# 111b 0 let us simplify matters somewhat by assuming

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Unformatted text preview: 9.1.1. The Green's function G( x) for the operator L of (9.1.1) satis es G( x) 2 C 0 (a b) (a b) L G( x) = 0 T (a ; ) ( + b) = lim 0 ! G( a) = G( b) = 0 G( x +) ; G x ( (9.1.12a) ; ) = 1: (9.1.12b) (9.1.12c) (9.1.12d) When viewed as a function of x, the Green's function has a unit jump in its rst derivative at the point . 4 Now, if we choose v(x) in (9.1.11c) as the Green's function G( x) so that the only (M = 2) discontinuity occurs at z1 = , we have u( ) = (G( ) Lu): (9.1.13a) If u(x) is chosen as y(x), the solution of (9.1.1), then y( ) = (G( ) Ly) = (G( ) r): (9.1.13b) The relation (9.1.13a) holds for any smooth function u(x) and not just the solution of (9.1.1). Since, for example, the collocation solution Y (x) 2 C 1 (a b), we can replace u in (9.1.13a) by Y to obtain Y ( ) = (G( ) LY ): (9.1.13c) e(x) = y(x) ; Y (x) (9.1.14a) Finally, letting denote the discretization error of the collocation solution, we subtract (9.1.13c) from (9.1.13b) to obtain e( ) = (G( ) Le) = Z b a G( x)Le(x)dx: (9.1.14b) Re...
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## This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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