教程案例成果

ARnn

AxbRn

xRn

xBx

（4-BRnn,fRn,xRn陣 Axb

x(0)(11

x)x(2)B)x(3) Bx(2)得到向量序列 x(1),x(2),x(3),…..x(k),.x(k1)Bx(k) (k0,1,2,

（4-如果對任意x(0)，

k時(shí)x(k)x*x(k)(x(k),x(k),,x(k))T

limx(k1)k

Blimx(k)k

x*x*Bx*三種基本的迭代一雅可比(Jacobi)迭代法。a11x1a12x2a1nxna21x1a22x2a2nxnan1x1an2x2annxnA(aij)nn非奇異，且aii0(i12,n)x1

1b x x 11 x 1bax ax

1b x

（4-nannn

x(kxi

bi

njnj

(kaa

(i1,2,, （4-x(kxi

x(k

bi

nnj

(kaa

（4-i(i1,2,,i 8x3x 20x1

2x

4x111x2x333 14xx

x12xx n nx(k1)

1b

ax(k)

(i1,2,,

（4-a aii

jj

x(k

1(20

3x(k

2x(k) 3x2

2x3

x(k

()

x(k) 2x1x24x3x(k

1(122x(k)x(k x(0) x(k1)1b ax(k)4x(1)4

0

x

ii

j 3j xx1

8，

x2x

33，

x3x

1241；1xx1

12033232.875

…，x(10)x(2)x2

13348

…，x(10)xx3

11411224

3

2…，x(10)233

(k1)x(k

321T(k

(k

j1j

(i1,2,,

a

x(k1)

1n

a11

11

(4- a22

a22

a22

an

(k)

ann

x(k1)Bx(k)

(4-

a1n

a2n A

ain

nn (4-Ddiag(a11,a22,,ann 21 L

jj

a1n

U

j1n

n1n Axb(DLU)xbDx(LU)xbDx(LU)xx D1(LU

xBJxf

Jx(k1)J

x(k

fJ

k

(4-

a1n

LU

ann

1na21

a2n

a21

a2n

22

1

ann

b1

ba11

a a

1f

b2 22

1 b ba a nn

ann

n二、高斯-賽德爾（Gauss-Seidel)迭代

x2x

時(shí)x 11

xk1,xk1,,xk

(kxix

xix些。故對1x(k1)1

1(203x(k23823

2x(k)xx

1(33

313

x(k)x(k1)

(k

(k (12343

x(k1)x

1(203x(k282

2x(k)xx

1(334x(k11

x(k)3x(k1)1(122x(k1)x(k 3 1x(1)1

x(0)0,0,0)T2 3 .523 30.52.50.25 31x(5)2.99984312x(5)2.00007223x(5)1.0000613

x(k

x(k

(k 1

(k

(k)a aii

aijxj

aijxj k0,1,2,;i1,2,,

（4-x(k1)

1b

nax(k)n

(i12, a aii

ji

a(ka

(k j

a a ax(kj

i1,2,, (k1)

x(k1)11

(k1)

x(k1)

2 2

(k x(k1)

nn

x(k) b 12

1n

1

x(k)

b2+ 2n +

0 (k) 0 nDx(k1)- 令x(k1)(DL)1Ux(k)(DL)1令 (DL)1 (DL)1 x(k1)Bx(k) （k 10

(k13x(k2(k3)x(k (5(k13x(k3)2x(kx(k31(k2)(k1(k2(k3)x(k2(5(k1(k3)x(k(k(k3x12)-代高格斯-代高格斯式

取初值x 方程組的近似解9)))

取初值x 方程組的近似解5)7)8)三.超松弛(SOR) ~(k

1

(k

(k

iaiii

aijxj

aijxj

(kxix

(ki

(1)x(k(k

(k

na(ka

(k (1)xi

aijx

aijx i

ii j

j 稱(4-14)為超松弛迭代法，簡稱為SOR迭代法，

x(k1)Bx(k) (4-

fDL三、迭代法的收斂性Ax

x(k1)Bx(k) 設(shè)

是(1)的解x(k

k若有l(wèi)imx(k kk注意.limx(k)x*limx(k)x*.k kx(k)

為任意向量范

x*Bx*fx(k1

x*Bx(k)BxB(x(k)x*) Bk1x(0)x*令ε(k)x(k) ，ε(k1)Bε(k)B2ε(k1)Bklimε(k

0

(k

limBk1ε(0)

k

k

k ε(0) x(0)

limBk1k

引

BRnn limBk

(B)1k證明:略

存在非奇異矩陣P使得B=P-1JP,Bk=P-1JkJ J

. . J

i

1Jir Jir s 迭代法收斂性的判定條定理4.1(充要條件 迭代x(k1)Bx(k)

x(0) .

(B)1證明

若||B||

||x(k)x*

||B||

||x(k

x(k1)

(4-||x(k

x*

||B||k1||B.

||

x(0)

(4-當(dāng)||B||1時(shí),方陣I-B非奇異，方程組(I-B)x=f設(shè)解為x*， x*-x(m)=B(x*-x(m-1)

||x*x(m)

B(x*

x(m1) x*

B

x*x(m2)

B

x*x(0)||x*x(0)||為常數(shù)，||B||<1,||x*-x(m)||→0,

m

B(m) Bx(k)x(k B

x(k1)x(k p

x(k

x*(x(k)x*)p

x(k

x(kp

p

)x(kp

p例 x12x22x3

2x

A 121 2 2x2x321 2 2 D1(LU)

10 0 detIBJ

3 (B)0 detI得

detI(DL)1Udet(DL)1D-LU

3424(2)2 進(jìn)一步得(BG)2

A(aij

n|aii||aijnjj

(i

等式成立，稱矩陣A為嚴(yán)格對角占優(yōu)陣 A為非奇異矩陣，det(A) 2

0

1

1 6 6

04 04 定理4.3（充分性條件）Ax中的A為嚴(yán)格對角占優(yōu)陣則Jacobi證（1）Jacobi0 0a

a1na

nn a2n D1(LU)

a22

|a|

(i1,2,,|aii

j A為嚴(yán)格對角占優(yōu)陣，|aii||aijj

(i1,2,,

所 1|a (i1,2,,|aii 即||

||

j

(D 的特征值 detI(DL)1Udet(DL)1det[(DL)U] det(DL)1 C(DL)U

anndet(C)det(DL)U （4-現(xiàn)在證明|| .用反證法，假設(shè)||1，又由|||aii||||aij|j j