用时间延迟在Python中求解颂歌

Question

任何人都可以给我一些建议，如何在python中解决一个在其中实现的时间延迟的颂歌？我似乎无法弄清楚如何使用scipy.integrate.odeint做到这一点。我要寻找的是：

# the constants in the equation
b = 1/50
d = 1/75
a = 0.8
G = 10 ** (-2)
tau = 0.5
u = [b, d, tau, a, G]
# enter initial conditions
N0 = 0.1
No0 = 10
w = [N0, No0]
def logistic(w, t, u):
    N, No = w
    b, d, tau, a, G = u
    dNdt = b * (No(t) - N(t) ) * (N(t) / No(t) ) - d * N(t - tau)
    dNodt = G * (a * No(t) - N(t) ) * (N(t) / No(t) )
    return [dNdt, dNodt]
# create timescale
# create timescale
stoptime = 1000.0
numpoints = 10000
t = np.linspace(0, stoptime, numpoints)
# in my previous code I would use scipy.integrate.odeint here to integrate my 
# equations, but with a time-delay that doesn't work (I think)
soln = ...

其中n（t），n（t -tau）等表示函数的时间参数。是否存在良好的库来解决这些类型的方程式？非常感谢！

Answer 1

我是jitcdde的作者，它可以求解延迟微分方程，并且主要类似于 scipy.ode。您可以使用pip3 install jitcdde安装它。据我所知，python的其他现有DDE库是破碎的，要么基于弃用的依赖项。

以下代码将整合您的问题：

from jitcdde import t, y, jitcdde
import numpy as np
# the constants in the equation
b = 1/50
d = 1/75
a = 0.8
G = 10**(-2)
tau = 0.5
# the equation
f = [    
    b * (y(1) - y(0)) * y(0) / y(1) - d * y(0, t-tau),
    G * (a*y(1) - y(0)) * y(0) / y(1)
    ]
# initialising the integrator
DDE = jitcdde(f)
# enter initial conditions
N0 = 0.1
No0 = 10
DDE.add_past_point(-1.0, [N0,No0], [0.0, 0.0])
DDE.add_past_point( 0.0, [N0,No0], [0.0, 0.0])
# short pre-integration to take care of discontinuities
DDE.step_on_discontinuities()
# create timescale
stoptime = 1000.0
numpoints = 100
times = DDE.t + np.linspace(1, stoptime, numpoints)
# integrating
data = []
for time in times:
    data.append( DDE.integrate(time) )