Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Constant phase is an energy-stationary state

Example

Let d≥1, T>0, ε,M,ν>0, W∈C2(R) be bounded below, and c∈R. On Td=Rd/(2πZ)d, consider the periodic equations ∂tϕ+u⋅∇ϕ=MΔμc, ∂tu+(u⋅∇)u+∇π=νΔu−ϕ∇μc, and ∇⋅u=0, with μc=ε−1W′(ϕ)−εΔϕ. The constant fields u=0, ϕ=c, and π=0 solve this system with constant chemical potential μc=ε−1W′(c). Their total energy is E(t)=(2π)dε−1W(c) for all t∈[0,T], so this is an energy-stationary state, meaning that its energy is constant in time; no local-minimum or stability claim is made. All quantities are dimensionless mathematical variables.

Facts & Assumptions

Given: The parameters, potential, constant c, and periodic equations in the example statement.

[F1]

Cancellation of capillary and transport energy exchange: for a classical periodic solution of these equations, E′(t)=−ν∫Td∣∇u∣2 dx−M∫Td∣∇μc∣2 dx.

Verification

1.1givenalgebra

For u=0 and ϕ=c, all time derivatives and spatial derivatives of u and ϕ vanish, while μc=ε−1W′(c) is constant and therefore ∇μc=Δμc=0. The phase equation has both sides zero; the momentum equation has every term zero; and ∇⋅u=0. Thus these fields are a classical periodic solution.

2.1givenF1step 1.1algebra∎

Their kinetic and gradient-energy terms vanish, while the potential term integrates to (2π)dε−1W(c) because the fundamental cube has volume (2π)d. Alternatively, [F1] gives E′(t)=0 because both dissipation integrands vanish. Hence the energy is constant and the state is energy-stationary.

Depends on

Dependency tree · 2 levels