Finite Element Analysis. Barna Szabó

Finite Element Analysis

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any perturbation of u by v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis

will increase pi left-parenthesis u right-parenthesis

. Therefore pi left-parenthesis u plus epsilon v right-parenthesis

is minimum at epsilon equals 0

and hence

(1.39) period vertical-bar times times times d pi left-parenthesis right-parenthesis plus plus u times times epsilon v times times d epsilon Subscript epsilon equals 0 Baseline equals 0 period

Therefore we have

(1.40)

where the last two terms are zero because v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis . Integrating the first term by parts,

integral Subscript 0 Superscript script l Baseline kappa u prime v Superscript prime Baseline d x equals ModifyingBelow kappa u prime left-parenthesis script l right-parenthesis v left-parenthesis script l right-parenthesis minus kappa u prime left-parenthesis 0 right-parenthesis v left-parenthesis 0 right-parenthesis With presentation form for vertical right-brace Underscript 0 Endscripts minus integral Subscript 0 Superscript script l Baseline left-parenthesis kappa u prime right-parenthesis prime v d x

and, substituting this into eq. (1.40), we get

(1.41) integral Subscript 0 Superscript script l Baseline left-parenthesis minus left-parenthesis kappa u Superscript prime Baseline right-parenthesis Superscript prime Baseline plus c u minus f right-parenthesis v d x equals 0 period

Since this holds for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis , the bracketed expression must be zero. In other words, the solution of the differential equation

(1.42)

minimizes the potential energy defined by eq. (1.38). This is the strong form of the problem.

Remark 1.2 The procedure in Example 1.2 is used in the calculus of variations for identifying the differential equation, known as the Euler⁹‐Lagrange¹⁰ equation, the solution of which maximizes or minimizes a functional. In this example the solution minimizes the potential energy on the space ModifyingAbove upper E With tilde left-parenthesis upper I right-parenthesis .

Remark 1.3 Whereas the strain energy is always positive, the potential energy may be positive, negative or zero.

1.3 Approximate solutions

The trial and test spaces defined in the preceding section are infinite‐dimensional, that is, they span infinitely many linearly independent functions. To find an approximate solution, we construct finite‐dimensional subspaces denoted, respectively, by upper S subset-of upper X , upper V subset-of upper Y and seek the function u element-of upper S that satisfies upper B left-parenthesis u comma v right-parenthesis equals upper F left-parenthesis v right-parenthesis for all v element-of upper V . Let us return to the introductory example described in Section 1.1 and define

u equals u Subscript n Baseline equals sigma-summation Underscript j equals 1 Overscript n Endscripts a Subscript j Baseline phi Subscript j Baseline comma v equals v Subscript n Baseline equals sigma-summation Underscript i equals 1 Overscript n Endscripts b Subscript i Baseline phi Subscript i Baseline

where ϕi ( i equals 1 comma 2 comma ellipsis n ) are basis functions. Using the definitions of k Subscript i j and m Subscript i j given in eq. (1.12), we write the bilinear form as

(1.43)

Similarly,

(1.44) upper F left-parenthesis v right-parenthesis identical-to integral Subscript 0 Superscript script l Baseline f v d x equals sigma-summation Underscript i equals 1 Overscript n Endscripts b Subscript i Baseline r Subscript i Baseline equals StartSet b EndSet Superscript upper T Baseline StartSet r EndSet

where ri is defined in eq. (1.12). Therefore we can write Скачать книгу