Table of Contents¶
Runge-Kutta methods for ODE integration in Python¶
I want to implement and illustrate the Runge-Kutta method (actually, different variants), in the Python programming language.
The Runge-Kutta methods are a family of numerical iterative algorithms to approximate solutions of Ordinary Differential Equations. I will simply implement them, for the mathematical descriptions, I let the interested reader refer to the Wikipedia page, or any good book or course on numerical integration of ODE.
- I will start with the order 1 method, then the order 2 and the most famous order 4.
- They will be compared on different ODE.
Preliminary¶
import numpy as np
import matplotlib.pyplot as plt
%load_ext watermark
%watermark
from scipy.integrate import odeint # for comparison
I will use as a first example the one included in the scipy documentation for this odeint function.
$$\theta''(t) + b \theta'(t) + c \sin(\theta(t)) = 0.$$
If $\omega(t) = \theta'(t)$, this gives $$ \begin{cases} \theta'(t) = \omega(t) \\ \omega'(t) = -b \omega(t) - c \sin(\theta(t)) \end{cases} $$
Vectorially, if $y(t) = [\theta(t), \omega(t)]$, then the equation is $y' = f(t, y)$ where $f(t, y) = [y_2(t), -b y_2(t) - c \sin(y_1(t))]$.
def pend(y, t, b, c):
return np.array([y[1], -b*y[1] - c*np.sin(y[0])])
We assume the values of $b$ and $c$ to be known, and the starting point to be also fixed:
b = 0.25
c = 5.0
y0 = np.array([np.pi - 0.1, 0.0])
The odeint function will be used to solve this ODE on the interval $t \in [0, 10]$, with $101$ points.
t = np.linspace(0, 10, 101)
It is used like this, and our implementations will follow this signature.
sol = odeint(pend, y0, t, args=(b, c))
plt.plot(t, sol[:, 0], 'b', label=r'$\theta(t)$')
plt.plot(t, sol[:, 1], 'g', label=r'$\omega(t)$')
plt.legend(loc='best')
plt.xlabel('t')
plt.grid()
plt.show()
Runge-Kutta method of order 1, or the Euler method¶
The approximation is computed using this update: $$y_{n+1} = y_n + (t_{n+1} - t_n) f(y_n, t_n).$$
The math behind this formula are the following: if $g$ is a solution to the ODE, and so far the approximation is correct, $y_n \simeq g(t_n)$, then a small step $h = t_{n+1} - t_n$ satisfy $g(t_n + h) \simeq g(t_n) + h g'(t_n) \simeq y_n + h f(g(t_n), t_n) + \simeq y_n + h f(y_n, t_n)$.
def rungekutta1(f, y0, t, args=()):
n = len(t)
y = np.zeros((n, len(y0)))
y[0] = y0
for i in range(n - 1):
y[i+1] = y[i] + (t[i+1] - t[i]) * f(y[i], t[i], *args)
return y
sol = rungekutta1(pend, y0, t, args=(b, c))
plt.plot(t, sol[:, 0], 'b', label=r'$\theta(t)$')
plt.plot(t, sol[:, 1], 'g', label=r'$\omega(t)$')
plt.legend(loc='best')
plt.xlabel('t')
plt.grid()
plt.show()
With the same number of points, the Euler method (i.e. the Runge-Kutta method of order 1) is less precise than the reference odeint method. With more points, it can give a satisfactory approximation of the solution:
t2 = np.linspace(0, 10, 1001)
sol2 = rungekutta1(pend, y0, t2, args=(b, c))
t3 = np.linspace(0, 10, 10001)
sol3 = rungekutta1(pend, y0, t3, args=(b, c))
plt.plot(t, sol[:, 0], label=r'$\theta(t)$ with 101 points')
plt.plot(t2, sol2[:, 0], label=r'$\theta(t)$ with 1001 points')
plt.plot(t3, sol3[:, 0], label=r'$\theta(t)$ with 10001 points')
plt.legend(loc='best')
plt.xlabel('t')
plt.grid()
plt.show()
Runge-Kutta method of order 2¶
The order 2 Runge-Method uses this update: $$ y_{n+1} = y_n + h f(t + \frac{h}{2}, y_n + \frac{h}{2} f(t, y_n)),$$ if $h = t_{n+1} - t_n$.
def rungekutta2(f, y0, t, args=()):
n = len(t)
y = np.zeros((n, len(y0)))
y[0] = y0
for i in range(n - 1):
h = t[i+1] - t[i]
y[i+1] = y[i] + h * f(y[i] + f(y[i], t[i], *args) * h / 2., t[i] + h / 2., *args)
return y
For our simple ODE example, this method is already quite efficient.
t4 = np.linspace(0, 10, 21)
sol4 = rungekutta2(pend, y0, t4, args=(b, c))
t = np.linspace(0, 10, 101)
sol = rungekutta2(pend, y0, t, args=(b, c))
t2 = np.linspace(0, 10, 1001)
sol2 = rungekutta2(pend, y0, t2, args=(b, c))
t3 = np.linspace(0, 10, 10001)
sol3 = rungekutta2(pend, y0, t3, args=(b, c))
plt.plot(t4, sol4[:, 0], label='with 11 points')
plt.plot(t, sol[:, 0], label='with 101 points')
plt.plot(t2, sol2[:, 0], label='with 1001 points')
plt.plot(t3, sol3[:, 0], label='with 10001 points')
plt.legend(loc='best')
plt.xlabel('t')
plt.grid()
plt.show()
Runge-Kutta method of order 4, "RK4"¶
The order 4 Runge-Method uses this update: $$ y_{n+1} = y_n + \frac{h}{6} (k_1 + 2 k_2 + 2 k_3 + k_4),$$ if $h = t_{n+1} - t_n$, and $$\begin{cases} k_1 &= f(y_n, t_n), \\ k_2 &= f(y_n + \frac{h}{2} k_1, t_n + \frac{h}{2}), \\ k_3 &= f(y_n + \frac{h}{2} k_2, t_n + \frac{h}{2}), \\ k_4 &= f(y_n + h k_3, t_n + h). \end{cases}$$
def rungekutta4(f, y0, t, args=()):
n = len(t)
y = np.zeros((n, len(y0)))
y[0] = y0
for i in range(n - 1):
h = t[i+1] - t[i]
k1 = f(y[i], t[i], *args)
k2 = f(y[i] + k1 * h / 2., t[i] + h / 2., *args)
k3 = f(y[i] + k2 * h / 2., t[i] + h / 2., *args)
k4 = f(y[i] + k3 * h, t[i] + h, *args)
y[i+1] = y[i] + (h / 6.) * (k1 + 2*k2 + 2*k3 + k4)
return y
For our simple ODE example, this method is even more efficient.
t4 = np.linspace(0, 10, 21)
sol4 = rungekutta4(pend, y0, t4, args=(b, c))
t = np.linspace(0, 10, 101)
sol = rungekutta4(pend, y0, t, args=(b, c))
t2 = np.linspace(0, 10, 1001)
sol2 = rungekutta4(pend, y0, t2, args=(b, c))
plt.plot(t4, sol4[:, 0], label='with 21 points')
plt.plot(t, sol[:, 0], label='with 101 points')
plt.plot(t2, sol2[:, 0], label='with 1001 points')
plt.legend(loc='best')
plt.xlabel('t')
plt.grid()
plt.show()
I also want to try to speed this function up by using numba.
from numba import jit
@jit
def rungekutta4_jit(f, y0, t, args=()):
n = len(t)
y = np.zeros((n, len(y0)))
y[0] = y0
for i in range(n - 1):
h = t[i+1] - t[i]
k1 = f(y[i], t[i], *args)
k2 = f(y[i] + k1 * h / 2., t[i] + h / 2., *args)
k3 = f(y[i] + k2 * h / 2., t[i] + h / 2., *args)
k4 = f(y[i] + k3 * h, t[i] + h, *args)
y[i+1] = y[i] + (h / 6.) * (k1 + 2*k2 + 2*k3 + k4)
return y
Both versions compute the same thing.
t2 = np.linspace(0, 10, 1001)
sol2 = rungekutta4(pend, y0, t2, args=(b, c))
sol2_jit = rungekutta4_jit(pend, y0, t2, args=(b, c))
np.linalg.norm(sol2 - sol2_jit)
Comparisons¶
methods = [odeint, rungekutta1, rungekutta2, rungekutta4]
markers = ['+', 'o', 's', '>']
def test_1(n=101):
t = np.linspace(0, 10, n)
for method, m in zip(methods, markers):
sol = method(pend, y0, t, args=(b, c))
plt.plot(t, sol[:, 0], label=method.__name__, marker=m)
plt.legend(loc='best')
plt.title("Comparison of different ODE integration methods for $n={}$ points".format(n))
plt.xlabel("$t = [0, 10]$")
plt.grid()
plt.show()
test_1(10)
test_1(20)
test_1(100)
test_1(200)
Comparisons on another integration problem¶
Consider the following ODE on $t\in[0, 1]$: $$ \begin{cases} y'''(t) = 12 y(t)^{4/5} + \cos(y'(t))^3 - \sin(y''(t)) \\ y(0) = 0, y'(0) = 1, y''(0) = 0.1 \end{cases} $$
It can be written in a vectorial form like the first one:
def f(y, t):
return np.array([y[1], y[2], 12 * y[0] ** (4/5.) + np.cos(y[1])**3 - np.sin(y[2])])
def test_2(n=101):
t = np.linspace(0, 1, n)
y0 = np.array([0, 1, 0.1])
for method, m in zip(methods, markers):
sol = method(f, y0, t)
plt.plot(t, sol[:, 0], label=method.__name__, marker=m)
plt.legend(loc='best')
plt.title("Comparison of different ODE integration methods for $n={}$ points".format(n))
plt.xlabel("$t = [0, 1]$")
plt.grid()
plt.show()
test_2(10)
test_2(50)
Consider the following ODE on $t\in[0, 3]$: $$ \begin{cases} y''''(t) = y(t)^{-5/3} \\ y(0) = 10, y'(0) = -3, y''(0) = 1, y'''(0) = 1 \end{cases} $$
It can be written in a vectorial form like the first one:
def f(y, t):
return np.array([y[1], y[2], y[3], y[0]**(-5/3.)])
def test_3(n=101):
t = np.linspace(0, 3, n)
y0 = np.array([10, -3, 1, 1])
for method, m in zip(methods, markers):
sol = method(f, y0, t)
plt.plot(t, sol[:, 0], label=method.__name__, marker=m)
plt.legend(loc='best')
plt.title("Comparison of different ODE integration methods for $n={}$ points".format(n))
plt.xlabel("$t = [0, 1]$")
plt.grid()
plt.show()
test_3(10)
test_3(50)
Our hand-written Runge-Kutta method of order 4 seems to be as efficient as the odeint method from scipy... and that's because odeint basically uses a Runge-Kutta method of order 4 (with smart variants).
Small benchmark¶
We can also compare their speed:
methods = [odeint, rungekutta1, rungekutta2, rungekutta4, rungekutta4_jit]
y0 = np.array([10, -3, 1, 1])
for n in [20, 100, 1000]:
print("\n")
t = np.linspace(0, 3, n)
for method in methods:
print("Time of solving this ODE for {} points with {} method...".format(n, method.__name__))
%timeit sol = method(f, y0, t)
- Well, that's disappointing, the Numba Jit version was NOT faster than the manual implementation...
- The order 1 method is simpler and so faster than the order 2, which itself is simpler and faster than the order 4 method.
- And we can check that the SciPy implementation is much faster than our manual implentations!
Conclusion¶
That's it for today, folks! See my other notebooks, available on GitHub.