Consider the ODE system $$ x' = -y, \qquad y' = x $$ with initial condition $$ x(0) = 1, \qquad y(0) = 0 $$ The exact solution is $$ x(t) = \cos(t), \qquad y(t) = \sin(t) $$ This solution is periodic. It also has a quadratic invariant $$ x^2(t) + y^2(t) = 1, \qquad \forall t $$
import numpy as np
from matplotlib import pyplot as plt
def ForwardEuler(h,T):
N = int(T/h)
x,y = np.zeros(N),np.zeros(N)
x[0] = 1.0
y[0] = 0.0
for n in range(1,N):
x[n] = x[n-1] - h*y[n-1]
y[n] = y[n-1] + h*x[n-1]
return x,y
h = 0.02
T = 4.0*np.pi
x,y = ForwardEuler(h,T)
plt.plot(x,y)
plt.axes().set_aspect('equal')
The phase space trajectory is spiralling outward.
Eliminate $y_n$ from first equation to get $$ x_n = \frac{x_{n-1} - h y_{n-1}}{1 + h^2} $$
def BackwardEuler(h,T):
N = int(T/h)
x,y = np.zeros(N),np.zeros(N)
x[0] = 1.0
y[0] = 0.0
for n in range(1,N):
x[n] = (x[n-1] - h*y[n-1])/(1.0 + h**2)
y[n] = y[n-1] + h*x[n]
return x,y
h = 0.02
T = 4.0*np.pi
x,y = BackwardEuler(h,T)
plt.plot(x,y)
plt.axes().set_aspect('equal')
The phase space trajectory is spiralling inward.
Eliminate $y_n$ from first equation $$ x_n = \frac{ (1-\frac{1}{4}h^2) x_{n-1} - h y_{n-1} }{1 + \frac{1}{4}h^2} $$
def Trapezoid(h,T):
N = int(T/h)
x,y = np.zeros(N),np.zeros(N)
x[0] = 1.0
y[0] = 0.0
for n in range(1,N):
x[n] = ((1.0-0.25*h**2)*x[n-1] - h*y[n-1])/(1.0 + 0.25*h**2)
y[n] = y[n-1] + 0.5*h*(x[n-1] + x[n])
return x,y
h = 0.02
T = 4.0*np.pi
x,y = Trapezoid(h,T)
plt.plot(x,y)
plt.axes().set_aspect('equal')
The phase space trajectory is exactly the unit circle.
Multiply first equation by $x_n + x_{n-1}$ and second equation by $y_n + y_{n-1}$ $$ (x_n + x_{n-1})(x_n - x_{n-1}) = - \frac{h}{2}(x_n + x_{n-1})(y_n + y_{n-1}) $$ $$ (y_n + y_{n-1})(y_n - y_{n-1}) = + \frac{h}{2}(x_n + x_{n-1})(y_n + y_{n-1}) $$ Adding the two equations we get $$ x_n^2 + y_n^2 = x_{n-1}^2 + y_{n-1}^2 $$ Thus the Trapezoidal method is able to preserve the invariant.