利用Richardson 外推法求 $$ \left\{\begin{array}{l} L u=-\frac{d}{d x}\left(p \frac{d u}{d x}\right)+r \frac{d u}{d x}+q u=f \quad a<x<b \\ u(a)=\alpha, u(b)=\beta \end{array}\right. $$ 的变步长差分方法d1matlab代码

以下是利用Richardson外推法求解二阶常微分方程的变步长差分方法的MATLAB代码：

function [x, u] = richardson(d, a, b, alpha, beta, p, q, r, f, tol)
% d: 初始步长
% a, b: 区间左右端点
% alpha, beta: 边值条件
% p, q, r, f: 方程参数
% tol: 容许误差

Nmax = 10000; % 最大迭代次数
x = a:d:b; % 网格点
N = length(x); % 网格数
u = zeros(1, N); % 数值解
u(1) = alpha; % 边界条件
u(N) = beta; % 边界条件
err = tol + 1; % 初始误差
n = 0; % 迭代次数

while err > tol && n < Nmax
    % 列出差分方程
    A = zeros(N-2, N-2);
    b = zeros(N-2, 1);
    for i = 2:N-1
        if i == 2
            A(i-1, i-1) = -2/d^2+p(x(i))/d+q(x(i));
            A(i-1, i) = 1/d^2-r(x(i))/(2*d);
            b(i-1) = f(x(i))-u(i-1)*(p(x(i))/d-r(x(i))/(2*d));
        elseif i == N-1
            A(i-1, i-2) = 1/d^2+r(x(i))/(2*d);
            A(i-1, i-1) = -2/d^2+p(x(i))/d+q(x(i));
            b(i-1) = f(x(i))-u(i+1)*(p(x(i))/d+r(x(i))/(2*d));
        else
            A(i-1, i-2) = 1/d^2+r(x(i))/(2*d);
            A(i-1, i-1) = -2/d^2+p(x(i))+q(x(i))*d^2;
            A(i-1, i) = 1/d^2-r(x(i))/(2*d);
            b(i-1) = f(x(i));
        end
    end
    
    % 求解线性方程组
    v = A \ b;
    
    % 利用 Richardson 外推法计算新的数值解
    u_new = zeros(1, N);
    u_new(1) = alpha;
    u_new(N) = beta;
    for i = 2:N-1
        u_new(i) = u(i) + (u(i)-u(i-1))/(4^(n+1)-1) + (v(i-1)-v(i-2))/(3*4^(n+1)*(d/2)^(2)-1);
    end
    
    % 计算误差
    err = max(abs(u_new-u));
    
    % 更新数值解
    u = u_new;
    
    % 更新步长和迭代次数
    d = d/2;
    n = n+1;
end

end

使用方法：

% 定义方程参数和边值条件
a = 0; b = 1;
alpha = 0; beta = 1;
p = @(x) 1+x.^2;
q = @(x) exp(x);
r = @(x) cos(x);
f = @(x) 2*x.^3;
tol = 1e-6;

% 初始步长为 1/10
d = 1/10;

% 求解方程
[x, u] = richardson(d, a, b, alpha, beta, p, q, r, f, tol);

% 绘制数值解图像
plot(x, u);