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Euler stepping, convergence and stability

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Start with the idea

A numerical solver repeatedly approximates the change over a short time interval.
Explicit Euler uses the slope at the beginning of each interval.
Smaller steps can reduce approximation error, but the scheme must also be stable for the equation being solved.

Symbols, units & horizon

n: time-step index
h: positive time-step length
t_n=nh: grid time
f(t,y): rate function defining the ODE
y_n: numerical state, not generally the exact solution at t_n
κ: positive decay speed
|1−κh|<1: stability condition for this scalar Euler decay example

When and why to use this

Use an exact solvable equation as a benchmark, compare several step sizes, and test stability. The ODE lab deliberately allows unstable steps so their consequences are visible.

A local slope is not constant over a finite interval, so replacing the full path with repeated tangent steps introduces discretisation error.
Under standard smoothness and stability conditions, explicit Euler has first-order global error over a fixed horizon.
For a decay equation, a step can be so large that the numerical solution oscillates or grows even though the exact solution decays.
This is an algorithmic effect, not newly discovered behaviour of the physical or financial model.

y_{n + 1} = y_{n} + h f (t_{n}, y_{n}), y^{'} = - κ y \Rightarrow y_{n + 1} = (1 - κ h) y_{n}

Calculus: derivation and arithmetic

Euler stepping, convergence and stability

Integrate y′=f(t,y) over one step: y(t+h)−y(t)=∫ₜᵗ⁺ʰ f(s,y(s))ds. Replace the integrand by its starting value to obtain Euler’s update.
For f=−κy, factor the update into (1−κh)y_n. Repeating n times gives y_n=(1−κh)ⁿy₀.
Decay of the numerical magnitude requires |1−κh|<1, equivalent to 0<κh<2. If 1<κh<2 the numerical path alternates sign even though its magnitude decays.

Work it by hand

For κ=2 and y₀=1, h=.25 gives successive values 1,.5,.25. The exact value at t=.5 is exp(−1)≈.3679. With h=1.1 the multiplier is −1.2, so the numerical magnitude grows.

An analogy to remember

Following a curved route using straight steps works better with appropriately short steps. A step that is too long can cross the destination and keep overshooting.

How this becomes a building block

ODE solvers support calibration, filtering and deterministic state evolution. Euler–Maruyama adds a Brownian increment to a similar update, but stochastic convergence notions differ. Solver error should be checked before interpreting a simulation result as model risk or economic evidence.

Python implementation

Self-contained teaching example. Python 3.10+; dependencies and input conventions are shown in the code and notation. Run in your own Python environment.

def euler(function,initial,start,end,steps):
    if not isinstance(steps,int) or steps<1 or end<=start: raise ValueError("Valid interval and positive steps required")
    h=(end-start)/steps
    t,y=start,initial
    path=[(t,y)]
    for i in range(steps):
        y=y+h*function(t,y)
        t=start+(i+1)*h
        path.append((t,y))
    return path

print(euler(lambda t,y:-2*y,1,0,.5,2))

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