Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 2 / 17C (Numerical Analysis)

A one-step or linear multistep method has order $p$ if, when the exact sufficiently smooth solution is substituted for the numerical values, its unscaled local truncation error is $O(h^{p+1})$, and this estimate is not generally $O(h^{p+2})$. For a stable method this corresponds to a global error of order $O(h^p)$ on a fixed time interval.

Expand the exact solution about $t_{n+1}=t$. The left side isSince $f(t,y(t))=y^{\prime }(t)$, the right side expands asThe $h{y}^{\prime }$ terms agree for every $\alpha ,\beta$, so the method is always consistent and is generically of order one.

The method is explicit exactly when $1+\alpha =0$. With $\alpha =-1$, its $h^2$ defect vanishes only for $\beta =3/2$, so a generic explicit member has order one; that exceptional explicit member has order two and cannot satisfy the next order condition.

For the unrestricted method to have order at least three, the $h^2$ and $h^3$ coefficients must agree:Solving givesFor these values, $1+2\alpha +\beta =1/2$, so the right-side $h^4$ coefficient is $1/12$, different from the left-side coefficient $1/24$. Hence the order is exactly three.

The first characteristic polynomial of the linear multistep method comes only from its left side:Its roots are zero and one; both lie in the closed unit disc, and the only unit-modulus root is simple. Thus the method satisfies the root condition for a multistep method and is zero-stable. Part (b) showed consistency for every $\alpha ,\beta$. The Dahlquist equivalence theorem states that a consistent linear multistep method is convergent exactly when it is zero-stable. Thereforeunder the usual Lipschitz hypotheses on $f$.