Forthelinearequation system:x=Bx+fwe adoptthefollowing steps:x(0) into the right side of thetaking the initial vector xequations,thengettingx(1) = Bx(0) + fand soon, we obtainx(2) = Bx(1) + fx(k+1) = Bx(k) + f (k = 0,1,2,..)This method called the iteration method, the above process isknownas theiterativeprocess.The iterative method produces a sequence: [x(*)]°.is, limx(k) = x, then the iterativeIf its limit exist, that is,kvomethod is convergent, otherwise, divergent
For the linear equation system: x = Bx + f This method called the iteration method, the above process is known as the iterative process. The iterative method produces a sequence: . 0 ( ) { } k x If its limit exist, that is, , then the iterative method is convergent, otherwise, divergent. ( ) * limx x k k = → (1) (0) x Bx f = + and so on, we obtain we adopt the following steps: x = Bx + f (2) (1) ( 1) ( ) k k x Bx f + = + (k = 0,1,2, ) taking the initial vector into the right side of the equations, then getting (0) x
If the iterative sequence ( x(*) is convergent: lim x(k) = x,k->o0then takinglimiton both sides oftheiterativeformatx(k+1) =Bx(k) +f ,we can get: x* = Bx* + fIt is equivalent to Ax" = b. Thus the vector sequence ( x()} isconvergent to the solution vector x *of the equations Ax =b.The calculation accuracy of this method is controllable.especially suitable for solving the equations system withlarge sparse coefficient matrix.Research Howto build theiterativeformat?eontents The convergence condition of the vector sequence?The convergence speed? The error estimation?
The calculation accuracy of this method is controllable, especially suitable for solving the equations system with large sparse coefficient matrix. Research contents How to build the iterative format? The convergence condition of the vector sequence? The convergence speed? The error estimation? . * * x B x f = + then taking limit on both sides of the iterative format ( 1) ( ) k k x B x f + = + If the iterative sequence is convergent: ( ) { } k x * It is equivalent to A x b = convergent to the solution vector of the equations . x* A x b = . Thus the vector sequence is ,we can get: ( ) { } k x ( ) * lim , k k x x → =
s2 The basic iterative methods1. The Jacobi MethodThe general form of the linear eguations system is:(a11X +a12X2 +...+ainXn =b)a21X1 +a22X2 +...+ a2nXn = b2aniX +an2X2 +. ±anXn = b,Setting au, ±O (i=l,2,...,n) , we can solve x, according tothe above formula:1[b - (a12x2 +... + ainxn)]X11an1[b2 -(a21x1 +a23X3 +... + a2nxn)]X2a22
1. The Jacobi Method §2 The basic iterative methods [ ( )] 1 1 1 2 2 1 1 1 1 n n b a x a x a x = − ++ The general form of the linear equations system is: 11 1 12 2 1 b1 a x a x a x + ++ n n = 21 1 22 2 2 b2 a x a x a x + ++ n n = n n nn n bn a 1 x1 + a 2 x2 ++ a x = [ ( )] 1 2 2 1 1 2 3 3 2 2 2 2 n n b a x a x a x a x = − + ++ 0 ( 1,2, , ) ii Setting a i n = , we can solve according to i x the above formula:
And so on,the linear eguations can be turned into:=-二(b-Zar,x) =x +二(b-Zar,x)?x.alal层1j=11(b-21La2x=x2+(b2X2a2,x,)a22a22厦j=1一(b, -Cax,=x,+=(b,-Zajx,)二x,aiaii万j=1之1(bnCamx)=x, +1(b,-Zamx,)x.-nannj=1Qj=1nnjtn
( ) 1 1 1 1 1 1 1 1 = = − n j j j j b a x a x And so on, the linear equations can be turned into: ( ) 1 2 1 2 2 2 2 2 = = − n j j j j b a x a x ( ) 1 1 = = − n j i j i ij j ii i b a x a x ( ) 1 1 = = − n j n j n nj j n n n b a x a x ( ) 1 1 1 1 11 1 = = + − n j j j b a x a x ( ) 1 1 2 2 22 2 = = + − n j j j b a x a x ( ) 1 1 = = + − n j i ij j ii i b a x a x ( ) 1 1 = = + − n j n nj j nn n b a x a x
Conducting iterative process of the equations:x(k+1) = x(k) +1(b, -Zayx()aiij=1(i=1,2,...,n;k =0,1,2,..)Setting D=diag(ai1,a22,",an)then the above formula is transformedinto the matrix form:x(k+1) = x(k) + D-1(b - Ax(k))x(+1) = x(k) - D-1 Ax(k) + D-1b1x(k+1) = D-1(D - A)x(k) + D-1b
( ) 1 1 ( 1) ( ) ( ) = + = + − n j k i i j j i i k i k i b a x a x x Conducting iterative process of the equations: (i = 1,2, ,n;k = 0,1,2, ) then the above formula is transformed into the matrix form: ( ) (k 1) (k ) 1 (k ) x = x + D b − Ax + − x x D Ax D b (k +1) (k ) −1 (k ) −1 = − + x D D A x D b (k 1) 1 (k ) 1 ( ) + − − = − + 11 22 ( , , , ) D diag a a a = nn Setting