Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 341 / 1 / b / Solution

For the Dahlquist test equation, put $z=h\lambda$. The amplification roots satisfyWe show that no root can reach the unit circle for $\mathrm{Re}z<0$. If $\xi =e^{i\theta }$, solving the quadratic in $z$ givesWrite $c=\cos \theta$, $u=\mathrm{Re}C=-3/4+4c-c^2$, and $v=\mathrm{Im}C=(c-4)\sin \theta$. Then $u\le 9/4$ andSince $9/2-u>0$, this gives $|C|+u\le 9/2$. Consequently $|\mathrm{Re}\sqrt{C}{|}^2=(|C|+u)/2\le 9/4$, so both possible $z$ have nonnegative real part. Equality can occur only at $\theta =0$, which gives $z=0$ or $z=3$.

The leading coefficient has zeros $(3\pm i\sqrt{5})/2$, both in the right half-plane. Hence the amplification roots vary continuously as a pair throughout the left half-plane. At $z=-1$ they are $1/3$ and $1/5$; neither can leave the unit disk without crossing the unit circle, which the preceding calculation excludes. The same conclusion extends to the imaginary axis, with strict inequality away from $z=0$. At zero the roots $1,1/7$ satisfy the simplicity requirement.

Thus the method is A-stable. Its third order does not contradict the usual second-order barrier, because it is a multiderivative multistep method.