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Transport of a translating rigid region: the rate integral vanishes for a travelling profile

Example

Assume ACω (The Axiom of Countable Choice (ACω)) for the actual Lebesgue transport and change-of-variables suppliers.

Let B:=B1(0)⊆R3 be the open unit ball (Open ball, closed ball and sphere in a metric space), a nonempty bounded open set and hence a material region in the sense of Material regions, densities and their mass; let v∈R3 be fixed, and let

ψ(t,X):=X+tv(t∈R, X∈R3)

be the uniform translation motion of Continuum motions, spatial velocity, material acceleration and swept regions (each ψt is the identity plus a translation, a C∞ diffeomorphism with DXψt=I3). Let h:R3→R be C1 (Ck Euclidean maps and diffeomorphisms) and define the travelling profile

f(t,x):=h(x−tv)((t,x)∈R×R3).

Then ∫ψt(B)f(t,x) dx is independent of t, while the transport integrand vanishes pointwise:

∂tf+div⁡(fv)=0,

so the transport theorem reproduces the constant rate 0.

Facts & Assumptions

Given: Countable choice. The unit ball B, the velocity v, the motion ψ(t,X)=X+tv, the C1 profile h and f(t,x)=h(x−tv).

[L1]

Transport theorem: for the motion ψ with spatial velocity w and a C1 f on a neighbourhood of the swept compact region, ddt∫ψt(B)f=∫ψt(B)(∂tf+div⁡(fw)); here w(t,ψ(t,X))=∂tψ(t,X)=v, so w≡v (The transport theorem for a C2 family of diffeomorphisms: ddt∫ψt(B)f=∫ψt(B)(∂tf+div⁡(fw)), Continuum motions, spatial velocity, material acceleration and swept regions).

[L2]

Change of variables: for the translation ψt (a C1 diffeomorphism with ∣det⁡Dψt∣=1), ∫ψt(B)f(t,x)dx=∫Bf(t,X+tv)dX=∫Bh(X)dX (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions, The Axiom of Countable Choice (ACω), Integral over a measurable subset).

[L3]

The affine map A(t,x):=x−tv has total derivative DA(t,x)(τ,z)=z−τv, directly from its exact increment. Since h is C1, The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a) gives f=h∘A of class C1 and Df(t,x)(τ,z)=Dh(x−tv)(z−τv). Derivatives: ∂tf(t,x)=−Dh(x−tv)v and ∇f(t,x)=(∇h)(x−tv), hence ∂tf+⟨∇f,v⟩2=0; moreover div⁡(fv)=⟨∇f,v⟩2+fdiv⁡v=⟨∇f,v⟩2 because v is constant (Euclidean spatial calculus: gradient, divergence and material derivative, Sums, scalar multiples, products and quotients: (f+g)′(c)=f′(c)+g′(c), (αf)′(c)=αf′(c), (fg)′(c)=f′(c)g(c)+f(c)g′(c), and (f/g)′(c)=(f′(c)g(c)−f(c)g′(c))/g(c)2 when g(c)≠0, A total derivative computes every directional derivative, and its matrix is the Jacobian, The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, The Lebesgue integral is linear on L1(μ)).

Verification

technique · direct; evaluate both sides of the transport identity for the explicit translating flow
1.1givenL2algebra

By [L2], ∫ψt(B)f(t,x)dx=∫Bh(X)dX for every t; the right-hand side does not depend on t, so the derivative is 0.

1.2givenL3algebra

By [L3], ∂tf+div⁡(fv)=(∂tf+⟨∇f,v⟩2)+fdiv⁡v=0, and the same conclusion holds for every point of the swept region.

2.1step 1.1step 1.2L1algebra∎

The transport theorem [L1] gives ddt∫ψt(B)f=∫ψt(B)(∂tf+div⁡(fv))=0, in agreement with step 1.1: both computations produce the constant rate.

Remarks

  • This example verifies the transport identity on a flow with constant velocity and non-constant density-like profile; it exercises the change-of-variables and material-derivative pieces without any physical claim.

Depends on

Dependency tree · 2 levels

Sources