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A globally Lipschitz alignment model with exact mean trajectory

Example

For $a \geq 0$ on $R^{d}$ , take $K (x, y) = a (y - x)$ . For $a > 0$ , each particle is pulled toward the current mean position; for $a = 0$ , every particle is stationary. It is globally Lipschitz, and its particle trajectories are explicit.

Facts & Assumptions

Given: $a \geq 0$ , $N \geq 1$ , and initial positions $x_{i} (0) \in R^{d}$ .

[F1]

A globally Lipschitz interaction kernel uses the two-coordinate bound in Globally Lipschitz mean-field interaction kernel.

[F2]

The equal-probability particle system is ${\dot{x}}_{i} = N^{- 1} \sum_{j} K (x_{i}, x_{j})$ (Equal-probability mean-field and weighted signed vortex particle systems).

[F3]

The empirical particle measure satisfies the exact weak equation (Exact weak mean-field identity for empirical particle measures).

[F4]

General characteristic probability solutions obey the $e^{2 L t}$ stability estimate (Dobrushin stability in 1-Wasserstein distance for mean-field flows).

Verification

Given: The data in the Example and $K (x, y) = a (y - x)$ .

1.1givenF1algebra

For all $x, y, x^{'}, y^{'}$ , $∣ K (x, y) - K (x^{'}, y^{'}) ∣ = a ∣ (y - y^{'}) - (x - x^{'}) ∣ \leq a (∣ x - x^{'} ∣ + ∣ y - y^{'} ∣)$ , so the global Lipschitz constant may be taken as $L = a$ in [F1].

2.1F2F3F4step 1.1algebra∎

Substitution into [F2] gives ${\dot{x}}_{i} = a (\overset{ˉ}{x} - x_{i})$ , where $\overset{ˉ}{x} = N^{- 1} \sum_{j} x_{j}$ . Summing over $i$ yields $\dot{\overset{ˉ}{x}} = 0$ , and hence $x_{i} (t) = \overset{ˉ}{x} (0) + e^{- a t} (x_{i} (0) - \overset{ˉ}{x} (0)) .$ For $a > 0$ , the formula tends to the constant mean as $t \to \infty$ ; for $a = 0$ , it equals $x_{i} (0)$ . Thus the empirical measure is an exact weak solution by [F3]; for probability solutions the stability factor is $e^{2 a t}$ by [F4].

Depends on

Used by

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Sources

François Golse, Mean-Field Limits in Statistical Dynamics, Lecture 1 (standard reference, not scraped)