数值分析课程设计--松弛迭代法中松弛因子.doc

安徽建筑大学数值分析 设计报告书题 目 松弛迭代法中松弛因子 院 系 数理系 专 业 信息与计算科学 班 级 信息班 学 号 12207210220 姓 名 穆海山 时 间 2013-12-102013-12-23 指导教师 刘华勇 题目：选用Jacobi迭代法、Gauss-Seidel迭代法和超松弛迭代法求解下面的方程组（考虑等于150）=考虑初值的变化和松弛因子的变化收敛效果的影响；对上述方程组还可以采用哪些方法求解？选择其中一些方法编程上机求解上述方程组，说明最适合的是什么方法；将计算结果进行比较分析，谈谈你对这些方法的看法。一、摘要本课程设计用matlab就线性方程组数值方法，Jacobi迭代法，Gauss-Seidel迭代法，超松弛法对所设计的问题进行求解，并编写程序在Matlab中实现，在文章中对各种迭代法进行了收敛性分析。接着用几种不同方法对线性方程组进行求解及结果分析，最后对此次课程设计进行了总结。关键词：线性方程组，迭代，Matlab，结果分析二、设计目的用熟悉的计算机语言编程上机求解线性方程组。三、理论基础对方程组 做等价变换 如：令 ，则 则，我们可以构造序列 若 同时：所以，序列收敛，与初值的选取无关则转化为矩阵形式 (1)令 或者 故迭代过程(1)化为 (2) 等价线性方程组为 称(2)式为解线性方程组(1)的Jacobi迭代法(J法)迭代矩阵 考虑迭代式(2) 即 将上式改为 (3) (4)超松弛迭代记 则 可以看作在前一步上加一个修正量。若在修正量前乘以一个因子有 对GaussSeidel迭代格式 迭代矩阵四、 程序代码及运算结果1.利用Jacobi迭代法求解：编制名为majacobifun.m的文件，内容如下：function x,n=jacobifun(A,b,x0,eps)%jacobifun为编写在雅可比迭代函数%A为线性方程组的系数矩阵%b为线性方程组的常数向量%x0为迭代初始向量%eps为解的精度%x为线性方程组的解%n为求出所需精度的解实际的迭代步数D=diag(diag(A);B=D(D-A);f=Db;x=B*x0+f;n=0;if max(abs(eig(B)=1 disp(Warning:不收敛！); return;endwhile norm(x-x0)=eps x0=x; x=B*x0+f; n=n+1;end2.利用Gauss-Seidel迭代法求解：编制名为G_Seidelifun.m的文件，内容如下：nction x,n=G_Seidelifun(A,b,x0,eps)%G_Seidelifun为编写在高斯-赛德尔迭代函数%A为线性方程组的系数矩阵%b为线性方程组的常数向量%x0为迭代初始向量%eps为解的精度%x为线性方程组的解%n为求出所需精度的解实际的迭代步数D=diag(diag(A);L=-tril(A,-1);U=-triu(A,1);G=(D-L)U;f=(D-L)b;x=G*x0+f;n=0;if max(abs(eig(G)=1 disp(Warning:不收敛！); return;endwhile norm(x-x0)=eps x0=x; x=G*x0+f; n=n+1;end3.利用SOR迭代法求解：编制名为SORfun.m的文件，内容如下：function x,n=SORfun(A,b,x0,w,eps)%SORfun为编写在松弛迭代函数%A为线性方程组的系数矩阵%b为线性方程组的常数向量%x0为迭代初始向量%w为松弛因子%eps为解的精度%x为线性方程组的解%n为求出所需精度的解实际的迭代步数if(w=2) error return;endD=diag(diag(A);L=-tril(A,-1);U=-triu(A,1);S=(D-w*L)(1-w)*D+w*U);f=w*(D-w*L)b);x=S*x0+f;n=0;if max(abs(eig(S)=1 disp(Warning:不收敛！); return;end while norm(x-x0)=eps x0=x; x=S*x0+f; n=n+1; end在Matlab命令窗口输入： clearw=1.5;eps=1e-6;n=150;A=diag(8*ones(1,n-1),-1)+diag(6*ones(1,n)+diag(ones(1,n-1),1);b(1)=10.5;b(n)=21;for i=2:n-1 b(i)=22.5;endb=b;x0=zeros(n,1);x1,n=jacobifun(A,b,x0,eps);x2,n=G_Seidelifun(A,b,x0,eps);x3,n=SORfun(A,b,x0,w,eps);plot(x1);hold onplot(x2,r);hold onplot(x3,g*)表1 Jacobi、Guass_seidel、SOR迭代结果表x0=zeros(n,1)x0=ones(n,1)w=1.5w=1.1w=1.5w=1.1雅可比高斯-赛德尔超松弛雅可比高斯-赛德尔超松弛不收敛！不收敛！n =n =n =n =n =n =n =n =084355663083353655x =x =x =x =x =x =x =x =1.0e+028 *1.0e+029 *1.0e+028 *1.0e+029 *1.7501.501.583301.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.502.2501.503.7501.37502.2501.37503.750202.250203.7500.502.2500.503.750302.250303.750002.250003.750402.250403.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.750002.250003.75-0.0001002.25-0.0001003.750.0002002.250.0002003.75-0.000300.00012.25-0.000300.00013.750.00060-0.00022.250.00060-0.00023.75-0.001200.00042.25-0.001200.00043.750.00240-0.00072.250.00240-0.00073.75-0.004900.00142.25-0.004900.00143.750.00970-0.00282.250.00970-0.00283.75-0.019500.00572.25-0.019500.00573.750.03890-0.01132.250.03890-0.01133.75-0.077600.02262.25-0.077600.02263.750.15460-0.0452.250.15460-0.0453.75-0.306800.08932.25-0.306800.08933.750.60390-0.17572.250.60390-0.17573.75-1.168700.342.25-1.168700.343.752.18170-0.63472.252.18170-0.63473.75-3.7401.0882.25-3.7401.0883.54.98660-1.45072.16674.98660-1.4507以下为方程组的解的图：图1 初值x0=0；.；0 松弛因子=1.5时图图2 初值x0=1；.；1 松弛因子=1.5时图图3 初值x0=1；.；1 松弛因子=1.1时图五、结果分析 通过以上数据测试可以分析出以下几点：1. 系数矩阵为150阶时，经判断Jacobi迭代是发散的、Gauss-Seidel、SOR两种迭代是收敛的，当初值一定时，Gauss-Seidel、SOR对应的迭代次数是逐渐减少的，也就是说SOR迭代比Gauss-Seidel迭代收敛更快。2. 当初值变大时，Gauss-Seidel、SOR对应的迭代次数是逐渐减少的，当初值相同，松弛因子 变小时，SOR对应的迭代次数