cite{KirschGrinberg} for a comprehensive understanding.
We also refer to Liu and Zhang \cite{LiuZhang2015} for a recent progress on the Factorization Method.
Recently, a type of direct sampling methods are proposed for inverse scattering problems,
e.g., Orthogonality Sampling by Potthast \cite{Potthast2010}, Direct Sampling Method by Ito et.al. \cite{ItoJinZou}, Single-shot Method by Li et.al. \cite{LiLiuZou},
Reverse Time Migration by Chen et.al. \cite{CCHuang}.
These direct sampling methods inherit many advantages of the classical ones, e.g., they are independent of any a priori information on the geometry and physical properties of
the unknown objects. The main feature of these direct sampling methods is that only inner product of the measurements with some suitably chosen functions is involved in the
computation of the indicator, thus is robust to noises and computationally faster than the classical sampling methods.
However, the theoretical foundation of the direct sampling methods is far less well developed than for the classical sampling methods.
In particular, there are no theoretical analysis of the indicators for the sampling points inside the scatterers.
In this paper, we propose a new direct sampling method for inverse acoustic scattering problems by using the scattering amplitudes.
We will study the behavior of our new indicator for the sampling points both outside and inside the scatterer.

We begin with the formulations of the acoustic scattering problems.  Let $k=\omega/c>0$ be the wave number of
a time harmonic wave where $\omega>0$ and $c>0$ denote the frequency and sound
speed, respectively. Let $\Omega\subset{\mathbb R}^n (n=2,\, 3)$ be an open and bounded domain with
Lipschitz-boundary $\partial\Omega$ such that the exterior ${\mathbb R}^n\backslash\overline{\Omega}$ is connected.
Furthermore, let the incident field $u^i$ be a plane wave of the form
\begin{eqnarray}\label{incidenwave}
u^i(x)\ =\ u^i(x,\hat{\theta}) = e^{ikx\cdot \hat{\theta}},\quad x\in{\mathbb R}^n\,,
\end{eqnarray}
where $\hat{\theta}\in S^{n-1}$ denotes the direction of the incident wave and $S^{n-1}:=\{x\in{\mathbb R}^n:|x|=1\}$ is the unit sphere in ${\mathbb R}^n$.
Then the scattering problem for the inhomogeneous medium is to find the total field $u=u^i+u^s$ such that
\begin{eqnarray}
\label{HemEqumedium}\Delta u + k^2 (1+q)u = 0\quad \mbox{in }{\mathbb R}^n,\\
\label{Srcmedium}\lim_{r:=|x|\rightarrow\infty}r^{\frac{n-1}{2}}\left(\frac{\partial u^{s}}{\partial r}-iku^{s}\right) =\,0
\end{eqnarray}
where $q\in L^{\infty}({\mathbb R}^n)$ such that $\Im (q)\geq 0$ and $q=0$ in ${\mathbb R}^n\backslash\overline{\Omega}$,
the Sommerfeld radiating condition \eqref{Srcmedium} holds uniformly with respect to all directions $\hat{x}:=x/|x|\in S^{n-1}$.
If the scatterer $\Omega$ is impenetrable, the direct scattering is to find the total field $u=u^i+u^s$ such that
\begin{eqnarray}
\label{HemEquobstacle}\Delta u + k^2 u = 0\quad \mbox{in }{\mathbb R}^n\backslash\overline{\Omega},\\
\label{Bc}\mathcal{B}(u) = 0\quad\mbox{on }\partial\Omega,\\
\label{Srcobstacle}\lim_{r:=|x|\rightarrow\infty}r^{\frac{n-1}{2}}\left(\frac{\partial u^{s}}{\partial r}-iku^{s}\right) =\,0,
\end{eqnarray}
where $\mathcal{B}$ denotes one of the following three boundary conditions:
\begin{eqnarray*}
(1)\,\mathcal{B}(u):=u\,\,\mbox{on}\ \partial \Omega;\qquad
(2)\,\mathcal{B}(u):=\frac{\partial u}{\partial\nu}\,\,\mbox{on}\ \partial \Omega;\qquad
(3)\,\mathcal{B}(u):=\frac{\partial u}{\partial\nu}+\lambda u\,\,\mbox{on}\ \partial \Omega
\end{eqnarray*}
corresponding, respectively, to the case when the scatterer $\Omega$ is sound-soft, sound-hard, and of impedance type.
Here, $\nu$ is the unit outward normal to $\partial\Omega$ and $\lambda\in L^{\infty}(\partial \Omega)$ is the (complex valued) impedance function such that $\Im(\lambda)\geq0$ almost everywhere on $\partial \Omega$.
Uniqueness of the scattering problems \eqref{HemEquobstacle}--\eqref{Srcmedium} and \eqref{HemEquobstacle}--\eqref{Srcobstacle}
can be shown with the help of Green's theorem, Rellich's lemma and unique continuation principle, see e.g., \cite{CK}. The proof of existence can be done
by variational approaches (cf. \cite{CK, Mclean} for the Dirichlet boundary condition and \cite{CC2014, LiuZhangHu} for other boundary conditions) or by integral equation methods
(cf.\cite{CK, KirschGrinberg,LiuZhangSiam,LiuZhang2012}).

Every radiating solution of the Helmholtz equation has the following asymptotic
behavior at infinity:
\begin{eqnarray}\label{0asyrep}
u^s(x,\hat{\theta})
=\frac{e^{i\frac{\pi}{4}}}{\sqrt{8k\pi}}\left(e^{-i\frac{\pi}{4}}\sqrt{\frac{k}{2\pi}}\right)^{n-2}\frac{e^{ikr}}{r^{\frac{n-1}{2}}}\left\{u^{\infty}(\hat{x},\hat{\theta})+\mathcal{O}\left(\frac{1}{r}\right)\right\}\quad\mbox{as }\,r:=|x|\rightarrow\infty
\end{eqnarray}
uniformly with respect to all directions $\hat{x}:=x/|x|\in S^{n-1}$, see, e.g., \cite{KirschGrinberg}.
The complex valued function $u^\infty=u^\infty(\hat{x},\hat{\theta})$ defined on the unit sphere $S^{n-1}$
is known as the scattering amplitude or far-field pattern with $\hat{x}\in S^{n-1}$ denoting the observation direction.
Then, the {\em inverse problem} we consider in this paper is to determine $\Omega$ from a knowledge of the scattering amplitude $u^{\infty}(\hat{x},\hat{\theta})$ for
$\hat{x},\hat{\theta}\in S^{n-1}$.
It is well known that the scatterer $\Omega$ can be uniquely determined by the scattering amplitude $u^\infty(\hat{x},\hat{\theta})$ for all $\hat{x}, \hat{\theta}\in S^{n-1}$ \cite{CK}.
What we interested, in this paper, is to design a direct sampling approach for shape reconstruction by using the far field measurements.

The indicator functional which will be used for inverse acoustic scattering problems is given as follows,
\begin{eqnarray}\label{IndicatorLiu}
I_{new}(z):=\Big|\int_{S^{n-1}}e^{-ik\hat{\theta}\cdot z}\int_{S^{n-1}}u^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})ds(\hat{\theta})\Big|,\quad z\in{\mathbb R}^n.
\end{eqnarray}
In practice, the indicator functional is given
by the following discrete form
\begin{eqnarray*}\label{IndicatorLiudiscrete}
I_{new}(z):=
\left|\left(
    \begin{array}{cccc}
      e^{-ik\hat{\theta}_1\cdot z}, e^{-ik\hat{\theta}_2\cdot z}, \cdots, e^{-ik\hat{\theta}_N\cdot z} \\
    \end{array}
  \right)
  \left(
    \begin{array}{cccc}
      u_{1,1}^\infty\, u_{1,2}^\infty\, \cdots\, u_{1,N}^\infty \\
      u_{2,1}^\infty\, u_{2,2}^\infty\, \cdots\, u_{2,N}^\infty \\
      \vdots\,\quad \vdots\,\quad \ddots\,\quad \vdots \\
      u_{N,1}^\infty\, u_{N,2}^\infty\, \cdots\, u_{N,N}^\infty \\
      \end{array}
  \right)\left(
           \begin{array}{c}
             e^{ik\hat{x}_1\cdot z} \\
             e^{ik\hat{x}_2\cdot z} \\
             \vdots \\
             e^{ik\hat{x}_N\cdot z} \\
           \end{array}
         \right)\right|,\quad z\in{\mathbb R}^n,
\end{eqnarray*}
where $u^{\infty}_{i,j}=u^\infty(\hat{x}_j,\hat{\theta}_i)$ for $1\leq i, j\leq N$ corresponding to $N$ observation directions $\hat{x}_j$ and $N$ incident directions $\hat{\theta}_i$.
Clearly, only matrix multiplication is involved in the computational implementation.

This paper is organized as follows.
The theoretical foundation of the proposed reconstruction scheme will be established in the next section.
With the help of the inf-criterion characterization obtained by using the factorization of the far field operator, we show a lower bound of
the indicator $I_{new}$ for the sampling points inside the scatterers.
The decay behavior of $I_{new}$ will then be studied for sampling points outside the scatterers.
A stability statement will also be established to reflect the important feature of the reconstruction scheme under consideration.
With the help of well known properties of the scattering amplitudes and the corresponding far field operator, some connections with other sampling methods
are then established.
Some numerical simulations in two dimensions will be presented in Section~\ref{sec3} to indicate the new sampling method is computationally efficient and robust
with respect to data noise.

\section{Theoretical foundation of the proposed sampling method}
\label{sec2}
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The aim of this section is to establish the mathematical basis of our sampling method.
First, we recall the far field operator $F:L^2(S^{n-1})\rightarrow L^2(S^{n-1})$ defined by
\begin{eqnarray}\label{ffoperator}
(Fg)(\hat{x})=\int_{S^{n-1}}u^{\infty}(\hat{x},\hat{\theta})\,g(\hat{\theta})\,ds(\hat{\theta})\,,\quad \hat{x}\in S^{n-1}\,.
\end{eqnarray}
The far field operator $F$ plays an essential role in the investigations of the inverse scattering problems, we refer to the monographs of Colton and Kress\cite{CK} and Kirsch and
Grinberg\cite{KirschGrinberg} for a survey on the state of the art of its properties and applications.

For all sampling point $z\in {\mathbb R}^n$, define a test function $\phi_z\in L^2(S^{n-1})$ by
\begin{eqnarray}
\label{phiz}\phi_z(\vartheta):=e^{-ikz\cdot\vartheta},\quad \vartheta\in S^{n-1}.
\end{eqnarray}
Then we may rewrite our indicator $I_{new}$ given by \eqref{IndicatorLiu} in a very simple form
\begin{eqnarray}\label{IndLiunewform}
I_{new}(z):=|( F\phi_z, \phi_z)|,\quad z\in{\mathbb R}^n.
\end{eqnarray}
Here and throughout this paper we denote by $(\cdot, \cdot)$ the inner product of $L^2(S^{n-1})$.

\subsection{Lower bound estimate of $I_{new}(z)$ for $z\in\Omega$}
For all $z\in {\mathbb R}^n$, define $A_z\subset L^2(S^{n-1})$ by
\begin{eqnarray*}
A_z:=\{\psi\in L^2(S^{n-1}):\,(\psi, \phi_z)=1\},
\end{eqnarray*}
where again $\phi_z$ is defined by \eqref{phiz}.
We recall that the fundamental solution $\Phi(x,z), x\in {\mathbb R}^n, x\neq z,$ of the Helmholtz equation is given by
\begin{eqnarray}\label{Phi}
\Phi(x,y):=\left\{
              \begin{array}{ll}
                \frac{ik}{4\pi}h^{(1)}_0(k|x-y|)=\frac{e^{ik|x-y|}}{4\pi|x-y|}, & n=3; \\
                \frac{i}{4}H^{(1)}_0(k|x-y|), & n=2,
              \end{array}
            \right.
\end{eqnarray}
where $h^{(1)}_0$ and $H^{(1)}_0$ are, respectively, spherical Hankel function and Hankel function of the first kind and order zero.

Considering the case of scattering by impenetrable scatterers, we have the inf-criterion for sampling points $z$ which is both necessary and sufficient.

\begin{lemma}\label{infcriobstacle}{\rm (See Theorem 1.20 and 2.8 of \cite{KirschGrinberg})}\,
Consider the inverse scattering by impenetrable scatterers. We assume that $k^2$ is not an eigenvalue of $-\Delta$ in $\Omega$ with respect to the boundary condition under consideration.
Then $z\in\Omega$ if, and only if,
\begin{eqnarray*}
\inf|\{|(F\psi, \psi)|:\,\psi\in A_z\}>0.
\end{eqnarray*}
Furthermore, for $z\in\Omega$ we have the estimate
\begin{eqnarray}\label{estimate}
\inf|\{|(F\psi, \psi)|:\,\psi\in A_z\}\geq \frac{c}{\|\Phi(\cdot,z)\|^2_{H^{1/2}(\partial\Omega)}},
\end{eqnarray}
for some constant $c>0$ which is independent of $z$.
\end{lemma}

Turning now the case of scattering by an inhomogeneous medium, the analogous results of Lemma \ref{infcriobstacle} is, to our knowledge, still not established. So we proceed by
studying the corresponding inf-criterion for inhomogeneous medium.
We first make the following general assumptions on the contrast function $q$.

\begin{assumption}\label{qassumption}
Let $q\in L^{\infty}({\mathbb R}^n)$ satisfy
\begin{enumerate}
  \item $q=0$ in ${\mathbb R}^n\backslash\overline{\Omega}$.
  \item $\Im(q)\geq 0$ and there exists $c_1>0$ with $1+\Re(q)\geq c_1$ for almost all $x\in\Omega$.
  \item $|q|$ is locally bounded below, i.e., for every compact subset $D\subset\Omega$ there exists $c_2>0$ (depending on $D$) such that $|q|\geq c_2$ for almost all $x\in\Omega$.
  \item There exists $t\in [0, \pi]$ and $c_3>0$ such that $\Re[e^{-it}q(x)]\geq c_3|q|$ for almost all $x\in\Omega$.
\end{enumerate}
\end{assumption}

For simplicity, we denote by $(\cdot,\cdot)_{\Omega}$ the inner product of $L^2(\Omega)$. A confusion with the inner product $(\cdot,\cdot)$ of $L^2(S^{n-1})$ is not expected.
We define the weighted space $L^2(\Omega, |q|dx)$ as the completion of $L^2(\Omega)$ with respect to the norm
$(\phi,\psi)_{L^2(\Omega, |q|dx)}=(\phi,|q|\psi)_{\Omega}$.
We say that $k^2$ is an interior transmission eigenvalue if there exists $(u,w)\in H^1_{0}(\Omega)\times L^2(\Omega, |q|dx)$ with  $(u,w)\neq (0,0)$ and a sequence $\{w_j\}$ in
$H^2(\Omega)$ with $w_j\rightarrow w$ in $L^2(\Omega, |q|dx)$ and $\Delta w_j+k^2w_j=0$ in $\Omega$ and
\begin{eqnarray*}
\int_{\Omega}[\nabla u\cdot\nabla\psi-k^2(1+q)u\psi]dx=k^2\int_{\Omega}qw\psi\,dx \quad \mbox{for\,all}\,\psi\in H^1(\Omega).
\end{eqnarray*}
We list now some of the results on the factorization of the far field operator for inhomogeneous medium \cite{KirschGrinberg}.

\begin{lemma}\label{fmedium}{\rm (See Theorems 4.5, 4.6 and 4.8 of \cite{KirschGrinberg})}\,
Let Assumption \ref{qassumption} hold and $F:L^2(S^{n-1})\rightarrow L^2(S^{n-1})$ be the far field operator defined by \eqref{ffoperator}. Then
\begin{enumerate}
  \item We have the following factorization of  the far field operator $F$
        \begin{eqnarray*}
        F=H^{\ast}TH.
        \end{eqnarray*}
        The operator $H:L^2(S^{n-1})\rightarrow L^2(\Omega)$ is defined by
        \begin{eqnarray*}
        (Hg)(x)=\sqrt{|q(x)|}\int_{S^{n-1}}g(\theta)e^{ikx\cdot\theta}ds(\theta),\quad x\in \Omega.
        \end{eqnarray*}
        The operator $T:L^2(\Omega)\rightarrow L^2(\Omega)$ is defined by
        \begin{eqnarray*}
        Tf=k^2(signq)(f+\sqrt{|q|}v_{\Omega}),\quad f\in L^2(\Omega),
        \end{eqnarray*}
        where $signq:=q/|q|$ and $v\in H^1_{loc}({\mathbb R}^n)$ is the radiating solution of $\Delta v+k^2(1+q)v=-k^2(signq)f$ in ${\mathbb R}^n$.
  \item Let $B_{\epsilon}(z)\subset \Omega$ be some closed ball with center $z$ and radius $\epsilon>0$ which is completely contained in $\Omega$. Choose a function $\chi\in C^{\infty}({\mathbb R})$
        with $\chi(t)=1$ for $|t|\geq \epsilon$ and $\chi(t)=0$ for $|t|\leq\epsilon/2$ and define $w_0\in C^{\infty}({\mathbb R}^3)$ by $w_0=\chi(|x-z|)\Phi(x,z)$ in ${\mathbb R}^n$. Now we set
        \begin{eqnarray}\label{w}
        w=\left\{
              \begin{array}{ll}
                -(\Delta w_0+k^2w_0)/\sqrt{|q|}, & \hbox{in $B_{\epsilon}(z)$;} \\
                0, & \hbox{in $\Omega\backslash\overline{B_{\epsilon}(z)}$.}
              \end{array}
            \right.
        \end{eqnarray}
        Then $w\in L^2(\Omega)$ and $\phi_z=H^{\ast}w$ where $\phi_z$ is again defined by \eqref{phiz}.
  \item Define $T_0:L^2(\Omega)\rightarrow L^(\Omega)$ by $T_0f=k^2(signq)f$ for $f\in L^2(\Omega)$. Then $T-T_0$ is compact and $\Re[e^{-it}T_0]$ is coercive, i.e.,
        \begin{eqnarray}\label{T0coercive}
        \Re[e^{-it}(T_0f,f)_{\Omega}]\geq c\|f\|^2_{L^2(\Omega)},\quad f\in L^2(\Omega)
        \end{eqnarray}
        for some constant $c>0$.
  \item Assume that $k^2$ is not an interior transmission eigenvalue. Then
       \begin{eqnarray}
       \Im(Tf, f)_{\Omega}>0\quad \forall\,\,f\in\overline{range(H)}, \,f\neq 0.
       \end{eqnarray}
\end{enumerate}
\end{lemma}

We will provide the inf-criterion for inverse scattering by inhomogeneous medium with the help of the following lemma.

\begin{lemma}\label{Tcoercive}
In addition to Assumption \ref{qassumption} assume that $k^2$ is not an interior transmission eigenvalue.
Then the middle operator $T:L^2(\Omega)\rightarrow L^2(\Omega)$ satisfy the coercivity condition, i.e., there exists a constant $c>0$ with
\begin{eqnarray}\label{Tproperty}
|(Tf, f)_{\Omega}|\geq c\|f\|^2_{L^2(\Omega)},\quad\forall\,\, f\in range(H).
\end{eqnarray}
\end{lemma}
\begin{proof}
If there exists no constant $c>0$ with \eqref{Tproperty} then there exists a sequence $\{f_j\}$ in $range(H)$ such that
\begin{eqnarray}\label{fj1}
\|f_j\|_{L^2(\Omega)}=1\quad\mbox{and}\quad (Tf_j, f_j)_{\Omega}\rightarrow 0,\quad\mbox{as} \quad j\rightarrow\infty.
\end{eqnarray}
Since the unit ball in $L^2(\Omega)$ is weakly compact there exists a subsequence which converge weakly to some $f\in\overline{range(H)}$.
We denote this subsequence again by $\{f_j\}$. By Lemma \ref{fmedium}(3), the difference operator $T-T_0$ is compact,
which implies that
\begin{eqnarray}\label{TT0}
(T-T_0)f_j\rightarrow (T-T_0)f\quad \mbox{in}\, L^2(\Omega),
\end{eqnarray}
and thus also
\begin{eqnarray}\label{TT02}
\big((T-T_0)(f-f_j), f_j\big)_{\Omega}\rightarrow 0\quad\mbox{as} \quad j\rightarrow\infty.
\end{eqnarray}
By linearity,
\begin{eqnarray*}
(Tf, f_j)_{\Omega}=(Tf_j, f_j)_{\Omega}+\big((T-T_0)(f-f_j), f_j\big)_{\Omega}+\big(T_0(f-f_j), f\big)_{\Omega}-\big(T_0(f-f_j), f-f_j\big)_{\Omega}.
\end{eqnarray*}
The left hand side converges to $(Tf, f)_{\Omega}$, the first three terms on the right hand side converge to zero.
By the definition of $T_0$ and the assumption that $\Im(q)\geq 0$, we deduce that $\Im (\big(T_0(f-f_j), f-f_j\big)_{\Omega})\geq 0$.
This fact combines the forth result of Lemma \ref{fmedium} implies that $f=0$.
Therefore, by using the third result of Lemma \ref{fmedium}, we have
\begin{eqnarray*}
c\|f_j\|^2_{L^2(\Omega)}
&\leq& \Re[e^{-it}(T_0f_j,f_j)_{\Omega}]\leq |e^{-it}(T_0f_j,f_j)_{\Omega}|\cr
&=&|(T_0f_j,f_j)_{\Omega}|\leq |\big((T_0-T)f_j,f_j\big)_{\Omega}|+|(Tf_j,f_j)_{\Omega}|
\end{eqnarray*}
which tends to zero as $j\rightarrow \infty$. Therefore, also $f_j\rightarrow 0$ which contradicts to the assumption \eqref{fj1}, i.e., $\|f_j\|_{L^2(\Omega)}=1$.
\end{proof}

Now the analogous result of Lemma \ref{infcriobstacle} for inhomogeneous medium, by using Theorem 1.16 from \cite{KirschGrinberg} and the previous two Lemmas
\ref{fmedium} and \ref{Tcoercive}, can be formulated as the following lemma.

\begin{lemma}\label{infcrimedium}
Consider the inverse scattering by inhomogeneous medium.
In addition to Assumption \ref{qassumption} assume that $k^2$ is not an interior transmission eigenvalue.
Then $z\in\Omega$ if, and only if,
\begin{eqnarray*}
\inf|\{|(F\psi, \psi)|:\,\psi\in A_z\}>0.
\end{eqnarray*}
Furthermore, for $z\in\Omega$ we have the estimate
\begin{eqnarray}\label{estimatemedium}
\inf|\{|(F\psi, \psi)|:\,\psi\in A_z\}\geq \frac{c}{\|w(\cdot,z)\|^2_{L^{2}(\Omega)}},
\end{eqnarray}
for some constant $c>0$ which is independent of $z$. Here $w$ is defined by \eqref{w}.
\end{lemma}

Lemmas \ref{infcriobstacle} and \ref{infcrimedium} are satisfactory from the theoretical point of view.
However, there is a major drawback with respect to the computational point of view since it is very time consuming to solve a minimization
problem for every sampling point $z$.
What is interesting is that the estimates \eqref{estimate} and \eqref{estimatemedium} given, respectively, in Lemmas \ref{infcriobstacle} and \ref{infcrimedium}
provide some insight to our indicator $I_{new}(z)$ for sampling points $z\in\Omega$.
Actually, a straightforward calculation shows that
\begin{eqnarray}\label{gama}
\gamma:=(\phi_z, \phi_z)=\int_{S^{n-1}}|\phi_z|^2ds=\int_{S^{n-1}}1ds=\left\{
                                                                    \begin{array}{ll}
                                                                      2\pi, & \hbox{$n=2$;} \\
                                                                      4\pi, & \hbox{$n=3$.}
                                                                    \end{array}
                                                                  \right.
\end{eqnarray}
This implies $\psi_z:=\phi_z/\gamma\in A_z$, and therefore, by noting the linearity of the far field operator $F$ and using the estimate \eqref{estimate} or \ref{estimatemedium}, we have
\begin{eqnarray*}
I_{new}(z)
&=&|( F\phi_z, \phi_z)|\cr
&=&\gamma|( F\psi_z, \phi_z)|\cr
&\geq&\gamma\inf|\{|(F\psi, \psi)|:\,\psi\in A_z\}\cr
&\geq& \frac{c\gamma}{M_z},\quad z\in\Omega
\end{eqnarray*}
for some constant $c>0$ which is independent of $z$. Here $M_z$ is defined by
\begin{eqnarray}\label{Mz}
M_z:=\left\{
             \begin{array}{ll}
               \|\Phi(\cdot,z)\|^{2}_{H^{1/2}(\partial\Omega)}, & \hbox{for the scattering by impenetrable scatterers;} \\
               \|w(\cdot,z)\|^2_{L^{2}(\Omega)}, & \hbox{for the scattering by inhomogeneous medium,}
             \end{array}
           \right.
\end{eqnarray}
with $w$ defined by \eqref{w}.
We formulate this result as the main result of this subsection.

\begin{theorem}\label{Ilowbound}
Under the assumptions of Lemmas \ref{infcriobstacle} and \ref{infcrimedium}, we have
\begin{eqnarray}
I_{new}(z)\geq \frac{c\gamma}{M_z},\quad z\in\Omega
\end{eqnarray}
for some constant $c>0$ which is independent of $z$. Here, $M_z$ is defined by \eqref{Mz}, $\gamma$ is defined by \eqref{gama}.
\end{theorem}

Theorem \ref{Ilowbound} provides a lower bound of the indicator $I_{new}$ for sampling points in $\Omega$.
We finally remark that the assumptions in Theorem \ref{Ilowbound} are only used for theoretical analysis.

\subsection{Resolution analysis for the sampling points go away from the boundary $\partial\Omega$}
This subsection is devoted to the study of the behavior of the indicator $I_{new}$ for sampling points outside the scatterer.

Let $Y_\alpha^\beta(\cdot)$ for $\alpha\in\mathbb{N}\cup\{0\}$ and $\beta=-\alpha,\ldots, \alpha$ ($\beta=\pm\alpha$ in two-dimensional case) be the
spherical harmonics which form a complete orthonormal system in $L^2(\mathbb{S}^{n-1})$ (cf. \cite{CK}).
In particular, we recall the spherical harmonics $Y_\alpha^\beta(\hat{x})$ of order $\alpha=0,1$, for $\hat{x}=(\hat{x}^l)_{l=1}^n\in\mathbb{S}^{n-1}$. In the three-dimensional case,
\begin{eqnarray*}
Y^0_0(\hat{x})=\sqrt{\frac{1}{4\pi}},\
Y^{-1}_1(\hat{x})=\sqrt{\frac{3}{8\pi}}(\hat{x}^1-i\hat{x}^2),\
Y^{0}_1(\hat{x})=\sqrt{\frac{3}{4\pi}}\hat{x}^3,\
Y^{1}_1(\hat{x})=\sqrt{\frac{3}{8\pi}}(\hat{x}^1+i\hat{x}^2).
\end{eqnarray*}
In the two-dimensional case, $Y_1^0$ does not exist and
\begin{eqnarray*}
Y^0_0(\hat{x})=\sqrt{\frac{1}{2\pi}},\quad
Y^{-1}_1(\hat{x})=\sqrt{\frac{1}{2\pi}}(\hat{x}^1-i\hat{x}^2),\quad
Y^{1}_1(\hat{x})=\sqrt{\frac{1}{2\pi}}(\hat{x}^1+i\hat{x}^2).
\end{eqnarray*}

\begin{lemma}\label{lemma_funkheck}
\begin{eqnarray*}\label{FunkHeckgeneralize}
\int_{S^{n-1}}e^{-ik\hat{x}\cdot p}ds(\hat{x}) &=& \mu_{0}f_{0}(k|p|),\quad p\in{\mathbb R}^n, \\
\int_{S^{n-1}}\hat{x} e^{-ik\hat{x}\cdot p}ds(\hat{x}) &=& \left\{
                                                 \begin{array}{ll}
                                                   0, & \hbox{$p=0$;} \\
                                                   \mu_{1}\hat{p}f_1(k|p|), & \hbox{$p\in{\mathbb R}^n, p\neq 0$,}
                                                 \end{array}
                                               \right.
\end{eqnarray*}
where
\begin{eqnarray*}
\hat{p}=p/|p|,\quad
\mu_{\alpha}=\left\{
               \begin{array}{ll}
                 \frac{2\pi}{i^{\alpha}}, & \hbox{$n=2$;} \\
                 \frac{4\pi}{i^{\alpha}}, & \hbox{$n=3$}
               \end{array}
             \right.
\qquad\mbox{and}\qquad
f_{\alpha}(t)=\left\{
               \begin{array}{ll}
                 J_{\alpha}(t), & \hbox{$n=2$;} \\
                 j_{\alpha}(t), & \hbox{$n=3$}
               \end{array}
             \right.
\end{eqnarray*}
with $J_{\alpha}$ and $j_{\alpha}$ being the Bessel functions and spherical Bessel functions of order $\alpha$, respectively.
\end{lemma}
\begin{proof}
The lemma follows by the well known Funk-Hecke formula
\begin{eqnarray*}\label{FunkHeckformula}
\int_{S^{n-1}}e^{-ikz\cdot\hat{x} }Y_\alpha^\beta(\hat{x})ds(\hat{x})=\mu_{\alpha}f_{\alpha}(k|z|)Y_\alpha^\beta(\hat{z}),
\end{eqnarray*}
and the fact that
\begin{eqnarray*}
\hat{x} = (\hat{x}^1, \hat{x}^2, \cdots, \hat{x}^n)
    = \left\{
        \begin{array}{ll}
          \sqrt{\pi/2}\Big(Y_1^1(\hat{x})+Y_1^{-1}(\hat{x}),\,\,iY_1^{-1}(\hat{x})-iY_1^{1}(\hat{x})\Big), & \hbox{$n=2$;} \\
          \sqrt{2\pi/3}\Big(Y_1^1(\hat{x})+Y_1^{-1}(\hat{x}),\,\, iY_1^{-1}(\hat{x})-iY_1^{1}(\hat{x}),\,\, \sqrt{2}Y_1^{0}(\hat{x})\Big), & \hbox{$n=3$.}
        \end{array}
      \right.
\end{eqnarray*}
\end{proof}

\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Bessel function $J_0$}]{
    \includegraphics[width=3in]{pic/Bessel0.pdf}}
  \subfigure[\textbf{Bessel function $J_1$}]{
    \includegraphics[width=3in]{pic/Bessel1.pdf}}
  \subfigure[\textbf{Spherical Bessel function $j_0$}]{
    \includegraphics[width=3in]{pic/SphericalBessel0.pdf}}
  \subfigure[\textbf{Spherical Bessel function $j_1$}]{
    \includegraphics[width=3in]{pic/SphericalBessel1.pdf}}
    \caption{Decay behaviors of $f_{\alpha}$ in two dimensions (a)-(b), and in three dimensions (c)-(d) for $\alpha=0,1$.}
\label{falphadecay}
\end{figure}

For both the scattering problems \eqref{HemEqumedium}-\eqref{Srcmedium} and \eqref{HemEquobstacle}-\eqref{Srcobstacle}, it is well known that the scattering amplitude
has the following form (cf. \cite{KirschGrinberg})
\begin{eqnarray*}
u^{\infty}(\hat{x},\hat{\theta})=\int_{\partial\Omega}\left\{u^s(y,\hat{\theta})\frac{\partial e^{-ik\hat{x}\cdot y}}{\partial\nu(y)}-\frac{\partial u^s}{\partial\nu}(y,\hat{\theta})e^{-ik\hat{x}\cdot y}\right\}ds(y),\quad \hat{x}\in S^{n-1}.
\end{eqnarray*}
Inserting this into our indicator \eqref{IndicatorLiu} yields
\begin{eqnarray}\label{Inewz2}
&&I_{new}(z)\cr
&=&\Big|\int_{S^{n-1}}\int_{S^{n-1}}\int_{\partial\Omega}\left\{u^s(y,\hat{\theta})\frac{\partial e^{-ik\hat{x}\cdot (y-z)}}{\partial\nu(y)}-\frac{\partial u}{\partial\nu}(y,\hat{\theta})e^{-ik\hat{x}\cdot (y-z)}\right\}ds(y)ds(\hat{x})e^{-ik\hat{\theta}\cdot z}ds(\hat{\theta})\Big|\cr
&=&\Big|\int_{S^{n-1}}\int_{\partial\Omega}\int_{S^{n-1}}\left\{-iku^s(y,\hat{\theta})\nu(y)\cdot\hat{x} e^{-ik\hat{x}\cdot (y-z)}-\frac{\partial u^s}{\partial\nu}(y,\hat{\theta})e^{-ik\hat{x}\cdot (y-z)}\right\}ds(\hat{x})ds(y)\cr
&& \qquad\qquad\qquad\qquad  e^{-ik\hat{\theta}\cdot z}ds(\hat{\theta})\Big|\cr
&:=&\Big|\int_{S^{n-1}}G(z, \hat{\theta})e^{-ik\hat{\theta}\cdot z}ds(\hat{\theta})\Big|
\end{eqnarray}
with
\begin{eqnarray*}
G(z,\hat{\theta})&:=&\int_{\partial\Omega}\Big\{-iku^s(y,\hat{\theta})\nu(y)\cdot \int_{S^{n-1}}\hat{x} e^{-ik\hat{x}\cdot (y-z)}ds(\hat{x})\cr
&&  \qquad\quad       -\frac{\partial u^s}{\partial\nu}(y,\hat{\theta})\int_{S^{n-1}}e^{-ik\hat{x}\cdot (y-z)}ds(\hat{x})\Big\}ds(y).
\end{eqnarray*}
By the well known Riemann-Lebesgue Lemma, we obtain that
\begin{eqnarray}\label{Inewz2}
I_{new}(z)\rightarrow 0\quad\mbox{as}\,|z|\rightarrow\infty.
\end{eqnarray}
From Lemma \ref{lemma_funkheck} we deduce that
\begin{eqnarray*}
G(z,\hat{\theta})=\int_{\partial\Omega}\left\{-ik\mu_1u^s(y,\hat{\theta})\nu(y)\cdot\frac{y-z}{|y-z|}f_1(k|y-z|)-\mu_0\frac{\partial u^s}{\partial\nu}(y,\hat{\theta})f_0(k|y-z|)\right\}ds(y).
\end{eqnarray*}
This means that $G(z,\hat{\theta})$ and thus also $I_{new}(z)$ are superpositions of the Bessel functions $f_0$ and $f_1$.
For large argument, we have the following asymptotic formulas for the Bessel and spherical Bessel functions
\begin{eqnarray*}
j_0(t)&=&\frac{\sin t}{t}\left\{1+O\Big(\frac{1}{t}\Big)\right\},\quad t\rightarrow\infty,\cr
j_1(t)&=&\frac{\cos t}{t}\left\{-1+O\Big(\frac{1}{t}\Big)\right\},\quad t\rightarrow\infty,\cr
J_0(t)&=&\frac{\cos t+\sin t}{\sqrt{\pi t}}\left\{1+O\Big(\frac{1}{t}\Big)\right\},\quad t\rightarrow\infty,\cr
J_1(t)&=&\frac{\cos t-\sin t}{\sqrt{\pi t}}\left\{-1+O\Big(\frac{1}{t}\Big)\right\},\quad t\rightarrow\infty.
\end{eqnarray*}
We refer the readers to Figure \ref{falphadecay} for a visual display of the behavior of these four functions.
Thus, we expect that the $I_{new}(z)$ decays as the sampling point $z$ goes away from the boundary $\partial\Omega$.
Actually, this phenomenon have been verified in a lot of the numerical simulations, see Section 3 in this paper.

We end this subsection by a stability statement, which reflects an important feature of the reconstruction scheme under consideration.
\begin{theorem}{\rm (Stability statement).}\label{stablity}
\begin{eqnarray}\label{stabilityestimate}
I_{new}(z)-I_{new}^{\delta}(z)\leq c\|u^{\infty}-u_{\delta}^{\infty}\|_{L^2(S^{n-1}\times S^{n-1})},\quad z\in{\mathbb R}^n.
\end{eqnarray}
where $I_{new}^{\delta}(z)$ is the indicator functional with $u^{\infty}$ replaced by $u_{\delta}^{\infty}$, $c$ is a constant independent of sampling point $z$.
\end{theorem}
\begin{proof}
\begin{eqnarray*}
&&I_{new}(z)-I_{new}^{\delta}(z)\cr
&:=&\Big|\int_{S^{n-1}}e^{-ik\hat{\theta}\cdot z}\int_{S^{n-1}}u^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})ds(\hat{\theta})\Big|\cr
&&-\Big|\int_{S^{n-1}}e^{-ik\hat{\theta}\cdot z}\int_{S^{n-1}}u_{\delta}^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})ds(\hat{\theta})\Big|\cr\vspace{2cm}
&\leq&\Big|\int_{S^{n-1}}e^{-ik\hat{\theta}\cdot z}\int_{S^{n-1}}[u^\infty(\hat{x},\hat{\theta})-u_{\delta}^\infty(\hat{x},\hat{\theta})]e^{ik\hat{x}\cdot z}ds(\hat{x})ds(\hat{\theta})\Big|\cr
&\leq&c\|u^\infty-u_{\delta}^\infty\|_{L^{2}(S^{n-1}\times S^{n-1})}.
\end{eqnarray*}
where we have used the Triangle Inequality in first inequality and the Cauchy-Schwarz Inequality in the second inequality.
\end{proof}

\subsection{Comparisons with other sampling methods}
In this subsection, we will study some connections with some other sampling methods. We mainly consider the Orthogonality Sampling Method proposed by Potthast in \cite{Potthast2010}
and the Reverse Time Migration proposed by Chen et.al. in \cite{CCHuang}.

In 2010, Potthast proposed a direct sampling method in \cite{Potthast2010} which he called Orthogonality Sampling based on the indicator
\begin{eqnarray}\label{IndicatorPotthast1}
I_{small}(z,\hat{\theta}):=\Big|\int_{S^{n-1}}u^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})\Big|,\quad z\in{\mathbb R}^n.
\end{eqnarray}
This indicator actually is an evaluation of the modulus of the $L^2(S^{n-1})$ inner product of the far field measurements $( u^\infty(\hat{x},d)$ and suitably chosen function
$e^{-ik\hat{x}\cdot z}$,
thus is very easy and simple for numerical simulation.
For small objects (whose size are much smaller than the wavelength), theoretical analysis shows that the indicator behaves like the Bessel functions.
Numerical examples (see also \cite{LiZou}) show the feasibility and effectiveness of the indicator $I_{small}$ for location reconstruction, in particular for small objects.
However, $I_{small}$ does not work for shape reconstruction, at least for extended scatterers.
To solve this difficulty, Potthast suggested in \cite{Potthast2010} the superpositions of $I_{small}$
with respect to the all the incident directions, i.e., by using the indicator
\begin{eqnarray}\label{IndicatorPotthast2}
I^{(\rho)}_{OSM}(z)
&:=&\int_{S^{n-1}} [I_{small}(z,\hat{\theta})]^{\rho} ds(\hat{\theta})\cr
&=&\int_{S^{n-1}}\Big|\int_{S^{n-1}}u^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})\Big|^{\rho} ds(\hat{\theta}),\quad z\in{\mathbb R}^n,
\end{eqnarray}
with $\rho=1$ or $\rho=2$.
Numerical examples show that it is indeed a good indicator for shape reconstruction for extended scatterers. However, no theoretical analysis is established for the behavior
of $I^{(\rho)}_{OSM}$.
As pointed out by Potthast in \cite{Potthast2010}, the modulus as used in \eqref{IndicatorPotthast2} before further integral is very important for reconstruction. Recall our
indicator \eqref{IndicatorLiu2}, that is,
\begin{eqnarray}\label{IndicatorLiu2}
I_{new}(z):=\Big|\int_{S^{n-1}}e^{-ik\hat{\theta}\cdot z}\int_{S^{n-1}}u^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})ds(\hat{\theta})\Big|,\quad z\in{\mathbb R}^n,
\end{eqnarray}
where the modulus is taken after the intergral but a weight function $e^{-ik\hat{\theta}\cdot z}$ is used in the integral w.r.t. $\hat{\theta}\in S^{n-1}$.
Another work close to our method is the Reverse Time Migration method which has been studied by Chen et.al. in \cite{CCHuang}.
The authors considered the point source as the incident field and the scattered field as the measurement. Based on similar arguments,
for the case of incident plane waves and far field measurements, one may consider the following indicator functional
\begin{eqnarray}\label{RTM}
I_{RTM}(z):=\Im\int_{S^{n-1}}e^{-ik\hat{\theta}\cdot z}\int_{S^{n-1}}u^\infty(\hat{x},\hat{\theta})e^{ik\hat{x}\cdot z}ds(\hat{x})ds(\hat{\theta}),\quad z\in{\mathbb R}^n.
\end{eqnarray}

In the following, we will show that
\begin{eqnarray}\label{IndicatorsRelations}
c I^{2}_{OSM}(z)\leq I_{RTM}(z)\leq I_{new}(z) \leq C \sqrt{I^{2}_{OSM}(z)},
\end{eqnarray}
for two positive constants $c$ and $C$ which are independent of the sampling point $z\in {\mathbb R}^n$.
This implies these indicators  $I^{2}_{OSM}(z)$ in \eqref{IndicatorPotthast2},  $I_{new}(z)$ in \eqref{IndicatorLiu2} and
$I_{RTM}(z)$ in \eqref{RTM} are equivalent in some sense.

To show \eqref{IndicatorsRelations}, we recall the reciprocity relation of the scattering amplitudes and some properties of the far field operator $F$ given by \eqref{ffoperator}.

\begin{lemma}\label{Fproperties}
(1) The scattering amplitude satisfies the reciprocity relation
\begin{eqnarray}\label{reciprocityrelation}
u^{\infty}(\hat{x},\hat{\theta})=u^{\infty}(-\hat{\theta}, -\hat{x})\quad\forall\, \hat{x},\hat{\theta}\in S^{n-1}.
\end{eqnarray}\\
(2) The far field operator satisfies
\begin{eqnarray}\label{FFastR}
F-F^{\ast}-\frac{i}{4\pi}\Big(\frac{k}{2\pi}\Big)^{n-2}F^{\ast}F=2iR,
\end{eqnarray}
where $F^{\ast}$ denotes the $L^2-$adjoint of $F$ and $R: L^2(S^{n-1})\rightarrow L^2(S^{n-1})$ is some self-adjoint non-negative operator.
The operator $R$ vanishes for the cases of Dirichlet or Neumann boundary conditions.
For the impedance boundary conditions, the operator $R$ is given by
\begin{eqnarray}\label{Rimp}
(Rh)(\hat{x}):=\int_{S^{n-1}}\Big(\int_{\partial\Omega}\Im(\lambda)u(y,\hat{\theta})\overline{u(y,\hat{x})}ds(y)\Big)h(\hat{\theta})ds(\hat{\theta}),\quad \hat{x}\in S^{n-1}.
\end{eqnarray}
For the case of inhomogeneous medium, the operator $R$ is given by
\begin{eqnarray}\label{Rmed}
(Rh)(\hat{x}):=\int_{S^{n-1}}\Big(\int_{\Omega}k^2\Im(q)u(y,\hat{\theta})\overline{u(y,\hat{x})}dy\Big)h(\hat{\theta})ds(\hat{\theta}),\quad \hat{x}\in S^{n-1},
\end{eqnarray}
where $u(\cdot,\hat{\theta})$ is the total field in $\Omega$ corresponding to the incident plane wave $u^i(\cdot,\hat{\theta})$ with incident direction $\hat{\theta}$.
Clearly, the operator $R$ also vanishes for the cases of $\Im(\lambda)=0$ or $\Im(q)=0$.
\end{lemma}
\begin{proof}
The proof of three-dimensional case can be found in Theorems 1.8, 2.5 and 4.4 of \cite{KirschGrinberg} (see also Theorem 2.1 in \cite{KirschLiuMMAS}).
By literally the same proofs, one can show the two-dimensional case with suitably modified coefficient. Thus we omit the proof here.
\end{proof}

By interchanging the roles of $\hat{x}$ and $\hat{\theta}$, with the help of the reciprocity relation \eqref{reciprocityrelation}, we have
\begin{eqnarray}\label{Fphiz2}
I^{(2)}_{OSM}(z)
&=&\int_{S^{n-1}}\Big|\int_{S^{n-1}}u^{\infty}(\hat{x},\hat{\theta})e^{ikz\cdot\hat{x}}ds(\hat{x})\Big|^2ds(\hat{\theta})\cr
&=&\int_{S^{n-1}}\Big|\int_{S^{n-1}}u^{\infty}(-\hat{\theta},-\hat{x})e^{ikz\cdot\hat{x}}ds(\hat{x})\Big|^2ds(\hat{\theta})\cr
&=&\int_{S^{n-1}}\Big|\int_{S^{n-1}}u^{\infty}(\hat{\theta},\hat{x})e^{-ikz\cdot\hat{x}}ds(\hat{x})\Big|^2ds(\hat{\theta})\cr
&=&\int_{S^{n-1}}\Big|\int_{S^{n-1}}u^{\infty}(\hat{x},\hat{\theta})e^{-ikz\cdot\hat{\theta}}ds(\hat{\theta})\Big|^2ds(\hat{x})\cr
&=&\|F\phi_z\|^2_{L^2(S^{n-1})}.
\end{eqnarray}

By using the operator identity \eqref{FFastR}, we have
\begin{eqnarray*}
\Im(Fg, g)=\frac{1}{8\pi}\Big(\frac{k}{2\pi}\Big)^{n-2}(Fg,Fg)+(Rg,g),\quad \forall\,\, g\in L^{2}(S^{n-1}).
\end{eqnarray*}
In particular, taking $g=\phi_z$ yields that
\begin{eqnarray}\label{lowb}
|(F\phi_z,\phi_z)|
&\geq& \Im(F\phi_z, \phi_z)\cr
&=& \frac{1}{8\pi}\Big(\frac{k}{2\pi}\Big)^{n-2}\|F\phi_z\|^2_{L^2(S^{n-1})}+(R\phi_z,\phi_z)\cr
&\geq& \frac{1}{8\pi}\Big(\frac{k}{2\pi}\Big)^{n-2}\|F\phi_z\|^2_{L^2(S^{n-1})},
\end{eqnarray}
where we have used the fact that $R$ is an non-negative operator.
On the other hand, by using the Cauchy-Schwartz inequality, we have
\begin{eqnarray}\label{upb}
|(F\phi_z,\phi_z)|^2
\leq \|F\phi_z\|^2_{L^2(S^{n-1})}\|\phi_z\|^2_{L^2(S^{n-1})}
= 2^{n-1}\pi\|F\phi_z\|^2_{L^2(S^{n-1})}.
\end{eqnarray}
Combing the previous two inequalities \eqref{lowb} and \eqref{upb} yields
\begin{eqnarray}
\frac{1}{8\pi}\Big(\frac{k}{2\pi}\Big)^{n-2} \|F\phi_z\|^2_{L^2(S^{n-1})}\leq \Im(F\phi_z, \phi_z)\leq |(F\phi_z,\phi_z)| \leq \sqrt{\pi}2^{\frac{n-1}{2}}\|F\phi_z\|_{L^2(S^{n-1})},
\end{eqnarray}
which actually implies \eqref{IndicatorsRelations} with $c=\frac{1}{8\pi}\Big(\frac{k}{2\pi}\Big)^{n-2}$ and $C=\sqrt{\pi}2^{\frac{n-1}{2}}$.

We have established the theory foundation of the reconstruction scheme by using the indicator $I_{new}$ in the previous subsections.
Thus similar results can be shown for the indicators $I^{(2)}_{OSM}$ in Orthogonal Sampling method and $I_{RTM}$ in Reverse Time Migration method.

Finally, we want to remark that similar indicators also been proposed in \cite{BaoHuangLiZhao} based on the idea of the MUSIC algorithm and in
\cite{BellisBonnetCakoni} by using the topological derivative of far field measurements-based $L^{2}$ cost functionals. The contributions of our paper
are new theory basis of such kind sampling methods and extensive numerical experiments shown in the next section.

\section{Numerical examples and discussions}
\label{sec3}
\setcounter{equation}{0}

Now we turn to present a variety of numerical examples in two dimensions to illustrate the applicability and effectiveness of our sampling method.
All the programs in our experiments are written in Matlab and run on a Core i5-5200U 2.2GHz PC.

There are totally seven groups of numerical tests to be considered, and they are
respectively referred to as {\bf Comparison, Dirichlet, OtherPhyPro, MixedType, MultiScalar, ResolutionLimit}  and {\bf HighResolution}.
The boundaries of the scatterers used in our numerical experiments are parameterized as follows:
\begin{eqnarray}
\label{circle}&\mbox{\rm Circle:}&\quad x(t)\ =(a,b)+r\ (\cos t, \sin t),\quad 0\leq t\leq2\pi,\\
\label{peanut}&\mbox{\rm Peanut:}&\quad x(t)\ =(a,b)+\ \sqrt{3\cos^2 t+1}(\cos t, \sin t),\quad 0\leq t\leq2\pi,\\
\label{pear}&\mbox{\rm Pear:}&\quad x(t)\ =(a,b)+(2+0.3\cos 3t)\ (\cos t, \sin t),\quad 0\leq t\leq2\pi,\\
\label{kite}&\mbox{\rm Kite:}&\quad x(t)\ =(a,b)+\ (\cos t+0.65\cos 2t-0.65, 1.5\sin t),\quad 0\leq t\leq2\pi,
\end{eqnarray}
with $(a,b)$ be the location of the scatter which may be different in different examples.

In our simulations, we used the boundary integral equation method to compute the scattering amplitudes $u_\Omega^\infty(\theta_j, \theta_l)$  with $\theta_j = 2\pi j/N$,
for $N$ equidistantly distributed incident directions and $N$ observation directions.
These data are then stored in the matrices $F_\Omega \in {\mathbb C}^{N \times N}$.
We further perturb
$F_\Omega$ by random noise using
\begin{eqnarray*}
F_{\Omega}^{\delta}\ =\ F_{\Omega} +\delta\|F_{\Omega}\|\frac{R_1+R_2 i}{\|R_1+R_2 i\|},
\end{eqnarray*}
where $R_1$ and $R_2$ are two $N \times N$ matrixes containing pseudo-random values
drawn from a normal distribution with mean zero and standard deviation one. The
value of $\delta$ used in our code is $\delta:=\|F_{\Omega}^{\delta} -F_{\Omega}\|/\|F_{\Omega}\|$ and so presents the relative error.

In the simulations, we used a grid $\mathcal{G}$ of $M\times M$ equally spaced sampling points on some rectangle $[-c,c]\times[-c,c]$.
For each point $z \in \mathcal{G}$, we define the indicator function
\begin{eqnarray*}
W(z)\ :=|\phi_z^{\ast}F_{\Omega}^{\delta}\phi_z|^{\rho},
\end{eqnarray*}
where $\phi_z=(e^{-ik\theta_1\cdot z}, e^{-ik\theta_2 \cdot z},\dots, e^{-ik\theta_{N}\cdot z})^\top \in {\mathbb C}^{N}$.
Clearly, the indicator is independent of any a priori information of the unknown scatterers. \\

{\bf Example Comparison.}
This example is designed to compare the reconstructions by using different sampling methods. We set the wave number $k=5$, $\rho=1$ and consider a sound soft kite shaped domain.
The research domain is $[-4,4]\times[-4,4]$ with $151\times 151$ equally spaced sampling points.
We take $N=64$, i.e., the scattering amplitude is collected in 64 observation directions and 64 incident directions.
Figure \ref{comparison} show the reconstruction by using the Factorization method, Orthogonal Sampling method, Reverse Time Migration method and Our novel method.
Clearly, the reconstructions are comparable to each other.
Different to the classical Factorization method, in the later three sampling methods, the indicator decays like the bessel functions as the sampling points away from
the boundary. This verifies our theory analyses and can be observed in later examples.
\begin{figure}[htbp]
  \centering
\subfigure[\textbf{Factorization method}]{
    \includegraphics[width=3in]{pic/FM.pdf}}
  \subfigure[\textbf{Orthogonal Sampling method}]{
    \includegraphics[width=3in]{pic/OSM.pdf}}
  \subfigure[\textbf{Reverse Time Migration method}]{
    \includegraphics[width=3in]{pic/RTM.pdf}}
  \subfigure[\textbf{Our novel method}]{
    \includegraphics[width=3in]{pic/Our.pdf}}
\caption{{\bf Example Comparison.}\, Reconstruction of Kite shaped domain with different sampling methods and $30\%$ noise. }
\label{comparison}
\end{figure}

\medskip

\noindent

To enhance the contrast, we take $\rho=2$ in the next five examples.

{\bf Example Dirichlet.}
This example is designed to check the validity of our method for scatterers with different shapes. For simplicity, we impose Dirichlet boundary condition on the
underlying scatterers. The other boundary conditions will be considered in the subsequent examples.
The same as previous example, we set $k=5$, $N=64$.
The research domain is $[-4,4]\times[-4,4]$ with $151\times 151$ equally spaced sampling points.
For the circle given in \eqref{circle}, we take the radius $r$ to be $2$.
Figures \ref{softkite}-\ref{softpear} show the reconstructions of kite, circle, peanut and pear shaped domain, respectively. All the results show that our method is very robust
to noise. Surprisingly, we observe that, even up to $90\%$ noise is added, both the location and shape of the underlying scatterer can be roughly reconstructed.
\begin{figure}[htbp]
  \centering
\subfigure[\textbf{Kite}]{
    \includegraphics[width=3in]{pic/SoftKite0.pdf}}
  \subfigure[\textbf{10\% noise}]{
    \includegraphics[width=3in]{pic/SoftKite1.pdf}}
  \subfigure[\textbf{30\% noise}]{
    \includegraphics[width=3in]{pic/SoftKite3.pdf}}
  \subfigure[\textbf{90\% noise}]{
    \includegraphics[width=3in]{pic/SoftKite9.pdf}}
\caption{{\bf Example Soft.}\, Reconstruction of Kite shaped domain with different noise.}
\label{softkite}
\end{figure}
\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Circle}]{
    \includegraphics[width=3in]{pic/SoftCircle0.pdf}}
  \subfigure[\textbf{10\% noise}]{
    \includegraphics[width=3in]{pic/SoftCircle1.pdf}}
  \subfigure[\textbf{30\% noise}]{
    \includegraphics[width=3in]{pic/SoftCircle3.pdf}}
  \subfigure[\textbf{90\% noise}]{
    \includegraphics[width=3in]{pic/SoftCircle9.pdf}}
\caption{{\bf Example Soft.}\, Reconstruction of a disk with different noise.}
\label{softcircle}
\end{figure}
\begin{figure}[htbp]
  \centering
\subfigure[\textbf{Peanut}]{
    \includegraphics[width=3in]{pic/SoftPeanut0.pdf}}
  \subfigure[\textbf{10\% noise}]{
    \includegraphics[width=3in]{pic/SoftPeanut1.pdf}}
  \subfigure[\textbf{30\% noise}]{
    \includegraphics[width=3in]{pic/SoftPeanut3.pdf}}
  \subfigure[\textbf{90\% noise}]{
    \includegraphics[width=3in]{pic/SoftPeanut9.pdf}}
\caption{{\bf Example Soft.}\, Reconstruction of Peanut shaped domain with different noise.
}\label{softpeanut}
\end{figure}
\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Pear}]{
    \includegraphics[width=3in]{pic/SoftPear0.pdf}}
  \subfigure[\textbf{10\% noise}]{
    \includegraphics[width=3in]{pic/SoftPear1.pdf}}
  \subfigure[\textbf{30\% noise}]{
    \includegraphics[width=3in]{pic/SoftPear3.pdf}}
  \subfigure[\textbf{90\% noise}]{
    \includegraphics[width=3in]{pic/SoftPear9.pdf}}
\caption{{\bf Example Soft.}\, Reconstruction of Pear shaped domain with different noise.
}\label{softpear}
\end{figure}

\medskip

\noindent

{\bf Example OtherPhyPro.}
For comparison, we consider the the other boundary conditions and the penetrable inhomogeneous medium. We take the pear shaped domain as an example.
Figure \ref{impedancepear} shows the results for impedance boundary conditions with different impedance functions $\lambda$.
Figure \ref{penetrablepear} shows the results for penetrable inhomogeneous medium with different contrast functions $q$.
From the reconstructions shown in Figures \ref{softpear}-\ref{penetrablepear}, we observe that the shape of the pear shaped domain can always be roughly captured,
it only changes slightly for different physical properties. This further show that our method is independent of the physical properties of the underlying scatterer.\\
\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{$\lambda=0$}]{
    \includegraphics[width=3in]{pic/ImpedancePear30.pdf}}
  \subfigure[\textbf{$\lambda=1$}]{
    \includegraphics[width=3in]{pic/ImpedancePear31.pdf}}
  \subfigure[\textbf{$\lambda=i$}]{
    \includegraphics[width=3in]{pic/ImpedancePear3i.pdf}}
  \subfigure[\textbf{$\lambda=1+i$}]{
    \includegraphics[width=3in]{pic/ImpedancePear31jiai.pdf}}
    \caption{{\bf Example OtherPhyPro.}\, Reconstruction of Pear shaped domain with different impedance conditions and $30\%$ noise.}
\label{impedancepear}
\end{figure}

\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{$q=0.5$}]{
    \includegraphics[width=3in]{pic/PenetrablePear33.pdf}}
  \subfigure[\textbf{$q=0.5+0.5i$}]{
    \includegraphics[width=3in]{pic/PenetrablePear32.pdf}}
  \subfigure[\textbf{$q=-0.5+0.5i$}]{
    \includegraphics[width=3in]{pic/PenetrablePear31.pdf}}
  \subfigure[\textbf{$q=-0.5$}]{
    \includegraphics[width=3in]{pic/PenetrablePear34.pdf}}
    \caption{{\bf Example OtherPhyPro.}\, Reconstruction of Pear shaped domain with different inhomogeneities and $30\%$ noise.}
\label{penetrablepear}
\end{figure}

\medskip

\noindent

{\bf Example MixedType.}
We consider the scattering by a scatterer with two disjoint components. We set the scatterer to be a combination of a peanut shaped domain centered at
$(-3,3)$ and a kite shaped domain centered at $(3,-3)$. Further, we impose different physical properties on each component.
The search domain is the rectangle $[-7,7]\times[-7,7]$ with $301\times 301$ equally spaced sampling points.
Figure \ref{multiscalar1} shows the reconstructions for scatterers with different physical property components.

\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Original domain}]{
    \includegraphics[width=3in]{pic/MixedType.pdf}}
  \subfigure[\textbf{Dirichlet Kite, Neumann Peanut}]{
    \includegraphics[width=3in]{pic/MixedTypeSoftKiteHardPeanutk10n128.pdf}}
  \subfigure[\textbf{Dirichlet Peanut, Neumann Kite}]{
    \includegraphics[width=3in]{pic/MixedTypeSoftPeanutHardKitek10n128.pdf}}
  \subfigure[\textbf{Dirichlet Peanut, Penetrable Kite}]{
    \includegraphics[width=3in]{pic/MixedTypeSoftPeanutPenetrableKitek10n128.pdf}}
    \caption{{\bf Example MixedType.}\, Reconstruction of mixed type scatterers with $30\%$ noise. We take $k=10$, $N=128$, $q=-0.5$.}
\label{mixedtype}
\end{figure}

\medskip

\noindent

{\bf Example MultiScalar.}
In this example, the underlying scatterer is a combination of a big pear shaped domain and a small disk with radius $r=0.1$ or $r=0.2$.
The Dirichlet boundary condition is imposed on the boundary  of the scatterer.
The research domain is $[-6,6]\times[-6,6]$ with $301\times 301$ equally spaced sampling points.
The scattering amplitude is collected in 256 observation directions and 256 incident directions.
We observe, from Figures \ref{multiscalar1} and \ref{multiscalar2}, that the boundary of the scatterer is clearly located.
Figures \ref{multiscalar1} and \ref{multiscalar2} also show that the resolution can be improved with higher wave number.
\\
\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Original domain}]{
    \includegraphics[width=2in]{pic/MultiscalarPearcircle.pdf}}
  \subfigure[\textbf{$k=5$}]{
    \includegraphics[width=2in]{pic/MultiscalarPearcircle1k5.pdf}}
  \subfigure[\textbf{$k=10$}]{
    \includegraphics[width=2in]{pic/MultiscalarPearcircle1k10.pdf}}
\caption{{\bf Example MultiScalar.}\, Reconstruction of a big pear shaped domain and a small disk with radius  $r=0.1$ and 30\% noise. }
\label{multiscalar1}
\end{figure}

\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Original domain}]{
    \includegraphics[width=2in]{pic/MultiscalarPearcircle2.pdf}}
  \subfigure[\textbf{$k=5$}]{
    \includegraphics[width=2in]{pic/MultiscalarPearcircle2k5.pdf}}
  \subfigure[\textbf{$k=10$}]{
    \includegraphics[width=2in]{pic/MultiscalarPearcircle2k10.pdf}}
\caption{{\bf Example MultiScalar.}\, Reconstruction of a big pear shaped domain and a small disk with radius  $r=0.2$ and 30\% noise. }
\label{multiscalar2}
\end{figure}

\medskip

\noindent

{\bf Example ResolutionLimit.}
As shown in Figure \ref{resolutionlimit}(a), the underling scatterer $\Omega$ is given as the union of two disjoint obstacles
$\Omega=\Omega_{1}\cup\Omega_{2}$ where $\Omega_{1}$ is a disk with radius $r=4$ and center $(a,b)=(-1,0)$, while $\Omega_{2}$ is
kite shaped domain centered at $(a,b)=(5,0)$. These two disjoint components are very close to each other.
Again, we impose Dirichlet boundary condition on the boundary $\partial \Omega$.
The search domain is the rectangle $[-7,7]\times[-7,7]$ with $301\times 301$ equally spaced sampling points.
To make the gap available, we have used 360 incident directions and 360 observation directions in our simulations.
The distance between these two scatterers is about 0.8.
The relation between the wave number $k$ and the probe wavelength $l$ is given by $kl=2\pi$.
We test two wave numbers $k=4$ and $k=8$, which means that the corresponding wavelengths are $l=2\pi/4\approx1.6$ and $l=2\pi/8\approx0.8$.
The results are shown in Figure \ref{resolutionlimit}, from which we observe that the gap appears clearly even if it is just about half a wavelength and gets more visible by
using a higher wave number.\\

\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{Original domain}]{
    \includegraphics[width=2in]{pic/ResolutionLimit360.pdf}}
  \subfigure[\textbf{$k=4$}]{
    \includegraphics[width=2in]{pic/ResolutionLimit360k4n30.pdf}}
  \subfigure[\textbf{$k=8$}]{
    \includegraphics[width=2in]{pic/ResolutionLimit360k8n30.pdf}}
\caption{{\bf Example ResolutionLimit.}\, Reconstruction of two objects that are close to each other with $30\%$ noise by using different wave numbers.}
\label{resolutionlimit}
\end{figure}

\medskip

\noindent

{\bf Example HighResolution.}
In the previous two Examples {\bf MultiScalar} and {\bf ResolutionLimit}, we have found that the resolution can be improved with the increase of the wave number $k$.
We will further verify this fact in this example. The second contribution of this example is to show that the resolution can also be improved by a higher power $\rho$.
We choose a kite shaped domain with Dirichlet boundary condition as an unknown object.
We set three different wave numbers $k=5,\,10$ and $15$ and three different powers $\rho=1,\,2$ and $\rho=8$. The corresponding observation and incident direction number $N=16*k$.
The research domain is $[-4,4]\times[-4,4]$ with $151\times 151$ equally spaced sampling points.
Figure \ref{highresolution} shows the corresponding reconstructions.
Obviously, the shadows are greatly reduced with the increase of the power $\rho$.
To our surprise, it seems that the indicator always takes its maximum on the boundary of the scatterer, which results in a shaper reconstruction of the boundary of the scatterer.
However, there is no general theory for this fact.
\\

\begin{figure}[htbp]
  \centering
  \subfigure[\textbf{$k=5,\,\rho=1$}]{
    \includegraphics[width=2in]{pic/HRk5-1.pdf}}
  \subfigure[\textbf{$k=5,\,\rho=2$}]{
    \includegraphics[width=2in]{pic/HRk5-2.pdf}}
  \subfigure[\textbf{$k=5,\,\rho=8$}]{
    \includegraphics[width=2in]{pic/HRk5-8.pdf}}
  \subfigure[\textbf{$k=10,\,\rho=1$}]{
    \includegraphics[width=2in]{pic/HRk10-1.pdf}}
  \subfigure[\textbf{$k=10,\,\rho=2$}]{
    \includegraphics[width=2in]{pic/HRk10-2.pdf}}
  \subfigure[\textbf{$k=10,\,\rho=8$}]{
    \includegraphics[width=2in]{pic/HRk10-8.pdf}}
  \subfigure[\textbf{$k=15,\,\rho=1$}]{
    \includegraphics[width=2in]{pic/HRk15-1.pdf}}
  \subfigure[\textbf{$k=15,\,\rho=2$}]{
    \includegraphics[width=2in]{pic/HRk15-2.pdf}}
  \subfigure[\textbf{$k=15,\,\rho=8$}]{
    \includegraphics[width=2in]{pic/HRk15-8.pdf}}
\caption{{\bf Example HighResolution.}\, Reconstruction by using different indicators $I^{\rho}_{new}$ with different wave numbers $k$ and different powers $\rho$. $30\%$ noise is
added to the far field data.}
\label{highresolution}
\end{figure}

In all the above examples, we observe that the reconstructions are rather satisfactory, with the consideration of the severe ill-posedness of the inverse scattering problems and the
fact that at least $30\%$ noise is added in the measurements (scattering amplitudes).

\section{Concluding remarks}
\label{sec4}
\setcounter{equation}{0}

In this paper we propose a new sampling method for shape identification in inverse acoustic scattering problems.
Both the theory foundation and numerical simulations are presented.
Only matrix multiplications are involved in the computation, thus
our method is very fast and robust against measurement noise from the numerical point of view.
The recovering scheme works independently of the physical properties of the underlying scatterers.
There might be several components with different physical properties, or with different scalar sizes, presented simultaneously.
Our method also allows us to distinguish two components of distance about one half of the wavelength, which is known to be challenging for numerical reconstruction.

However, the theory foundation is still less developed than the classical sampling method, e.g., the factorization method \cite{KirschGrinberg}. In our numerical simulations,
we have observed that the indicator always takes its maximum on the boundary of the scatterer. However, there are still no theory analysis on this fact.
Similar techniques can also be applied to inverse scattering of elastic waves or electromagnetic waves, which shall be addressed in a forthcoming work.

\section*{Acknowledgements}

The research of X. Liu was supported by the NNSF of China under grants 11571355, 61379093 and 91430102.

redpajama_set_name: RedPajamaArXiv


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text:
Q: Submit Form with Input still selected I have this simple form for which I am attempting to submit.  It has 3 input:text fields which are populated with data from a calculation depending on which field you are entering information in.  I can hit enter and it works fine because I am still focused in the field but the form won't submit if I hit the submit button.  Here is the html.
<form id="pr">

    Pressure Conversion Calculator

    Psi
    MPa
    Bar

    <input id="psi" name="psi" type="text"  size="7" value=""/>
    <input id="mpa" name="mpa" type="text"  size="7" value="" />
    <input id="bar" name="bar" type="text"  size="7" value="" />

    <input type="submit" name="button" id="submit" value="Calculate" class="submitbutton"/>
    <input type="button" name="clear" id="clear" value="Clear Fields" class="submitbutton"/>

and then the jquery
$('#submit').click(function(event){
event.preventDefault();

    if($('#psi').is(':focus')){
        pressure = parseInt($('#psi').val());
        bar = pressure * 0.0690;
        $('#bar').val(bar.toFixed());
        mpa = pressure * 0.006894759086775369;
        $('#mpa').val(mpa.toFixed());
    }

    else if($('#mpa').is(':focus')){
        pressure = parseInt($('#mpa').val());
        bar = pressure * 10;
        $('#bar').val(bar.toFixed());
        psi = pressure * 145.0377;
        psi = Math.round(psi/1000)*1000;
        $('#psi').val(psi);
    }

     else if($('#bar').is(':focus')){
        pressure = parseInt($('#bar').val());
        mpa = pressure * 0.1;
        $('#mpa').val(mpa.toFixed());
        psi = pressure * 14.500;
        psi = Math.round(psi/1000)*1000;
        $('#psi').val(psi);
    }

    else{}

});

I am kind of new to this and am just trying to work out all of the bugs.  Any help would be appreciated.

A: Use a regular button (remove type="submit"), then submit the form in your code:
document.forms[0].submit();

redpajama_set_name: RedPajamaStackExchange


text:
Peggy West Lowman
March 3, 1932 - January 15, 2018
Peggy West Lowman, 85, formerly of Altavista died Monday, January 15, 2018 at Blue Ridge Manor in Martinsville. She was the loving wife of the late Robert Preston Lowman, Jr. for sixty years. She was born March 3, 1932 in Campbell County, a daughter of the late Luther James West, Sr. and Elsie Eades West. She was a member of First Baptist Church and retired from the Campbell County School System. She is survived by two sons, Robert P. Lowman, III and his wife, Sharon of Hilton Head, SC and James Scott Lowman and his wife, Susan of Altavista; two daughters, Dana L. Pennington and her husband, Mike of Martinsville and Stacy L. Wheeler and her husband, Zach of Thomasville, GA; a sister-in-law, Janet West of Altavista; four grandchildren, Derek Pennington and his wife, Peiching of China, Michelle McKnight and her husband, Jon of Knoxville, TN, Allison Wheeler of Thomasville, GA, and Nicholas Lowman of Altavista; and three great grandchildren, Jieling Pennington, Sarah McKnight, and Emily McKnight. She was preceded in death by a brother, Luther James West, Jr. A funeral service will be conducted at 1:00 p.m., Saturday, January 20, 2018 at Finch and Finch Chapel, Altavista by Chaplain Jeff Rhea with interment to follow in Altavista Memorial Park. The family will receive friends prior to the service from noon until 12:45 p.m., at Finch and Finch Funeral and Cremation Service, Altavista and other times at the residence of Janet West, 1265 Lynch Rd., Altavista. The family suggests that those wishing to make memorials consider Mountain Valley Hospice, 430 East Church St., Ste. 13, Martinsville, VA 24112 or Lynchburg Grows, 1339 Englewood St., Lynchburg, VA 24501.
Peggy West Lowman, 85, formerly of Altavista died Monday, January 15, 2018 at Blue Ridge Manor in Martinsville. She was the loving wife of the late Robert Preston Lowman, Jr. for sixty years. She was born March 3, 1932 in Campbell County, a... View Obituary & Service Information
The family of Peggy West Lowman created this Life Tributes page to make it easy to share your memories.
Peggy West Lowman, 85, formerly of Altavista died Monday, January...
Send flowers to the Lowman family.

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A guy is driving around and sees a sign in front of a house: "Talking Dog For Sale."
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"He's a liar. He didn't do any of that crap."


The Registered Metal Crusher Unit Owners Association had submitted a representation to the Technical Committee for consideration of their demands.
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Agra is the one of the prominent destinations of the World Tourism map with three heritage monuments - The Taj Mahal, Red Fort & Fatehpur Sikri. It is more just a dacadent city of graveyards and stones, but it is a vibrant centre of Culture, Art and Religious philosphies that have enriched mankind and shaped humen thought over centuries.
Agra was once the capital of the Mughal empire and even today it seems to linger in the past . Not surprising , for the Mughal emperors with their passion for building, endowed the city with some of the finest structures in the world . It is very easy to slip away here through the centuries into the grandeur and intrigues of the Mughal court.
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This bird paradise acquired its name –Keoladeo because of the existence of the Keoladeo temple of Lord Shiva. After the name of this temple the lake is called 'Ghana' Keoladeo. Ghana being the vernacular synonym for dense forest. It was known as the best duck shooting resort during the British reign, but was declared a reserve for birds in 1956 and later upgraded to a National Park. UNESCO has listed it as a world heritage site.
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One of the benchmarks of successful public relations is the ability to attract media attention for the organization subjects. Media attention is important in a number of cases, for example: to communicate crisis responses, to build a good reputation among the public, to create publicity for key events, etc. And what a better way to do so than to create a powerful press release that gets picked up by the media?
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823 press releases issued in 2012 from 30 of the largest Dutch corporations were analysed with quantitative content analysis. News articles were sampled from Dutch news media – three newspapers in their print and online version (De Telegraaf, De Volkskrant, De Financieele Dagblad), two largest Dutch news websites and the press agency ANP. Comparing the source data with the media data turns out that only 220 press releases were picked up by the media. Which results in ignoring the other 75% press releases. This leads us to the question – what makes one press release more successful than others?
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It turns out that within the field of business news only a selection of news factors improves the press releases' chance for succession in news media. The presence of controversy, negative consequences, surprise and elite organization is found to impact the newsworthiness of a press release positively. Mentioning positive consequences on the other hand leads to a negative influence. Which reinforces the saying among journalist that actually "Bad news is good news". The inclusion of incentives hardly improved the chance for media attention. The actual message itself – the bare facts presented in the press release – of only some topics, for example, business news, was also a substantive factor in agenda building. For example, general topics like financial performance and employees and management may be interpreted as news factors in the field of business news. The paper suggests that this is because their aim is to cater for a specific audience: stock owners.
What is your opinion on the topic? Do you think that the news factors' agenda building power would vary in different context, for example culture, sports, and politics? How do you think the mix of influential news factors would differ from field to field?
For more information check out the article in Public Relations Review and get acquainted with the details of the research. For further information you may contact the corresponding author p.h.j.schafraad@uva.nl (P. Schafraad).


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text:
Zgornje Bitnje je naselje u slovenskoj Općini Kranju. Zgornje Bitnje se nalazi u pokrajini Gorenjskoj i statističkoj regiji Gorenjskoj.

Stanovništvo
Prema popisu stanovništva iz 2002. godine naselje je imalo 1,228 stanovnika.

Izvor

Naselja u Općini Kranj

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text:
Q: Does Android NDK support "-z execstack" flag? I'm tring to build a executable with android ndk to test shellcode. I set 'LOCAL_CFLAGS := -z execstack' to make the stack executable but it does not work. Doesn't android ndk support '-z execstack' option?
Here's the code:
18 include $(CLEAR_VARS)
 19 LOCAL_MODULE := shellcode-test
 20 LOCAL_SRC_FILES := src/shellcode-test.c
 21 LOCAL_CFLAGS := -z execstack -Wl,--execstack
Check result: only 'RW', no 'E'
$ aarch64-linux-gnu-readelf -l shellcode-test
...
GNU_STACK      0x0000000000000000 0x0000000000000000 0x0000000000000000
                 0x0000000000000000 0x0000000000000000  RW     0x10
...

A: Even if you get your code to compile with an executable stack, the platform won't load it as executable. Executable stacks are a big security risk.

A: Resolved.
Android NDK forces noexecstack options when build executables or libraries. So '-z execstack' in my Android.mk is overridden by NDK's default option:

./build/core/build-binary.mk:LOCAL_CFLAGS += -Wa,--noexecstack
./build/core/build-binary.mk:LOCAL_LDFLAGS += -Wl,-z,noexecstack


text:
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Addresses the participation of veterans in STEM (science, technology, engineering, and mathematics) fields, including by...


When you're in a relationship with someone, it's important to work as a team. But when your partner struggles to balance their loyalties to you and to their family, you might feel anything but united. It's common to feel hurt or betrayed if your partner doesn't back you up when their family criticizes or judges you. Not seeing eye-to-eye on how to handle family conflicts can damage your relationship with your partner, so it's important to know how to cope with this situation. You can learn to do so by communicating better with your partner, setting boundaries with your partner's family, and learning to stand up for yourself.
Choose a good time to have the talk. Discussing your partner's family is a sensitive subject, so you'll want to bring up the topic when your partner is in a receptive mood. Avoid discussing this when they are angry, tired, or stressed. Instead choose a time when you are both in a good mood and relatively relaxed.
It may even be a good idea to have the discussion when you are engaging in a mutual activity side-by-side rather than face-on to alleviate some of the tension. Consider bringing up the topic when you are driving or folding laundry. Say, "Baby, I've been wanting to talk to you about your family. Sometimes, I feel like they are very judgmental of me, and you don't seem to have my back."
Also, keep in mind that your partner may need time to process what you have to say, so you might need to break up the discussion into a few conversations over time. Try to be sensitive to your partner's needs and give them time.
Tell your partner how their family is making you feel. Be honest about what is bothering you. Your partner may not realize how their family's behavior affects you.
Try to keep your tone neutral during this conversation, even if you're frustrated. Your partner might get defensive if you're visibly angry.
Tell your partner you need them to stand up for you. Often it's best for your partner to handle issues that arise with their own family. Let them know that you need support from them.
Avoid blaming them for not supporting you in the past. Try to focus on what you need from them moving forward.
Avoid talking about anyone's character. If you make any kind of personal attacks on your partner's family, your partner may instinctively jump to their side. Keep everything strictly factual when you make your case. Refer to specific events that have happened, and avoid making character judgments.
Similarly, avoid using "always" and "never" statements. These statements are rarely true, and they often lead to an argument.
Remember that your partner loves their family, and it's natural for them to feel loyal to their family members.
Brainstorm solutions with your partner. Your partner knows their family best, and they may have some good insights for how to handle them. Work as a team to come up with some ideas for avoiding conflict and hurt feelings at future family get-togethers.
For instance, you and your partner could sit down and evaluate what's happening and how to approach the situation based on the unique personalities of their family members. Perhaps, your partner knows a way to handle a certain family member that could help in your interactions. They might say, "Aunt Sarah is judgmental of everyone I date. It may be better if we just ignore any comments from her."
You could even try to come up with some dialogue and rehearse what each of you might say in certain situations. This may make it easier for your partner to step in when you need them.
Consider attending couples counseling. If you and your partner are struggling to reach an agreement on how to handle family conflict, couples counseling can help you understand each other better. A good counselor can teach you communication strategies and help you create solutions that work for both of you.
You might suggest, "Sweetie, I can see you are having trouble standing up to your family. I think it would be helpful if we saw a therapist who can help us figure out how to manage this situation. Does that sound good to you?"
If you feel like the conflict is affecting your relationship, take a moment to remember all the things you appreciate about your partner that have nothing to do with their family. Write these down and read over the list now and then.
For example, if you only see your partner's family on holidays and special occasions, you might not fret so much about any tension because you don't have to deal with it all that often."
Discuss boundaries with your partner. Sit down with your partner and come up with some reasonable limits together. Think about what both of you can do to minimize conflict and maintain family peace.
For instance, one of your boundaries might be that your partner's family members can't stay the night when they come to visit.
Another boundary might be not allowing the family to factor in on certain couple decisions such as having babies, practicing a certain religion, or deciding where you live.
Have your partner communicate your family's boundaries to your partner's family. Your partner's family members will need to know about the new rules you and your partner have decided on. Your partner should communicate these boundaries to their family members, and you can both enforce them as needed. Be kind and polite, but firm. However, if someone insults you over the new boundaries, then stand up for yourself.
Make sure your partner's family knows the reasoning behind your boundaries, too.
Maintain your boundaries. You may need to remind your partner's family about your boundaries now and then. It takes time for people to learn new behaviors when they've gotten used to acting a certain way.
If your boundaries are violated, you need to reinforce them by saying, "Remember, we decided that we are not having children, Mom/mother-in-law. Can you please support our decision even if you don't like it?"
Be confident and assertive. Remember that you're an adult, too. Dealing with older family members, such as your partner's parents, can make you feel like a child again, but it shouldn't. If you feel like you're being harassed or judged negatively, it's your right to stand up for yourself.
Being assertive doesn't have to mean being disrespectful. You can stand up for yourself while still being respectful and kind.
For instance, you can assertively say, "I know you don't understand my culture, but it's important for me and Henry to celebrate this holiday. I respect your beliefs, and I'd appreciate it if you did the same for me."
Talk with your partner's family. If you're having trouble with a particular person, try bringing up the issue with them yourself. Taking the initiative to solve the problem will show your maturity, and they may respect you more for speaking up.
It's much better to talk through problems when they happen than to let them go unaddressed for years. Say, "When you talk over me like that Josie, it makes me feel like I don't have a voice. I'd really appreciate it if you let me finish talking before sharing your opinion."
Deflect unwanted advice or comments. If your partner's family frequently offers you unsolicited advice or criticism, prepare a few noncommittal responses to redirect the conversation. Practice these responses before you need to use them. This will help you stay calm and collected in the moment.
If you're talking to someone older, a good way to handle unsolicited advice is to respond politely with something like, "How interesting!" or "What a neat story!" For instance, if your partner's mother tells you that you should feed your kids differently, ask her how she fed her children when they were growing up.
Consider limiting your contact with your partner's family. If you really can't resolve your conflicts with your partner's family, the best course of action might be to limit your contact with them. Skipping family events may be the best way to maintain peace and avoid straining your relationship with your partner. However, if you don't want to skip family events, then you can also set a time limit for how long you will be there.
Limiting contact is probably a good idea if your partner's family is outright abusive or disrespectful to you. People who act like this are unlikely to change their behavior.
My boyfriend's son has a girlfriend who is evil, mean and hurts me every chance she gets. She is a drunk and lies manipulates. I cannot keep dealing with her. What should I do?
Tell your boyfriend or boyfriend's son about her. If they refuse to talk to her, then avoid her at all costs. If she is putting her hands on you, you might need police help.
My husband's family is always trying to raise our 2-year-old son, and it drives me insane! The days I do need his family's help to watch him, they always shut me down. What can I do?
It's not that you set out the plan how your children should be raised, and everybody else is simply under your control when it comes to executing that plan. Everybody has their own views, ideas, experiments and experiences on how to raise children. So basically, it's either use them to help and allow them to do their own thing, or not use their help and do everything yourself. Another way to look at it is this: my house, my rules. When the kid is at your place, he follows your rules; when he's at their place, he follows their rules.