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TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generated
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Exact weak mean-field identity for empirical particle measures

Statement

Let x1,…,xN solve the equal-probability system of Equal-probability mean-field and weighted signed vortex particle systems on [0,T], and put μN(t)=N−1∑iδxi(t). Then for every φ∈Cc1(Rd), ∫φdμN(t)−∫φdμN(0)=∫0t∫∇φ(x)⋅(∫K(x,y)dμN(s)(y))dμN(s)(x)ds. Thus the empirical measure satisfies the same weak transport equation exactly, with no particle quadrature error in this identity.

Facts & Assumptions

Given: A classical particle solution, the particle equations, and a test function φ∈Cc1(Rd).

[F1]

The equal-probability particle system is x˙i=N−1∑jK(xi,xj) and its empirical law is N−1∑iδxi (Equal-probability mean-field and weighted signed vortex particle systems).

[F2]

The chain rule differentiates t↦φ(xi(t)) (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Proof

1.1F1F2algebra

By [F2] and [F1], ddt1N∑iφ(xi(t))=1N2∑i,j∇φ(xi(t))⋅K(xi(t),xj(t)). The double sum is finite, so integration in time gives the left side of the Statement as its time integral.

2.1F1step 1.1algebra∎

Expanding both empirical integrals on the right of the Statement gives exactly the same finite double sum as in step 1.1. This proves the identity for every t and includes repeated particle positions without a distinctness assumption.

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Sources