text:
Face Up 21 Strategy Card
It seemed to me that we just kind of knew what we were supposed to do.
He had just been out with his father to dinner in New York when I interviewed him: up. The Review of the National Policy toward Gambling concluded that these hotels"represent the base online of the island's tourist industry." Ey far, the most interesting aspect of Puerto Rico's experience with casino gambling is its success in limiting gambling by its resident population. 21 - later he denied on the witness stand that Vallon and Schepps were the two guys who jumped in the automobile and told him to"go fast" or they would"blow my roof off," or that he had ever told anybody that"Vallon swiped me over the head with a gun and told me to drive on" after the shooting.
In this underhand manner they conducted their business for about twenty years, without once attempting to open a fashionable skin-game (chart). During a lengthened residence at Nice last winter, I had joined a committee which was formed there for the purpose of having public gambling at Monte Carlo suppressed.
Blackjack - ball is an exclusion but what are members of the Club.
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An almost similar evolution leads us from the root ghar, to yearn after, through the notions of desiring, concupiscentia, to the graciousness of the CJiarites, by stages each one of which marks a gradual refinement of the purely sexual longing (card). He attends to every department; decides what should be done when any perplexing question arises, or convokes the three other directors to consult with him: face.
Court of Common Pleas held that if the defendant wishes to ascertain the nature of the Unsoundness, he should take out a summons for that purpose ie): play. Percentage of Lottery Players Who Would Play Lottery Games Less Often Due to New Gambling Scenarios While low, these percentages are much higher than reported in Connecticut prior to Foxwoods, they would play each of the different lottery games less often. DISPUTED CLAIMS: No internal DON authority exists to adjudicate disputed claims or enforce settlement of private claims made against members:
Summing up our results for the old civilisations, we find women slightly more brachycephalic and slightly more variable than their male comrades. Indeed, the moral tale was hardly a form. Free - croix provide general government services to the proposed gaming facility. In Spain, the lands of nobles were long exempted from direct taxation. Simulcasting and Common Pool Wagering wagering. Alcohol comes in three basic forms: beer, wine, and distilled When you have a drink, the alcohol is absorbed, distributed through the body, and gradually eliminated.
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text:
Fifty Years Later, PARKLAND Plays The Assassination Of JFK In A Minor Key
"I filmed a murder."
Since I'm from Boston, I guess it goes without saying that I'm a big fan of the Kennedys. Full disclosure: a few members of the family went to my high school (as did Teddy back in the day) and Mike Kennedy, whose father Michael died in a tragic skiing accident our freshman year, was a member of my graduating class. I certainly wouldn't say we were friends so much as we were friendly - I think we played freshman soccer together and we were in one or two of the same classes, but that's about it. Though even before I had ever met a Kennedy, I was already familiar with the legends. This is a family of folk heroes, respected across the country and downright beloved in their home state of Massachusetts where even the cloud of a possible manslaughter isn't enough to deter the public from voting them into office.
Conspiracy theories absolutely fascinate me and the JFK assassination is pretty much the brass ring of American historical puzzles. In the case of something like the supposed alien crash at Roswell,* I think the compelling factor is not necessarily the idea of the government cover up, but instead the idea of conclusive evidence of extraterrestrial life. It's one of those big "what if" questions that humanity has pondered for as long as we've properly grasped the structure of the universe and that will continue to fill our dreams and imaginations until the day comes that we make finally first contact. But because alien life is still such an unknown (Would they look anything like us? Would they be bellicose or benevolent?) it still feels like more of a curiosity, an amorphous idea upon which everyone can imprint their own values or beliefs. The JFK assassination is different because Kennedy wasn't just a known quantity, he was rock star, a pop culture icon that captured the hearts and minds of the American public. He was polarizing to be sure and there were plenty of people out there who outright hated the guy, but that only added to his mystique. Either way, he certainly wasn't someone who inspired ambivalence and that kind of public loss demands an explanation. Sadly, the subsequent death of Oswald robbed America of those answers, leaving behind so many unknowns that the mystery of that day has endured for half a century.
The violent and unexpected murder of someone as inspirational as Kennedy, whether you're a White House staffer or the owner of a local dress shop, is shocking to say the least and potentially traumatic if you had the misfortune to actually witness the event and/or its immediate aftermath. That's the focus of Parkland, an ensemble drama which follows those left in the wake of the President's death in Dallas, Texas on November 22, 1963. There's the swarm of Secret Service agents and presidential aides, the doctors at nearby Parkland Hospital, the Oswald family, the local FBI office that could have arrested Oswald days before, and finally Abraham Zapruder, the man who happened to be filming the Presidential motorcade when shots rang out from the Texas Book Depository. The film has an absolutely stacked cast and it seemed like Jamie and I spent the whole first half hour saying, "Hey it's that guy!" Seriously, at one point Jackie Earl Haley showed up as a hospital priest who blesses the body and then walks away, never to be seen again. At a certain point, the whole thing started to feel disorienting and seriously threatened to pull me out of drama of the story itself. Is that Tom Welling as a Secret Service agent? What's Mark Duplass doing in this movie? Oh, hi Rory Cochrane! What's happening again? Oh. Right.
I love the idea of focusing on the ancillary figures of that day while folks like Jackie and Lyndon Johnson remain out of focus or partially out of frame. Sadly, the film suffers from the biggest potential trap of most ensemble dramas, namely that the glut of characters causes the film to spread itself too thin across the various storylines until none of them really pop to command your attention. The emergency room scenes are well executed and appropriately intense because of the circumstances, but I don't really care about any of the hospital staff. The same goes for the Secret Service agents who rush LBJ to Air Force One and then have to find an appropriate place for the casket as well as the local FBI agents grappling with their own innocent mistake. It's the Zapruder and Oswald stories that offer the only truly engaging character work. To be honest, Zapruder's story isn't even all that interesting, but Paul Giamatti is so damn good in the role that you almost don't notice that his entire plot consists of developing a role of film.
The Oswald stuff really stands apart, due mostly to the great work by Jacki Weaver as Lee Harvey's mother Margueritte and James Badge Dale as his brother Robert. (When is Dale is gonna graduate from incredible supporting actor to carrying his own movie? I think it's time.) Margueritte always maintained that her son was actually a secret agent acting under orders from the U.S. government, and whether she truly believed it or simply clung to that explanation in order to cope with Oswald's horrific act is never really addressed. Regardless of your interpretation, Weaver's steely determination suggests a woman who refuses to be done in by her seemingly tenuous grip on reality. But your heart really goes out to Bob Oswald, a nice guy who is equal parts shocked and angered and then, before he can properly wrap his head around his own emotions, watches as his brother is gunned down in a police station on national television. Oswald had no pallbearers and there was no church service because nobody wanted to take the body, a drama we saw play out again this year following the death of one of the two Boston Marathon bombers. You just can't help but sympathize with the Robert Oswald, as any of us could suddenly be thrust into the same situation without warning. How would you handle it if someone close to you suddenly committed a horrifying act of violence? Do you renounce them? Do you try to understand and perhaps even forgive? These are the questions plaguing the families of James Holmes and the Tsarnaev brothers, and I hope I never have to answer them myself.
I was in a large meeting at work yesterday and one of the presenters asked how many people in the audience remembered exactly where they were when President Kennedy was shot. Only two people raised their hands, which surprised me since, at the ripe old age of 30, I'm generally on the younger end of the employee spectrum. The number of people who can answer that question is only going down and thankfully we haven't seen the successful assassination of a U.S. President in the 50 years since JFK. I find that to be a rather miraculous fact and I'd be lying if I said I haven't spent the past five years quietly preparing myself for the news that some bigoted asshole has attempted/succeeded in killing Barack Obama for being a Socialist Muslim Hitler. Anger and vitriol on both sides of the political aisle continues to bubble over each and every day, as evidenced by yesterday's Senate debate, where the adoption of simply majority rule was labeled "the nuclear option." But even if we are lucky enough to dodge the literal bullet of an assassin's rifle for another 50 years, we will continue to rack up plenty of those "where were you then?" moments. For me, (so far) it's 9/11 and the Marathon Bombing. For my kids? I shudder to think. But at least for all those moments of tragedy, there are also moments of joy and I look forward to telling my children about seeing the Red Sox win their first World Series in 86 years as well as the election of this country's first African American President.
And so today my thoughts are with the entire Kennedy family. JFK is one of my very favorite Presidents along with FDR, Lincoln, Clinton and Obama. (What can I say, I have a type.) While I wish I could have been alive to see Jack in his prime, I'd rather he was alive today so that I wouldn't have to write this piece.
*Some would also include the veracity of the moon landing here, but those people are crazy people. It happened. Science says so.
Title: Parkland
Director: Peter Landesman
Starring: Paul Giamatti, Zac Efron, Marcia Gay Harden, Billy Bob Thornton, James Badge Dale, Jacki Weaver, Ron Livingston, Mark Duplass, Tom Welling, David Harbour, Rory Cochrane, Jackie Earl Haley, Jeremy Strong
Viewing Method: Redbox DVD
Labels: assassination, jacki weaver, james badge dale, jfk, kennedy, lee harvey oswald, parkland, paul giamatti, roswell, zac efron

redpajama_set_name: RedPajamaCommonCrawl


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The LERU Summer School 2015 brought together over 40 PhD researchers from 19 European universities to discuss about knowledge exchange. We were welcomed in the wonderful venue of the University of Oxford and attended several informative courses ranging from networking to IP rights, over consultancy and start-ups. The study programme was excellently organised and included theoretical, practical and leisure parts. However, the greatest added value of this Summer School undoubtedly lied in the inspiring interdisciplinary contact with other researchers. The variety of discussions covering topics such as artificial robotic intelligence, experiments on mice, social behaviour and archaeological sites brings one out of the everyday life and provides new inputs, perspectives and a boost of creativity. The exchange was so fruitful that a reunion of the LERU Summer School group 2015 is already on planning with the aim of ensuring a long-lasting interdisciplinary knowledge exchange.


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text:
\section{Introduction and main results}
\subsection{Direct electrostatic and elasticity problems}
\label{1.1}
The {\it Poisson} and {\it Navier} (also known as {\it Lam\'e}\,) partial differential equations of elliptic type are commonly used for direct problems in electrostatics and elasticity theory. Let $\Omega$ be a bounded domain in $\mathbb{R}^n$, $n=2,3,\dots$; the cases $n=2,3$ are the most physically interesting.
It is assumed throughout the paper that   $\Omega$ has a   piecewise smooth boundary $\Gamma=\partial \Omega$.

In the direct formulation of the electrostatic problem, the Poisson equation   for a real valued function $u(x)$,  called
the {\it electric potential distribution},
\begin{equation}
\label{Poisson}
\Delta u = f(x),
\end{equation}
is solved in  the domain $\Omega$
  with a known distribution of the {\it electric charge density}, $-f(x)$, and with definite boundary conditions set on $\Gamma$. The boundary conditions may be formulated either in the form of potential values ({\it Dirichlet conditions})
\begin{equation}
\label{dir:elec}
u|_{\Gamma}=\phi_1
\end{equation}
or in terms of the electric field ({\it Neumann conditions}),
\begin{equation}
\label{neum:elec}
(\nabla u, \nu)|_{\Gamma} =\phi_2.
\end{equation}
Here $\nu=(\nu_1,\dots,\nu_n)$ is the unit outer normal vector to $\Gamma$ and $(\cdot,\cdot)$ is the Euclidean scalar product in ${\mathbb R}^n$.
Different parts of   $\Gamma$ may have different types of boundary conditions, and at any part of $\Gamma$  only one boundary condition may be set (which may be a linear combination of Dirichlet and Neumann conditions), so that the
problem is not overconstrained.

\smallskip

The direct formulation of the elasticity problem is described by the Navier equation
\begin{equation}
\label{Navier}
\Delta U +\alpha\operatorname{grad} \operatorname{div} U =F(x),
\end{equation}
for a vector-valued function $U:\Omega \to \mathbb{R}^n$, called
the {\it displacement field}.
Here $F(x)= -\frac{2(1+\mu)}{E}\mathcal{F}(x)$, where $\mathcal{F}$ is  the distribution of {\it body forces}, $\mu$ is the  {\it Poisson's ratio}, $E$ is the {\it Young's modulus} and parameter $\alpha$ is related to the Poisson's ratio by  formula
$\alpha=\frac{1}{1-2\mu}.$   The body $\Omega$ is assumed to be elastically isotropic.
The equation \eqref{Navier} is solved for the known distribution of body forces in  $\Omega$  and the boundary conditions at  $\Gamma$  defined for displacements  ({\it Dirichlet conditions})
\begin{equation}
\label{dir:elast}
U|_\Gamma=\Phi_1
\end{equation}
or for  traction forces ({\it Neumann conditions}),
\begin{equation}
\label{neum:elast}
(\sigma, \nu)=\Phi_2.
\end{equation}
Here $\sigma$ is a $(0,2)$--tensor (called the {\it stress tensor}), whose components are  related to the components of the displacement gradient through Hooke's law:
$$\sigma_{ij}(U)=(\alpha-1)\, \delta_{ij}\, \operatorname{div} U+
\partial U_i/\partial x_j + \partial U_j/\partial x_i,\,\,\,\, $$
$i,j=1,\dots,n$,  where $\delta_{ij}$ is the Kronecker symbol. Note
that the Hooke's law is given above in dimensionless form
corresponding to the unit value of the shear modulus. The scalar product
$(\sigma, \nu)$ is  a vector in ${\mathbb R}^n$ with the components
$$\sum_{j=1}^n \sigma_{ij}(U)\,\nu_j, \,\,\,i=1,\dots, n.$$ At any
part of $\Gamma$ the boundary condition can be specified for
the displacement, or for the traction force, or for a linear combination
between displacements and traction forces. As in the  direct
electrostatic problem,  only one boundary condition can be
assigned at any part of $\Gamma$. The attempt to define
simultaneously two different types of boundary conditions at the
same part of  $\Gamma$ (i.e. to impose the {\it Cauchy conditions}
\cite[chapter 6]{MF} corresponding to the overconstraining of the
system) may lead to the loss of the solution.

The properties of the direct electrostatic  and elasticity problems have been studied intensively for almost two centuries. It is well-known that  problems \eqref{Poisson} and \eqref{Navier} have unique solutions
under the Dirichlet boundary conditions \eqref{dir:elec} and \eqref{dir:elast}, respectively. For Neumann boundary conditions,  solutions exist under
additional  assumptions $\int_\Omega f dx =\int_{\Gamma} \phi_2 ds$ and $\int_\Omega F dx = \int_{\Gamma} \Phi_2 ds$,
respectively, and are unique up to additive constants  (see \cite{Ja, TG}).

There are various analytical and numerical methods to find solutions of the boundary value problems for Poisson and Navier equations. Most of the numerical methods developed for these problems are based on finite difference approximations  \cite{Hi, MG, Sa, St},  finite element analysis \cite{Ba, Sa, CS}
and Fourier transform  \cite{Du, Kh}.The finite element method has become a dominant approach to solving the elasticity problems,  with the exception of the microelasticity analysis for strain interactions in microstructures, where the Fourier transform is still used intensively. All major numerical techniques are still used for the electrostatic (or magnetostatic) and electromagnetic  problems.

\subsection{Inverse problems for Poisson and Navier equations}
\label{sec:unique}
In the present paper we consider the following inverse electrostatic and elasticity problems:  the  charge distributions or  the internal body forces are not known, and the states of the system, i.e., the functions $f$ or  $\mathcal{F}$, should be determined from the boundary conditions \eqref{dir:elec}--\eqref{neum:elec} or, respectively,  \eqref{dir:elast}--\eqref{neum:elast}.   Note that it is assumed that both Dirichlet and Neumann data are obtained from the boundary measurements. Such problems have attracted much interest in the recent years among physicists and engineers (see section \ref{sec:disc} and references therein).

The problems described above can be viewed as examples of inverse problems of potential theory \cite{Isak3}. These problems are  different from the
Calder\'on's inverse conductivity and elasticity problems, for which the coefficients of the left--hand sides of the equations, rather than the right--hand sides, are unknown and have to be determined from the boundary data (see, for example,  \cite{Ca, Uh1, Uh2,  AMR,  Isak2}).

The inverse electrostatic problem formulated above is  closely related to  the  inverse gravimetry problem\footnote{We thank Leonid Polterovich for bringing this link  to our attention.} that has important applications to geophysics and has been intensively studied for many years (see, for instance,  \cite{Isak1, Isak2,  FM} and references therein).

In the present subsection we collect some general results on uniqueness of solutions of inverse problems for Poisson and Navier equations.  Essentially, they are well-known (see, for example, \cite{BSB}). We present their proofs  in subsection \ref{sec:harm} for the sake of completeness.

Let, as before, $\Omega \subset \mathbb{R}^n$ be a Euclidean domain with piecewise smooth boundary $\Gamma$.
Consider the following overdetermined boundary value problem for the Poisson equation:
\begin{equation}
\label{problem}
 \Delta u = f, \quad x \in \Omega,
\end{equation}
$$u|_{\Gamma} =0, \quad \left.(\nabla u, \nu)\right|_{\Gamma}=0.
$$
Let $H(\Omega)$ be the space of {\it harmonic functions} on $\Omega$.
Denote by $Z(\Omega)$ its orthogonal complement in $L^2(\Omega)$.
We have the following
\begin{theorem}
\label{harm} A nonzero solution of problem \eqref{problem} exists
if and only if  $f \in Z(\Omega)$.
\end{theorem}

Let $V \subset L^2(\Omega)$ be a linear subspace. We say that the inverse electrostatics problem  possesses a {\it uniqueness property} for charge distributions in $V$ if
for any  two solutions  $u$ and $w$ of the Poisson equations $\Delta u =f$ and $\Delta w= g$ in $\Omega$  with $f, g \in V$, the equalities
$u|_{\Gamma}=w|_{\Gamma}$ and $ (\nabla u,\nu)|_{\Gamma} =(\nabla w, \nu)|_{\Gamma}$ imply $f \equiv g$. Since $V$ is a linear subspace of $L^2(\Omega)$ and the Poisson equation is also linear, this is equivalent to saying that for any nonzero $f \in V$,  problem \eqref{problem} does not have
a solution.  Therefore, Theorem \ref{harm} implies the following
\begin{cor}
\label{cor1}
The inverse electrostatics problem possesses a uniqueness property for charge distributions in a linear subspace $V(\Omega) \subset L^2(\Omega)$ if and only if $V(\Omega) \cap Z(\Omega) =0.$
\end{cor}
\begin{remark} It follows immediately from Corollary \ref{cor1} that the inverse electrostatics problem possesses a uniqueness property if $V(\Omega)
\subset H(\Omega)$. For instance, this is true if $V(\Omega)$ is the space of {\it linear functions} on~$\Omega$.
\end{remark}
Similar results hold for the inverse elasticity problem. Consider an
overdetermined problem for the Navier equation:
\begin{equation}
\label{problem2}
\Delta U + \alpha \, \operatorname{grad} \operatorname{div} U = F, \quad x \in \Omega,
\end{equation}
$$U|_{\Gamma}=0,  \quad  \left. (\sigma, \nu)
\right|_{\Gamma}=0.
$$
Let
$$\mathcal{L}=\Delta  + \alpha \, \operatorname{grad} \operatorname{div} $$
be the Navier operator acting on vector-valued functions $U:\Omega \to
{\mathbb R}^n$.
Denote by $\mathcal{H}(\Omega)$ the kernel of $\mathcal{L}$ (i.e., the
analogue of harmonic functions for the Navier operator) and by $\mathcal{Z}(\Omega)$ its orthogonal
complement in $L^2(\Omega, {\mathbb R}^n)$.
\begin{theorem}
\label{harm2}  A nonzero solution of  problem
\eqref{problem2} exists if and only if $f \in \mathcal{Z}(\Omega)$.
\end{theorem}
Let $\mathcal V(\Omega) \subset L^2(\Omega, \mathbb{R}^n)$ be a linear subspace. We say that  the inverse elasticity problem possesses a uniqueness property for internal stress
distributions in $\mathcal V$ if  for any  two solutions  $U$ and $W$ of the Navier equations $\mathcal{L}U=F $ and $\mathcal{L}W=G$ in $\Omega$  with $F, G \in \mathcal V$, the equalities
$U|_{\Gamma}=W|_{\Gamma}$ and $ (\sigma_U,\nu)|_{\Gamma} =(\sigma_W, \nu)|_{\Gamma}$ imply $F \equiv G$. Here
 $\sigma_U$ and $\sigma_W$ denote the stress tensors associated with $U$ and $W$, respectively.
Since the Navier equation and the space $\mathcal V(\Omega)$ are linear,  Theorem~\ref{harm2} immediately implies
\begin{corollary}
\label{cor2}
The inverse elasticity problem possesses a uniqueness property for internal stress distributions in a linear subspace $\mathcal V(\Omega) \subset L^2(\Omega,\mathbb{R}^n)$
if and only if $\mathcal V(\Omega) \cap \mathcal{Z}(\Omega)=0$.
\end{corollary}

\subsection{Checkered distributions}
\label{sec:check}
In practical applications one can often assume that the distributions of electric charge as well as of the internal stress  have  a certain structure.
For example, the geometry of the charge density distribution in the electronic component can be dictated by the structure of the component. Such structures often consist of elements with rectangular shape (see, for instance, \cite{LK, XCS}). Let us also note that similar structures appear naturally in  geophysics \cite{Tsch}. This motivates the following definition.

We say that the set $\Pi \subset \mathbb{R}^n$ is a {\it box} if  $\Pi
=[a_1,b_1)\times \dots \times [a_n,b_n)$, $a_i<b_i, i=1,\dots, n$.
Denote by  $V_c(\Pi) \subset L^2(\Pi)$ a linear subspace generated
by the characteristic functions of all boxes contained in $\Pi$.
Elements of $V_c(\Pi)$ are called {\it checkered functions}.
Equivalently, a function $f\in L^2(\Pi)$ is checkered if $\Pi$ can
be represented as a  finite union of disjoint boxes,
$\Pi=\Pi_1\sqcup\dots \sqcup\Pi_N$, such that $f|_{\Pi_i} \equiv
{\rm const}$, $i=1,\dots N$ (such a
representation is clearly not unique).
Note that the subspace $V_c(\Pi)$ is dense in $L^2(\Pi)$.
\begin{theorem} \label{main}  Let $u$ and $w$ be solutions of the Poisson equations $\Delta u =f$ and $\Delta w=g$ in the interior of the box $\Pi \subset \mathbb{R}^n$  with $f, g \in V_c(\Pi)$.  If $u|_{\partial \Pi}=w|_{\partial \Pi}$ and $ (\nabla u,\nu)|_{\Gamma} =(\nabla w, \nu)|_{\partial \Pi}$, then $f \equiv g$.
\end{theorem}
 In other words, the inverse electrostatics problem on $\Pi$ possesses a uniqueness property for electric charge distributions given by checkered functions.

\begin{remark}
In the context of the inverse gravimetry problem,  the right--hand side of equation \eqref{Poisson} should be understood as  the mass density and the function $u$ as the gravitational potential. Therefore, Theorem~\ref{main} can be reformulated as follows: the inverse gravimetry problem possesses
a uniqueness property for mass distributions given by checkered functions.  To our knowledge, distributions of this type have not been previously studied  for the inverse gravimetry problem. Uniqueness results for other types of mass distributions
could be found in \cite[Corollary 4.2.3]{Isak1} and  \cite[Theorem 2.1]{Isak3}.
\end{remark}

An analogue of Theorem \ref{main}  holds also for the inverse elasticity problem. Denote by
$\mathcal V_c(\Pi) \subset L^2(\Pi, {\mathbb R}^n)$ a linear
subspace generated by functions $F=(f_1,f_2,\dots,f_n)$, where $f_i
\in V_c(\Pi)$, $i=1,\dots, n$.
\begin{theorem} \label{main2} Let $U$ and $W$ be solutions of the Navier equations  $\Delta U +\alpha \operatorname{grad} \operatorname{div} U=F$ and
$\Delta W +\alpha \operatorname{grad} \operatorname{div} W=G$   in the interior of the box $\Pi \subset \mathbb{R}^n$  with $F, G \in \mathcal V_c(\Pi)$.  Suppose that $U|_{\partial \Pi}=W|_{\partial\Pi}$ and $ (\sigma_U,\nu)|_{\partial \Pi} =(\sigma_W, \nu)|_{\partial \Pi}$, where $\sigma_U$ and $\sigma_W$ are the stress tensors associated with $U$ and $W$, respectively.  Then $F \equiv G$.
\end{theorem}
In other words, the inverse elasticity  problem on $\Pi$ possesses a uniqueness property for  internal stress distributions with components given by
checkered functions.

Theorems \ref{main} and \ref{main2} are proved using Corollaries \ref{cor1} and \ref{cor2}, see section \ref{sec:checkered}.

\subsection{Discussion}
\label{sec:disc}
The interest in  the  inverse problems considered in the present paper arises from a number of practical applications. For example, in microelectronics, the observation of the internal voltage distribution in a device can be very important for the testing and diagnostic of devices under development  \cite{LK, BS}.

The inverse elasticity problem naturally appears in the analysis of {\it residual stresses} \cite{Wi1, Wi2}.  These  stresses are produced in the materials as a result of non-uniform deformation during forming, heat treatment and welding processes. The effect of a residual stress field is similar to the effect of an internal force distribution, and one can be converted into the other.  Modern experimental methods, such as Scanning Probe Microscopy \cite{KB, GAT}, can be used to obtain data on the electric potential and electric field at the surfaces of the component \cite{Pr}.
Digital image correlation \cite{CRS} can be applied to study the displacement distribution. These methods allow to obtain the Cauchy boundary conditions for electrostatic or elastic problems corresponding to real objects or components with high accuracy and fine resolution. The important question is to which extent such information can be used to find the charge (and the potential) or the internal force distributions inside the body,  and whether the corresponding inverse problems  have unique solutions. For simple distributions of internal charges or residual stresses the inverse problems can be solved easily (for example,  for a 2-D distribution of charges in a thin layer or 1-D distribution of residual stresses with a single significant stress component).
However, for general charge and internal stress distributions the issue becomes quite difficult.  While Theorems \ref{main} and \ref{main2} give a complete mathematical solution of the inverse electrostatics and elasticity problems for checkered distributions, from the viewpoint of practical applications these results are far from satisfactory, see section \ref{sec:chal}.

\subsection{Non-uniqueness of solutions: an example}
\label{sec:example}
One may ask whether
the analogues of Theorems \ref{main} and \ref{main2} hold for other, non-checkered,
electric charge and internal stress distributions.  Below we provide an example of a natural class
of distributions for which the solutions of the inverse problems are not unique.  Similar examples are well-known for the
inverse gravimetry problem (see \cite{Isak3}).

Let $S=S(r_1,r_2)$ be a {\it spherical layer} centered at the
origin, that is $S(r_1,r_2)=\{r_1\le |x|<r_2\}$ for some $r_2>r_1 \ge 0$.
Denote by $V_\sigma(S) \subset L^2(S)$ the linear subspace
generated by characteristic functions of  spherical
layers centered at the origin. In other words, $f\in V_\sigma(S)$ if and only if there exists a decomposition of
$S$ into a disjoint union of spherical layers $S=S_1\sqcup\dots \sqcup S_N$, such that $ f|_{S_i} \equiv {\rm const}$, $i=1,\dots N$.
We also denote by  $\mathcal V_\sigma(S) \subset
L^2(S,{\mathbb R}^n)$ the linear subspace of vector
functions whose components belong to $V_\sigma(S)$.
\begin{theorem}
\label{spherical} Let $S \subset \mathbb{R}^n$ be a spherical layer. Then

(i) $V_\sigma (S) \cap Z(S) \neq ~\{0\}$ and
(ii) $\mathcal V_\sigma(S) \cap \mathcal{Z}(S) \neq \{0\}.$
\end{theorem}
Theorem \ref{spherical} is proved in subsection \ref{sec:sph}. Together with Corollaries \ref{cor1} and \ref{cor2},  it immediately implies
\begin{corollary} The solutions of the inverse electrostatics and elasticity problems are not unique in $V_\sigma(S)$ and $\mathcal V_\sigma(S)$, respectively.
\end{corollary}

\subsection{Plan of the paper}

Section \ref{sec:checkered} is devoted to the proof of Theorems~\ref{main} and \ref{main2}. In subsection \ref{discr} an auxiliary discretization of the checkered functions is constructed.  In subsection \ref{sec:exp} we introduce a family of harmonic functions given by complex exponentials,
that are used to show that there are no nonzero checkered functions orthogonal to the space of harmonic functions.
Theorem \ref{main} then follows from Theorem \ref{harm}. In subsection
\ref{sec:elast} the above arguments are modified in order to prove Theorem \ref{main2}. Theorems \ref{harm}  and \ref{harm2} as well as Theorem \ref{spherical} are proved in section \ref{sec:other}.

\section{Inverse  problems for checkered distributions}
\label{sec:checkered}
The goal of this section is to prove  Theorems \ref{main} and \ref{main2}. We present the proofs in
three dimensions, which is the most interesting case for
applications. A similar argument works in any dimension $n\ge~2$.

\subsection{Discretization of checkered functions}
\label{discr}
Let $\Omega$ be a box as defined in section \ref{sec:check}.
For any $f \in V_c(\Omega)$, let us construct a function $\tilde f$
supported on a finite number of points. Consider an arbitrary box
$\Pi=[\beta_1^-,\beta_1^+) \times [\beta_2^-,\beta_2^+) \times
[\beta_3^-,\beta_3^+) \subset \mathbb{R}^3$. Set
\begin{equation}
\label{rule} \Upsilon(\chi_\Pi)=\sum_{\sigma_1,\sigma_2,\sigma_3 \in
\pm} \sigma_1\sigma_2\sigma_3 \mathbf{1}_{(\beta_1^{\sigma_1},
\beta_2^{\sigma_2}, \beta_3^{\sigma_3})}
\end{equation}
 Here
$\mathbf{1}_{(x,y,z)}$ is a function that takes value $1$ at the
point $(x,y,z)$ and vanishes elsewhere. The function
$\Upsilon(\chi_{\Pi})$ is supported on the vertices of $\Pi$ and
takes values $\pm 1$ at each vertex. The map $\Upsilon$ can be then
extended by linearity to the whole space $V_c(\Omega)$.

Given a function $f\in V_c(\Omega)$, set $\tilde f = \Upsilon(f)$.
Denote by $\tilde V_c(\Omega)$ the space of functions supported on
finite subsets of $\Omega$.

\begin{prop}
\label{ups} The map $\Upsilon:V_c(\Omega) \to \tilde V_c(\Omega)$ is
injective. Moreover,  there exists a constructive procedure to
recover $f\in V_c(\Omega)$ from the function $\Upsilon(f)=\tilde f$.
\end{prop}

To prove Proposition \ref{ups} we need an  auxiliary lemma below.

 Let $\{(x_l,y_l,z_l)\}_{l=1}^N$ be the
collection of vertices of all the boxes appearing in some
representation of $f$ as a linear combination of characteristic
functions of boxes. We say that a point $(a,b,c) \in \Omega$ is a
{\it node} of the function $f$ if $a=x_i, b=y_j, c=z_k$ for some $1
\le i,j,k \le N$.
A node $v$ is {\it interesting} if
$\tilde f(v) \neq 0$.   We also call a node $v$ {\it artificial} if  there exists a neighborhood of $v$ in which $f$ does not change its value across  a plane passing through $v$ and parallel to one of the coordinate planes.
It is easy to check that all artificial nodes are not interesting (and, therefore, artificial nodes can not be determined from $\tilde f$), but the
converse is not necessarily true.

\begin{example} Let $\Omega$ be a cube with side $2$ centered at the origin $v=(0,0,0)$.
Let $f$ be a restriction to $\Omega$ of a function which is identically equal to $1$ in
the positive and the negative octants, and vanishes elsewhere. Then $v$ is not an artificial node, but at the same time  $\tilde f(v)=0$ by \eqref{rule}
and, hence,  $v$ is  not interesting.
\end{example}

\begin{remark} One could view the difference between artificial and non-artificial nodes as follows. Let us  colour $\Omega$ in such a way that points $x,y \in \Omega$ have the same colour if and only
if $f(x)=f(y)$.  Then $\Omega$ can be represented as a disjoint union of sets $\Omega=\sqcup_{j=1}^J \Omega_j$, such that all points in $\Omega_j$,$ j=1,\dots J$,  have the same colour,
and the points in $\Omega_i$ and $\Omega_k$, $i\neq k$ have different colours.
Each $\Omega_j$ is a not necessarily connected union of boxes.  A node is not artificial if it is a vertex of one of the sets  $\Omega_j$, and artificial otherwise.
\end{remark}

Let $\operatorname{supp} \tilde f=\{(p_l,q_l,s_l)\}_{l=1}^M$ be the set of
interesting nodes. We say that a point $(a,b,c) \in \Omega$ is a
 {\it marked node} if $a=p_i, b=q_j, c=s_k$ for some $1 \le i,j,k \le
 M$.

 Note that the properties of being
a marked node or an interesting node do not depend on the choice of
the representation of $f$.
\begin{lemma}
\label{marked} The set of all marked nodes contains the set of
all non-artificial nodes.
\end{lemma}
\begin{proof} Without loss of generality, suppose that the node
$(0,0,0)$  is not marked. This means that among interesting nodes
there are either no points with $x=0$, or with $y=0$, or with $z=0$.
In each case, the corresponding plane (say, $x=0$) does not contain
interesting nodes. Let us show that the function $f$ does not change
its value across this plane. This would mean that
all nodes contained in this plane are  artificial, including
$(0,0,0)$.

Consider a  decomposition of $\Omega$ into boxes,  such that the set of all their vertices coincides with the set of all nodes of $f$ (this could be achieved by
constructing planes through each node parallel to the coordinate planes).  It follows from the definition of a node that $f$ is constant on each of these boxes.  Take one of the corner nodes belonging to the plane  $x=0$ (i.e. a node lying on one of the edges of $\Omega$). At each such node
at most two boxes meet. Therefore, if this  node is not
interesting, the values of $f$ at the boxes adjacent to it are equal and
hence the node is artificial.   Note that by  formula \eqref{rule}, the  total contribution of these two boxes to the value of $\tilde f$ at any other node lying on the plane $x=0$  is zero.
Let us throw away these two boxes and pick another node where at most two of the remaining boxes meet.  Again, the value of $\tilde f$ at this node is zero and hence
the values of $f$ at the boxes adjacent to it  coincide. Therefore, this node is also artificial. We repeat the procedure until all boxes adjacent to the plane $x=0$ are
thrown away.  At each step we get artificial nodes only. This
completes the proof of the lemma.
\end{proof}
Let us now prove Proposition \ref{ups}. The proof is based on a similar inductive argument as above.  We start at a corner box,
on which  formula \eqref{rule} allows us to reconstruct  in an unambiguous way the value of $f$   from the value of $\tilde f$ on the corresponding corner vertex. We remove that box, move to an adjacent one
and repeat the procedure.  A similar  approach will be used again in the proof of Proposition \ref{prop3}.
\begin{proof} As follows from Lemma \ref{marked}, knowing $\operatorname{supp} \tilde f$
allows us to construct a decomposition of $\Omega$ into boxes, whose
vertices include all non-artificial nodes. We know the values of
$\tilde f$ at each vertex of these boxes. Let us now reconstruct the
value of $f$ at each of the boxes using the following inductive
procedure. Start with a vertex that is also a vertex of $\Omega$,
and take the box that contains it (there is a unique box with this
property). Since there are no other boxes containing this vertex, by
\eqref{rule}, the value of $\tilde f$ at this vertex determines the
value of $f$ at the box. We subtract the contribution of this box to
$\tilde f$, throw away this box and take one of the new corner
vertices, at which at most two of the remaining boxes  meet.  At each step of this procedure we determine the value of
$f$ on the corner box, and reduce the number of boxes by one. Since
the number of boxes is finite, eventually we will determine the
value of $f$ on each box.
\end{proof}

\subsection{Exponential functions}
\label{sec:exp}
 Let
$$e=e(x)=e(\alpha,\Theta,\Psi;x)=e^{\alpha(\Theta,x)+i\alpha(\Psi,x)}$$
be a function of the variable $x\in \mathbb{R}^3$, depending on the
parameters $0\neq \alpha \in \mathbb{R}, \Theta\in \mathbb{R}^3$,
$\Psi \in \mathbb{R}^3$, such that $(\Theta, \Psi)=0$,
$|\Theta|=|\Psi|=1$. It is easy to check that $e(x) \in
H(\mathbb{R}^3)$.

We say that a pair of vectors $(\Theta,\Psi)$ is {\it admissible} if
the plane it generates is not  orthogonal to any of the coordinate
axes. Set
\begin{equation}
\label{int:vol}
 P_f(\alpha, \Theta, \Psi)=(f,
e(\alpha,\Theta,\Psi;x))=\int_\Omega f(x)
e^{\alpha(\Theta,x)+i\alpha(\Psi,x)}\,dx.
\end{equation}
\begin{lemma}
\label{lemma:prod} Let $\{v_j\}$ be the set of interesting nodes of
$f\in V_c(\Omega)$. Then, for any $\alpha \neq 0$ and any admissible
pair $(\Theta, \Psi)$ we have:
\begin{equation}
\label{prod} P_f(\alpha,\Theta,\Psi)=C \sum_{v_j} \tilde
f(v_j)\,e(\alpha,\Theta, \Psi;v_j),
\end{equation}
where
\begin{equation}
\label{const} C=C(\alpha,\Theta, \Psi)=\frac{1}{\alpha^3}
\prod_{l=1}^3 \frac{1}{\Theta_l+i\Psi_l}.
\end{equation}
\end{lemma}
Note that the constant $C$ is well-defined for any admissible pair
$(\Theta, \Psi)$.
\begin{proof} The result follows from \eqref{rule} by a direct computation of the triple
integral \eqref{int:vol}.
\end{proof}
Since any function $e(x)$ is harmonic, the right-hand side in
\eqref{prod} can be computed using the boundary data $\phi_1,\phi_2$
of problem \eqref{problem} by Green's formula:
$$P_f(\alpha, \Theta, \Psi)=\int_{\Gamma} \left(e(x)\phi_2
-\frac{\partial e(x)}{\partial n}\,\phi_1 \right)\, ds
$$
\begin{prop}
\label{prop3} Knowing the value of $P_f(\alpha, \Theta, \Psi)$ for
any $\alpha \neq 0$ and any admissible pair $(\Theta, \Psi)$, one
can reconstruct the function $\tilde f$.
\end{prop}
\begin{proof}
Let $K$ be the convex hull of $\operatorname{supp} \tilde f$. It is easy to see
that $K$ is a convex polyhedron; let $w_j$ be its vertices. Then, for
any $\Theta \in \mathbb{R}^3$ and any $\Psi$ chosen in such a way
that the pair $(\Theta, \Psi)$ is admissible,  we have:
\begin{equation}
\label{limsup} \limsup_{\alpha \to \infty}
\frac{\log|P_f(\alpha,\Theta,\Psi)|}{\alpha}=\max_j(w_j,\Theta)=\max_{y\in K}
(y,\Theta),
\end{equation}
where the first equality follows from \eqref{prod} and the second
one from a well-known fact that the maximum of a linear functional
on a convex polyhedron is attained at one of the  vertices. At the
same time, a convex set $K$ can be represented as the intersection
of its supporting half-spaces:
\begin{equation}
\label{326}
K=\cap_{\Theta \in \mathbb{R}^3} \{ x \in \mathbb{R}^3\,|\,
(x,\Theta) \le \max_{y\in K} (y, \Theta)\}.
\end{equation}
Therefore, by \eqref{limsup}, we can recover $K$ and, in particular,
all its vertices $w_j$, $j=1,\dots, N$.

In order to recover the values $\tilde f(w_j)$ we use the following
procedure. Let $\Theta(j)$ be an external unit normal vector to a plane passing through $w_j$ and not intersecting the convex set $K$ (external means here that $\Theta(j)$ points to the half-space not containing $K$). One can easily check that in this case
\begin{equation}
\label{thetacond} (\Theta(j),w_j)>(\Theta(j),w_k)
\end{equation}
 for all $k \neq j$, $k=1,\dots,N$.

Choose $\Psi(j)$ in such a
way that the pair $(\Theta(j), \Psi(j))$ is admissible. We have:
\begin{equation}
\label{327}
\tilde f(w_j)=\lim_{\alpha \to
\infty}\frac{P_f(\alpha,\Theta(j),\Psi(j))}{C(\alpha,\Theta(j),\Psi(j))\,
e(\alpha_,\Theta(j),\Psi(j),w_j)}.
\end{equation}
This allows us to  determine the values of $\tilde f$ at all vertices
$w_j$, $j=1,\dots,N$. Using Lemma \ref{lemma:prod} we can subtract
the contributions of these nodes from $P_f(\alpha,\Theta,\Psi)$ and
repeat the procedure. Since at each step the number of nodes
decreases, the number of steps will be finite and at the end we will
recover all elements of $\operatorname{supp} \tilde f$ and the values of $\tilde
f$ at each of these points.
\end{proof}
Combining Propositions \ref{prop3} and \ref{ups} we obtain the proof
of  Theorem \ref{main}.
\begin{remark} The results of this section generalize in a straightforward way to any dimension $n\ge 2$.
Note that in dimension $n=2$ the admissibility assumption can be
omitted, because for any orthogonal nonzero vectors $\Theta, \Psi
\in {\mathbb R}^2$, the denominators in \eqref{const} are
automatically nonzero.
\end{remark}

\subsection{Proof of Theorem \ref{main2}}
\label{sec:elast}
Let us indicate how the proof of Theorem \ref{main}  can be modified in order to
prove Theorem \ref{main2}. Let $F=(f_1,f_2,f_3) \in \mathcal V_c(\Omega)$
and let $\widetilde F = (\tilde f_1, \tilde f_2, \tilde f_3)$ be its
discretization in the sense of section \ref{discr}. We say that $v_j
\in \Omega$ is a node of $F$ if it is a node of one of the
functions $f_i$, $i=1,2,3$. As before, consider a harmonic function
$$e:=e(\alpha,\Theta,\Psi;x)=e^{\alpha(\Theta,x)+i\alpha(\Psi,x)}.$$ It
is easy to check that
$$\operatorname{curl}(e(\alpha,\theta,\Psi;x),0,0)=(0,\alpha(\Theta_3+i \Psi_3)e,
-\alpha (\Theta_2+i\Psi_2)e) \in \mathcal{H}(\Omega).$$ This follows from
the fact that $e$ is harmonic and that $\operatorname{div} \operatorname{curl} =0$. Similarly,
$\operatorname{curl}(0,e(\alpha,\theta,\Psi;x),0) \in \mathcal{H}(\Omega)$.

Set
$$
\mathcal P_F^1(\alpha, \Theta, \Psi)=(F,
\operatorname{curl}(e(\alpha,\theta,\Psi;x),0,0)),$$ $$ \mathcal P_F^2(\alpha, \Theta,
\Psi)=(F, \operatorname{curl}(0,e(\alpha,\theta,\Psi;x),0))$$
(now $(\cdot,\cdot)$ means the natural inner product in $L_2(\Omega,\mathbb{R}^3)$).
Theorem \ref{main2} can be now deduced from Proposition \ref{ups}
and the following analogue of Proposition \ref{prop3}:
\begin{prop}
\label{prop4} Knowing the value of $\mathcal P^l(\alpha, \Theta, \Psi)$,
$l=1,2$,  for any $\alpha \neq 0$ and any admissible (in the sense
of section \ref{sec:exp}) pair $(\Theta, \Psi)$ one can reconstruct
the function $\widetilde F$.
\end{prop}
\begin{proof}
Similarly to  Lemma \ref{lemma:prod}, we have:
\begin{multline}
\label{prod2}\mathcal P_F^1(\alpha,\Theta,\Psi)=
\frac{1}{\alpha^2}\prod_{l=1}^3 \frac{1}{\Theta_l+i\Psi_l}
\sum_{v_j} e(v_j)(\tilde f_2(v_j)(\Theta_3+i\Psi_3) - \\ \tilde
f_3(v_j)(\Theta_2+i\Psi_2)).
\end{multline}
When choosing the unit vector $\Psi$, we will make sure
that, apart from the admissibility condition,
the following condition is satisfied:
$$\Theta_3 \Psi_2 -
\Theta_2 \Psi_3 \neq 0.$$ This condition guarantees that if $$\tilde
f_2(v_j)(\Theta_3+i\Psi_3) - \\ \tilde
f_3(v_j)(\Theta_2+i\Psi_2)=0,$$ this automatically implies $\tilde
f_2(v_j)=\tilde f_3(v_j)=0$, and so no term in the sum \eqref{prod2}
may ``accidentally'' vanish.  Therefore, the contribution of each
interesting node will be taken into account. Arguing in the same way
as in the proof of Proposition \ref{prop3} we can recover the convex
hull of $\operatorname{supp} \tilde f_2\cup\operatorname{supp} \tilde f_3$.

Taking $\mathcal P_F^2(\alpha,\Theta,\Psi)$  instead of
$\mathcal P_F^1(\alpha,\Theta,\Psi)$ in the argument above we recover the
convex hull of $\operatorname{supp} \tilde f_1\cup\operatorname{supp} \tilde f_3$. Taking a
union of these two sets, we recover the convex hull of $\operatorname{supp}
\widetilde F$.

Let $w_j$ be a vertex of the convex hull of $\operatorname{supp} \widetilde F$.
Choose a  unit vector $\Theta(j)$ satisfying \eqref{thetacond} as in
the proof of Proposition \ref{prop3}. Consider two admissible pairs
$(\Theta(j), \Psi^1(j))$ and $(\Theta_j, \Psi^2(j))$. Using
\eqref{prod2} we can calculate
$$\tilde f_2(w_j)(\Theta_3(j)+i\Psi_3^k(j)) - \tilde
f_3(w_j)(\Theta_2(j)+i\Psi_2^k(j)), \quad k=1,2.$$  We obtain a system
of two linear equations on $\tilde f_3(w_j)$ and $\tilde f_2(w_j)$.
Clearly, we can choose the admissible pairs $(\Theta(j), \Psi^1(j))$ and
$(\Theta_j, \Psi^2(j))$ in such a way that the determinant of this
system is nonzero. Thus, we can compute $\tilde f_3(w_j)$ and
$\tilde f_2(w_j)$.

Applying the same argument to $\mathcal P_F^2(\alpha,\Theta,\Psi)$, we
compute $\tilde f_1(w_j)$. Therefore, we have computed $\widetilde
F(w_j)$, and this can be done for any vertex of the convex hull of
$\operatorname{supp} \widetilde F$. As in the proof of Proposition \ref{prop3}, we
subtract the contributions of these nodes from
$\mathcal P_F^1(\alpha,\Theta,\Psi)$ and $\mathcal P_F^2(\alpha,\Theta,\Psi)$, and
repeat the argument. The process will stop after a finite number of
steps because the number of nodes of $\tilde F$ is finite, and it
decreases at each step. This completes the proof of Proposition
\ref{prop4} and of Theorem \ref{main2}.
\end{proof}
\begin{remark} Instead of using the curl in the proof of Proposition
\ref{prop4}, we could  take $\operatorname{grad} e(\alpha,\theta,\Psi;x)$.
Clearly, $$\operatorname{grad} e(\alpha,\theta,\Psi;x)~\in~\mathcal{H}(\Omega).$$ In this
case, for each $\Theta(j)$ we need to consider three admissible
pairs $(\Theta(j), \Psi(j;k))$, $k=1,2,3$, in order to get a system
of three linear equations on $\tilde f_k(w_j)$, $k=1,2,3$. The rest
of the proof goes along the same lines as above. The advantage of
this approach is that it works in any dimension, while the curl is
defined only in dimension three.
\end{remark}
\subsection{Computational challenges}
\label{sec:chal} It is convenient to prove Theorems \ref{main} and
\ref{main2} using the exponential functions introduced in subsection
\ref{sec:exp}. In principle, our proof could be presented as an
algorithm that allows to reconstruct in a unique way the solutions
of the Poisson and Navier equations from the corresponding boundary
values.  However, numerical implementation of our approach faces
serious computational difficulties that we describe below. For
simplicity, a $2$--dimensional example is presented.

We have tested  the developed algorithm for the solution of the inverse electrostatic
problem
on a rectangle $\Omega \subset \mathbb{R}^2$ with boundary $\Gamma$,
containing two rectangular charged areas (Fig. 1). The boundary
conditions corresponding to this problem were obtained using the Green
function method implemented numerically. In other words, we computed
a solution $u$ of the equation $\Delta u =f$ on the whole plane
using Green's function , and calculated numerically its values
as well as the values of its normal derivative on $\Gamma$. Here $f$
is the characteristic function of the total charged area.
\begin{figure}
\begin{center}
\includegraphics[width=6cm]{figure1.pdf}
\caption{Geometry of the test problem. Gray areas represent charge
areas. Thick black line represents the convex envelope of charge
areas. Thin gray line represents the boundary $\Gamma$ of the domain
$\Omega$.}
 \end{center}
\end{figure}

It was found that if $P_f(\alpha,\Theta,\Psi)$  is calculated directly
using the integration over $\Omega$,  then the procedure based on
\eqref{limsup} and \eqref{326} produces the convex hull of the
charged areas with high accuracy. However, when
$P_f(\alpha,\Theta,\Psi)$  is calculated using the integration over
the boundary,  the algorithm based on \eqref{limsup} produces the
convex hull occupying the whole $\Omega$. Such a drastic difference
is the result of small numerical errors in the boundary conditions
determined by the numerical solution of the direct problem,  and also
in the numerical integration over the boundary.

Strong sensitivity to numerical errors
arises from a specific nature of exponential functions. When $\alpha$ is large, these functions
are rapidly increasing in one direction and rapidly oscillating in the orthogonal direction.
As a result, small errors in boundary conditions are multiplied by large factors and the convex
hull estimate based on \eqref{326} and \eqref{327} becomes distorted. For small  values of $\alpha$, the
value of $\log |P_f(\alpha,\Theta,\Psi)|/\alpha$  obtained using the integration over $\Omega$ is close to the one
obtained using the integration over $\Gamma$. However, as $\alpha$ increases these two values diverge.

In order to verify the validity of the computational model,  calculations were performed for the constant charge density: $f \equiv 1$ on $\Omega$.
In this case, the precise value of  $P_f(\alpha, \Theta,\Psi)$ can be computed analytically. For all  values of $\alpha$, the analytical equations produced the
same  $\log |P_f(\alpha,\Theta,\Psi)|/\alpha$  values,  no matter if  the integration was performed  over $\Omega$ or its boundary $\Gamma$.
However, in the  numerical analysis of this problem,  the area and
boundary integration were producing close values for small values  of $\alpha$ and were diverging
for  large $\alpha$.

In practical problems based on experimental data,  boundary conditions are always  obtained with some
errors. Therefore, as the above analysis show, in order to produce an algorithm for the solution of the inverse electrostatics problem
that is numerically implementable, one needs to modify our approach. One possibility would be to find a set of harmonic functions
exhibiting good  behavior from the numerical viewpoint, which could  replace the exponentials in the proof of Theorem \ref{main}.

\section{Necessary and sufficient conditions for uniqueness of solutions}
\label{sec:other}
\subsection{Proofs of uniqueness criteria}
\label{sec:harm}
The goal of this subsection is to prove Theorems \ref{harm} and \ref{harm2}.
Let us start with Theorem \ref{harm}.
To prove necessity, suppose that  $u$ is a solution of problem
\eqref{problem} and let $h \in H(\Omega)$. Then
$$\int_\Omega f \cdot h = \int_\Omega \Delta u \cdot h = \int_\Omega
u \cdot \Delta h =0.$$ Note that the boundary terms in the
integration by parts disappear since $\phi_1=\phi_2=0$.

To prove sufficiency, denote by $G_1(x,y)$ the Green's function of
the Dirichlet boundary value problem in $\Omega$:
$$\Delta_x
G_1(x,y)=\delta(x-y), \,\,\, G_1(x,y)|_{x \in \partial \Omega}=0,$$
and by $G_2(x,y)$ the Green's function for the corresponding Neumann
boundary value problem:
$$\Delta_x
G_2(x,y)=\delta(x-y)-\frac{1}{|\Omega|},\,\,\, (\operatorname{grad}_x G_2(x,y),
\nu)|_{x \in
\partial \Omega}=0.$$
Note that the integral over $\Omega$ of the right--hand
side of the equation above is zero, which is necessary for the existence of a solution of the Neumann problem with zero boundary conditions.

It follows from the definitions of $G_1$ and $G_2$ that
$$(G_1-G_2)(x,y)+ \frac{x_1^2}{2|\Omega|}$$ is a harmonic function of $x$.
Therefore, the assumption $f \in Z$ implies
\begin{equation*}
u(y):=\int_\Omega f(x)G_1(x,y)\,dx = \int_\Omega f(x) G_2(x,y)\, dx -\int_\Omega f(x)\frac{x_1^2}{2|\Omega|}\, dx
\end{equation*}
for all $y \in \Omega$. Note that the term $$\int_\Omega
f(x)\frac{x_1^2}{2|\Omega|}\,dx$$ is constant and hence of no
importance for the Neumann boundary value problem.
Let us also remark that  $\int_\Omega f(x) dx =0$ since $f \in Z(\Omega)$.
It is easy to check that, by the properties of $G_1$ and $G_2$, the function $u$
constructed above is a solution of problem \eqref{problem}.
This completes the proof of Proposition \ref{harm}. \qed

The proof of Theorem \ref{harm2}  is completely analogous. We note
that the existence of Green's functions for the Navier equation with
either Dirichlet or Neumann boundary conditions is well-known~---
see, for instance \cite[section 7.12]{S}.
\subsection{Proof of Theorem \ref{spherical}}
\label{sec:sph} \noindent {\it Proof of (i).}
Let $A,B,S$ be three
spherical layers centered at the origin such that $A\cup B=S$ and
$A~\cap~B=~\emptyset$. Consider the following linear combination of
characteristic functions of the sets $A$ and $B$:
$$u=\operatorname{Vol}(B)\, \chi_A-\operatorname{Vol}(A)\,\chi_B.$$
By the mean value theorem for harmonic functions we immediately have
$u\in Z(S)$, and this completes the proof of part (i) of the
proposition.

\bigskip

\noindent {\it Proof of (ii)} In order to prove the second part of
the proposition we note that a function $U$ lying in the kernel of
the Navier operator $\mathcal L$ is {\it biharmonic}. Indeed, let $\mathcal L(U) =
\Delta U + \alpha \operatorname{grad}\, \operatorname{div} U =0$. Taking the divergence on both
sides we get $\operatorname{div} (\Delta U)=0$. Here we took into account that $\operatorname{div}
\operatorname{grad} = \Delta$ and that the Laplacian commutes with the divergence.
At the same time, applying the Laplacian to $\mathcal L(U)$ we get
$$\Delta^2\,U + \alpha \Delta\,\operatorname{grad}\,\operatorname{div} U=0.$$ But since $\operatorname{div}
(\Delta U)=0$, the second term vanishes, because $$\Delta \,
\operatorname{grad}\,\operatorname{div} U = \operatorname{grad} \, \operatorname{div} \, \Delta U.$$ Hence, $\Delta^2 U = 0$
and $U$ is biharmonic.

It is well-known that a real valued biharmonic function $f(x)$
satisfies the following mean-value property (see, for example,
\cite{K}):
\begin{equation}
\label{meanvalue} \int_{B(x,r)} f(y) dy = \omega_n r^n f(x) +
\frac{\omega_n \, r^{n+2}}{2(n+2)}\Delta f(x),
\end{equation}
where $B(x,r)$ is a ball of radius $r$ centered at $x \in {\mathbb
R}^n$ and $\omega_n$ is the volume of the unit ball in $\mathbb{R}^n$.

Consider now a  ball $B=B(r_3)$ centered at the origin. Let us
represent it as a union of three sets $B(r_1)\cup S(r_1,r_2) \cup
S(r_2,r_3)$, $r_3 > r_2 > r_1 >0$. Let $F_{a,b}$ be a piecewise
constant vector-valued function taking the values $a$ and $b$ on
$S(r_2,r_3)$ and $S(r_1,r_2$), respectively, and the value one on
$B(r_1)$. Clearly, $F_{a,b} \in V_\sigma$ for all $a,b \in {\mathbb
R}$. Let us show that for any triple $0<r_1 < r_2 < r_3$ there
exists a choice of parameters $a, b$ such that $F_{a,b} \in \mathcal{Z}(B)$.
It can be deduced from the mean-value formula \eqref{meanvalue},
that the inclusion $F_{a,b} \in \mathcal{Z}(B)$ holds if the following system
of equations is satisfied:
$$a(r_3^n-r_2^n) + b (r_2^n-r_1^n) +r_1^n=0,\,\,a(r_3^{n+2}-r_2^{n+2}) + b (r_2^{n+2}-r_1^{n+2})
+r_1^{n+2}=0.
$$
One may check that the determinant of this system does not vanish
for $r_3 > r_2 > r_1 >0$. Indeed, if $r_3 \neq r_2$, the only
positive roots of the determinant considered as a polynomial in
$r_1$ are $r_1=r_2$ and $r_1=r_3$ (there are no other positive roots
because, as can be easily verified, the derivative with respect to
$r_1$ has only one positive root). Hence, there always exists a
unique solution $a,b$ of the system above, and the corresponding
function $F_{a,b} \in \mathcal{Z}(B)$.

This completes the proof of Theorem \ref{spherical}.
\begin{remark}
The proof of part (i) of Theorem \ref{spherical} shows that the
intersection $V_\sigma(S) \cap \mathcal{Z}(S)$  is in fact quite large.
Indeed, it is easy to show that for any partition of $S$ into $k$
concentric spherical layers,  there exists a $(k-1)$--dimensional
linear subspace of functions in $V_\sigma \cap \mathcal{Z}(S)$.

The proof of part (ii) goes through without changes if instead of a
ball $B(r_3)$ one takes a spherical layer
$S=S(r_3,r_0)=S(r_0,r_1)\cup S(r_1,r_2) \cup S(r_2,r_3)$.  It follows
from the proof that for any partition of $S$ into $k$ concentric
spherical layers, there exists a $(k-2)$--dimensional linear space
of functions in $\mathcal V_\sigma \cap \mathcal{Z}(S)$.
\end{remark}
\subsection*{Acknowledgments} The authors are grateful to Victor Isakov and Leonid Polterovich for  useful discussions.
Research of Leonid Parnovski is supported by EPSRC  grant EP/F029721/1.
Research of Iosif Poltero\-vich  is supported by NSERC, FQRNT and Canada Research Chairs program.

redpajama_set_name: RedPajamaArXiv


Posted on 02.08.2018 | by LEATRICE D.
Video igaming will be reasonably some sort of innovative sphere for Bangladesh. Games need recently been constructed because 2002, mostly individually. Having said that, with 2014, several That vendors possess going to build up video tutorial online games vendor.
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that the control that men seek over women is broken when women take control of the power that they have. P 55 The concept of socialization is that our actions are driven/learned by culture. The photograph chosen, which appeared on the Boston Globes website and was captured by John Tlumacki, exposes the emotion of a three-year-old boy saying goodbye to his father before deployment, while including references to military masculinity and gender roles. Ramsay is occupied with matronly duties, such as knitting socks and running errands. These expectations can have negative effects on a person and how they grow. At first glance, the viewer sees the obvious separation between the civilians and the Military Police Soldiers through physical distance. In Sweetheart of the Song Tra Bong, the character Mary Anne Bell appears to be apparently similar to the character Martha from The Things They Carried, in that both are young women who have relations with a soldier in the war.
tags: Gender Powerful Essays 1480 words (4.2 pages) Preview - Many of the short stories within the collection The Things They Carried, written by Tim OBrien, can be compared and contrasted in regards to elements including theme, characters, and technique. Wearing T-shirts and jeans or playing with sporting equipment. However, the reader is kept in the dark that Dimmesdale is the childs father until latter part of the novel. Unlike Hamlet, King Lear, and Macbeth, which are set against a backdrop of affairs of state and reverberate with suggestions of universal human concerns, Othello is set in a private world and focuses on the passions and personal lives of its major figures. From the 1700s to modern day society, women have held more of an equal role in society with the rights of suffrage, education, and reproductive rights given to them through Roe. Two authors, Judy Mann of The Difference and Bernard Lefkowitz of Our Guys face the issue of gender roles and stereotypes, and how they affect our lives today. With time, society continuously creates new definitions and expectations for one another, and we are all expected to comply with modern day terms. tags: Violence, Creation of Roles, Modern Society Powerful Essays 1598 words (4.6 pages) Preview - Gender affects every aspect of our life, from how we feel about ourselves and set our goals in educational, recreational and work opportunities as well as the the nature.


text:
Q: sum group values based on factor from second data frame I have two data frames - one containing a list of all samples (sample ID, area etc)(sampleid.df) and a second a list of counts (obs.count) made in each sample. What I need is a table which lists ALL samples and provides a sum for all obs.measur (by sampleid) - in addition samples returning no counts should get a 0. I tried aggregate and merge but the difference in length was an issue. Any ideas/suggestions much appreciated.
sampleid.df <- data.frame(sampleid = 1:20)
sampleid.df$sampleid <- as.factor(sampleid.df$sampleid)

obs.count  <- data.frame(counts = c(sample(1:50,15,rep=TRUE)), sampleid = c(sample(1:20,15,rep=TRUE)))

A: You could try merge with option all=TRUE so that the output ("df1") will have NA for "sampleids" from "sampleid.df" that does not have a match in the same column from "obs.count".  Then use aggregate with option na.action=na.pass (by default, the option is na.action=na.omit) to include the NA rows as well.  Using na.rm=TRUE within sum remove the NAs but if a particular "sampleid" have only NAs, it will return NA.
df1 <- merge(sampleid.df, obs.count, all=TRUE)
res <- aggregate(counts~sampleid, df1, sum, na.rm=TRUE, na.action=na.pass)
res$counts[is.na(res$counts)] <- 0

redpajama_set_name: RedPajamaStackExchange


Did we say $99.9 million? We meant $233 million.
Attorneys who represented Eloise Cobell, the lead plaintiff in the Indian Trust Fund case, are seeking to more than double their fee request for their role in winning the $3.4 billion dollar settlement.
According to an article by Mike Scarcella in The National Law Review, Dennis Gingold and lawyers from the firm Kilpatrick Townsend and Stockton have filed a fee petition in federal court in Washington D.C. that seeks to increase their originally agreed upon fee cap of $100 million to $233 million.
Former Senator Byron Dorgan, D-North Dakota described the request as a "cruel irony" in an article by Rob Capriccioso of Indian Country Today. In his role as Chairman of the Senate Committee on Indian Affairs Dorgan has been a long time supporter of American Indians and a proponent of settling the trust fund case.
"It's a profound disappointment to me that the lawyers are trying to go beyond the agreement reached by Congress," said Dorgan.
Dorgan said that the "nearly one-quarter of a billion dollars" the lawyers are now requesting would have "never gotten through Congress in the first place" had the lawyers been clear with their intentions from the outset.
Capriccioso reports that a Justice Department spokeswoman said it's safe to assume that the agency will continue to argue that the funds included in the settlement should go directly to the plaintiffs, and not to the lawyers in the case.


text:
Reuters Events | Renewables
CSP Espanol
Blockchain in Energy
Innovation slashes UK offshore costs to ease pain of new CCL rules
The UK's removal of Climate Change Levy (CCL) exemptions for renewable energy projects puts further pressure on developers to improve on the latest record-low offtake prices of GBP114-119/MWh through large scale deployment of the most advanced turbine and grid technology.
Siemens' latest flagship offshore wind turbine, the SWT-7.0-154
UK offshore load factors accelerate on tech breakthroughs: 10-year study
Milestone for three offshore wind farms in the UK
Rising offshore plant capacities lead to greater data-based O&M gains
Monopiles to remain dominant offshore foundation in Europe: consultant
UK Chancellor George Osborne announced in July that renewable energy projects would no longer receive Climate Change Levy exemption certificates (LECs) which have provided generators with around GBP4/MWh in extra revenue.
The Conservative government's move comes amid wide-ranging cuts across UK State departments to meet deficit reduction targets and follows significant cost reductions in a wind power market dominated by European firms.
Renewable electricity has been exempt from the CCL since its introduction in 2001 and the UK had installed 5 GW of offshore wind capacity by June 2015, some 1,450 turbines, according to industry group RenewableUK.
As developers install more projects in UK waters, advancements in technology allowed ScottishPower Renewables to announce it had lowered the cost of offshore power to £119/MWh for its 714 MW East Anglia ONE project in the latest contract round held earlier this year.
By securing a Contract for Difference (CfD) at this price, the company has sliced the cost of offshore power by over £20/MWh compared with CfDs secured in 2014, the company said.
In the same contract round, Mainstream Renewable Power announced a CfD at GBP114/MWh for its 448 MW Neart na Gaoithe project in Scotland, with additional savings made on introducing Siemens' new Offshore Transmission Module (OTM).
Offshore developers will certainly miss the revenues from CCL exemptions, but on-going technological advancements will allow the offshore sector to close in on conventional generation costs.
Other support mechanisms for offshore wind also remain in place, including the Renewable Obligation Scheme for which suppliers pay generators to meet green energy requirements. In contrast, onshore wind developers will not be eligible for the renewable obligation scheme from April 2016, a decision which was announced by the government in June and sent shockwaves through the sector.
Below, the UK offshore project pipeline as at June 2015, according to RenewablesUK industry group.
Slicing costs
The cost reductions on East Anglia ONE are driven by a supply contract with Germany's Siemens for 102 advanced direct drive 7 MW turbines, representing the largest UK turbine deal ever, ScottishPower Renewables said.
"East Anglia ONE will be the most cost effective offshore wind farm ever delivered. Selecting the turbine supplier will be the single largest agreement for East Anglia ONE, and the most significant in terms of achieving important cost reduction goals," Keith Anderson, CEO of ScottishPower Renewables, said in a company statement.
"We are confident that we have selected the most efficient offshore wind turbines in the world," he said.
According to Robert Norris, spokesman for RenewableUK, the location of the East Anglia ONE project has also reduced project costs.
"East Anglia ONE is relatively close to the coast (approximately 50km) for a Round 3 project and water depth is shallow (30 – 40m). The combination of all these factors drives down costs significantly," he told Wind Energy Update.
ScottishPower Renewables aims to start construction of East Anglia ONE in 2017. The first turbines are to be installed by 2019 and the project is scheduled to be fully operational by 2020.
Source: European Wind Energy Association
Transition talks
The UK's Climate Change Levy exemption was introduced to stimulate demand for renewable energy and applies to energy supplied to businesses.
The government has argued that the value of the LECs for renewables projects will fall sharply in the coming years and be "negligible by the early 2020s" as renewable energy capacity rises faster than eligible business demand.
"As a result it is hard for generators to factor in the exemption into their long term investment plans," the UK Treasury said on its website.
However, Dr Gordon Edge, Director of Policy at RenewableUK, called the levy cut a "punitive measure" for the entire clean energy sector.
Fergus Ewing, Scotland's Energy Minister has also spoken out against the Government's decision, calling it an "anti-business" move, which he believes is creating uncertainty in the investment community.
In reaction to the government's move, UK utilities said they would contact their individual renewable energy suppliers to discuss any impact on contracts and they were awaiting clarity from the government on how the change in law will be implemented.
"Whilst the announcement was unexpected and disappointing, we remain committed to independent generators and to supplying the UK's leading companies with renewable electricity," Robert Groves, CEO of UK supplier Smartest Energy said in a statement.
"We are working through the detail required to transition our business away from LECs and have spoken to our customers. As you would expect there are still many questions unanswered but we will move forward as quickly as possible."
Despite CCL cuts being heavily opposed by many in the UK's renewables sector, developers and equipment suppliers are working hard to innovate and a suite of major UK projects are under development.
Siemens is currently working on several "levers" which the company hopes will bring even greater cost cuts to the offshore wind energy sector, and may even benefit East Anglia ONE, according to Michael Hannibal, CEO of Siemens' Offshore Wind Power and Renewables Division.
The company is seeking to produce technologies by 2020 which drive down offshore power-generation costs to below Eur100/MWh.
In addition to turbine improvements, Hannibal has said that innovating the vital grid connections will make offshore wind more cost and energy efficient.
A major innovation in this area is Siemens Offshore Transmission Module (OTM), which can replace the high voltage alternating current (AC) transformer platforms currently in use. The system is around a third of the size and weight of a conventional platform and Siemens estimates it can reduce the cost of offshore substations by up to 40%.
On announcing the record low CfD for the Neart na Gaoithe project, Mainstream renewable Power said the OTM's would provide an "expected 30% project saving against the offshore substation system CAPEX costs as well as reducing the operating costs significantly."
Mainstream will also use High Wind's Boom Lock system to install the wind turbine components and this will reduce installation time and provide "savings due to reduced cost for installation vessels, installation crews and related costs," the company said.
Sharing gains
UK offshore activity remains high and ScottishPower Renewables announced August 7 an agreement with Sweden's Vattenfall to independently develop offshore wind farm projects within the East Anglia Zone.
The agreement followed changes to agreements with the Crown Estate, the U.K sovereign land agency, which replaced zone development agreements with project specific contracts.
While the two companies will develop projects in East Anglia independently, they are to share offshore wind measuring equipment, zone wide survey data and other key interfaces.
"ScottishPower Renewables intends to focus on developing projects in the southern area of the zone, with Vattenfall developing projects in the northern section of the zone. The overall capacity remains at up to 7.2 GW of potential, with each developer working on proposals for around half of the overall capacity," the UK firm said in a statement.
In addition to the East Anglia ONE project, ScottishPower Renewables is developing the East Anglia THREE project with a capacity of up to 1.2 GW, for which a planning submission is expected by the end of the year.
Vattenfall intends to take forward the East Anglia FOUR project in the northern area, and both companies are discussing further projects in the area with The Crown Estate.
Powering forward
The consistent rollout of new UK offshore power projects in areas such as East Anglia provides a platform for new technology providers to introduce the latest technologies to help drive down project costs.
The government argues that the CCL exemption will become less and less important for renewable energy projects, and other UK measures such as the UK Renewable Obligation will continue to provide support for offshore power.
David Casale, Director at Turquoise, a merchant bank specialising in energy and the environment, considers the removal of the CCL exemption good news for the industry.
"Clearly in the short term, removing a subsidy is never going to be met with open arms. However, as in wider society, we in the clean energy industry need to be weaned off of the subsidy fix," he said.
According to Casale, everyone must be willing to pay the necessary costs of a low carbon economy, including consumers, but secure in the knowledge that there are no hidden schemes. Casale has called the CCL such a scheme and said, "I for one say good riddance."
PV Operations Europe 2021
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*Research note* Wind Energy Operations & Maintenance Costs
Long term contracts have led to a higher cost profile for asset owners and contracted asset owner, due to the "partial wrap" or "full wrap" component warranty for blades, gearboxes, and generators. However, the global COVID-19 pandemic has had some material impacts on this trend with OpEx budgets that were rapidly slashed. Has this affected you?
*Research Note* Wind Turbine OEMs: Reducing your bill of materials cost opens new markets
One of the most important tools they have to open new market is to reduce the bill of material costs. Drawing on an example from Nordex Acciona, the following research note takes a look at their global product sales pipeline, bill of materials reduction trends, global turbine price trends as well as buying patterns.
*Research Note* Global offshore wind energy market & commercial outlook
The global offshore wind energy market is undergoing a period of explosive growth. The market could grow to a total of 584GW installed if the industry finds a way to unlock the US $1.432 trillion in investment required to develop the proposed capacity additions.
*Research note* Offshore wind energy technology & supply chain outlook
Facing a period of tremendous growth, the offshore wind energy sector must continue to expand the supply chain to gain maximum global leverage and fulfil its potential. Understanding technology growth patterns and how to position yourself in this changing market landscape will be critical.
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Knitting Like Crazy: A Pyramid O' Yarn?
First off, work has been crazy busy. No time for knitting during my morning break or lunch. It is sad. I dream of the day when I can knit at work again. There is not much to show on the knitting front. I have been working a row here and there on my sock, and there is some secret birthday knitting that is taking place, but I can't show you that. Soon though.
Anyway, this weekend, Alex and I headed over to Chocolate Fest in Long Grove. You may think I wanted to go because of the chocolate, but the truth is, I had a coupon for 20% off non-sale yarn at I'd Rather Be Knitting (btw, if you go to their website, I think the coupon is good for the same thing during Strawberry Fest, which is in June some time...just saying). There was some purchasing for the Secret Pal, but certainly I cannot show you that. There was some purchases for me too, of course!!
Here we have some Mountain Colors Bearfoot in Marias Falls. I think it is a little darker than the picture shows. I am imagining something lacey with it.
It's Kathmandu Aran. I have never knit with it before, but it is soft and nice and pretty and cheap! I am imagining another shrug or some kind of bag with this.
Hey look! It's a pyramid of yarn!
Chocolate Fest was a lot of fun. We walked around, ate some chocolate fondue, people watched. Alex even walked around the knitting store like he almost cared about the yarn. I swear, it is just a matter of time before that boy starts knitting.
That's all. The rest of this week brings more working late, more American Idol, and more secret birthday knitting.
This post was brought to you by the letter Y and the number 12.


The perfect way to have a complete overview of the museums or monuments of your interests in short time, avoiding waiting in line and guided by a professional guide that will focus on the highlights of each site.
-Park Guell, Sagrada Familia, Casa Batlló, Casa Milá, Palau Güell, Casa Lleó Morera, Museu Picasso, Fundació Miró, Hospital de Sant Pau, Museu d'Història de Barcelona (MUHBA Plaça del Rei), Museu Nacional d'Art de Catalunya-MNAC, Museu del Modernisme, Torre Bellesguard, Catedral de Barcelona, Sinagoga Medieval.
Time spent at each museum/monument: 1hour 15 min approx.
Transportation (taxi, bus and/or metro or chauffeured car/van) is needed to get to each site and it is not included in the price of the tour.
Entrance fees to monuments/museums are not included in the price of the tour.