(2) can be written as Dx(i+1

師大 Successive Over Relaxation (SOR) Method Symmetric Successive Over Relaxation (SSOR) Method

The single-step method (SSM)

(D + L)x⁽ⁱ⁺¹⁾= −U x⁽ⁱ⁾+ b can be written in the form

x⁽ⁱ⁺¹⁾ = x⁽ⁱ⁾+{−D⁻¹Lx⁽ⁱ⁺¹⁾−D⁻¹U x⁽ⁱ⁾+D⁻¹b−x⁽ⁱ⁾} := x⁽ⁱ⁾+ν⁽ⁱ⁾. (1) Consider a general form of (1)

x⁽ⁱ⁺¹⁾= x⁽ⁱ⁾+ ων⁽ⁱ⁾ (2)

with constant ω. (2) can be written as

Dx⁽ⁱ⁺¹⁾ = Dx⁽ⁱ⁾− ωLx⁽ⁱ⁺¹⁾− ωU x⁽ⁱ⁾+ ωb − ωDx⁽ⁱ⁾. Then

x⁽ⁱ⁺¹⁾ = (D + ωL)⁻¹[(1 − ω)D − ωU ] x⁽ⁱ⁾+ ω(D + ωL)⁻¹b.

(3) Hence the iteration matrix

Tω = (D + ωL)⁻¹[(1 − ω)D − ωU ] .

師大 Successive Over Relaxation (SOR) Method Symmetric Successive Over Relaxation (SSOR) Method

These methods is called for

ω < 1: under relaxation, ω = 1: single-step method,

ω > 1: over relaxation. (In general: relaxation methods.)

師大 Successive Over Relaxation (SOR) Method Symmetric Successive Over Relaxation (SSOR) Method

Let A be symmetric and A = D + L + L^T. The idea is in fact to implement the SOR formulationtwice,one forwardandone backward, at each iteration. That is, SSOR method defines

(D + ωL)x^(k−12) = (1 − ω)D − ωL^T x^(k−1)+ ωb (4) (D + ωL^T)x^(k) = [(1 − ω)D − ωL] x^(k−12)+ ωb. (5) Define

 Mω: = D + ωL,

Nω: = (1 − ω)D − ωL^T. Then from the iterations (4) and (5), it follows that

x_i+1 = M_ω^−TN_ω^TM_ω⁻¹N_ω
x_i+ ω M_ω^−TN_ω^TM_ω⁻¹+ M_ω^−T
b

≡ T (ω)x_i+ M (ω)⁻¹b.

師大 Successive Over Relaxation (SOR) Method Symmetric Successive Over Relaxation (SSOR) Method

But

((1 − ω)D − ωL) (D + ωL)⁻¹+ I

= (−ωL − D − ωD + 2D)(D + ωL)⁻¹+ I

= −I + (2 − ω)D(D + ωL)⁻¹+ I

= (2 − ω)D(D + ωL)⁻¹, Thus

M (ω)⁻¹ = ω D + ωL^T
−1

(2 − ω)D(D + ωL)⁻¹, then the splitting matrix is

M (ω) = 1

ω(2 − ω)(D + ωL)D⁻¹ D + ωL^T
. The iteration matrix is

T (ω) = (D + ωL^T)⁻¹[(1 − ω)D − ωL] (D + ωL)⁻¹(1 − ω)D − ωL^T .