Finite Element Analysis. Barna Szabó

Finite Element Analysis - Barna Szabó


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multiplying eq. (1.133) by a test function v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis and integrating by parts:

      (1.134)integral Subscript 0 Superscript script l Baseline left-parenthesis kappa u prime v Superscript prime Baseline plus c u v right-parenthesis d x equals minus integral Subscript 0 Superscript script l Baseline mu StartFraction partial-differential squared u Over partial-differential t squared EndFraction v d x period

      We now introduce u equals upper U left-parenthesis x right-parenthesis upper T left-parenthesis t right-parenthesis where upper U element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis, upper T element-of upper C squared left-parenthesis 0 comma infinity right-parenthesis. This is known as separation of variables. Therefore we get

      (1.135)upper T integral Subscript 0 Superscript script l Baseline left-parenthesis kappa upper U prime v prime plus c upper U v right-parenthesis d x equals minus StartFraction partial-differential squared upper T Over partial-differential t squared EndFraction integral Subscript 0 Superscript script l Baseline mu upper U v d x

      (1.136)StartFraction integral Subscript 0 Superscript script l Baseline left-parenthesis kappa upper U prime v Superscript prime Baseline plus c upper U v right-parenthesis d x Over integral Subscript 0 Superscript script l Baseline mu upper U v d x EndFraction equals minus StartFraction 1 Over upper T EndFraction StartFraction partial-differential squared upper T Over partial-differential t squared EndFraction equals omega squared period

      Since the functions on the left are independent of t, the function T depends only on t, both expressions must equal some positive constant denoted by omega squared. That constant has to be positive because the expression on the left holds for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis and if we select v equals upper U then the expression on the left is positive.

      The function upper T left-parenthesis t right-parenthesis satisfies the ordinary differential equation

      (1.137)StartFraction partial-differential squared upper T Over partial-differential t squared EndFraction plus omega squared upper T equals 0

      the solution of which is

      (1.138)upper T equals a cosine left-parenthesis omega t right-parenthesis plus b sine left-parenthesis omega t right-parenthesis

      where ω is the angular velocity (rad/s). Alternatively ω is written as omega equals 2 pi f where f is the frequency (Hz).

      To find ω and U we have to solve the problem

      (1.139)integral Subscript 0 Superscript script l Baseline left-parenthesis kappa upper U prime v Superscript prime Baseline plus c upper U v right-parenthesis d x minus omega squared integral Subscript 0 Superscript script l Baseline mu upper U v d x equals 0 for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis

      which will be abbreviated as

      There are infinitely many solutions called eigenpairs left-parenthesis omega Subscript i Baseline comma upper U Subscript i Baseline right-parenthesis, i equals 1 comma 2 comma ellipsis comma infinity. The set of eigenvalues is called the spectrum. If Ui is an eigenfunction and α is a real number then alpha upper U Subscript i is also an eigenfunction. In the following we assume that the eigenfunctions have been normalized so that

upper D left-parenthesis upper U Subscript i Baseline comma upper U Subscript i Baseline right-parenthesis identical-to integral Subscript 0 Superscript script l Baseline mu upper U Subscript i Superscript 2 Baseline d x equals 1 period StartLayout 1st Row 1st Column upper B left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis minus omega Subscript i Superscript 2 Baseline upper D left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis 2nd Column equals 0 2nd Row 1st Column upper B left-parenthesis upper U Subscript j Baseline comma upper U Subscript i Baseline right-parenthesis minus omega Subscript j Superscript 2 Baseline upper D left-parenthesis upper U Subscript j Baseline comma upper U Subscript i Baseline right-parenthesis 2nd Column equals 0 period EndLayout

      Subtracting the second equation from the first we see that if omega Subscript i Baseline not-equals omega Subscript j then Ui and Uj are orthogonal functions:

      (1.141)upper D left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis identical-to integral Subscript 0 Superscript script l Baseline mu upper U Subscript i Baseline upper U Subscript j Baseline d x equals 0

      and hence upper B left-parenthesis upper U Subscript i Baseline comma upper U Subscript j Baseline right-parenthesis equals 0.

      Importantly, it can be shown that any function f element-of <hr><noindex><a href=Скачать книгу