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$$
\begin{align}
u'(t) &= au(t)
\tag{1}\\
u(0) &= I
\tag{2}
\end{align}
$$
Here,


The \( \theta \) rule, $$ u^{n+1} = \frac{1  (1\theta) a\Delta t}{1 + \theta a\Delta t}u^n, \quad n=0,1,\ldots,N1 $$ contains the Forward Euler (\( \theta=0 \)), the Backward Euler (\( \theta=1 \)), and the CrankNicolson (\( \theta=0.5 \)) schemes.
def solver(I, a, T, dt, theta):
"""Solve u'=a*u, u(0)=I, for t in (0,T]; step: dt."""
dt = float(dt) # avoid integer division
N = int(round(T/dt)) # no of time intervals
T = N*dt # adjust T to fit time step dt
u = zeros(N+1) # array of u[n] values
t = linspace(0, T, N+1) # time mesh
u[0] = I # assign initial condition
for n in range(0, N): # n=0,1,...,N1
u[n+1] = (1  (1theta)*a*dt)/(1 + theta*dt*a)*u[n]
return u, t
# Set problem parameters
I = 1.2
a = 0.2
T = 8
dt = 0.25
theta = 0.5
from solver import solver, exact_solution
u, t = solver(I, a, T, dt, theta)
import matplotlib.pyplot as plt
plt.plot(t, u, t, exact_solution)
plt.legend(['numerical', 'exact'])
plt.show()
Exact solution of the scheme: $$ u^n = A^n,\quad A = \frac{1  (1\theta) a\Delta t}{1 + \theta a\Delta t}\thinspace .$$
Key results:
Only the Backward Euler scheme is guaranteed to always give qualitatively correct results.