ODE solvers — Euler's method (derivation, global error): Question bank

The whole page tests one thing: do you know which error is which, when the method behaves, and why the geometry is a straight step along a curving path. No arithmetic grinding here — that lives in the worked-example pages.

True or false — justify

Euler's method is second order because its local truncation error is

O (h^{2})

False. The local error of one step is

O (h^{2})

, but the global error at a fixed endpoint accumulates over

N \propto 1/ h

steps, losing one power to give

O (h)

— so Euler is first order.

Halving the step

h

roughly halves the global error.

True. Global error

\propto h^{1}

, so scaling

h

\frac{1}{2}

scales the error by about

\frac{1}{2}

. (If it quartered, the method would be second order.)

If we always start each step exactly on the true curve, the only error left is the local truncation error.

True — that is the very definition of LTE: the error of a single step assuming a perfect start. Global error is what you get when you don't restart on the curve each time and mistakes compound.

Making

h

as small as your computer allows always gives the most accurate answer.

False. Shrinking

h

cuts truncation error but multiplies the number of steps, so round-off error accumulates and cost explodes. There is a sweet spot, not a limit at zero.

For a linear ODE

y^{'} = λ y

with

λ < 0

, Euler's approximation always decays like the true solution.

False. Euler gives

y_{n + 1} = (1 + hλ) y_{n}

; decay only happens if

∣1 + hλ ∣ < 1

. Too large an

h

makes this factor

\geq 1

in magnitude and the numerical solution grows or oscillates — a stability failure, not an accuracy one. See Numerical Stability.

The global error bound

\frac{h M}{2 L} (e^{L (x_{n} - x_{0})} - 1)

says error grows without limit as we integrate further.

True in the worst case: for a fixed

h

, pushing the endpoint

x_{n}

further makes the

e^{L (x_{n} - x_{0})}

factor larger, so accumulated error can grow exponentially along the interval.

Euler's method needs a closed-form solution of the ODE to run.

False. It needs only the slope function

f (x, y)

and a starting point. The whole point is it works precisely when no closed form exists.

y^{''} = 0

everywhere (the true solution is a straight line), Euler's method is exact.

True. The discarded term is

\frac{h ^{2}}{2} y^{''} (ξ) = 0

, so there is nothing to throw away — the tangent step lands exactly on the line.

Spot the error

y_{n + 1} = y_{n} + h f (x_{n + 1}, y_{n + 1})

— that's Euler's update."

Wrong — that uses the slope at the new point, which is Backward (implicit) Euler and requires solving for

y_{n + 1}

. Forward Euler uses

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

, the slope at the current point.

"Global error

= N \times

LTE

= \frac{b - x _{0}}{h} \cdot \frac{h ^{2}}{2} y^{''} = \frac{( b - x _{0} ) h}{2} y^{''}

, so the error is

O (h^{2})

The algebra is right up to the last claim:

\frac{( b - x _{0} ) h}{2} y^{''}

is linear in

h

, i.e.

O (h)

, not

O (h^{2})

. The student misread

h^{1}

h^{2}

"The step formula is

y_{n + 1} = y_{n} + f (x_{n}, y_{n})

The step size

h

is missing. Without it the update takes a full unit-length rise regardless of how far

h

actually is — it must be

y_{n + 1} = y_{n} + h f (x_{n}, y_{n})

N = (b - x_{0}) / h

steps, and since each new step is independent, errors just add and stay

O (h^{2})

total."

Two mistakes: errors don't stay

O (h^{2})

because

N

grows as

h

shrinks, and they aren't independent — earlier errors are amplified by later steps, which is exactly where the

e^{L (x - x_{0})}

factor comes from.

"To get the tangent at

(x_{n}, y_{n})

we need to know

y (x)

near there."

No — the ODE hands us the slope directly as

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

. That is the whole trick: we get the tangent's slope without knowing the curve.

"Since LTE

= \frac{h ^{2}}{2} y^{''} (ξ)

with

ξ

unknown, we can't say anything useful about the error."

We can: bounding

∣ y^{''} ∣ \leq M

on the interval turns the unknown

ξ

into a usable inequality

∣ LTE ∣ \leq \frac{h ^{2}}{2} M

, which drives the whole global bound.

Why questions

Why does global error lose exactly one power of

h

compared to local error?

Because reaching a fixed endpoint takes

N = (b - x_{0}) / h

steps, and

N

grows like

1/ h

. Multiplying the per-step

O (h^{2})

O (1/ h)

steps kills one power:

O (h^{2}) \cdot O (h^{- 1}) = O (h)

Why does the error bound contain an exponential

e^{L (x - x_{0})}

rather than a plain sum?

Because an error made early doesn't just sit there — it gets fed back through

f

at every later step, and nearby solutions can diverge at rate

L

(the Lipschitz constant). The worst-case compounding of such growth is exponential.

Why do we use a forward difference

\frac{y _{n + 1} - y _{n}}{h}

and not the exact derivative?

A computer cannot take the limit

h \to 0

that defines the derivative. Keeping

h

finite gives a usable, computable approximation — the price is the truncation error we then analyse.

Why is Taylor's theorem the right tool for the error, rather than just guessing?

Taylor's theorem is exact: the remainder term

\frac{h ^{2}}{2} y^{''} (ξ)

is not an approximation but the precise thing we discard. That lets us name and bound the error instead of hand-waving. See Taylor Series Expansion.

Why does the Lipschitz condition on

f

matter for Euler at all?

It guarantees nearby solutions can't diverge faster than rate

L

, which is exactly what keeps accumulated errors bounded. Without it, the

e^{L (x - x_{0})}

factor — and the whole error guarantee — could fail to exist.

Why does a small step keep us close to a curving true solution even though we walk in straight lines?

Over a tiny horizontal distance

h

, the curve barely bends, so its tangent and the curve nearly coincide. The gap per step is

O (h^{2})

— shrinking

h

makes each straight step hug the curve tighter.

Edge cases

What does Euler do when

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

at a step (slope is flat)?

The update becomes

y_{n + 1} = y_{n}

: the point stays put for that step. It's a perfectly valid step — a horizontal tangent — not an error.

For

y^{'} = - 2 y

y (0) = 1

with

h = 0.5

, Euler gives

y_{1} = 0

then stays at

0

. What went wrong?

The factor

1 + hλ = 1 + 0.5 (- 2) = 0

collapses the value to exactly zero, and the true

e^{- 1} \approx 0.37

is badly missed. The step is too big for the fast decay — an instability, curable by shrinking

h

What happens to Euler if the true solution has a vertical asymptote (blows up) inside the interval?

Euler keeps taking finite tangent steps and produces finite numbers even where the true solution is infinite, so it silently gives nonsense past the blow-up. The bound assumed

∣ y^{''} ∣ \leq M

, which fails there.

h

is chosen so that

x_{0} + N h

overshoots the target

b

, what's the danger?

The final step lands past

b

, so you evaluate the solution at the wrong point. Choose

h

so that

N = (b - x_{0}) / h

is a whole number, or take a smaller final step.

What is the smallest meaningful number of steps,

N = 1

A single Euler step from

x_{0}

b = x_{0} + h

: it's just one tangent jump. Global and local error coincide here because there's nothing to accumulate yet.

Does Euler behave differently if the slope function

f

depends only on

x

(i.e.

y^{'} = g (x)

Then Euler reduces to the left-endpoint rectangle rule for the integral

\int g d x

— a familiar first-order quadrature. It confirms the

O (h)

order from a different angle.

Connections

Euler's Method — Derivation & Global Error — the parent this bank drills
Taylor Series Expansion — source of the exact remainder term
Numerical Stability — the decay/blow-up edge cases
Backward Euler & Implicit Methods — the implicit look-alike trap
Lipschitz Continuity — why the exponential error factor exists
Runge-Kutta Methods — higher-order fixes for the order-one limit
Finite Difference Approximations — the forward-difference derivation route