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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 17D / c / Solution

From the same value

y_{n}

, the two trial steps are

y_{F} = (I + h M) y_{n}, y_{B} = (I - h M)^{- 1} y_{n} .

The exact step and the two approximations expand as

e^{h M} y_{n} y_{F} y_{B} = (I + h M + \frac{1}{2} h^{2} M^{2} + O (h^{3})) y_{n}, = (I + h M) y_{n}, = (I + h M + h^{2} M^{2} + O (h^{3})) y_{n} .

Thus their leading local errors have opposite signs, and the Milne device for forward and backward Euler estimates either magnitude by

E_{n} = \frac{1}{2} ∥ y_{B} - y_{F} ∥ = \frac{1}{2} h^{2} ∥ M^{2} y_{n} ∥ + O (h^{3}) .

Use backward Euler as the accepted step because it is A-stable. Reject a trial if

E_{n}

exceeds the prescribed local tolerance; otherwise accept

y_{B}

and choose, with a safety factor

s < 1

,

h_{new} = s h (\frac{tol}{E _{n}})^{1/2} .

The small steps resolve the initial fast transient. Once its amplitude has decayed, the error estimator permits much larger steps, while the accepted backward-Euler evolution remains stable. This controls the error without paying the forward-Euler stability restriction throughout the integration.

Solved by gpt-5.6-sol high.

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