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Solutions of projected contact dynamics are unique under the inherited choice assumptions

Statement

Assume Countable Choice and Dependent Choice. Let d∈N and T>0, let C⊆Rd be nonempty, closed, and convex, and let f:[0,T]×C→Rd satisfy ∥f(t,x)−f(t,y)∥2≤L∥x−y∥2 for all t,x,y, with L≥0. Any two absolutely continuous solutions of the projected dynamics of Projected contact dynamics and explicit projection Euler step with the same initial state agree on [0,T]. For d=0, extend the solution convention directly: C=R0={0}, NC(0)={0}, and a solution is the sole map x(t)=0.

Facts & Assumptions

Given: Countable Choice, Dependent Choice, the set and forcing, and two absolutely continuous solution maps in the Statement.

[A1]

The Axiom of Countable Choice (ACω) is inherited through [F4]'s absolutely-continuous FTC.

[F1]

For d≥1, Projected contact dynamics and explicit projection Euler step: a solution satisfies f(t,x(t))−x˙(t)∈NC(x(t)) for almost every t.

[F2]

The convex normal-cone operator is maximal monotone and its resolvent is projection: if nx∈NC(x) and ny∈NC(y), then ⟨nx−ny,x−y⟩≥0.

[F3]

Gronwall's integral inequality with variable and constant coefficients: if continuous u≥0 satisfies u(t)≤u(0)+∫0t2Lu(s) ds, then u(t)≤u(0)e2Lt.

[F4]

The product of two absolutely continuous functions is absolutely continuous and Fundamental theorem of calculus for absolutely continuous functions: products of absolutely continuous real functions are absolutely continuous with the ordinary product derivative almost everywhere, and the integral of that derivative is the endpoint difference.

[F6]

The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: R0 is the singleton zero vector space, with zero inner product and norm.

Proof

1.1F6given

If d=0, then C={0} and both solution maps are identically zero. Coordinatewise absolute continuity and differentiation are vacuous, and the inclusion is 0∈NC(0)={0}, so this agrees with the extended convention and uniqueness holds.

1.2F1F2given

Assume d≥1 and let x,y be the two solutions and define nx=f(t,x)−x˙ and ny=f(t,y)−y˙ at almost every time where both inclusions hold. By [F1] these belong to NC(x) and NC(y); [F2] gives ⟨nx−ny,x−y⟩≥0.

2.1A1A2F1F2F3F4F5step 1.2givenalgebra∎

The function u(t)=∥x(t)−y(t)∥22 is absolutely continuous by [F4]. At almost every t, [F1] and step 1.2 give u′(t)=2⟨x˙−y˙,x−y⟩=2⟨f(t,x)−f(t,y),x−y⟩−2⟨nx−ny,x−y⟩≤2Lu(t), using the Lipschitz hypothesis and [F5]. Since u(0)=0, [F4] integrates this differential inequality to u(t)≤∫0t2Lu(s) ds. Applying [F3] gives u(t)=0 for every t, so x=y. This argument uses no choices beyond the two axioms already required by [F4].

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