Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 67 / 2

Use the coefficient $2-\alpha$ in the numerator, as verified in the PDF; the converted TeX lost its $-\alpha$. Multiplying the method by $2+\alpha$ gives the characteristic polynomials of a linear multistep methodFor every admissible real $\alpha$, $\rho (1)=0$ and ${\rho }^{\prime }(1)=2\alpha =\sigma (1)$, giving consistency of a numerical method. Factorization givesThe second root is $r=(2-\alpha )/(2+\alpha )$. The root condition for a multistep method requires $|r|\le 1$ and forbids a repeated root at one. For real $\alpha \ne -2$,At $\alpha =0$ both roots equal one, violating zero-stability. For $\alpha >0$ the second root lies strictly inside the unit disk. The Dahlquist equivalence theorem therefore givesTo find the order rather than only consistency, expand the exact-solution residual through the generating function:Thus its unnormalized local truncation error begins with $(\alpha /3)h^3{y}^{(3)}$ when $\alpha >0$, and the global method order is exactly two for every convergent parameter. For comparison, $\alpha =0$ has formal order three but is not convergent because of its double unit root:These orders assume a smooth solution, the usual local Lipschitz condition and starting approximations of the required accuracy.

Only $\alpha >0$ needs consideration, by the preceding zero-stability analysis. On the Dahlquist test equation, all roots of the amplification polynomialmust satisfy the unit-disk root condition if the method is A-stable. Evaluate it at $\zeta =-1$:Take a real negative $z<-4/\alpha$. Then $P_z(-1)<0$, whereas the leading coefficient $2+\alpha -z$ is positive and $P_z(\zeta )\to +\infty$ as $\zeta \to -\infty$. The intermediate value theorem gives a real root strictly below $-1$. Its numerical mode grows in modulus, contradicting absolute stability at this negative test parameter. ConsequentlyThis proof detects an unstable root directly; the second-order value alone would not exclude A-stability. Equivalently the limiting derivative polynomial has an exterior root $-\alpha -\sqrt{1+{\alpha }^2}$, explaining the instability for large negative $z$.