Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 341 / 2 / a

The nodes $0,1/2,1$ have Lagrange interpolation polynomialsThe coefficients obtained from $a_{ij}={\int }_0^{c_i}{\ell }_j(s)ds$ and $b_j={\int }_0^1{\ell }_j(s)ds$ reproduce the given Butcher tableau. In particular, the second row's last entry is $-1/24$, which is missing from the TeX transcription. Thus this is the three-stage Lobatto IIIA method.

One applicable collocation theorem is that $s$-stage Lobatto IIIA collocation has order $2s-2$. Alternatively, the fourth-order conditions for a Runge-Kutta method give a direct verification. With $e=(1,1,1{)}^T$, $c=Ae$ and $C=\mathrm{diag}(c)$, the eight required Butcher order conditions arePowers of $c$ here are componentwise. Substitution satisfies all eight. The stability function computed from the stages isIts expansion satisfies $R(z)-e^z=-z^5/720+O(z^6)$, so even the scalar linear problem fails the fifth-order condition. HenceThe Butcher order condition theorem equates order $p$ with all conditions for rooted trees through order $p$; the displayed conditions are its complete specialization through order four.