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A transverse backward boundary hit has a certified time and trace error bound

Statement

Let J=[a,b] be a compact bracket, let y be an exact incoming characteristic on J, and let g(t,x) be a boundary guard that is Lg-Lipschitz in its state argument on a tube containing the exact path and a dense reconstruction U. Suppose h(t):=g(t,y(t)) has a root τ∈J and a declared orientation σ∈{−1,1} with σh′(t)≥ν>0 throughout J. If sup⁡t∈J∥y(t)−U(t)∥2≤η,∣g(τ^,U(τ^))∣≤ε,τ^∈J, then the matched first-hit time satisfies ∣τ^−τ∣≤Δ:=Lgη+εν. If the boundary value β(t,x) is Lipschitz in time and state with constants Ltβ,Lxβ on the same tube and ∥y′(t)∥2≤M, then ∣β(τ^,U(τ^))−β(τ,y(τ))∣≤LtβΔ+Lxβ(η+MΔ). This is a local one-hit bound. It gives no conclusion at grazing, unresolved, or simultaneous events.

Facts & Assumptions

Given: A smooth mode characteristic, a boundary guard and orientation, a matched transverse first-hit bracket, a dense-output error bound, a reported guard residual, and the stated boundary-data Lipschitz bounds.

[F1]

Smooth hybrid ODE modes, guards, resets, and deterministic event priority: a smooth mode follows its ODE until a selected guard event; the event orientation is part of the guard data.

[F2]

The derivative of a guard along an ODE mode is its Lie derivative: along y′=b(t,y), the guard derivative is h′(t)=∂tg(t,y(t))+Dxg(t,y(t))b(t,y(t)).

[F3]

A transverse hybrid event time is stable under dense-output and root error: dense state error η, guard residual ε, guard state-Lipschitz constant Lg, and transverse margin ν imply ∣τ^−τ∣≤(Lgη+ε)/ν on the matched bracket.

[F4]

A first-order initial value problem is equivalent to its Volterra integral equation: a mode trajectory increment equals the oriented integral of its vector field.

[F5]

For a≤b and f:[a,b]→Rm integrable when a<b, ∥∫abf∥2≤∫ab∥f∥2; for a<b, ∥f∥2 is integrable: for ordered endpoints c≤d, ∥∫cdf∥2≤∫cd∥f∥2; reversed oriented integrals must first be rewritten with ordered endpoints.

Proof

1.1givenalgebra

If a=b, then τ^=τ=a, so the time error is zero and hence at most Δ. The stated bounds give ∥U(a)−y(a)∥2≤η; the state-Lipschitz bound for β gives ∣β(a,U(a))−β(a,y(a))∣≤Lxβη≤LtβΔ+Lxβ(η+MΔ). Here Δ≥0 from the error and residual bounds, ν>0, and the nonnegative Lipschitz constant Lg, while M,Ltβ,Lxβ≥0. Assume a<b below.

1.2givenF1F2F3

By [F2], the transversality hypothesis is the mode derivative condition required by [F3]. Apply [F3] to this matched bracket with the stated η,ε,Lg,ν; it yields the unique transverse root and the displayed bound ∣τ^−τ∣≤Δ.

2.1givenF4F5step 1.2algebra∎

By [F4], the increment is the oriented integral from τ to τ^. Set c=min⁡(τ,τ^) and d=max⁡(τ,τ^); changing orientation only negates the vector integral, so [F5] gives ∥y(τ^)−y(τ)∥2≤∫cd∥y′(r)∥2 dr≤M(d−c)=M∣τ^−τ∣≤MΔ. The triangle inequality with the dense-output bound gives ∥U(τ^)−y(τ)∥2≤η+MΔ. Apply the time and state Lipschitz inequalities for β to obtain the second displayed bound.

Remarks

An inflow/exit implementation must use a certified first-hit bracket and return unresolved when the bracket or orientation is not certified. This local result does not propagate errors through resets or an event list; those require the full finite-separated-event hypotheses of Finite separated transverse events give an event-aligned hybrid trajectory error bound.

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