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

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.

Use the stability function $R(z)=(z^2+6z+12)/(z^2-6z+12)$. Its denominator vanishes only at $3\pm i\sqrt{3}$, both in the open right half-plane. For $z=x+iy$,This is nonnegative whenever $x\le 0$, so $|R(z)|\le 1$ throughout the closed left half-plane. The internal stage equations are also solvable there, since $\det (I-zA)=1-z/2+z^2/12$. Therefore the method is A-stable.

The inequality is strict in the open left half-plane, while $|R(iy)|=1$. Moreover $R(z)\to 1$ as $|z|\to \infty$, so the method is not L-stable. These conclusions follow directly from the definitions of A-stability and L-stability, rather than from the order of the method.

For algebraic stability of a Runge-Kutta method, the weights must be nonnegative andmust be a positive semidefinite matrix. Here all weights are positive, but direct calculation givesIn particular $e_1^{T}M{e}_1=-1/36<0$; equivalently, its eigenvalues are $0,\pm \sqrt{3}/36$. Therefore the method is not algebraically stable. This does not contradict its A-stability: algebraic stability is a stronger condition designed to guarantee nonlinear B-stability through the Runge-Kutta contractivity identity.