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A state-Lipschitz vector field makes the Picard operator a contraction when
Statement
On the invariant curve ball of A bounded vector field makes the Picard operator preserve a sufficiently short closed curve ball, suppose has state-Lipschitz constant . Then
In particular, if , the Picard operator is a contraction.
Facts & Assumptions
Given: Curves in the invariant ball and a state-Lipschitz constant .
For an integrable vector-valued function on with , (For and integrable when , ; for , is integrable).
On every compact time-state cylinder the state-variable inequality holds with one finite constant (Local Lipschitz continuity in the state variable, locally uniform in time and parameters).
Proof
Subtracting the two Picard images and applying [L1] and [L2] gives for every in the cylinder.
Taking the supremum and using gives the displayed estimate; if this is a contraction, while gives contraction constant .
Depends on
- Local Lipschitz continuity in the state variable, locally uniform in time and parameters
- The Picard operator and Picard iterates on a closed ball of continuous curves
- A bounded vector field makes the Picard operator preserve a sufficiently short closed curve ball
- 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
Used by
Dependency tree · two levels
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Sources
- Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, Ch. 2 (standard reference, not scraped)
- Jiri Lebl, Basic Analysis I, Section 6.3 (standard reference, not scraped)