概述
一、高斯法和Doolitle法
高斯法和Doolitle法都比较简单，其实Doolitle可以直接用maltab里面的lu命令来求，下面提一下迭代法
二、Jacobi迭代
1.算法实现
2.收敛条件
3.matlab实现
根据以上的算法可以写出一个简单的jacobi迭代如下：
function x = jacobi(a,b,k)
n=length(b);
d=diag(a);
r=a-diag(d);
x=zeros(n,1);
for j = 1:k
x = (b-r*x)./d;
end以上代码是最简单的写法，但是做题的时候一般要求把迭代过程给显示，这样写的话就不够完善，所以有以下拓展版。
function r = jacobi(A,B,varargin)
sizeA=size(A);
sizev=size(varargin);
if sizev(2) == 0
rol = 0.000001;
n = 1000;
x = zeros(sizeA(1),1);
elseif sizev(2) == 1
rol = varargin{1};
n = 1000;
x = zeros(sizeA(1),1);
elseif sizev(2) == 2
rol = varargin{1};
n = varargin{2};
x = zeros(sizeA(1),1000);
elseif sizev(2) == 3
rol = varargin{1};
n = varargin{2};
x = varargin{3};
else
error("输入参数过多");
end
for i = 2:n
for j = 1:sizeA(2)
sum1=0;
for k = 1:sizeA(1)
if j == k
continue;
end
sum1 = sum1 - x(k,i-1)*A(j,k)/A(j,j);
end
x(j,i)=B(j)/A(j,j)+sum1;
end
if any(abs(x(:,i)-x(:,i-1))>rol) == 0
break;
end
end
r = x;
end三、Gauss-Seidel迭代法
1.算法实现
2.收敛条件
3.matlab实现
function r = Gauss_Seidel(A,B,varargin)
sizeA=size(A);
sizev=size(varargin);
if sizev(2) == 0
rol = 0.000001;
n = 1000;
x = zeros(sizeA(1),1);
elseif sizev(2) == 1
rol = varargin{1};
n = 1000;
x = zeros(sizeA(1),1);
elseif sizev(2) == 2
rol = varargin{1};
n = varargin{2};
x = zeros(sizeA(1),1000);
elseif sizev(2) == 3
rol = varargin{1};
n = varargin{2};
x = varargin{3};
else
error("输入参数过多");
end
for i = 2:n
for j = 1:sizeA(2)
sum1=0;
for k = 1:j
if j == k
continue;
end
sum1 = sum1 - x(k,i)*A(j,k)/A(j,j);
end
for k = j+1:sizeA(1)
sum1 = sum1 - x(k,i-1)*A(j,k)/A(j,j);
end
x(j,i)=B(j)/A(j,j)+sum1;
end
if any(abs(x(:,i)-x(:,i-1))>rol) == 0
break;
end
end
r = x;
end四、逐次超松弛(SOR)迭代法
1.算法实现
2.收敛条件
3.matlab实现
function r = SOR(A,B,w,varargin)
sizeA=size(A);
sizev=size(varargin);
if sizev(2) == 0
rol = 0.000001;
n = 1000;
x = zeros(sizeA(1),1);
elseif sizev(2) == 1
rol = varargin{1};
n = 1000;
x = zeros(sizeA(1),1);
elseif sizev(2) == 2
rol = varargin{1};
n = varargin{2};
x = zeros(sizeA(1),1000);
elseif sizev(2) == 3
rol = varargin{1};
n = varargin{2};
x = varargin{3};
else
error("输入参数过多");
end
for i = 2:n
for j = 1:sizeA(2)
sum1=0;
for k = 1:j
if j == k
sum1 = sum1 - w*(1-1/w)*x(k,i-1);
continue;
end
sum1 = sum1 - w*x(k,i)*A(j,k)/A(j,j);
end
for k = j+1:sizeA(1)
sum1 = sum1 - w*x(k,i-1)*A(j,k)/A(j,j);
end
x(j,i)=w*B(j)/A(j,j)+sum1;
end
if any(abs(x(:,i)-x(:,i-1))>rol) == 0
break;
end
end
r = x;
end最后
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