Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 68 / 3

The first characteristic polynomial factors asAlso $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the order conditions for a linear multistep method hold for every $\alpha$. Convergence of the full linear multistep method recurrence requires, in addition, the root condition for a multistep method: all polynomial roots of $\rho$ lie in the closed unit disk and its polynomial roots on the unit circle are simple.

For $-1<\alpha <1$, write $\alpha =\cos \theta$ with $0<\theta <\pi$. The other polynomial roots are $e^{\pm i\theta }$, distinct from each other and from 1, so the root condition for a multistep method holds. For $|\alpha |>1$, the two quadratic polynomial roots are real and reciprocal, and one has modulus greater than one. At $\alpha =1$, $\rho =(w-1{)}^3$; at $\alpha =-1$, $\rho =(w-1)(w+1{)}^2$. Both endpoints violate simplicity. Consistency of a numerical method plus zero-stability therefore givesThe conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha =1$ the polynomials contain $(w-1{)}^2$, and at $\alpha =-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the common-factor cancellation defect in a multistep recurrence, so those parameter values are not added to the interval.

Apply the exponential-symbol order criterion for a multistep method, which follows by substituting the bilateral shift operator $e^{h{\partial }_t}$ into the exact-solution residual. Direct expansion givesAll coefficients through degree three vanish. If $\alpha \ne -5$, the fourth-degree coefficient is nonzero, so the normalized local truncation error has exact order three. At $\alpha =-5$, the fourth-degree coefficient vanishes and the fifth-degree coefficient is $1/10$, which is nonzero. ThusThis is the formal order of the full recurrence. The order-four member is not zero-stable, so it is not convergent. All the convergent members from part (a) have order three. At the degenerate value $\alpha =1$, the full recurrence still has formal order three, while cancellation of its common factor gives Backward Euler method of order one; that cancellation changes the method rather than its order calculation.

For $|\alpha |>1$, an exterior polynomial root of $\rho$ persists by continuity for sufficiently small negative real $z$ in the stability polynomial $\rho (w)-z\sigma (w)$. Hence these parameters cannot be A-stable.

For $-1\le \alpha <1$, the principal polynomial root at $w=1,z=0$ is simple. Let $s(z)=\log w(z)$ denote its analytic logarithm near zero. Using the expansion in part (b), and ${\rho }^{\prime }(1)=2(1-\alpha )$, solve the polynomial root equation to obtainIndeed the residual at $s=z$ is $-(\alpha +5)z^4/12$, and the correction in $s$ cancels it by multiplication with $2(1-\alpha )$. Choose $z=-C{\epsilon }^4/2+i\epsilon$. This lies strictly in the left half-plane, butfor small positive $\epsilon$. Thus $|w(z)|=e^{\mathrm{Re}s(z)}>1$, proving linear instability within the A-stability domain. This is a direct principal-root obstruction to A-stability, rather than an appeal to an order-barrier theorem alone.

Finally, at $\alpha =1$, the full stability polynomial isIt retains a double unit polynomial root for every negative $z$, allowing growing parasitic solutions. ThereforeThe canceled Backward Euler method recurrence at $\alpha =1$ is A-stable, but its admissible solutions and starting relations differ from those of the original method.