Questions: Euler's Method for ODEs (Error Analysis)

Question 1

You use Euler's method to solve an ODE over [0, 1] with step size h = 0.1 and find the global error at x = 1 is approximately 0.08. If you repeat the computation with h = 0.05, what global error do you expect?

• A. Approximately 0.04 — halving the step size halves the global error for a first-order method
• B. Approximately 0.002 — halving the step size reduces the global error by a factor of 4 for a first-order method
• C. Approximately 0.08 — the global error is determined by the ODE, not the step size
• D. Approximately 0.016 — halving h reduces global error by a factor of 5 due to error cancellation

Answer: (A)

Euler's method is first-order globally: global error = O(h). Halving h halves the global error. Option B describes a second-order method (like classical Runge-Kutta), where halving h reduces error by a factor of 4. The first-order relationship O(h) is what defines Euler's method as first-order — it's the practical consequence of having O(h²) local error accumulated over O(1/h) steps.

Question 2

Why is Euler's method called 'first-order' when its local truncation error at each step is O(h²)?

• A. Because the method was first published in the first volume of Euler's collected works
• B. Because the global error is O(h¹) — one power lower than the local error — due to accumulation of O(1/h) local errors over the integration interval
• C. Because Euler's method uses only first-order Taylor expansion terms to compute f(x, y)
• D. Because the method is only accurate to first decimal place regardless of step size

Answer: (B)

The 'order' of a method refers to its global error, not its local truncation error. Local error per step is O(h²), but over [0, T] there are N = T/h steps, so total accumulated error is approximately N × O(h²) = (T/h) × O(h²) = O(h). This one-order reduction is why global order = local order − 1 for Euler's method. The confusingly similar terminology (local truncation error vs. method order) is a common source of error.

Question 3

The global error of Euler's method is the same order as its local truncation error: both are O(h²).

• T. True
• F. False

Answer: False

Local truncation error is O(h²), but global error is O(h) — one order lower. This is because local errors accumulate over approximately T/h steps. Each step contributes O(h²) error, and (T/h) × O(h²) = O(h). The method is thus called 'first-order globally,' not second-order. Confusing local and global error is one of the most common misconceptions in numerical ODE analysis.

Question 4

Halving the step size in Euler's method approximately halves the global error at the end of the integration interval.

• T. True
• F. False

Answer: True

Because Euler's method has O(h) global error, the relationship is linear: cutting h in half cuts global error in half. This contrasts with higher-order methods: classical fourth-order Runge-Kutta has O(h⁴) global error, so halving h reduces error by a factor of 16. The linear relationship in Euler's method makes it expensive for high accuracy — to reduce error by a factor of 1000, you need 1000× as many steps.

Question 5

Explain why the global error of Euler's method is O(h) rather than O(h²), despite each individual step having local truncation error of O(h²). What happens to the local errors over the course of the integration?

Model answer: Local truncation error at each step is O(h²) because the Taylor expansion of the true solution includes a (h²/2)y'' term that Euler's method omits. But over an integration interval [0, T] with step size h, there are N = T/h steps. If these local errors simply add without interaction, total error ≈ N × O(h²) = (T/h) × O(h²) = T × O(h) = O(h). Additionally, errors from early steps propagate and perturb subsequent steps, but for Lipschitz-continuous ODEs, this propagation does not cause exponential blowup — it remains bounded, and the net effect is still O(h) globally.

The practical implication is that to achieve global error ε with Euler's method, you need h ~ ε, meaning N ~ T/ε steps. For ε = 10⁻⁶, that's a million steps over [0,1]. This cost motivates higher-order methods: fourth-order Runge-Kutta achieves the same accuracy with only N ~ T/ε^(1/4) steps, requiring far fewer steps for the same precision.