Finite Element Analysis. Barna Szabó

Finite Element Analysis

lies in X and satisfies:(1.151) The normed linear spaces, , the linear functional F and the bilinear form B satisfy the respective properties listed in sections A.1.1 and A.1.2.

2 Finite element spaces. The finite‐dimensional subspaces , () are defined and it is assumed that there are such that the sequence of functions ûi () converges in the space X to , that is:(1.152) The functions ûi are not the finite element solutions in general.

3 The finite element solution. The finite element solution satisfies:(1.153)

4 The stability criterion. The finite element method is said to be stable if(1.154) for all possible . The necessary and sufficient condition for a finite element method to be stable is that for every there is a so that(1.155) where is a constant, independent of i, or for every there is a so that this inequality holds. This inequality is known as the Babuška‐Brezzi condition, usually abbreviated to “the BB condition”. This condition was formulated by Babuška in 1971 [9] and independently by Brezzi in 1974 [29].

If the Babuška‐Brezzi condition is not satisfied then there will be at least some u Subscript upper E upper X Baseline element-of upper X for which double-vertical-bar u Subscript upper E upper X Baseline minus u Subscript i vertical-bar upper F upper E Baseline double-vertical-bar Subscript upper X Baseline right-arrow with stroke 0 as i right-arrow infinity even though there may be for which as . Examples are presented in [6]. In general it is difficult, or may even be impossible, to separate those u Subscript upper E upper X for which the method works well from those for which it does not. The Babuška‐Brezzi condition guarantees that the condition number of the stiffness matrix will not become too large as i increases.

Remark 1.15 Any implementation of the finite element method must be shown to satisfy the Babuška‐Brezzi condition otherwise there will be some input data for which the method will fail even though it may work well for other input data. The formulation based on the principle of virtual work satisfies the Babuška‐Brezzi condition.

Exercise 1.23 Show that the finite element method based on the principle of virtual work satisfies the Babuška‐Brezzi condition.

1.8.1 The mixed method

Consider writing eq. (1.5) in the following form:

(1.156) kappa u Superscript prime Baseline minus upper F equals 0

(1.157) minus upper F Superscript prime Baseline plus c u equals f

and assume that the boundary conditions are u left-parenthesis 0 right-parenthesis equals u left-parenthesis script l right-parenthesis equals 0 .

In the following we will use the one‐dimensional equivalent of the notation introduced in sections A.2.2 and A.2.3. Multiply eq. (1.156) by upper G element-of upper L squared left-parenthesis upper I right-parenthesis and eq. (1.157) by v element-of upper H Superscript 1 Baseline left-parenthesis upper I right-parenthesis , integrate by parts and sum the resulting equations to obtain:

(1.158) integral Subscript 0 Superscript script l Baseline left-parenthesis kappa StartFraction d u Over d x EndFraction upper G minus upper F upper G right-parenthesis d x plus integral Subscript 0 Superscript script l Baseline left-parenthesis upper F StartFraction d v Over d x EndFraction plus c u v right-parenthesis d x equals integral Subscript 0 Superscript script l Baseline upper T v d x period

We define the bilinear form:

(1.159) upper B left-parenthesis u comma upper F semicolon v comma upper G right-parenthesis equals Overscript def Endscripts integral Subscript 0 Superscript script l Baseline left-parenthesis kappa u prime upper G minus upper F upper G right-parenthesis d x plus integral Subscript 0 Superscript script l Baseline left-parenthesis upper F v prime plus c u v right-parenthesis d x

and the linear form

(1.160) upper F left-parenthesis v right-parenthesis equals Overscript def Endscripts integral Subscript 0 Superscript script l Baseline f v d x period

The problem is now stated as follows: Find u Subscript upper E upper X Baseline element-of upper H 0 Superscript 1 Baseline left-parenthesis upper I right-parenthesis , upper F Subscript upper E upper X Baseline element-of upper L squared left-parenthesis upper I right-parenthesis such that

(1.161) upper B left-parenthesis u Subscript upper E upper X Baseline comma upper F Subscript upper E upper X Baseline semicolon v comma upper G right-parenthesis equals upper F left-parenthesis v right-parenthesis f o r a l l v element-of upper H 0 Superscript 1 Baseline left-parenthesis upper I right-parenthesis comma upper G element-of upper L squared left-parenthesis upper I right-parenthesis period

The finite element problem is formulated as follows: Find u Subscript upper F upper E Baseline element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis where is a subspace of upper H 0 Superscript 1 Baseline left-parenthesis upper I right-parenthesis and upper F Subscript upper F upper E Baseline element-of upper V left-parenthesis upper I right-parenthesis where is a subspace of Скачать книгу