Well-posedness of linearized Taylor equations in magnetohydrodynamics
aa r X i v : . [ m a t h . A P ] M a y WELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONSIN MAGNETOHYDRODYNAMICS
ISABELLE GALLAGHER, DAVID G´ERARD-VARET
Abstract.
This paper is a first step in the study of the so-called Taylor model, introducedby J.B. Taylor in [20]. This system of nonlinear PDE’s is derived from the viscous incom-pressible MHD equations, through an asymptotics relevant to the Earth’s magnetic field.We consider here a simple class of linearizations of the Taylor model, for which we showwell-posedness. Presentation of the model and main result
Introduction.
The concern of this paper is the so-called Taylor model, derived by J.B.Taylor in 1963. The general motivation behind this model is the understanding of the dynamoeffect in the Earth. By dynamo effect, we mean the generation of magnetic energy by theflow of liquid iron in the Earth’s core. This dynamical process has been recognized since thefirst half of the twentieth century, and sustains the magnetic field of the Earth, despite Jouledissipation. We refer to [16, 10] for an introduction to the subject.A standard model in dynamo theory is the so-called incompressible MHD system, which isobtained after coupling and simplifying the Navier-Stokes and Maxwell equations (see [16],[2]). The resulting system reads(1.1) ρ ( ∂ t u + u · ∇ u ) + ∇ p + ρ Ω e × u − µ ∆ u = µ − curl B × B∂ t B = curl ( u × B ) + η ∆ B div u = div B = 0 . The first line corresponds to the Navier-Stokes equation, that describes the evolution of thefluid velocity u and pressure p . The density ρ and viscosity µ are constant. The equationis written in a frame rotating with the Earth, which is responsible for the Coriolis forcingterm ρ Ω e × u , with Ω the angular speed of the Earth and e = e is taken as the rotationaxis. Finally, as one is describing a conducting fluid, one must take into account the Laplaceforce µ − curl B × B exerted by the magnetic field B on the fluid ions, with µ the magneticpermeability constant.The second line is the so-called induction equation, that describes the evolution of themagnetic field. It can be written ∂ t B = curl E , where the electric field E = u × B − η rot B isdeduced from Ohm’s law in a moving medium (see [16] for details). Finally, the divergencefree constraints on u and B correspond to the incompressibility of the fluid and the absenceof magnetic monopole respectively.With regards to the dynamo problem, the MHD system has been the matter of manyworks, see [18, 1, 12, 14, 15] among many. Most of them focus on linear studies : namely, bylinearizing (1.1) around ( u = u ( x ) , B = 0), one is left with(1.2) ∂ t b = curl ( u × b ) + η ∆ b where u is given. In other words, the retroaction of the magnetic field on the fluid is neglected,and one tries to determine which fluid flows allow for the growth of magnetic perturbations b . Date : August 15, 2018.
This amounts to establishing the existence of unstable spectrum for the operator at the right-hand side of (1.2). However, this spectral problem turns out to be difficult. Roughly, to bea dynamo field, u must exhibit some kind of complexity. For instance, if u has too manysymmetries, there is only stable spectrum : this is the point of several antidynamo theorems,see [8, 1]. Also, if one looks for fast dynamos, meaning with a lower bound on the growthrate independent of the magnetic diffusivity η , then the field u must exhibit some lagrangianchaos: we refer to [21, 7] for more on fast dynamos.Hence, a good understading of the Earth’s magnetic field through explicit analytical calcu-lations seems out of reach. Unfortunately, numerical simulation of the MHD system is also avery challenging problem, due to the presence of many small parameters : in a dimensionlessform, (1.1) becomes(1.3) ∂ t u + u · ∇ u + ∇ pε + e × uε − Eε ∆ u = Λ εθ curl B × B∂ t B = curl ( u × B ) + 1 θ ∆ B div u = div B = 0 . The dimensionless parameters ε, E,
Λ and θ are the Rossby, Ekman, Elsasser and magneticReynolds numbers respectively. Typical values for the Earth’s core are ε ∼ − , Λ = O (1) , εθ ∼ − , E ∼ − . We refer to [9] for more.Due to these small values and underlying small-scale phenomena, a direct computationof the solution is not possible. It is thus tempting to simplify (1.3), notably neglecting theinertia and viscous dissipation of the fluid. Proceeding formally, we obtain(1.4) e × u + ∇ p = Λ θ curl B × B∂ t B = curl ( u × B ) + 1 θ ∆ B div u = div B = 0 . This system was introduced formally and briefly discussed by J.B. Taylor in [20]. Its mathe-matical analysis is the subject of the present paper. Let us stress that other ”degeneracies” ofthe MHD system have been recently considered, in link to the magnetic relaxation problemintroduced by K. Moffatt. We refer to [17], [3].1.2.
The Taylor model.
Our long term goal is to understand better and possibly justifythe asymptotics that leads from (1.3) to (1.4). In geophysical contexts, a huge litteraturehas been devoted to evolution equations with a linear skew-symmetric penalization [5, 11].The peculiarity of the present problem is its genuine nonlinear character, originating in thepenalization of the nonlinear Laplace force F L := curl B × B in (1.3). We shall only discusshere the limit Taylor system. For simplicity, we normalize all constants to 1 and thus consider(1.5) e × u + ∇ p = curl B × B∂ t B = curl ( u × B ) + ∆ B div u = div B = 0 . We assume that this system is set in a domain Ω with an impermeable boundary: u · n | ∂ Ω = 0(we do not discuss the boundary conditions on B for the time being).Let us start with general comments on the first equation (1.5a). Time is only a parameterthere, so that we omit it temporarily from notations. This equation involves naturally theCoriolis operator C u := P ( e × u ), where P := Id − ∇ ∆ − div is the Leray projector onto ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 3 divergence free vector fields. It defines a skew-symmetric operator over the space L σ (Ω)of L divergence free fields tangent at ∂ Ω. The balance equation (1.5a) implies the
Taylorconstraint :(1.6) P F L ∈ Range( C ) . Under this constraint, for a given B such that P F L belongs to L σ (Ω), any solution u of (1.5a)can be written u = u m + u g , where • u m belongs to the orthogonal of ker C , and is uniquely determined by P F L . In partic-ular, it can be expressed in terms of B , possibly in an implicit way. Such a field u m will be called magnetostrophic . • u g is any field in ker C . It satisfies(1.7) e × u + ∇ p = 0 , div u = 0 , u · n | ∂ Ω = 0 . Such a field u g will be called geostrophic .Note that, by skew-symmetry of the Coriolis operator and by the Taylor constraint (1.6), P F L must be orthogonal to ker C :(1.8) Z Ω P F L · u g = Z Ω F L · u g = 0 , for all geostrophic fields u g . Inserting the decomposition u = u m + u g into the induction equation (1.5b), we get(1.9) ∂ t B = curl ( u m × B ) + curl ( u g × B ) + ∆ B . As u m = u m ( B ), the first term at the right hand-side can be seen a nonlinear functionalof B . The other term is more delicate, as the geostrophic field u g is a priori not determined.The idea is that the term curl ( u g × B ) should be a kind of Lagrange multiplier associatedwith the Taylor constraint (1.6). From this point of view, a parallel can be drawn withthe incompressible Navier-Stokes: the term curl ( u g × B ) would correspond to the pressuregradient, whereas the Taylor constraint would correspond to the incompressibility condition.Let us for instance consider the case of the ball Ω = B (0 , u g have a simple characterization. It can be shown that they are exactlythose of the form u g = (0 , U θ ( r ) ,
0) in cylindrical coordinates ( r, θ, z ). Hence, condition (1.8)amounts to(1.10) Z (F L ) θ dzdθ = 0along any cylinder r = r in cylindrical coordinates. Moreover, if F L is regular enough, (1.10)is equivalent to the original Taylor constraint (1.6). See [20] for more details.Again, the term curl ( u g × B ) should allow to preserve (1.10) through time. We can write0 = ∂ t Z (F L ) θ dzdθ = ∂ t Z (curl B × B ) θ dzdθ = Z (curl ∂ t B × B + curl B × ∂ t B ) θ dzdθ and substitute for ∂ t B using equation (1.9). This formal manipulation is performed by Taylorin [20]. He finds an elliptic equation of second order on U θ , whose coefficients and sourcedepend on B (the source is explicit in terms of B and u m ). Such an equation can be seen as ananalogue of the Poisson equation on pressure which is derived from the incompressible Navier-Stokes system. Following [20], one may hope to invert this elliptic equation and express inthis way u g in terms of B . Eventually the evolution equation (1.9) may make sense !Of course, above computations and remarks lack mathematical justification. The presentpaper aims at taking a little step forward. ISABELLE GALLAGHER, DAVID G´ERARD-VARET
The Taylor model in the torus.
To avoid technicalities due to boundaries, we con-sider the Taylor system (1.5) in the torus T . The space L σ (Ω) is now replaced by thespace L σ ( T ) of L divergence-free fields with zero mean over T . The Taylor constraint canbe made explicit in this setting. We assume that F L is smooth enough, and take the curl ofequation (1.5a). We find(1.11) ∂ u = curl F L , which is solvable if and only if:(1.12) Z T curl F L ( · , x ) dx = 0 . Under this condition, curl F L has a unique antiderivative with zero mean in x , that is(1.13) u m ( · , x ) = Z x curl F L ( · , y ) dy − Z T Z x curl F L ( · , y ) dy . This field u m is a solution of (1.11), but also of equation (1.5a). Indeed, it follows from (1.11)that ∂ div u m = 0, and as u m has zero mean in x , div u m = 0. Hence, the relation ∂ z u m =curl F L can be written curl ( e × u m ) = curl F L , which is the same as (1.5a). More generally,any solution of (1.5a) is of the form u = u m + u g , where u g is any element in the kernel ofthe Coriolis operator. It is well-known that these elements are the u g = u g ( t, x h ), with x h :=( x , x ), satisfying(1.14) div h u g, h := ∂ u g, + ∂ u g, = 0 and Z T u g ( · , x h ) dx h = 0 . Clearly, u m is orthogonal to any field u g of the previous kind in L σ ( T ). Hence, according tothe terminology of the previous paragraph, we have identified the Taylor constraint (1.12),the magnetostrophic field u m , and the geostrophic fields u g . A straightforward computationshows that (1.12) is equivalent to(1.15) Z T F L, ( · , x ) dx = 0 , P h Z T F L, h ( · , x ) dx = 0where P h is the two-dimensional Leray projector over L σ ( T ). By the identity curl B × B = B · ∇ B + ∇| B | , it can also be written(1.16) Z T B · ∇ h B dx = 0 , P h Z T B · ∇ h B h dx = 0 . It is equivalent to (1.8) as well.One still needs to make sense of (1.9), notably of the Lagrange multiplier curl ( u g × B ).We follow the approach initiated by Taylor in the case of the ball. Assuming that B, u m , u g are smooth enough and satisfy (1.9), we derive an evolution equation for B · ∇ B . We write ∂ t ( B · ∇ B ) = B · ∇ ∂ t B + ∂ t B · ∇ B = B · ∇ (curl ( u g × B )) + (curl ( u g × B )) · ∇ B + F where F = B · ∇ (curl ( u m × B ) + ∆ B ) + (curl ( u m × B ) + ∆ B ) · ∇ B is a nonlinear functional of u m and B , that can be seen by (1.13) as a nonlinear functionalof B alone. Using the relation(1.17) curl ( a × b ) = b · ∇ a − a · ∇ b , valid for all divergence-free vector fields a, b , and after a few simplifications, we obtain(1.18) ∂ t ( B · ∇ B ) + u g · ∇ ( B · ∇ B ) = ( B · ∇ ) u g + F .
ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 5
The role of u g is to preserve the Taylor constraint through time. In the form (1.16), thisamounts to the system of equations (cid:0) P h (cid:1) (cid:18) − Z T ( B · ∇ ) dx u g + u g · Z T ∇ ( B · ∇ B ) dx (cid:19) = (cid:0) P h (cid:1) Z T F dx which splits into(1.19) − Z T ( B · ∇ ) dx u g, h + u g, h · ∇ h Z T ∇ ( B · ∇ B h ) dx + ∇ h p = Z T F h dx , div u g, h = 0for some p = p ( x h ) and(1.20) − Z T ( B · ∇ ) dx u g, = Z T F dx . This set of equations, where t is only a parameter, can be seen an analogue of the Poissonequation for the pressure in Navier-Stokes, or an analogue of the second order equation de-rived by Taylor in the case of the ball, cf the previous paragraph. Roughly, the system (1.19),that is satisfied by the horizontal part of the geostrophic field u g, h , looks like a Stokes equa-tion, whereas the equation (1.20) satisfied by u g, looks like a Poisson equation. The maindifference is that the usual two-dimensional Laplacian operator is replaced by Z T ( B · ∇ ) dx .Moreover, the Stokes-like equation contains a zero order term. This makes unclear the op-timal conditions for which these equations are well-posed. Still, we can state the followingresult. Proposition 1.1.
Let B = B ( x ) be smooth and divergence-free over T . Assume that (1.21) for all x h ∈ T , B h ( x h , · ) has non-constant direction.Then, for sup x h ∈ T (cid:12)(cid:12)(cid:12)(cid:12)Z T ( B · ∇ B h )( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) small enough, and for any smooth F = F ( x ) with zeromean over T , equations (1.19) and (1.20) have unique smooth solutions u g, h and u g, withzero mean over T .Proof. We first consider (1.19). One can associate to it the variational formulation(1.22) Z T Z T ( B h · ∇ h u g ) · ( B h · ∇ h ϕ ) dx dx h + Z T Z T ( B · ∇ B h ) dx · ( u g · ∇ h ϕ ) dx h = Z T Z T F h dx · ϕ dx h for all ϕ in H σ ( T ), that is the set of 2d solenoidal vector fields in H ( T ) with zero mean.Assumption (1.21) is equivalent to the strict Cauchy-Schwarz inequality(1.23) | Z T ( B B )( · , x ) dx | < Z T | B ( · , x ) | dx Z T | B ( · , x ) | dx . uniformly in x h . It follows that for any f = f ( x h ) with zero mean Z T Z T | B h · ∇ h f | dx dx h ≥ c Z T |∇ h f | dx h so that the first term in (1.22) is coercive over H σ ( T ). Under the smallness assumption of theproposition, the whole left hand-side of (1.22) is coercive, which yields a unique solution u g, h by the Lax Milgram lemma. The smoothness of u g, h follows from classical elliptic regularity,as the principal symbol of the operator R T ( B · ∇ ) dx is uniformly elliptic under (1.23).The case of (1.20) is similar and easier: the existence of a unique smooth solution u g, isobtained under the single assumption (1.21), as there is no zero order term. (cid:3) ISABELLE GALLAGHER, DAVID G´ERARD-VARET
Proposition 1.1 opens the way towards a local well-posedness result of the Taylor sys-tem(1.5). Indeed, if the initial datum B is smooth, satisfies (1.21) and ifsup x h ∈ T (cid:12)(cid:12)(cid:12)(cid:12)Z T ( B · ∇ B , h )( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) is small enough(for instance if it is zero), then the right-hand side of (1.9) is well-defined and smooth atinitial time, and one may hope to solve the equation at least for short time. Note furthermorethat (1.5) is (formally) dissipative: we find12 ddt k B k L ( T ) + k∇ B k L ( T ) = − Z T curl ( u × B ) · B = Z T u · (curl B × B ) = Z T u · ( e × u + ∇ p ) = 0 . Still, this energy decay is not enough to build strong solutions, because it does not providea control of higher order derivatives. The current paper, devoted to a linearized analysisof (1.5), can be seen as a first step in the study of these derivatives.1.4.
Linearization and statement of the main result.
From now on, we study the well-posedness of simple linearizations of the Taylor model in T . With Proposition 1.1 in mind,we consider reference states of the form ( u = 0 , B ) with(1.24) B ( x ) := ( B ( x ) , B ( x ) , B has zero mean and non-constant direction, meaning:(1.25) ∀ η ∈ S , k B h · η k L ( T ) > . This last assumption is made coherently with Proposition 1.1. Note that Z T B · ∇ B h = 0, sothat B also satisfies the smallness assumption of the proposition. Let us mention that thecouple ( u = 0 , B ) is not a solution of the source free Taylor model (1.5): one should add aforcing term f = − ∆ B at the right hand-side of (1.5b). But this is a usual approximation,reminiscent of the study of shear flow stability in fluid dynamics.The linearized system is then(1.26) e × u + ∇ p = curl b × B + curl B × b∂ t b = curl ( u × B ) + ∆ b div u = div b = 0where b is now the magnetic perturbation. The main result of this paper is the followingwell-posedness result: Theorem 1.
Assume that B is a zero-mean smooth function of the form (1.24) , satisfyingassumption (1.25) . If b belongs to L σ ( T ) , system (1.26) has a unique solution ( u, b ) suchthat b ∈ C ( R + , L σ ( T )) ∩ L loc ( R + , H σ ( T )) , u ∈ L loc ( R + , H − σ ( T )) , satisfying for some constant C and for all t ≥ : (1.27) k b ( t ) k L + Z t k∇ b ( s ) k L ds ≤ C k b k L exp( Ct ) . The rest of the paper is dedicated to the proof of this theorem. As usual, the keypoint isto establish an a priori estimate of type (1.27) for smooth solutions of (1.26). The existenceand uniqueness of a solution follows then from standard arguments. But the derivation ofthis a priori bound is far from obvious. The difficulty comes from the so-called inductionterm curl ( u × B ) in (1.26b). As will be seen below, one can express u = u [ b ] as a linear ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 7 operator in b , second order in variables ( x , x ). It follows that one can write curl ( u × B )as curl ( u × B ) = L B b , where L B is a third order operator in ( x , x ). The principal part ofthis operator L B is shown to be skew-symmetric, but second order terms remain: one mayfind fields b = b ( x ) such that Z T L B b · b ≥ c B k∇ b k L where roughly, the constant c B > B . In particular, if B islarge, this term can not be absorbed in a standard energy estimate by the term coming fromthe laplacian in b . Hence, the linear system (1.26) may be ill-posed, with growth similar tothe one of the backward heat equation.The point of the paper is to show that such instability does not occur. It is based on acareful normal form argument, annihilating the second order symmetric part by the thirdorder skew-symmetric one. 2. Preliminaries and reductions
Computation of the linear Taylor constraint and the magnetostrophic field.
In order to prove Theorem 1, we first need to compute curl ( u × B ) in terms of b . The approachis the same as in the nonlinear analysis of Paragraph 1.3. We first focus on equation (1.26a),which amounts to ∂ u = curl (curl b × B + curl B × b )and is solvable if and only if(2.1) Z T curl (curl b × B + curl B × b ) dx = 0 . Under this constraint, (1.26a) has a unique solution u m with zero mean in x . Let(2.2) ∂ − := P Z x P , where P f := f − Z T f dx is the projection onto functions with zero mean in x . Then:(2.3) u m = u m, [ b ] + u m, [ b ] , with u m, [ b ] := ∂ − curl (curl b × B ) , u m, [ b ] := ∂ − curl (curl B × b ) . It is easily seen that b curl ( u m, [ b ] × B ) is skew-adjoint with respect to the L scalarproduct. Moreover, relying on the identity (1.17), we can write:(2.4) curl ( u m, [ b ] × B ) = B h · ∇ h ∂ − ( B h · ∇ h curl b − curl b · ∇ B ) − ∂ − ( B h · ∇ h curl b − curl b · ∇ B ) · ∇ B = B h · ∇ h ∂ − B h · ∇ h (cid:0) ∇ h (cid:1) × b + R m, b . Again, b B h · ∇ h ∂ − B h · ∇ h (cid:0) ∇ h (cid:1) × b is skew-adjoint with respect to the L scalar product.It is a third operator in variables x , x , regularizing in x thanks to ∂ − . The remainingoperator R m, is also skew-adjoint (because the total operator is), second order in x , x ,bounded in x . As regards b → curl ( u m, [ b ] × B ), we use again identity (1.17), and write(2.5) curl ( u m, [ b ] × B ) = B h · ∇ h ∂ − ( b · ∇ curl B − (curl B ) h · ∇ h b ) − ∂ − ( b · ∇ curl B − (curl B ) h · ∇ h b ) · ∇ B .
We recall that B depends only on x , so that the vertical component of curl B is zero. In theright-hand side, only the term b → − B h · ∇ h ∂ − (curl B ) h · ∇ h b is second order in x , x (and ISABELLE GALLAGHER, DAVID G´ERARD-VARET regularizing in x ), all other terms are first order in x , x (and regularizing in x ). We cansplit this second order term into self-adjoint and skew-adjoint terms: we end up with(2.6) curl ( u m, [ b ] × B ) = − (cid:0) B h ·∇ h ∂ − (curl B ) h ·∇ h b − (curl B ) h ·∇ h ∂ − B h ·∇ h b (cid:1) + R m, b where the operator R m, gathers a second order skew-adjoint operator and a first orderoperator in x , x . Eventually, we can write(2.7) curl ( u m × b ) = A m b + C m b + R m b where A m and C m are respectively third and second order operators, with A m skew-adjointand C m self-adjoint, defined by(2.8) A m b := B h · ∇ h ∂ − B h · ∇ h (cid:0) ∇ h (cid:1) × b C m b := − B h · ∇ h ∂ − (curl B ) h · ∇ h b + 12 (curl B ) h · ∇ h ∂ − B h · ∇ h b and where R m is the sum of a skew-adjoint operator of second order in x , x and of a firstorder operator in x , x (both bounded in x ). In particular(2.9) kR m b k L ( T ) ≤ C k b k H ( T ) , (cid:12)(cid:12)(cid:12) Z T R m b · b dx (cid:12)(cid:12)(cid:12) ≤ C k b k H ( T ) k b k L ( T ) , for all smooth fields b with zero mean.2.2. Computation of the geostrophic field.
Like in the nonlinear Taylor system, anysolution u of (1.26a) reads u = u m + u g , where u m is defined in (2.3), and u g is in the kernelof the Coriolis operator. As before, u g = u g ( t, x h ), with div h u g, h = 0. Then,(2.10) ∂ t b = curl ( u m × B ) + curl ( u g × B ) + ∆ b . The term curl ( u g × B ) is there to preserve the linear Taylor constraint (2.1) through time.To determine u g , we proceed exactly as in the nonlinear case, see the lines around (1.19)and (1.20). Here, we get(2.11) (cid:0) P h (cid:1) ( − ∆ B ) u g = (cid:0) P h (cid:1) Z T F dx where F = B h · ∇ h (curl ( u m × B ) + ∆ b ) + (curl ( u m × B ) + ∆ b ) · ∇ B and ∆ B := Z T ( B h · ∇ h ) dx . The proof of Proposition 1.1, on the well-posedness of (1.19)-(1.20), applies to (2.11). Underassumption (1.25), if b is smooth (which implies that u m is smooth), (2.11) has a uniquesmooth solution with zero mean: u g = (cid:0) P h (cid:1) ( − ∆ B ) − Z T F dx . Note that P h and ∆ B commute, since B only depends on x . Note also that ∆ B is asecond order operator, uniformly elliptic by assumption (1.25). Hence, ( − ∆ B ) − gains twoderivatives in the Sobolev scale. Taking this into account, and using the decomposition (2.7),we can write(2.12) u g = u highg [ b ] + u lowg [ b ]where u highg and u lowg are both quasigeostrophic and satisfy(2.13) u highg [ b ] = (cid:0) P h (cid:1) ( − ∆ B ) − Z T B h ( x ) · ∇ h A m b ( · , x ) dx , ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 9 and where u lowg satisfies the estimate(2.14) k u lowg [ b ] k H s ( T ) ≤ C k b k H s +1 ( T ) for all s ≥ b with zero mean. It follows that(2.15) (cid:13)(cid:13)(cid:13) curl (cid:16) u lowg [ b ] × B (cid:17)(cid:13)(cid:13)(cid:13) H s ( T ) ≤ C k b k H s +2 ( T ) . The operator curl ( u lowg [ · ] × B ) therefore is a second order operator, but one notes that if b satisfies (2.1), one has Z T (curl b × B + curl B × b ) · v g = 0 for all geostrophic fields v g . With the choice v g = u lowg , we therefore obtain that Z T curl ( u lowg × B ) · b = Z T u lowg · ( B × curl b )= Z T u lowg · (curl B × b )so that(2.16) (cid:12)(cid:12)(cid:12)(cid:12)Z T curl ( u lowg × B ) · b (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k b k H ( T ) k b k L ( T ) . Hence, the operator curl ( u lowg [ · ] × B ) will act as a first order operator in an L energy estimate.Eventually, as regards u highg [ b ], we can notice that Z T ( B h · ∇ h A m b ) h dx = −∇ ⊥ h Z T (cid:0) ( B h · ∇ h ) ∂ − ( B h · ∇ h ) b (cid:1) dx where we have noted ∇ ⊥ h = (cid:16) − ∂ ∂ (cid:17) so that it is a 2d divergence-free field. Thus, there is noneed for P h in (2.13), and we find(2.17) u highg [ b ] = ( − ∆ B ) − Z T B h ( x ) · ∇ h A m b ( · , x ) dx . We can then writecurl ( u highg × B ) = B h · ∇ h u highg − u highg · ∇ B = Π A m b − u highg · ∇ B where Π is the self-adjoint bounded operator defined by(2.18) Π f := B h · ∇ h ( − ∆ B ) − Z T B h · ∇ h f dx . Eventually, to mimic decomposition (2.7), we write(2.19) curl ( u g × B ) = A g b + C g b + R g b where(2.20) A g b := Π A m b + A m Π b , C g b := −A m Π b − u highg [ b ] · ∇ B , R g b := curl ( u lowg [ b ] × B ) . Note that A g , like A m , is skew-adjoint. In contrast C g is not self-adjoint, contrary to C m .Also, by (2.15)-(2.16) we have for all smooth b with zero mean(2.21) kR g b k L ( T ) ≤ C k b k H ( T ) and (cid:12)(cid:12)(cid:12) Z T R g b · b dx (cid:12)(cid:12)(cid:12) ≤ C k b k H ( T ) k b k L ( T ) for all smooth b satisfying (2.1) . Fourier transform.
From the two previous paragraphs, we can reformulate the lin-earized Taylor system (1.26) in terms of b only. The evolution equation on b is(2.22) ∂ t b = A b + C b + R b + ∆ b with A = A m + A g , C = C m + C g , R = R m + R g , see (2.8)-(2.9),(2.20)-(2.21). Moreover, the solution b should satisfy the linear Taylor con-straint and divergence-free constraints, namely(2.23) (cid:0) P h (cid:1) Z T (curl b × B + curl B × b ) = 0 , div b = 0 . The point is to establish the a priori estimate (1.27) for smooth solutions b of (2.22)-(2.23).This is easier after taking a horizontal Fourier transform, since B does not depend on x h .Given horizontal Fourier modes ξ := ( ξ , ξ ) ∈ (2 π Z ) , we write b ( t, x ) = e iξ · x h b b ( t, ξ, x ) . Is is easily seen that the zero mode b b ( t, , x ) satisfies the heat equation, so that it decaysexponentially. We can therefore focus on the case ξ = 0. From now on we omit the dependenceof b b on ξ which is fixed and simply write b b ( t, x ).We introduce the notation, recalling that ξ ⊥ := (cid:16) − ξ ξ (cid:17) ,(2.24) β ξ ( x ) := B h ( x ) · ξ , β ′ ξ ⊥ ( x ) := B ′ h ( x ) · ξ ⊥ , e ( x ) := β ξ ( x ) k β ξ k L ( T ) · Let us stress that: ∃ c, C > , c | ξ | ≤ k β ξ k L ( T ) ≤ C | ξ | , ∀ ξ ∈ (2 π Z ) , the lower bound coming from (1.25). We denote by Π e the orthogonal projection over R e in L ( T ). NamelyΠ e f := (cid:18)Z T e f dx (cid:19) e = 1 k β ξ k L ( T ) (cid:18)Z T β ξ f dx (cid:19) β ξ . To lighten notation, we will also call Π e the orthogonal projection over R e in L ( T ) ,acting component-wise: b b (Π e b b , Π e b b , Π e b b ).With such notations, we find that b Π = − Π e and therefore ∂ t b b = b A b b + b C b b + b R b b + ( ∂ − | ξ | ) b b , with(2.25) b A b b := (cid:16)d A m − Π e d A m − d A m Π e (cid:17) b b , (2.26) b C b b := − β ξ ∂ − β ′ ξ ⊥ b b + 12 β ′ ξ ⊥ ∂ − β ξ b b + d A m Π e b b − i k β ξ k L ( T ) Z T β ξ (cid:16)d A m b b (cid:17) dx B ′ , where(2.27) d A m b b := − iβ ξ ∂ − β ξ (cid:0) ξ (cid:1) × b b , and with, for all smooth b with zero mean(2.28) k b R ˆ b k L ( T ) ≤ C | ξ | k b b k L ( T ) and (cid:12)(cid:12)(cid:0) b R b b | b b (cid:1)(cid:12)(cid:12) ≤ C | ξ | k b b k L ( T ) for all smooth b satisfying (2.1).See (2.9)-(2.21). Here, ( | ) is the usual scalar product on L ( T , C ). ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 11
The operator b A is skew-selfadjoint, and therefore satisfies,(2.29) ( b A b b | b b ) = 0 . Furthermore, one can notice that(2.30) Π e d A m Π e = 0due to the fact that for all V ∈ R : Z T β ξ · d A m ( β ξ V ) dx = − i Z T β ξ ∂ − β ξ dx (cid:0) ξ (cid:1) × V = 0 . Hence, we can write in a condensed way:(2.31) b A = Π ⊥ e d A m Π ⊥ e , Π ⊥ e = Id − Π e . As regards the operator b C , we find(2.32) k b C b b k L ( T ) ≤ C (cid:16) | ξ | k b b k L ( T ) + | ξ | k Π e b b k L ( T ) (cid:17) where the O ( | ξ | ) comes from the term d A m Π e in the expression for b C . We now state thefollowing useful lemma. Lemma 2.1.
Let b satisfy the Taylor constraint (2.1) . Then (2.33) Π e b b = β ξ k β ξ k L ( T ) Z T (cid:16) i b b B ′ − i | ξ | − β ′ ξ b b (cid:0) ξ (cid:1) (cid:17) dx . In particular (cid:13)(cid:13) Π e b b (cid:13)(cid:13) L ≤ C | ξ | k b b k L , while (2.34) d A m Π e b b = d A m β ξ k β ξ k L ( T ) Z T i b b B ′ dx ! . Proof.
We notice that the Fourier transform of the Taylor constraint (2.1) can be written Z T (cid:0) iβ ξ b b + B ′ b b (cid:1) dx = i (cid:0) ξ (cid:1) b p . This implies on the one hand thatΠ e b b = β ξ k β ξ k L ( T ) Z T (cid:0) i b b B ′ + (cid:0) ξ (cid:1) b p (cid:1) dx and also b p = 1 | ξ | Z T (cid:0) β ξ b b h · ξ − i b b B ′ h · ξ (cid:1) dx . The formula follows from the divergence free condition b b h · ξ = i b b ′ and an integration byparts. (cid:3) It follows from this lemma and (2.32) that(2.35) k b C b b k L ( T ) ≤ C | ξ | k b b k L ( T ) . Reduction to large horizontal frequencies.
Let us write an L energy estimate onthe equation satisfied by b b . From (2.28),(2.29) and (2.35), we deduce(2.36) 12 ddt k b b ( t ) k L ( T ) + | ξ | k b b ( t ) k L ( T ) + k ∂ b b ( t ) k L ( T ) ≤ C (1 + | ξ | ) k b b ( t ) k L ( T ) . Gronwall’s lemma gives therefore directly that k b b ( t ) k L ( T ) ≤ k b b (0) k L ( T ) exp (cid:0) C (1 + | ξ | ) t (cid:1) so from now on we may restrict our attention to the case when | ξ | ≫
1: we introduce ε := 1 | ξ | ≪ , η := ε ξ ∈ S and express equation (2.22) in terms of ε and η . Similarly to β ξ and β ′ ξ ⊥ , we define(2.37) β η ( x ) := B h ( x ) · η , β ′ η ⊥ ( x ) := B ′ h ( x ) · η ⊥ . Note that(2.38) δ := min η h ∈ S k β η k L ( T ) > . Instead of the operators b A = b A ( ξ, ∂ ) , b C = b C ( ξ, ∂ ) , b R = b R ( ξ, ∂ ) , we introduce A = A ( η, ∂ ) := ε b A (cid:16) ηε , ∂ (cid:17) , C = C ( η, ∂ ) := ε b C (cid:16) ηε , ∂ (cid:17) , and R = R ( η, ∂ ) := ε b R (cid:16) ηε , ∂ (cid:17) . We also drop the b on b b , denoting b instead. We have notably(2.39) A b = Π ⊥ e A m Π ⊥ e b , and(2.40) C b := C m b + A m β η k β η k L ( T ) Z T i b B ′ dx ! − i k β η k L ( T ) Z T β η ( A m b ) dx B ′ where(2.41) A m b := − iβ η ∂ − β η ( η ) × b, C m b = − β η ∂ − β ′ η ⊥ b + 12 β ′ η ⊥ ∂ − β η b . Note that the second term at the right-hand side of (2.40) comes from (2.34). We get(2.42) ∂ t b = Abε + Cbε + Rbε − ε b + ∂ b . Operators A and C are independent of ε . Moreover, ( Ab | b ) = 0, where ( | ) is the usual scalarproduct over L ( T ; C ). The remainder R is bounded uniformly in ε , and (2.28) implies that(2.43) 1 ε | ( Rb | b ) | ≤ Cε k b k L ( T ) for all smooth b satisfying the Fourier version of the Taylor constraint (cid:18) Id − | ξ | − ξ ⊗ ξ (cid:19) Z T (cid:0) β η b + εb B ′ (cid:1) dx = 0 . So, for ε small enough, it can be controlled by the term − ε k b k L ( T ) coming from the diffusion.The obstacle to estimate (1.27) is therefore the term ε ( Cb | b ). ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 13
General strategy.
To prove Theorem 1, we shall resort to a normal form argument.In this section we present the method in a formal way. We denote generically by O ( ε α ) anoperator which may depend on on ε and η , but is uniformly bounded by ε α in operator normover L ( T ). The idea is to change unknown by defining d := (Id + εQ ) b with Q = O (1). We then expect that b = (Id + εQ ) − d = (Id − εQ ) d + O ( ε ) d and ∂ t d = (Id + εQ ) ∂ t b + O ( ε ) d . It follows that d should satisfy an equation of the type ∂ t d − ( ∂ − ε ) d = 1 ε Ad + 1 ε (cid:0) C + [ Q, A ] + R (cid:1) d + O (cid:16) ε (cid:17) d . The idea is to take Q as a solution of the homological equation [ A, Q ] = C . We refer to[6, 4, 13] for applications of this strategy. Nevertheless, solving this equation is difficult inour case. To explain why, let us consider a simplified version where we neglect all termscoming from the geostrophic part curl ( u g × b ). This means we consider the equation[ A m , Q ] = C m with A m and C m defined in (2.41).If the matrix ( η ) × were invertible, and if the function β η were not vanishing anywhereon T , then a natural candidate for Q would be: Q = − i β ′ η ⊥ β η (( η ) × ) − which satisfies formally [ A m , Q ] = C m .Unfortunately, trying to make this kind of construction rigorous, we face several difficulties:(1) The matrix ( η ) × is not invertible.(2) The full expression of A and C involves additional terms, related to the geostrophicfield u g , notably the orthogonal projection Π e .(3) For any x , there is some η ∈ S such that β η ( x ) = 0. From a different perspective,one may say that the multiplication by β η is not a bounded operator over L ( T ).Hence, the normal form cannot be applied directly, and we need additional arguments to over-come the issues just mentioned. Roughly, the first difficulty, related to the kernel of ( η ) × willbe handled thanks to the divergence-free constraint, which makes b ( t, x ) almost orthogonalto this kernel (up to a power of ε ). The second one will be handled taking advantage of theTaylor constraint, notably through the identity of Lemma 2.1, which now reads(2.44) Π e b = ε β η k β η k L ( T ) Z T (cid:16) i b b B ′ − i β ′ η b ( η ) (cid:17) dx . Eventually, the last problem will be overcome by a spectral truncation. This means weshall only perform the construction of Q in a finite-dimensional setting, projecting on the lowmodes of the skew-self-adjoint operator A . The high modes (that as we will show correspondto high frequencies in x ), will be controlled, and discarded, thanks to the presence of theoperator ∂ − in C . Normal form argument
Using the divergence-free constraint.
Let Π η the orthogonal projection in C overthe vector ( η, t , and Π ⊥ η := Id − Π η . We shall use the divergence free constraint on b , whichnow reads(3.1) iη · b h + ε∂ b = 0to show that we can somehow restrict to the control of Π ⊥ η b . More precisely we have thefollowing result. Proposition 3.1.
For all b, c in L , divergence free in the sense of (3.1) , we have (cid:12)(cid:12)(cid:12) ( Cb | c ) − ( C Π ⊥ η b | Π ⊥ η c ) (cid:12)(cid:12)(cid:12) ≤ C ε k b k L ( T ) k c k L ( T ) . Proof.
Let us start by proving that(3.2) k C Π η b k L ( T ) ≤ Cε k b k L ( T ) . From (3.1), Π η b = − ( ε∂ (Π ⊥ η b ) ) (cid:18) η (cid:19) . From (2.40), we get(3.3) C Π η b := − β η ∂ − β ′ η ⊥ Π η b + 12 β ′ η ⊥ ∂ − β η Π η b = ε (cid:18) β η ∂ − (cid:16) β ′ η ⊥ ∂ (Π ⊥ η b ) (cid:17) − β ′ η ⊥ ∂ − (cid:16) β η ∂ (Π ⊥ η b ) (cid:17)(cid:19) (cid:18) η (cid:19) We notice that ∂ − ( β η ∂ Π ⊥ η b ) = β η Π ⊥ η b − Z T ( β η Π ⊥ η b ) − ∂ − (cid:16) β ′ η Π ⊥ η b (cid:17) so that(3.4) k ∂ − ( β η ∂ Π ⊥ η b ) k L ( T ) ≤ C k b k L ( T ) and the same with β ′ η ⊥ instead of β η . Inequality (3.2) follows. Then to end the proof, writing( Cb | c ) = ( C Π η b | c ) + ( C Π ⊥ η b | Π η c ) + ( C Π ⊥ η b | Π ⊥ η c ) , it suffices to prove that (cid:12)(cid:12) ( C Π ⊥ η b | Π η c ) (cid:12)(cid:12) ≤ Cε k b k L ( T ) k c k L ( T ) . We have ( C Π ⊥ η b | Π η c ) = − ε (cid:16) C Π ⊥ η b | ∂ (Π ⊥ η c ) ( η ) (cid:17) = ε (cid:16) ∂ C Π ⊥ η b | (Π ⊥ η c ) ( η ) (cid:17) . Using again (3.4), it is easily seen that k ∂ C Π ⊥ η b k L ( T ) ≤ C k Π ⊥ η b k L ( T ) . The result followsfrom the Cauchy-Schwarz inequality. (cid:3) ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 15
Spectral analysis of A . In this paragraph we diagonalize the operator A , recalling Ab := i Π ⊥ e β η ∂ − β η Π ⊥ e b × (cid:18) η (cid:19) . The following result holds.
Proposition 3.2.
Given η ∈ S , recall (3.5) e := β η / k β η k L ( T ) and for all k ∈ Z ∗ define (3.6) µ k := k β η k L ( T ) ikπe k := β η,k / k β η,k k L ( T ) , with β η,k ( x ) := β η ( x ) exp (cid:16) µ k Z x β η ( y ) dy (cid:17) . Then, the set (Φ ± k ) k ∈ Z defined by Φ − k := (cid:18) η (cid:19) e k , Φ +2 k := 1 √ (cid:18) η ⊥ i (cid:19) e k , Φ +2 k +1 := 1 √ (cid:18) η ⊥ − i (cid:19) e k is an orthonormal basis of L ( T ) satisfying A Φ − k = A Φ +0 = A Φ +1 = 0 , A Φ +2 k = µ k Φ +2 k , A Φ +2 k +1 = − µ k Φ +2 k +1 , k ∈ Z ∗ . Proof.
We start by considering the compact operator D := Π ⊥ e β η ∂ − β η Π ⊥ e which clearly sends e to 0 and Π ⊥ e L ( T ) to itself. Let µ = 0 be an eigenvalue of D in Π ⊥ e L ( T ), and f an associate eigenfunction. Then since Π ⊥ e f = f this means that f must solve, in Π ⊥ e L ( T ), the equation(3.7) β η ∂ − β η f = µf + αβ η , for some α ∈ C . Moreover Π ⊥ e f = f means that β η f has zero average with respect to x , so that u := ∂ − ( β η f )satisfies ∂ u = β η f . Hence, u must satisfy β η u = µβ η ∂ u + αβ η hence u ( x ) = exp (cid:16) µ Z x β η ( y ) dy (cid:17) − Z T exp (cid:16) µ Z x β η ( y ) dy (cid:17) dx , and α = − Z T exp (cid:16) µ Z x β η ( y ) dy (cid:17) dx . But u (0) = u (1) thereforeexp (cid:16) µ Z T β η ( y ) dy (cid:17) = 1which implies that µ = µ k where for k ∈ Z ∗ ,µ k := 12 ikπ Z T β η ( x ) dx . Finally we have f ( x ) = 2 ikπ k β η k L β η ( x ) exp (cid:16) ikπ k β η k L Z x β η ( y ) dy (cid:17) . It follows that the family ( e k ) k ∈ Z ∗ defined in (3.6) is an orthonormal basis of Π ⊥ e L ( T ),while ( e k ) k ∈ Z is an orthonormal basis of L ( T ). Finally to recover an orthonormal basis ofeigenfunctions of A in ( L ( T )) , we use the fact that (cid:18) η ⊥ ± i (cid:19) , (cid:18) η (cid:19) is a basis of eigenvectors of the operator i · × (cid:18) η (cid:19) , where the two first are associated withthe eigenvalues ± (cid:3) Recalling that Π η is the projector onto (cid:18) η (cid:19) remark thatΠ η b = X k ∈ Z ( b | Φ − k )Φ − k and Π ⊥ η b = X k ∈ Z ( b | Φ + k )Φ + k . Reduction to a finite dimensional setting.
In order to build up our normal form,we need to reduce to a finite dimensional setting, cf the discussion in paragraph 2.5. Roughly,Proposition 3.1 will help us to get rid of the infinite dimensional subspacevect( { Φ − k , k ∈ Z } ) ⊂ ker A .
The point is then to be able to restrict to vect( { Φ k + , | k | ≤ N ) } , for some (possibly large) N .We first define spectral projectors on low and high modes of A :Π ♭N b := N +1 X k = − N ( b | Φ + k ) = X | k |≤ N (cid:0) ( b | Φ +2 k )Φ +2 k + ( b | Φ +2 k +1 )Φ +2 k +1 (cid:1) , Π ♯N := Π ⊥ η − Π ♭N . Let us collect some properties of these spectral projectors.
Lemma 3.3.
For any divergence free vector field in the sense of (3.1) b in L ( T ) , thefollowing holds for all integers n , under Assumption (2.38) : k ∂ n Π ♭N b k L ( T ) . N n + δ k Π ♭N b k L ( T ) , k Π ♭N ∂ n Π ♯N b k L ( T ) . N n + δ k Π ♯N b k L ( T ) , and k Π ♭N ∂ n b k L ( T ) . N n + δ k b k L ( T ) . Proof.
The third relation follows easily from the first two and from the inequality k Π ♭N ∂ n b k L ( T ) ≤ k Π ♭N ∂ n Π ♭N b k L ( T ) + k Π ♭N ∂ n Π ♯N b k L ( T ) ≤ k ∂ n Π ♭N b k L ( T ) + k Π ♭N ∂ n Π ♯N b k L ( T ) . To fix ideas, we assume that N is odd. To prove the first inequality, we write ∂ Π ♭N b = N +1 X k = − N ( b | Φ + k ) ∂ Φ + k = X | k |≤ N (cid:16) ( b | Φ +2 k ) ∂ Φ +2 k + ( b | Φ +2 k +1 ) ∂ Φ +2 k +1 (cid:17) . Recalling (3.6) we know that(3.8) ∂ e k ( x ) = 1 k β η k L ( T ) (cid:0) β ′ η ( x ) + 1 µ k β η ( x ) (cid:1) exp (cid:16) µ k Z x β η ( y ) dy (cid:17) ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 17 and thanks to (2.38) we have ∀| k | ≤ N , | µ k | . Nδ · It follows that ∀| k | ≤ N , k ∂ e k k L ( T ) . Nδ so by the Cauchy-Schwarz inequality k ∂ Π ♭N b k L ( T ) . N δ k Π ♭N b k L ( T ) . The argument is identical for higher derivatives.For the second inequality, we write(3.9) Π ♭N ∂ Π ♯N b = N +1 X j = − N X k [ − N, N +1] ( b | Φ + k )( ∂ Φ + k | Φ + j )Φ + j = − N +1 X j = − N X k [ − N, N +1] ( b | Φ + k )(Φ + k | ∂ Φ + j )Φ + j = − X | j |≤ N X | k | >N (cid:16) ( b | Φ +2 k )( e k | ∂ e j )Φ +2 j + ( b | Φ +2 k +1 )( e k | ∂ e j )Φ +2 j +1 (cid:17) , noticing that by construction (Φ +2 k | Φ +2 kj +1 ) = 0.By the definitions (3.5) and (3.6), setting µ = + ∞ (that is µ = 0), we get( e k | ∂ e j ) = 1 k β η k L ( T ) Z T β η ( x ) (cid:0) β ′ η ( x ) + 1 µ j β η ( x ) (cid:1) × exp (cid:16)(cid:0) µ k − µ j (cid:1) Z x β η ( y ) dy (cid:17) dx = 1 k β η k L ( T ) Z T β η ( x ) (cid:0) β ′ η ( x ) + 1 µ j β η ( x ) (cid:1) × exp (cid:16) iπ ( k − j ) 1 k β η k L ( T ) Z x β η ( y ) dy (cid:17) dx = 1 k β η k L ( T ) ( β ′ η + 1 µ j β η | e j − k ) . Hence k Π ♭N ∂ Π ♯N b k L ( T ) ≤ k β η k L ( T ) X | j |≤ N (cid:12)(cid:12)(cid:12) X | k | >N ( b | Φ +2 k )( β ′ η + 1 µ j β η | e j − k ) (cid:12)(cid:12)(cid:12) + 1 k β η k L ( T ) X | j |≤ N (cid:12)(cid:12)(cid:12) X | k | >N ( b | Φ +2 k +1 )( β ′ η + 1 µ j β η | e j − k ) (cid:12)(cid:12)(cid:12) ≤ k β η k L ( T ) X | j |≤ N k β ′ η + 1 µ j β η (cid:13)(cid:13) L ( T ) × (cid:16)(cid:13)(cid:13) X | k | >N ( b | Φ +2 k ) e j − k (cid:13)(cid:13) L ( T ) + (cid:13)(cid:13) X | k | >N ( b | Φ +2 k +1 ) e j − k (cid:13)(cid:13) L ( T ) (cid:17) so finally k Π ♭N ∂ Π ♯N b k L ( T ) ≤ k β η k L ( T ) X | j |≤ N (cid:13)(cid:13) β ′ η + 1 µ j β η (cid:13)(cid:13) L ( T ) X k [ − N, N +1] | ( b | Φ + k ) | ≤ CN δ k Π ♯N b k L ( T ) . The argument is identical for higher order derivatives. (cid:3)
Remark 3.4.
The decay of the scalar product ( e k | ∂ e j ) as | k − j | goes to infinity could bespecified thanks to stationary phase theorems. For instance, the term k β η k L ( β ′ η | e k − j ) in theright hand-side of (3.10) is proportional to an integral of the form Z T Ψ ′′ ( x ) exp (cid:0) i ( k − j )Ψ( x ) (cid:1) dx with Ψ( x ) := Z x β η ( y ) dy . The behaviour of this integral depends on the stationarypoints of Φ. For instance, if β η does not vanish, the integral will behave like | k − j | − n forall n , because Φ has no stationary point. A contrario , if β η has a (say single) zero of order m , then Φ has a critical point of order 2 m , and then, according to [19], Chapter VIII.1.3, onehas (cid:12)(cid:12)(cid:12)(cid:12)Z T Ψ ′′ ( x ) exp (cid:0) i ( k − j )Ψ( x ) (cid:1) dx (cid:12)(cid:12)(cid:12)(cid:12) . C | k − j | − m m +1 . The key proposition to be able to neglect the high modes is the following:
Lemma 3.5.
For any divergence-free vector field in the sense of (3.1) b in L ( T ) , thefollowing holds: (cid:12)(cid:12)(cid:12) ( C Π ⊥ η b | Π ⊥ η b ) − ( C Π ♭N b | Π ♭N b ) (cid:12)(cid:12)(cid:12) ≤ ǫ ( N ) δ k b k L ( T ) where ǫ ( N ) goes to zero as N → + ∞ .Proof. We decompose( C Π ⊥ η b | Π ⊥ η b ) = ( C Π ♭N b | Π ♭N b ) + ( C Π ⊥ η b | Π ♯N b ) + ( C Π ♯N b | Π ♭N b ) . We must show that the last two terms go to zero as N → + ∞ . They are very similar, so wefocus on ( C Π ⊥ η b | Π ♯N b ). We first consider the magnetostrophic part, recalling the decomposi-tion (2.40). We have( C m Π ⊥ η b | Π ♯N b ) = 12 X j ∈ Z , | k | >N (cid:16) ( b | Φ +2 j )( b | Φ +2 k ) + ( b | Φ +2 j +1 )( b | Φ +2 k +1 ) (cid:17) ( C m e j | e k ) . We write k β η k L ( C m e j | e k ) = − Z T (cid:18) ∂ − (cid:0) β ′ η ⊥ β η exp (cid:0) µ j Z x β η (cid:1)(cid:1)(cid:19) β η exp (cid:0) − µ k Z x β η (cid:1) + 12 Z T (cid:18) ∂ − (cid:0) β η exp (cid:0) µ j Z x β (cid:1)(cid:1)(cid:19) β ′ η ⊥ β η exp (cid:0) − µ k Z x β η (cid:1) =: I jk + I jk . ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 19
An integration by parts yields (noticing that k = 0 by construction) I jk = 12 Z T β ′ η ⊥ β η exp (cid:0) µ j Z x β η (cid:1) (cid:18) ∂ − (cid:0) β η exp (cid:0) − µ k Z x β η (cid:1)(cid:1)(cid:19) = − µ k Z T β ′ η ⊥ β η exp (cid:0) µ j Z x β η (cid:1) (cid:18) exp (cid:0) − µ k Z x β η (cid:1) − Z T exp (cid:0) − µ k Z x β η (cid:1)(cid:19) = − µ k β ′ η ⊥ | e k − j ) + µ k (cid:18)Z T exp (cid:0) − µ k Z x β η (cid:1)(cid:19) ( β ′ η ⊥ | e − j )Similarly, for j ∈ Z ∗ , I jk = µ j β ′ η ⊥ | e k − j ) − µ j (cid:18)Z T exp (cid:0) µ j Z x β η (cid:1)(cid:19) ( β ′ η ⊥ | e k )This implies(3.10) 12 k β η k L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X j ∈ Z ∗ , | k | >N (cid:16) ( b | Φ +2 j )( b | Φ +2 k ) + ( b | Φ +2 j +1 )( b | Φ +2 k +1 ) (cid:17) ( C m e j | e k ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:18) X j ∈ Z ∗ , | k | >N (cid:12)(cid:12)(cid:12)(cid:12) k − j (cid:12)(cid:12)(cid:12)(cid:12) | ( β ′ η ⊥ | e k − j ) | b j b k + X j ∈ Z ∗ , | k | >N b k | k | | ( β ′ η ⊥ | e − j ) | b j + X j ∈ Z ∗ , | k | >N b j | j | | ( β ′ η ⊥ | e k ) | b k (cid:19) with b ℓ := (cid:12)(cid:12) ( b | Φ +2 ℓ ) (cid:12)(cid:12) + (cid:12)(cid:12) ( b | Φ +2 ℓ +1 ) (cid:12)(cid:12) .To treat the first term at the right-hand side of (3.10), we split the sum over j , distin-guishing between | j | ≤ N and | j | > N . One has on the one hand X < | j |≤ N , | k | >N (cid:12)(cid:12)(cid:12)(cid:12) k − j (cid:12)(cid:12)(cid:12)(cid:12) | ( β ′ η ⊥ | e k − j ) | b j b k ≤ X < | j |≤ N , | k | >N b j | j | | ( β ′ η ⊥ | e k − j ) | b k ≤ (cid:18) X | k | >N b k (cid:19) / (cid:18) X < | j |≤ N b j | j | (cid:19)(cid:18) X | k ′ |≥ N | ( β ′ η ⊥ | e k ′ ) | (cid:19) / by Young’s inequality, so X < | j |≤ N , | k | >N (cid:12)(cid:12)(cid:12)(cid:12) k − j (cid:12)(cid:12)(cid:12)(cid:12) | ( β ′ η ⊥ | e k − j ) | b j b k ≤ C ǫ ⊥ ( N ) (cid:18) X | k | >N b k (cid:19) / (cid:18) X < | j |≤ N b j (cid:19) / , with ǫ ⊥ ( N ) := (cid:18) X | ℓ |≥ N | ( β ′ η ⊥ | e ℓ ) | (cid:19) / . On the other hand X | j | > N , | k | >N (cid:12)(cid:12)(cid:12)(cid:12) k − j (cid:12)(cid:12)(cid:12)(cid:12) | ( β ′ η ⊥ | e k − j ) | b j b k ≤ X | j | > N , | k | >N (cid:18) | k | + 1 | j | (cid:19) ( β ′ η ⊥ | e k − j ) b j b k ≤ C k β ′ η ⊥ k L ( T ) (cid:18) (cid:18) X | k | >N b k | k | (cid:19) (cid:18) X | j | > N | b j | (cid:19) / + (cid:18) X | j | > N b j | j | (cid:19) (cid:18) X | k | >N | b k | (cid:19) / (cid:19) ≤ C ′ N / X | ℓ | > N | b ℓ | . The second term at the right-hand side of (3.10) is bounded by X j ∈ Z ∗ , | k | >N b k | k | | ( β ′ η ⊥ | e − j ) | b j ≤ C k β ′ η ⊥ k L ( T ) (cid:18) X | k |≥ N b k | k | (cid:19) (cid:18) X j ∈ Z ∗ | b j | (cid:19) / ≤ C ′ N / X ℓ ∈ Z ∗ | b ℓ | . The third term at the right-hand side of (3.10) is bounded by : X j ∈ Z ∗ , | k | >N b j | j | | ( β ′ η ⊥ | e k ) | b k ≤ X j ∈ Z ∗ b j | j | (cid:18) X | k |≥ N | ( β ′ η ⊥ | e k ) | (cid:19) / (cid:18) X | k |≥ N | b k | (cid:19) / ≤ C ǫ ⊥ ( N ) X ℓ ∈ Z ∗ | b ℓ | . Notice that P ℓ ∈ Z | b ℓ | ≤ k b k L ( T ) . Gathering all bounds, we find12 X j ∈ Z ∗ , | k | >N (cid:16) ( b | Φ +2 j )( b | Φ +2 k ) + ( b | Φ +2 j +1 )( b | Φ +2 k +1 ) (cid:17) ( C m e j | e k ) ≤ ǫ ∗ ( N ) δ k b k L ( T ) where ǫ ∗ ( N ) goes to zero as N → + ∞ . We still have to examine the case j = 0. We write12 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X | k | >N (cid:0) ( b | Φ +0 )( b | Φ +2 k ) + ( b | Φ +1 )( b | Φ +2 k +1 ) (cid:1) ( C m e | e k ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C k b k L ( T ) (cid:18) X | k | >N b k (cid:19) / (cid:18) X | k | >N | ( C m e | e k ) | (cid:19) / ≤ ǫ ( N ) δ k b k L ( T ) with ǫ ( N ) := (cid:18) X | k | >N | ( C m e | e k ) | (cid:19) / → N → + ∞ . Note that there is a priori no rate of convergence for this term ǫ ( N ), contrary to the otherterms which could be quantified: this is related to the fact that β η e is not mean free.Eventually,(3.11) (cid:12)(cid:12)(cid:12) ( C m Π ⊥ η b | Π ♯N b ) (cid:12)(cid:12)(cid:12) ≤ ǫ ( N ) δ k b k L ( T )ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 21 for some ǫ ( N ) going to zero as N goes to infinity.To conclude the proof of the lemma, we still have to study the contribution of the othertwo terms involved in the definition of the operator C , see (2.40), namely C g, b := A m β η k β η k L ( T ) Z T i b B ′ dx ! , C g, b = − i k β η k L ( T ) Z T β η ( A m b ) dx B ′ . We first consider( C g, Π ⊥ η b | Π ♯N b ) = − k β η k L ( T ) (cid:18)Z T i (Π ⊥ η b ) β ′ η ⊥ dx (cid:19) (cid:16) A m (cid:0) β η (cid:0) η ⊥ (cid:1)(cid:1) | Π ♯N b (cid:17) . It is clear that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k β η k L ( T ) (cid:18)Z T i (Π ⊥ η b ) β ′ η ⊥ dx (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Cδ k b k L ( T ) . Then, (cid:16) A m (cid:0) β η (cid:0) η ⊥ (cid:1)(cid:1) | Π ♯N b (cid:17) = X k [ − N, N +1] (cid:16) A m (cid:0) β η (cid:0) η ⊥ (cid:1)(cid:1) | Φ + k (cid:17) ( b | Φ + k )so that (cid:12)(cid:12)(cid:12)(cid:16) A m (cid:0) β η (cid:0) η ⊥ (cid:1)(cid:1) | Π ♯N b (cid:17)(cid:12)(cid:12)(cid:12) ≤ C δ ( N ) (cid:18) X k [ − N, N +1] | ( b | Φ + k ) | (cid:19) / ≤ C δ ( N ) k b k L ( T ) with δ ( N ) := (cid:16) X | k |≥ N (cid:12)(cid:12)(cid:12)(cid:16) A m (cid:0) β η (cid:0) η ⊥ (cid:1)(cid:1) | Φ + k (cid:17)(cid:12)(cid:12)(cid:12) (cid:17) −−−−−→ N → + ∞ . Hence: | ( C g, Π ⊥ η b | Π ♯N b ) | ≤ δ ( N ) k b k L ( T ) for some δ ( N ) going to zero as N goes to infinity.The treatment of ( C g, Π ⊥ η b | Π ♯N b ) follows the same lines as the previous one, and we omitit for brevity: we get | ( C g, Π ⊥ η b | Π ♯N b ) | ≤ ǫ ( N ) k b k L ( T ) for some ǫ ( N ) going to zero as N goes to infinity. Putting these last two estimates togetherwith (3.11) ends the proof of the lemma. (cid:3) Construction of the matrix Q . The aim of this paragraph is to prove the followingresult.
Proposition 3.6.
There exist two finite dimensional operators
Q, T : Π ♭N L ( T ) → Π ♭N L ( T ) such that [Π ♭N Q Π ♭N , A ] = − Π ♭N C Π ♭N + εT , where C m is defined in (2.41) and where for all smooth b satisfying the Taylor constraint, (3.12) (cid:12)(cid:12) ( T b | b ) L (cid:12)(cid:12) ≤ C k b k L . Proof.
We are going to construct Q , and show that T = 1 ε Π ⊥ η Π e C m Π e Π ⊥ η . We notice indeed that as Π ⊥ η and Π e commute, we have C m Π e Π ⊥ η b = C m Π ⊥ η Π e b . Then if b satisfies the linearized Taylor constraint, which amounts to identity (2.44), then weget easily | ( C m Π e Π ⊥ η b | Π e Π ⊥ η b ) | ≤ Cε k b k L . So now let us prove that there exists Q satisfying: for all | i | , | j | ≤ N ,( AQ Φ + i − QA Φ + i | Φ + j ) = − ( C Φ + i | Φ + j ) + (cid:16) C m Π e Φ + i | Π e Φ + j (cid:17) . Let ( λ k ) k ∈ Z the family of eigenvalues of A associated to (Φ + k ) k ∈ Z . We recall from Proposi-tion 3.2 that λ = λ = 0 , λ k = µ k , λ k +1 = − µ k , k ∈ Z ∗ . The last equality reads(3.13) ( λ j − λ i )( Q Φ + i | Φ + j ) = − ( C Φ + i | Φ + j ) + (cid:16) C m Π e Φ + i | Π e Φ + j (cid:17) . A necessary and sufficient condition for the existence of Q is that the right-hand side is zerowhen λ i = λ j . There are three cases to consider: • The first case is of course when i = j . We compute (cid:0) C Φ + i | Φ + i (cid:1) = (cid:0) C m Φ + i | Φ + i (cid:1) + A m β η k β η k L ( T ) Z T i Φ + i, B ′ dx ! | Φ + i ! − i k β η k L ( T ) Z T β η (cid:0) A m Φ + i (cid:1) dx B ′ | Φ + i ! . In the case i = 0 or i = 1, we have Φ + i = Π e Φ + i . Moreover, we know from (2.30)that Π e A m Π e = 0. It implies that the second term vanishes. Finally, Z T β η (cid:0) A m Φ + i (cid:1) dx = i √ k β η k L ( T ) Z T β η ∂ − β η dx = 0so thatif i ∈ { , } , ( C Φ + i | Φ + i ) = ( C m Φ + i | Φ + i ) = ( C m Π e Φ + i | Π e Φ + i ) . This means that right-hand side of (3.13) is zero.In the case i = 2 k , k ∈ Z ∗ , we find (cid:0) C m Φ +2 k | Φ +2 k (cid:1) = −ℜ Z T β η ∂ − ( β ′ η ⊥ e k ) e k = ℜ Z T e k β ′ η ⊥ β η β η ∂ − ( β η e k )= ℜ Z T e k β ′ η ⊥ β η Π ⊥ e β η ∂ − ( β η Π ⊥ e e k ) + ℜ Z T e k β ′ η ⊥ β η Π e β η ∂ − ( β η e k )= ℜ µ k Z T β ′ η ⊥ β η | e k | + ℜ k β k L ( T ) Z T β ′ η ⊥ e k dx Z T β ∂ − β η e k = ℜ k β k L ( T ) Z T β ′ η ⊥ e k dx Z T β ∂ − β η e k . ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 23
As regards the second term, A m β η k β η k L ( T ) Z T i Φ + i, B ′ dx ! | Φ + i ! = 1 k β η k L ( T ) Z T β ′ η ⊥ √ e k dx Z T β η √ ∂ − β η e k dx = − k β η k L ( T ) Z T β ′ η ⊥ √ e k dx Z T β η √ ∂ − β η e k dx . As regards the third term, − i k β η k L ( T ) Z T β η (cid:0) A m Φ +2 k (cid:1) dx B ′ | Φ +2 k ! = − k β η k L ( T ) Z T β η √ ∂ − β η e k dx Z T β ′ η ⊥ √ e k dx . We find ( C Φ +2 k | Φ +2 k ) = 0 as expected. In the same way, ( C Φ +2 k +1 | Φ +2 k +1 ) = 0. • the second case is when ( i, j ) ∈ { (0 , , (1 , } , which implies λ = λ = 0. Wecompute (cid:0) C Φ +0 | Φ +1 (cid:1) = (cid:0) C m Φ +0 | Φ +1 (cid:1) + A m β η k β η k L ( T ) Z T i Φ +0 , B ′ dx ! | Φ +1 ! − i k β η k L ( T ) Z T β η (cid:0) A m Φ +0 (cid:1) dx B ′ | Φ +1 ! . We argue exactly as in the first case to conclude that( C Φ +0 | Φ ) = ( C m Φ +0 | Φ +1 ) = ( C m Π e Φ +0 | Π e Φ +1 )which means that the right-hand side of (3.13) is zero when i = 0 , j = 1. By the samecomputation, it is also true when i = 1 , j = 0. • the last case is when ( i, j ) ∈ { (2 k, − k + 1) , ( − k + 1 , k ) } for some k ∈ Z ∗ , whichimplies λ i = λ j = µ k . Note thatΦ +2 k = 1 √ e k (cid:16) η ⊥ i (cid:17) , Φ + − k +1 = Φ +2 k = 1 √ e − k (cid:16) η ⊥ − i (cid:17) , with (cid:16)(cid:16) η ⊥ i (cid:17) | (cid:16) η ⊥ − i (cid:17)(cid:17) = 0 . This last orthogonality property implies easily that (cid:0) C m Φ +2 k | Φ + − k +1 (cid:1) = 0 . Then, A m β η k β η k L ( T ) Z T i Φ +2 k, B ′ dx ! | Φ + − k +1 ! = 1 k β η k L ( T ) Z T β ′ η ⊥ √ e k dx Z T β η √ ∂ − β η e k dx
34 ISABELLE GALLAGHER, DAVID G´ERARD-VARET while − i k β η k L ( T ) Z T β η (cid:0) A m Φ +2 k (cid:1) dx B ′ | Φ + − k +1 ! = − k β η k L ( T ) Z T β η √ ∂ − β η e k dx Z T β ′ η ⊥ √ e k dx . Finally, ( C Φ +2 k | Φ + − k +1 ) = 0. In the same way, ( C Φ + − k +1 | Φ +2 k ) = 0.This ends the proof of the proposition. (cid:3) Remark 3.7.
From the relation (3.13), one infers that(3.14) (cid:12)(cid:12)(cid:12) ( Q Φ + i | Φ + j ) (cid:12)(cid:12)(cid:12) ≤ CN , ∀| i | , | j | ≤ N From this estimate, we deduce that the operator norm of Q from Π ♭N L to itself is boundedby:(3.15) k Q k ≤ X ij (cid:12)(cid:12)(cid:12) ( Q Φ + i | Φ + j ) (cid:12)(cid:12)(cid:12) / ≤ C N . Conclusion.
We have now all the elements to establish the stability estimate (1.27).Following the discussion of Paragraph 2.5, we introduce the new unknown d := (Id + εQ N ) b , Q N := Π ♭N Q Π ♭N , with Q constructed in Proposition 3.6. From the estimate (3.15), we deduce that thereexists C > ε k Q N k ≤ CεN . Hence, for εN ≪
1, the operator (Id + εQ N ) is invertible, with(Id + εQ N ) − = Id − εQ N + O ( ε N ) . It follows that there exists
C > − CεN ) k b k L ( T ) ≤ k d k L ( T ) ≤ (1 + CεN ) k b k L ( T ) . As b satisfies equation (2.42), we obtain(3.17) ∂ t d = Adε + ([ Q N , A ] + C ) dε + Rdε − ε d + ∂ d + O ( N ε ) d + ε [ Q N , ∂ ](Id + εQ N ) − d , where O ( N ε ) stands for an operator whose operator norm is controlled by C N ε . As regardsthe commutator [ Q n , ∂ ], we invoke Lemma 3.3 together with (3.15) to get k [ Q N , ∂ ] k ≤ CN N / = C N / . Thanks to Proposition 3.6, we get ∂ t d = Adε + ( C + εT − Π ♭N C Π ♭N ) dε + Rdε − ε d + ∂ d + O (cid:0) N ε + εN / (cid:1) d . ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 25
We then multiply by d and perform an energy estimate:(3.18) 12 ddt k d k L ( T ) + 1 ε k d k L ( T ) + k ∂ d k L ( T ) ≤ ε ℜ (( Cd | d ) − ( C Π ♭N d | Π ♭N d ))+ 1 ε ℜ ( T d | d ) + 1 ε ℜ ( Rd | d ) + C (cid:18) N ε + εN / (cid:19) k d k L ( T ) . Equivalently(3.19) 12 ddt k d k L ( T ) + 1 ε k d k L ( T ) + k ∂ d k L ( T ) ≤ ε ℜ (( Cd | d ) − ( C Π ♭N d | Π ♭N d ))+ 1 ε ℜ ( T b | b ) + 1 ε ℜ ( Rb | b ) + C (cid:18) N ε + εN / (cid:19) k d k L ( T ) . By (2.43) | ( Rb | b ) | ≤ C ε k b k L ( T ) ≤ C ′ ε k d k L ( T ) and by (3.12) | ( T b | b ) | ≤ C k b k L ( T ) ≤ C ′ k d k L ( T ) . Thus, the energy estimate on d reduces to12 ddt k d k L ( T ) + 1 ε k d k L ( T ) + k ∂ d k L ( T ) ≤ ε ℜ (( Cd | d ) − ( C Π ♭N d | Π ♭N d ))+ C (cid:0) N ε + N / (cid:1) k d k L ( T ) . Eventually, to estimate the first term at the right-hand side, we combine Proposition 3.1 andLemma 3.5 to get (cid:12)(cid:12)(cid:12) ( Cb | b ) − ( C Π ♭N b | Π ♭N b ) (cid:12)(cid:12)(cid:12) ≤ ( Cε + δ ( N )) k b k L ( T ) which gives (cid:12)(cid:12)(cid:12) ( Cd | d ) − ( C Π ♭N d | Π ♭N d ) (cid:12)(cid:12)(cid:12) ≤ (cid:0) Cε + δ ( N ) + εN (cid:1) k d k L ( T ) . We end up with12 ddt k d k L ( T ) + k d k L ( T ) ε + k ∂ d k L ( T ) ≤ C (cid:18) δ ( N ) ε + N ε + εN / (cid:19) k d k L . Taking N large enough so that C δ ( N ) ≤ , one can absorb the first term at the right-handside by the diffusion term at the left-hand side. This N being fixed, for small enough ε , allremaining terms can be absorbed as well. This leads to the estimate k d ( t ) k L ( T ) + 1 ε Z t k d ( t ′ ) k L ( T ) dt ′ + Z t k ∂ d ( t ′ ) k L ( T ) dt ′ ≤ C k d (0) k L . Using (3.16) this implies that k b ( t ) k L ( T ) + 1 ε Z t k b ( t ′ ) k L ( T ) dt ′ + Z t k ∂ d ( t ′ ) k L ( T ) dt ′ ≤ C k b (0) k L , and finally since Lemma 3.3 together with (3.15) imply that k [ Q N , ∂ ] k ≤ CN N / = C N / , we have k ∂ b k L ( T ) ≤ k ∂ d k L ( T ) + CεN / k d k L ( T )6 ISABELLE GALLAGHER, DAVID G´ERARD-VARET so for ε small enough, k b ( t ) k L ( T ) + 1 ε Z t k b ( t ′ ) k L ( T ) dt ′ + Z t k ∂ b ( t ′ ) k L ( T ) dt ′ ≤ C k b (0) k L . We notice that there is a global control on b in L , with no exponential loss in t . The expo-nential appearing on the right-hand side of (1.27) in Theorem 1 is due to the contrinbutionof low horizontal frequencies as explained in Section 2.4. The end of the proof of the theoremconsists in noticing that the velocity is obtained as a second order operator with respect to b (see Sections 2.1 and 2.2). Theorem 1 is proved. Acknowledgements
The authors are very grateful to Laure Saint-Raymond for multiple discussions at the earlystage of this work. They acknowledge the support of ANR Project Dyficolti ANR-13-BS01-0003-01.
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ELLPOSEDNESS OF LINEARIZED TAYLOR EQUATIONS 27 (Isabelle Gallagher)
Universit´e Paris Diderot, Sorbonne Paris Cit´e, Institut de Math´ematiquesde Jussieu-Paris Rive Gauche, UMR 7586, F- 75205 Paris, France (David G´erard-Varet)(David G´erard-Varet)