# 32a and a u u then ku u k1 ku k2 1 n 8u 2 s0 a c

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Unformatted text preview: the Finite Element Method Proof. Using the triangular inequality ku ; U k1 = ku ; V + V ; U k1 ku ; V k1 + kW k1 (7.3.4a) where W = U ; V: (7.3.4b) Using (7.3.2b) and (7.3.4b) kW k2 A (U ; V W ) = A (U W ) ; A (V W ): 1 Using (7.3.1a) with V replaced by W to eliminate A (U W ), we get kW k2 (f W ) ; A (V W ): 1 Adding the exact Galerkin equation (7.2.8) with v replaced by W kW k2 (f W ) ; (f W ) + A(u W ) ; A (V W ): 1 Adding and subtracting A(V W ) and taking an absolute value kW k2 j(f W ) ; (f W )j + jA(u ; V W )j + jA(V W ) ; A (V W )j: 1 Now, using the continuity condition (7.3.2a) with u replaced by u ; V and v replaced by W , we obtain kW k2 j(f W ) ; (f W )j + ku ; V k1 kW k1 + jA(V W ) ; A (V W )j: 1 Dividing by kW k1 1 f ku ; V k + j(f W ) ; (f W )j + jA(V W ) ; A (V W )j g: kW k1 1 kW k1 kW k1 Combining the above inequality with (7.3.4a), maximizing the inner product ratios over W , and choosing C as the larger of 1 + = or 1= yields (7.3.3). N Remark 1. Since the error estimate (7.3.3) is valid for all V 2 S0 it can be written in the...
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