Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 2 / 17D (Numerical Analysis)

For the test equation $y^{\prime }=\lambda y$, write one numerical step asThe linear stability domain isA method is A-stable when $\{z:\Re z\le 0\}\subseteq \mathcal{S}$.

Forward Euler has $R(z)=1+z$, soThis disk does not contain the whole left half-plane, so forward Euler is not A-stable. Backward Euler has $R(z)=(1-z{)}^{-1}$, soIt contains the left half-plane, and backward Euler is A-stable.

A differential equation is stiff when it contains rapidly decaying modes on time scales much shorter than those of interest, forcing an explicit method to take very small steps for stability rather than accuracy.

Hereso the decay rates are $1$ and $100$. For a negative real eigenvalue, forward Euler requirestherefore the fast mode imposesFor backward Euler the amplification factors arewhose moduli are at most one for every $h\ge 0$. ThusThis is the stiff two-mode linear system.

From the same value $y_n$, the two trial steps areThe exact step and the two approximations expand asThus their leading local errors have opposite signs, and the Milne device for forward and backward Euler estimates either magnitude by

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 $s<1$,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.