{c}

A Lobatto IIIA method collocates at the <Lobatto quadrature> nodes, including both endpoints of the time interval. Its three-stage version has
$$
A=\begin{pmatrix}0&0&0\\5/24&1/3&-1/24\\1/6&2/3&1/6\end{pmatrix},
\qquad b=(1/6,2/3,1/6)^T,
\qquad c=(0,1/2,1)^T.
$$
The <Butcher order conditions> give order four. Its <stability function> is
$$
R(z)=\frac{1+z/2+z^2/12}{1-z/2+z^2/12},
$$
which is <A-stable> but not <L-stable>. It is not <algebraically stable>: the first diagonal entry of the defining <matrix> is $2b_1a_{11}-b_1^2=-1/36$. This distinguishes linear <A-stability> from nonlinear <algebraic stability>.

Lobatto IIIA method

The tableau is the three-stage <Lobatto IIIA method>. For its stage <matrix> $A$, weights $b$, nodes $c$ and $C=\operatorname{diag}(c)$, the <Butcher order conditions> through order four are
$$
\begin{gathered}
b^T\mathbf1=1,\quad b^Tc=\frac12,\quad b^Tc^{\circ2}=\frac13,\quad b^TAc=\frac16,\\
b^Tc^{\circ3}=\frac14,\quad b^TCAc=\frac18,\quad b^TAc^{\circ2}=\frac1{12},\quad b^TA^2c=\frac1{24},
\end{gathered}
$$
where $c^{\circ j}$ means coordinatewise powers. Direct substitution verifies all eight identities. Equivalently, this is a <collocation Runge-Kutta method> at the three <Lobatto quadrature> nodes, whose weights integrate cubic <polynomials> exactly.

Order five would require $b^Tc^{\circ4}=1/5$, but here $b^Tc^{\circ4}=5/24$. Therefore the method has \b[exactly order four].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 2 / a / Solution

{c}

An $s$-node Lobatto quadrature rule includes both endpoints and is exact for <polynomials> of degree at most $2s-3$. On $[-1,1]$, its interior nodes are the zeros of the derivative of the degree-$(s-1)$ <Legendre polynomial>. The three-node rule on $[0,1]$ has nodes $0,1/2,1$ and weights $1/6,2/3,1/6$. Integrating the interpolation basis from zero to each node gives the three-stage <Lobatto IIIA method>.

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 Incoming links: Lobatto quadrature