Question

如果有更多实施经验的人会帮助我发现我当前代码中的逻辑缺陷，我会感激不尽。在过去的几个小时里，我一直坚持使用以下RK4函数的各种步长的实现和测试来解决Lotka-Volterra Differential equation。

我尽最大努力确保代码的可读性并注释掉关键步骤，因此下面的代码应该清楚。

import matplotlib.pyplot as plt
import numpy as np

def model(state,t):
    """
    A function that creates an 1x2-array containing the Lotka Volterra Differential equation

    Parameter assignement/convention:
    a   natural growth rate of the preys
    b   chance of being eaten by a predator
    c   dying rate of the predators per week
    d   chance of catching a prey 
    """

    x,y = state     # will corresponding to initial conditions  
                    # consider it as a vector too 

    a = 0.08
    b = 0.002
    c = 0.2
    d = 0.0004

    return np.array([ x*(a-b*y) , -y*(c - d*x) ]) # corresponds to [dx/dt, dy/dt]


def rk4( f, x0, t):
    """ 
    4th order Runge-Kutta method implementation to solve x' = f(x,t) with x(t[0]) = x0.

    INPUT:
        f     - function of x and t equal to dx/dt. 

        x0    - the initial condition(s).  
                Specifies the value of x @ t = t[0] (initial).  
                Can be a scalar or a vector (NumPy Array)

                Example: [x0, y0] = [500, 20]

        t     - a time vector (array) at which the values of the solution are computed at.
                t[0] is considered as the initial time point 
                the step size h is dependent on the time vector, choosing more points will 
                result in a smaller step size. 


    OUTPUT:
        x     - An array containing the solution evaluated at each point in the t array.

    """

    n = len( t )
    x = np.array( [ x0 ] * n )      # creating an array of length n 

    for i in xrange( n - 1 ):
        h = t[i+1]- t[i]            # step size, dependent on time vector

        # starting below - the implementation of the RK4 algorithm:
        # for further informations visit http://en.wikipedia.org/wiki/Runge-Kutta_methods

        # k1 is the increment based on the slope at the beginning of the interval (same as Euler)
        # k2 is the increment based on the slope at the midpoint of the interval 
        # k3 is AGAIN the increment based on the slope at the midpoint 
        # k4 is the increment based on the slope at the end of the interval

        k1 = f( x[i], t[i] )
        k2 = f( x[i] + 0.5 * h * k1, t[i] + 0.5 * h ) 
        k3 = f( x[i] + 0.5 * h * k2, t[i] + 0.5 * h )  
        k4 = f( x[i] + h * k3, t[i] + h  )

        # finally computing the weighted average and storing it in the x-array
        t[i+1] = t[i] + h
        x[i+1] = x[i] + h * ( ( k1 + 2.0 * ( k2 + k3 ) + k4 ) / 6.0 )

    return x


################################################################
# just the graphical output

# initial conditions for the system

x0 = 500
y0 = 20


# vector of times
t = np.linspace( 0, 200, 150 )

result = rk4( model,[x0,y0], t )


plt.plot(t,result)

plt.xlabel('Time')
plt.ylabel('Population Size')
plt.legend(('x (prey)','y (predator)'))
plt.title('Lotka-Volterra Model')
plt.show()

目前的输出看起来还不错。在一个小间隔然后去疯狂＃39;。奇怪的是，当我选择更大的步长而不是小的步长时，代码似乎表现得更好，这表明我的实现必定是错误的，或者我的模型可能已关闭。我自己无法发现错误。

输出（错误）：

current output by the RK4 algorithm

这是所需的输出，可以通过使用Scipys集成模块之一轻松获得。请注意，在时间间隔[0,50]上，模拟似乎是正确的，然后每一步都会变得更糟。

correct desired output

Answer 1

不幸的是，你遇到了我偶尔遇到的陷阱：你的初始x0数组包含整数，因此，所有得到的x[i]值将在计算后转换为整数。 / p>

为什么？因为int是您初始条件的类型：

x0 = 500
y0 = 20

当然，解决方案是明确地使它们成为float：

x0 = 500.
y0 = 20.

那么为什么scipy在为它提供整数起始值时能正确执行？在开始实际计算之前，它可能会将它们转换为float。例如，您可以这样做：

x = np.array( [ x0 ] * n, dtype=np.float)

然后你仍然可以安全地使用整数初始条件而没有问题。至少在这种情况下，转换是在函数内完成一次性的，如果你再次使用它半年（或者，其他人使用它），你就不能再陷入陷阱了。

需要帮助修复我的RK4实现

1 个答案: