= Solution
The coefficients of this <linear multistep method> are $\alpha_0=b$, $\alpha_1=-a$, $\alpha_2=1$ and $\beta_0=\beta_1=0$, $\beta_2=c$. Substitution into the <order of a numerical method> conditions gives
$$
1-a+b=0,\qquad 2-a-c=0,\qquad 2-\frac a2-2c=0.
$$
The second equation gives $c=2-a$; substituting this into the third gives $-2+3a/2=0$. Consequently there is exactly one second-order choice,
$$
\boxed{a=\frac43,\qquad b=\frac13,\qquad c=\frac23.}
$$
The resulting scheme is the <BDF2 method>,
$$
3Y_{n+2}-4Y_{n+1}+Y_n=2h f(x_{n+2},Y_{n+2}).
$$
For a $C^4$ exact solution, <Taylor's theorem> in the original normalization gives the exact leading order defect
$$
\begin{aligned}
T_n(h)
&=\frac{y(x_n+2h)-\frac43y(x_n+h)+\frac13y(x_n)}h-\frac23y'(x_n+2h)\\
&=-\frac29h^2y'''(x_n)+O(h^3).
\end{aligned}
$$
In particular the coefficient $-2/9$ is nonzero: the method has formal order exactly two, rather than three. Bounded third derivatives already suffice for the uniform estimate $T_n(h)=O(h^2)$; the stronger smoothness above only makes the displayed remainder immediate. Multiplying the recurrence by $3/2$ instead gives the common derivative normalization with leading defect $-h^2y'''(x_n)/3$.
The first of the <characteristic polynomials of a linear multistep method> is
$$
\rho(z)=z^2-\frac43z+\frac13=(z-1)\left(z-\frac13\right).
$$
Both roots are simple and lie in the closed unit disk, so the <root condition for a multistep method> proves <zero-stability>.
The <Dahlquist equivalence theorem> states that, for fixed-coefficient <linear multistep methods> on the usual class of <initial value problems> with a right-hand side continuous in $x$ and uniformly <Lipschitz continuous> in $y$, <convergence of a numerical method> for every such problem and every convergent set of starting values is equivalent to <consistency of a numerical method> and <zero-stability>. The statement is on fixed finite intervals where the exact solution exists, with implicit updates well defined. Its order estimate says that a <zero-stable> method with $T_n(h)=O(h^p)$ and starting errors $O(h^p)$ has <global error> $O(h^p)$ under the corresponding solution smoothness and <Lipschitz continuity> assumptions; merely convergent starting values do not ensure the order-$p$ rate.
For this application, sufficient assumptions are a unique solution $y\in C^3([x_0,x_0+T])$, a uniform <Lipschitz continuity> bound $L$ in $y$ on a neighborhood of its graph, and the nearby solution branch of each implicit update. For sufficiently small $h$, the map
$$
z\longmapsto\frac43Y_{n+1}-\frac13Y_n+\frac23h f(x_{n+2},z)
$$
is a <contraction mapping> with factor at most $2hL/3<1$ on a suitable small closed neighborhood. The <Banach fixed-point theorem> supplies local solvability of the nearby update. With starting errors
$$
|Y_0-y(x_0)|+|Y_1-y(x_0+h)|=O(h^2),
$$
the order estimate gives
$$
\max_{0\leq nh\leq T}|Y_n-y(x_n)|\leq C_T\left(|Y_0-y(x_0)|+|Y_1-y(x_0+h)|+\max_n|T_n(h)|\right)=O(h^2).
$$
Thus the <BDF2 method> is second-order convergent under these standard hypotheses. Exact $Y_0=y(x_0)$ together with one <Euler method> step for $Y_1$ already gives the required starting error when $y''$ is bounded, since only one starting step is taken.
The question's assumption of <continuity> of $f$ alone does not establish uniqueness, the needed solution smoothness, or this uniform convergence estimate. The added <Lipschitz continuity> condition is a sufficient hypothesis for the theorem used here; its failure does not imply that every particular <initial value problem> is nonunique or that its numerical approximations cannot converge.
Back to article page
