Finite Element Analysis. Barna Szabó

Finite Element Analysis

left-parenthesis upper I right-parenthesis"/> such that

(1.162) upper B left-parenthesis u Subscript upper F upper E Baseline comma upper F Subscript upper F upper E 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 S Superscript 0 Baseline left-parenthesis upper I right-parenthesis comma upper G element-of upper V left-parenthesis upper I right-parenthesis period

We now ask: In what sense will left-parenthesis u Subscript upper F upper E Baseline comma upper F Subscript upper F upper E Baseline right-parenthesis be close to ? The answer is that there is a constant C, independent of the finite element mesh and , such that

(1.163)

provided, however, that upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis and upper V left-parenthesis upper I right-parenthesis were properly selected.

For example, let S be the space defined in eq. (1.61) with p Subscript k Baseline equals 1 , k equals 1 comma 2 ellipsis upper M left-parenthesis normal upper Delta right-parenthesis . The space upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis has the dimension . For upper V left-parenthesis upper I right-parenthesis consider three choices:

1 is the set of functions which are constant on each finite element. has the dimension .

2 is the space S defined in (3.11) with , (dimension ).

3 is the set of functions which are linear on every element and discontinuous at the nodes (dimension ).

For these choices of upper V left-parenthesis upper I right-parenthesis the mixed formulation leads to systems of linear equations with 2 upper M left-parenthesis normal upper Delta right-parenthesis minus 1 , and 3 upper M left-parenthesis normal upper Delta right-parenthesis minus 1 unknowns, respectively. In the cases upper V equals upper V 1 and upper V equals upper V 3 , a constant C exists such that the inequality (1.163) holds for all u Subscript upper E upper X , and upper F Subscript upper E upper X . In the case of upper V equals upper V 2 , however, such a constant does not exist. This means that no matter how large C is, there exist some u Subscript upper E upper X Baseline element-of upper H 0 Superscript 1 Baseline left-parenthesis upper I right-parenthesis and upper F Subscript upper E upper X Baseline element-of upper L squared left-parenthesis upper I right-parenthesis and mesh normal upper Delta so that the inequality (1.163) is not satisfied. On the other hand, there will be u Subscript upper E upper X Baseline element-of upper H 0 Superscript 1 Baseline left-parenthesis upper I right-parenthesis and upper F Subscript upper E upper X Baseline element-of upper L squared left-parenthesis upper I right-parenthesis for which the inequality is satisfied and therefore the finite element solutions will converge to the underlying exact solution.

1.8.2 Nitsche's method

Nitsche's method¹⁶ allows the treatment of essential boundary conditions as natural boundary conditions. This has certain advantages in two and three dimensions. An outline of the algorithmic aspects of the method is presented in the following. For additional details we refer to [51].

Consider the problem:

(1.164) minus u Superscript double-prime Baseline plus c u equals f left-parenthesis x right-parenthesis comma x element-of left-parenthesis 0 comma script l right-parenthesis

with the boundary conditions u prime left-parenthesis 0 right-parenthesis equals 0 and u left-parenthesis script l right-parenthesis equals modifying above u with caret Subscript script l . However, at x equals script l we substitute the natural boundary condition:

(1.165)Скачать книгу