RK4 Python解释 [英] RK4 Python Explanation

问题描述

我想将在网上找到的RK4的实现用于某些事情，但是我很难理解在网上找到的实现.例如:

I'd like to use an implementation of RK4 I found online for something, but I'm having a bit of difficulty wrapping my head around the implementations I have found online. For example:

def rk4(f, x0, y0, x1, n):
    vx = [0] * (n + 1)
    vy = [0] * (n + 1)
    h = (x1 - x0) / float(n)
    vx[0] = x = x0
    vy[0] = y = y0
    for i in range(1, n + 1):
        k1 = h * f(x, y)
        k2 = h * f(x + 0.5 * h, y + 0.5 * k1)
        k3 = h * f(x + 0.5 * h, y + 0.5 * k2)
        k4 = h * f(x + h, y + k3)
        vx[i] = x = x0 + i * h
        vy[i] = y = y + (k1 + k2 + k2 + k3 + k3 + k4) / 6
return vx, vy

有人可以帮我了解参数的确切含义吗?如果可能的话，我希望有一个更笼统的解释，但是，如果更加具体，则更易于解释，我将专门针对理想的弹簧系统使用它.

Could someone please help me understand what exactly the parameters are? If possible, I'd like a more general explanation, but, if being more specific makes it easier to explain, I'm going to be using it specifically for an ideal spring system.

推荐答案

您在此处询问参数:

def rk4(f, x0, y0, x1, n):
    ...
    return vx, vy

• f 是ODE函数，对于微分方程 y'(x)= f(x，y)声明为 def f(x，y)(x))，
• (x0，y0)是初始点和值
• x1 是积分间隔的结束时间 [x0，x1]

• n 是子间隔或积分步骤的数量

  • f is the ODE function, declared as def f(x,y) for the differential equation y'(x)=f(x,y(x)),
  • (x0,y0) is the initial point and value,
  • x1 is the end of the integration interval [x0,x1]

  • n is the number of sub-intervals or integration steps

    vx，vx 是计算出的样本点， vy [k] 近似于 y(vx [k]).

    vx,vx are the computed sample points, vy[k] approximates y(vx[k]).

    不能将其用于spring系统，因为该代码仅适用于标量 v .您需要对其进行更改以与 numpy 一起进行矢量操作.

    You can not use this for the spring system, as that code only works for a scalar v. You would need to change it to work with numpy for vector operations.

    def rk4(func, x0, y0, x1, n):
        y0 = np.array(y0)
        f = lambda x,y: np.array(func(x,y))
        vx = [0] * (n + 1)
        vy = np.zeros( (n + 1,)+y0.shape)
        h = (x1 - x0) / float(n)
        vx[0] = x = x0
        vy[0] = y = y0[:]
        for i in range(1, n + 1):
            k1 = h * f(x, y)
            k2 = h * f(x + 0.5 * h, y + 0.5 * k1)
            k3 = h * f(x + 0.5 * h, y + 0.5 * k2)
            k4 = h * f(x + h, y + k3)
            vx[i] = x = x0 + i * h
            vy[i] = y = y + (k1 + 2*(k2 + k3) + k4) / 6
        return vx, vy
    

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