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

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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...
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