Unformatted text preview: 2)(LY ) ( )
2 (9.1.17b) 00 i i i where
2 (x 1 x ).
Using (9.1.17) in (9.1.16) yields
i i i; e( )= i Z xi i xi where ;1 (x ; 1)(x ; 2)g( x)dx
i 2 (x
= i 1 i; 1
g( x) = 2 G( x) (Ly) ( ) ; (LY ) ( )]:
00 00 i i x)
(9.1.18b) where e was de ned in (9.1.15). We bound (9.1.18a) as
i je ( )j Z xi i Since jx ; ij xi ;1 jx ; 1jjx ; 2jjg( x)jdx:
i i j h , j = 1 2, we have
i je ( )j h3 jjg( )jj
i i i1 6 2 (x
= i; 1 x)
i (9.1.19a) GL Y
GL y x i-1 ξ i,1 ξ ξ i,2 x i Figure 9.1.3: Functions GLy, GLY , and GPr on a subinterval (x 1 x ) containing the
point x = .
i; i where jjf ( )jj = i;max i jf (x)j:
i1 x x (9.1.19b) x Typically, one subinterval contains the point where G is discontinuous. The analysis leading to (9.1.19) cannot be used on this subinterval since G( x) 2 C 1(x 1 x ).
There are several ways of showing that je ( )j increases from O(h3) to O(h2) on this
subinterval. We'll choose one which restricts to lie between 1 and 2. In this case,
we interpolate GLy and GLY by piecewise constant func...
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This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.
- Spring '14
- The Land