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Entropy solution orbits are strongly continuous in L1

Statement

Assume Countable Choice and Dependent Choice (The Axiom of Countable Choice (ACω), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain) for the heat-kernel, L1 completeness and vanishing-viscosity extraction interfaces used below. Let n≥1, let f ⁣:R→Rn be C1, and let u0∈L1(Rn)∩L∞(Rn). Then the orbit t↦Stu0 is continuous from [0,∞) to L1(Rn): for every t≥0, ∥Stu0−Ssu0∥1→0 as s→t, and in particular lim⁡t↓0∥Stu0−u0∥L1(Rn)=0. If additionally f(0)=0 and f is globally Lipschitz, its extension is a strongly continuous semigroup of L1 contractions on all of L1 (The entropy solution semigroup on L1∩L∞, Kruzhkov entropy solutions, The space Lp(μ) as the quotient by null functions).

Facts & Assumptions

Given: Countable and Dependent Choice, n≥1, a C1 flux, with the extra global Lipschitz and f(0)=0 hypotheses only for the extension, a datum u0∈L1∩L∞ with M=∥u0∥∞, and the entropy solution semigroup St of The entropy solution semigroup on L1∩L∞.

[F1]

Semigroup and contraction: St+s=StSs, each St is order-preserving, and ∥Stu0−Stv0∥1≤∥u0−v0∥1 for all t≥0 and u0,v0∈L1∩L∞; when f(0)=0 and f is globally Lipschitz the maps extend to order-preserving L1 contractions on all of L1, agreeing with the flow on L1∩L∞ (The entropy solution semigroup on L1∩L∞).

[F2]

Every any-time contraction, for data in L1∩L∞, holds for every t because the solutions have Lloc1-continuous representatives: ∥Stu0−Stv0∥1≤∥u0−v0∥1 (Global L1 contraction from the local estimate, Existence of bounded Kruzhkov entropy solutions).

[F3]

Finite propagation: if u0 vanishes almost everywhere outside B(0,R), the entropy solution vanishes almost everywhere outside B(0,R+Lt) for almost every t, where L is a Lipschitz constant of the flux on the common range; with the Lloc1-continuous representative this support statement upgrades to every t (Finite propagation for scalar conservation laws, Open ball, closed ball and sphere in a metric space).

[F4]

The strong local L1 trace: for every compact K, ess sup⁡0<t<δ∫K∣u(t,x)−u0(x)∣ dx→0 as δ↓0; and monotone convergence controls the tails of an L1 function over increasing balls (Kruzhkov entropy solutions, Monotone convergence for the integral, The space Lp(μ) as the quotient by null functions).

Proof

technique · direct
1.1F2F3F4

Continuity at time zero. Fix R>0, put u0R=u01B(0,R) and vR(t)=Stu0R; the datum u0R is compactly supported, so by [F3] the solution vR is supported in B(0,R+Lt) for every t, where L is a Lipschitz constant of f on the common range [−∥u0∥∞,∥u0∥∞]. By [F2], ∥Stu0−vR(t)∥1≤∥u0−u0R∥1=∥u0∥L1(B(0,R)c) for every t. Hence for 0<t<1/L (any t>0 when L=0) the exterior B(0,R+1)c is contained in B(0,R+Lt)c and ∥Stu0−u0∥1≤∥Stu0−u0∥L1(B(0,R+1))+2∥u0∥L1(B(0,R)c), because on B(0,R+1)c both Stu0 and u0 have L1 norms bounded by ∥u0∥L1(B(0,R)c) (for Stu0 combine the contraction bound with vR(t)=0 there). The first term tends to 0 as t↓0 by the local L1 continuity of the chosen representative [F2] and its trace [F4], and the tail term tends to 0 as R→∞ by monotone convergence. Therefore lim⁡t↓0∥Stu0−u0∥1=0.

2.1F1F2step 1.1

Continuity at every time. Let s,t≥0 with t>s. By the semigroup law and the contraction estimate of [F1], ∥Stu0−Ssu0∥1=∥Ss(St−su0)−Ssu0∥1≤∥St−su0−u0∥1, and the right side tends to 0 as t↓s by step 1.1 applied to the fixed datum u0. The case s<t is symmetric, so the orbit is continuous at every t≥0.

3.1F1step 2.1∎

Strong continuity on the closure. If f(0)=0 and f is globally Lipschitz, the extension of [F1] is defined on all of L1, and L1∩L∞ is dense in L1 (the closure appearing in the statement). For u0∈L1 and u0m=(−m)∨(u0∧m)∈L1∩L∞ with u0m→u0 in L1, the contraction property gives ∥Stu0−Stu0m∥1≤∥u0−u0m∥1 for every t, so ∥Stu0−Ssu0∥1≤2∥u0−u0m∥1+∥Stu0m−Ssu0m∥1→0 by first making the two approximation errors small with a fixed large m and then taking s→t, by step 2.1 applied to each u0m. Hence the extended semigroup is strongly continuous on all of L1, which is the closure of L1∩L∞.

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