matlab - 为 DE 系统实现 Runge-Kutta

Question

我正在尝试实现 Runge-Kutta Method for Systems of DEs在 MATLAB 中。我没有收到 correct answers ，我不确定代码或我用来运行它的命令是否有问题。

这是我的代码:

function RKSystems(a, b, m, N, alpha, f)
    h = (b - a)/N;
    t = a;
    w = zeros(1, m);

    for j = 1:m
        w(j) = alpha(j);
    end

    fprintf('t = %.2f;', t);
    for i = 1:m
        fprintf(' w(%d) = %.10f;', i, w(i));
    end
    fprintf('\n');

    k = zeros(4, m);
    for i = 1:N
        for j = 1:m
           k(1, j) = h*f{j}(t, w);
        end

        for j = 1:m
            k(2, j) = h*f{j}(t + h/2, w + (1/2)*k(1));
        end

        for j = 1:m
            k(3, j) = h*f{j}(t + h/2, w + (1/2)*k(2));
        end

        for j = 1:m
            k(4, j) = h*f{j}(t + h, w + k(3));
        end

        for j = 1:m
            w(j) = w(j) + (k(1, j) + 2*k(2, j) + 2*k(3, j) + k(4, j))/6;
        end

        t = a + i*h;

        fprintf('t = %.2f;', t);
        for k = 1:m
            fprintf(' w(%d) = %.10f;', k, w(k));
        end
        fprintf('\n');

    end 
end

我正在尝试在 this problem 上测试它.
这是我的命令和输出:

>> U1 = @(t, u) 3*u(1) + 2*u(2) - (2*t^2 + 1)*exp(2*t);

>> U2 = @(t, u) 4*u(1) + u(2) + (t^2 + 2*t - 4)*exp(2*t);

>> a = 0; b = 1; alpha = [1 1]; m = 2; h = 0.2; N = (b - a)/h;

>> RKSystems(a, b, m, N, alpha, {U1 U2});

t = 0.00; w(1) = 1.0000000000; w(2) = 1.0000000000;

t = 0.20; w(1) = 2.2930309680; w(2) = 1.6186020410;

t = 0.40; w(1) = 5.0379629523; w(2) = 3.7300162165;

t = 0.60; w(1) = 11.4076339762; w(2) = 9.7009491301;

t = 0.80; w(1) = 27.0898576892; w(2) = 25.6603894354;

t = 1.00; w(1) = 67.1832886708; w(2) = 67.6103125539;

Answer 1

所以......这是我将如何做到这一点，我很难阅读您的代码片段中发生的事情，但我希望这对您有所帮助。此外，matlab 内置了 rk 求解器，我建议您熟悉这些求解器。

%rk4 solver
dt = .2;
t = 0:dt:1;
u = zeros(2,numel(t));
u(:,1) = 1;

for jj = 2:numel(t)
    u_ = u(:,jj-1);
    t_ = t(jj-1);
    fa = rhs(u_,t_);
    fb = rhs(u_+dt/2.*fa,t_+dt/2);
    fc = rhs(u_+dt/2.*fb,t_+dt/2);
    fd = rhs(u_+dt.*fc,t_+dt);
    u(:,jj) = u(:,jj-1) + dt/6*(fa+2*fb+2*fc+fd);
end
disp([t;u]')

和 rhs.m 如下:

function dudt = rhs(u,t)
dudt = [(3*u(1) + 2*u(2) - (2*t^2 + 1)*exp(2*t));
        (4*u(1) + u(2) + (t^2 + 2*t - 4)*exp(2*t))];

这将返回以下内容:

>> rkorder4
     0    1.0000    1.0000
0.2000    2.1204    1.5070
0.4000    4.4412    3.2422
0.6000    9.7391    8.1634
0.8000   22.6766   21.3435
1.0000   55.6612   56.0305

或者，您可以使用以下内容调用 ode45:

dt = .2;
t = 0:dt:1;
rhs=@(t,u)[(3*u(1) + 2*u(2) - (2*t^2 + 1)*exp(2*t));
        (4*u(1) + u(2) + (t^2 + 2*t - 4)*exp(2*t))];

[ts,us]=ode45(@(t,u) rhs(t,u),t,[1 1],[]);
disp([ts us]);

这给了你:

                   0   1.000000000000000   1.000000000000000
   0.200000000000000   2.125018862140859   1.511597928697035
   0.400000000000000   4.465156492097592   3.266022178547346
   0.600000000000000   9.832481410895053   8.256418221678395
   0.800000000000000  23.003033457636558  21.669270713784108
   1.000000000000000  56.738351357036301  57.106230777717208

这与您从我的代码中得到的非常接近。可以通过减小时间步长来增加两者之间的一致性， dt .它们总是在 t 的低值时最一致，两者之间的差异随着 t 的增加而增加。这两种实现也与分析解决方案非常接近。