Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 341 / Section A / 1

Substitute the exact solution and expand every value about $t_n$. The coefficient of $h^q{y}^{(q)}(t_n)$ in the defect is zero for $q=0,\dots ,p$ exactly whenThese are the order conditions for a linear multistep method, specialized to the two nonzero derivative coefficients. The first failed identity determines the leading local truncation error.

For $s=1$, imposing order two givesThe method is therefore the trapezoidal ruleIts first characteristic polynomial is $\rho (\zeta )=\zeta -1$, so it satisfies the root condition for a multistep method.

For $s=2$, the four conditions through order three giveand henceHereso the root condition for a multistep method again holds. Both methods are consistent and zero-stable, and the Dahlquist equivalence theorem therefore proves that both are convergent.

A highest-order method in this family has order $s+1$. The Second Dahlquist barrier says that an irreducible A-stable multistep method has order at most two, so $s+1\le 2$ and hence $s=1$. Part b then leaves only the trapezoidal rule. Its amplification factor isand $|R(z)|\le 1$ whenever $\mathrm{Re}z\le 0$. Thus the trapezoidal rule is the unique highest-order A-stable method of the stated form, apart from representations containing removable common factors.