NAVIER-STOKES EQUATIONS IN THIN SPHERICAL DOMAINS

To appear in Contemporary Mathematics, AMS, 1997.

NAVIER-STOKES EQUATIONS IN THIN SPHERICAL DOMAINS

R. Temam 1,2 and M. Ziane 1

November 19, 2003

Abstract. Our aim in this article is to give a mathematical justification for the primi-tive equations of the atmosphere and the ocean which are known to be the fundamental

equations of meteorology and oceanography [LTW4], [P]. These equations are based on anapproximation of the Navier-Stokes equations which we justify here by considering the 3D

incompressible Navier-Stokes equations (NSE) in thin spherical shells: we show how these

equations approximate the primitive equations when the thickness converges to zero.We also show that for initial data belonging to large sets that contain the physically

relevant cases, the 3D Navier-Stokes equations have global smooth solutions. Then, we prove

that for volume forces belonging to large sets, all the Leray-Hopf weak solutions are eventuallysmooth.

0. Introduction

The primary purpose of this paper is to give a mathematical justification for some of theapproximations used to derive the primitive equations of the atmosphere and the ocean[LTW1,2,3,4]. Allthough, the atmosphere is a compressible fluid we will consider hereas a model the incompressible Navier-Stokes equations. This model contains the maindifficulties and the case of the compressible Navier-Stokes equations can be treated withsimilar techniques and will appear elswhere.

In this work, we study a geophysical model: the atmosphere is an incompressible fluidoccupying a thin layer around the earth. The equations discribing its motion are the 3Dincompressible Navier-Stokes equations (NSE) in thin spherical shells. Our purpose is toprove the validity of the 2D approximation of flows in thin spherical layers, which justifiesthe study of geophysical flows on the sphere or more generally on 2D manifolds; see, forinstance, Avez and Bamberger [AB], and Ebin and Marsden [EM]. We also derive the

1991 Mathematics Subject Classification. 34C35, 35Q30, 76D05.Key words and phrases. Navier Stokes Equations, Hydrostatic equation, Global existence and regu-

larity, Thin domains, Geophysics.1 Laboratoire d’Analyse Numerique. Universite de Paris-Sud Orsay, France,2 Department of Mathematics & The Institute for Scientific Computing and Applied Mathematics,

Indiana University, Bloomington, Indiana 47405, USA

E-mail: [email protected]

Typeset by AMS-TEX

2 R. TEMAM AND M. ZIANE

hydrostatic equation used in the primitive equations (see [LTW1]). Our approach is basedpartially on some techniques that we developed in [TZ] in studying the Navier-Stokesequations in thin rectangular domains.

We consider the Navier-Stokes equations (NSE) in a thin spherical shell Ωε = x ∈R

3, a < |x| < a+εa, where 0 < ǫ < 1 is a nondimensional small parameter, and establisha number of different results valid for ε sufficiently small. First, we show that, for initialdata belonging to large sets that contain the physically relevant cases, the Navier-Stokesequations have global, in time, smooth solutions. Then, we prove that, for volume forcesbelonging to large sets, all the Leray-Hopf weak solutions are eventually smooth, i.e., aftera period of time. Finally, we show that the averages in the radial direction of the strongsolutions of the NSE on the thin spherical shells converge as the thickness ε → 0 to thesolution of the NSE on the sphere. Furthermore the dominant part of the conservationof momentum equation in the radial part direction is precisely the hydrostatic equationappearing in the primitive equation.

With different or related motivations, the theory of partial differential equations in thindomains has been the object of much studies in a recent past; e.g. Babin and Vishik [BV],Ciarlet [Ci], Ghidaglia and Temam [GT], Hale and Raugel [HR1], [HR2], Mardsen, Ratiuand Raugel [MRR], Raugel and Sell [RS1]-[RS3], Temam and Ziane [TZ] and the referencestherein.

The Mathematical Setting of the Problem

We consider the Navier-Stokes equations of viscous incompressible fluids, namely

∂u

∂t− ν∆u + (u · ∇)u + ∇p = f in Ωǫ × (0,∞),(0.1)

div u = 0 in Ωǫ × (0,∞),(0.2)

u(·, 0) = u0(·) in Ωǫ.(0.3)

Here u = (u1, u2, u3) is the velocity vector at point x = (x1, x2, x3) and time t, andp = p(x, t) is the pressure; ν > 0 is the kinematic viscosity and f = f(x, t) representsvolume forces.

Equations (0.1)-(0.3) are supplemented with the free boundary conditions:

(0.4) u · ~n = 0 and curl u × ~n = 0 on ∂Ωε.

The first boundary condition in (0.4) is the non penetration condition, while the secondone means that the tangential component of the stress tensor applied to the normal to∂Ωε vanishes on ∂Ωε, i.e.,

(σ · ~n)τ = 0 on ∂Ωε.

Here σ = σ(u) is the stress tensor with components

σij(u) = 2νǫij(u) − νpδij , ǫij(u) =1

2

(

∂ui

∂xj+

∂uj

∂xi

)

, ∀ i, j,

NAVIER-STOKES EQUATIONS 3

where δij is the Kronecker symbol.

We denote by Hs(Ωǫ), s ∈ R, the Sobolev space constructed on L2(Ωǫ) and L2(Ωǫ) =

(L2(Ωǫ))3, H

s(Ωǫ) = (Hs(Ωǫ))3. We also denote by Hs

0(Ωǫ) the closure in the space Hs(Ωǫ)of C∞

0 (Ωǫ), the space of infinitely differentiable functions with compact support in Ωǫ. Wewill eventually introduce Sobolev spaces on the sphere Sa of radius a (see Section 5), butso far all function spaces are usual Sobolev spaces defined on a domain of R

3.

For the mathematical setting of the Navier-Stokes equations, we consider as usual theHilbert space Hǫ, which is a closed subspace of L

2(Ωǫ) (see e.g. [T1]).

Hε =

u ∈ L2(Ωǫ); div u = 0; u · ~n = 0 on ∂Ωǫ

.

Another useful space is Vǫ, a closed subspace of H1(Ωǫ), which is defined as follows:

Vε = u ∈ H1(Ωǫ) ∩ Hε; u · ~n = 0 on ∂Ωǫ.

The scalar product on Hǫ or L2(Ωε) is denoted by (·, ·)ǫ, that on Vǫ is denoted by ((·, ·))ǫ,

and the associated norms are denoted by | · |ǫ and || · ||ǫ respectively. We denote by Aǫ theStokes operator defined as an isomorphism from Vǫ onto the dual V ′

ǫ of Vǫ, by

(0.7) ∀v ∈ Vǫ, < Aǫu, v >V ′

ǫ ,Vǫ= ((u, v))ǫ.

In (0.7) we have set

((u, v))ε =

3∑

i,j=1

∫

Ωε

∂ui

∂xj

∂vi

∂xjdx;

note that ||u||ε = 0 implies that u is a constant vector and if u ∈ Vε, we have u · ~n = 0,i.e., u is tangent to ∂Ωε, and must be 0. Hence || · ||ε is a norm on Vε equivalent to theusual norm on H

1(Ωε) and we have a Poincare inequality |u|ε ≤ c(ε)||u||ε, ∀u ∈ Vε. Thedependence of c(ε) on ε will be made more explicit in Lemma 2.1.

The operator Aǫ is extended to Hǫ as a linear unbounded operator. The domain ofAǫ in Hǫ is denoted by D(Aǫ). The space D(Aǫ) can be fully characterized using theregularity theory for the Stokes operator; see e.g. [Ca], [Gh], and [So]. Here we give thecharacterization of the domain of the Stokes operator:

D(Aε) =

u ∈ H2(Ωǫ) ∩ Vε; curl u × ~n = 0 and u · ~n = 0 on ∂Ωε

.

We also recall the Leray’s projector Pǫ, which is the orthogonal projector of L2(Ωǫ) onto

Hǫ. Using the Leray projector, the Stokes operator can be defined as follows:

(0.8) Aǫu = Pǫ(−∆u), for u ∈ D(Aǫ).


Let bǫ be the continuous trilinear form on Vǫ defined by:

(0.9) bǫ(u, v, w) =

3∑

i,j=1

∫

Ωǫ

ui∂vj

∂xiwj dx, u, v, w ∈ H

1(Ωǫ).

We denote by Bǫ the bilinear mapping from Vǫ into V ′ǫ defined for (u, v) ∈ Vǫ × Vǫ by

< Bǫ(u, v), w >V ′

ǫ ,Vǫ= bǫ(u, v, w), ∀w ∈ Vǫ,

and we setBǫ(u) = Bǫ(u, u).

We assume in this work that the data ν, u0 and f satisfy

(0.10) ν > 0, u0 ∈ Hǫ (or Vǫ), f ∈ L∞(0, +∞; Hǫ).

The system of equations (0.1)–(0.4) can be written as a differential equation in V ′ǫ :

(0.11)

u′ + νAǫu + Bǫ(u) = f,

u(0) = u0,

where u′ denotes the derivative (in the distribution sense) of the function u with respectto time. We recall now the classical result of existence of solutions to problem (0.11); see[CF], [FGT], [KL], [La], [Li], [Le], [T1,2], etc....

Theorem 0.1. For u0 ∈ Hǫ, there exists a solution (not necessarily unique) u = uǫ toproblem (0.11) such that:

(0.12) uε ∈ L2(0, T ; Vǫ) ∩ L∞(0, T ; Hǫ), ∀T > 0.

Moreover, if u0 ∈ Vǫ, then there exists Tǫ = Tǫ(Ωǫ, ν, u0, f) > 0, and a unique solution uε

to problem (0.11) on [0, Tǫ], such that:

(0.13) uε ∈ L2(0, Tǫ; D(Aǫ)) ∩ L∞(0, Tǫ; Vǫ).

The solution uε which satisfies (0.13) is called the strong solution of (0.11). We recallthat the global existence (in time) of the strong solutions and the uniqueness of solutionsto problem (0.11) are still open problems.

The Main results:We are given a function R0 from (0, 1] into R such that

(0.14) limǫ→0

ǫqR0(ǫ) = 0,

for some q < 12 . We prove that if ǫ is sufficiently small and f, u0 satisfy

(0.15) |A1/2ǫ u0|2ǫ + |f |2ǫ ≤ R2

0(ǫ)

then the maximal time of existence of the strong solution to the 3D-Navier-Stokes equationsin the spherical domain Ωε is infinite. More precisely, we prove


Theorem A. Assume that the function R0 is given satisfy (0.14) for some 0 < q < 12 .

Then there exists ǫ1 = ǫ1(ν, a, q, R0) such that if uǫ0 ∈ Vǫ and f ǫ ∈ H are given satisfying

(0.15), then there exists a strong solution uε of (0.1)-(0.4) defined for all times; i.e. Tε =+∞ in Theorem 0.1 and

uε ∈ C0([0,∞); Vε) ∩ L2(0, T ; D(Aε)), ∀ T > 0.

We also prove that the averages Mǫuǫ in the radial direction1 of the solutions of the 3D-

Navier-Stokes equations on Ωε converge to the solution of the 2D-Navier-Stokes equationson the sphere Sa. More precisely, we prove

Theorem B. Let (uε0)ε>0 ∈ Vε and (fε)ε>0 ∈ Hε(resp. L∞(0,∞; Hε)), satisfying (0.14),

(0.15), and assume that there exists g ∈ H0 and v0 ∈ V0, such that,

limε→0

Mεfε = g in H0-weak.

limε→0

Mεuε0 = v0 in H0-weak, |A1/2

ǫ Muǫ0|ǫ is bounded.

Then, for all T > 0, there exists ε1 = ε1(g, v0, ν, T, R0) such that

limε→0

Mεuε = v in C([0, T ]; H0) ∩ L2(0, T ; V0).

where v is the unique solution of the Navier-Stokes equations on the sphere Sa.

This article is layed out as follows: in Section 1 we introduce the average operatorsMz = Mǫ and study some of its properties which are related to the Stokes operator.Section 2 is devoted to the determination of the dependence on ε of the constants of someSobolev type inequalities, while Section 3 deals with some estimates of the nonlinear term.In Section 4 we derive the a priori estimates on the solutions of the NSE, and in Section 5,we prove Theorem A and an eventual regularity result. Finally, in Section 6, we prove the2D-limit stated in Theorem B. The article ends with an Appendix recalling some classicalformulas and expressions from vector analysis needed in this work.

1. The average operator

We will use the spherical coordinate system (r, θ, ϕ), where θ ∈ (0, π) is the colatitudeand ϕ ∈ (0, 2π) is the longitude; ~er, ~eθ, ~eϕ is the local orthonormal basis associated withthese coordinates. For a real valued function Ψ(r, θ, ϕ) defined on Ωε = x ∈ R

3, a <|x| < a + εa, we consider the average operator

(1.1) (MΨ)(θ, ϕ) =1

εa

∫ a+εa

a

rΨ(r, θ, ϕ)dr.

1See the precise definition of Mǫ in Section 1.


For a vector field u = (ur, uθ, uϕ), we write

(1.2)Mu = (0, Muθ, Muϕ), Nu = u − Mu

r,

Nuθ = uθ −Muθ

r, Nuϕ = uϕ − Muϕ

r;

it is clear from the definition of the average operator that

(1.3)

∫ a+εa

a

r Nuθ dr = 0, and

∫ a+εa

a

r Nuϕ dr = 0.

We chose in (1.1) to average Ψ with respect to the measure r dr instead of dr for thefollowing reason: thanks to (1.1) and (1.2), if u ∈ Hε, i.e. ur = 0 on ∂Ωε and div u = 0 in

Ωε, thenMu

r· ~n = 0 and Nu · ~n = 0 and also

(1.4) divMu

r= 0 in Ωε and divNu = 0 on ∂Ωε.

Hence

(1.5) if u ∈ Hε thenMu

r∈ Hε and Nu ∈ Hε.

Now we give a lemma that will provide us useful test functions in Hε. These testfunctions which will be used for deriving the a priori estimates on Muε and Nuε are notAεMuε, AεNuε but some related functions Aε(Muε/r), AεNuε. We now define Aε andgive some properties of Aε and Aε

Lemma 1.1. If u ∈ D(Aε), then

(1.6) Aεu = curl curl u ∈ Hε and Aεu = curl (r2curl u) ∈ Hε,

(1.7)

Mu

r∈ D(Aε) and Nu ∈ D(Aε), and thus

AεMu

rand AεNu ∈ Hε.

Proof. By (0.8), (A.5) and since div u = 0,

Aεu = Pεcurl curl u.

Hence Aεu = curl curl u follows if we show that curl curl u ∈ Hε; since this vector isdivergence free, it suffices to show that ~er · curl curl u = 0 on ∂Ωε. But, by (A.6),

(1.8) ~er · curl curl u = curl u · curl~er − div (~er × curl u).


We notice that curl~er = 0 (see (A.10)) and, by (A.11) and (A.16),

(1.9) div (~er × curl u) =a

rdiv 2 (~er × curl u).

For u ∈ D(Aε), ~er × curl u vanishes on ∂Ωε and therefore by (1.8) and (1.9) ~er · curl curl uvanishes on ∂Ωε too.

Then, having defined Aεu as curl (r2curl u), we want, for (1.6), to show that Aεu ∈ Hε.Since this vector is divergence free, we need to show that ~er · curl r2curl u = 0 on ∂Ωε, andthis is done as for curl curlu :

~er · curl (r2curl u) = r2curl u · curl~er − div (r2~er × curl u)

= −div (r2~er × curl u)

= −2ardiv 2 (~er × curl u)

= 0 on ∂Ωε.

For (1.7) we only need to show e.g. thatMu

r∈ D(Aε). Then, because of (1.4) we only

need to show that ~er × curlMu

rvanishes on ∂Ωε. We see with (1.2), (A.10), (A.14) that

(1.10) curlMu

r=

a

r2curl 2Mu~er

and therefore ~er × curlMu

rvanishes everywhere. The lemma is proved.

Remark 1.1. For u ∈ D(Aε), we have

AεMu

r= r2Aε

Mu

rand AεNu = r2AεNu + ∇r2 × curl Nu.

Moreover, for u, v ∈ D(Aε),

< Aεu, v > =

∫

Ωε

curl r2curl u · v dx

= (by (A.7))

=

∫

Ωε

r2curl u · curl v dx.

The last quantity defines a weighted H1−product on Vε:

(1.11) ((u, v))ε,r =

∫

Ωε

r2curl u · curl v dx;


the corresponding norm || · ||ε,r is equivalent to || · ||ε, uniformly with respect to ε for, say,0 < ε ≤ 1 :

(1.12) a2||u||2ε ≤ ||u||2ε,r ≤ 4a2||u||2ε.

We also define a weighted L2−product on Hε as

(1.13) (u, v)ε,r =

∫

Ωε

r2u · v dx, u, v ∈ Hε.

We have similarly

(1.14) a2|u|2ε ≤ |u|2ε,r ≤ 4a2|u|2ε.

Now we prove a weighted H1 orthogonality property for the scalar product ((·, ·))ε,r

defined in (1.11).

Lemma 1.2. For u and v ∈ D(Aε), we have

(1.15) ((Mu

r, Nv))ε,r = (

Mu

r, AεNv)ε = (Aε

Mu

r, Nv)ε = 0.

Moreover,

(1.16) (u, AεNv)ε = ((Nu, Nv))ε,r, and

(

u, AεMv

r

)

ε

=

((

Mu

r,Mv

r

))

ε,r

Proof. It is clear that (1.16) follows from (1.2) and (1.15). The three quantities in (1.15)

are equal because of (A.6), the integrals on ∂Ωε vanishing sinceMu

rand Nv are in D(Aε)

(see Remark 1.1). Hence, it suffices to show that one of these quantities vanish. For this

we write with (1.10), (A.10) and since

∫ a+εa

a

r Nvϕ dr = 0 and

∫ a+εa

a

r Nvθdr = 0 :

(1.17)

∫

Ωε

r curlMu

r· r curlNv dx =

=

∫

Sa

∫ a+εa

a

ar curl 2 Mu

(

∂

∂θ(sin θ Nvϕ) − ∂Nvθ

∂ϕ

)

dr dθ dϕ = 0,

and (1.15) follows.


2. Sobolev inequalities in thin spherical shells and applications.

One of the basic tools in our study of nonlinear partial differential equations in thindomains is the knowledge of the exact dependence on the thickness of the domain of theconstants appearing in the Sobolev and related inequalities; we now derive estimates onsuch constants for Ωε. From now on we assume that ε ≤ 1/8.

2.1. Poincare’s inequality in thin spherical shells.

Lemma 2.1. We have (for 0 < ε ≤ 1/2),

(2.1) |Nu|2ε ≤ 4ε2a2|curl Nu|2ε, ∀u ∈ Vε,

which implies

(2.2) |Nu|2ε ≤ 4ε2||Nu||2ε,r, ∀u ∈ Vε,

Proof. By density, it is enough to prove (2.1) for smooth functions. Let Ψ ∈ C(Ωε) be areal continuous function. We write for any ξ, η in [a, a + εa]:

(2.2) ξ2Ψ2(x′, ξ) + η2Ψ2(x′, η) = 2ξηΨ(x′, ξ)Ψ(x′, η) +(

ξΨ(x′, ξ)− ηΨ(x′, η))2

= 2ξηΨ(x′, ξ)Ψ(x′, η) +

(∫ ξ

η

∂(rΨ)

∂r(x′, r)dr

)2

,

with x′ = x/|x|. We fix ξ and integrate with respect to η to obtain

(2.3) aεξ2Ψ2(x′, ξ) +

∫ a+εa

a

η2Ψ2(x′, η)dη = 2ξΨ(x′, ξ)

∫ a+εa

a

ηΨ(x′, η)dη

+

∫ a+εa

a

(∫ ξ

η

∂(rΨ)

∂r(x′, r)dr

)2

dη.

We apply (2.3) with Ψ = ur and ξ = a, observing that Ψ(x′, a) = 0. Then we apply (2.3)with Ψ = Nuθ and Ψ = Nuϕ; in each of these cases we obtain

(2.4)

∫ a+εa

a

η2Ψ2(x′, η)dη ≤∫ a+εa

a

(∫ ξ

η

∂(rΨ)

∂r(x′, r)dr

)2

dη.

Using the Cauchy-Schwarz inequality, we find∫ a+εa

a

η2Ψ2(x′, η)dη ≤∫ a+εa

a

|ξ − η| dη

∫ a+εa

a

∣

∣

∂

∂r(rΨ)

∣

∣

2dr ≤ ε2a2

∫ a+εa

a

∣

∣

∂

∂r(rΨ)

∣

∣

2dr

≤ 2ε2a2

∫ a+εa

a

∣

∣

∂

∂rΨ

∣

∣

2r2dr + 2ε2a2

∫ a+εa

a

∣

∣Ψ∣

∣

2dr(2.5)

≤ 2ε2a2

∫ a+εa

a

r2∣

∣

∂

∂rΨ

∣

∣

2dr + 2ε2

∫ a+εa

a

∣

∣rΨ∣

∣

2dr,


and since ε ≤ 1/2, we have

(2.6)

∫ a+εa

a

r2Ψ2(θ, ϕ, r)dr ≤ 4ε2a2

∫ a+εa

a

∣

∣

∂

∂rΨ

∣

∣

2r2dr.

We then integrate with respect to θ and ϕ, and add the relations (2.6) for Ψ = ur, Nuθ,and Nuϕ; using finally

(2.7)

∫

Ωε

∣

∣

∂

∂rNu

∣

∣

2dx ≤ ||Nu||2ε = |curl Nu|2ε, ∀u ∈ Vε,

we conclude the lemma.

Using the integration by parts formula (A.7), we have

||Nu||2ε =

∫

Ωε

|curl Nu|2 dx = (AεNu, Nu)ε ≤ |AεNu|ε|Nu|ε

≤ 2εa|AεNu|ε||Nu||ε.

Furthermore, using Remark 1.1 and the integration by parts formula (A.7), we can write

(2.8)

∫

Ωε

r2∣

∣curlNu∣

∣

2dx = (AεNu, Nu)ε = (rAεNu, rNu)ε + (∇r2 × curl Nu, Nu)ε

≤ 2a|AεNu|ε,r|Nu|ε + 2||Nu||ε,r|Nu|ε≤ 4εa

∣

∣AεNu∣

∣

ε,r||Nu||ε,r + 4ε||Nu||2ε,r,

and for ε ≤ 1/8, we have

||Nu||ε,r ≤ 8εa|AεNu|ε,r.

Hence,

Corollary 2.1. For every u ∈ D(Aε), and for ε ≤ 1/8:

|curlNu|2ε ≤ 4ε2a2|AεNu|2ε,

(2.9)||Nu||2ε,r ≤ 4ε2|AεNu|2ε,r,

|Nu|2ε ≤ 256ε4a2|AεNu|2ε,r.

The next proposition which will be used throughout the article follows promptly from(1.6), (1.10), Remark 1.1 and Corollary 2.1.


Proposition 2.1. For f ∈ Hε, and u ∈ D(Aε), we have

1

2|Mf |ε ≤ a

∣

∣

Mf

r

∣

∣

ε≤ |Mf |ε,

∣

∣rcurl (Mu

r)∣

∣

ε= a|curl 2Mu/r|ε,

|AεMu

r|ε = r2|Aε

Mu

r|ε,

a2

2|AεNu|ε ≤ |AεNu|ε ≤ 8a2|AεNu|ε, for ε ≤ 1/8.

Sobolev-type inequalities.

In order to determine the dependence of the constants on ε, we will use our previousresults on parallelepipeds [TZ]. To that end, we divide the domain Ωε into a finite numberof isometric overlapping subdomains, each one of them is C∞−diffeomorphic to a paral-lelpiped; the diffeomorphisms are defined with the help of appropriate spherical coordinatesystems and their Jacobians are bounded from above and below independently of ε. Hence,we can transfer the inequalities from the parallelepiped to the domain Ωε without changingthe dependence of the constants on ε.

More precisly, let Qε =(π

4,3π

4

)

×(

0,3π

2

)

× (a, a + aε), and let Tε be the mapping

(2.10)Tε : Qε −→ Ωε

(θ′, ϕ′, r) 7→ (x1, x2, x3) = (r sin θ′ cos ϕ′, r sin θ′ sin ϕ′, r cos θ′).

Note that Tε is a C∞−diffeomorphism from Qε onto Ωε = Tε(Ωε); its jacobian Jac Tε isbounded independently of ε, for 0 < ε ≤ 1. In fact

(2.11)a2

2≤ |Jac Tε| ≤ 4a2.

It is clear that Ωε is the union of eight overlapping subdomains Ωiε, i = 1, . . . , 8, where

each Ωiε is the image of Ωε under some rotation. Thus

(2.12) |u|Lp(Ωε) ≤8

∑

i=1

|u|Lp(Ωiε) ≤ 8|u|Lp(Ωε), 1 ≤ p ≤ ∞.

Furthermore, since the Sobolev imbeddings are independent of the coordinate system, i.e.,on the position of the domain (up to an isometry), it is sufficient to derive the dependence

of the constants on ε for the subdomain Ωε, it will be the same for all the Ωiε, i = 1, . . . , 8,

and then, thanks to (2.12), we obtain the dependence of the constants for the domain Ωε.

For a function u ∈ Vε, we set for (x1, x2, x3) = Tε(r, θ′, ϕ′) ∈ Ωε

(2.13) v(r, θ′, ϕ′) = Nu(x1, x2, x3).


Due to (2.6) the Poincare inequality of Lemma 2.1 extends to the domain Ωε and then tothe domain Qε via Tε; it reads for the function v = Nu and thanks to (2.11)

(2.14) |v|2L2(Qε) ≤ 16ε2a2∣

∣

∂v

∂r

∣

∣

2

L2(Qε).

Moreover, for u ∈ D(Aε), we have

(2.15)∂

∂r(rvθ′) =

∂

∂r(rvϕ′) = vr = 0, for r = a or r = a + aε.

Hence, by integration by parts, Cauchy-Schwarz inequality and (2.14), we can write

(2.16)∣

∣

∂v

∂r

∣

∣

2

L2(Qε)≤ 16ε2a2

∣

∣

∂2v

∂r2

∣

∣

2

L2(Qε).

Now we are ready to prove the following (Hereafter the value of the constant c0 may notbe the same at different places):

Lemma 2.2. (Agmon’s inequality). There exist an absolute constant c0 independent of εsuch that

(2.17) |Nu|L∞(Ωε) ≤ c0|Nu|1

4

ε |AεNu|3

4

ε ≤ c0a− 3

2 |Nu|1

4

ε |AεNu|3

4

ε,r, ∀ u ∈ D(Aε),

and

(2.18) |Nu|L∞(Ωε) ≤ c0ε1

2 a− 1

2 |AεNu|ε,r, ∀ u ∈ D(Aε).

Proof. Thanks to Proposition 2.1 and Corollary 2.1, it is clear that (2.18) follows from(2.14). In order to prove (2.17), we will use the version of Agmon’s inequality establishedin [TZ]. We set y1 = θ′, y2 = ϕ′, y3 = r, and, by [TZ], we write for v ∈ H2(Qε),

(2.19)

∣

∣v∣

∣

L∞(Qε)≤ c0

∣

∣v∣

∣

1

4

L2(Qε)

2∏

i=1

(

∣

∣

∂2v

∂y2i

∣

∣

L2(Qε)+

∣

∣

∂v

∂yi

∣

∣

L2(Qε)+

∣

∣v∣

∣

L2(Qε)

)

1

4

×(

∣

∣

∂2v

∂y23

∣

∣

L2(Qε)+

1

εa

∣

∣

∂v

∂y3

∣

∣

L2(Qε)+

1

ε2a2

∣

∣v∣

∣

L2(Qε)

)

1

4

,

where c0 is an absolute constant. Thus, for the function v in (2.13) (v = Nu Tε), wehave, thanks to the Poincare inequalities (2.14) and (2.16),

(2.20)∣

∣v∣

∣

L∞(Qε)≤ c0|v|

1

4

L2(Qε)|∆v|3

4

L2(Qε).

Returning to the variables (r, θ′, ϕ′) and using (2.11), we obtain

(2.21) |Nu|L∞(Ωε) ≤ c0|Nu|1

4

L2(Ωε)|AεNu|

3

4

L2(Ωε).

Hence

|Nu|L∞(Ωiε) ≤ c0|Nu|

1

4

L2(Ωiε)|AεNu|

3

4

L2(Ωiε)

, i = 1, . . . , 8,

and, by (2.12), we have

|Nu|L∞(Ωε) ≤ c0|Nu|1

4

ε |AεNu|3

4

ε .

Finally, using |AεNu|ε ≤ 1

a2|AεNu|ε,r and Corollary 2.1, we conclude the lemma.


Lemma 2.3. (Ladyzhenskaya’s inequality). There exists a constant c0 independent of ε,such that

(2.22)∣

∣Nu∣

∣

L6(Ωε)≤ c0||Nu||ε ≤ c0

a||Nu||ε,r, ∀ u ∈ Vε.

Proof. Thanks to the version of Ladyzhenskaya inequality obtained in [TZ], there existsan absolute constant c0 (independent of ε) such that for all v ∈ H1(Qε),

(2.23)∣

∣v∣

∣

L6(Qε)≤ c0

2∏

i=1

(

∣

∣v∣

∣

L2(Qε)+

∣

∣

∂v

∂yi

∣

∣

L2(Qε)

)1

3

×(

1

εa

∣

∣v∣

∣

L2(Qε)+

∣

∣

∂v

∂r

∣

∣

L2(Qε)

)1

3

,

Now we proceede as in the proof of Lemma 2.2, we take for v the function in (2.13) anduse the Poincare inequality (2.14) to obtain

(2.24)∣

∣v∣

∣

L6(Qε)≤ c0|v|H1(Qε).

Using the change of variables (r, θ′, ϕ′) = T−1ε (x1, x2, x3), (2.11) and (2.12), inequality

(2.22) follows promptly.

Now interpolating between (2.2) and (2.22), we obtain

Lemma 2.4. For 2 ≤ q ≤ 6, there exists a constant c0(a, q) independent of ε such that

(2.25)∣

∣Nu∣

∣

2

Lq(Ωε)≤ c0(a, q)ε

6−q

q ||Nu||2ε,r, ∀ u ∈ Vε.

A pseudo-orthogonality inequality.

We will use the Poincare inequality (2.1) to establish a pseudo-orthogonality inequality.

Because of the geometry of the domain, AεNu and Aε

(Mu

r

)

are not orthogonal in Hε as

in the flat case [TZ]. Instead the pseudo-orthogonality inequality below, will replace thecorresponding orthogonality identity used in [TZ] and will be sufficient for our purpose.

Lemma 2.5. There exists an absolute constant c0 independent of ε such that for all u ∈D(Aε), we have

(2.26)(

Aεu, AεNu)

ε=

∣

∣r AεNu∣

∣

2

ε+ Iε(u) =

∣

∣AεNu∣

∣

2

ε,r+ Iε(u),

with

(2.27) |Iε(u)| ≤ c0ε|AεNu|ε,r

[

|AεMu

r|ε,r + |AεNu|ε,r].


Similarly

(2.28)

(

Aεu, AεMu

r

)

ε

=∣

∣r AεMu

r

∣

∣

2

ε+ Jε(u) =

∣

∣AεMu

r

∣

∣

2

ε,r+ Jε(u),

with

(2.29) |Jε(u)| ≤ c0ε|AεMu

r|ε,r

[

|AεNu|ε + |AεMu

r|ε,r

]

.

Proof. (i) We write

(2.30) (Aεu, AεNu)ε = (AεNu, AεNu)ε + (AεMu

r, AεNu)ε,

and thanks to Remark 1.1, we see that

(AεNu, AεNu)ε = |AεNu|2ε,r + (AεNu,∇r2 × curl Nu)ε,

and with (2.9)

(2.31) |(AεNu,∇r2 × curl Nu)ε| ≤ c0ε|AεNu|2ε,r.

Furthermore, thanks again to Remark 1.1,

(AεMu

r, AεNu)ε = (Aε

Mu

r, r2AεNu)ε + (Aε

Mu

r,∇r2 × curl Nu)ε,

(AεMu

r, r2AεNu)ε =

∫

Ωε

r2

a2r curl curl

Mu

rr curl curl Nu dx

+

∫

Ωε

a2 − r2

a2r curl curl

Mu

r· r curl curl Nu dx.(2.32)

Thanks to (1.3), the first integral in the right hand side of (2.32) vanishes. Then, usingthe Cauchy-Schwarz inequality and a < r < a + εa, we can bound the second integral by

c0ε∣

∣AεMu

r

∣

∣

ε,r|AεNu|ε,r. Finally, we note that

(2.33) |(AεMu

r,∇r2 × curl Nu)ε| ≤ c0ε|Aε

Mu

ru|ε,r|AεNu|ε,r.

Combining (2.31), (2.32) and (2.33), we obtain (2.27).

The proof of (2.28) follows the same lines as above. We just note that ∇r2×curlMu

r=

0, and

(2.34) Jε(u) =

(

r AεMu

r, r AεNu

)

ε

.


3. Estimates on the trilinear form

We will use systematically an alternative expression of the trilinear form; we note that

(u · ∇)u = u × curlu +1

2∇(|u|2) and

∫

Ωε

∇(|u|2) · w dx = 0, for w ∈ Hε, and write

(3.1) bǫ(u, v, w) =

∫

Ωε

u × curl v · w dx, ∀u, v, w ∈ Vε

This section gives some estimates on the form b = bε. First we give an orthogonalityproperty of the trilinear form bε related to the operator Aε, which is similar to the classicalone on the sphere with the 2D Stokes operator.

Proposition 3.1. For all u ∈ D(Aε), we have

(3.2) bε

(

Mu

r,Mu

r, Aε

Mu

r

)

= 0.

Proof. We write

(3.3)

b

(

Mu

r,Mu

r, curl r2curl

Mu

r

)

=

∫

Ωε

(

Mu

r× curl

Mu

r

)

· curl r2curlMu

rdx

= (by integration by parts using (A.7))

=

∫

Ωε

r2curl

(

Mu

r× curl

Mu

r

)

· curlMu

rdx

= (using (A.4))

=

∫

Ωε

curl

(

r2 Mu

r× curl

Mu

r

)

· curlMu

rdx

−∫

Ωε

[

∇r2 ×(

Mu

r× curl

Mu

r

)]

· curlMu

rdx.

Now note that

(3.4) ∇r2 ×(

Mu

r× curl

Mu

r

)

=

(

∇r2 · curlMu

r

)

Mu

r−

(

∇r2 · Mu

r

)

curlMu

r,

and, since ∇r2 = 2r~er, ~er · Mu = 0 and curlMu

r=

a

r2curl 2Mu~er, we have

(3.5) ∇r2 ×(

Mu

r× curl

Mu

r

)

=2a

r2(curl 2Mu) Mu,


and

(3.6)

[

∇r2 ×(

Mu

r× curl

Mu

r

)]

· curlMu

r= 0.

Moreover, since div 2Mu = 0, we have

(3.7) curl

(

r2 Mu

r× curl

Mu

r

)

=a2

r2div 2(Mu curl 2 Mu)~er =

a2

r2[∇2(curl 2 Mu) ·Mu]~er.

Hence,

∫

Ωε

curl

(

r2 Mu

r× curl

Mu

r

)

· curlMu

rdx = −

∫

Ωε

a3

r4(∇2(curl 2 Mu) · Mu)curl 2 Mu dx

= −∫

Ωε

a3

2r4∇2|curl 2 Mu|2 · Mu dx(3.8)

=

∫

Ωε

a3

2r2|curl 2 Mu|2div 2 Mu dx = 0.

We now consider certain terms which vanish in the cylindrical case but do not vanishhere because of the curvature. We are able however to bound them in a convenient way.

Lemma 3.1. For 0 < q < 12 , there exists a constant c(a, q) independent of ε such that for

every u ∈ D(Aε),

∣

∣b(Mu

r,Mu

r, AεNu

)∣

∣ ≤ c(a, q)ε1+q∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

ε,r|Aε

Mu

r|ε,r|AεNu|ε,r,(3.9)

∣

∣b(

Nu,Mu

r, Aε

Mu

r

)∣

∣ ≤ c(a, q)ε3

2 ||Nu||ε,r|AεMu

r|2ε,r,(3.10)

∣

∣b(Mu

r, Nu, Aε

Mu

r

)∣

∣ ≤ c(a, q)ε1+q||Nu||ε,r|AεMu

r|2ε,r.(3.11)

Proof. (i) Proof of (3.9). First we show that

b

(

Mu

r,Mu

r, curl (r2curl Nu)

)

=

∫

Ωε

a2 − r2

a2[curl Nu · ~er]

[

a2

r2(∇2curl 2 Mu) · Mu

]

dx

− 2

∫

Ωε

a(a2 − r2)

a2

curl 2 Mu

r2Mu · curl Nu dx.(3.12)

We have

(3.13) B1 = b

(

Mu

r,Mu

r, curl (r2curl Nu)

)

=

∫

Ωε

(

Mu

r× curl

Mu

r

)

·curl r2curl Nu dx.


Integrating by parts using formula (A.6), the boundary integral vanishes thanks to (A.2)and (curl Nu) × ~n = 0 on ∂Ωε. Hence,

(3.14) B1 =

∫

Ωε

r2curlNu · curl

(

Mu

r× curl

Mu

r

)

dx,

and formula (A.4) yields

(3.15)

B1 =

∫

Ωε

curl Nu · curl

[

r2

(

Mu

r× curl

Mu

r

)]

dx

− 2

∫

Ωε

curl Nu ·(

∇r2 ×(

Mu

r× curl

Mu

r

))

dx.

Using (3.7) and (3.10), we can write

(3.16) B1 =

∫

Ωε

[curlNu ·~er]a2

r2(∇2curl 2 Mu) ·Mu dx−2a

∫

Ωε

curl 2 Mu

r2Mu ·curl Nu dx.

Thanks to (1.3), we have

(3.17)

∫

Ωε

[curl Nu · ~er](∇2curl 2 Mu) · Mu dx = 0, and

∫

Ωε

(curl 2Mu)Mu · curl Nu dx = 0.

Thus (3.12) follows.

Now since a < r < a + εa, we have by Holder’s inequality with1

p+

1

p∗=

1

2,

(3.18) |B1| ≤ 2ε∣

∣

a2

r2∇2curl 2 Mu

∣

∣

L2(Ωε)|Mu|Lp(Ωε)|curlNu · ~er|Lp∗

(Ωε)

+ 2ǫa∣

∣

curl 2 Mu

r2

∣

∣

ε|Mu|Lp(Ωε)|curl Nu · ~er|Lp∗

(Ωε).

Thanks to the Sobolev embedding H1(S2) ⊂ Lp(S2), for 2 ≤ p < ∞, we have (see [A])

(3.19)∣

∣Mu∣

∣

Lp(Ωε)≤ c(a, p)ε

1

p− 1

2

∣

∣Mu∣

∣

H1(Ωε), (c(a, p) independent of ε),

and with (2.25) we obtain,

(3.20)|B1| ≤ c(a, p)ε

1

p+ 3

p∗

∣

∣

1

r2∇2curl 2 Mu

∣

∣

ε|curlMu|ε|curl curl Nu|ε

≤ c(a, p)ε1

p+ 3

p∗ |curlMu

r|ε,r|Aε

Mu

r|ε,r|AεNu|ε,r.


Inequality (3.9) follows promptly.

(ii) Proof of (3.10). We write

b

(

Nu,Mu

r, curl r2curl

Mu

r

)

=

∫

Ωε

(

Nu × curlMu

r

)

· curl r2curlMu

rdx

=1

a2

∫

Ωε

r2

(

Nu × curlMu

r

)

· curl r2curlMu

rdx(3.21)

+

∫

Ωε

a2 − r2

a2

(

Nu × curlMu

r

)

· curl r2curlMu

rdx

Note that

r2

[

Nu × curlMu

r

]

= aNuϕ curl 2 Mu~eθ − aNuθ curl 2 Mu~eϕ,(3.22)

curl

(

r2curlMu

r

)

=a

r sin θ

∂

∂ϕcurl 2 Mu~eθ −

a

r

∂

∂θcurl 2 Mu~eϕ.(3.23)

Now using

∫ a+εa

a

r Nuθdr = 0 and

∫ a+εa

a

r Nuϕ dr = 0, we find that the first integral in

the right hand side of (3.21) vanishes. Therefore,(3.24)

b

(

Nu,Mu

r, curl r2curl

Mu

r

)

=

∫

Ωε

a2 − r2

a2

(

Nu × curlMu

r

)

· curl r2curlMu

rdx.

Now, since a < r < a + εa, we have by Holder’s inequality

(3.25)∣

∣b

(

Nu,Mu

r, curl r2curl

Mu

r

)

∣

∣ ≤ ε|Nu|L∞(Ωε)

∣

∣curlMu

r

∣

∣

ε

∣

∣curl r2curlMu

r

∣

∣

ε,

and according to Lemma 2.2, we have

(3.26)∣

∣b

(

Nu,Mu

r, curl r2curl

Mu

r

)

∣

∣ ≤ c(a, p)ε3

2 |curlMu

r|ε,r|Aε

Mu

r|ε,r|AεNu|ε,r.

(iii) Proof of (3.11). First we write(3.27)

b

(

Mu

r, Nu, curl r2curl

Mu

r

)

=

∫

Ωε

(Mu

r× curl Nu

)

· curl (r2curlMu

r) dx

=

∫

Ωε

r2

a2

(Mu

r× curl Nu

)

· curl (r2curlMu

r) dx

+

∫

Ωε

a2 − r2

a2

(Mu

r× curl Nu

)

· curl (r2curlMu

r) dx.


As above, we can show, using (1.3), that the first integral in the right hand side of (3.27)is zero. Furthermore, with Holder’s inequality, we have

(3.28) |b(

Mu

r, Nu, curl r2curl

Mu

r

)

| ≤ ε∣

∣

Mu

r

∣

∣

Lp(Ωε)|Nu|Lp∗

(Ωε)

∣

∣curl (r2curlMu

r)∣

∣

ε,

and just as in the proof of (3.3), we obtain (3.5).

Lemma 3.2. For 0 < q < 12 , there exists a constant c(a, q) independent of ε such that for

all u ∈ D(Aε), we have

|b(Nu, Nu, AεNu)| ≤ c(a, q)ε1/2||Nu||ε,r|AεNu|2ε,r,(3.29)

∣

∣b(

Nu,Mu

r, AεNu

)∣

∣ ≤ c(a, q)ε1/2∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

ε,r|AεNu|2ε,r,(3.30)

∣

∣b(Mu

r, Nu, AεNu

)∣

∣ ≤ c(a, q)εq∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

ε,r|AεNu|2ε,r,(3.31)

∣

∣b(

Nu, Nu, AεMu

r

))∣

∣ ≤ c(a)||Nu||3/2ε,r |AεNu|/12

ε,r |AεMu

r|ε,r.(3.32)

Proof. The proof is based on Lemmas 2.1-2.5 and Holder’s inequalities, and follows thelines of the proof of Lemma 3.1; we skip the details.

4. A priori estimates

First of all we introduce the following notations; we set

(4.1)

α(ε) = |Mf

r|ε, β(ε) = |Nf |ε

a(ε) =∣

∣

∣

∣

Mu0

r

∣

∣

∣

∣

ε,r=

∣

∣rcurlMu0

r

∣

∣

ε,

b(ε) = ||Nu0||ε,r = |rcurlNu0|ε,R2

0(ε) = a2(ε) + b2(ε) + α2(ε) + β2(ε),

B2ε = b0(ε) + β2(ε).

We have, according to Lemma 1.2,

(4.2) ||u0||2ε,r = a2(ε) + b2(ε),∣

∣f∣

∣

2

ε= α2(ε) + β2(ε),

and

(4.3) ||u0||2ε,r ≤ R20(ε).

Now, given σ > 1, we have the following classical fact which is a consequence of Theorem0.1 (see e.g. [Le 1,2] and [Li]):

(4.4) ∃ Tσ(ε) > 0, such that ||u(t)||2ε,r ≤ σR20(ε), ∀ 0 ≤ t ≤ Tσ(ε).

Here [0, Tσ(ε) ) is the maximal interval on which (4.4) holds. It is clear that if Tσ(ε) < ∞,then

(4.5) ||u(Tσ(ε))||2ε,r = σR20(ε).


A priori estimates for Nu.

Thanks to Lemma 1.1, we can multiply (0.11) with Aε = curl (r2curl Nu), and obtainwith Lemmas 1.2 and 2.5

(4.6)1

2

d

dt||Nu||2ε,r + ν|AεNu|2ε,r = (Nf, AεNu) − b(u, u, AεNu) − Iε(u).

We write

(4.7) b(u, u, AεNu) = b(Mu

r,Mu

r, AεNu) + b(Nu,

Mu

r, AεNu)

+ b(Mu

r, Nu, AεNu) + b(Nu, Nu, AεNu),

and with Lemmas 2.5, 3.1 and 3.2, we have, for 0 < t < Tσ(ε),

1

2

d

dt||Nu||2ε,r + ν|AεNu|2ε,r ≤ |Nf |ε|AεNu|ε + c0νε|AεNu|ε,r|Aε

Mu

r|ε,r

+ c0νε|AεNu|2ε,r + c(q, a)εq||u||ε,r|AεNu|2ε,r(4.8)

+ c(q, a)ε1+q∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

ε,r|AεNu|ε,r|Aε

Mu

r|ε,r,

where q < 12 , and c(q, a) is a constant independent of ε. For ε ≤ 1

4c0, we have

c0νε|AεNu|ε,r|AεMu

r|ε,r + c0νε|AεNu|2ε,r ≤ ν

4|AεNu|2ε,r + c0νε|Aε

Mu

r|2ε,r,

and, by Proposition 2.1, we also have, for ε ≤ 1/8,

|Nf |ε|AεNu|ε ≤ 8a2|Nf |ε|AεNu|ε ≤ 3a|Nf |ε|AεNu|ε,r ≤ ν

4|AεNu|2ε,r +

9a2

ν|Nf |2ε.

Hence

d

dt||Nu||2ε,r + ν|AεNu|2ε,r ≤ c0a

2

ν|Nf |2ε + c0νε|Aε

Mu

r|2ε,r

+ c(q, a)εq||u||ε,r|AεNu|2ε,r

+ c(q, a)ε1+q∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

ε,r|AεNu|ε,r|Aε

Mu

r|ε,r,

where c0 is a numerical constant (we denote all numerical constants by c0). Now, with(4.3) and (4.4), we can write for 0 < t < Tσ(ε),

(4.9)

d

dt||Nu||2ε,r + ν|AεNu|2ε,r ≤ c0a

2

ν|Nf |2ε + c0νε2|Aε

Mu

r|2ε,r

+ c(q, a)εq√

σR0(ε)|AεNu|2ε,r

+ c(q, a)ε1+q√

σR0(ε)|AεNu|ε,r|AεMu

r|ε,r.


At this stage, we use assumption (0.14) with 0 < q < 12 . We note that assumption (0.14)

is not restrictive, since physically, R0(ε) can be assumed to go to zero when ε goes to zero,which is the case when f and u0 are independent of ε. Thanks to (0.14), we can chooseε1 = ε1(ν, a, q, σ, R0) > 0, such that

(4.10)√

σc(q, a)εqR0(ε) ≤ν

2, ∀ε, 0 < ε ≤ ε1.

Therefore, for 0 < t < Tσ(ε) and 0 < ε < ε1, we have

(4.11)d

dt||Nu||2ε,r + ν|AεNu|2ε,r ≤ 18a2

ν|Nf |2ε + c0νε2|Aε

Mu

r|2ε,r.

Similarly, with Lemma 1.1, we can multiply (0.11) by AεMu

r= curl

(

r2curlMu

r

)

, and

obtain

(4.12)1

2

d

dt

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r+ ν

∣

∣AεMu

r

∣

∣

2

ε,r=

(

Mf

r, Aε

Mu

r

)

− b

(

u, u, AεMu

r

)

− Jε(u).

We write

(4.13)b(

u, u, AεMu

r

)

= b(Mu

r,Mu

r, Aε

Mu

r

)

+ b(Mu

r, Nu, Aε

Mu

r

)

+ b(

Nu, Nu, AεMu

r

)

+ b(

Nu,Mu

r, Aε

Mu

r

)

,

and with Proposition 3.1, Lemmas 2.5, 3.1 and 3.2, we have

d

dt

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r+ ν

∣

∣AεMu

r

∣

∣

2

ε,r≤ a2

2ν

∣

∣

Mf

r

∣

∣

2

ε+ c0νε|AεNu|ε,r|Aε

Mu

r|ε,r

+ c0νε2|AεMu

r|2ε,r

+ c(q, a)ε1+q||Nu||ε,r|AεMu

r|2ε,r(4.14)

+ c(a)ε3

2 ||Nu||ε,r|AεNu|ε,r|AεMu

r|ε,r

+ c(a)||Nu||3

2

ε,r|AεNu|1

2

ε,r|AεMu

r|ε,r.

Thanks to Holder’s inequality and (4.10), we infer from (4.14)

(4.15)d

dt

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r+ ν|Aε

Mu

r|2ε,r ≤ a2

2ν

∣

∣

Mf

r

∣

∣

2

ε+ c0νε2|AεNu|2ε,r.

Adding (4.11) and (4.15), we obtain

d

dt||u||2ε,r+ν|Aε

Mu

r|2ε,r+ν|AεNu|2ε,r ≤ c0a

2

ν|f |2ε+c(a, q)νε2

[

|AεNu|2ε,r+|AεMu

r|2ε,r

]

.


Now, by integration between 0 and t, we obtain for ε2 <1

4c(a, q)ν

(4.16)

∫ t

0

|AεMu

r|2ε,r(s)ds ≤ 2

ν(a2

0(ε) + b20(ε)) +

c0a2t

ν2|f |2ε

≤ c0R20(ε)

ν(1 +

a2t

ν), 0 < t < Tσ(ε), 0 < ε < ε1

and

(4.17)

∫ t

0

|AεNu|2ε,r(s)ds ≤ 2

ν(a2

0(ε) + b20(ε)) +

c0a2t

ν2|f |2ε

≤ c0R20(ε)

ν(1 +

a2t

ν), 0 < t < Tσ(ε), 0 < ε < ε1.

Now we return to (4.11) and infer from Poincare’s inequality

(4.18)d

dt||Nu||2ε,r +

ν

4ε2||Nu||2ε,r ≤ c0a

2|Nf |2εν

+ c(a, q)νε2|AεMu

r|2ε,r,

and Gronwall’s lemma yields

(4.19)

||Nu||2ε,r(t) ≤ b20(ε) exp(− νt

4ε2) +

c0ε2a2

ν2|Nf |2ε + 2c(a, q)νε2

∫ t

0

|AεMu

r|2ε,r(s)ds

≤ (with (4.16))

≤ b20(ε) exp(− νt

4ε2) +

c0ε2a

ν2|Nf |2ε + 2c(a, q)νε2R2

0(ε)ν(1 +a2t

ν).

Integrating (4.18) between 0 and t, we have

(4.20)

∫ t

0

||Nu||2ε,r(s)ds ≤ 4ε2

νb20(ε) +

36ε2a2t

ν2|Nf |2ε + c(a, q, ν)ε2 R2

0(ε)

ν(1 +

a2t

ν).

Combining (4.17), (4.19) and (4.20), we have thanks to Cauchy-Schwarz inequality

∫ t

0

||Nu||3ε,r|AεNu|ε,rds ≤ sup0≤s≤t

||Nu||2ε,r(s)

(∫ t

0

||Nu||2ε,rds

)1

2

(∫ t

0

|AεNu|2ε,rds

)1

2

≤ c(a, q, ν)εR40(ε)(1 + t)2.

We collect the a priori estimates on Nu in the following:


Lemma 4.1. Let uε0 ∈ Vε and fε ∈ Hε, 0 < ε < 1, satisfying (0.14) and (0.15) for some

0 < q < 12 . Then, for any σ > 1, there exists ε1 = ε1(σ, a, q, ν, R0)) > 0 such that for all ε,

0 < ε ≤ ε1, there exists Tσ(ε) > 0 and a positive constant c5, independent of ε, such thatfor 0 < ε ≤ ε1 and 0 < t < Tσ(ε), we have

||Nu||2ε,r(t) ≤ b20(ε) exp(− νt

4ε2) + c(a, q, ν)ε2R2

0(ε)(1 +a2t

ν),(i)

∫ t

0

||Nu||2ε,r(s)ds ≤ 4ε2

νb20(ε) + c(a, q, ν)ε2R2

0(ε)(1 +a2t

ν),(ii)

∫ t

0

|AεNu|2ε,r(s)ds ≤ c0R20(ε)

ν(1 +

a2t

ν)(iii)

∫ t

0

||Nu||3ε,r|AεNu|ε,rds ≤ c(a, q, ν)εR40(ε)(1 + t)2.(iv)

A priori estimates on Mu.

Using (4.14), we write

d

dt

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r+ ν|Aε

Mu

r|2ε,r ≤ a2

2ν

∣

∣

Mf

r

∣

∣

2

ε+ c0νε2|AεNu|2ε,r

+ c(a, q, ν)||Nu||3

2

ε,r|AεNu|1

2

ε,r|AεMu

r|ε,r.

Thanks to Holder’s inequality, we have

(4.22)

d

dt

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r+ ν|Aε

Mu

r|2ε,r ≤ a2

2ν

∣

∣

Mf

r

∣

∣

2

ε+ c0νε2|AεNu|2ε,r

+ c(a, q, ν)||Nu||3ε,r|AεNu|ε,r.

and with the Poincare inequality

(4.23)∣

∣r curlMu

r

∣

∣

2

ε≤ 1

λ(a)|Aε

Mu

r|2ε,r,

with λ(a) =λ1

a2, where λ1 is the first eigenvalue of the Laplace-Beltrami operator on the

unit sphere, we can write

(4.24)

d

dt

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r+

ν

λ(a)|Aε

Mu

r|2ε,r ≤ a2

ν|Mf

r|2ε + c0νε2|AεNu|2ε,r

+ c(a, q, ν)||Nu||3ε,r|AεNu|ε,r.


Therefore, Gronwall’s lemma yields

(4.25)

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r(t) ≤ a2

0(ε) exp(−νλ(a)t) +a2

2λ(a)ν2α2(ε)

+ c0νε2

∫ t

0

|AεNu|2ε,rds + c(a, q)ν

∫ t

0

||Nu||3ε,r|AεNu|ε,rds.

Hence, thanks to (4.16) and (4.21), we obtain

(4.26)

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r(t) ≤ a2

0(ε) exp(−νλ(a)t) +a2

2λ2(a)ν2α2(ε)

+ c(a, q, ν)ε2R20(ε)(1 + t) + c(a, q, ν)εR4

0(ε)(1 + t)2.

Moreover, by integration of (4.24) between 0 and t,

∫ t

0

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r(s)ds ≤ a2

0(ε)

2λ(a)ν+

α2(ε)t

a2λ2(a)ν2+ c(a, q, ν)ε2R2

0(ε)(1 + t)

+ c(a, q, ν)εR40(ε)(1 + t)2.

We collect the a priori estimates on Mu in the following:

Lemma 4.2. Let uε0 ∈ Vε and fε ∈ Hε, 0 < ε < 1, satisfying (0.14) and (0.15) for some

0 < q < 12 . Then, there exists ε1 = ε1(ν, a, q, σ, R0) > 0 such that for all ε, 0 < ε ≤ ε1,

there exists Tσ(ε) > 0 and a positive constant c(a, q, ν, σ), independent of ε, such that for0 < ε ≤ ε1 and 0 < t < Tσ(ε):

(i)∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r(t) ≤ a2

0(ε) exp(−νλ(a)t) +a2α2(ε)

2λ2(a)ν2+ c(a, q, ν)εR4

0(ε)(1 + t)2,

(ii)

∫ t

0

|AεMu

r|2ε,r(s)ds ≤ c1R

20(ε)(1 +

a2t

ν),

(iii)

∫ t

0

∣

∣

∣

∣

Mu

r

∣

∣

∣

∣

2

ε,r(s)ds ≤ a2

0(ε)

λ(a)ν+

α2(ε)t

a2λ2(a)ν2+ c(a, q, ν)εR4

0(ε)(1 + t)2.

5. Global existence and regularity of strong solutions

In this section we study the behavior of the maximal time of existence of strong solutionsof the 3D-Navier-Stokes equations in the thin spherical shells. First we establish the globalexistence of strong solutions for initial data belonging to large sets. Then, we give aneventual regularity result for Leray-Hopf weak solutions when the volume forces belong tolarge sets.

5.1. Global regularity.

First we prove the following auxiliary result that will show that the maximal time ofexistence of strong solutions is bounded from below independently of ε.


Proposition 5.1. Let uε0 ∈ Vε and fε ∈ Hε satisfying (0.14) and (0.15), for some q <

1

2,

limε→0

ε2q(∣

∣A1

2

ε u0

∣

∣

2

ε+

∣

∣f∣

∣

2

ε) = 0.

Then

(5.1) limε→0

ε1/2−q Tσ(ε) = +∞.

Proof. According to Lemmas 4.1 and 4.2, we have, for 0 < t ≤ Tσ(ε)

(5.2)

||u(t)||2ε,r ≤ a20(ε) exp(−λ(a)νt) + b2

0(ε) exp(− νt

4ε2)

+a2

λ(a)ν2α2(ε) +

[

c(a, ν, q))ε1−2qε2qR20(ε)(1 + t)2

]

R20(ε).

Now we fix σ by setting

(5.3) σ = 8 max(

1,a2

λ(a)ν2

)

.

We infer from (5.2) that

(5.4) ||u(t)||2ε,r ≤ σ

4R2

0(ε) +

[

c(a, ν, q))ε1−2qε2qR20(ε)(1 + t)2

]

R20(ε), 0 ≤ t ≤ Tσ(ε).

Now suppose that Tσ(ε) < ∞. We have, according to (4.5),

(5.5) σR20(ε) ≤

σ

4R2

0(ε) +

[

c(a, ν, q))ε1−2qε2qR20(ε)(1 + Tσ(ε))2

]

R20(ε).

We make the obvious assumption R20(ε) 6= 0, and obtain

(5.6)3σ

4≤

[

c(a, q, ν)ε1−2q(1 + Tσ(ε))2]

ε2qR20(ε).

Hence

ε1/2−qTσ(ε) >

√

3σ

4c(a, ν, q)

(

1

εqR0(ε)− 1

)

,

and (5.1) follows promptly.

Now we go to the second step and establish that the maximal time of existence of thestrong solutions of the 3D- Navier-Stokes equations in thin domains is infinite for largeinitial data. More precisely, we prove


Theorem 5.1. Let uε0 ∈ Vε and fε ∈ Hε satisfying (0.14) and (0.15), for some q < 1

2 ,Then, there exists ε1(ν, a, q, R0) such that for 0 < ε < ε1, the strong solution uε of (0.1)-(0.4) exists for all times; i.e. Tε = +∞ in Theorem 0.1 and

(5.7) uε ∈ C0([0,∞); Vε) ∩ L2(0, T ; D(Aε)), ∀ T > 0.

Proof. First recall that σ is fixed to be σ = 8 max(1,a2

λ2(a)ν2). We set

(5.8) B2ε = b2

0(ε) + R20(ε) and K2

ε = a20(ε) +

4a2α2(ε)

λ2(a)ν2+ B2

ε .

Note that

(5.9) B2ε ≤ σ

4R2

0(ε) and K2ε ≤ 3σ

4R2

0(ε).

Now we fix ε and choose ǫ4(ν, a, q, R0) so that for 0 < ε ≤ ε4(ν, q, q),

(5.10. i) c(a, q, ν)ε2qR20(ε) ≤

1

32,

(5.10. ii) exp(−νλ(a)εq−1/2) ≤ 1

4,

(5.10. iii) exp(−νεq−1/2

4ε2) ≤ 1

4,

(5.10. iv) ε1/2−qTσ(ε) > 1.

The existence of ε4 is obvious, since the left hand sides of (i)-(iii) go to zero when ε goesto zero; and, by Proposition 5.1, the left hand side of (iv) goes to infinity as ε → 0.

Recall that

(5.11) ||Nu||2ε,r(t) ≤ b0(ε)2 exp(− νt

4ε2) + c(a, q, ν)ε2R2

0(ε)(1 + t), 0 ≤ t < Tσ(ε),

(5.12) ||Mu

r||2ε,r(t) ≤ a0(ε)

2 exp(−νλ1(a)t) +a2α2(ε)

2λ2(a)ν2

+ c(a, q, ν)ε2qR40(ε)(1 + t)2ε1−2q, 0 ≤ t < Tσ(ε),

where the constant c(a, q, ν) appearing in (5.11) and (5.12) is the same as the one appearingin (5.10. i).

We set tε = εq−1/2, 0 < ε ≤ ε4. Note that (5.10. iv) implies that tε < Tσ(ε), andaccording to (5.11), (5.12) together with (5.10), we have

||Nu||2ε,r(tε) ≤b0(ε)

2

4+

1

4R2

0(ε) =1

4B2

ε ,

||Mu

r||2ε,r(tε) ≤

a0(ε)2

4+

1

4K2

ε ≤ 1

2K2

ε .


Now we use an induction argument to show that

(Hn) Tσ(ε) > ntε, and ||Nu||2ε,r(ntε) ≤1

2B2

ε , ||Mu

r||2ε,r(ntε) ≤

1

2K2

ε .

The proof of (H1) is given above. Now assuming that (Hn) holds, our goal is to prove(Hn+1); we write (5.11) and (5.12) when the initial data are given at time t0 = ntε. Wehave, for ntε ≤ t < Tσ(ε)

(5.13)||Nu||2ε,r(t) ≤ ||Nu||2ε,r(ntε) exp(−ν(t − ntε)

4ε2)

+ c(a, q, ν)ε2R40(ε)(1 + t − ntε), 0 ≤ t < Tσ(ε),

(5.14)||Mu

r||2ε,r(t) ≤ ||Mu

r||2ε,r(ntε) exp(−νλ1(a)(t − ntε)) +

a2α2(ε)

2λ2(a)ν2

+ c(a, q, ν)ε2qR40(ε)(1 + t − ntε)

2ε1−2q, 0 ≤ t < Tσ(ε),

Hence, if Tσ(ε) ≤ (n + 1)tε, we have thanks to (Hn)

||Nu||2ε,r(Tσ(ε)) ≤ B2

ε

2+

1

4K2

ε ≤ σ

8R2

0(ε) +3σ

16R2

0(ε) =5σ

16R2

0(ε),

||Mu

r||2ε,r(tε) ≤

K2ε

2+

1

4K2

ε ≤ 9σ

16R2

0(ε),

and

||u||2ε,r(Tσ(ε)) ≤ 14σ

16R2

0(ε) < σR20(ε).

This contradicts the definition of Tσ(ε). Thus

Tσ(ε) > (n + 1)tε,

and we are able to set t = (n + 1)tε in (5.13) and (5.14). We obtain, using (5.10) (i)-(iv)and (Hn)

||Nu||2ε,r((n + 1)tε) ≤B2

ε

2exp(−νεq−1/2

ε2) + c(a, q, ν)ε2R2

0(ε)(1 + εq−1/2)

≤ 1

8B2

ε +1

4R2

0(ε) ≤1

2B2

ε ,

and

||Mu

r||2ε,r((n + 1)tε) ≤

K2ε

2exp(−νλ1(a)εq−1/2) +

a2α2(ε)

2λ2(a)ν2

+ c(a, q, ν)ε2qR40(ε)(1 + εq−1/2)2ε1−2q

≤ 1

8K2

ε +1

8K2

ε +1

8K2

ε <1

2K2

ε .

Therefore, (Hn+1) is established, and as a consequence of it, we have

Tσ(ε) = ∞, 0 < ε ≤ ε4.

The proof of Theorem 5.1 is complete.


5.2. Eventual regularity.

In this subsection we show that for volume forces belonging to large sets all Leray-Hopf’s weak solutions become regular after a period of time. More precisely we prove thefollowing:

Theorem 5.2. Let fε ∈ Hε, 0 < ε < 1, be such that

(5.15) |f ǫ|2ǫ ≤ R20(ǫ),

R0 satisfying (0.14). Then, there exists ε1(ν, a, q) such that: for 0 < ε < ε1, there existsT0(ε) such that any Leray-Hopf’s type weak solution uε of (0.1)-(0.4) is a strong solutionon [T0(ε), +∞), i.e.,

(5.16) uε ∈ C0([T0(ε),∞); Vε) ∩ L2(T0(ε), T ; D(Aε)), ∀ T > T0(ε).

Proof. We use the energy inequality

(5.17)1

2

d

dt|uε|2ε + ν||uε||2ε ≤

∣

∣

Mfε

r

∣

∣

ε

∣

∣

Muε

r

∣

∣

ε+ |Nfε|ε|Nuε|ε.

Recall the Poincare inequalities

(5.18)|Mu

r|ε ≤ c0||

Mu

r||ε,r

|Nu|ε ≤ c0ε||Nu||ε,r,

where c0 is an absolute constant. We infer from (1.12), (5.17) and (5.18)

(5.19)1

2

d

dt|uε|2ε +

ν

4a2||uε||2ε,r ≤ ν

8a2

∣

∣

Muε

r

∣

∣

2

ε,r+

ν

8a2|Nuε|2ε +

8c20a

2

ν|f |2ε.

Henced

dt|uε|2ε +

ν

2a2||uε||2ε,r ≤ 16c2

0a2

ν|f |2ε,

and

(5.20)1

t

∫ t

0

||uε(s)||2ε,r ds ≤ 64c20a

4

ν2|f |2ε +

4a2

νt|uε(0)|2ε.

There exists t0 = t0(ε) such that

(5.21)4a2

νt|uε(0)|2ε ≤ 64c2

0a4

ν2|f |2ε, for t ≥ t0(ε),

and, according to (5.20), there exists T0(ε) such that

(5.22) ||uε(T0(ε))||2ε,r ≤ c20a

4

ν2|f |2ε .

Thanks to (0.14), (5.15) and (5.22), we have

||uε(T0(ε))||2ε,r + |f |2ε ≤ R20(ǫ).

The condition (0.15) of Theorem 5.1 holds when the initial condition are given at timeT0(ε). We can apply Theorem 5.1 and conclude the proof.


6. The 2D-limit

In this section we establish the convergence of the average, in the radial direction, of thestrong solutions of the three-dimensional Navier-Stokes equations in thin spherical domainsto the strong solution of the two-dimensional Navier-Stokes equations on a sphere, whenthe thickness of the domain goes to zero.

First we introduce the Navier-Stokes equations on the sphere S2a. For simplicity, we use

v = vi ∂∂xi = (vi) to denote a vector field on S2

a. The 2D Navier-Stokes equations on the

sphere S2a are written as follows:

(6.1)

vt + ∇v v + ∇2q − ν∆2v = g,

div 2v = 0,

v|t = 0 = v0,

where ∇2 is the tangential gradient, div 2v, is the tangential divergence, ∆2 is the tangentialLaplace-Beltrami operator (see Appendix), and the covariant derivative ∇v v is defined asfollows:

(6.2) ∇v v = vkvi;k

∂

∂xi.

We introduce the function spaces H0 and V0 defined as follows

H0 = L2(S2)/R =

ζ ∈ L2(S2a);

∫

S2a

ζ dS2a = 0

,

V0 = H1(S2a)/R.

Here the function spaces are defined on the manifold S2 (see [A]). Let (fε)ε>0 ∈ Hε (orL∞(0,∞; Hε)) and (uε

0)ε>0 ∈ Vε such that, for some q < 12 , we have (0.15) and

(6.3) limε→0

ǫqR0(ǫ) = 0,

and assume that there exists g ∈ H0(or L∞(0,∞; H0)) and v0 ∈ H0 such that:

limε→0

Mfε = g in H0-weak.(6.4)

limε→0

Muε0 = v0 in H0-weak,

∣

∣A1

2

ε Muε0

∣

∣

εis bounded.(6.5)

First note that (6.3) implies (0.14). Let T > 0 be given and fixed. Thanks to Theorem5.1, there exists ε1(ν, v0, g) such that T < Tσ(ε) for 0 < ε ≤ ε1, and according to Lemma4.2, and Proposition 2.1, we have

Muε is bounded in L∞(0, T ; V0) independently of 0 < ε ≤ ε1.(6.6)

Muε is bounded in L2(0, T ; V0) independently of 0 < ε ≤ ε1.(6.7)

Muε is bounded in L2(0, T ; H0) independently of 0 < ε ≤ ε1.(6.8)


Hence, there exists a sequence (εn)n∈N, with limn→∞ εn = 0, and a function v∗ such that

limεn→0

Mεnuεn = v∗ in L2(0, T ; H0) − strong.(6.9)

limεn→0

Mεnuεn = v∗ in L2(0, T ; V0) − weak.(6.10)

Now we write the weak formulation of the equation satisfied byMεu

ε

r: for v defined on

S2, we havev

r∈ Hε, ∀ε > 0, and

Mv

r= a(1 +

ǫ

2)v

r. Hence

d

dt

(

Muε

r,Mv

r

)

ε

+ ν

(

curl curlMuε

r,Mv

r

)

ε

+ b

(

Muε

r,Muε

r,Mv

r

)

=

(

Mf

r,Mv

r

)

ε

+ ν

(

curl curl Nu,Mv

r

)

ε

− b

(

Muε

r, Nu,

Mv

r

)

(6.11)

− b

(

Nu,Muε

r,Mv

r

)

− b

(

Nu, Nu,Mv

r

)

Now note that Muε and Mv are defined on the sphere S2a, and

(6.12)

(

Mεuε

r,Mv

r

)

ε

= ǫa(Muε, Mv)0,

(6.13)

(

curl curlMεu

ε

r,Mv

r

)

ε

=ε

a + εa(curl 2Mεu

ε, curl 2Mv)0,

and

(6.14)

(

curl curl Nu,Mv

r

)

ε

=

(

a2 − r2

a2curl Nu, curl

Mv

r

)

ε

,

so that

(6.15)

∣

∣

∣

∣

(

curl curl Nu,Mv

r

)

ε

∣

∣

∣

∣

≤ ε|curl Nu|ε∣

∣curlMv

r

∣

∣

ε≤ ε2

a|curl Nu|ε|curl 2

Mv

r|0.

Moreover, we have∣

∣

∣

∣

1

εb(Nu, Nu,

Mv

r)

∣

∣

∣

∣

≤ 1

ε|Nu|2L4(Ωε) |curl

Mv

r|ε

≤ c0ε1

2 |curlNu|2ε |curl 2Mv|0≤ (with Lemma 4.1)(6.16)

≤[

c0ε1

2 b20(ε) exp(− νt

4ε2) + c(a, ν, q)ε2R2

0(ε)(1 + t)

]

|curl 2Mv|0,


and under the assumptions H1, there exists a constant K independent of ε such that

(6.17)

∣

∣

∣

∣

1

εb(Nu, Nu,

Mv

r)

∣

∣

∣

∣

≤ Kε1

2−q

[

exp(− νt

4ε2) + c(a, ν, q)(1 + T )

]

|curl 2Mv|0,

and

(6.18) limn→∞

∣

∣

∣

∣

1

εnb(Nu, Nu,

Mv

r)

∣

∣

∣

∣

= 0.

Now we claim that

(6.19) limn→∞

∣

∣

∣

∣

1

εnb(

Mεuε

r, Nu,

Mv

r)

∣

∣

∣

∣

= 0 and limn→∞

∣

∣

∣

∣

1

εnb(Nu,

Mεuε

r,Mv

r)

∣

∣

∣

∣

= 0.

For note that, with (1.3), we have

(6.20) b(Mεu

ε

r, Nu,

Mv

r) = b(

a2 − r2

a2

Mεuε

r, Nu,

Mv

r).

Hence

(6.21)

∣

∣

∣

∣

1

εb(

Mεuε

r, Nu,

Mv

r)

∣

∣

∣

∣

≤ |Mεuε

r|L4(Ωε) |curl Nu|ε |

Mv

r|L4(Ωε)

≤ c0ε1

2 |curl Nu|ε|curl 2Mεuε|0|curl 2Mv|0

≤ (with H1 and Lemma 4.2)

≤ c1(ν, T, g, v0)ε1

2−q|curl 2Mv|0, q <

1

2.

Similarly, with (1.3), we write

b(Nu,Mεu

ε

r,Mv

r) = b(

a2 − r2

a2Nu,

Mεuε

r,Mv

r),

and as above, we obtain

(6.22)

∣

∣

∣

∣

1

εb(Nu,

Mεuε

r,Mv

r)

∣

∣

∣

∣

≤ c1(ν, T, g, v0)ε1

2−q|curl 2Mv|0, q <

1

2.

This proves (6.18).

Now we write the formulation for Mεuε (defined on the fixed domain S2)

(6.23)d

dt(Mεu

ε, Mv)0 +ν

1 + ε(∆2Mεu

ε, Mv)0 +1

1 + εb0(Mεu

ε, Mεuε, Mv) = (Mf, Mv)0 + fn,


where fn satisfies, thanks to (6.17), (6.21) and (6.22)

(6.24) |fn| ≤ c0(T, ν, g, v0)ε1

2−q

n |curl 2Mv|0, limn→∞

fn = 0, q <1

2.

Taking into account (6.6)-(6.10) and (6.24), it is straightforward to pass to the limit in(6.23) (see Chapter III of [T1]). We have

(6.25)

d

dt(v∗, v)0 + ν(curl 2v

∗, curl 2v)0 + b0(v∗, v∗, v) = (g, v)0,

v∗(·, 0) = v0.

Finally, the uniqueness of solutions to (6.1)-(6.3) implies that v∗ = v.

Now we are ready to prove the following:

Theorem 6.1. Let (uε0)ε>0 ∈ Vε and (fε)ε>0 ∈ Hε(resp. L∞(0,∞; Hε)), and assume that

there exists g ∈ H0 and v0 ∈ H0, such that , for some q < 12 , we have limε→0 εq(||uε

0|| +|fε|) = 0, and

limε→0

Mεfε = g in H0-weak.

limε→0

Mεuε0 = v0 in H0 − weak.

Moreover, we assume that∣

∣curl 2Muε0

∣

∣

0is bounded. for 0 < ε ≤ 1. Then, for all T > 0,

there exists ε1 = ε1(g, v0, ν) such that

(6.26) limε→0

Mεuε = v in C([0, T ]; H0) ∩ L2(0, T ; V0).

where v is the unique solution to (6.1)-(6.3).

Proof. Thanks to the estimates derived above, the proof is now classical see [TZ] and [T1]for the details.

The limit of the radial equation and the hydrostatic equation.

Let Mv be a smooth scalar function on the unit sphere. It is easy to see that Mv/r ~er

belongs to Hε. Hence, setting

Muεr =

1

εa

∫ a+εa

a

ruεr(r, θ, ϕ) dr,

we have

d

dt

(

Muεr

r,Mv

r

)

ε

+ ν

(

curl 2Muε

r

r, curl 2

Mv

r

)

ε

+ b

(

Muε

r,Muε

r,Mv

r

)

=

(

Mfr

r,Mv

r

)

ε

+ ν

(

curl curl Nu,Mv

r

)

ε

− b

(

Muε

r, Nu,

Mv

r

)

− b

(

Nu,Muε

r,Mv

r

)

− b

(

Nu, Nu,Mv

r

)

.


Thanks to (6.15)-(6.19), we can show with the methods used above that the limit ur ofMuε

r as ε → 0 is solution of

(6.27)∂ur

∂t− curl 2curl 2ur +

1

a2|v|2ur = gr,

where v is the solution of (6.1)-(6.3) and gr is the limit of Mfεr .

In order to obtain the hydrostatic equation, we take f = −ρger where ρ is the densityand g is the gravity constant. Since f is not in Hε we absorb its gradient part in thepressure term as ∂

∂r (p + ρgr), using the Helmholtz decomposition of f.Now let Φ be a smooth scalar function in Ωε; we multiply equation (0.1) with NΦ~er.

We then pass to the limit in the resulting equation after dividing by ε. Using the resultsand methods above, we obtain thanks to (6.15)-(6.19) and (6.27), the hydrostatic equation

∂p

∂r= −ρg.

Remark 6.1. In order to obtain the primitive equations of the atmosphere, the linearizationof the spherical metric is used in [LTW1]. This amounts to replacing r2 sin θdθdϕdr bya2 sin θdθdϕdz, where z = r − a is the distance of a point to the surface of the earth; see[LTW1] and [P]. With the results obtained in this article and our previous results [TZ], thelinearization procedure can be fully justified at least for the incompressible Navier-Stokesequations.

A detailed discussion on both the hydrostatic approximation and the linearization ofthe metric will appear elsewhere. The Boussinesq approximation will be also discussed.

A. Appendix

We collect in this appendix some standard formulas from vector analysis. For ~A, ~B, ~Cvectors in R

3, we have

( ~A × ~B) × ~C = ( ~A · ~C) ~B − ( ~B · ~C) ~A( A.1)

~A × ( ~B × ~C) = ( ~A · ~C) ~B − ( ~A · ~B) ~C.

~A · ( ~B × ~C) = ( ~A × ~B) · ~C = ( ~C × ~A) · ~B = ~C · ( ~A × ~B)( A.2)

= ~B · ( ~C × ~A) = ( ~B × ~C) · ~A.

For Ψ a scalar function, and A and B vector fields, we have

(A.3) div (Ψ ~A) = ∇Ψ · ~A + Ψdiv ~A

(A.4) curl (Ψ ~A) = (∇Ψ) × ~A + Ψcurl ~A.


(A.5) curl curl ~A = −∆ ~A + ∇div ~A.

(A.6) div ( ~A × ~B) = ~B · (curl ~A) − ~A · curl ~B

(A.7)

∫

Ω

~B · curl ~Adx =

∫

Ω

~A · curl ~B +

∫

∂Ω

( ~A × ~B) · ~n dσ.

The Laplace-Beltrami operator of a scalar function Ψ, in spherical coordinates is ex-pressed as

(A.8) ∇2Ψ =∂2f

∂r2+

2

r

∂Ψ

∂r+

1

r2 sin θ

∂

∂θ( sin θ

∂Ψ

∂θ) +

1

r2 sin θ

∂2Ψ

∂ϕ2,

and its gradient is given by

(A.9) ∇Ψ =∂Ψ

∂r~er +

1

r

∂Ψ

∂θ~eθ +

1

r sin θ

∂Ψ

∂ϕ~eϕ.

We recall also the expressions of the curl , the divergence and the Laplacian of a vectorfield written in spherical coordinates, u = ur~er + uθ~eθ + uϕ~eϕ

curl u =1

r sin θ

[

∂

∂θ(sin θuϕ) − ∂uθ

∂ϕ

]

~er +

[

1

r sin θ

∂ur

∂ϕ− 1

r

∂

∂r(ruϕ)

]

~eθ(A.10)

+

[

1

r

∂

∂r(ruθ) −

1

r

∂ur

∂θ

]

~eϕ

and its divergence reads

(A.11) div u =1

r2

∂(r2ur)

∂r+

1

r sin θ

∂

∂θ(uθ sin θ) +

1

r sin θ

∂uϕ

∂ϕ.

The Laplacian of a vector field written in spherical coordinates is

(A.12)

(∆u)r = ∇2ur −2ur

r2− 2

r2

∂uθ

∂θ− 2

cot θ

r2uθ −

2

r2 sin θ

∂uϕ

∂ϕ,

(∆u)θ = ∇2uθ +2

r2

∂ur

∂θ− uθ

r2 sin 2θ− 2 cos θ

r2 sin 2θ

∂uϕ

∂ϕ,

(∆u)ϕ = ∇2uϕ − uϕ

r2 sin 2θ+

2

r2 sin θ

∂ur

∂ϕ+

2 cos θ

r2 sin 2θ

∂uθ

∂ϕ,

where ∇2ur, ∇2uθ and ∇2uϕ are as in (A.8).


Finally we recall the expressions of some tangential operators on the sphere S = Sa =x ∈ R

3, |x| = a centred at the origin, of radius r = a. We use the spherical coordinatesθ, ϕ as above

For p a scalar function defined on Sa, the tangential gradient is given by

(A.13) grad 2 p =1

a

∂p

∂θ~eθ +

1

a sin θ

∂p

∂ϕ~eϕ.

We also define curl 2p as a tangent vector function

(A.14) curl 2p = ∇p × ~er =1

a sin θ

∂p

∂ϕ~eθ −

1

a

∂p

∂θ~eϕ.

The Laplace-Beltrami of a scalar function p is

(A.15) ∆2 p =1

a2 sin θ

[

∂

∂θ

(

sin θ∂p

∂θ

)

+1

sin θ

∂2p

∂ϕ2.

For a tangent vector field v defined on Sa, v = vθ~eθ + vϕ~eϕ, the tangential divergenceis expressed by

(A.16) div 2 v =1

a sin θ

∂

∂θ(vθ sin θ) +

1

a sin θ

∂vϕ

∂ϕ,

and the curl 2v is the scalar function defined by

(A.17) curl 2v =1

a sin θ

∂

∂θ(vϕ sin θ) − 1

a sin θ

∂vθ

∂ϕ.

The tangential Laplacian of the vector field v is given by

(A.18)

∆2 v =

[

∆2 vθ −2 cos θ

a2 sin2 θ

∂vϕ

∂ϕ− vθ

a2 sin2 θ

]

~eθ

+

[

∆2 vϕ +2 cos θ

a2 sin2 θ

∂vθ

∂ϕ− vϕ

a2 sin2 θ

]

~eϕ,

where ∆2 vθ and ∆2 vϕ are as in (A.15).

Acknowledgment

This work was partially supported by the National Science Foundation under GrantNSF-DMS-9400615, by the Office of Naval Research under grant NAVY-N00014-96-1-0425,and by the Research Fund of Indiana University.


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