Stiff equations — implicit methods, backward Euler: Worked examples

Before anything, a one-line refresher of the two amplification factors — the numbers that get raised to the power $n$ :

We call $z = hλ$ the stability variable. Every example below is really asking: where does this $z$ land, and what does each $R$ do there?

The scenario matrix

Every case a stiff-ODE problem can hand you, and the example that hits it:

Cell	Case class	What is special	Example
A	$λ < 0$ , small $∣ z ∣$	both methods stable, mild step	Ex 1
B	$λ < 0$ , $z$ in FE danger band ( $- 2 < z < - 1$ )	FE oscillates but bounded	Ex 2
C	$λ < 0$ , $z < - 2$	FE blows up, BE fine	Ex 3
D	Boundary $z = - 2$ exactly	FE marginal ( $∣ R ∣ = 1$ )	Ex 4
E	Degenerate $λ = 0$ ( $z = 0$ )	no decay, both give constant	Ex 5
F	$λ > 0$ (true growth)	neither should force $\to 0$ ; check both faithfully grow	Ex 6
G	Complex $λ$ (oscillatory decay)	left half-plane, magnitude test	Ex 7
H	Nonlinear $f$ → Newton needed	implicit solve by iteration	Ex 8
I	Real-world word problem (two time scales)	choose method + step	Ex 9
J	Exam twist: pick largest stable $h$	invert the stability test	Ex 10

The stability variable $z = hλ$ lives on a number line (real $λ$ ) or in a plane (complex $λ$ ). This picture shows the safe zones of each method — refer back to it constantly:

Figure — Stiff equations — implicit methods, backward Euler

Worked examples

Worked example Example 7 — Cell G: complex

$λ$ , oscillatory decay $y^{'} = λ y$ with $λ = - 1 + 4 i$ , $h = 0.5$ . Is backward Euler stable? (magnitude test)

Forecast: does $Re λ < 0$ guarantee BE stability?

$z = hλ = 0.5 (- 1 + 4 i) = - 0.5 + 2 i$ . Why this step? Now $z$ is a point in the plane, not on a line.
$1 - z = 1 - (- 0.5 + 2 i) = 1.5 - 2 i$ ; its magnitude $∣1 - z ∣ = 1. 5^{2} + 2^{2} = 2.25 + 4 = 6.25 = 2.5$ . Why this step? $∣ R_{BE} ∣ = \frac{1}{∣1 - z ∣}$ , so we need the size of the denominator.
$∣ R_{BE} ∣ = \frac{1}{2.5} = 0.4 < 1$ → stable. Why this step? Magnitude below 1 means the oscillation decays, matching $∣ e^{λ t} ∣ = e^{- t} \to 0$ .

Verify: $Re λ = - 1 < 0$ , and A-stability guarantees $∣ R_{BE} ∣ < 1$ for the whole left half-plane — confirmed numerically ( $0.4$ ). Forward Euler here: $∣1 + z ∣ = ∣0.5 + 2 i ∣ = 0.25 + 4 = 4.25 = 2.06 > 1$ → FE unstable even though the true solution decays. ✓ (See Region of absolute stability.)

Worked example Example 8 — Cell H: nonlinear

$f$ , one Newton solve Backward Euler on $y^{'} = - y^{2}$ , $y_{0} = 2$ , $h = 0.5$ . Find $y_{1}$ using Newton's method for root finding.

Forecast: the implicit equation is quadratic — will Newton land in a couple of steps?

Set up $g (y) = y - y_{0} - h (- y^{2}) = y - 2 + 0.5 y^{2} = 0$ . Why this step? Backward Euler is $y_{1} = y_{0} + h f (y_{1})$ ; moving all terms to one side gives a root problem.
Derivative $g^{'} (y) = 1 + h \cdot 2 y = 1 + y$ (since $h = 0.5$ ). Why this step? Newton needs $g^{'}$ ; here $g^{'} = 1 - h \partial f / \partial y = 1 - 0.5 (- 2 y) = 1 + y$ .
Start $y^{(0)} = y_{0} = 2$ . $g (2) = 2 - 2 + 0.5 (4) = 2$ , $g^{'} (2) = 3$ . $y^{(1)} = 2 - 2/3 = 1.3333$ . Why this step? One Newton step: subtract $g / g^{'}$ .
$g (1.3333) = 1.3333 - 2 + 0.5 (1.7778) = 0.2222$ , $g^{'} = 2.3333$ . $y^{(2)} = 1.3333 - 0.09524 = 1.2381$ . Why this step? Iterate; residual shrinking fast (quadratic convergence).
One more: $g (1.2381) = 1.2381 - 2 + 0.5 (1.5330) = 0.00460$ , $g^{'} = 2.2381$ , $y^{(3)} = 1.2381 - 0.002056 = 1.23605$ .

Verify: exact root of $0.5 y^{2} + y - 2 = 0$ is $y = \frac{- 1 + 1 + 4}{1} = \frac{- 1 + 5}{1} = 1.23607$ (positive root). Newton reached $1.23605$ in three steps — matches. And $g (1.23607) = 0$ . ✓

Worked example Example 9 — Cell I: real-world two-time-scale problem

A chemical mixes: fast species decays like $e^{- 500 t}$ , slow like $e^{- t}$ , model $y^{'} = - 500 y$ (transient) coupled with a slow drift. To integrate to $t = 5$ after the fast part is dead, which method and roughly how many steps?

Forecast: guess the step-count ratio between FE and BE.

Fastest rate $∣ λ ∣ = 500$ , so FE needs $h < \frac{2}{500} = 0.004$ . Why this step? FE step is capped by the fastest scale for stability, even after that scale is gone.
Steps for FE to reach $t = 5$ : $\frac{5}{0.004} = 1250$ steps. Why this step? Total time ÷ max step.
BE has no stability cap; the slow scale (rate 1) allows $h \approx 0.1$ for accuracy → $\frac{5}{0.1} = 50$ steps. Why this step? A-stability lets us pick $h$ for accuracy alone.

Verify: FE $1250$ vs BE $50$ steps — a $25 \times$ saving, and the ratio $\approx ∣ λ_{fast} ∣/∣ λ_{slow} ∣ \times (accuracy factor)$ scales with the stiffness ratio $500$ . ✓

Worked example Example 10 — Cell J: exam twist, largest stable

$h$ "For $y^{'} = - 40 y$ , find the largest step $h$ for which forward Euler is stable, and confirm backward Euler has no such limit."

Forecast: invert the stability inequality — what's the cutoff?

FE stable needs $∣1 + hλ ∣ < 1$ , i.e. $- 1 < 1 - 40 h < 1$ . Why this step? Plug $λ = - 40$ into the test and read off the window.
Left inequality $1 - 40 h > - 1 \Rightarrow 40 h < 2 \Rightarrow h < 0.05$ . Right inequality $1 - 40 h < 1$ is automatic for $h > 0$ . Why this step? The binding constraint is $h < 2/∣ λ ∣$ .
So the supremum is $h = 0.05$ (strict; at $h = 0.05$ , $∣ R ∣ = 1$ , marginal — see Ex 4). Why this step? Equality is the boundary, not stable.
Backward Euler: $∣1/ (1 - hλ) ∣ = ∣1/ (1 + 40 h) ∣ < 1$ for all $h > 0$ since $1 + 40 h > 1$ . Why this step? Denominator always exceeds 1 for decaying $λ$ — no upper bound on $h$ .

Verify: $2/∣ λ ∣ = 2/40 = 0.05$ matches. At $h = 0.05$ : $R_{FE} = 1 - 2 = - 1$ (marginal). BE at even $h = 10$ : $1/ (1 + 400) = 0.0025 < 1$ still stable. ✓

Recall Quick self-test across the matrix

Which cell blows up forward Euler but leaves backward Euler untouched? ::: Cell C ( $z < - 2$ ), e.g. Ex 3. At $z = - 2$ exactly, what does forward Euler do? ::: Oscillates with constant magnitude $∣ R ∣ = 1$ — not stable (Cell D). When is "unstable growth" actually the correct behaviour? ::: When $λ > 0$ (Cell F): the true solution grows, so both methods should too. For complex $λ$ , what replaces " $z < 0$ "? ::: The magnitude test $∣ R ∣ < 1$ , e.g. $∣1/ (1 - z) ∣ < 1$ over the whole left half-plane (Cell G).

Connections

4.8.26 Stiff equations — implicit methods, backward Euler (Hinglish)
Forward Euler method
Region of absolute stability
A-stability and L-stability
Trapezoidal / Crank–Nicolson method
Newton's method for root finding
Runge–Kutta methods
Eigenvalues and time scales of linear systems