text:
Northumberland schoolboy named UK's Young Neighbour of the Year after being inspired by Marcus Rashford
Noah Macdonald from Wylam won the award for his fundraising efforts to help those in need, his efforts inspired by Marcus Rashford's free school meals campaign
Mike KellyReporter
The UKs Young Neighbour of the Year, Noah Macdonald (Image: Handout)
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A 10-year-old Northumberland schoolboy been named the UK's Young Neighbour of the Year for his efforts to raise money for those in need.
Noah Macdonald from Wylam was inspired by Marcus Rashford's free school meals campaign which aimed to ensure that children from low income families did not go hungry.
The Ovingham Middle School pupil organised his first charity walk to raise money for Fenham Association of Residents (FAR) and to raise awareness of the issue of food poverty.
Go here for more Northumberland news and updates from Northumberland Live
Based in Newcastle, FAR aims to break down the social isolation that people on low incomes often experience by providing a range of opportunities for learning and recreational activities, to build self-confidence and develop new skills.
Noah walked a local eight mile loop twice across one weekend with some of his friends to raise awareness for his campaign.
His mum, Dee, helped publicise his fundraising campaign on social media, and donations came in from the UK as well as abroad.
Earlier this month Noah embarked on a 10 mile charity walk through sleet and rain from his home in Wylam to Newcastle, raising an impressive £620.
The funds will be used to buy Christmas presents which will be handed out to young refugees in the area by North East Solidarity and Teaching (NEST) – a student-led organisation based at Newcastle University which aims to support those fleeing from conflict.
Noah said: "I'm so happy to win this award. People like Marcus Rashford really inspire me to help others - it's not fair that some children don't have enough food or won't get any Christmas presents.
"I think if we all work together we can make a big difference. My Grandmother helped to set up a charity in 1991 which helps people in our local community and it has really shown me how important it is to think of other people."
It was his mother, Dee, who nominated Noah for the Co-op Insurance and Neighbourhood Watch UK's Young Neighbour of the Year award.
She said: " Noah is a kind and caring young man with a social conscience. He wants fairness and does believe he can help to change.
"I'm so proud of what he's achieved this year and am delighted that he's getting recognition for his efforts."
Noah was inspired by Marcus Rashford's free school meals campaign (Image: Handout)
Charles Offord, managing director of Co-op Insurance, said: " The next generation is in very safe hands if Noah's story is anything to go by.
"It was heart-warming to see the hundreds of entries we had illustrating the amazing work of young people all over the UK but his passion and motivation really stood out."
John Hayward-Cripps, CEO of Neighbourhood Watch, said: "We are absolutely thrilled to crown Noah as our Young Neighbour of the Year – his passion for helping others really shone through and we hope his work will inspire many other young people to get out in their community and help those who are less fortunate."
The Co-op Insurance and Neighbourhood Watch Neighbour of the Year awards, now in its fourth year, launched in 2018 to shine a light on the UK's unsung heroes who are carrying out work in their communities to improve the lives of others.
Winners are chosen in the categories of Neighbour, Young Neighbour and Community of the Year.
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Sam Fender urges people to buy The Big Issue as high streets empty due to Covid surge
Covid: Boris Johnson 'considering three options including full lockdown' to slow spread of Omicron

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Logitech is preparing the first release of their Gaming Software for Mac OS X (LGS). Just as with their gaming gear, they put the software through multiple phases of internal testing. But no Logitech product can be approved for release until its been field tested under real-world conditions by its most intensive users. That's where you come in.
If interested, please click the link below to join this public beta which is starting right now. Also, to see this and other open projects for OnlineBeta please visit www.onlinebeta.com/open.


Most of the population of western Ireland was engaged in agriculture of some form. However in the towns there were store owners, blacksmiths, stone masons, chimney sweeps, car men, thatchers, and labourers of all sorts. Near the coasts there were fishermen.
SKETCHES OF THE PESANTRY IN MAYO"
These three images show the shepherd, a young woman carrying fleece and a young woman spinning the yarn.
"Irish Granite Industry. Making setts "
Charles Lardner, the godfather of Joseph Walsh, was a stone cutter.
This card is not post marked. However, someone has written "1908" in pencil on the back of the card.
An Irish Thatcher: One of the characteristic features of the Irish countryside is the old-style farm house with its thatched roof. The style of thatching varies throughout the country. Along the west coast the thatch is bound down securely with a network of ropes, fastened to a row of pegs set in the walls, or to a line of stones along the eave. Inland thatching is bound with cords or scollops. Thatch was a cheap from of roof in the old days.
In the days before the automobile an important member of the community was the blacksmith here shown shoeing a horse.


State regulators are considering the estimate from an industry analyst that the Atlantic Coast Pipeline will increase costs for Dominion Energy ratepayers by more than $1 billion.
CHARLOTTESVILLE, Va. (WVIR) - Virginia regulators are considering the estimate from a natural gas industry analyst that the Atlantic Coast Pipeline will increase costs for Dominion Energy ratepayers by more than $1 billion.
The analysis suggests the proposed pipeline will cost Dominion customers $1.61 billion to $2.36 billion.
The energy company wanted the State Corporation Commission to throw out that report, which was filed in response to Dominion's Integrated Resource Plan.
Monday, March 12, the commission ordered Dominion to include the analysis in its final long-term plan which is due May 1.


The British Library is one of the world's greatest research libraries and is regarded as the international institution. It holds around 170 million items from different nations in several languages in both formats print and digital. Library include items such as books, manuscripts, music, newspapers, drawings, sound, journals etc (Lankes, 2011). The institution act as non-departmental public body sponsored by the Department for culture,media and sport. It was established in 1973 in part of London. Since then library has seen great transformation in its services and strategic priorities. Management continues to work on delivering high quality and innovative services for the users. Digital modernization was an important phase in the services of British Library (Goodchild, 2012). It has now become an important part of its strategic goals and objectives. This research is undertaken to talk about one of the component of digital world that is digital scholarship for the scholars.
What is the impact of digital modernization on the services of British Library?
The significance of this research study is that it will give an clear insight view about the existing digital research trends. It will also determine which digital tools, content and services are needed to be prioritized by British Library. Apart from that it will help in identifying the need and expectations of the library users. Scope of the study is that it will further open new opportunities for British Library for bringing more digital changes in the future.
Data Collection is the most vital section of the research. It is very important that valid and relevant data is to be collected for evaluation. It is required to be collected from valid and authentic sources. There are two types of data primary and secondary (Sapsford and Jupp, 2006). Primary data is new and fresh in nature and on getting used once become secondary to others. For this study, primary data will be collected from users of British Library through survey method. Secondary information is available from books, journals, online sources, Annual reports related to the British Library (Punch, 2009). Survey method will be applied because it is the most effectual approach of collecting substantial amount of information.
Digital library is a focused collection of digital objects such as audio material, visual material, text, video material etc. All these stored in the form of electronic media formats. These libraries varies in size and scope. These are maintained by individuals, groups or organizations. The electronic content may be stored locally or accessed remotely through computer networks (Tang and et.all., 2012). It is a kind of information retrieval system. It is the most advanced medium of gaining knowledge and learning for the users. These users include researchers, scholars, students, experts etc. It is used widely because it provides lot of convenience and simplicity as the user can have the data in a very quick span of time (Hutchinson and et. all., 2013). This type of library is a kind of gift for the people as they can avail the knowledge sitting in any part of the world. It is essential that these kind of approaches should be encouraged on wide level.
Digital scholarship is an highly advanced method of offering scholarship for the users. It make use of digital tools and techniques to offer learning and knowledge to the scholars. Several concerns have been risen related to regulation of digital scholarship across many universities across the world. The increasing demand of the scholarly work has left many Universities to think how they will manage scholarship in future (Xiaolin, 2011). There is a growing interest of people in performing scholarly and research activities. In that context, digital scholarship has come highly into the picture. They are using digital scholarship in order to make money and staying ahead and also for the protection of their brand. People are also finding lot of familiarity with this approach because people these days have become techno-friendly. They are more into computers, laptops, gadgets, smart phones etc (Zhang, 2010). These things are making their life easy and comfortable. Hence in such case invention of digital scholarship adds higher value to their life.
The self-reflection is considered to be the most important part of research work. It is through self-reflection that the researcher is able to contribute significantly to the research work. The reflective statement helps in identifying one's weaknesses and strengths that in-turn helps in conducting research with efficiency. The research on digital modernization has helped me in gaining various valuable insights. I am glad to acquire knowledge about development of digital libraries. Moreover, it also helped me in gaining awareness about concepts and development of digital library. The research added to my knowledge base of manner in which digital services of British Library helped in improving life of people. I feel that in today's world of digital technology the every aspect of human life is turning digital. I am surprised to see the manner in which digital services of British library has increase the level of convenience among people. Library is meant for gaining knowledge about various fields. In traditional times, individuals used to spend most of time in reading books in library. However, in present world due to busy schedule it has been found that reading habits of people are declining on continuous basis.
It can be therefore said that the digital library provides a platform to inculcate reading habit among people. I am glad to know that digital services of library provides an opportunity to people for maintaining their reading habits within comfort zone of their home. I have heard of online libraries and services offered by them. However, I have never availed the services provided by them due to lack of awareness. It is through present research that awareness has been created in me regarding presence of digital library within the country. I am surprised to see the manner in which digital modernization leads to improvement in services of British library. I can say that the digital modernization has changed the life style of people. The conduct of research also proves to be highly significant for me. I am able to equip skills related to the conduct of research. This in turn helps me in development of skills that are highly valuable for my personality development. The academic research is considered to be part and parcel of researchers' life.
I am really glad to acquire skills related to manner in which research can be conducted with efficiency. Moreover, the skills acquired will help me in my future endeavours. I will be able to conduct the research work in future with high proficiency. I can say that the skills and learning developed as a part of research work support me for long-term. It is through the present skills acquired that I will be able to understand various concepts related to research work. Moreover, I can claim that the issues that I am capable of solving issues that arises as a part of research work. I can therefore say the present research has laid down a firm foundation in me for conduct of future research work.
To identify the awareness and adoption of digital tools and content on the research practices of British library users.
The British library as the name suggest is the national library of United Kingdom. Library is the place that aims at increasing individual's knowledge by providing reading material of varied range. The British library is well-renowned library across the world with collection of approximately 170 million items (Garibay, Gutiérrez & Figueroa, 2010). The library has maintained collection of books, journals, manuscripts, journals, magazines, sound-recordings, music-recordings, videos, play-scripts, patents, databases, maps, stamps, prints and drawings. The collection consists of items in different languages and formats. Moreover, the library has occupied the largest library across world measured through the number of items (Hutchinson and et.all., 2013). As the renowned library, British library tends to receive copy of all books that are published in United Kingdom and Ireland.
The library has attracted large number of readers from various parts of world. The library owes its existence to 1973 when it was created as a declaration of British Library Act 1972. The library was considered as a part of British museum prior to its creation (Goodchild, M. F. (2012)). The library has been retaining its image of national library since its creation. It has changed the way of operations with the time. In present time, it has also adopted digital methods to offer services to readers (About Us, British Library, n.d.). It can be said that the library is highly professional and satisfies demand of all its readers. The report is highly focused on identifying the impact of digital modernization on the services of the institution. It tries to identify whether users of BL are satisfied from their digital content and services. It will also identify the awareness and adoption of digital tools and content on the research practices of British library users.
This section shows the application of research tools and techniques that are applied for the study. It is important that all the methods must be applied with appropriate tools and approaches along with proper justification and logic (Sapsford and Jupp, 2006).
Research philosophy is related with examining the thought process of the people. There are four type of philosophies realism, positivism, pragmatism and interpretivism. The present report will follows interpretivism philosophy (Punch, 2009). It will work towards understanding the impact of Digital modernization on the services of the British Library. It will help in investigating the phenomenon in more natural environment. It helps in identifying the thoughts of the people and their actions.
It outlines the framework carried by the study. Research approach helps in bringing effectiveness throughout the study. There are two types of inductive approach and deductive approach. For this study inductive approach will be applied. It is related with discovery of new theory from the information (Olsen, 2011). This approach is applied because this research work is focused on exploring new phenomenon and it also look at the previously researched phenomenon from a different view point. Further this study is associated with qualitative research.
Research design reflects the quality of the study. It helps in collecting, interpreting and analyzing the data related to the subject. Selection of a suitable design offers knowledge in more conceptual manner. Different types of research design are descriptive, exploratory, casual, case study, experimental etc (McBurney and White 2009). For this study a combination of descriptive and exploratory design will be applied. Descriptive design is applied because the study is related with an specific entity that is British Library. Here the researcher wants to examine a specific behavior as it occurs in the environment. Further this design demands collection of data through survey (Erik and Marko 2011). Exploratory design is applied because the researcher has observed something about the British Library and is making efforts to gain more knowledge about it.
It states the manner under which the data is to be presented and analyzed. There are two research types qualitative and quantitative. For this study qualitative type will be used as the study is more subjective in nature (Cohen, Manion & Morrison, 2007). It requires evaluation of the data by using qualitative methods. Further it examines the behavior of the consumers in certain environment.
Sampling activity aids in selection of persons who can provide some valuable data for the research. It helps in making the data collection task easy and effortless. There are two methods of sampling probability and non probability (Sapsford and Jupp, 2006). For this study combination of convenience and random sampling is applied which belongs to both the types. Convenience sampling is adopted because it helps in collecting the data as per the convenience of the researcher. For this study researcher will collected the data through survey approach by visiting the British Library (Punch, 2009). On visiting the library through random sampling approach he will select the respondents for the survey. A sample size of 50 respondents have been kept for the survey.
There can be some ethical issues with the research work, hence researcher will follow all the ethical guidelines during the study. All the data and information will be collected from valid and authentic sources. Act of plagiarism will be avoided to the higher extent. During the survey process identity of the respondents will be kept secretive until being told to disclose. Opinions and responses of the participants will be respected (McBurney and White, 2009). There should not be any harm to their feelings and sentiments. Researcher will also try to avoid any kind of subjectivity within the report.
From the above study it can be concluded that this research proposal will prove to be an extinguishing research for the readers. Researcher can say that the skills and learning developed as a part of research work support support for him. He will gain lot of new skills and learning from this report. It will help him in his career development. There are certain ethical and limitation issues which are to be taken into consideration by the researcher. This study will successfully identify the impact of digital modernization on the services of British Library.
Garibay, C., Gutiérrez, H., & Figueroa, A. 2010. Evaluation of a digital library by means of quality function deployment (QFD) and the Kano model.
Goodchild, M. F. 2012. The Alexandria digital library project review, assessment, and prospects.Trends in Information Management (TRIM),1(1).
Hutchinson and et.all., 2013. The international children's digital library: a case study in designing for a multilingual, multicultural, multigenerational audience.
Lankes, R. D. 2011.The atlas of new librarianship. MIT Press.
McBurney, D. and White, T. 2009. Research Methods. Cengage Learning.


text:
Das Pawlodarski Traktorny Sawod, dt. Pawlodarer Traktorenwerk, kurz PTS oder PTZ (, kurz ПТЗ) war ein zunächst sowjetischer und später kasachischer Hersteller von Traktoren. Das 1966 gegründete Unternehmen saß in Pawlodar, Kasachstan und wurde 1998 erstmals für zahlungsunfähig erklärt. Verschiedene Nachfolgegesellschaften und Investoren versuchten bis etwa 2006, eine neue Traktorenproduktion am alten Standort aufzubauen, was letztlich scheiterte. Mindestens zu sowjetischen Zeiten trug das Werk den Beinamen W. I. Lenin.

Unternehmensgeschichte

Das Werk wurde 1966 als Pawlodarski Traktorny Sawod imeni W. I. Lenin (russisch Павлодарский тракторный завод им. В. И. Ленина) gegründet, die Traktorenproduktion startete am 12. August 1968. Gebaut wurden Traktoren vom Typ DT-75, die zusätzlich die Bezeichnung "Kasachstan" erhielten. Das Fahrzeug war ursprünglich im Wolgogradski Traktorny Sawod entwickelt worden und wird dort auch heute noch gebaut.

Die in den 1960er- und 1970er-Jahren gebauten Traktoren hatten noch die ursprüngliche runde Kabine, wurden blau lackiert und trugen den Schriftzug "Казахстан" an der Motorhaube. Erst Mitte der 1980er-Jahre wurde mit dem DT-75ML eine modernisierte Version mit rechteckiger Kabine und abweichender Farbgebung eingeführt. 1984 war das Jahr mit dem höchsten Ausstoß in der Unternehmensgeschichte – 55.000 Traktoren wurden gebaut.

Mit dem Zerfall der Sowjetunion wurde auch das Werk in Pawlodar von schweren wirtschaftlichen Problemen getroffen. 1997 wurden noch 1879 Traktoren gebaut. 1998 wurde das Unternehmen erstmals für bankrott erklärt.

In den folgenden Jahren firmierte das Werk mehrfach um und wurde auch mehrmals verkauft. Es bestand der Plan, Traktoren der Typen MTZ-80 und MTZ-82 aus dem Minski Traktorny Sawod in Lizenz zu fertigen. Der Plan scheiterte aber, auch aufgrund diverser Fehlinvestitionen. 2006 erklärte der damalige Premierminister Kasachstans, Danial Achmetow, dass die Regierung keine Finanzmittel zur Rettung des Unternehmens bereitstellen würde.

2007 gründete ein Unternehmer auf Grundlage des alten Werks eine angeblich lizenzierte Ersatzteilproduktion für Belarus-Traktoren. Wenig später wurde er von einem kasachischen Gericht wegen Betrugs zu sieben Jahren Haft verurteilt. Die verbliebenen Maschinen wurden als Bezahlung für die Gläubiger genutzt. Seitdem sind keine weiteren Aktivitäten anderer Firmen bekannt, die eine Traktorenproduktion in Pawlodar anstreben.

Produkte
Das Werk fertigte in seiner Geschichte im Wesentlichen nur ein einziges Traktorenmodell. Vom Kettentraktor DT-75 wurden zwischen 1968 und 1995 etwa 862.300 Exemplare in Pawlodar gebaut. Die Fertigung lief um 1998 aus.

Zusätzlich wurde nach dem Zerfall der Sowjetunion ein neuer Kettenschlepper projektiert. Das Fahrzeug mit der Bezeichnung T-95.4 sollte ab 2001 gefertigt werden, erreichte jedoch nie auch nur annähernd die gleichen Stückzahlen wie der DT-75.

Ob tatsächlich je Traktoren des Minski Traktorny Sawods in Lizenz gebaut wurden, ist unklar.

Einzelnachweise

Weblinks

 Webseite zur Geschichte des Unternehmens (russisch)
 Fotogalerie mit Abbildungen von in Pawlodar gebauten DT-75-Traktoren (russisch)

Ehemaliger Traktorenhersteller
Gegründet 1966
Aufgelöst 2006
Pawlodar
Produzierendes Unternehmen (Kasachstan)

redpajama_set_name: RedPajamaWikipedia


Personal Capital is one of the growing number of investment platforms available. But it may be even better described as an account aggregator since it also connects virtually all of your financial accounts, including your bank accounts, investment accounts, credit cards and even your mortgage.
On top of that, they also provide investment advice to help you manage and grow your investments. It's professional investment management, but without the high fees typically associated with that service.
5 Should You Choose Personal Capital to Help You Manage Your Investments?
Personal Capital provides such a wide range of services that you'll feel like you're getting a very personal touch – and often you literally are.
The platform is free to use and can also provide you with a dedicated investment advisor if you have over $100k in your portfolio.
They offer what they refer to as The Big Picture view of your entire financial profile. This is also referred to as the "360 degree View of Your Financial Life". It includes all of your financial accounts, and considers the impact that each has as a part of your overall financial situation. This is almost unique among web based investment platforms.
Personal Capital also incorporates an Integrated Investment Portfolio feature, where they import all of your investment accounts into the platform to give you a comprehensive look at your asset allocation. This is incredibly valuable if you have your investments spread out over several accounts, with various investment brokers.
The have a unique custom feature referred to as The "You Index". This tool specifically monitors the performance of your equity investments, including both funds and individual stocks. However, it does not include cash-type accounts or any holdings of individual bonds, options or alternative investments.
Like most investment platforms now, Personal Capital allows you to incorporate mobile apps. The platform can be downloaded on Apple iPhone, iPad and Adroid apps, and the performance level will be comparable to the desktop version.
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Avoiding mutual funds, which are tax inefficient.
Tax allocation. For individuals with both taxable accounts and retirement accounts, this means placing the right investments in the right accounts. High yielding stocks and fixed income generally go into tax deferred or exempt accounts. REITs pay non-qualified dividends and should also be owned in retirement accounts.
Tax loss harvesting. Because they use individual securities, they can realize losses to offset gains or provide a deduction.
It's always a major plus when your investment advisor "has your back" on the tax front.
Based in San Francisco, Personal Capital's co-founder and CEO is Bill Harris, who was formerly the CEO at both PayPal and Intuit (makers of Quicken and TurboTax).
The platform works by customizing your investment portfolio based on your goals, preferences and risk tolerance. Once they learn about your entire financial situation and goals, they then determine how much growth you need to reach your goals. A portfolio will be designed that incorporates an appropriate risk and return profile. They also account for unique circumstances, such as savings goals like buying a house. For the most part, however you want your portfolio customized, they'll make it happen.
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Here's how it works: Numerous government programs, most notably Social Security benefits and the income thresholds for tax brackets, are indexed for inflation. But inflation can be measured in a number of ways. The tax code, for instance, uses CPI-U (Consumer Price Index – Urban), which measures prices for consumers in urban areas, to adjust the income cutoffs for different tax brackets. Social Security uses CPI-W, which is like CPI-U but only counts prices paid by urban wage-earners, not all consumers.
Various deficit-reduction frameworks, including Bowles-Simpson, Domenici-Rivlin and the Gang of Six plan, would convert all programs using CPI-U or CPI-W to a third measure — called C-CPI-U, or chained CPI. Most inflation measures, including CPI-U and CPI-W, track the price of a certain basket of goods. That basket could include, say, a year's supply of propane. When propane costs go up, CPI-U and CPI-W include that as an increase in the cost of living.
But some people would just stop using propane if its price went up. They'd switch to electric heating, or a geothermal system, or a wood stove. So their actual heating costs wouldn't go up as much as CPI-U and CPI-W would suggest. Chained CPI attempts to take "substitution effects" like this into account. Thus, its number generally rises more slowly than other metrics.
The group getting the biggest tax hike is families making between $30,000 and $40,000 a year. Their increase is almost six times that faced by millionaires. That's because millionaires are already in the top bracket, so they're not being pushed into higher marginal rates because of changing bracket thresholds. While a different inflation measure might mean that the cutoff between the 15 percent and 25 percent goes from $35,000 to $30,000, the threshold for the top 35 percent bracket is already low enough that all millionaires are paying it. Some of their income is taxed at higher rates because of lower thresholds down the line, but as a percentage of income that doesn't amount to a whole lot.
All told, chained CPI raises average taxes by about 0.19 percent of income. So, taken all together, it's basically a big (5 percent over 12 years; more, if you take a longer view) across-the-board cut in Social Security benefits paired with a 0.19 percent income surtax. You don't hear a lot of politicians calling for the drastic slashing of Social Security benefits and an across-the-board tax increase that disproportionately hits low earners. But that's what they're sneakily doing when they talk about chained CPI.
That's why watchdog groups like the Center for Budget and Policy Priorities argue that the only fair way to do chained CPI would be to pair it with an increase in Social Security benefits, and to exempt Supplemental Security Income, which provides support for impoverished elderly, disabled and blind people. Otherwise, it's just a typical "raise taxes, cut benefits" plan, and an arguably regressive one at that.


text:
\section{Introduction}

The plane billiard consists in the free motion of a point particle
on a connected bounded region in ${\cal R}^2$, being reflected elastically
at the boundary. The billiard defines a 2-dimensional discrete dynamical
system.

Depending on the boundary, this discrete dynamical system may have
different dynamical behaviour.
For instance, if it is a circle or an ellipse, it is known that
the system is integrable, and the phase space is ordered by
invariant curves (figure~\ref{fig:circ}).

A quite different situation appears when the components of the boundary have
negative curvature. The system is, then, ergodic and almost every orbit is
dense on the phase space (as in Sinai billiards).

The first example of an ergodic billiard with convex boundary was given by
Bunimovich \cite{kn:bun}. The boundary is the circular stadium, composed by
joining two half-circles by means of two straight segments of equal length.

A convex generalization is the elliptical stadium, composed by two
half-ellipses, with half axes $a\geq 1$ and $b=1$, joined by two straight
segments of equal length $2h$. For $a=1$, ie, for the circular stadium
billiard,
$h>0$ implies the existence of a positive Lyapunov exponent almost everywhere.
This will not be the case for $a\neq 1$ and small values
of $h$, because the structure of the phase space of the elliptical billiard
is not as degenerate as that of the circular billiard (figure~\ref{fig:circ}).

V. Donnay, in \cite{kn:don}, proved that for the billiard on the elliptical
stadium, if $1<a<\sqrt 2$ and $h$ is sufficiently big, then there is
a positive Lyapunov exponent almost everywhere. Also that $h$ must grow as
$a$ increases to $\sqrt 2$. And he gives a challenge: ''One could try to
calculate bounds on these lengths.''

In \cite{kn:can}, Canale and Markarian studied this problem numerically and,
varying $a$, found a value $h(a)$ such that if $0\leq h<h(a)$,
the billiard does not seem to be chaotic and, if $h>h(a)$, it seems ergodic.

In this article we give a partial answer to Donnay's challenge. We prove that
if
$1<a<\sqrt{4-2\sqrt 2}$, then $h>2a^2\sqrt{a^2-1}$ assures the positiveness of
a
Lyapunov exponent. To do so, we construct a cone-field, eventually strictly
invariant under the derivative of the billiard map, and apply Wojtkowski's
Theorem \cite{kn:woj}. The magic number $\sqrt{4-2\sqrt 2}$ is related to
the structure of bifurcation of  hyperbolic caustic periodic orbits of the
elliptical billiard (see figure~\ref{fig:bif}).

The existence of a positive Lyapunov exponent, meaning sensitive dependence on
initial conditions, does not automatically imply ergodicity. However, in this
example, as the numerical experiments suggest, ergodicity (and K-property)
can be proved. In fact, the additional conditions required on the Fundamental
Theorem of Sinai and Chernov are satisfied by this system (see \cite{kn:mark}
and \cite{kn:liv}).

\begin{figure}
\vspace{4truecm}  \special{psfile=fig1.ps}
\caption{Phase space of the circular, the elliptical and the
         circular stadium billiards}
\label{fig:circ}
\end{figure}

\section{Billiards}
Let $\Gamma$ be an oriented plane closed curve, or a finite set of
(topologically) closed curves on the plane, and $Q$ the region enclosed by
$\Gamma$. The billiard problem consists in the free motion of a point particle
on $Q$, being reflected elastically at the impacts on $\Gamma$. At any instant,
the state of the particle is determined by its position and
velocity. Since the motion is free on $Q$, all the motion is determined
either by two consecutive points of reflection at $\Gamma$ or by the point of
reflection and the direction of motion immediately after each reflection.

\begin{figure}
\putfigurewithtex fig2.ps fig2.tex  4.5 16
\caption{Coordinates for the elliptical billiard}
\label{fig:eli}
\end{figure}

Let $t\in [0, L)$ be a parameter for $\Gamma$ and the direction of motion be
determined by the angle $\theta$ with the tangent to the boundary. The billiard
defines a map $T$ from the annulus ${\cal A}=[0, L) \times (0,\pi)$
into itself. Let $(t_0,\theta_0)$ and $(t_1,\theta_1) \in {\cal A}$ be such
that
$T(t_0,\theta_0)=(t_1,\theta_1)$. As usual, we will call $t$ either the
value of the parameter or the point $\Gamma(t)$ at the boundary.
If $\Gamma$ is $C^k, k \geq 2$, in some neighborhoods of $t_0$ and $t_1$, then
$T$ is a $C^{k-1}$-diffeomorphism in some neighborhoods of $(t_0,\theta_0)$ and
$(t_1,\theta_1)$ \cite{kn:str}. $T$ also preserves the measure
$d\mu = \sin \theta d\theta \frac{ds}{dt}dt$ \cite{kn:bir}, where $s$ is the
arclenght parameter for $\Gamma$.

$({\cal A},\mu, T)$ defines a discrete dynamical system, whose orbits are
given by $\{ (t_n,\theta _n)=T^n(t_0,\theta_0), n \in Z\}$.

The polygonal on $Q$, whose vertices are $\Gamma (t_n), n \in Z $, is often
called the trajectory of the particle. Because of the obvious correspondence
between orbits and trajectories, we will call both of them either orbits or
trajectories.

\subsection{The Elliptical Billiard}

Given an orthogonal coordinate system in ${\cal R}^2$, let $\Gamma$ be the
ellipse $\frac{x^2}{a^2}+y^2=1$, with $a>1$. We will use the angle $\varphi$
between the oriented tangent and the $x$-axis as a parameter for $\Gamma$
(figure~\ref{fig:eli}). We will call half-ellipses the pieces of $\Gamma$
corresponding to $0\leq \varphi \leq \pi$ and $\pi \leq \varphi \leq 2\pi$.

Each trajectory has a conic caustic, confocal with $\Gamma$. If a segment of a
given trajectory cuts the segment joining the two foci, all the other segments
of this trajectory will cut it and the caustic will be a hyperbola. If it
passes by  one focus, the trajectory will always pass by the foci, thus having
the two foci as a degenerate caustic. Otherwise, the caustic will be an ellipse
(see, for instance \cite{kn:cha}).

The phase space of the elliptical billiard is the annulus
${\cal A}=[0,2\pi)\times (0,\pi)$, with coordinates $\varphi$ and $\theta$.
As $\Gamma$ is $C^{\infty}$, the map $T:{\cal A} \rightarrow {\cal A}$ is a
$C^{\infty}$-diffeomorphism, where
$(\varphi_1,\theta_1)=T(\varphi_0,\theta_0)$ is given by:
\begin{eqnarray}
a\tan \varphi_1 &=&\frac{a\tan \gamma -a\tan \varphi_0}
{1+(a\tan \gamma )(a\tan \varphi_0)} \ ,
\  \ a\tan \gamma = \frac{2a\tan (\varphi_0+\theta_0)}
{1-(a\tan (\varphi_0+\theta_0))^2} \nonumber \\
\theta_1&=& \varphi_1 -(\varphi_0+\theta_0)
\end{eqnarray}
and the derivative of $T$ is (see, for instance \cite{kn:str}):
\begin{equation}
DT_{(\varphi_0,\theta_0)}=\frac{1}{R_1\sin \theta_1}\left(
\begin{array}{cc}
l_{01}-R_0\sin\theta_0&l_{01}\\
l_{01}-R_0\sin\theta_0-R_1\sin\theta_1&l_{01}-R_1\sin\theta_1
\end{array}\right),
\label{eq:derivelipse}
\end{equation}
where $l_{01}$ is the distance between $(x(\varphi_0), y(\varphi_0))$ and
$(x(\varphi_1), y(\varphi_1))$, and
$R_{i}=a^2/(a^2\sin^2 \varphi_{i}+\cos^2 \varphi_{i})^{3/2}$
is the radius of curvature at $\varphi_{i}$, $i=0, 1$.

$T$ has no fixed points and has exactly four two-periodic orbits, being
two elliptic
(both eigenvalues of $DT^2$ have moduli one) and two hyperbolic (the moduli
of the eigenvalues of $DT^2$ are one smaller and the other greater than one).

The function $F(\varphi,\theta)=\frac{\cos
^2\theta-\varepsilon^2\cos^2\varphi}{1-\varepsilon^2\cos^2\varphi}$, where
$\varepsilon=\frac{\sqrt{a^2-1}}{a}$ is the eccentricity of the ellipse
$\Gamma$, is a first integral
for $T$, ie, $F$ is constant on orbits under $T$ (see, for instance,
\cite{kn:lev}). Physically, $F$ may be interpreted as the product of angular
momenta about the two foci \cite{kn:berry2}. Geometrically,
$\sqrt{\mid F(\varphi_0,\theta_0)\mid}$ measures the length of the minor axis
of the conical caustic \cite{kn:fel}.
So, for every $a>1$, the phase space ${\cal A}$ is foliated by
the level curves of $F$ (figure~\ref{fig:circ}).

Let us call $\cal E$=\{ trajectories with elliptical caustic\} and
$\cal H$=\{ trajectories with hyperbolic caustic\}.
The obvious correspondence between trajectories on the configuration space
$Q$ and orbits on the phase space ${\cal A}$ gives  the following geometrical
interpretation:
\begin{itemize}
\item trajectories over the major axis $\longleftrightarrow$ period 2,
hyperbolic
orbits , $F=0$
\item trajectories over the minor axis $\longleftrightarrow$ period 2, elliptic
orbits, $F=1-a^2$
\item trajectories that pass by the foci $\longleftrightarrow$ saddle
connection,
$F=0$
\item trajectories on ${\cal E}$ $\longleftrightarrow$ orbits with $F>0$
\item trajectories on ${\cal H}$ $\longleftrightarrow$ orbits with $1-a^2<F<0$.
\end{itemize}
Because of this interpretation, we will also call $\cal E$ the
set of orbits with $F>0$ and $\cal H$ the set of orbits with $1-a^2<F<0$.

Poncelet's theorem \cite{kn:cha} assures that trajectories that
share the same caustic have the same dynamical behaviour.
And to each conical caustic, or equivalently, to each integral curve $F=k$, we
can associate a rotation number (see, for instance,\cite{kn:pal}).

\begin{itemize}
\item
For $k>0$ (trajectories on ${\cal E}$) Jacobi coordinates (see,
for instance \cite{kn:cha}) give a natural conjugation between ${\cal E}$ and
the trajectories of the circular billiard with $\theta \neq \pi /2$. It
follows that, to each $k>0$ we can associate a rotation number
$\rho(k) \in (0,1)$, such that:
\begin{itemize}
 \item $\rho = \frac{n}{p} \in {\cal Q}, (n,p)\equiv 1$ corresponds to
    periodic orbits of period $p$,
    with $T^p(\varphi_0, \theta_0) = $ \break $(\varphi_0 + 2n\pi, \alpha_0)
\cong
    (\varphi_0 , \alpha_0)$.
 \item $\rho \in {\cal R} \setminus {\cal Q}$ corresponds to dense orbits
  (dense on the level curve).
\end{itemize}

\item
For $1-a^2<k<0$ (trajectories on ${\cal H}$) notice
that every periodic orbit has an even period and $(0,\pi/2)$ and
$(\pi,\pi/2)$ are elliptic fixed points of the integrable, measure
preserving diffeomorphism $T^2$, with
eigenvalues
$$\lambda_j =e^{2\pi i\nu_j (a)},\:\hbox{where}\:
\nu_j(a)= \frac{1}{\pi}
          \arctan \frac{\pm 2 \sqrt{a^2-1}}{2-a^2},\,\,\,
j=1,2.$$
Notice that for $a=\sqrt 2$, $\lambda_j=\pm 1$.

So, to every level curve it is associated a rotation number $\tau$
such that
$$ \nu (a)=\frac{1}{\pi}
            \arctan \frac{2 \sqrt{a^2-1}}{a^2-2}<\tau <1.$$
\begin{itemize}
  \item  $\tau=\frac{n}{p} \in {\cal Q}, (n,p)\equiv 1$, corresponds to
    periodic orbits of period $2p$, crossing $2n$ times the minor axis of
    $\Gamma$.
 \item $\tau \in {\cal R} \setminus {\cal Q}$ corresponds to dense orbits
  (dense on the level curve).
\end{itemize}
\end{itemize}

Notice that $\nu (a)$ is strictly decreasing, $\nu (a) \longrightarrow 1$,
as $a \longrightarrow 1^+$ and $\nu (a) \longrightarrow 0^+$, as $a
\longrightarrow +\infty$.
This indicates the first difference between the trajectories on ${\cal E}$ and
on ${\cal H}$: for each $a>1$, we have periodic orbits of every period
on ${\cal E}$, but not on ${\cal H}$. For instance, orbits of period 4
exist only for $a>\sqrt{2}$, or there is no period 8 orbits if
$a<\sqrt{4-2\sqrt 2}$. On figure~\ref{fig:bif}, we show the bifurcation
diagram for periodic orbits on ${\cal H}$.

\begin{figure}
\putfigurewithtex fig3.ps fig3.tex  8 16
\caption{Diagram of bifurcation of periodic orbits on ${\cal H}$}
 \label{fig:bif}
\end{figure}

Another difference lies on where the trajectory touches the conical caustic.
The tangency to the caustic on $\cal E$ always occurs inside $Q$ (ie, inside
the
original ellipse) and so, inside the segment of trajectory that joins two
reflections at $\Gamma$. However, on $\cal H$, the tangency may occur outside
$Q$, ie, you do not necessarily have the point of tangency between two
reflections at $\Gamma$. It may occurs even at infinity, if a segment of
trajectory is contained on an asymptote of the caustic hyperbola.

The results below give conditions to allow tangencies on $Q$ for trajectories
on ${\cal H}$.
\begin{lemma} \label{lem:D}
Suppose that a trajectory $\{T^n(\varphi_0,\theta_0)\}\in{\cal H}$ has two
consecutive reflections at the same half-ellipse. If $1<a<\sqrt 2$, then the
tangency between the segment joining those two reflections and the hyperbolic
caustic occurs inside the billiard.
\end{lemma}
proof:
Suppose that those two consecutive reflections are at $\varphi_0\in
[0,\pi/2]$ and $\varphi_1\in [\pi/2,\pi]$ (the other cases are symmetrical).
We will call $\overline{\varphi_0\varphi_1}$ the segment joining
$\Gamma(\varphi_0)$ to $\Gamma(\varphi_1)$.

For a fixed $\varphi_0$, if $\theta_0=\pi/2-\varphi_0$, then
$\overline{\varphi_0\varphi_1}$ touches the caustic at its vertex (and so,
inside $Q$).
Suppose that there is a $\theta$ such that $\varphi_1\in [\pi/2,\pi]$, and
$\overline{\varphi_0\varphi_1}$ touches the caustic outside $Q$. Then there is
a $\tilde\theta_0$ such that $\varphi_1\in [\pi/2,\pi]$, and
$\overline{\varphi_0\varphi_1}$ has the tangency exactly at the boundary.
But this means that either $\Gamma(\varphi_0)$ or $\Gamma(\varphi_1)$ is the
intersection of the hyperbolic caustic and the elliptical boundary and, since
they form an orthogonal family, either $\tilde\theta_0$ or $\theta_1 = \pi/2$.

But if $1<a<\sqrt 2$, $\varphi_0\in [k\pi/2,(k+1)\pi/2]$ and $\theta_0=\pi/2$,
then $\varphi_1\in [(k+2)\pi/2,(k+3)\pi/2]$ (\cite{kn:don}), and this implies
that $\varphi_0$ and $\varphi_1$ are not on the same half-ellipse.

\begin{lemma} \label{lem:DD}
Suppose that a trajectory $\{T^n(\varphi_0,\theta_0)\}\in{\cal H}$ has two
consecutive reflections, $\varphi_0$ and $\varphi_1$, at the same half-ellipse.
If $1<a<\sqrt{4-2\sqrt 2}$, then the tangency between each one of the segments
$\overline{\varphi_{-1}\varphi_0}$ and $\overline{\varphi_0\varphi_1}$
and the hyperbolic caustic occurs inside the billiard.
\end{lemma}
proof:
$\{T^n(\varphi_0,\theta_0)\}\in{\cal H}$. So,
$F(\varphi_0,\theta_0)=k_0<0$. Fix this hyperbolic caustic $k_0$. Call
$\theta(\varphi)$ the solution of $F(\varphi,\theta(\varphi))=k_0$. The piece
of trajectory $(\varphi_{-1},\theta(\varphi_{-1}))$, $(0,\theta(0))$,
$(\varphi_1,\theta(\varphi_1))$ is such that $0$ and $\varphi_1\in [0,\pi]$,
and $\varphi_{-1}$ and $0\in [\pi,2\pi]$. So, by lemma~\ref{lem:D}, and since
$\sqrt{4-2\sqrt 2}<\sqrt 2$, $\overline{0\varphi_1}$ and
$\overline{\varphi_{-1}0}$ touch the hyperbola inside $Q$.
Suppose that for a given $\varphi_0\in [0,\pi/2]$,
$F(\varphi_0,\theta_0)=k_0$, $\varphi_1\in [\pi/2,\pi]$
and $\overline{\varphi_{-1}\varphi_0}$
touches the hyperbola outside $Q$. Then, there exists a
$\tilde{\varphi_0}\in [0,\pi/2]$, with
$F(\tilde{\varphi_0},\theta(\tilde{\varphi_0}))=k_0$,
$\varphi_1\in [\pi/2,\pi]$, and $\overline{\varphi_{-1}\tilde{\varphi_0}}$
has the tangency exactly at the boundary. Since $\varphi_1\in [\pi/2,\pi]$ and
$a<\sqrt 2$, $\theta(\varphi_{-1})=\pi/2$.

Fix now the configuration $(\varphi_{-1},\pi/2)\mapsto
(\varphi_0,\theta_0)\mapsto (\varphi_1,\theta_1)$, $\varphi_{-1}\in
[3\pi/2,\pi]$, $\varphi_0\in [0,\pi/2]$, $\varphi_1\in [\pi/2,\pi]$.
There exists a $k_1$, $0>k_1>k_0$, such that $F(\varphi_0,\theta_0)=k_1$, and
the piece of trajectory is $(\varphi_{-1},\pi/2)\mapsto
(\varphi_0,\theta_0)\mapsto (\pi,\theta_1)$. Completing this trajectory, we
see that it has period 8 and crosses the minor axis 6 times,
or has $\tau =3/4$ that exists only if $a>\sqrt{4-2\sqrt 2}$.

\begin{corollary} \label{coro:DDD}
Suppose that a trajectory $\{T^n(\varphi_0,\theta_0)\}\in{\cal H}$ has two
consecutive reflections $\varphi_0$ and $\varphi_1$ at the same half-ellipse.
If $1<a<\sqrt{4-2\sqrt 2}$, then the tangency between each one of the segments
$\overline{\varphi_{-1}\varphi_0}$, $\overline{\varphi_0\varphi_1}$ and
$\overline{\varphi_1\varphi_2}$ and the hyperbolic caustic occurs inside the
billiard.
\end{corollary}
proof: Apply lemma~\ref{lem:DD} twice.

\subsection{ The Billiard on the Elliptical Stadium}
\label{estadio}

The boundary of the elliptical stadium is a curve $\Gamma$ constructed by
joining two half-ellipses, with major axes $a>1$ and minor axes $b=1$, by two
segments of equal length $2h$, as shown in figure~\ref{fig:estadio}.
\begin{figure}
\putfigurewithtex fig4.ps fig4.tex 2.5 16
 \caption{Elliptical stadium}
 \label{fig:estadio}
\end{figure}

$\Gamma$ is a Jordan curve, convex, but not strictly convex, globally $C^1$
but not $C^2$, and piecewise $C^{\infty}$. We parametrize $\Gamma$ by the angle
$\varphi$ on the half-ellipses and by the arclenght parameter on the straight
parts.

The billiard map associated to $\Gamma$
$$
\begin{array}{cccc}
S:&[0,2\pi+4h]\times (0,\pi)&\longrightarrow&[0,2\pi+4h]\times (0,\pi)\\
&(t_0,\theta_0)&\longmapsto&(t_1,\theta_1)
\end{array}$$
is piecewise $C^{\infty}$, but globally only $C^0$.

The derivative of $S$ is: if $t_0$ and $t_1$ belong to the elliptical part,
then $DS_{(t_0,\theta_0)}$ is equal to (\ref{eq:derivelipse});\\
if $t_0$ belongs to the elliptical part and $t_1$ to the straight part, then
\begin{equation}
DS_{(t_0,\theta_0)}=\frac{1}{\sin \theta_1}\left(
\begin{array}{cc}
k_0 l_{01}-\sin\theta_0 & l_{01}\\
-k_0\sin\theta_1 & -\sin\theta_1
\end{array}\right);
\label{eq:derivestadio,er}
\end{equation}
if $t_0$ belongs to the straight part and $t_1$ belongs to the elliptical part,
then
\begin{equation}
DS_{(t_0,\theta_0)}=\frac{1}{\sin \theta_1}\left(
\begin{array}{cc}
-\sin\theta_0 & l_{01}\\
-k_1\sin\theta_1 & k_1 l_{01}-\sin\theta_1
\end{array}\right);
\label{eq:derivestadio,re}
\end{equation}
and if $t_0$ and $t_1$ belong to the straight part, then
\begin{equation}
DS_{(t_0,\theta_0)}=\left(
\begin{array}{cc}
-1 & l_{01}/\sin\theta_1\\
 0 & -1
\end{array}\right),
\label{eq:derivestadio,rr}
\end{equation}
where $k_i$ is the curvature of the ellipse at $\varphi_i$ and $l_{01}$ is the
length of the segment $\overline{t_0t_1}$.

It is easy to see that for every $(t_0,\theta_0)$, with $t_0\in (0,\pi)$
(resp. $(\pi +2h, 2\pi +2h)$), there exists an $n=n(t_0,\theta_0)$ such that
$S^n(t_0,\theta_0)=(t_n,\theta_n)$, with $t_n \in (\pi +2h, 2\pi +2h)$
(resp. $(0,\pi)$), ie, if we begin in a half-ellipse, we will reach the other
one on a finite number of iterations.

This allows us to construct a restricted map:
$$
\begin{array}{cccc}
 R:&{\cal B}&\longrightarrow&{\cal B} \\
&(\varphi_0,\theta_0)&\longmapsto&(\varphi_1,\theta_1)
\end{array}
$$
where ${\cal B}=\{[0,\pi ]\bigcup \, [\pi+2h, 2\pi+2h ]\} \times (0,\pi)$,
$\varphi_1$ is the next reflection at a half-ellipse and $\theta_1$ is the
direction of motion immediately after the reflection at $\varphi_1$.

Notice that while the reflections remain at the same half-ellipse, $R$ is just
$T$, the diffeomorphism associated to the elliptical billiard. So, for those
reflections, the orbit will stay on an integral curve of the elliptical
billiard.
\begin{figure}
\putfigurewithtex fig5.ps fig5.tex 4 16
\caption{Integral curves and caustics in the half-ellipses}
 \label{fig:meia}
\end{figure}

\section{Main Tools}

\subsection{Cone-fields and consequences}
\label{cones}

Let $\Gamma =\{\gamma(t)=(x(t), y(t)), t\in[0, L)\}$ be a $C^k$-piecewise
Jordan
curve in $R^2$, $k\geq 1$, $Q$ the region of $R^2$ bounded by $\Gamma$ and
${\cal A}$ the annulus $[0, L)\times (0,\pi)$.

Let $T:{\cal A}\rightarrow{\cal A}$ be the local $C^{k-1}$ measure preserving
diffeomorphism associated to the billiard on $Q$.

\begin{definition}
For each $(t,\theta)\in{\cal A}$, the Lyapunov exponents of $T$ at
$(t,\theta)$ are given by:
\begin{eqnarray*}
\lambda_{+}(t,\theta)&=&\lim_{n\rightarrow \infty}\frac{1}{n} \ln
\parallel DT^n(t,\theta)\parallel \nonumber \\
\lambda_{-}(t,\theta)&=&-\lambda_{+}(t,\theta)
\end{eqnarray*}
\end{definition}

\begin{definition}
Given
two linearly independent vector fields $X_1(t,\theta)$ and
$X_2(t,\theta)$ in the tangent space $T_{(t,\theta)}{\cal A}\approx R^2$,
a {\bf cone} in $T_{(t,\theta)}{\cal A}$ is defined by
$$C(t,\theta)=\{rX_1(t,\theta)+sX_2(t,\theta), r. s\geq 0\}. $$
Its interior is
$$int(C(t,\theta))=\{rX_1(t,\theta)+sX_2(t,\theta), r.s>0 \:\hbox{or}\:\:
r=s=0\}. $$
A {\bf measurable cone-field} is a family of cones $\{C(t,\theta)\}\subset TA$
defined $\mu$-almost everywhere, and such that the vectors $X_1(t,\theta)$ and
$X_2(t,\theta)$ vary measurably with $(t,\theta)$.
\end{definition}

{\bf THEOREM} (Wojtkowski,\cite{kn:woj})
{\em
Let $C(t,\theta)$ be a
measurable cone field such that for almost every $(t,\theta)$,
\samepage
$$DT(C(t,\theta))\subset C(T(t,\theta))$$
\samepage
and for almost every $(t,\theta)$ there exists a $k(t,\theta)$ for which
\samepage
$$DT^{k(t,\theta)}(C(t,\theta))\subset intC(T^{k(t,\theta)}(t,\theta)). $$
\samepage
Then the Lyapunov exponents $\lambda_{+}(t,\theta)$ are positive for almost
every $(t,\theta)$.
}

(Such a cone-field is called eventually strictly invariant.)

Our goal is to construct a measurable cone-field, eventually strictly
invariant under the derivative $DS$ of the map associated to the billiard on
the elliptical stadium. But $S$ is not differentiable at
$t=0$, $\pi$, $\pi +2h$, $2\pi +2h$.

For $\alpha \in [0,2\pi+4h]$, let
\begin{eqnarray}
W_{\alpha}&=&\{(t_0,\theta_0)\in [0,2\pi+4h]\times (0,\pi)\:
\hbox{such that}\:
\exists n\in Z,\: S^n(t_0,\theta_0)=(\alpha,\theta_n)\} \nonumber \\
&=&\bigcup_{n\in Z}S^n(\{(\alpha,\theta),\theta\in (0,\pi)\}) \nonumber
\end{eqnarray}
ie, the images of a vertical line on $[0,2\pi+4h]\times (0,\pi)$.
$\mu (W_{\alpha})=0$ because $\mu (\{(\alpha,\theta),\theta \in (0,\pi)\})=0$
and $S$ is measure preserving. Let $W=W_0\bigcup W_{\pi}\bigcup
W_{\pi+2h}\bigcup W_{2\pi +2h}$. Then $\mu (W)=0$ and we will work on
$[0,2\pi+4h]\times (0,\pi)\setminus W$.

On the other hand, on the straight part of the elliptical stadium, $DS$ is
just a translation and an inversion. So, it is enough to construct
a cone-field strictly invariant under $DR$, the derivative of the restricted
map $R$, on ${\cal B}\setminus (W\bigcap {\cal B})$.

Finally, if a given trajectory, between $(\varphi_0,\theta_0)$ and
$(\varphi_1,\theta_1)=R(\varphi_0,\theta_0)$ hits $q$ times the straight part
of the elliptical stadium, then

\begin{equation}
DR_{(\varphi_0,\theta_0)}=\frac{(-1)^q}{R_1\sin \theta_1}\left(
\begin{array}{cc}
l_{01}-R_0\sin\theta_0 & l_{01} \\
l_{01}-R_0\sin\theta_0-R_1\sin\theta_1& l_{01}-R_1\sin\theta_1
\end{array}\right).
\end{equation}
So, the image under $DR$ of a cone $C(\varphi_0,\theta_0)$ can be inverted
when arriving at $(\varphi_1,\theta_1)$.
But, as
$-C=\{r(-X_1)+s(-X_2),\: rs\geq 0\}=\{rX_1+sX_2,\: rs\geq 0\}=C$, all that will
interest us is the slope of the boundaries of the cone.

\subsection{A little bit of optics}

We are going to follow Donnay ideas \cite{kn:don}, giving an optical
interpretation of the tangent space.

{}From now on, we are going to suppose that $\Gamma$ is strictly convex
(although the elliptical stadium is not strictly convex, we are going
to reduce our problem to the elliptical part of the boundary, looking only to
the restricted map $R$).

\begin{definition}
A {\bf pencil of rays} is a $C^1$ curve in the phase space ${\cal A}$,
given by $\eta:(-\epsilon,\epsilon)\longrightarrow {\cal A}$,
$\eta (\sigma)=(\varphi(\sigma),\theta(\sigma))$.
The {\bf base point} of the pencil is $\eta(0)=(\varphi_0,\theta_0)$, the {\bf
span} is $\eta '(0)=(\alpha,\beta)\in T_{(\varphi_0,\theta_0)}{\cal A}$ and the
{\bf slope} is $\beta/\alpha$.

To each $\sigma\in (-\epsilon,\epsilon)$, we can associate a straight line
$r(\sigma)$, passing by $\Gamma(\varphi(\sigma))$, with slope
$\tan(\varphi(\sigma)+\theta(\sigma))$. The {\bf core} of the pencil is the
straight line $r_0=r(0)$. Let $P(\sigma)=r(\sigma)\bigcap r_0,\: \sigma\neq
0$. A pencil $\eta$ {\bf focus} at $P_0=\lim_{\sigma \rightarrow 0}
P(\sigma)$, if the limit exists.
\end{definition}

If $P_0=(x,y)$ exists, it is given by the equations
\begin{eqnarray*}
\sec^2(\varphi_0+\theta_0)(1+\frac{\beta}{\alpha})(x-x_0)&
=&(\tan(\varphi_0+\theta_0)
 \cos\varphi_0-\sin\varphi_0)R_0 \nonumber \\
y-y_0&=&\tan(\varphi_0+\theta_0)(x-x_0),
\end{eqnarray*}
 where $(x_0, y_0)=\Gamma(\varphi_0),\: \theta_0=\theta(0)$ and $R_0$ is the
radius of curvature of $\Gamma$ at $\varphi_0$.

So $P_0$ depends only on the base point $(\varphi_0,\theta_0)$ and the slope
$\beta/\alpha$.
{}From the point of view of focusing properties, we may identify
pencils with the same base point and slope to either the linear pencil
$\eta(\sigma)=(\varphi_0+\alpha\sigma,\theta_0+\beta\sigma)$ or to the point
$(\varphi_0,\theta_0,\alpha,\beta)\in T{\cal A}$. We can also define and
calculate the focusing distance
$$f=dist(P_0, (x_0, y_0))=\frac{R_0\sin\theta_0}{1+\frac{\beta}{\alpha}}.$$

Notice that $R_0\sin\theta_0$ is the length of the intersection of the core of
the pencil with the half-osculating circle at $\varphi_0$ (the half-osculating
circle is the circle with radius $R_0/2$, tangent to $\Gamma$ at $\varphi_0$).

\subsection{Pencils of rays and beams of trajectories}

When we give a pencil of rays, we give a collection of initial conditions of
the billiard ball problem inside $\Gamma$. So, to a pencil of rays there
corresponds a beam of trajectories.

We may have to check if a given beam of trajectories
$\eta(\sigma)=(\varphi(\sigma),\theta(\sigma))
\approx (\varphi_0+\alpha\sigma,\theta_0+\beta\sigma)$ focuses forward and
backward, after and before the reflections at $\Gamma(\varphi(\sigma))$.
The forward focusing distance is given by
\begin{equation}
f_s=\frac{R_0\sin\theta_0}{1+\frac{\beta}{\alpha}}
\label{eq:fs}
\end{equation}
and the backward,
\begin{equation}
f_e=\frac{R_0\sin\theta_0}{1-\frac{\beta}{\alpha}}
\label{eq:fe}
\end{equation}
because going backward is the same as going forward for the pencil
$$\overline{\eta}(\sigma)=(\varphi(\sigma),\pi-\theta(\sigma))
\approx
(\varphi_0+\alpha\sigma,\pi-(\theta_0+\beta\sigma))
=(\varphi_0+\alpha\sigma,\pi-\theta_0-\beta\sigma)\ . $$

To compare the image of a pencil with another pencil, we will use Donnay's
Focusing Lemma \cite{kn:don}, rewritten as:

{\bf FOCUSING LEMMA}
{\em Let $T:{\cal A}\rightarrow
{\cal A}$ be the map associated to the billiard problem at $\Gamma$. Let
$(\varphi_0,\theta_0)$, $(\varphi_1,\theta_1)$ $\in{\cal A}$,
$(\varphi_1,\theta_1)=T(\varphi_0,\theta_0)$.
Let $\eta(\sigma)$ be a pencil with base point $(\varphi_0,\theta_0)$,
focusing forward at $P_0$ and $\xi(\sigma)$, a pencil with base point
$(\varphi_1,\theta_1)$, focusing backward at $P_1$. If
$P_0$ appears before $P_1$ when going from
$\Gamma(\varphi_0)$ to $\Gamma(\varphi_1)$ along the trajectory, then
$$\mbox{slope}(T\circ\eta(\sigma))>\mbox{slope}(\xi(\sigma)). $$}

\section{Construction of a cone-field on TB}

\subsection{Zones of the phase space B} \label{zonasB}

Take $R:{\cal B}\rightarrow {\cal B}$, the restricted map defined in
subsection~\ref{estadio}.
${\cal B}={\cal B}_1\bigcup{\cal B}_2$,
${\cal B}_1=[0,\pi] \times (0,\pi)$,
${\cal B}_2=[\pi+2h, 2\pi+2h]\times (0,\pi)
\approx [\pi, 2\pi]\times (0,\pi)$,
each one corresponding to the reflections at one half-ellipse.
Let $B={\cal B}\setminus (W\bigcap {\cal B})$, $W$
the measure zero set defined in subsection~\ref{cones}.
$B=B_1 \bigcup B_2$,
$B_i={\cal B}_i\setminus (W\bigcap {\cal B}_i),\:i=1, 2. $

Let  $B_1=U_1\bigcup \, M_1 $ where
\begin{eqnarray*}
U_1 &=&
\{ (\varphi_0,\theta_0)\in  B_1
\hbox{ such that } R(\varphi_0,\theta_0) \in B_2 \hbox{ and }
R^{-1}(\varphi_0,\theta_0) \in B_2\}\\
M_1 &=&
\{ (\varphi_0,\theta_0)\in
B_1 \: \hbox{such that}\:\hbox{either}\: R(\varphi_0,\theta_0) \in
B_1 \: \hbox{or} \:
R^{-1}(\varphi_0,\theta_0) \in B_1\}
\end{eqnarray*}

The boundary of $U_1$ is composed by two curves
$\varphi\mapsto\theta_{\pm}(\varphi)$ such that
$R(\varphi,\theta_{+}(\varphi))=(\pi,\theta_1)$ and
$R^{-1}(\varphi,\theta_{-}(\varphi))=(0,\theta_1)$
with equation:
$$\tan(\varphi+\theta_{\pm}(\varphi))=\frac{\tan\varphi}
{1\mp\sqrt{1+a^2\tan\varphi}}. $$
Notice that this boundary is contained on $W$.

By symmetry, we define $U_2$ and $M_2$, with $B_2=U_2 \bigcup\, M_2. $

\subsection{Caustic, Vertical and Horizontal Pencils}
\label{sec:caus}

If $(\varphi_0,\theta_0) \in M_1$ (resp. $M_2$), there exist $m, n \in Z,\, m
\leq 0\leq n, m\neq n$ such that $R^{m}(\varphi_0,\theta_0)$,
$R^{m+1}(\varphi_0,\theta_0)$, $\ldots$, $R^{n-1}(\varphi_0,\theta_0)$, $
R^{n}(\varphi_0,\theta_0) \in M_1$ (resp. $M_2$) \cite{kn:don}.
And $R^{i}=T^{i}, m\leq i\leq n$, because $M_1\subset B_1$
(resp. $M_2\subset B_2$). So, there is a piece
of the orbit of $(\varphi_0,\theta_0)$ that behaves exactly as it would
on the elliptical billiard. It stays on an integral curve $F=k$ of the
elliptical billiard and this piece of trajectory  has a caustic.

\begin{definition}
The caustic pencil at $(\varphi_0,\theta_0)$ is a curve
$\eta:(-\epsilon,\epsilon)\rightarrow B_1$,
$\eta(\sigma)=(\varphi(\sigma),\theta(\sigma))$
and $F(\eta(\sigma))=F(\varphi_0,\theta_0)$
or, by identification, a tangent vector at $(\varphi_0,\theta_0)$ to the
integral curve $F=F(\varphi_0,\theta_0)$.
\end{definition}

Clearly, a caustic pencil focuses forward and backward at the point of tangency
of its core with the conic caustic. Also, caustic pencils are transformed by
$T$ on caustic pencils.

\begin{lemma} \label{coro:tangelipse}
On the elliptical billiard, with $a<\sqrt 2$, choose a trajectory on ${\cal
E}$ such that $\varphi_0\in [\pi/2,\pi]$, $\varphi_0+\theta_0\geq \pi$ and
$\varphi_1\in[\pi, 2\pi]$. The point $P_0$ of tangency of
$\overline{\varphi_0\varphi_1}$ with the elliptical caustic is contained in
the interior of the half-osculating circle at $\varphi=\pi$.
\end{lemma}
proof: Let us fix one elliptical caustic. To each $(\varphi,\theta)$,
$\varphi\in [\pi/2,\pi]$, $\varphi+\theta\geq\pi$, we associate $P(\varphi)$,
the point of tangency with the caustic. Let $\tilde{\varphi_0}$ be such that
$P(\tilde{\varphi_0})$ is the upper vertex of the elliptical caustic.
Ordering the elliptical caustic via the counter clockwise orientation, we
have that $P(\tilde{\varphi_0})\leq P(\varphi)\leq P(\pi)$.
It is clear that $P(\tilde{\varphi_0})$ is contained in the interior of the
half-osculating circle at $\varphi=\pi$.
The integral curve associated to this caustic has a minimum at $\varphi=\pi$.
So the maximum of the forward focusing distance for the caustic pencil is
$f_{s}=\frac{R_{\pi}\sin\theta}{1+0}=a^2\sin\theta$.

\begin{definition}
The vertical pencil is defined by
$v(\sigma)=(\varphi_0,\theta_0+\sigma)$ or by the vector $(0, 1) \in
T_{(\varphi_0,\theta_0)}B$.
\end{definition}

The vertical pencil $v(\sigma)$ focuses forward and backward at
$\Gamma(\varphi_0)$, ie, at the boundary of the billiard, since $\alpha=0,\:
\beta=1$ and $f_{s, e}= \alpha\frac{R_0\sin\theta_0}{\alpha\pm \beta}=0$.

\begin{lemma} \label{pro:vert}
Let $a<\sqrt{2}$, $h>0$ and $(\varphi_1,\theta_1)=R(\varphi_0,\theta_0)$. Let
$v_0(\sigma)$ and $v_1(\sigma)$ be the vertical pencils at
$(\varphi_0,\theta_0)$ and $(\varphi_1,\theta_1)$, respectively. Let
$\eta(\sigma)=(\varphi_1+\alpha\sigma,\theta_1+\beta\sigma)$ be the
linearization of the pencil $R(v_0(\sigma))$. Then $\alpha\neq 0$ and
$\frac{\beta}{\alpha}>0$.
\end{lemma}
proof:
$(\alpha,\beta)=DR_{(\varphi_0,\theta_0)}(0, 1)=\frac{(-1)^n}{R_1\sin\theta_1}
(l_{01}, l_{01}-R_1\sin\theta_1). $
So, $\alpha\neq 0$ and $\frac{\beta}{\alpha}=
\frac{l_{01}-R_1\sin\theta_1}{l_{01}}$.

If $a<\sqrt 2$, the half-osculating circles of an ellipse with half axes $a$
and 1 are totally contained in the ellipse \cite{kn:don}.
So, $l_{01}>R_1\sin\theta_1$ and $\frac{\beta}{\alpha}>0$.

Notice that, geometrically, this lemma shows that $R$ deviates the vertical to
the right.

\begin{definition}
The horizontal pencil is defined by $h(\sigma)=(\varphi_0+\sigma,\theta_0)$
or by the vector $(1, 0)\in T_{(\varphi_0,\theta_0)}B$.
\end{definition}

The horizontal pencil focuses forward and backward at a distance
$R_0\sin\theta_0$, ie, at the intersection of the core with the half-osculating
circle at $\varphi_0$.

\subsection{ A cone field on TB} \label{coneB}

\begin{definition}
For each $(\varphi,\theta)\in B$, define the measurable cone field
$\{C(\varphi,\theta)\}\subset TB$ by
\begin{itemize}
\item
if $(\varphi,\theta)\in U_1$ or $U_2$, $C(\varphi,\theta)=\{rX_v+sX_h,\,
r. s\geq 0\}$,
\item if $(\varphi,\theta)\in M_1$ or $M_2$, $C(\varphi,\theta)=\{rX_v+sX_c,\,
r. s\geq 0\}$,
\end{itemize}
where $X_h =(1, 0)$ is the horizontal pencil at $(\varphi,\theta)$,
$X_v =(0, 1)$ is the vertical pencil at $(\varphi,\theta)$ and
$X_c =(1, -\frac{\partial F/\partial \varphi}{\partial F/\partial \theta})$,
where $F$ is the first integral of the elliptical billiard, is the caustic
pencil at $(\varphi,\theta)$.
\end{definition}

Remember that the elliptical stadium is defined by two parameters: $a$ and $h$.
The two half-ellipses have minor axes equal to 1 and major axes equal to
$a>1$, and the straight parts have length $2h$.

\begin{theorem} \label{teo:coneB}
If $1<a<\sqrt{4-2\sqrt{2}}$ and $h>2a^2\sqrt{a^2-1}$, then
$\{C(\varphi,\theta)\}$
is eventually strictly invariant under $DR$.
\end{theorem}

outline of the proof: If $(\varphi_0,\theta_0)\in M_i,\:i=1, 2$, the trajectory
associated to $(\varphi_0,\theta_0)$ has at least two consecutive reflections
at
the same half-ellipse, being $\varphi_0$ one of them. So, the orbit remains
on the integral curve $F=F(\varphi_0,\theta_0)$ and the trajectory has,
while staying on the same half-ellipse, a conic caustic.
Taking $$M(i, {\cal E})=\{(\varphi_0,\theta_0)\:\hbox{such that the caustic is
an ellipse}\}$$ and $$M(i, {\cal H})=\{(\varphi_0,\theta_0)\:\hbox{such that
the caustic is a hyperbola}\}, $$ we have $B_i=(M(i, {\cal E}))\bigcup
(M(i, {\cal H}))\bigcup U_i$, $i=1, 2$.

Our aim is to prove that for almost every $(\varphi_0,\theta_0)\in B$, with
$(\varphi_n,\theta_n)=R^n(\varphi_0,\theta_0)$,
we will have $DR^n(C(\varphi_0,\theta_0))\subset C(\varphi_n,\theta_n)$
and, for almost every $(\varphi_0,\theta_0)\in B$, we can find a
$k=k(\varphi_0,\theta_0)$ such that
$DR^k(C(\varphi_0,\theta_0))\subset intC(\varphi_k,\theta_k). $

The cones $C(\varphi,\theta)$ are bounded by the vertical pencil and by either
the horizontal or the caustic pencil. If $a<\sqrt 2$, the vertical pencil is
deviated to the right under $DR$ (lemma~\ref{pro:vert}). Since
$det(DR_{(\varphi_0,\theta_0)})>0$, all we have to do is to compare the
slope of the image of $X_c(\varphi_0,\theta_0)$ or $X_h(\varphi_0,\theta_0)$,
under $DR$, with the slope of $X_c(R(\varphi_0,\theta_0))$ or
$X_h(R(\varphi_0,\theta_0))$, depending on where $R(\varphi_0,\theta_0)$ is.
We have to look for the point of tangency with a conical caustic or for the
intersection with a half-osculating circle, forward or backward, in order to
apply Donnay's Focusing Lemma. All the possible situations are analysed in 7
propositions in the appendix, and we conclude that if
$1<a<\sqrt{4-2\sqrt{2}}$ and $h>2a^2\sqrt{a^2-1}$, the forward focusing point
appears before the backward one.

\section{ Main Results}

Take $S:{\cal A}\longrightarrow{\cal A}$ the map associated to the billiard on
the elliptical stadium.

${\cal A}=[0,2\pi+4h]\times (0,\pi)={\cal B}_1\bigcup {\cal F}_1\bigcup {\cal
B}_2\bigcup{\cal F}_2$, where: ${\cal B}_1=[0,\pi] \times (0,\pi)$,
${\cal F}_1=[\pi+h,\pi+2h] \times (0,\pi)$,
${\cal B}_2=[\pi+2h, 2\pi+2h]\times (0,\pi)$
and ${\cal F}_2=[2\pi+2h, 2\pi+4h] \times (0,\pi)$, the ${\cal B}_i$
corresponding to the reflections at the elliptical part and the ${\cal F}_i$
corresponding to the reflections at the straight part.

Let $A={\cal A}\setminus (W\bigcap {\cal A})$, where $W$ is
the measure zero set defined in subsection~\ref{cones}.
$A=B_1\bigcup F_1\bigcup B_2\bigcup F_2$,
$B_i={\cal B}_i\setminus (W\bigcap {\cal B}_i),\:i=1, 2$,
$F_i={\cal F}_i\setminus (W\bigcap {\cal F}_i),\:i=1, 2. $

As in subsection~\ref{zonasB}, let
$U_i
=\{ (t_0,\theta_0)\in  B_i \:\hbox{such that}\:S(t_0,\theta_0) \,\hbox{and} \,
S^{-1}(t_0,\theta_0) \not\in B_i\}$ and
$M_i=\{ (t_0,\theta_0)\in B_i \: \hbox{such that}\:\hbox{either}\:
S(t_0,\theta_0) \in B_i \: \hbox{or} \:S^{-1}(t_0,\theta_0) \in B_i\}$.

\begin{corollary} \label{coro:coneA}
For each $(t,\theta)\in A$, define the measurable cone field $\{C(t,\theta)\}
\subset TA$ by
\begin{itemize}
\item if $(t,\theta)\in M_1\bigcup M_2$, $C(t,\theta)=\{rX_v+sX_c,\,
r. s\geq 0\}$.
\item
if $(t,\theta)\in U_1\bigcup U_2$, $C(t,\theta)=\{rX_v+sX_h,\, r. s\geq 0\}$.
\item if $(t,\theta)\in F_1\bigcup F_2$, $C(t,\theta)=DS(C(S^{-1}(t,\theta)))$.
\end{itemize}
If $1<a<\sqrt{4-2\sqrt{2}}$ and $h>2a^2\sqrt{a^2-1}$, then $\{C(t,\theta)\}$
is eventually strictly invariant under $DS$.
\end{corollary}

Applying Wojtkowski's theorem, it follows that for those values of $a$ and $h$,
the Lyapunov exponents are non-vanishing $\mu$-almost everywhere. In
(\cite{kn:liv},\S 14. A), Liverani and Wojtkowski proved that the strictly
invariance of the cone-field implies the non-contraction property of its
vectors. And also that the other conditions on the dynamical behaviour of the
singularity curves (the vertical lines on the phase space, given by $W_{0}$,
$W_{\pi}$, $W_{\pi + 2h}$ and $W_{2 \pi +2h}$) have been satisfied. All those
additional conditions allow to prove that the system is not only chaotic, ie,
has non-vanishing Lyapunov exponents, but also ergodic and (see \cite{kn:mark},
sec. 4) is a K-system.

So, we have:
\begin{theorem}
\label{teo:main}
If $a<\sqrt{4-2\sqrt 2}$, then $\forall h>2a^2\sqrt{a^2-1}$, the map
associated to the elliptical stadium has non-vanishing Lyapunov exponents
$\mu$-almost everywhere, is ergodic and has the K-property.
\end{theorem}

\section{ Final Remarks}

Actually, the numerical simulations suggest much more about the dynamical
behaviour, as we present bellow.
\begin{figure}
 \vspace{5truecm}  \special{psfile=fig6.ps}
 \caption{$a=1.07$ and $h=0.01$, $h=0.1$, $h=0.815$}
\end{figure}

1) $2a^2\sqrt{a^2-1}$ does not seem to be an optimal lower bound for $h$. The
phase space seems "chaotic" for values of $h$ smaller than this bound, at
least for $a<\sqrt{4-2\sqrt 2}$.

The dependence on $\sqrt{a^2-1}$ seems reasonable, since it measures the focal
distance of the half ellipses. What seems to be unnecessarily big is the
factor $2a^2$ that is clearly an overestimation for $h$, used on the proof
of proposition~\ref{pro:caso3}, to assure that the focusing points of
orbits from $M(1, {\cal H})$ to $M(2, {\cal H})$ are separated by a reflection
at the straight part.

In any case, the lower bound curve $h(a)=2a^2\sqrt{a^2-1}$ is in agreement with
what we expected for $a\sim 1$. For the circular stadium billiard, or $a=1$,
$h>0$ implies ergodicity. This will not be the case for $a\neq 1$, because the
structure of the phase space of the elliptical billiard is not as degenerate as
that of the circular billiard. The vertical tangent at $a=1$ reflects this
fact.

2) $\forall a>1$ and $\forall h>0$, it seems that a neighbourhood of the
boundaries $\theta=0$ and $\theta=\pi$ is invariant. As shown in
proposition~\ref{pro:caso2}, the behaviour of the cone field there is analogous
to
the behaviour of the cone field constructed by Donnay \cite{kn:don} for the
circular stadium. And this, perhaps, invariant neighbourhood is a chaotic zone
of positive measure, even for small $h$'s (figure~\ref{fig:uma orbita}).

3)  $\forall a>1$ and $\forall h>0$, some initial conditions near $\theta=0$
(or $\theta=\pi$), after a finite number of iterations, come near $\theta=\pi$
(resp. $\theta=0$), showing that the trajectory changes orientation in
relation to the orientation of the boundary of the billiard. This indicates
that the invariant manifolds of the hyperbolic 2-periodic orbit
$\{ (\pi/2,\pi/2), (3\pi/2+2h,\pi/2)\}$ (which have corners) cross and that
there are homoclinic points and, then, non-integrability
(figure~\ref{fig:uma orbita}).

\begin{figure}
 \vspace{5truecm}  \special{psfile=fig7.ps}
 \caption{32,000 points of a unique orbit and the unstable manifold,
        for  a=1. 07 and h=0.15}
 \label{fig:uma orbita}
\end{figure}

4) Canale and Markarian \cite{kn:can} have proved the existence of symmetric
periodic orbits for the elliptical stadium and have pointed out that for
$a>1$, $h>0$, there exists a 4-periodic orbit, which is elliptic if
$h<\sqrt{a^2-1}$ and hyperbolic if $h>\sqrt{a^2-1}$.
The numerical simulations they caried out show that, while it is elliptic, it
is
encircled by elliptic islands of positive measure and, fixing $a<\sqrt 2$
and increasing $h$, those elliptic islands seem to be the last  to
disappear, among all those that can be observed, at least, for small values of
$h$. For $a>\sqrt 2$, the elliptic islands remain, even if $h$ is very big.

5) In the construction of the cone field, as in Donnay's proof \cite{kn:don},
it
becomes clear why the lower bound for $h$ must increase to infinity as $a$
approaches $\sqrt 2$. As $a$ approaches ${\sqrt 2}^{\: +}$, there is a
trajectory on $M\bigcap {\cal H}$ that has two reflections at the same
half-ellipse and cuts the minor axis of the half-ellipse closer and closer to
the center of the half-ellipse, ie, with a direction near the direction of
the asymptote of the hyperbolic caustic. This means that the focusing
distance of the caustic pencil goes to infinity and so, we need to put the
half-ellipses farther and farther apart to assure that the caustic pencil will
have enough space  to focus and to open again.

This also shows  that $h(a)=2a^2\sqrt{a^2-1}$ can not be the lower bound for
$a\sim \sqrt 2$. $h(a)$ must be determined by pieces. We believe
that the determination of $h(a)$ is closely related to the structure of
bifurcation of periodic orbits with hyperbolic caustic on the elliptical
billiard (figure~\ref{fig:bif}).

redpajama_set_name: RedPajamaArXiv


text:
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text:
Showing http://www.firstamendmentcenter.org/news.aspx?id=12219 captured on Apr 23, 2010
Suit seeking to open sealed court cases dismissed
WATERBURY, Conn. — A lawsuit seeking access to sealed state court cases has been dismissed by a federal judge, who ruled that Connecticut's top judicial officials don't have the authority to vacate Superior Court judges' sealing orders.
The Connecticut Law Tribune and the Hartford Courant filed suit earlier this year seeking information about the sealed cases.
At issue were about 10,000 sealed civil cases and an unknown number of Level 1 or "super-sealed" cases — files that are so secret their existence cannot officially be acknowledged.
Using a little-known quirk in court procedure, court officials reportedly super-sealed cases involving such defendants as saxophonist Clarence Clemons and Boston Celtics star Vin Baker.
The practice was first reported by The Connecticut Law Tribune last December. It drew criticism from editorial boards, legislators and many lawyers, in part because the rich, famous and powerful were among its beneficiaries.
U.S. District Judge Gerard L. Goettel ruled in Waterbury this week that the newspapers were essentially asking two top state court administrators to overturn sealing orders entered by the Superior Court judges who presided over the cases. Goettel said the defendants, Chief Court Administrator Joseph Pellegrino and state Supreme Court Chief Justice William Sullivan, lacked the authority to vacate the sealing orders.
The news outlets have argued that they were not seeking to have any orders overturned or cases unsealed. Their lawyers said they merely sought the disclosure of "docket sheets," a limited catalog of activity in each case. With those, they said, they could decide whether and how to contest the sealing of a particular case file.
The state had argued that the lawsuit amounted to an attempted "end run" around the judges. Lawyers for both newspapers said they would likely appeal.
"The judge is basically saying that neither of these judges has the authority to set aside the other judges' sealing orders," Hartford Courant lawyer Ralph G. Elliot said, "but the problem is there isn't a shred of evidence that any of the judges issued sealing orders in the first place because we don't have access to these files."
Connecticut Law Tribune lawyer Daniel J. Klau said the publications were trapped in a Catch-22.
"If you accept Judge Goettel's decision as correct," Klau said, "there's nothing that any newspaper can do, because you can't intervene in a case where you don't know the docket number. And you can't intervene in 10,000 cases where you know the docket number and no other information. So for all intents and purposes, there's no remedy, according to Judge Goettel."
Maureen Cox — a lawyer with Carmody & Torrance in Waterbury, the firm that represented the state in the case — said the judicial department was pleased that the ruling recognized "a balance between the public's right to access and the rights of parties to litigation to privacy."
The state's estimate of the number of Level 1 cases has fluctuated. By its count, 54 cases were on Level 1 status in May; the state has said most of those have since been moved to Level 2 status: The files remain sealed, but the party names are in the captions.
2nd Circuit revives newspapers' challenge to sealed court files
Three-judge panel rules that court docket sheets should generally be public, but sends case back to federal judge in Connecticut for further litigation. 06.09.04
Connecticut newspapers seek secret court 'X-Files'
Suit in federal court asks for files on cases including E Street Band member paternity suit, Boston Celtics player family-court matter. 08.21.03
Newspaper seeks court records in clergy sex-abuse case
Washington Post files motions in two Oklahoma counties for documents related to former priest now in prison for molesting teen-age boy. 06.01.02
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Congratulations Jan for your INPhINIT diploma!
Written by Irene Anton on July 4, 2018 . Posted in Awards and scholarships, General News, Group activities.
Jan Grzelak attended the award ceremony from La Caixa Fellows in Caixa Forum (Barcelona) and received a diploma for his excellence fellowship supported by La Caixa to perform his PhD at the N&N group on the topic "Mesoporous magnetic nanorods for theranostics". Two other ICMAB researchers have also been granted with by a PhD grant from La Caixa and also assisted the ceremony.
Written by Irene Anton on July 31, 2017 . Posted in Awards and scholarships, General News, Group activities.
Jan Grzelak defended his Master thesis at the University of Warsaw on the 27th of July 2017. His supervisors Anna Laromaine and Martí Gich came to Poland for the defence.
Jan was enrolled in a Master's in Nanostructure Engineering and the title of the thesis was 'Synthesis of supports for catalysis with palladium nanoparticles templated by bacterial cellulose'.
He got the highest mark (excellent) for the defence and the final mark from the whole studies was 4.88/5. He will receive a diploma with distinction! Next academic year Jan will be back to the N&N group to do his PhD with a grant from La Caixa.
We welcome Jan in the N&N group!
Written by Anna May on November 16, 2016 . Posted in News, Welcome and farewell.
Jan Grzelak obtained a Bachelor's degree in Nanostructure Engineering at the University of Warsaw (Poland). Then he started studying the Master program at the same university, also in Nanostructure Engineering. Now he is doing his second and last year of Master with an Erasmus program at the UAB, on Advanced Nanoscience and Nanotechnology.
Jan will join our N&N group at ICMAB, and will be supervised by Martí Gich and Anna Laromaine. He will be working on alumina tubes with palladium nanoparticles, using bacterial cellulose as a template for the synthesis.
We hope you enjoy your stay with us, Jan!


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It is indeed a time for outrage. The far right is exerting considerable political influence in most Western countries to the point where rhetoric and ideological approaches to welfare and society appear indistinguishable. Critical thinking seems to be absent in many school curricula: see for example creationism still taught in faith schools.


Sergio Sette Camara topped an eventful FIA Formula 2 Championship free practice session at Baku, Azerbaijan, the Carlin driver setting a 1:57.136 to beat Campos Vexatec Racing's Luca Ghiotto and MP Motorsport's Ralph Boschung to the top of the timing boards.
The Trident pair of Santino Ferrucci and Arjun Maini cut through the breezy conditions first to lead the field out onto the track, beginning a series of exploratory laps as the drivers got to grips with the Baku City Circuit. Ferrucci set the opening pace, followed by Nyck de Vries (PERTAMINA PREMA Theodore Racing) across the line.
A yellow flag was called out shortly after, as Artem Markelov of RUSSIAN TIME stranded his car while braking too deep into turn 4 to prematurely end his session and prompt a virtual safety car early on. Roy Nissany (Campos Vexatec Racing) produced a second yellow immediately after, going off at the second corner before rejoining.
During the interruptions, Sean Gelael and teammate De Vries headed the timing boards, before Sette Camara ended the first fifteen minutes on top of the pile. His Carlin stablemate Lando Norris assumed the lead of the session to bring the pace below the two minute mark, before Ghiotto popped up to wrest control of the times.
Embroiled in a fight for dominance, Sette Camara and Ghiotto traded fastest times amid further fleeting yellow flags, as the field explored the limits of the track throughout the session; at different stages, Alexander Albon (DAMS) and George Russell (ART Grand Prix) stopped on the exit of the pitlane, while De Vries and Tadasuke Makino (RUSSIAN TIME) went off at the first corner before returning to the circuit.
Sette Camara reclaimed first position after Boschung briefly set the quickest time, holding the advantage into the final stages of the session. A planned virtual safety car period in the final five minutes was hastened by Maximilian Günther's stopped BWT Arden, before a final dash for laptimes in the dying minutes of the session.
While Ghiotto grabbed his fastest lap of the session, he was unable to overcome Sette Camara and settled for second ahead of Boschung. Albon and Russell atoned for their earlier stoppages to finish fourth and fifth, as Norris claimed the sixth best time of the session. Makino was seventh, ahead of a wild final lap from Fuoco, who narrowly missed the barriers on the exit of turn 18. Maini and Nirei Fukuzumi (BWT Arden) completed the top 10.
Qualifying for Saturday's Feature Race will take place today at 3pm, meaning that the drivers will have to quickly tame the Baku City Circuit as they battle for pole position.


text:
FaceTime Bloomberg features a full-fledged avatar based on an iOS-like interface and a full-fledged interface
by Elizabeth Pascka Latim | Tue, 24 Jan 2023 at 08:15:11 | Tech
Apple's long-awaited mixed reality headset is a company's ambitious attempt to create a 3D version of the iPhone operating system that would stand out from competitors' products. The roughly $3,000 gadget, likely called Reality Pro, will demonstrate a new approach to virtual meetings and immersive video, stumbling in the Meta-dominated VR industry.
Please lend a hand.
Apple enters an uncertain marketplace with a premium product. The company's technology development team of around 1,000 people has spent more than seven years on the project, and the business expects it to become a new revenue source, especially considering that sales growth will stagnate this year.
But virtual reality has proved to be a challenge for the greatest technology kingpins. Although some analysts forecast the industry to surpass $100 billion by the end of the decade, headsets remain considered niche goods—and Meta has lost billions in its efforts.
Apple's aim is to introduce something new. Among those familiar with the product, eye and hand tracking will be the most compelling selling point. These include advanced video conferencing and FaceTime-based meeting rooms. The headset will also be able to display immersive video content, connect to an external display for an iPhone and iPad.
The headset will include several external cameras that can monitor the user's hands and sensors in the eye-reading device. This will enable a person to select an item on the screen, be it a button, a program icon, or a list. It's quite simple to operate a headset that relies on a handheld controller.
Apple's new gadget will combine virtual and augmented reality. With virtual reality, users can see the image and content while wearing glasses. Additionally, augmented reality superimposes digital content on the real world.
Apple will offer special lenses for eyeglasses that are located inside the case itself.
Apple expects the device to have the Digital Crown, a Apple Watch-like wheel that allows you to toggle between VR and AR.
In virtual reality, the FaceTime software will realistically show the user's face and entire body. These avatars will allow two people — each with an Apple headset — to communicate immersively. The technology differs from the virtual meeting rooms on the Meta headset, which creates a more cartoon-like user avatar.
Because of the enormous processing power required for this feature, the headset will only support realistic avatars during one-on-one video chats. However, additional users will appear as an icon or Memoji, Apple's customized emoji.
When making an iPhone FaceTime call, use Animoji. Photographer: David Paul Morris/Bloomberg
Apple intends to market the device this spring, although the schedule may change. This will enable the company to discuss the product at the annual Software Developers Conference in June before releasing it later this year, according to Bloomberg sources.
Apple is already working on a cheaper version, which may be released in late 2024 or early 2025, that will cost less than $1,500.
Apple expects to sell about a million copies of its new headset in its first year, compared to more than 200 million units of the iPhone, the company's biggest earner in Cupertino. However, the company does not intend to immediately make a profit on the initial version – even at a high price – indicating that it intends to continue developing the platform.
Apple has discussed with a dozen media partners whether or not it will be possible to use the new gadget in any way. Apple is also working on updating its own Apple TV+ content to work with the headset in 2020. NextVR, a streaming company, has developed sports content for the new device.
Apple intends to make the headset a video viewing experience that will make viewers feel like they're watching a film in another world, such as the desert or outer space. However, the headset's speakers may not be able to support it, and users may need to wear AirPods to enjoy the full surround sound experience.
The device will also have a variety of productivity options, including the ability to act as a Mac external monitor. With this feature, users will be able to see the computer display in virtual reality while still controlling it using a trackpad or mouse and a physical keyboard.
The headset's operating system, called xrOS, will have many of the same features as the iPhone and iPad but in a 3D environment. These include the Safari web browser, Photos, Messages, and the Calendar app, as well as apps for company services such as the App Store for installing third-party software, Apple TV+, and podcasts.
Apple users should already know the headset's main interface, which will be similar to that of the iPhone and iPad, with a main screen that may be changed. App icons will include weather, calendar appointments, and stock market numbers.
When users need to type in the air with their hands, they can use the Siri voice assistant or the iPhone, Mac, or iPad keyboard. However, you don't need an iPhone to work, according to the company.
Third-party games are expected to become a popular offering from third-party developers. In 2017, ARKit and other tools were released to assist in the creation of the iPhone's augmented reality experience. This provided a framework for programmers to write programs, games, and services.
According to trademark applications filed by Apple, the device will include a modified M2 chip found in previous Macs, as well as a dedicated processor for graphics and mixed reality.
Apple decided to make an external battery that is placed in the user's pocket and connected via a cable to solve a second issue: the device overheated when placed on the user's face. Another improvement is the inclusion of a cooling fan, similar to high-end Macs.
The Meta Quest headset can last up to two hours on a single battery, comparable to competitors' devices. The battery is large, measuring about the size of two iPhone 14 Pro Max stacked on top of each other, or about 15.2 cm high and 1.27 thick.
Meta's batteries are placed on the back of the headset in a manner that helps balance the device on the person's head. Some testers found the product to be bulky.
Other problems might arise due to the relatively short battery life — about 20 hours less than Apple's new MacBook Pro — and the fact that users will need to purchase many batteries and replace them often.
Apple recognized these difficulties within the company and attempted to establish realistic expectations for the product. According to the company, the device has an advantage of encouraging consumers to visit Apple retail stores to try out the product or acquire another item, such as an iPad or AirPods.
The initial headset will be made of aluminum and glass, and will resemble Apple's $550 AirPods Max. The product will have a curved screen on the front, speakers on the sides, and a special headband that allows the device to be placed on the head, removing it from the competition's usually plastic design.
Apple likes to make its headsets unique: the iPod had a wheel, the iPhone and iPad had a multi-touch approach, and the Apple Watch had the Digital Crown. Now, the company is hoping that its next product will follow the lead with its sci-fi interface.
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Open YEAR 'ROUND, the "Huh" Resort offers spectacular views of the 1800-acre Thunder Lake . Located just minutes from Sugar Camp and Three Lakes the cabin is winterized for year round fun! Located on the snowmobile trails the Huh Resort is great for fishing, ice fishing, boating and snowmobiling.