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

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: 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...
View Full Document

Ask a homework question - tutors are online