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A bouncing-ball impact has a transverse event-time and reset-error bound

Example

Let q be height and v velocity, with mode field q′=v, v′=−2 before and after impacts, initial state (q(0),v(0))=(1,0), downward guard g(q,v)=q of orientation σ=−1, and time-independent reset R(q,v)=(0,−v/2). On the event bracket [1,51/50], use the approximate incoming reconstruction U(t)=(1−t2+1/25,−2t). It has uniform state error η=1/25 from the exact incoming solution y(t)=(1−t2,−2t). The reported guard root is τ^=26/25, and the exact impact time is τ=1. This example isolates the event-time and reset estimates; it does not claim that U is a residual-certified reconstruction from the exact initial state.

Facts & Assumptions

Given: A smooth hybrid mode and reset, the transverse event-time estimate, and the mistimed-reset estimate.

[F1]

Smooth hybrid ODE modes, guards, resets, and deterministic event priority: a mode follows its ODE until an oriented guard crossing, then applies its reset and changes mode.

[F2]

A transverse hybrid event time is stable under dense-output and root error: if the exact guard has signed derivative at least ν>0, the dense state error is at most η, and the reported guard residual is at most ε, then ∣τ^−τ∣≤(Lgη+ε)/ν.

[F3]

Reset and smooth-mode flow propagate a transverse event-time error: with incoming error η, event-time error Δ, reset constants LR,LR,t, and incoming/outgoing field bounds M−,M+, the aligned post-reset error is at most LRη+(LRM−+LR,t+M+)Δ.

Verification

1.1givenF1algebra

The exact incoming solution is y(t)=(1−t2,−2t), so its guard h(t)=1−t2 has the unique root τ=1 on the bracket. For t∈[1,51/50], −h′(t)=2t≥2, so the downward crossing is transverse with margin ν=2.

2.1givenstep 1.1algebra

The dense guard is 1−t2+1/25, whose root in the bracket is τ^=26/25 and whose residual there is ε=0. For x≥0, both sides are nonnegative and (1+x/2)2−(1+x)=x2/4≥0, so 1+x≤1+x/2. Taking x=1/25 gives 0≤τ^−τ≤1/50.

3.1givenF2step 1.1step 2.1algebra

On the bracket, ∥U(t)−y(t)∥2=1/25. Since g(q,v)=q, ∣g(q1,v1)−g(q2,v2)∣=∣q1−q2∣≤∥(q1,v1)−(q2,v2)∥2, so Lg=1; the reported guard residual is zero. Apply [F2] with ν=2 to obtain ∣τ^−τ∣≤(1/25)/2=1/50, agreeing with the explicit bound in step 2.1.

4.1givenF3step 1.1step 3.1algebra

The reset is defined on all of R2, is LR=1/2-Lipschitz in state, and has LR,t=0. On the incoming bracket, ∥U(t)−y(t)∥2=1/25 and ∥f−(t,y(t))∥2=4t2+4<3. Step 3.1 gives ∣τ^−τ∣≤1/50. The outgoing field (q,v)↦(v,−2) is globally 1-Lipschitz in state. The exact outgoing path through τ^ and the reported reset state lie in [−1/10,1/10]×[4/5,6/5], where the field norm is less than 3; by local existence and continuity both outgoing solutions remain in the bounded convex rectangle [−1/5,1/5]×[3/4,5/4] for a short common forward interval. Thus the reset and outgoing tube hypotheses of [F3] hold with M−=M+=3 and L+=1. By [F3], the aligned post-reset error is at most 12⋅125+(12⋅3+3)150=11/100.

5.1givenstep 2.1step 4.1algebra∎

Directly, the exact outgoing state at τ^=1+Δ is (Δ−Δ2,1−2Δ) and the reported reset state is (0,1+Δ). Their difference has components bounded by 1/50 and 3/50, so its Euclidean norm is at most (1/50)2+(3/50)2<11/100. This verifies the reset estimate for this instance.

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