1 as stated the gmres procedure does not calculate

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Unformatted text preview: computational requirements when M becomes too large. One remedy is to restart the orthogonalization after a xed number (M ) steps using the last computed solution x(M ) as an initial guess. This procedure tends to converge slowly when A is not positive de nite. 4. A second technique for restricting the size of M is to truncate the Arnoldi process. With this approach, only the last k vectors are made orthogonal. The rst M ; k vectors can be discarded and need not be stored. The only remaining detail is the solution of the least-squares problem min k e1 ; HM ck2: c Since H is a Hessenberg matrix, it seems reasonable to transform it to upper triangular form. We'll do this by using plane rotations (Givens matrices) of the form 21 3 6 ... 7 6 7 6 7 1 6 7 6 7 ci si 6 7 row i (9.4.16a) 7 i=6 ;si ci 6 7 row i + 1 6 7 1 6 7 6 ... 7 4 5 1 with c2 + s2 = 1: i i (9.4.16b) 9.4. Krylov Subspace Methods 65 We multiply e1 ; HM c on the left by a sequence of rotations i with the coe cients ci and si designed to eliminate hi+1 i , i = 1 2 : : : M . If M were ve, we would have 2 3 23 h11 h12 h13 h14 h15 6 h21 h22 h23 h24 h25 7 607 6 7 67 6 67 h32 h33 h34 h35 7 7 HM = 6 g = 6 0 7: 6 607 h43 h44 h45 7 6 7 67 4 5 405 h54 h55 h65 0 We multiply HM and g by 2 3 c1 s1 6 ;s1 c1 7 6 7 6 7 1 1=6 7 4 15 1 with c1 = p 2h11 2 s1 = p 2h21 2 h11 + h21 h11 + h21 to obtain 2 (1) (1) (1) (1) (1) 3 2 3 h11 h12 h13 h14 h15 c1 6 6 ;s1 7 h(1) h(1) h(1) h(1) 7 6 6 7 22 23 24 25 7 (1) 6 h32 h33 h34 h35 7 (1) = 6 0 7 : 6 7 6 HM = 6 g 607 7 h43 h44 h45 7 6 7 6 4 407 5 h54 h55 5 0 h65 The process continues with the parameters for rotation i satisfying h(iii;1) ci = q (i;1) si = q (i;hi+1 i : (9.4.16c) (hii )2 + h2+1 i (hii 1) )2 + h2+1 i i i At the end of M (= 5) rotations, we have 2 (5) (5) (5) (5) (5) 3 h11 h12 h13 h14 h15 23 6 h(5) h(5) h(5) h(5) 7 1 6 22 23 24 25 7 6 6 27 (5) (5) (5) 7 h33 h34 h35 7 H(5) = 6 g(5) = 6 .. 7 : 6 67 M (5) h(5) 7 6 7 4.5 h44 45 7 6 (5) 5 4 6 h55 0 In the general case, let QM = M M ;1 : : : 1 (9.4.17a) 66 Solution Techniques for Elliptic Problems and M RM = H(M ) = QM HM Since the matrices i, (M gM = gM ) = QM e1: (9.4.17b) i = 1 2 : : : M , and QM are orthogonal, min k e1 ; HM ck2 = min kgM ; RM ck2: c c The solution of this problem is obtained by deleting the last row of RM and gM and solving the...
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