数值分析9.ppt

?解线性方程组的迭代法 ( Iterative Methods ) ?雅可比迭代法?赛德尔迭代?迭代法的矩阵表示《数值分析》9 ?????????????yyu xuxuyu yxuu yy xx? sin ),1( 0)1,()0,(),0( 1,0,0 ( i, j = 1, ···, n ) 0 4 1,,1 1,,1?????????jiji ijjijiuuuuu记 X= [u 11 ···u 1,nu 21 ···u 2,n ······ u n ,1 ···u n,n] T AX = b 0 5 10 15 0 5 10 15 nz = 64 0 5 10 15 0 5 10 15 nz = 118 A=LU ?( L + U – I ) 边值问题: 2/18 ????????????????13 15 810 7 9 321 321 321xxx xxx :系数矩阵主对角元均不为零?????????????????????????????????????????? 15 / 13 10 /8 9/70 15 /1 15 /1 10 /10 10 /1 9/19/10 )0(3 )0(2 )0(1 )1(3 )1(2 )1(1x x x x x x计算格式 X (1)= B X (0) + f ??????????????15 /)13 ( 10 /)8( 9/)7( 213 312 321xxx xxx xxx取X (0) = ??????????0 0 0 4/18 X * X (0) 0 0 0 X (1) X (2) X (3) 计算格式: X (k+1) = BX (k)+f准确解 X (4) ········ 5/18 || X (4) –X (3) || 1 = , 005 .0 || || || || 1 )4( 1 )3()4(??X XX反例(不收敛) ????????????????13 15 7 9 810 321 321 321xxx xxx xxx - - - - - - ······ ······ ······ - - 6/18 ??????????????????? nn nn nn nn nnbxaxaxa bxaxaxa bxaxaxa??????????? 2211 22222 121 11212 1 11雅可比迭代法(i = 1,2, …n ; k =1,2, ……) 取初始向量 X (0) =[x 1 (0) x 2 (0) ···x n (0)] T, 迭代计算?????????? nij kj ij ij kj iji ii kixaxaba x 1 )( 11 )( )1( ] [ 1 i nj j ijbxa???1(i = 1,2, …,n) 7/18 分裂: AX = b 任取x (0), 迭代计算产生向量序列:若* lim )(XX kk???则X*是方程组 AX = b 的解迭代矩阵迭代计算格式( k =0,1,2, ······ ) X (k+1) = BX (k) + fX (1),X (2), ···,X (k), ······ splitting D + ( A – D ) X = BX + f DX = ( D – A ) X + b ? 8/18 迭代法适用于解大型稀疏方程组(万阶以上的方程组,系数矩阵中零元素占很大比例,而非零元按某种模式分布) 背景: 电路分析、边值问题的数值解和数学物理方程问题: (1) 如何构造迭代格式? (2) 迭代格式是否收敛? (3) 收敛速度如何? (4) 如何进行误差估计? 9/18 i nj j ijbxa???1(i = 1,2, …,n)高斯-赛德尔迭代法??????????? nij kj ij ij kj iji ii kixaxaba x 1 )( 11 )1( )1(] [ 1(i = 1,2, …n ; k =1,2, ……) 取初始向量 x (0) =[x 1 (0) x 2 (0) ···x n (0)] T, 迭代计算 10/18