向后差分中verlet方法的增幅

Question

我在使用向后差分的简谐振子上增加振荡时遇到了麻烦。这是我在 Scilab

中的代码

function [x] = back(h, tf)
k = 2;
m = 1;
i = 2;

t(i - 1) = 0;
x(i - 1) = 10;
v(i - 1) = 0;
t(i) = t(i - 1) + h
v(i) = v(i - 1) - h * (k / m) * x(i - 1)

while t(i) < tf
    t(i + 1) = t(i) + h
    x(i + 1) = x(i - 1) - 2 * (k / m) * v(i) * h
    i = i + 1
end

plot(t, x, 'b');

endfunction

Answer 1

我不太确定你想要达到什么目的，如果你的数学是正确的也不是。但假设你想解决以下数值问题：

//coefficients of:
k = 2.;
m = 1.;

// with an initial condition of:
t(1) = 0.;
x(1) = 10.;
v(1) = 0.;

// time paramters:
N = 50;
tf = 10;
h = tf / 50.;


for ii = 2:N
    t(ii) = t(ii - 1) + h;
    x(ii) = x(ii - 1) - 2 * (k / m) * v(ii - 1) * h
    v(ii) = v(ii - 1) - h * (k / m) * x(ii - 1)
    disp(x(ii))
end


plot(t, x, 'b');

将导致：

这似乎不对，但无论如何。请再次检查你的数学。

Answer 2

根据您的代码，我认为您正在尝试实现 velocity-Verlet scheme。这是一个简单振荡器的微分方程的实现：

function [x] = back(h, tf)
  k = 2;
  m = 1;

  t = 0:h:tf;
  x(1) = 10;
  v(1) = 0;

  for i=2:length(t)
    x(i) = x(i - 1) + v(i - 1) * h - k / m * x(i-1) * h^2 / 2;
    v(i) = v(i - 1) - k / m * (x(i) + x(i-1)) * h / 2;
  end

  plot(t, x, 'b');
endfunction

[x] = back(0.01, 10)

向后差分中verlet方法的增幅

Amplitude increasing in the verlet method in backward difference

numerical-methods

differential-equations

scilab