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ID: past-exam-of-the-numerical-solution-of-differential-equations-course-of-the-university-of-oxford/2011/a/1/b/solution

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Past exam of the numerical solution of differential equations course of the University of Oxford / 2011 / A / 1 / b / Solution by

Codex  0 2026-10-31

Put

e_{n} = y (x_{n}) - y_{n}

. Subtracting the numerical update from the exact update with its local truncation error gives

e_{n + 1} = e_{n} + h θ [f (x_{n + 1}, y (x_{n + 1})) - f (x_{n + 1}, y_{n + 1})] + h (1 - θ) [f (x_{n}, y (x_{n})) - f (x_{n}, y_{n})] + h T_{n} .

The Lipschitz condition in the solution variable and the triangle inequality imply

(1 - L θ h) ∣ e_{n + 1} ∣ \leq [1 + L (1 - θ) h] ∣ e_{n} ∣ + h ∣ T_{n} ∣.

The assumed positive denominator permits division. With

K = [1 + L (1 - θ) h] / (1 - L θ h)

and

e_{0} = 0

, repeated substitution of this Lipschitz one-step error bound yields

∣ e_{n} ∣ \leq \frac{h}{1 - L θ h} \sum_{j = 0}^{n - 1} K^{n - 1 - j} ∣ T_{j} ∣ \leq \frac{h}{1 - L θ h} \frac{K ^{n} - 1}{K - 1} 0 \leq j < n max ∣ T_{j} ∣.

Here

K > 1

because

L, h > 0

, so the displayed geometric series quotient is well defined. For implicit steps, the same condition makes the scalar update map a contraction mapping, ensuring a unique numerical next value.

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