线性方程组迭代解法

1、实验六：线性方程组迭代解法 1）实验目的 熟悉Matlab编程； 学习线性方程组迭代解法的程序设计算法2）实验题目1.研究解线性方程组Ax=b迭代法收敛速度。A为20阶五对角距阵 要求：(1)选取不同的初始向量x0 及右端向量b，给定迭代误差要求，用雅可比迭代和高斯-赛德尔迭代法求解，观察得到的序列是否收敛？若收敛，记录迭代次数，分析计算结果并得出你的结论。(2)用SOR迭代法求解上述方程组，松弛系数取1< <2的不同值,在 时停止迭代.记录迭代次数，分析计算结果并得出你的结论。2.给出线性方程组，其中系数矩阵为希尔伯特矩阵：，假设若取分别用雅可比迭代法及SOR迭代（）求解，比较计

2、算结果。3）实验原理与理论基础1.雅克比（Jacobi）迭代法算法设计： 输入矩阵a与右端向量b及初值x(1,i); 按公式计算得 2.高斯赛得尔迭代法算法设计：1. 输入矩阵a与右端向量b及初值x(1,i). 2. （i = 1, 2, n） 3.超松驰法算法设计：输入矩阵a与右端向量b及初值x(1,i)。 ,4）实验内容第一题实验程序：1.雅克比迭代法：function =yakebi(e)%输入矩阵a与右端向量b。for i=1:20 a(i,i)=3;endfor i=3:20 for j=i-2 a(i,j)=-1/4; a(j,i)=-1/4; endendfor i=2:20 f

3、or j=i-1 a(i,j)=-1/2; a(j,i)=-1/2; endendb=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2;k=1;n=length(a);for i=1:n x(1,i)=1;%数组中没有第0行。endwhile k>=1 for i=1:n m=0; %此步也可以用ifj=i条件判定一下。 for j=1:(i-1) m=m+a(i,j)*x(k,j); end for j=(i+1):n m=m+a(i,j)*x(k,j); end x(k+1,

4、i)=(b(i)-m)/a(i,i); end l=0; %判定满足条件使循环停止迭代。 for i=1:n l=l+abs(x(k+1,i)-x(k,i); end if l<e break end k=k+1;end%输出所有的x的值。 x(k+1,:)k 2.高斯赛德尔迭代法：function =gaoshisaideer(e)for i=1:20 a(i,i)=3;endfor i=3:20 for j=i-2 a(i,j)=-1/4; a(j,i)=-1/4; endendfor i=2:20 for j=i-1 a(i,j)=-1/2; a(j,i)=-1/2; endend

5、b=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2;k=1;n=length(a);for i=1:n x(1,i)=0;%数组中没有第0行。endwhile k>=1 for i=1:n p=0;q=0; for j=1:(i-1) p=p+a(i,j)*x(k+1,j); end for j=(i+1):n q=q+a(i,j)*x(k,j); end x(k+1,i)=(b(i)-q-p)/a(i,i); end l=0; %判定满足条件使循环停止迭代。 for i=1:

6、n l=l+abs(x(k+1,i)-x(k,i); end if l<e break end k=k+1;end%输出所有的x的值。 x(k+1,:)k3.SOR迭代法程序：function =caosongci(e,w)for i=1:20 a(i,i)=3;endfor i=3:20 for j=i-2 a(i,j)=-1/4; a(j,i)=-1/4; endendfor i=2:20 for j=i-1 a(i,j)=-1/2; a(j,i)=-1/2; endendb=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.

7、5 1.5 1.5 1.5 1.5 1.7 2.2;k=1;n=length(a);for i=1:n x(1,i)=0;%数组中没有第0行。endwhile k>=1 if w>=2|w<=1 '请重新输入w的值，w在1与2之间' break end for i=1:n p=0;q=0; for j=1:(i-1) p=p+a(i,j)*x(k+1,j); end for j=i:n q=q+a(i,j)*x(k,j); end x(k+1,i)=x(k,i)+w*(b(i)-q-p)/a(i,i); end l=0; %判定满足条件使循环停止迭代。 for

8、 i=1:n l=l+abs(x(k+1,i)-x(k,i); end if l<e break end k=k+1;end%输出所有的x的值。 x(k+1,:)k第二题实验程序：1.雅克比迭代法：function X = p211_1_JJ(n)Hn = GET_Hn(n);b = GET_b(n);temp = 0;X0 = zeros(1, n);X_old = zeros(1, n);X_new = zeros(1, n);disp('Now Jacobi method!');disp('Start with the vector that (0, 0,

9、0, .)T');for i = 1:n for k = 1:n X_old = X_new; temp = 0; for j = 1:n if(j = i) temp = temp + Hn(i, j) * X_old(j); end end X_new(i) = (b(i) - temp) / Hn(i, i); endendX = X_new;end2.SOR迭代法：function X = p211_1_SOR(n, w)Hn = GET_Hn(n);b = GET_b(n);temp01 = 0;temp02 = 0;X0 = zeros(1, n);X_old = zero

10、s(1, n);X_new = zeros(1, n);disp('Now Successive Over Relaxtion method!');disp('Start with the vector that (0, 0, 0, .)T');for i = 1:n for k = 1:n X_old = X_new; temp01 = 0; temp02 = 0; for j = 1:n if(j < i) temp01 = temp01 + Hn(i, j) * X_new(j); end if(j > i) temp02 = temp02 +

11、 Hn(i, j) * X_old(j); end end end X_new(i) = w * (b(i) - temp01 - temp02) / Hn(i, i) + X_old(i);endX = X_new;end5）实验结果第一题实验结果：1.雅克比迭代法：输入：>> b=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2;yakebi(0.00001)结果：ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.997

12、0 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 122.高斯赛德尔迭代法：此时初值全取1；输入：>> b=2.2 1.7 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.7 2.2;>> gaoshisaideer(0.00001)结果：ans = Column

13、s 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k =14此时初值全取1；输入：>> b=2.5 1.9 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.9 2.5;gaoshisaideer(0.0

14、0001)结果：ans = Columns 1 through 12 1.0969 1.0707 1.0219 1.0103 1.0039 1.0016 1.0006 1.0003 1.0001 1.0001 1.0001 1.0001 Columns 13 through 20 1.0003 1.0006 1.0016 1.0039 1.0103 1.0219 1.0707 1.0969k = 143.SOR迭代法：>> caosongci(0.00001,1.1)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9

15、989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 11>> caosongci(0.00001,1.2)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20

16、0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 12>> caosongci(0.00001,1.3)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 15>> caoso

17、ngci(0.00001,1.4)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 19>> caosongci(0.00001,1.5)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9

18、989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 25>> caosongci(0.00001,1.6)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20

19、0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 34>> caosongci(0.00001,1.7)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 47>> caoso

20、ngci(0.00001,1.8)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 73>> caosongci(0.00001,1.9)ans = Columns 1 through 12 0.9793 0.9787 0.9941 0.9970 0.9

21、989 0.9995 0.9998 0.9999 1.0000 1.0000 1.0000 1.0000 Columns 13 through 20 0.9999 0.9998 0.9995 0.9989 0.9970 0.9941 0.9787 0.9793k = 150第二题实验结果：1.雅克比迭代法： >> p211_1_JJ(6) Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans = 2.4500 1.1036 0.6265 0.4060 0.2831 0.2071>> p211_1_JJ

22、(8)Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans =2.7179 1.4101 0.8524 0.5809 0.4221 0.3198 0.2497 0.1995 >> p211_1_JJ(10)Now Jacobi method!Start with the vector that (0, 0, 0, .)Tans = Columns 1 through 9 2.9290 1.6662 1.0517 0.7423 0.5554 0.4315 0.3445 0.2807 0.2325 Column 10

23、 0.19512.SOR迭代法：n=6, =1,1.25,1.5的时候>> p211_1_SOR(6, 1)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = 2.4500 1.1036 0.6265 0.4060 0.2831 0.2071>> p211_1_SOR(6, 1.25)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = 3.062

24、5 0.2310 0.8704 0.3389 0.3141 0.2097>> p211_1_SOR(6, 1.5)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans = 3.6750 -1.1009 2.0106 -0.3994 0.7670 -0.0384与n=8, =1,1.25,1.5的时候>> p211_1_SOR(8, 1)Now Successive Over Relaxtion method!Start with the vector that (

25、0, 0, 0, .)Tans =2.7179 1.4101 0.8524 0.5809 0.4221 0.3198 0.2497 0.1995>> p211_1_SOR(8, 1.25)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans =3.3973 0.4887 1.0898 0.5062 0.4501 0.3203 0.2573 0.2042>> p211_1_SOR(8, 1.5)Now Successive Over Relaxtion method!Start with the vector that (0, 0, 0, .)Tans =4.0768 -0.9424 2.2923 -0.2753 0.9252 0.0578 0.4071 0.1275与n