Y Almog

B Helffer

On the stability of laminar flows between plates

Consider a two-dimensional laminar flow between two plates, so that (

We prove that the flow is linearly stable in the large Reynolds number limit, in two different cases:

(nearly Couette flows),

We assume either no-slip or fixed traction force conditions on the plates, and an arbitrary large (but much smaller than the Reynolds number) period in the x 1 direction.

Introduction

Consider the incompressible Navier-Stokes equations in the two-dimensional pipe D = R × (-1, 1)

∂ t v -ǫ∆v + v • ∇v = -∇p in R + × D v = v b î1 on R + × ∂D , (1.1)
where î1 = (1, 0) and v = (v 1 , v 2 ). We require periodicity in the x 1 direction, i.e., for all (t, x 1 , x 2 ) ∈ R + × D we have

v(t, x 1 + L, x 2 ) = v(t, x 1 , x 2 ) , (1.2)
for some L > 0, which may be arbitrary large, but must satisfy L ≪ ǫ -1 as ǫ → 0. An initial condition on v at t = 0 must be placed as well. The vector v = (v 1 , v 2 ) denotes the fluid velocity field which belongs, for all T > 0, to

W T,L = {u ∈ L 2 (0, T ; H 2 loc (D, R 2 )) | div u = 0 , u(t, x 1 + L, x 2 ) = u(t, x 1 , x 2 ) } ,
where the divergence is taken in the spatial coordinates. Here u ∈ H k loc (D)) for some k ∈ N means that for any φ ∈ C ∞ 0 (R), it holds that (x 1 , x 2 ) → φ(x 1 )u(x 1 , x 2 ) ∈ H k (D). Similarly, u ∈ H k 0,loc (D) means that for any

φ ∈ C ∞ 0 (R), (x 1 , x 2 ) → φ(x 1 )u(x 1 , x 2 ) ∈ H k 0 (D). Recall that v • ∇v = v 1 ∂v ∂x 1 + v 2 ∂v ∂x 2 .
The pressure p belongs, for all T > 0, to

Q T,L = {p ∈ L 2 (0, T ; H 1 loc (D)) | ∇p(t, x 1 + L, x 2 ) = ∇p(t, x 1 , x 2 ) } .
The trace of v on the boundary is constant on each connected component of R + ×∂D:

{v b (-1), v b (1)} ∈ R 2 .
The parameter ǫ > 0 denotes the inverse of the flow's Reynolds number Re > 0.

Beyond the above no-slip boundary condition we shall also consider a prescribed constant traction force on the boundary, i.e.,

∂v 1 ∂x 2 = s b , v 2 = 0 , (1.3)
where s b (±1) ∈ R denote the prescribed traction force on ∂D = R × {±1}.

We consider the stability of a stationary pair (v, p) where the flow v is of the form, for (x 1 , x 2 ) ∈ D, v(x 1 , x 2 ) = U(x 2 ) î1 .

For such flows, (1.1) is satisfied for (v, p) if and only if there exists some constant a and p 0 such that, for (x 1 , x 2 ) ∈ D, We shall not confine ourselves, however, in the sequel to such unperturbed velocity fields and discuss a more general class of motions (cf. also [12, p. 154]), which can be obtained if we add a non-uniform body force b(x 1 , x 2 ) = b(x 2 ) î2 to the right-hand-side of (1.1). Such a generalization can be useful if one attempts to examine the stability of a flow in an arbitrary 2D cross-section but uniform in the longitudinal direction. The linearized operator associated with (1.1), at the flow (v, p), assumes the form (u, q) → T 0 (u, q) := Tu -∇q (1.5) where

U ′′ (x 1 , x 2 ) = a ; p(x 1 , x 2 ) = ǫ a x 1 + p 0 . ( 1
Tu = -ǫ ∆u + U ∂u ∂x 1 + u 2 U ′ î1 , in D , (1.6)
where q ∈ Q L given by

Q L = {p ∈ H 1 loc (D) | ∇p(• + L, •) = ∇p(•, •) and
(0,L)×(-1,+1)

p(x 1 , x 2 )dx 1 dx 2 = 0 } , and u = (u 1 , u 2 ) belongs to either

W D = u ∈ W u 1 ∂D = 0 , u 2 ∂D = 0 , or to W S = u ∈ W ∂u 1 ∂x 2 ∂D = 0 , u 2 ∂D = 0 , where W = {u ∈ H 2 loc (D, R 2 ) | div u = 0 ; u(• + L, •) = u(•, •)} .
Remark 1.1.

• Note that when no-slip boundary conditions, introduced in (1.1), are applied, then u ∈ W D . Otherwise, if we select (1.3) instead, then u ∈ W S .

• Note that that for any p ∈ Q L there exists A ∈ R and p such that p(x 1 , x 2 ) = Ax 1 + p(x 1 , x 2 ) with p satisfying the periodicity condition p(• + L, •) = p(•, •).

We now attempt to define a spectral problem for T 0 . We seek an estimate for the solution (u, q) for some Λ ∈ C and F in a suitable space of the equation T 0 (u, q) -Λu = F .

(1.7)

To this end we need to define the function space in which the solution should reside, and then to formulate an effective spectral problem involving only u, so that q is recovered in a final step directly from u.

The local stability of the flow (1.4) has been addressed mostly by physicists and and engineers [START_REF] Drazin | Hydrodynamic stability[END_REF][START_REF] Joseph | Stability of fluid motions I[END_REF][START_REF] Yaglom | Hydrodynamic Instability and Transition to Turbulence[END_REF][START_REF] Schmid | Stability and transition in shear flows[END_REF]. Poiseuille flow (U(x) = 1x 2 ), which falls outside the scope of this work, and Couette flow (U(x) = x), have received some special attention. Thus, for Couette flows it has been established, using a mix of numerical and analytical techniques that Couette flow is always stable [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF][START_REF] Romanov | Stability of plane-parallel Couette flow[END_REF], and established numerically [START_REF] Orszag | Accurate solution of the Orr-Sommerfeld stability equation[END_REF] that Poiseuille flow looses its stability for ǫ -1 ≈ 5772. In [START_REF] Shkalikov | Spectral portraits of the Orr-Sommerfeld operator at large Reynolds numbers[END_REF] a survey of results published in Russian is presented where the locus of part of the spectrum (but not its left margin) for Couette and Poiseuille flows is approximated in the limit ǫ → 0.

It should be emphasized that experimental observations (cf. [START_REF] Chapman | Subcritical transition in channel flows[END_REF]) conclude that Couette and Poiseuille flows loose their stability for Reynolds numbers that are much lower than ǫ -1 = 5772. It is commonly believed that these instabilities arise due to finite, though small, initial conditions. Thus, it has been established in [START_REF] Bedrossian | On the stability threshold for the 3D Couette flow in Sobolev regularity[END_REF] that unbounded Couette flows (in R instead of [-1, 1] as is presently considered), assuming a period L = 1, are finitely stable for O(ǫ 3/2 ) initial data. From a different perspective, in [START_REF] Gérard-Varet | Phénomène d'amortissement dans les équations d'Euler[END_REF], Sobolev stability of a Prandtl boundary layer is studied, leading to a similar problem in D = R × R + , for sufficiently small period (L ≪ 1).

To obtain non-linear stability one needs in [START_REF] Bedrossian | On the stability threshold for the 3D Couette flow in Sobolev regularity[END_REF] to use semigroup estimates (and not only the locus of the eigenvalues as in [START_REF] Shkalikov | Spectral portraits of the Orr-Sommerfeld operator at large Reynolds numbers[END_REF]), associated with time dependent equation u t -T 0 (u, q) = 0 in W D or W S .

Unlike the unbounded Couette flow in [START_REF] Bedrossian | On the stability threshold for the 3D Couette flow in Sobolev regularity[END_REF], the semigroup associated with more general laminar flows in [-1, 1] is not explicitly known. In the present contribution we thus consider, in the limit ǫ → 0, velocity fields U ∈ C 4 |U ′ (x)| , and velocity fields for which U ′′ = 0 in [-1 , 1]. We prove that these laminar flows are stable and provide C(R + ; L(L 2 )) estimates for their associated semigroup norm. We believe that these linear estimates would be useful when considering the nonlinear stability of these flows in a bounded interval.

The rest of this contribution is arranged as follows. In the next section we formulate the spectral problem by using Hodge decomposition. Since the standard Hodge decomposition [START_REF] Girault | Finite element approximation of the Navier-Stokes equations[END_REF] is not a direct sum in this periodic setup, we define a space of zero-flux perturbations, and then formulate our main results in this space. In Section 3 we present the problem in terms of the stream function and its Fourier coefficients and arrive at the Orr-Sommerfeld operator.

In the next section we consider the inviscid problem (where ǫ = 0) in Fourier space. Section 5 includes some resolvent estimates obtained for one-dimensional Schrödinger operators on the entire real line and for their Dirichlet realization in (-1, 1). In Section 6 we consider the same Schrödinger operator on (-1, 1) and R + but this time in a Sobolev space of functions satisfying certain orthogonality conditions (cf. [START_REF] Shkalikov | Spectral portraits of the Orr-Sommerfeld operator at large Reynolds numbers[END_REF]). Section 7 provides inverse estimates for the Orr-Sommerfeld operator for the fixedtraction problem, whereas Section 8 provides the same estimates for the no-slip realization.

In Section 9 we prove the main results. Finally in the appendix we bring some auxiliary estimates obtained for Airy functions and generalized Airy functions.

Spectral problem formulation and main results

We now amend the spectral question presented in (1.7) to a more standard spectral problem. To this end we use a variant of Hodge decomposition adapted to our periodic setting (see [START_REF] Girault | Finite element approximation of the Navier-Stokes equations[END_REF]Theorem 3.4] for the standard case), which allows us to eliminate q.

Hodge theory

Let H = {u ∈ L 2 loc (D, R 2 ) | u(• + L, •) = u(•, •)} , (2.1)
where the scalar product for the Hilbert space H is given by

H × H ∋ (u, v) → (0,L)×(-1,1) ū • v dx 1 dx 2 ,
and the two closed subspaces of H

H curl = {u ∈ L 2 loc (D, R 2 ) | curl u = 0 ; u(• + L, •) = u(•, •)} ,
and

H div = {u ∈ L 2 loc (D, R 2 ) | div u = 0 ; u • n| ∂D = 0 ; u(• + L, •) = u(•, •)} .
We have Lemma 2.1.

H div ∩ H curl = span(1, 0) .

Proof. Let u ∈ H div ∩ H curl , and û = (û 1 , û2 ) denote its partial Fourier transform with respect to x 1 , i.e., û(n,

x 2 ) = 1 L L 0 u(x 1 , x 2 )e -i2πnx 1 /L dx 1 . (2.
2)

It can be easily verified that, for any n ∈ Z,

i 2π L nû 2 (n, x 2 ) -d dx 2 û1 (n, x 2 ) = 0 i 2π
L nû 1 (n, x 2 ) + d dx 2 û2 (n, x 2 ) = 0 û2 (n, -1) = û2 (n, +1) = 0.

From the above we can conclude that û(n, x 2 ) = 0 for n = 0, û2 (0, x 2 ) = 0 and û1 (0, x 2 ) = Const.

With the above in mind, we introduce

H 0 div := {u ∈ H div , u, (1, 0) = 0}. (2.3)
The orthogonality condition reads (0,L)×(-1,+1)

u 1 (x 1 , x 2 ) dx 1 dx 2 = 0 . (2.4)
We can now prove the following Hodge decomposition: It can be easily verified, by using the Lax-Milgram Lemma, that there exists a unique solution for (2.6). Similarly, by using (see Remark 1.1) the ansatz

φ c = Ax 1 + φc , (2.8)
there where φc (•, •) = φc (• + L, •), it follows that there exists a unique, solution of (2.5). (Note that the Neumann condition in the second line of (2.5) is satisfied in H -1/2 loc (∂D) sense.) Equivalently, we can say that φc is the unique periodic solution, orthogonal to the constant function, of (-1,+1)×(0,L) (∇ φcu) • ∇v dx 1 dx 2 = 0 , for every L-periodic v ∈ H 1 ((-1, +1) × (0, L)) .

Clearly, ∇φ c ∈ H curl , ∇ ⊥ φ d ∈ H 0 div , u -∇φ c -∇ ⊥ φ d , (1, 0) = 0, and, by the periodicity of φ d and ∇φ c , it holds that ∇φ c , ∇ ⊥ φ d = 0 .

Finally we set u = v + ∇φ c + ∇ ⊥ φ d to obtain that v ∈ H curl ∩ H 0 div and hence, by Lemma 2.1, v ≡ 0.

Zero flux solution

Since W # ⊂ H 0 div for # ∈ {D, S}, we need to introduce the following spaces, as the domain of the operator in the spectral formulation

W 0 D = u ∈ W 0 u 1 ∂D = 0 , u 2 ∂D = 0 , (2.9a)
and

W 0 S = u ∈ W 0 ∂u 1 ∂x 2 ∂D = 0 , u 2 ∂D = 0 , (2.9b)
where

W 0 = {u ∈ H 2 loc (D, R 2 ) | div u = 0 ; u(• + L, •) = u(•, •), u, (1
, 0) H = 0} . While it seems that the orthogonality requirement (2.4) is an unjustified limitation of the perturbation space, we show in the sequel how one can solve the stability problem in W # from its solution in W 0 # . Hence, (2.4) does not result in any loss of generality.

Let then P : H → H 0 div denote the orthogonal projection on H 0 div , given explicitly for some u ∈ H by ∇ ⊥ φ d , where φ d is the solution of (2.6). Rewriting (1.7)

in the form

Tu -Λu -F = ∇q , we observe that ∇q ∈ H curl . Then, projecting on H 0 div , we may write, for u ∈ W 0

♯ P (T -Λ)u -F) = 0 .
With the above in mind, we now define as an unbounded operator on H 0 div whose domain is W 0 ♯ (which is clearly dense in H 0 div )

T ♯ P := P T .

(2.10)

By this definition we have (T ♯ P -Λ)u = P F , (2.11) which appears to be a proper formulation of the resolvent equation.

Proposition 2.3. T ♯ P is semi-bounded on H 0 div and has compact resolvent. Furthermore, e -t T ♯ P ≤ e 1 2 U ′ ∞ t .

(2.12)

Proof. Let u ∈ W 0 # . As u ⊥ H curl we have

ℜ u, T ♯ P u = ℜ u, Tu = ǫ ∇u 2 2 + ℜ u 2 , U ′ u 1 ≥ ǫ ∇u 2 2 - 1 2 U ′ ∞ u 2 2 , (2.13)
verifying, thereby, semi-boundedness. More precisely, the resolvent set of T ♯ P contains {λ ∈ C | ℜΛ < - 1 2 U ′ ∞ }. The semigroup estimate (2.12) is then a consequence of the Hille-Yosida theorem. The compactness of the resolvent is proved by observing that W 0 # is compactly embedded in H 0 div .

Remark 2.4.

• Suppose that for some Λ 0 ∈ R and C > 0, we have

sup Λ≤Λ 0 Λ∈ρ(T # P ) (T # P -Λ) -1 ≤ C .
Then, by the compactness of the resolvent the spectrum is discrete, and hence it holds that sup ℜΛ≤Λ 0 (T # P -Λ) -1 ≤ C .

• It results from (2.13) that, for any F ∈ H and # ∈ {S, D}, it holds that

sup ℜΛ≤-1 2 U ′ ∞-ǫ (T # P -Λ) -1 F 1,2 ≤ 1 ǫ F 2 .
(2. [START_REF] Gérard-Varet | Phénomène d'amortissement dans les équations d'Euler[END_REF] If Λ is in the resolvent set of T ♯ P , we recover u by u = (T ♯ P -Λ) -1 P F .

(2.15)

Once we have derived u we can obtain q in the following manner Proposition 2.5. Let (u, Λ, F) ∈ W 0 # × C × H satisfy (2.15). Then, there exists a unique q ∈ Q L such that (1.7) holds for (u, q, Λ, F).

Proof. From (2.11) it follows that

(T -Λ)u -F = G ∈ H curl .
It remains to prove the existence of a unique q G ∈ Q L satisfying ∇q G = G .

(2. [START_REF] Gohberg | Operator Theory: Advances and Applications[END_REF] This, however, easily follows from the proof of Lemma 2.2, i.e., one obtain q G as the unique solution of (2.5) with G in the place of u.

Corollary 2.6. If Λ is in the resolvent set of T ♯ P , then for any F ∈ H there exists a unique pair (u 0 , q 0 ) ∈ W 0 # × Q L such that (1.7) holds. We use the term "the zero flux solution of (1.7)" for this solution.

Remark 2.7.

• From the proof of Proposition 2.5 we learn, in addition, that if Λ ∈ σ(T ♯ P ) and u Λ is a corresponding eigenfunction in W 0 # then there exists q Λ such that (u Λ , q Λ ) satisfies (1.7) with F = 0. We cannot exclude, at the moment, the possibility that Λ is not a simple eigenvalue.

• The proof shows also that, for any Λ ∈ ρ(T ♯ P ), the map F → (u 0 , q 0 ) (as defined in the corollary) is continuous from H onto W 0 # × Q L .

General solution

Once the zero flux solution of (1.7) has been found in W 0 # × Q L , we attempt to solve the problem more generally in W # × Q L . Proposition 2.8. Let Λ ∈ ρ(T ♯ P ), then for any F and any γ ∈ R, there exists a unique pair

(u γ , q γ ) ∈ W # × Q L satisfying (1.7) and u γ , (1, 0) = γ .
(2.17)

Moreover, (u γ , ∇q γ ) continuously depends on γ in

H 2 ((0, L) × (-1, 1])) × L 2 ((0, L) × (-1, 1)) .
Proof.

Fixed traction

In this case, we observe that γ î1 ∈ W S . Let F ∈ H, and (u 0 , q 0 ) ∈ W 0 S × Q L satisfy (1.7) for some Λ ∈ ρ(T S ). Set then

u γ = u 0 + γ L î1 , q γ = q 0 + Λ γ L x 1 .
It can be easily verified that (u γ , q γ , Λ, F ) satisfy both (1.7) and (2.17).

No-slip

Let again F ∈ H, and (u

0 , q 0 ) ∈ W 0 D × Q L satisfy (1.7) for some Λ ∈ ρ(T D ). Then, set q = q 0 + ǫγx 1 ; ũ = ũ0 + γV (x 2 ) î1 , where V ∈ W D satisfies V ′′ (x 2 ) = - 3 2L . As V (x 2 ) = 3 4L (1 -x 2
2 ), we have

1 -1 V (x 2 ) dx 2 = 1 L .
and hence ũ, (1, 0) = γ. Clearly, T(ũ) -Λũ -∇q = F -ΛγV î1 .

To construct (u γ , q γ ) we introduce (û, q) ∈ W 0

D × Q L satisfying T(û) -Λû -∇q = V î1 .
The existence of (û, q) follows from Proposition 2.5. We then set (u γ , q γ ) = (u 0 , q 0 ) + Λγ(û, q) + γ(ǫx 1 , V î1 ) .

(2.18)

Clearly, (u γ , q γ ) is a solution of (1.7) satisfying (2.17). It must also be unique by Proposition 2.5.

Note that the above non-uniqueness results from the periodic setup we have chosen. If instead we had selected to work in L 2 (D) (an infinite pipe) the solution of (1.7) would have been unique, since the perturbation flux must vanish at infinity, and hence must be zero on each cross section. Nevertheless, as will become evident in the sequel the periodic setting offers significant simplification over the infinite pipe and is not necessarily less effective as an approximation of a more realistic setup.

Averaging with respect to x 1

We begin by defining the projection p : L 2 ((0, L) × (-1, 1)) → L 2 ((0, L) × (-1, 1)) by

pu(x 1 , x 2 ) = 1 L L 0 u(s, x 2 ) ds , (2.19)
and then extend it to Π :

L 2 ((0, L) × (-1, 1); R 2 ) → L 2 ((0, L) × (-1, 1); R 2 ) by writing Πu = (pu 1 , pu 2 ) for u = (u 1 , u 2 ) . (2.20)
We first show Lemma 2.9. Π is a projection on H 0 div . Moreover for any # ∈ {D, S}, we have

Π W 0 # ⊂ W 0 # . Proof. Let u ∈ H 0 div . Then u = ∇ ⊥ φ d where φ d is a solution of (2.6
). We may then write, using the periodicity of φ d ,

Πu = ∂ x 2 (pφ d ) î1 .
(2.21)

Obviously, div Πu = 0, and the orthogonality of Πu to (1, 0) in H follows from

1 -1 ∂ x 2 (pφ d ) dx 2 = 0 .
Hence Πu ∈ H 0 div . It can now be easily verified that Π W 0 # ⊂ W 0 # .

Lemma 2.10. P commutes with Π.

Proof. Let F ∈ H. Then, by the proof of Lemma 2.2

F = ∇ ⊥ φ d + ∇φ c ,
where φ d is a solution of (2.6) and φ c is a solution of (2.5).

Clearly,

Π F = ∂ x 2 (pφ d ) + A) î1 + ∂ x 2 (p φc ) î2 ,
where A is given by (2.7) (with F instead of u).

Hence Π F = ∇ ⊥ (pφ d ) + A î1 + ∇(p φc ) ,
and by uniqueness of the Hodge decomposition we obtain

P Π F = (∂ x 2 pφ d ) î1 .
Next, we compute ΠP F. Observing that

P F = ∇ ⊥ φ d , we get ΠP F = (∂ x 2 pφ d ) î1 = P Π F . (2.22)
We can now prove the following commutation result Lemma 2.11. For any # ∈ {S, D}, T ♯ P commutes with Π.

Proof.

We simply observe that for all u ∈ W 0 # T Πu = Π Tu , and use the commutation of P and Π.

An immediate consequence follows

Proposition 2.12. For any # ∈ {D, S} it holds that Π e -t T # P = e -t T # P Π .

(2.23)

Main results

Throughout this work we assume that:

Assumption 2.13. U ′ does not vanish in [-1, +1], or m := inf x∈[-1,1] |U ′ (x)| > 0 . (2.24)
The statement of the main results below involves the spectral properties of the complex Airy operator on R +

L + = - d 2 dx 2 + ix , (2.25a) defined on D(L + ) = {u ∈ H 2 (R + ) ∩ H 1 0 (R + ) | xu ∈ L 2 (R + ) } . (2.25b)
We denote its leftmost eigenvalue [START_REF] Almog | The Stability of the Normal State of Superconductors in the Presence of Electric Currents[END_REF] by ν 1 . We further set

J m (U) = min(|U ′ (-1)|, |U ′ (1)|) (2.25c)
We also need below

δ 2 (U) := U ′′ 1,∞ , (2.26)
where

u 1,∞ = u ∞ + u ′ ∞ .
Finally we define, for any r > 1 and k ≥ 2,

S k r = {v ∈ C k ([-1, +1]), inf x∈[-1,+1] |v ′ (x)| ≥ r -1 and v k,∞ ≤ r} , (2.27)
and then set for convenience of notation S r = S 4 r .

(2.28)

For U ∈ S r , ǫ > 0 and L > 0, we recall that T S P := T S P (U, ǫ, L) is defined in (2.9b) and (2.10) (where ǫ appears in the definition of T and L is the x 1 periodicity). For β > 0, we introduce

Ω(β) := {(ǫ, L) ∈ (0, 1] × R + , (Lǫ) -1 ≥ β/(2π)} .
Theorem 2.14. The following statements hold for any r > 1.

1. For all δ > 0, there exist β 0 > 0 and C > 0 such that, for any U ∈ S r satisfying

inf x∈[-1,1] |U ′′ (x)| ≥ 1/r , (2.29)
any (ǫ, L) ∈ Ω(β 0 ), and any t > 0, it holds that e -t T S P (U,ǫ,L) ≤ C ǫ -2 e -ǫt(π 2 /4-δ) .

(2.30a)

Furthermore, for all δ > 0, there exist Υ > 0, β 0 > 0 and C > 0 such that for all all (ǫ, L) ∈ Ω(β 0 ) , t > 0 , and any U ∈ S r satisfying (2.29) we have

e -t T S P (U,ǫ,L) (I -Π) ≤ C L 1 3 - δ ǫ -5 3 - δ e -ǫΥ[Lǫ] -2/3 t , (2.30b)
where Π is given by (2.19).

2. For all δ > 0, there exist δ > 0, β 0 > 0 and C > 0 such that, for any U ∈ S r satisfying δ 2 (U) < δ , (2.30a) holds true for all (ǫ, L) ∈ Ω(β 0 ) and t > 0 .

Moreover, for all Υ < J 2/3 m ℜν 1 , there exist β 0 > 0, δ > 0 and C > 0 such that, for any (ǫ, L) ∈ Ω(β 0 ), U ∈ S r satisfying δ 2 (U) < δ , and t > 0 , we have

e -t T S P (U,ǫ,L) (I -Π) ≤ CL 2/3 ǫ -4 3 e -ǫΥ[Lǫ] -2/3 t . (2.31)
For the case # = D, we first define, for some θ > 0, the operator L θ whose differential operator is given by (2.25a) and its domain by

D(L θ ) = { u ∈ H 2 (R + ) | e -θ• , u = 0 , xu ∈ L 2 (R + ) } .
We show later (see Proposition 6.4 and Corollary 6.7) that L θ is a closed operator and that

μm := inf θ≥0 (inf ℜσ(L θ ) + θ 2 2 ) . (2.32)
is finite and positive. For U ∈ S r , ǫ > 0 and L > 0, we recall that

T D P := T D P (U, ǫ, L)
is defined in (2.9a) and (2.10).

Theorem 2.15. For r > 1, the following properties hold.

1. For all δ > 0, there exist β 0 > 0 and C > 0 such that for any U ∈ S r satisfying (2.29), any (ǫ, L) ∈ Ω(β 0 ) and any t > 0 it holds that e -t T D P (U,ǫ,L) ≤ Cǫ -2 e -ǫt(π 2 /4-δ) .

(2.33a)

Furthermore, for all δ > 0, there exist Υ > 0, β 0 > 0 and Ĉ > 0 such that, for any U ∈ S r satisfying (2.29), (ǫ, L) ∈ Ω(β 0 ) and t > 0 , we have

e -t T D P (U,ǫ,L) (I -Π) ≤ ĈL 1 3 - δ ǫ -5 3 - δ e -ǫΥ[Lǫ] -2/3 t , (2.33b)
where Π is given by (2.19).

2.

For any δ > 0, there exist δ > 0, C > 0, and β 0 > 0, such that for any U ∈ S r satisfying δ 2 (U) ≤ δ , (ǫ, L) ∈ Ω(β 0 ), and t > 0 , (2.33a) holds true.

Moreover, for all Υ < J 2/3 m μm , there exist δ 0 > 0, β 0 > 0 and C > 0, such that for all U ∈ S r satisfying δ 2 (U) ≤ δ 0 , (ǫ, L) ∈ Ω(β 0 ), and t > 0 it holds that

e -t T D P (U,ǫ,L) (I -Π) ≤ C L 2/3 ǫ -4 3 e -ǫΥ[Lǫ] -2/3 t . (2.34)
3 The Orr-Sommerfeld operator

We focus attention in the sequel on T ♯ P and its resolvent.

Stream Function

When considering a two-dimensional incompressible fluid flow, it is customary to introduce a stream function, i.e., to let u = ∇ ⊥ ψ. Its introduction is again related to Hodge decomposition theory.

Lemma 3.1. Let u ∈ H 0 div . Then, there exists a unique ψ ∈ H 1 0,loc (D, R 2 )) such that ψ(x 1 + L, x 2 ) = ψ(x 1 , x 2 ) and u = ∇ ⊥ ψ. If in addition, u ∈ W 0 # , then ψ ∈ H 3 loc (D, R 2 )) and ψ satisfies ∂ x 2 ψ = 0 on ∂D if # = D and ∂ 2 x 2 ψ = 0 on ∂D if # = S.
Proof. Existence and uniqueness of ψ follow from the proof of Lemma 2.2. In particular, for any u ∈ H 0 div we have ψ = φ d where φ d is a solution of (2.6). The second part of the lemma is immediate.

We next substitute u = ∇ ⊥ ψ into (1.6) and take the curl of the ensuing equation, which leads to the following equation, in the distributional sense,

P # Λ,ǫ ψ = curl F in D , (3.1)
with # ∈ {D, S},

P # Λ,ǫ := -ǫ∆ 2 + U ∂ ∂x 1 ∆ -U ′′ ∂ ∂x 1 -Λ ∆ . (3.2)
We treat P # Λ,ǫ as an unbounded operator on L 2 per (D), where

L 2 per (D) := {u ∈ L 2 loc (D) , u(x 1 + L, x 2 ) = u(x 1 , x 2 )} , is equipped with the L 2 ([-1, 1] × [0, L]) norm.
Note that additional regularity is needed while attempting to use results obtained on L 2 per (D) for spectral problem for T # P . We shall obtain the necessary regularity at a later stage. Similarly, we introduce for k = 1, 2 and s ≥ 0

H s per (D, C k ) := {u ∈ H s loc (D, C k ) , u(x 1 + L, x 2 ) = u(x 1 , x 2 )} ,
and assign to it the

H 1 ([-1, 1] × [0, L]) norm.
In the interest of brevity we write in the sequel H s per (D, C) = H s per (D). For no-slip boundary conditions we take

D(P D Λ,ǫ ) = {ψ ∈ H 4 per (D), ψ = 0 and ∂ x 2 ψ = 0 on ∂D} .
For fixed traction, the domain of P S Λ,ǫ is given by

D(P S Λ,ǫ ) = {ψ ∈ H 4 per (D), ψ = 0 and ∂ 2 x 2 ψ = 0 on ∂D} .
We can now make the following statement Lemma 3.2. The operator P # Λ,ǫ is invertible for each # ∈ {S, D} and Λ ∈ ρ(T # P ).

Proof. Let ψ ∈ D(P # Λ,ǫ ) and f ∈ L 2 per (D) satisfy P # Λ,ǫ ψ = f for some # ∈ {S, D} and Λ ∈ ρ(T # P ). Let F denote the unique vector field in H 0 div satisfying curl

F = f . As (T -Λ)∇ ⊥ ψ -F ∈ H curl , it follows that ∇ψ 2 ≤ (T # P -Λ) -1 F 2 ≤ (T # P -Λ) -1 f 2 .
Due to the periodicity of our function spaces and the fact that the coefficients of the differential operator P # Λ,ǫ and the associated boundary conditions do not depend on x 1 , it is natural to consider the operator in a Fourier space. Hence, we introduce

L 2 per (D) ∋ ψ → (F ψ) ∈ ℓ 2 ( 2π L Z) ⊗ L 2 (-1, +1) satisfying (F ψ)(α n , x 2 ) = 1 L L 0 e -iαnx 1 ψ(x 1 , x 2 ) dx 1 , (3.3)
for α n = 2πn/L (n ∈ Z).

We then obtain the Hilbertian sum

F P # Λ,ǫ F -1 = ǫ ⊕ n∈Z\{0} B # λ,αn,βn ⊕ B # Λ ,
where

B # Λ = - d 4 dx 4 - Λ d 2 dx 2 , in which Λ = Λ ǫ . (3.4)
The unbounded operator B # λ,α,β on L 2 (-1, +1), which is commonly referred to as the Orr-Sommerfeld operator is given by

φ → B # λ,α,β φ = (L β -βλ) d 2 dx 2 -α 2 φ -iβU ′′ φ , (3.5)
in which

β = α ǫ -1 , (3.6)
L β = - d 2 dx 2 + iβU , (3.7)
and, for β = 0,

λ = β -1 ( Λ -α 2 ) . (3.8)
In the sequel, unless stated otherwise, we consider β and α as independent parameters.

We now define two different realizations associated with the differential operator appearing in (3.5). The domain of B S λ,α,β , corresponding to the prescribed traction force boundary condition, is given by

D(B S λ,α,β ) = {u ∈ H 4 (-1, 1) , u(±1) = 0 and u ′′ (±1) = 0 } , (3.9)
whereas the operator B D λ,α,β , corresponding to the no-slip condition, is defined on

D(B D λ,α,β ) = {u ∈ H 4 (-1, 1) , u(±1) = 0 and u ′ (±1) = 0 }. (3.10)
The domain of B # Λ is similarly defined by (3.9) for # = S and (3.10) for # = D.

Inverse estimates

The Orr-Sommerfeld operator given by (3.5) has extensively been studied in the Physics literature [START_REF] Drazin | Hydrodynamic stability[END_REF][START_REF] Yaglom | Hydrodynamic Instability and Transition to Turbulence[END_REF][START_REF] Schmid | Stability and transition in shear flows[END_REF]. Very few rigorous studies, however, address its spectrum (cf. [START_REF] Romanov | Stability of plane-parallel Couette flow[END_REF] in the Couette case U ′′ = 0) and none, to the best of our knowledge provide estimates for it inverse norm (B # λ,α,β ) -1 in L(L 2 (-1, +1)). Assuming that Λ ∈ ρ(T # P ), the inverse of P Λ,ǫ is bounded. To estimate its norm, one needs a proper uniform bound of (B # λ,α,β ) -1 for all α > 0 and (B # Λ ) -1 . Hence we write,

(P # Λ,ǫ ) -1 ≤ ǫ -1 max sup n∈Z\{0} (B # λn,αn,βn ) -1 , (B # Λ ) -1 ,
where, for n ∈ Z \ {0}, Λ = ǫ Λ,

λ n = β -1 n ( Λ -α 2 n ) = ǫα -1 n ( Λ -α 2 n ) = α -1 n Λ -ǫα n .
Clearly, for any ǫ > 0 and Λ ∈ R,

sup n∈Z\{0} (B # λn,αn,βn ) -1 ≤ sup α≥α 1 λ=α -1 ǫ( Λ-α 2 ) (B # λ,α,αǫ -1 ) -1 ≤ sup β≥β 1 sup α≥α 1 λ=β -1 ( Λ-α 2 ) (B # λ,α,β ) -1 , with β 1 (ǫ, L) = ǫ -1 α 1 = 2π/(Lǫ) .
Consequently, for any Λ0 ∈ R and ǫ > 0, we have, the following inequality sup

ℜΛ≤ǫ Λ0 Λ∈ρ(T # P ) (P # Λ,ǫ ) -1 ≤ ǫ -1 max sup β≥β 1 sup α≥0 ℜλ≤β -1 ( Λ0 -α 2 ) (B # λ,α,β ) -1 , sup ℜ Λ≤ Λ0 (B # Λ ) -1 . (3.11) Note that B # λ,α,β = B # λ,-α,β
and hence, it is sufficient to consider α ≥ 0 in the above. We emphasize that the supremum with respect to λ, β and α of (B # λ,α,β ) -1 is obtained while ignoring the dependence of β on α. Note also that β 1 = β 1 (ǫ) tends to +∞ as Lǫ → 0. Hence we shall attempt to obtain a bound on (B # λ,α,β ) -1 for large values of β, since we are interested in the small ǫ limit.

Throughout this work we recall that we always assume Assumption 2.13. Without any loss of generality we can assume that

U ′ > 0 on [-1, +1] .
(3.12) Indeed, the case U ′ < 0 can similarly be treated after applying the transformation x → -x.

In view of (3.11), we attempt to obtain, in the large β limit, a bound on B -1 λ,α,β . To this end, we introduce

A λ,α def = (U + iλ) - d 2 dx 2 + α 2 + U ′′ . (3.13)
We define A λ,α , for ℜλ = 0 or when ℜλ = 0 and ℑλ ∈ [U(-1), U(1)], on

D(A λ,α ) = H 2 (-1, 1) ∩ H 1 0 (-1, 1) . (3.14)
In the case ℜλ = 0 and ℑλ ∈ [U(-1), U(1)] we set

D(A iν,α ) = H 2 ((-1, 1); |U -ν| 2 dx) ∩ H 1 0 (-1, 1) , (3.15)
where ν ∈ [U(-1), U(1)].

Note that A λ,α = A λ,-α and hence we consider it only in the case α ≥ 0 . One can formally obtain A λ,α from B λ,α,β by dividing it by β and taking the limit β → ∞ which corresponds to the limit α -1 ǫ → 0. This is why it has been commonly referred to as the "inviscid operator" in [START_REF] Drazin | Hydrodynamic stability[END_REF]. We note that the formal limit of the Orr-Sommerfeld operator as β → ∞ is very different from that of the Schrödinger operator -d 2 dx 2 + iβ(Uν) (ν = ℑλ). In the latter case, we expect the resolvent to be small away from the set were U = ν. This fact was used in [START_REF] Almog | The Stability of the Normal State of Superconductors in the Presence of Electric Currents[END_REF][START_REF] Almog | On a Schrödinger operator with a purely imaginary potential in the semiclassical limit[END_REF], for instance, to obtain resolvent estimates via localization techniques. For B λ,α,β , the best one can expect is that v = A λ,α φ would be small outside a close neighborhood of the set where U = ℑλ. We note that A λ,α raises considerable interest independently of the viscous operator B # λ (cf. [START_REF] Drazin | Hydrodynamic stability[END_REF] and [START_REF] Lin | Some recent results on instability of ideal plane flows[END_REF]).

The inviscid operator

We consider here the inviscid operator A λ,α associated with the differential operator (3.13), whose domain of definition is given either by (3.14) where ℜλ = 0 or ℑλ ∈ [U(-1), U(1)], or by (3.15) in the case λ = iν for ν ∈ [U(-1), U(1)]. While the spectrum of A λ,α has been studied in the context of inviscid (Euler) flows and their stabilty [START_REF] Drazin | Hydrodynamic stability[END_REF][START_REF] Lin | Some recent results on instability of ideal plane flows[END_REF], it appears that its inverse norm has not been estimated in the literature. The purpose of this section is therefore to offer a systematic study of A -1 λ,α , with emphasis on the limit ℜλ → 0 for ν ∈ [U(-1), U(1)].

The case ℜλ = 0 : preliminaries

We begin by showing that A λ,α is a closed operator on L 2 (-1, 1). In the cases where D(A λ,α is given by (3.14) the proof is standard and will therefore be omitted.

V := {u ∈ C ∞ ([-1, +1]), s.t. u(-1) = u(1) = 0}
is dense in D(A iν,α ) under the graph norm.

Proof. Let {u n } ∞ n=1 denote a sequence in V such that

u n → û in L 2 (-1, +1) and Au n → v in L 2 (-1, +1) (4.1)
To prove the proposition we need to show that û ∈ D(A iν,α ) and that A iν,α û = v. Let U(x ν ) = ν. We use Hardy's inequality for weighted Sobolev spaces associated with the intervals (-1, x ν ) and (x ν , 1) separately. Set then, for k ∈ {1, 2}

W k 1,ν (x ν , 1) := {u ∈ H 1 (x ν , 1) , (U -ν)u (k) ∈ L 2 (x ν ,
v ∈ W 2 1,ν (x ν , 1) , v L 2 (xν ,1) ≤ 2 (x -x ν )v ′ L 2 (xν ,1) . (4.2)
Note that (4.2) follows by extension from Hardy's inequality in (0, +∞), given for any v ∈ W 2 1,0 (0, +∞)) by

v L 2 (0,+∞) ≤ 2 x v ′ L 2 (0,+∞) . (4.3)
Hence, for any u ∈ W 2 1,ν (x ν , 1) (and hence for the restriction to [x ν , 1] of any u ∈ V)

u ′ L 2 (xν ,1) ≤ 2 (x -x ν )u ′′ L 2 (xν ,1) ≤ C m (U -ν)u ′′ L 2 (xν ,1) . (4.4)
From (4.1) and (4.4) we deduce that u n is a Cauchy sequence in H 1 (x ν , 1) and in H 1 (-1, x ν ). Hence there are two corresponding limits u + ∈ H 1 (x ν , 1) and u -∈ H 1 (-1, x ν ), with û/(-1,xν) = u -and û/(xν,1) = u + . By continuity we have u

-(-1) = 0, u + (1) = 0 and u -(x ν ) = u + (x ν ). This shows that û ∈ H 1 0 (-1, +1). Finally it is clear that v = A iν,α û in D ′ (-1, +1) and hence û ∈ D(A iν,α ).
The density argument is a consequence, after localization, of Proposition 2.1 in [START_REF] Bolley | Sur une classe d'opérateurs elliptiques et dégénérés à une variable[END_REF].

The cases x ν = ±1 are easier, since one can apply Hardy's inequality only once, in (-1, 1).

Consider the case of Couette flow U = x in (-1, 1), which is one of the most popular examples of uniaxial flows ( [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF][START_REF] Romanov | Stability of plane-parallel Couette flow[END_REF][START_REF] Bedrossian | Stability of the Couette flow at high Reynolds numbers in 2D and 3D[END_REF]. In this case

A c iν,α def = (x -ν) - d 2 dx 2 + α 2 .
We can construct in some cases the explicit solution of the inhomogeneous problem

A c iν,α u = f . (4.5)
For example, when α = 0, ν ∈ (-1, +1) and f ≡ 1. It can be easily verified that

u ν (x) = A 1 (x -ν) + A 2 + (x -ν) log |x -ν| , (4.6a)
where

A 1 = -1 2 (1 -ν) log |1 -ν| + (1 + ν) log |1 + ν| , A 2 = 1 2 (1 -ν) log |1 -ν| -(1 + ν) log |1 + ν| . (4.6b)
It can be easily verified that u ν satisfies (4.5).

While it seems at a first glance that A c iν,0 :

D(A c iν,0 ) → L 2 (-1, 1
) is injective for all ν ∈ R, it is wrong for ν ∈ (-1, +1). It has indeed a non-trivial solution of the form

ϕ ν (x) = x-ν 1-ν -1 if ν ≤ x , -x-ν 1+ν -1 if x < ν . While ϕ ν ∈ H 2 (-1, 1
) for all ν ∈ (-1, 1), it does satisfy (4.8) in the sense of distributions.

We now look at the injectivity of A c iν,α when ν ∈ (-1, +1). We have already explicitly obtained a non trivial element in KerA c iν,0 . More generally, we prove Lemma 4.2. For all ν ∈ (-1, 1), α ≥ 0, it holds that Dim KerA c iν,α = 1 . (4.7)

Proof. We observe that if u is in the kernel, we have u ∈ H 1 0 (-1, +1) and

- d 2 dx 2 + α 2 u = c δ(x -ν)
, where c is the jump u ′ undergoes through x = ν. Since δ(xν) belongs to H -1 (-1, +1), we may use the Lax-Milgram Lemma for the Dirichlet problem to obtain a unique solution u 1 ∈ H 1 0 (-1, +1) for

- d 2 dx 2 + α 2 u 1 = δ(x -ν) .
It follows that u = cu 1 .

Using [10, Proposition 3.1] and the same argument as in the proof of [10, Theorem 3.1] one can show that A c iν,α , which is a bounded operator from the Hilbert space D(A c iν,α ) into L 2 (-1, +1), is a Fredholm operator of index 1 for ν ∈ (-1, +1). With the aid of Lemma 4.2, we can then conclude the surjectivity of A c iν,α for any ν ∈ (-1, +1) or more precisely the existence of a right inverse. We note that surjectivity of A c iν,α follows from the surjectivity A iν,α we prove in the sequel.

Remark 4.3. When f ∈ C 1 ([-1, 1]
), we can prove (4.7) in the following alternative manner. Recall the solution of A c iν,0 u ν = 1 we have obtained in (4.6). If f (ν) = 0, we can solve (4.5) by dividing it by (x-ν). When f (ν) = 0 we write u = f (ν)u ν + ũ , to obtain A c iν,α ũ = f , with f (x) := f (x)f (ν)α 2 (xν)u ν , which can be easily solved as f (ν) = 0. It is not clear, however, how to extend by density the above solution to any f ∈ L 2 (-1, +1).

In the next subsection, we obtain the surjectivity of A iν,α (for any U satisfying (3.12)) via a non-explicit compactness argument.

Construction of a right inverse of A iν,α

We begin by establishing the surjectivity of A iν,α Lemma 4.4. Suppose that (3.12) holds, and that λ = iν for some ν ∈ [U(-1), U(1)]. Then, for any v ∈ L 2 (-1, 1) and α ≥ 0 there exists some φ ∈ D(A λ,α ) satisfying A λ,α φ = v. Furthermore, there exists C > 0, such that for all ν, α and v ∈ L 2 , φ satisfies

φ 1,2 + [1 + α 2 ] -1 (U -ν)φ ′′ 2 ≤ C v 2 , (4.8)
where φ 1,2 denotes the norm of φ in the Sobolev space H 1 (-1, 1).

Proof. Let κ > 0. With (3.13) in mind, we can use the following alternative form of A λ,α φ = v, which is valid wherever U = ν,

-(U -ν) 2 φ U -ν ′ ′ + α 2 (U -ν)φ = v .
(4.9)

We look first for some

φ κ ∈ H 2 (-1, +1) ∩ H 1 0 (-1, +1) ⊂ D(A iν,α ) satisfying the regularized equation -[(U -ν) 2 + κ 2 ] φ κ U -ν + iκ ′ ′ + α 2 (U -ν -iκ)φ κ = v . (4.10)
We may set

w κ = (U -ν + iκ) -1 φ κ to obtain -([(U -ν) 2 + κ 2 ]w ′ κ ) ′ + α 2 [(U -ν) 2 + κ 2 ]w κ = v
Taking the inner product with w κ then yields

[(U -ν) 2 + κ 2 ] 1/2 w ′ κ 2 2 + α 2 [(U -ν) 2 + κ 2 ] 1/2 w κ 2 2 = w κ , v . (4.11)
This immediately implies, by the Lax-Milgram Lemma, that w κ (and hence also φ κ ) uniquely exists in H 2 (-1, 1) ∩ H 1 0 (-1, 1). Since ν ∈ [U(-1), U(1)], there exists x ν ∈ [-1, 1] such that U(x ν ) = ν. Let m > 0 be defined by (2.24). Clearly, we have

(U -ν) 2 + κ 2 ≥ m 2 (x -x ν ) 2 . Hence [(U -ν) 2 + κ 2 ] 1/2 w ′ κ 2 2 ≥ m 2 (x -x ν )w ′ κ 2 2 = m 2 (x -x ν )w ′ κ 2 L 2 (-1,xν ) + (x -x ν )w ′ κ 2 L 2 (xν ,1)
.

Upon translation we apply Hardy's inequality (4.3), and the fact that w κ (1) = 0 (permitting the extension w k = 0 in (1, +∞) so that

w k ∈ H 1 (x ν , ∞)), it holds that (x -x ν )w ′ κ 2 L 2 (xν ,1) ≥ 1 4 w κ 2 L 2 (xν ,1) .
A similar bound can be established on (-1, x ν ). Hence, by (4.11),

w κ 2 ≤ 4 m 2 v 2 . (4.12)
Substituting once again into (4.11) yields, in addition,

[(U -ν) 2 + κ 2 ] 1/2 w ′ κ 2 ≤ 2 m v 2 . (4.13)
We note that (4.12) and (4.13) do not imply convergence of w κ in L 2 (-1, 1) to (Uν) -1 φ where φ is a solution of (3.13) (with λ = iν). As a matter of fact it is expected that, in most cases, (Uν) -1 φ will be unbounded in L 2 . We thus use (4.13) and (4.12) to obtain that

φ ′ κ 2 = ([U -ν + iκ]w κ ) ′ 2 ≤ [(U -ν) 2 + κ 2 ] 1/2 w ′ κ 2 + U ′ w κ 2 ≤ C v 2 . (4.
14) It follows, either by Poincaré's inequality or by (4.12), that {φ κ } ∞ k=1 is bounded in a ball of size Ĉ v 2 in H 1 0 (-1, 1). By weak compactness, there exists a sequence κ n > 0 tending to 0 such that φ κn converges weakly in H 1 0 (-1, 1) to a limit φ 0 ∈ H 1 0 (-1, 1) as n → +∞ in the same ball, and hence

φ 0 H 1 0 (-1,1) ≤ Ĉ v . (4.15)
It remains to establish that φ 0 is a weak solution of (3.13) for λ = iν. We thus write (4.10) in its weak form for some ψ ∈ C ∞ 0 (-1, 1)

ψ ′ , (U -ν -iκ)φ ′ κ -ψ ′ , (U -ν -iκ) (U -ν + iκ) U ′ φ κ + α 2 ψ, (U -ν -iκ)φ κ = ψ, v .
(4.16) Letting κ = κ n and then taking the limit n → +∞ yields by the above established weak convergence in

H 1 that ψ ′ , (U -ν)φ ′ 0 -ψ ′ , U ′ φ 0 + α 2 ψ, (U -ν)φ 0 = ψ, v .
Consequently, φ 0 is a weak solution of (3.13) satisfying (3.12). Finally, since for all x = x ν we have (Uν)[φ ′′ 0α 2 φ 0 ] = U ′′ φ 0 , we easily obtain from (4.15) that

(U -ν)φ ′′ 0 2 ≤ C(1 + α 2 ) v 2 .
The lemma is proved. Remark 4.5. As in the case of a Couette flow, one can show that the index of A iν,α as a Fredholm operator from the Hilbert space D(A iν,α ) (equipped with the graph norm) into L 2 (-1, 1) is one for ν ∈ [U(-1), U(1)]. Since the multiplication by U ′′ is a compact operator from D(A iν,α ) into L 2 (-1, 1) we may conclude from Fredholm theory that Ãν,α = (Uν) -d 2 dx 2 + α 2 and A iν,α have the same index. Then, we observe that the multiplication operator by (U(x)ν)/(xx ν ) has index 0, and hence Ãν,α and A c iν,α again have the same index. Consequently, the index of A iν,α equals to one, and by Lemma 4.4 it holds that dim kerA iν,α = 1. We note that one can construct a direct proof of injectivity with much greater difficulty.

We have proved above, the surjectivity of A = A iν,α . By Fredholm theory, there is a natural right inverse E right iν,α which associates with v the solution φ of Aφ = v which is orthogonal to KerA iν,α in D(A iν,α ) for the scalar product

(φ, ψ) → φ , ψ ν,α := Aφ, Aψ L 2 + φ, ψ L 2 .
Note that φ , ψ ν,α coincides with the ordinary L 2 product whenever ψ ∈ KerA iν,α .

Employing the estimates of Lemma 4.4 we now prove:

Proposition 4.6. Suppose that (3.12) holds, and that λ = iν for some ν ∈ [U(-1), U(1)].

Then, there exists C > 0, such that for all ν, α

E right iν,α L(L 2 ,H 1 0 ) + E right iν,α L(L 2 ,D(A iν,α )) ≤ C . (4.17)
Proof. We first observe that the solution constructed in Lemma 4.4 satisfies

φ α,ν ≤ C v 2
To obtain E right iν,α v we now need to subtract π α,ν φ, where π α,ν is the orthogonal projector from D(A iν,α ) onto KerA iν,α . Obviously,

E right iν,α v 2 α,ν = φ -π α,ν φ 2 α,ν ≤ φ 2 α,ν ≤ C v 2 2 ,
establishing, thereby, a bound on the right inverse in L(L 2 , D(A iν,α )). To estimate

E right iν,α in L(L 2 , H 1
0 ) we observe that by (4.4) there exists C 0 > 0 such that for any α ∈ R, ν ∈ [U(-1), U(1)], and φ ∈ D(A iν,α )

φ 1,2 ≤ C 0 φ α,ν .

Nearly Couette velocity fields

We now attempt to estimate A -1 λ,α in the case where ℜλ = 0. We shall begin with the case where δ 2 (U), defined in (2.26), is small. To this end we introduce

I(φ, λ) = 1 2 φ ′ 2 2 + U ′′ (U -ν)φ (U -ν) 2 + µ 2 , φ , (4.18a)
and

γ m (λ, U) = inf φ∈H 1 0 (-1,1)\{0} I(φ, λ) φ 2 2 . (4.18b)
The following result proves that under suitable assumptions on U, the infimum of γ m (λ) over C \ J , where

J = {λ ∈ C | ℜλ = 0 , ν ∈ [U(-1), U (1)
]} is positive.

Lemma 4.7. For any r > 1, there exists δ 0 > 0 and γ 0 > 0 such that for any

U ∈ S 3 r satisfying δ 2 (U) ≤ δ 0 , we have inf λ∈C\J γ m (λ, U) ≥ γ 0 > 0 . (4.19)
Proof. Writing I(φ, λ) in the form

I(φ, λ) = 1 2 φ ′ 2 2 + ℜ φ , U ′′ φ (U + iλ) .
we attempt to estimate the second term on the right hand side. For some ν 0 > 0 to be chosen later, we consider two different cases depending on the size of |λ|.

Case I: |µ| 2 + |ν| 2 < ν 0 . Integration by parts yields, accounting for the Dirichlet condition φ satisfies at

x = ±1, φ, U ′′ φ U + iλ = U ′′ U ′ |φ| 2 , U ′ U + iλ = - U ′′ |φ| 2 U ′ ′ , log (U + iλ) .
We estimate the right-hand-side as follows

U ′′ |φ| 2 U ′ ′ , log (U + iλ) ≤ C(ν 0 ) δ 2 (U)[ φ ′ 2 φ ∞ + φ 2 ∞ ] . (4.20)
Sobolev's embedding and Poincaré's inequality then yield that for some C(ν 0 , r) > 0,

ℜ φ, U ′′ φ U + iλ ≤ C(ν 0 , r) δ 2 (U) φ ′ 2 2 . (4.21)
Case II:

|µ| 2 + |ν| 2 ≥ ν 0 . As U ∈ S r we conclude that there exists C(r) > 0 such that ℜ φ, U ′′ φ U + iλ ≤ φ, U ′′ φ U + iλ ≤ C |ν 0 | φ 2 2 ≤ Ĉ(r) ν 0 φ ′ 2 2 . (4.22)
Poincaré's inequality was applied to obtain the last estimate. We can now set ν 0 = 8 Ĉ in (4.22) and δ 2 (U) ≤ δ 0 = 1 8 C(ν 0 ,r) in (4.22), to obtain

I(φ, λ) ≥ 1 4 φ ′ 2 2 .
The lemma is proved by using Poincaré's inequality once again.

If δ 2 (U) is not small we may still show Lemma 4.8. For any r > 1, there exists γ 0 ∈ R such that for any U ∈ S 3 r we have

inf λ∈C\J γ m (λ, U) ≥ γ 0 .
Proof. Indeed, we obtain by (4.20), Poincaré's inequality and Sobolev's embeddings, that for any ε > 0 and any ν 0 > 0 there exists C ε,ν 0 such that, for any λ ∈ C \ J and φ ∈ H 1 0 (-1, 1),

I(φ, λ) ≥ 1 2 φ ′ 2 2 -C φ ′ 3/2 2 φ 1/2 2 ≥ 1 2 -ε φ ′ 2 2 -C ε,ν 0 φ 2 2 .
For |λ| > ν 0 , semiboundedness of I follows immediately from (4.22).

We now obtain an estimate for A -1 λ,α which is neither singular as |µ| → 0 (unlike (4.49)), nor does it necessitate any assumption on the injectivity of A iν,α . Instead, we simply assume (4.19). We also observe that m ≥ 1/r for any U ∈ S 2 r where m is defined in (2.24). Proposition 4.9. For any r > 1 and p > 1, there exists a constant C such that for any U ∈ S 2 r satisfying (4.19) we have:

1. For all v ∈ H 1 (-1, 1), sup |ℜλ|≤(2r) -1 \{ℜλ=0} 0≤α 1 log (|ℜλ| -1 ) A -1 λ,α v 1,2 ≤ C v ∞ . (4.23)
2. For all v ∈ W 1,p (-1, 1),

sup |ℜλ|≤(2r) -1 \{ℜλ=0} 0≤α A -1 λ,α v 1,2 ≤ C ( v ′ p + v ∞ ). (4.24) 3. For all v ∈ L p (-1, +1), sup |ℜλ|≤(2r) -1 \{ℜλ=0} 0≤α |ℜλ| 1/p A -1 λ,α v 1,2 ≤ C v p . (4.25)
Proof.

Proof of (4.23). Let v ∈ H 1 (-1, 1) and φ = A -1 λ,α v for some λ = µ + iν with 0 < |µ| ≤ m/2. We begin by observing that

ℜ φ, v U -ν + iµ = φ ′ 2 2 + α 2 φ 2 2 + ℜ φ, U ′′ φ U + iλ , (4.26)
and note that by (4.19), which is assumed to hold here,

ℜ φ, v U -ν + iµ = 1 2 φ ′ 2 2 + α 2 ||φ 2 + I(φ, λ) ≥ 1 2 φ ′ 2 2 . (4.27)
The left hand side of (4.27) can be estimated as follows

ℜ φ, v U + iλ ≤ φ ∞ v ∞ 1 U + iλ 1 ≤ C log (m|µ| -1 ) φ ∞ v ∞ . (4.28)
To obtain (4.28), we consider two different cases.

We first apply the following computation, valid whenever ν ∈ [U(-1), U(1)], and 0 < |µ| ≤ m/2,

1 U +iλ 1 = +1 -1 1 √ (U (x)-ν) 2 +µ 2 dx ≤ m -1 1 -1 1 √ |x-xν| 2 +|µ/m| 2 dx ≤ m -1 2m µ -2m µ 1 √ 1+y 2 dy ≤ C m -1 log (m|µ| -1 ) . (4.29)
In the case ν < U(-1) we may write

1 U +iλ 1 = +1 -1 1 √ (U (x)-ν) 2 +µ 2 dx ≤ +1 -1 1 √ (U (x)-U (-1)) 2 +µ 2 dx ≤ m -1 +1 -1 1 √ (x+1) 2 +(µ/m) 2 dx ,
and proceed as before. The case U(1) > ν is similar. Hence (4.28) is proved in both cases.

Sobolev's embedding and Poincaré's inequality then yield

φ, v U + iλ ≤ C log (m|µ| -1 ) φ ′ 2 v ∞ . (4.30)
Together with (4.27), it implies for 0

< |µ| ≤ m/2 , φ 1,2 ≤ C log (m|µ| -1 ) v ∞ , (4.31)
and hence also (4.23).

Proof of (4.24).

Let |ν| > 2 max(|U(1)|, |U(-1)|). Then we have ℜ φ, v U + iλ ≤ C φ 2 v 2 . (4.32) If |ν| ≤ 2 max(|U(1)|, |U(-1)|) we integrate by parts to obtain φ, v U + iλ = -log (U + iλ), φ v U ′ ′ .
Use of Hölder's inequality and Sobolev embedding lead, for every p > 1, to the conclusion that there exists

C p > 0 such that φ v U ′ ′ , log (U + iλ) ≤ C p ( φ ∞ v ′ p + φ ′ p v ∞ ) .
where we have used the fact that the L q -norm (q being the Hölder conjugate) of log (U + iλ) can be uniformly bounded for

(|ℜλ|, |ℑλ|) ∈ [0, m/2] × [0, 2 max(|U(1)|, |U(-1)|)] .
Combining the above with (4.32), and making use of Sobolev embedding once again yield

φ, v U + iλ ≤ C p ( v ′ p φ ∞ + v ∞ φ ′ p ) . (4.33)
By (4.27) we obtain that

1 2 φ ′ 2 2 ≤ C φ 1,p min( v ′ p , v ∞ ) , (4.34)
from which (4.24) easily follows for 1 < p ≤ 2, and then for p > 2 by the inequality

v ′ 2 ≤ 2 1-2/p v ′ p .
Proof of (4.25). We first observe that

φ, v U + iλ ≤ 1 U + iλ q φ ∞ v p , (4.35)
where

q = p/(p -1). If ν ∈ [U(-1), U (1)
], we estimate the right-hand-side as follows

1 U +iλ q ≤ m -1 1 -1 dx ([(x-xν ) 2 +|µ/m| 2 ] q/2 1/q ≤ m -1 ∞ -∞ dx [x 2 +|µ/m| 2 ] q/2 1/q = m -1 ( |µ| m ) 1 q -1 ∞ -∞ dx [x 2 +1] q/2 1/q = C q m -1 p |µ| -1/p . (4.36)
If ν < U(-1), we write

1 U + iλ q ≤ m -1 1 -1 dx [(x + 1) 2 + |µ/m| 2 ] q/2 1/q
and proceed as before. The case ν > U(-1) is similar. Thus, without any restriction on ν, we get 1

U + iλ q ≤ Ĉq |µ| -1/p . (4.37)
Substituting the above into (4.35) then yields

φ, v U + iλ ≤ C|µ| -1/p φ ∞ v p . (4.38)
Combining the above with (4.27), we obtain

1 2 φ ′ 2 2 ≤ C|µ| -1/p φ ′ 2 v p , (4.39)
from which (4.25) easily follows

We shall need in Section 8 the following immediate consequence of Proposition 4.9.

Corollary 4.10. For any r > 1 there exists C > 0 such that for all v ∈ L 2 (-1, 1), U ∈ S r satisfying (4.19), λ ∈ C \ J and α ≥ 0 it holds that

A -1 λ,α v 1,2 ≤ C (1 ± x) 1/2 v U + iλ 1 . (4.40) Proof. Let v ∈ L 2 (-1, 1) and φ = A -1 λ,α v. By (4.27) we have that 1 2 φ ′ 2 2 ≤ ℜ φ, v U + iλ . ( 4

.41)

As φ ∈ H 1 0 (-1, 1) we may write

ℜ φ, v U + iλ = ℜ φ -φ(±1), v U + iλ ≤ φ ′ 2 (1 ± x) 1/2 v U + iλ 1 .
Combining the above with (4.41) yields (4.40) with the aid of Poincaré's inequality.

In the next lemma we address the optimality of (4.25).

Lemma 4.11.

Let U ∈ C 3 ([-1, 1]
) satisfy (3.12). Then, (4.25) is optimal, i.e., there exists a sequence

{λ k , v k } ∞ k=1 ∈ C N × [L p (-1, 1)] N and α ∈ R + such that v k p = 1, µ k = ℜλ k → 0, and lim inf k→∞ |µ k | 1/p A -1 λ k ,α v k 1,2 > 0 . (4.42)
Proof. We prove (4.42) for α = 0. Consider then A λ,0 for some λ = µ + iν with µ = 0 and ν ∈ (U(-1), U(1)). Let (φ, v) satisfy

A λ,0 φ = v , with v ∈ L p (-1, 1) and v p = 1 .
We may rewrite this equation in the form

(U + iλ) 2 φ U + iλ ′ ′ = v . Integrating once yields (U(t) + iλ) 2 φ U + iλ ′ (t) = A 1 + t -1 v(τ )dτ .
Integrating again leads to

φ(x) U(x) + iλ = A 1 x -1 dt (U(t) + iλ) 2 + x -1 1 (U(t) + iλ) 2 t -1 v(τ )dτ dt .
The Dirichlet boundary condition at x = 1 is then satisfied through the requirement that A 1 satisfies

A 1 1 -1 dt (U(t) + iλ) 2 + 1 -1 1 (U(t) + iλ) 2 t -1 v(τ )dτ dt = 0 .
Making use of Fubini's Theorem, we finally obtain

φ(x) = (U(x) + iλ) x -1 v(t) x t dτ (U(τ ) + iλ) 2 dt -A 1 x -1 dτ (U(τ ) + iλ) 2 , (4.43a)
where

A 1 = 1 -1 v(t) 1 t dτ (U (τ )+iλ) 2 dt 1 -1 dτ (U (τ )+iλ) 2 .
(4.43b)

We now write

x t dτ (U(τ ) + iλ) 2 = - 1 U ′ (U + iλ) x t - x t U ′′ (τ ) dτ |U ′ (τ )| 2 (U(τ ) + iλ) . (4.44)
It can be verified (as in the proof of (4.29)) that for some positive, independent of µ, constants C 1 and C,

x t U ′′ (τ )dτ |U ′ (τ )| 2 (U(τ ) + iλ) ≤ C 1 1 1 |U(τ ) + iλ)| -1 dτ ≤ C 1 + log |µ| .
Consequently, as µ → 0 we have

x -1 v(t) x t dτ (U(τ ) + iλ) 2 dt - x -1 v(t) 1 U ′ (U + iλ) x t dt ≤ C | log |µ| | v 1 . (4.45) We seek a sequence {λ k , v k } with v k ∈ L p (-1, 1
) and µ k > 0 a decreasing sequence tending to 0 for which (4.42) holds. Let then

λ k = µ k + iν .
For sufficiently large k we have that x ν + 2µ k < 1. We then define v k by

v k (x) = µ -1/p k x ∈ [x ν , x ν + µ k ] 0 otherwise . Clearly v k p = 1, and v k 1 = µ 1-1/p k
. By (4.45) we then have as µ k tends to 0 ,

x -1 v k (t) x t dτ (U(τ ) + iλ k ) 2 dt = x -1 v k (t) 1 U ′ (U + iλ k ) x t dt+O(|µ k | 1-1/p | log µ k |) . (4.46) We now write, for (x ν + 1)/2 < x ≤ 1, x -1 v k (t) 1 U ′ (U +iλ k ) x t dt = = 1 U ′ (x)(U (x)+iλ k ) µ 1-1/p k -µ -1/p k xν +µ k xν 1 U ′ (t)(U (t)+iλ k ) dt . As xν +µ k xν dτ U ′ (U + iλ k ) - xν +µ k xν dτ J ν [J ν (τ -x ν ) + iµ k ] ≤ C µ k 1 + log µ -1 k , where J ν = U ′ (x ν ), we obtain, for (x ν + 1)/2 < x ≤ 1, x -1 v k (t) 1 U ′ (U + iλ k ) x t dt = - µ -1/p k J 2 ν log [J ν (τ -x ν ) + iµ k ] xν +µ k xν + O(µ 1-1/p k ) .
Consequently,

x -1 v k (t) 1 U ′ (U + iλ k ) x t dt = - µ -1/p k J 2 ν log (1 -iJ ν ) + O(µ 1-1/p k | log µ k |) . (4.47)
Substituting the above into (4.43a) and (4.43b) yields, for all (x ν + 1)/2 < x ≤ 1,

φ k (x) = -(U(x)+iλ k ) µ -1/p k J 2 ν log (1-iJ ν ) 1 x (U(τ ) + iλ k ) -2 dτ 1 -1 (U(τ ) + iλ k ) -2 dτ +O(µ 1-1/p k | log µ k |) ,
where

φ k := A -1 λ k ,0 v k . Clearly, for all (x ν + 1)/2 < x ≤ 1, lim k→∞ 1 x (U(τ ) + iλ k ) -2 dτ = 1 x (U(τ ) -ν) -2 dτ > 0 ,
the convergence being uniform in x. We now prove (4.42) by establishing that lim inf

k→∞ |µ k | 1/p φ k L 2 ((xν +1)/2,1])) > 0 ,
which will immediately imply lim inf k→∞ |µ k | 1/p φ k L 2 (-1,1) > 0 and consequently (4.42).

To this end we need to prove that

1 -1 (U(τ ) + iλ k ) -2 dτ ≤ C . (4.48)
We now use (4.44) together with the fact that

U ∈ C 3 ([-1, 1]) to obtain that 1 -1 dτ (U(τ ) + iλ k ) 2 = - 1 U ′ (U + iλ k ) - U ′′ log (U + iλ k ) |U ′ | 3 1 -1 + 1 -1 U ′′ |U ′ | 3 ′ log (U+iλ k ) dτ . Since U ′ > 0 and |U(±1) + iλ k )| ≥ C(1 ± x ν )
, for some C > 0, we can conclude (4.48) from the above.

Non-vanishing U ′′

We dedicate this subsection to the case when U ∈ C 2 ([-1, +1] and U ′′ = 0 which may result from a combination of non-vanishing pressure gradient and relative velocity between the plates at x 2 = ±1. We begin by establishing the following, rather straightforward, result.

Proposition 4.12. Suppose that U ′′ = 0 on [-1, +1], then, for any λ ∈ C for which ℜλ = 0 and α ≥ 0, A λ,α is invertible. Moreover, for any r > 1 there exists C > 0, such that, for any λ with ℜλ = 0, α ≥ 0, and U ∈ S 2 r satisfying (2.29), it holds that

A -1 λ,α + d dx A -1 λ,α ≤ C 1 + |ℜλ| 1/2 |ℜλ| . (4.49) Proof. For a pair (φ, v) such that v = A λ,α φ, with λ = µ + iν, we write, ℑ φ, v U -ν + iµ = -µ U ′′ φ (U -ν) 2 + µ 2 , φ . (4.50)
Consequently, since U ′′ = 0 (hence has constant sign) we obtain

φ U -ν + iµ 2 ≤ C |µ| v 2 . (4.51)
Let ν 0 > 0 be chosen at a later stage and consider the following two cases. In the case |µ| 2 + |ν| 2 < ν 0 , we immediately deduce from (4.51) that there exists

C(ν 0 ) such that φ 2 ≤ C(ν 0 ) |µ| v 2 . (4.52)
In the case

|µ| 2 + |ν| 2 ≥ ν 0 , as ℜ φ, v U -ν + iµ = φ ′ 2 2 + α 2 φ 2 2 + U ′′ (U -ν)φ (U -ν) 2 + µ 2 , φ , (4.53)
we can use (4.51) once again to obtain that, for ν 0 ≥ 2 sup |U|, there exists

C > 0 such that φ ′ 2 2 + α 2 - C ν 0 φ 2 2 ≤ C |µ| v 2 2 .
Since φ ∈ H 1 0 (-1, 1) we can use Poincaré's inequality to obtain for sufficiently large ν 0 that there exists C > 0 such that, for any α ≥ 0 (and |λ| ≥ ν 0 ),

φ 1,2 ≤ C |µ| 1/2 v 2 , (4.54)
which, combined with (4.52) yields (4.49).

Once injectivity of A λ,α is established, we may apply Fredholm theory to prove its surjectivity. By the compactness of the multiplication with U ′′ from D(A λ,α ) into L 2 (-1, +1), we can conclude, as in Remark 4.5, that the index of A λ,α is the same as the index of (U + iλ)

-d 2 dx 2 + α 2 . Since for µ = 0, U + iλ = 0 on [-1, +1], it follows that the indices of A λ,α and -d 2 dx 2 + α 2 from H 2 (-1, 1) ∩ H 1 0 (-1, 1) onto L 2 (-1, 1
) are the same. Consequently, the index of A λ,α is 0 and surjectivity follows from injectivity.

It should be noted that (4.9) is unsatisfactory. Clearly, it is significantly inferior to (4.23)- (4.25), where a bound of O(|ℜλ| -1/2 ) for A -1 λ,α is obtained. We seek, therefore, a better estimate for A -1 λ,α that will be applicable in Sections 7 and 8.

Proposition 4.13. Let r > 1 and p > 1. There exist µ 0 > 0 and C > 0 such that for all v ∈ W 1,p (-1, 1) and U ∈ S 3 r satisfying (2.29) we have

sup |ℜλ|≤µ 0 \{ℜλ=0} 0≤α 1 log (|ℜλ| -1 ) A -1 λ,α v 1,2 ≤ C v ∞ , (4.55a)
sup

|ℜλ|≤µ 0 \{ℜλ=0} 0≤α A -1 λ,α v 1,2 ≤ C ( v ′ p + v ∞ ) , (4.55b)
and sup

|ℜλ|≤µ 0 \{ℜλ=0} 0≤α |ℜλ| 1/p A -1 λ,α v 1,2 ≤ C v p . (4.55c)
Proof. In the case where ν ∈ [U(-1), U(1)] we (uniquely) select

x ν ∈ [-1, 1] where U(x ν ) = ν. Otherwise if ν > U(1) (ν < U(-1)) we set x ν = 1 (x ν = -1).
Step 1: For p > 1 and

ℜλ = 0 define N ± m,p by v → N ± m,p (v, λ) := min (1 ± •) 1/2 v U + iλ 1 , v 1,p .
We prove that there exists C > 0 such that, for all ε > 0 and 0

< |µ| ≤ 1 it holds that |φ(x ν )| ≤ C ε -1/2 N ± m,p (v, λ) + (|µ| 1/2 + ε 1/2 ) φ ′ 2 , (4.56
)

for all pairs (φ, v) ∈ D(A λ,α ) × W 1,r (-1, 1) satisfying A λ,α φ = v. As |φ(x)| 2 ≥ 1 2 |φ(x ν )| 2 -|φ(x) -φ(x ν )| 2 ,
we may use (4.50) to obtain

|ℑ φ, v U -ν + iµ | ≥ |µ| |U ′′ | (U -ν) 2 + µ 2 , 1 2 |φ(x ν )| 2 -|φ(x) -φ(x ν )| 2 . (4.57)
We note that, for any p > 1, there exists C > 0 such that

φ, v U +iλ = φ ′ v + φ v′ , log (U + iλ) ≤ C ( φ ′ 2 v ∞ + φ ∞ v ′ p ) ≤ C φ ′ 2 v 1,p .
The constant C depends on a uniform bound of log(U + iλ) 2 and log(U + iλ) p/(p-1) . On the other hand, observing that

|φ(x)| = |φ(x) -φ(±1)| ≤ φ ′ 2 (1 ± x) 1 2 ,
we may conclude that

φ, v U + iλ ≤ φ ′ 2 (1 ± •) 1/2 v U + iλ 1
and hence, there exists C > 0, such that

φ, v U + iλ ≤ C φ ′ 2 N ± m,p (v, λ) . (4.58)
Substituting the above into (4.57) yields

|µ| 2 |φ(x ν )| 2 |U ′′ | 1/2 U + iλ 2 2 ≤ |µ| sup |U ′′ | φ -φ(x ν ) U + iλ 2 2 + C φ ′ 2 N ± m,p (v, λ) .
We now observe, as in (4.36) (but with a lower bound in mind), that, for some positive C and Ĉ (note that |µ| ≤ 1), it holds

|U ′′ | 1/2 U + iλ 2 2 ≥ 1 C 1 (x -x ν ) 2 + µ 2 1 ≥ 1 Ĉ|µ| .
Hence, for another constant C > 0, we get

|φ(x ν )| 2 ≤ C |µ| φ -φ(x ν ) U + iλ 2 2 + φ ′ 2 N ± m,p (v, λ) . (4.59)
To estimate the first term on the right-hand-side of (4.59) we first observe that for some C > 0 we have

1 U(x) + iλ ≤ C |x -x ν | , ∀x ∈ (-1, +1) .
Then we notice that for any w ∈ H 1 0 (R + ) we have by (4.3) and some integration by parts

(xw) ′ 2 2 = xw ′ 2 2 ≥ 1 4 w 2 2 .
Using a constant smaller than 1/4 on its right-hand-side, the above inequality can can be extended in an interval (a, b)

to the space {w ∈ H 1 (a, b) , v(a) = 0} . We thus apply it to w = (φ -φ(x ν ))/(x -x ν ) in (-1, x ν ) and (x ν , 1) to obtain φ -φ(x ν ) U + iλ 2 2 ≤ C φ ′ 2 2 ,
which when substituted into (4.59) readily yields (4.56) via Cauchy's inequality. Note that, for ν ∈ [U(-1), U(1)], (4.56) is trivial as φ(x ν ) = 0 .

Step 2: Let d = min(1x ν , 1 + x ν ). We prove that for any A > 0, and

d 1 > 0, there exists C A,d 1 and µ A,d 1 such that, for α 2 ≤ A, x ν ∈ (-1, 1), |µ| ≤ µ A,d 1 , and d ≥ d 1 , φ 1,2 ≤ C A,d 1 N ± m,p (v, λ) . (4.60) holds for any pair (φ, v) satisfying A λ,α φ = v. Let χ ∈ C ∞ 0 (R, [0, 1]) satisfy χ(x) = 1 |x| < 1/2 0 |x| > 1 . Let χ d (x) = χ((x -x ν )/d) and set φ = ϕ + φ(x ν )χ d . (4.61)
Note that by the choice of d, ϕ satisfies also the boundary condition at ±1 . It can be easily verified that

A λ,α ϕ = v + φ(x ν ) (U + iλ)(χ ′′ d -α 2 χ d ) -U ′′ χ d .
By construction we have that w = (Uν) -1 ϕ ∈ H 2 (-1, 1), and hence we can rewrite the above equality (using (3.13) twice) in the form

-(U -ν) 2 ϕ U -ν ′ ′ + α 2 (U -ν)ϕ = v + φ(x ν ) (U -ν)(χ ′′ d -α 2 χ d ) -U ′′ χ d + iµ(φ ′′ -α 2 φ) = (U -ν)v U + iλ + φ(x ν ) (U -ν)(χ ′′ d -α 2 χ d ) -U ′′ χ d + iµ U ′′ φ U + iλ .
Taking the scalar product with w and integrating by parts then yield

(U -ν)w ′ 2 2 + α 2 ϕ 2 2 = ϕ, v U + iλ -w, φ(x ν )U ′′ χ d + φ(x ν ) ϕ, χ ′′ d -α 2 χ d + iµ w, U ′′ φ U + iλ . (4.62)
As in the proof of (4.58), the first term on the right-hand side is estimated as follows

ϕ, v U + iλ ≤ ϕ ′ 2 (1 + •) 1/2 v U + iλ 1 ≤ C( φ ′ 2 + d -1/2 |φ(x ν )|)N ± m,p (v, λ) . ( 4
.63) To estimate the second term on the right-hand-side, we note that by Hardy's inequality, we have

w 2 ≤ C ϕ ′ 2 ≤ Ĉ φ ′ 2 + 1 d 1/2 |φ(x ν )| . (4.64) From (4.64) we get | w, φ(x ν )U ′′ χ d | ≤ Č d 1 2 |φ(x ν )| φ ′ 2 + 1 d 1/2 |φ(x ν )| . (4.65)
Then, we write for the third term on the right-hand-side of (4.62), using integration by parts and the upper bound on

α 2 | ϕ, χ ′′ d -α 2 χ d | ≤ ϕ ′ 2 χ ′ d 2 + C ϕ 2 ≤ Ĉ 1 d 1/2 ϕ ′ 2 + ϕ 2 .
Consequently, by (4.61),

|φ(x ν )| | ϕ, χ ′′ d -α 2 χ d | ≤ ≤ C|φ(x ν )| 1 d 1/2 φ ′ 2 + 1 d 1/2 |φ(x ν )| + ( φ 2 + d 1/2 |φ(x ν )|) .
Hence, using Poincaré's inequality,

|φ(x ν )| | ϕ, χ ′′ d -α 2 χ d | ≤ C 1 d 1/2 |φ(x ν )| φ ′ 2 + 1 d 1/2 |φ(x ν )| ,
from which we conclude the existence of C such that for any ε 1 ∈ (0, 1), we have

|φ(x ν )| | ϕ, χ ′′ d -α 2 χ d | ≤ C ε 1 φ ′ 2 2 + 1 ε 1 d |φ(x ν )| 2 . (4.66)
To estimate the last term on the right-hand-side of (4.62), we use (4.64), (4.37), and (4.64), to obtain

w, U ′′ φ U + iλ ≤ C w 2 φ ∞ 1 U + iλ 2 ≤ C|µ| -1/2 φ ∞ φ ′ 2 + 1 d 1/2 |φ(x ν )|
(4.67) Substituting (4.67) together with (4.63), (4.65), and (4.66) into (4.62) yields that there exists C > 0 such that for every 0

< ε 1 < 1 it holds that (U -ν)w ′ 2 2 +α 2 ϕ 2 2 ≤ C (ε 1 +|µ| 1/2 ) φ ′ 2 2 + 1 ε 1 d |φ(x ν )| 2 +ε -1 1 N ± m,p (v, λ) 2 (4.68)
By Hardy's inequality (4.2), Poincaré's inequality, and (4.56) we obtain, for 0 < |µ| ≤ 1, ε ∈ (0, 1), and ε 1 ∈ (0, 1), that

w 2 ≤ C |µ| 1/4 + ε 1/2 1 + ε 1/2 + |µ| 1/2 [ε 1 d] 1/2 φ ′ 2 + [ε 1 dε] -1/2 N ± m,p (v, λ) .
Selecting ε = ε 2 1 then yields, for any |µ| ≤ µ A,d 1 , and ε 1 ∈ (0, 1),

w 2 ≤ C |µ| 1/4 (1 + |µ| 1/4 [ε 1 d] -1/2 ) + d -1/2 ε 1/2 1 φ ′ 2 + C[ε 3 1 d] -1/2 N ± m,p (v, λ) .
As d ≥ d 1 we then write

w 2 ≤ C |µ| 1/4 (1 + |µ| 1/4 [ε 1 d 1 ] -1/2 ) + d -1/2 1 ε 1/2 1 φ ′ 2 + C[ε 3 1 d 1 ] -1/2 N ± m,p (v, λ) .
(4.69) Recalling the definitions of φ, ϕ and w, we immediately conclude that

φ ′ 2 ≤ (U -ν)w ′ 2 + U ′ w 2 + Cd -1/2 |φ(x ν )| .
which together with (4.56) gives

φ ′ 2 ≤ (U -ν)w ′ 2 + U ′ w 2 + C [d 1/2 1 ε 1 ] -1 N ± m,p (v, λ) + d -1 2 1 (|µ| 1 2 + ε 1 ) φ ′ 2 . ( 4
φ ′ 2 ≤ C(d 1 ) ε -3/2 1 N ± m,p (v, λ) + (|µ| 1 4 + |µ| 1 2 ǫ -1 2 1 + ε 1/2 1 ) φ ′ 2
Hence, we can choose first ǫ 1 and then µ A,d 1 > 0 such that (4.60) follows for |µ| ≤ µ A,d 1 and d ≥ d 1 .

Step 3: We prove (4.60) under the assumption that

α 2 ≥ C U := 1 4 C 4 0 δ 2 (U) 4 , (4.71) with C 0 = 2 sup |µ|≤1 ν∈[U (-1),U (1)] log (U + iλ) 2 .
(4.72)

We recall (4.26) which reads

ℜ φ, v U + iλ = φ ′ 2 2 + α 2 φ 2 2 -ℜ U ′′ φ, φ U + iλ .
For the last term we have, using Poincaré's inequality and Sobolev's embeddings

U ′′ φ, φ U +iλ ≤ U ′′ U ′ |φ| 2 ′ , log (U + iλ) ≤ log (U + iλ) 2 φ ∞ 2 U ′′ U ′ ∞ φ ′ 2 + U ′′ U ′ ′ ∞ φ 2 ≤ C 0 δ 2 (U) φ ′ 3/2 2 φ 1/2 2
, where C 0 is given by (4.72).

Consequently, by (4.58)

φ ′ 2 2 + α 2 φ 2 2 ≤ C 0 δ 2 (U) φ ′ 3/2 2 φ 1/2 2 + C φ ′ 2 N ± m,p (v, λ) . Using Young's inequality we obtain 1 8 φ ′ 2 2 ≤ 1 4 C 4 0 δ 2 (U) 4 -α 2 φ 2 2 + ĈN ± m,p (v, λ) 2 (4.73)
Hence, for α 2 ≥ C U (4.60) follows immediately from the above inequality in conjunction with Poincaré's inequality.

Step 4: We prove that there exist d 0 > 0, µ 0 > 0 and C > 0 such that, for all

d ≤ d 0 , ν ∈ [U(-1), U(1)], α ≥ 0 and |µ| ≤ µ 0 , φ 1,2 ≤ CN ± m,p (v, λ) . (4.74) holds for any pair (φ, v) such that A λ,α φ = v.
Without any loss of generality we can assume that d = 1x ν . As

(U -U(1))(-φ ′′ + α 2 φ) -U ′′ φ = v -(U(1) -ν + iµ)(-φ ′′ + α 2 φ)
or equivalently, by (3.13),

-(U -U(1)) 2 φ U -U (1) ′ ′ + α 2 (U -U(1))φ = (U -U (1))v U +iλ + (U(1) -ν + iµ) U ′′ φ U +iλ .
Taking the scalar product with (U -U(1)) -1 φ and integrating by parts then yield

(U -U(1)) φ U -U(1) ′ 2 2 + α 2 φ 2 2 = φ, v U + iλ + (U(1) + iλ) φ U -U(1) , U ′′ φ U + iλ . (4.75)
For the first term on the right-hand-side of (4.75) we use (4.58).

Next, we estimate the second inner product on the right-hand-side of (4.75) by splitting the domain of integration in two sub-intervals: (1 -2d, 1) and (-1, 1 -2d).

The integral over (1 -2d, 1). To estimate the integral over (1 -2d, 1) we use the identity

1 [U(1) -U](U + iλ) = 1 U(1) + iλ 1 U + iλ + 1 U(1) -U , to obtain that (U(1) + iλ) 1 1-2d U ′′ |φ| 2 [U(1) -U](U + iλ) dx = 1 1-2d U ′′ |φ| 2 U + iλ dx + 1 1-2d U ′′ |φ| 2 U(1) -U dx . (4.76) As 1 1-2d U ′′ |φ| 2 U + iλ dx = - U ′′ U ′ |φ| 2 | log (U + iλ) 1-2d - 1 1-2d U ′′ U ′ |φ| 2 ′ | log (U + iλ)| dx ,
we may conclude, having in mind that µ and d are bounded, that

1 1-2d U ′′ |φ| 2 U +iλ dx ≤ C U ′′ U ′ 1,∞ d log d φ ′ 2 2 + 1 1-2d (|φ| |φ ′ | + |φ| 2 ) | log (U + iλ)| dx ≤ Ĉ d 1/2 φ ′ 2 2 .
(4.77) Furthermore, employing Hardy's inequality and Cauchy-Schwarz inequality yields

1 1-2d U ′′ |φ| 2 U -U(1) dx ≤ C d 1/2 φ ′ 2 φ ∞ .
Substituting the above inequalities into (4.76) yields

(U(1) + iλ) 1 1-2d U ′′ |φ| 2 [U(1) -U](U + iλ) dx ≤ Cd 1/2 φ ′ 2 2 . (4.78)
The integrals over (-1, 1 -2d). We now estimate the integrals over [-1, 1 -2d] for the inner products on the right-hand-side of (4.75). To this end we write, using Hardy's inequality (4.2) and lower bounds of |U + iλ|,

1-2d -1 U ′′ |φ| 2 [U(1) -U](U + iλ) dx ≤ C φ ′ 2 φ ∞ 1-2d -1 1 |U + iλ| 2 1/2 ≤ Ĉ (max(d, |µ|)) -1/2 φ ′ 2 2 . (4.79)
Returning to the estimate of the right hand side of (4.75), we use (4.79), and the fact that |U(1)-ν| ≤ Ĉd (following from by definition of d) together with Poincaré's inequality, to obtain

|U(1) -ν| 1-2d -1 U ′′ |φ| 2 [U(1) -U](U + iλ) dx ≤ Cd 1/2 φ ′ 2 2 .
and

|µ| 1-2d -1 φ (v + U ′′ φ) [U(1) -U](U + iλ) dx ≤ C|µ| 1/2 φ ′ 2 2 .
Combining the above and (4.78) yields

(U(1) -ν + iµ) φ U -U(1) , U ′′ φ U + iλ ≤ C[d 1/2 + |µ| 1/2 ] φ ′ 2 2 . ( 4
(U -U(1)) φ U -U(1) ′ 2 2 ≤ C [d 1/2 + |µ| 1/2 ] φ ′ 2 2 + φ ′ 2 N ± m,p (v, λ) .
From Hardy's inequality in the form (4.2) we then conclude

φ U -U(1) 2 2 ≤ C([d 1/2 + |µ| 1/2 ] φ ′ 2 2 + φ ′ 2 N ± m,p (v, λ)) .
Combined with the following straightforward observation

φ ′ 2 ≤ (U -U(1)) φ U -U(1) ′ 2 + U ′ φ U -U(1) 2 , (4.81) this yields, φ ′ 2 2 ≤ C([d 1/2 + |µ| 1/2 ] φ ′ 2 2 + φ ′ 2 N + m,p (v, λ)) .
Hence, there exists d 0 > 0 and µ 0 > 0 such that (4.74) holds for all d ≤ d 0 and |µ| ≤ µ 0 .

Step 5: We prove that there exist C > 0 and µ 0 > 0 such that (4.74) holds for all ν ∈ R \ [U(-1), U(1)] and |µ| ≤ µ 0 .

Without any loss of generality we assume ν > U(1). We begin by rewriting

A λ,α φ = v in the form -(U -ν) 2 φ U -ν ′ ′ + α 2 (U -ν)φ = v -iµ v + U ′′ φ U + iλ = (U -ν)v -iµU ′′ φ U + iλ .
Taking the inner product with φ/(Uν) on the left yields

(U -ν) φ U -ν ′ 2 2 + α 2 φ 2 2 = φ, v U + iλ -iµ φ U -ν , U ′′ φ U + iλ . (4.82) Let φ(x) = φ(x) x ∈ [-1, 1] 0 |x| > 1 . Let Û = U(x) x ∈ [-1, 1] U(1) + U ′ (1)(x -1) x > 1 .
Let xν > 1 denote the unique zero of Ûν. We may now use Hardy's inequality to obtain the existence of C > 0 such that for all

ν ∈ R \ [U(-1), U(+1)], all φ ∈ H 1 0 (-1, +1), (U -ν) φ U -ν ′ 2 2 = ( Û -ν) φ Û -ν ′ 2 L 2 (-1,xν ) ≥ 1 C φ Û -ν 2 L 2 (-1,xν ) = 1 C φ U -ν 2 2 . (4.83)
Next we use the analog of (4.81)

φ ′ 2 2 ≤ 2 (U -ν) φ U -ν ′ 2 2 + 2 U ′ φ U -ν 2 2
, which leads, together with (4.83), to

φ ′ 2 ≤ Ĉ (U -ν) φ U -ν ′ 2 .
(4.84)

On the other hand we have from (4.82) and (4.37)

(U -ν) φ U -ν ′ 2 2 ≤ φ ′ 2 N ± m,p (v, λ) + |µ| φ U -ν 2 φ ∞ 1 U +iλ 2 ≤ C φ ′ 2 N ± m,p (v, λ) + |µ| 1/2 φ U -ν 2 .
Combining the above with (4.83) yields first

(U -ν) φ U -ν ′ 2 ≤ Č (N ± m,p (v, λ) + |µ| 1 2 (U -ν) φ U -ν ′ 2 ) , hence, for sufficiently small |µ| , (U -ν) φ U -ν ′ 2 ≤ 2 Č N ± m,p (v, λ) . (4.85)
Finally, using (4.84) once again leads to the existence of C > 0 and

µ 0 > 0 such that if |µ| ≤ µ 0 φ ′ 2 ≤ C N ± m,p (v, λ) , (4.86)
and we obtain (4.74) in this case as well.

Step 6: Prove (4.55). By (4.74) (established in steps 4 and 5 for d ≤ d 0 ) and (4.60) (proved in steps 2 and 3 for d ≥ d 1 = d 0 ), there exist C > 0 and µ 0 > 0 such that for |µ| ≤ µ 0 we have

φ 1,2 ≤ C N ± m,p (v, λ) , (4.87)
As N ± m,p (v, λ) ≤ v 1,p , we can immediately conclude (4.55b). To conclude (4.55a,c) we first observe that

N ± m,p (v, λ) ≤ 2 v U + iλ 1 and then use Hölder inequality v U + iλ 1 ≤ v p 1 U + iλ q ,
valid for any 1 < p ≤ ∞, together with (4.37), and (4.29).

For later reference we also need the following estimate which can be deduced from the proofs of Propositions 4.12 and 4.13. Proposition 4.14. Let r > 1. Then, there exists C > 0 such that, for all v ∈ L 2 (-1, 1), λ such that ℜλ = 0, and U ∈ S 3 r satisfying (2.29) it holds that

A -1 λ,α v 1,2 ≤ C (1 ± x) 1/2 v U + iλ 1 . (4.88)
Proof. Let µ 0 be as in the statement of Proposition 4.13. Let λ = µ + iν, v ∈ L 2 (-1, 1), and φ = A -1 λ,α v. For |µ| < µ 0 , (4.88) is an immediate result of (4.87).

Consider then the case |µ| ≥ µ 0 . By (4.50) and (4.58), we obtain that

|µ| min x∈[-1,1] |U ′′ | φ (U -ν) 2 + µ 2 , φ ≤ φ, v U -ν + iµ ≤ C φ ′ 2 N ± m,p (v, λ) .
Consequently, as |µ| > µ 0 , there exists C(µ 0 ) > 0 such that

φ U + iλ 2 2 ≤ C φ ′ 2 N ± m,p (v, λ) . (4.89)
Then, we use (4.53) to establish that

φ ′ 2 2 + α 2 φ 2 2 ≤ φ, v U -ν + iµ + φ U + iλ 2 φ 2 ,
from which we conclude, with the aid of (4.89), that

φ ′ 2 2 + α 2 φ 2 2 ≤ C φ ′ 2 N ± m,p (v, λ) 1/2 φ 2 + φ ′ 2 N ± m,p (v, λ) .
Using Poincaré's inequality we can now establish (4.88).

Some Schrödinger operators and their resolvents

In this section we derive several refinements of estimates obtained in [START_REF] Henry | On the semi-classical analysis of Schrödinger operators with purely imaginary electric potentials in a bounded domain[END_REF][START_REF] Almog | On a Schrödinger operator with a purely imaginary potential in the semiclassical limit[END_REF] for the resolvent of L β = -d 2 dx 2 + iβU, (as in (3.7)) defined over different domains. As in the rest of this contribution, we are assuming (2.24). These estimates will be used in Sections 7 and 8.

The entire real line

We begin by stating the following result on R.

Proposition 5.1. For Ũ ∈ C 1 (R), let L β,R be given by L β,R = - d 2 dx 2 + iβ Ũ , (5.1)
with domain

D( L β,R ) = {u ∈ H 2 (R) | xu ∈ L 2 (R) } . (5.2)
Then, for all positive Υ, m, M, C, and ǫ, there exist β 0 > 0 and Ĉ, such that for all β ≥ β 0 and Ũ ∈ C 1 (R) satisfying

0 < m ≤ Ũ′ (x) ≤ M for all x ∈ R | Ũ′ (x) -Ũ′ (y)| ≤ C|x -y| ǫ , for all x, y ∈ R s.t. |x -y| ≤ 1 , (5.3)
it holds that sup

ℜλ≤Υβ -1/3 (1 + β 1/3 |ℜλ|) ( L β,R -βλ) -1 + β -1/3 sup ℜλ≤Υβ -1/3 d dx ( L β,R -βλ) -1 ≤ Ĉ β 2/3 , (5.4)
and sup

ℜλ≤Υβ -1/3 ( Ũ -ℑλ)( L β,R -βλ) -1 ≤ Ĉ β . (5.5)
Proof.

The estimation of the first term on the left-hand-side of (5.4), can be obtained, for

-1 < β 1/3 µ < Υ (with µ = ℜλ) as in the proof of [21, Theorem 1.1 (ii)].
The difference is that the interval is infinite and that we impose here less regularity on the potential. To accommodate C 1,ǫ potential in the proof in [START_REF] Henry | On the semi-classical analysis of Schrödinger operators with purely imaginary electric potentials in a bounded domain[END_REF], we construct a partition of unity composed of intervals of size β -ρ 2 ), and select ρ ∈

( 2 3(1+ǫ) , 2 3 ) instead of ρ ∈ ( 1 3 , 2
3 ) (as in p. 16, line 6 in [START_REF] Henry | On the semi-classical analysis of Schrödinger operators with purely imaginary electric potentials in a bounded domain[END_REF]). The remaining details are skipped. We now observe that

ℜ u, ( L β,R -βλ)u = u ′ 2 2 -βµ u 2 2 , (5.6)
For µ ≤ -β -1/3 , we deduce

u 2 ≤ C β|µ| ( L β,R -βλ)u 2 ≤ 2C β 2/3 (1 + β 1/3 |µ|) ( L β,R -βλ)u 2 ,
which gives the estimate of the first term on the left-hand-side of (5.4) for

β 1/3 µ ≤ -1.
To estimate the second term on the right-hand-side of (5.4), we return to (5.6) to conclude from the bound of ( L β,Rβλ) -1 , we have just obtained, that

u ′ 2 ≤ C β 1/3 ( L β,R -βλ)u 2 .
(5.7)

Finally, to prove (5.5) we use the identity

ℑ ( Ũ -ν)u, ( L β,R -βλ)u = β ( Ũ -ν)u 2 2 + ℑ Ũ′ u, u ′ ,
to obtain with the aid of (5.4) that

β ( Ũ -ν)u 2 2 ≤ ( L β,R -βλ)u 2 ( Ũ -ν)u 2 + C β ( L β,R -βλ)u 2 2 ,
from which (5.5) easily follows.

A Dirichlet problem

We now obtain some resolvent estimate for the Dirichlet realization L D β of L β in (-1, 1). Proposition 5.2. For any r > 1 and Υ < J 2/3 m ℜν 1 , there exist C > 0 and β 0 > 0 such that for all U ∈ S 2 r and β ≥ β 0 sup

ℜλ≤Υβ -1/3 (L D β -βλ) -1 + β -1/3 d dx (L D β -βλ) -1 + β -2/3 d 2 dx 2 (L D β -βλ) -1 ≤ C β 2/3 , (5.8)
and sup

ℜλ≤Υβ -1/3 (U -ℑλ)(L D β -βλ) -1 ≤ C β .
(5.9)

Furthermore, for every 1 < p < 2 there exists C p > 0 such that for all f ∈ L 2 (-1, 1)

sup

ℜλ≤Υβ -1/3 d dx (L D β -βλ) -1 f p ≤ C p β 2+p 6p f 2 .
(5.10)

Proof. By [21, Theorem 1.1] we have, under the assumptions of the proposition,

(L D β -βλ) -1 ≤ C β 2/3 . (5.11)
We next observe that for any

u ∈ D(L D β ) ℜ u, (L D β -βλ)u = u ′ 2 2 -βµ u 2 2 ,
which together with (5.11) yields

u ′ 2 ≤ C β 1/3 (L D β -βλ)u 2 .
(5.12)

To complete the proof of (5.8) we write, with µ = ℜλ,

-ℜ u ′′ , (L D β -βλ)u = u ′′ 2 2 -βµ u ′ 2 2 -β ℑ U ′ u, u ′ .
From (5.11) and (5.12) we then obtain that

u ′′ 2 ≤ C (L D β -βλ)u 2 ,
completing, thereby, the proof of (5.8). We establish (5.9) in the same way we have established (5.5).

It remains to prove (5.10). Let f ∈ L 2 (-1, 1) and u ∈ D(L D β ) satisfy

(L D β -βλ)u = f . (5.13) Let η and η denote C ∞ functions such that η(t) = 0 |t| < 1/2 1 |t| > 1 , and
η(t) = 1 -η(t) 2 . (5.14) Let x ν ∈ [-1, 1] be such that U(x ν ) = ν (otherwise, if U(x) = ν for all x ∈ [-1, 1],
we arbitrarily set

x ν = -2). Let η ν (x) = η(β 1/3 (x -x ν ))1 R + (x -x ν ) . (5.15) As ℜ η 2 ν (U -ν)u, (L D β -βλ)u = η ν |U -ν| 1/2 u ′ 2 2 + ℜ η ν (U ′ η ν + 2(U -ν)η ′ ν )u, u ′ -µβ η ν |U -ν| 1/2 u 2 2 ,
we obtain, observing that

|(U -ν)η ′ ν | is uniformly bounded, η ν |U -ν| 1/2 u ′ 2 2 ≤ (U -ν)u 2 f 2 + C β 2/3 η ν |U -ν| 1/2 u 2 2 + u 2 u ′ 2 . (5.16) Furthermore, since ℑ η 2 ν u, (L D β -βλ)u = β η ν |U -ν| 1/2 u 2 2 + 2ℑ η ′ ν u, η ν u ′ , (5.17)
we obtain

η ν |U -ν| 1/2 u 2 2 ≤ 1 β ( u 2 f 2 + β 1/3 u 2 u ′ 2 ,
and hence, by (5.8),

η ν |U -ν| 1/2 u 2 ≤ C β 5/6 f 2 .
Substituting the above into (5.16) yields, with the aid of (5.8) and (5.9),

η ν |U -ν| 1/2 u ′ 2 ≤ C β 1/2 f 2 .
Setting η - ν (x) = η ν (-x), we obtain in a similar manner

η - ν |U -ν| 1/2 u ′ 2 ≤ C β 1/2 f 2 ,
and hence, with the aid of (5.8), we can conclude that

|U + iλ| 1/2 u ′ 2 ≤ η ν |U -ν| 1/2 u ′ 2 + η - ν |U -ν| 1/2 u ′ 2 + Cβ -1/6 u ′ 2 ≤ Ĉβ -1/2 f 2 .
We can now conclude, assuming first |µ| ≥ Υβ -1/3 /2, with the aid (4.36),

u ′ p ≤ |U + iλ| 1/2 u ′ 2 |U + iλ| -1/2 2p 2-p ≤ Cβ -2+p 6p f 2 .
(5.18)

Otherwise, if |µ| < Υβ -1/3 /2, we rewrite (5.13) in the form

(L D β -β[Υβ -1/3 + iν])u = f -(Υβ -1 3 -µ)β u to obtain from (5.18) that u ′ p ≤ Cβ -2+p 6p ( f 2 + β 2/3 u 2 ) .
The proof of (5.10) can now be completed using (5.8).

We can now deduce the following corollary Corollary 5.3. For any r > 1 and Υ < J

2/3

m ℜν 1 , there exist C > 0 and β 0 > 0 such that for all U ∈ S 2 r , β ≥ β 0 , 2 < p ≤ ∞, and f ∈ L 2 (-1, 1) it holds that sup

ℜλ≤Υβ -1/3 (L D β -βλ) -1 f p ≤ Cβ -3p+2 6p f 2 . (5.19) Proof. Let f ∈ L 2 (-1, 1) and set v = (L D β -βλ) -1 f
. By (5.8), there exist C > 0 and β 0 such that, for

β ≥ β 0 v 2 ≤ C β 2/3 f 2 , and v ′ 2 ≤ Cβ -1/3 f 2 .
(5.20)

Consequently, using the interpolation inequality

v p ≤ v 2 p 2 v 1-2 p
∞ , and the Sobolev embedding

v ∞ ≤ √ 2 v 1 2 2 v ′ 1 2 2 , we can conclude that v p ≤ 2 p-2 2p v p+2 2p 2 v ′ p-2 2p 2 ≤ Cβ -3p+2 6p f 2 .
For later reference we also need the following refined estimate. Proposition 5.4. For any r > 1 and Υ < ℜν 1 , , there exist C > 0 and β 0 > 0 such that, for all U ∈ S r , β ≥ β 0 , and f ∈ L ∞ (-1, 1), sup

ℜλ≤ΥJ 2/3 m β -1/3 (L D β -βλ) -1 f 2 ≤ C β 5/6 f ∞ .
(5.21)

Proof. Let η be given by (5.14) and for any

x ∈ [-1, 1] , η x (t) = η(β 1/3 (t -x))1 [-1,1] (t) . Let further η x,2 (t) = η(β 1/3 (t -x)/2)1 [-1,1] (t) , implying that η x η x,2 = η x .
Step 1: We prove that, for any ν ∈ [U(-1), U(1)], there exist C > 0 and β 0 such that, for all

β ≥ β 0 , x ∈ [-1, 1], f ∈ L ∞ (-1, 1), ℜλ ≤ Υβ -1 3 , (U -ν)η x v 2 2 ≤ C β 2 η x f 2 2 + C β 1/3 |x -x ν | + 1 β 2/3 η x,2 v 2 2 , (5.22)
where

x ν ∈ [-1, +1] is chosen so that U(x ν ) = ν and v = (L D β -βλ) -1 f . (5.23)
Clearly,

ℑ η 2 x (U -ν)v, (L D β -βλ)v = β (U -ν)η x v 2 2 + ℑ [η 2 x (U -ν)] ′ v, v ′ , which implies (U -ν)η x v 2 2 ≤ C β 2 η x f 2 2 + C β sup t |η ′ x (t)(U(t) -ν)| η x v ′ 2 η x,2 v 2 . (5.24)
To estimate η x v ′ 2 we use the identity

ℜ η 2 x v, (L D β -βλ)v = (η x v) ′ 2 2 -η ′ x v 2 2 -µβ η x v 2 2 , (5.25)
from which we easily conclude, using the fact that β

1 3 µ is bounded by assumption, that η x v ′ 2 2 ≤ C(β -2/3 η x f 2 2 + β 2/3 η x,2 v 2 2 ) . (5.26)
Finally we note that

C -1 |x -x ν | ≤ |U(x) -ν| ≤ C|x -x ν | . (5.27) Suppose first that |x -x ν | < 3β -1/3 .
Since by (5.27) it holds that (sup t |η ′ x (t)(U(t)ν)|) ≤ C we may use (5.24) to obtain

(U -ν)η x v 2 2 ≤ C β 2 η x f 2 2 + C β η x v ′ 2 η x,2 v 2 .
which together with (5.26) yields (5.22). Suppose now that |xx ν | ≥ 3β -1/3 . This time we have by (5.27) that (sup

t |η ′ x (t)(U(t) -ν)|) ≤ Cβ 1 3 |x -x ν |
, and hence we get from (5.24)

(U -ν)η x v 2 2 ≤ C β 2 η x f 2 2 + C |x -x ν | β 2/3 η x,2 v 2 η x v ′ 2 .
(5.28)

To estimate the last term on the right-hand-side of (5.28), we use (5.25) once again and get instead of (5.26)

η x v ′ 2 2 ≤ C β 2/3 η x,2 v 2 2 + (U -ν)η x v 2 η x f U -ν 2 , ( 5.29)
or, alternatively, for any δ > 0 ,

η x v ′ 2 ≤ C β 1/3 η x,2 v 2 + δ β 1/2 |x -x ν | 1/2 (U -ν)η x v 2 + |x -x ν | 1/2 δβ 1/2 η x f U -ν 2 .
Substituting the above into (5.28) then yields

(U -ν)η x v 2 2 ≤ C 1 β 2 η x f 2 2 + |x -x ν | β 1/3 η x,2 v 2 2 + + |x -x ν | 3/2 δβ 7/6 η x,2 v 2 η x f U -ν 2 + δ|x -x ν | 1/2 β 1/6 η x,2 v 2 (U -ν)η x v 2 .
Observing that

η x f U -ν 2 ≤ C|x -x ν | -1 η x f 2 , yields (5.22
) by choosing a sufficiently small value of δ.

To proceed to the next step, we need to define, yet, two additional β dependent cutoff functions. Let then for s ≥ 2, η s and ηs in C ∞ (R, [0, 1]) satisfy

η s (t) = 1 |t -x ν | ≤ sβ -1/3 0 |t -x ν | ≥ (s + 1)β -1/3 and ηs = 1 -η 2 s .
We further require that there exists C and β 0 such that for any s ≥ 2 and

β ≥ β 0 η ′ s ∞ ≤ Cβ 1/3 ; η ′′ s ∞ ≤ Cβ 2/3 .
Step 2: We prove that there exist s 0 > 0, and C > 0 such that, for all s ≥ s 0 , there exists

β s such that β ≥ β s ηs v 2 2 ≤ C(β -5/3 f 2 ∞ + s -1 v 2 2 ) , (5.30)
for any pair (f, v) satisfying (5.23). By (5.22) we have

η x v 2 2 ≤ C |x -x ν | 2 β 2 η x f 2 2 + C |x -x ν |β 1/3 η x,2 v 2 2 .
(5.31)

We now integrate the above inequality with respect to x over (-1, x νsβ -1/3 /2) ∪ (x ν + sβ -1/3 /2, 1). By changing the order of integration we obtain that for all s > 4

sβ -1/3 /2<|x-xν | 1 [-1,1] (x) η x v 2 2 dx = -1,1 |v| 2 (t) dt sβ -1/3 /2<|x-xν | 1 [-1,1] (x)η 2 (β 1/3 (x -t)) dx ≥ -1,1 |η s v| 2 (t) dt sβ -1/3 /2<|x-xν | 1 [-1,1] (x)η 2 (β 1/3 (x -t)) dx ≥ β -1/3 ηs v 2 2 . Note that sβ -1/3 /2<|x-xν | 1 [-1,1] (x)η 2 (β 1/3 (x-t)) dx = β -1 3 s/2<|τ +β 1/3 (t-xν )| η 2 (τ )dτ ≥ β -1 3 ,
for any t in the support of ηs . We rewrite the above in the form

ηs v 2 2 ≤ β 1 3 sβ -1/3 /2<|x-xν | 1 [-1,1] (x) η x v 2 2 dx . (5.32) As η x f 2 2 ≤ 2β -1/3 f 2 ∞ we have sβ -1/3 /2<|x-xν | 1 [-1,1] (x) 1 |x-xν| 2 β 2 η x f 2 2 dx ≤ Cβ -7/3 f 2 ∞ sβ -1/3 /2<|x-xν | 1 [-1,1] (x) 1 |x-xν | 2 dx ≤ Ĉβ -2 f 2 ∞ . (5.33)
Finally, we have for all s ≥ 16 ,

sβ -1/3 /2<|x-xν | 1 [-1,1] (x) 1 |x-xν|β 1/3 η x,2 v 2 2 dx ≤ -1,1 |v| 2 (t) dt sβ -1/3 /2<|x-xν | 1 [-1,1] (x) η 2 (β 1/3 (x-t)/2) |x-xν|β 1/3 dx ≤ β -1/3 sβ -1/3 /4<|t-xν | 1 [-1,1] (t) log |t-xν |+2β -1/3 |t-xν |-2β -1/3 |v(t)| 2 dt ≤ Cβ -1/3 s v 2 2 .
Note that η(

β 1/3 (x -t)/2) vanishes for sβ -1/3 /2 < |x -x ν | (for r ≥ 16) and |t -x ν | < sβ -1/3 /4 . Note further that 0 < log |t -x ν | + 2β -1/3 |t -x ν | -2β -1/3 ≤ log 1 + 8/s 1 -8/s ≤ C s .
Combining the above with (5.33), (5.32), and (5.31) easily yields (5.30).

Step 3: We now prove (5.21).

Writing (L D β -βλ)(η s v) = η s f + 2η ′ s v ′ + η ′′ s v
, we obtain from (5.8) and the definitions and properties of the cut-off functions η s and η x ,

η s v 2 2 ≤ C(β -4/3 η s f 2 2 + β -2/3 η xs v ′ 2 2 + ηs/2 v 2 2 ) , (5.34)
where x s = x ν + s+1 2 β -1/3 . We now use (5.29) (with x = x s ) together with (5.22) to obtain that

η xs v ′ 2 2 ≤ C β 2/3 η xs,2 v 2 2 + (U -ν)η xs v 2 η xs f U -ν 2 ≤ Ĉ β 2/3 η xs,2 v 2 2 + s -1 β 1/3 β -1 η xs f 2 + sβ -1/3 η xs,2 v 2 η xs f 2 ,
from which we can conclude, using the inequality ηs/2 ≥ η xs,2 for s > 16, that

η xs v ′ 2 2 ≤ C(β 2/3 ηr/2 v 2 2 + β -2/3 η xs f 2 2 ) .
Substituting the above into (5.34), yields that, for some C > 0, it holds:

η s v 2 2 ≤ C s (β -5/3 f 2 ∞ + ηs/2 v 2 2 ) .
Using (5.30) (used with r replaced by s 2 and for s large enough) then yields

η s v 2 2 ≤ C(β -5/3 f 2 ∞ + s -1 v 2 2 ) .
Combining the above with again (5.30) yields (5.21), by choosing a sufficiently large value of s.

L 1 estimates

In this subsection we establish new L 1 estimates for the resolvent of L β,R , defined by (5.1) and (5.2). We first observe that the proof of Proposition 5.4 can be applied to the entire real line case, replacing the estimates of resolvent for the Dirichlet problem (L D ββλ) -1 by the corresponding estimates of the resolvent ( L β,Rβλ) -1 for ℜλ ≤ β -1 3 Υ. Hence, we may state the following Lemma 5.5. For any r > 1, any Υ > 0 and a > 0 there exists C > 0 and β 0 > 0 such that, for all U ∈ S r , g ∈ L 2 (R) ∩ L ∞ (R) and β ≥ β 0 , sup

ℜλ≤Υβ -1/3 ( L β,R -βλ) -1 g L 2 (-a,a) ≤ C β 5/6 g ∞ . (5.35)
We continue this subsection with the following L 1 estimate Lemma 5.6. For any r > 1 and Υ > 0, there exist C > 0 and β 0 > 0 such that, for all

β ≥ β 0 , U ∈ S 2 r , ℜλ ≤ ΥJ 2/3 m β -1/3 , and any g ∈ L ∞ (R) supported in [-1, 1], we have ( L β,R -βλ) -1 g L 1 (-1,1) ≤ C min(β -5/6 g 2 , β -1 log β g ∞ ) .
(5.36)

Proof. Let Υ denote a fixed positive value and

ℜλ ≤ Υβ -1/3 . Let further g ∈ L ∞ (R), supported on [-1, 1], and v ∈ D( L β,R ) satisfy ( L β,R -βλ)v = g .
(5.37) By (5.4)-(5.5) and (4.37), we have

v L 1 (-1,1) ≤ ( U -ν + iβ -1/3 ) -1 L 2 (-1,+1) ( U -ν + iβ -1/3 )v 2 ≤ Cβ -5/6 g 2 ,
which proves the first inequality of (5.36). Let x ν ∈ R satisfy U (x ν ) = ν. Let η and η ν be given by (5.14) and (5.15) and

ζ ν (x) = η ν (x) η(|x -x ν |/6) .
Taking the inner product of (5.37) with ζ 2 ν v yields for the imaginary part (see (5.17) with η ν instead of ζ ν )

β ζ ν | U -ν| 1/2 v 2 2 + 2ℑ ζ ′ ν v, (ζ ν v) ′ = ζ ν v, ζ ν g .
(5.38) By (5.35) we have that

ζ ′ ν v 2 ≤ C β 1/2 g ∞ . (5.39) To estimate (ζ ν v) ′ 2 we use the identity (ζ ν v) ′ 2 2 -ℜλ β ζ ν v 2 2 -ζ ′ ν v 2 2 = ℜ ζ ν v, ζ ν g
, to obtain from (5.4), (5.35), and (5.39)

(ζ ν v) ′ 2 2 ≤ C β g 2 ∞ + ζ ν v 1 g ∞ .
Substituting the above into (5.38) yields

ζ ν | U -ν| 1/2 v 2 ≤ C 1 β g ∞ + β -1/2 ζ ν v 1/2 1 g 1/2 ∞ .
(5.40)

As ζ ν v 1 ≤ ζ ν | U -ν| 1/2 v 2 1 (xν + 1 2 β -1/3 ,xν +6) | U -ν| -1/2 2 ≤ C(log β) 1 2 ζ ν | U -ν| 1/2 v 2 , (5.41)
we obtain from (5.40) that

ζ ν | U -ν| 1/2 v 2 ≤ C (log β) 1 2 β g ∞ .
Then, using (5.41) once again yields

ζ ν v 1 ≤ C log β β g ∞ ,
from which we easily conclude that

1 (xν ,xν +3) v 1 ≤ ζ ν v 1 + Cβ -1/6 1 (xν ,xν +β -1/3 ) v 2 ≤ Č log β β g ∞ .
In a similar manner we obtain that

1 (xν -3,xν ) v 1 ≤ Č log β β g ∞ .
This proves (5.36) in the case where (-1, +1) ⊂ (x ν -3, x ν + 3).

It remains to prove (5.36) in the case |x ν | ≥ 2 . By (5.5) and the fact that g is supported on [-1, 1], we have that

v L 1 (-1,1) ≤ √ 2 v L 2 (-1,1) ≤ C ( U -ν)v L 2 (-1,1) ≤ C β g 2 ≤ C β g ∞ .
The lemma is proved A similar statement can be proved in the Dirichlet case.

Lemma 5.7. For any r > 1 and Υ ∈ (0, ℜν 1 ), there exist C > 0 and β 0 > 0 such that, for all β ≥ β 0 , U ∈ S 2 r and ℜλ ≤ J 2/3 m Υβ -1/3 , and for any g ∈ L ∞ (-1, 1) we have

(L D β -βλ) -1 g 1 ≤ C min(β -5/6 g 2 , β -1 log β g ∞ ) .
(5.42)

The proof is similar to the proof of Lemma 5.6 and is therefore skipped.

6 No-slip resolvent estimates 6.1 A no-slip Schrödinger operator

We begin by providing a short explanation of the difficulties arising when the no-slip boundary condition (3.10) is prescribed. Complete details will be given in Section 8.

In the zero-traction case, estimating φ ∈ D(B S λ,α,β ) satisfying B S λ,α,β φ = f for some f ∈ L 2 (-1, 1), we may write, by (3.5),

- d 2 dx 2 + iβU -βλ) (-φ ′′ + α 2 φ) = iβU ′′ φ + f .
Since by (3.9), it holds that -φ ′′ + α 2 φ satisfies a Dirichlet condition at ±1, one can now use, for instance, (5.8) and (5.21) to obtain

-φ ′′ + α 2 φ 2 ≤ Cβ 1/6 φ ∞ + β -2/3 f 2 .
Such an estimate is particularly useful in the case α ≫ β 1/6 , but also in other cases (detailed in Section 8). Similar estimates can be obtained for v = A λ,α φ and ṽ = (U + iλ) -1 v.

If we now consider the same problem in the no-slip case the above approach is inapplicable. Thus, for φ ∈ D(B D λ,α,β ) satisfying B D λ,α,β φ = f , we can no longer use neither (5.8) nor (5.21), as -φ ′′ + α 2 φ does not satisfy a Dirichlet condition at x = ±1. However, integration by parts easily yields that for all φ ∈ D(B D λ,α,β ) we have e ±αx , -φ ′′ + α 2 φ = 0 .

If we consider v or ṽ instead of -φ ′′ + α 2 φ we can still obtain similar orthogonality conditions (see (8.22) and (8.32c)). These conditions read

v , ζ ± L 2 (-1,+1) = 0 , (6.1)
where ζ -and ζ + are linearly independent, β dependent, and belong to H 1 (-1, +1). With (3.5) in mind, we let L ζ β be the differential operator -d 2 /dx 2 + iβU with domain

D(L ζ β ) = {u ∈ H 2 (-1, 1) | ζ ± , u = 0 } . (6.2)
For convenience we require that ζ ± satisfy

ζ + (1) ζ -(1) ζ + (-1) ζ -(-1) = 1 0 0 1 . (6.3)
Note that ζ ± = e ±αx do not satisfy the above requirement, and we shall therefore need to replace them by a pair of proper linear combinations of them [START_REF] Shkalikov | Spectral portraits of the Orr-Sommerfeld operator at large Reynolds numbers[END_REF] (a more detailed explanation is brought in Section 8). We seek resolvent estimates for L ζ β in the following. In the absence of a Dirichlet boundary condition, it seems reasonable to approximate the solution of

(L ζ β -βλ)v = g , (6.4)
by a sum of a solution in R of the inhomogeneous equation and a linear combination of two independent approximate solutions of the homogeneous equation whose coefficients will be determined by the above integral conditions. Using affine approximations of U in (-1, +1) or extensions outside of (-1, +1), the approximate solutions can be described by a pair of dilated and translated Airy functions in (-1, +∞) and (-∞, +1).

The solution in R. We now explain our construction of an approximate inverse in R by defining first a natural C 1 -extension Ũ of U outside of [-1, +1], satisfying (5.3), by

Ũ (x) =      U(x) x ∈ [-1, 1] U(1) + U ′ (1)(x -1) x > 1 U(-1) + U ′ (-1)(x + 1) x < -1 .
We note that Ũ satisfies the conditions of Proposition 5.1. We also extend g by g(x) = g(x) x ∈ [-1, 1] 0 otherwise ,

and set u = Γ (-1,1) ( L β,R -βλ) -1 g . (6.5)
where L β,R is defined by (5.1) and (5.2) and Γ (-1,1) denotes the restriction to (-1, 1).

Boundary terms. To obtain the boundaries effect, we replace U(x) by its affine approximation at ±1 and consider the L 2 ∩ L 1 solutions ψ ± of the approximate problems

(-d 2 /dx 2 + iβ[U(-1) + J -(x + 1)] -βλ)ψ -= 0 in (-1, +∞) +∞ -1 ψ -(x) dx = (J -β) -1/3 , (6.6) and (-d 2 /dx 2 + iβ[U(1) + J + (x -1))] -βλ)ψ + = 0 in (-∞, 1) 1 -∞ ψ + (x) dx = (J + β) -1/3 , (6.7)
with

J ± = U ′ (±1).
Except, perhaps for some particular values of λ, the above solutions are unique, and ψ ± rapidly decays as x → ∓∞, but their existence (due to the additional integral condition) could depend on (β, λ, J ± ) as is clarified below. We express ψ ± using Airy functions. Having in mind the definition of the generalized Airy functions [35, eq. (39)] or [30, Lemma 2] (for more details see [START_REF] Drazin | Hydrodynamic stability[END_REF]Appendix] or our short review in Appendix A). These solutions are given, assuming that the denominator does not vanish, by

ψ -(x) = e iπ/6 Ai (J -β) 1/3 e iπ/6 (1 + x) + iJ -1 -(λ -iU(-1)) A 0 iβ 1/3 J -2/3 - [λ -iU(-1)] , (6.8a)
and

ψ + (x) = -e iπ/6 Ai (J + β) 1/3 e iπ/6 (1 -x) + iJ -1 + ( λ + iU(1)) A 0 iβ 1/3 J -2/3 + [ λ + iU(1)] . (6.8b)
where A 0 is the holomorphic extension to C of

x → A 0 (x) = e iπ/6 +∞ x
Ai (e iπ/6 t) dt . (6.9)

Much of the properties of A 0 are recalled (mainly from Wasow's paper [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF]) in Appendix A. It has been established in [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF] (see also Appendix A) that The zeroes of z → A 0 (iz) are located in the sector arg z ∈ ( π 6 , π 2 ). Let

S λ = { z | A 0 (iz) = 0 } ,
and further define ϑ r 1 := inf z∈S λ

ℜz . (6.10)

In addition, we prove in Appendix A.2 that S λ = ∅ and(relying on [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF]) that ϑ r 1 > 0 . It follows that the denominators in (6.8b) and (6.8a) do not vanish, if

ℜλ < ϑ r 1 β -1/3 J 2/3 m ,
where J m is given by (2.25c).

The functions ψ ± are not exact solutions of (-d 2 /dx 2 + iβUβλ)ψ = 0 and hence we must introduce a correction term. We thus consider

g ± = (-d 2 dx 2 + iβ(U + iλ))ψ ± for x ∈ (-1, 1) 0 otherwise , (6.11)
and then introduce ṽ± = Γ (-1,1) ( L β,Rβλ) -1 g ± . (6.12)

This correction term can be estimated as follows Lemma 6.1. For any r > 1 and Υ < ϑ r 1 , there exist C and β 0 such that, for all U ∈ S 2 r , λ ∈ C satisfying

β 1/3 ℜλ ≤ ΥJ 2/3 m , (6.13)
and β ≥ β 0 , we have

(U + iλ)ṽ ± 2 + β -1/3 ṽ± 2 ≤ Cβ -5/6 . ( 6

.14)

Proof.

A simple computation shows that:

g ± = iβ [U -U(±1) -J ± (x ∓ 1))] ψ ± in (-1, 1) . Let λ ± = µ -i(U(±1) -ν) . (6.15)
We note that ψ ± (x) = e i π 6 Ψ (J ± β)

1 3 J -1 ± λ ± ((J ± β) 1 3 (1 ∓ x)) (6.16)
where Ψ λ is defined in the appendix (see (A.42)). Using translation, dilation, and (A.43a), we can conclude that, under the assumptions of Lemma 6.1, it holds, for k ∈ [0, 4], that

(1 ∓ x) k ψ ± 2 ≤ C [1 + |λ ± |β 1/3 ] 1-2k 4 β -(1+2k)/6 . (6.17) Hence, as |g ± (x)| ≤ Cβ(1 ∓ x) 2 |ψ ± (x)|, we have g ± 2 ≤ C β 1/6 [1 + |λ ± |β 1/3 ] -3/4 . (6.18)
By (5.4) and (5.5) we have

(U + iλ)ṽ ± 2 + β -1/3 ṽ± 2 ≤ Cβ -5/6 [1 + |λ ± |β 1/3 ] -3/4 , (6.19)
establishing thereby (6.14).

We are now ready for introducing a solution of (6.4) in the form

v = A + (ψ + -ṽ+ ) + A -(ψ --ṽ-) + u . (6.20)
We observe that (L ββλ)v = g for any pair (A -, A + ). Therefore, one can attempt to find two linear forms g → A -(g) and g → A + (g) such that v belongs to the domain of L ζ β , hence is the solution of (6.4). This is the object of the next lemma. Lemma 6.2. Let θ > 0 and C ζ > 0, and suppose that ζ -, ζ + ∈ H 1 (-1, 1) satisfy the conditions (6.3),

ζ ± ∞ ≤ C ζ , (6.21
)

and ζ ′ ± 2 ≤ θ β 1/6 . (6.22)
Let further r > 1 and Υ < ϑ r 1 . Then, there exist β 0 > 0 and θ 0 > 0 such that for all β ≥ β 0 , 0 < θ ≤ θ 0 , U ∈ S 2 r and λ ∈ C satisfying

β 1/3 ℜλ ≤ J 2/3 m Υ , (6.23)
(6.4) and (6.20) hold true with v ∈ D(L β ζ ) and A ± = A ± (g) denoting a pair of linear forms A ± (g) : L ∞ (-1, 1) → C. Furthermore, there exists C > 0 such that, for all β ≥ β 0 , U ∈ S 2 r and g ∈ L ∞ (-1, +1), we have

|A ± (g)| ≤ C min(β -1/2 g 2 , β -2/3 ln β g ∞ ) . (6.24)
Proof. In view of the discussion preceding the statement of the lemma, it remains to show the existence of A ± (g) satisfying (6.24).

Taking the inner product of (6.20) in L 2 (-1, +1), first by ζ + and then by ζ - while having (6.1) in mind yields the following system

ζ + , (ψ + -ṽ+ ) ζ + , (ψ --ṽ-) ζ -, (ψ + -ṽ+ )
ζ -, (ψ -ṽ-)

A + (g) A -(g) = ζ + , u ζ -, u . (6.25)
We now write

ζ ± , ψ ± = 1, ψ ± + ζ ± -1, ψ ± . (6.26)
For the first term on the right-hand-side we have

1, ψ -= (J -β) -1/3 - ∞ 1 ψ -(x) dx .
The integral on the other side can be estimated as follows: we first write

∞ 1 ψ -(x) dx ≤ (1 + x) 3 ψ -1 .
Then, using (A.43b), (6.16) and dilation, we obtain for all s ∈ [0, 3],

(1 ∓ x) s ψ ± 1 ≤ C [1 + |λ ± |β 1/3 ] -s/2 β -(s+1)/3 . (6.27)
The above estimate for s = 3 yields,

∞ 1 ψ -(x) dx ≤ Cβ -4/3 .
A similar estimate can be obtained for 1, ψ + . Consequently we have

1, ψ ± = (J ± β) -1/3 [1 + O(β -1 )] . (6.28)
For the second term on the right-hand-side of (6.26) we use the fact that, for all x ∈ [-1, 1], we have by (6.3)

|ζ ± (x) -1| ≤ [1 ∓ x] 1/2 ζ ′ ± 2 .
We obtain, using (6.27) with s = 1 2 and (6.22),

| ζ ± -1, ψ ± | ≤ ζ ′ ± 2 [1 ∓ x] 1/2 ψ ± 1 ≤ C(Υ)θ 0 β -1/3 . (6.29)
Furthermore, by (4.36) and (6.19), we have

| ζ ± , ṽ± | ≤ 2 (U -ν + iβ -1/3 ) -1 2 (U -ν + iβ -1/3 )ṽ ± 2 ≤ C β -2/3 [1 + |λ ± |β 1/3 ] -3/4 .
(6.30)

By the above, (6.28), and (6.29), there exists C > 0 and β 0 , such that, for any β ≥ β 0 and any λ satisfying ℜλ ≤ Υβ -1/3 , we have

| ζ ± , (ψ ± -ṽ± , ) -(J ± β) -1/3 | ≤ C(θ 0 β -1/3 + β -2/3 ) . (6.31)
As ζ ± (∓1) = 0, we obtain as in (6.29)

| ζ ± , ψ ∓ | ≤ C θ 0 β -1/3 . (6.32)
Furthermore, as in (6.30) we obtain that

| ζ ± , ṽ∓ | ≤ C β -2/3 [1 + |λ ± |β 1/3 ] -3/4 . (6.33)
Substituting the above, together with (6.31) and (6.32) into (6.25) then yields, for θ 0 small enough, and β large enough, the invertibility of (6.25) together with the estimate

|A ± (g)| ≤ | ζ ± , u |(J ± β) 1/3 [1 + Cθ 0 ] + Cθ 0 β -1/3 | ζ ∓ , u | . (6.34)
By (5.36) we obtain that

| ζ ± , u | ≤ C min(β -1/2 g 2 , β -2/3 log β g ∞ ) .
Substituting the above into (6.34) yields (6.24).

Remark 6.3. We may replace in Lemma 6.2, L β,R by L D β in (6.5) and (6.12), but under the condition Υ < J 2/3 m ℜν 1 . In this case, we have by (6.20) (with L D β instead of L β,R ) and having in mind the Dirichlet condition at ±1 ,

v(±1) = A ± ψ ± (±1) + A ∓ ψ ∓ (±1) . By (A.35) we have |ψ ± (±1)| ≤ C[1 + |λ ± | 1/2 β 1/6 ] ,
and by (A.43c)

|ψ ∓ (±1)| ≤ Cβ -4/3 [1 + |λ ± |β 1/3 ] -2 .
Combining the above with (6.24) yields that

|v(±1)| ≤ C [1 + |λ ± | 1/2 β 1/6 ] β -1/2 g 2 .

A no-slip Schrödinger in R +

In the previous subsection we have considered a space of functions satisfying the orthogonality condition (6.1). We have assumed that the functions spanning the orthogonal space ζ + and ζ -satisfy the bound

ζ ′ ± 2 ≤ θβ 1/6 ,
where θ ∈ (0, θ 0 ] for some sufficiently small θ 0 > 0 .

Of particular interest is the example ζ ± = e -α(1∓x) (or a proper linear combination of them satisfying (6.3)). In this case, we have

(e -α(1∓x) ) ′ 2 ∼ = α/2 ,
for sufficiently large α. Consequently, as long as α ≪ β 1/3 , Lemma 6.2 is applicable in this case. We, however, need to consider also the case where α ∼ β 1/3 , or even α ≫ β 1/3 . These cases can, nevertheless, be treated using localization techniques as in [START_REF] Henry | Spectral instability for even non-selfadjoint anharmonic oscillators[END_REF][START_REF] Almog | On a Schrödinger operator with a purely imaginary potential in the semiclassical limit[END_REF]. To this end we have to consider a localized version of L ζ near x = ±1. This subsection is devoted therefore to the study of the ensuing linearized operator. We begin by establishing a proper spectral formulation for the problem.

Proposition 6.4. Let, for some θ > 0,

L θ = - d 2 dx 2 + ix , (6.35a)
be defined on

D(L θ ) = { u ∈ H 2 (R + ) | e θ , u = 0 , xu ∈ L 2 (R + ) } . (6.35b)
where e θ (x) := e -θx .

Then, L θ is a closed operator with non empty resolvent set and compact resolvent. Moreover L θ has index 0.

Before proceeding to the proof of the proposition we establish the following H 1 estimate of any v ∈ D(L θ ) in term of the L 2 -norm of L θ v and v. Lemma 6.5. There exists some constant C(θ) such that, for any λ ∈ C and any v ∈ D(L θ ), we have

v ′ 2 + |v(0)| ≤ C(θ)[(1 + |ℜλ| 1/2 sign ℜλ) v 2 + (L θ -λ)v 2 ] . (6.36) Proof. Let v ∈ D(L θ ) and g ∈ L 2 (R + ) satisfy (L θ -λ)v = g . (6.37)
Taking the inner product with v yields

v ′ 2 2 -λ v 2 2 + i xv, v + v ′ (0)v(0) = v, g . (6.38)
To obtain an estimate for the fourth term on the left-hand-side of (6.38) we need an effective bound on v ′ (0). Integration by parts yields, with the aid of the fact that e θ , v = 0 ,

-e θ , v ′′ = v ′ (0) + θ v(0) .
Taking the inner product of (6.37) with e θ , we then obtain

v ′ (0) + θv(0) + i xe θ , v = e θ , g , (6.39)
from which we conclude that

|v(0)v ′ (0)| ≤ θ |v(0)| 2 + 1 2θ 3/2 |v(0)| ( v 2 + θ √ 2 g 2 ) . (6.40) As |v(0)| 2 ≤ v ′ 2 v 2 , (6.41)
we obtain, using (6.40) that

|v(0)v ′ (0)| ≤ θ 2 v ′ 2 v 2 + 1 2θ 3/2 |v(0)| ( v 2 + θ √ 2 g 2 ) . (6.42)
Combining the real part of (6.38) with (6.42) yields

v ′ 2 2 -µ v 2 2 ≤ θ 2 v ′ 2 v 2 + v 2 g 2 + 1 2θ 3/2 |v(0)|( v 2 + θ √ 2 g 2 ) . (6.43)
Consequently, we obtain (6.36).

Proof of Proposition 6.4.

Step1: We prove that L θ is a closed operator.

Let

{(v n , L θ v n )} ∞ n=1 ∈ [D(L θ )] N ×[L 2 (R + )] N converge, as n → ∞, in L 2 (0, +∞)× L 2 (0, +∞) to (v ∞ , g ∞ ).
We need to establish that v ∞ ∈ D(L θ ). The orthogonality of v ∞ with e θ immediately follows from the L 2 convergence. From (6.36) (with λ = 0) we conclude that v n is a Cauchy sequence in H 1 (R + ) and hence must converge to v ∞ in the H 1 (R + ) norm.

Let χ ∈ C ∞ (R) be supported on [ 1 2 , +∞) and satisfy χ = 1 for x > 1. Clearly,

- d 2 dx 2 + ix (χv n ) = χg n + 2χ ′ v ′ n + χ ′′ v n .
Since we can smoothly extend χv n to H 2 (R), it follows from (5.4) and (5.5) (with β = 1 and Ũ = x) that χv n and χxv n are Cauchy sequences in H 2 (R) and in L 2 (R) respectively. Hence, its limit χv ∞ satisfies χv ∞ ∈ H 2 (R) and χxv ∞ ∈ L 2 (R). By the

H 1 (R + ) convergence of {v n } ∞ n=1 it follows that v ∞ is a weak solution of v ′′ ∞ = -g ∞ + ixv ∞ .
Since the right-hand-side is in L 2 (R + ), it follows that v ∞ ∈ H 2 (R + ) and hence L θ is closed.

Step 2: We prove that L θλ has index 0.

Let L : H 2 (R + )∩L 2 (R + ; x 2 dx) → L 2 (R + ) be associated with the same differential operator as L θ . Clearly, L is a Fredholm operator of index 1. Indeed, it is clearly surjective (we can find a unique solution satisfying a Dirichlet condition at x = 0) and it is easy to see that the kernel has dimension 1 (span{Ai(e iπ/6 •)}. Consequently, for any λ ∈ C, Lλ has index 1. We now observe that L θ is obtained by imposing a single orthogonality condition in the domain. Hence the index of L θλ equals 0.

Step 3: We show that ρ(L θ ) = ∅ . We prove that there exists μ < 0 such that for all ℜλ < μ the operator L θλ is injective. Combined with the above zero index property, it would yield that the resolvent set contains the half plane ℜλ < μ. The injectivity follows from (6.41) and (6.43), by which there exist µ 0 and C > 0 such that for all ℜλ < µ 0 , we have

v 2 + v ′ 2 + |v(0)| ≤ C g 2 . (6.44)
Finally, the compactness of the resolvent follows from the fact that D(L θ ) is compactly embedded in L 2 (R + ).

The previous proof has also shown to us that L θ is semi-bounded. The next proposition provides a a more explicit lower bound for the spectrum as a function of θ. Proposition 6.6. For all θ ∈ R + , we have

µ 0 (θ) = inf ℜσ(L θ ) ≥ - 1 2 min(θ 2 , θ -2 ) . (6.45)
Proof. Suppose that for some positive θ0 there exist λ 0 ∈ σ(L θ0 ) and

v 0 ∈ D(L θ0 ) such that (L θ0 -λ 0 )v 0 = 0 . (6.46)
Since v 0 is an L 2 solution of the complex Airy equation in R + it is expressible, up to a multiplicative constant, in the form v 0 (x) = Ai (e iπ/6 (x + iλ 0 )) . (6.47a)

The orthogonality condition for v 0 reads

F (λ 0 , θ0 ) = 0 , (6.47b)
where

F (λ, θ) = R + e θ (x)
Ai (e iπ/6 (x + iλ)) dx . (6.47c)

Step 1: We prove that the set {θ ∈ (0, +∞)

, ∃λ ∈ σ(L θ ) with ℜλ < ℜν 1 } is open.
We use the implicit function theorem. If indeed F (λ 0 , θ0 ) = 0, we get after integration by parts that ∂F ∂λ (λ 0 , θ0 ) = -iAi (e i2π/3 λ 0 ) = 0 .

Hence there exists a neighborhood of θ0 and a C 1 -solution λ(θ) in this neighborhood such that λ( θ0 ) = λ 0 .

Step 2: Let ε < ℜν 1 . Consider the set Σ(ε) := {(λ, θ) ∈ C × (0, +∞) , ℜλ < ε and F (λ, θ) = 0} , which can be described as a countable (or finite) union of simple analytic curves denoted by {λ k (θ)} k∈K , each with an interval of definition (θ k , θ * k ). Let further µ k (θ) = ℜλ k (θ). We prove that for all k ∈ K and θ ∈ (θ k , θ * k )

µ k (θ) + θ 2 /2 ≥ (µ k (θ k ) + θ 2 k /2) exp - θ θ k v k (•, τ ) 2 2 |v k (0, τ )| 2 dτ , (6.48)
where v k (x, θ) := Ai (e iπ/6 (x + iλ k (θ))) .

Let k ∈ K. For convenience of notation we set

λ(θ) = λ k (θ) , µ(θ) = ℜλ(θ) and v(θ) = v k (θ) .
By (6.47b,c) we have that

F (λ(θ), θ) = 0 , ∀θ ∈ Σ(0) . (6.49)
Differentiating this identity with respect to θ yields

e 2πi/3 dλ dθ (θ) I 1 (θ) -I 2 (θ) = 0 , (6.50a)
where

I 1 (θ) = R + e θ (x) Ai ′ (e iπ/6 (x + iλ(θ))) dx , (6.50b)
and

I 2 (θ) = R + xe θ (x)
Ai (e iπ/6 (x + iλ(θ))) dx . (6.50c)

Integration by parts yields, in conjunction with (6.49),

I 1 (θ) = -e -iπ/6
Ai (e 2πi/3 λ(θ)) . (6.51)

We now write, with the aid of (6.49) and Airy's equation

I 2 (θ) = R + (x + iλ(θ))e θ (x) Ai (e iπ/6 (x + iλ(θ))) dx = e -iπ/6 R + e θ (x) Ai ′′ (e iπ/6 (x + iλ(θ))) dx .
Integration by parts and (6.51) then yield

I 2 (θ) = -e -iπ/3 Ai ′ (e 2πi/3 λ(θ)) + iθAi (e 2πi/3 λ(θ)) .
Substituting the above, together with (6.51) into (6.50) yields

dλ dθ (θ) = - ∂ x v(0, θ) v(0, θ) -θ . (6.52)
Taking the inner product of (6.46) with v(•, θ) we obtain for the real part

∂ x v(•, θ) 2 2 -µ(θ) v(•, θ) 2 2 + ℜ{v(0, θ)∂ x v(0, θ)} = 0 ,
where µ(θ) = ℜλ(θ).

Combining the above with (6.52) then yields

dµ dθ (θ) + θ = ∂ x v(•, θ 2 2 |v(0, θ)| 2 -µ(θ) v(•, θ 2 2 |v(0, θ)| 2 .
We then have on the branch

d(µ + θ 2 /2) dθ + v(•, θ) 2 2 |v(0, θ)| 2 (µ + θ 2 /2) > 0 .
Solving in [ θ0 , θ * 0 ] yields (6.48).

Step 3: We prove that along every curve in Σ( 0)

ℜλ(θ) = µ(θ) ≥ -θ -2 /2 . (6.53)
From (6.43) with g = 0, and (6.41) we obtain that

1 2 v ′ 2 2 -µ(θ) v 2 2 ≤ 1 2θ 3/2 v ′ 1/2 2 v 3/2 2 .
The above, in conjunction with Young's inequality yields µ(θ) ≥ -θ -2 /2 which is precisely (6.53). In particular it implies that lim inf

θ→+∞ inf F (λ,θ)=0 ℜλ ≥ 0 . (6.54)
Step 4: We prove (6.45).

If inf θ inf F (λ,θ)=0 ℜλ ≥ 0 then (6.45) readily follows. Hence, we can assume that there exists (λ 0 , θ0 ) such that ℜλ 0 < 0 and F (λ 0 , θ0 ) = 0. We then look at {λ k (θ)} k∈K inside Σ(-ε) where ℜλ 0 < -ε < 0. By (6.54), all branches exit Σ(-ε) for sufficiently large θ. We now observe that F (λ(0), 0) = 0 ⇒ µ(0) > 0, .

Indeed, as F (λ, 0) = e -iπ 6 A 0 (iλ) we can apply Corollary A.4. Hence, these branches must lie outside Σ(0) for θ ∈ [0, θ inf ) for sufficiently small θ inf > 0. Assume that ε is chosen small enough so that

-ε + 1 2 θ 2 inf > 0 . Consider any branch λ(θ) in Σ(-ε) with θ in some interval [θ(ε), θ * (ε)] ⊂ (0, +∞) such that ℜλ(θ(ε)) = ℜλ(θ * (ε)) = -ε
We can then use (6.48) to obtain

µ(θ) + θ 2 /2 ≥ (µ(θ(ε)) + θ(ε) 2 /2) exp - θ θ(ε) v 2 2 |v(0)| 2 dθ = (-ε + θ(ε) 2 /2) exp - θ θ(ε) v 2 2 |v(0)| 2 dθ ≥ (-ε + θ 2 inf /2) exp - θ θ(ε) v 2 2 |v(0)| 2 dθ > 0 . (6.55)
This completes the proof of (6.45). Corollary 6.7. It holds that

μm := inf θ∈R + inf λ∈σ(L θ ) ℜλ + θ 2 2 > 0 . (6.56)
The proof of (6.56) with the non strict inequality follows immediately from (6.45). Note now that by (6.55), taking into account that µ(θ) + θ 2 2 ----→ θ→+∞ +∞ and that µ(0) > 0 we can conclude the strict inequality in (6.56).

The adjoint operator

We note that D(L θ ) is not dense in L 2 (R + ). We thus introduce H 0 = D(A), and then define

D(A * ) = {v ∈ H s.t. D(A) ∋ u → v,
Au extends as a continuous linear map on H 0 } , (6.57) and set A * v to be the unique (by Riesz theorem) y ∈ H 0 for which v, Au = y, u .

The standard definition is recovered when H

0 = H. Note that (A -λI) * = A * -λ Π H 0 ,
where Π H 0 is the projector on H 0 . We further note that A * A is an unbounded operator on H 0 .

In the particular case A = L θ , H 0 is the orthogonal complement in L 2 (R + ) of e θ . Hence, Π H 0 = I -P θ , where, for any u ∈ L 2 (R + ),

P θ u(x) = 1 2θ e θ , u e θ .
We next provide a more explicit representation of (L θ ) * . Lemma 6.8. We have

D((L θ ) * ) = H 2 0 (R + ) ∩ L 2 (R + ; x 2 dx) , (6.58a)
and for any v ∈ D((L θ ) * ),

(L θ ) * v = (I -P θ ) - d 2 dx 2 -ix v . (6.58b)
Proof. The proof is reminiscent of the analysis of selfadjointness for (1D)-problems in [START_REF] Reed | Methods of Modern Mathematical Physics[END_REF]. Let φ ∈ C ∞ 0 (R + ) and then set u

= (I -P θ )φ ∈ D(L θ ). Let v ∈ D(L * θ )
, where D(L * θ ) is defined by (6.57). From the definition we deduce that the distribution

C ∞ 0 (R + ) ∋ φ → v, L θ (I -P θ )φ L 2 (R + )
should extend as a continuous linear map on L 2 (R + ). We then observe that

v, L θ (I -P θ )φ = - d 2 dx 2 -ix v, φ -v, - d 2 dx 2 + ix e θ e θ , φ .
The second term on the right hand side defines a linear form on L 2 (R + ). Hence, from (6.57) we get that φ → (-

d 2 /dx 2 -ix)v, φ is a distribution in L 2 (R + ). Hence, it holds that (-d 2 /dx 2 -ix)v ∈ L 2 (R + ). We can thus conclude that v ∈ H 2 (R + ), xv ∈ L 2 (R + ).
We now compute v, L θ u using integration by parts to obtain v, -

d 2 dx 2 + ix u = -u ′ (0)v(0) + u(0)v ′ (0) + - d 2 dx 2 -ix v, u .
To conform with (6.57) u → -u ′ (0)v(0) + u(0)v ′ (0) must be a continuous map on (I -P θ )L 2 . This, however, is possible only if v(0) = v ′ (0) = 0 (consider the sequence u n = χ n -P θ χ n with χ n (x) = √ nχ(nx)), leading thereby to (6.58a). Consequently

for any v ∈ D(L * θ ), we have v, - d 2 dx 2 + ix u = - d 2 dx 2 -ix v, u .
Having in mind that v,

L θ u = L * θ v, u , (6.59)
leads to

L * θ v = (I -P θ ) - d 2 dx 2 -ix v .
We can then extend (6.59) by density to any u ∈ H 0 . Proposition 6.9. The eigenfunctions of L θ are complete in (I -P θ )L 2 (R + ) .

Proof. We take a similar approach to the one in [START_REF] Almog | On the spectrum of non-selfadjoint Schrödinger operators with compact resolvent[END_REF].

Step 1: By the semi-boundedness of L θ and (6.58a) there exists c 0 > 0 and μ0 ∈ R -such that for all u ∈ D(L θ )

c 0 -u ′′ + xu 2 2 ≤ 2c 0 xu 2 2 + u 2 2,2 ≤ -u ′′ + (ix -μ0 )u 2 2 .
(6.60)

Step 2: We now show that the resolvent of L θ is in S p for every p > 3/2 , where S p denotes the Schatten of order p.

By the Max-Min principle the singular values (µ

n ) n∈N * of the operator (L θ -μ0 ) -1 satisfy for k ∈ N µ -2 k+1 = max U k ∈H k 0 min u∈U ⊥ k ∩D(L θ ) -u ′′ + (ix -μ0 )u 2 2 u 2 2 .
Let further

κ -2 k+1 = max U k ∈(L 2 (R + )) k min u∈U ⊥ k ∩D(L θ ) -u ′′ + xu 2 2 u 2 2
.

By (6.60) we have, for n ∈ N * ,

κ -2 n ≤ c 0 µ -2 n .
Finally, let κ-2 k+1 = max

U k ∈L 2 (R + ) k min u∈U ⊥ k ∩H 2 (R + )∩L 2 (R + ;x 2 dx) -u ′′ + xu 2 2 u 2 2 .
In view of the additional constraint embedded in D(L θ ) we have

κ n ≤ κn .
By the Max-Min principle the κ-2 n are eigenvalues of

A N := (-d 2 /dx 2 + x) 2
defined on

D(A N ) = {u ∈ H 4 (R + ) ∩ L 2 (R + ; x 4 dx) | u ′′ (0) = (-u ′′ + xu) ′ (0) = 0 } . Let λ = α 2
, where α > 0, denote an eigenvalue of A N . (Note that λ = 0 is an eigenvalue.) Let u α denote the corresponding eigenfunction. As

- d 2 dx 2 + x 2 -α 2 = - d 2 dx 2 + x + α - d 2 dx 2 + x -α ,
we easily obtain that, up to a product by an arbitrary constant,

u α = - 1 2α Ai (x + α) + A 1 Ai (x -α) ,
where A 1 has to be determined from the requirement u α ∈ D(A N ). It can now be easily verified that α ∈ σ(A N ) if and only if

δ(α) := - Ai ′ (α) Ai (α) + Ai ′ (-α) Ai (-α) = 0 . (6.61) Let {ω n } ∞ n=1 ⊂ R -denote the zeroes of Airy's function Ai (x)
. By computation of its derivative δ(α) is a monotone increasing for α ∈ (-ω n , -ω n+1 ) and tends to ±∞ at the edges. Consequently, there is precisely one solution of (6.61) in (-ω n , -ω n+1 ).

As -ω n ∼ n 2/3 we may conclude from the foregoing discussion that κ-1 n ∼ n 2/3 as well. Consequently, there exists C > 0 such that, for sufficiently large n ,

µ n ≤ C n 2/3 .
As a result, for all p > 3/2 it holds that

∞ n=1 µ p n < ∞ .
Step 3: We complete the proof of the proposition.

We take a similar approach to the one in [START_REF] Almog | On the spectrum of non-selfadjoint Schrödinger operators with compact resolvent[END_REF]. By (6.41) and ( 6.43) we have, for sufficiently large -ℜλ,

(L θ -λ) -1 ≤ C(θ) |λ| . Let v = (L θ -λ) -1 g for some g ∈ L 2 (R + ).
From the imaginary part of (6.38) we learn that

-ℑλ v 2 2 + x 1/2 v 2 2 = ℑ v, g -ℑ{v ′ (0)v(0)} .
With the aid of (6.36) and ( 6.42) we then obtain that

-ℑλ v 2 2 ≤ C(θ)[(1 + |µ| 1/2 ) v 2 2 + g 2 2 ]
. Hence, there exists Ĉ such that if -ℑλ ≥ Ĉ (1 + |µ| 1/2 ), then

(L θ -λ) -1 2 ≤ C(θ) |λ| .
From the foregoing discussion we may conclude that every direction where π/2 < arg λ < 2π is a direction of minimal growth for (L θλ) -1 . Following the arguments of the proof of [2, Theorem 16.4] (cf. also [START_REF] Gohberg | Operator Theory: Advances and Applications[END_REF]Theorem X.3.1 ] or [13, Corollary XI.9.31])) we can conclude that the eigenspace of L θ is given by Proof. We begin the proof by applying Rouché's Theorem, in the large θ limit, to the holomorphic functions θF (λ, θ) and Ai (e 2i π 3 λ) inside a disk of radius r > 0 centered at ν 1 and containing no other eigenvalue of L + . As λ → Ai (e 2i π 3 λ) has a unique zero in this disk, Rouché's Theorem would show the same for the zeros of F (•, θ). It is therefore necessary to compare the two functions for λ ∈ ∂B(ν 1 , r). We thus write

D(L θ ) = (I -P θ )L 2 (R + ).
θF (λ, θ) -Ai (e i2π/3 λ) = θ R + e -θx [Ai (e i2π/3 λ + e iπ/6 x) -Ai (e i2π/3 λ)] dx . (6.63)
We bound the right-hand-side in the following manner θ R + e -θx |Ai (e i2π/3 λ + e iπ/6 x) -Ai (e i2π/3 λ)| dx

≤ θ R + e -θx x 1/2 Ai ′ (e iπ/6 (. + iλ) L 2 (0,x) dx ≤ C θ 1/2
Ai ′ (e iπ/6 (. + iλ) 2 . (6.64)

From this, we obtain the existence of r 0 > 0 and C > 0 such that, for any r ∈ (0, r 0 ] and any λ ∈ ∂B(ν 1 , r), we have

|θF (λ, θ) -Ai (e 2iπ/3 λ)| ≤ C rθ 1 2
|Ai (e 2iπ/3 λ)| .

It follows from Rouché's Theorem that for sufficiently large θ, F (λ, θ) has a unique zero in B(ν 1 , r).

At this stage we have obtained lim sup

θ→+∞ µ 0 (θ) ≤ ℜν 1 .
Using the arguments as above and supposing now that r < |λν 1 | < R and ℜλ ≤ ℜν 1 , we can establish that

θF (λ, θ) Ai (e i2π/3 λ) -1 ≤ C(R, r) θ 1/2 . (6.65)
Consequently, we obtain that, there exists θ 1 (R, r) such that, for all θ > θ 1 (R, r) F (λ, θ) does not vanish in (B(ν

1 , R) \ B(ν 1 , r)) ∩ {ℜλ ≤ ℜν 1 }.
To complete the proof we need yet to establish that there exists R 0 > 0, and θ 2 (R 0 ) > 0 such that for all θ > θ 2 (R 0 ) we have that inf

ℜλ≤ℜν 1 |λ-ν 1 |>R 0 [θ + (-λ) 1/2 ]F (λ, θ)
Ai (e i2π/3 λ) > 0 . (6.66)

To this end we set as in (A.6)

F (λ, θ) Ai (e i2π/3 λ) = R + e -θx e -(-λ) 1/2 x dx + R + e -θx w(x) dx .
To bound the second term we use (A.14) and (A.17) (with µ 0 = ℜν 1 ) together with Sobolev embeddings to obtain

w 2 ∞ ≤ w 2 w ′ 2 ≤ C|λ| -3/2 . Hence, [θ + (-λ) 1/2 ]F (λ, θ)
Ai (e i2π/3 λ) -1 ≤ C|λ| -1/4 , (6.67) from which (6.66) easily follows.

6.3

No-slip operator on (-1, 1) for large α

Consider L ζ β , defined in (6.2), with ζ ± = z ± , where z ± ∈ C 2 (-1, 1) is the solution of -z ′′ ± + α 2 z ± = 0 for x ∈ (-1, 1) z ± (±1) = 1 and z ± (∓1) = 0 . (6.68)
An immediate computation gives

z + (x) = sinh α(1 + x) sinh 2α , (6.69)
and a similar formula for z -.

We attempt to obtain a resolvent estimate for L ζ β in the case α ≥ θ 1 β 1/3 where θ 1 > 0 . If we try to use the arguments of Subsection 6.1 we would encounter a problem while attempting to use (6.29). It can be verified from (6.69) that

z ± -e -α(1∓x) ∞ ≤ Ce -2α ≤ Ce -2θ 1 β 1/3 . (6.70)
Then, one can deduce in the same manner that for some C > 0, C 1 ∈ (0, 1), and sufficiently large β,

z ′ ± 2 ≥ α 1/2 (1 -Ce -2α ) ≥ C 1 θ 1/2
1 β 1/6 . Thus, the error introduced by (6.29) is not necessarily small and one needs an alternate route for the estimation of (L ζ βλ) -1 . Since for α ≥ θ 1 β 1/3 we need to consider, in the next section, only the case ζ ± ≈ z ± , we focus attention here on the resolvent of L ζ β in that case. Thus, we no longer approximate ζ ± near x = ±1 by 1, as in Subsection 6.1, and use instead the approximation ζ ± ≈ e -α(1∓x) as observed in (6.70). Note that L ζ β depends on α through the orthogonality conditions appearing in the definition of its domain. Consequently, we need to renormalize ψ ± from (6.8b) and (6.8a) in a manner that would suit the approximation used for ζ ± . For some θ > 0, the renormalization factor will be defined by

ω ± (β, λ, θ) := F ( λ± , 0) F ( λ± , θJ -1/3 ± ) , (6.71)
where, (see (6.15) for the definition of λ ± ),

λ± = β 1/3 J -2/3 ± [λ -iU(±1)] = β 1/3 J -2/3 ± λ ± . (6.72) We now define ψ ±,θ = ω ± (β, λ, θ)ψ ± , (6.73)
where ψ ± was introduced in (6.8b)-(6.8a).

The above normalization provides the approximation ζ ± , ψ ±,θ ∼ (J ± β) -1 3 , in the limit β → ∞, as in Subsection 6.1 (see below (6.84)).

We similarly introduce with the notation of (6.11) and (6.12)

g ±,θ = ω ± (β, λ, θ)g ± and ṽ±,θ = ω ± (β, λ, θ)ṽ ± . ( 6

.74)

We can now state: Proposition 6.11. Let r > 1, θ 1 > 0 and κ > 0 . Then, there exist β 0 > 0 and C(κ) > 0 such that, for all U ∈ S 2 r , β ≥ β 0 and θ = αβ -1/3 ≥ θ 1 , sup

ℜλ≤(μ 0 (θ)-κ)β -1/3 (L ζ β -βλ) -1 ≤ C(κ)β -2/3 , (6.75)
where μ0 (θ) = min(J

2/3 -µ 0 (J -1/3 - θ), J 2/3 + µ 0 (J -1/3 + θ)) .
(6.76) Remark 6.12. In the sequel we apply Proposition 6.11 with θ 1 = θ 0 where θ 0 is defined in the statement of Lemma 6.2.

Proof.

The proof goes along similar lines to the proof of Lemma 6.2. Let λ ∈ C satisfy

β 1/3 ℜλ ≤ Υ(θ, κ) := μ0 (θ) -κ .
Furthermore, let the pair (g, v) in L 2 (-1, 1) × D(L ζ β ) satisfy the relation

(L ζ β -βλ)v = g .
We assume, as in (6.20),

v = A + (g)(ψ +,θ -ṽ+,θ ) + A -(g)(ψ -,θ -ṽ-,θ ) + u , (6.77)
where u is given by (6.5), and then estimate A ± (g) in the relevant regime of α values. We first estimate the renormalization factor. We note that

ω ± (β, λ, θ) = e -iπ/6 A 0 (i λ± ) F ( λ± , θJ -1/3 ± ) ,
and from (A.44) which reads, sup ℜ λ≤Υ( θ)

A 0 (i λ)

F ( λ, θ) ≤ C(κ)(1 + θ) ,
we obtain that there exist C(κ) > 0 and β 0 (κ), such that for all θ ≥ θ 1 , ℜλ ≤ Υ(θ, κ)β -1/3 , and β ≥ β 0 (κ) we have

|ω ± (β, λ, θ)| ≤ C θ . (6.78)
We can now use (6.17), (6.18) and (6. [START_REF] Helffer | From resolvent bounds to semigroup bounds[END_REF]) to obtain that

ψ ±,θ 2 ≤ C θ [1 + |λ ± | 1/2 β 1/6 ] 1/2 β -1/6 , (6.79)
g ±,θ 2 ≤ C θ β 1/6 [1 + |λ ± | 1/2 β 1/6 ] 3/2 , (6.80) and (
U -ν + iβ -1/3 )ṽ ±,θ 2 + β -1/3 ṽ±,θ 2 ≤ C θ β -5/6 [1 + |λ ± | 1/2 β 1/6 ] 3/2 .
(6.81)

We note that (6.25) remain valid in the case θ > 0, i.e.,

z + , (ψ +,θ -ṽ+,θ ) z + , (ψ -,θ -ṽ-,θ ) z -, (ψ +,θ -ṽ+,θ )
z -, (ψ -,θṽ-,θ )

A + A - = z + , u z -, u , (6.82)
We now write z ± , ψ ±,θ = e -α(1±x) , ψ ±,θ + z ±e -α(1±x) , ψ ±,θ .

Since by (6.78) and (A.43b) we have that

ψ ±,θ 1 ≤ C(κ)θ ψ ± 1 ≤ Cθβ -1/3 , (6.83)
we can easily deduce using (6.70) that for sufficiently large β

|z ± -e -α(1±x) , ψ ±,θ | ≤ Ĉe -2θβ 1/3 ψ ±,θ 1 ≤ Čβ -2 3 e -θ 1 β 1/3 .
Furthermore, as

+∞ -1 e -α(1+x) Ai (J -β) 1/3 e iπ/6 (1 + x) + iJ -1 -(λ -iU(-1)) dx = (J -β) -1/3 R + e -θJ 1/3 -ξ Ai e iπ/6 ξ + iJ -2/3 - β 1/3 (λ -iU(-1)) dξ = (J -β) -1/3 F ( λ± , θJ -1/3 ± ) ,
we have that

e -α(1+x) , ψ -,θ = (J -β) -1/3 -1 1,∞ e -α(1+x) , ψ -,θ .
Since by (6.83) we have

| 1 1,∞ e -α(1+x) , ψ -,θ | ≤ Ce -2θβ 1/3 ψ -,θ 1 ≤ Čβ -1 3 e -θ 1 β 2/3 ,
and hence we obtain

e -α(1+x) , ψ -,θ = (J -β) -1/3 [1 + O(e -θ 1 β 1/3 )] , (6.84)
and a similar estimate can be obtained for e -α(1-x) , ψ +,θ . We now write

| z + , ψ -,θ | ≤ ψ -,θ L 1 (-1,0) e -θβ 1/3 + ψ -,θ L ∞ (0,1) z + L 1 (0,1) ,
and then use (6.83), (6.78), and (A.43c) to obtain that

| z + , ψ -,θ | ≤ Cβ -5/3 . (6.85) Bounds for z ± , u . Set, λ =    λ if |ℜλ| > max 1, U ′ ∞ θ β -1/3 max 1, U ′ ∞ θ β -1/3 + iℑλ if |ℜλ| ≤ max 1, U ′ ∞ θ β -1/3 . (6.86) As β 1 3 |ℜ λ| ≥ 1 and |ℜ λ| ≥ |ℜλ|, it can be verified that 1 + |ℜλ| 2 β 2/3 ≤ 2 |ℜ λ| 2 β 2/3 ,
and hence 1 +

β 2/3 |λ ± | 2 ≤ 2 β 2/3 |U(±1) + i λ| 2 . (6.87)
We now verify that

z 2 -(U + i λ) -2 ∞ = |U(-1) + i λ| -2 (6.88) Indeed, we have [z 2 -|U + i λ| -2 ] ′ = z 2 -|U + i λ| -4 -2θβ 1/3 [(U -ℑλ) 2 + |ℜ λ| 2 ] + 2U ′ (U -ℑλ) = z 2 -|U + i λ| -4 -2θβ 1/3 (U -ℑλ) + U ′ 2θβ 1/3 2 + |U ′ | 2 2θβ 1/3 -2θβ 1/3 |ℜ λ| 2
, which is non positive by (6.86). Hence the maximum of z 2 -|U + i λ| -2 is obtained at x = -1. A similar inequality can be established for z 2 + |U + i λ| -2 . Combining (6.87) and (6.88) yields

z ± (U + i λ) -1/2 1 ≤ z 1/2 ± 1 z 1/2 ± (U + i λ) -1/2 ∞ ≤ C β -1/6 θ |β 1/3 (U(±1) + i λ)| -1/2 ≤ Ĉ β -1/6 θ[1 + β 1/3 |λ ± |] 1/2 . (6.89)
Note that the replacement of λ by λ avoids the burden of a vanishing denominator.

We now write

| z ± , u | ≤ z ± (U + i λ) -1/2 1 (U + i λ) 1/2 u ∞ . (6.90)
To estimate (U +i λ) 1/2 u ∞ in the right hand side of (6.90) , we first obtain a bound for |U -ν| 1/2 u ∞ . Thus, integration by parts yields for all (x,

x 0 ) ∈ [-1, 1] 2 (U -ν)|u| 2 x x 0 = 2ℜ x x 0 (U -ν)ūu ′ dt - x x 0 U ′ |u| 2 dt .
From which we conclude, by integrating the above for x ∈ (-1, +1) and Cauchy-Schwarz inequalities, that

|U -ν| 1/2 u 2 ∞ ≤ C( |U -ν| u 2 u ′ 2 + u 2 2 + |U -ν| 1/2 u 2 2 )
. By (5.4) and (5.5) we then have

|U -ν| 1/2 u 2 ∞ ≤ C β 4/3 g 2 2 .
We now write

(U + i λ) 1/2 u 2 ∞ ≤ 2 |U -ν| 1/2 u 2 ∞ + (|ℜλ| + 2β -1/3 ) u 2 ∞ ≤ C(β -4/3 g 2 2 + (|ℜλ| + 2β -1/3 ) u 2 u ′ 2 .
By (5.4) we then have

(|ℜλ| + 2β -1/3 ) u 2 ≤ C β g 2 ,
and hence, using(5.4) once again to estimate u ′ 2 , we may conclude that

(U + i λ) 1/2 u ∞ ≤ C β 2/3 g 2 . (6.91)
Combining the above with (6.89) and (6.90) yields

| z ± , u | ≤ C θ[1 + |λ ± |β 1/3 ] 1/2 β -5/6 g 2 .
(6.92)

Bounds for z ± , ṽ±,θ .

The estimation of z ± , ṽ±,θ follows a similar path to that of z ± , u . We begin by writing

| z ± , ṽ±,θ | ≤ z ± (U + i λ) -1/2 1 (U + i λ) 1/2 ṽ±,θ ∞ .
Since ṽ±,θ , given by (6.74) satisfies the same problem as u with g replaced by g ±,θ we may conclude as in (6.91) that

(U + i λ) 1/2 ṽ±,θ ∞ ≤ C β 2/3 g ±,θ 2 .
Consequently, by (6.80) and (6.89)

| z ± , ṽ±,θ | ≤ C β -2/3 [1 + |λ ± |β 1/3 ] 5/4 . (6.93)
In a similar manner we can obtain that

| z ∓ , ṽ±,θ | ≤ C β -2/3 [1 + |λ ± |β 1/3 ] 3/4 [1 + |λ ∓ |β 1/3 ] 1/2 ≤ Cβ -2/3 . (6.94)
As in (6.31) we can now write, in view of (6.93) and

| z ± , (ψ ±,θ -ṽ±,θ ) -(J ± β) -1/3 | ≤ Cβ -2/3 .
Combining the above with (6.85), (6.93), and (6.94) yields

|A ± (g)| ≤ Cβ 1/3 | z ± , u | + Cβ -2/3 | z ∓ , u | .
The above, together with (6.92) yields

|A ± (g)| ≤ Cβ -1/2 θ 1 [1 + |λ ± |β 1/3 ] 1/2 + β -1 1 [1 + |λ ∓ |β 1/3 ] 1/2 g 2 .
As |λ +λ -| = |U(1) -U(-1)|, we obtain for sufficiently large β

|A ± (g)| ≤ Ĉβ -1/2 θ 1 [1 + |λ ± |β 1/3 ] 1/2 g 2 .
(6.95)

Combining the above with (6.77), (6.79), and (6.81) yields (6.75).

7 Zero traction Orr-Sommerfeld operator

A short reminder

We recall for the commodity of the reader that B S λ,α,β is defined by (3.5) and (3.10) i.e.

B λ,α,β := (L β -βλ) d 2 dx 2 -α 2 -iβU ′′ , with domain D(B S λ,α,β ) = { u ∈ H 4 (-1, 1) ∩ H 1 0 (-1, 1) | u ′′ ∈ H 1 0 (-1, 1) } .
Here andL D β is the Dirichlet realization of L β in (-1, +1). Finally we recall that the inviscid operator A λ,α associated with U is defined by

L β = - d 2 dx 2 + iβU ,
A λ,α = (U + iλ) - d 2 dx 2 + α 2 + U ′′ with domain D(A λ,α ) = H 2 (-1, +1) ∩ H 1 0 (-1, +1).
7.2 The case U ′′ = 0 .

We now prove Proposition 7.1. For all r > 1 and δ > 0 there exist positive β 0 , Υ, and C such that, for any β ≥ β 0 and U ∈ S r satisfying (2.29), it holds that sup

ℜλ≤Υβ -1/3 0≤α (B S λ,α,β ) -1 + d dx • (B S λ,α,β ) -1 ≤ C β -1 2 -δ . (7.1)
Proof. Let λ = µ + iν. Let further 0 < Υ < ℜν 1 and suppose µ ≤ Υβ -1/3 . Let φ ∈ D(B S λ,α,β ) and f ∈ L 2 (-1, +1) satisfy

B S λ,α,β φ = f (7.2) and v := A λ,α φ . (7.3)
We note that v ∈ D(L D β ) and, defining g by

g := (L D β -λβ)v , (7.4)
we have

g = -(U + iλ)f + 2U ′ (φ (3) -α 2 φ ′ ) + U ′′ (φ ′′ -α 2 φ) -(U ′′ φ) ′′ . (7.5)
By (5.19), there exist C > 0 and β 0 > 0 such that, for β ≥ β 0 and p ∈ (2, +∞] we have

v p ≤ Ĉβ -3p+2 6p g 2 . (7.6)
As

ℜ (U ′′ ) -1 (φ ′′ -α 2 φ), B S λ φ = (U ′′ ) -1/2 (φ (3) -α 2 φ ′ ) 2 2 + + ℜ (U ′′ ) -1 ′ (φ ′′ -α 2 φ), φ (3) -α 2 φ ′ -βµ φ ′′ -α 2 φ 2 2 , (7.7)
we easily obtain that

φ (3) -α 2 φ ′ 2 ≤ C( f 2 + |µ| 1/2 β 1/2 φ ′′ -α 2 φ 2 ) . (7.8)
We now write,

(L D β -βλ)(φ ′′ -α 2 φ) = iβU ′′ φ + f . (7.9)
With the aid of (5.8) and (5.21) we then obtain

φ ′′ -α 2 φ 2 ≤ C(β 1/6 φ ∞ + β -2/3 f 2 ) , (7.10)
Substituting the above into (7.8) then yields

φ (3) -α 2 φ ′ 2 ≤ C( f 2 + |µ| 1/2 β 1/2 φ ′′ -α 2 φ 2 ) . (7.11)
To bound (U ′′ φ) ′′ 2 we first use the fact that

φ ′′ , φ ′′ -α 2 φ = φ ′′ 2 2 + α 2 φ ′ 2 2 . (7.12)
Then, by (7.10) we obtain that

φ ′′ 2 ≤ C(β 1/6 φ ∞ + β -2/3 f 2 ) , (7.13) and (
U ′′ φ) ′′ 2 ≤ C(β 1/6 φ ∞ + β -2/3 f 2 + φ 1,2 ) . (7.14)
We continue the proof by considering B S λ,α,β in a few different regimes of λ values.

Case 1: Bounded |λ| Suppose first that |ν| ≤ C 0 ; 0 < |µ| ≤ Υβ -1/3 , (7.15a,b)
where C 0 ∈ R + . The value of C 0 to be selected above will be determined in a later stage. Using (7.10) , (7.11), and (7.14), we obtain from (7.5) that

g 2 ≤ C( f 2 + (|µ| 1/2 β 2/3 + β 1/6 )[ φ ∞ + φ 1,2 ]) ,
or by Sobolev's embedding

g 2 ≤ C( f 2 + (|µ| 1/2 β 2/3 + β 1/6 ) φ 1,2 ) . (7.16) (7.
3), we may apply Proposition 4.13 to the pair (φ, v) to conclude, by (4.24), that for any q > 1 there exists C > 0 such that

φ 1,2 ≤ C( v ′ q + v ∞ ) . (7.17)
Similarly, by (4.25), for any p > 2 there exists C > 0 such that

|µ| 1/2 β 2/3 φ 1,2 ≤ |µ| 1/2-1/p β 2/3 v p . (7.18)
Hence,

g 2 ≤ C f 2 + |µ| 1/2-1/p β 2/3 v p + β 1/6 ( v ′ q + v ∞ ) .
We may now use (7.6) to obtain, for p > 2,

|µ| 1/2-1/p β 2/3 v p ≤ C|µ| 1 2 -1 p β 2 3 -3p+2 6p g 2 ≤ ĈΥ 1 2 -1 p β 2 3 -p-2 6p -3p+2 6p g 2 = ĈΥ 1 2 -1
p g 2 Applying (7.6) (which is valid for p = +∞ as well) once again yields

β 1 6 v ∞ ≤ Cβ 1 6 -1 2 g 2 = Cβ -1 3 g 2 .
Finally, we apply (5.10) (with q = p) to the pair (v, g) satisfying (7.4)) to conclude, for 1 < q < 2, that

β 1/6 v ′ q ≤ C p β 1 3r g 2 .
Combining the above we then obtain, for sufficiently small Υ and β -1 ,

g 2 ≤ C f 2 . (7.19)
From (5.10) and (7.19) we now get, for any p ∈ (1, 2),

v ′ p ≤ Cβ -2+p 6p g 2 ≤ Ĉβ -2+p 6p f 2 .
From which we deduce, for p sufficiently close to 1 and sufficiently small δ > 0, the existence of C δ > 0 such that

v ′ p + v ∞ ≤ C δ β -1 2 -δ f 2 .
We now return to (7.17) to conclude that

φ 1,2 ≤ Ĉδ β -1 2 -δ f 2 . (7.20)
Hence (7.1) is proven in Case 1 for sufficiently small Υ > 0.

Case 2: ℜλ unbounded negative Next, consider the case where (7.15a) is kept in place but instead of (7.15b), µ < -Υβ -1/3 is assumed. In this case we return to (7.7) and use the positivity of -βµ for the last term on its right hand side. In a similar manner to the one used to derive (7.8) we establish the existence of β(Υ) > 0 such that for all β ≥ β(Y ),

φ ′′ -α 2 φ 2 ≤ C Υβ 2/3 f 2 .
Since φ ∈ H 1 0 (-1, 1) it can be easily verified that

φ 1,2 ≤ 4 π φ ′′ -α 2 φ 2 , (7.21)
and hence for any Υ > 0, there exist C and β 0 such that, for all α ≥ 0 and all In this case we write

β ≥ β 0 φ 1,2 ≤ Cβ -2/3 f
-ℑ φ ′′ -α 2 φ, B S λ φ = β φ ′′ -α 2 φ, (ν -U)(φ ′′ -α 2 φ) + φ ′′ -α 2 φ, U ′′ φ As φ ∈ H 1 0 (-1, 1) we have that φ 2 ≤ 4 π 2 φ ′′ -α 2 φ 2 ,
and hence, under (7.22),

| φ ′′ -α 2 φ, (ν-U)(φ ′′ -α 2 φ) + φ ′′ -α 2 φ, U ′′ φ | ≥ (C 0 -U ∞ -4π -2 U ′′ ∞ ) φ ′′ -α 2 φ 2 2 . Selecting C 0 > U ∞ + 4π -2 U ′′ ∞ we then obtain φ ′′ -α 2 φ 2 ≤ C β f 2 .
We can now conclude from (7.21) that

φ 1,2 ≤ C β f 2 .
As (4.23)-(4.25) are not valid for µ = 0, we need to establish (7.20) for the case µ = 0. Let B S iν,α φ = f . Then, we may write by (7.7) that

φ ′′ + α 2 φ 2 ≤ C f 2 .
Then, as B S iν+β -1 ,α φ = φ ′′ + α 2 φ + f , we may conclude (7.1) from (7.20).

The nearly Couette case

We now proceed to consider the nearly Couette case addressed by both Theorems 2.14 and 2.15. Let U ∈ C 4 ([-1, 1]) satisfy (2.24) and recall the definition of δ 2 (U) from (2.26), δ 2 (U) := U ′′ 1,∞ . We next recall from (2.28), for some r > 1,

S r = {v ∈ C 4 ([-1, 1]) | , inf x∈[-1,1] v ′ ≥ 1/r and v 4,∞ ≤ r} .
We shall consider the case where δ 2 (U) is small. Note that in contrast with the assumptions of Proposition 7.1, U ′′ may change its sign. Proposition 7.3. For any r > 1 and Υ < ℜν 1 , there exist δ ∈ (0, 1 4 ), and positive β 0 and C, such that, for all U ∈ S r satisfying 0 < δ 2 (U) ≤ δ and all β ≥ β 0 , sup

β 1/3 ℜλ≤J 2/3 m Υ 0≤α (B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β 5/6 . (7.23)
Proof.

Note that by (2.28)

r ≥ m ≥ 1 r . (7.24)
As in the proof of Proposition 7.1 we separately consider different regimes of λ ∈ C.

Case 1: ℜν 1 /2 ≤ J -2/3 m β 1/3 µ ≤ Υ or β 1/3 µ ≤ -J 2/3 m ℜν 1 /2 Let f ∈ L 2 (-1, 1
). Using the definitions (7.2)-( 7.3) we further set

ṽ = v U + iλ , Clearly (L D β -βλ)ṽ = f + U ′′ φ U + iλ ′′ = h . (7.25)
We now write v 2 ≤ (U + iλ)(L D ββλ) -1 h 2 By (5.8) and (5.9) we then have

v 2 ≤ C β h 2 . (7.26)
Next, we turn to estimate h 2 . Clearly, by (7.25),

h 2 ≤ f 2 + C(r) φ ′′ |U + iλ| 2 + φ ′ 1 + 1 |U + iλ| 2 2 + U ′′ φ |U + iλ| 3 2 + φ 1 + 1 |U + iλ| 2 2 . (7.27)
where we have use that fact that

1 |U + iλ| ≤ 1 2 1 + 1 |U + iλ| 2 .
For the third term in the right-hand-side of (7.27) we have

φ ′ 1 + 1 |U + iλ| 2 2 ≤ Cβ 2/3 φ ′ 2 . (7.28)
We now turn to estimate the second term on the right-hand-side of (7.27). Here we write

φ ′′ U + iλ 2 ≤ Cβ 1/3 φ ′′ 2 .
Then, we use (7.13) (which remains valid in the nearly-Couette case) to obtain

φ ′′ U + iλ 2 ≤ C(r)(β 1/2 φ ∞ + β -1/3 f 2 ) .
Sobolev embedding then yields

φ ′′ U + iλ 2 ≤ C(r)(β 1/2 φ 1,2 + β -1/3 f 2 ) . (7.29)
Finally, we turn to estimate the last term on the right-hand-side of (7.27). Suppose first that for some

x ν ∈ [-1, 1] we have U(x ν ) = ν. Clearly, as |U(x) + iλ| ≥ m(x -x ν ) + 1 2 i J 2/3 m ℜν 1 β -1/3 (7.30)
it holds by (7.24) that

U ′′ φ |U + iλ| 3 2 ≤ Cδ φ |(x -x ν ) + i J 2/3 m ℜν 1 β -1/3 | 3 2
.

Hence, we obtain

U ′′ φ |U + iλ| 3 2 ≤ Ĉδβ 5/6 φ ∞ . (7.31)
Substituting the above together with (7.29) and (7.28) into (7.27) yields with the aid of Sobolev embedding and Poincare inequality

h 2 ≤ f 2 + C(r)(δβ 5/6 + β 2 3 ) φ 1,2 ,
which for β ≥ β 0 (δ) gives

h 2 ≤ f 2 + Ĉ(r) δ β 5/6 φ 1,2 . (7.32)
By (7.26) we then have

v 2 ≤ C β f 2 + C(r) δ β -1/6 φ 1,2 . (7.33)
The proof can now be completed, for

ℜν 1 /2 ≤ J -2/3 m β 1/3 µ ≤ Υ or β 1/3 µ ≤ -J 2/3
m ℜν 1 /2 , by using (4.25) with p = 2. In the case where U(x)ν = 0 on [-1, +1], we may still apply the previous arguments by replacing (7.30) with

|U(x) + iλ| ≥ m(x ± 1) + 1 2 iJ 2/3 m ℜν 1 β -1/3 (7.34)
where ± denotes + for ν < inf U(x) andfor ν > sup U(x) .

Case 2:

β 1/3 |µ| < J 2/3 m ℜν 1 /2 Let s = β 1/3 J 2/3 m ℜν 1 2 -µ ,
and then write

B S λ+sβ -1/3 ,α φ = f -sβ 2/3 (φ ′′ -α 2 φ) . Since ℜ(λ + sβ -1/3 ) = β -1/3 J 2/3
m ℜν 1 /2 , we can use (7.23) to obtain that

φ 1,2 ≤ C(β -5/6 f 2 + β -1/6 φ ′′ -α 2 φ 2 ) . (7.35)
We now recall (7.9)

(L D β -βλ)(φ ′′ -α 2 φ) = iβU ′′ φ + f ,
and then use (5.8) and (5.21) to establish that

φ ′′ -α 2 φ 2 ≤ C(β -2/3 f 2 + δβ 1/6 φ ∞ ) (7.36)
Substituting the above into (7.35) yields

φ 1,2 ≤ C(β -5/6 f 2 + δ φ ∞ ) .
Using Sobolev embedding yields (7.23) , for sufficiently small δ and β 1/3 µ ≤ Υ.

No slip Orr-Sommerfeld operator

In contrast with the prescribed traction condition, the auxiliary function v = A λ,α φ, does not satisfy, if φ ∈ D(B D λ,α,β ), a Dirichlet boundary condition. Consequently, special attention must be given to the behaviour of v near the boundary through (6.20) for instance. Since φ satisfies a Dirichlet boundary condition, we expect that the rapidly decaying boundary terms ψ ± -ṽ± in (6.20) should have a negligible contribution to φ compared with that of u (i.e., A -1 λ,α (ψ ± -ṽ± ) 1,2 ≪ A -1 λ,α u 1,2 ). The next subsection is dedicated to the establishment of such estimates.

Preliminaries

We recall that J ± = U ′ (±1) and that ϑ r 1 and ψ ± are respectively defined by (6.10), (6.6) and (6.7). The next lemma holds true under the assumptions of either Proposition 4.9 or Proposition 4.13. Lemma 8.1. For any r > 1 and Υ < ϑ r 1 , there exist positive constants C and β 0 such that, for all β ≥ β 0 , λ ∈ C for which ℜλ ≤ β -1/3 J 2/3 ± Υ, and U ∈ S r satisfying either (2.29) or (4. [START_REF] Helffer | From resolvent bounds to semigroup bounds[END_REF], it holds that

A -1 λ,α (U + iλ)ψ ± 1,2 ≤ C [1 + β 1/3 |λ ± |] -1/4 β -1/2 . (8.1)
where

λ ± = µ + i(ν -U(±1)) . (8.2)
Proof.

Let φ = A -1 λ,α (U + iλ)ψ ± . We write

φ U + iλ , (U + iλ)ψ ± ≤ (φ -φ(±1))ψ ± 1 . (8.3) As |φ(x) -φ(±1)| ≤ φ 1,2 (1 ∓ x) 1/2 , ∀x ∈ (-1, 1) , we may write (φ -φ(±1))ψ ± 1 ≤ φ 1,2 (1 ∓ x) 1/2 ψ ± 1 .
It follows from (6.27) (with s = 1 2 ) that for some positive C

(1 ∓ x) 1/2 ψ ± 1 ≤ C [1 + |λ ± |β 1/3 ] -1/4 β -1/2 . (8.4)
In the case where (4. [START_REF] Helffer | From resolvent bounds to semigroup bounds[END_REF]) is satisfied, we may use (4.40) so that

φ 1,2 ≤ C (1 ∓ x) 1/2 ψ ± 1 ,
which, combined with (8.4) yields (8.1).

In the case where (2.29) is satisfied, we may use (4.88) for v = (U + iλ)ψ ± to obtain that

A -1 λ,α (U + iλ)ψ ± 1,2 ≤ C (1 ∓ x) 1/2 ψ ± 1 .
Then, we apply (8.4) to obtain (8.1).

We shall also need the following Lemma 8.2. Let r > 1 and Υ < ϑ r 1 . Let further (λ, λ) ∈ C 2 satisfy λ -λ ∈ R, ℜλ and ℜ λ are in (-∞, β -1/3 J 2/3 ± Υ) and |ℜλ -ℜ λ| ≤ 2ϑ r 1 β -1/3 . Then, there exist positive C and β 0 such that, for all β ≥ β 0 , and all U ∈ S r satisfying either (2.29) or (4. [START_REF] Helffer | From resolvent bounds to semigroup bounds[END_REF], it holds that

A -1 λ,α (U + i λ)Γ [-1,1] ( L β,R -β λ) -1 ψ± (λ) 1,2 ≤ C [1 + |λ ± |β 1/3 ] -1 β -7/6 , (8.5)
where

ψ± (x, λ) = ψ ± (x, λ) x ∈ [-1, 1] 0 |x| > 1 .
Proof.

For later reference we note that the requirement set above that |λ -λ| ≤ 2ϑ r 1 β -1/3 implies the existence of C > 0 such that

1 C [1 + |λ ± |β 1/3 ] ≤ [1 + | λ± |β 1/3 ] ≤ C[1 + |λ ± |β 1/3 ] and 1 C (1 + β 1/3 |µ|) ≤ (1 + β 1/3 |μ|) ≤ C(1 + β 1/3 |µ|) ,
where μ = ℜ λ.

Step 1: We prove that there exist positive C, and β 0 such that for all β > β 0 and k ∈ {0, 1, 2}

(1 ∓ x) k ( L β,R -β λ) -1 ψ± (λ) 2 ≤ Cβ -(5/6+k/2) [1 + β 1/3 |λ ± |] -3/4-k/2 . (8.6)
For convenience of notation we prove (8.6) only for ( L β,Rβ λ) -1 ψ+ (λ). The proof for ( L β,Rβ λ) -1 ψ-(λ) can be obtained in a similar manner.

Let

u k = (1 -x) k ( L β,R -β λ) -1
ψ+ , for k ∈ {0, 1, 2, 3}. For convenience we also set u k ≡ 0 for all k ≤ -1.

The case k = 0 . For k = 0, we observe from (5.4) and (6.17), that we have

u 0 2 ≤ C β 5/6 (1 + β 1/3 |µ|) [1 + β 1/3 |λ + |] 1/4 . (8.7)
We note that (8.6) for k = 0 does not follow from (8.7). Some additional estimates for large values β 1/3 |ν| should be obtained to this end. For k ∈ {1, 2, 3}, we now write:

( L β,R -β λ)u k = (1 -x) k ψ± (λ) -2ku ′ k-1 + k(k -1)u k-2 . (8.8)
By (5.4) it holds that

u k 2 ≤ C β 2/3 (1 + β 1/3 |µ|) ( (1 -x) k ψ± (λ) 2 + u ′ k-1 2 + (k -1) u k-2 2 ) . (8.9) As ℜ u k-1 , ( L β,R -β λ)u k-1 = u ′ k-1 2 2 -β μ u k-1 2 2 , (8.10)
we obtain, as μ ≤ β -1/3 J

2/3 ± Υ, that u ′ k-1 2 ≤ C β 1/3 u k-1 2 + β -1/3 (1 -x) k-1 ψ± (λ) 2 + (k -1) u ′ k-2 2 + (k -1)(k -2) u k-3 . (8.11)
Substituting the above into (8.9) yields, for k = 1,

u 1 2 ≤ Cβ -2 3 (1 + β 1 3 |µ|) -1 (1 -x) ψ± (λ) 2 + β 1 3 u 0 2 + β -1 3 ψ± (λ) 2 .
Using (6.17) for k = 0 and k = 1, which holds as

β 1/3 ℜλ ≤ Υ < J 2/3 + ϑ r 1 , yields u 1 2 ≤ C β 2/3 (1 + β 1/3 |µ|) [1 + β 1/3 |λ + |] 1/4 β -1/2 + β 1/3 u 0 2 . (8.12)
From (8.7) and (8.12) we then get

u 1 2 ≤ C β -7/6 (1 + β 1/3 |µ|) -1 [1 + β 1/3 |λ + |] 1/4 . (8.13)
Using (8.13) we can now complete the proof of (8.6) for k = 0 . To this end we write

[β -1/3 + |λ + |] u 0 2 ≤ (β -1/3 + |µ|) u 0 2 + [U -U(1)]u 0 2 + (U -ν)u 0 2 .
From (5.5) and (6.17), we deduce that

(U -ν)u 0 2 ≤ Cβ -1 ψ± (λ) 2 ≤ Ĉβ -7/6 [1 + β 1/3 |λ + |] -3/4 .
On the other hand, we have, by (8.13)

[U -U(1)]u 0 2 ≤ C u 1 2 ≤ Ĉβ -7 6 (1 + β 1/3 |µ|) -1 [1 + β 1/3 |λ + |] 1/4 ≤ Ĉβ -7 6 [1 + β 1/3 |λ + |] 1/4 .
Together with (8.7), we obtain

u 0 2 ≤ C β 5/6 [1 + β 1/3 |λ + |] -3/4 , (8.14)
which proves (6.11) for k = 0 .

The case k = 1 . By (8.10) we may write, instead of (8.11),

u ′ 0 2 ≤ C β 1/3 [1 + β 1/3 |λ + |] 1/2 u 0 2 + β -1/3 [1 + β 1/3 |λ + |] -1/2 ψ± (λ) 2 .
Hence, by (6.17), it holds that

u ′ 0 2 ≤ C β 1/2 [1 + β 1/3 |λ + |] -1/4 , (8.15)
which, when substituted into (8.9) for k = 1 , yields

u 1 2 ≤ C β 7/6 (1 + β 1/3 |µ|) [1 + β 1/3 |λ + |] -1/4 .
Substituting into (8.11) yields, for k = 2 ,

u ′ 1 2 ≤ C β 5/6 [1 + β 1/3 |λ + |] -1/4 , (8.16)
and hence, by (8.9) for k = 2 ,

u 2 2 ≤ C β 3/2 (1 + β 1/3 |µ|) [1 + β 1/3 |λ + |] -1/4 . (8.17)
As above we now write

[β -1/3 + |λ + |] u 1 2 ≤ (β -1/3 + |µ|) u 1 2 + C u 2 2 + (U -ν)u 1 2 , (8.18)
to obtain from (8.17), (8.16), (5.5), and (6.17) that

u 1 2 ≤ C β 7/6 [1 + β 1/3 |λ + |] -5/4 . (8.19)
Proposition 8.3. For every r > 1 and Υ < ϑ r 1 , there exist positive δ, β 0 , α 0 , and C such that, for all U ∈ S r satisfying 0 < δ 2 (U) ≤ δ (where δ 2 is given by (2.26)) and β ≥ β 0 , it holds that

sup ℜλ≤ΥJ 2/3 m β -1/3 0≤α≤α 0 β 1/3 (B D λ,α,β ) -1 + d dx (B D λ,α,β ) -1 ≤ Cβ -5/6 . ( 8
-(U + iλ)ζ ′′ ± + (α 2 [U + iλ] + U ′′ )ζ ± = 0 in (-1, 1) ζ ± (±1) = 1 and ζ ± (∓1) = 0 . (8.22)
Note that the differential operator on the left-hand-side is identical with that of A λ,α given by (3.13). We now write

ζ ± (x) = ζ 0 ± (x) + exp{-α(1 ∓ x)}η ± (x) (8.23)
in which η± (x) = η(1 ∓ x), where η is given by (5.14), and ζ 0 ± ∈ D(A λ,α ).

ζ 0 ± and ζ ± estimates We begin by establishing L ∞ and W 1,q estimates for ζ 0 ± . To this end we first write

A λ,α ζ 0 ± = -U ′′ η± + 2α(U + iλ)η ′ ± + η′′ ± (U + iλ) exp{-α(1 ∓ x)} . (8.24)
Note that since the derivatives of η are supported on [-1/2, 1/2] we have, since

|µ| is bounded, 2α(U+iλ)η ′ ± +η ′′ ± (U+iλ) exp{-α(1∓•)} 1,∞ ≤ C(1+|ν|)(1+α 2 )e -α/2 ≤ C(1+|ν|) .
Note further that, for all 1 ≤ q ≤ 2 , we have

U ′′ η± exp{-α(1 ∓ •)} q ≤ C(1 + α) -1/q . and U ′′ η± exp{-α(1 ∓ •)} 1,q ≤ C(1 + α) 1-1/q .
Consequently, for all 1 ≤ q ≤ 2 , we have

A λ,α ζ 0 ± q ≤ C(1 + |ν| + (1 + α) -1/q ) ≤ Ĉ(1 + |ν|) . (8.25a) and A λ,α ζ 0 ± 1,q ≤ C(1 + |ν| + α 1-1/q ) . (8.25b)
Observing that

ℜ ζ 0 ± , (U + iλ) -1 A λ,α ζ 0 ± = 1 2 |(ζ 0 ± ) ′ | 2 + α 2 |ζ 0 ± | 2 + I(ζ 0 ± , λ) ,
we deduce from Lemma 4.7 that, for sufficiently small δ there exist γ 0 > 0 and C > 0 such that, for any α ≥ 0 and any λ ∈ C \ J , (recall that |ℜλ| > J

2/3 m β -1/3 ϑ r 1 /2 in this step) ℜ ζ 0 ± , (U + iλ) -1 A λ,α ζ 0 ± ≥ 1 2 (ζ 0 ± ) ′ 2 2 + (γ 0 + α 2 ) ζ 0 ± 2 2 ≥ C(1 + α) 1/2 (ζ 0 ± ) ′ 3/2 2 ζ 0 ± 1/2 2 . (8.26)
Consequently, by (4.33) (which is applied with q = p, φ = ζ 0 ± and v = A λ,α ζ 0 ± ) and (8.25), for any 1 < q < 2 there exists C q > 0 such that

(ζ 0 ± ) ′ 3/2 2 ζ 0 ± 1/2 2 ≤ C q (1 + α) -1/2 ( (A λ,α ζ 0 ± ) ′ q ζ 0 ± ∞ + A λ,α ζ 0 ± ∞ (ζ 0 ± ) ′ q ) ≤ C q (1 + |ν|) (1 + α) -1/q+1/2 (ζ 0 ± ) ′ 2 .
(We use the fact that q ≤ 2 and Poincaré's inequality to obtain the last inequality.)

We may thus conclude that

(ζ 0 ± ) ′ 1/2 2 ζ 0 ± 1/2 2 ≤ C q (1 + |ν|) (1 + α) -1/q+1/2 .
Hence, for q = 3 2 , we get

ζ 0 ± ∞ ≤ (ζ 0 ± ) ′ 1/2 2 ζ 0 ± 1/2 2 ≤ C q (1 + |ν|) (1 + α) -1/6 , which finally implies ζ 0 ± ∞ ≤ C (1 + |ν|) . (8.27)
By (8.26) we also have that

ℜ ζ 0 ± , (U + iλ) -1 A λ,α ζ 0 ± ≥ 1 2 (ζ 0 ± ) ′ 2 2 ,
and hence

(ζ 0 ± ) ′ 2 ≤ C q (1 + |ν|) (1 + α) -1/q+1 . For q = 3
2 we may thus conclude

(ζ 0 ± ) ′ 2 ≤ C(1 + |ν|)(1 + α) 1 3 . (8.28)
As (8.27) and (8.28) are unsatisfactory for large values of ν, we use the fact that

ℜ (U -ν) 2 ζ 0 ± , (U + iλ) -1 A λ,α ζ 0 ± = ((U -ν)ζ 0 ± ) ′ 2 2 -U ′ ζ 0 ± 2 2 + α 2 (U -ν)ζ 0 ± 2 2 -ℜ (U -ν) 2 ζ 0 ± , (U + iλ) -1 U ′′ ζ 0 ± .
Assuming |ν| ≥ 2|U| ∞ , we first obtain, for any ε > 0

ε||(U -ν)ζ 0 ± 2 2 + 4 ε A λ,α ζ 0 ± || 2 2 ≥ ((U -ν)ζ 0 ± ) ′ 2 2 -C (U -ν)ζ 0 ± 2 ζ 0 ± 2 .
Poincaré's inequalit then yields, for sufficiently small ε > 0 We can thus consider the problem

ν ζ 0 ± 1,2 ≤ C( A λ,α ζ 0 ± 2 + ζ 0 ± 2 ) ,
- d 2 dx 2 + iβ(U + iλ) ṽ = h ,
where ṽ satisfies (8.32c) and h is given by (see (7.25))

h = f + U ′′ φ U + iλ ′′ .
By (8.32) we may now use Lemma 6.2 to conclude from (6.20) (applied for (g, u, v) = (h, ũ, ṽ)) that ṽ admits the decomposition .33) where A ± (h) and ṽ± respectively satisfy (6.24) and (6.14), and ũ is given by

ṽ = A + (h)(ψ + -ṽ+ ) + A -(h)(ψ --ṽ-) + ũ . ( 8
ũ = Γ [-1,1] ( L β,R -βλ) -1 h , ( 8
h 2 ≤ f 2 + C (δ β 5/6 + β 2 3 ) φ 1,2 + φ ′′ U + iλ 2 .
which leads for β ≥ β 0 (δ) to

h 2 ≤ f 2 + Ĉ δ β 5/6 φ 1,2 + φ ′′ U + iλ 2 . (8.36)
To bound the last term on the right-hand-side we write, recalling that |µ| ≥ J

2/3 m β -1/3 ϑ r 1 /2 in this step, φ ′′ U + iλ 2 ≤ Cβ 1/3 φ ′′ 2 .
(8.37)

Then, we use the fact that by (7.3) and (3.13)

φ ′′ -α 2 φ 2 ≤ ṽ 2 + U ′′ (U + iλ) -1 φ 2 .
As in the proof of (7.31), we get

U ′′ (U + iλ) -1 φ 2 ≤ Cδβ 1/6 φ ∞ .
We now obtain a bound for the norm of ṽ by estimating each term appearing in the right hand side of (8.33). By (6.14) we get for ṽ± , ṽ± ≤ Cβ -1 2 , By (6.17) (for k = 0) we get for ψ ± ,

ψ ± ≤ C β -1 6 [1 + |λ ± |β 1/3 ] 1 4 , (8.38)
and by (6.24) we get for A ± (h)

|A ± (h)| ≤ C β -1 2 h 2 . (8.39)
Together with (8.35) for ũ, we then have

ṽ 2 ≤ Cβ -2/3 [1 + |λ ± |β 1/3 ] 1 4 h 2 ,
where λ ± is given by (6.15). Since (7.12) remains valid for no-slip conditions we may conclude that

φ ′′ 2 ≤ C(β -2/3 [1 + |λ ± |β 1/3 ] 1 4 h 2 + δβ 1/6 φ ∞ ) .
Since |λ ± | is bounded in this step, we obtain

φ ′′ 2 ≤ C(β -7/12 h 2 + δβ 1/6 φ ∞ ) . ( 8
v = v o + v + + v -, where v ± = A ± (h)(U + iλ)(ψ ± -ṽ± ) , (8.43)
and

v o = (U + iλ)Γ [-1,1] ũ = (U + iλ)Γ [-1,1] ( L β,R -βλ) -1 h . (8.44) Set further φ ± = A -1 λ,α v ± ; φ o = A -1 λ,α v o . Clearly, φ = φ o + φ + + φ -. (8.45)
To estimate φ ± we write

φ ± = A ± (h)A -1 λ,α (U + iλ)ψ ± -A ± (h)A -1 λ,α (U + iλ)ṽ ± . (8.46)
For the first term in the right hand side, we use (8.1) to obtain

A ± (h)A -1 λ,α (U + iλ)ψ ± 1,2 ≤ Cβ -1 2 |A ± (h)| .
Using (4.25) with p = 2 and (6.14), yields for the second term

A ± (h)A -1 λ,α (U + iλ)ṽ ± 1,2 ≤ Cβ -1 |A ± (h)| .
Hence using (8.39), we obtain

φ ± 1,2 ≤ C(β -1 f 2 + δβ -1/6 φ 1,2 ) . (8.47)
To estimate φ 0 we first recall that by (8.34) and (8.44)

v o = (U + iλ)Γ [-1,1] ũ = (U + iλ)Γ [-1,1] ( L β,R -βλ) -1 h .
By (5.4) and (5.5) we then have

v o 2 ≤ C β h 2 ,
and hence by (8.41)

v o 2 ≤ Cβ -1 f 2 + Cδβ -1/
/2 ≤ β 1/3 J -2/3 m ℜλ ≤ Υ or -3r ≤ β 1/3 J -2/3 m ℜλ ≤ -ϑ r 1 /2 .
Step 2: We prove (8.21) for sufficiently small α 0 and λ satisfying

β 1/3 J -2/3 m |ℜλ| < ϑ r
1 /2 and |ℑλ| ≤ 3r.

Let z ± ∈ C 2 ([-1, 1]) be given by (6.68). Note that

z ± ∞ = 1 ; z ± 1,2 ≤ C(1 + α 1/2 ) ≤ C(1 + α 1/2 0 β 1/6
g = f + iβU ′′ φ and v = -φ ′′ + α 2 φ , that -φ ′′ + α 2 φ = B + (ψ + -ṽ+ ) + B -(ψ --ṽ-) + Γ (-1,1) ( L β,R -βλ) -1 g , (8.52)
where

B ± = A ± (f + iβU ′′ φ) , and
g := γ R (f + iβU ′′ φ)(x) = (f + iβU ′′ φ)(x) x ∈ [-1, 1] 0 otherwise .
By (6.24) and the fact that U ′′ ∞ ≤ δ in this nearly Couette case, it holds that

|B ± | ≤ C(β -1/2 f 2 + δβ 1/3 log β φ ∞ ) . ( 8
B ± ṽ± 2 ≤ C(β -1 f 2 + δβ -1/6 log β φ ∞ ) . ( 8
( L β,R -βλ) -1 γ R (f + iβU ′′ φ) 2 ≤ C(β -2/3 f 2 + δβ 1/6 φ ∞ ) .
Combining the above with (8.52) and (8.54) then yields

w 0 2 ≤ C(β -2/3 f 2 + δβ 1/6 φ ∞ ) , (8.55)
where

w 0 = -φ ′′ + α 2 φ -B + ψ + -B -ψ -.
As in the proof of Proposition 7.3, we use the result of the previous step by considering λ + Υ 0 β -1/3 for a suitable value of Υ 0 . We choose Υ 0 such that

J 2/3 m ϑ r 1 2 ≤ β 1/3 ℜλ + Υ 0 ≤ J 2/3 m Υ . (8.56)
We now write

B D λ+Υ 0 β -1/3 ,α φ = f -Υ 0 β 2/3 (w 0 + B + ψ + + B -ψ -) ,
and introduce the following decomposition of φ

φ = χ 0 -Υ 0 β 2/3 (B + χ + + B -χ -) , (8.57)
where

χ 0 = (B D λ+Υ 0 β -1/3 ,α ) -1 (f -Υ 0 β 2/3 w 0 ) and χ ± = (B D λ+Υ 0 β -1/3 ,α ) -1 ψ ± .
For convenience we set λ = λ + Υ 0 β -1/3 .

We may now apply (8.21) (with λ replaced by λ) to obtain, with the aid of (8.55

) that χ 0 2 ≤ C(β -5/6 f 2 + δ φ ∞ ) . (8.58)
Estimate of χ ± .

We seek an estimate of χ + 1,2 . To this end we repeat the same procedure applied in step 1. For convenience of notation we consider only χ + in the following. The same estimates for χ -can be obtained in a similar manner. Let then

w+ = -χ ′′ + + α 2 χ + + U ′′ U + i λ χ + , and
H + := - d 2 dx 2 + iβ(U + i λ) w+ . (8.59)
It can be easily verified that

H + = ψ + + U ′′ χ + U + i λ ′′ .
and that ζ ± , w+ = 0. Consequently, we can use Lemma 6.2 with λ replaced by λ. With the notation ψ ± = ψ ± (λ), ψ+ = ψ ± ( λ) ,and v± = ṽ± ( λ) (as defined in (6.12)) Lemma 6.2 yields

w+ = B+ ( ψ+ -v+ ) + B-( ψ--v-) + ( L β,R -β λ) -1 H+ ,
where B± = A ± (H + ) and

H+ (x) = H + (x) x ∈ (-1, 1) 0 |x| ≥ 1 .
By (8.42) with χ + instead of φ and ψ + replacing f , it holds that

U ′′ χ + U + i λ ′′ 2 ≤ C β -5/12 [1 + |λ ± | 1/4 β 1 12 ] + δβ 5/6 χ + 1,2 .
Combining the above with (8.65) leads to

A -1 λ,α (U+i λ)Γ [-1,1] ( L β,R -β λ) -1 U ′′ χ + U + i λ ′′ 1,2 ≤ C β -5/4 [1+|λ ± | 1/4 β 1 12 ]+δ χ + 1,2 ,
which, together with (8.64), yields

φo 1,2 ≤ C β -7/6 + δ χ + 1,2 .
Combining the above with (8.62) and (8.63) yields, for δ > 0 small enough,

χ + 2 ≤ Cβ -7/6 .
In a similar manner we obtain

χ -2 ≤ Cβ -7/6 .
Substituting the above, (8.58), and (8.53) into (8.57) yields that (8.21) holds for

|ℜλ| ≤ β -1/3 J 2/3 m ϑ r 1 /2 .
Step 3: We prove (8.21) for λ ∈ C satisfying ℜλ ≤ -3r. As f = B D λ,α,β φ and

-ℜ φ, B D λ,α,β φ = φ ′′ 2 2 -(βµ -α 2 ) φ ′ 2 2 -βµα 2 φ 2 2 -βℑ U ′ φ, φ ′ , we can conclude that β(3r φ ′ 2 2 -U ′ ∞ φ ′ 2 φ 2 ) ≤ φ 2 f 2 . Since φ ∈ H 1 0 (-1, 1) it holds that φ ′ 2 ≥ π 2 φ 2 ,
and hence (recall that U ′ ∞ < r and 3r > 3 > 2/π) we can conclude that there exists C > 0 such that φ

′ 2 ≤ C β f 2 , (8.66)
completing, thereby, the proof of (8.21) for ℜλ ≤ -3r.

Step 4: We prove (8.21) for λ ∈ C satisfying |ℑλ| ≥ 3r.

An integration by parts yields

ℑ φ, B D λ,α,β φ = βν φ ′ 2 2 + α 2 φ 2 2 -β Uφ ′ , φ ′ + α 2 Uφ, φ + ℜ U ′ φ, φ ′ + U ′′ φ, φ ,
and as

ℜ U ′ φ, φ ′ = - 1 2 U ′′ φ, φ , we obtain that β |ν| -U ∞ - 1 2 U ′′ ∞ φ ′ 2 2 + α 2 β(|ν| -U ∞ ) φ 2 2 ≤ φ 2 f 2 . As |ν| -U ∞ - 1 2 U ′′ ∞ ≥ |ν| - 3 2 r ,
we establish (8.66) whenever |ν| ≥ 3r. The proposition is proved.

8.2.2 The case α ≥ α # β 1/6
We separately treat the case # = D and the case # = S.

Proposition 8.4. For any r > 1 and any κ > 0 there exist β 0 > 0, α D > 0 and C such that for all β ≥ β 0 and U ∈ S r sup

ℜλ≤β -1/3 (μ 0 (β -1/3 α)-κ) α D β 1/6 ≤α) (1 + α) (B D λ,α,β ) -1 + d dx (B D λ,α,β ) -1 ≤ C β -5/6 . (8.67) Proof. Let z ± ∈ C 2 ([-1, 1
]) be given by (6.68). By (6.75), (7.9), and (8.51), we have

-φ ′′ + θ 2 β 2/3 φ 2 ≤ C(β 1/3 φ 2 + β -2/3 f 2 ) , (8.68)
where θ = αβ -1/3 . Hence,

φ ′ 2 2 + θ 2 β 2/3 φ 2 2 = -φ ′′ + θ 2 β 2/3 φ, φ ≤ C(β 1/3 φ 2 2 + β -2/3 f 2 φ 2 ) . (8.69)
As θ ≥ α D β -1/6 , we obtain that for sufficiently large α D and β,

φ 2 ≤ C θ 2 β 4/3 f 2 , which implies α φ 2 ≤ C θβ f 2 ≤ C α D β 5 6 f 2 .
Returning to (8.69), we get

φ ′ 2 2 ≤ β -2 3 f 2 φ 2 ≤ C θ 2 β 2 f 2 2 , hence φ ′ 2 ≤ C θβ f 2 ≤ C α D β 5 6
f 2 Remark 8.5. Using the definition of μ0 from (6.76) yields that

μ0 (β -1/3 α)β 2/3 + α 2 = β 2/3 min J 2/3 + µ 0 (J -1/3 + θ) + θ 2 , J 2/3 -µ 0 (J -1/3 - θ) + θ 2 , where θ = β -1/3 α. Let θ ± = J -1/3 ± θ.
Using the definition of μm from (6.56) we then conclude

J 2/3 ± µ 0 (J -1/3 ± θ) + θ 2 = J 2/3 ± (µ 0 (θ ± ) + θ 2 ± ) ≥ J 2/3 m μm . Consequently we obtain that μ0 (β -1/3 α)β 2/3 + α 2 ≥ β 2/3 J 2/3
m μm . By the foregoing discussion, we may conclude from (8.67) that sup

ℜλ≤β -1/3 (J 2/3 m μm-κ-β -2 3 α 2 ) α D β 1/6 ≤α (1 + α) (B D λ,α,β ) -1 + d dx (B D λ,α,β ) -1 ≤ C β 5/6 . (8.70)
A similar estimate holds true also for B S λ,α,β . Proposition 8.6. For any r > 1 and Υ < ℜν 1 there exist positive β 0 , α S and C such that for all β ≥ β 0 and U ∈ S r such that

sup ℜλ≤ΥJ 2/3 m β -1/3 α S β 1/6 ≤α (1 + α) (B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ Cβ -5/6 . (8.71)
Proof. To prove (8.71) we note that by (7.9) we have

φ ′′ -α 2 φ 2 ≤ C(β 1/3 φ 2 + β -2/3 f 2 ) .
Consequently,

φ ′ 2 2 + α 2 φ 2 2 ≤ C(β 1/3 φ 2 2 + β -2/3 f 2 φ 2 )
. from which (8.71) follows as in the D-case.

β -1/3 α1 ≤α≤ α0 β 1/3 (B # λ,α,β ) -1 + d dx (B # λ,α,β ) -1 ≤ Cβ -5/6 , (8.72)
and

sup ℜλ≤ΥJ 2/3 m β -1/3 d(ℑλ,[U (-1),U (1)])≥νβ -1/3 0≤α≤ α1 (B # λ,α,β ) -1 + d dx (B # λ,α,β ) -1 ≤ Cβ -5/6 . (8.73) Proof. Let f ∈ L 2 (-1, 1), φ ∈ D(B # λ,α,β ) and h satisfy B # λ,α,β φ = f and h = f + U ′′ φ U + iλ ′′ .
Proof of ( Proof of (8.72).

Case 1: |ℜλ| > J 2/3 m β -1/3 min(ϑ r 1 , ℜν 1 )/2. To prove (8.72) for # = D we repeat the same procedure used in the proof of Proposition 8.3 to establish (8.48) with δ = r (note that we always have δ 2 (U) ≤ r and recall that v 0 is defined by (8.44))

v o 2 ≤ C β f 2 + Cβ -1/6 φ 1,2 . (8.75)
In the case # = S we repeat the same steps as in the proof of Proposition 7.3 to establish (7.33) with δ = r. For convenience we use the notation v o instead of v (which is defined by (7.3)).

We now estimate φ

o = A -1 λ,α v o .
To this end we rewrite (4.27) in the form

ℜ φ o U -ν + iµ , v o ≥ 1 2 φ ′ o 2 2 + (α 2 + γ m (λ, U)) φ o 2 2 .
The left-hand-side can be bounded by (4.38) (for p = 2) to obtain

1 2 φ ′ o 2 2 + (α 2 + γ m (λ, U)) φ o 2 ≤ Cβ 1/6 φ o ∞ v o 2 .
Since by Lemma 4.8, γ m (λ, U) ≥ γ 0 > -∞ we pick α ∈ R such that

α 2 ≥ sup(1, -2γ 0 ) . Then we write α φ o 2 ∞ ≤ α φ ′ o 2 φ o 2 ≤ 1 2 ( φ ′ o 2 2 + α 2 φ o 2 2 ) ≤ Cβ 1/6 φ o ∞ v o 2 . ( 8
φ o 1,2 ≤ Ĉ √ α β -5/6 f 2 + Ĉ √ α φ 1,2 . (8.78)
In the case # = S we have φ 0 = φ and hence (8.72) immediately follows. In the case # = D we continue as in the proof of Proposition 8.3 to establish (8.47) with δ = r, or explicitly,

φ ± 1,2 ≤ C(β -1 f 2 + β -1/6 φ 1,2 ) .
The above, combined with (8.78) and (8.45) yields for α ≥ α1 with α1 large enough and β ≥ β 0 with β 0 large enough

φ 1,2 ≤ C 1 √ α + β -1/6 β -5/6 f 2 , (8.79)
from which (8.72) readily follows.

Case

2: |ℜλ| ≤ min(ℜν 1 , ϑ r 1 ) 2 -1 J 2 3 m β -1 3 .
For # = S we use (7.36) for δ = r, i.e.,

φ ′′ -α 2 φ 2 ≤ C(β -2/3 f 2 + rβ 1/6 φ ∞ ) . (8.80)
Then, as

B S λ+sβ -1/3 ,α,β φ = f -sβ 2/3 (φ ′′ -α 2 φ
) . we may use (8.78), having in mind that φ 0 = φ, to obtain that

φ 1,2 ≤ Ĉ √ α β -5/6 ( f 2 + β 2/3 φ ′′ -α 2 φ 2 ) .
Substituting (8.80) into the above yields (8.72).

For # = D we first obtain (8.55) with δ = r, which implies

w 0 2 ≤ C(β -2/3 f 2 + rβ 1/6 φ ∞ ) , (8.81)
where

w 0 = -φ ′′ + α 2 φ -B + ψ + -B -ψ -, and
B ± = A ± (f + iβU ′′ φ) . As B D λ+Υ 0 β -1/3 ,α,β φ = f -Υ 0 β 2/3 (w 0 + B + ψ + + B -ψ -)
, where Υ 0 satisfies (8.56), we can write as in (8.57) that

φ = χ 0 -Υ 0 β 2/3 (B + χ + + B -χ -) , (8.82)
where

χ 0 = (B D λ+Υ 0 β -1/3 ,α,β ) -1 (f -Υ 0 β 2/3 w 0 ) and χ ± = (B D λ+Υ 0 β -1/3 ,α ) -1 ψ ± .
The estimation of χ 0 can be done with the aid of (8.79) and (8.81), yielding

χ 0 1,2 ≤ C 1 √ α + β -1/6 β -5/6 ( f 2 + φ ∞ ) . (8.83)
To estimate χ + (or χ -) we write

χ + = χ 1 + + χ 2 + ,
where

χ 1 + = A -1 λ,α (U + i λ)Γ [-1,1] ( L β,R -β λ) -1 U ′′ χ + U + i λ ′′ . As U ′′ χ + U + i λ ′′ 2 ≤ C β -1/3 + β 5/6 χ + 1,2 ,
we can conclude from (5.4), (5.5), and (8.77) that

χ 1 + 1,2 ≤ C √ α (β -7/6 + χ + 1,2 ) .
The estimation of χ 2 + in the proof of Proposition 8.3 does not involve δ at all, but only r and hence we can conclude by (8.63) and (8.64) that

χ 2 + 1,2 ≤ C(β -7/6 + β -1/6 χ + 1,2 ) .
Consequently we obtain that for sufficiently large α χ + 1,2 ≤ Cβ -7/6 .

A similar estimate holds for χ -, and hence we can conclude (8.72) from (8.83), (8.82), and (8.53).

8.3.2

The case U ′′ = 0 Lemma 8.8. Let r > 1 and δ ∈ (0, 1 2 ]. Then, there exist β 0 > 0, Υ > 0, and C > 0 such that for all β ≥ β 0 and U ∈ S r satisfying (2.29), it holds that

sup ℜλ≤ΥJ 2 3 m β -1/3 0≤α≤ α1 (B D λ,α,β ) -1 + d dx (B D λ,α,β ) -1 ≤ C β 1/2 - δ , (8.84)
where α1 is the same as in Proposition 8.7.

Proof.

Let φ ∈ D(B D λ,α,β ), α ≤ α1 , f = B D λ,α,β φ and v D ∈ H 2 (-1, +1) defined by v D = A λ,α φ + (U + iλ)[φ ′′ (1) ψ+ + φ ′′ (-1) ψ-] , (8.85)
where ψ± = ψ ± Θ ± /ψ ± (±1) (8.86) in which ψ ± is defined in (6.8) and Θ ± (x) = 1η(1 ∓ x), with η given by (5.14). With u(x, λ) given by (A.4), we have

ψ± (x, λ) = u (J ± β) 1/3 (1 ∓ x), iJ -2/3 ± β 1/3 λ ± . (8.87)
Hence we may thus use all the estimates derived for u in the proof of Proposition A.1.

Note that v D ∈ H 1 0 (-1, 1) and hence we may introduce

g D := (L D β -βλ)v D .
We have

g D = (U + iλ)(-f + φ ′′ (1)ĝ + + φ ′′ (-1)ĝ -) -(U ′′ φ) ′′ -2U ′ ṽ′ D -U ′′ ṽD , (8.88) wherein ĝ± = - d 2 dx 2 + iβU -βλ ψ± , and ṽD = v D -U ′′ φ U + iλ = -φ ′′ + α 2 φ + φ ′′ (1) ψ+ + φ ′′ (-1) ψ-.
We note that

(L D β -βλ)ṽ D -iβU ′′ φ = -f + φ ′′ (1)ĝ + + φ ′′ (-1)ĝ -. (8.89)
As in the proof of Proposition 7.1 (see in particular (7.7)) we can integrate by parts to obtain

-ℜ (U ′′ ) -1 ṽD , (L D β -βλ)ṽ D -iβU ′′ φ = (U ′′ ) -1/2 ṽ′ D 2 2 + + ℜ (U ′′ ) -1 ′ ṽD , ṽ′ D -βµ ṽD 2 2 + βℜ φ ′′ (1) ψ+ + φ ′′ (-1) ψ-, iφ . (8.90)
We begin the estimation by obtaining a bound for the last term on the right-handside.

Estimate of βℜ φ ′′ (1) ψ+ + φ ′′ (-1) ψ-, iφ .

We first write

φ(x) = 1 x (ξ-x)φ ′′ (ξ) dξ = φ ′′ (1) 1 x (ξ-x) ψ+ (ξ) dξ+ 1 x (ξ-x)[φ ′′ (ξ)-φ ′′ (1) ψ+ (ξ)] dξ . Let w + (x) = 1 x (ξ -x) ψ+ (ξ) dξ .
It can be easily verified that w ′ + (±1) = 0 and w + (1) = 0 . Hence,

ℜ φ ′′ (1) ψ+ , iφ ′′ (1)w + = |φ ′′ (1)| 2 ℜ w ′′ + , iw + = 0 .
We then obtain for x ∈ (-1, +1), using the fact that ψ+ ψ-≡ 0 ,

1 x (ξ -x)[φ ′′ (ξ) -φ ′′ (1) ψ+ (ξ)] dξ = 1 x (ξ -x)[ṽ D (ξ) + α 2 φ(ξ)] dξ ≤ C( ṽ′ D 2 + α 2 φ ′ 2 )(1 -x) 5/2 .
We thus get, as α ≤ α1 ,

|ℜ φ ′′ (1) ψ+ , iφ | ≤ |φ ′′ (1)|( ṽ′ D 2 + α2 1 φ ′ 2 ) (1 -x) 5/2 ψ+ 1 .
Using (8.87), (A.32), translation and dilation (see the proof of (6.15)-(6.16)) we may conclude that

(1 -x) s ψ+ 1 ≤ Cβ -(s+1)/3 [1 + |λ + | 1/2 β 1/6 ] -(s+1) , ∀s ≤ 3 ,
and since α ≤ α1 we can determine that

β| ℜ φ ′′ (1) ψ+ , iφ | ≤ C [1 + |λ ± |β 1/3 ] -7/4 β -1/6 |φ ′′ (1)|( ṽ′ D 2 + φ ′ 2 ) . ( 8
D 2 2 ≤ C ṽD 2 ( f 2 + |φ ′′ (1)| ĝ+ 2 + |φ ′′ (-1)| ĝ-2 ) + CΥβ 2/3 ṽD 2 2 + C[1 + |λ ± |β 1/3 ] -2 β -1/6 |φ ′′ (1)|( v ′ D 2 + φ ′ 2 ) . (8.92)
By (8.53), which is valid here with δ = r and Remark 6.3, we have that (8.99)

|φ ′′ (±1)| ≤ C[1 + |λ ± | 1/2 β 1/6 ](β -1/6 f 2 + β 1/3 log β φ ∞ ) . ( 8
Substituting the above into (8.97) yields, with the aid of (8.98)

g D 2 ≤ C( f 2 + (Υ 1/2 β 1/2 + β 1/6 log β) φ D 1,2 ) .
By (4.24) and (4.25) we then have for any q > 1 and p > 2

g D 2 ≤ C f 2 + Υ 1/2-1/p β 3p+2 6p
v D p + β 1/6 log β( v ′ D q + v D ∞ ) . (8.100) By (7.6) there exists C > 0 such that for all p > 2 (including p = ∞) it holds that

v D p ≤ Cβ -3p+2 6p g D 2 .
(8.101)

Similarly, by (5.10), for all 1 < q < 2, there exists C q > 0 such that We can now conclude. Proposition 8.9. Let r > 1 and δ > 0. Then, there exist β 0 > 0, Υ > 0, and C > 0 , such that for all β ≥ β 0 and U ∈ S r satisfying (2.29), it holds that sup

v ′ D q ≤ C q β -2+q
ℜλ+β -1 α 2 ≤ΥJ 2/3 m β -1/3 0<α (B D λ,α,β ) -1 + d dx (B D λ,α,β ) -1 ≤ C β 1/2- δ , (8.103)
The proof follows immediately by combining (8.70), (8.72), and (8.84) for sufficiently small valeue of Υ > 0.

Semigroup estimates

In this section we prove Theorems 2.14 and 2.15. To estimate (T ♯ P (U, ǫ, L) -Λ) -1 we seek therefore a bound for (P ♯ Λ,ǫ ) -1 , for Λ in a suitable region of C. We later derive the properties of the semi-group e -tT ♯ P from these resolvent estimates.

Preliminaries

Let Λ0 ∈ R and ǫ > 0. We seek a bound for the L(L 2 per , H 1 per ) norm of (P # Λ,ǫ ) -1 in the domain in ℜΛ ≤ Λ 0 with Λ 0 = ǫ Λ0 . Recall that P ♯ Λ,ǫ depends on L > 0 through the periodicity condition appearing in the definition of its domain.

As in Section 3 we can rewrite (3.11), which provides an L(L 2 per ) bound for (P ♯ Λ,ǫ ) -1 to the following L(L 2 per , H 1 per ) estimate Then there exists β0 (Υ) such that for all β ≥ β0 (Υ) we have S 1 (β, Λ0 ) ⊂ S 2 (β, Υ). By (7.1) we have that for sufficiently small Υ > 0 there exists β ′ 0 (Υ) > 0 and C > 0 such that sup (ℜλ,α)∈S 2 (β,Υ)

[ (1 + α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β -1/3+ δ , (9.8)
for all β ≥ β ′ 0 (Υ). For such sufficiently small Υ we may then write, for all β ≥ β0 := max( β0 (Υ), β ′ 0 (Υ)), that sup (9.9) On the other hand, it follows from (8.71) in a similar manner that for any Λ0 > 0 there exist α S > 0, β ′′ 0 > 0, and C > 0, such that for all β ≥ β ′′ 0 we have sup Combining the above with (9.7) for α M = α S yields that for any δ ∈ (0, 1 3 ) and Λ0 > 0 there exists C > 0 and ǫ 0 > 0 such that for all L > 0 and ǫ > 0 satisfying Lǫ ≤ ǫ 0 we have sup Hence we obtain that for any Λ0 and δ ∈ (0, 1 3 ) there exists ǫ 0 > 0 and C > 0 such that for all L > 0, ǫ > 0 satisfying 0 < Lǫ ≤ ǫ 0 it holds that To estimate the resolvent for Λ ∈ C satisfying -U ′ ∞ /2ǫ ≤ ℜΛ ≤ ǫ(π 2 /4 -δ), we introduce ω = -U ′ ∞ /2ǫ and use the resolvent identity (T S P -Λ) -1 = (T S P + ω + iℑΛ) -1 + (ω + ℜΛ)(T S P -Λ) -1 (T S P + ω + iℑΛ) -1 , (9.17 We now apply (9.21) to S(t) = e -t T S P (U,ǫ,L) . By (2.12) it follows that (9.19) holds for M = 1 and ω = -U ′ ∞ /2. By (9.16) and (9.18) we have that (9.20) holds for ω = ǫ(π 2 /4 -δ) and r(ω) = ǫ 2 /C. Consequently by (9.21) , we obtain that for any δ, there exist C > 0 and ǫ 0 > 0 such that for all ǫ ∈ (0, 1] and L > 0 satisfying 0 < Lǫ ≤ ǫ 0 we have e -t T S P (U,ǫ,L) ≤ C ǫ 2 e -ǫt (π 2 /4-δ) . This completes the proof of (2.30a). Consequently, as above, we may apply (I -Π) to the resolvent identity (9.17) to establish that, for ǫ ∈ (0, 1], L > 0 such that Lǫ ≤ ǫ 0 , sup ℜΛ≤ǫΥβ 1 (ǫ,L) 2/3

(I -Π)(T S P (U, ǫ, L) -Λ) -1 ≤ CL 1/3-δ ǫ -5/3-δ . (9.28)

As by (2.12) e -t T S P (U,ǫ,L) (I -Π) ≤ e t U ′ ∞/2 , We may apply, as above, [19, Proposition 2.1] to S(t) = e -tT S P (I -Π) to establish (2.30b).

Proof of Part 2. The proof of part 2 is obtained in the same manner. For the first statement leading to (2.30a), we replace Proposition 9.1 by the following result Proposition 9.3. For any 0 < Λ0 < π 2 /4 and C > 2/π + 4/π 2 there exists ǫ 0 > 0 and δ > 0, such that for all positive ǫ and L for which 0 < Lǫ ≤ ǫ 0 and U ∈ S r satisfying δ 2 (U) < δ , it holds that Proof. As the estimate (9.12) of B S 0 ( Λ0 ) is unaffected by the change of U in this part, we have only to prove that (9.11) still holds under the new assumptions. Recall that (9.11) follows from (9.10), which remains valid in this part, and (9.9) which needs to be remanded to fit this case. To this end we use (7.23) The proof of (2.30a) proceeds from here in the same manner as in Part 1.

We may conclude (2.30b) by following, once again, the same steps as those used in the proof of part 1 of the theorem. We use again (7.23) to establish, for Υ < J

Proof of Theorem 2.15

Since the proof is similar to the proof of Theorem 2.14 we address here only it main ingredients.

Proof of Part 1 For the first part of the Theorem we use (8.70) and (8.103) to establish that for any Λ0 and δ ∈ (0, 1 3 ) there exists ǫ 0 > 0 and C > 0 such that for all positive L and ǫ satisfying 0 < Lǫ ≤ ǫ 0 we have B D * ( Λ0 , ǫ, L) ≤ C β 1 (ǫ, L) -1/3+ δ , (9.33)

which is similar to (9.11) in the case # = S.

To bound B D 0 ( Λ0 ) we use integration by parts to establish for any φ ∈ D(B D Λ )

-ℜ B D Λ φ, φ = ||φ ′′ || 2 -ℜ Λ ||φ ′ || 2 .
As φ ′ ∈ H 1 0 (-1, 1) we obtain, for ℜ Λ ≤ π 2 4 ,

-ℜ B D Λ φ, φ ≥ π 2 4 -ℜ Λ ||φ ′ || 2 .
Upon the above inequality we use the fact that φ ∈ H 1 0 to obtain the equivalent of (9.12) for all 0 < Λ0 < π 2 /4 , In a similar manner we obtain the estimates for k ∈ {3, 4} . We note in particular that x 4 w 1 2 + (x 4 w 1 ) ′ 2 ≤ C |λ| 9/4 , and hence, by Sobolev embeddings,

B D 0 ( Λ0 ) = sup ℜ Λ≤ Λ0 (B D Λ ) -1 + d dx (B D Λ ) -1 ≤ 2 π + 4 π 2 1 (π 2 /4 -Λ0 ) . ( 9
x 4 w 1 ∞ ≤ C |λ| 9/4 .
Combining the above with (A.6) and (A.20) yields, for k ∈ [0, 4] ,

x k u ∞ ≤ C |λ| k 2
.

(A.30)

Weighted L 1 estimates. Recall that in deriving (A.2) we needed to establish that, for k = 0, . . . , 4, it holds that

x k u L 2 ≤ C < λ > -2k+1 4 . (A.31)
By interpolation it is enough to treat the case when s is an integer. Then we use Hölder inequality and (A.31) for k = s and k = s + 1 to obtain, for s ≤ 3,

x s u L 1 ≤ [x + i < λ > -1 2 ]x s u 2 [x + i < λ > -1 2 ] -1 2 ≤ x s+1 u 2 + < λ > -1 2 x s u 2 [x + i < λ > -1 2 ] -1 2 ≤ C < λ > -s+1 2 .
(A.32)

We then recover (A.3) by using (A.16).

A.2 Definition of A 0 and the locus of its zeroes.

Let A 0 : C → C be given (see (6.9)) by the holomorphic extension to C of R ∋ z → A 0 (z) = e iπ/6 +∞ z

Ai (e iπ/6 t) dt .

To use the results of Wasow [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF] (see [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF]Eq. (39)]) and justify this holomorphic extension we observe the following relation Combining the above with (A.45) and (A.46) yields (A.44).

. 4 )

 4 Depending on the boundary condition (no-slip or fixed traction force), U should satisfy an inhomogeneous Dirichlet condition U(±1) = v b (±1) or an inhomogeneous Neumann condition U ′ (±1) = s b (±1).

  ([-1, 1]) satisfying U ′ = 0 in [-1, 1], of two different types: nearly Couette flows, so that sup x∈[-1,1] |U ′′ (x)| + sup x∈[-1,1] |U ′′′ (x)| ≪ min x∈[-1,1]

Lemma 2 . 2 .-

 22 H 0 div and H curl are orthogonal subspaces of H andH = H curl ⊕ H 0 div . Proof. Let u ∈ H. Let further φ c ∈ H 1 loc ( D) and H 1 loc ( D) denote the weak solutions of  c (•, •) = φ c (• + L, •) + A L in D , ∆φ d = curl u in D φ d = 0 on ∂D φ d (•, •) = φ d (• + L, •) in D .

  .70) Substituting (4.69) and (4.68) into (4.70) yields

Proposition 6 . 10 .

 610 Let µ 0 (θ) = inf λ∈σ(L θ ) ℜλ. Then, lim θ→+∞ µ 0 (θ) = ℜν 1 , (6.62)where ν 1 the left most eigenvalue of L + the Dirichlet realization of L in R + .

. 21 ) 2 . 1 :- 2 /3 m β 1 / 3 3 m β - 1 /3 ϑ r 1 / 2 ,

 21212133112 Proof. Let φ ∈ D(B D λ,α,β ) and f ∈ L 2 (-1, 1) satisfy B D λ,α,β φ = f . Let λ = µ + iν. Let further v be given by (7.3), and set ṽ = (U + iλ)-1 v. Without any loss of generality we select Υ > ϑ r 1 /Step We prove (8.21) for λ ∈ C satisfying either ϑ r 1 /2 ≤ J ℜλ ≤ Υ or -3r ≤ ℜλ ≤ -J 2/and |ℑλ| ≤ 3r. Let ζ ± ∈ C 2 ([-1, 1]) satisfy the problem

8. 3 Proposition 8 . 7 .

 387 Strictly convex/concave flows 8.3.1 Large α or |ℑλ| If we assume large α or |ν| we may obtain resolvent estimates for all U ∈ C 4 ([-1, 1]), without the necessity to assume any further restrictions on U as in the nearly Couette case or in the case U ′′ = 0 . Let r > 1, Υ < min(ℜν 1 , ϑ r 1 ), and # ∈ {S, D}. Then, there exist α0 > 0, α1 > 0, ν > 0, β 0 > 0, and C such that, for all β ≥ β 0 and U ∈ S r satisfying (4.23)-(4.25), it holds that sup ℜλ≤ΥJ 2/3 m

φ

  = φ D + φ+ + φ-,(8.98) whereφ D = A -1 λ,α v D ; φ± = A -1 λ,α [U + iλ]φ ′′ (±1) ψ± . By (8.1) and (8.93) we have φ± 1,2 ≤ C[1 + |λ ± | 1/2 β 1/6 ] -1/2 (β -2/3 f 2 + β -1/6 log β φ 1,2 ) ,and hence, by (8.98),φ± 1,2 ≤ C[1 + |λ ± | 1/2 β 1/6 ] -1/2 (β -2/3 f 2 + β -1/6 log β φ D 1,2 ) .

6q g D 2 .

 2 (8.102) Substituting(8.101) and (8.102) into (8.100) yields, choosing Υ > 0 small enough and β 0 large enough, the existence of C > 0 such that for β ≥ β 0g D 2 ≤ C f 2 .Using (4.24) and (4.25) once again upon (8.101) and (8.102) and the above inequality readily verifies (8.84).

For

  ♯ ∈ {S, D}, let F = (F 1 , F 2 ) ∈ H 1 loc ( D, R 2 ) ∩ H, Λ ∈ C, and u ∈ W 0 ♯ satisfy (T ♯ P (U, ǫ, L) -Λ)u = P F .(9.1)Recall that by writing u = ∇ ⊥ ψ for some ψ ∈ D(P ♯ Λ,ǫ ) we have established in (3.1) thatP ♯ Λ,ǫ ψ = curl F in D .(9.2)This gives, for Λ ∈ ρ(T ♯ P ), u = ∇ ⊥ (P ♯ Λ,ǫ ) -1 curl F = (T ♯ P (U, ǫ, L) -Λ) -1 P F . (9.3)

1 perS 1

 11 ) ≤ ǫ -1 max(B ♯ * ( Λ0 , ǫ, L), B ♯ 0 ( Λ0 )) , λ,α,β ) -1 , (9.5)in which β 1 (ǫ, L) = (2π)/(Lǫ), andB ♯ 0 ( Λ0 ) = sup ℜ Λ≤ Λ0 1 (P S Λ,ǫ ) -1 estimatesEstimation of B S * ( Λ0 , ǫ, L). We begin by showing that for all δ ∈ (0, 1/3), Λ0 > 0 and α M > 0, there exist β0 > 0 and C > 0, such that, for all β ≥ β0 sup0≤α≤α M β 1/6 ℜλ≤β -1 Λ0 (1 + α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β -1/3+ δ . (β, Λ0 ) = {(ℜλ, α) ∈ R 2 | ℜλ ≤ β -1 Λ0 ; 0 ≤ α ≤ α M β 1/6 } , and S 2 (β, Υ) = {(ℜλ, α) ∈ R 2 | ℜλ ≤ β -1/3 Υ ; 0 ≤ α ≤ α M β 1/6 } .

(

  ℜλ,α)∈S 1 (β, Λ0 )[ (1 + α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β -1/3+ δ ,which is precisely (9.7). Consequently, for all L > 0, ǫ > 0 such that Lǫ ≤ 2π β0 we have supβ≥β 1 (ǫ,L) sup 0≤α≤α M β 1/6 ℜλ≤β -1 Λ0 (1 + α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β 1 (ǫ, L) -1/3+ δ .

α S β 1/6 ≤α ℜλ≤β - 1 Λ0( 1 +

 11 α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β -5/6 . (9.10)As a result, for all L > 0 and ǫ > 0 such that Lǫ ≤ ǫ 0 := 2π/(max(β ′′ 0 , β0 )) (implying that β 1 (ǫ, L) ≥ max(β ′′ 0 , β0 )) it holds that supβ≥β 1 (ǫ,L) sup α S β 1/6 ≤α ℜλ≤β -1 Λ0 (1 + α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β 1 (ǫ, L) -5/6 .

( 1 +

 1 α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 ≤ C β 1 (ǫ, L) -1/3+ δ .We now observe thatB S * ( Λ0 , ǫ, L) ≤ sup β≥β 1 (ǫ,L) sup 0≤α ℜλ≤β -1 Λ0 (1 + α)(B S λ,α,β ) -1 + d dx (B S λ,α,β ) -1 .

4 - Λ d 2 dx 2 . 1 π 2 12 ) 9 . 1 .

 42121291 B S* ( Λ0 , ǫ, L) ≤ C β 1 (ǫ, L) -1/3+ δ . As for any φ ∈ D(B S Λ ) we have φ ′′ ∈ H 1 0 (-1, 1), we easily obtain thatd 2 dx 2 (B S Λ ) -1 ≤ /4 -ℜ Λ .Since D(B S Λ ) ⊂ H 1 0 (-1, 1) we obtain that for all 0 < Λ0 < π 2In conclusion, we have established the following Proposition For any 0 < Λ0 < π 2 /4 and C > 2/π + 4/π 2 there exists ǫ 0 > 0, such that for all positive L and ǫ for which 0 < Lǫ ≤ ǫ 0 and U ∈ S r satisfying (2.29), it holds that sup

≤

  (T S P + ω + iℑΛ) -1 L(H) +|ω + ℜΛ| (T S P -Λ) -1 L(H 1 per (D,C 2 ),H) (T S P + ω + iℑλ) -1 L(H,H 1 per (D,C 2 )). With the aid of (9.14) and (2.14) we then obtain that for each δ > 0 there exist C > 0 and ǫ 0 > 0 such that for all ǫ ∈ (0, 1], L > 0 satisfying 0< ǫL ≤ ǫ 0 sup -U ′ ∞/2-ǫ≤ℜΛ≤ǫ(π 2 /4-δ) (T S P (U, ǫ, L) -Λ) use [19, Proposition 2.1] (or[START_REF] Helffer | Spectral theory and its applications[END_REF] Proposition 13.31]), which we repeat here for the benefit of the reader Proposition 9.2. Let S(t) be a strongly continuous semigroup, defined on a Hilbert space H, which satisfies for some M ≥ 1 and ω ∈ R, S(t) ≤ M e -ωt .(9.19)Let -A : D(A) → H denote the generator of S(t). Suppose further that for some ω > ω there exists r(ω) > 0 such that sup ℜz≤ω (Az)

  λ,α,β ) -1 ≤ C β 1 (ǫ, L) -2/3 . (9.30)Combining the above with (9.10) yieldsB S * ( Λ0 , ǫ, L) ≤ C β 1 (ǫ, L) -2/3, which together with (9.4) yields (9.36).

  λ,α,β ) -1 ≤ C β 1 (ǫ, L) -2/3 , (9.31) Together with (8.71), (9.31) yields, as in Part 1, supℜΛ≤ǫΥβ 1 (ǫ,L) 2/3 (I -Π)(T S P (U, ǫ, L) -Λ) -1 ≤ CL 2/3 ǫ -4 3 .(9.32)from which (2.30b) is verified by following the same route used in Part 1.

. 34 ) 9 . 4 .

 3494 Combining the above with (9.33) and (9.4) we may conclude that Proposition For any 0 < Λ0 < π 2 /4 and C > 2/π + 4/π 2 there exists ǫ 0 > 0, such that for all positive L and ǫ > 0 for which 0 < Lǫ ≤ ǫ 0 and U ∈ S r satisfying (2.29), it holds thatsup ℜΛ≤ǫ Λ0 (P D Λ,ǫ ) -1 L(L 2 per ,H 1 per ) ≤ C ǫ(π 2 /4 -Λ0 ) .we use (A.28), (A.21), and (A.12) to obtainx 2 w 1 2 ≤ C |λ| 9/4 , (A.29)which can be used, together with (A.20) and (A.6) to obtain (A.2) for k = 2 .

Lemma A. 2 .0

 2 A 0 (z) = -ψ(-e iπ/6 z) . (A.[START_REF] Sjöstrand | Resolvent estimates for non-selfadjoint operators via semigroups. Around the research of Vladimir Maz[END_REF] where ψ is the holomorphic extension of the real functionR ∋ x → ψ(x) := Ai(-t)dt .(A.34)The case |λ| ≥ R.If |λ| ≥ R we may use (6.67) which reads[θ + (-λ) 1/2 ]F (λ, θ) Ai (e i2π/3 λ) -1 ≤ C |λ| -1/4 ,to obtain for R large enough with the aid of (A.1), (A.33), and (A.35) that,A 0 (iλ) F (λ, θ) ≤ A 0 (iλ) Ai (e i2π/3 λ) Ai (e i2π/3 λ) F (λ, θ) ≤ C θ + |λ| 1/2 |λ| 1/2 .

  Proposition 4.1. Let U satisfy Assumption 2.13 and ν ∈ [U(-1), U(1)] . Then A iν,α , whose domain is given by(3.15), is closed as an unbounded operator on L 2 (-1, +1) and the space

  Cβ -5/6 f 2 ,which is precisely(8.21) established in this step for all λ ∈ C such that |ℑλ| ≤ 3r, and either ϑ r 1

	6 φ 1,2 .	(8.48)
	By (4.25) (with p = 2), and the definition of φ o we obtain
	φ o 2 ≤ C(β -5/6 f 2 + δ φ 1,2 ) .	(8.49)
	Substituting (8.49) and (8.47) into (8.45) yields,for sufficiently small δ
	φ 1,2 ≤

  ) is still valid, and in view of (8.50) and (8.51), we can apply Lemma 6.2, assuming that α 0 is small enough, with ζ ± replaced by z ± , to obtain for

	) .	(8.50)
	By (3.10), (6.68), and two integrations by parts, it holds that
	z ± , -φ ′′ + α 2 φ = 0 .	(8.51)
	As (7.9

  .91)

	Estimate of ṽ′ D .
	Next we obtain from (8.90) and (8.91) that
	ṽ′

  By(5.21) and(8.89) we have thatṽD 2 ≤ C β 1/6 φ ∞ + β -2/3 [ f 2 + |φ ′′ (1)| ĝ+ 2 + |φ ′′ (-1)| ĝ-2 ] . ≤ C(β 1/6 φ ∞ + β -2/3 f 2 ) . ≤ C( f 2 + (Υ 1/2 β 1/2 + β 1/6 log β) φ 1,2) . (8.96) We now combine (8.88), (8.96), (8.95), (8.93), and (8.94) to obtain thatg D 2 ≤ C( f 2 + (Υ 1/2 β 1/2 + β 1/6 log β) φ 1,2 ) . (8.97)Proof of (8.84) We continue as in the proof of Proposition 8.3. We first write, in view of (8.85)

	Hence, by (8.94) and (8.93) we have that
	ṽD 2 Substituting the above into (8.95) yields
	ṽ′ D 2
		.93)
	Furthermore, by (6.18) we have that
	ĝ± 2 ≤ Cβ -1/6 [1 + |λ ± | 1/2 β 1/6 ] -1 ,	(8.94)

which, when substituted into (8.92), yields, with the aid of Sobolev's embeddings,

ṽ′ D 2 ≤ C( f 2 + β 1/6 log β φ 1,2 ) + Υ 1/2 β 1/3 ṽD 2 .

(8.95)

Estimate of ṽD .

The case k = 2 We briefly repeat the same argument as in the case k = 1. By (8.10) with k = 2 we may write (instead of using (8.11))

From which we obtain, with the aid of (8. [START_REF] Helffer | From resolvent bounds to semigroup bounds[END_REF]) and (8.15)

Substituting (8.20) into (8.9) with k = 2 then yields with the aid of (8.14)

Substituting the above, (8.20), and (8.14) into (8.11), with k = 3 yields

From (8.9) with k = 3 we then obtain

As in (8.18) and (8.19) we can now obtain that

Combining the above with (8.19) and (8.14) yields (8.12).

Step 2: We prove (8.5). We first observe that by interpolation (8.6) holds for any k ∈ [0, 2]. If the conditions of (4.9) are met, we may now obtain (8.5) from (4.19) as in the proof of (8.1).

Otherwise if the assumptions of Proposition 4.13 are met, we may conclude (8.1) from (4.88).

Nearly Couette flows

We begin by considering the case where the flow U is nearly linear, as in Subsection 7.3.

8.2.1

The case 0 ≤ α ≤ α0 β 1/3

Let B D λ,α,β be defined by (3.5). We can now state and prove Moreover, we have that

(8.60)

We now follow the arguments of the previous step with (χ + , ψ+ , H + , w+ ) respectively replacing (φ, f, h, ṽ). We then reach by the equivalent of (8.42)

from which, using (6.17), we get

Hence, by (8.60) for k = 0, we get

As above we set

where

H+ .

(8.62b) By (8.1) and (8.61) we have

Note here that as |λ -λ| = Υ 0 β -1/3 we have used that, for some C > 1,

Moreover, by (8.61), (6.14), and (4.25) (with p = 2) it holds that

Combining the above yields

Next, we estimate φo 1,2 . By (8.5) we have

Furthermore, by (5.5), (5.4), and (4.25), we have

Proof of (2.30b).

Let Π be defined by (2.19). We have (see (2.21)) that Πu = Π∇ ⊥ ψ = ∇ ⊥ pψ .

We also have from (9.2)

and from (9.3)

We can thus conclude, as in the proof of (2.30a) but with Λ0 being replaced by Υβ 1 (ǫ, L)

Consequently, we obtain, using the fact that β

We also note that (Ip)curl F = curl (I -Π)F .

As in the proof of (2.30a) we can use (9.8) and (9.10) to establish that for all δ ∈ (0, 1/3) there exist Υ, ǫ 0 > 0 and C such that for all positive L and ǫfor which 0

As in the proof of (2.30b) we need yet to establish a bound on ∇(I -Π)u 2 for Λ < - We may now proceed as in the proof of (2.30a) to establish (2.33a).

To prove (2.33b) we use (8.70) together with (8.103) once again to obtain

From the above, we can obtain as in the proof of (2.30b) that sup

We can now establish (2.33b) in the same manner as in the proof of (2.30b).

Proof of Part 2

As in the proof of Part 2 of Theorem 2.14 we establish from (9.34), (8.21), and (8.70) that Proposition 9.5. For any 0 < Λ0 < π 2 /4 and C > 2/π + 4/π 2 there exists ǫ 0 > 0 and δ > 0, such that for all ǫ ∈ (0, 1] and L > 0 for which 0 < Lǫ ≤ ǫ 0 and U ∈ S r satisfying δ 2 (U) < δ , it holds that

We then continue as in the proof of (2.33a).

To establish (2.33b) in Part 2 we use (8.21) and (8.70) to show that

Then we can continue in the same manner as in the proof of (2.33b) in the first part of the theorem.

Acknowledgments: Y. Almog was partially supported by NSF grant DMS-1613471. Both authors would like to thank Pierre Bolley and Nader Masmoudi for for some fruitful discussions with B. Helffer.

A Basic properties of the Airy function and Wasow's results on A 0

A.1 Airy function properties

In this subsection, we summarize some of the basic properties of Airy function Ai (z), and the generalized Airy function A 0 (z), that are being used throughout this work (see [START_REF] Abramowitz | Handbook of Mathematical Functions[END_REF] for details) and establish some new inequalities satisfied by these functions. We recall that Airy function is the unique solution of

on the line such that u(x) tends to 0 as x → +∞ and Ai (0) = 1/ 3

. Standard ODE theory shows that Airy function is entire and strictly decreasing on R + , but has an infinite number of zeros in R -. Airy function satisfies different asymptotic expansions as |z| → ∞ depending on arg z. We bring two of them here

The following moment estimates are needed in Subsection 6.1.

Proposition A.1. Let < λ >:= 1 + |λ| 2 . For every µ 0 > 0 there exists C > 0 such that, for any λ with ℜλ ≤ µ 0 , we have,

Proof.

If |λ| ≤ 3µ 0 then, by (A.1a), all the estimates of the proposition are satisfied for some C = C(µ 0 ). Hence, we can consider from now on the case where |λ| > 3µ 0 . Note that by interpolation, it is sufficient to consider k ∈ Z.

Proof of (A.2) for k = 0 . Let

which is well-defined for

since the denominator can vanish only when arg λ = π/3 . It can be easily verified that

Let further w = ue λ (x) with e λ (x) := e -(-λ) 1/2 x , (A. [START_REF] Almog | Superconductivity near the normal state in a half-plane under the action of a perpendicular electric current and an induced magnetic field[END_REF] where (-λ) 1 2 is well-defined for λ ∈ V(µ 0 ) as

Note that for λ ∈ V(µ 0 )

implying, for any µ 0 > 0, the existence of C(µ 0 ) > 0, such that, for all λ ∈ V(µ 0 ), it holds that

Substituting into (A.5) yields

where L + is associated with the differential operator -d 2 /dx 2 + ix and is defined on the domain [17, §5] that, for any μ ∈ R + , there exist ν 0 (μ) > 0 and C(μ) such that

(A.10)

In addition, we have

Denote by {ν n } ∞ n=1 the eigenvalues of L + , and recall that they are located on the ray arg λ = π/3 (see [3, §2.2]). Observing that V(µ 0 ) does not contain any eigenvalue (a consequence of the fact that arccos 1/3 > π/3) and combining the above with (A.10)-(A.11) yield the existence of C(µ 0 ) > 0 such that sup λ∈V(µ 0 )

Hence, for λ ∈ V(µ 0 ),

Using (A.7), we obtain that

and hence

By the above, (A.6), and (A.8), we thus have

To prove (A.2) for k = 0, we need to establish yet an upper bound for Ai (e 2i π 3 λ), for λ ∈ V(µ 0 ). This is an immediate consequence of (A.1a). We observe indeed that for any ǫ > 0, arg λ ∈ (-π 2 + ǫ, π 2ǫ) as |λ| → +∞ with ℜλ ≤ µ 0 . This implies that, for any ǫ > 0 , there exists λ 0 (ǫ, µ 0 ) > 0 such that for |λ| ≥ λ 0 (ǫ) and ℜλ ≤ µ 0 , arg λ exp 2iπ 3 ∈ (-5π 6ǫ, π 6 + ǫ). Consequently, for any µ 0 > 0, there exists a constant C(µ 0 ), such that for all λ ∈ V(µ 0 )

|Ai

Together with (A.15), (A.16) yields (A.2) for k = 0 and λ ∈ V(µ 0 ).

Proof of (A.2) for k = 1. We begin by deriving an estimate of xw. To this end we observe that

Following the same procedure applied in the previous proof, we need an estimate of w ′ . To achieve this end, we first observe that

which leads to the estimate

Implementing (A.13) and (A.14) leads to

We observe in addition that

Proceeding as in the proof of (A.2) for k = 0, we obtain (A.2) for k = 1 .

Proof of (A.2) for k = 2 . In this case the approximation of u by e λ is unsatisfactory. To improve it, in light of (A.9), we solve in H 2 (R + ) the problem

We look for f λ in the form f λ = p λ e λ which means that p λ must satisfy

We search for a polynomial solution. A simple computation leads to

We now write u = e λ + w = e λ + f λ + w 1 , (A.20)

to obtain

In order to prove (A.2) (k = 2) we observe first that

Hence, it remains necessary to obtain an estimate of x 2 w 1 in L 2 . Here we use the fact that (A.9) is similar to (A.21), the only difference being that the right-hand-side is given by -ixf λ instead of -ixe λ . Note that xf λ 2 is much smaller than xe λ 2 as |λ| → ∞. By (A.12), (A.21) and (A.22) (for ℓ = 1), we then have

which is significantly smaller than the bound provided by (A.14) for w. We continue as in the case k = 1. We first use the identity

to conclude with the aid of (A.21), (A.22), (A.23), and recalling that µ ≤ µ 0 ,

Then we write

Combining (A.24), (A.21), and (A.12) yields

From (A.26) and (A.23) we get in addition

Next, we write

to conclude from (A.25) and (A.26) that

(A.28)

Upon writing

It has been proved by Wasow in [35, Section 3] that the zeroes of ψ are all located in the sector | arg z| < π 6 .

Proposition A.3. The zeroes of ψ belongs to -π/6 < arg z < π/6 . Moreover ψ has no real zeroes.

Sketch of the proof. To establish that result is a combination of the argument principle and Rouché's. The change of arg ψ is estimated along the path Γ = Γ 1 ∪Γ 2 ∪Γ 3 where, for some R > 0

Since Ai (z) is real and positive for z ∈ R + it follows that arg ψ does not change along Γ 3 . Along Γ 2 one uses (A.1), with a bound on the remainder. Finally, to estimate ∆ arg ψ along Γ 1 one uses the power series of Ai (t), for 0 < |t| < 9 , and (A.1) for |t| > 9. The tails of the ensuing power series of ℜψ and ℑψ are Leibniz series with terms of alternating sign and decreasing moduli. Thus, one may truncate the series into finite sums, and the remainders can be easily estimated. Once the above procedure is applied, one can establish that |∆ arg ψ| < 2π and hence that ∆ arg ψ = 0 along Γ. Since ψ(z) = 0 ⇒ ψ(z) = 0, the first statement of the proposition follows.

The second statement is proved in [35, p. 199].

Corollary A.4. Let A 0 (iz) = 0. Then, π/6 < arg z < π/2 .

We continue with the following result stated in [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF] (Eq. ( 35)) which allows us to obtain additional information on the location of the zeroes of A 0 . It is also serves as a useful tool in some of the proofs in Subsection 6.1.

Lemma A.5. Let ψ(z) = ψ(-z). For any 0 < δ < π there exists C δ > 0 and r 0 ( δ) > 0 such that for all |z| > r 0 ( δ) in the sector | arg z| < π -δ it holds that

The proof is a rather standard application of the method of steepest descent method [START_REF] Miller | Applied asymptotic analysis[END_REF]Chapter 4] and is therefore being skipped.

A.3 Asymptotic of the zeroes

In Subsection 6.1 we also need to establish the following lemma about the asymptotic behavior of the zeroes of A 0 . A similar statement for ψ is made in [35, § 3] without a clear proof.

Proposition A.6. Let S 0 denote the set of points λ ∈ C satisfying A 0 (iλ) = 0. Then, for any R > 0, S 0 ∩ B(0, R) is finite and its cardinality tends to +∞ as R tends to +∞. In particular S 0 is non empty. Moreover, for any ǫ > 0, there exists R such that, for λ ∈ S 0 ∩ B(0, R) c ,

Proof.

We apply Jensen's formula [START_REF] Lang | Introduction to complex hyperbolic spaces[END_REF]Theorem 1.7] to ψ(z) = ψ(-z).

In the above N ψ (R, 0) is the Nevanlinna counting function,

where we have used the fact that all zeroes of ψ are simple. We recall from [START_REF] Wasow | On small disturbances of plane Couette flow[END_REF] that none of them is real, and all the zeroes of ψ′ = -Ai lie on the negative real axis. Note that if S 0 = ∅ then N ψ (R, 0) ≡ 0 for all R > 0. Recall from the definition that ψ(0) = 1/3 and hence if we show that the first term on the right-hand-side of (A.37) is unbounded as R → ∞, we may conclude the first statement of the proposition. It follows from (A.35) that for any C 0 ∈ (0, 8 9 ) there exist δ > 0 and r 0 > 0 such that for all R > r 0 we have

Indeed, using Lemma A.5, we eastablish the existence for any δ > 0 of r 0 ( δ) and C such that, for all R ≥ r 0 ( δ),

To estimate the integral for | arg z| ∈ (π -δ, π) we write, owing to the concavity of

(A.39) Since by (A.34) we have, for any 0 < δ < π/3 and R > R 0 ,

where R 0 is fixed, but sufficiently large so that Ai (se iθ ) obeys (A.1b) for all s > R 0 . For the first term on the right-hand side there exists C(R 0 ) > 0 such that

For the second term we use (A.1b) to obtain the rather crude estimate for R

Consequently, for every

This implies, by (A.39), the existence, for any 0

Combining the above with (A.38) yields, by fixing δ small enough, the existence of R 3 and Ĉ0 > 0 such that for all R > R 3

The proof shows that the above bound holds true for any Ĉ0 ∈ (0, 8 9 ) for R 3 ( Ĉ0 ) large enough. Hence, we get a lower bound for N ψ (R, 0) which implies the first statement of the proposition.

To prove (A.36) we notice that for any δ > 0, it holds by (A.35) that there exists r 0 > 0 such that ψ cannot have any zeroes, for |z| > r 0 in the sector | arg z| < π -δ. We then observe that A 0 (iλ) = -ψ(e 2iπ/3 z) .

We can now immediately draw the following conclusion.

Corollary A.7.

inf ℜS 0 = ϑ r 1 > 0 .

(A.41)

A.4 Normalized Airy functions

We complete the appendix with some corollaries of (A.2)-(A.3) and (A.35) needed in Subsection 6.1 and with other estimates needed in Subsection 6.3.

Proposition A.8. Let Ψ λ ∈ L 2 (R + ) be defined by

Then, for any δ1 > 0 there exists C > 0 such that for all ℜλ ≤ ϑ r 1 -δ1

for k ∈ [0, 4] , (A.43a)

Proof. Let |λ| > 3ϑ r 1 (the proof for |λ| ≤ 3ϑ r 1 follows by continuity as the denominator is bounded away from zero). As A 0 (iλ) = -ψ(e i2π/3 λ) we obtain from (A.35) that for any δ > 0 and | arg λ + 2π/3| < π

ℜ{(e i2π/3 λ) 3/2 } , which combined with (A.2) and (A.3) yields (A.43a,b) . The proof of (A.43c) follows from (A.30).

We also need, in Subsection 6.3 the following estimate Lemma A.9. Let κ > 0 and, for θ > 0, Υ(θ) = µ 0 (θ)κ where µ 0 (θ) := inf ℜσ(L θ ) .

Let F (λ, θ) be defined by (6.47c). There exists C(κ) > 0 such that, for all θ > 0, sup ℜλ≤Υ(θ)

A 0 (iλ) F (λ, θ) ≤ C(κ)(1 + θ) .

(A.44)

Proof. Let R > 0 be determined later.

The case |λ| < R.

Let θ 1 (κ) satisfy sup We may now use the fact that Υ(θ) < ℜν