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Barotropic Kelvin and Bernoulli with exact hypotheses — formal worked solution

Example

Let I=R, Ω=R3, P(q)=q for q>0, ρref=1, ρ(t,x)=1, V=0, and u(t,x)=(1,0,0). These fields solve the barotropic Euler system. The enthalpy is h(q)=log⁡q. Define the Bernoulli quantity by B=∣u∣2/2+h(ρ)+V; here it is the constant 1/2. The closed loop X(t,s)=(cos⁡(2πs)+t,sin⁡(2πs),0) is material and has circulation 0 for every t.

Facts & Assumptions

Given: Use the classical fields, positive density, equation of state, potential and closed material loop in [A1].

[A1]

I=R, Ω=R3, P(q)=q, ρref=1, ρ=1, V=0, u=(1,0,0), and X(t,s)=(cos⁡(2πs)+t,sin⁡(2πs),0) for 0≤s≤1.

[F1]

Barotropic Kelvin and Bernoulli with exact hypotheses: for classical positive-density barotropic Euler with conservative V, h(q)=∫ρrefqP′(r)/r dr; steady Bernoulli is constant on streamlines, and a closed C1 material loop has constant circulation under the stated mixed-derivative condition.

Verification

1.1givenA1F1algebra

The fields have the required smoothness and positive density. Since ρ is constant, ρt+div⁡(ρu)=0; also ut=0, (u⋅∇)u=0, ∇(P(ρ))=0, and ∇V=0. Thus both barotropic Euler equations hold pointwise. The enthalpy formula in [F1] gives h(q)=∫1qdr/r=log⁡q, so h(1)=0 and ∣u∣2/2+h(1)+V=1/2.

2.1givenA1F1algebra∎

The loop is closed because its endpoint values at s=0,1 agree, and Xt=(1,0,0)=u(t,X); its spatial and mixed derivatives are continuous. Its tangent is Xs=(−2πsin⁡(2πs),2πcos⁡(2πs),0), so Γ(t)=∫01u⋅Xs ds=−2π∫01sin⁡(2πs) ds=0. This verifies the theorem's circulation conclusion directly.

Depends on

Dependency tree · 2 levels