Unformatted text preview: ven by the values of where
f ( !) = g( ) (cf. (9.2.22)). With = (MJ ), these eigenvalues are shown
as points A and B on Figure 9.2.4. The larger value of at the points labeled A
corresponds to (M! ) (Item 2). 22 Solution Techniques for Elliptic Problems
g(λ,ρ ( MJ )) 1 ∼
f (λ,ω) A A A C
f (λ,1) f (λ,ω) f (λ,0) B λ
1 B B Figure 9.2.4: Functions f ( !) and g( (MJ )) vs. . 4. Setting ! = 0 gives f ( 0) as the line = 1. If ! < 0, then the values of at
the intersection points (not shown in Figure 9.2.4) would exceed unity, and the
iteration (9.2.21) would diverge.
5. Setting ! = 1 in (9.2.23) gives f ( 1) = , which is the Gauss-Seidel method.
6. As seen in Figure 9.2.4, the maximum eigenvalue of M! can be reduced further
by choosing ! > 1. The minimum real solution for occurs at ! = ! when f ( !)
is tangent (at point C ) to g( (MJ )).
Let us rewrite (9.2.22) as ;! 1=2 + ! ; 1 = 0: 9.2. Basic Iterative Solution Methods 23 This is a quadratic equation in 1=2 thus,
= ! =2
(! =2)2 + 1 ; ! : (9.2.24a) The point of tangency (C ) occurs at ! when
(~ =2)2 + 1 ; ! = 0
1, we see that 1 !+ 2 and 2 !; < 1. We'll show in a moment that
the interesting range of ! is 0 2] thus, the appropriate value of ! is !+. We'll drop the
+ subscript and simply let !=
1+ 1; 2 (9.2.24b) Substituting (9.2.24b) into (9.2.24a) yields
~ = (!) = (~ =2)2 = ! ; 1:
~ (9.2.24c) Values of ! < ! produce real values of while values of ! > ! yield complex values
of . In this latter case, we may use (9.2.24a) to show that j j=!;1 ! > !:
~ We're now able to state our main result. Theorem 9.2.6. Let A be a symmetric and consistently-ordered matrix, then the optimal
relaxation parameter satis es !OPT =
and (M! ) = (h p p22
1 + 1 ; (MJ ) ! (MJ )=2 + (! (MJ )=2)2 + 1 ; !
!;1 i2 (9.2.25a) if 0 ! !OPT : (9.2.25b)
if !OPT < ! 24 Solution Techniques for Elliptic Problems
ρ( Mω )
1 ρ( M J ) = 0.95 0.9 ω
1 ω OPT 2 Figure 9.2.5: Spec...
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