# ODE with periodic solution¶

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 $$x^2(t) + y^2(t) = 1, \qquad \forall t$$

In :
import numpy as np
from matplotlib import pyplot as plt


## Forward Euler scheme¶

In :
def ForwardEuler(h,T):
N = int(T/h)
x,y = np.zeros(N),np.zeros(N)
x = 1.0
y = 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

In :
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.

## Backward Euler scheme¶

$$x_n = x_{n-1} - h y_n, \qquad y_n = y_{n-1} + h x_n$$

Eliminate $y_n$ from first equation to get $$x_n = \frac{x_{n-1} - h y_{n-1}}{1 + h^2}$$

In :
def BackwardEuler(h,T):
N = int(T/h)
x,y = np.zeros(N),np.zeros(N)
x = 1.0
y = 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

In :
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.

## Trapezoidal scheme¶

$$x_n = x_{n-1} - \frac{h}{2}(y_{n-1} + y_n), \qquad y_n = y_{n-1} + \frac{h}{2}(x_{n-1} + x_n)$$

Eliminate $y_n$ from first equation $$x_n = \frac{ (1-\frac{1}{4}h^2) x_{n-1} - h y_{n-1} }{1 + \frac{1}{4}h^2}$$

In :
def Trapezoid(h,T):
N = int(T/h)
x,y = np.zeros(N),np.zeros(N)
x = 1.0
y = 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

In :
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.