Finite Element Analysis. Barna Szabó

Finite Element Analysis

alt="ModifyingAbove upper S With tilde left-parenthesis upper I right-parenthesis"/>, denoted by u Superscript black star

. We then seek u overbar element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis

such that

(1.79)

for all v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis . Observe that the solution u equals u overbar plus u Superscript black star is independent of the choice of .

We denote the global numbers of the basis functions that are unity at x equals 0 and x equals script l by K and L respectively. For instance, in Example 1.6 upper K equals 1 and upper L equals 4 . It is advantageous to define u Superscript black star in terms of phi Subscript upper K Baseline left-parenthesis x right-parenthesis and phi Subscript upper L Baseline left-parenthesis x right-parenthesis :

(1.80) u Superscript black star Baseline equals modifying above u with caret 0 phi Subscript upper K Baseline left-parenthesis x right-parenthesis plus modifying above u with caret Subscript script l Baseline phi Subscript upper L Baseline left-parenthesis x right-parenthesis

as indicated in Fig. 1.6. When Dirichlet boundary condition is prescribed on only one of the boundary points then this expression is modified to include the term corresponding to that point only.

On substituting eq. (1.80) into eq. (1.79) the second term on the right‐hand side of eq. (1.79) can be written as

integral Subscript 0 Superscript script l Baseline left-parenthesis kappa left-parenthesis u Superscript black star Baseline right-parenthesis prime v prime plus c u Superscript black star Baseline v right-parenthesis d x equals sigma-summation Underscript i equals 1 Overscript upper N Subscript u Baseline Endscripts b Subscript i Baseline left-parenthesis c Subscript i upper K Baseline plus c Subscript i upper L Baseline right-parenthesis

where Nu is the number of unconstrained equations, that is, the number of equations prior to enforcement of the Dirichlet boundary conditions. (For instance, in Example 1.6 upper N Subscript u Baseline equals 7 .) The coefficients c Subscript i upper K , c Subscript i upper L are elements of the assembled coefficient matrix.

Geometric representation of recommended choice of the function u★ in one dimension.

Figure 1.6 Recommended choice of the function

in one dimension.

Since v element-of upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis , we have b Subscript upper K Baseline equals b Subscript upper L Baseline equals 0 and therefore the Kth and Lth rows of matrix left-bracket upper C right-bracket are multiplied by zero and can be deleted. The Kth and Lth columns of matrix are multiplied by modifying above u with caret 0 and modifying above u with caret Subscript script l respectively, summed and the resulting vector is transferred to the right‐hand side. The resulting coefficient matrix has the dimension N which is Nu minus the number of Dirichlet boundary conditions. The number N is called the number of degrees of freedom. It is the maximum number of linearly independent functions in upper S Superscript 0 Baseline left-parenthesis upper I right-parenthesis .

Remark 1.7 In order to avoid having to renumber the coefficient matrix once the rows and columns corresponding to φK and φL were eliminated, all elements in the Kth and Lth rows and columns can be set to zero, with the exception of the diagonal elements, which are set to unity. The corresponding elements on the right hand side vector are set to û₀ and ûℓ. This is illustrated by the following example.

Example 1.7 Consider the problem

minus u Superscript double-prime Baseline plus 4 u equals 0 comma u left-parenthesis 0 right-parenthesis equals 1 comma u left-parenthesis 1 right-parenthesis equals 2

the exact solution of which is

u equals StartFraction exp left-parenthesis 2 right-parenthesis minus 2 Over exp left-parenthesis 2 right-parenthesis minus exp left-parenthesis negative 2 right-parenthesis EndFraction exp left-parenthesis minus 2 x right-parenthesis plus StartFraction 2 minus exp left-parenthesis negative 2 right-parenthesis Over exp left-parenthesis 2 right-parenthesis <hr><noindex><a href=

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