python - Hindmarsh-Rose 模型的相空间轨迹

Question

我正在尝试绘制 Hindmarsh-Rose 模型的相空间轨迹。我已经实现了一个 RK4 积分器来解决以下方程组:

到目前为止，我编写的代码如下。

import numpy as np
import matplotlib.pyplot as plt

def RK4(f, x0, t):
    
    dt   = t[2] -t[1]  #time span 
    N    = len(t)
    X    = np.empty((len(t), len(x0)))
    X[0] = x0
    
    for i in range(1, N):
        
        k1 = f(X[i-1], t[i-1])
        k2 = f(X[i-1] + dt/2*k1, t[i-1] + dt/2)
        k3 = f(X[i-1] + dt/2*k2, t[i-1] + dt/2)
        k4 = f(X[i-1] + dt*k3, t[i-1] + dt)
        
        X[i] = X[i-1] + dt/6*(k1 + 2*k2 + 2*k3 + k4)
        
    return X


def hindmarsh(X, t):
    
    a  = 3.0
    c  = 1.0
    d  = 5.0
    s  = 4.0
    x0 = - 1.6
    
    # Bifurcation parameters
    
    b   = 3.09
    I   = 3.2
    eps = 0.001
      
    x,y,z = X
    
    dxdt  = y - (a * x**3) + (b * x**2) + I - z 
    dydt  = c - (d * x**2) - y 
    dzdt  = eps * ( (s * (x - x0)) - z) 
    
    return np.array([dxdt, dydt, dzdt])


T     = np.linspace(0,100,10000)

Y     = [0.03, 0.03, 3]

param = RK4( hindmarsh, Y, T )

ax = plt.axes(projection='3d')

zline = param[2]
yline = param[1]
xline = param[0]


ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')

ax.plot3D(xline, yline, zline)

但是，我没有在相空间中获得轨道，如下图所示，而是在相空间中得到一条直线。我将不胜感激有关如何获得下面的情节的任何提示。

Answer 1

param有形状(len(T), len(Y)) ，所以时间在第一维，x,y,z 在第二维。你会得到正确的情节

zline = param[:,0]
xline = param[:,1]
yline = param[:,2]