我需要以极性形式给出的以下非线性系统的相图...
\ dot {r} = 0.5 *(r-r ^ 3)\ dot {\ theta} = 1
我知道如何在Mathematica中做到这一点...
field1 = {0.5*(r - r^3), 1};
p1 = StreamPlot[Evaluate@TransformedField["Polar" -> "Cartesian", field1, {r, \[Theta]} -> {x, y}], {x, -3, 3}, {y, -3, 3}, Axes -> True, StreamStyle -> Gray, ImageSize -> Large];
Show[p1, AxesLabel->{x,y}, ImageSize -> Large]
如何在Python中使用pyplot.quiver来做同样的事情?
只是非常幼稚的实现,但可能会有所帮助...
import numpy as np
import matplotlib.pyplot as plt
def dF(r, theta):
return 0.5*(r - r**3), 1
X, Y = np.meshgrid(np.linspace(-3.0, 3.0, 30), np.linspace(-3.0, 3.0, 30))
u, v = np.zeros_like(X), np.zeros_like(X)
NI, NJ = X.shape
for i in range(NI):
for j in range(NJ):
x, y = X[i, j], Y[i, j]
r, theta = (x**2 + y**2)**0.5, np.arctan2(y, x)
fp = dF(r, theta)
u[i,j] = (r + fp[0]) * np.cos(theta + fp[1]) - x
v[i,j] = (r + fp[0]) * np.sin(theta + fp[1]) - y
plt.streamplot(X, Y, u, v)
plt.axis('square')
plt.axis([-3, 3, -3, 3])
plt.show()