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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 leastsquares 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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 Spring '14
 JosephE.Flaherty

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