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Global L1 contraction from the local estimate

Statement

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the analytic prerequisites used below.

Let n≥1, T>0, let f ⁣:R→Rn be C1 and Lipschitz on the common essential range of the solutions below with constant L≥0, and let u,v be bounded Kruzhkov entropy solutions on ΠT in the sense of Kruzhkov entropy solutions, with initial data u0,v0∈L1(Rn)∩L∞(Rn). Then for almost every t∈(0,T), ∥u(t,⋅)−v(t,⋅)∥1≤∥u0−v0∥1. If u,v have representatives continuous in Lloc1 on [0,T], the same inequality holds for every t∈[0,T] (Open ball, closed ball and sphere in a metric space, The space Lp(μ) as the quotient by null functions).

Facts & Assumptions

Given: Countable Choice, n≥1, T>0, f Lipschitz with constant L≥0 on the common essential range of u and v, bounded Kruzhkov entropy solutions u,v on ΠT with ∣u∣,∣v∣≤M almost everywhere, and initial data u0,v0∈L1(Rn)∩L∞(Rn). Set L∗:=sup⁡∣z∣≤M∣f′(z)∣<∞; in the proof use L∗, irrespective of the given constant on the common essential range.

[F1]

Since f∈C1, L∗<∞, and coordinatewise FTC gives f(b)−f(a)=∫abf′(z) dz and ∣f(b)−f(a)∣≤L∗∣b−a∣ for a,b∈[−M,M] (The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a)). Local L1 contraction with this interval constant: for every centre x0 and radius R>0, for almost every t∈(0,T) with L∗t<R, ∫B(x0,R−L∗t)∣u(t,x)−v(t,x)∣ dx≤∫B(x0,R)∣u0(x)−v0(x)∣ dx, with an exceptional null set depending on x0,R (Local L1 contraction for two entropy solutions, Kruzhkov entropy solutions).

[F2]

Monotone convergence for integrals of nonnegative functions over increasing sets: if 0≤gm↑g pointwise then ∫gm↑∫g; in particular the integrals of a fixed nonnegative function over the balls B(0,ρ)↑Rn increase to its integral over Rn, finite or infinite (Monotone convergence for the integral).

[F3]

Almost-everywhere assertions concern equivalence classes: a countable union of null sets in (0,T) is null, and members of L1 are defined up to modification on null sets (The space Lp(μ) as the quotient by null functions, Open ball, closed ball and sphere in a metric space).

Proof

technique · direct
1.1F1

Ball estimates along an exhausting sequence. For m≥1 put Rm=L∗T+m, so that Rm>L∗t for every t∈(0,T) and Rm−L∗t=m+L∗(T−t)>0. Applying [F1] with centre 0 and radius Rm gives, for every m, an exceptional null set Nm⊆(0,T) such that for all t∈(0,T)∖Nm ∫B(0,Rm−L∗t)∣u(t,x)−v(t,x)∣ dx≤∫B(0,Rm)∣u0(x)−v0(x)∣ dx≤∥u0−v0∥1.

2.1F2F3step 1.1

Intersection and monotone limit. The set N=⋃m≥1Nm is null by [F3]. Fix t∈(0,T)∖N, so that the estimates of step 1.1 hold for every m. The balls B(0,Rm−L∗t) increase to Rn as m↑∞, hence the integrals of the fixed nonnegative function ∣u(t,⋅)−v(t,⋅)∣ over them increase to ∫Rn∣u(t,x)−v(t,x)∣ dx, while ∫B(0,Rm)∣u0−v0∣↑∥u0−v0∥1 by monotone convergence. Passing to the limit in step 1.1 gives ∥u(t,⋅)−v(t,⋅)∥1≤∥u0−v0∥1<∞ for every t∈(0,T)∖N, which is the almost-everywhere assertion and shows that the slice integrals are finite for almost every t.

3.1F2step 2.1∎

The every-time assertion under Lloc1 continuity. Assume now that u and v have representatives on [0,T] such that tj→t implies u(tj)→u(t) and v(tj)→v(t) in L1(K) for every compact K⊆Rn (with one-sided sequences at t=0,T). Fix t∈[0,T] and choose tj∈(0,T)∖N with tj→t, possible because N is null. For every fixed ball B(0,ρ), step 2.1 gives ∫B(0,ρ)∣u(tj)−v(tj)∣≤∥u(tj)−v(tj)∥1≤∥u0−v0∥1, and ∣u(tj)−v(tj)∣→∣u(t)−v(t)∣ in L1(B(0,ρ)) because u(tj)→u(t) and v(tj)→v(t) there; the inequality ∣∥a∥1−∥b∥1∣≤∥a−b∥1 gives convergence of these integrals directly. Hence ∫B(0,ρ)∣u(t)−v(t)∣≤lim inf⁡j∫B(0,ρ)∣u(tj)−v(tj)∣≤∥u0−v0∥1. Letting ρ↑∞ and using monotone convergence once more gives ∥u(t)−v(t)∥1≤∥u0−v0∥1.

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