數(shù)值分析11對(duì)稱正定矩陣的收斂性完整

1、1邊值問(wèn)題數(shù)值解算例邊值問(wèn)題數(shù)值解算例對(duì)稱正定矩陣的收斂性對(duì)稱正定矩陣的收斂性超松馳迭代算法超松馳迭代算法分塊矩陣的塊迭代分塊矩陣的塊迭代數(shù)值分析 112 . 0)1(, 0)0()1 , 0(0yyxxyy例例1 1 常微分方程邊值問(wèn)題常微分方程邊值問(wèn)題 在在x1=0.1, x2=0.2,x9=0.9 處的數(shù)值解處的數(shù)值解解解: 令令 h = 0.1, 記記 yj=y(xj) ( j = 1,2,9),將將)9 , 2 , 1(2)(211 jhyyyxyjjjj代入微分方程代入微分方程,整理得整理得 yj-1 + (2 h2) yj yj+1 = xj h2 ( j= 1,2,9) 2/1

2、83yj = xj h2 + yj-1 + yj+1/ (2 h2) , ( j= 1,2,9) 2922219212222112112hxhxhxyyyhhh yj-1 + (2 h2) yj yj+1 = xj h2 高斯高斯-賽德?tīng)柕袷劫惖聽(tīng)柕袷?212)(1)1(12)1(hxyyhyjkjkjkj 3/184程序程序h=0.1;x=0:h:1;y=zeros(size(x);r1=h*h;r=2-r1;er=1;k=0;while e0.0001 er=0; for j=2:10 s=(y(j-1)+y(j+1) + r1*x(j)/r; er=max(er,abs(s-y(

3、j);y(j)=s; end k=k+1end00.10.20.30.40.50.60.70.80.9100.010.020.030.040.050.060.070.08準(zhǔn)確解準(zhǔn)確解: y(x)=sin x/sin 1 - x- y(x)o yj4/185正方形區(qū)域上第一邊值問(wèn)題正方形區(qū)域上第一邊值問(wèn)題 yyuxuxuyuyxuuyyxx sin), 1(0)1 ,()0 ,(), 0(1,0, 0yshxshyxu sin),( 準(zhǔn)確解準(zhǔn)確解: :O1x1y5/186取正整數(shù)取正整數(shù)n, ,記記 對(duì)區(qū)域做網(wǎng)格剖分對(duì)區(qū)域做網(wǎng)格剖分: : 11 nhxi = i h ( i = 0，1，n+1

4、)yj = j h ( j = 0，1，n+1)在點(diǎn)在點(diǎn)(xi，yj ) 處記處記 uij = u(xi ,yj) ,五點(diǎn)差分格式五點(diǎn)差分格式02221,1,2, 1, 1 huuuhuuujiijjijiijji)(411,1, 1, 1 jijijijij iuuuuu整理整理)(41)(1,)1(1,)(, 1)1(, 1)1(,kjikjikjikjikjiuuuuu 6/187邊界條件邊界條件: u0, j = 0 (j = 1, , n); un, j = ( j = 1, , n); ui,0 = 0 ( i = 1, , n); ui,n = 0 ( i = 1, , n) j

5、hsin結(jié)點(diǎn)數(shù)結(jié)點(diǎn)數(shù)n2 102 202 402迭代次數(shù)迭代次數(shù) 91 303 978CPU時(shí)間時(shí)間(s) 0.14 1.843 24.6720誤差誤差 0.0028 5.6995e-4 6.6671e-4高斯高斯- -賽德?tīng)柕▽?shí)驗(yàn)賽德?tīng)柕▽?shí)驗(yàn):7/188定理定理4.4 方程組方程組 Ax=b中中, 若若A是實(shí)對(duì)稱正定矩是實(shí)對(duì)稱正定矩陣陣,則則Gauss-Seidel迭法收斂迭法收斂證明證明: 由由 A = D L LT BG-S = (D L)-1LT設(shè)設(shè)為為BG-S的任一特征值的任一特征值, x為其特征向量為其特征向量,則則(D L)-1LT x =x LT x =(D L)x A

6、正定正定,故故 p = xTDx0, 記記 xTLTx = a , 則有則有xTLTx =xT(D L)xxTAx=xT(D L LT)x=p a a =p 2a 08/1891)2(2222222 aappaapapa 所以所以, 迭代矩陣迭代矩陣BG-S的譜半徑的譜半徑(BG-S) 1,從而從而當(dāng)方程組當(dāng)方程組 Ax=b的系數(shù)矩陣的系數(shù)矩陣A 是實(shí)對(duì)稱正定矩是實(shí)對(duì)稱正定矩陣時(shí)陣時(shí),Gauss-Seidel迭代法收斂迭代法收斂apaxLDxxLxTTT )( 稱稱 R= ln(B) 為迭代法的漸近收斂速度為迭代法的漸近收斂速度9/1810)()1()1()(1)1()(bUxLxDxxkkk

7、k )1(1)1(11)()1()( nijkjijijkjijiiikikixaxabaxx (i=1,2, n; k = 1,2,3, )超松馳超松馳(SOR)迭代法迭代法Gauss-Seidel迭代格式迭代格式 nijkjijijkjijiiikixaxabax1)(11)1()1(110/1811定理定理4.8 若若A是對(duì)稱正定矩陣是對(duì)稱正定矩陣,則當(dāng)則當(dāng)02時(shí)時(shí)SOR迭代法解方程組迭代法解方程組 Ax = b 是收斂的是收斂的定理定理4.9 若若A是嚴(yán)格對(duì)角占優(yōu)矩陣是嚴(yán)格對(duì)角占優(yōu)矩陣,則當(dāng)則當(dāng)00.0005 er=0;k=k+1; for i=1:3 s=0;t=x(i);x(i)=

8、0; for j=1:3 s=s+A(i,j)*x(j); end x(i)=(1-w)*t+w*(b(i)-s)/A(i,i); er=max(abs(x(i)-t),er); endendkk=10 x= 1.1999 1.3999 1.5999=1.2,只需只需k=612/1813塊迭代法簡(jiǎn)介塊迭代法簡(jiǎn)介設(shè)設(shè) ARnn, xRn, bRn將方程組將方程組A x = b中系數(shù)矩陣中系數(shù)矩陣A分塊分塊 rrrrrrrrBBBXXXAAAAAAAAA2121212222111211其中其中, AiiRnini, AijRninj , xiRni, biRni13/1814將將A分解分解, A

9、= DB LB UB Jacobi塊迭代塊迭代 DB X(k+1) = (LB + UB)X(k) + b ijkjijikiiiXABXA)()1(i=1,2, r(2)Gauss-Seidel塊迭代塊迭代 DB X(k+1) = LB X(k+1)+ UBX(k) + b rijkjijijkjijikiiiXAXAbXA1)(11)1()1(i=1,2, r14/1815 626050410100141010014001100410010141001014654321xxxxxx塊迭代求解塊迭代求解 DIIDA 4114114DX1 = x1 x2 x3TX2 = x4 x5 x6Tb1

10、 = 0 5 0Tb2 = 6 2 6TDX1 X2 = b1X1+ DX2=b2DX1(k+1)= b1+X2(k)DX2(k+1)=b2 + X1(k)15/1816 yyuxuxuyuyxuuyyxx sin), 1(0)1 ,()0 ,(), 0(1,0, 0( i, ,j = 1,n )041, 11, 1 jijiijjijiuuuuu邊值問(wèn)題邊值問(wèn)題:051015051015nz = 641,1, 1, 14 jijijiijjiuuuuu)(1,)(1,)1(, 1)1()1(, 14kjikjikjikijkjiuuuuu ( i, ,j = 1,n ; k = 1,2,3,

11、 ）保留三對(duì)角塊保留三對(duì)角塊16/1817 0)1 ,(, 0)0 ,(0), 1 ,(, 0), 0 ,(0), 1(, 0), 0(1,0),sin()sin()sin(32yxuyxuzxuzxuzyuzyuzyxzyxuuuzzyyxx 取正整數(shù)取正整數(shù)n, h=1/(n+1)離散點(diǎn)離散點(diǎn)xi = i h yj = j h zm = m h (i, j, m = 0,1,n+1)用七點(diǎn)差分格式計(jì)算求解，用用七點(diǎn)差分格式計(jì)算求解，用slice命令繪四維圖命令繪四維圖17/181805101520250510152025nz = 135ijmmjimjimjimjimjimjimjimjimjifhuuuhuuuhuuu 21,1,2, 1, 1,2, 1, 1222)(61)(1,)1(1,)(, 1,)1(, 1,)(, 1)