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$$ \begin{align} u'(t) &= -au(t) \label{ode}\\ u(0) &= I \label{initial:value} \end{align} $$

Here,

- \( t\in (0,T] \)
- \( a \), \( I \), and \( T \) are prescribed parameters
- \( u(t) \) is the unknown function
- The ODE \eqref{ode} has the initial condition \eqref{initial:value}

- Mesh in time: \( 0= t_0 < t_1 \cdots < t_N=T \)
- Assume constant \( \Delta t = t_{n}-t_{n-1} \)
- \( u^n \): numerical approx to the exact solution at \( t_n \)

The \( \theta \) rule, $$ u^{n+1} = \frac{1 - (1-\theta) a\Delta t}{1 + \theta a\Delta t}u^n, \quad n=0,1,\ldots,N-1 $$ contains the Forward Euler (\( \theta=0 \)), the Backward Euler (\( \theta=1 \)), and the Crank-Nicolson (\( \theta=0.5 \)) 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,...,N-1
u[n+1] = (1 - (1-theta)*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:

- Stability: \( |A| < 1 \)
- No oscillations: \( A>0 \)
- \( \Delta t < 1/a \) for Forward Euler (\( \theta=0 \))
- \( \Delta t < 2/a \) for Crank-Nicolson (\( \theta=1/2 \))

Only the Backward Euler scheme is guaranteed to always give qualitatively correct results.