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Finite speed of propagation for linear symmetric hyperbolic systems: cone domination and independence of the extension

Statement

Assume Countable Choice ACω (The Axiom of Countable Choice (ACω)), carrying the hypotheses of the analytic suppliers used below.

Let T>0, let Aμ:(−T,T)×R3→Sym⁡m(R) (μ=0,1,2,3) be smooth with A0(t,x)≥aI for some constant a>0 and all (t,x), and let ∥Ai(t,x)∥≤b for the operator norm and all i,t,x. Let C:(−T,T)×R3→Mm(R) be smooth and bounded by K.

(a) Put λ:=3 b/a+1; then λA0(t,x)+ξiAi(t,x)≥0 as a quadratic form for every (t,x) and every ξ∈R3 with ∣ξ∣≤1.

(b) Let w∈C1((−T,T)×R3;Rm) solve A0∂tw+Ai∂iw+Cw=0, and let x0∈R3, 0<t0<T, and suppose w(0,x)=0 for every ∣x−x0∣≤λt0. Then w(t,x)=0 for every (t,x)∈{(t,x):0≤t≤t0, ∣x−x0∣≤λ(t0−t)}.

(c) [independence of the extension] If w1,w2 are two solutions of two such systems whose complete coefficient families, including C, agree on an open set containing the truncated cone D={(t,x):0≤t≤t0,∣x−x0∣≤λ(t0−t)}, and if w1(0,x)=w2(0,x) for ∣x−x0∣≤λt0, then w1=w2 on D.

Facts & Assumptions

Given: Countable Choice and The coefficients, their stated bounds and the C1 solutions, and the point and time determining the compact cone D.

[L2]

The divergence theorem applies on bounded piecewise C1 domains (Divergence for finite piecewise C1 presentations).

[L3]

Continuous nonnegative E satisfying E(t)≤c∫0tE(s) ds vanishes by Gronwall (Gronwall's integral inequality with variable and constant coefficients).

Proof

technique · direct
1.1givenalgebra

(Correct symbol domination.) For every ∣ξ∣≤1, ∥∑iξiAi∥≤∑i∣ξi∣∥Ai∥≤b∥ξ∥1≤3 b. Hence for every z one has zT(λA0+ξiAi)z≥(λa−3b)∣z∣2=a∣z∣2≥0. This proves (a).

1.2givenalgebra

(Energy-current inequality on the compact cone.) Put jμ=wTAμw. Symmetry and the equation give ∂μjμ=wT(∂tA0+∑i∂iAi−C−CT)w. On compact D let M=sup⁡D∥∂tA0+∑i∂iAi∥+2K<∞. Thus ∂μjμ≤M∣w∣2≤(M/a)j0 throughout D.

2.1step 1.1step 1.2L2algebra

(Truncated cone flux.) For 0<τ<t0, apply the divergence theorem to Dτ={(s,x):0≤s≤τ,∣x−x0∣≤λ(t0−s)}. Base flux is −E(0) and top flux is E(τ), where E(s)=∫∣x−x0∣≤λ(t0−s)j0(s,x) dx. The lateral outward unit normal is (λ,ξ)/1+λ2, so its flux is nonnegative by step 1.1. Step 1.2 gives E(τ)≤E(0)+(M/a)∫0τE(s) ds. The domains avoid the cone vertex, so their boundaries are piecewise C1 as required.

3.1step 2.1L3algebra

(Vanishing and the vertex.) E is continuous and nonnegative; zero initial data give E(0)=0. Gronwall therefore gives E(s)=0 for 0≤s<t0. Positivity of A0 and continuity of w imply w vanishes on each closed top ball, since a nonzero value would give positive integral on a neighborhood. Letting s increase to t_0 and using continuity also gives w(t0,x0)=0. This proves (b).

4.1step 3.1givenalgebra∎

(Independence.) On the common neighborhood of D, subtract the two identical linear equations. The difference w1−w2 solves the homogeneous system with the common matrices Aμ and the common zeroth-order coefficient C; it vanishes on the base. The preceding argument uses only coefficients on D and gives equality throughout D. This proves (c).

Depends on

Dependency tree · 2 levels

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