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Barotropic Kelvin and Bernoulli with exact hypotheses

Statement

Let I⊆R be an open time interval, let Ω⊆R3 be open, and let P:(0,∞)→R be C2. Suppose u:I×Ω→R3 has continuous first time, first spatial, and second spatial derivatives; ρ:I×Ω→(0,∞) has continuous first time and first spatial derivatives; and V:Ω→R is time-independent and C2. Assume pointwise

∂tρ+div⁡(ρu)=0,∂tu+(u⋅∇)u=−ρ−1∇(P(ρ))−∇V.

For any reference density ρref>0, define h(q)=∫ρrefqP′(r)/r dr for q>0. For a steady velocity field, a streamline is a C1 curve x:J→Ω, where J⊆R is an interval, satisfying x′(s)=u(x(s)) with u(x) denoting its time-independent spatial value. If u and ρ are independent of time, every streamline has ∣u∣2/2+h(ρ)+V constant along it. If in addition curl⁡u=0 on a connected component of Ω, this quantity is constant throughout that component.

For any closed C1 material loop X:J×[0,1]→Ω satisfying X(t,0)=X(t,1), Xt(t,s)=u(t,X(t,s)), with Xs and Xts continuous, the circulation Γ(t)=∫01u(t,X(t,s))⋅Xs(t,s) ds is constant on J. Independently of the Euler assumptions, every C2 vector field v on an open subset of R3 obeys the identity

(v⋅∇)v=∇(∣v∣2/2)−v×curl⁡v.

No claim is made for shocks, nonbarotropic pressure, nonconservative body forces, or viscous circulation. Under SI interpretation, u is in metres per second, ρ in kilograms per cubic metre, P in pascals, and h,V in square metres per second squared.

Facts & Assumptions

Given: Use the positive density, C2 equation of state, conservative time-independent potential, classical pointwise equations, steady-flow hypothesis where stated, and material-loop regularity in the Statement. The result is a conditional deduction from the stated smooth model.

[F1]

Formal parameterized fluid systems and observables: the barotropic Euler equations are ρt+div⁡(ρu)=0 and ut+(u⋅∇)u=−ρ−1∇(P(ρ))−∇V, with positive density, P∈C2, and time-independent V∈C2; divergence, gradient, curl and convective derivative use Cartesian coordinates.

Proof

1.1givenalgebra

In coordinates, ((v⋅∇)v)i=∑jvj∂jvi and ∂i(∣v∣2/2)=∑jvj∂ivj. Let ϵijk be the alternating symbol with ϵ123=1. Using (v×curl⁡v)i=∑j,k,l,mϵijkϵklmvj∂lvm=∑jvj∂ivj−∑jvj∂jvi gives the asserted identity component by component.

2.1givenF1step 1.1algebra

The fundamental theorem of calculus and the chain rule give ∇h(ρ)=(P′(ρ)/ρ)∇ρ=ρ−1∇(P(ρ)). For a steady solution, ut=0, so dotting the momentum equation with u and applying step 1.1 yields u⋅∇(∣u∣2/2+h(ρ)+V)=0, because u⋅(u×curl⁡u)=0. Along a streamline x′(s)=u(x(s)), the chain rule makes the derivative of this quantity zero, proving streamline constancy. If curl⁡u=0, step 1.1 and the momentum equation give ∇(∣u∣2/2+h(ρ)+V)=0. A differentiable function with zero gradient is constant along every polygonal path; in an open connected subset of R3, polygonally reachable points form a nonempty relatively open and relatively closed set (each point has a ball contained in the open set). Connectedness therefore gives constancy on that component.

3.1givenF1step 2.1algebra∎

By the assumed continuous mixed derivatives, differentiate under the integral sign and commute Xts=∂sXt. The chain rule and Xt=u(t,X) give ∂t(u(t,X)⋅Xs)=(∂tu+(u⋅∇)u)(t,X)⋅Xs+∂s(∣u(t,X)∣2/2). The momentum equation and step 2.1's enthalpy-gradient identity turn this into −∂s(h(ρ(t,X))+V(X))+∂s(∣u(t,X)∣2/2). Integrating in s gives the endpoint difference of ∣u∣2/2−h(ρ)−V, which is zero because X(t,0)=X(t,1) and all fields are single-valued. Hence dΓ/dt=0 on J.

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