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Wave energy need not be conserved through an open boundary

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice (ACω)) for the Lebesgue and Riemann integral bridge used below. The claim refuted is that the total energy of a classical wave solution is automatically constant whenever the domain is bounded, without any hypothesis on the boundary flux. Witness: let c>0, let Ω=(0,1)⊆R and let F∈Cc2(R) have F′ nonzero somewhere in (0,1); put

u(x,t):=F(x−ct),

the right-moving packet. Then u solves □cu=0 on R×R, but its energy in the fixed interval,

EΩ(t)=c2∫01F′(x−ct)2 dx,

equals c2∫01F′(x)2 dx>0 at t=0 and is 0 for all sufficiently large t, once the packet has left the interval. The decrease is exactly the boundary flux: with q=−c2utux=c3F′2,

ddtEΩ(t)=q(0,t)−q(1,t),

so the energy lost through the right endpoint is accounted for, and Conservation of total wave energy in three admissible settings may not be invoked on a domain with an open boundary without the vanishing-flux hypothesis.

Homogeneous-boundary comparison on the half-line. If v∈C2([0,∞)×I) solves vtt−c2vxx=0 on an open time interval I, and for each compact J⊆I there is RJ<∞ with vx=vt=0 for x≥RJ and t∈J, then E+(t):=12∫0∞(vt2+c2vx2) dx is constant under either v(0,t)=0 for every t∈I or vx(0,t)=0 for every t∈I (the homogeneous Dirichlet and Neumann comparisons for the linear case of the cited problem). No nonlinear potential term is asserted.

Facts & Assumptions

Given: Countable Choice; c>0, Ω=(0,1), F∈Cc2(R) with F′ nonzero somewhere in (0,1), u(x,t)=F(x−ct), and the fields e=12(ut2+c2ux2), q=−c2utux of Wave energy density, energy flux and total energy; for the last part a solution v on [0,∞)×I with the stated support hypothesis.

[F1]

The local balance: ∂te+∂xq=fut, hence ∂te=−∂xq for a classical solution. (The local wave-energy conservation law)

[F3]

Differentiation under the integral sign: if x↦f(x,t) is integrable for every t, t↦f(x,t) is differentiable for almost every x, and the t-derivative is dominated on the time interval by a fixed integrable function, then F(t)=∫f(x,t) dx is differentiable with F′(t)=∫∂tf(x,t) dx. (Differentiation under the integral sign)

[F6]

The support of f is the closure of {f≠0}; for the translate, supp⁡F(⋅−ct)=ct+supp⁡F. (The support of a function on Rn and its compactly supported Riemann integral)

Proof

1.1givenF1F2algebra

The packet and its flux: by the chain rule [F2], ut=−cF′(x−ct) and ux=F′(x−ct), so utt=c2F′′(x−ct), uxx=F′′(x−ct) and □cu=0; hence [F1] holds and the energy density and flux are e=c2F′(x−ct)2 and q=−c2utux=c3F′(x−ct)2.

2.1givenstep 1.1F6algebra

Positive initial energy and late vanishing: since F′ is continuous and nonzero somewhere in (0,1), there are a subinterval of (0,1) on which F′2≥m>0 and hence EΩ(0)=c2∫01F′(x)2 dx>0; and by [F6] the support of x↦u(x,t) is ct+supp⁡F, so for every t>(1−min⁡supp⁡F)/c the packet is disjoint from [0,1] and EΩ(t)=0.

2.2givenstep 1.1F3F4algebra

The flux identity: e and ∂te are continuous on [0,1] and bounded on compact time intervals, so [F3] gives EΩ′(t)=∫01∂te(x,t) dx=−c3∫01(F′2)′(x−ct) dx, and by [F4] the Darboux fundamental theorem applies to the continuous function x↦F′2(x−ct) on [0,1], whose Lebesgue integral equals that Darboux integral, giving ∫01(F′2)′(x−ct) dx=F′2(1−ct)−F′2(−ct); therefore EΩ′(t)=c3(F′2(−ct)−F′2(1−ct))=q(0,t)−q(1,t).

3.1step 2.1step 2.2algebra

Conclusion for the open boundary: by steps 2.1 and 2.2 the energy EΩ is positive at t=0, zero for all large t, and its rate of change is exactly the difference of the outward fluxes at the two endpoints; so EΩ is not constant, the drop is accounted for by the flux through the right endpoint, and automatic conservation cannot be inferred without controlling the boundary flux.

4.1givenstep 1.1F2F3F4F5∎

The half-line comparison: fix a compact J⊆I and R≥RJ; for t∈J the integrand of E+ vanishes for x≥RJ, so E+(t)=12∫0R(vt2+c2vx2) dx, and [F3] with the domination constant sup⁡[0,R]×J∣vtvtt+c2vxvxt∣ gives E+′(t)=∫0R(vtvtt+c2vxvxt) dx=c2∫0R(vtvxx+vxvxt) dx=c2∫0R∂x(vtvx) dx, using vtt=c2vxx and the product rule [F2]; by [F4], ∫0R∂x(vtvx) dx=vt(R,t)vx(R,t)−vt(0,t)vx(0,t); the upper endpoint term is zero by the support hypothesis, and the lower endpoint term is zero because in the Dirichlet case the trace t↦v(0,t) is identically zero and differentiable with derivative vt(0,t), while in the Neumann case vx(0,t)=0 directly; hence E+′(t)=0 for every interior t∈J and, E+ being continuous on J with vanishing derivative there, [F5] makes E+ constant on J; as J is an arbitrary compact subinterval of I, E+ is constant on I.

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