Unformatted text preview: tral radius (M! ) as a function of ! for SOR iteration.
Proof. With the various symmetries (cf. Strikwerda 5], Section 13.3), it su ces to
consider positive values of . With this choice and our interest in calculating the spectral
radius of M! , we clearly want the positive sign in (9.2.24a). Thus, in summary, we have 1. = 1 when ! = 0,
2 + 1 ; ! , 0 < ! ! , and
2. = ! =2 + (! =2)
3. j j = ! ; 1, ! < !.
The values of increase with thus, the spectral radius (M! ) occurs when =
(MJ ), as given by (9.2.25b). The minimum value of (M! ) occurs at ! corresponding to
= (MJ ), as given by (9.2.25a). The spectral radius (M! ) is displayed as a function
of ! in Figure 9.2.5.
Several corollaries follow from Theorem 9.2.6 and we'll list one. Corollary 9.2.1. Under the conditions of Theorem 9.2.6, the SOR method converges
for ! 2 (0 2) when (MJ ) < 1. 9.2. Basic Iterative Solution Methods 25 Proof. From (9.2.25) (M!OP T ) = !OPT ; 1 = p22
1 + 1 ; (MJ ) Thus, 0 < (M!OP T ) < 1 when 0 < (MJ ) < 1.
Additional inspection of (9.2.25b) reveals that d (M! )=d! is non-positive when ! <
!OPT and unity when ! > !OPT (Figure 9.2.5). Since (M! ) = 1 at ! = 0 2, we conclude
that (M! ) < 1 for 0 < ! < 2.
An examination of (9.2.25) and Figure 9.2.5 reveals that !OPT 2 (1 2). Increasing
(MJ ) increases !OPT and, in particular, !OPT ! 2 as (MJ ) ! 1. From Figure
9.2.5, we also see that overestimating !OPT by a given amount increases (M! ) less than
underestimating it by the same amount.
Example 9.2.8. Let us solve the problem of Example 9.2.4 using SOR iteration. With
J = K = 3 and x = y = 1=4, we obtain the Gauss-Seidel method (9.2.16) as
Ujk +1) = 1 (Uj(+1 k + Uj(;+1) + Uj( k)+1 + Uj( k+1) ):
4 With relaxation, we compute
Ujk +1) = !Ujk +1) + (1 ; !)Ujk ) : We'll explicitly eliminate the intermediate variable to obtain the SOR method as
Ujk +1) = ! (Uj(+1 k + Uj(;+1) + Uj( k)+1 + Uj( k+1) ) + (1 ; !)Ujk ) :
4 Using (9.2...
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