Constant-step BDF2 is second-order convergent with a consistent start and selected roots

Given: A $C^3$ exact solution, compact tube and Lipschitz bound, a constant positive grid, prescribed BDF2 branch balls with a unique root per step, and starting values satisfying the stated $O(h^2)$ bound.

Variable-step BDF2 with an explicit local implicit-branch rule: BDF2 selects the unique root in each prescribed closed branch ball and makes no convergence assertion by itself.

Variable-step BDF2 is second-order convergent under a uniform ratio bound: under its $C^3$ solution, compact-tube, state-Lipschitz, ratio $0<r_{\min }\le r_n\le r_{\ast }<1+\sqrt{2}$, and $O(H^2)$ starting-pair hypotheses, BDF2 has maximum error $O(H^2)$ for sufficiently small $H$, provided there are fixed $\eta ,\epsilon >0$ such that every preceding pair within $\eta$ of the exact history has exactly one step root in $\bar{B}(y(t_{n+1}),\epsilon )\subset K$, and the prescribed branch ball contains exactly this same unique root.

First-order systems, initial value problems, and solutions on intervals: the exact solution satisfies $y^{\mathrm{\prime }}=f(t,y)$ on its interval.

Multivariable Taylor formula with a Lagrange remainder along a line segment is scalar-valued and is applied separately to each component of the $C^3$ solution. The supplier is scalar-valued. For a vector map $F:U\subset {\mathbb{R}}^s\to {\mathbb{R}}^d$, apply it separately to each real component $F_i$; its intermediate points ${\theta }_i$ may differ. If the absolute values of all order-$(k+1)$ component partial derivatives on the segment are at most $M$, each component remainder is at most $M{C}_{s,k}\mathrm{\parallel }v{\mathrm{\parallel }}_2^{k+1}$, where $C_{s,k}={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k+1}1\mathrm{/}\alpha !$ is a fixed finite constant, since $\mathrm{\mid }{v}^{\alpha }\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2^{k+1}$. Hence the Euclidean remainder is at most $\sqrt{d}M{C}_{s,k}\mathrm{\parallel }v{\mathrm{\parallel }}_2^{k+1}$. In one time variable the max-norm bound is $M\mathrm{\mid }h{\mathrm{\mid }}^{k+1}\mathrm{/}(k+1)!$. No common intermediate point for all components is asserted. Thus $y(h)=y(0)+h{y}^{\mathrm{\prime }}(0)+O(h^2)$ and $y^{\mathrm{\prime }}(h)=y^{\mathrm{\prime }}(0)+O(h)$ uniformly in the fixed finite dimension. The same componentwise degree-two formula gives the cubic constant-step residual used below.

Backward Euler step equation and algebraic residual: an admissible BE endpoint satisfies $z=y_0+hf(h,z)$ and its update residual is $z-y_0-hf(h,z)$.

Implicit Runge–Kutta stage equations and update: for one stage with $A=b=c=(1)$, the stage equation is $Z=y_0+hf(h,Z)$ and the update is $Z$ when a stage is selected.

Small-step implicit Runge–Kutta stages have a unique branch and controlled residual error: on an invariant product ball, if $q=h\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}L<1$, there is a unique stage root there and every approximate stage vector in the ball has error at most its update residual divided by $1-q$.

The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$: $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2={\sum }_{i<d}{v}_i^2$.

Squaring is monotone on the nonnegatives: if $a,b\ge 0$, then $a\le b$ whenever $a^2\le b^2$.

The tube, Lipschitz, smoothness, $O(h^2)$ starting-pair, branch-ball, and unique-root hypotheses above hold uniformly as $h\to 0$. For the BE startup, $B_{\mathrm{B}\mathrm{E}}$ contains $y(h)$ and the exact BE root, and its stage map is invariant and $q$-contractive with $q=h{L}_2\le 1\mathrm{/}2$.