Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 69 / 2

For the linear multistep method, let $\rho (\zeta )={\zeta }^2-12\zeta /7+5/7$ and $\sigma (\zeta )=4{\zeta }^2/7-2/7$. The exponential-symbol order criterion for a multistep method compares $\rho (e^z)$ with $z\sigma (e^z)$. Direct expansion givesEquivalently, substitution of exact smooth values into the recurrence leaves $-2h^3{y}^{\prime \prime \prime }(t_n)/21+O(h^4)$, while the coefficients of $1,h,h^2$ cancel. Therefore $\boxed{\text{order }2}$; its unnormalized step defect is of order three. The nonzero cubic term excludes order three.

Factor the first characteristic polynomial:The unit root is simple and the other root is strictly inside the unit disk, so the method is zero-stable. Also $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2/7$, giving numerical consistency. The Dahlquist equivalence theorem therefore makes it convergent.

The mechanism can also be seen directly. The zero-stable homogeneous recurrence has bounded propagators because its two error modes are $1$ and $(5/7{)}^n$. Variation of constants for the perturbed recurrence bounds the error by a fixed multiple of its two starting errors and the sum of the step defects, plus $O(h)$ times a sum of errors from the Lipschitz vector field. Absorb the current implicit error term for sufficiently small $h$, then apply the discrete Gronwall inequality. On a fixed time interval the sum of $O(h^3)$ exact-start defects is $O(h^2)$, giving second-order numerical convergence with second-order starting values. The implicit step is locally well defined, for example by contraction when $4hL/7<1$ for a Lipschitz constant $L$.

For the Dahlquist test equation, write $z=h\lambda$. The amplification roots solveA root on $|\zeta |=1$ would require the boundary-locus valueIts denominator cannot vanish on that circle. Setting $\zeta =e^{i\theta }$ and simplifying the real part givesThus no amplification root can cross the unit circle while $\mathrm{Re}z<0$. The leading coefficient is also nonzero throughout that half-plane. At $z=-1$, the roots are $(6\pm \sqrt{3})/11$, both strictly inside the unit disk. Connect any other left-half-plane point to $-1$: continuity of the roots, without a boundary crossing or a root escaping through infinity, keeps both roots inside.

On the imaginary axis a boundary root can occur only at $\theta =0$, which gives $z=0$; there the roots $1$ and $5/7$ satisfy the simple-unit-root condition. Hence $\boxed{\text{the method is A-stable}}$, including the usual boundary root convention at zero.