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Reset and smooth-mode flow propagate a transverse event-time error
Statement
Let be an exact event time and its reported time, with . On the interval between these times, suppose the incoming exact mode extension and dense reconstruction satisfy , and . Assume . Suppose the reset is defined on a neighborhood containing , , and the coordinate segments joining them, with Lipschitz bounds in state and in time there. Let start from at time , and let start from at time . Assume both outgoing solution pieces through lie in a common convex tube where the outgoing field is bounded by and is -Lipschitz in state. For while both outgoing solutions remain in a common convex tube on which the field is -Lipschitz, we have For while both outgoing solutions remain in the tube,
Facts & Assumptions
Given: The exact and reported event times, incoming tube/error and field bounds, reset Lipschitz bounds, and a common outgoing mode tube.
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 at the event.
A Lipschitz ODE mode flow is stable to its initial state on a tube: same-mode solutions with initial error at a common time differ later by at most .
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.
For and integrable when , ; for , is integrable: the norm of a vector integral is at most the integral of the norm.
Proof
If , the state increment and its oriented integral are both zero. If , the incoming mode extension satisfies by [F3] on the nondegenerate interval between them. For either order of , orientation and [F4] give . Hence .
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 .
If , the earlier-reset displacement is zero. Otherwise, at , the trajectory that reset earlier has followed the outgoing field for at most time. By [F3] and [F4], its displacement is at most . Adding this to step 2.1 gives the first displayed inequality.
From the common time 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.
Depends on
- Smooth hybrid ODE modes, guards, resets, and deterministic event priority
- A first-order initial value problem is equivalent to its Volterra integral equation
- For $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$, $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$; for $a<b$, $\lVert f\rVert_2$ is integrable
- A Lipschitz ODE mode flow is stable to its initial state on a tube
Used by
Dependency tree · two levels
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Sources
- Ernst Hairer, course notes, Chapter III: Équations Différentielles Ordinaires (standard reference, not scraped)