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Cleanup differential equations examples. #22043

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Jan 5, 2022
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Cleanup differential equations examples. #22043

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20 changes: 13 additions & 7 deletions examples/animation/double_pendulum.py
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Expand Up @@ -12,7 +12,6 @@
from numpy import sin, cos
import numpy as np
import matplotlib.pyplot as plt
import scipy.integrate as integrate
import matplotlib.animation as animation
from collections import deque

Expand All @@ -22,13 +21,13 @@
L = L1 + L2 # maximal length of the combined pendulum
M1 = 1.0 # mass of pendulum 1 in kg
M2 = 1.0 # mass of pendulum 2 in kg
t_stop = 5 # how many seconds to simulate
t_stop = 2.5 # how many seconds to simulate
history_len = 500 # how many trajectory points to display


def derivs(state, t):

def derivs(t, state):
dydx = np.zeros_like(state)

dydx[0] = state[1]

delta = state[2] - state[0]
Expand All @@ -51,7 +50,7 @@ def derivs(state, t):
return dydx

# create a time array from 0..t_stop sampled at 0.02 second steps
dt = 0.02
dt = 0.01
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Please check if this is really necessary. This doubles the number of frames and thus likely the size of the generated file.

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I decreased the end time by a factor of two as well; the shorter time step does appear to improve the accuracy.

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@timhoffm timhoffm Dec 23, 2021 •
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But that means that you're only going half the path from before. Please check if that's still a reasonable path length. I'm afraid that might be a bit short.

I can imagine that the shorter time step improves accuracy. But does that matter here? Alternatively, one could also render only every second step, but that adds additional compelity to the code and I'm not sure that's worth it.

Of course, one could also change to another example like a regular single pendulum. That's simpler to calculate and one could have a continuous cyclic animation with, a period of e.g. 2s.

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I did check, the path length is fine.

t = np.arange(0, t_stop, dt)

# th1 and th2 are the initial angles (degrees)
Expand All @@ -64,8 +63,15 @@ def derivs(state, t):
# initial state
state = np.radians([th1, w1, th2, w2])

# integrate your ODE using scipy.integrate.
y = integrate.odeint(derivs, state, t)
# integrate the ODE using Euler's method
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This makes the example longer and IMHO a little harder to understand. Does this outweigh the benefit of not depending on scipy in this example?

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The only other use of scipy in the docs can also easily be removed (I have another patch ready for that) so we could entirely drop the dependency on scipy for building the docs, which may or may not be a useful thing (e.g., with the yearly release cycle of cpython, I fear that we're always going to have a small period between having scipy wheels available after the new cpython release, and scipy is significantly harder to build from source (e.g., due to fortran sources) than other dependencies, even numpy).

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I think dropping SciPy would be pretty helpful if we don't need it. I think its a pretty heavy dependency to carry around for a couple of examples.

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We have oscillated back and forth on having a scipy documentation dependencies a couple of times. I do not have a strong preference either way.

y = np.empty((len(t), 4))
y[0] = state
for i in range(1, len(t)):
y[i] = y[i - 1] + derivs(t[i - 1], y[i - 1]) * dt

# A more accurate estimate could be obtained e.g. using scipy:
#
# y = scipy.integrate.solve_ivp(derivs, t[[0, -1]], state, t_eval=t).y.T

x1 = L1*sin(y[:, 0])
y1 = -L1*cos(y[:, 0])
Expand Down
40 changes: 18 additions & 22 deletions examples/mplot3d/lorenz_attractor.py
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Expand Up @@ -18,45 +18,41 @@
import matplotlib.pyplot as plt


def lorenz(x, y, z, s=10, r=28, b=2.667):
def lorenz(xyz, *, s=10, r=28, b=2.667):
"""
Given:
x, y, z: a point of interest in three dimensional space
s, r, b: parameters defining the lorenz attractor
Returns:
x_dot, y_dot, z_dot: values of the lorenz attractor's partial
derivatives at the point x, y, z
Parameters
----------
xyz : array-like, shape (3,)
Point of interest in three dimensional space.
s, r, b : float
Parameters defining the Lorenz attractor.

Returns
-------
xyz_dot : array, shape (3,)
Values of the Lorenz attractor's partial derivatives at *xyz*.
"""
x, y, z = xyz
x_dot = s*(y - x)
y_dot = r*x - y - x*z
z_dot = x*y - b*z
return x_dot, y_dot, z_dot
return np.array([x_dot, y_dot, z_dot])


dt = 0.01
num_steps = 10000

# Need one more for the initial values
xs = np.empty(num_steps + 1)
ys = np.empty(num_steps + 1)
zs = np.empty(num_steps + 1)

# Set initial values
xs[0], ys[0], zs[0] = (0., 1., 1.05)

xyzs = np.empty((num_steps + 1, 3)) # Need one more for the initial values
xyzs[0] = (0., 1., 1.05) # Set initial values
# Step through "time", calculating the partial derivatives at the current point
# and using them to estimate the next point
for i in range(num_steps):
x_dot, y_dot, z_dot = lorenz(xs[i], ys[i], zs[i])
xs[i + 1] = xs[i] + (x_dot * dt)
ys[i + 1] = ys[i] + (y_dot * dt)
zs[i + 1] = zs[i] + (z_dot * dt)

xyzs[i + 1] = xyzs[i] + lorenz(xyzs[i]) * dt

# Plot
ax = plt.figure().add_subplot(projection='3d')

ax.plot(xs, ys, zs, lw=0.5)
ax.plot(*xyzs.T, lw=0.5)
ax.set_xlabel("X Axis")
ax.set_ylabel("Y Axis")
ax.set_zlabel("Z Axis")
Expand Down