使用scipy.integrate.odeint求解odes系统(具有不断变化的常数！)? [英] Solving a system of odes (with changing constant!) using scipy.integrate.odeint?

问题描述

我目前有一个随时间变化的常数的odes系统.例如

I currently have a system of odes with a time-dependent constant. E.g.

def fun(u, t, a, b, c):
    x = u[0]
    y = u[1]
    z = u[2]
    dx_dt = a * x + y * z
    dy_dt = b * (y-z)
    dz_dt = -x*y+c*y-z
    return [dx_dt, dy_dt, dz_dt]

常量为"a"，"b"和"c".我现在有一个要在每个时间步中插入的"a"列表，在使用scipy ode求解器时，我想在每个时间步中插入...可以吗?

The constants are "a", "b" and "c". I currently have a list of "a"s for every time-step which I would like to insert at every time-step, when using the scipy ode solver...is this possible?

谢谢！

推荐答案

是的，这是可能的.在a为常数的情况下，我猜您调用了scipy.integrate.odeint(fun, u0, t, args)，其中fun的定义与您的问题相同，u0 = [x0, y0, z0]是初始条件，t是需要解决的时间点序列ODE和args = (a, b, c)是传递给fun的额外参数.

Yes, this is possible. In the case where a is constant, I guess you called scipy.integrate.odeint(fun, u0, t, args) where fun is defined as in your question, u0 = [x0, y0, z0] is the initial condition, t is a sequence of time points for which to solve for the ODE and args = (a, b, c) are the extra arguments to pass to fun.

如果a取决于时间，则只需重新考虑a作为一个函数，例如(给定常量a0):

In the case where a depends on time, you simply have to reconsider a as a function, for example (given a constant a0):

def a(t):
    return a0 * t

然后，您将不得不修改fun，该计算将在每个时间步计算导数，以将先前的更改考虑在内:

Then you will have to modify fun which computes the derivative at each time step to take the previous change into account:

def fun(u, t, a, b, c):
    x = u[0]
    y = u[1]
    z = u[2]
    dx_dt = a(t) * x + y * z # A change on this line: a -> a(t)
    dy_dt = b * (y - z)
    dz_dt = - x * y + c * y - z
    return [dx_dt, dy_dt, dz_dt]

最后，请注意u0，t和args保持不变，您可以再次调用scipy.integrate.odeint(fun, u0, t, args).

Eventually, note that u0, t and args remain unchanged and you can again call scipy.integrate.odeint(fun, u0, t, args).

关于这种方法的正确性的一句话.数值积分逼近的性能会受到影响，我不知道具体如何实现(没有理论保证)，但这是一个简单的示例，可以工作:

A word about the correctness of this approach. The performance of the approximation of the numerical integration is affected, I don't know precisely how (no theoretical guarantees) but here is a simple example which works:

import matplotlib.pyplot as plt
import numpy as np
import scipy as sp
import scipy.integrate

tmax = 10.0

def a(t):
    if t < tmax / 2.0:
        return ((tmax / 2.0) - t) / (tmax / 2.0)
    else:
        return 1.0

def func(x, t, a):
    return - (x - a(t))

x0 = 0.8
t = np.linspace(0.0, tmax, 1000)
args = (a,)
y = sp.integrate.odeint(func, x0, t, args)

fig = plt.figure()
ax = fig.add_subplot(111)
h1, = ax.plot(t, y)
h2, = ax.plot(t, [a(s) for s in t])
ax.legend([h1, h2], ["y", "a"])
ax.set_xlabel("t")
ax.grid()
plt.show()

我希望这会对您有所帮助.

I Hope this will help you.

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