数值分析SOR迭代

博客围绕数值分析中的SOR迭代展开,虽未给出具体内容,但可知聚焦于该迭代方法在数值分析领域的相关探讨。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

function [X,Number_of_iteration] = SOR_iteration(A,B,P,delta,max1,W)
% Input  - A is an N x N nonsingular matrix
%        - B is an N x 1 matrix
%        - P is an N x 1 matrix: the initial guess
%        - delta is the tolerance for P
%        - max1 is the maximum number of iterations
%        - W is the relax factor
% Output - X is an N x 1 matrix: the SOR approximation to the 
%        - solution of the AX = B
N = length(B);
count = 0;
for k = 1 : max1
    for j = 1 : N
        if j == 1 
            X(1) = (1 - W) * P(1) - (B(1) - (A(1,(2:N)) * P(2:N))) / (A(1,1) / W);
        elseif j == N
            X(N) = (1 - W) * P(N) - (B(N) - (A(N,(1:N-1)) * X(1:N-1)')) / (A(N,N) / W);
        else
            %X contains the kth approxiamation and P the (k-1)st
            X(j) = (1 - W) * P(j) - (B(j) - A (j,1:j-1) * X(1,j-1)' - A(j,(j+1:N)) * P(j+1 : N)) / (A(j,j) / W);
        end
    end
    count = count + 1;
    err = abs(norm(X'-P));
    reletive_err = err/(norm(X)+eps);
    P = X';
        if(err < delta) || (reletive_err < delta)
            break
        end
end
X = X';
Number_of_iteration = count;

 

评论 1
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值