Differential Equations Solutions 128

Differential Equations Solutions 128 - 138 Chapter 23....

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138 Chapter 23. Solutions: Case Study: Finite Diferences and Finite Elements since we can compute the integral of a sum as the sum of the integrals and then move the constants outside the integrals. Therefore, a ( u h ,v h )= a ( u h , M 2 X j =1 v j φ j ) = M 2 X j =1 v j a ( u h j ) = M 2 X j =1 v j ( f,φ j ) =( f, M 2 X j =1 v j φ j ) f,v h ) . (b) We compute a ( φ j j Z 1 0 ( φ ± j ( t )) 2 d t = Z ( j +1) h ( j 1) h ( φ ± j ( t )) 2 d t =2 Z jh ( j 1) h 1 h 2 d t = 2 h , and a ( φ j j +1 Z 1 0 φ ± j ( t ) φ ± j +1 ( t )d t = Z ( j +1) h φ ± j ( t ) φ ± j +1 ( t )d t = Z
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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