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Finite propagation speed for the wave equation

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)). Let n≥1, c>0, x0∈Rn, t0>0 and let u∈C2 solve □cu=f on a neighbourhood of the closed backward cone K−(x0,t0)={(x,t):0≤t≤t0, ∣x−x0∣≤c(t0−t)} (Forward and backward wave cones, domain of dependence and influence). If f=0 on K−(x0,t0) and u(⋅,0)=ut(⋅,0)=0 on the base ball Bct0(x0), then u≡0 on K−(x0,t0); in particular u(x0,t0)=0. Data and source vanishing in a backward cone control the whole cone: the source term is included, in the sharp form of the enrichment row thm-finite-propagation-for-forced-waves.

Facts & Assumptions

Given: ACω; a C2 function u solving □cu=f on a neighbourhood of the closed cone K−=K−(x0,t0), with f=0 on K− and u(⋅,0)=ut(⋅,0)=0 on Bct0(x0); the density e=12(ut2+c2∣Du∣2)≥0 of Wave energy density, energy flux and total energy and E(t)=∫Bc(t0−t)(x0)e(x,t) dx for 0<t<t0.

[F1]

Cone energy identity: for 0<t1<t2<t0, with ℓ≥0 on the lateral surface, ∫K(t1,t2)fut=E(t2)−E(t1)+∫t1t2∫∂Bc(t0−t)(x0)ℓ dS dt. (The energy identity on a truncated wave cone)

[F2]

Dominated convergence: if fk→f pointwise and ∣fk∣≤g with ∫g<∞, then ∫fk→∫f. (Dominated convergence)

[F3]

A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere; a continuous nonnegative function with vanishing integral on an open ball vanishes identically there. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

[F4]

On an open convex set, a C1 function with vanishing gradient is constant; the open cone int⁡K−={0<t<t0, ∣x−x0∣<c(t0−t)} is convex, being the increasing union of the convex frusta K(t1,t2). (Vanishing gradient and time derivative force constancy on convex sets, Truncated wave cones: convexity, piecewise C1 presentation and outward normals)

Proof

1.1givenF2algebraF5

The energy tends to zero at the base: for every t∈(0,t0) the integral E(t) over the ball Bc(t0−t)(x0) is finite and E(t)≥0 because e≥0 is continuous on the compact set K− and hence bounded there; as t↓0, the functions x↦1Bc(t0−t)(x0)(x)e(x,t) converge pointwise on Bct0(x0) to 1Bct0(x0)(x)e(x,0) by continuity of e up to t=0, and they are dominated by the constant sup⁡K−e<∞, so [F2] gives E(t)→∫Bct0(x0)e(x,0) dx=12∫Bct0(x0)(ut(x,0)2+c2∣Du(x,0)∣2)dx=0, the last equality because both Cauchy data vanish on the base ball.

2.1givenstep 1.1F1algebra

Monotonicity and vanishing of the energy: for 0<t1<t2<t0 the frustum K(t1,t2) lies in K−, where f=0, so [F1] gives E(t2)−E(t1)=−∫t1t2∫∂Bc(t0−t)(x0)ℓ dS dt≤0 because ℓ≥0; thus E is nonincreasing on (0,t0), with E≥0 and E(t)→0 as t↓0 by step 1.1, so E(t)=0 for every t∈(0,t0).

3.1givenstep 2.1F3algebra

Vanishing of the derivatives on the open cone: fix t∈(0,t0); by step 2.1 e(⋅,t)≥0 has vanishing integral over the open ball Bc(t0−t)(x0), so [F3] and continuity give e(x,t)=0 for every x in that ball, and hence ut(x,t)=0 and Du(x,t)=0 there; letting t vary gives ut=Du=0 on the open cone int⁡K−.

4.1givenstep 3.1F4algebra∎

Constancy and conclusion: the open cone int⁡K− is convex [F4], so the vanishing-gradient lemma [F4] makes u constant on it; the constant is 0 because u is continuous on a neighbourhood of the closed cone and u(⋅,0)=0 on the base ball, so evaluating along points of the open cone tending to a base point gives u≡0 on int⁡K−; finally K− is the closure of int⁡K− (each point of the base, of the lateral surface or the vertex is a limit of interior points), so continuity gives u≡0 on K−, and in particular u(x0,t0)=0.

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