Quentin Vila

Time-Asymptotic Study of a Viscous Axisymmetric Fluid without Swirl

Keywords: fluid mechanics, incompressible Navier-Stokes equations, vorticity equation, longtime behaviour, asymptotic expansion, axisymmetry, axisymmetry without swirl

We study the long-time behaviour of axisymmetric solutions without swirl for the threedimensional Navier-Stokes equations in the whole space. Assuming that the initial vorticity is sufficiently localised, we compute explicitly the leading terms in the asymptotic expansion of the solution, both for the vorticity and the velocity field. In particular, we identify optimal temporal decay rates depending on the spatial localisation of the initial data. Our approach relies on accurate L p -L q estimates for the linearised evolution equation and its Taylor expansion in self-similar variables.

Introduction

In absence of nonconservative forces, the Navier-Stokes equations for homogeneous and incompressible fluids can be expressed in dimensionless variables as the system ∂ t u + (u.∇)u = -∇P + ∆u div u = 0 (1.1) whose unknown is the pair (u, P ), where u is the (dimensionless) velocity field and P the (dimensionless) pressure field of the fluid. The present paper supposes the studied fluid to be axisymmetric, meaning that the fluid is evolving in the three-dimensional real space in such a way that its motion is invariant by rotation around the vertical axis. Considering the cylindrical coordinates system (r, θ, z) and its associated basis

e r =    cos θ sin θ 0    , e θ =    -sin θ cos θ 0    , e z =    0 0 1    ,
it means that the cylindrical components of the velocity u r = u • e r , u θ = u • e θ and u z = u • e z are independant of θ: u r (t, x) = u r (t, r, θ, z) = u r (t, r, z) for all x = (r cos θ, r sin θ, z) ∈ R 3 and likewise for u θ and u z . The tangential component of the velocity field u θ is named the swirl. In this paper we also make the hypothesis that the swirl is identically equal to zero, so u(t, r, θ, z) = u r (t, r, z)e r + u z (t, r, z)e z .

To a certain extent, the axisymmetric fluid without swirl is comparable to a two-dimensional fluid admitting an axial symmetry.

The first mathematical results in axisymmetric fluid mechanics date back to 1968, when Ladyzhenskaya [START_REF] Aleksandrovna | Unique solvability in the large of the three-dimensional Cauchy problem for the Navier-Stokes equations in the presence of axial symmetry[END_REF] and Ukhovskii and Yudovich [START_REF] Ukhovskii | Axially symmetric flows of ideal and viscous fluids filling the whole space[END_REF] tackle the uniqueness problem of the threedimensional Navier-Stokes equations by adding this hypothesis. Comes later the study of Leonardi, Málek, Nečas and Pokorný [START_REF] Leonardi | On axially symmetric flows in R 3[END_REF], who show the global existence of a zero-swirl axisymmetric solution to (1.1) from any such initial data in H 2 (R 3 ), then the one of Abidi [START_REF] Abidi | Résultats de régularité de solutions axisymétriques pour le système de Navier-Stokes[END_REF] who improved the result for an initial data in H 1 2 (R 3 ). Follows finally the recent paper of Gallay and Šverák [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF], suggesting to work in some two-dimensional-like L p spaces equipped with scale-invariant norms.

The present article examines in detail what can be said about the long-time behaviour of the axisymmetric fluid when the initial vorticity is well localised. Using tools specific to axisymmetry without swirl, we determine the asymptotic expansions of the fluid's vorticity and velocity under essentially optimal hypotheses. The interest for the long-time behaviour of the Navier-Stokes flow in the litterature is substantial, and intensified in the end of the last century. Starting in the 1980's with the papers of Kato [START_REF] Kato | Strong L p -solutions of the Navier-Stokes equation in R m , with applications to weak solutions[END_REF], Schonbek [START_REF] Elena | L 2 Decay for Weak Solutions of the Navier-Stokes Equations[END_REF][START_REF] Elena | Large time behaviour of solutions to the Navier-Stokes equations[END_REF] and Wiegner [START_REF] Wiegner | Decay results for Weak Solutions of the Navier-Stokes Equations on R n[END_REF], a series of important works have addressed the time-decay problem for fluids in R 3 , or R n , among which can be cited [START_REF] Carpio | Asymptotic behavior for the vorticity equations in dimension two and three[END_REF][START_REF] Carpio | Large-time behavior in incompressible Navier-Stokes equations[END_REF][START_REF] Miyakawa | On optimal decay rates for weak solutions to the Navier-Stokes equations in R n[END_REF][START_REF] Fujigaki | Asymptotic Profiles of Nonstationary Incompressible Navier-Stokes Flows in the Whole Space[END_REF][START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF][START_REF] Brandolese | Asymptotic behavior of the energy and pointwise estimates for solutions to the Navier-Stokes equations[END_REF][START_REF] Brandolese | Space-time decay of Navier-Stokes flows under rotations[END_REF][START_REF] Okabe | Navier-Stokes flow in the weighted Hardy space with applications to time decay problem[END_REF][START_REF] Brandolese | Annihilation of slowly-decaying terms of Navier-Stokes flows by external forcing[END_REF]. For an extensive overview on the subject, we invite the reader to look at the introduction of the most recent ones [START_REF] Okabe | Navier-Stokes flow in the weighted Hardy space with applications to time decay problem[END_REF][START_REF] Brandolese | Annihilation of slowly-decaying terms of Navier-Stokes flows by external forcing[END_REF] and the references therein, or to refer to the chapter by Brandolese and Schonbeck in [START_REF]Handbook of Mathematical Analysis in Mechanics of Viscous Fluids[END_REF].

The hypothesis of axisymmetry without swirl encourages the consideration of the vorticity ω, which is the curl of the velocity ω = rot(u), as the core of the problem. Indeed, in this context, the vorticity boils down to a real-valued function ω θ by writing that ω = ω θ e θ where ω θ = ∂ z u r -∂ r u z . Given the Navier-Stokes equations (1.1) and the zero-swirl axisymmetry hypothesis, ω θ can be found as the solution of the equation

       ∂ t ω θ + u.∇ω θ - u r r ω θ = ∆ω θ - ω θ r 2 on ]0, +∞[×Ω ω θ (t, 0, z) = 0 ∀(t, 0, z) ∈ [0, +∞[×∂Ω ω θ (0, r, z) = ω 0 (r, z) ∀(r, z) ∈ Ω (1.2)
where Ω is the open half plane Ω = ]0, +∞[×R and where ω θ = 0 at r = 0 due to the regularity of u. Here, in cylindrical coordinates, u • ∇ = u r ∂ r + u z ∂ z and ∆ = ∂ 2 r + 1 r ∂ r + ∂ 2 z . One notable advantage of working with the vorticity is that the pressure P is not present any more in the equation, yet no generality is lost thanks to the Biot-Savart law. Deeper discussions about the relevance to work in vorticity can also be found in [START_REF] Gallay | Invariant Manifolds and the Long-Time Asymptotics of the Navier-Stokes and Vorticity Equations on R 2[END_REF][START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF].

Following [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF], our study will be carried out using the scale-invariant norms

w L p (Ω) = Ω |w(r, z)| p drdz 1 p if 1 p < ∞ ; w L ∞ (Ω) = ess sup Ω |w|,
according to the two-dimensional Lebesgue measure drdz on the half plane Ω instead of the threedimensional Lebesgue measure on R 3 . Up to a multiplicative factor, w L p (Ω) is the norm of (r, z) → r -1 p w(r, z) in the usual L p (R 3 , rdrdθdz) space. Within this framework, it is known thanks to [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF] that the zero-swirl axisymmetric problem (1.2) is well-posed for any initial data ω 0 in L 1 (Ω) = L 1 (Ω, drdz) and that moreover, unless ω 0 is trivial, the L 1 -norm of the solution ω θ (t) L 1 (Ω) = Ω |ω θ (t, r, z)|drdz is a decreasing function of time and tends to zero as t tends to infinity. We here go even further, naming I(t) = Ω r 2 ω θ (t, r, z) drdz and J(t) = Ω r 2 z ω θ (t, r, z) drdz (1.3) then stating the following theorem.

Theorem 1.1. Consider ω θ the solution of (1.2) with initial data ω 0 ∈ L 1 (Ω, drdz). If r 2 ω 0 L 1 (Ω) < ∞ then I(t) is constantly equal to I 0 = I(0) and the relation Remarks.

ω θ (t,
1. Theorem 1.1 is the concatenation of the two central results of this paper, propositions 3.5 and 3.9, and of remark 3.10. [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF], but with the additional hypothesis that ω 0 is nonnegative. Let us note especially that (1.4) gives for p = 1 the general estimate ω θ (t) L 1 (Ω) C ω0 t -1 , where C ω0 is a positive constant that depends only on the initial data ω 0 .

Expansion (1.4) already appears in

3. Expansion (1.5) is a new result for zero-swirl axisymmetry. Its terms could have been obtained from the general three-dimensional study developed in [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF], but (1.5) is here deduced from much weaker assumptions thanks to the specific context of zero-swirl axisymmetry. We discuss this point more precisely in appendix A. [START_REF] Brandolese | Space-time decay of Navier-Stokes flows under rotations[END_REF]. The technical lemmas that lead to theorem 1.1 give more information than just showing the terms of the asymptotic expansion: they provide estimates on the L p -norms of the vorticity for all the intermediate hypotheses on its initial moments. Indeed, proposition 3.4 treats the cases where ω 0 ∈ L 1 (Ω) is such that r α ω 0 L 1 (Ω) < ∞ with 0 α 2, proposition 3.6 the cases where moreover r α (r 2 + r|z| + |z|)ω 0 L 1 (Ω) < ∞, 0 < α 1, proposition 3.11 the cases where r α (r 3

+ rz 2 + z 2 )ω 0 L 1 (Ω) < ∞, 0 < α < 1,
and proposition 4.1 treats the case where (r 4 + r 2 z 2 + rz 2 )ω 0 L 1 (Ω) < ∞.

These increasingly precise hypotheses on the finite moments of ω 0 are the right amount of localisation to ask in order to refine our observation of the vorticity's behaviour. This echoes the well-known fact that there exists a link between the spatial and the temporal decays at infinity, for both the velocity (cf [START_REF] Elena | Large time behaviour of solutions to the Navier-Stokes equations[END_REF][START_REF] Carpio | Large-time behavior in incompressible Navier-Stokes equations[END_REF][START_REF] Miyakawa | On optimal decay rates for weak solutions to the Navier-Stokes equations in R n[END_REF][START_REF] Fujigaki | Asymptotic Profiles of Nonstationary Incompressible Navier-Stokes Flows in the Whole Space[END_REF][START_REF] Brandolese | Asymptotic behavior of the energy and pointwise estimates for solutions to the Navier-Stokes equations[END_REF][START_REF] Brandolese | Space-time decay of Navier-Stokes flows under rotations[END_REF][START_REF] Okabe | Navier-Stokes flow in the weighted Hardy space with applications to time decay problem[END_REF]) and the vorticity (cf [START_REF] Gallay | Invariant Manifolds and the Long-Time Asymptotics of the Navier-Stokes and Vorticity Equations on R 2[END_REF][START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF]). Theorem 1.1 has a translation in terms of velocity. When ω θ is solution of (1.2), u is the solution of the system

                 ∂ t u + (u • ∇)u = -∇P + ∆u on ]0, +∞[×R 3 ∂ r u r + 1 r u r + ∂ z u z = 0 on ]0, +∞[×R 3 u r (0, r, θ, z) = u 0 (r, z) • e r ∀(r, z) ∈ Ω, ∀θ ∈ [0, 2π[ u θ (0, r, θ, z) = 0 ∀(r, z) ∈ Ω, ∀θ ∈ [0, 2π[ u z (0, r, θ, z) = u 0 (r, z) • e z ∀(r, z) ∈ Ω, ∀θ ∈ [0, 2π[ (1.6)
subject to u 0 • e θ = 0, which ensures that u remains axisymmetric with zero swirl at all times, as explained in [START_REF] Joseph | Vorticity and Incompressible Flow[END_REF]. Let us introduce the vector fields

u G1 = - 1 r ∂ z ϕ e r + 1 r ∂ r ϕ e z and u G2 = -∂ z u G1 (1.7)
where ϕ is the Stokes stream function of

G 1 : (r, z) → r 16 √ π exp(-r 2 +z 2 4 ), ϕ(r, z) = - r 4 √ π ∂ r 1 √ r 2 + z 2 √ r 2 +z 2 0 e -σ 2 4 dσ , (1.8)
and let us adapt notations (1.3) for the velocity: I 0 = 1 2π R 3 r rot(u 0 (r, z)) rdrdθdz and J ∞ is the finite limit of J(t) = 1 2π R 3 rz rot(u(t, r, z)) rdrdθdz when t → +∞. The study carried out through this paper leads then to the following theorem, giving information on the velocity's decay rate in L p (R 3 ) 3 for 1 < p ∞. Theorem 1.2. Consider u the solution of (1.6) with initial data u 0 such that (r, z)

→ 1 r rot(u 0 (r, z)) is in L 1 (R 3 ). If r rot(u 0 ) L 1 (R 3 ) < ∞ then u(t, r, z) = I 0 u G1 r √ t , z √ t t -3/2 + o t→+∞ t 3 2p -3 2 (1.9) holds in L p (R 3 ) 3 for every p ∈ ]1, +∞]. If moreover (r 2 + r|z| + |z|) rot(u 0 ) L 1 (R 3 ) < ∞, then u(t, r, z) = I 0 u G1 r √ t , z √ t t -3/2 + J ∞ u G2 r √ t , z √ t t -2 + o t→+∞ t 3 2p -2
(1.10)

holds in L p (R 3 ) 3 for every p ∈ ]1, +∞].
This theorem gives for example another insight into the paper of Miyakawa and Schonbek [START_REF] Miyakawa | On optimal decay rates for weak solutions to the Navier-Stokes equations in R n[END_REF] in the case of axisymmetry without swirl, namely the fact that, taking p = 2, the solutions to the Navier-Stokes system decreasing faster than t → t -5/4 are exactly those such that I 0 = J ∞ = 0. Naming

E = 1 r rot(u 0 ) ∈ L 1 (R 3 ) (r 2 + r|z| + |z|) rot(u 0 ) L 1 (R 3 ) < ∞, I 0 = J ∞ = 0
the set of all such zero-swirl axisymmetric solutions (to be exact, the set of all initial vorticity data corresponding to such solutions), one can show by following the developments in [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF] that the space E is a smooth invariant submanifold of codimension 2 of the weighted space {w ∈ L

1 (R 3 ) | ( 1 r + r 2 + r|z| + |z|)w L 1 (R 3 ) < ∞}.
This paper is written as follows. The second section analyses the semigroup associated with the linearised vorticity equation and its behaviour at large times. We give step by step the first two terms of its expansion, together with new L p -L q estimates on the successive remainders. Considering the integral version of (1.2), these technical estimates will be fundamental to deal with both the linear and the non-linear components of the vorticity. The third section initiates the asymptotic study of the vorticity until its second order, following quite the same steps as for the semigroup. Adding increasingly strong hypotheses on the localisation of the initial data ω 0 , we refine the bound on the decay rate of ω θ . The final section discusses the consequences of this study on different levels. After speaking briefly about the emergence of the first resonant term, we state some corollary on the velocity's decay rate and prove theorem 1.2. We then write the first two asymptotics when the swirl is nonzero. In the appendices, we develop some argument proving the local existence of the moments of ω θ then we establish a specific Grönwall lemma, both of them will be used several times throughout section 3.

Notations

B will be reserved in this article to denote the Euler Beta function. Remember that, in particular, B(1/2, 1/2) = π. For more details upon the properties of the B function, see for example [START_REF] William | Asymptotics and Special Functions[END_REF].

C will denote any generic positive constant. Constants named C α or C ω0 will be constants depending in their contexts on an α parameter or on the initial data ω 0 of the vorticity equation (1.2).

L p (Ω) will denote the Lebesgue space L p (Ω, drdz) with respect to the two dimensional Lebesgue measure on Ω = R * + × R. L p (R 3 ), on the contrary, will refer to the usual L p (R 3 , rdrdθdz) space with the three-dimensional Lebesgue measure. div * will be the pseudo-divergence operator acting on a function w = w r e r +w z e z (in cylindrical coordinates) as div * w = ∂ r w r + ∂ z w z .

The Semigroup associated with the linearised equation

Before anything, some crucial results are useful to have in mind. Focus on (1.2) and remark that the equation can also be expressed as

       ∂ t ω θ + div * (u ω θ ) = ∆ω θ - ω θ r 2 on ]0, +∞[×Ω ω θ (t, 0, z) = 0 ∀(t, 0, z) ∈ [0, +∞[×∂Ω ω θ (0, r, z) = ω 0 (r, z) ∀(r, z) ∈ Ω (2.1) by writing div * (u ω θ ) = ∂ r (u r ω θ ) + ∂ z (u z ω θ ) and using that div u = ∂ r u r + 1 r u r + ∂ z u z = 0. The differential operator ∆ -1
r 2 is the generator of a semigroup (S(t)) t 0 of bounded linear operators on L p (Ω, drdz) which is strongly continuous when 1 p < ∞ (the calculation can be found in [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF]), defined by S(0) = id and

S(t)[ω 0 ] : (r, z) -→ r 4πt 5/2 Ω K rρ t e -(r-ρ) 2 +(z-ζ) 2 4t ρ 2 ω 0 (ρ, ζ) dρdζ (2.2)
for any t > 0, where

K(τ ) = 1 τ √ 4π π -π e -τ sin 2 ( φ 2 ) cos(φ)dφ. (2.
3)

The properties of the function K will be discussed in the next subsection. The Duhamel formula says then that (2.1) is equivalent to the integral equation

ω θ (t) = S(t)ω 0 - t 0 S(t -s) [div * (u(s) ω θ (s))] ds t 0, (2.4)
to be still considered on the open half-plane Ω = {(r, z) ∈ R 2 | r > 0} and with the Dirichlet boundary condition ω θ = 0 at r = 0. Theorem 1.1 from [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF] tells that for an initial data ω 0 in L 1 (Ω), this equation admits one unique global solution

ω θ ∈ C 0 ([0, +∞[, L 1 (Ω)) ∩ C 0 (]0, +∞[, L ∞ (Ω))
which is infinitely differentiable on ]0, +∞[×Ω. Because the solution of (2.4) is unique, it also satisfies for every t 0 > 0 the relation

ω θ (t) = S(t -t 0 )ω θ (t 0 ) - t t0 S(t -s) div * (u(s) ω θ (s)) ds ∀t t 0 . (2.5)
Indeed, following for a duration t -t 0 the solution stemming from ω θ (t 0 ), which exists, one necessarily gets back to the solution ω θ at time t. Equation (2.4) consists of two very different terms to deal with, one linear and the other nonlinear, both of which involving the semigroup (S(t)) t 0 . Having a good idea as to how the semigroup behaves at infinity is a key point to the asymptotic study of ω θ .

The Semigroup kernel

The function K defined by (2.3) has the following properties. Proposition 2.1. K is infinitely differentiable on [0, +∞[ and the functions τ → (1+τ

) 3/2+i K (i) (τ ) are all in L ∞ ([0, +∞[) for i ∈ {0, 1, 2, 3}.
This point will be crucial for propositions 2.3, 2.4 and 2.7, which give different kinds of estimates on the semigroup (S(t)) t 0 .

Proof. Let us rewrite K as

K(τ ) = √ π τ e -τ /2 I 1 τ 2 (2.6)
where

I n (τ ) = 1 π π 0 e τ cos φ cos(nφ)dφ (2.7)
denotes the modified Bessel function of order n, n ∈ N. The desired properties of K stem from the properties of I n for n ∈ {1, 2, 3, 4}.

To begin with, writing the exponential in (2.7) as a series and exchanging the series with the integral shows that I 1 can be expressed as

I 1 (τ ) = k 0 a k (2k+1)! τ 2k+1 where a k = 1 2 × 3 4 ×...× 2k+1 2k+2 . Hence K is infinitely differentiable on [0, +∞[.
Let us then look at the behaviour of K, K , K and K at infinity. The asymptotic expansions of the modified Bessel functions are given by

I n (τ ) = 1 √ 2πτ e τ 1 - (4n 2 -1) 8τ + (4n 2 -1)(4n 2 -9) 128τ 2 + O τ -3 as τ → +∞, (2.8)
see for example the book of Franck Olver [22, chapter 7], and in particular

I 1 (τ ) = 1 √ 2πτ e τ (1 + O(τ -1
)) so K(τ ) ∼ τ -3/2 at infinity. Moreover, one gets by differentiating (2.7) that I n (τ ) = I n+1 (τ ) + n τ I n (τ ) so successively

I 1 (τ ) = I 2 (τ ) + 1 τ I 1 (τ ), I 1 (τ ) = I 3 (τ ) + 3 τ I 2 (τ ), and I 1 (τ ) = I 4 (τ ) + 6 τ I 3 (τ ) + 3 τ 2 I 2 (τ ).
Calculating the derivatives of K from (2.6) involves these three last formulae, which combined with (2.8) lead finally to

K (i) (τ ) = O(τ -3/2-i ) as τ → +∞ for i ∈ {1, 2, 3}.

Estimates on the semigroup

Let us write some L p (Ω) -L q (Ω) estimates on the moments of the semigroup (S(t)) t 0 . Some of them already exist in [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF], and can be stated the following way.

Proposition 2.2. Choose 1 p q ∞ and -1 α β 2.

(i) There exists a positive constant C such that for all time t > 0

r α S(t)ω 0 L q (Ω) C r β ω 0 L p (Ω) × t α-β 2 + 1 q -1 p (2.9)
provided that (r, z) → r β ω 0 (r, z) belongs to L p (Ω).

(ii) If -1 α β 1, then there exists a positive constant C such that for all time t > 0

r α S(t)div * w L q (Ω) C r β w L p (Ω) × t -1 2 + α-β 2 + 1 q -1 p (2.10) provided that (r, z) → r β w(r, z) belongs to (L p (Ω)) 2 .
The constants in (2.9) and (2.10) are actually independent of p, q, α, β and ω 0 . The constraints -1 α β 2 and -1 α β 1 suggested respectively for (2.9) and (2.10) are optimal as long as one wishes to get time-homogeneous bounds. Proposition 2.2 is sufficient to deal with the existence, uniqueness and regularity to (2.1). Proposition 2.2 is sufficient as well to give the decay rates of the vorticity when the initial data ω 0 is such that rω 0 is in L 1 (Ω), using inequality (2.10) up to its limit β = 1 (this will be the object of lemma 3.2). However, in order to go further into the asymptotic expansion of ω θ , we shall need estimates on the r-moments of S(t) higher than 2 as well as some of its r-and-z-moments that proposition 2.2 does not handle.

In practice, the following inequalities will be required in the next section; some concerning the semigroup strictly speaking, like in (2.9),

r α z S(t)ω 0 L q (Ω) C r α (r + |z|) ω 0 L p (Ω) × t 1 q -1 p , (2.11) r 2+α S(t)ω 0 L q (Ω) C r 2 (t α 2 + r α ) ω 0 L p (Ω) × t 1 q -1 p , (2.12
)

r α z 2 S(t)ω 0 L q (Ω) C (r 2 t α 2 + r α z 2 ) ω 0 L p (Ω) × t 1 q -1 p , (2.13)
and the others concerning the semigroup acting upon the div * operator, like in (2.10),

r 2 S(t)div * w L q (Ω) C r( √ t + r)w L p (Ω) × t -1 2 + 1 q -1 p , (2.14)
r α z S(t)div * w L q (Ω) C (r t α 2 + r α |z|) w L p (Ω) × t -1 2 + 1 q -1 p , (2.15)
r 2+α S(t)div * w L q (Ω) C (r t 1+α 2 + r 2+α ) w L p (Ω) × t -1 2 + 1 q -1 p , (2.16
)

r α z 2 S(t)div * w L q (Ω) C (r t 1+α 2 + r α z 2 ) w L p (Ω) × t -1 2 + 1 q -1 p (2.17)
where α lies in [0, 1]. These inequalities can be proven simultaneously as consequences of the following statement.

Proposition 2.3. Choose 1 p q ∞ and α , γ 0. Define for t ∈ ]0, +∞[ and (r, z) ∈ Ω

N (t, r, z) = r √ t β + r √ t β z √ t γ 1 + r √ t α .
(2.18)

(i) If -1 α β, β 2
, there exists a positive constant C α ,γ such that for all time t > 0

r α+α z γ S(t)ω 0 L q (Ω) C α ,γ N (t)ω 0 L p (Ω) × t α+α +γ 2 + 1 q -1 p (2.19) provided that (r, z) → (r β + r β+α + r β |z| γ + r β +α |z| γ )ω 0 (r, z) ∈ L p (Ω).
(ii) If -1 α β, β 1, there exists a positive constant C α ,γ such that for all time t > 0

r α+α z γ S(t)div * w L q (Ω) C α ,γ N (t)w L p (Ω) × t -1 2 + α+α +γ 2 + 1 q -1 p (2.20) provided that (r, z) → (r β + r β+α + r β |z| γ + r β +α |z| γ )w(r, z) ∈ (L p (Ω)) 2 .
If one wishes to know explicitly how the constant in (2. [START_REF] Joseph | Vorticity and Incompressible Flow[END_REF]) and (2.20) depends on α and γ, they may write the optimal majorations ∀δ > 0, ∀x 0 and conclude while reading the proof coming below that C α ,γ can be taken as

(1 + x) δ max(1, 2 δ-1 ) (1 + x δ ) (2.
C α ,γ = C 0 × max(1, 2 α -1 ) max(1, 2 γ-1 ) max 1, 2α e α /2 max 1, 2γ e γ/2
, where C 0 is determined by the bounds on K and its derivatives (cf proposition 2.1), by the value of exp(-1 8 (r 2 + z 2 )) L 1 (Ω) = 4π, and by the maximum value of 2δ e δ/2 for δ between 0 and 5.

Proof. Let us write from (2.2) that for all time t > 0

r α+α z γ S(t)ω 0 (r, z) = 1 4πt 5/2 Ω r α z γ e -X r α+1 ρ 2 K rρ t ω 0 (ρ, ζ) dρdζ where we note X = X(t, r -ρ, z -ζ) = 1 4t ((r -ρ) 2 + (z -ζ) 2
) for the clarity of the demonstration.

Using an integration by part, write as well that, for all time t > 0,

r α+α z γ S(t)[div * w](r, z) = - 1 4πt 3 Ω r α z γ e -X r α (A r w r + A z w z ) dρdζ
where w = (w r , w z ) and the terms A r and A z have the following expressions:

A r (t, r, z, ρ, ζ) = rρ 2 2 √ t (r -ρ)K rρ t + r 2 ρ 2 2 √ t K rρ t + 2rρ √ tK rρ t and A z (t, r, z, ρ, ζ) = rρ 2 2 √ t (z -ζ)K rρ t .
In both cases, cut the factor r α z γ e -X into four parts according to (2.21) then (2.22):

r α |z| γ C(|r -ρ| α + ρ α )(|z -ζ| γ + |ζ| γ ) C(t α 2 + ρ α )(t γ 2 + |ζ| γ )e X 4 .
Afterwards, thanks to proposition 2.1 and (2.22), let us show that the factor r α+1 ρ 2 K( rρ t ) is less than or equal to the quantity

C( ρ √ t ) β e X 4 t α+3 2
as follows.

• If (α + β -1 0 and 0 r 2ρ) or (α + β -1 0 and 0 ρ < 2r) then

r α+1 ρ 2 K rρ t = r α+β-1 2 ρ β-α+1 2 t β-α-3 2 × rρ t 3+α-β 2 K rρ t C ρ √ t β e X 4 t α+3 2
since 0 3 + α -β 3 (cf proposition 2.1) and applying (2.21) with δ = 0.

• If α + β -1 0 and 0 2ρ r then • If finally α + β -1 0 and 0 2r ρ then

r α+1 ρ 2 K rρ t = r α+β-1 ρ β t 2β-4 2 × rρ t 2-β K rρ t C ρ √ t β e X 4 t α+3
r α+1 ρ 2 K rρ t = ρ β ρ -(α+β-1) t -2α-2 2 × rρ t α+1 K rρ t C ρ √ t β e X 4 t α+3 2 since 0 α + 1 3 2 and ρ |ρ -r| + r |ρ -r| + ρ 2 so ρ 2|ρ -r| 4 √ t X. Taking β instead of β gives that r α+1 ρ 2 K( rρ t ) C( ρ √ t ) β e X 4 t α+3 2
for -1 α β 2 as well. Let us treat the factors r α |A r | and r α |A z | in the same way when -1 α β 1, noticing that they are sums of terms of the form r α+1+i ρ 1+j X k/2 t 1-i-j 2

K (i) ( rρ t ) where i, j, k are in {0, 1}. • If (α + β + i -j 0 and 0 r 2ρ) or (α + β + i -j 0 and 0 ρ < 2r) then r α+1+i ρ 1+j t i+j-1 2 X k 2 K (i) rρ t = r α+β+i-j 2 ρ β-α-i+j 2 t β-α-3 2 X k 2 × rρ t 2+α+i+j-β 2
K (i) rρ t C ρ √ t β e X 4 t α+3 2 since 0 2 + α + i + j -β 3 + 2i.
• If α + β + i -j 0 and 0 2ρ r then

r α+1+i ρ 1+j t i+j-1 2 X k 2 K (i) rρ t = r α+β+i-j ρ β t 2β+i-j-3 2 X k 2 × rρ t 1+j-β K (i) rρ t C ρ √ t β e X 4 t α+3 2 since 0 1 + j -β 3+2i 2
and r |r -ρ| + ρ |r -ρ| + r 2 so r 2|r -ρ| 4 √ t X.

• If finally α + β + i -j 0 and 0 2r ρ then

r α+1+i ρ 1+j t i+j-1 2 X k 2 K (i) rρ t = ρ β ρ -(α+β+i-j) t -2α-i+j-3 2 X k 2 × rρ t α+i+1 K (i) rρ t C ρ √ t β e X 4 t α+3 2 since 0 α + i + 1 3+2i 2 and ρ |ρ -r| + r |ρ -r| + ρ 2 so ρ 2|ρ -r| 4 √ t X.
Let us take β instead of β and get that

r α |A r | + r α |A z | C( ρ √ t ) β e X 4 t α+3 2
as well. We have now shown that on the one hand

|r α+α z γ S(t)ω 0 (r, z)| C α ,γ 4π Ω e -X/2 N (t, ρ, ζ) |ω 0 (ρ, ζ)| dρdζ × t α+α +γ 2 -1
and on the other hand

|r α+α z γ S(t)div * w(r, z)| C α ,γ 4π Ω e -X/2 N (t, ρ, ζ) |w(ρ, ζ)| dρdζ × t -1 2 + α+α +γ 2 -1 ,
which concludes the proof thanks to Young's inequality for convolution.

Let us remark eventually that all of the estimates (2.11) to (2.17) are indeed obtained from proposition 2.3 with well choosen values of α, α , β, β and γ. Incidentally, let us see that proposition 2.2 is also contained in proposition 2.3, corresponding to the cases when β = β and α = γ = 0.

Approximation of the semigroup at first order

The goal is now to specify the behaviour of the semigroup (S(t)) t 0 at large times. We know from (2.9) that S(t)ω 0 L 1 (Ω)

C r β ω 0 L 1 (Ω) t -β/2 as long as β ∈ [0, 2]. In particular, when (r, z) → r 2 ω 0 (r, z) is in L 1 (Ω), it appears that the equivalent

S(t)ω 0 ∼ t→+∞ I 0 r 16 √ π e -r 2 +z 2 4t t -5/2 (2.23)
holds whenever I 0 = Ω r 2 ω 0 (r, z) drdz = 0. This will be the object of lemma 2.5. Hence, as long as I 0 is nonzero, t → S(t)ω 0 decreases in L 1 (Ω) exactly as t → t -1 up to a multiplicative constant. One thing to notice here is the preponderant role of the radial coordinate in axisymmetry, for until now all the hypotheses made relate to the r-moments of ω 0 and no additional spatial decay has been asked so far along the vertical axis.

Let us introduce the Gaussian-like function

G 1 (r, z) = r 16 √ π e -r 2 +z 2 4
(2.24)

on Ω and the notation

S 1 (t)ω 0 : (r, z) → S(t)ω 0 (r, z) - Ω ρ 2 ω 0 (ρ, ζ) dρdζ G 1 r √ t , z √ t t -2 (2.25)
when (1 + r 2 )ω 0 L 1 (Ω) < ∞ and t > 0. Despite its denomination, chosen because S 1 gives the first order approximation of S at large times, the family of operators (S 1 (t)) t>0 is not a semigroup and is not strongly continuous at t = 0. We now state some estimates on (S 1 (t)) t>0 comparable to those from proposition 2.3.

Proposition 2.4. Choose 1 p q ∞ such that p < 2 and -1 α β 2.

(i) If 3 -2 p < β 2 then there exists a positive constant C such that for all time t > 0

r α S 1 (t)ω 0 L q (Ω) C 1 p + β+1 2 -2 r β (r + |z|) ω 0 L p (Ω) × t -1 2 + α-β 2 + 1 q -1 p (2.26) provided that (r, z) → r 2 ω 0 (r, z) ∈ L 1 (Ω) and (r, z) → r β (r + |z|)ω 0 (r, z) ∈ L p (Ω).
(ii) If 2 -2 p < β 1 then there exists a positive constant C such that for all time t > 0

r α S 1 (t)div * w L q (Ω) C 1 p + β+2 2 -2 r β (r + |z|) w L p (Ω) × t -1+ α-β 2 + 1 q -1 p (2.27) provided that (r, z) → r 2 div * w(r, z) ∈ L 1 (Ω) and (r, z) → r β (r + |z|)w(r, z) ∈ (L p (Ω)) 2 .
The constants in (2.26) and (2.27) can be shown to be independent of p, q, α, β and ω 0 . The proof follows the same procedure than the one of proposition 2.3, with the difference that we shall conduct this one in the self-similar space-and-time coordinates (r √ t, z √ t).

Proof. Let us write for all t > 0 and (r, z), (ρ, ζ) ∈ Ω the first order asymptotic expansion with integral remainder

K rρ √ t e -X(t) = √ π 4 e -1 4 (r 2 +z 2 ) + +∞ t 1 2s 3/2 A 0 e -X(s) ds (2.28)
where we shall use the notation

X(t) = 1 4 ((r -ρ √ t ) 2 +(z -ζ √ t ) 2
) and where A 0 denotes the quantity

A 0 = rρK rρ √ s + 1 2 ρ r - ρ √ s + ζ z - ζ √ s K rρ √ s . (2.29)
From that expansion and the expression of the semigroup (2.2) in the self-similar coordinates (r √ t, z √ t), one gets that on the one hand

r α S 1 (t)ω 0 (r √ t, z √ t) = 1 8πt 2 Ω +∞ t 1 s 3/2 r α+1 ρ 2 A 0 e -X(s) ω 0 (ρ, ζ) ds dρdζ (2.30)
for all time t > 0, and on the other hand after integrating by parts

r α S 1 (t)[div * w](r √ t, z √ t) = - 1 8πt 2 Ω +∞ t 1 s 2 r α (A r w r + A z w z )e -X(s) ds dρdζ (2.31)
where w = (w r , w z ) and

A r = rρ 2 2 √ s r -ρ √ s A 0 + ∂ ρ (rρ 2 A 0 ) √ s A z = rρ 2 2 √ s z -ζ √ s A 0 + ∂ ζ (rρ 2 A 0 ) √ s.
(2.32)

The rest of the demonstration uses the arguments already encountered in the proof of proposition 2.3. The calculation gives that rρ 2 A 0 , A r and A z are sums of terms of the form

r 1+i ρ 1+j s j-1 2 ρ k r - ρ √ s k ζ 1-k z - ζ √ s k K (i) rρ √ s with j, k ∈ {0, 1} and i, k , k ∈ {0, 1, 2}.
It is important to note here that the terms with j = 0 appear only in A r and A z , when β 1. Then see that according to (2.22)

ρ k r - ρ √ s k |ζ| 1-k z - ζ √ s k (ρ + |ζ|) r - ρ √ s k z - ζ √ s k C(ρ + |ζ|)e X(s) 4
for every k ∈ {0, 1} and k , k ∈ {0, 1, 2}.

Afterwards, thanks to proposition 2.1 and to (2.22), let us show that the factors of the form as follows. For i ∈ {0, 1, 2} and j ∈ {0, 1} corresponding to the factors of interest, meaning that j = 0 only when β 1,

r α × r 1+i ρ 1+j s 1-j 2 K (i) ( rρ
• if (α + β + i -j 0 and 0 r 2 ρ √ s ) or (α + β + i -j 0 and 0 ρ √ s < 2r) then r α+1+i ρ 1+j s j-1 2
K (i) rρ √ s = r α+β+i-j 2 ρ β-α-i+j 2 s β-α-i+j-4 4 × rρ √ s 2+α+i+j-β 2 K (i) rρ √ s Cρ β s 2-β 2 e X(s) 4 since 0 2 + α + i + j -β 3 + 2i, • if α + β + i -j 0 and 0 2 ρ √ s r then r α+1+i ρ 1+j s j-1 2
K (i) rρ √ s = r α+β+i-j ρ β s β-2 2 × rρ √ s 1+j-β K (i) rρ √ s Cρ β s 2-β 2 e X(s) 4 since 0 1 + j -β 3+2i 2
and r |r -

ρ √ s | + ρ √ s |r -ρ √ s | + r 2 so r 2|r -ρ √ s | 4 X(s),
• and if α + β + i -j 0 and 0 2r

ρ √ s then r α+1+i ρ 1+j s j-1 2
K (i) rρ √ s = ρ β s β-2 2 ρ √ s -(α+β+i-j) × rρ √ s α+i+1 K (i) rρ √ s Cρ β s 2-β 2 e X(s) 4 since 0 α+i+1 3+2i 2 and ρ √ s | ρ √ s -r|+r | ρ √ s -r|+ ρ 2 √ s so ρ √ s 2| ρ √ s -r| 4 X(s).
It is only left to write, using first a Fubini inversion between the time and space integrals and then Young's inequality for convolution, that on the one hand

r α S 1 (t)ω 0 (r √ t, z √ t) L q (Ω) C t 2 Ω +∞ t s -β+1 2 e -X(s) 2 ρ β (ρ + |ζ|) |ω 0 (ρ, ζ)| ds dρdζ L q (Ω) C t 2 +∞ t s -β+1 2 Ω e -X(s) 2 ρ β (ρ + |ζ|) |ω 0 (ρ, ζ)| dρdζ L q (Ω) ds C t 2 +∞ t s -β+1 2 +1-1 p ds r β (r + |z|)ω 0 (r, z) L p (Ω)
and on the other hand

r α S 1 (t)div * w(r √ t, z √ t) L q (Ω) C t 2 Ω +∞ t s -β+2 2 e -X(s) 2 ρ β (ρ + |ζ|) |w(ρ, ζ)| ds dρdζ L q (Ω) C t 2 +∞ t s -β+2 2 Ω e -X(s) 2 ρ β (ρ + |ζ|) |w(ρ, ζ)| dρdζ L q (Ω) ds C t 2 +∞ t s -β+2 2 +1-1 p ds r β (r + |z|)w(r, z) L p (Ω) ,
and proof is made with a last change of variables on the left hand sides of these two last inequalities.

We can now prove the statement made in (2.23), which is true in every L p (Ω) for 1 p ∞. The demonstration will be performed again in the self-similar coordinates (z √ t, z √ t).

Lemma 2.5. Take ω 0 ∈ L 1 (Ω) such that (r, z) → r 2 ω 0 (r, z) ∈ L 1 (Ω). Remember that (S(t)) t 0 is the semigoup defined by (2.2) and I 0 = Ω r 2 ω 0 (r, z) drdz. Then

S(t)ω 0 (r, z) = I 0 r 16 √ π e -r 2 +z 2 4t t -5/2 + o t→+∞ t 1 p -2 (2.33) in L p (Ω) for every p ∈ [1, +∞]. Proof. Let us recall that G 1 : Ω → R : (r, z) → r 16 √ π exp(-r 2 +z 2 4
) and S 1 :

ω 0 → S(t)ω 0 - I 0 G 1 ( • √ t )t -2 . The goal is to show that t 2 S 1 (t)ω 0 (• √ t) L p (Ω)
tends to zero as t goes to infinity for p in [1, +∞].

Let us consider for any a > 1 the indicator function χ a of the truncated subset [ 1 a , a] × [-a, a] of Ω, and the quantity

I a 0 = a 1/a a -a ρ 2 ω 0 (ρ, ζ) dρdζ = Ω ρ 2 χ a (ρ, ζ)ω 0 (ρ, ζ
) dρdζ which tends towards I 0 as a goes to infinity. Thanks to estimate (2.26) with α = 0 and β = 2, we see that

t 2 S 1 (t)[χ a ω 0 ](• √ t) L p (Ω) = t 2-1 p S 1 (t)[χ a ω 0 ] L p (Ω) (2.26) 2C r 2 (r + |z|)χ a ω 0 L 1 (Ω) × t -1 2 ----→ t→+∞ 0.
We see as well that

t 2-1 p S(t)[(1 -χ a )ω 0 ] L p (Ω) C r 2 ω 0 (1 -χ a ) L 1
(Ω) using (2.9) with α = 0 and β = 2, then writing

t 2 S 1 (t)ω 0 (r √ t, z √ t) = t 2 S(t)[ω 0 -χ a ω 0 ](r √ t, z √ t) + t 2 S 1 (t)[χ a ω 0 ](r √ t, z √ t) + (I a 0 -I 0 )G 1 (r, z)
we finally have lim sup

t→+∞ t 2 S 1 (t)ω 0 (• √ t) L p (Ω) lim sup t→+∞ t 2 S(t)[(1 -χ a )ω 0 ](• √ t) L p (Ω) + lim sup t→+∞ t 2 S 1 (t)[χ a ω 0 ](• √ t) L p (Ω) + lim sup t→+∞ (I a 0 -I 0 )G 1 L p (Ω) (2.9) C r 2 ω 0 (1 -χ a ) L 1 (Ω) + 0 + (I a 0 -I 0 )G 1 L p (Ω)
which tends to zero as a → +∞.

Approximation of the semigroup at second order

Let us suppose briefly that I 0 = Ω r 2 ω 0 (r, z)drdz vanishes. In this case S 1 (t) = S(t) for every t > 0, and estimate (2.26) taken with α = 0 and q = p = 1 implies that the semigroup decreases in L 1 (Ω) at least like t → t -β/2 when (r, z)

→ (1 + r β + zr β-1 )ω 0 (r, z) is in L 1 (Ω), if β 3.
The case β = 3 corresponds to the second term in the asymptotic approximation of S:

S(t)ω 0 ∼ t→+∞ J 0 rz 32 √ π e -r 2 +z 2 4t t -7/2 (2.34) if I 0 = 0 but J 0 = Ω r 2 z ω 0 (r, z) drdz = 0
. This is the object of lemma 2.8 below.

Let us introduce on Ω the Gaussian-like function

G 2 (r, z) = -∂ z G 1 (r, z) = rz 32 √ π e -r 2 +z 2 4 , (2.35)
and the notation

S 2 (t)ω 0 : (r, z) → S(t)ω 0 (r, z) - Ω ρ 2 ω 0 (ρ, ζ) dρdζ G 1 r √ t , z √ t t -2 - Ω ρ 2 ζ ω 0 (ρ, ζ) dρdζ G 2 r √ t , z √ t t -5/2 (2.36) defined when (1 + r 2 + r 2 |z|)ω 0 L 1 (Ω) < ∞
S(t -s) 1 s 2 G 1 • √ s (r, z) = 1 t 2 G 1 r √ t , z √ t (2.38)
and

S(t -s) 1 s 5/2 G 2 • √ s (r, z) = 1 t 5/2 G 2 r √ t , z √ t (2.39)
hold for all positive times t s > 0. Other notable properties are the values of their r 2 and r 2 z-moments:

Ω r 2 G 1 (r, z) drdz = 1 and Ω r 2 z G 1 (r, z) drdz = 0 (2.40) while Ω r 2 G 2 (r, z) drdz = 0 and Ω r 2 z G 2 (r, z) drdz = 1. (2.41)
Let us now see the estimates that hold for (S 2 (t)) t>0 , similar to the ones of proposition 2.4 upon S 1 . Once again, the constants in proposition 2.7 can be shown independent of p, q, α, β and ω 0 . Proposition 2.7. Choose 1 p q ∞ such that p < 2 and -1 α β 2.

(i) If 3 -2 p < β 2 then there exists a positive constant C such that for all time t > 0

r α S 2 (t)ω 0 L q (Ω) C 1 p + β+1 2 -2 r β (r 2 + z 2 ) ω 0 L p (Ω) × t -1+ α-β 2 + 1 q -1 p (2.42) provided that (r, z) → r 2 (1 + |z|)ω 0 (r, z) ∈ L 1 (Ω) and (r, z) → r β (r 2 + z 2 )ω 0 (r, z) ∈ L p (Ω).
(ii) If 2 -2 p < β 1 then there exists a positive constant C such that for all time t > 0

r α S 2 (t)div * w L q (Ω) C 1 p + β+2 2 -2 r β (r 2 + z 2 ) w L p (Ω) × t -3 2 + α-β 2 + 1 q -1 p (2.43) provided that (r, z) → r 2 (1 + |z|)div * w(r, z) ∈ L 1 (Ω) and (r, z) → r β (r 2 + z 2 )w(r, z) ∈ (L p (Ω)) 2 .
Proof. The proof is strictly identical to the one of proposition 2.4 once taken into account the following modification and its consequences. The asymptotic expansion with integral remainder (2.28) has to be considered here at order two

K rρ √ t e -X(t) = √ π 4 e -1 4 (r 2 +z 2 ) + zζ √ π 8 √ t e -1 4 (r 2 +z 2 ) + +∞ t 1 2s 3/2 1 √ t - 1 √ s A 0 e -X(s) ds,
(2.44) A 0 being now the quantity

A 0 = r 2 ρ 2 K rρ √ s + rρ 2 ρ r -ρ √ s + ζ z -ζ √ s K rρ √ s + 1 4 ρ r -ρ √ s + ζ z -ζ √ s 2 K rρ √ s - 1 2 ρ 2 + ζ 2 K rρ √ s
instead of (2.29). Equalities (2.30) and (2.31) change accordingly, meaning that their left hand side members involve here S 2 (t) instead of S 1 (t) and that the integrands have to be multiplied by the factor (

1 √ t -1 √ s ), which is less than 1 √ t .
Last consequence to take into account, A r and A z such as defined in (2.32) and rρ 2 A 0 are now sums of terms of the form

r 1+i ρ 1+j s j-1 2 ρ k r - ρ √ s k ζ 2-k z - ζ √ s k K (i) rρ √ s with j ∈ {0, 1}, k ∈ {0, 1, 2}, i, k , k ∈ {0, 1, 2, 3}
, and must be bounded by Cs

2-β 2 ρ β (ρ 2 + ζ 2 )e X(s) 2 instead of Cs 2-β 2 ρ β (ρ + |ζ|)e X(s)
2 . Using this bound inside of (2.44), together with a Fubini integral permutation and Young's inequality for convolution, leads to the desired conclusion.

We can now come back to (2.34), which is true in every L p (Ω), 1 p ∞. Lemma 2.8. Take ω 0 ∈ L 1 (Ω) such that (r, z) → (r 3 + r 2 |z|)ω 0 (r, z) ∈ L 1 (Ω). Remember that (S(t)) t 0 is the semigoup defined by (2.2), I 0 = Ω r 2 ω 0 (r, z) drdz and J 0 = Ω r 2 z ω 0 (r, z) drdz. Then

S(t)ω 0 = I 0 r 16 √ π e -r 2 +z 2 4 t -5/2 + J 0 rz 32 √ π e -r 2 +z 2 4 t -7/2 + o t→+∞ t 1 p -5 2 (2.45) in L p (Ω) for every p ∈ [1, +∞].
Proof. The goal is to prove that t

J a 0 = a 1/a a -a ρ 2 ζ ω 0 (ρ, ζ) dρdζ = Ω ρ 2 ζ
χ a ω 0 dρdζ which tends towards J 0 as a goes to infinity. Let us use (2.26) with α = 0 and β = 2 to show that

t 5 2 S 1 (t)[(1 -χ a )ω 0 ](• √ t) L p (Ω) = t 5 2 -1 p S 1 (t)[(1 -χ a )ω 0 ] L p (Ω) 2C r 2 (r + |z|)(1 -χ a )ω 0 L 1 (Ω)
for any positive time t, and use (2.42) with α = 0 and β = 2 to see that

t 5 2 S 2 (t)[χ a ω 0 ](• √ t) L p (Ω) = t 5 2 -1 p S 2 (t)[χ a ω 0 ] L p (Ω) (2.42) 2C r 2 (r 2 + z 2 )χ a ω 0 L 1 (Ω) × t -1 2 ----→ t→+∞ 0.
Let us write eventually for every a > 1

t 5 2 S 2 (t)ω 0 (r √ t, z √ t) = t 5 2 S 1 (t)[(1-χ a )ω 0 ](r √ t, z √ t)+t 5 2 S 2 (t)[χ a ω 0 ](r √ t, z √ t)+(J a 0 -J 0 )G 2 (r, z) so lim sup t→+∞ t 5 2 S 2 (t)ω 0 (• √ t) L p (Ω) lim sup t→+∞ t 5 2 S 1 (t)[(1 -χ a )ω 0 ](• √ t) L p (Ω) + lim sup t→+∞ t 5 2 S 2 (t)[χ a ω 0 ](• √ t) L p (Ω) + lim sup t→+∞ (J a 0 -J 0 )G 2 L p (Ω) (2.26) C r 2 (r + |z|)ω 0 (1 -χ a ) L 1 (Ω) + 0 + (J a 0 -J 0 )G 2 L p (Ω)
which tends to zero as a → +∞, concluding the proof.

Time-asymptotic expansion of the vorticity

In [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF], Gallay and Šverák show important results about the axisymmetric vorticity when the swirl is zero: for any nontrivial initial data ω 0 ∈ L 1 (Ω), the L 1 -norm of the vorticity t → ω θ (t) L 1 (Ω) decreases to zero and for every p in ]1, +∞]

lim t→0 t 1-1 p ω θ (t) L p (Ω) = lim t→+∞ t 1-1 p ω θ (t) L p (Ω) = 0. (3.1)
As we said in the introduction, they also show that when the initial data ω 0 is nonnegative and has a finite impulse I 0 = Ω r 2 ω 0 (r, z)drdz < ∞, the vorticity has the same first-order asymptotic expansion as the linear semigroup S. The goal of this section is to refine these results by reinforcing gradually the hypotheses on the localisation of the initial data, proving in particular that the nonnegativity of ω 0 is never required. Let us start by introducing the Biot-Savart law for axisymmetric fluids, which gives crucial estimates between u and ω θ .

The Axisymmetric Biot-Savart law

As stated in the introductory section, working with the vorticity or directly with the velocity is equivalent given that to any solution ω of the vorticity equation (1.2) corresponds only one solution u to the Navier-Stokes equations (1.1) whose curl is ω. In the context of axisymmetry without swirl, the Biot-Savart law can be written as

u r = -1 r ∂ z ψ u θ = 0 u z = 1 r ∂ r ψ where ψ(r, z) = 1 4π Ω 2π 0 rρ cos θ ω θ (z -ζ) 2 + r 2 + ρ 2 -2rρ cos θ dθdρdζ, (3.2)
see for example [START_REF] Fukumoto | Motion and expansion of a viscous vortex ring. Part 1. A higher-order asymptotic formula for the velocity[END_REF]. The reader will then find the following proposition in [START_REF] Gallay | Remarks on the Cauchy problem for the axisymmetric Navier-Stokes equations[END_REF].

Proposition 3.1. Consider u defined from ω θ by the axisymmetric Biot-Savart law (3.2) and choose 1 p < q ∞.

(i) If 1 < p < q < ∞ such that 1 p = 1 q + 1 2
and ω θ ∈ L p (Ω) then there exists a positive constant C such that for all time t 0

u(t) L q (Ω) C ω θ (t) L p (Ω) . (3.3)
More generally, if 1 < p < q < ∞ and 0 α β 2 such that 0 β -α < 1 and

1 p = 1 q + 1+α-β 2
and if (r, z) → r β ω θ (r, z) ∈ L p (Ω) then there exists a positive constant C such that for all time t 0 r α u(t) L q (Ω) C r β ω θ (t) L p (Ω) .

(3.4)

(ii) If 1 p < 2 < q ∞ and ω θ ∈ L p (Ω) ∩ L q (Ω) then there exists a positive constant C such that for all time t 0

u(t) L ∞ (Ω) C ω θ (t) σ L p (Ω) ω θ (t) 1-σ L q (Ω) where σ = p 2 q -2 q -p ∈ ]0, 1[ . (3.5)
Remark. The combination of (3.1) and (3.5) ensures the existence of two positive constants C and C ω0 such that

u(t) L ∞ (Ω) C ω θ (t) 1 2 L 1 (Ω) C ω0 t -1 1 2 ∀t > 0, (3.6)
and the fact that t → ω θ (t) L 1 (Ω) is nonincreasing gives the existence of a positive constant

C ω0 such that u(t) L ∞ (Ω) C ω0 t -1 2 ∀t > 0. (3.7)
This bound will be useful in what comes next, and already tells something about how u vanishes at infinity.

Decay rates for the vorticity

Let us show first that a finite initial impulse r 2 ω 0 L 1 (Ω) < ∞ implies that the second radial moment r 2 ω θ (t) L 1 (Ω) is bounded and that ω θ (t) L p (Ω) decreases like t → t 1 p -2 ; this is lemma 3.3 below. This result requires the intermediate study of the vorticity's decay rates when only its first radial moment rω 0 L 1 (Ω) is initially supposed finite.

Lemma 3.2. Take ω θ the solution of (2.1) with initial condition ω 0 ∈ L 1 (Ω).

If (r, z) → rω 0 (r, z) is in L 1 (Ω) then there exists a positive constant C ω0 such that for every q ∈ [1, +∞] and all time t > 0

ω θ (t) L q (Ω) C ω0 (1 + t) -1 2 t 1 q -1 (3.8) rω θ (t) L q (Ω) C ω0 t 1 q -1 . (3.9)
Proof. Suppose that rω 0 L 1 (Ω) < ∞. We show first the local existence of rω θ (t) L 1 (Ω) , then the fact that it does not blow up in finite time and eventually that it is bounded on [0, +∞[. We shall come afterwards to inequalities (3.8) and (3.9). First of all, a fixed point argument shows that there exists a time T > 0 such that ω θ is continuous from [0, T ] into the space

L 1 1 (Ω) = {w : Ω → R | w L 1 (Ω) + rw L 1 (Ω) < ∞}.
This standard argument will be used several times in the next lemmas, we thus provide some details in appendix B for the reader's convenience.

We then expose why rω θ (t) L 1 (Ω) does not blow up in finite time, using a Grönwall argument. Let us write for t 0 and t 0 > 0

rω θ (t 0 + t) L 1 (Ω) (2.5) rS(t)ω θ (t 0 ) L 1 (Ω) + t0+t t0 rS(t 0 + t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.9)(2.10) C rω θ (t 0 ) L 1 (Ω) + C t0+t t0 (t 0 + t -s) -1 2 u(s) L ∞ (Ω) rω θ (s) L 1 (Ω) ds (3.10) (3.7) C rω θ (t 0 ) L 1 (Ω) + C t 0 (t -s) -1 2 (t 0 + s) -1 2 rω θ (t 0 + s) L 1 (Ω) ds
and conclude using some Grönwall-type lemma that rω θ (t) L 1 (Ω) is finite and well defined for every time t in [0, +∞[. For example, lemma 7.1.1 from [START_REF] Henry | Geometric Theory of Semilinear Parabolic Equations[END_REF] gives exactly the desired conclusion if we observe that (t 0 + s) -1 2 is bounded by the constant t -1 2 0 . Let us now come back to line (3.10) and give a bound to u(s) L ∞ (Ω) by using inequality (3.6) instead of (3.7), then u(s) L ∞ (Ω) C ω θ (s)

1 2 L 1 (Ω) s -1
2 and inequality (3.10) becomes

rω θ (t 0 + t) L 1 (Ω) C rω θ (t 0 ) L 1 (Ω) + C 2 t0+t t0 (t 0 + t -s) -1 2 ω θ (s) 1 2 L 1 (Ω) s -1 2 rω θ (s) L 1 (Ω) ds
for any t 0, C 2 being some positive constant depending on ω 0 L 1 (Ω) . Let us remember here that t → ω θ (t) L 1 (Ω) decreases to zero and consider t 0 large enough so that ω θ (t 0 ) L 1 (Ω) (2C 2 π) -2 ; the presence of π being due to the particular value of the Euler Beta function

B( 1 2 , 1 2 ) = 1 0 (1 - s) -1
2 s -1 2 ds = π appearing right below. Indeed, for every t > 0 we can now write

rω θ (t 0 + t) L 1 (Ω) C rω θ (t 0 ) L 1 (Ω) + C 2 ω θ (t 0 ) 1 2 L 1 (Ω) t0+t 0 (t 0 + t -s) -1 2 s -1 2 ds sup [t0,t0+t] rω θ L 1 (Ω) C rω θ (t 0 ) L 1 (Ω) + 1 2 sup [t0,t0+t] rω θ L 1 (Ω)
and so sup [t0,t0+t] rω θ L 1 (Ω) C+ 1 2 sup [t0,t0+t] rω θ L 1 (Ω) . This proves that there exists a positive constant C ω0 such that rω θ (t) L 1 (Ω) C ω0 for any t in [0, +∞[. We can now prove inequalities (3.8) and (3.9), which are consequences of this uniform bound on rω θ (t) L 1 (Ω) as shown below. To establish (3.8), let us recall first that ω θ (t) L q (Ω) C ω0 t 1 q -1 as a consequence of (3.1). Referring then to (2.9) and (2.10) with α = 0 and β = 1 2 , let us write for t > 0

ω θ (t) L 1 (Ω) (2.4) S(t)ω 0 L 1 (Ω) + t 0 S(t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.9)(2.10) C r 1 2 ω 0 L 1 (Ω) t -1 4 + C t 0 (t -s) -3 4 u(s) L ∞ (Ω) r 1 2 ω θ (s) L 1 (Ω) ds (3.7) Ct -1 4 + C t 0 (t -s) -3 4 s -1 2 ds = C(1 + B(1/4, 1/2)) t -1 4 .
Use straightaway the decay rate obtained this way on ω θ (t) L 1 (Ω) in the same calculation (referring this time to (2.9) with α = 0, β = 1 and to (2.10) with α = 0, β = 3 4 ),

ω θ (t) L 1 (Ω) (2.4) S(t)ω 0 L 1 (Ω) + t 0 S(t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds
(2.9)(2.10)

C rω 0 L 1 (Ω) t -1 2 + C t 0 (t -s) -7 8 u(s) L ∞ (Ω) r 3 4 ω θ (s) L 1 (Ω) ds (3.6) Ct -1 2 + C t 0 (t -s) -7 8 ω θ (s) 1 2 L 1 (Ω) s -1 2 r 3 4 ω θ (s) L 1 (Ω) ds Ct -1 2 + C t 0 (t -s) -7 8 s -1 8 s -1 2 ds = C(1 + B(1/8, 3/8)) t -1 2 ,
to get the case q = 1 of (3.8). Finally, another iterative argument gives all of the remaining cases in three steps. Let us proceed to the following calculations three times, successively for q ∈ [1, 2[, p = 1, then q ∈ [ 4 3 , 4[, p = 4 3 and at last q ∈ [3, +∞], p = 3:

ω θ (t) L q (Ω) (2.5) S(t/2)ω θ (t/2) L q (Ω) + t t/2
S(t -s)div * (u(s)ω θ (s)) L q (Ω) ds

(2.9)(2.10)

C rω θ (t/2) L 1 (Ω) t 1 q -3 2 + C t t/2 (t -s) 1 q -1 p -1 2 u(s) L ∞ (Ω) ω θ (s) L p (Ω) ds (3.7) Ct 1 q -3 2 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 ω θ (s) L p (Ω) ds Ct 1 q -3 2 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 s 1 p -3 2 ds C 1 + 1 1/2 (1 -s) 1 q -1 p -1 2 s 1 p -2 ds t 1 q -3 2 and rω θ (t) L q (Ω) (2.5) rS(t/2)ω θ (t/2) L q (Ω) + t t/2 rS(t -s)div * (u(s)ω θ (s)) L q (Ω) ds (2.9)(2.10) C rω θ (t/2) L 1 (Ω) t 1 q -1 + C t t/2 (t -s) 1 q -1 p -1 2 u(s) L ∞ (Ω) rω θ (s) L p (Ω) ds (3.7) Ct 1 q -1 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 rω θ (s) L p (Ω) ds Ct 1 q -1 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 s 1 p -1 ds C 1 + 1 1/2 (1 -s) 1 q -1 p -1 2 s 1 p -3 2 ds t 1 q -1
for t > 0. Note that the constants in the computations only depend on the initial data ω 0 .

Remark. Under the hypothesis of rω 0 L 1 (Ω) < ∞, (3.8) is an refinement of (3.1). It allows us to replace (3.7) by the following estimate

u(t) L ∞ (Ω) C(1 + t) -1/2 t -1/2 ∀t > 0, (3.11)
obtained directly thanks to (3.8) combined with (3.5). This bound will be used exclusively to prove the next lemma, before being superseded by a better one.

Lemma 3.3. Take ω θ the solution of (2.1) with initial condition

ω 0 ∈ L 1 (Ω). If (r, z) → r 2 ω 0 (r, z) is in L 1 (Ω)
then there exists a positive constant C ω0 such that for every q ∈ [1, +∞] and all time t > 0

ω θ (t) L q (Ω) C ω0 (1 + t) -1 t 1 q -1 (3.12) rω θ (t) L q (Ω) C ω0 (1 + t) -1 2 t 1 q -1 (3.13) r 2 ω θ (t) L q (Ω) C ω0 t 1 q -1 . (3.14)
Proof. Suppose that r 2 ω 0 L 1 (Ω) < ∞. As for lemma 3.2, we shall show first the local existence of the quantity r 2 ω θ (t) L 1 (Ω) , then the fact that it does not blow up in finite time and eventually that it is bounded on [0, +∞[. We shall afterwards concentrate on inequalities (3.12) to (3.14). Note that, in particular, rω 0 L 1 (Ω) < ∞ so the conclusions of lemma 3.2 hold. First of all, a fixed point argument shows that there exists a time T > 0 such that ω θ is continuous from [0, T ] into

{w : Ω → R | w L 1 (Ω) + r 2 w L 1 (Ω) < ∞},
read for example the development in appendix B but adapted with the norm

ω θ X T = sup 0<t T t 1/4 ω θ (t) L 4/3 (Ω) + t 1/4 r 2 ω θ (t) L 4/3 (Ω)
on X T and using (2.14) together with an interpolation inequality when adapting equations (B.1) and (B.2).

Let us now show that r 2 ω θ (t) L 1 (Ω) does not blow up in finite time. Let us write for t 0 that

rω θ (t) L 1 (Ω) (2.5) rS(t/2)ω θ (t/2) L 1 (Ω) + t t/2 rS(t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.9)(2.10) C r 2 ω θ (t/2) L 1 (Ω) t -1 2 + C t t/2 (t -s) -1 2 u(s) L ∞ (Ω) rω θ (s) L 1 (Ω) ds (3.11)(3.9) C r 2 ω θ (t/2) L 1 (Ω) t -1 2 + C t t/2 (t -s) -1 2 (1 + s) -1 2 s -1 2 ds C r 2 ω θ (t/2) L 1 (Ω) + 1 1/2 (1 -s) -1 2 s -1 ds t -1 2 , (3.15)
which leads for t 0 and t 0 > 0 to

r 2 ω θ (t 0 + t) L 1 (Ω) (2.5) r 2 S(t)ω θ (t 0 ) L 1 (Ω) + t0+t t0 r 2 S(t 0 + t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.9)(2.14) C r 2 ω θ (t 0 ) L 1 (Ω) + C t0+t t0 u(s) L ∞ (Ω) rω θ (s) L 1 (Ω) , ds + C t0+t t0 (t 0 + t -s) -1 2 ru(s) L 4 (Ω) rω θ (s) L 4/3 (Ω) ds (3.15)(3.4)(3.9) C r 2 ω θ (t 0 ) L 1 (Ω) + C t0+t t0 (1 + s) -1 2 s -1 2 ( r 2 ω θ (s/2) L 1 (Ω) + 1)s -1 2 ds + C t0+t t0 (t 0 + t -s) -1 2 s -1 2 ds
where in the last line (3.4) is used with α = β = 1 and p = 4, q = 4 3 . Translating the integration variable s by t 0 , using that (t 0 + s) -1/2 min(1, t 0 ) -1/2 (1 + s) -1/2 and in the first integrand that

r 2 ω θ (s/2) L 1 (Ω) sup [0,t0] r 2 ω θ L 1 (Ω) + sup [t0,s] r 2 ω θ L 1 (Ω)
, one gets for every t t 0 > 0 that sup

[t0,t] r 2 ω θ L 1 (Ω) C ω0,t0 + C t0 t 0 (1 + s) -3 2 sup [t0,s] r 2 ω θ L 1 (Ω) ds.
Grönwall's inequality applied to the map t → sup s∈[t0,t] r 2 ω θ (s) L 1 (Ω) therefore ensures that t → r 2 ω θ (t) L 1 (Ω) is well-defined and bounded on [0, +∞[. This sets estimate (3.14) when q = 1.

We can now thanks to this uniform bound prove inequalities (3.12) to (3.14). To establish (3.12) and (3.13), one has to recall first that ω θ (t) L q (Ω) C ω0 t 1 q -1 due to (3.1) and rω θ (t) L q (Ω) C ω0 t 1 q -1 due to (3.9). Then, let us write for t > 0

ω θ (t) L 1 (Ω) (2.5) S(t/2)ω θ (t/2) L 1 (Ω) + t t/2 S(t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.9)(2.10) C r 2 ω θ (t/2) L 1 (Ω) t -1 + C t t/2 (t -s) -1 2 u(s) L ∞ (Ω) ω θ (s) L 1 (Ω) ds (3.11)(3.8) Ct -1 + C t t/2 (t -s) -1 2 (1 + s) -1 2 s -1 2 (1 + s) -1 2 ds C 1 + 1 1/2 (1 -s) -1 2 s -3 2 ds t -1
to get the case q = 1 of (3.12), and let us remember calculation (3.15) which gives the case q = 1 of (3.13) once we know that r 2 ω θ L 1 (Ω) is bounded on [0, +∞[. Finally, a three-steps iterative argument leads to all of the remaining cases. Let us proceed to the following computations three times, successively for q ∈ [1, 2[, p = 1 then q ∈ [ 4 3 , 4[, p = 4 3 and at last q ∈ [3, +∞], p = 3:

ω θ (t) L q (Ω) (2.5) S(t/2)ω θ (t/2) L q (Ω) + t t/2 S(t -s)div * (u(s)ω θ (s)) L q (Ω) ds (2.9)(2.10) C r 2 ω θ (t/2) L 1 (Ω) t 1 q -2 + C t t/2 (t -s) 1 q -1 p -1 2 u(s) L ∞ (Ω) ω θ (s) L p (Ω) ds (3.7) Ct 1 q -2 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 ω θ (s) L p (Ω) ds Ct 1 q -2 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 s 1 p -2 ds C 1 + 1 1/2 (1 -s) 1 q -1 p -1 2 s 1 p -5 2 ds t 1 q -2 and rω θ (t) L q (Ω) (2.5) rS(t/2)ω θ (t/2) L q (Ω) + t t/2 rS(t -s)div * (u(s)ω θ (s)) L q (Ω) ds (2.9)(2.10) C r 2 ω θ (t/2) L 1 (Ω) t 1 q -3 2 + C t t/2 (t -s) 1 q -1 p -1 2 u(s) L ∞ (Ω) rω θ (s) L p (Ω) ds (3.7) Ct 1 q -3 2 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 rω θ (s) L p (Ω) ds Ct 1 q -3 2 + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 s 1 p -3 2 ds and r 2 ω θ (t) L q (Ω) (2.5) r 2 S(t/2)ω θ (t/2) L q (Ω) + t t/2 r 2 S(t -s)div * (u(s)ω θ (s)) L q (Ω) ds (2.9)(2.14) C r 2 ω θ (t/2) L 1 (Ω) t 1 q -1 + C t t/2 (t -s) 1 q -1 p u(s) L ∞ (Ω) rω θ (s) L p (Ω) ds + C t t/2 (t -s) 1 q -1 p -1 2 u(s) L ∞ (Ω) r 2 ω θ (s) L p (Ω) ds (3.7) Ct 1 q -1 + C t t/2 (t -s) 1 q -1 p s -1 2 rω θ (s) L p (Ω) ds + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 r 2 ω θ (s) L p (Ω) ds Ct 1 q -1 + C t t/2 (t -s) 1 q -1 p s -1 2 s 1 p -3 2 ds + C t t/2 (t -s) 1 q -1 p -1 2 s -1 2 s 1 p -1 ds,
where the constants only depend on the initial data ω 0 .

Remark. In particular, we deduce from the Biot-Savart law that if r 2 ω 0 (r, z) L 1 (Ω) < ∞ then there is some positive constant C such that for all time t > 0

u(t) L ∞ (Ω) C(1 + t) -1 t -1/2 . (3.16)
This estimate is given by inequalities (3.5) and (3.12) put together, and it is the best one can get concerning the supremum of u as long as the impulse I 0 is not zero.

Lemmas 3.2 and 3.3 are the most direct way to reach the case where (r, z) → (1 + r 2 )ω 0 (r, z) is in L 1 (Ω), but one could consider all of the intermediate cases:

Proposition 3.4. Take ω θ the solution of (2.1) with initial condition ω 0 ∈ L 1 (Ω). Take α in [0, 2]. If (r, z) → r α ω 0 (r, z) is in L 1 (Ω) then there exists a positive constant C ω0,α such that for every q ∈ [1, +∞] and all time t > 0

ω θ (t) L q (Ω) C ω0,α (1 + t) -α 2 t 1 q -1
(3.17)

r α ω θ (t) L q (Ω) C ω0,α t 1 q -1 . (3.18)
Proof. The arguments presented in the proofs of lemmas 3.2 and 3.3 are applicable in the same order, following the proof of lemma 3.2 when α ∈ [0, 1] and the proof of lemma 3.3 when α ∈ ]1, 2].

First-Order asymptotics

In this subsection, we show first (1.4) then we study how close ω θ (t) and its first asymptotic term are the one to the other. Due to the symmetry of the problem, when both ω 0 and (r, z) → r 2 ω 0 (r, z) are in L 1 (Ω) the vorticity has a constant impulse:

I(t) = Ω r 2 ω θ (t, r, z) drdz = Ω r 2 ω 0 (r, z) drdz = I 0 ∀t 0. (3.19) Indeed, remembering that ∆ = ∂ 2 r + 1 r ∂ r + ∂ 2 z in (2.1) implies that İ(t) = Ω r 2 ∂ t ω θ (t) = Ω r 2 ∂ 2 r + ∂ 2 z + 1 r ∂ r - 1 r 2 ω θ (t) - Ω r 2 div * (u(t)ω θ (t))
where after several integrations by parts, the first integral vanishes and the second is equal to

Ω 2r u r (t)ω θ (t).
Writing therefore İ(t) = Ω 2r u r (t)ω θ (t), then substituting u r with its expression by the Biot-Savart law (3.2) shows that İ(t) is zero, because the integrand under the double integral obtained that way is odd with respect to permuting the variables (r, z) and (ρ, ζ) in Ω × Ω.

Remark. The existence of İ(t) = d dt I(t) is ensured by the fact that

ω θ ∈ C 1 (]0, +∞[, L 1 2 (Ω)), where L 1 2 (Ω) = L 1 (Ω, ( 1+r
2 )drdz). The reader can refer to the end of appendix B for more details.

Proposition 3.5. Take ω θ the solution of (2.1) with initial data

ω 0 ∈ L 1 (Ω). If (r, z) → r 2 ω 0 (r, z) is in L 1 (Ω) then ω θ (t, r, z) = I 0 r 16 √ π e -r 2 +z 2 4t t -5/2 + o t→+∞ t 1 p -2 (3.20) in L p (Ω) for every p ∈ [1, +∞].
Proof. Let r 2 ω 0 L 1 (Ω) be finite, ω 0 ∈ L 1 (Ω). The goal is to show that, in the self-similar coordinates (r √ t, z √ t), the quantity t 2 ω θ (t, • √ t, • √ t) -I 0 G 1 tends to zero in any L p (Ω) when t tends to infinity (remembering that G 1 is given by (2.24)).

Let us fix t 0 > 0, then for any p in [1, +∞] and any t > 2t 0 we can write

t 2 ω θ (t, • √ t, • √ t) -I 0 G 1 L p (Ω) (2.5) t 2 S(t -t 0 )[ω θ (t 0 )](• √ t) - t t -t 0 2 I 0 G 1 • √ t √ t -t 0 L p (Ω) + I 0 1 - t 0 t -2 G 1 • 1 - t 0 t -1 2 -I 0 G 1 L p (Ω) + t/2 t0 t 2 S(t -s)div * (u(s) ω θ (s)) • √ t L p (Ω) ds + t t/2 t 2 S(t -s)div * (u(s) ω θ (s)) • √ t L p (Ω)
ds where three of those four terms tend to zero as t tends to infinity and one will have to be dominated by a decreasing function of t 0 . The convergence of the second term is trivial. The first one can be written with a change of variables as

( t t-t0 ) 2-1 p (t -t 0 ) 2 S(t -t 0 )[ω θ (t 0 )](• √ t -t 0 ) -I 0 G 1 L p (Ω)
, which tends to zero according to lemma 2.5 and the fact that I(t 0 ) = I 0 (recall (3.19)). The fourth one satisfies

t t/2 t 2 S(t -s)div * (u(s) ω θ (s)) • √ t L p (Ω) ds (2.10) C t t/2 t 2-1 p (t -s) -1 2 u(s) L ∞ (Ω) ω θ (s) L p (Ω) ds (3.16)(3.12) C t t/2 t 2-1 p (t -s) -1 2 s -3 2 s 1 p -2 ds C 1 1/2 (1 -s) -1 2 s 1 p -7 2 ds t -1
so converges to zero as well, and the third term can be bounded as

t/2 t0 t 2 S(t -s)div * (u(s) ω θ (s)) • √ t L p (Ω) ds (2.10) C t/2 t0 t 2-1 p (t -s) 1 p -2 u(s) L ∞ (Ω) rω θ (s) L 1 (Ω) ds (3.16)(3.13) C t/2 t0 2 2-1 p (1 + s) -1 s -1 2 (1 + s) -1 2 ds C +∞ 0 (1 + s) -3 2 ds t -1 2 0
where the constants do not depend on t 0 . Finally,

lim sup t→+∞ t 2 ω θ (t, • √ t, • √ t) -I 0 G 1 L p (Ω) 0 + 0 + C t -1 2 0 + 0
for any t 0 > 0, which ends the proof.

Let us now specify the o(t 1 p -2 ) in proposition 3.5 by reinforcing the hypotheses on ω 0 . For this, let us remark that another consequence of

İ(t) = - Ω r 2 div * (u(t, r, z)ω θ (t, r, z)) drdz = 0 (3.21)
is that for every t s 0

S 1 (t -s)div * (u(s)ω θ (s)) = S(t -s)div * (u(s)ω θ (s)), (3.22)
where S 1 is defined by (2.25). This means that the difference ω θ (t)

-I 0 G 1 ( • √ t )t -2
between the vorticity and its first asymptotic term satisfies

ω θ (t) -I 0 G 1 • √ t t -2 = (2.4)(3.22) S 1 (t)ω 0 - t 0 S 1 (t -s)div * (u(s)ω θ (s)) ds. (3.23)
Proposition 3.6. Take ω θ the solution of (2.1) with initial condition

ω 0 ∈ L 1 (Ω). Take α in ]0, 1]. If (r, z) → r α (r 2 + r|z| + |z|)ω 0 (r, z) is in L 1 (Ω)
then there exists a positive constant C ω0,α such that for every q ∈ [1, +∞] and all t > 0

ω θ (t, r, z) - I 0 r 16 √ π e -r 2 +z 2 4t t -5 2 L q (Ω) C ω0,α t 1 q -2-α 2 . (3.24)
Before showing proposition 3.6, we need a lemma certifying that some of the finite moments asked to ω 0 stay finite through the evolution of ω θ . Lemma 3.7. Take ω θ the solution of (2.1) with initial condition

ω 0 ∈ L 1 (Ω). Take α in [0, 1]. If (r, z) → r α (r + |z|)ω 0 (r, z) is in L 1 (Ω) then t → r α (r + |z|) ω θ (t, r, z) L 1 (Ω) is bounded on [0, +∞[.
Proof. Let r α (r + |z|) ω 0 L 1 (Ω) be finite, where 0 α 1. The strategy of this proof will be similar to those of lemmas 3.2 and 3.3. r 1+α ω 0 L 1 (Ω) being supposed finite, we already have access to the conclusions of proposition 3.4, especially the fact that by combining (3.5) and (3.17) we know that u(t) L ∞ (Ω) C(1 + t) -1+α 2 t -1/2 for all time t > 0.

First of all, a fixed point argument shows that there exists a time T > 0 such that ω θ is continuous from [0, T ] into

{w : Ω → R | w L 1 (Ω) + rw L 1 (Ω) + r α zw L 1 (Ω) < ∞},
read for example the development in appendix B adapted with the norm

ω θ X T = sup 0<t T t 1/4 ω θ (t) L 4/3 (Ω) + t 1/4 rω θ (t) L 4/3 (Ω) + t 1/4 r α zω θ (t) L 4/3 (Ω)
on X T and using (2.15) together with interpolations inequalities when adapting equations (B.1) and (B.2).

Let us then show that r α zω θ (t) L 1 (Ω) is well-defined and bounded on [0, +∞[, by referring to lemma C.1 in appendix and using the estimate u(s) L ∞ (Ω) C(1 + s) -1+α 2 s -1/2 as follows. For every t 0 and t 0 > 0,

r α zω θ (t 0 + t) L 1 (Ω) (2.5) r α zS(t)ω θ (t 0 ) L 1 (Ω) + t0+t t0 r α zS(t 0 + t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.11)(2.15) C r α (r + |z|) ω θ (t 0 ) L 1 (Ω) + C t0+t t0 (t 0 + t -s) α-1 2 u(s) L ∞ (Ω) rω θ (s) L 1 (Ω) ds + C t0+t t0 (t 0 + t -s) -1 2 u(s) L ∞ (Ω) r α zω θ (s) L 1 (Ω) ds (3.9)
C + C t0+t t0 (t 0 + t -s) α-1 2 (1 + s) -1+α 2 s -1 2 ds + C t0+t t0 (t 0 + t -s) -1 2 (1 + s) -1+α 2 s -1 2 r α zω θ (s) L 1 (Ω) ds where t0+t 0 (t 0 +t-s) α-1
2 (1+s) -1+α 2 s -1 2 ds 10 and where, after the change of variable s → t 0 +s in the last integral, we can write (1+t 0 +s) -1+α 2 (1+s) -1 2 and (t 0 +s) -

1 2 min(1, t 0 ) -1 2 (1+s) -1 2 . Hence r α zω θ (t 0 + t) L 1 (Ω) C + C t 0 (t -s) -1 2 (1 + s) -1 r α zω θ (t 0 + s) L 1 (Ω) ds,
and we can apply lemma C.1. Let us note that the constants only depend on ω 0 and the choice of t 0 , but not on α.

Proof of proposition 3.6. Let us recall the definitions of G 1 and S 1 , given by (2.24) and (2.25). Using lemma 3.7 and knowing (3.22), let us carry out the following calculation, starting from (3.23):

ω θ (t) -I 0 G 1 • √ t t -2 L q (Ω) (3.23) S 1 (t)ω 0 L q (Ω) + t/2 0 S 1 (t -s)div * (u(s)ω θ (s)) L q (Ω) ds + t t/2 S(t -s)div * (u(s)ω θ (s)) L q (Ω) ds (2.26)(2.27)(2.10) 2C α r α (r 2 + r|z|) ω 0 L 1 (Ω) t 1 q -4+α 2 + t/2 0 2C α (t -s) 1 q -4+α 2 u(s) L ∞ (Ω) r α (r + |z|) ω θ (s) L 1 (Ω) ds + t t/2 C(t -s) -1 2 u(s) L ∞ (Ω) ω θ (s) L q (Ω) ds (3.16)(3.12) C t 1 q -4+α 2 + C t 2 1 q -4+α 2 +∞ 0 (1 + s) -1 s -1 2 ds + C t t/2 (t -s) -1 2 s -1 s 1 q -3+α 2 ds C t 1 q -2-α 2
for any t > 0. Let us note that the last constant depends on the initial data ω 0 and on the quantity 1 α .

Considering the limit case α = 1 in proposition 3.6, in particular r 2 zω 0 (r, z) L 1 (Ω) and r(r + |z|) ω 0 (r, z) L 1 (Ω) are finite so the moment J(t) = Ω r 2 z ω θ (t, r, z) drdz of the vorticity is bounded.

Corollary 3.8. Take ω θ the solution of (2.1) with initial condition

ω 0 ∈ L 1 (Ω). If (r, z) → (r 2 |z| + r 2 + r|z|)ω 0 (r, z) is in L 1 (Ω)
, then J is bounded on [0, +∞[. To be more precise, J is differentiable and tends to the finite limit

J ∞ = J 0 + +∞ 0 Ω (2rz u r (t, r, z) + r 2 u z (t, r, z)) ω θ (t, r, z) drdzdt (3.25)
when time tends to infinity.

Proof. It has to be said first that ω θ is differentiable in time into L 1 (Ω, (1 + r 2 + r 2 |z|)drdz), for reasons given in remark B.1. Then J is differentiable and, referring to the vorticity equation (2.1) and using that div * (uω θ ) = ∂ r (u r ω θ ) + ∂ z (u z ω θ ), a direct calculation gives its evolution for all time t 0:

J(t) = Ω r 2 z ∂ t ω θ (t) = - Ω r 2 z div * (u(t)ω θ (t)) = Ω (2rz u r + r 2 u z ) ω θ . (3.26)
Hence | J(t)| 3 u(t) L ∞ (Ω) r(r+|z|) ω θ (t) L 1 (Ω) , and from lemma 3.7 and estimate (3.16) follows directly the existence of a positive constant C ω0 such that

| J(t)| C ω0 (1 + t) -1 t -1/2 , (3.27)
so J is bounded on [0, +∞[ and the expression of J ∞ (3.25) holds.

Second-Order asymptotics

At that point, we want to express the next term in the asymptotic expansion of ω θ . The object of the next propositions is to write for all (t, r, z

) in ]0, +∞[ * × Ω ω θ (t, r, z) = I 0 r 16 √ π e -r 2 +z 2 4t t -5/2 + J(t)rz 32 √ π e -r 2 +z 2 4t t -7/2 + ω θ (t, r, z) (3.28)
where the remainder ω θ (t, r, z) is a o(t

1 p - 5
2 ) in every L p (Ω) when t goes to infinity (this is proposition 3.9) and even a O(ln(t)t 1 p -3 ) if enough spatial localisation is asked on the initial data (this is proposition 4.1). Before moving towards the demonstration of these results, there are a few things to say about ω θ .

Given that ω θ defined by (3.28) is a difference between C ∞ maps, it is itself in C ∞ (]0, +∞[×Ω). Moreover, given that ω θ satisfies the vorticity equation (2.1) and that (t, r, z)

→ G 1 ( r √ t , z √ t )t -2 and (t, r, z) → G 2 ( r √ t , z √ t )t -5
2 are solutions to the linearised equation (2.37), then for every t > 0

∂ t ω θ = ∆ ω θ - ω θ r 2 -div * (uω θ ) - J(t) t 5/2 G 2 • √ t (3.29)
which is equivalent to the integral formulation

ω θ (t) = S 2 (t)ω 0 - t 0 S(t -s) div * (u(s)ω θ (s)) + J(s) s 5/2 G 2 • √ s ds (3.30)
where S 2 is the family of operators defined as (2.36) and G 1 and G 2 are the Gaussian-like functions (2.24) and (2.35).

Let us pay special attention to the various possible ways to write the integrand in equation (3.30), because this will be helpful in further calculation. First, let us remember that (t, r, z)

→ G 2 ( r √ t , z √ t )t -5
2 can be taken out of the semigroup, as noted in (2.39), and especially that

t1 t0 S(t - s)[ J(s)G 2 ( • √ s )s -5 2 ] ds = (J(t 1 ) -J(t 0 ))G 2 ( • √ t )t -5 2
, whatever 0 t 0 t 1 t are. Let us notice that it can also be put inside the div * by writing J(s)

s 5/2 G 2 r √ s , z √ s = div * J(s) 32 √ π s 2 -z √ s -r √ s e -1 4s (r 2 + z 2 ) . (3.31)
Finally, in view of the fact that Ω r 2 div * (u(s)ω θ (s)) = 0, that Ω r 2 zdiv * (u(s)ω θ (s)) = -J(s) and that Ω r 2 zG 2 = 1 (see respectively (3.21), (3.26) and (2.41)), the relation

S(t -s) div * (u(s)ω θ (s)) + J(s) s 5/2 G 2 • √ s = S 2 (t -s) div * (u(s)ω θ (s)) + J(s) s 5/2 G 2 • √ s (3.
32) holds on Ω for any t s > 0. Obviously, (3.32) remains true by substituting S 1 for S 2 .

This being said, let us start the study of ω θ under the hypotheses of proposition (3.6) when α = 1. Proposition 3.9. Take ω θ the solution of (2.1) with initial data Let us fix any t 0 > 0 and see that, given property (2.39) of G 2 , (3.30) can as well be written as

ω 0 ∈ L 1 (Ω). If (r, z) → (r 3 + r 2 |z| + r|z|)ω 0 (r, z) is in L 1 (Ω) then ω θ (t,
ω θ (t) = S(t -t 0 )ω θ (t 0 ) -I 0 G 1 • √ t t -2 -J(t 0 ) G 2 • √ t t -5 2 - t t0 S(t -s) div * (u(s)ω θ (s)) + J(s) s 5/2 G 2 • √ s ds
for every time t > t 0 . Therefore, for any p in [1, +∞] and t > 2t 0 , t

5 2 ω θ (t, • √ t, • √ t) L p (Ω)
is less than or equal to the sum of the six terms The convergence of T 2 and T 3 to zero is obvious when p = +∞, and is a straightforward application of the Lebesgue dominated convergence theorem when p < ∞. The first term T 1 equals ( t t-t0 )

T 1 (t, t 0 , p) = t 5 2 S 2 (t -t 0 )[ω θ (t 0 )](• √ t) L p (Ω) , T 2 (t, t 0 , p) = I 0 √ t 1 - t 0 t -2 G 1 • 1 - t 0 t -1 2 -I 0 √ t G 1 L p (Ω) , T 3 (t, t 0 , p) = J(t 0 ) 1 - t 0 t -5 2 G 2 • 1 - t 0 t -1 2 -J(t 0 ) G 2 L p (Ω) , T 4 (t, t 0 , p) = t/2 t0 t 5 2 S 1 (t -s)div * u(s) ω θ (s) + J(s) 32 √ π s 2 -z √ s -r √ s e -1 4s (r 2 + z 2 ) • √ t L p (Ω)
5 2 -1 p (t -t 0 ) 5 2 S 2 (t -t 0 )[ω θ (t 0 )](• √ t -t 0 ) L p (Ω)
with a change of variables, and so converges to zero according to lemma 2.8. The fifth one satisfies T 5

(2.10)

C t t/2 t 5 2 -1 p (t -s) -1 2 u(s) L ∞ (Ω) ω θ (s) L p (Ω) ds (3.16)(3.12) C t t/2 t 5 2 -1 p (t -s) -1 2 s -3 2 s 1 p -2 ds C 1 1/2 (1 -s) -1 2 s 1 p -7 2 ds t -1 2 ,
and T 6 Ct -1 2 given (3.27), so both tend to zero. Finally, thanks to lemma 3.7 the fourth term can be bounded as + 0 + 0 for any t 0 > 0, which ends the proof.

T 4 (2.27) 2C t/2 t0 t 5 2 -1 p (t -s) 1 p -5 2 u(s) L ∞ (Ω) r(r + |z|)ω θ (s) L 1 (Ω) ds + 2C t/2 t0 t 5 2 -1 p (t -s) 1 p -5 2 | J(s)| s 2 r(r + |z|) -z √ s -r √ s e -r
Remark 3.10. Within the scope of this article, each time the quantity J(t) appears in the expansion of ω θ it can be replaced by its limit J ∞ . This is because, in view of (3.27), for all t > 0

|J(t) -J ∞ | +∞ t | J(s)|ds C +∞ t s -3 2 ds = 2Ct -1 2 so the difference between J(t)G 2 ( • √ t )t -5 2 and J ∞ G 2 ( • √ t )t -5 2 is a O(t 1 p -3
) in every L p (Ω); and it appears that in propositions 3.9, 3.11 and 4.1 the precision of the expansions never exceeds a O(t 1 p -3 ). This remark, together with the previous proposition, conclude the proof of theorem 1.1.

The same way we did in proposition 3.6 concerning the difference ω θ -I 0 G 1 ( • √ t )t -2 , the remainder ω θ can be shown less than a o(t

1 q -5
2 ) by assuming more localisation on the initial vorticity ω 0 . Proposition 3.11. Take ω θ the solution of (2.1) with initial condition ω 0 ∈ L 1 (Ω). Take α in ]0, 1[. If (r, z) → r α (r 3 +rz 2 +z 2 )ω 0 (r, z) is in L 1 (Ω) then there exists a positive constant C ω0,α such that for every q ∈ [1, +∞] and all t > 0

ω θ (t) L q (Ω) C ω0,α t 1 q -5 2 -α 2 (3.34)
where ω θ is given by (3.28).

Lemma 3.12. Take ω θ the solution of (2.1) with initial condition

ω 0 ∈ L 1 (Ω). Take α in [0, 1]. If (r, z) → r α (r 2 + z 2 )ω 0 (r, z) is in L 1 (Ω)
then there exists a positive constant C ω0 such that for all t 0

r α (r 2 + z 2 ) ω θ (t) L 1 (Ω) C ω0 (1 + t) α/2 . (3.35)
Proof. Let r α (r 2 + z 2 ) ω 0 L 1 (Ω) be finite, where 0 α 1. Let us show first that the quantity

r α (r 2 + z 2 ) ω θ (t) L 1 (Ω) exists locally, then that t → (1 + t) -α/2 r α (r 2 + z 2 ) ω θ (t) L 1 (Ω)
is bounded on [0, +∞[ and thus does not blow up in finite time. Given that r 2 ω 0 L 1 (Ω) < ∞, we have access to the conclusions of lemma 3.3. First of all, a fixed point argument shows that there exists a time T > 0 such that ω θ is continuous from [0, T ] into

{w : Ω → R | w L 1 (Ω) + r 2+α w L 1 (Ω) + r α z 2 w L 1 (Ω) < ∞}.
read for example the development in appendix B adapted with the norm

ω θ X T = sup 0<t T t 1/4 ω θ (t) L 4/3 (Ω) + t 1/4 r 2+α ω θ (t) L 4/3 (Ω) + t 1/4 r α z 2 ω θ (t) L 4/3 (Ω)
on X T and using (2.16) and (2.17) together with an interpolation inequality when adapting equations (B.1) and (B.2).

Let us now show (3.35) with the use of lemma C.1. Let us write for every t 0 and t 0 > 0

r α (r 2 + z 2 ) ω θ (t 0 + t) L 1 (Ω) (2.5) r α (r 2 + z 2 )S(t)ω θ (t 0 ) L 1 (Ω) + t0+t t0 r α (r 2 + z 2 )S(t 0 + t -s)div * (u(s)ω θ (s)) L 1 (Ω) ds (2.12)(2.13)(2.16)(2.17) C r 2 ω θ (t 0 ) L 1 (Ω) t α 2 + C r α (r 2 + z 2 ) ω θ (t 0 ) L 1 (Ω) + C t0+t t0 (t 0 + t -s) α 2 u(s) L ∞ (Ω) rω θ (s) L 1 (Ω) ds + C t0+t t0 (t 0 + t -s) -1 2 u(s) L ∞ (Ω) r α (r 2 + z 2 ) ω θ (s) L 1 (Ω) ds (3.16)(3.13) C(1 + t α 2 ) + C t0+t t0 (t 0 + t -s) α 2 (1 + s) -1 s -1 2 (1 + s) -1 2 ds + C t0+t t0 (t 0 + t -s) -1 2 (1 + s) -1 s -1 2 r α (r 2 + z 2 ) ω θ (s) L 1 (Ω) ds where t -α 2 × t 0 (t -s) α 2 (1 + s) -3 2 s -1
2 ds 4 and where, after the change of variable s → t 0 + s in the last integral, (1

+ t 0 + s) -1 (1 + s) -1 and (t 0 + s) -1 2 min(1, t 0 ) -1 2 (1 + s) -1 2 . Thus r α (r 2 + z 2 ) (t 0 + t) α 2 ω θ (t 0 + t) L 1 (Ω) C t -α 2 0 + 1 + 4C + C t 0 (t -s) -1 2 (1 + s) -3 2 r α (r 2 + z 2 ) (t 0 + s) α 2 ω θ (t 0 + s) L 1 (Ω)
ω θ (t) L q (Ω) (3.30) S 2 (t)ω 0 L q (Ω) + t/2 0 S 2 (t -s)div * u(s)ω θ (s) + J(s) 32 √ π s 2 -z √ s -r √ s e -1 4s (r 2 + z 2 ) L q (Ω) ds + t t/2 S(t -s)div * (u(s)ω θ (s)) L q (Ω) ds + t t/2 | J(s)| S(t -s) G 2 • √ s s -5 2 L q (Ω)
ds so using respectively (2.42), (2.43), (2.10) and (3.27) in the four terms

ω θ (t) L q (Ω) 2C α r α (r 3 + rz 2 ) ω 0 L 1 (Ω) t 1 q -5+α 2 + t/2 0 2C α (t -s) 1 q -5+α 2 u(s) L ∞ (Ω) r α (r 2 + z 2 ) ω θ (s) L 1 (Ω) ds + t/2 0 2C(t -s) 1 q -5+α 2 | J(s)| s 2 r α (r 2 + z 2 ) z √ s r √ s e -1 4s (r 2 + z 2 ) L 1 (Ω) ds + t t/2 C(t -s) -1 2 u(s) L ∞ (Ω) ω θ (s) L q (Ω) ds + t t/2 (1 + s) -1 s -1 2 ds G 2 • √ t L q (Ω) t -5 2 .
Using again estimate (3.27) and then finally (3.16), (3.12), lemma 3.7 and lemma 3.12, one can state that

ω θ (t) L q (Ω) Ct 1 q -5+α 2 + C t 2 1 q -5+α 2 t/2 0 (1 + s) α 2 -1 s -1 2 ds + C t 2 1 q -5+α 2 t/2 0 (1 + s) -1 s -1 2 s 2 s 4+α 2 ds + C t t/2 (t -s) -1 2 s -2+α 2 s 1 q -2 ds + C t t/2 s -2+α 2 ds t 1 q t -5 2 C t 1 q -5 2 -α 2
for any positive time t, where the last constant depends on the initial data ω 0 and on the quantity 4 Additional Results and concluding remarks

The First Resonant term

In the previous section we have shown that the first terms in the vorticity's asymptotic expansion decrease in L p as negative powers of t. This had already been highlighted in [START_REF] Carpio | Asymptotic behavior for the vorticity equations in dimension two and three[END_REF][START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF], or in the case of the velocity by the works of Wiegner [START_REF] Wiegner | Decay results for Weak Solutions of the Navier-Stokes Equations on R n[END_REF], Carpio [START_REF] Carpio | Large-time behavior in incompressible Navier-Stokes equations[END_REF], Miyakawa and Schonbek [START_REF] Elena | L 2 Decay for Weak Solutions of the Navier-Stokes Equations[END_REF][START_REF] Elena | Large time behaviour of solutions to the Navier-Stokes equations[END_REF][START_REF] Fujigaki | Asymptotic Profiles of Nonstationary Incompressible Navier-Stokes Flows in the Whole Space[END_REF][START_REF] Miyakawa | On optimal decay rates for weak solutions to the Navier-Stokes equations in R n[END_REF] and more recently Brandolese [START_REF] Brandolese | Asymptotic behavior of the energy and pointwise estimates for solutions to the Navier-Stokes equations[END_REF][START_REF] Brandolese | Space-time decay of Navier-Stokes flows under rotations[END_REF] for instance. However, when addressing the higher order asymptotics, Gallay and Wayne observe in [START_REF] Gallay | Invariant Manifolds and the Long-Time Asymptotics of the Navier-Stokes and Vorticity Equations on R 2[END_REF][START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF] what they call resonances: interactions between the eigenvalues of the linear operator Λ = ∆ + 1 2 x • ∇ + 1, which is the one that appears when they consider the vorticity equation (2.1) in the self-similar variables (ln(1 + t), x √ 1+t ). In particular, in the two-dimensional case, they show explicitly the presence of a ln(t) factor in the third-order asymptotic term. This is also something we observe. When (r, z) → (r 4 + r 2 z 2 + rz 2 )ω 0 (r, z) is in L 1 (Ω) appears the first tangible manifestation of the nonlinearity, namely a bound that does not decreases merely as a negative power of t. This is the reason why, in general, proposition 3.11 does not hold for α = 1 and has to be replaced by the following result. Proposition 4.1. Take ω θ the solution of (2.1) with initial condition ω 0 ∈ L 1 (Ω). If (r, z) → (r 4 + r 2 z 2 + rz 2 )ω 0 (r, z) is in L 1 (Ω) then there exists a positive constant C ω0 such that for every q ∈ [1, +∞] and all t > 0

ω θ (t) L q (Ω) C ω0 (1 + ln(1 + t)) t 1 q -3 (4.1)
where ω θ is given by (3.28).

Proof. The proof is identical to the one of proposition 3.11, with the only difference that having α = 1 in the last calculation leads to consider the integrals t/2 0 (1+s) -1 2 s -1 2 ds and t/2 0 (1+s) -1 ds which grow towards infinity like ln(1 + t).

This article ends before the study of the third-order asymptotic term, which is the first resonant term in the long-time expansion of the three-dimensional fluid.

Decay rates for the velocity

To deduce theorem 1.2 from what has been seen in the previous section, as well as some decay rates for the velocity, one has to invoke estimate (3.4) and the fact that the velocity fields corresponding to the vorticities (t, r, z)

→ I 0 G 1 ( r √ t , z √ t )t -2 and (t, r, z) → J(t)G 2 ( r √ t , z √ t )t -5/2 are respectively (t, r, z) → I 0 u G1 ( r √ t , z √ t )t -3/2 and (t, r, z) → J(t)u G2 ( r √ t , z √ t )t -2
, where u G1 and u G2 are defined in (1.7). Propositions 3.4, 3.6, 3.11 and 4.1 then lead to the following result, where I 0 and J(t) denote respectively Ω r 2 rot(u 0 (r, z))drdz and Ω r 2 z rot(u(t, r, z))drdz.

Corollary 4.2. Take u the solution of (1.6) with initial data u 0 such that rot(u 0 )•e θ = ω 0 ∈ L 1 (Ω). Take β 0.

(i) If β ∈ ] 1 2 , 2] and (r, z) → r β ω 0 (r, z) is in L 1 (Ω) then for every q ∈ [2, +∞] there exists a positive constant C such that for all time t > 0 u(t) L q (R 3 ) C t (ii) If β ∈ ]2, 3] and (r, z) → (r β +r β-1 |z|+r β-2 |z|) ω 0 (r, z) is in L 1 (Ω) then for every q ∈ [2, +∞] there exists a positive constant C such that for all time t > 0

u(t) -I 0 u G1 • √ t t -3 2 L q (R 3 ) C t 3 2q -1 2 -β 2 . ( 4.3)
(iii) If β ∈ ]3, 4[ and (r, z) → (r β +r β-2 z 2 +r β-3 z 2 ) ω 0 (r, z) is in L 1 (Ω) then for every q ∈ [2, +∞] there exists a positive constant C such that for all time t > 0

u(t) -I 0 u G1 • √ t t -3 2 -J(t)u G2 • √ t t -2 L q (R 3 ) C t 3 2q -1 2 -β 2 .
(4.4)

(iv) If (r, z) → (r 4 + r 2 z 2 + rz 2 ) ω 0 (r, z) is in L 1 (Ω) then for every q ∈ [2, +∞] there exists a positive constant C such that for all time t > 0

u(t) -I 0 u G1 • √ t t -3 2 -J(t)u G2 • √ t t -2 L q (R 3 )
C ln(1 + t) t The first and second order asymptotic terms can still be calculated, for example from the general three-dimensional formulae given by Gallay and Wayne in [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF], but the vorticity has to be assumed continuous in time into the weighted space L 2 (R 3 , (1 + |x|) m dx) 3 for some m > 7 2 . Let us suppose this is the case. The vorticity's expansion in [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF] then shows that the first two asymptotics of the tangential vorticity ω θ are the same as when the swirl is zero: in L 2 (R 3 ), where I 0 = Ω r 2 ω θ (0, r, z)drdz and J(t) = Ω r 2 z ω θ (t, r, z)drdz. The radial and vertical vorticities, however, present now nontrivial terms which contribute to the second-order asymptotics as in L 2 (R 3 ), where K 0 = Ω r 2 z ω r (0, r, z)drdz and L 0 = Ω r(2 -z 2 ) ω z (0, r, z)drdz.

ω θ (t,
ω r (t,
A natural arising question is whether it would be easy to follow a similar approach to what has been done in the present paper when the swirl is not supposed zero, in order to weaken the hypotheses taken on ω by [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF]. to get that ω θ (t 0 ) belongs to L 1 (Ω).

Our assumption is thus weaker than the one from [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF] for any positive time. Let us notice finally that due to the symmetry of our study we do not need every moment of order three of ω 0 to be finite, while (r

B Appendix: A Fixed Point argument

The goal of this appendix is to show that under the hypothesis of (r, z) → (1+r)ω 0 (r, z) belonging to L 1 (Ω), there exists a time T > 0 such that the solution t → ω θ (t) to the integral vorticity equation (2.4) is continuous from [0, T ] into the space L 1 1 (Ω) = {w : Ω → R | w L 1 (Ω) + rw L 1 (Ω) < ∞}. The strategy developed is similar to the two-dimensional case, see for example [START_REF] Ben-Artzi | Global solutions of two-dimensional Navier-Stokes and Euler equations[END_REF][START_REF] Kato | The Navier-Stokes equation for an incompressible fluid in R 2 with a measure as the initial vorticity[END_REF]. We rewrite here, up to the appropriate adaptations, the fixed-point argument that can be found in [10, prop. 4.1]. Let ω 0 belong to L 1 1 (Ω). Define for any T > 0 the function space

X T = ω θ ∈ C 0 ]0, T ], L 4/3 (Ω) ω θ X T < ∞
equipped with the norm ω θ X T = sup 0<t T t 1/4 ω θ (t) L 4/3 (Ω) + t 1/4 rω θ (t) L 4/3 (Ω) .

Let us see that, given (2.9), there exists some positive constant C > 0 independent of T such that S(•)ω 0 X T C (1 + r)ω 0 L 1 (Ω) . In particular, t → S(t)ω 0 belongs to X T whatever T is. Moreover, it also follows from (2.9) that S(•)ω 0 X T -→ 0 as T tends to 0 for the following reasons. By density, for any ε > 0 there exists ω ε 0 such that (1 + r)ω ε 0 L 4/3 (Ω) < ∞ and (1 + r)(ω 0ω ε 0 ) L 1 (Ω) < ε. Hence, using (2.9) twice with β = α = 0 and twice with β = α = 1, C ω 0 -ω ε 0 L 1 (Ω) + r(ω 0 -ω ε 0 ) L 1 (Ω) +t 1/4 ω ε 0 L 4/3 (Ω) + t 1/4 rω ε 0 L 4/3 (Ω) 2ε and so lim T →0 S(•)ω 0 X T = 0. This will be useful latter.

Given ω θ in X T and p ∈ [1, 4 3 ], and taking u the velocity field obtained from ω θ via the axisymmetric Biot-Savart law (3.2), define the map

(t 0 -s) β-1 f L ∞ ([0,t0]) ds + 1 2 f L ∞ ([t0,T [) so f L ∞ ([t0,T [) 2a + 2b 1 β t β 0 f L ∞ ([0,t0]) 2a + 2ab 1 β t β 0 k 0 c k t kβ 0 and f L ∞ ([0,T [) f L ∞ ([0,t0]) + f L ∞ ([t0,T [) aC b,β,γ .
The crucial point of this lemma is that the constant C b,β,γ can be chosen independently of T . The restriction β 1 can be easily removed, but the proof would be longer.

2 since 0 2 -β 3 2

 23 and r |r -ρ| + ρ |r -ρ| + r 2 so r 2|r -ρ| 4 √ t X.

√s

  ) are less than or equal to the quantity Cρ β s

and ] 0

 0 ,1]×R |ω θ (t 0 , r, z)| drdz ]0,1]×R drdz (1 + |z|) m (1 + |z|) m ω θ (t 0 ) L ∞ (Ω)

1 2 + r m+ 1 2 + r 1 2

 11 |z| m )ω 0 L 2 (Ω) involves all the moments of order m of ω 0 .

t 1 / 4 S

 14 (t)[ω 0 -ω ε 0 ] L 4/3 (Ω) + t 1/4 rS(t)[ω 0 -ω ε 0 ] L 4/3 (Ω) +t 1/4 S(t)ω ε 0 L 4/3 (Ω) + t 1/4 rS(t)ω ε 0 L 4/3 (Ω)

Aω θ : t -→ t 0 S 1 γ-β ds if β = 1 .if β = 1 .

 0111 (t -s)div * (u(s)ω θ (s)) ds which is continuous from ]0, T ] into the space {w : Ω → R | w L p (Ω) + rw L p (Ω) < ∞} . UsingLemma C.1. Consider f in L ∞ loc ([0, T [, [0, +∞[) where 0 < T +∞. If there are a 0, b 0 and γ > β > 0 such that ∀t ∈ [0, T [ f (t) a + b t 0 (t -s) β-1 (1 + s) -γ f (s) ds (C.2)with β 1, then there exists a positive constant C b,β,γ such that f L ∞ ([0,T [) a C b,β,γ .Proof. Let us first exhibit one dominating function for f that does not depend on f itself. This is lemma 7.1.1 of[START_REF] Henry | Geometric Theory of Semilinear Parabolic Equations[END_REF], if we use that trivially (1 + s) -γ 1. Denoting by A the linear operatorAg : t -→ b t 0 (t -s) β-1 (1 + s) -γ g(s) ds on L ∞ loc ([0, T [),and remembering that B denotes the Euler Beta function, a recurrence gives that on [0, T [ ∀n ∈ N f (t) a n k=0 c k t kβ + A n+1 f (t) where c 0 = 1, c n+1 = c n bB(β, nβ + 1) and ∀n ∈ N * A n f (t) b d n t 0 (t -s) nβ-1 f (s) ds where d 1 = 1, d n+1 = d n bB(β, nβ). Given that B(β, nβ + 1) and B(β, nβ) are both O(n -β ) when n tends to infinity, A n f (t) tends to zero for every t in [0, T [ and c k t k has an infinite radius of convergence. Thus f (t) a k 0 c k t kβ for all t in [0, T [. Let us now give a uniform bound to f on [0, T [ which does not depend on T . Set t 0= (2bB(β, 1 -β)) 1 γ-β if β < 1, and t 0 = ( 2b γ-β ) Let us see that in the first case t t0 (t -s) β-1 (1 + s) -γ ds t t0 (t -s) β-1 s -β t β-γ 0 ds B(β, 1 -β) t β-γ s) β-1 (1 + s) -γ ds t t0 (t -s) β-1 s -β t β-γTherefore, in both cases, either T t 0 and obviously f L ∞ ([0,T [) a k 0 c k t kβ 0 , or T > t 0 and f L ∞ ([t0,T [) a + b t0 0 (t -s) β-1 (1 + s) -γ f (s) ds + b t t0 (t -s) β-1 (1 + s) -γ f (s) ds a + b t0 0

  If moreover (r 3 + r 2 |z| + r|z|)ω 0 L 1 (Ω) < ∞, then J(t) has a finite limit J ∞ as t → +∞ and

	ω θ (t, r, z) =	rI 0 16 √ π	e -r 2 +z 2 4t	t -5/2 +	rzJ ∞ 32 √ π	e -r 2 +z 2 4t	t -7/2 + o t→+∞	t	1 p -5 2	(1.5)
	holds in L
		r, z) =	rI 0 16 √ π	e -r 2 +z 2

4t t -5/2 + o t→+∞ t 1 p -2 (1.4) holds in L p (Ω, drdz) for every p ∈ [1, +∞]. p (Ω, drdz) for every p ∈ [1, +∞].

  and t > 0. Like S 1 , the family of operators (S 2 (t)) t>0 is not a semigroup nor is it strongly continuous at t = 0 on any L p (Ω).Remark 2.6. G 1 and G 2 defined by (2.24) and (2.35) are such that the first term (2.23) and the second term (2.34) in the asymptotic expansion of the semigroup are both self-similar solutions to the linearised vorticity equation ∂ t ω θ = ∆ω θ -

	ω θ r 2 ,	(2.37)
	meaning that the equalities

  5/2 S 2 (t)ω 0 (• √ t) L p (Ω) tends to zero when t goes to infinity for every p in [1, +∞]. Let us recall the definitions of G 1 , S 1 , G 2 and S 2 , respectively in (2.24), (2.25), (2.35) and (2.36). Let us consider for any a > 1 the indicator function χ a of the truncated subset [ 1

a , a] × [-a, a]

of Ω, and the quantity

  tends to zero in L p (Ω) as t goes to infinity. Let us recall for this the definitions of G 1 , S 1 , G 2 and S 2 from (2.24), (2.25), (2.35), and (2.36).

	r, z) =	rI 0 16 √ π	e -r 2 +z 2 4t	t -5/2 +	rzJ(t) 32 √ π	e -r 2 +z 2 4t	t -7/2 + o t→+∞	t	1 p -5 2	(3.33)
	in L p (Ω) for every p ∈ [1, +∞].
	Proof. In view of (3.28), the goal is to show that t	5 2 ω θ (t, • √	t, • √	t)

  | J(s)| × G 2 L p (Ω) ds,among which five tend to zero when t goes to infinity and one will have to be dominated by a decreasing function of t 0 . Let us remember what has been discussed right above between equation (3.30) and relation(3.32) to understand why the integrands of T 4 and T 6 can be written as they are.

	T 5 (t, p) =	t t/2	t	5 2	S(t -s)div * (u(s) ω θ (s)) • √	t	L p (Ω)	ds,
		t
	T 6 (t, p) =
		t/2
								ds,

  ds which is a case covered by lemma C.1. Let us note that the constants only depend on ω 0 and on the choice of t 0 .Proof of proposition 3.11. Be sure to have in mind the definitions of G 1 , G 2 and S 2 given by (2.24), (2.35) and (2.36) respectively. Starting from (3.30) while remembering (3.32) and (3.31), let us write for all t > 0

Acknowledgment

Many thanks to Thierry Gallay for his support, his advice, his comments and suggestions. The author wants to thank as well the reviewer for their detailed feedback and their helpful remarks.

Proof. When q = +∞, one only has to apply inequality (3.5) together with the propositions named right above, the L ∞ -norms being rigorously the same for u(t) on (R 3 , rdrdθdz) and (Ω, drdz).

When 2 q < ∞, a straightforward generalisation of those propositions is required. Let us consider α ∈ [0, 1] such that α β and 1 q α < 1 q + 1. Estimate (3.4) with 1 p = 1 q + 1+ 1 q -α 2 gives that f (t) L q (R 3 ) = (2π)

1 q r 1 q f (t) L q (Ω) C r α rotf (t) L p (Ω) (4.6) where f (t) is u(t), u(t) -I 0 u G1 ( • √ t )t -3/2 or u(t) -I 0 u G1 ( • √ t )t -3/2 -J(t)u G2 ( • √ t )t -2 depending on the case. Estimate (3.4) requires that p is not chosen equal to 1, which is possible by taking α > 3 q -1 given that β > 3 q -1 and that 1 q + 1 > 3 q -1. Finally, the proof is achieved by using that

when β 2 (by interpolation from proposition 3.4),

• and

the demonstrations of the last three inequalities being the same calculations as for propositions 3.6 and 3.11 as long as α ∈ [0, 1].

Eventually, let us prove theorem 1.2.

Proof of theorem 1.2. Let us keep the same notations as in the previous proof, considering here only the cases when β = 2 and

. Propositions 3.5 and 3.9 are still true with a radial weight r α when 0 α 2, meaning that

The demonstrations of (4.7) and (4.8) are similar to the original ones (respectively (3.20) and (3.33)). Inequality (4.6) can thus be applied with 3 q -1 < α 2, so q can be taken in all ]1, +∞[ . This concludes the proof, remembering that the case q = +∞ is directly included in propositions 3.5 and 3.9 and remembering remark 3.10 about the presence of J ∞ in (1.10).

Axisymmetry with swirl

When the swirl is nonzero, the information of the Navier-Stokes equations is contained in the pair (u θ , ω θ ) through the vorticity-stream formulation

where ψ is the stream function, u = -1 r ∂ z ψ e r + u θ e θ + 1 r ∂ r ψ e z and ω = -∂ z u θ e r + ω θ e θ + 1 r ∂ r (ru θ )e z , see for example. The tangential vorticity ω θ is thus linked to the velocity in the axial plane (u r , u z ), and the swirl u θ is linked to the vorticity (ω r , ω z ) in the axial plane.

A Appendix: On The Hypotheses of theorem 1.1

We come back in this appendix to the comparison between the hypotheses we made for the second point of our theorem (1.5) and those needed for the second-order asymptotic development from [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF]. Applying theorem 4.2 of [START_REF] Gallay | Long-time asymptotics of the Navier-Stokes and vorticity equation on R 3[END_REF] in the zero-swirl axisymmetric case requires that the initial vorticity ω(0) = ω 0 e θ satisfies (r

. This implies in particular that (r

< ∞ for all positive time t > 0 (see [12, prop 2.5]); we can therefore consider any instant t 0 > 0 as the initial time and see that our assumption (considered at t 0 > 0) is a consequence of that.

What we assume in theorem 1.

then Young's inequality to write (1+r) C (r

where the constant C only depends on m > 7 2 . Similar computations hold for r 3 ω θ (t 0 ) L 1 (Ω) and rzω θ (t 0 ) L 1 (Ω) , however ω θ (t 0 ) L 1 (Ω) has to be cut in two parts. Indeed, one has to write using Hölder's inequality that

(2.10) with β = α = 0, Hölder's inequality and then (3.3), see that

for t in ]0, T ]. The same computation gives that t 1-1 p rAω θ (t) L p (Ω) C ω θ 2 X T , using (2.10) with β = α = 1. This entails, by choosing p = 4 3 , that Aω θ belongs to X T and that Aω θ X T C 1 ω θ 2 X T for some positive constant C 1 . More generally, the Lipschitz estimate

holds for every ω θ , ω θ in X T , stemming again from computations similar to (B.1). Consider now the operator A : X T → X T : ω θ → S(•)ω 0 -Aω θ . Fix R > 0 such that 2C 1 R < 1 and denote by B R the closed ball of radius R centered at the origin in X T . One can now chose T > 0 small enough for S(•)ω 0 X T 1 2 R to hold, so that A maps B R into B R and is a strict contraction there and thus has a unique fixed point in B R thanks to the Banach fixed point theorem. By construction, this unique fixed point ω θ is the unique solution of (2.4), which we already know.

To achieve the proof, it only remains to show that t → Aω θ (t) is continuous at 0 for the norm

. Indeed, (S(t)) t 0 is a strongly continuous semigroup on L 1 1 (Ω) and we already said that Aω θ is continuous from ]0, T ] into L 1 1 (Ω). Let us note that ω θ X T -→ 0 as T goes to zero, since one can take T > 0 as small as desired and, from the moment that T is little enough, the Banach fixed point theorem applies in the ball B R of radius R = 2 S(•)ω 0 X T which tends to zero as seen above. Consequently, remembering that t 1-1 p (1+r)Aω θ (t) L p (Ω) C ω θ 2 X T for all t ∈ ]0, T ] and taking p = 1 implies that Aω θ (t) L 1 1 (Ω) -→ 0 when t goes to zero. Therefore

Remark B.1. These arguments can be extended to show that ω θ is also differentiable from ]0, T ] into L 1 1 (Ω). This results from the analyticity of the semigroup (S(t)) t 0 and the fact that A defined above is a locally Lipschitz operator (see (B.2)), as developed for example in [14, §3.3]. The same way, when saying in the proof of lemma 3.

). The same is true as well on L 1 (Ω, (1 + r 2 + r 2 |z|)drdz). The hypothesis made here (C.1) differs from the usual sufficient condition asked for Grönwall's inequality by the presence of the factor (t -s) β-1 in the integrand. This hypothesis is partially covered in the Dan Henry's book [START_REF] Henry | Geometric Theory of Semilinear Parabolic Equations[END_REF], but lemmas 7.1.1 and 7.1.2 from [START_REF] Henry | Geometric Theory of Semilinear Parabolic Equations[END_REF] do not not rigorously include the cases we need here.

C Appendix: A Grönwall lemma