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Conserved energy of a travelling wave packet

Example

Assume Countable Choice for the Lebesgue measure and multidimensional volume assertions below (The Axiom of Countable Choice (ACω)). Let c>0 and let F∈Cc2(R) be compactly supported, and put

u(x,t):=F(x−ct)(x,t∈R),

a right-moving travelling packet. Then u is a classical solution of □cu=0 (Wave equation, Cauchy data and wave speed) and for every t∈R its total energy (Wave energy density, energy flux and total energy) is finite, independent of t, and splits equally between its kinetic and potential parts:

E(t)=∫R12(ut2+c2ux2) dx=c2∫RF′(s)2 ds,Ekin(t)=Epot(t)=12c2∫RF′(s)2 ds.

The equal split is the signature of a nondispersive packet: pointwise ut=−cF′(x−ct) and ∣Du∣=∣F′(x−ct)∣, so both densities equal 12c2F′(x−ct)2 and the energy density is c2F′(x−ct)2.

The finite-energy statement is genuinely one-dimensional. In dimension n≥2 the profile u(x,t)=F(ω⋅x−ct) with a unit vector ω is still a classical solution and still satisfies ut=−cF′ and Du=F′ω pointwise, so the two densities still split equally; but the density c2F′(ω⋅x−ct)2 then depends on x only through the single variable ω⋅x−ct, and whenever F′≢0 it is bounded below by a positive constant on a slab of infinite n-dimensional measure, so the total energy over Rn is +∞. The conserved finite total energy computed here is therefore the energy of a one-dimensional packet.

Facts & Assumptions

Given: Countable Choice; c>0 and F∈Cc2(R), and u(x,t)=F(x−ct) on R×R; write F2(s):=F′(s)2 for the continuous compactly supported density profile.

[F1]

Chain rule: D(G∘H)(a)=DG(H(a))∘DH(a) for composable totally differentiable maps. (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a))

[F2]

Change of variables: for a C1 diffeomorphism T:U→V of open sets and a continuous compactly supported f:V→R, ∫Vf dλn=∫U(f∘T) ∣det⁡DT∣ dλn. (The published Riemann change-of-variables theorem already gives the Lebesgue formula for continuous compactly supported integrands)

[F3]

The energy density and flux of a C2 function are e=12(ut2+c2∣Du∣2) and q=−c2utDu; the wave operator is □c=∂t2−c2Δ. (Wave energy density, energy flux and total energy, Wave equation, Cauchy data and wave speed)

[F4]

The nonnegative Lebesgue integral is monotone and positively homogeneous; a continuous compactly supported function is bounded and supported in a set of finite measure. (Monotonicity and nonnegative homogeneity of the nonnegative integral, The support of a function on Rn and its compactly supported Riemann integral)

Verification

1.1givenF1F3algebra

Derivatives and the pointwise split: by [F1], ut=−cF′(x−ct), utt=c2F′′(x−ct), ux=F′(x−ct) and uxx=F′′(x−ct), so □cu=c2F′′(x−ct)−c2F′′(x−ct)=0 and u is a classical solution with continuous second derivatives [F3]; moreover ut2=c2F′(x−ct)2 and ∣Du∣2=F′(x−ct)2, so ekin=epot=12c2F2(x−ct) and e=c2F2(x−ct).

2.1givenstep 1.1F2F4algebra

The total energy: for each t, E(t)=∫Rc2F2(x−ct) dx=c2∫RF2(y) dy by [F2] applied to the diffeomorphism T(x)=x+ct of R, whose derivative is 1; the value is finite because F2=F′2 is continuous with compact support, so it is bounded by a constant and vanishes outside a bounded interval, and [F4] bounds its integral by the constant times the finite length of that interval; similarly Ekin(t)=Epot(t)=12c2∫RF2(y) dy by [F4]. Hence E(t) is finite, independent of t, and equals c2∫RF′(s)2 ds, with the equal split Ekin=Epot=12c2∫RF′(s)2 ds.

3.1givenstep 1.1F4F5algebra∎

The multidimensional caution: for n≥2, given a unit vector ω and u(x,t)=F(ω⋅x−ct), the same chain rule gives ut=−cF′(ω⋅x−ct), Du=F′(ω⋅x−ct)ω and utt=c2F′′, Δu=F′′∣ω∣2=F′′, so u is again a classical solution and the densities again satisfy ekin=epot=12c2F′(ω⋅x−ct)2; but if F′≢0 then F′2≥m>0 on some nondegenerate interval I=(a,b) by continuity. Choose an orthogonal O with Oe0=ω: take O=I if ω=e0, and otherwise set v=(e0−ω)/∣e0−ω∣ and O=I−2vvT; direct multiplication gives OTO=I and Oe0=ω. For each R>0, the rotated box ctω+O(I×(−R,R)n−1) lies in the slab {x:ω⋅x−ct∈I}. By [F5] its measure is (b−a)(2R)n−1. The density is at least c2m throughout it, so [F4] gives ERn(t)≥c2m(b−a)(2R)n−1 for every R; letting R→∞ proves ERn(t)=+∞. Thus the finite total energy of the Example cannot be extended beyond n=1.

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