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Exact Reynolds and Mach nondimensionalization

Statement

Let I⊆R and Ω⊆R3 be open, let ρ0,μ,U,L,c>0, and set ν=μ/ρ0, Re⁡=UL/ν, and Ma⁡=U/c. Define T=Ut/L, X=x/L, I∗={Ut/L:t∈I}, and Ω∗={x/L:x∈Ω}. For the incompressible branch, let u:I×Ω→R3 have continuous first time and first and second spatial partial derivatives, and let p:I×Ω→R have continuous first spatial partial derivatives. Write u∗:I∗×Ω∗→R3 and p∗:I∗×Ω∗→R for the corresponding fields with the same stated regularity. Set u(t,x)=Uu∗(T,X) and p(t,x)=pref+ρ0U2p∗(T,X), where pref is constant. Then the original system is equivalent under this invertible change of variables to

∂Tu∗+(u∗⋅∇X)u∗=−∇Xp∗+Re⁡−1ΔXu∗,div⁡Xu∗=0.

For the barotropic branch, let P:(0,∞)→R have two continuous derivatives with P′(ρ0)=c2>0. Let u:I×Ω→R3 have continuous first time and first and second spatial partial derivatives, let ρ:I×Ω→(0,∞) have continuous first time and spatial partial derivatives, and let any potential V have continuous spatial partial derivatives through order two. Write u∗:I∗×Ω∗→R3, ρ∗:I∗×Ω∗→(0,∞), and p∗:I∗×Ω∗→R for the rescaled fields with the corresponding stated regularity. Set ρ=ρ0ρ∗ and P(ρ)=P(ρ0)+ρ0c2p∗; if a potential is present, set V=U2V∗ with V∗:Ω∗→R. Then the exact transformed equations are

∂Tρ∗+div⁡X(ρ∗u∗)=0,∂Tu∗+(u∗⋅∇X)u∗=−Ma⁡−2(ρ∗)−1∇Xp∗−∇XV∗.

These are coefficient identities under positive changes of scale. They do not assert a limit as either dimensionless ratio tends to zero or infinity, an error estimate, uniqueness, or identical solutions when geometry or rescaled data differ.

In the optional SI interpretation, [ρ0]=kg m−3, [μ]=Pa s, [U]=[c]=m s−1, [L]=m, and [ν]=m2 s−1; Re⁡ and Ma⁡ are dimensionless.

Statement references

The characteristic Reynolds ratio and similarity discussion: Stewart, MIT 8.09, Chapter 6, §6.7, printed pp. 133–134.

Independent Reynolds and Mach parameter treatment: MIT 16.01 Unified Engineering, F5 lecture notes, pp. 1–4.

The formal systems and derivative conventions: Formal parameterized fluid systems and observables.

Facts & Assumptions

Given: Positive constants ρ0,μ,U,L,c, open domains I,Ω, and regular fields u,p,ρ as in the statement. Use Cartesian derivatives and the SI dimensional interpretation only to identify the ratios as dimensionless.

[A1]

On the arbitrary open sets I,Ω allowed here, define the original constant-density incompressible system by the pointwise equations ∂tu+(u⋅∇)u+ρ0−1∇p−νΔu=0 and div⁡u=0. These are exactly the equations of Formal parameterized fluid systems and observables whenever I is a nonempty open interval and Ω is nonempty; the present definition also permits disconnected or empty open sets, with equations on an empty product holding vacuously.

[A2]

On those same open sets, define the original barotropic system by the pointwise equations ∂tρ+div⁡(ρu)=0 and ∂tu+(u⋅∇)u=−ρ−1∇(P(ρ))−∇V. This likewise agrees with Formal parameterized fluid systems and observables on its nonempty interval scope and extends the same formulas to disconnected or empty open sets. If no potential is present, use V=0 and V∗=0.

[F1]

The gradient, divergence, Laplacian, and componentwise convective derivative in these systems use the Cartesian coordinate formulas of Formal parameterized fluid systems and observables, evaluated locally on the open sets here.

[F2]

Let g:A→R and f:B→R with g[A]⊆B. If x∈A and g(x)∈B are limit points of their respective domains and g,f are differentiable there, then f∘g is differentiable at x and (f∘g)′(x)=f′(g(x))g′(x) (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f∘g is differentiable at c with (f∘g)′(c)=f′(g(c)) g′(c)).

Proof

1.1givenA1A2F1F2

Every point of I×Ω, when this product is nonempty, has a neighborhood J×B⊆I×Ω with J a nonempty open interval and B an open ball. On this neighborhood [A1] and [A2] agree with the supplier systems, and each coordinate slice and its positive rescaling are open intervals, so both limit-point hypotheses in [F2] hold. From t=LT/U, x=LX, and u=Uu∗(T,X), applying that one-variable chain rule componentwise gives ∂tu=(U2/L)∂Tu∗, ∇xu=(U/L)∇Xu∗, Δxu=(U/L2)ΔXu∗, (u⋅∇x)u=(U2/L)(u∗⋅∇X)u∗, and div⁡xu=(U/L)div⁡Xu∗. The pressure substitution gives ρ0−1∇xp=(U2/L)∇Xp∗ because the gradient of pref is zero.

1.2givenA2F1F2

With ρ=ρ0ρ∗ and u=Uu∗, the time derivative and divergence terms in the mass equation both acquire the common factor ρ0U/L; dividing by this positive factor gives ∂Tρ∗+div⁡X(ρ∗u∗)=0.

1.3givenA2F1F2algebra

The left side of the barotropic momentum equation acquires the factor U2/L. The pressure gradient satisfies ρ−1∇xP(ρ)=(c2/L)(ρ∗)−1∇Xp∗ by the stated pressure scaling, while ∇xV=(U2/L)∇XV∗. Dividing by U2/L gives the displayed momentum equation and the coefficient c2/U2=Ma⁡−2.

2.1step 1.1A1algebra

Substituting the identities of step 1.1 into [A1] and dividing by U2/L>0 yields the displayed incompressible equations, since ν/(UL)=Re⁡−1. Reversing the same invertible substitutions proves the converse.

3.1givenA1A2step 1.1step 2.1step 1.2step 1.3∎

Since U,L,ρ0,μ,c are positive, every scaling map and every division above is invertible. The local identities hold at every point of the open product, regardless of its connected components. If the product is empty, its rescaled product is also empty, all the pointwise equations hold vacuously, and the unique empty fields correspond under the same substitutions. Thus the equations transform exactly in both directions on every allowed domain; this algebra alone supplies no uniqueness, limiting, or error estimate.

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