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Reset and smooth-mode flow propagate a transverse event-time error

Statement

Let τ be an exact event time and τ^ its reported time, with Δ=∣τ^−τ∣. On the interval J between these times, suppose the incoming exact mode extension y− and dense reconstruction U− satisfy ∥U−(t)−y−(t)∥2≤η, and ∥f−(t,y−(t))∥2≤M−. Assume η,M−,M+,LR,LR,t,L+≥0. Suppose the reset is defined on a neighborhood containing (τ,y−(τ)), (τ^,U−(τ^)), and the coordinate segments joining them, with Lipschitz bounds LR in state and LR,t in time there. Let y+ start from R(τ,y−(τ)) at time τ, and let y~+ start from R(τ^,U−(τ^)) at time τ^. Assume both outgoing solution pieces through t0=max⁡{τ,τ^} lie in a common convex tube where the outgoing field is bounded by M+ and is L+-Lipschitz in state. For s≥t0 while both outgoing solutions remain in a common convex tube on which the field is L+-Lipschitz, we have ∥y~+(t0)−y+(t0)∥2≤LRη+(LRM−+LR,t+M+)Δ. For s≥t0 while both outgoing solutions remain in the tube, ∥y~+(s)−y+(s)∥2≤eL+(s−t0)[LRη+(LRM−+LR,t+M+)Δ].

Facts & Assumptions

Given: The exact and reported event times, incoming tube/error and field bounds, reset Lipschitz bounds, and a common outgoing mode tube.

[F1]

Smooth hybrid ODE modes, guards, resets, and deterministic event priority: the incoming and outgoing pieces solve their mode ODEs and the state is reset by R(t,x) at the event.

[F2]

A Lipschitz ODE mode flow is stable to its initial state on a tube: same-mode solutions with initial error e at a common time differ later by at most eL+Δte.

[F3]

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

Proof

1.1givenF1F3F4algebra

If a=b, the state increment and its oriented integral are both zero. If a≠b, the incoming mode extension satisfies y−(b)−y−(a)=∫abf−(t,y−(t)) dt by [F3] on the nondegenerate interval between them. For either order of a,b, orientation and [F4] give ∥y−(b)−y−(a)∥2≤M−∣b−a∣. Hence ∥U−(τ^)−y−(τ)∥2≤∥U−(τ^)−y−(τ^)∥2+∥y−(τ^)−y−(τ)∥2≤η+M−Δ.

2.1givenF1step 1.1algebra

Apply the reset's state and time Lipschitz bounds to the two reset inputs in step 1.1. The reset states differ by at most LRη+(LRM−+LR,t)Δ.

3.1givenF1F3F4step 2.1algebra

If Δ=0, the earlier-reset displacement is zero. Otherwise, at t0, the trajectory that reset earlier has followed the outgoing field for at most Δ time. By [F3] and [F4], its displacement is at most M+Δ. Adding this to step 2.1 gives the first displayed inequality.

4.1givenF2step 3.1∎

From the common time t0 onward, both curves solve the same outgoing mode ODE and remain in the common tube. Apply [F2] with the initial error from step 3.1 to obtain the second displayed inequality.

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