Modified Euler (Heun's method): Question bank

The one picture to keep in your head: two slopes drawn at the two ends of a step, and a single walk taken along their average direction.

True or false — justify

Heun's method is always more accurate than plain Euler for the same

h

Usually, not always. For smooth problems it is far better (

O (h^{2})

O (h)

), but on a very stiff or badly-scaled ODE with too-large

h

, the predictor can overshoot so wildly that the corrected step is worse than Euler's — accuracy order only describes the limit as

h \to 0

Heun's method evaluates the function

f

exactly twice per step.

True for basic Heun: once for

k_{1} = f (x_{0}, y_{0})

and once for

k_{2} = f (x_{1}, y_{1}^{(p)})

. The iterated corrector variant re-evaluates

f

again for each extra sweep, so it costs more.

The predictor value

y_{1}^{(p)}

is the final answer for the step.

False.

y_{1}^{(p)}

is only a scratch guess whose sole purpose is to let us compute the end-slope

k_{2}

; it is thrown away and replaced by the corrected

y_{1}

Heun's method is a Runge–Kutta method.

True. It is RK2 — a 2-stage Runge–Kutta scheme with equal weights

\frac{1}{2}, \frac{1}{2}

on its two slopes.

f (x, y)

depends only on

x

(not on

y

), Heun's method reduces to the trapezoidal rule for a plain integral.

True. When

f = f (x)

, the predicted

y

never enters

f

, so

y_{1} = y_{0} + \frac{h}{2} [f (x_{0}) + f (x_{1})]

— exactly the Trapezoidal Rule applied to

\int f d x

Doubling the step size

h

roughly doubles Heun's global error.

False. Global error is

O (h^{2})

, so doubling

h

multiplies error by about

4

, not

2

. (Doubling does roughly double plain Euler's

O (h)

error.)

The

\frac{h}{2}

in the corrector is a rounding convenience with no deeper meaning.

False. The

\frac{1}{2}

is the average of two slopes: each slope contributes half the step. Replace it with

h

and you take a double-length step — a genuine error, not a style choice.

Heun's local (per-step) error and global error have the same order.

False. Local error is

O (h^{3})

per step; accumulating over

\sim 1/ h

steps drops one power, giving global

O (h^{2})

. This one-power drop is generic to one-step methods.

Spot the error

A student writes the corrector as

y_{1} = y_{0} + \frac{h}{2} [f (x_{0}, y_{0}) + f (x_{1}, y_{0})]

. What's wrong?

The end-slope is evaluated at the old

y_{0}

instead of the predicted

y_{1}^{(p)} = y_{0} + h k_{1}

. Ignoring how

y

moved across the step destroys the second-order accuracy — you lose the very "look-ahead" that defines Heun.

A student computes

k_{2} = f (x_{0}, y_{0} + h k_{1})

— same

x_{0}

, moved

y

. Fix it.

The end-slope must be sampled at the end point

x_{1} = x_{0} + h

, so

k_{2} = f (x_{0} + h, y_{0} + h k_{1})

. Advancing

y

but not

x

evaluates

f

at a point that never exists on the step.

A student uses the corrected

y_{1}

back inside

k_{1}

before finishing the step. Why is this circular?

k_{1} = f (x_{0}, y_{0})

is fixed by the start of the step and is known before anything else. Feeding a later value into it means you no longer have a start-slope — the method loses its predictor and stops being Heun.

Someone writes

y_{1} = y_{0} + h f (x_{0}, y_{0}) + \frac{h}{2} f (x_{1}, y_{1}^{(p)})

. Diagnose.

They kept a full Euler step

h k_{1}

and added a half end-slope — a mix of Euler and Heun. The correct total weight on the two slopes must sum to

h

: it should be

\frac{h}{2} k_{1} + \frac{h}{2} k_{2}

In Example 3 a student iterates the corrector but keeps re-using the very first predictor

y^{(p)} = 1.1

for

k_{2}

each sweep. What's the mistake?

Iterating means re-predicting the end-slope from the latest corrected

y

(e.g.

1.11

), so

k_{2}

updates each sweep. Freezing

k_{2}

at the original guess just repeats the first corrector and converges to nothing new.

Why questions

Why do we predict

y_{1}

before we can correct it?

The trapezoidal corrector needs

f (x_{1}, y_{1})

, but

y_{1}

is the unknown we are solving for. A cheap Euler predictor breaks this chicken-and-egg deadlock with a first guess.

Why average the two slopes instead of just using the end-slope

k_{2}

Using only

k_{2}

would over-correct — the true slope varies across the whole interval, and the mean of entry and exit slopes matches the interval's behaviour to one order higher than either endpoint alone.

Why is Heun called a "predictor–corrector" method rather than just "Euler twice"?

The two stages play different roles: the predictor makes a rough forward guess, the corrector uses that guess to build a better-averaged step. See Predictor-Corrector Methods — this asymmetry is the defining feature.

Why does Heun match the true Taylor series up to the

h^{2}

term but fail at

h^{3}

Averaging start and end slopes reproduces

y^{'}

and

y^{''}

exactly, but the

h^{3}

term needs

y^{'''}

(curvature-of-curvature) which two slope samples cannot capture. Getting

h^{3}

right is what pushes you to RK4. See Taylor Series Methods.

Why does halving

h

help Heun more than it helps Euler?

Heun's error scales as

h^{2}

, Euler's as

h

; halving

h

divides them by

4

and

2

respectively — the higher power rewards refinement more. This is the whole point of Order of Accuracy and Step Size.

Why can iterating the corrector improve the answer, yet never reach the exact solution?

Iteration solves the implicit trapezoidal equation more precisely, removing predictor error — but the trapezoidal rule itself still has an

O (h^{3})

discretisation error that no amount of iterating can erase.

Why is Heun exact when the true solution is a straight line (

y^{''} = 0

A straight solution has constant slope, so

k_{1} = k_{2}

and their average equals the true slope everywhere. The trapezoid is exact whenever the integrand is constant, so no error is introduced.

Edge cases

What does Heun do when

f (x_{0}, y_{0}) = 0

(flat start slope)?

The predictor stays put (

y_{1}^{(p)} = y_{0}

), but

k_{2} = f (x_{1}, y_{0})

can be nonzero, so the corrector still moves. This is exactly why Heun "notices" curvature that plain Euler misses when it starts flat (see Example 2).

If the step size is

h = 0

, what does the formula give?

y_{1} = y_{0}

and

x_{1} = x_{0}

— no step is taken. Both predictor and corrector collapse to identity; the method is consistent (a zero step changes nothing).

Suppose

f (x, y)

is linear in

y

, say

f = a y + b (x)

. Is the predictor still needed?

Yes in practice, but note the corrector becomes an explicit linear equation you could solve directly for

y_{1}

. Heun sidesteps that by predicting — trading exactness for a fixed, cheap recipe.

What happens to the corrector if the predictor massively overshoots (huge

h

on a fast-decaying ODE)?

k_{2}

is sampled at a wildly wrong

y

, so the averaged slope is corrupted and

y_{1}

can be worse than Euler's — a stability failure. Order-2 accuracy is a small-

h

promise, not a large-

h

guarantee.

For a purely oscillatory problem like

y^{'} = - y

over one step, does Heun's answer sit above or below the true decay?

Heun's averaged slope slightly under-decays for large

h

because the trapezoid rule over/undershoots curved integrands; for small

h

the discrepancy is tiny (

O (h^{3})

per step) and Heun tracks the true exponential closely.

If two different smooth ODEs share the same

y_{0}

k_{1}

and

k_{2}

at a step, do they take the same Heun step?

Yes — Heun's step depends only on

y_{0}

and the two slope samples, not on the underlying formula. Identical inputs to the recipe give an identical output, by construction.

Connections

Euler's Method — the predictor is one plain Euler step.
Trapezoidal Rule — the corrector is this rule in disguise (and equals it when $f = f (x)$ ).
Runge-Kutta Methods — Heun is RK2; RK4 fixes the $h^{3}$ gap.
Order of Accuracy and Step Size — why $O (h^{2})$ rewards halving $h$ .
Predictor-Corrector Methods — Heun is the simplest member.
Taylor Series Methods — the tool that proves Heun is second order.