text:
EuroLex/E/Tranquillizer

tranquillizer

 * Original language: English
 * Original form and meaning: n. 'a drug used to diminish anxiety'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: < róandi(lyf); Romanian: < tranchilizant; Russian: < trankvilizator; Bulgarian: < trankvilizator;

,

title: EuroLex/E/Tranquillizer


text:
EuroLex/E/Transmitter

transmitter

 * Original language: English
 * Original form and meaning: n. 2 'a set of equipment used to generate and transmit electromagnetic waves carrying messages, signals, etc.', 3 'a neurotransmitter (biol.)'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: < sendir, senditœki (2); Romanian: < transmitator (2); Russian: transmitter, beg20c, meaning: 2, status: restricted use: technical; Bulgarian: transmiter, mid20c, meaning: 2, status: restricted use: technical; Albanian: < transmetues;

,

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: EuroLex/E/Transmitter


text:
EuroLex/E/Transporter

transporter

 * Original language: English
 * Original form and meaning: n. 2 'a vehicle used to transport other vehicles or large pieces of machinery etc. by road'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: transportor, meaning: the same as in English, see above, status: word comes from other source than English: French; Russian: transporter, mid20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English (technical) / word comes from other source than English: French; Albanian: transportier < mjet transportues;

,

title: EuroLex/E/Transporter


text:
EuroLex/E/Trapper

trapper

 * Original language: English
 * Original form and meaning: n. 1 'a person who traps wild animals, esp. to obtain furs'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: trapeur, 20c, via French, meaning: the same as in English, see above, status: known as a foreignism, used only with reference to British or American contexts / not (or no longer) recognized as English; Bulgarian: traper, beg20c, meaning: the same as in English, see above, status: known as a foreignism, used only with reference to British or American contexts; Albanian: traper, end19c, meaning: the same as in English, see above, status: restricted use: regional;

,

title: EuroLex/E/Trapper


text:
User talk:Annahoj

 Hello Annahoj, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Abd 14:17, 11 March 2011 (UTC)

Your pages
Please explain to me why you have created several hundreds of pages, some of which appear to be blank. TeleComNasSprVen 00:31, 12 March 2011 (UTC)

title: User talk:Annahoj


text:
EuroLex/E/Trash

trash

 * Original language: English
 * Original form and meaning: n. +6 'a style of music'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Russian: trésh, 1990s, meaning: the same as in English, see above, status: restricted use: youth, modern; Bulgarian: trash, end20c, meaning: the same as in English, see above, status: restricted use: youth, modern/technical; Greek: trash, 1990s, meaning: the same as in English, see above, status: restricted use: technical, youth, modern;

,

title: EuroLex/E/Trash


text:
EuroLex/E/Traveller's cheque

traveller's cheque

 * Original language: English
 * Original form and meaning: n. 'a cheque for a fixed amount that may be cashed on signature, usu. internationally'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: < ferðatékki; Romanian:  traveller's cheque [= English], 1990s, meaning: the same as in English, see above, status: restricted use: technical; Russian: trevler-chek, end20c, meaning: the same as in English, see above, status: restricted use: technical; Bulgarian: travel chek < putnicheski chek, end20c, meaning: the same as in English, see above, status: restricted use: technical, slang; Albanian: < cek udhetues (rend.); Greek: traveller's (tsek), 1990s, meaning: the same as in English, see above, status: restricted use: technical;

,

title: EuroLex/E/Traveller's cheque


text:
EuroLex/E/Trawl (net)

trawl (net)

 * Original language: English
 * Original form and meaning: n. 2 'a large wide-mouthed fishing-net dragged by a boat along the bottom of the sea'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: troll > botnvarpa (creat.), end19c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Romanian: traul [trawl], 1960s, meaning: the same as in English, see above, status: not (or no longer) recognized as English (technical); Russian: tral, mid20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English (technical); Bulgarian: tral, beg20c, via Russian, meaning: the same as in English, see above, status: not (or no longer) recognized as English (technical); Albanian: tral, mid20c, meaning: the same as in English, see above, status: restricted use: technical;

,

title: EuroLex/E/Trawl (net)


text:
EuroLex/E/Trawler

trawler

 * Original language: English
 * Original form and meaning: n. 1 'a boat used for trawling'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trollari < togari; botnvörpungur (creat.), end19c, meaning: the same as in English, see above, status: restricted use: obsolete; Romanian: trauler [trawlər], 1950s, meaning: the same as in English, see above, status: not (or no longer) recognized as English (technical); Russian: trauler, end19c, meaning: the same as in English, see above, status: fully accepted, but still marked as English (technical); Bulgarian: trauler, mid20c, via Russian, meaning: the same as in English, see above, status: not (or no longer) recognized as English (technical) / traler, beg20c, meaning: the same as in English, see above, status: not (or no longer) recognized as English;

,

title: EuroLex/E/Trawler


text:
EuroLex/E/Trekking

trekking

 * Original language: English
 * Original form and meaning: n. +1a 'a long-distance walk', +3 'a touristic expedition'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Russian: trekking, end20c, meaning: +3, status: restricted use: technical; Bulgarian: treking, end20c, meaning: +3, status: restricted use: technical, modern; Greek: trekking/treking = pezoporia, 1990s, meaning: +1a, status: restricted use: technical;

,

title: EuroLex/E/Trekking


text:
EuroLex/E/Trench (coat)

trench (coat)

 * Original language: English
 * Original form and meaning: n. 2 'a loose belted raincoat'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trenchkápa < rykfrakki [thrεns-], end20c, meaning: the same as in English, see above, status: restricted use + word comes from other source than English; Romanian: trenci/trencicot [trent∫/trent∫ kot/= English], 1930s, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Bulgarian: trenchkot, beg20c, meaning: the same as in English, see above, status: restricted use: archaic; Greek: trents kot, mid20c, meaning: the same as in English, see above, status: restricted use;

,

title: EuroLex/E/Trench (coat)


text:
EuroLex/E/Trend

trend

 * Original language: English
 * Original form and meaning: n. 1 'a general direction and tendency (esp. of events, fashion, or opinion etc.)'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trend [= English], 1980s, meaning: the same as in English, see above, status: restricted use: slang; Romanian: trend [= English], 1970s, meaning: the same as in English, see above, status: word is known mainly to bilinguals and felt to be English > restricted use: technical; Russian: trend, end20c, meaning: the same as in English, see above, status: restricted use: modern; Bulgarian: trend, end20c, meaning: the same as in English, see above, status: restricted use: technical (economy); Albanian: trend, mid20c, meaning: the same as in English, see above, status: restricted use: regional; Greek: < tasi (mean.);

,

title: EuroLex/E/Trend


text:
EuroLex/E/Trick

trick

 * Original language: English
 * Original form and meaning: n./cp. 1 'an action or scheme undertaken to fool, outwit, or deceive', 2 'an optical or other illusion', 3 'a special technique', 4b 'an unusual action learnt by an animal'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trikk [th rihk], mid20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English (colloquial) / trix [th rixs], 1930s, via Danish, meaning: 1,3, status: fully accepted, but still marked as English (colloquial); Romanian: < truc, the same as in English, see above, status: word comes from other source than English: French/Italian; Russian: < tryuc, the same as in English, see above, status: word comes from other source than English: French; Bulgarian: trik, beg20c, meaning: 1,2,3, status: fully accepted, but still marked as English; Albanian: trik, beg20c, meaning: the same as in English, see above, status: restricted use: regional; Greek: trik, beg20c, meaning: 1,2,3, status: fully accepted, but still marked as English;

,

title: EuroLex/E/Trick


text:
EuroLex/E/Trigger

trigger

 * Original language: English
 * Original form and meaning: n./cp. 2 'an event, occurrence, etc., that sets off a chain reaction' (electronic, medicine)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: trigger [trigər], 1980s, meaning: the same as in English, see above, status: restricted use: technical; Russian: trigger, mid20c, meaning: the same as in English, see above, status: restricted use: technical; Bulgarian: triger, mid20c, meaning: the same as in English, see above, status: restricted use: technical;

,

title: EuroLex/E/Trigger


text:
EuroLex/E/Trim 1

trim1

 * Original language: English
 * Original form and meaning: v. 1 'make neat or of the required size or form, esp. by cutting away irregular or unwanted parts; shear (a dog)', 4 'adjust the balance of (a ship or aircraft) by the arrangement of its cargo etc.', 5 'arrange (sails) to suit the wind', +8 'transport coal', +9 'train (a dog)', +10 (refl.) 'become/remain fit', +11 'undertake physical exercise in order to keep fit'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trimma [th rim:a], 1970s, meaning: 4, status: not (or no longer) recognized as English (technical); +11, status: not (or no longer) recognized as English;

,

title: EuroLex/E/Trim 1


text:
EuroLex/E/Trim 2

trim2

 * Original language: English
 * Original form and meaning: adj./n. '2 'in good order' (nautical), +6, 'the position of a ship in water', +7 'unused filming material', +8 'the act of becoming or keeping fit by physical exercise'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trimm [th rim:], 1970s, meaning: +8, status: not (or no longer) recognized as English;

,

title: EuroLex/E/Trim 2


text:
EuroLex/E/Trimmer

trimmer

 * Original language: English
 * Original form and meaning: n. 3 'an instrument for clipping etc.', 5 'a small capacitator (etc.) used to tune a radio set' (electronic)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: trimer [trimer], 1960s, meaning: 5, status: restricted use: technical; Russian: trimmer, mid20c, meaning: 5, status: restricted use: technical; Bulgarian: trimer, mid20c, meaning: 5, status: restricted use: technical; Albanian: trimer, end20c, meaning: 5, status: restricted use: technical;

,

title: EuroLex/E/Trimmer


text:
EuroLex/E/Trip

trip

 * Original language: English
 * Original form and meaning: n./cp. 1 'a journey or excursion made for pleasure', 4 'a drug-induced experience, esp. from taking LSD', +6 'enthusiasm'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: tripp < voyage [th rihp], 1970s, meaning: 4, status: fully accepted, but still marked as English; Bulgarian: trip, end20c, meaning: 4, status: restricted use: technical; Greek: trip, end20c, meaning: 4, status: restricted use: slang;

,

title: EuroLex/E/Trip


text:
EuroLex/E/Trolley

trolley

 * Original language: English
 * Original form and meaning: n./cp. 3 'a wheel attached to a pole etc. used for collecting current from an overhead electric wire to drive a vehicle', 4b 'a trolley bus'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Bulgarian: trolei, mid20c, via Russian, meaning: 4b, status: not (or no longer) recognized as English (colloquial); Greek: trolei, 20c, meaning: the same as in English, see above, status: not (or no longer) recognized as English;

,

title: EuroLex/E/Trolley


text:
EuroLex/E/Trolleybus

trolleybus

 * Original language: English
 * Original form and meaning: n. 1 'a bus powered by electricity obtained from an overhead cable by means of trolley-wheel'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: troleibuz [trolejbuz], mid20c, via French, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Russian: troleibus, mid20c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Bulgarian: troleibus/trolei, mid20c, via Russian, meaning: the same as in English, see above, status: fully accepted, but still marked as English / not (or no longer) recognized as English (colloquial); Albanian: trolejbus, mid20c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Greek: troleibas < trolei, 1950s, meaning: the same as in English, see above, status: restricted use: obsolete;

,

title: EuroLex/E/Trolleybus


text:
EuroLex/E/Truck

truck

 * Original language: English
 * Original form and meaning: n. 2 'a big lorry', +8 'a pivoted chassis of a railway or lorry', +9 'a forklift truck'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: trukkur [th ryhkyr], mid20c, meaning: 2, status: not (or no longer) recognized as English; Romanian: truc [truk], mid20c, via French/German, meaning: +8, status: fully accepted, but still marked as English (technical);

,

title: EuroLex/E/Truck


text:
EuroLex/E/Truck (system)

truck (system)

 * Original language: English (not an English word, but an Anglicism nevertheless)
 * Original form and meaning: n. 'the payment of workers in the form of goods or vouchers'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: truck [= English], 20c, meaning: the same as in English, see above, status: restricted use: technical;

,

title: EuroLex/E/Truck (system)


text:
EuroLex/E/Truism

truism

 * Original language: English
 * Original form and meaning: n. 'an obviously true or hackneyed statement'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: truism [truism], beg20c, via French, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Russian: tryuizm, beg20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English; Bulgarian: truizum, beg20c, meaning: the same as in English, see above, status: restricted use: technical; Albanian: truizem, beg20c, meaning: the same as in English, see above, status: restricted use: regional;

,

title: EuroLex/E/Truism


text:
EuroLex/E/Trust

trust

 * Original language: English
 * Original form and meaning: n. 8c 'an organized association of several companies for the purpose of reducing or defeating competition etc.'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: trust [trust], beg20c, via French, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Russian: trest, mid20c, meaning: the same as in English, see above, status: restricted use: technical; Bulgarian: trust, beg20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English (technical); Albanian: trust, mid20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English; Greek: trast, end20c, meaning: the same as in English, see above, status: restricted use: technical;

,

title: EuroLex/E/Trust


text:
EuroLex/E/Tube

tube

 * Original language: English
 * Original form and meaning: n. 2 'a soft metal or plastic cylinder sealed at one end and having a screw cap at the other, for holding a semi-liquid substance ready for use (toothpaste etc.)', 4a 'an underground railway system, esp. the one in London'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: túpa [th u:pa], beg20c, meaning: 2, status: word comes from other source than English: Danish; 4a, status: known as a foreignism, used only with reference to British or American contexts; Romanian: tub [tub], beg20c, meaning: the same as in English, see above, status: word comes from other source than English: French/Italian; Russian: tyubik, beg20c, meaning: 2, status: not (or no longer) recognized as English; Bulgarian: tuba, beg20c, via German, meaning: 2, status: word comes from other source than English: French; Albanian: tubet, beg20c, via Italian, meaning: the same as in English, see above, status: fully accepted, but still marked as English;

,

title: EuroLex/E/Tube


text:
EuroLex/E/Tunnel

tunnel

 * Original language: English
 * Original form and meaning: n. 1,2 'an artificial underground passage built by humans or animals'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: tunel [tunel], mid19c, via French, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Russian: tunnel, mid19c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Bulgarian: tunel, end19c, via German, meaning: the same as in English, see above, status: fully accepted, but still marked as English; Albanian: tunel, 20c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Greek: tunel, beg20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English;

,

title: EuroLex/E/Tunnel


text:
EuroLex/E/Tuner

tuner

 * Original language: English
 * Original form and meaning: n. 2 'a device for tuning a radio receiver', 3 'an electric device for tuning a guitar etc.', +4 'FM radio'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: tuner [= English], end20c, meaning: 3, status: restricted use: technical; Russian: tyuner, end20c, meaning: 2,+4, status: restricted use: technical/modern; Bulgarian: tuner, mid20c, meaning: 2, status: restricted use: technical; Greek: tuner/tiuner, end20c, meaning: +4, status: word is known mainly to bilinguals and felt to be English (technical);

,

title: EuroLex/E/Tuner


text:
EuroLex/E/Tupperware

Tupperware

 * Original language: English
 * Original form and meaning: n./cp. 'tupperware, a range of plastic containers for food' (original ™)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: ™ only; Greek: taper, mid20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English;

,

title: EuroLex/E/Tupperware


text:
EuroLex/E/Turf

turf

 * Original language: English
 * Original form and meaning: n. 1a 'a layer of grass etc. with earth and matted roots, as the surface of grassland', 3a 'horse-racing', 3b 'the racecourse for horse racing'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: turf [turf], end19c, via French, meaning: 3a,b, status: not (or no longer) recognized as English (technical); Russian: torf, via German, meaning: 1a, status: not (or no longer) recognized as English;

,

title: EuroLex/E/Turf


text:
EuroLex/E/Tweed

tweed

 * Original language: English
 * Original form and meaning: n. 1 'a rough-surfaced woollen cloth, usu. of mixed flecked colours, originally produced in Scotland'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: tweed/tvíd [th vi:t], end19c, meaning: the same as in English, see above, status: fully accepted, but still marked as English; Romanian: tweed [= English], 1960s, meaning: the same as in English, see above, status: fully accepted, but still marked as English (technical); Russian: tvid, mid20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English; Bulgarian: tuid, mid20c, meaning: the same as in English, see above, status: restricted use: technical; Greek: tuid, beg20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English (technical);

,

title: EuroLex/E/Tweed


text:
Talk:SPIR608 Political Simulations and Gaming/2011/BRIC

Card game being developed. Not like Red Guard! as no alliances more like simplified Modern SocietyLeutha 14:51, 11 March 2011 (UTC)


text:
EuroLex/E/Twinset

twinset

 * Original language: English
 * Original form and meaning: n. 'a matching combination of jumper and cardigan'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: -

,

title: EuroLex/E/Twinset


text:
EuroLex/E/Twist 1

twist1

 * Original language: English
 * Original form and meaning: v. 10 'dance the twist'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: tvista [th vista], 1960s, meaning: the same as in English, see above, status: fully accepted, but still marked as English (colloquial, archaic); Russian: tvistovat, mid20c, meaning: the same as in English, see above, status: restricted use: obsolete;

,

title: EuroLex/E/Twist 1


text:
EuroLex/E/Twist 2

twist2

 * Original language: English
 * Original form and meaning: n. 12 'the act of twisting (gymnastics)', 15 'a dance with a twisting movement of the body'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: tvist [thvist], 1960s, meaning: 15, status: fully accepted, but still marked as English (colloquial, archaic); Romanian: twist [= English], 1960s, meaning: the same as in English, see above, status: restricted use: obsolete; Russian: tvist, mid20c, meaning: the same as in English, see above, status: restricted use: obsolete; Bulgarian: tsvist, 20c, meaning: 12, status: restricted use: technical; tuist, 1960s, meaning: 15, status: restricted use: obsolete; Albanian: tuist, mid20c, meaning: 15, status: fully accepted, but still marked as English; Greek: tuist, 1960s, meaning: the same as in English, see above, status: restricted use: obsolete;

,

title: EuroLex/E/Twist 2


text:
EuroLex/E/Two-step

two-step

 * Original language: English
 * Original form and meaning: n. 'a round dance in march or polka time'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Icelandic: tústep [thustεhp], beg/mid20c, via Danish, meaning: the same as in English, see above, status: restricted use: obsolete; Russian: tustep, beg20c, meaning: the same as in English, see above, status: restricted use: obsolete;

,

title: EuroLex/E/Two-step


text:
EuroLex/E/Typhoon

typhoon

 * Original language: English
 * Original form and meaning: n. 'a tropical storm in the western Pacific'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: * DEA = Dictionary of European Anglicisms by Manfred Görlach (2001), Oxford: OUP.; ** CODEE = The Concise Oxford Dictionary of English Etymology by T.F. Hoad (1986), Oxford: Clarendon.

 * Information on Other Languages: Romanian: taifun [tajfun], end19c, via German, meaning: the same as in English, see above, status: not (or no longer) recognized as English; tifon [tifon], end19c, via French, meaning: the same as in English, see above, status: not (or no longer) recognized as English (archaic); tifone [tifone], mid19c, via Italian, meaning: the same as in English, see above, status: not (or no longer) recognized as English (archaic); Russian: taifun, 19c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Bulgarian: taifun, beg20c, meaning: the same as in English, see above, status: fully accepted, but still marked as English; Albanian: taifun, end19c, meaning: the same as in English, see above, status: not (or no longer) recognized as English; Greek: tyfonas, 20c, meaning: the same as in English, see above, status: not (or no longer) recognized as English;

,

title: EuroLex/E/Typhoon


text:
User:Fem.tm7/HW5

Problem Description

 * 1) Solve the general 1-D model for an elastic bar with data set 1b (G1DM1.0/D1b) using the weight function with polynomial, Fourier, and exponential basis functions until convergence of $$\displaystyle u^h (0.5) \rightarrow O(10^{-6}) $$.
 * 2) Plot u, uh, and convergence (error versus n).

Given
For G1DM1.0/D1 and D1b, the governing partial differential equation (PDE) and given conditions are,

 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial}{\partial x} \left[ a_2(x) \frac{\partial u}{\partial x} \right ] + f(x,t) = \bar{m} \frac{\partial ^s u}{\partial t ^s} (x,t)$$ (5.1.1)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

 * where
 * $$\displaystyle \Omega = [{0,1}]$$

 * $$\displaystyle a_2 (x) = 2 + 3x$$

 * $$\displaystyle f(x) = 5x$$

 * $$\displaystyle \bar{m} \frac{\partial ^s u}{\partial t ^s} (x,t) = 0 $$ (static case)

The natural and essential boundary conditions are respectively defined below for data set D1b.

 * {| style="width:100%" border="0"

$$ n(1) a_2(1) \frac{\partial u}{\partial x}(1)=12 \qquad \Rightarrow \qquad \frac{\partial u}{\partial x}(1)= \frac{12}{5}$$ (5.1.2)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }
 * {| style="width:100%" border="0"

$$\displaystyle u(0)=4$$ (5.1.3)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Weak Solution
The weak form must first be constructed such that the weight function, $$\displaystyle w(x)$$, vanishes on the essential boundary (x=0). First, the governing equation (5.1.1) and natural boundary condition (5.1.2) are multiplied by the weight function and integrated over the domain. The restriction on the weight function is that $$\displaystyle w(0)=0$$.

 * {| style="width:100%" border="0"

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) + f(x) \right] dx = 0 \qquad and \qquad w(x)\left( \frac{du}{dx} - \frac{12}{5} \right )_{x=1} = 0 $$     (5.1.4)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

The first equation can be expanded as

 * {| style="width:100%" border="0"

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) \right] dx + \int_0^1 w(x) f(x) dx= 0 \qquad \forall w. $$ (5.1.5)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Integrating by parts yields

 * {| style="width:100%" border="0"

$$ \int_0^1 w(x) \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) dx = \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=1} - \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=0} - \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx. $$     (5.1.6)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Recall that weight function must vanish at the essential boundary. Then by combining equations 5.1.5 and 5.1.6, the following expression is determined.

 * {| style="width:100%" border="0"

$$ \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx = \left( w(x) a_2(x) \frac{du}{dx} \right)_{x=1} + \int_0^1 w(x)f(x)dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.1.7)
 * style="width:10%; |
 * style="width:10%; |
 *
 * }

Then substitute the natural boundary condition (equation 5.1.2) and the given conditions for $$\displaystyle a_2(x)$$ and $$\displaystyle f(x)$$ into the previous equation. The result is the weak form for G1DM1.0/D1b.

 * {| style="width:100%" border="0"

$$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = 12w(1) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.1.8)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Exact Solution
The governing partial differential equation (PDE):
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right] + 5x = 0{\text{ }}\forall {\text{x}} \in \left] {0,1} \right[

$$     (5.1.9) The natural and essential B.C.:
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac\left( 1 \right) = \frac{5}

$$     (5.1.10)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

u\left( 0 \right) = 4

$$     (5.1.11) Then
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int {\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right]} dx = \int { - 5xdx}

$$     (5.1.12)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( {2 + 3x} \right)\frac + {C_1} = - \frac{5}{2}{x^2}{\text{ where }}{C_1}{\text{ is a constant}}

$$     (5.1.13) By the natural B.C.
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{C_1} = - \frac{2}

$$     (5.1.14) Then
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac = \frac

$$     (5.1.15)
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

u = - \frac{5}{x^2} + \frac{5}{9}x + \frac\log \left( {3x + 2} \right) + {C_2}{\text{ where }}{C_2}{\text{ is a constant}}

$$     (5.1.16) By the essential B.C.
 *
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{C_2} = 4 - \frac\log \left( 2 \right)

$$     (5.1.17) Then
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

u = - \frac{5}{x^2} + \frac{5}{9}x + \frac\log \left( {3x + 2} \right) + 4 - \frac\log \left( 2 \right)

$$     (5.1.18)
 *

 * }

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 0 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:

 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$

 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.

\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$

\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

\displaystyle

\tilde f({b_1}) = {b_1}(1)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x)}(5x)dx}=41/3

$$

\displaystyle

\tilde f({b_2}) = {b_2}(1)h + \int\limits_0^1 {{b_2}(5x)dx}=1*12 + \int\limits_0^1  {{(x)^2}(5x)dx}=53/4

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {41/3} \\ {53/4} \\  {.} \\  {.} \\  {.} \end{array}} \right]

$$

\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$

\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$

\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$

\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x)dx}=4

$$

\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x)*(2+3x)*1dx}=4

$$

\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x)*(2+3x)*2(x)dx}=17/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&4 &.&. \\ 4&17/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$

\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$

\displaystyle

K_{FF}d_F=F_F $$

\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.5840*10^{-6})$$. The following plots were generated:

For Fourier Basis
The family of Fourier basis functions is of the form
 * $$\displaystyle F_F = \left\{ 1, cos(x-1), sin(x), cos(2x)-1, sin(2x),..., cos\left( \frac{nx}{2}\right)-1, sin\left(\frac{nx}{2}\right) \right\}$$.

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

 * {| style="width:100%" border="0"

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$
 * style="width:10%; |
 * style="width:10%; |
 * }

 * where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then equation 5.1.11 for the unconstrained case can be rewritten as

 * {| style="width:100%" border="0"

$$ \mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$
 * style="width:10%; |
 * style="width:10%; |
 * }

This agrees with Vu-Quoc's notes. Now the matrices must be determined. From the derivation of the polynomial basis case above, it holds that $$\displaystyle d_0 = 4 $$, and $$\displaystyle \mathbf{d}_F = \mathbf{K}_{FF}^{-1} \mathbf{F}_F$$.

 * For n=6,

 * $$\displaystyle

\mathbf{F}_F = \tilde f(b_i) = {b_i}(1)h + \int_0^1 {{b_i}(5x)dx} $$

 * {| style="width:100%" border="0"

$$ \mathbf{F}_F = \begin{bmatrix} \tilde f(b_1) \\ \tilde f(b_2) \\ \tilde f(b_3) \\ \tilde f(b_4) \\ \tilde f(b_5) \\ \tilde f(b_6) \end{bmatrix} \qquad \Rightarrow \qquad \mathbf{F}_F = \begin{bmatrix} -6.108 \\ 11.603 \\ -18.991 \\ -13.089 \\ -27.25 \\ 3.422 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The stiffness matrix is determined,

 * $$\displaystyle

\mathbf{K}_{FF} = \tilde k(b_i,b_j) = \int_0^1 {b_i}'(2+3x){b_j}'dx $$

 * {| style="width:100%" border="0"

$$ \mathbf{K}_{FF} = \begin{bmatrix} 1.144 & -1.361 & 3.256 & -0.541 & 3.701 & 2.321 \\ -1.361 & 2.356 & -4.187 & 2.4 & -5.846 & -0.067 \\ 3.256 & -4.187 & 9.512 & -2.35 & 11.619 & 5.465 \\ -0.541 & 2.4 & -2.35 & 4.488 & -5.473 & 5.35 \\ 3.701 & -5.846 & 11.619 & -5.473 & 16.813 & 2.205 \\ 2.321 & -0.067 & 5.465 & 5.35 & 2.205 & 14.687 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

Now, the matrix $$\displaystyle \mathbf{d}_F$$ can be found.

 * {| style="width:100%" border="0"

$$ \mathbf{d}_{F} = \mathbf{K}_{FF}^{-1} \mathbf{F}_{F} \qquad \Rightarrow \qquad \mathbf{d}_{F} = \begin{bmatrix} 17.629 \\ 23.592 \\ 0.383 \\ -9.94 \\ -0.952 \\ 1.177 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

 * OR (simplified)

 * {| style="width:100%" border="0"

$$\displaystyle u^h(x)= 17.629*cos(x) \ + \ 23.592*sin(x) \ + \ 0.383*cos(2x) \ - \ 9.94*sin(2x) \ - \ 0.952*cos(3x) \ + \ 1.177*sin(3x) \ - \ 13.06 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * }

For Exponential Basis
The exponential basis begins to converge as n increases to around n=3, but then quickly begins to diverge. At n=10, the solution begins to diverge at the right end, and when n=20, it is clear that the solution is not stable.

Problem Description
Rework the previous problem for data set 1, i.e. G1DM1.0/D1.
 * Note: The difference between data set 1 and data set 1b is where the boundary conditions are applied. Specifically, the essential boundary condition for D1 is applied at x=1 while the natural boundary condition is applied at x=0. The opposite is true for data set 1b.

The essential and natural boundary conditions are respectively defined below for data set D1.
 * $$ \Gamma_g = \left \{ 1 \right \} \quad \text{and} \quad \Gamma_h = \left \{ 0 \right \}$$

 * $$ n(0) a_2(0) \frac{\partial u}{\partial x}(0)=-12 \qquad \Rightarrow \qquad \frac{\partial u}{\partial x}(0)= 6$$
 * $$\displaystyle u(1)=4$$

Weak Solution
The weak form must first be constructed such that the weight function, $$\displaystyle w(x)$$, vanishes on the essential boundary (x=0). First, the governing equation (5.1.1) and natural boundary condition (5.1.2) are multiplied by the weight function and integrated over the domain. The restriction on the weight function is that $$\displaystyle w(1)=0$$.

 * {| style="width:100%" border="0"

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) + f(x) \right] dx = 0 \qquad and \qquad w(x)\left( \frac{du}{dx} - \frac{12}{5} \right )_{x=0} = 0 $$     (5.2.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

The first equation can be expanded as

 * {| style="width:100%" border="0"

$$ \int_0^1 w(x) \left[ \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) \right] dx + \int_0^1 w(x) f(x) dx= 0 \qquad \forall w. $$ (5.2.2)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Integrating by parts yields

 * {| style="width:100%" border="0"

$$ \int_0^1 w(x) \frac{d}{dx} \left(a_2(x) \frac{du}{dx} \right) dx = \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=1} - \left( w(x)a_2(x) \frac{du}{dx} \right )_{x=0} - \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx. $$     (5.2.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Recall that weight function must vanish at the essential boundary. Then by combining equations 5.1.5 and 5.1.6, the following expression is determined.

 * {| style="width:100%" border="0"

$$ \int_0^1 \frac{dw}{dx}a_2(x) \frac{du}{dx} dx =- \left( w(x) a_2(x) \frac{du}{dx} \right)_{x=0} + \int_0^1 w(x)f(x)dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.2.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Then substitute the natural boundary condition (equation 5.1.2) and the given conditions for $$\displaystyle a_2(x)$$ and $$\displaystyle f(x)$$ into the previous equation. The result is the weak form for G1DM1.0/D1.

 * {| style="width:100%" border="0"

$$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = -12w(0) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.2.5)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Exact Solution
The governing partial differential equation (PDE):
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right] + 5x = 0{\text{ }}\forall {\text{x}} \in \left] {0,1} \right[

$$     (5.2.6) The natural and essential B.C.:
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac\left( 0 \right) = 6

$$     (5.2.7)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

u\left( 1 \right) = 4

$$     (5.2.8) Then
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int {\frac{\partial }\left[ {\left( {2 + 3x} \right)\frac} \right]} dx = \int { - 5xdx}

$$     (5.2.9)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left( {2 + 3x} \right)\frac + {C_1} = - \frac{5}{2}{x^2}{\text{ where }}{C_1}{\text{ is a constant}}

$$     (5.2.10) By the natural B.C.
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{C_1} = - 12

$$     (5.2.11) Then
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac = \frac

$$     (5.2.12)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

u = - \frac{5}{x^2} + \frac{5}{9}x - \frac\log \left( {3x + 2} \right) + \frac{5}{9} + {C_2}{\text{ where }}{C_2}{\text{ is a constant}}

$$     (5.2.13) By the essential B.C.
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{C_2} = 4 + \frac\log \left( 5 \right)-\frac{10}{9}+\frac{5}{12]}

$$     (5.2.14) Then
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

u = - \frac{5}{x^2} + \frac{5}{9}x - \frac\log \left( {3x + 2} \right) + \frac{5}{12} + \frac\log \left( 5 \right)+4-\frac{5}{9}

$$     (5.2.15)
 *

 * }

For Polynomial Basis
The corresponding family $${{\bar{F}}_{P}}=\left\{ {{{\bar{b}}}_{j}} \right\}$$ of polynomial basis function satisfying the Constraint Breaking Solution (CBS) for $$\Gamma_g = \left \{ 1 \right \}$$ is:

$${{F}_{P}}=\left\{ {{\left( x-1 \right)}^{j}},\text{ }j=0,1,2,\ldots \right\}$$

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$

where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then the discrete weak form is simplified to:

\mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$

The first constant, $$\displaystyle d_0$$, is determined from the formulation of the CBS. It is known that:

 * $$\displaystyle u^h(\beta) = d_0 b_0(\beta)=g$$

 * $$\displaystyle d_0= \frac{g}{b_0(\beta)}=4/1=4$$

Now the matrices used to determine the rest of the coefficients are determined as follows. Note that the number of terms needed in the approximate solution, $$\displaystyle n$$, is determined by the convergence interval given. The first few terms of each matrix computation is shown. A Matlab script was written to iterate these computation for increasing values of $$\displaystyle n$$ until the desired accuracy was reached.

\displaystyle

{F_F} = \left\{ {{F_i},i = 1,2,...,n} \right\} = \left\{ {\tilde f\left( \right),i = 1,2,...,n} \right\}

$$

\displaystyle

\tilde f({b_i}) = {b_i}(\alpha )h + \int\limits_\alpha ^\beta {{b_i}fdx}

$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 2*6=12

$$

\displaystyle

\tilde f({b_1}) = {b_1}(0)h + \int\limits_0^1 {{b_1}(5x)dx}=1*12+\int\limits_0^1  {{(x-1)}(5x)dx}=-77/6

$$

\displaystyle

\tilde f({b_2}) = {b_2}(0)h + \int\limits_0^1 {{b_2}(5x)dx}=-1*12 + \int\limits_0^1  {{(x-1)^2}(5x)dx}=149/12

$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {-77/6} \\ {149/12} \\  {.} \\  {.} \\  {.} \end{array}} \right]

$$

\displaystyle

{K_{FE}} = \left\{ {{K_{0j}};j = 1,...,n} \right\} = \left\{ {\tilde k\left( {{b_0},{b_j}} \right),j = 1,...,n} \right\}

$$

\displaystyle

\tilde k({b_i},{b_j}) = \int\limits_0^1 {b{'_i}{a_2}b{'_j}dx}

$$

\displaystyle

\tilde k({b_0},{b_1}) = \int\limits_0^1 {b{'_0}{a_2}b{'_1}dx} = \int\limits_0^1 {b{'_0}(2+3x)b{'_1}dx}  = 0 = \tilde k({b_0},{b_2}) = \tilde k({b_0},{b_3}) = 0

$$

\displaystyle

\therefore {K_{EF}} = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

And：

\displaystyle

\because {K_{FE}} = {K_{EF}}^T = \left[ {\begin{array}{ccccccccccccccc} 0 \\ 0 \\  . \\ .\\ . \end{array}} \right]

$$

\displaystyle

{K_{FF}} = \left[ {{K_{ij}};i,j = 1,2,...} \right] = \tilde k({b_i},{b_j})

$$

\displaystyle

\tilde k({b_1},{b_1}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_1}dx} = \int\limits_0^1 {1*(2+3x)*1dx}=7/2

$$

\displaystyle

\tilde k({b_1},{b_2}) = \int\limits_0^1 {b{'_1}{2+3x}b{'_2}dx} = \int\limits_0^1 {1*(2+3x)*2(x-1)dx}=-3

$$

\displaystyle

\tilde k({b_2},{b_1}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_1}dx} = \int\limits_0^1 {2(x-1)*(2+3x)*1dx}=-3

$$

\displaystyle

\tilde k({b_2},{b_2}) = \int\limits_0^1 {b{'_2}{2+3x}b{'_2}dx} = \int\limits_0^1 {2(x-1)*(2+3x)*2(x-1)dx}=11/3

$$

\displaystyle \therefore {K_{FF}} = \begin{bmatrix} 7/2&-3 &.&. \\ -3&11/3  & .&.\\ . &.  & .&.\\ . &.  & .&. \end{bmatrix} $$

The matrices are then solved for the coefficients:

\displaystyle

Kd = F $$

\displaystyle

F = {F_F} - K_{FE}g = {F_F} $$

\displaystyle

K_{FF}d_F=F_F $$

\displaystyle

d_F=K_{FF}^{-1} F_F $$

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

As stated above a Matlab script was written to iterate $$\displaystyle n$$ until the desired accuracy was reached. At $$\displaystyle n=6$$ the error was $$\displaystyle  O(3.4091*10^{-6})$$. The following plots were generated:

For Fourier Basis
The family of Fourier basis functions is of the form
 * $$\displaystyle F_F = \left\{ 1, cos(x-1)-1, sin(x-1), cos[2(x-1)]-1, sin[2(x-1)],..., cos[\frac{n}{2}(x-1)]-1, sin[\frac{n}{2}(x-1) \right\}$$.

Now that the family of basis functions are determined, we recall that the discrete weak form is defined by

 * {| style="width:100%" border="0"

$$ \mathbf{\tilde{d}} \cdot \mathbf{b}( \beta ) = g \qquad \text{and} \qquad \mathbf{\tilde{c}} \cdot \left[ \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} + \mathbf{\tilde{K}} \mathbf{\tilde{d}} - \mathbf{\tilde{F}} \right] = 0 $$
 * style="width:10%; |
 * style="width:10%; |
 * }

 * where the expression $$\displaystyle \mathbf{\tilde{M}} \mathbf{\tilde{d}}^{(s)} = 0 $$ for the static case. Then equation 5.1.11 for the unconstrained case can be rewritten as

 * {| style="width:100%" border="0"

$$ \mathbf{K} \mathbf{d} = \mathbf{F} \qquad \Rightarrow \qquad \mathbf{K} \mathbf{d} = \mathbf{F}_F - \mathbf{K}_{FE} g. $$
 * style="width:10%; |
 * style="width:10%; |
 * }

This agrees with Vu-Quoc's notes. Now the matrices must be determined. From the derivation of the polynomial basis case above, it holds that $$\displaystyle d_0 = 4 $$, and $$\displaystyle \mathbf{d}_F = \mathbf{K}_{FF}^{-1} \mathbf{F}_F$$.

 * For n=6,

 * $$\displaystyle

\mathbf{F}_F = \tilde f(b_i) = {b_i}(0)h + \int_0^1 {{b_i}(5x)dx} $$

 * {| style="width:100%" border="0"

$$ \mathbf{F}_F = \begin{bmatrix} \tilde f(b_1) \\ \tilde f(b_2) \\ \tilde f(b_3) \\ \tilde f(b_4) \\ \tilde f(b_5) \\ \tilde f(b_6) \end{bmatrix} \qquad \Rightarrow \qquad \mathbf{F}_F = \begin{bmatrix} -5.718 \\ -10.89 \\ -17.724 \\ -12.275 \\ -25.274 \\ -3.282 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The stiffness matrix is determined,

 * $$\displaystyle

\mathbf{K}_{FF} = \tilde k(b_i,b_j) = \int_0^1 {b_i}'(2+3x){b_j}'dx $$

 * {| style="width:100%" border="0"

$$ \mathbf{K}_{FF} = \begin{bmatrix} 0.764 & 1.117 & 2.305 & 0.885	& 3.06	& -0.773 \\ 1.117 & 2.736 & 3.675 & 3.819 & 5.929 & 2.855 \\ 2.305 & 3.675 & 7.137 & 3.437 & 10.079 & -1.18 \\ 0.885 & 3.819 & 3.437 & 6.863 & 7.19 & 8.293 \\ 3.06 & 5.929 & 10.079 & 7.19 & 16.154 & 2.415 \\ -0.773 & 2.855 & -1.18 & 8.293 & 2.415 & 15.346 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

Now, the matrix $$\displaystyle \mathbf{d}_F$$ can be found.

 * {| style="width:100%" border="0"

$$ \mathbf{d}_{F} = \mathbf{K}_{FF}^{-1} \mathbf{F}_{F} \qquad \Rightarrow \qquad \mathbf{d}_{F} = \begin{bmatrix} -26.379 \\ 1.214 \\ 8.796 \\ -3.542 \\ -1.074 \\ 0.991 \end{bmatrix} $$
 * style="width:10%; |
 * style="width:10%; |
 * }

The approximate solution, $$\displaystyle u^h(x)$$ is defined as,

 * $$\displaystyle u^h(x)= d_0b_0+\sum_{j=1}^{n}d_jb_j(x) \qquad \text{where} \quad d_0=4 \ \text{and} \ b_0=1$$.

 * OR (simplified)

 * {| style="width:100%" border="0"

$$\displaystyle u^h(x)= - 26.379*cos(x-1) \ + \ 1.214*sin(x-1) \ + \ 8.796*cos(2x-2) \ - \ 3.542*sin(2x-2)] \ - \ 1.074*cos(3x-3) \ + \ 0.991*sin(3x-3) \ + \ 22.657 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * }

For Exponential Basis
The exponential basis function is given by:
 * {| style="width:100%" border="0"

$$ \displaystyle {u^{h}}(x)={d_{j}}\left (e^{j(x-1)}-1 \right ) $$       (5.2.exp.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle \frac{{u^{h}}(x)}{dx} ={d_{j}}\left (e^{j(x-1)} \right ) $$       (5.2.exp.2) The general form of the stiffness matrix becomes:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle K=\int_{0}^{1}i*e^{i(x-1)}*(2+3x)*j*e^{j(x-1)}dx $$       (5.2.exp.3) The general form of the force matrix becomes:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle F=12(e^{i(0-1)}-1)+\int_{0}^{1} 5x*i(e^{i(x-1)}-1)dx $$       (5.2.exp.4) Using MatLab, the K and F matrix with N=10 is:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

 * {| style="width:100%" border="0"

$$ \displaystyle K=\begin{bmatrix} 1.716166179 &	2.633475288	& 3.170329089	& 3.512453499	&3.74690156	&3.917139098	&4.046398014	&4.147965319	&4.229930538	&4.297494435\\ 2.633475288	&4.227105451	&5.268680249	&5.995042496	&6.528565163	&6.93668231	&7.258939308	&7.519876512	&7.735489984	&7.916648746\\ 3.170329089	&5.268680249	&6.744422808	&7.834278196	&8.670852887	&9.332921967	&9.869837922	&10.31398664	&10.68747581	&11.00590793\\ 3.512453499	&5.995042496	&7.834278196	&9.248909746	&10.3699133	&11.27981477	&12.03298442	&12.66663799	&13.20708952	&13.67346515\\ 3.74690156	&6.528565163	&8.670852887	&10.3699133	&11.74980705	&12.89248331	&13.85413531	&14.67454391	&15.38264829	&15.99999816\\ 3.917139098	&6.93668231	&9.332921967	&11.27981477	&12.89248331	&14.24996774	&15.4082711	&16.40815817	&17.27999802	&18.04687424\\ 4.046398014	&7.258939308	&9.869837922	&12.03298442	&13.85413531	&15.4082711	&16.7499948	&17.91999794	&18.94921795	&19.86159138\\ 4.147965319	&7.519876512	&10.31398664	&12.66663799	&14.67454391	&16.40815817	&17.91999794	&19.24999918	&20.42906542	&21.48148136\\ 4.229930538	&7.735489984	&10.68747581	&13.20708952	&15.38264829	&17.27999802	&18.94921795	&20.42906542	&21.74999987	&22.93628804 \end{bmatrix} $$       (5.2.exp.5)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle F=\begin{bmatrix} -194.4238\\ 556.4552\\ -794.5674\\ 388.5998\\ 355.4029\\ -550.4317\\ 202.6984\\ 13.1421\\ 77.9578\\ -68.7552\\ -267.2934\\ 294.8667\\ 224.2897\\ -420.1257 150.8844 \end{bmatrix} $$       (5.2.exp.5) Solving for the general solution with n=6 gives an error of 2.83E-06 at u(.5) and a general solution of:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle u^{h}=-167.2599*e^{1*(x-1)}	487.6591e^{2*(x-1)}	-808.7098e^{3*(x-1)}	770.2175e^{4*(x-1)}	-393.7902e^{5*(x-1)}	83.8821e^{6*(x-1)}+32.0012 $$       (5.2.exp.5) The boundary condition u(1)=4 is kept and the exact solution is very close to the trial solution:
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

1. Explain how the basis functions satisfy the constraint breaking solution (CBS).
The following are the basis functions:

 * {| style="width:100%" border="0"

$$\displaystyle \sum_{i=1}^m \prod_{k=1,i\ne k}^m \frac{(x-x_k)}{(x_i-x_k)} {m=2,4,6,8....}$$ (5.3.1)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

3. Solve the following equation using these basis functions and matlab.

 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial}{\partial x} \left[ a_2(x) \frac{\partial u}{\partial x} \right ] + f(x,t) = \bar{m} \frac{\partial ^s u}{\partial t ^s} (x,t)$$ (5.3.2)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

with boundary conditions:

 * {| style="width:100%" border="0"

$$ n(1) a_2(1) \frac{\partial u}{\partial x}(1)=12 \qquad \Rightarrow \qquad \frac{\partial u}{\partial x}(1)= \frac{12}{5}$$ (5.3.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle u(0)=4$$ (5.3.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

4. Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

 * {| style="width:100%" border="0"

$$\displaystyle u_m^h$$ vs $$\displaystyle u$$ (5.3.5)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

and,

 * {| style="width:100%" border="0"

$$\displaystyle u_m^h(0.5)-u(0.5)$$ vs $$\displaystyle m$$ (5.3.6)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Part 1
These basis functions have the property where at each node there is a single function equal to one and the rest are zero, as seen in figure on part 2. So, all that needs to be done is set the coefficient of the single nonzero basis function to zero to meet the constraint breaking solution.

Part 2
Using the following Matlab code for m=2:

With the Following Plot:

For M=4:

With Plot:

For m=6:

With Plot:

For m=8:

With Plot for m=8:

To get a better picture of the properties of these 8 basis functions, the following is a plot where the basis functions are only evaluated at the nodes:

Part 3
Starting with the weak form of equation 5.3.3 (see derivation of weak form in problem 5.1):
 * {| style="width:100%" border="0"

$$ \int_0^1 \frac{dw}{dx}(2+3x) \frac{du}{dx} dx = 12w(1) + \int_0^1 w(x)5x dx \qquad \forall w \quad with \quad w(0)=0. $$     (5.1.8)
 *

 * }

Now approximate the solution by:
 * {| style="width:100%" border="0"

$$\displaystyle u\simeq u^h = \sum_{i=1}^md_i\prod_{k=1,k\neq i}^m \frac{(x-x_k)}{(x_i-x_k)}=\sum_{i=1}^md_i*b_i(x) $$     (5.3.7)
 *

 * }

and approximate the weighting function by:

 * {| style="width:100%" border="0"

$$\displaystyle w\simeq w^h = \sum_{i=1}^mc_i\prod_{k=1,k\neq i}^m \frac{(x-x_k)}{(x_i-x_k)}\sum_{i=1}^mc_i*b_i(x) $$     (5.3.8)
 *

 * }

Where bi(x) has been substituted in for the basis functions for ease of finding the solution.

Substituting Equation 5.3.7 and 5.3.8 into 5.1.8 and using linearity of the integral operator gives after some rearrangement:

 * {| style="width:100%" border="0"

$$\displaystyle \sum_{i=1}^m[\int_0^1(\frac{d}{dx}c_ib_i(x))(2+3x)(\frac{d}{dx}d_ib_i(x)-12c_ib_i(1)-\int_0^1c_ib_i(x)5xdx] $$     (5.3.9)
 *

 * }

Factoring out the constant ci:

 * {| style="width:100%" border="0"

$$\displaystyle \sum_{i=1}^mc_i[\int_0^1\frac{d}{dx}b_i(x)(2+3x)(\frac{d}{dx}d_ib_i(x)-12b_i(1)-\int_0^1b_i(x)5xdx] $$     (5.3.10)
 *

 * }

To solve for the unknown coefficients simultaneously, we will move to write 5.3.10 in tensor form.

 * {| style="width:100%" border="0"

$$\displaystyle \vec C=\begin{bmatrix} \vec C_E \\ \vec C_F\end{bmatrix}=\begin{bmatrix} 0 \\ \vec C_F\end{bmatrix} $$     (5.3.11)
 *

 * }

The zero in equation comes from satisfying the constraint breaking solution where the coefficient at the natural boundary condition for the only nonzero basis function coefficient on the weighting function is zero. Hence the subscript E in equation 5.3.11 for essential. Looking at the coefficients for uh:

 * {| style="width:100%" border="0"

$$\displaystyle \vec d=\begin{bmatrix} \vec d_E \\ \vec d_F\end{bmatrix}=\begin{bmatrix} g \\ \vec d_F\end{bmatrix} $$     (5.3.12)
 *

 * }

Likewise here, g is the value at the essential boundary condition. Since there is only one nonzero basis function here, the coefficient (since the basis function is one) must be equal to the essential boundary condition value. The subscript F can be thought of as free and will be determined.

Now defining the following

 * {| style="width:100%" border="0"

$$\displaystyle \tilde k(b_i,b_j)=\int_0^1(\frac{d}{dx}b_i)(2+3x)(\frac{d}{dx}b_j)dx $$     (5.3.13)
 *

 * }

This shows how the far left integral term of equation 5.3.10 can be written in components of a matrix. Additionally defining:

 * {| style="width:100%" border="0"

$$\displaystyle K_{EE}=[\tilde k(b_1,b_1)] $$     (5.3.14)
 *

 * }

Where subscript one signifies the first basis functions.
 * {| style="width:100%" border="0"

$$\displaystyle K_{EF}=[\tilde k(b_1,b_j)] $$     (5.3.15)
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle K_{FE}=[\tilde k(b_i,b_1)] $$     (5.3.16)
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle K_{FF}=[\tilde k(b_{remaining i},b_{remaining j})] $$     (5.3.17)
 *

 * }

Giving:

 * {| style="width:100%" border="0"

$$\displaystyle \mathbf K=\begin{bmatrix} \mathbf K_{EE} \mathbf K_{EF} \\ \mathbf K_{FE} \mathbf K_{FF}\end{bmatrix} $$     (5.3.18)
 *

 * }

Lastly, combining the constants into:
 * {| style="width:100%" border="0"

$$\displaystyle \mathbf F = [\tilde f_i] $$     (5.3.19)
 *

 * }

Where,

 * {| style="width:100%" border="0"

$$\displaystyle \tilde f_i=12b_i(1)+\int_0^1b_i(x)5xdx $$     (5.3.20)
 *

 * }

Putting it all together:

 * {| style="width:100%" border="0"

$$\displaystyle \vec C \cdot \mathbf K\mathbf d=\begin{bmatrix}0\\\vec C_F\end{bmatrix}\cdot \Big (\begin{bmatrix} \mathbf K_{EE} \mathbf K_{EF} \\ \mathbf K_{FE} \mathbf K_{FF}\end{bmatrix}\begin{bmatrix}\mathbf g\\\vec d_F\end{bmatrix}-\begin{bmatrix}\mathbf F_E\\ \mathbf F_F\end{bmatrix} \Big )=0 $$     (5.3.21) Now, our weighting function is arbitrary so what is in parenthesis corresponding to CF must equal zero. Writing this out explicitly for four basis functions gives the following:
 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle \begin{bmatrix}K_{11}K_{12}K_{13}K_{14}\\K_{21}K_{22}K_{23}K_{24}\\K_{31}K_{32}K_{33}K_{34}\\K_{41}K_{42}K_{43}K_{44}\end{bmatrix}\begin{bmatrix}d_1\\d_2\\d_3\\d_4\end{bmatrix}=\begin{bmatrix}f_1+r_1\\f_2\\f_3\\f_4\end{bmatrix} $$     (5.3.22)
 *

 * }

In order to solve four our unknowns d2-d4, in terms of our knowns, we rearrange to get the following:

 * {| style="width:100%" border="0"

$$\displaystyle \begin{bmatrix}K_{22}K_{23}K_{24}\\K_{32}K_{33}K_{34}\\K_{42}K_{43}K_{44}\end{bmatrix}\begin{bmatrix}d_2\\d_3\\d_4\end{bmatrix}=\begin{bmatrix}f_2-K_{21}d_1\\f_3-K_{31}d_1\\f_4-K_{41}d_1\end{bmatrix} $$     (5.3.23)
 *

 * }

This solution method was applied in Matlab to give the following:

m=2

m=4

m=6

m=8

Part 4
The following are the resulting plots from Part 3: m=2

m=4

m=6

m=8

This Matlab code was used to plot uh-u vs m:

Giving:

Problem Description
Solve G1D1M1.D/D1d, for$$\frac{d}((2 + 3x)\frac) + 5x = 0$$, using weak form with LIBF (m=4, 6, 8, ..), until convergence of $${u^h}(0.5)to{\rm O}({10^{ - 6}})$$.
 * 1) Explain how LIBF are used as CBS
 * 2) Plot all LIBF used
 * 3) Use Matlab quad or WolfAlpha to integrate
 * 4) Plot $${u^h}$$ vs $$u$$, $${u^h}(0.5) - u(0.5)$$ vs m

Solution
(1)
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{L_{i,m}}({x_j}) = {\delta _{i,m}} = \left\{ {\begin{array}{ccccccccccccccc} {1,forj = i} \\ {0,forj \ne i} \end{array}} \right.

$$ So for series of LIBF, at the essential boundary condition, there will be only one term does not equal to zero, the other terms equal to zero. So the CBS has been satisfied.
 *
(Eq 5.4.1)
 * }

(2) Basis function:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{L_{i,m}}(x) = \prod\limits_{\begin{array}{ccccccccccccccc} {k = 1} \\ {k \ne i} \end{array}}^m {\frac}

$$     (Eq 5.4.2)
 *

 * }

Figs of the basis functions

m=4:

m=6:

m=8：

(3) For weak form
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

w(x) = \sum\limits_{i = 1}^m {{c_i} \cdot {L_{i,m}}}

$$     (Eq 5.4.3)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

u(x) = \sum\limits_{i = 1}^m {{b_i} \cdot {L_{i,m}}}

$$     (Eq 5.4.4) To boundary condition
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

w(1) = 0

$$     (Eq 5.4.5)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{\alpha _m} = 0

$$     (Eq 5.4.6)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

u(1) = 4

$$     (Eq 5.4.7)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{b_m} = 4

$$     (Eq 5.4.8) The weak form is:
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\left. {w(2 + 3x)\frac} \right|_0^1 - \int\limits_0^1 {\frac(2 + 3x)\fracdx} + \int\limits_0^1 {w \cdot 5x}  = 0

$$     (Eq 5.4.9)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

- w(0)(2 + 3*0)\frac(0) - \int\limits_0^1 {\frac(2 + 3x)\fracdx} + \int\limits_0^1 {w \cdot 5x}  = 0

$$     (Eq 5.4.10)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

- w(0)*12 + \int\limits_0^1 {w \cdot 5x} = \int\limits_0^1 {\frac(2 + 3x)\fracdx}

$$     (Eq 5.4.11)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow \sum\limits_i {{C_i}\left[ {\sum\limits_j  - {F_i}} \right]}  = 0

$$     (Eq 5.4.12) Where
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{K_{ij}} = \int\limits_0^1 {{L_{i,m}}'(2 + 3x){L_{j,m}}'}dx

$$     (Eq 5.4.13)
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{F_i} = \int\limits_0^1 {{L_{i,m}} \cdot 5x}dx + w(0)*12

$$     (Eq 5.4.14) In this static case
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

Kd = Fwhere{F_F} - {K_{FE}}g

$$     (Eq 5.4.15) Solve for d
 *

 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{u^h} = \sum\limits_{i = 1}^m  \cdot {L_{i.m}}

$$     (Eq 5.4.16)
 *

 * }

Plot $${u^h}$$ vs $$u$$

m=4:

m=6:

m=8：

Calculate for error

Error vs m :

Problem 5.5.1 and 5.5.2
'''5.5.1 For the disc problem find node information : nodes numbers and their coordinates, Element numbers and their respective node numbers. 5.5.2 For the same disc problem generate 3 meshes with triangle elements by changing number of elements. ''' Eml5526.s11.team7.Srilalithkumar 16:03, 22 March 2011 (UTC)

Problem 5.5.3: CalculiX CrunchiX (CCX) for Dummies
Adopted from CCX User's Manual and Mechanical Hack's website. This walk through provides the basic overview of CCX. Experimentation and actual development of input files will be required on one's on.

Note: Previous installation of Calculix in Windows additionally installed CCX, so one can immediately begin to follow this example.

CCX is computational tool based on the information the user puts into the input file. CCX is constructed from FORTRAN77 and C programing languages, with a definite emphasis on FORTRAN. In relation to Calculix, CCX is considered the main subroutine of the total program Calculix.

CCX uses what it calls an "input deck" which specifies what will be the input. This input deck is made up of two parts, the model and step portions. Inside of these portions keywords are used. These keywords can be thought of as commands. These keywords belong either to the model or the step portion, and the model portion MUST be used before the step portion. To get a visual, look at the following figure built off of Mechanical Hack's website:

The *WORD are the keywords, or commands. The *HEADING command allows for data to be entered on the following line for the benefit of the user. The next set of commands describes the material that will be analyzed with CCX. This data can either be entered through *INCLUDE, which copies the data from the specified file (who must be in the designated path for CCX) to the input, or it can be entered explicitly as seen in the following code from the test folder on the Calculix website:

A search of the users manual provides additional methods of inputting data if needed. The remaining parts of this model section adds the necessary attributes relating to the physics of the model (boundary condition, physical properties, etc). This can be seen with the *BOUNDARY command, which allows for the proper formatting of the data being read from the input FIXED_123.BOU to be read as boundary conditions. Next the *MATERIAL is used to allow for assignment through NAME=EL of a name to the model being created. The keyword *ELASTIC allows for assignment of the elastic properties of the material. There are a multitude of additional keywords to completely specify the material of the problem. These are in the user's manual as well, and an example of them can be seen here:

Now that the material has been specified the keyword step can be used. Step allows for a quasi-static analysis. For example, in the picture a dynamic load is applied, so a step in time can be used to find the final results. End step signifies the end of the step, and there is a limit to the number of steps the program is capable of running.

With the basics of the input file created, the results are found by clicking on Tools, Calculate:

This splits the screen in two and shows the command portion of Calculix. If the input file was created correctly, the following "Job finished, Exit code:0" will be seen as well as the output.

Additionally output can be specified to a file for post processing with CGX (as seen in the figure above with the *NODE FILE and *EL FILE commands. If this happens, the data is sent to a .frb file which is post processed simply by Tools->Post Process, or Shift-F10.

Problem 5.6: Quadratic Lagrangian Element Basis Functions (QLEBF)
Plot three figures, similar to those in lecture 30, for the quadratic Lagrangian element basis functions. Consider the case where the distance between nodes is not equal.

Solution

Comments:
 * The first three quadratic basis functions are defined as:
 * $$\displaystyle b_1(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}$$,
 * $$\displaystyle b_2(x)=\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)}$$,
 * $$\displaystyle b_3(x)=\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}$$.
 * where $$\displaystyle x_N$$ represents the x value of the global node 'N'. Subsequent basis functions (i.e. $$\displaystyle b_4, \ b_5, \ b_6, \ \text{etc.}$$) follow the same form.

 * Each element is not discretized uniformly. In other words each node is not equidistant from adjacent nodes. The x-position for each node was arbitrarily chosen and are given as:
 * $$\displaystyle \qquad x_1=1, \qquad x_2=1.75, \qquad x_3=3, \qquad x_4=3.5, \qquad \text{and} \qquad x_5=5$$.

 * The sum of all basis functions at any point along 'x' is one.
 * The first three plots of Figure 5.6 illustrate the global representation. The last plot of Figure 5.6 shows the case for the individual element $$\displaystyle \omega^{(1)}$$ and is defined between global nodes 1 and 3. A second element, say $$\displaystyle \omega^{(2)}$$, would then be defined between global nodes 3 and 5.
 * While each element contains three nodes, each node is assigned a "local" node number from 1 to 3. Specifically for the case of element 1 in Figure 5.6, the global node numbers are the same as the local node number. However, for element 2, global node 3 corresponds to local node 1, global node 4 corresponds to local node 2, and global node 5 corresponds to local node 3.
 * Since it can easily be confusing when correlating global node numbers and local node numbers, the relationship between the two can be generalized using the equation below.

 * $$\displaystyle I = i+(3-1)(e-1) \qquad \text{(for the case of 3 nodes per element)}$$
 * where $$\displaystyle I$$ is the global node number, $$\displaystyle i$$ is the local node number, and $$\displaystyle e$$ is the element number.

Problem Description
Solve the general 1-D model for an elastic bar with data set 1b (G1DM1.0/D1b) using the Lagrangian Interpolation Basis Function until convergence of $$\displaystyle u^h (0.5) \rightarrow O(10^{-6}) $$.
 * 1. Explain how the basis functions satisfy the constraint breaking solution (CBS).
 * 2.Plot these basis functions.
 * 3.Use matlab to integrate.
 * 4.Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

Solution
The Lagrangian Element Basis Function (LEBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:

 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LEBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:

 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$

 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LEBF). In other words the basis function is equal to the kronicker delta at each node.

 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$

 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

A Matlab script was written to plot the LEBFs at each node for each element. The Matlab script is given followed by the plots. The first plot shows each of the LEBFs plotted across the entire domain. The second plot is more informative as the plot is divided into the elemental domains and lines corresponding to the values zero and one are present. The legend states the node number, $$n$$, and element number, $$e$$, of each LEBF.

As with the CBS, the coefficient at the essential boundary condition is equal to the essential boundary condition because the basis function is equal to one at the essential boundary condition:

 * $$g = \mathbf{d_{1}^{(1)}}\mathbf{b_{1}^{(1)}}(0)=\mathbf{d_{1}^{(1)}}*1=\mathbf{d_{1}^{(1)}}=4$$

The element matrices are formed as in problems 5.1 and 5.2 as follows:

 * $$\displaystyle \mathbf{F}^{e}_{F}=b_{i}^{(1)}(\alpha_{i}^{e} )h + \int_{\omega^{e}}^{.} {{b_{i}^{(1)}}fdx}$$

\displaystyle

h = {a_2}(\alpha )\frac = (2+3\alpha)\frac = 5*12/5=12

$$

 * $$\displaystyle \mathbf{F_{F}^{(1)}}=\mathbf{F}^{(1)}_{2}=b_{2}^{(1)}(.25)h + \int_{0}^{.25} {{b_{2}^{(1)}}fdx}=1*12+\int_{0}^{.25} 4x(5x)=12.104167$$

$$ \displaystyle

\therefore {F_F} = \left[ {\begin{array}{ccccccccccccccc} {12.104167} \end{array}} \right]

$$

 * $$\displaystyle \mathbf{K}^{e}_{ij}=\int_{\omega^{e}}^{.} b_{i}^{e'}(x^e)a_2(x^e)b_{j}^{e'}(x^e)dx$$

\displaystyle

\mathbf{K^{(1)}_{FE}} = \left\{ {\mathbf{K^{(1)}_{1j}};j = 2,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{12}}= \int\limits_0^{.25} b_{1}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}-4*(2+3x)*4=-32(.25)-24(.25)^2=-9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} -9.5 \end{bmatrix} $$

\displaystyle

\mathbf{K^{(1)}_{FF}} = \left\{ {\mathbf{K^{(1)}_{ij}};i,j = 2,...,n} \right\}

$$

\displaystyle

\mathbf{K^{(1)}_{22}}= \int\limits_0^{.25} b_{2}^{(1)'}a_2b_{2}^{(1)'}=\int\limits_0^{.25}4*(2+3x)*4=32(.25)+24(.25)^2=9.5

$$

\displaystyle \therefore \mathbf{K^{(1)}_{FF}} = \begin{bmatrix} 9.5 \end{bmatrix} $$

 * $$\mathbf{K^{(1)}} \mathbf{d^{(1)}} = \mathbf{F_{F}^{(1)}} - \mathbf{K_{FE}^{(1)}} g$$

 * $$\mathbf{d_{2}^{(1)}} = \mathbf{K}_{FF}^{(1)-1}( \mathbf{F_{F}^{(1)}} - \mathbf{K_{FE}^{(1)}} g)$$

 * $$\therefore \mathbf{d_{2}^{(1)}} = 5.5208$$

Now the process is repeated for each element. The matrices are assembled into global matrices and all of the coefficients can be determined. The approximate solution is formed for each element by taking the coefficients times the basis functions. The approximate solutions for each element make up the global approximate solution.

 * $$\displaystyle \mathbf{K}_{ij} = \sum_{e=1}^{nel}\mathbf{K}^{e}_{ij}$$

 * $$\displaystyle \mathbf{F}_{i} = \sum_{e=1}^{nel}\mathbf{F}^{e}_{i}$$

 * $$\displaystyle u^{he}(x)= \sum_{j=1}^{n}d_jb_j(x),\quad \quad u^{h}(x)=\sum_{j=1}^{nel}u^{he}(x) $$

A Matlab script was written to generate the plots. The results and script are given below:

Problem Description
Solve the general 1-D model for an elastic bar with data set 1 (G1DM1.0/D1) using the Linear Lagrangian element basis function (LLEBF) with uniform discretization(equidistant element nodes, nel=4, 6, 8).
 * 1.Explain how the basis functions satisfy the constraint breaking solution (CBS).
 * 2.Plot these basis functions.
 * 3.Use matlab to integrate.
 * 4.Make Plots of u^h vs u and u^h(.5)-u(.5) vs m

Explain how LLEBF are used as CBS
This problem is similar to problem 5.7 except switch the location of natural and essential boundary conditions.

The Lagrangian Element Basis Function (LEBF) is given by the following equation and generates a polynomial of degree $$\displaystyle n-1$$:

 * $$L_{i,n}\left (x\right )=\prod_{k=1}^{n}\frac{x-x_{k}}{x_{i}-x_{k}},k\neq i$$

For a linear function the number of nodes is 2 ($$\displaystyle n=2$$). For uniform discretization with $$\displaystyle nel=4$$ the domain $$\displaystyle \Omega$$ will be subdivided into 4 equal length elements each having two nodes. The LEBFs for the first element $$\displaystyle n=1$$ over the elemental domain $$\displaystyle \omega^{(1)}=[0,.25]$$ can be written as follows:

 * $$L_{1,2}^{(1)}\left (x\right)=b_{1}^{(1)}=\frac{(x-x_2)}{x_1-x_2}=\frac{(x-.25)}{0-.25}=-4(x-.25)$$

 * $$L_{2,2}^{(1)}\left (x\right)=b_{2}^{(1)}=\frac{(x-x_1)}{x_2-x_1}=\frac{(x-0)}{.25-0}=4(x)$$

Recall the constraint breaking solution (CBS) states the basis functions must satisfy $$\displaystyle \left\{ b_i(x), i=0,1,2,...\right\}$$ such that $$\displaystyle b_0(\beta) \neq 0$$ and $$\displaystyle b_i(\beta) = 0$$ for $$\displaystyle i=1,2,...,n$$.

It can be seen that these basis functions for the first element satisfy the CBS at each node. For the first node, the basis function has a value of one at that node and a value of zero at every other node (there is only one other node for the linear LEBF). In other words the basis function is equal to the kronicker delta at each node.

 * $$b_{1}^{(1)}(\beta^{(1)}_1)=L_{1,2}^{(1)}\left (0\right)=b_{1}^{(1)}(0)=-4(0-.25)=1$$

 * $$L_{1,2}^{(1)}\left (.25\right)=b_{1}^{(1)}(.25)=-4(.25-.25)=0$$

Similarly for the second node, the basis function has a value of one at that second node and a value of zero at the first node.

Therefore, LLEBF satifies CBS

title: User:Fem.tm7/HW5


text:
User talk:Caprice/Clue Train

Abd, at the top of your personal user page, you have prominently posted this template:

 In view of your avowed commitment to Scholarly Ethics, please complete your as-yet-unfinished responses to the following incompleted discussion thread:

No rangeblocks
No rangeblocks just for one editor, Abd. Precisely why the Wikimedia Foundation cautions against such practice is because it affects a wide range of other users along with just Moulton, and is primarily used as a last resort against frequent IP-address socking. Do you understand what it is, and how it could adversely affect other users other than just Moulton? TeleComNasSprVen 01:24, 9 March 2011 (UTC)

 * Sigh. TCNSV, I have not asked for Range Blocks. Did you notice what was done the last time, Moulton was blocked, a few days ago? Massive range blocks. Basically, TCNSV, you are clueless. Range blocks will be used if a custodian thinks them necessary. I'm monitoring activity and may inform or advise. I'm not asking for your advice, but you are free to offer it to the custodians. Just as I am. --Abd 03:55, 9 March 2011 (UTC)

 * Then monitor, and don't give him any ideas with doling out unnecessary extraneous comments. Like this one it gives the appearance that you support rangeblocks, chastising SB_Johnny for not taking the appropriate measures levied against him. TeleComNasSprVen 06:37, 9 March 2011 (UTC)

There, you've asked for a rangeblock! It will have adverse affects on other users sharing Moulton's range. Why are you so opposed to blocking Draubb's school whilst at the same time advocating for a rangeblock of someone else? Your analysis often seems inconsistent with your practice. TeleComNasSprVen 02:17, 11 March 2011 (UTC)

 * The disruption caused by Draubb is trivial, compared to that caused by Moulton. If you want Draubb's school blocked, go to RCA and request it. Show the problem. Look like the vengeful idiot that you are. *Draubb is not blocked. You think Draubb's school IP should be blocked? Without the user being blocked? Why?

 * Something is seriously off, TCNSV, this makes no sense at all. Moulton is *blocked.* If he revert wars, he's making it impossible to tolerate his block evasion. First of all, I've not asked for a range block, yet, only if Moulton makes it necessary. I've asked for a block of a single IP. I don't think you have any concept of how to handle a situation, except by going to meta and implying massive vandalism, or the risk of same, when nothing is happening. That really took the cake.

 * Stay away from my Talk page, you are a waste of time. --Abd 05:50, 11 March 2011 (UTC)

Truthout Issue
Note: This thread was originally posted on Darklama's page.

 * I have an idea, Abd. Why don't you go running to Darkmama and ask her to dictate what passes for truth. —Palomino of Certainty 17:09, 9 March 2011 (UTC)
 * My position, Moulton, is that custodians should not be placed in the position of determining "truth," as to "outside reality." They need only look at the edits themselves, or at the edits and verified complaints, possibly. "Dictate" is your reframing for any determination regarding a situation, when, on wikis, generally, nobody "dictates," but people do act for themselves, and sometimes structure has been set up that does, collectively, dictate. Management (i.e, ownership) may also dictate, but technically, they cannot compel wiki users to do anything. They may merely assert legal control. Get it straight, Moulton, you've been confused about all this for years. --Abd 18:44, 9 March 2011 (UTC)

 * by the way, I brought this to Darklama instead of RCA, because Darklama had "touched" your edit, so I wouldn't revert it, for example. I am, generally, only acting with reverts to enforce a block, or to prevent outing, but not to tendentiously edit with respect to other editors in good standing. --Abd 18:46, 9 March 2011 (UTC)

 * According to a complaint from JWS, Custodians are placed in the awkward position of having to give a valid reason for a block. They aren't supposed to just make up a haphazard (and demonstrably false) reason out of whole cloth.  Do you agree with JWS that the reason for a block has to at least be a true statement on its own, whether or not it's also a statement that, if true, would amount to a violation of some policy?  —Moulton 19:23, 9 March 2011 (UTC)
 * You're asking me on DL's talk page? She may remove this if she wants. Custodians may block, giving no reason at all. However, if they do this more than infrequently, their competence to continue might be questioned. "Making up a haphazard and demonstrably false reason out of whole cloth, by definition, would indicate serious bias, and if your condition is true, the custodian would properly be deysopped. However, making an error in judgment, without the intention implied in your conditions, is not necessarily a problem, because it's a wiki and almost everything can be fixed. You claim that nothing can be fixed ("Moulton's law") but that is an error in judgment. Nothing can be fixed if no significant elements of the community, particularly the administrative community, wants to fix it. They can, on a wiki, look the other way. They are not employees of the "state." They have no specific obligations, they are volunteers in everything they do. The solution to a community with administrators who "oppress" by looking the other way is more administrators who won't look the other way, but, as you know, if an oligarchy has developed, it resists change. Still, with individual admins who demonstrate clear bias in use of tools, it's quite possible to desysop. It's just difficult, not for the inexperienced or highly reactive, and possibly not worth the effort and cost. --Abd 20:26, 9 March 2011 (UTC)
 * Yes, it's astonishingly difficult. Look at how hard it was for Charles Ainsworth to bring charges against FeloniousMonk for his abuses on Wikipedia.  And look how hard it was for ArbCom to adjudicate the case.  Now, as you may know, it was FeloniousMonk who called upon Jimbo to come here to block me, per Jimbo's threat to do just that, over the issue of the silly song parodies on my personal blog lampooning IDCab. Now Jimbo unilaterally declared that those song parodies, posted off-wiki on my own obscure and little-read blog amounted to a violation of Civility here on Wikiversity.  Don't you think the community here should decide whether silly song parodies, posted off-wiki on an obscure and little-noted personal blog are a sufficient cause of action for the God-King to come galumphing in here, against the will of the community, to personally execute blocks without due process?  Is it an objectively true statement that those song parodies, posted off-wiki amount to a violation of the local Civility policy?—Moulton 21:20, 9 March 2011 (UTC)

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