Transport in Rotating Fluids

Peter Constantin

Department of Mathematics

The University of Chicago

11/14/02

Dedicated to Professor M. I. Vishik with deep admiration.

Abstract. We consider uniformly rotating incompressible Euler and Navier-Stokes equations. We study the suppression of vertical gradients of Lagrangian displacement (”vertical” refers to the direction of the rotation axis). We employ a formalism that relates the total vorticity to the gradient of the back-to-labels map (the inverse Lagrangian map, for inviscid flows, a diffusive analogue for viscous flows). We obtain bounds for the vertical gradients of the Lagrangian displacement that vanish linearly with the maximal local Rossby number. Consequently, the change in vertical separation between fluid masses carried by the flow vanishes linearly with the maximal local Rossby number.

1 Introduction

Consider a container filled with water and rotated at a constant angular velocity Ω. The equations ([2]) are

∂_{t}u + u · ∇u − νΔu + ∇π + 2Ωẑ × u = 0  (1)

with

∇ · u = 0  (2)

Δ = ∇^{2} is the Laplacian. We use Cartesian coordinates, the unit vector ẑ = e_{3} is the rotation axis. The vector u is the relative velocity, a function of three space variables and time, representing the velocity of the fluid recorded by an ideal observer attached to the container and rotating thus with it, at uniform angular velocity. The function π contains the physical pressure and the centrifugal force. The Taylor-Proudman theorem ([2]) states that, if one neglects viscosity and inertia, then the only time independent solutions are two-dimensional.

Neglecting viscosity brings us to the three dimensional incompressible Euler equations

∂_{t}u + u · ∇u + ∇π + 2Ωẑ × u = 0.  (3)

Neglecting inertia brings about the linear equation

∂_{t}u + 2Ωẑ × u + ∇π = 0, ∇ · u = 0,  (4)

and time independent solutions of (4) obey ∂_{3}π = 0, u_{1} = −(2Ω)^{−1}∂_{2}π, u_{2} = (2Ω)^{−1}∂_{1}π. Incompressibility demands then ∂_{3}u_{3} = 0 and standard boundary conditions require then u_{3} = 0, so indeed, the solutions are two dimensional. G.I. Taylor showed experimentally that, when motions are slow and steady, the motion of the fluid is organized in sheets that remain parallel to the axis of rotation. Two fluid elements which are initially on a line parallel to the rotation axis, will remain so. If these elements are a certain distance apart, they will remain at the same distance apart ([2]). In Taylor’s and in more recent [1] experiments the rotation axis coincides with the direction of gravity, so we refer to the direction of the axis as “vertical”. Thus, for slow, steady inviscid motions, there is no vertical transport.

The main objective of this paper is the quantitative study of the suppression of vertical transport in the presence of inertia.

Extensive mathematical studies, ([5], [6] and more recent works), based on the averaging of the interaction of the fast waves of (4) with inertia show that two dimensional structures emerge in the limit Ω → ∞.

In the present work we employ a formalism that relates the total vorticity to the gradient of the back-to-labels map (the inverse Lagrangian map, for inviscid flows, a diffusive analogue for viscous flows ([3], [4])). In the presence of rotation the total vorticity decomposes in local vorticity (curl of relative velocity) and Ωe_{3}. The ratio between the magnitude of the relative vorticity and Ω is the local Rossby number ρ. Rotation dominated flows have small Rossby numbers. For solutions of the inviscid equations (3), the Lagrangian displacement λ(a, t) = X(a, t) − a obeys the time independent differential equation (see 89):

(∂_{a_{3}} + \frac{1}{2}ρ_{0}(a)ξ_{0}(a) · ∇_{a}) λ(a, t) =

= \frac{1}{2} (ρ_{t}(a)ξ(a, t) − ρ_{0}(a)ξ_{0}(a))

where

ρ_{0}(a) = \frac{|ω_{0}(a)|}{Ω}, ξ_{0}(a) = \frac{ω_{0}(a)}{|ω_{0}(a)|},

ρ_{t}(a) = \frac{|ω(X(a, t), t)|}{Ω}, ξ(a, t) = \frac{ω(X(a, t), t)}{|ω(X(a, t), t)|}

and ω is the relative vorticity. Note that, in the limit ρ → 0 we recover

∂_{a_{3}}λ_{3}(a, t) = 0,

that is, the Taylor-Proudman theorem. For small local Rossby numbers we obtain bounds for vertical gradients of the Lagrangian displacements that vanish linearly with ρ.

2 Maps, Velocity and Vorticity

We consider flow in an open domain R with smooth boundary ∂R and we impose natural boundary conditions: the relative velocity is tangent to the boundary: u(x, t) · n(x) = 0, for x ∈ ∂R. We denote by x = (x_{1}, x_{2}, x_{3}) or a = (a_{1}, a_{2}, a_{3}) Cartesian coordinates of independent variables. The third Cartesian direction ẑ = e_{3} is singled out. The fluid can be described employing Lagrangian or Eulerian descriptions. While doing this, we always keep the Cartesian coordinates for independent variables. The relative Lagrangian path map X(a, t), defined for a ∈ R, and associated to the relative velocity u(x, t) is the solution of the familiar ordinary differential equation

∂_{t}(X(a, t)) = u(X(a, t), t)  (5)

with initial data

X(a, 0) = a.  (6)

The rotating Euler equations (3) are thus

\frac{d}{dt}(∂_{t}X(a, t)) + (∇_{x}π)(X(a, t), t) + 2Ωẑ × u(X(a, t), t) = 0  (7)

There is no difference between \frac{d}{dt} and ∂_{t}: the labels a are held fixed. We use both notations to emphasize the operation at hand. We take one of the directions a = (a_{1}, a_{2}, a_{3}), say a_{α} and differentiate the Lagrangian map in the direction a_{α} to obtain the vector ∂_{α}(X(a, t)). We take the scalar product of the equation (7) with this vector, and using the chain rule, we get:

\frac{d}{dt} (∂_{t}X · ∂_{α}X) + ∂_{α}(π − \frac{1}{2} |∂_{t}X|^{2}) + 2Ω (ẑ; ∂_{t}X; ∂_{α}X) = 0.  (8)

where (u; v; w) denotes the determinant of the matrix whose columns are u, v and respectively, w. We compute (ẑ; ∂_{t}X; ∂_{α}X) = ∂_{t} (ẑ; X; ∂_{α}X) − (ẑ; X; ∂_{α}∂_{t}X) = ∂_{t} (ẑ; X; ∂_{α}X) − ∂_{α} (ẑ; X; ∂_{t}X) + (ẑ; ∂_{α}X; ∂_{t}X), and thus, using the antisymmetry of the determinant, we get

2 (ẑ; ∂_{t}X; ∂_{α}X) = ∂_{t} (ẑ; X; ∂_{α}X) − ∂_{α} (ẑ; X; ∂_{t}X) .  (9)

Inserting (9) in (8) we get

\frac{d}{dt} \{(∂_{t}X · ∂_{α}X) + Ω (ẑ; X; ∂_{α}X)\} + ∂_{α}ρ = 0  (10)

where

ρ = π − \frac{1}{2}|∂_{t}X|^{2} − Ω (ẑ; X; ∂_{t}X) .  (11)

Integrating in time, from time zero to time t, and remembering that time integration commutes with label derivative, we obtain

{(∂_{t}X · ∂_{α}X) + Ω (ẑ; X; ∂_{α}X)}

= {(∂_{t}X · ∂_{α}X) + Ω (ẑ; X; ∂_{α}X)}_{|t=0} − ∂_{α} ∫_{0}^{t} ρ.  (12)

The initial data area X(a, 0) = 0 and ∂_{t}X(a, 0) = u_{0}(a). The equation (12) is therefore

{∂_{t}X + Ωẑ × X} · ∂_{α}X = {u_{0}(a) + Ωẑ × a} · e_{α} − ∂_{α}q  (13)

where

q = ∫_{0}^{t} ρ  (14)

and e_{α} = (δ_{iα}) is the unit vector associated with the direction α. We consider the “back-to-labels” map, (the inverse Lagrangian map)

A(x, t) = X^{−1}(x, t)  (15)

We read (14) at a = A(x, t), we multiply it by (∂_{i}A^{α})(x, t) and sum. We use the chain rule, and obtain:

u^{i}(x, t) + Ω (ẑ; x; e_{i}) =
∂_{i}A^{α}(x, t) {u_{0}^{α}(A(x, t), t) + Ω (ẑ; A(x, t); e_{α})} − ∂_{i}r(x, t),

where

r(x, t) = q(A(x, t), t).  (16)

Re-arranging the expressions a bit, we write

u^{i}(x, t) = (∂_{i}A^{α}(x, t))u_{0}^{α}(A(x, t), t) − ∂_{i}r(x, t) +
Ω {(ẑ; A(x, t), ∂_{i}A(x, t) − (ẑ; x; e_{i})} .  (17)

This is the Weber formula for rotating Euler equations. Obviously, the “back-to-labels” map A obeys the equation

∂_{t}A(x, t) + u(x, t) · ∇A(x, t) = 0  (18)

which follows by time differentiation from the statement that A(x, t) is the label corresponding to x, a = A(X(a, t), t). One can verify, by direct calculation, that if A(x, t) solves the equation (18) with initial data

A(x, 0) = x  (19)

and with velocity computed from (17), then the velocity satisfies the rotating Euler equations. Indeed, writing Γ_{0}(u, ∇) = ∂_{t} +u·∇, and applying Γ_{0}(u, ∇) to (17) we obtain, using (18)

Γ_{0}(u, ∇)u^{i} = −Ω (ẑ; u; e_{i}) − (∂_{i}u^{j})(∂_{j}A^{α}(x, t)) {u_{0}^{α}(A) + Ω (ẑ; A(x, t); e_{α})}

−∂_{i}(Γ_{0}(u, ∇)r) + (∂_{i}u^{j})∂_{j}r.  (20)

Now insering the definition (17) in the equation above, we have

Γ_{0}(u, ∇)u^{i} = −Ω (ẑ; u; e_{i}) − (∂_{i}u^{j})u^{j} − ∂_{i}(Γ_{0}(u, ∇)r) − Ω(∂_{i}u^{j}) (ẑ; x; e_{j}) (21)

We notice that

(∂_{i}u^{j}) (ẑ; x; e_{j}) = ∂_{i} (ẑ; x; u) − (ẑ; e_{i}; u)  (22)

so that the equation (21) reads

Γ_{0}(u, ∇)u^{i} = −2Ω (ẑ; u, e_{i}) − ∂_{i}π  (23)

with

π = Γ_{0}(u, ∇)r + \frac{1}{2}|u|^{2} + Ω (ẑ; x; u)  (24)

The obtained equation (23) is just (3), so we have proved therefore:

Theorem 1. Any twice continuously differentiable solution of the active vector system (17, 18) solves the rotating Euler equations (3) with pressure π given in (24). Viceversa, if u is a twice differentiable solution of the rotating Euler equations, then it obeys (17) with A determined by (18).

The Navier-Stokes equations (1) in a rotating frame admit a somewhat similar treatment. One considers a map A(x, t) that obeys

(∂_{t} + u · ∇ − νΔ) A(x, t) = 0  (25)

together with the initial condition (19). It is convenient to denote

Γ_{ν}(u, ∇) = ∂_{t} + u · ∇ − νΔ  (26)

the operator of advection by u and diffusion with diffusivity ν. One starts with an ansatz like (17),

u^{i} = (∂_{i}A^{α})v^{α} + Ω {(ẑ; A; ∂_{i}A) − (ẑ; x, e_{i})} − ∂_{i}r  (27)

which is exactly (17) except that instaed of u_{0}(A(x, t)) we have now v(x, t). Note that, in the inviscid case, the function v_{0}(x, t) = u_{0}(A(x, t)) obeys Γ_{0}(u, ∇)v_{0} = 0. The viscosity introduces corrections to this equation. We apply Γ_{ν}(u, ∇) to (27) and compute the various pieces separately. First, we get

Γ_{ν}(u, ∇) {(∂_{i}A^{α}) v^{α}} =
(∂_{i}A^{α})Γ_{ν}(u, ∇)v^{α} − 2ν(∂_{j}∂_{i}A^{α})∂_{j}v^{α} − (∂_{i}u^{j}) (∂_{j}A^{α}) v^{α}.  (28)

Secondly, we have

Γ_{ν}(u, ∇)(∂_{i}r) = ∂_{i} (Γ_{ν}(u, ∇)r) − (∂_{i}u^{j})(∂_{j}r)  (29)

Subtracting (29) from (28) we get

Γ_{ν}(u, ∇) {(∂_{i}A^{α}) v^{α} − ∂_{i}r} = −∂_{i} (Γ_{ν}(u, ∇)r) +

(∂_{i}A^{α})Γ_{ν}(u, ∇)v^{α} − 2ν(∂_{j}∂_{i}A^{α})∂_{j}v^{α} − (∂_{i}u^{j}) {(∂_{j}A^{α}) v^{α} − ∂_{j}r} . (30)

Let us give a temporary name to the term involving Ω in (27):

U^{i} = Ω {(ẑ; A; ∂_{i}A) − (ẑ; x; e_{i})} .  (31)

Then, using (27) we write (30) as

Γ_{ν}(u, ∇)u = (∂_{i}A^{α})Γ_{ν}(u, ∇)v^{α} − 2ν(∂_{j}∂_{i}A^{α})∂_{j}v^{α} − ∂_{i} (Γ_{ν}(u, ∇)r)

−(∂_{i}u^{j})u^{j} + (∂_{i}u^{j})U^{j} + Γ_{ν}(u, ∇)U^{i}  (32)

Now we compute Γ_{ν}(u, ∇)U^{i}:

Γ_{ν}(u, ∇)U^{i} = −Ω {(∂_{i}u^{j})(ẑ; A; ∂_{j}A) + (ẑ; u; e_{i}) + 2ν(ẑ; ∂_{j}A, ∂_{j}∂_{i}A)} (33)

Using

∂_{i} (−Ω(ẑ; x; u)) = −Ω(ẑ; x; (∂_{i}u^{j})e_{j}) − Ω(ẑ; e_{i}; u)

and noticing that two terms Ω(∂_{i}u^{j})(ẑ; A; ∂_{j}A) cancel, we deduce that

(∂_{i}u^{j})U^{j} + Γ_{ν}(u, ∇)U^{i} = −Ω∂_{i}(ẑ; x; u) − 2Ω(ẑ; u; e_{i}) − 2Ων(ẑ; ∂_{j}A; ∂_{i}∂_{j}A). (34)

Therefore, we have arrived at

Γ_{ν}(u, ∇)u = −∂_{i} \{Γ_{ν}(u, ∇)r + \frac{1}{2}|u|^{2} + Ω(ẑ; x; u)\} − 2Ω(ẑ; u; e_{i})+

(∂_{i}A^{α})Γ_{ν}(u, ∇)v^{α} − 2ν(∂_{j}∂_{i}A^{α})(∂_{j}v^{α}) − 2Ων(ẑ; ∂_{j}A; ∂_{i}∂_{j}A). (35)

Now the last piece in (35) vanishes if v solves the equation

Γ_{ν}(u, ∇v) = 2νC_{j;β}^{α}v_{j}^{α} + 2Ων(ẑ; ∂_{j}A; C_{j;β}^{.})  (36)

Here

C_{j;β}^{α} = (∇A)_{kβ}^{−1}∂_{j}∂_{k}A^{α} = −Γ_{βγ}^{α}(∂_{j}A^{γ})  (37)

is related to the Riemann-Christoffel symbol Γ_{βγ}^{α}. Thus we have proved

Theorem 2. Assume that A solves (25) with u given by (27) and with v evolving according to (36). Then u solves the rotating Navier-Stokes equation (1). Vice-versa, if u solves the rotating Navier-Stokes equations and if we solve the linear equation (25) and then the linear inhomogeneous equation (36), then (27) holds.

Let us derive the analogue of (27) for the vorticity. We will take the curl of (27) and express the vorticity ω = ∇ × u in terms of the diffusive backto-labels map A associated to the relative velocity, and the virtual vorticity ζ = ∇_{A} ×v. This object is a generalization of the Cauchy invariant ζ_{0}(x, t) = ω_{0}(A(x, t)), to which it reduces if ν = 0. Thus, ζ is defined by

ζ^{α}(x, t) = \frac{1}{2}ϵ_{αβγ}(∇A)_{jβ}^{−1}(∂_{j}v^{γ}).  (38)

We compute the curl of (27); we start with the curl of (∂_{i}A^{α})v^{α}:

ϵ_{ijk}∂_{j} {(∂_{k}A^{α})v^{α}} = ϵ_{ijk}(∂_{k}A^{α})∂_{j}v^{α}.

Using the convenient notation v_{;β}^{α} = (∇A)_{jβ}^{−1}∂_{j}v^{α}, we obtain

ϵ_{ijk}∂_{j} \{(∂_{k}A^{α})v^{α}\} = ϵ_{ijk}(∂_{k}A^{α})(∂_{j}A^{β})v_{;β}^{α} = \frac{1}{2}ϵ_{ijk}(∂_{k}A^{α})(∂_{j}A^{β})(v_{;β}^{α} − v_{;α}^{β})

The last equality holds because the contribution of \frac{1}{2}(v_{;β}^{α}+v_{;α}^{β}) vanishes. Thus

ϵ_{ijk}∂_{j} \{(∂_{k}A^{α})v^{α}\} = \frac{1}{2}ϵ_{ijk}(∂_{k}A^{α})(∂_{j}A^{β})ϵ_{βαγ}ζ^{γ}

and consequently

ϵ_{ijk}∂_{j} \{(∂_{k}A^{α})v^{α}\} = \frac{1}{2}ϵ_{ijk}(∂_{j}A; ∂_{k}A; ζ)  (39)

Now we compute the curl of (31):

ϵ_{ijk}∂_{j}U^{k} = Ωϵ_{ijk} {(ẑ; ∂_{j}A; ∂_{k}A; ) + (ẑ; A; ∂_{j}∂_{k}A) − (ẑ; e_{j}; e_{k})}

which gives

ϵ_{ijk}∂_{j}U^{k} = Ωϵ_{ijk} {(ẑ; ∂_{j}A; ∂_{k}A; ) − (ẑ; e_{j}; e_{k})} .  (40)

Adding (39) to (40 we deduce

ω^{i} = \frac{1}{2}ϵ_{ijk} \{(∂_{j}A; ∂_{k}A; ζ) + 2Ω [(∂_{j}A; ∂_{k}A; ẑ) − (e_{j}; e_{k}; ẑ)]\} .  (41)

This states that the total vorticity ω +2Ωẑ obeys the Cauchy invariance ([3], [4])

ω + 2Ωẑ = C(∇A; ζ + 2Ωẑ)  (42)

with respect to the trajectories of the relative velocity. Here we used the notation C(M; v) = \frac{1}{2}ϵ_{ijk}(M_{.j}; M_{.k}; v).

3 Lagrangian Transport

We will look more closely now at the term involving Ω in the expression (41). We start by writing

ω = C(∇A; ζ) + 2ΩR(ℓ),  (43)

Here

ℓ(x, t) = A(x, t) − x  (44)

is the Eulerian-Lagrangian displacement, and

(R(ℓ))_{i} = \frac{1}{2}ϵ_{ijk} \{(ẑ; ∂_{j}ℓ; ∂_{k}ℓ) + (ẑ; e_{j}; ∂_{k}ℓ) + (ẑ; ∂_{j}ℓ; e_{k})\}  (45)

Using ẑ = e_{3}, the term depending linearly on ℓ is

\frac{1}{2}ϵ_{ijk} \{(e_{j}; ∂_{k}ℓ; e_{3}) + (∂_{j}ℓ; e_{k}; e_{3})\} = ϵ_{ijk}(e_{j}; ∂_{k}ℓ; e_{3}) =

{−δ_{k1}δ_{j2}δ_{i3}(∂_{2}ℓ; e_{1}; e_{3}) + δ_{k1}δ_{j3}δ_{i2}(∂_{3}ℓ; e_{1}; e_{3}) + δ_{k2}δ_{j1}δ_{i3}(∂_{1}ℓ; e_{2}; e_{3})

−δ_{k2}δ_{j3}δ_{i1}(∂_{3}; e_{2}; e_{3})} = {(∂_{2}ℓ_{2})e_{3} − (∂_{3}ℓ_{2})e_{2}} +

{(∂_{1}ℓ_{1})e_{3} − (∂_{3}ℓ_{1})e_{1}} = {−∂_{z}ℓ + (∇ · ℓ)e_{3}}  (46)

The quadratic term is

\frac{1}{2}ϵ_{ijk}(∂_{j}ℓ; ∂_{k}ℓ; e_{3}) = \frac{1}{2}ϵ_{ijk}ϵ_{αβγ}(∂_{j}ℓ_{α})(∂_{k}ℓ_{β})δ_{γ3} =

= ϵ_{ijk}(∂_{j}ℓ_{1})(∂_{k}ℓ_{2}) = ∇ℓ_{1} × ∇ℓ_{2}.  (47)

Putting together (46) and (47) we have thus

R(ℓ) = −∂_{z}ℓ + (∇ · ℓ)ẑ + (∇ℓ_{1} × ∇ℓ_{2})  (48)

The detailed expression is:

R(ℓ) = ( \begin{matrix}−∂_{3}ℓ_{1} + (∂_{2}ℓ_{1})(∂_{3}ℓ_{2}) − (∂_{3}ℓ_{1})(∂_{2}ℓ_{2}) \\ −∂_{3}ℓ_{2} + (∂_{3}ℓ_{1})(∂_{1}ℓ_{2}) − (∂_{1}ℓ_{1})(∂_{3}ℓ_{2}) \\ −∂_{3}ℓ_{3} + (∇ · ℓ) + (∂_{1}ℓ_{1})(∂_{2}ℓ_{2}) − (∂_{2}ℓ_{1})(∂_{1}ℓ_{2}) \end{matrix} )  (49)

Let us consider the case of the rotating three dimensional Euler equations. In this case the back-to-labels map is the inverse Lagrangian path map. The Eulerian-Lagrangian displacement ℓ(x, t) is related to the Lagrangian displacement

λ(a, t) = X(a, t) − a  (50)

by

λ(a, t) + ℓ(x, t) = 0, a = A(x, t).  (51)

As long as the evolution is smooth, the incompressibility constraint (∇·u = 0) implies

Det(∇A) = 1  (52)

In view of ∇A = 1 + (∇ℓ), one has therefore

∇ · ℓ + (∂_{2}ℓ_{2})(∂_{3}ℓ_{3}) − (∂_{3}ℓ_{2})(∂_{2}ℓ_{3}) + (∂_{1}ℓ_{1})(∂_{3}ℓ_{3}) − (∂_{3}ℓ_{1})(∂_{1}ℓ_{3}) +

(∂_{1}ℓ_{1})(∂_{2}ℓ_{2}) − (∂_{2}ℓ_{1})(∂_{1}ℓ_{2}) = 0  (53)

Now comes the main observation concerning R(ℓ): The only terms in (49) that are not explicitly multiples of some component of ∂_{z}ℓ are found in the third component; but, in view of (53), (∇ · ℓ) + (∂_{1}ℓ_{1})(∂_{2}ℓ_{2}) − (∂_{2}ℓ_{1})(∂_{1}ℓ_{2}) is a quadratic expression involving exclusively multiples of ∂_{z}ℓ. Therefore one can factor

R(ℓ) = −M(∇ℓ)∂_{z}ℓ,  (54)

where the matrix M(∇ℓ) is:

M(∇ℓ) = ( \begin{matrix}1 + ∂_{2}ℓ_{2} & −∂_{2}ℓ_{1} & 0 \\ −∂_{1}ℓ_{2} & 1 + ∂_{1}ℓ_{1} & 0 \\ −∂_{1}ℓ_{3} & −∂_{2}ℓ_{3} & 1 + ∂_{1}ℓ_{1} + ∂_{2}ℓ_{2} \end{matrix} )  (55)

Returning to (43) we obtain the relation

M(∇ℓ)∂_{z}ℓ = \frac{1}{2Ω} \{C(∇A; ζ) − ω\}  (56)

Let us consider the matrix

N (∇ℓ) = ( \begin{matrix}1 + ∂_{1}ℓ_{1} & ∂_{2}ℓ_{1} & 0 \\ ∂_{1}ℓ_{2} & 1 + ∂_{2}ℓ_{2} & 0 \\ 0 & 0 & 1 \end{matrix} )  (57)

and multiply both sides of (56) by (57) from the left. Let us denote the right hand side by the single letter s (for small).

s = \frac{1}{2Ω}N (∇ℓ) \{C(∇A; ζ) − ω\}  (58)

We obtain thus

D_{2}(∇ℓ)∂_{z}ℓ_{1} = s_{1},  (59)

D_{2}(∇ℓ)∂_{z}ℓ_{2} = s_{2},  (60)

and

(1 + t_{2}(∇ℓ))∂_{z}ℓ_{3} − (∂_{1}ℓ_{3})(∂_{3}ℓ_{1}) − (∂_{2}ℓ_{3})(∂_{3}ℓ_{2}) = s_{3}.  (61)

We use the notations

D_{2}(∇ℓ) = (1 + ∂_{1}ℓ_{1})(1 + ∂_{2}ℓ_{2}) − (∂_{1}ℓ_{2})(∂_{2}ℓ_{1}),  (62)

d_{2}(∇ℓ) = (∂_{1}ℓ_{1})(∂_{2}ℓ_{2}) − (∂_{1}ℓ_{2})(∂_{2}ℓ_{1}),  (63)

t_{2}(∇ℓ) = ∂_{1}ℓ_{1} + ∂_{2}ℓ_{2}.  (64)

From definitions obviously

D_{2}(∇ℓ) = 1 + t_{2}(∇ℓ) + d_{2}(∇ℓ).  (65)

Also, the information contained in (53) can be written as

t_{2}(∇ℓ) + d_{2}(∇ℓ) + (1 + t_{2}(∇ℓ))∂_{z}ℓ_{3} = (∂_{1}ℓ_{3})(∂_{3}ℓ_{1}) + (∂_{2}ℓ_{3})(∂_{3}ℓ_{2}) (66)

Substituting (66) in (61) we obtain

t_{2}(∇ℓ) + d_{2}(∇ℓ) = −s_{3}  (67)

Using (65) we obtain Proposition 1. Let ω be the relative vorticity ω = ∇ × u of a solution of the rotating three dimensional Euler equations (3), let A be the inverse Lagrangian map associated to the relative velocity u, let ζ be the Cauchy invariant (ζ(x, t) = ω_{0}(A(x, t)) , with ω_{0} = ω_{|t=0}), and let D_{2}(∇ℓ) be the determinant of the block (1, 2) of ∇A, i.e.

D_{2}(∇ℓ) = (∂_{1}A_{1})(∂_{2}A_{2}) − (∂_{1}A_{2})(∂_{2}A_{1}).  (68)

Then one has

D_{2}(∇ℓ) = 1 + \frac{1}{2Ω} \{ω_{3} − (∂_{1}A; ∂_{2}A; ζ)\} ,  (69)

and the equations (59, 60, 61).

Let us assume now that

sup_{t∈[0,T]} ∥ω(·, t)∥_{L^{∞}(dx)} = M < ∞.  (70)

Let us assume also that on a time interval I we have

sup_{t∈I} ∥∇ℓ(·, t)∥_{L^{∞}(dx)} ≤ g.  (71)

Clearly then

|s_{3}| ≤ (1 + 2g + 3g^{2})\frac{M}{Ω}  (72)

and

|s_{j}| ≤ (1 + 2g)(1 + 2g + 3g^{2})\frac{M}{Ω}  (73)

j = 1, 2 hold on this time interval. We deduce from (69) that

D_{2}(∇ℓ) ≥ 1 − (1 + 2g + 3g^{2})\frac{M}{Ω}  (74)

holds on the time interval I. Using (59, 60) we deduce that

|∂_{z}ℓ_{j}| ≤ (1 + 2g)(1 + 2g + 3g^{2}) (1 − (1 + 2g + 3g^{2})\frac{M}{Ω} )^{−1} \frac{M}{Ω}  (75)

holds for j = 1, 2 on the time interval I_{i}. From (65, 74) we deduce that

1 + t_{2}(∇ℓ) ≥ 1 − 2g^{2} − (1 + 2g + 3g^{2})\frac{M}{Ω}  (76)

holds on the same interval. From (61, 76) and from (75) we deduce now that

|∂_{z}ℓ_{3}| ≤ C_{g} \frac{M}{Ω}  (77)

with

C_{g} = (1 − 2g^{2} − (1 + 2g + 3g^{2})\frac{M}{Ω} )^{−1} ×

[1 + 2g(1 + 2g) (1 − (1 + 2g + 3g^{2})\frac{M}{Ω} )^{−1}] (1 + 2g + 3g^{2})\frac{M}{Ω}  (78)

In order to simplify the exposition, let us assume for example, that

g ≤ \frac{1}{4}  (79)

Let us introduce the maximal local Rossby number

ρ = \frac{M}{Ω} = \frac{sup_{0≤t≤T} ∥ω(·, t)∥_{L^{∞}(dx)}}{Ω} .  (80)

The most stringent condition of invertibility, 1 > 2g^{2} + (1 + 2g + 3g^{2})ρ is satisfied if the Rossby number is small enough. Let us assume that the Rossby number satisfies

ρ ≤ \frac{1}{4}  (81)

Then we deduce

Proposition 2. Let a solution of the rotating Euler equation (3) satisfy

sup_{t∈I} ∥∇ℓ(·, t)∥_{L^{∞}} ≤ \frac{1}{4}  (82)

on a time interval I and assume that the local Rossby number ρ (80) does not exceed 1/4 (81). Then

sup_{t∈I} ∥∂_{z}ℓ∥_{L^{∞}(dx)} ≤ 14ρ  (83)

holds on the interval I.

In order to bound the vertical derivative of the direct Lagrangian map X(a, t) we recall that

(∇_{a}X)(A(x, t), t) = (∇A(x, t))^{−1}  (84)

and in particular

\frac{∂X}{∂a_{3}} (A(x, t), t) = (∇A_{1}(x, t)) × (∇A_{2}(x, t)).  (85)

Expressing this in terms of the displacement ℓ we arrive at

\frac{∂X}{∂a_{3}} (A(x, t), t) = e_{3} + ( \begin{matrix}(∂_{2}ℓ_{1})(∂_{3}ℓ_{2}) − (1 + ∂_{2}ℓ_{2})(∂_{3}ℓ_{1}) \\ (∂_{1}ℓ_{2})(∂_{3}ℓ_{1}) − (1 + ∂_{1}ℓ_{1})(∂_{3}ℓ_{2}) \\ t_{2}(∇ℓ) + d_{2}(∇ℓ) \end{matrix} )  (86)

In view of (65, 67, 69, 72, 75) we deduce

Proposition 3. Let a solution of the rotating Euler equation (3). Consider the Lagrangian map X(a, t) associated to the relative velocity. Then

\frac{∂X_{3}}{∂a_{3}} (a, t) = 1 + \frac{1}{2Ω} (ω_{3}(X(a, t), t) − ζ(a) · \frac{∂X_{3}}{∂a} (a, t))  (87)

holds. If the displacement satisfies (82) on a time interval I and the local Rossby number ρ does not exceed 1/4 then

sup_{t∈I} ∥∂_{a_{3}} (X(a, t) − a) ∥_{L^{∞}(da)} ≤ 9ρ  (88)

holds on the interval I.

Comparing (86) to (49) employing (66), and then recalling the definition (50) of the Lagrangian displacement we have

Theorem 3. The Lagrangian displacement λ(a, t) of a solution of the rotating Euler equations (3) for local Rossby number less than two obeys the differential equation

∂_{a_{3}}λ(a, t) + \frac{1}{2}ρ_{0}(a)ξ_{0}(a) · ∇_{a}λ(a, t) =

= \frac{1}{2} (ρ_{t}(a)ξ(a, t) − ρ_{0}(a)ξ(a, 0))  (89)

ρ_{t}(a) = \frac{|ω(X(a, t), t)|}{Ω}.  (90)

and

ξ(a, t) = \frac{ω(X(a, t), t)}{|ω(X(a, t), t)|}  (91)

Let us consider the motion of a rotating incompressible ideal fluid for a duration T, and assume that in this period of time the flow is smooth, and so,

∫_{0}^{T} ∥∇u(·, t)∥_{L^{∞}} < ∞  (92)

Notice that, from the equation

\frac{∂_{t}(∇_{a}X)}{∂t}(a, t) = (∇u(X(a, t), t))(∇_{a}X)(a, t)  (93)

it follows immediately that

∥\frac{∂X}{∂a} (a, t)∥_{L^{∞}(da)} ≤ e^{∫_{t_{0}}^{t} ∥∇u(·,t)∥_{L^{∞}(dx)}dt}  (94)

holds.

Theorem 4. Consider a smooth (92) solution of the rotating Euler equations (3), defined on a time interval [0, T]. Consider, at time t_{0}, two particles P = (a, b, c) and Q = (a, b, c + d) separated by a vertical segment of length d. (Vertical refers to the direction of the rotation axis e_{3} = ẑ). Consider the Lagrangian evolution of the particles X(P, t), X(Q, t). Then, the separation between the two particles obeys

|X(P, t) − X(Q, t) + d| ≤ \frac{ρd}{2} (1 + e^{∫_{t_{0}}^{t} ∥∇u(·,t)∥_{L^{∞}(dx)}dt})  (95)

for all t_{0} ≤ t ≤ T.

Ideed, (89) can be written as

∂_{a_{3}}λ(a, t) = \frac{1}{2Ω} (ω(X(a, t), t) − ω_{0}(a) · ∇_{a}X(a, t))  (96)

and the statement follows by integrating ∫_{c}^{c+d} da_{3} both sides of (96) and holding a_{1} = a, a_{2} = b. The right hand side is the bounded using the definition of ρ and the inequality (94).

For a lower bound we need to restrict the duration of time. Let us denote

g(t) = ∥∇ℓ(·, t)∥_{L^{∞}}.  (97)

In view of the equation

(∂_{t} + u · ∇)(∇ℓ) + ∇u + (∇ℓ)∇u = 0  (98)

obeyed by the gradient of the displacement, we deduce that

g(t) ≤ (e^{∫_{t_{0}}^{t} ∥∇u(·,t)∥_{L^{∞}(dx)}dt} − 1)  (99)

holds on any interval I = [t_{0}, t_{0} + τ] where ℓ(x, t_{0}) = 0.

If wish to ensure that the Lagrangian displacement obeys (82) on the interval of time I = [t_{0}, t_{0} + τ], then, if (92) holds, it is enough to require

∫_{t_{0}}^{t_{0}+τ} ∥∇u(·, t)∥_{L^{∞}(dx)}dt ≤ log \frac{5}{4}.  (100)

Now we are in position to prove a lower bound for the vertical separation of Lagrangian trajectories in terms of the maximal local Rossby number:

We consider two fluid masses (sets) Σ_{j}(t), j = 1, 2. We define their vertical separation as

δ(t) = inf{|z_{1} − z_{2}| ; (x, y, z_{j}) ∈ Σ_{j}(t)}  (101)

and distance d(t)

d(t) = inf{|(x_{1}, y_{1}, z_{1}) − (x_{2}, y_{2}, z_{2}) ; (x_{j}, y_{j}, z_{j}) ∈ Σ_{j}(t)}  (102)

Obviously, d(t) ≤ δ(t).

Theorem 5. Let Σ_{j}(t) be two sets carried by the flow of a smooth (92) solution of the rotating Euler equations (3). Assume that at some time t_{0}, the distance between the two sets is positive, d(t_{0}) > 0. Consider a time interval I = [t_{0}, t_{0} + τ] in which (82) holds. Then

δ(t) ≥ (1 + 14ρ)^{−1}d(t_{0})  (103)

holds for t ∈ [t_{0}, t_{0} + τ]

Indeed, if two points are in the respective sets, (x, y, z_{j}) ∈ Σ_{j}(t), then, considering the back-to-labels map starting from time t_{0}, ℓ(x, t_{0}) = 0 we have by assumption A(x, y, z_{j}, t) ∈ Σ_{j}(t_{0}), and, in view of (83) we have

d(t_{0}) ≤ |A(x, y, z_{1}, t) − A(x, y, z_{2}, t)| = |z_{1} − z_{2} + ℓ(x, y, z_{1}, t) − ℓ(x, y, z_{2}, t)| ≤

|z_{1} − z_{2}|(1 + 14ρ).

Taking the infimum produces (103). The bounds above hold in great generality; they become most effective when ρ → 0. It is quite easy to exhibit at least some situations in which this limit is well defined mathematically.

Proposition 4. Consider the rotating Euler system (3) with spatially periodic boundary conditions, u(x + L_{1}, y + L_{2}, z + L_{3}, t) = u(x, y, z, t). Assume that the initial data u(x, y, z, 0) = u_{0}(x, y, z) belongs to the space H^{s} of divergence-free square integrable functions having s > \frac{5}{2} square integrable derivatives. Then, there exists a time T and a constant M_{1}, depending only on the norm of the initial data in H^{s}, such that, for any Ω > 0, the solution of (3) with initial data u_{0} exists on the time interval [0, T] and obeys

sup_{0≤t≤T} ∥∇u(·, t)∥_{L^{∞}(dx)} ≤ M_{1}.  (104)

By restricing the duration of time further (see 100),

M_{1}T ≤ log \frac{5}{4}  (105)

we guarantee (82) on the time interval [0, T]. Fixing thus u_{0} and T, we may let Ω → ∞ and consequently ρ ≤ \frac{M_{1}}{Ω} → 0. Assuming that

Ω ≥ 4M_{1}  (106)

we have the conditions for the bounds (83, 88) to hold.

Theorem 6. Let u_{0} ∈ H^{s}, s > \frac{5}{2} and T > 0 be fixed as above satisfying (104, 105). Then, for each Ω satisfying (106), consider the inverse and direct Lagrangian displacements

ℓ(x, y, z, t) = A(x, y, z, t) − (x, y, z)

and

λ(a_{1}, a_{2}, a_{3}, t) = X(a_{1}, a_{2}, a_{3}, t) − (a_{1}, a_{2}, a_{3}).

They obey

∥∂_{z}ℓ(·, t)∥_{L^{∞}(dx)} ≤ 14ρ  (107)

and

∥∂_{a_{3}}λ(·, t)∥_{L^{∞}(da)} ≤ 9ρ  (108)

with ρ = Ω^{−1} sup_{0≤t≤T} ∥ω(·, t)∥_{L^{∞}(dx)} ≤ Ω^{−1}M_{1}.

Let Ω_{j} → ∞ be an arbitrary sequence and let X^{j}(a_{1}, a_{2}, a_{3}, t) denote the Lagrangian paths associated to Ω_{j}. Then, there exists a subsequence (denoted for convenience by the same letter j) an invertible map X(a_{1}, a_{2}, a_{3}, t), and a periodic function of two variables λ(a_{1}, a_{2}, t) such that

lim_{j→∞} X^{j}(a_{1}, a_{2}, a_{3}, t) = X(a_{1}, a_{2}, a_{3}, t)

holds uniformly in a, t and

X(a_{1}, a_{2}, a_{3}) = (a_{1}, a_{2}, a_{3}) + λ(a_{1}, a_{2}, t)

The proof of the second part follows easily from the Arzeli-Ascola theorem and inequalities of the form

|X^{(j)}(a_{1}, a_{2}, b_{3}, t) − X^{(j)}(a_{1}, a_{2}, c_{3}, t) + c_{3} − b_{3}| ≤ 9ρ_{j}|b_{3} − c_{3}|

Acknowledgment Research supported in part by NSF DMS-0202531 and by the ASCI Flash Center at the University of Chicago under DOE contract B341495.

References

[1] Ch. N. Baroud, B.B. Plapp, Z-S She, H. L. Swinney, Phys. Rev. Lett. 88, 114501-1 (18 March 2002).

[2] S. Chandrasekhar, Hydrodynamic and Hydromagnetic stability, Oxford University Press, Oxford, (1961).

[3] P. Constantin, An Eulerian-Lagrangian approach for incompressible fluids: local theory, Journal of the Americal Mathematical Society, 14 (2001), 263-278

[4] P. Constantin, An Eulerian-Lagrangian approach to the Navier-Stokes equations, Commun. Math. Phys 216 (2001), 663-686.

[5] P. Embid, A. Majda, Averaging over fast gravity waves for geophysical flows with arbitrary potential vorticity. Comm. Partial Differential Equations 21 (1996), 619-658.

[6] A. Babin, A. Mahalov, B. Nicolaenko, Global splitting, integrability and regularity of 3D Euler and Navier-Stokes equations for uniformly rotating fluids. European J. Mech. B Fluids 15 (1996), 291–300.