Finite Element Analysis. Barna Szabó

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Название Finite Element Analysis
Автор произведения Barna Szabó
Жанр Физика
Серия
Издательство Физика
Год выпуска 0
isbn 9781119426462



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u comma v right-parenthesis minus upper F left-parenthesis v right-parenthesis equals 0"/> in the following form:

      (1.45)StartSet b EndSet Superscript upper T Baseline left-parenthesis left-parenthesis left-bracket upper K right-bracket plus left-bracket upper M right-bracket right-parenthesis StartSet a EndSet minus StartSet r EndSet right-parenthesis equals 0 period

      Since this must hold for any choice of StartSet b EndSet, it follows that

      (1.46)left-parenthesis left-bracket upper K right-bracket plus left-bracket upper M right-bracket right-parenthesis StartSet a EndSet equals StartSet r EndSet

StartLayout 1st Row 1st Column upper B left-parenthesis u comma v right-parenthesis equals 2nd Column upper F left-parenthesis v right-parenthesis for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis 2nd Row 1st Column upper B left-parenthesis u Subscript n Baseline comma v right-parenthesis equals 2nd Column upper F left-parenthesis v right-parenthesis for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis period EndLayout

      Subtracting the second equation from the first we have,

      (1.47)upper B left-parenthesis u minus u Subscript n Baseline comma v right-parenthesis identical-to upper B left-parenthesis e comma v right-parenthesis equals 0 for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis period

      Proof: Let e equals u minus u Subscript n and let v be an arbitrary function in upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis. Then

vertical-bar vertical-bar vertical-bar vertical-bar plus plus ev Subscript upper E left-parenthesis upper I right-parenthesis Superscript 2 Baseline identical-to one half upper B left-parenthesis e plus v comma e plus v right-parenthesis equals one half upper B left-parenthesis e comma e right-parenthesis plus upper B left-parenthesis e comma v right-parenthesis plus one half upper B left-parenthesis v comma v right-parenthesis period

      The first term on the right is double-vertical-bar e double-vertical-bar Subscript upper E left-parenthesis upper I right-parenthesis Superscript 2, the second term is zero on account of Theorem 1.3, the third term is positive for any v not-equals 0 in upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis. Therefore double-vertical-bar e double-vertical-bar Subscript upper E left-parenthesis upper I right-parenthesis is minimum.

      Theorem 1.4 states that the error depends on the exact solution of the problem u Subscript upper E upper X and the definition of the trial space ModifyingAbove upper S With tilde left-parenthesis upper I right-parenthesis.

      The finite element method is a flexible and powerful method for constructing trial spaces. The basic algorithmic structure of the finite element method is outlined in the following sections.

      The standard polynomial space of degree p, denoted by script upper S Superscript p Baseline left-parenthesis upper I Subscript st Baseline right-parenthesis, is spanned by the monomials 1 comma xi comma xi squared comma ellipsis comma xi Superscript p defined on the standard element

      (1.49)upper I Subscript st Baseline equals StartSet xi vertical-bar negative 1 less-than xi less-than 1 EndSet period

      The choice of basis functions is guided by considerations