# 35 pcg algorithm 2 rather than compute m1 it is more e

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: +1) uses the same manipulations. 4. Storage is needed for A, M (possibly in factored form), x( ) , p( ) , r( ), z( ) , and Ap( ) . Relative to the conjugate gradient procedure, additional storage is needed for M and z( ) . Storage costs are still modest relative to those of direct methods. 5. In addition to the matrix multiplication (Ap( ) ) and the two inner products per step required by the conjugate gradient method, a linear equations solution (Mz( +1) = r( +1) ) is required 44 Solution Techniques for Elliptic Problems 6. Using (9.3.13) for the conjugate gradient method ~~ ~ (p( ) )T Ap( ) = 0 (~( ) )T ~( ) = 0 rr 6= : Using (9.3.18a, 9.3.18b), we obtain the \orthogonality conditions" (r( ) )T M;1 r( ) = 0 (p( ) )T Ap( ) = 0 6= (9.3.20) 7. When M is positive de nite (r( ) )T z( ) = (r( ))T M;1 r( ) > 0: Thus, values of can always be obtained and the procedure does not fail. Let us select some preconditionings, beginning with some choices based on iterative strategies. It will be convenient to write the basic xed-point strategy (9.2.2) in the form ^ ^ Mx( +1) = Nx( ) + b (9.3.21a) ^^ A=M;N (9.3.21b) where Comparing (9.3.21) with (9.2.10b), (9.2.15b), and (9.2.21d), we have Jacobi iteration: ^ MJ = D ^ NJ = L + U (9.3.22) Gauss-Seidel iteration: ^ MGS = D ; L ^ NGS = U (9.3.23) SOR Iteration: 1 ^ M! = ! (D ; !L) ^ N! = 1 ; ! D + U: ! (9.3.24) 9.3. Conjugate Gradient Methods 45 Recall that D is the diagonal part, L is the negative of the lower traingular part, and U is the negative of the upper triangular part of A (cf. (9.2.8)). Let us also include symmetric successive over relaxation (SSOR) in our study. As discussed in Section 9.2, SSOR takes two SOR sweeps with the unknowns placed in reverse order on the second sweep. Using (9.3.24) and (9.3.21), the rst step of the SOR procedure is (D ; !L)x( +1=2) = (1 ; !)D + !U]x( ) + !b: (9.3.25a) Reversing the sweep direction on the second step yields (D ; !U)x( +1) = (1 ; !)D + !L]x( +1=2) + !b: (9.3.25b) The intermediate solution x( +1=2) can be eliminated to obtain a scheme...
View Full Document

## This document was uploaded on 03/16/2014 for the course CSCI 6840 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online