23 to obtain au u 12kruk2 12 kuk2 0 0 kuk2 1 where

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Unformatted text preview: 1 where = (1=2) max(1 ). Thus, (7.2.2) is satis ed with s = 1 and A(u v) is coercive (H 1-elliptic). Continuity and coercivity of the strain energy reveal the nite element solution U to be nearly the best approximation in S N Theorem 7.2.1. Let A(v u) be symmetric, continuous, and coercive. Let u 2 H01 satisfy N (7.1.1a) and U 2 S0 H01 satisfy (7.1.2). Then ku ; U k1 ku ; V k1 N 8V 2 S0 (7.2.4a) with and satisfying (7.2.1) and (7.2.2). Remark 1. Equation (7.2.4a) may also be expressed as ku ; U k1 C infN ku ; V k1: V 2S0 Thus, continuity and H 1-ellipticity give us a bound of the form (7.1.5). Proof. cf. Problem 2 at the end of this section. The bound (7.2.4) can be improved when A(v u) has the form (7.1.1c). (7.2.4b) 6 Analysis of the Finite Element Method Theorem 7.2.2. let A(v u) be a symmetric, continuous, and coercive bilinear form u 2 H01 minimize I w] = A(w w) ; 2(w f ) 1 8w 2 H0 (7.2.5) N 1 and S0 be a nite-dimensional subspace of H0 . Then N 1. The minimum of I W ] and A(u ; W u ; W ), 8W 2 S0 , are achieve...
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