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Zero wave energy means a spatial constant, fixed by the displacement datum

Example

Assume the Axiom of Countable Choice. Let c>0 and let u be a classical solution of the homogeneous equation □cu=0 on Rn×(0,T) with Cauchy data (u0,u1) and differentiable displacement u0, so Du0 in the initial-energy hypothesis is defined, with E(t)=E(0) for every t∈(0,T) — the sharp form of conservation in the senses of Conservation of total wave energy in three admissible settings — and with E(0)=12∫Rn(u12+c2∣Du0∣2) dx=0. Then u1=0 and Du0=0, so the displacement datum u0 is constant on Rn; the energy seminorm sees only (ut,Du) and cannot fix that constant, and the evolution keeps it: u(⋅,t)≡u0 for every t.

In particular every constant displacement with zero initial velocity, u(x,t)≡k for a fixed k∈R, is a genuine classical solution of zero energy. Thus "zero energy" is strictly weaker than "zero solution": the displacement datum is what fixes the residual constant (Energy uniqueness for the wave Cauchy problem(ii)).

Facts & Assumptions

Given: ACω; a classical homogeneous solution u with Cauchy data (u0,u1) and differentiable displacement u0, so Du0 in the initial-energy hypothesis is defined, conserved total energy in the sharp form E(t)=E(0) for t∈(0,T) and E(0)=12∫(u12+c2∣Du0∣2)=0; the density e=12(ut2+c2∣Du∣2)≥0 of Wave energy density, energy flux and total energy.

[F1]

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

[F2]

On an open convex set, a C1 function with vanishing gradient is constant; Rn is convex. (Vanishing gradient and time derivative force constancy on convex sets)

[F3]

In each setting of the conservation theorem the energy is constant on the interval; the hypothesis of this Example records the sharp form E(t)=E(0) for all t∈(0,T). (Conservation of total wave energy in three admissible settings)

Verification

1.1givenF1F2F3algebra

Vanishing at positive times: sharp conservation gives E(t)=E(0)=0 for each t∈(0,T). The nonnegative continuous density therefore vanishes everywhere by [F1], so ut(⋅,t)=Du(⋅,t)=0. By [F2], each spatial slice is constant.

2.1givenstep 1.1F4algebra

Time constancy and the data: for each fixed x, the function t↦u(x,t) has derivative zero on (0,T), so [F4] makes it constant there. Together with step 1.1 this gives one constant k on all space-time. The Cauchy limits then give u0(x)=k and u1(x)=0 at every x, hence Du0=0 and u(⋅,t)≡u0. This derives pointwise data vanishing without inferring it from an almost-everywhere statement at t=0.

3.1givenstep 2.1algebra∎

The converse check: the constant displacement u(x,t)≡k has ut=Du=0, so e≡0 and E≡0, and □cu=0 trivially; hence the zero-energy solutions are exactly the constant displacements, and that constant is precisely the initial displacement datum, which the energy cannot see.

Depends on

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