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A state-Lipschitz vector field makes the Picard operator a contraction when $L h < 1$

Statement

On the invariant curve ball of A bounded vector field makes the Picard operator preserve a sufficiently short closed curve ball, suppose $F$ has state-Lipschitz constant $L \geq 0$ . Then

$d_{\infty} (T x, T y) \leq L h d_{\infty} (x, y) .$

In particular, if $L h < 1$ , the Picard operator is a contraction.

Facts & Assumptions

Given: Curves $x, y$ in the invariant ball and a state-Lipschitz constant $L$ .

[L1]

For an integrable vector-valued function on $[a, b]$ with $a \leq b$ , ${∥ \int_{a}^{b} f ∥}_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ (For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable).

[L2]

On every compact time-state cylinder the state-variable inequality holds with one finite constant $L$ (Local Lipschitz continuity in the state variable, locally uniform in time and parameters).

Proof

technique · direct

1.1givenL1L2

Subtracting the two Picard images and applying [L1] and [L2] gives $∥ (T x) (t) - (T y) (t) ∥_{2} \leq L ∣ t - t_{0} ∣ d_{\infty} (x, y)$ for every $t$ in the cylinder.

2.1step 1.1algebra∎

Taking the supremum and using $∣ t - t_{0} ∣ \leq h$ gives the displayed estimate; if $L h < 1$ this is a contraction, while $L = 0$ gives contraction constant $0$ .

A state-Lipschitz vector field makes the Picard operator a contraction when $L h < 1$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A state-Lipschitz vector field makes the Picard operator a contraction when Lh<1Lh<1

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A state-Lipschitz vector field makes the Picard operator a contraction when $L h < 1$