111b 0 let us simplify matters somewhat by assuming

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online