Also let z0 a and z b consider j v lu z m b v

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: sum of integrals over the subintervals (z0 z1), (z1 z2), : : : , (z 1 z ). For simplicity, we'll also assume that A = B = 0 and that u(x) and v(x) satisfy these conditions. Using (9.1.1), we integrate (9.1.9) by parts to obtain M; (v LU ) = M Z b ;v u ; (pv) u + qvu]dx + 0 a 0 0 X M j 3 =1 fv(x)u (x) + p(x)v(x)u(x)g jj;1 : 0 z z Integrating the rst term in the integrand by parts once more (v LU ) = Z b v ; (pv) + qv]udx + 00 a 0 X M j =1 fv(x)u (x) ; v (x)u(x) + p(x)v(x)u(x)g jj;1 : 0 z 0 z We can write this result in a simpler form by using the inner product notation (9.1.8) and by de ning the jump in a function q(x) at a point z as q(x)] = = lim q(z + ) ; q(z ; ): 0 x z (9.1.10) ! With this notation, we have (v Lu) = (L v u) ; T X1 M; j =1 vu ; v u + pvu] j 0 0 z (9.1.11a) where L is called the adjoint operator and satis es T L v = v ; (pv) + qv: T 00 (9.1.11b) 0 Let us simplify matters somewhat by assuming that u 2 C 1(a b) and v 2 C 0 (a b). Then, (9.1.11a) becomes X1 M; j =1 v u] j = (v Lu) ; (L v u): T 0 z (9.1.11c) De nition...
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