Civil Engineering Reference
In-Depth Information
Since the convergence only depends on the spectral radius of the iteration matrix,
without loss of generality we can assume that
b
=
0. Moreover, observe that
x
k
+
1
=
x
k
since
Ax
k
0.
(1) Let
A
be positive definite. For every
x
0
−
b
=
in the unit sphere
S
n
−
1
n
:
={
x
∈ R
;
x
=
1
}
,
(1.17) implies
x
1
Ax
1
x
0
Ax
0
<
1
.
In view of the continuity of the mapping
x
0
→
x
1
and the compactness of the
sphere
S
n
−
1
, there exists
β<
1 with
x
1
Ax
1
x
0
Ax
0
≤
β<
1
,
(
1
.
18
)
∈
S
n
−
1
. Since the quotient on the left-hand side of (1.18) does not
change if
x
0
for all
x
0
0, (1.18) holds for all
x
0
is multiplied by a factor
=
=
0. It follows
by induction that
x
k
Ax
k
≤
β
k
x
0
Ax
0
,
and thus
x
k
Ax
k
0. By the definiteness of
A
,wehavelim
k
→∞
x
k
0.
(2) Let
A
be indefinite. Without loss of generality, consider the iteration
(1.9) for the homogeneous equation with
b
→
=
0. Then there exists
x
0
=
=
0 with
=
f(x
0
)<
0. Now (1.17) implies
f(x
k
)
≤
f(x
k
−
1
)
and
f(x
k
)
α
:
≤
α<
0
,k
=
0
,
1
...,
and we conclude that
x
k
→
0.
In carrying out the relaxation methods (1.9) and (1.14), the components of the
vectors are recomputed in the order from
i
=
1to
i
=
n
. Obviously, we can also
run through the indices in the reverse order. In this case, the roles of the submatrices
L
and
U
are reversed. In the
symmetric
SOR
method
, SSOR method for short, the
iteration is performed alternately in the forward and backward directions. Thus,
each iteration step consists of two half steps:
Dx
k
+
1
/
2
=
ω
[
Lx
k
+
1
/
2
+
Ux
k
1
)Dx
k
,
+
b
]
−
(ω
−
(
1
.
19
)
Dx
k
+
1
=
ω
[
Lx
k
+
1
/
2
+
Ux
k
+
1
1
)Dx
k
+
1
/
2
.
+
b
]
−
(ω
−
By the remark following (1.16), we conclude that the assertion of the theorem also
holds for the symmetric iteration process.
The SSOR method has two advantages. The computational effort in perform-
ing
k
SSOR cycles is not the same as that for 2
k
cycles of the SOR method, but
only to
k
+
1
/
2 cycles. Moreover, the associated iteration matrix
M
−
1
−
ω)(D
−
ωU)
−
1
D(D
−
ωL)
−
1
=
ω(
2
(
1
.
20
)
in the sense of (1.1) is symmetric; cf. Problem 1.10.
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