## Phase portraits of a system of ODEs

Posted February 21, 2013 at 09:00 AM | categories: math, ode | tags: | View Comments

Updated February 27, 2013 at 02:56 PM

Matlab post An undamped pendulum with no driving force is described by

$$ y'' + sin(y) = 0$$

We reduce this to standard matlab form of a system of first order ODEs by letting \(y_1 = y\) and \(y_2=y_1'\). This leads to:

\(y_1' = y_2\)

\(y_2' = -sin(y_1)\)

The phase portrait is a plot of a vector field which qualitatively shows how the solutions to these equations will go from a given starting point. here is our definition of the differential equations:

To generate the phase portrait, we need to compute the derivatives \(y_1'\) and \(y_2'\) at \(t=0\) on a grid over the range of values for \(y_1\) and \(y_2\) we are interested in. We will plot the derivatives as a vector at each (y1, y2) which will show us the initial direction from each point. We will examine the solutions over the range -2 < y1 < 8, and -2 < y2 < 2 for y2, and create a grid of 20 x 20 points.

import numpy as np import matplotlib.pyplot as plt def f(Y, t): y1, y2 = Y return [y2, -np.sin(y1)] y1 = np.linspace(-2.0, 8.0, 20) y2 = np.linspace(-2.0, 2.0, 20) Y1, Y2 = np.meshgrid(y1, y2) t = 0 u, v = np.zeros(Y1.shape), np.zeros(Y2.shape) NI, NJ = Y1.shape for i in range(NI): for j in range(NJ): x = Y1[i, j] y = Y2[i, j] yprime = f([x, y], t) u[i,j] = yprime[0] v[i,j] = yprime[1] Q = plt.quiver(Y1, Y2, u, v, color='r') plt.xlabel('$y_1$') plt.ylabel('$y_2$') plt.xlim([-2, 8]) plt.ylim([-4, 4]) plt.savefig('images/phase-portrait.png')

>>> ... >>> >>> >>> >>> >>> >>> ... ... ... >>> >>> <matplotlib.text.Text object at 0xbcfd290> <matplotlib.text.Text object at 0xbd14750> (-2, 8) (-4, 4)

Let us plot a few solutions on the vector field. We will consider the solutions where y1(0)=0, and values of y2(0) = [0 0.5 1 1.5 2 2.5], in otherwords we start the pendulum at an angle of zero, with some angular velocity.

from scipy.integrate import odeint for y20 in [0, 0.5, 1, 1.5, 2, 2.5]: tspan = np.linspace(0, 50, 200) y0 = [0.0, y20] ys = odeint(f, y0, tspan) plt.plot(ys[:,0], ys[:,1], 'b-') # path plt.plot([ys[0,0]], [ys[0,1]], 'o') # start plt.plot([ys[-1,0]], [ys[-1,1]], 's') # end plt.xlim([-2, 8]) plt.savefig('images/phase-portrait-2.png') plt.show()

>>> ... ... ... [<matplotlib.lines.Line2D object at 0xbc0dbd0>] [<matplotlib.lines.Line2D object at 0xc135050>] [<matplotlib.lines.Line2D object at 0xbd1a450>] [<matplotlib.lines.Line2D object at 0xbd1a1d0>] [<matplotlib.lines.Line2D object at 0xb88aa10>] [<matplotlib.lines.Line2D object at 0xb88a5d0>] [<matplotlib.lines.Line2D object at 0xc14e410>] [<matplotlib.lines.Line2D object at 0xc14ed90>] [<matplotlib.lines.Line2D object at 0xc2c5290>] [<matplotlib.lines.Line2D object at 0xbea4bd0>] [<matplotlib.lines.Line2D object at 0xbea4d50>] [<matplotlib.lines.Line2D object at 0xbc401d0>] [<matplotlib.lines.Line2D object at 0xc2d1190>] [<matplotlib.lines.Line2D object at 0xc2f3e90>] [<matplotlib.lines.Line2D object at 0xc2f3bd0>] [<matplotlib.lines.Line2D object at 0xc2e4790>] [<matplotlib.lines.Line2D object at 0xc2e0a90>] [<matplotlib.lines.Line2D object at 0xc2c7390>] (-2, 8)

What do these figures mean? For starting points near the origin, and small velocities, the pendulum goes into a stable limit cycle. For others, the trajectory appears to fly off into y1 space. Recall that y1 is an angle that has values from \(-\pi\) to \(\pi\). The y1 data in this case is not wrapped around to be in this range.

Copyright (C) 2013 by John Kitchin. See the License for information about copying.