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Quadratic interpolation gives the variable-step BDF2 formula and cubic residual

Statement

Let hn−1,hn>0, rn=hn/hn−1, and let y be an exact solution with three continuous derivatives near tn−1,tn,tn+1, where hj=tj+1−tj. The derivative at tn+1 of the quadratic interpolant through (tn+1,yn+1), (tn,yn), and (tn−1,yn−1) is

(1+2rn)yn+1−(1+rn)2yn+rn2yn−1hn(1+rn).

Equating this derivative to f(tn+1,Yn+1) defines the variable-step BDF2 formula

(1+2rn)Yn+1−(1+rn)2Yn+rn2Yn−1hn(1+rn)=f(tn+1,Yn+1).

For an exact vector solution y′=f(t,y) whose third derivative is bounded, the unscaled residual

τn=(1+2rn)y(tn+1)−(1+rn)2y(tn)+rn2y(tn−1)−hn(1+rn)y′(tn+1)

has norm ∥τn∥∞=O(hn3) uniformly whenever rn stays in a fixed compact subinterval of (0,∞); the formula and remainder hold componentwise.

Facts & Assumptions

Given: Distinct ordered nodes tn−1<tn<tn+1, their positive step sizes, and the linear multistep coefficient convention in Linear multistep methods, consistency, order, and the root condition.

[F1]

Multivariable Taylor formula with a Lagrange remainder along a line segment applies separately to every real coordinate of y on the interval between the nodes. For a component with ∣yi(3)∣≤M, its degree-two remainder at displacement δ has absolute value at most M∣δ∣3/6; the intermediate points may differ by component and node. Taking the maximum of these finitely many component bounds gives the same max-norm estimate, without a common vector remainder point.

Proof

1.1givenconstructalgebra

Set s=(t−tn+1)/hn. The interpolation nodes become 0,−1,−(1+1/rn). Their Lagrange basis derivatives at s=0 are respectively (1+2rn)/(1+rn), −(1+rn), and rn2/(1+rn), as follows by differentiating the three quadratic basis products. Since ds/dt=1/hn, the derivative formula in the statement follows.

1.2F1algebra

Apply [F1] separately to each component at tn+1 through degree two. In the weighted combination S=(1+2rn)y(tn+1)−(1+rn)2y(tn)+rn2y(tn−1), the constant coefficient is zero; the coefficient of y′ is (1+rn)hn; and the coefficient of y′′ is −12(1+rn)2hn2+12rn2(hn+hn−1)2=0.

2.1F1step 1.2algebra∎

Let M bound the absolute third derivatives of every component. By [F1], the max norm of the remaining weighted remainders is at most (M/6)((1+rn)2hn3+rn2(hn+hn−1)3). On the fixed compact ratio interval, hn−1=hn/rn makes this at most Chn3 with a uniform finite C. The canceled terms in step 1.2 leave exactly these remainders after subtracting the exact derivative term, so ∥τn∥∞≤Chn3.

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