Recipe

Find Euler's stability threshold for a stiff equation

Use when: the equation locally reduces to $y'=-k(y-g(x))$ (or exactly $y'=-ky$ ) near an equilibrium, with $k$ large and $g(x)$ slowly varying, and you need the largest Euler step size that stays stable

Goal: the stability bound $h<2/k$ , plus a way to check whether a specific candidate $h$ satisfies it

Taught in Stiffness, informally

Quick recipe

Identify the local decay rate $k$
Apply Euler's method to the local linear equation
Solve $|1-hk|<1$ for $h$
Check a specific candidate step size against the bound

1
Write the equation in (or linearize it near an equilibrium into) the form $y'=-k(y-g(x))$ , treating $g(x)$ as roughly constant over one step; $k$ is the coefficient multiplying $(y-g(x))$ .

Example

For $y'=-50(y-\cos x)$ , $k=50$ directly — no linearization needed since the equation is already in this form.

Why: Euler's stability behavior near an equilibrium of this kind depends only on $k$ , not on the slowly varying $g(x)$ term riding along with it.
2
Substituting the local form into $y_{n+1}=y_n+hf(y_n)$ gives $y_{n+1}=(1-hk)y_n$ , measuring $y$ relative to the (locally frozen) equilibrium.

Example

With $k=50$ : $y_{n+1}=(1-50h)y_n$ .

Why: this is Euler's own update rule, applied to the simplest equation that captures the fast-decaying component's behavior in isolation.
3
The iteration stays bounded exactly when $|1-hk|<1$ , i.e. $0<h<2/k$ .

Example

$k=50$ : $h<2/50=0.04$ .

⚡ Shortcut

For any equation that locally reduces to $y'=-ky$ , the bound is always exactly $h<2/k$ — there's no need to redo the algebra above each time. A larger $k$ (a faster-decaying component) always forces a smaller stable step, in exact inverse proportion.

Why: $|1-hk|>1$ makes each step multiply the error by something bigger than $1$ in magnitude, so the numerical solution grows — and since $1-hk$ can be negative, oscillates in sign — without bound.
4
Compute $|1-hk|$ for the proposed $h$ directly; it's stable exactly when that number is under $1$ .

Example

$k=50$ , $h=0.03$ : $|1-0.03(50)|=|1-1.5|=0.5<1$ — stable. $k=50$ , $h=0.045$ : $|1-0.045(50)|=|1-2.25|=1.25>1$ — unstable.
- $|1-hk|<1$ — stable, safe to use.
- $|1-hk|\ge1$ — unstable; the step must be shrunk below $2/k$ .
Why: this is the same stability condition from Step 3, just evaluated numerically instead of solved symbolically — useful when you're handed a candidate $h$ rather than asked to find the threshold from scratch.

Common mistakes

Using the equation's overall behavior (how $g(x)$ varies) to judge the step size, instead of the fast local rate $k$ that actually governs stability.
Computing $h<2/k$ but then picking $h$ equal to (rather than strictly below) the threshold and calling it safe — $h=2/k$ sits exactly on the boundary, where $|1-hk|=1$ oscillates without decaying.
Confusing accuracy with stability — a step size can satisfy $h<2/k$ (stable, not blowing up) while still being far too large to track the solution's shape accurately.
Forgetting the absolute value in $|1-hk|<1$ and checking only $1-hk<1$ , which holds for every positive $h$ and misses the actual constraint entirely.