python - 如何使用 matplotlib 生成逻辑方程 K = 1 的向量场图？

Question

我正在学习 Strogatz 的《非线性动力学和混沌》，但在第 2 章练习 2.8.1 中遇到了障碍。 (教育者旗帜:我已经毕业了，所以这不是为了一个类，我只是想回到微分方程的数值求解)这是一个非常简单的微分方程，我可以在给定不同初始条件的情况下绘制单独的解曲线但我尝试使用 quiver 或 streamplot 将各个解决方案叠加在矢量场之上。

我的问题是理解如何转换矢量场图以找到 dy/dx 形式的类似问题 here转向 Strogatz 书中主要讨论的 dx/dt 形式。

鉴于逻辑函数中定义的 x 向量只是一维，我很难推理出如何表达 quiver 或 streamplot 中的 u 和 v 流，因为问题似乎只有 u 流。这可能非常简单，并且经过深思熟虑，但任何指导或帮助将不胜感激!

到目前为止，我有以下内容:

# 2.8.1
# Plot the vector field and some trajectories for xdot = x(1-x) given 
# some different initial conditions for the logistic equation with carrying 
# capacity K = 1

# dx/dt = x(1-x)  

# Imports:
from __future__ import division
from scipy import *
import numpy as np
import pylab
import matplotlib as mp
from matplotlib import pyplot as plt  
import sys
import math as mt

def logistic(x,t):
    return np.array([x[0]*(1-x[0])])

def RK4(t0 = 0, x0 = np.array([1]), t1 = 5 , dt = 0.01, ng = None):  
    tsp = np.arange(t0, t1, dt)
    Nsize = np.size(tsp)
    X = np.empty((Nsize, np.size(x0)))
    X[0] = x0

    for i in range(1, Nsize):
        k1 = ng(X[i-1],tsp[i-1])
        k2 = ng(X[i-1] + dt/2*k1, tsp[i-1] + dt/2)
        k3 = ng(X[i-1] + dt/2*k2, tsp[i-1] + dt/2)
        k4 = ng(X[i-1] + dt*k3, tsp[i-1] + dt)
        X[i] = X[i-1] + dt/6*(k1 + 2*k2 + 2*k3 + k4)
    return X

def tplot():
    t0 = 0
    t1 = 10
    dt = 0.02
    tsp = np.arange(t0,t1,dt)
    X = RK4(x0 = np.array([2]), t1 = 10,dt = 0.02, ng = logistic)
    Y = RK4(x0 = np.array([0.01]), t1 = 10,dt = 0.02, ng = logistic)
    Z = RK4(x0 = np.array([0.5]), t1 = 10,dt = 0.02, ng = logistic)
    P = RK4(x0 = np.array([3]), t1 = 10,dt = 0.02, ng = logistic)
    Q = RK4(x0 = np.array([0.1]), t1 = 10,dt = 0.02, ng = logistic)
    R = RK4(x0 = np.array([1.5]), t1 = 10,dt = 0.02, ng = logistic)
    O = RK4(x0 = np.array([1]), t1 = 10,dt = 0.02, ng = logistic)
    pylab.figure()
    pylab.plot(tsp,X)
    pylab.plot(tsp,Y)
    pylab.plot(tsp,Z)
    pylab.plot(tsp,P)
    pylab.plot(tsp,Q)
    pylab.plot(tsp,R)
    pylab.plot(tsp,O)
    pylab.title('Logistic Equation - K=1')
    pylab.xlabel('Time')
    pylab.ylabel('Xdot')
    pylab.show()

print tplot()

image here

Answer 1

要根据导数(如dx/dt)绘制斜率，您可以首先找到dx/dt，然后使用固定的dt 计算dx。然后，在每个感兴趣的 (t, x) 处，绘制从 (t,x) 到 (t+dt, x+dx) 的小线段。

以下是方程式的示例dx/dt = x(1-x)。 (Strogatz 图片没有箭头，所以我也将其删除。)

import numpy as np
import matplotlib.pyplot as plt

times = np.linspace(0, 10,  20)
x = np.linspace(0 ,2, 20)
T, X = np.meshgrid(times, x)   # make a grid that roughly matches the Strogatz grid  

dxdt = X*(1-X)            # the equation of interest
dt = .5*np.ones(X.shape)  # a constant value (.5 is just so segments don't run into each other -- given spacing of times array
dx = dxdt * dt            # given dt, now calc dx for the line segment

plt.quiver(T, X, dt, dx, headwidth=0., angles='xy', scale=15.)
plt.show()