Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 63 / 2 / 2

Use the stability function just calculated. Its denominator has its two zeros at $3\pm i\sqrt{3}$, so there is no pole in the closed left half-plane. Put $z=x+iy$, $N=1+z/2+z^2/12$ and $D=1-z/2+z^2/12$. A direct calculation yieldsFor $x\le 0$ this is nonnegative, and therefore $|R(z)|\le 1$. The method is A-stable. Equality holds on the imaginary axis, consistently with the absence of numerical damping for those scalar oscillatory modes. Since $R(z)\to 1$ as $z\to \infty$, the A-stability established here does not imply strong damping of very stiff decaying modes.