使用极坐标在Python中绘制相像 [英] Plotting phase portraits in Python using polar coordinates

问题描述

我需要以极性形式给出的以下非线性系统的相图...

I need a phase portrait of the following nonlinear system given in polar form...

\dot{r} = 0.5*(r - r^3)
\dot{\theta} = 1

\dot{r} = 0.5*(r - r^3)
\dot{\theta} = 1

我知道如何在Mathematica中做到这一点...

I know how to do it in Mathematica...

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做同样的事情?

How can I do the same using pyplot.quiver in Python?

推荐答案

更正上一个答案:

• 从 x=r*cos(theta) 得到 dx = dr*cos(theta)-r*sin(theta)*dtheta = x*dr/ry*dtheta，
• 从 y = r * sin(theta)中获得 dy = dr * sin(theta)+ r * cos(theta)* dtheta = y * dr/r + x *dtheta,
• 一个人可以使用numpy的矢量化操作来避免所有循环

def dF(r, theta):
    return 0.5*r*(1 - r*r), 1+0*theta

X, Y = np.meshgrid(np.linspace(-3.0, 3.0, 30), np.linspace(-3.0, 3.0, 30))
R, Theta = (X**2 + Y**2)**0.5, np.arctan2(Y, X)
dR, dTheta = dF(R, Theta)
C, S = np.cos(Theta), np.sin(Theta)
U, V = dR*C - R*S*dTheta, dR*S+R*C*dTheta

plt.streamplot(X, Y, U, V, color='r', linewidth=0.5, density=1.6)
plt.axis('square')
plt.axis([-3, 3, -3, 3])
plt.show()

这给出了下面的图.使用streamplot的density选项来增加绘图线的密度.

This gives the plot below. Use the density option of streamplot to increase the density of plot lines.

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