On pseudospectral bound for non-selfadjoint operators and its ... - arXiv

On pseudospectral bound for non-selfadjoint operators and its application to stability of Kolmogorov flows arXiv:1710.05132v1 [math.AP] 14 Oct 2017

Slim Ibrahim∗

Yasunori Maekawa†

Nader Masmoudi‡

Abstract We study the stability of the Kolmogorov flows which are stationary solutions to the two-dimensional Navier-Stokes equations in the presence of the shear external force. We establish the linear stability estimate when the viscosity coefficient ν is sufficiently small, where 1 the enhanced dissipation is rigorously verified in the time scale O(ν− 2 ) for solutions to the linearized problem, which has been numerically conjectured and is much shorter than the usual viscous time scale O(ν−1 ). Our approach is based on the detailed analysis for the resolvent problem. We also provide the abstract framework which is applicable to the resolvent estimate for the Kolmogorov flows. Keywords Navier-Stokes equations · enhanced dissipation · nearly inviscid flows Mathematics Subject Classification (2010) 35Q30 · 35P15 · 47A10 · 76D09

1

Introduction

For nearly-inviscid fluids, turbulent phenomena often occur at transient time scales that are much smaller than the viscous time scale. Describing the fluid, by means of simple solutions, for such long transient times helps to understand turbulence. This is of course of great interest both physically and mathematically. But finding such solutions and estimating their basin of attraction are in general not easy tasks both experimentally and theoretically. To investigate this phenomena let us consider the two dimensional incompressible Navier-Stokes equations in the domain M = T2 or M = R2 , ∂t U + (U · ∇)U + ∇P = ν∆U + F ,

t > 0,

(x, y) ∈ M .

(1.1)

Here U = (U1 , U2 ) : M2 × (0, ∞) → R2 is the velocity field of a fluid, P : M2 × (0, ∞) → R is the pressure field, and ν > 0 is the viscosity coefficient. The vector field F describes a given external force. Setting the vorticity Ω as Ω = rot U = ∂ x U2 − ∂y U1 , one can rewrite (1.1) in the vorticity form ∂t Ω + (U · ∇)Ω = ν∆Ω + rot F . (1.2) ∗

Department of Mathematics and Statistics, University of Victoria, PO BOX 1700 STN CSC, Victoria BC, Canada. E-mail: [email protected] † Department of Mathematics, Kyoto University, Kitashirakawa Oiwakecho, Sakyo-ku, Kyoto 606-8502, Japan. E-mail: [email protected] ‡ Department of mathematics, New York University in Abu Dhabi, Saadyiat Island, Abu Dhabi, UAE. E-mail: [email protected]

1

Recall that the velocity field can be formally recovered from its vorticity using the Biot-Savart law: U = KBS ∗ Ω . (1.3) 1 (−y,x) when M = R2 , and ∗ denotes the convolution Here the kernel KBS is given by KBS (x, y) = 2π x2 +y2 with respect to the spatial variables. In the sequel, we will review two important examples of solutions to (1.2), the Kolmogorov flow and the Lamb-Oseen vortex, and explain how the study of their stability is related to spectral problems for non-self adjoint operators. The Kolmogorov flow, which is the main object of this paper, is an explicit stationary solution to (1.1) with a shear sourcing term F = (aν sin y, 0), a ∈ R, and is given by

U a (x, y) = a(sin y, 0) ,

Ωa (x, y) = −a cos y .

(1.4)

By Iudovich [15] these solutions are known to be globally stable for initial perturbations in Sobolev class with zero mean condition for the streamfunctions; see also Marchioro [20]. By changing the length of the periodicity (e.g., for x) the detailed bifurcation analysis has also been done, and there are a lot of important works in this direction; see, for example, [23, 15, 21, 1, 26, 24]. As a closely related subject of this paper, there are also explicit solutions having the similar forms to (1.4) when F = 0, but instead, the initial data is chosen as in (1.4). Indeed, in this case one can check that U a (x, y, t) = ae−νt (sin y, 0) solves (1.1) with F = 0. These solutions describe a quasi-steady state of the fluid, and are exact steady solutions to the Euler equations when ν = 0. These quasi-steady solutions are known as “bar-states” or also as the Kolmogorov flows, and they qualitatively match the maximum entropy solutions found in [7, 22, 27]. Both numerical and experimental evidences [27] claim that solutions to (1.1) rapidly approach barstates on time scale O( √1ν ) for high Reynolds number. Note that the time scale O( √1ν ) is much shorter than the scale O( 1ν ) which is the scale for the linear Stokes equation (and thus, heat equation in this problem) with the viscosity ν. The aim of this paper is to study this enhanced dissipation in view of the stability analysis of the steady Kolmogorov flows (1.4). Expanding solutions to (1.2) around (1.4) yields ∂t ω = Lν,a ω − (KBS ∗ ω · ∇)ω

(1.5)

where we have set Ω = Ωa,1 + ω, and the linearized operator Lν,a is given by Lν,a ω = ν∆ω − a sin y ∂ x (I + ∆−1 )ω .

(1.6)

We note that the linearized operator around the bar-state has the similar form but becomes timedependent as Lν,a (t)ω = ν∆ω − ae−νt sin y ∂ x (I + ∆−1 )ω .

(1.7)

Showing that the solution ω to (1.5) decays rapidly within a nontrivial time scale t  O( 1ν ) is a challenging mathematical problem, even in the linear case. In studying the flows generated by (1.6) or (1.7) the main difficulty comes from the presence of the non-local term in these linearized operators. In [2], Beck and Wayne proved the stability and enhanced dissipation of the bar-states for the model linear problem by removing the nonlocal term ∆−1 from (1.7). Their method is based on hypocoercivity arguments developed by Villani [25], and provide the decay in the time scale O( √1ν ) for solutions to the model linear problem in a suitable invariant 2

subspace. However, it is not clear how to extend their argument in the presence of the nonlocal term ∆−1 . Moreover, beside the nonlocality, the presence of ∆−1 in (1.6) or (1.7) leads to an additional difficulty in view of the symmetry of the operator. Indeed, although the operator sin y∂ x is antisymmetric in the standard L2 space, sin y∂ x (I + ∆−1 ) is not. Very recently, the full evolution operator (1.7) and the corresponding nonlinear problem were studied in details by Lin and Xu [18], and the enhanced dissipation is verified at some time scale o( 1ν ) for Qω, called the non-shear part of the solution ω in [18], where Q is the projection to the orthogonal complement of the kernel of − sin y∂ x (I + ∆−1 ) in L2 . The core idea in [18] is to use the Hamiltonian structure of the operator − sin y∂ x (I + ∆−1 ) = JL with J = − sin y∂ x and L = I + ∆−1 that naturally leads to the use of the weighted L2 space hL·, ·iL2 in which JL becomes antisymmetric. Then one can apply the RAGE Theorem for the estimate of the group etJL and the argument of Constantin, Kiselev, Ryzhik, and Zlatoˇs [8]. Note that the inner product hL·, ·iL2 was a key tool also in obtaining the global stability of the Kolmogorov flows with arbitrary amplitude a; cf. [15]. The argument and the result of [18] are verified without any change also for the stability problem of the steady Kolmogorov flows (1.5) - (1.6). However, for a deeper quantitative point of view, the spectral property of Lν,a requires a further study. Indeed, the argument in [18] kQω(t/ν)k provides little information on the required smallness of ν to achieve the smallness of kω(0)k 2L2 , L ν,a which depends on t in an implicit way, even for the linear solution ω(t) = etL ω(0). In particukQω(t/νβ )k lar, the question whether or not the smallness of kω(0)k 2 L2 holds for some β ∈ (0, 1), as solved L

in [2] with β = 12 for the model problem to (1.7), has been a challenging problem; see also Remark 1.2 below. In this paper we will establish some resolvent estimate on the imaginary axis of the resolvent parameters for the linearization (1.6) around the steady Kolmogorov flows. Our resolvent estimate is related to the pseudospectrum as in the work by Gallagher, Gallay, and Nier [11] of the spectral analysis for large skew-symmetric perturbations of the Harmonic oscillator. As a main result, we will verify the enhanced dissipation in a time scale O( √1ν ) for the linear flow ν,a etL ω0 ; see (1.10) below. In particular, our result gives an affirmative answer to the problem numerically conjectured in [2]. We expect that the similar enhanced dissipation will be true also for the linear flow generated by the evolution operator Lν,a (t) in (1.7), which is still under investigation due to an obstacle from the time dependence of the operator. The nonlinear problem (1.5) can also be handled based on the linear estimates of this paper, but here we focus only on the linear problem. By rescaling time as t 7→ νt, one can rewrite the evolution problem ∂t ω = Lν,a ω as, by relabeling the variable and the unknown again as t and ω respectively, ∂t ω = ∆ω −

a sin y∂ x (I + ∆−1 )ω . ν

(1.8)

This problem is viewed in the more abstract form ∂t ω = (A − αΛ)ω ,

(1.9)

where α > 0 is a large positive parameter, A is a dissipative operator, and Λ has a Hamiltonian structure. It will be worthwhile investigating the spectral property for such operators in the abstract level, which is handled in Section 2. The problem (1.8) for the Kolmogorov flows is discussed in Section 3, and we will show the key estimate for the resolvent with a rate on α = aν 3

(Theorems 3.11 and 3.13), and then for the semigroup (Theorem 3.14). In the original variables, ν,a our result in particular provides the bound for the semigroup etL such as ν,a

kQetL ω0 kL2 ≤ Ce−c

√

aν t

1 t≥ √ , aν

kQω0 kL2 ,

(1.10)

see Corollary 3.15. Here C and c are positive constants independent of t, a, ω0 , and sufficiently 1 small ν > 0. This implies the enhanced dissipation in the time scale O(ν− 2 ) for 0 < ν  1, that kQω(t/νβ )k is much shorter than o( 1ν ) and provides a decay of kω(0)k 2 L2 for β = 21 in the linear problem. L

Remark 1.1 It should be emphasized that the semigroup estimate (1.10) is in fact new even for the model problem considered by [2] in which the nonlocal operator ∆−1 is dropped, though our result does not handle the time-dependent operator as in (1.7). More precisely, the argument used in [2] provides the semigroup bound of the model problem in a weighted H 1 space whose norm has a dependence in ν. In particular, the norm introduced in [2] involves the term 1 ν− 4 kMcos y f kL2 , in addition to the usual L2 norm. As a result, the estimate obtained in [2] shows 1 that kMcos y ω(t)kL2 becomes small in a time scale O(ν− 2 ) for the model problem, while in order to achieve the dissipation in the usual L2 norm it seems that one needs to take a slightly longer 1 time scale, e.g., O(ν− 2 | log ν|). Our result (1.10) gives the dissipation in the L2 norm exactly in 1 a time scale O(ν− 2 ). Our proof is based on the detailed resolvent analysis and is very different from the approach in [2]. Next let us briefly mention a topic which is closely related to the present work as another example of (1.9): the asymptotic stability problem of the Lamb-Oseen vortex. By working on M = R2 , it is known that there exists a family of self-similar solutions to the vorticity equation (1.2) given by Ω(x, y, t) =

x y γ G( √ , √ ), νt νt νt

and

γ x y U(x, y, t) = √ V G ( √ , √ ) , νt νt νt

(1.11)

1 −|ξ| /4 1 ξ (1 − e−|ξ| /4 ). The constant γ = where the profiles are G(ξ) = 4π e and V G (ξ) = 2π |ξ|2 R Ω(x, y, t) dx dy is the circulation at infinity of the flow. By the significant work of Gallay R2 and Wayne [14] it is known that this solution is the only forward self-similar solution to (1.1) in R2 with an integrable vorticity. This solution is called the Lamb-Oseen vortex. It is well known that, through a suitable similarity transformation, the asymptotic stability of the LambOseen vortex is equivalent with the two dimensional stability of the Burgers vortex, which is a stationary solution to the three dimensional Navier-Stokes equations in the presence of the axisymmeric linear strain. The reader is referred to a recent review article [13] by Gallay and the second author of this paper about the research on the stability of the Burgers vortex. The two dimensional linearized problem for the Burgers vortex with circulation α is given by ⊥

2

∂τ ω = (L − αΛ)ω ,

τ > 0,

ξ ∈ R2 ,

2

(1.12)

where Lω = ∆ω + 12 ξ · ∇ω + ω, and Λω = (V G · ∇)ω + (KBS ∗ ω · ∇)G. Here ∆ and ∇ are now about the variables ξ = (ξ1 , ξ2 ). In the weighted L2 space L2 (R2 ; dξ ), the operator −L is G nonnegative self-adjoint with compact resolvent, and Λ becomes antisymmetric as proved in [14]. Hence the linear analysis falls into the analysis of the operator of the form (1.9). In the 4

space L2 (R2 , G−1 dξ) with zero mass condition, we have −L ≥ 12 , and thus, the antisymmetry of Λ provides 21 spectral gap for L − αΛ for any α. This yields the linear stability with a uniform estimate in α. However, this simple argument does not provide further informations for the fast rotation case |α|  1, at the time when numerical and experimental evidence suggest that the basin of attraction should be α-dependent, at least “away” from the kernel of the operator Λ. In [19] the second author of this paper verifies a behavior of the pseudospectral bound but without the information on the rate about α. On the other hand, in [11] and Deng [9] simplified model operators are studied in details, where the main simplification is dropping the nonlocal term (KBS ∗ w, ∇)G, and the optimal dependence on α of the pseudospectral bound that decays 1 like |α| 3 is obtained for these model operators. The same result is proved for the full linearized operator L − αΛ in Deng [10] but in a smaller subspace than the orthogonal complement of Ker Λ. Very recently, Li, Wei and Zhang [17] gave a sharp pseudospectral bound as well as the spectral bound of L − αΛ in the orthogonal complement of Ker Λ, and this result is applied to the nonlinear problem by Gallay [12]. Remark 1.2 In Li, Wei, and Zhang [17] the key elegant idea is to introduce the wave operator which converts the original skew-symmetric operator Λ, containing a nonocal term that leads to an essential difficulty, into a skew-symmetric operator for which the nonlocal operator is removed and hence the approach of [11] is applied. In [17] it is announced without details that this approach for the Lamb-Oseen operator can be applied also for the estimate of the enhanced dissipation around the Kolmogorov flows. We note that our approach for (1.10) or Theorem 1.3 below is different and independent of [17], and in particular, does not rely on the construction of the wave operator. To summarize, the above two examples of the Kolmogorov flows and the Lamb-Oseen vortex show that to measure the basin of attraction, it is important to obtain a pseudospectral bound as sharp as possible for the operator in the abstract form given in (1.9). We also note that the enhanced dissipation is one of the important subjects in fluid mechanics, and recently, significant progress has been achieved around some class of simple flows such as the Couette flow; see, e.g., [3, 4, 6, 5]. This paper consists of two parts. The first one is an abstract result, in which the spectral properties of some class of non self-adjoint operators are established. The other one is the application of the abstract result to the linearized operator for the Kolmogorov flows. As for the abstract part, we consider the operator in a Hilbert space X of the form Lα = A − αΛ ,

(1.13)

where A is positive self-adjoint with compact resolvent, α ∈ R, and Λ is a densely defined ˆ by the relation closed linear operator relatively compact to A. For later use we set Λ ˆ. Λ = iΛ

(1.14)

We denote by DX (A) the domain of A in X. We are interested in the spectral property of Lα ˆ it is natural to introduce the for large |α|. Since the effect of α is absent for functions in Ker Λ orthogonal projections ˆ
⊥ , Q : X → Y := Ker Λ (1.15) 5

where K ⊥ denotes the orthogonal complement space in X for a given closed subspace K. We are interested in the estimate of QetLα for large α. Since the semigroup etLα is expressed in terms of the resolvent of Lα the problem is reduced to the estimate of Q(iλ − Lα )−1 when iλ belongs to the resolvent set of Lα . When the invariance QA ⊂ AQ holds, which will be assumed in this paper, the estimate of Q(iλ − Lα )−1 is reduced to the resolvent analysis of the operator QLα in Y, which is realized as DY (QLα ) = DY (QA) := DX (A) ∩ Y , ˆ , QLα u = QAu − iαQΛu u ∈ DY (QLα ) .

(1.16)

Indeed, we have Q(iλ − Lα )−1 f = (iλ − QLα )−1 f for f ∈ Y when QA ⊂ AQ. In order to obtain the estimate of QetLα or etQLα , the following quantity plays an essential role:  −1 −1 (1.17) ΨY (α; QLα ) = sup k(iλ − QLα ) kY→Y . λ∈R

The quantity (1.17) was introduced in [11], where the basic pseudospectral property and the relation with the semigroup estimate are also presented. For convenience we call (1.17) the pseudospectral bound of QLα . In our framework the operator Λ is not necessarily antisymmetric, but instead, is assumed to possess a Hamiltonian structure; see Assumption 2 in Section 2. This structural assumption is of course motivated by the application to the Kolmogorov flows. There are two theorems in the abstract part. The first one is a pseudospectral bound without a concrete dependence on α (Theorem 2.4). The argument in Theorem 2.4 shares some common features with the argument in [18]. While the result of [18] is based on the RAGE theorem, our argument is much more elementary, though Theorem 2.4 dose not necessarily give a stronger result than [18]. The second result of the abstract part has a concrete dependence on α (see ˆ The key additional condition is Assumption Theorem 2.9), under additional assumptions on Λ. ˆ with µ ∈ R by allowing a presence of the 4 which imposes some coercive estimate for µ − Λ term yielding a “loss of derivative” but with a small factor in front. This derivative loss with a small prefactor is controlled by the smoothing effect of A at the end, and this balance determines the rate in α for the pseudospectral bound. In Assumption 4 we also introduce a structural conˆ dition on A of the form −A = T ∗ T and impose some bounds related to the commutator [T, Λ]. This type of condition is motivated by the work of [25, 11] and it is useful also in this paper to achieve the resolvent estimate with a better dependence in α. As an application of the abstract result, we study in Section 3 the rescaled version of the linear operator (1.6), i.e., the problem (1.8). By taking the Fourier series in x, the key is to analyze the operator only in the y variable in the space L2 (T): ˆl, Lα,l = Al − iαlΛ

ˆ l = Msin y (I + A−1 Λ l ).

ˆ l has a nontrivial Here Al = ∂2y − l2 , α = aν , Msin y f = sin y f , and l ∈ Z \ {0}. The operator Λ kernel only when l = ±1 which is spanned by the constant functions, and thus, the projection Ql : L2 (T) → L2 (T) is defined by Z 2π 1 Ql f = f for |l| ≥ 2 , Ql f = f − f dy for |l| = 1 . 2π 0 ˆ l with The main effort is to check the coercive estimates described in Assumption 4 for µ − Λ µ ∈ R which is essential to achieve the pseudospectral bound with a rate in α. We shall verify 6

ˆ l . The main difficulty Assumption 4 by analyzing the ODE corresponding to the operator µ − Λ −1 ˆ here is the presence of the term Al in Λl , which makes the problem nonlocal and also leads to ˆ l , 0 when l = ±1. This loss of invariance some lack of invariance, namely the fact that (I −Ql )Λ ˆ l , and gives rise to an additional nonlocality coming is due to the absence of the symmetry of Λ from the projection Ql . Therefore, we have to deal with two nonlocalities; the one in A−1 l and the one in Ql . For a given µ ∈ R the point y ∈ T satisfying sin y = µ is called a critical point of this problem. The difficulty coming from the nonlocality of A−1 l is significant when the critical points are degenerate, and this corresponds to the case when |µ| is around 1 in the analysis of ˆ l . The core part of the analysis is Lemma 3.8 which deals with this singularity. The key µ−Λ idea is to use a contradiction argument, which enables us to focus on the functions concentrating around the critical points, for which the nonlocal term essentially becomes a small order since the operator A−1 l has a smoothing effect. On the other hand, the influence of the projection Ql ˆ l , for I − Ql is the projection becomes relevant only when µ is closed to 0 in the analysis of µ − Λ ˆ l . As a result, these two kinds of difficulty related to A−1 and to Ql appear in to the kernel of Λ l different parameter regimes of µ, and thus we can handle them separately. After establishing the ˆ l , which are stated in Proposition 3.10, the resolvent estimate for key coercive bounds of µ − Λ Lα,l is obtained in Theorem 3.13 by applying the abstract result in Section 2 and also by using a specific property of the trigonometric functions. For convenience, it will be worthwhile stating our resolvent estimate for the Kolmogorov flow in this introductory section: Theorem 1.3 There exist C, M, α0 > 0 such that the following statement holds for all α ∈ R with |α| ≥ α0 . Let λ ∈ R and l ∈ Z \ {0} with |l| ≥ 1. Then  λ 1 C    if 1 − | | > ,  1 2 1  λ 3  αl 3 (1 − | 2 |αl| |αl| |)   αl    C 1 M λ  −1 if − ≤1−| |≤ , (1.18) k(iλ − Ql Lα,l ) kYl →Yl ≤  1 1 1   αl 2 2 2  |αl| |αl| |αl|    λ M C    if 1 − | | < − 1 .   |αl| (| λ | − 1) αl |αl| 2 αl Here Yl = Ql L2 (T). The estimate (1.18) actually gives more detailed information on the spectrum of QLα,l than the pseudospectral bound defined by (1.17), and seems to be considerably sharp in view of the degeneracy of the critical points. In fact, we observe that the critical points become degenerate 1 when | αlλ | ∼ 1, and (1.18) claims that the rate is O(|αl|− 2 ) around this case. When | αlλ | is less than 2 1, the critical points are nondegenerate and the rate is improved as O(|αl|− 3 ). Note that these 1 2 rates, O(|αl|− 2 ) and O(|αl|− 3 ), depending on the degeneracy of the critical points are compatible with the result in [11]. Finally, if | αlλ | is larger than 1, the critical points are no longer present, resulting in the rate O(|αl|−1 ). This paper is organized as follows. In Section 2 we discuss the problem in an abstract framework. Section 3 is devoted to the study of the linearized problem for the Kolmogorov flows. The main results in Section 3 are Theorem 3.14 and its Corollary 3.15 for the estimate ν,a of the semigroup {etL }t≥0 .

7

2

Abstract result

In this section we establish the abstract result in obtaining the resolvent estimate for the operator (1.13), by taking into account the application to the stability of the Kolmogorov flows. In fact, to prove the estimate stated in Theorem 1.3 requires a rather complicated and long argument, and thus, the abstract result is useful in understanding the basic strategy. First we state the basic assumption on A. Assumption 1 The operator A : DX (A) ⊂ X → X is self-adjoint in X with compact resolvent, and −A is positive and satisfies h−Au, uiX ≥ kuk2X ,

u ∈ DX (A) .

(2.1)

Remark 2.1 One can extend the result of this section to more general class of A such that −A is m-sectorial satisfying the positivity Reh−Au, uiX ≥ kuk2X + C| ImhAu, uiX | with compact ˆ But for simplicity we focus on the case resolvent, by slightly modifying the assumption on Λ. when A is self-adjoint. ˆ Next we state the conditions on the relation between A and Λ. ˆ is a densely defined closed operator and is relatively compact to −A in X, Assumption 2 (i) Λ ˆ ∗ ), and there exists a positive constant C such that QA ⊂ AQ, DX (A) ⊂ DX (Λ ˆ , uiX | + | Imh−Au, Λui ˆ X | ≤ Ch−Au, uiX , |hΛu

u ∈ DX (A) .

(ii) There exist closed symmetric operators B1 , B2 , and positive constants c1 and C such that ˆ = B1 B2 , Ker Λ ˆ = Ker B2 , and Λ 1

kB2 uk2X ≤ Ck(−A) 2 uk2X , hu , B2 uiX ≥

c1 kuk2X

, 1 2

Reh−Au, B2 uiX ≥ c1 k(−A) uk2X

u ∈ DX (A) ,

(2.2)

u ∈ DX (A) ∩ Y ,

(2.3)

u ∈ DX (A) ∩ Y .

(2.4)

ˆ has a property similar to a closed symmetric operRemark 2.2 Assumption 2 (ii) states that Λ ˆ is closed symmetric then it suffices to take B1 = Λ ˆ and B2 = Q. ator. Indeed, if Λ ˆ the range of Λ, ˆ i.e, Ran Λ ˆ = { f ∈ X | f = Λg ˆ for some g ∈ DX (Λ)}. ˆ Let us denote by Ran Λ ˆ ⊂ DX (Λ ˆ ∗ ). Assumption 3 (i) Ker Λ ˆ ∩ Ran Λ ˆ = {0}. (ii) Ker Λ ˆ does not possess eigenvalues in R \ {0}. (iii) Λ Remark 2.3 (1) Assumption 3 (i) is imposed due to a technical reason; indeed, it is imposed so that formal computations are justified. Thus, for applications to concrete problems this condition can be relaxed to some extent. ˆ is closed symmetric then Assumption 3 (ii) holds. Indeed, it suffices to use the or(2) If Λ ˆ ∗ ⊕ Ran Λ ˆ ∗∗ = Ker Λ ˆ ∗ ⊕ Ran Λ; ˆ then for any f we have the thogonal decomposition X = Ker Λ 8

ˆ n. corresponding decomposition f = ϕ + ψ, and ψ is approximated by {ψn } such that ψn = Λφ ˆ ˆ Then for any u ∈ Ker Λ ∩ Ran Λ we have hu, f iX = hu, ϕiX + hu, ψn iX + hu, ψ − ψn iX = hu, ψ − ψn iX → 0

n → ∞.

Hence u = 0. First we state the abstract result of the spectral behavior of QLα with α ∈ R in the limit |α| → ∞, but without any information on the rate of convergence. Theorem 2.4 Suppose that Assumptions 1, 2, and 3 hold. Let σY (QLα ) be the set of the spectrum of QLα , α ∈ R, in Y. Then we have lim

sup

|α|→∞ µ∈σY (QLα )

Re µ = −∞ ,

(2.5)

and lim sup k(iλ − QLα )−1 kY→Y = 0 .

|α|→∞ λ∈R

(2.6)

Moreover, for sufficiently large |α| the set {ζ ∈ C | Re ζ > −1} is contained in the resolvent set of Lα in X, and we have lim sup kQ(iλ − Lα )−1 kX→X = 0 .

|α|→∞ λ∈R

(2.7)

Proof. The proof consists of several steps. Without loss of generality we may assume that α > 0. Step 1: The operator QA in Y is a closed linear operator with compact resolvent. This follows directly from the invariance QA ⊂ AQ and Assumption 1. We denote by A|Y the restriction of A to Y with the domain DY (A|Y ) = DX (A) ∩ Y. Step 2: σY (QLα ) ⊂ {µ ∈ C | Re µ < 0}. Let c1 > 0 be the number in (2.4). We have already seen that −A|Y is bounded from below in Y ˆ is relatively compact to A by the assumption, and has a compact resolvent in Y. Then, since Λ ˆ we see that QLα = A|Y − iαQΛ is also a closed operator with compact resolvent in Y. Thus, the spectrum of QLα consists of isolated eigenvalues with finite multiplicities. Let µ ∈ C be an eigenvalue of QLα in Y and let u ∈ DX (A) ∩ Y be an associated eigenfunction such that kukX = 1. Note that −µu = −QLα u holds. By taking the inner product with B2 u, we have ˆ , B2 uiX −µhu , B2 uiX = h−QAu , B2 uiX + iαhQΛu = h−QAu , B2 uiX + iαhB1 B2 u , QB2 uiX .

(2.8)

ˆ = B1 B2 by Assumption 2 (ii). Moreover, we verify that (I − Q)B2 = 0 Here we have used Λ ˆ = Ker B2 . Hence we have QB2 = B2 , which implies since B2 is closed symmetric and Ker Λ hB1 B2 u , QB2 uiX = hB1 B2 u , B2 uiX and h−QAu , B2 uiX = h−Au , QB2 uiX = h−Au , B2 uiX . Therefore, since B1 is closed symmetric in X, the real part of (2.8) yields −(Re µ)hu , B2 uiX = Reh−Au , B2 uiX 1

≥ c1 k(−A) 2 uk2X ≥ c1 . 9

(2.9)

Here we have used the assumptions (2.4) and (2.1) with kukX = 1. Hence, (2.3) yields Re(µ) ≤ −

c1 < 0. hu, B2 uiX

(2.10)

Thus, σY (QLα ) ⊂ {µ ∈ C | Re µ < 0} for all α ≥ 0. Step 3: The spectral limits (2.5) and (2.6) hold. By Step 2 it suffices to show lim sup k(iλ − QLα )−1 kY→Y = 0 .

(2.11)

α→∞ λ∈R

Suppose that (2.11) does not hold. Then there exist δ > 0, {αn } ⊂ R+ , {λn } ⊂ R, and fn ∈ Y with k fn kX = 1 such that αn → ∞, and k(iλn − QLα )−1 fn kX ≥ δ. Set un = (iλn − QLα )−1 fn , which solves ˆ n = fn . iλn un − QAun + iαn QΛu

(2.12)

By taking the inner product with B2 un in the above equation, we obtain iλn hun , B2 un iX + h−Aun , B2 un iX + iαn hB1 B2 un , B2 un iX = h fn , B2 un iX .

(2.13)

Here we have used QB2 = B2 as observed in Step 2. Since B1 and B2 are symmetric, the real part of (2.13) yields Reh−Aun , B2 un iX = Reh fn , B2 un iX , and then the assumptions (2.2), (2.3), and (2.4) imply 1

c1 h−Aun , un iX ≤ Reh fn , B2 un iX ≤ k fn kX kB2 un kX ≤ Ck(−A) 2 un kX . Thus we obtain the uniform bound 1

sup k(−A) 2 un kX < ∞ .

(2.14)

n 1

Now we recall that, since −A is positive self-adjoint with compact resolvent, (−A) 2 also has a compact resolvent ( [16, Theorem V-3.49]). Since (2.14) implies the uniform bound of 1 k(−A) 2 un kX , {un } is compact in X, and thus, in Y. Then there exists a subsequence of {un }, 1 denoted again by {un }, which strongly converges to some u∞ ∈ Y and satisfies k(−A) 2 u∞ kX ≤ 1 supn k(−A) 2 un kX < ∞. By the strong convergence we have ku∞ kX ≥ δ, so u∞ ∈ Y is nontrivial. Let us go back to (2.12), and take the inner product with un . Then we have i

λn 1 ˆ n , un iX = 1 h fn , un iX kun k2X + h−Aun , un iX + ihΛu αn αn αn

and the imaginary part of this identity yields the bound |

λn ˆ n , un iX | + | Imh fn , un iX | ≤ Ck(−A) 12 un k2X + k fn kX kun kX | kun k2X ≤ |hΛu αn
1 ≤ C sup k(−A) 2 un k2X + 1 < ∞ . n

10

Since kun kX ≥ δ we have the uniform bound sup | n

λn | < ∞. αn

Set µn = αλnn . By taking a suitable subsequence we may assume that µn converges to some µ∞ ∈ R. For any ϕ ∈ DX (A) we have from (2.12), iµn hun , ϕiX −

i ˆ ∗ QϕiX = 1 h fn , ϕiX , hun , AQϕiX + ihun , Λ αn αn

and by taking the limit n → ∞, we have ˆ ∗ QϕiX = 0 , iµ∞ hu∞ , ϕiX + ihu∞ , Λ and thus, ˆ ∗ ϕiX = −µ∞ hu∞ , ϕiX + hu∞ , Λ ˆ ∗ (I − Q)ϕiX . hu∞ , Λ ˆ ⊂ DX (Λ ˆ ∗ ) by the assumption, and thus, Λ ˆ ∗ (I − Q) defines a Note that (I − Q) : X → Ker Λ bounded operator in X by the closed mapping theorem. Then, since DX (A) is dense in X this ˆ ∗ ), which implies u∞ ∈ DX (Λ) ˆ and identity holds for all ϕ ∈ DX (Λ ˆ ∞ = −µ∞ u∞ + (I − Q)Λu ˆ ∞. Λu ˆ ∞ ∈ DX (Λ) ˆ and This shows Λu ˆ 2 u∞ = −µ∞ Λu ˆ ∞. Λ

(2.15)

ˆ ∞ , 0. Thus, −µ∞ is an Since we have shown that u∞ ∈ Y and u∞ , 0, we conclude that Λu ˆ eigenvalue of Λ in X. Then, by the assumption of the theorem µ∞ must be 0, which implies ˆ ∞ ∈ Ker Λ. ˆ Thus, Λu ˆ ∞ ∈ Ran Λ ˆ ∩ Ker Λ, ˆ and by the assumption we conclude that Λu ˆ ∞ = 0, Λu
⊥ ˆ ˆ i.e., u∞ ∈ Ker Λ. On the other hand, we have also seen u∞ ∈ Y = Ker Λ , and hence, u∞ = 0. This is a contradiction, and (2.11) must hold. Step 4: σ(Lα ) ⊂ {ζ ∈ C | Re ζ ≤ −1} and (2.7) holds. Let ζ ∈ C satisfy Re ζ > −1 and let f ∈ X. Let v ∈ DY (QLα ) be the unique solution to (ζ − QLα )v = Q f ,

(2.16)

which is well-defined for all sufficiently large |α| by (2.5). Let w ∈ DX (A) be the solution to (ζ − A)w = −P(ζ − Lα )v + P f ,

P = I − Q,

ˆ by the invariance QA ⊂ AQ and v ∈ Y. That where the term −P(ζ − Lα )v coincides with −iαPΛv is, we have the following formula for w: ˆ − QLα )−1 Q f + (ζ − A)−1 P f . w = −iα(ζ − A)−1 PΛ(ζ 11

(2.17)

Note that (2.1) implies that ζ ∈ C satisfying Re ζ > −1 belongs to the resolvent set of A, and thus the above formula is well-defined. Moreover, from the invariance and the above equation, we get: 0 = Q(ζ − A)w = (ζ − A)Qw . ˆ Then u = v + w solves Hence we have Qw = 0, that is, w ∈ Ker Λ. (ζ − Lα )u = (ζ − Lα )v + (ζ − A)w = Q f + P(ζ − Lα )v + (ζ − A)w = Qf + Pf = f . Hence u ∈ DX (Lα ) is the solution to the resolvent problem, and the above construction also implies the uniqueness. Moreover, we have from the construction that Q(ζ − Lα )−1 f = (ζ − QLα )−1 Q f ,

f ∈ X.

(2.18)

Hence, (2.7) holds by (2.6). The proof is complete.

Remark 2.5 From (2.17) and (2.18) we have the formula Q(ζ − Lα )−1 = (ζ − QLα )−1 Q , ˆ − QLα )−1 Q + (ζ − A)−1 P . P(ζ − Lα )−1 = −iα(ζ − A)−1 PΛ(ζ

(2.19)

Theorem 2.4 and its proof do not give any information on the rate of convergence for |α| → ∞. To obtain a rate we make further assumptions as follows. Assumption 4 There exist C > 0, τ ∈ (0, ∞], m0 ≥ 1, γ j > 0, and bounded nonnegative functions h j : [m0 , ∞) × R → [0, ∞), j = 1, 2, such that the following statements hold. ˆ ⊂ DX (A). (i) Ker Λ (ii) (a) It follows that 1

1

kB2 (−A)− 2 ukX ≤ Ck(−A)− 2 ukX ,

kB2 ukX ≤ CkukX ,

(b) For all µ ∈ R and m ≥ m0 it follows that   1 ˆ 2X + m−2γ1 h21 (m, µ)k(−A) 21 uk2X k(−A)− 2 uk2X ≤ C m2 k(µ + Λ)uk

u ∈ X.

(2.20)

if |µ| ≥ τ and u ∈ DX (A) , (2.21)

− 12



1 2

ˆ 2X + m−2γ1 h21 (m, µ)k(−A) uk2X k(−A) uk2X ≤ C m2 kQ(µ + Λ)uk



if |µ| < τ and u ∈ DX (A) ∩ Y .

(2.22)

(iii) (a) There exists a densely defined closed operator T : DX (T ) → X such that −A ⊂ T ∗ T , B1 T B2 u ∈ X for any u ∈ DX (A), and 1

| ImhT u, B1 T B2 uiX | ≤ CkB2 ukX k(−A)− 2 ukX , 12

u ∈ DX (A) .

(2.23)

(b) [T, B1 ]B2 u ∈ X for any u ∈ DX (A), and for all µ ∈ R and m ≥ m0 it follows that   ˆ 2X + m−2γ2 h22 (m, µ)k(−A) 21 uk2X if |µ| ≥ τ and u ∈ DX (A) , k[T, B1 ]B2 uk2X ≤ C m2 k(µ + Λ)uk (2.24)   ˆ 2X + m−2γ2 h22 (m, µ)k(−A) 21 uk2X k[T, B1 ]B2 uk2X ≤ C m2 kQ(µ + Λ)uk if |µ| < τ and u ∈ DX (A) ∩ Y .

(2.25)

ˆ in Remark 2.6 (1) The condition (ii) (b) states quantitatively the absence of eigenvalues of Λ R \ {0}. To be precise let us compare the conditions (2.21)-(2.22) with Assumption 3. Assume ˆ ⊂ DX (A) holds. Then (2.21)-(2.22) imply that Λ ˆ does not possess eigenvalues in that Ker Λ R \ {0}, otherwise one can take the limit m → ∞ in (2.21)-(2.22) for the eigenvalue −µ and the ˆ 2 ⊂ DX (A), then eigenfunction u, which leads to a contradiction. Moreover, if in addition Ker Λ ˆ ∩ RanΛ ˆ there exists f ∈ DX (A) ∩ Y such that u = Λ ˆ f . Then (2.22) with µ = 0 for any u ∈ Ker Λ ˆ ∩ Ran Λ ˆ = {0}. implies f = 0, which leads to u = 0. Thus, in this case we also have Ker Λ 2 ˆ As a conclusion, under Assumption 4 (ii) and the condition Ker Λ ⊂ DX (A), Assumption 3 is automatically satisfied. (2) The case τ = ∞ means that the conditions (2.22) and (2.25) hold for all µ ∈ R. But since in actual applications Q can be nonlocal, then in the case µ is away from 0 the conditions (2.21) and ˆ is closed symmetric then Q in (2.22) and (2.25) is automatically (2.24) are easier to check. If Λ ˆ = 0 in this case. dropped since PΛ Remark 2.7 (1) In fact, one can obtain the pseudospectral bound with some rate without assuming (iii) of Assumption 4. However, as in the work of [25] the condition (iii) is useful in obtaining the pseudospectral bound with a better rate in α. (2) If B2 = I then (2.23) is automatically satisfied by the symmetry of B1 . In other words, the estimate (2.23) is valid when B2 is a smooth enough perturbation from the identity operator. Remark 2.8 The key idea of Assumption 4 is that we reduce the whole analysis to several ˆ This idea is useful in actual applications. Indeed, when A and Λ ˆ are coercive estimates of Λ. ˆ is lower than A − iαΛ, ˆ and hence, the analysis of (pseudo)differential operators, the order of Λ ˆ ˆ In principle, the operator Λ itself is expected to be simpler than the combined operator A − iαΛ. ˆ The A plays a role of recovering the regularity which was lost in the coercive estimates for Λ. ˆ functions h j in the assumption describe the degeneracy of the operator µ + Λ, which leads to an essential influence to the resolvent estimate for the full operator Lα . The next theorem provides the information on the convergence rate, once we know the behavior of the functions h j . Theorem 2.9 Suppose that Assumptions 1, 2, and 4 hold. Then for any α ∈ R the set {ζ ∈ C | Re ζ ≥ 0} is contained in the resolvent set of Lα in X and of QLα in Y. Moreover, for sufficiently large |α| the following resolvent estimate holds. λ kQ(iλ − Lα )−1 kX→X = k(iλ − QLα )−1 kY→Y ≤ CF(α, ) , α 13

(2.26)

where F(α, µ) =

 m2

inf

m1 ,m2 ≥m0

1 2 α

−γ  21 m1
43 m81 m21 m22 m21 m2 2 h2 (m2 , µ) −2γ1 2 + + 4 + + + m1 h1 (m1 , µ) α α α2 |α|

(2.27)

and m0 is given in Assumption 4. Here C is independent of α and λ. Remark 2.10 To make F(α, µ) small we will take to estimate by the simpler function e µ) = F(α,

m1 +m21 |α|

small and m1 ≤ m2 . In fact, it suffices

 21  m
4 m2 m2 m2 m−γ2 h (m , µ) 2 2 1 3 −2γ1 2 1 2 1 2 inf + + + m1 h1 (m1 , µ) . m1 ,m2 ≥m0 α α2 |α|

(2.28)

e we first choose m2 so that m2+γ2 = |α|h2 (m2 , µ) holds, which gives the Then, to evaluate F, 2 m2 m2

−γ2 h

m2 m

(m ,µ)

balance α1 2 2 = 1 2 |α|2 2 for any m1 . With this choice of m2 , the number m1 is chosen so that m2 m2 m1
34 1 2 + α1 2 2 + m−2γ h1 (m1 , µ) is minimized. 1 α Proof of Theorem 2.9. It suffices to consider the case α > 0. Set µ = αλ and let τ ∈ (0, ∞] be the number in Assumption 4. For simplicity of notations we write h j instead of h j (m j , µ), j = 1, 2. (i) The case |µ| < τ. From the definition, we have for u ∈ DX (A) ∩ Y: ˆ + µ) Q(Lα − iλ) = QA − iαQ(Λ and ˆ + µ)uk2X . ˆ + µ
uiX = hQAu, Λ ˆ + µ
uiX + iαkQ(Λ h(QLα − iλ)u, Λ

(2.29)

By taking the imaginary parts of both sides, we obtain ˆ + µ
uk2X ≤ | Imh(QLα − iλ)u, Q Λ ˆ + µ
uiX | + | Imh−QAu, Λ ˆ + µ
uiX | αkQ Λ ˆ + µ
ukX + | Imh−Au, Λui ˆ X| ≤ k(QLα − iλ)ukX kQ(Λ Here we have used that for u ∈ DX (A) ∩ Y, we have QAu = AQu = Au and the self-adjointness ˆ X , we use −Au = T ∗ T u of A, which gives Imh−QAu, µuiX = 0. For the mixing term Imh−Au, Λui and Assumption 4 (iii) (a) to obtain ˆ X | = | ImhT u, B1 T B2 uiX + ImhT u, [T, B1 ]B2 uiX | | Imh−Au, Λui 1

≤ CkB2 ukX k(−A)− 2 ukX + kT ukX k[T, B1 ]B2 ukX 1

1

= CkB2 ukX k(−A)− 2 ukX + k(−A) 2 ukX k[T, B1 ]B2 ukX .

(2.30)

1

Here we have used kT uk2X = k(−A) 2 uk2X by −A = T ∗ T . Thus we have arrived at ˆ
uk2X ≤ C k(QLα − iλ)uk2X + CkB2 ukX k(−A)− 12 ukX + k(−A) 21 ukX k[T, B1 ]B2 ukX . αkQ µ + Λ α (2.31)

14

Let us estimate k[T, B1 ]B2 ukX . Fix m2 ≥ m0 . We have from (2.25) and (2.31), ˆ 2X + Cm−2γ2 h22 k(−A) 21 uk2X k[T, B1 ]B2 uk2X ≤ Cm22 kQ(µ + Λ)uk 2  2 Cm2 1 1 2 − 12 2 k(QLα − iλ)ukX + kB2 ukX k(−A) ukX + k(−A) ukX k[T, B1 ]B2 ukX ≤ α α 1 2 2 + Cm−2γ h2 k(−A) 2 uk2X 2 ≤

Cm22 Cm22 1 2 k(QL − iλ)uk + kB2 ukX k(−A)− 2 ukX α X 2 α α  m4  1 2 2 + C 22 + m−2γ h2 k(−A) 2 uk2X . 2 α

(2.32)

1

Next we estimate k(−A)− 2 uk2X . Fix m1 ≥ m0 . We apply (2.22) and then use (2.32), which gives 1 ˆ 2X + Cm−2γ1 h21 k(−A) 21 uk2X k(−A)− 2 uk2X ≤ Cm21 kQ(µ + Λ)uk 1  2 Cm1 1 1 2 − 12 2 k(QLα − iλ)ukX + kB2 ukX k(−A) ukX + k(−A) ukX k[T, B1 ]B2 ukX ≤ α α 1 1 2 + Cm−2γ h1 k(−A) 2 uk2X 1

Cm41 Cm21 1 2 1 2 k(QL − iλ)uk + kB2 uk2X + Cm−2γ h1 k(−A) 2 uk2X α X 1 2 2 α α  2 Cm21 Cm Cm22 1 1 2 2 2 + k(−A) ukX kB2 ukX k(−A)− 2 ukX k(QLα − iλ)ukX + 2 α α α 1  m4   2 1 2 2 + C 22 + m−2γ h2 k(−A) 2 uk2X 2 α 4  m2 m 2 1 3 Cm1 m41 1 1 2 2 2 2 2 k(−A) ukX kB2 ukX ≤ 2 k(QLα − iλ)ukX + C 2 kB2 ukX + C 3 α α α2  m2 m2 m2 m−γ2 h  1 2 2 1 2 2 + C 12 2 + 1 2 + m−2γ h 1 k(−A) ukX . 1 α α Cm2 m4 ≤ 2 1 k(QLα − iλ)uk2X + C 21 kB2 uk2X α α   m2 m2 m2 m−γ2 h 1 2 2 1 2 2 h + C 12 2 + 1 2 + m−2γ 1 k(−A) ukX . 1 α α ≤

By the interpolation inequality, (2.33), and the boundedness of B2 in X, 1

1

kuk2X ≤ k(−A) 2 ukX k(−A)− 2 ukX  Cm2 m41 1 1 2 ≤ Ck(−A) 2 ukX k(QL − iλ)uk + C kB2 uk2X α X α2 α2  m2 m2 m2 m−γ2 h   21 1 2 1 2 2 uk2 + C 12 2 + 1 2 + m−2γ h k(−A) X 1 1 α α Cm21 Cm1 1 1 ≤ k(QLα − iλ)ukX k(−A) 2 ukX + kB2 ukX k(−A) 2 ukX α α  m2 m2 m2 m−γ2 h  12 1 2 1 2 + C 12 2 + 1 2 + m−2γ h1 k(−A) 2 uk2X 1 α α 15

(2.33)

≤

Cm1 1 k(QLα − iλ)ukX k(−A) 2 ukX α  m8 m2 m2 m2 m−γ2 h  12 1 2 2 1 2 2 + C 41 + 1 2 2 + 1 2 + m−2γ h 1 k(−A) ukX . 1 α α α

(2.34)

1

Next we estimate k(−A) 2 uk2X . We observe that the following identity holds: Reh(QLα − iλ)u, B2 uiX = Reh(Lα − iλ)u, QB2 uiX = RehAu , B2 uiX . ˆ , B2 uiX = ImhB1 B2 u , B2 uiX = 0, and Imhu , B2 uiX = 0 Here we have used QB2 = B2 , ImhΛu since B1 and B2 are symmetric. By (2.4) we have Reh−Au , B2 uiX ≥ c1 h−Au , uiX , which gives 1

k(−A) 2 uk2X ≤ Ck(QLα − iλ)ukX kB2 ukX ,

u ∈ DX (A) ∩ Y .

(2.35)

Thus we conclude from (2.34) and (2.35) that kuk2X

  m
4 m8 m2 m2 m2 m−γ2 h 2 1 3 −2γ1 2 1 2 1 1 2 + m1 h1 k(QLα − iλ)uk2X , + 4 + + ≤C 2 α α α α

u ∈ DX (A) ∩ Y . (2.36)

Note that (2.36) is valid for any m1 , m2 ≥ m0 and α > 0. (ii) The case |µ| ≥ τ. In this case we drop the projection Q in (2.29) and use the identity ˆ + µ
uiX = hAu, Λ ˆ + µ
uiX + iαk(Λ ˆ + µ)uk2X , h(Lα − iλ)u, Λ

u ∈ DX (A) ,

which gives by taking the imaginary parts, ˆ + µ
uk2X ≤ k(Lα − iλ)ukX k(Λ ˆ + µ
ukX + | Imh−Au, Λui ˆ X| . αk Λ Thus, as in (2.30), we obtain ˆ + µ
uk2X ≤ C k(Lα − iλ)uk2X + CkB2 ukX k(−A)− 21 ukX + k(−A) 12 ukX k[T, B1 ]B2 ukX , αk Λ α u ∈ DX (A) .

(2.37)

1

Then the estimates of the terms [T, B1 ]B2 u and (−A)− 2 u are obtained in the same manner as in the case (i), and we have k[T,

B1 ]B2 uk2X

Cm22 Cm22 1 2 ≤ 2 k(Lα − iλ)ukX + kB2 ukX k(−A)− 2 ukX α α  m4  1 2 2 + C 22 + m−2γ h2 k(−A) 2 uk2X , 2 α

(2.38)

Cm21 m41 2 k(L − iλ)uk + C kB2 uk2X α X α2 α2  m2 m2 m2 m−γ2 h  1 2 −2γ1 2 1 2 1 2 + + m1 h1 k(−A) 2 uk2X . +C 2 α α

(2.39)

and 1

k(−A)− 2 uk2X ≤

16

Then u is estimated by using the interpolation inequality as in (2.34), kuk2X ≤

Cm1 1 k(Lα − iλ)ukX k(−A) 2 ukX α 1  m8 m2 m2 m2 m−γ2 h 1 2 −2γ1 2 2 1 2 1 2 1 + + m1 h1 k(−A) 2 uk2X . + C1 4 + 2 α α α

(2.40)

ˆ ⊂ Next we observe that AP, where P = I − Q, is a bounded operator by the assumption Ker Λ DX (A), which gives from QA ⊂ AQ, 1

k(−A) 2 uk2X = h−Au, uiX = h−APu, PuiX + h−AQu, QuiX ≤ C2 kPuk2X + h−AQu, QuiX . Hence, for sufficiently large α so that  m8

C1C2 41 α

−γ  m21 m22 m21 m2 2 h2 1 −2γ1 2 2 ≤ , + + + m h 1 1 2 α α 2 1

(2.41)

the estimate (2.40) yields kuk2X

Cm21 Cm1 1 k(Lα − iλ)ukX k(−A) 2 QukX ≤ 2 k(Lα − iλ)uk2X + α α  m8 m2 m2 m2 m−γ2 h 1 1 2 −2γ1 2 2 1 2 1 2 1 + + m1 h1 k(−A) 2 Quk2X . +C 4 + 2 α α α

(2.42)

Since h−AQu, QuiX is estimated as (2.35) but with u replaced by Qu, we obtain the estimate like (2.36) such as kuk2X

Cm21 ≤ 2 k(Lα − iλ)uk2X α   m
4 m8 m2 m2 m2 m−γ2 h 2 1 3 −2γ1 2 1 2 1 1 2 + 4 + + + m1 h1 k(QLα − iλ)uk2X , +C 2 α α α α

u ∈ DX (A) . (2.43)

This is the desired estimate in the case |µ| ≥ τ. Note that (2.43) is valid for any m1 , m2 ≥ m0 and α > 0 satisfying the condition (2.41). Such a set is not empty since the functions h j are bounded and γ j are strictly positive. Now we recall that σ(Lα ), σY (QLα ), and σPX (A|PX ) consist only of discrete eigenvalues. Moreover, we have from QA ⊂ AQ, σ(Lα ) = σY (QLα ) ∪ σPX (A|PX ) , (2.44)
and the formula (2.19) holds for all ζ ∈ C \ σ(Lα ) . Indeed, let ζ ∈ C be in the resolvent of QLα in Y and also of A|PX in PX. Let us show that ζ − Lα is injective in X, which shows that ζ belongs to the resolvent of Lα since σ(Lα ) in X consists of eigenvalues. If u ∈ DX (A) satisfies ˆ = ΛQu. ˆ (ζ − Lα )u = 0, then (ζ − QLα )Qu = 0 by the invariance QA ⊂ AQ and Λu Then Qu = 0 by the assumption, and therefore, (ζ − A)Pu = 0, which also gives Pu = 0 by the assumption. Thus u = 0, and we have shown the inclusion σ(Lα ) ⊂ σY (QLα ) ∪ σPX (A|PX ) . 17

On the other hand, let ζ belong to the resolvent set of Lα in X. Since σPX (A|PX ) ⊂ σ(Lα ) ˆ ⊂ DX (A), ζ is also a resolvent of A|PX in PX. If holds by the assumptions QA ⊂ AQ and Ker Λ ˆ u ∈ DX (A)∩Y satisfies (ζ −QLα )u = 0 then by setting v ∈ DX (A)∩PX as v = −iα(ζ −A|PX )−1 PΛu we see that w = u + v solves from QA ⊂ AQ, ˆ − iαPΛu ˆ = 0. (ζ − Lα )w = (ζ − Lα )u + (ζ − Lα )v = (ζ − QLα )u + iαPΛu Since ζ is a resolvent of Lα in X, w = 0. This implies u = Qw = 0. Hence ζ is a resolvent of QLα in Y. Thus (2.44) holds. By arguing as in Step 2 of the proof of Theorem 2.4, we can show that σY (QLα ) ⊂ {ζ ∈ C | Re ζ < 0} for all α. Therefore, we observe from σ(A) ⊂ {ζ ∈ C | Re ζ ≤ −1} and (2.44) that σ(Lα ) ⊂ {ζ ∈ C | Re ζ < 0} for all α. In particular, iR belongs to the resolvent set of Lα and also of QLα . Let λ ∈ R. If | αλ | < τ then (2.36) gives the estimate for the resolvent (iλ − QLα )−1 in Y. If | αλ | ≥ τ then (2.43) yields the estimate of the resolvent (iλ − Lα )−1 in X as long as (2.41) is satisfied. The estimate of the resolvent (iλ − QLα )−1 in Y is then obtained from the formula (iλ − QLα )−1 f = Q(iλ − Lα )−1 f ,

f ∈Y,

and by using the inequality kQukX ≤ kukX . As a summary, we have −γ  m2 1 m1
43 m81 m21 m22 m21 m2 2 h2 −2γ1 2 2 1 −1 + 4 + + + m1 h1 , k(iλ − QLα ) kY→Y ≤ C 2 + α α α α2 α

(2.45)

as long as α is large enough so that the set of (m1 , m2 ) satisfying (2.41) is not empty. The proof is complete.

3

Application to Kolmogorov flow

In this section we study the spectral property of the operator (1.8) related to the linearization of the Kolmogorov flow. Set Z 2π 2 2 2 2 X = L0 (T ) = {ω ∈ L (T ) | ω(x, y) dx = 0 a.e. y ∈ T} . 0

Let A be the realization of ∆ = ∂2x + ∂2y in L02 (T2 ), i.e., D(A) = W 2,2 (T2 ) ∩ L02 (T2 ) ,

Aω = ∆ω ,

ω ∈ D(A) .

Next let us denote by Mg the multiplication operator with the multiplier g, i.e., Mg f = g f . We ˆ the realization of −i∂ x Msin y (I + A−1 ) in L2 (T2 ) which is given by denote by Λ 0 ˆ = {ω ∈ L02 (T2 ) | ∂ x Msin y ω ∈ L02 (T2 )} , D(Λ)

ˆ = −i∂ x Msin y (I + A−1 )ω , Λω

ˆ . ω ∈ D(Λ)

Since i∂ x is realized as a self-adjoint operator in L02 (T2 ) and Msin y (I + A−1 ) is bounded in L02 (T2 ), ˆ is a closed operator. Moreover, since H 1 (T2 ) ∩ L2 (T2 ) ⊂ D(Λ), ˆ it is densely the operator Λ 0 2 2 defined in L0 (T ). We are interested in the spectral property of ˆ, Lα = A − iαΛ

D(Lα ) = D(A) . 18

(3.1)

Let us denote by Y the closed subspace of L02 (T2 ) defined by Z 2π 2 2 Y = {ω ∈ L0 (T ) | (Pl ω)(x, y) dy = 0 , |l| = 1 , for all x ∈ T} ,

(3.2)

0

where 1 (Pl ω)(x, y) = 2π

Z

2π

ω(s, y)e−ils ds eilx ,

l ∈ Z.

0

The orthogonal projection from L02 (T2 ) to Y is denoted by Q. We observe that Y is an invariant space under the action of A and that Y is the orthogonal complement of {a cos x + b sin x | a, b ∈ C} in X . The spaces L02 (T2 ) and Y are diagonalized as L02 (T2 ) = ⊕l∈Z\{0} Pl L02 (T2 ) ,

Y = ⊕l∈Z\{0} Pl Y ,

and each of Pl L02 (T2 ) and Pl Y is identified with L2 (T) and Yl respectively, where    if l , ±1 , L2 (T) R Yl =   { f ∈ L2 (T) | 2π f dy = 0} if l = ±1 .

(3.3)

(3.4)

0

The orthogonal projection from L2 (T) to Yl is denoted by Ql . Since Pl L02 (T2 ) and Pl Y are invariant spaces for Lα , the operator Lα is also diagonalized as Lα = ⊕l∈Z\{0} Lα |Pl L02 (T2 ) ,

(3.5)

where Lα |Pl L02 (T2 ) is the restriction of Lα to the invariant subspace Pl L02 (T2 ), which is identified with Lα,l in L2 (T) defined as follows: ˆl, Lα,l = Al − iαlΛ

D(Lα,l ) = W 2,2 (T) ,

(3.6)

Al = ∂2y − l2 , D(Al ) = W 2,2 (T) , ˆ l = Msin y (I + A−1 ˆ l ) = L2 (T) . Λ D(Λ l ),

(3.7)

where

It is straightforward to see the following result. Proposition 3.1 Let |l| ≥ 1. (i) −Al is positive self-adjoint in L2 (T) and satisfies Assumption 1 in L2 (T). Moreover, the invariance Ql Al ⊂ Al Ql holds. ˆ l )⊥ = Yl , Ran Λ ˆ l ∩ Ker Λ ˆ l = {0}, and Λ ˆ l satisfies Assumption 2 (i). (ii) (Ker Λ ˆ l ∩ Ker Λ ˆ l = {0} with l = ±1, since Proof. We give a proof only for the statement Ran Λ ˆ l ∩ Ker Λ ˆ l with l = ±1. Then, since the other statements are easy to check. Let f ∈ Ran Λ ˆ ±1 = {Const.}, there exists a constant c and a function g ∈ L2 (T) such that f = c = Λ ˆ l g. Ker Λ

c −1 −1 ˆ l , we have sin y I + A g = c. However, I + Al g = By the definition of Λ cannot belong l sin y to L2 (T) if c , 0. Hence, we must have c = 0, that is, f = 0. The proof is complete. The following corollary immediately follows from the above proposition. 19

Corollary 3.2 (i) −A is positive self-adjoint in L02 (T2 ) and satisfies Assumption 1 in L02 (T2 ). Moreover, the invariance QA ⊂ AQ holds. ˆ = { f = a sin x + b cos x , a, b ∈ C}, (Ker Λ) ˆ ⊥ = Y, Ran Λ ˆ ∩ Ker Λ ˆ = {0}, and Λ ˆ (ii) Ker Λ satisfies Assumption 2 (i).

3.1

Estimate without rate

In this subsection we aim to apply Theorem 2.4. Let us first check Assumption 2 (ii) for Lα,l in L2 (T). We observe that ˆ l = Msin y B2,l , Λ


B2,l = I + A−1 , l

(3.8)

ˆ l = Ker B2,l without and B2,l is bounded self-adjoint in L2 (T). We can also check that Ker Λ −1 difficulty. The operator B2,l is positive in Yl . To see this we set φ = Al f for f ∈ Yl , which satisfies k∂y φk2L2 + l2 kφk2L2 = −h f , φiL2 ≤ k f kL2 kφkL2 .

(3.9)

Since f, φ ∈ Yl we see k∂y φk2L2 ≥ kφk2L2 if l = ±1, and k∂y φk2L2 ≥ 0 if |l| ≥ 2. Thus we have kφkL2

 −1    2 k f k L2 ≤   l−2 k f kL2

if l = ±1 , if |l| ≥ 2 .

(3.10)

Hence, for |l| ≥ 1 and f ∈ Yl , h f , B2,l f iL2 = k f k2L2 + h f , φiL2 ≥ k f k2L2 − k f kL2 kφkL2 1 1 ≥ k f k2L2 − k f k2L2 = k f k2L2 . 2 2

(3.11)

Since (2.4) is also not difficult to check, Assumptions 1 and 2 are satisfied by the above operators. To apply Theorem 2.4 it remains to show ˆ l ) = [−1, 1]. Moreover, Λ ˆ l in L2 (T) does not have eigenvalues in C \ {0}. Proposition 3.3 σ(Λ Remark 3.4 Proposition 3.3 and its Corollary 3.5 below are not new, and have been proved in a more general framework; see Lin and Xu [18, Lemmas 2.4, 5.1]. We give a proof here just for the convenience of the reader. Proof of Proposition 3.3. We first observe that the spectrum of Msin y in L2 (T) consists of the 2 essential spectrum and is [−1, 1]. Since Msin y A−1 l is compact in L (T), the essential spectrum ˆ l coincides with the one of Msin y , and thus, is [−1, 1]. Hence it suffices to consider the of Λ ˆ l . Suppose that f ∈ L2 (T) and µ ∈ C \ {0} satisfies existence of eigenvalues of Λ ˆlf = µf . Λ

20

(3.12)

We first consider the case µ ∈ R \ {0}. Then we have (sin y − µ) f + sin y A−1 l f = 0, and thus, f+

sin y A−1 f = 0, sin y − µ l

y < Sµ ,

(3.13)

2,2 1+δ where S µ = {θ ∈ T | sin θ = µ}. Note that φ = A−1 (T) function for some l f ∈ W (T) is a C δ > 0 by the Sobolev embedding inequality. From (3.13) we see that f ∈ C 1+δ (T \ {S µ }), and also (3.13) implies that φ(yµ ) = 0 for yµ ∈ S µ , otherwise f cannot be in L2 (T). By the bootstrap argument and (3.13), we see that f is smooth in T \ {S µ }. Thus φ is smooth in T \ {S µ } and solves the ODE

(Msin y − µ)Al φ + Msin y φ = 0 . By the identity Msin y = Msin y − µ + µ, we have (Msin y − µ)(Al + 1)φ + µφ = 0, and thus, −(Al + 1)φ +

µ φ = 0, µ − sin y

y ∈ T \ {S µ } .

(3.14)

Case (i) µ ≥ 0. When 0 ≤ µ < 1 let yµ , zµ ∈ S µ be the points such that yµ ∈ [ 12 π, π] and zµ ∈ [2π, 25 π] (they are uniquely determined). When µ ≥ 1 we simply take yµ = 12 π and zµ = 52 π. Then µ − sin y ≥ 0 for y ∈ (yµ , zµ ), and we obtain Z zµ Z zµ µ ¯ |φ|2 dy = 0 . (3.15) (−Al − 1)φ φ dy + yµ yµ µ − sin y Note that the second integral converges due to the regularity φ ∈ C 1+δ (T) and φ(yµ ) = φ(zµ ) = 0 when 0 ≤ µ ≤ 1. As for the first integral, the integration by parts and the condition φ(yµ ) = φ(zµ ) = 0 when 0 ≤ µ ≤ 1 and the periodicity of φ when µ > 1 yield Z zµ Z zµ ¯ (−Al − 1)φ φ dy = |∂y φ|2 + (l2 − 1)|φ|2 dy , yµ

yµ

thus we have Z

zµ

|∂y φ| + (l − 1)|φ| dy + 2

2

Z

zµ

2

yµ

yµ

µ |φ|2 dy = 0 . µ − sin y

(3.16)

Hence φ = 0 in [yµ , zµ ]. When µ ≥ 1 we clearly have φ = 0 on [ 12 π, 52 π]. When 0 ≤ µ < 1 we 1 see from φ ∈ C 1+δ (T) that φ(yµ ) = φ0 (yµ ) = 0. However, since the singularity of is first µ − sin y order when 0 ≤ µ < 1 it is easy to see that any C 1+δ solution φ to the ODE (3.14) satisfying φ(yµ ) = φ0 (yµ ) = 0 must be trivial. Thus, we have f = 0 in T. Case (ii) µ < 0. The argument is the same as above, and we omit the details. Case (iii) µ < R. Since −µ f + Msin y (I + A−1 l ) f = 0, by taking the inner product with B2,l f , we have −µh f, B2,l f iL2 + hMsin y B2,l f, B2,l f iL2 = 0. The imaginary part of this equality gives (Im µ)h f, B2,l f iL2 = 0, and thus, from the definition of B2,l , we observe that f = 0 if |l| ≥ 2 and f = constant if |l| = 1. On the other hand, if |l| = 1 and f = constant then (I+A−1 l ) f = B2,l f = 0, which gives −µ f = 0. Thus f = 0 since µ , 0. The proof is complete.

21

ˆ l in L2 (T) is easily translated to Λ ˆ in L2 (T2 ). Indeed, The above result for Λ 0 ˆ = B1 B2 , Λ

B1 = −i∂ x Msin y ,

B2 = (I + A−1 ) ,

(3.17)

ˆ = Ker B2 . The and B1 is closed symmetric and B2 is bounded self-adjoint in L02 (T2 ), and Ker Λ operator B2 is positive in Y, for so is B2,l in Yl for each l ∈ Z \ {0} with a uniform lower bound in l. Proposition 3.3 therefore implies ˆ = R. Moreover, Λ ˆ in L2 (T2 ) does not have eigenvalues in C \ {0}. Corollary 3.5 σ(Λ) 0 ˆ l is Proof. Let f ∈ L02 (T2 ) and set fl (y) = (Pl f )(x, y)e−ilx . If ζ ∈ C and Im ζ , 0 then ζ − lΛ −1 ˆ l ) fl satisfies invertible for any l ∈ Z \ {0} and ωl = (ζ − lΛ kωl kL2 (T) ≤

C k fl kL2 (T) if |l| ≥ 2 , | Im ζ|

kωl kL2 (T) ≤ C(| Im ζ|)k fl kL2 (T) if |l| = 1 .

Here C is independent of l and C(| Im ζ|) depends only on | Im ζ| (the concrete dependence of C(| Im ζ|) on | Im ζ| is not needed in the argument below). Moreover, from ζωl −lMsin y B2,l ωl = fl , we also have klMsin y ωl kL2 (T) ≤ klMsin y A−1 l ωl kL2 (T) + |ζ| kωl kL2 (T) + k fl kL2 (T) ≤ C(ζ)k fl kL2 (T) P with C(ζ) depending only on ζ and independent of l. Hence, ω(x, y) = l∈Z\{0} ωl (y)eilx satisfies ˆ ω ∈ L02 (T2 ) and ∂ x Msin y ω ∈ L02 (T2 ). Clearly ω solves (ζ − Λ)ω = f . The uniqueness is also ˆ in L2 (T2 ). shown by taking the Fourier series in x. Thus ζ belongs to the resolvent set of Λ 0 ˆ ⊂ R. Since σ(lΛ ˆ l ) = [−l, l], we conclude that σ(Λ) ˆ = R. If ζ ∈ R is This shows σ(Λ) ˆ and ω ∈ D(Λ) ˆ is an eigenfunction then there exists l ∈ Z \ {0} such that an eigenvalue of Λ −ilx ˆ l )ωl = 0, the number ζ must be an ωl (y) = (Pl ω)(x, y)e is nontrivial. Since ωl satisfies (ζ − lΛ l ˆ l in L2 (T), which is a contradiction. The proof is complete. eigenvalue of Λ We can now apply Theorems 2.4, which yields the following result. Theorem 3.6 Let Lα be as in (3.1). Then lim sup k(iλ − QLα )−1 kY→Y = lim sup kQ(iλ − Lα )−1 kX→X = 0 .

|α|→∞ λ∈R

|α|→∞ λ∈R

(3.18)

A similar result holds also for Lα,l for each l ∈ Z \ {0}.

3.2

Estimate with rate

Theorem 3.6 does not give any estimates on the convergence rate. To apply Theorem 2.9 we ˆ l = Msin y (I + A−1 ) in L2 (T). Note that focus on the study of Λ l −Al = T l∗ T l ,

T l = ∂y − l

and therefore, [T l , B1 ] = Mcos y . 22

ˆ is obtained by the diagonalization Λ ˆ = ⊕l∈Z\{0} lΛ ˆ l . To simplify the notation we The result for Λ use the symbols u and v as scalar functions on T in this subsection (i.e., in this subsection u and v do not mean velocity fields). ˆ l as stated in Assumption 4. The main Our goal is to show the coercive estimates of Λ difficulty comes from the degeneracy of the critical points, i.e., the case when (sin y)0 |y=yµ = cos yµ vanishes for the point yµ ∈ sin−1 µ. This is the case µ = ±1. More precisely, the difficulty is to show (2.21) uniformly in a neighborhood of µ = ±1, rather than on the exact points µ = ±1. Below we divide the regime of µ into three parts. The first part, discussed in Lemma 3.7, corresponds to the case |µ| ≥ 1, though we can take |µ| slightly below 1 depending on the value m in (2.21). The second part, which is the core part of this section and discussed in Lemma 3.8, is 21 ≤ |µ| < 1. The last part is |µ| < 21 and will be treated in Lemma 3.9, where the critical point is nondegenerate, while we need to handle the additional nonlocality due to the presence of the projection Ql when |l| = 1. Lemma 3.7 There exist κ ∈ (0, 1) and C > 0 such that the following statements hold for all 1 δ ∈ (0, 100 ) and l ∈ Z \ {0}. If µ ∈ R satisfies 1 − κδ2 ≤ |µ| ≤ 1 + κδ2 then 1 ˆ l )uk2 2 + δ4 k(−Al ) 21 uk2 2
, δ2 kuk2L2 + kMcos y B2,l uk2L2 + k(−Al )− 2 uk2L2 ≤ C δ−2 k(µ − Λ L L

u ∈ H 1 (T) , (3.19)

while if |µ| > 1 then (|µ| − 1)2 kB2,l uk2L2 + (|µ| − 1)kMcos y B2,l uk2L2 1 ˆ l )uk2 2 , + |µ|(|µ| − 1)k(−Al )− 2 uk2L2 ≤ Ck(µ − Λ L

u ∈ H 1 (T) .

(3.20)

ˆ l )u for u ∈ H 1 (T), i.e., Proof. Set f = (µ − Λ (µ − Msin y )u − Msin y A−1 l u = f

(3.21)

ˆ l . Setting v = A−1 u and using Msin y = Msin y − µ + µ , (3.21) is also written by the definition of Λ l as (µ − Msin y )(Al + 1)v − µv = f .

(3.22)

1 ) and take κ > 0 sufficiently Note that (Al + 1)v = B2,l u by the definition. Below we fix δ ∈ (0, 100 small. Taking the inner product with (Al + 1)v and by considering the real part, we obtain Z 2π µ − sin y 1 |(Al + 1)v|2 dy + k∂y vk2L2 + (l2 − 1)kvk2L2 = Reh f , (Al + 1)viL2 . (3.23) µ µ 0

If 0 < 1 − κδ2 ≤ µ ≤ 1 + κδ2 then we have Z 2π 1 − sin y |(Al + 1)v|2 dy + k∂y vk2L2 + (l2 − 1)kvk2L2 µ 0 Z 1 1 − µ 2π = |(Al + 1)v|2 dy + Re h f , (Al + 1)viL2 µ µ 0 2 Z 2π κδ 1 ≤ |(Al + 1)v|2 dy + k f kL2 k(Al + 1)vkL2 . µ 0 µ 23

(3.24)

Thus, we have 2π

Z 0

(1 − sin y)|(Al + 1)v|2 dy + k∂y vk2L2 + (l2 − 1)kvk2L2 ≤ Cκδ k(Al + 2

Next we compute Z 2π

1)vk2L2

(1 − sin y)|(Al + 1)v| dy =

+ Ck f kL2 k(Al + 1)vkL2 .

Z

2

|y− 12 π|≥δ

0

(3.25)

Z

. . . dy +

Z

. . . dy |y− 21 π| 0 is small enough but independently of µ and δ, then   2 2 2 2 2 3 2 −2 2 δ k(Al + 1)vkL2 + k∂y vkL2 + (l − 1)kvkL2 ≤ C δ kukL∞ + δ k f kL2 . From Al v = u we finally obtain   1 ˆ l )uk2 2 + δ3 kuk2L∞ . δ2 kuk2L2 + k(−Al )− 2 uk2L2 ≤ C δ−2 k(µ − Λ L

(3.27)

The argument is the same as above for the case −1 − κδ2 ≤ µ ≤ −1 + κδ2 < 0, and we have (3.27) also in this case. The details are omitted here. The estimate of kMcos x B2,l uk2L2 follows from (3.25) and (3.27) by the inequality cos2 y = (1 − sin2 y) ≤ 2(1 ± sin y) . 1

Then it suffices to apply the interpolation inequality kuk2∞ ≤ CkukH 1 kukL2 and kukH1 ≤ k(−Al ) 2 ukL2 to obtain (3.19). Estimate (3.20) for the case |µ| > 1 easily follows from the identity (3.23). Indeed, (3.23) is written from (Al + 1)v = B2,l u, Z 2π
1 1 |µ| 1 2 1− sin y |B2,l u|2 dy + k∂y vk2L2 + (l2 − 1)kvk2L2 = Reh f , B2,l uiL2 , (1 − )kB2,l ukL2 + |µ| |µ| 0 µ µ which gives (3.20) for |µ| > 1. The details are omitted here. The proof is complete. The coercive estimate for |µ| < 1 − κδ2 is more delicate, especially when κδ2 < |µ ± 1| ≤ o(1) as δ → 0 due to the degeneracy around the points such that (sin y)0 = cos y = 0 and the nonlocality. To overcome the difficulty we apply a contradiction argument. The contradiction argument is useful since it enables us to focus on the functions which concentrate around the critical points, by which the nonlocality is reduced since the nonlocal operator has a smoothing effect and thus becomes a small perturbation of the local operator for such functions. The following lemma, which requires a long proof, is the core result of this subsection. 24

Lemma 3.8 Let κ ∈ (0, 1) be the number in Lemma 3.7. There exists C > 0 such that if 1 ), l ∈ Z \ {0}, and µ ∈ R with 12 ≤ |µ| < 1 − κδ2 , then δ ∈ (0, 100 1

k(−Al )− 2 uk2L2 +

  1 δ6 2 −2 2 ˆ 2 kA−1 uk ≤ C δ k(µ − Λ )uk + h−A u , ui l l L , L2 L2 1 − |µ| l 1 − |µ|

u ∈ H 2 (T) , (3.28)

and   ˆ l )uk2 2 + δ2 (1 − |µ|)h−Al u , uiL2 , kMcos y B2,l uk2L2 ≤ C δ−2 k(µ − Λ L

u ∈ H 2 (T) .

(3.29)

The proof consists of several steps. We first consider (3.28). ˆ l and Al preserve the real valued, without loss of Proof of (3.28). Since µ is a real number and Λ generality it suffices to show (3.28) for real valued functions. We make use of a contradiction argument. Suppose that the estimate   1 δ6 2 −1 2 −2 2 − 21 ˆ kA ukL2 ≤ C δ lim sup k(µ + i − Λl )ukL2 + h−Al u , uiL2 , k(−Al ) ukL2 + 1 − |µ| 1 − |µ| l ↓0 δ ∈ (0,

1 ), 100

l ∈ Z \ {0} ,

1 ≤ |µ| < 1 − κ|δ|2 , 2

u ∈ H 2 (T; R) (3.30)

1 ), {ln } ⊂ Z \ {0}, {n } ⊂ (0, ln−4 δ10 does not hold. Then there exist {δn } ⊂ (0, 100 n ), {µn } ⊂ (−1 + 1 1 2 2 2 κδn , − 2 ] ∪ [ 2 , 1 − κδn ), and {un } ⊂ H (T; R) such that lim n = 0, n→∞

lim δn = δ∞ ∈ [0,

n→∞

1 ], 100

lim ln = l∞ ∈ {±∞} ∪ Z \ {0} ,

n→∞

1 1 lim µn = µ∞ ∈ [−1 + κδ2∞ , − ] ∪ [ , 1 − κδ2∞ ] , n→∞ 2 2 and 1 kA−1 un k2L2 = 1 , 1 − |µn | ln   δ6n 2 ˆ 2 + lim δ−2 k(µ + i − Λ )u k h−A u , u i 2 n n ln n L ln n n L = 0. n n→∞ 1 − |µ| 1

k(−Aln )− 2 un k2L2 +

(3.31)

Note that we can actually take each n as small as we want. Set ˆ fn = δ−1 n (µn + in − Λln )un ,

vn = A−1 ln un .

(3.32)

Since un is real valued, so is vn , and vn satisfies (µn + in − Msin y ) Aln + 1)vn − (µn + in )vn = δn fn . 1

The normalized condition in (3.31) implies k(−Aln ) 2 vn k2L2 + integration by parts, k∂y vn k2L2 + ln2 kvn k2L2 + 25

1 kv k2 1−|µn | n L2

1 kvn k2L2 = 1 . 1 − |µn |

(3.33)

= 1, and thus, from the

(3.34)

Since supn kvn kH1 < ∞, we may assume that, after taking a suitable subsequence, {vn } converges to v∞ weakly in H 1 (T; R), and thus, strongly in L2 (T; R). Let S µn = {y ∈ T | sin y = µn } be the set of critical points. Since the terms with the factor n are negligible because of the smallness of n , we have from (3.33), µn vn (yµ ) + δn fn (yµ ) = 0

(modulo n ) ,

yµ ∈ S µn .

(3.35)

1. Case δ∞ = 0 and µ∞ = ±1. Below we focus on the case δ∞ = 0 and µ∞ = ±1; the other case is in fact easier to handle. Assume that µ∞ = 1. The proof for the case µ∞ = −1 is similar. Our first goal is to show that vn must concentrate around a small neighborhood of the critical points. Step 1: Behavior of µn − sin y around y = π2 . For large n there exist unique yn,1 ∈ (0, 12 π) and yn,2 ∈ ( 12 π, π) such that µn = sin yn, j , j = 1, 2. 1 Note that lim yn, j = π and by the Taylor expansion and (sin y)0 = cos y = 0 for y = 21 π, we n→∞ 2 have 1 π  π  |1 − µn | = |yn, j − |2 1 + O(|yn, j − |2 ) . (3.36) 2 2 2 Again by the Taylor expansion of sin y around yn, j , we see sin y = sin yn, j + cos yn, j (y − yn, j ) −

1 1 sin yn, j (y − yn, j )2 − cos yn, j (y − yn, j )3 + O(|y − yn, j |4 ) , 2 6

and we also have π π − yn, j + O(| − yn, j |3 ) , 2 2 π = 1 + O(| − yn, j |2 ) . 2

cos yn, j = sin yn, j Note that

π min{|y − yn,1 | , |y − yn,2 |} < | − yn, j | 2

if y ∈ (yn,1 , yn,2 ) .

Hence,   π 1 π µn − sin y = − ( − yn, j ) + (y − yn, j ) + O(| − yn, j |3 ) (y − yn, j ) , 2 2 2

y ∈ (yn,1 , yn,2 ).

Thus we have 1 π |µn − sin y| ≥ | − yn, j | min{|y − yn,1 | , |y − yn,2 |} 4 2 1 1 ≥ |1 − µn | 2 min{|y − yn,1 | , |y − yn,2 |} , 4

y ∈ (yn,1 , yn,2 ) .

On the other hand, if y ∈ (0, yn,1 ] then from the monotonicity of cos y in (0, π2 ), Z yn,1 1π 1 1 µn − sin y = cos s ds ≥ cos yn,1 (yn,1 − y) ≥ | − yn,1 | |yn,1 − y| ≥ |1 − µn | 2 |yn,1 − y| . 2 2 2 y 26

The similar bound holds for y ∈ (yn,2 , π) with yn,1 replaced by yn,2 . Hence we conclude that 1 1 |µn − sin y| ≥ |1 − µn | 2 min{|y − yn,1 | , |y − yn,2 |} , 4

y ∈ (0, π) .

(3.37)

Step 2: Estimate of kvn k2L2 . We take the inner product with vn in (3.33), which gives from the integration by parts, Z 2π
− (µn + in − sin y) |∂y vn |2 + (ln2 − 1)|v|2 dy 0 Z 2π + cos y ∂y vn v¯n dy − (µn + in )kvn k2L2 = δn h fn , vn iL2 . 0

The real part of this identity and the integration by parts yield Z 2π Z 1 2π 2 2 2
2 (µn − sin y) |∂y vn | + (ln − 1)|v| dy + µn kvn kL2 − sin y |v|2 dy = −δn h fn , vn iL2 2 0 0 Then, since 0 < 1 − µn  1, we have Z   1 2 kvn kL2 + (µn − sin y) |∂y vn |2 + (ln2 − 1)|vn |2 dy 4 y κ00 | π2 − yn,1 |} for sufficiently small κ00 > 0. Then we have from (3.37) and (3.36), Z In,1,1


2 Z vn (y) − vn (yn,1 )
2 µn − sin y C vn (y) − vn (yn,1 ) dy ≤ π dy (µn − sin y)2 + n2 | 2 − yn,1 | In,1,1 |y − yn,1 |
2 Z Cκ00 | π2 − yn,1 | vn (y) − vn (yn,1 ) ≤ π dy | 2 − yn,1 | |y − yn,1 |2 In,1,1 ≤ Cκ00 k∂y vn k2L2

31

( by the Hardy inequality ) . (3.54)

Next we see again from (3.37) and (3.36), Z Z π2
2 µn − sin y µn − sin y vn (y) − vn (yn,1 ) dy ≥ (v (y) − vn (yn,1 ))2 dy 2 2 2 + 2 n π (µ − sin y) +  (µ − sin y) 00 n n In,1,2 yn,1 +κ ( 2 −yn,1 ) n n Z π2 2 |vn (y)| ≥ −C dy yn,1 +κ00 ( π2 −yn,1 ) |µn − sin y| Z π2 1 2 dy − C|vn (yn,1 )| yn,1 +κ00 ( π2 −yn,1 ) |µn − sin y| C|vn (yn,1 )|2 C 2 kv k − π 2 n L ((yn,1 , 2 )) κ00 | π2 − yn,1 |2 κ00 | π2 − yn,1 | Ckvn k2L2 ((yn,1 ,yn,2 )) C|vn (yn,1 )|2 ≥− . − 1 κ00 |1 − µn | κ00 |1 − µn | 2 ≥−

(3.55)

Combining (3.54) and (3.55), we obtain F1 ≥ −

Ckvn k2L2 ((yn,1 ,yn,2 )) κ00 |1

− µn |

−

C|vn (yn,1 )|2 κ00 |1

− µn |

1 2

− Cκ00 k∂y vn k2L2 .

As for the term F2 , we have from (3.37) and (3.36), Z 1 |vn (y) − vn (yn, j )| dy |F2 | ≤ C|vn (yn,1 )| π In,1 | 2 − yn,1 | |y − yn, j | π 1 ≤ C|vn (yn,1 )| | − yn,1 |−1 |In,1 | 2 k∂y vn kL2 2 (by the Cauchy-Schwarz inequality and the Hardy inequality) π 1 ≤ C|vn (yn,1 )| | − yn,1 |− 2 k∂y vn kL2 2 2 C|vn (yn,1 )| ≤ + Cκ00 k∂y vn k2L2 . 1 00 κ |1 − µn | 2

(3.56)

(3.57)

Finally, the term G in (3.52) is estimated from (3.36) and (3.37) as |G| ≤ C

|vn (yn,1 )|2 |vn (yn,1 )|2 ≤ C . 1 | π2 − yn,1 | |1 − µn | 2

(3.58)

Collecting (3.56), (3.57), and (3.58), we have Ckvn k2L2 ((yn,1 ,yn,2 )) C|vn (yn,1 )|2 µn − sin y 2 |v | dy ≥ − − Cκ00 k∂y vn k2L2 . (3.59) − 1 2 + 2 n 00 |1 − µ | 00 (µ − sin y) κ n n κ |1 − µn | 2 In,1 n R µn −sin y The term I (µn −sin |v |2 dy in (3.50) is estimated exactly in the same way, and thus we y)2 +n2 n n,2 obtain Z 2π Ckvn k2L2 ((yn,1 ,yn,2 )) X C|vn (yn, j )|2 µn − sin y 2 µn |v | dy ≥ − − − Cκ00 k∂y vn k2L2 . (3.60) 1 2 + 2 n 00 |1 − µ | 00 (µ − sin y) κ 2 n n 0 n j=1,2 κ |1 − µn | Z

32

Finally we estimate the last term in the right-hand side of (3.49): Z 2π fn δn Re vn dy µn + in − sin y 0 Z π2 Z π fn fn = δn Re (vn − vn (yn,1 )) dy + δn Re (vn − vn (yn,2 )) dy π µn + in − sin y 0 µn + in − sin y 2 Z π Z π2 fn fn dy + δn vn (yn,2 ) Re dy + δn vn (yn,1 ) Re π µn + in − sin y 0 µn + in − sin y 2 Z 2π fn + δn Re vn dy µn + in − sin y π 5 X Jk . = k=1

By using the lower bound (3.37) and the Hardy inequality the terms J1 and J2 are estimated as |J1 | + |J2 | ≤ C

δn |1 − µn | 2 1

k fn kL2 k∂y vn kL2 ,

(3.61)

while J5 is simply estimated as |J5 | ≤ Cδn k fn kL2 kvn kL2 .

(3.62)

Next we estimate J3 and J4 . It suffices to consider J3 , for the estimate of J4 is similar. Note that the contribution from Im fn = n δ−1 n un is negligible in the estimate. Hence we have Z π2 (µn − sin y) fn J3 ≈ δn vn (yn,1 ) dy (3.63) 2 2 0 (µn − sin y) + n Let c > 0 be a sufficiently small number but independent of n such that yn,1 + c

δ2n 1

|1−µn | 2

≤

long as 1 − µn ≥ κδ2n as we have assumed in this lemma. Then let βn ∈ ( 13 , 3) be such that sin(yn,1 +

cδ2n |1 − µn | 2 1

) − µn = µn − sin(yn,1 − βn

cδ2n |1 − µn | 2 1

),

and set the interval IIn as IIn = (yn,1 − βn

cδ2n |1 − µn |

1 2

, yn,1 +

π ) ⊂ [0, ] , 2 |1 − µn | cδ2n

1 2

for which we have from the symmetry Z µn − sin y cos y dy = 0 . 2 2 IIn (µn − sin y) + n Then we observe that Z π2 Z fn (yn,1 )
(µn − sin y) fn (µn − sin y) cos y fn dy = − dy 2 2 2 2 cos yn,1 0 (µn − sin y) + n IIn (µn − sin y) + n cos y Z (µn − sin y) fn + dy 2 2 (0, π2 )\IIn (µn − sin y) + n = J3,1 + J3,2 . 33

π 2

as

The term J3,1 is decomposed as Z 1 1
(µn − sin y) cos y − fn dy J3,1 = 2 2 cos yn,1 IIn (µn − sin y) + n cos y Z
1 (µn − sin y) cos y + fn − fn (yn,1 ) dy , 2 2 cos yn,1 IIn (µn − sin y) + n and thus, C C 1 1 |IIn | 2 k fn kL2 (IIn ) + |IIn | 2 k∂y fn kL2 (IIn ) 1 |1 − µn | |1 − µn | 2 Cδn Cδn ≤ k fn kL2 + k∂y fn kL2 (IIn ) . 3 5 |1 − µn | 4 |1 − µn | 4

|J3,1 | ≤

The estimate of δ2n k∂y fn k2L2 (IIn ) is similar to (3.42) due to the choice of IIn . Hence we obtain  C  Cδn 1 2 2 kB 2 2 2 2 k f k + |1 − µ | |J3,1 | ≤ u k + δ k∂ u k + k∂ v k n L n 2,ln n L (IIn ) y n L . n y n L 3 5 |1 − µn | 4 |1 − µn | 4 Finally J3,2 is estimated as 1

C |1 − µn | 2 21 µn − sin y 2 ((0, π )\II ) k fn k L2 ≤ k ( ) k fn kL2 |J3,2 | ≤ Ck L 1 n 2 (µn − sin y)2 + n2 δ2n |1 − µn | 2 C ≤ k fn k L 2 . 1 δn |1 − µn | 4 Thus by using 1 − µn ≥ κδ2n , we have arrived at
|J3 | ≤ Cδn |vn (yn,1 )| |J3,1 | + |J3,2 |  δ3n |vn (yn,1 )|  2 2 2 2 k f k + δ kB u k + k∂ u k + k∂ v k ≤C n L n 2,ln n L (IIn ) y n L y n L . 1 1 |1 − µn | 4 |1 − µn | 2

(3.64)

By the definition of IIn , the estimate of δn kB2,ln un kL2 (IIn ) is the same as (3.47): δ2n kB2,ln un k2L2 (IIn ) ≤

Cδ6n C 1 k(−Aln ) 2 un k2L2 + Cκ0 + Cδ4n + 0 2 k fn k2L2 . |1 − µn | κ

(3.65)

Since J4 is estimated similarly as J3 , collecting (3.61), (3.62), (3.64), and (3.65), we have Z 2π fn δn Re vn dy µn + in − sin y 0 (3.66) X |vn (yn, j )|2 Cδ6n C 1 2 2 0 00 2 ≤C + k(−An ) 2 un kL2 + 0 2 k fn kL2 + Cκ + Cκ k∂y vn kL2 . 00 |1 − µ | 12 |1 − µ | κ n κ n j=1,2 Substituting the estimates (3.60) and (3.66) into (3.49), we obtain for a small but fixed κ00 > 0, k∂y vn k2L2 + (ln2 − 1)kvn k2L2 ≤C

kvn k2L2 ((yn,1 ,yn,2 )) κ00 |1 − µn |

+C

X |vn (yn, j )|2 κ00 |1 − µn | 2

1

j=1,2

+

(3.67) Cδ6n C 1 k(−Aln ) 2 un k2L2 + 0 2 k fn k2L2 + Cκ0 . |1 − µn | κ 34

The key point of (3.67) is that k∂y vn k2L2 is estimated in terms of the norms of vn only around the critical points, modulo small order terms as n → ∞. Step 5: inf n |1 − µn |−1 kvn k2L2 ((yn,1 ,yn.2 )) > 0. If |1 − µn |−1 kvn k2L2 ((yn,1 ,yn,2 )) → 0 for some subsequence {vn }, then we see from (3.46) and (3.67) that this subsequence satisfies lim sup n→∞

k∂y vn k2L2

+

(ln2

−


1)kvn k2L2

 C X |v (y )|2  Cκ0 n n, j 0 ≤ lim sup 00 + Cκ ≤ 00  1 κ j=1,2 |1 − µn | 21 κ n→∞

by taking κ0 > 0 small enough. Then by (3.39) this subsequence must satisfy lim sup |1 − µn |−1 kvn k2L2  1 , n→∞

since k fn kL2 → 0 as n → ∞. This contradicts with the normalized condition (3.31). Hence we conclude that inf n |1 − µn |−1 kvn k2L2 ((yn,1 ,yn.2 ) > 0. Step 6: Estimate of ∂y vn in (rn , yn,1 ). Let rn ∈ [0, 12 ] be the number in Step 2. In view of the sign of the function µ − sin y, we can expect a better bound of vn in the interval (rn , yn,1 ), compared with the estimates in the interval (yn,1 , yn,2 ). To see this we set ψn = vn − vn (yn,1 ), which satisfies (∂2y − ln2 + 1)ψn − (µn + in )

vn (yn,1 ) ψn = (µn + in ) + (ln2 − 1)vn (yn,1 ) µn + in − sin y µn + in − sin y (3.68) fn . + δn µn + in − sin y

Integrating over (rn , yn,1 ) after multiplying by ψn , we obtain from the integration by parts, Z yn,1 Z yn,1 |ψn |2 2 2 2 dy |∂y ψn | + (ln − 1)|ψn | dy + (µn + in ) µn + in − sin y rn rn Z yn,1 Z yn,1 vn (yn,1 )ψn 2 = −(µn + in ) dy − (ln − 1)vn (yn,1 ) ψn dy − ψn (rn )∂y ψn (rn ) (3.69) µn + in − sin y rn rn Z yn,1 fn ψn − δn dy . µn + in − sin y rn By recalling that n is taken small enough, we have Z yn,1 vn (yn,1 )ψn (µn + in ) dy µn + in − sin y rn Z yn,1   Z yn,1 −4|1−µn | 12 |ψn | |ψn | ≤ C|vn (yn,1 )| dy + dy , 1 µn − sin y rn yn,1 −4|1−µn | 2 µn − sin y 1

and since the Taylor expansion implies µn − sin y ≥ c(yn,1 − y)2 for y ∈ (rn , yn,1 − 4|1 − µn | 2 )

35

where c > 0 is a constant independent of n, we get Z yn,1 −4|1−µn | 21 Z yn,1 −4|1−µn | 21 |vn (yn,1 )| + |vn | |ψn | dy ≤ C dy µn − sin y |yn,1 − y|2 rn rn Z yn,1 −4|1−µn | 21 C|vn (yn,1 )|
1 1 ≤ +C dy 2 kvn kL2 ((rn ,yn,1 )) 1 4 (yn,1 − y) |1 − µn | 2 rn C|vn (yn,1 )| Ckvn kL2 ((rn ,yn,1 )) ≤ + . 1 3 |1 − µn | 2 |1 − µn | 4 On the other hand, we have from (3.37), Z yn,1 Z yn,1 |ψn | C |ψn | dy ≤ dy 1 1 1 |1 − µn | 2 yn,1 −4|1−µn | 2 |yn,1 − y| yn,1 −4|1−µn | 2 µn − sin y C ≤ k∂y ψn kL2 ((rn ,yn,1 )) 1 |1 − µn | 4 ( by the Cauchy-Schwarz inequality and the Hardy inequality) C = k∂y vn kL2 ((rn ,yn,1 )) . 1 |1 − µn | 4 Thus we obtain from (3.46) and the normalized condition (3.34), Z yn,1 vn (yn,1 )ψn (µn + in ) dy µn + in − sin y rn  |vn (yn,1 )|  |vn (yn,1 )| kvn kL2 ((rn ,yn,1 )) 2 + + k∂ v k ≤C (3.70) y n L ((rn ,yn,1 )) 1 1 1 |1 − µn | 4 |1 − µn | 4 |1 − µn | 2 |vn (yn,1 )| ≤C . 1 |1 − µn | 4 Similarly, the H¨older inequality and the Hardy inequality yield Z yn,1 ψn fn ψn δn dy ≤ δn k fn kL2 k kL2 ((rn ,yn,1 )) µn + in − sin y µn − sin y rn ψn 1 ≤ Cδn k fn kL2 |1 − µn |− 2 k kL2 ((rn ,yn,1 )) yn,1 − y δn ≤C k fn kL2 k∂y ψn kL2 ((rn ,yn,1 )) . (3.71) 1 |1 − µn | 2 On the other hand, (3.46) and (3.40) imply 3

|ψn (rn )∂y ψn (rn )| = |vn (rn ) − vn (yn,1 )| |∂y vn (rn )| ≤ C|1 − µn | 4 . Finally we observe from ψn = vn − vn (yn,1 ) that Z yn,1 Z yn,1 2 2 2 |ψn | dy − (ln − 1)vn (yn,1 ) ψn dy − (ln − 1) rn rn Z yn,1 Z yn,1 2 2 2 = −(ln − 1) |vn | dy + (ln − 1)vn (yn,1 ) vn dy rn

rn

Z 2 ≤ (ln − 1)vn (yn,1 )

1 yn,1 −4|1−µn | 2

vn dy +

Z

yn,1 1

yn,1 −4|1−µn | 2

rn

36

 vn dy ,

(3.72)

and (3.39) together with the normalized condition (3.34) implies 1

Z

yn,1 −4|1−µn | 2

(ln2 − 1)vn (yn,1 )

vn dy rn 1

Z ≤ |vn (yn,1 )|

yn,1 −4|1−µn | 2 rn

1

Z yn,1 −4|1−µn | 2  12   21 1 2 2 dy (ln − 1) (µn − sin y)|vn |2 dy µn − sin y rn 1

Z ≤ C|vn (yn,1 )|

yn,1 −4|1−µn | 2

 21  1 1 2 2 2
2 dy (l − 1) |1 − µ | + δ k f k 2 n n n n L (yn,1 − y)2 rn  12 |vn (yn,1 )|  2 ≤C (l − 1)|1 − µ | , n n 1 |1 − µn | 4

while Z (ln2

− 1)vn (yn,1 )

yn,1

1

1 yn,1 −4|1−µn | 2

vn dy ≤ C(ln2 − 1)|vn (yn,1 )| |1 − µn | 4 kvn kL2 ≤ C(ln2 − 1)|1 − µn |

again from the normalized condition (3.34). Hence we obtain Z yn,1 Z 2 2 − (ln − 1)vn (yn,1 ) ψn dy − (ln − 1) rn

|vn (yn,1 )| |1 − µn | 4 1

yn,1

|ψn |2 dy

rn

1 |vn (yn,1 )|  2 2 2 2 2 . ≤C (l − 1)|1 − µ | + (l − 1) |1 − µ | n n n n 1 |1 − µn | 4

(3.73)

Thus, the real part of (3.69) and the estimates (3.70)-(3.73) together with ∂y ψn = ∂y vn imply k∂y vn k2L2 ((rn ,yn,1 )) ≤ C



δ2n
|vn (yn,1 )|  3 k fn k2L2 + |1 − µn | 4 + 1 + (ln2 − 1)|1 − µn | . 1 |1 − µn | |1 − µn | 4

(3.74)

Step 7: Rescaling and limiting process Set wn (ξ) =

π 1 vn ( + |1 − µn | 2 ξ) , 2 |1 − µn | 1

1 4

ξ ∈ [−2, 2] .

(3.75)

Let cn > 0 be the number such that yn,1 =

π 1 − |1 − µn | 2 cn , 2

yn,2 =

π 1 + |1 − µn | 2 cn . 2

Note that µn = sin yn, j = sin

π 1 π 1 π − ( − yn, j )2 + ( − yn, j )4 + · · · , 2 2 2 4! 2

which gives π (yn, j − )2 = 2(1 − µn ) + O(|1 − µn |2 ) . 2 37

(3.76)

Hence we see from (yn, j − π2 )2 = (1 − µn )c2n by its definition, √ 1 cn = 2 + O(|1 − µn | 2 ) .

(3.77)

In particular, we have for all large n ,

1 ≤ cn ≤ 2

lim cn =

√

n→∞

2.

(3.78)

In virtue of Step 5, we have k∂ξ wn k2L2 (−2,2))

+

kwn k2L2 ((−2,2))

inf kwn k2L2 ((−cn ,cn )) n

+

kvn k2L2

≤ 1, |1 − µn | = inf |1 − µn |−1 kvn k2L2 ((yn,1 ,yn,2 )) > 0 .

≤

k∂y vn k2L2

(3.79)

n

That is, the sequence {wn } is uniformly bounded in H 1 ((−2, 2)), and thus, we may assume that wn weakly converges in H 1 ((−2, 2)) to some w∞ ∈ H 1 ((−2, 2)) and strongly converges in L2 ((−2, 2)) as well as in C η ([−2, 2]) for some η > 0. Moreover, by the uniform lower bound in (3.79), the limit w∞ is nontrivial. The direct computation shows that wn satisfies   2 2 (µn + in − sinn ξ)∂ξ wn = |1 − µn | (µn + in − sinn ξ)(ln − 1)wn + (µn + in )wn + δn gn , (3.80) ξ ∈ (−2, 2) . Here we have set π 1 sinn ξ = sin( + |1 − µn | 2 ξ) , 2

π 1 1 gn (ξ) = |1 − µn |− 4 fn ( + |1 − µn | 2 ξ) . 2

On the points ξ = ±cn we have 1

1

wn (−cn ) = |1 − µn |− 4 vn (yn,1 ) ,

wn (cn ) = |1 − µn |− 4 vn (yn,2 ) .

(3.81)

By (3.46) and the hypothesis (3.31) we have √ |w∞ (± 2)| = | lim wn (±cn )| n→∞   δ3n C 0 12 2 ≤ lim sup C k∂y un kL2 + Cκ + 0 k fn kL2 1 κ n→∞ |1 − µn | 2 0 ≤ Cκ , where κ0 > 0 can be taken arbitrary small and C is independent of κ0 . Thus we have √ w∞ (± 2) = 0 . Next we see sinn ξ = sin

π 1 1 − |1 − µn |ξ2 + |1 − µn |2 ξ4 · · · 2 2 4!

and π 1 1 1 µn = sin yn,2 = sin( + |1 − µn | 2 cn ) = 1 − |1 − µn |c2n + |1 − µn |2 c4n + · · · 2 2 4! 38

(3.82)

which gives   1 2 2 µn + in − sinn ξ = |1 − µn |(ξ − cn ) 1 + |1 − µn | qn (ξ) . 2

(3.83)

Here qn is a smooth function on [−2, 2] satisfying the uniform bound sup k n

dk qn kL∞ ((−2,2)) < ∞ dξk

for k = 0, 1, 2 .

Thus (3.80) is written as   1 2 2 (ξ − cn ) 1 + |1 − µn | qn ∂2ξ wn 2   1 = |1 − µn |(ln2 − 1)(ξ2 − c2n ) 1 + |1 − µn | qn wn + (µn + in )wn + δn gn . 2

(3.84)

Note that, from 1 − µn ≥ κδ2n and k fn kL2 → 0 as n → ∞, δn kgn kL2 ((−2,2)) ≤

Cδn |1 − µn | 2

1

k fn kL2 → 0

n → ∞.

(3.85)

(i) When lim sup |1 − µn |(ln2 − 1) = ∞: n→∞

In this case we may assume that lim |1 − µn |(ln2 − 1) = ∞ by taking a suitable subsequence. n→∞

Then we divide both sides of (3.84) by |1 − µn |(ln2 − 1) and consider the weak formulation with arbitrary test function ϕ ∈ C0∞ ((−2, 2)): 1
 −1 2 2 (ξ − c h∂ w , ∂ ) 1 + |1 − µ |q ξ n ξ n n ϕ iL2 ((−2,2)) n |1 − µn |(ln2 − 1) 2   1 1 2 2 h(µn + in )wn + δn gn , ϕiL2 ((−2,2)) . = h (ξ − cn ) 1 + |1 − µn |qn wn , ϕiL2 ((−2,2)) + 2 |1 − µn |(ln2 − 1) Note that 1
 h∂ξ wn , ∂ξ (ξ2 − c2n ) 1 + |1 − µn |qn ϕ iL2 ((−2,2)) 2 k

is uniformly bounded in n and |1 − µn | k ddξqkn kL∞ ((−2,2)) → 0 as n → ∞ for k = 0, 1, 2. Then, by taking the limit n → ∞ in the above weak formulation and using (3.85), we obtain the identity 1 2 (ξ − 2)w∞ = 0 in (−2, 2), i.e., w∞ = 0, which is a contradiction. 2 (ii) When lim sup |1 − µn |(ln2 − 1) < ∞: n→∞

In this case we may assume that lim |1 − µn |(ln2 − 1) = d∞ ∈ [0, ∞) by taking a suitable subn→∞ sequence. Then, by considering the weak formulation for (3.84) with arbitrary test function φ ∈ C0∞ ((−2, 2)) as above, we verify that the limit w∞ ∈ H 1 ((−2, 2)) satisfies, in the sense of distributions, √ (ξ2 − 2)∂2ξ w∞ = d∞ (ξ2 − 2)w∞ + 2w∞ , ξ ∈ (−2, 2) , w∞ (± 2) = 0 . (3.86) 39

√ We can show from (3.86) that w∞ ∈ H 1 ((−2, 2)) is √ smooth except for the points ξ = ± 2, 1 and thus (3.86) is satisfied √ pointwise in (−2, 2) \ {± 2}. Moreover, since w∞ ∈ H ((−2, 2)) and vanishes at ξ = ± 2, the equation (3.86) together with the Hardy inequality implies that w∞ ∈ H 2 ((−2, 2)). In particular, w∞ ∈ C 1+η ((−2, 2)) for some η > 0. Our aim is to show that w∞ = 0 in (−2, 2), for it leads to the contradiction. The difficulty is that the polynomial ξ2 − 2 satisfies (3.86) at least when d∞ = 0, therefore we need to derive some additional estimate for w∞ . The key is the estimate of ∂y vn outside the interval (yn,1 , yn,2 ), which is obtained in Step 6. Indeed, (3.75) implies that Z −cn Z −cn π 1 1 2 2 |(∂y vn )( + |1 − µn | 2 ξ)|2 dξ |∂ξ wn | dξ = |1 − µn | 2 −2 −2 Z yn,1 = |∂y vn |2 dy , 1 π 2 2 −2|1−µn |

and hence, (3.74) and (3.46) together with the assumption lim |1 − µn |(ln2 − 1) = d∞ ∈ [0, ∞) n→∞ lead to Z −cn 1 lim sup |∂ξ wn |2 dξ ≤ Cd∞ κ0 2 . n→∞

−2

Since the number κ0 > 0 is taken arbitrary small we have lim sup k∂ξ wn kL2 ((−2,−cn )) = 0. Hence, n→∞ √ for any a ∈ (−2, − 2), we see that k∂ξ w∞ kL2 ((−2,a)) ≤ lim inf k∂y wn kL2 ((−2,a)) ≤ lim sup k∂y wn kL2 ((−2,−cn )) = 0 . n→∞

n→∞

√ That is, ∂ξ w∞ = 0 for ξ ∈ (−2, − 2). Since w∞ ∈ C 1+η ((−2, 2)) we conclude that √ ∂ξ w∞ (− 2) = 0 .

(3.87)

√ Note that the singularity (ξ2 − 2)−1 at ξ = − 2 is first order. Then it is√easy to show √that the solution w∞ ∈ H 2 ((−2, 2)) to (3.86) satisfying the initial condition w∞ (− 2) = ∂ξ w∞ (− 2) = 0 must be trivial, i.e., w∞ = 0 in (−2, 2). This contradicts with kw∞ kL2 ((−2,2)) > 0. The proof is complete for the case 1. 2. Case δ∞ > 0 or 21 ≤ |µ∞ | < 1. First we exclude the possibility δ∞ > 0. Indeed, in this case 1 we have k(−Aln ) 2 un kL2 → 0 and inf n |1 − µn | > 0 by (3.31), which implies 1

k(−Aln ) 2 vn k2L2 +

1 1 kvn k2L2 ≤ Ck(−Aln ) 2 un k2L2 → 0 . |1 − µn |

This contradicts with the normalized condition in (3.34). Hence it remains to consider the case δ∞ = 0 and 21 ≤ |µ∞ | < 1. It suffices to consider the case 12 ≤ µ∞ < 1. The argument is similar to the case 1 above, but since µn is uniformly away from the degenerate critical points ±1, the problem is easier than the case |µ∞ | = 1, and in particular, we do not need Step 7 of the case 1 above, where a rescaling argument is used. So we only give the sketch of the proof here. First we do not need Step 1 and Step 2 of the case 1. As for the estimate of |vn (yn, j )|2 discussed in 40

Step 3 of the case 1, we take zn, j in this case as zn,1 ∈ (yn,1 , yn,1 +κ0 2 δ2n ) and zn,2 ∈ (yn,2 −κ0 2 δ2n , yn,2 ) which satisfy | f (zn,1 )|2 ≤ | f (zn,2 )|2 ≤

C κ0 2 δ2n C κ0 2 δ2n

k fn k2L2 ((yn,1 ,yn,1 +κ0 2 δ2 )) ≤ n

k fn k2L2 ((yn,2 −κ0 2 δ2 ,yn,2 )) ≤ n

C κ0 2 δ2n C κ0 2 δ2n

k fn k2L2 , k fn k2L2 .

Then (3.41) is replaced by
|vn (yn, j )|2 ≈ δ2n | f (yn, j )|2 ≤ Cδ2n | f (yn, j ) − fn (zn, j )|2 + | fn (z, j )|2 C ≤ Cκ0 2 δ4n k∂y fn k2L2 ((yn, j ,zn, j )) + 0 2 k fn k2L2 , κ and from kµn + in − sin ykL∞ ((yn, j ,zn, j )) ≤ Cκ0 2 δ2n , (3.42) is in this case replaced by   δ2n k∂y fn k2L2 ((yn, j ,zn, j )) ≤ C kB2,ln un k2L( (yn, j ,zn, j ))2 + κ0 4 δ4n k∂y un k2L2 + k∂y vn k2L2 . Hence we obtain 1

1

|vn (yn, j )|2 ≤ Cκ0 2 δ2n kB2,ln un k2L2 ((yn, j ,zn, j )) + Cκ0 6 δ6n k(−Aln ) 2 un k2L2 + Cκ0 2 δ2n k(−Aln )− 2 un k2L2 +

C k f k2 . 2 n L2 0 κ

The term δ2n kB2,ln un k2L2 ((yn, j ,zn, j )) has the same estimate as in (3.47). Hence we have arrived at 1

|v(yn, j )|2 ≤ Cδ6n k(−Aln ) 2 un k2L2 + Cκ0 +

C k fn k2L2 , κ0 2

(3.88)

1

as in (3.46) of the case 1. The estimate of k(−Aln ) 2 vn k2L2 is obtained as in Step 4 above. The key R 2π R 2π µ −sin y fn n 2 v dy |v | dy and δ Re identity is (3.49), and we decompose the integral 0 (µn −sin n n 2 2 y) +n 0 µn +in −sin y n into the small interval around the critical points yn, j and the interval away from the critical points as done in Step 4. From the estimate near the critical points the term κ00 k∂y vn k2L2 appears by the use of the Hardy inequality, while the term k∂y vn kL2 does not appear in the estimate away from the critical points. In these computations the condition |µ∞ | < 1 is crucial, for we can use the fact that the critical point is nondegenerate uniformly in n. In the end we can show the bounds as in (3.60) and (3.66), and by the smallness of κ00 , we obtain the estimate like (3.67), that is, k∂y vn k2L2 + (ln2 − 1)kvn k2L2 ≤

C C C 1 2 2 6 2 u k2 + kv k |v (y )| + Cδ k(−A ) k f k2 + Cκ0 . 2 + 2 n n n, j l n n n L L 2 n L2 0 κ00 κ00 κ (3.89)

Note that the factor |1−µn | is dropped by the condition |µ∞ | < 1. The estimates (3.88) and (3.89) yield for 0 < κ0  κ00  1, k∂y vn k2L2 + (ln2 − 1)kvn k2L2 ≤

C Cκ0 C 6 C 1 2 kv k + + 00 δn k(−Aln ) 2 un k2L2 + 00 0 2 k fn k2L2 + Cκ0 . (3.90) 2 n L 00 00 κ κ κ κ κ 0

1

Note that by the convergence condition (3.31) we can make the term Cκ + κC00 δ6n k(−Aln ) 2 un k2L2 + κ00 C k f k2 as small as we want by taking κ0 small enough and then by taking n large enough. κ00 κ0 2 n L2 41

Then, in virtue of the normalized condition in (3.31), we conclude that inf n kvn k2L2 > 0 and l∞ < ∞, which in particular implies that there exists a subsequence, denoted again by {vn }, which is bounded in H 1 (T) and converges strongly in C η (T) for some η > 0 to a nontrivial limit v∞ ∈ H 1 (T). Let y∞, j be the limit of yn, j , j = 1, 2. The limit v∞ also satisfies from (3.88), |v∞ (y∞, j )|2 = lim |vn (yn, j )|2 ≤ Cκ0 . n→∞

Since κ0 > 0 is taken arbitrary small, we obtain v∞ (y∞, j ) = 0 ,

j = 1, 2 .

(3.91)

We can also show that the limit v∞ is smooth in (0, 2π) \ {y∞, j } and satisfies (µ∞ − Msin y )(Al∞ + 1)v∞ − µ∞ v∞ = 0 ,

y ∈ (0, 2π) \ {y∞, j } .

(3.92)

∞ v ∈ L2 (T), and thus, v∞ ∈ H 2 (T) by the By (3.91), we see from the Hardy inequality that µ∞µ−sin y ∞ equation (3.92). Since v∞ is nontrivial, the function u∞ = Al∞ v∞ ∈ L2 (T) is also nontrivial. This ˆ l∞ in L2 (T) and the function u∞ is an eigenfunction to implies that µ∞ , 0 is an eigenvalue of Λ µ∞ . This contradicts with Proposition 3.3. The proof of (3.28) is complete.

Proof of (3.29). Again we will use the contradiction argument. Suppose that there exist {δn } ⊂ 1 1 1 2 2 2 ), {ln } ⊂ Z \ {0}, {n } ⊂ (0, ln−4 δ10 (0, 100 n ), {µn } ⊂ (−1 + κδn , − 2 ] ∪ [ 2 , 1 − κδn ), and {un } ⊂ H (T; R) such that lim n = 0, n→∞

lim δn = δ∞ ∈ [0,

n→∞

1 ], 100

lim ln = l∞ ∈ {±∞} ∪ Z \ {0} ,

n→∞

1 1 lim µn = µ∞ ∈ [−1 + κδ2∞ , − ] ∪ [ , 1 − κδ2∞ ] , n→∞ 2 2 and kMcos y B2,ln un k2L2 = 1 ,   2 2 ˆ 2 k(µ + i − Λ )u k + δ (1 − |µ |)h−A u , u i lim δ−2 n n ln n L 2 n ln n n L = 0. n n

(3.93)

n→∞

Note again that we can actually take each n as small as we want. As in the previous lemma, set ˆ fn = δ−1 n (µn + in − Λln )un ,

vn = A−1 ln un .

(3.94)

Since un is real valued, so is vn , and vn satisfies (µn + in − Msin y ) Aln + 1)vn − (µn + in )vn = δn fn .

(3.95)

From (3.28) and the condition 1 − |µn | ≥ κδ2n , we have lim k∂y vn k2L2 +

n→∞


1 kvn k2L2 = 0 , 1 − |µn |

(3.96)

which is essential in the proof below. Moreover, we may assume that lim k∂y Mcos y B2,ln un kL2 = ∞ ,

n→∞

42

δ∞ = 0 .

(3.97)

Otherwise, Mcos y B2,ln un converges strongly in L2 (T) for a subsequence, while the condition 1 lim k(−Aln )− 2 un kL2 = 0 implies that Mcos y B2,ln un converges to 0 weakly in L2 (T). Hence Mcos y B2,ln un

n→∞

converges to 0 strongly in L2 (T), which contradicts with lim kMcos y B2,ln un k2L2 = 1. We also note n→∞

that we may assume µn ≥ 21 for all n, for the case µn ≤ − 12 is handle in the same manner. We go back to the equation (3.95), and set u˜ n = Mcos y (Aln + 1)vn = Mcos y B2,ln un . Let us recall that yn, j ∈ S µn are the critical points, sin yn, j = µn , such that yn,1 ∈ (0, π2 ) and yn,2 ∈ ( π2 , π). Then (3.95) gives the identity µn vn (yn, j ) + δn fn (yn, j ) = 0

(modulo n ) .

(3.98)

Step 1: Estimate of k˜un k2L2 . We have from (3.95), µn + in fn k˜un k2L2 = Reh vn , Mcos y u˜ n iL2 + δn Reh , Mcos y u˜ n iL2 µn + in − sin y µn + in − sin y Z 2π Z 2π µn − sin y µn − sin y ≈ µn vn u˜ n cos y dy + δn fn u˜ n cos y dy 2 2 (µn − sin y) + n (µn − sin y)2 + n2 0 0 = I + II . (3.99) We decompose I and II as I = µn

Z 0

π 2

+

π

Z π 2

+

2π

Z


π

=

X

Ik ,

II = δn

Z 0

k=1,2,3

π 2

+

π

Z π 2

+

Z

2π


π

=

X

IIk ,

k=1,2,3

and I3 and II3 are away from the critical points and thus easy to estimate as follows: |I3 | ≤ Ckvn kL2 k˜un kL2 ≤ Ckvn kL2 → 0 as n → ∞ , |II3 | ≤ Cδn k fn kL2 k˜un kL2 ≤ Cδn k fn kL2 → 0 as n → ∞ .

(3.100)

Next we consider the estimate of I1 + II1 . The other term I2 + II2 is handled in the same manner. We first take κ0 > 0 and bn ∈ ( 14 , 4) such that Z µn − sin y π 1 π 0 0 cos y dy = 0 , T = [y − κ b δ , y + κ δ ] ⊂ [0, − ( − yn,1 )] n n,1 n n n,1 n 2 2 2 8 2 T n (µn − sin y) + n (3.101) by the symmetry of the integrand. Here the number κ0 > 0 is taken small enough so that at least T n ⊂ [0, π2 − 18 ( π2 − yn,1 )] is satisfied . The required smallness of κ0 at this stage is, however, independent of n because of the condition 1 − µn ≥ κδ2n and the fact 1 − µn = O(| π2 − yn,1 |2 ), which gives δn ≤ O(| π2 − yn,1 |). Later we will take κ0 > 0 even smaller. Then I1 + II1 is decomposed as Z Z I1 + II1 = + = III + IV . Tn

[0, π2 ]\T n

43

We observe that cos y C µn − sin y ≤ ≤ cos y , 2 2 (µn − sin y) + n µn − sin y |y − yn,1 |

π y ∈ (0, ) 2

(3.102)

holds. Then the term IV is estimated as Z Z µn − sin y µn − sin y |IV| ≤ |µn vn u˜ n cos y dy| + |δn f u˜ cos y dy| 2 2 2 2 n n [0, π2 ]\T n (µn − sin y) + n [0, π2 ]\T n (µn − sin y) + n Z
µn − sin y vn − vn (yn,1 ) u˜ n cos y dy| ≤ |µn 2 2 π [0, 2 ]\T n (µn − sin y) + n Z µn − sin y + |µn vn (yn,1 ) u˜ cos y dy| + Ck fn kL2 k˜un kL2 2 2 n [0, π2 ]\T n (µn − sin y) + n C ≤ Ck∂y vn kL2 k˜un kL2 + |vn (yn,1 )| k˜un kL2 + Ck fn kL2 k˜un kL2 . (3.103) 1 (κ0 δn ) 2 As for III, we see from (3.98), (3.102), and the Hardy inequality,   Z µn − sin y µ v + δ f − µ v (y ) − δ f (y ) u ˜ cos y dy |III| ≈ n n n n n n n,1 n n n,1 n 2 2 T n (µn − sin y) + n Z
µn − sin y ≤ µ v − µ v (y ) u ˜ cos y dy n n n n n,1 n 2 2 T n (µn − sin y) + n Z
µn − sin y + δ f − δ f (y ) u ˜ cos y dy n n n n n,1 n 2 2 T n (µn − sin y) + n Z

2 µn − sin y δ f − δ f (y ) u − u (y ) cos y dy ≤ Ck∂y vn kL2 k˜un kL2 + n n n n n,1 n n n,1 2 2 T n (µn − sin y) + n Z
2 µn − sin y + un (yn,1 ) δ f − δ f (y ) cos y dy n n n n n,1 2 2 T n (µn − sin y) + n ≤ Ck∂y vn kL2 k˜un kL2 + Cδn |T n | k∂y fn kL2 (Tn ) k∂y un kL2 (Tn ) k cos ykL∞ (Tn ) 1

+ C|u(yn,1 )|δn |T n | 2 k∂y fn kL2 (Tn ) k cos ykL∞ (Tn ) 1

≤ Ck∂y vn kL2 k˜un kL2 + Cκ0 δ2n |1 − µn | 2 k∂y fn kL2 (Tn ) k∂y un kL2 (Tn ) 1

3

(3.104)

1

+ Cκ0 2 |u(yn,1 )|δn2 |1 − µn | 2 k∂y fn kL2 (Tn ) . 1

Here we have used k cos ykL∞ (Tn ) ≤ C|1 − µn | 2 . The term δn k∂y fn kL2 (Tn ) is computed from (3.95) as δn k∂y fn kL2 (Tn ) ≈ k∂y µn − Msin y )(un + vn ) − µn ∂y vn kL2 (Tn ) ≤ k˜un kL2 (Tn ) + k(µn − Msin y )∂y un kL2 (Tn ) + Ck∂y vn kL2 (Tn ) 1

≤ k˜un kL2 (Tn ) + Cδn |1 − µn | 2 k∂y un kL2 + Ck∂y vn kL2 (Tn ) ≤C.

(3.105)

1

Here we have used kµn − sin ykL∞ (Tn ) ≤ Cδn k cos ykL∞ (Tn ) ≤ Cδn |1 − µn | 2 . Hence we have from 1 δn (1 − µn ) 2 k∂y un kL2 = o(1) as n → ∞ by the hypothesis, 1

1

1

|III| ≤ Cκ0 2 |u(yn,1 )|δn2 |1 − µn | 2 + o(1) , 44

n → ∞.

(3.106)

Collecting (3.103), (3.104), and (3.106), we arrived at 1

1

1

1

|I1 + II1 | ≤ C|vn (yn,1 )|(κ0 δn )− 2 + Cκ0 2 |un (yn,1 )|δn2 |1 − µn | 2 + o(1) ,

n → ∞.

(3.107)

−1

The similar bound is obtained for I2 + II2 . Thus it suffices to show |vn (yn,1 )|δn 2 → 0 and 1

1

|un (yn,1 )|δn2 |1 − µn | 2 ≤ C as n → ∞, for this gives a contradiction with the normalized condition on u˜ n by taking κ0 > 0 small enough. Step 2: Estimate of |un (yn,1 )|. Let ζn,1 ∈ [yn,1 − δn , yn,1 ] be the point such that |un (ζn,1 )|2 ≤

C kuk2 2 . δn L ([yn,1 −δn ,yn,1 ])

Observe that 1 1 1 |1 − µn | 2 ≤ | cos y| ≤ C|1 − µn | 2 , C

y ∈ [yn,1 − δn , yn,1 ] ,

holds. Then C kMcos y un k2L2 δn (1 − µn )
C C ≤ kMcos y (un + vn )k2L2 + kvn k2L2 ≤ . δn (1 − µn ) δn (1 − µn )

|un (ζn,1 )|2 ≤

(3.108)

Thus 1

−1

1

|un (yn,1 )| ≤ |un (yn,1 ) − un (ζn,1 )| + |un (ζn,1 )| ≤ Cδn2 k∂y un kL2 + Cδn 2 (1 − µn )− 2 , that is, 1

1

1

|un (yn,1 )|δn2 (1 − µn ) 2 ≤ Cδn (1 − µn ) 2 k∂y un kL2 + C ≤ C ,

(3.109)

as desired. Step 3: Estimate of |v(yn,1 )|2 δ−1 n . The argument is similar to Step 3 in the proof of (3.28). For sufficiently small κ0 > 0 as above, we set Ten as Ten = [yn,1 , yn,1 + κ0 2 δn ] ⊂ T n . We take zn,1 ∈ Ten so that | f (zn,1 )|2 ≤

C C 2 k f k ≤ k fn k2L2 . n 2 e 2 2 L ( T ) 0 0 n κ δn κ δn

Then
µ2n |vn (yn,1 )|2 ≈ δ2n | f (yn,1 )|2 ≤ Cδ2n | f (yn,1 ) − f (zn,1 )|2 + | f (zn,1 )|2
C ≤ Cδ2n κ0 2 δn k∂y fn k2L2 (Te ) + 0 2 k fn k2L2 . n κ δn 45

Then from (3.105) for the estimate of δ2n k∂y fn k2L2 (Tn ) , µ2n |vn (yn,1 )|2 ≤ Cκ0 2 δn +

C δ k f k2 , 2 n n L2 0 κ

that is, 02 |µn vn (yn,1 )|2 δ−1 n ≤ Cκ +

C k fn k2L2 . κ0 2

(3.110)

From Step 1 - Step 3 above, we have arrived at 1

1 = k˜un k2L2 ≤ Cκ0 2 +

C 3 κ0 2

k fn k2L2 + o(1) ,

n → ∞,

which leads to a contradiction since k fn kL2 → 0 and κ0 > 0 can be taken arbitrary small. The proof of (3.29) is complete. This concludes the proof of Lemma 3.8.

Finally we consider the case |µ| < 12 . The proof is similar to the case 2 in the proof of Lemma 3.8. The only difference is the influence of the projection Ql when l = ±1, which yields an additional nonlocal term in the limit equation when we preform the contradiction argument. Lemma 3.9 Let κ ∈ (0, 1) be the number in Lemma 3.7. There exists C > 0 such that if 1 ), l ∈ Z \ {0}, and µ ∈ R with |µ| < 21 , then δ ∈ (0, 100   − 12 2 −1 2 −2 2 6 ˆ k(−Al ) ukL2 + kAl ukL2 ≤ C δ kQl (µ − Λl )ukL2 + δ h−Al u , uiL2 , (3.111) and kMcos y B2,l uk2L2

  −2 2 2 ˆ ≤ C δ kQl (µ − Λl )ukL2 + δ h−Al u , uiL2 ,

(3.112)

for all u ∈ H 2 (T) ∩ Yl . Here Ql : L2 (T) → Yl is the orthogonal projection on Yl . Again the proof consists of several steps. We first consider (3.111). Proof of (3.111). The proof is very similar to the proof of (3.28) and is based on the contradiction argument. Again it suffices to consider real valued functions. Suppose that the estimate   1 2 −2 2 6 ˆ 2 uk ≤ C δ lim sup kQ (µ + i − Λ )uk + δ h−A u , ui k(−Al )− 2 uk2L2 +kA−1 l l l L , l L2 L2 ↓0 (3.113) 1 1 2 δ ∈ (0, ) , l ∈ Z \ {0} , |µ| < , u ∈ H (T; R) ∩ Yl 100 2 1 1 1 does not hold. Then there exist {δn } ⊂ (0, 100 ), {ln } ⊂ Z \ {0}, {n } ⊂ (0, ln−4 δ10 n ), {µn } ⊂ (− 2 , 2 ), and {un } ⊂ H 2 (T; R) ∩ Yln such that lim n = 0, n→∞

1 ], n→∞ 100 1 1 lim µn = µ∞ ∈ [− , ] , n→∞ 2 2 lim δn = δ∞ ∈ [0,

lim ln = l∞ ∈ {±∞} ∪ Z \ {0} ,

n→∞

46

and 1

2 k(−Aln )− 2 un k2L2 + kA−1 ln un k L 2 = 1 ,   −2 2 6 ˆ lim δn kQln (µn + in − Λln )un kL2 + δn h−Aln un , un iL2 = 0 .

(3.114)

n→∞

Note again that we can take each n small as we want. We first observe that δ∞ = 0, otherwise 1 we have k(−Aln ) 2 un kL2 → 0(n → ∞) due to the second condition in (3.114), from which we easily reach a contradiction to the normalized condition in (3.114). Moreover, if |l∞ | , 1 then the situation is exactly the same as the case 2 in the proof of Lemma 3.8, for Ql = I when |l| , 1 1 is first order when |µ| < 21 . Therefore, it remains to consider the and the singularity of µ−sin y case |l∞ | = 1 and |µ∞ | ≤ 12 . Let us focus on the case l∞ = 1 and 0 ≤ µ∞ ≤ 21 ; the other cases are handled in the same manner. Then we may assume that ln = 1 for all n by taking a subsequence if necessary, though we often keep the notation ln for convenience. Set ˆ fn = δ−1 n (µn + in − Λln )un ,

vn = A−1 ln un .

(3.115)

Since un is real valued, so is vn , and vn satisfies (µn + in − Msin y ) Aln + 1)vn − (µn + in )vn = δn fn .

(3.116)

It is convenient to introduce the value 1 ϑn = 2π

2π

Z

δn fn dy ,

0

which gives δn fn = δn Qln fn + ϑn . Note that Aln + 1 = ∂2y + ln2 − 1 = ∂2y . Thus, ϑn is computed from the condition vn ∈ Yln as 1 ϑn = − 2π

Z

2π

sin y ∂2y vn

0

1 dy = 2π

2π

Z

sin y vn dy .

(3.117)

0

Then (3.116) is written as (µn + in − Msin y ) Aln + 1)vn − (µn + in )vn − ϑn = δn Qln fn .

(3.118)

and the trace relation of (3.35) is now written as µn vn (yµ ) + ϑn = −δn (Qln fn )(yµ )

(modulo n ) ,

(3.119)

where yµ is any critical point. A key difference from the proof of (3.28) is that the role of µn vn (yn, j ), where yn, j is the critical point, has to be replaced by µn vn (yn, j ) + ϑn , and similarly, the |µ v (y )|2 role of fn is replaced by Qln fn . Then the estimate of n n n,1j obtained in Step 3 in the proof of |1−µn | 2

(3.28), is replaced by 1

|µn vn (yn, j ) + ϑn |2 ≤ Cδ6n k(−Aln ) 2 un k2L2 + Cκ0 + Cδ4n +

47

C kQl fn k2 2 , κ0 2 n L

(3.120)

where κ0 > 0 is arbitrary sufficiently small number. The proof of (3.120) is exactly the same as the proof of (3.46), and note that the factor 1 − µn is not needed here, in virtue of the condition |µn | ≤ 21 . Similarly, the estimate of δ2n kB2,ln un k2L2 ((yn, j ,zn, j )) stated in (3.47) is also valid as follows. 1

δ2n kB2,ln un k2L2 ((yn, j ,zn, j )) ≤ Cδ6n k(−Aln ) 2 un k2L2 + Cκ0 + Cδ4n +

C kQln fn k2L2 . 2 0 κ

(3.121)

To estimate k∂y vn k2L2 we use Aln + 1)vn −

δn Qln fn (µn + in )vn + ϑn = . µn + in − sin y µn + in − sin y

(3.122)

By taking the inner product with vn and looking at the real part of it, we obtain the equality of the form (3.49). Thus we have Z k∂y vn k2L2

≤−

3π 2


µn − sin y µn vn + ϑn vn dy − δn Re 2 2 (µn − sin y) + n

− π2

Z

3π 2

− π2

Qln fn vn dy . µn + in − sin y (3.123)

) rather than over (0, 2π) for convenience in view of Here the integral is written over (− π2 , 3π 2 the positions of the critical points µn = sin yn when |µn | < 12 . To estimate the first term in R 3π the right-hand side of (3.123), we divide the integral − 2π dy into the part near the critical points 2 R R R dy. The integral dy near the critical points dy and the part away from the critical parts c Icr Icr Icr becomes the computation of the principal value as in Step 4, and is built upon the decomposition as follows. Z
µn − sin y µn vn + ϑn vn dy 2 2 Icr (µn − sin y) + n Z Z
µn − sin y µn − sin y µn vn − µn vn (yn, j ) vn dy + (µn vn (yn, j ) + ϑn ) v dy . = 2 2 2 2 n Icr (µn − sin y) + n Icr (µn − sin y) + n Then we can show from the Hardy inequality and also from kvn kH1 ≤ Ck∂y vn kL2 for vn ∈ H 1 (T)∩ Yln with |ln | = 1 that Z
µn − sin y | µn vn + ϑn vn dy| ≤ Cκ0 k∂y vn k2L2 (3.124) 2 2 Icr (µn − sin y) + n with κ0 > 0 small enough by taking the size of the interval Icr suitably small. Here the fact that the critical point yn (sin yn = µn ) is nondegenarate, cos yn , 0, is crucial. For the integral over µn −sin y c Icr there is no singularity from the term (µn −sin , and thus, we have y)2 + 2 n

Z | c Icr


µn − sin y C µn vn + ϑn vn dy| ≤ 0 kvn k2L2 . 2 2 (µn − sin y) + n κ

(3.125)

R 3π Q fn Note that we can allow κ0 in the denominator. The term δn Re − 2π µn +ilnn−sin v dy is handled in y n 2 P5 the same way as in the estimate of k=1 Jk in Step 4 of the proof of (3.28); the term fn there is simply replaced by Qln fn in this case. The key point is that we need to decompose the integral 48

again into the part around the critical points and the part away from the critical points, and the size of the interval for the decomposition is in this case chosen depending on δn ; see the argument for J3 in Step 4 of the proof of Lemma 3.8. By using the interpolation inequality |vn (yn, j )|2 ≤ kvn k2L∞ ≤ Ck∂y vn kL2 kvn kL2 for vn ∈ H 1 (T) ∩ Yln and |µ|n < 21 , one can show Z |δn Re

3π 2

− π2

Qln fn vn dy| µn + in − sin y

(3.126)

C C 1 ≤ 0 2 kQln fn k2L2 + Cκ0 k∂y vn k2L2 + 0 kvn k2L2 + Cδ6n k(−Aln ) 2 un k2L2 + Cκ0 . κ κ Here the number κ0 > 0 is taken small enough. The details are omitted here. Collecting (3.124), (3.125), and (3.126), by taking κ0 > 0 small enough (but independent of n) we have k∂y vn k2L2 ≤

C C 1 kvn k2L2 + Cδ6n k(−Aln ) 2 un k2L2 + 0 2 kQln fn k2L2 . 0 κ κ

(3.127)

Since
1 lim sup kvn k2L2 + δ6n k(−Aln ) 2 un k2L2 + kQln fn k2L2 ≤ lim sup kvn k2L2 n→∞

n→∞

holds by the hypothesis of the contradiction argument, we must have inf n kvn k2L2 > 0, otherwise the normalized condition in (3.114) is not satisfied due to the bound (3.127). Hence we conclude that {vn } is bounded in H 1 (T; R) ∩ Yln , and up to a subsequence, it strongly converges to a nontrivial limit v∞ ∈ H 1 (T; R) ∩ Yl∞ with l∞ = 1 in the topology of C η (T) for some η > 0. By taking the limit in the equality −h∂y vn , ∂y (µn − in − Msin y )ϕiL2 = µn hvn , ϕiL2 + δn h fn , ϕiL2 = µn hvn , ϕiL2 + δn hQln fn , ϕiL2 + δn

2π

Z

2π

Z

ϕ dy ,

fn dy 0

0

for any ϕ ∈ C ∞ (T) , we see that −h∂y v∞ , ∂y (µ∞ − Msin y )ϕi

L2

= µ∞ hv∞ , ϕi + lim (δn L2

n→∞

Z

2π

2π

Z

ϕ dy ,

fn dy) 0

ϕ ∈ C ∞ (T) .

0

(3.128) From (3.117) we have Z 2π Z δn fn dy = 0

2π

Z sin y vn dy →

0

2π

sin y v∞ dy =: c∞

(n → ∞) .

(3.129)

0

Let S µ∞ be the set of the critical points for µ∞ , i.e., S µ∞ = {θ ∈ T | sin θ = µ∞ }. Then, from (3.128) and (3.129), the limit v∞ is shown to be smooth in T \ S µ∞ , and (µ∞ − sin y)∂2y v∞ = µ∞ v∞ + c∞ ,

y ∈ T \ S µ∞ .

We observe that (3.120) implies for any y∞ ∈ S µ∞ , 1

|µ∞ v∞ (y∞ ) + c∞ | ≤ lim sup Cδ6n k(−Aln ) 2 un k2L2 + Cκ0 + n→∞

49

C 2
kQ f k ≤ Cκ0 , l n n L2 2 κ0

(3.130)

which gives by letting κ0 → 0, µ∞ v∞ (y∞ ) + c∞ = 0 ,

c∞ =

Z

2π

v∞ sin y dy ,

y∞ ∈ S µ∞ .

(3.131)

0 ∞ +c∞ ∈ L2 (T) by the Hardy inequality, and thus, (3.131) implies the regularity ∂2y v∞ ∈ Then µµ∞∞v−sin y L2 (T). Hence we see that v∞ ∈ H 2 (T; R) ∩ Y1 is nontrivial and satisfies Z 2π 2 (µ∞ − sin y)∂y v∞ = µ∞ v∞ + c∞ , c∞ = v∞ sin y dy , y ∈ T. (3.132)

0

Our goal is to show v∞ must be trivial, which leads to a contradiction. Recall that we are assuming that 0 ≤ µ∞ ≤ 12 and v∞ is real-valued. If µ∞ = 0 then c∞ = 0 by (3.131), thus ∂2y v∞ = 0 in T. Since v∞ ∈ H 2 (T) ∩ Y1 we have v∞ = 0, which is a contradiction. Next we consider the case 0 < µ∞ ≤ 21 . Let y∞, j ∈ S µ∞ , j = 1, 2, be such that y∞,1 ∈ ( π2 , π) and yn,2 ∈ (2π, 5π ). Then µ∞ − sin y ≥ 0 for y ∈ (y∞,1 , y∞,2 ). Thus we see 2 Z y∞,2 Z y∞,2 (µ∞ v∞ + c∞ )2 2 dy , ∂y v∞ (µ∞ v∞ + c∞ ) dy = µ∞ − sin y y∞,1 y∞,1 which makes sense by the condition (3.131). The integration by parts and (3.131) imply Z y∞,2 Z y∞,2 (µ∞ v∞ + c∞ )2 2 |∂y v∞ | dy + µ∞ dy = 0 . (3.133) µ∞ − sin y y∞,1 y∞,1 Therefore, we conclude that µ∞ v∞ + c∞ = 0 on [y∞,1 , y∞,2 ]. Set w∞ = ∂y v∞ ∈ H 1 (T), which then satisfies w∞ = 0 on [y∞,1 , y∞,2 ]. Moreover, from (3.132) and also from c∞ = −µ∞ v(y∞, j ), we see Z y µ∞ v∞ + c∞ µ∞ ∂y w∞ = = w∞ dz , y > y∞,2 , w∞ (y∞,2 ) = 0 . (3.134) µ∞ − sin y µ∞ − sin y y∞,2 Then it is easy to see that w∞ = 0 for y ∈ (y∞,2 , y∞,2 + τ) for some τ > 0, and thus, w∞ = 0 for all y > y∞,2 . Hence w∞ = 0, i.e., v∞ is a constant. Since v∞ ∈ Y1 we must have v∞ = 0, which is a contradiction. The proof of (3.111) is complete. Proof of (3.112). The proof is again very similar with the proof of (3.29) and is based on a contradiction argument. Then the problem is reduced to the analysis of the sequence {˜un }, where u˜ n = Mcos y B2,ln un , un ∈ H 2 (T) ∩ Yln . As in proof of (3.111), it suffices to consider the case |ln | = 1 and δ∞ = 0, and without loss of generality we may assume that ln = 1 for all n. From the hypothesis of the contradiction argument, we have the convergence   2 2 ˆ 2 lim δ−2 k(µ − Λ )u k + δ h−A u , u i (3.135) n ln n L 2 ln n n L = 0 , n n n→∞

and therefore, from (3.111) we have lim kvn k2H1 = 0 .

n→∞

(3.136)

Then we can apply the same argument as in the proof of (3.29), for the argument there relies only on (3.135) and (3.136). The only difference from the proof of (3.29) is that, here we use 50

the identities (3.118), (3.119), and the estimates (3.120), (3.121), by taking into account the presence of the projection Qln . We omit the details. The proof of Lemma 3.9 is complete.

Let us set          h1 (m, µ) =        

κ 1 − |µ| ≥ 2 m κ κ if − 2 ≤ 1 − |µ| < 2 , m m κ if |µ| > 1 + 2 , m

1

m−1 (1 − |µ|)− 2

if

1 0

(3.137)

and κ if 1 − |µ| ≥ 2 m κ κ if − 2 ≤ 1 − |µ| < 2 , m κ m if |µ| > 1 + 2 , m

 1   (1 − |µ|) 2       m−1 h2 (m, µ) =         0

(3.138)

Then h j are bounded in m and µ. Moreover, h j (m, −µ) = h j (m.µ) holds. From this symmetry and by Lemmas 3.7, 3.8, and 3.9, we immediately obtain Proposition 3.10 There exist C, m0 > 0 such that for all m ≥ m0 , µ ∈ R, and l ∈ Z \ {0},   1 2 2 − 21 2 2 −4 2 ˆ 2 k(−Al ) ukL2 ≤ C m k(µ + Λl )ukL2 + m h1 (m, µ) k(−Al ) ukL2 , (3.139) 1 κ u ∈ H 2 (T) , ≤ |µ| ≤ 1 + 2 , 2   m 1 1 ˆ l )uk2 2 + m−4 h21 (m, µ) k(−Al ) 2 uk2 2 , k(−Al )− 2 uk2L2 ≤ C m2 kQl (µ + Λ L L (3.140) 1 u ∈ H 2 (T) ∩ Yl , |µ| < . 2  1 ˆ l )uk2 2 + m−2 h22 (m, µ) k(−Al ) 2 uk2 2 , kMcos y B2,l ukL2 ≤ C m2 k(µ + Λ L L (3.141) 1 κ 2 u ∈ H (T) , ≤ |µ| ≤ 1 + 2 , 2  m 1 2 2 −2 2 2 ˆ l )uk 2 + m h2 (m, µ) k(−Al ) 2 uk 2 , kMcos y B2,l ukL2 ≤ C m kQl (µ + Λ L L (3.142) 1 2 u ∈ H (T) ∩ Yl , |µ| < . 2 On the other hand, if |µ| > 1 then C ˆ l )uk2 2 , k(µ + Λ L 2 (|µ| − 1) C ˆ l )uk2 2 , ≤ k(µ + Λ L (|µ| − 1) C ˆ l )uk2 2 , ≤ k(µ + Λ L |µ| (|µ| − 1)

kuk2L2 ≤ kMcos y B2,l uk2L2 1

k(−Al )− 2 uk2L2

51

u ∈ H 1 (T) , u ∈ H 1 (T) , u ∈ H 1 (T) .

(3.143) (3.144) (3.145)

Note that the constants C and m0 in Proposition 3.10 are independent of l ∈ Z \ {0}. Let us ˆ l , and thus, it is convenient to introduce recall that Lα,l is defined as Lα,l = Al − iαlΛ α˜ = α(l) ˜ = αl .

(3.146)

We are interested in the estimate of k(iλ − QLα,l )−1 kYl →Yl by applying Theorem 2.9. In particular, the dependence of the estimate on α˜ is important. For this purpose let us introduce the function F(α, ˜ µ) as F(α, ˜ µ) =

 m2

inf

m1 ,m2 ≥m0

1 α˜ 2

 21 m1
43 m81 m21 m22 m21 m−1 2 h2 (m2 , µ) −4 2 + + 4 + + + m1 h1 (m1 , µ) . (3.147) α˜ α˜ α˜ 2 α˜

Here h j are defined by (3.137) and (3.138). Our aim is to obtain the upper bound for F(α, ˜ µ). 3

1

Case 1: 1 − |µ| ≥ κ 4 α˜ − 2 . In this case we first take m2 as  1 1 3 2 . m2 = α(1 ˜ − |µ|) 3

(3.148)

1

This choice of m2 together with 1 − |µ| ≥ κ 4 α˜ − 2 satisfies the condition 1 − |µ| ≥ κm−2 2 , and then, 1 we have h2 (m2 , µ) = (1 − |µ|) 2 . Thus it follows that m21 m22 m21 m−1 2 h2 (m2 , µ) = α˜ 2 α˜

for any m1 > 0 .

With this choice of m2 we next take m1 as m1 = (

α˜ 1 )6 . 1 − |µ|

(3.149)

Then we have 3

κ κ κ4 2 1 = 1 =( 1 )3 κ2 ≤ 1 . 2 2 m1 (1 − |µ|) α˜ 3 (1 − |µ|) 3 α˜ 2 (1 − |µ|) 1

−2 Thus, h1 (m1 , µ) = m−1 by its definition, and we can also check the balance 1 (1 − |µ|)

m21 m−1 2 h2 (m2 , µ) 2 = m−4 1 h1 (m1 , µ) . α˜ Let us now compute the size of each term in the right-hand side of (3.147) for m1 and m2 defined 3 1 as (3.148) and (3.149): note that we use the lower bound 1 − |µ| ≥ κ 4 α˜ − 2 assumed in the case 1. 4 m21 m1 43 2 − 65 − 10 − 16
3 −1 − 61 9 (1 − |µ|)− 9 ≤ α ≤ ( ) = α ˜ (1 − |µ|) = α ˜ ˜ κ , α˜ 2 α˜ m2 2 1
4 1 1
4 ( 1 )4 = α˜ − 3 (1 − |µ|)− 3 ≤ α˜ − 2 κ− 4 = κ−1 α˜ −2 , α˜ m21 m22 1 2 1 1 = α˜ 3 + 3 −2 (1 − |µ|)− 3 + 3 = α˜ −1 . 2 α˜

52

Thus, in the case 1 we have  m2 1 2 α˜

2 m1
43 m81 m21 m22 m21 m−1 1 2 h2 (m2 , µ) −4 2 + + 4 + + + m1 h1 (m1 , µ) ≤ C α˜ − 2 . 2 α˜ α˜ α˜ α 1

(3.150)

Here C is also independent of l. 3 1 3 1 Case 2: −κ 4 α˜ − 2 ≤ 1 − |µ| < κ 4 α˜ − 2 . In this case we take m1 and m2 as 1

1

m1 = m2 = κ 4 α˜ 4 .

(3.151)

Then we see from κ ∈ (0, 1), 3 1 κ4 κ κ1− 2 . > ≥ 1 − |µ| = 1 1 m2j α˜ 2 α˜ 2

Hence h1 (m1 , µ) = 1 and h2 (m2 , µ) = m−1 2 for this choice of m1 and m2 . We can also check that there exist C, C 0 > 0 depending only on κ such that m21 m22 1 m21 m22 m21 m−1 2 h2 (m2 , µ) 0 −4 2 ≤ C m1 h1 (m1 , µ) ≤ C 2 . ≤ C α˜ 2 α˜ α˜ Let us compute the size of each term as in the case 1: m21 m1 4 1 3
4 1 ≤ ( ) 3 = κ 4 α˜ − 4 3 = κ 3 α˜ −1 , 2 α˜ α˜ 2 m 1
4 1 ( 1 )4 = κ 2 α˜ − 2 = κ2 α˜ −2 , α˜ m21 m22 = κα˜ −1 . α˜ 2 Thus, also in the case 2 we have  m2 1 α˜ 2

2 m1
43 m81 m21 m22 m21 m−1 1 2 h2 (m2 , µ) −4 2 + + 4 + + + m h (m , µ) ≤ C α˜ − 2 . 1 1 1 2 α˜ α˜ α˜ α˜ 1

3

(3.152)

1

Case 3: 1 − |µ| < −κ 4 α˜ − 2 . In this case let us take 1

m1 = m2 = α˜ 4 . Then we have 1+

κ 3 1 1 = 1 + κα˜ − 2 < 1 + κ 4 α˜ − 2 < |µ| . 2 mj

Thus, h j (m j , µ) = 0 by the definition, and m21 m1
34 ≤ = α˜ −1 , 2 α˜ α˜ m2 m2 m2 ( 1 )4 ≤ 1 2 2 = α˜ −1 . α˜ α˜ 53

(3.153)

Hence we obtain  21  m2 m1
43 m81 m21 m22 m21 m−1 2 h2 (m2 , µ) 1 −4 2 − 12 + + + + + m h (m , µ) ≤ C α ˜ , 1 1 1 α˜ 2 α˜ α˜ 4 α˜ 2 α

(3.154)

as desired. As a summary, we have arrived at the upper bound of F(α, ˜ µ) such that 1

1

F(α, ˜ µ) ≤ C|α| ˜ − 2 = C|αl|− 2 ,

(3.155)

where C is independent of α, µ, and l. Then Theorem 2.9 implies the following result. Theorem 3.11 The following statements hold for all sufficiently large |α|. (1) Let |l| ≥ 2 and let Lα,l be the operator in L2 (T) as in (3.6). Then 1

sup Re ζ ≤ −c |αl| 2 ,

(3.156)

ζ∈σ(Lα,l )

and sup k(iλ − Lα,l )−1 kL2 (T)→L2 (T) ≤ λ∈R

C 1

|αl| 2

.

(3.157)

Here c and C are independent of α and l. The result is also true for the case |l| = 1 by replacing Lα,l and L2 (T) by QLα,l and Yl , respectively. (2) Let Lα be the operators in X as in (3.1). Then the set {ζ ∈ C | Re ζ > −1} is contained in the resolvent set of Lα in X, and we have sup kQ(iλ − Lα )−1 kX→X = sup k(iλ − QLα )−1 kY→Y ≤ λ∈R

λ∈R

C 1

|α| 2

.

(3.158)

In particular, the bound (3.158) implies the spectral bound for QLα in Y such that sup

ζ∈σY (QLα )

1

Re ζ ≤ −c|α| 2 .

(3.159)

Here c > 0 is independent of α. ˆ l . The statement (1) follows from Theorem 2.9 and the estimate Recall that Lα,l = Al − iαlΛ (3.155) for Fl (α, ˜ µ) with α˜ = αl. Note that Ql = I when |l| ≥ 2. The statement (2) follows from the statement (1) and the diagonalization of Lα . The statement (1) is stronger than (2), for (1) gives the decay on αl rather than on α.

3.3

Improved estimate away from the degenerate critical points

In view of the degeneracy of the critical points the resolvent estimate for Lα,l with the rate 1 O(|αl|− 2 ) obtained in Theorem 3.11 is considered to be optimal around the degenerate critical points, i.e., when | αlλ | is close to 1. On the other hand, when | αlλ | is away from 1 the natural rate is 2 O(|αl|− 3 ). The aim of this section is to achieve this improvement by using the specific property ˆ rather than the abstract result of Theorem 2.9 which is based on the estimate for F(α, of Λ, ˜ µ) with α˜ = αl. The key is the following lemma. 54

Lemma 3.12 There exists C > 0 such that (i) for all l ∈ Z \ {0} with |l| ≥ 2, µ ∈ R with |µ| < 1, and u ∈ L2 (T),  C  2 C − 12 2 2 ˆ l )uk2 2 , kuk2L2 ≤ µ k(−A ) uk + kM B uk k(µ + Λ l cos y 2,l L2 + L2 L 2 1−µ (1 − µ2 )2 (ii) for l = ±1, µ ∈ R with |µ| < 1, and u ∈ L2 (T), Z 2π 1 2 u dy|2 kukL2 ≤ | 2π 0  C  C 1 ˆ l )uk2 2 . + k(−Al )− 2 Ql uk2L2 + kMcos y B2,l uk2L2 + kQl (µ + Λ L 1 − |µ| (1 − |µ|)2 In particular, if u ∈ Yl in addition, then  C  C 1 ˆ l )uk2 2 . kuk2L2 ≤ k(−Al )− 2 uk2L2 + kMcos y B2,l uk2L2 + kQl (µ + Λ L 1 − |µ| (1 − |µ|)2

(3.160)

(3.161)

(3.162)

Here C is independent of µ, l, and u. Proof. We first consider the estimate of kB2,l uk2L2 for u ∈ Yl . From cos2 y + sin2 y = 1 we have kB2,l uk2L2 = kMcos y B2,l uk2L2 + kMsin y B2,l uk2L2 = kMcos y B2,l uk2L2 + kMsin y B2,l uk2L2 , and for µ ∈ R, kMsin y B2,l uk2L2 = kµu + Msin y B2,l u − µuk2L2 = kµu + Msin y B2,l uk2L2 + µ2 kuk2L2 − 2µhµu + Msin y B2,l u, uiL2
2 −1 ˆ l )uk2 2 + µ2 kB2,l uk2 2 + kA−1 = k(µ + Λ l uk 2 − 2hB2,l u, Al uiL2 L

L

L

ˆ l )u, uiL2 . − 2µh(µ + Λ −1 2 2 2 Setting v = A−1 l u, we have hB2,l u, Al uiL2 = hAl v + v, viL2 = −k∂y vkL2 − (l − 1)kvkL2 . Thus we have arrived at the identity 2 2 −1 2
ˆ l )uk2 2 + µ2 kB2,l uk2 2 + 2k∂y A−1 kB2,l uk2L2 = kMcos y B2,l uk2L2 + k(µ + Λ l ukL2 + (2l − 1)kAl ukL2 L L ˆ l )u, uiL2 . − 2µh(µ + Λ

(3.163) When |µ| < 1 the identity (3.163) gives 2 2 −1 2
ˆ l )uk2 2 + µ2 2k∂y A−1 (1 − |µ|2 )kB2,l uk2L2 = kMcos y B2,l uk2L2 + k(µ + Λ uk + (2l − 1)kA 2 l l ukL2 L L ˆ l )u, uiL2 − 2µh(µ + Λ 1

ˆ l )uk2 2 + Cµ2 k(−Al )− 2 uk2 2 ≤ kMcos y B2,l uk2L2 + k(µ + Λ L L ˆ − 2µh(µ + Λl )u, uiL2 . ˆ l )u, uiL2 = hQl (µ + Λ ˆ l )u, uiL2 , which yields Since u ∈ Yl we have h(µ + Λ ˆ l )u, uiL2 ≤ C|µ| kQl (µ + Λ ˆ l )ukL2 kB2,l ukL2 , µh(µ + Λ 55

(3.164)

for B2,l is invertible in Yl . Hence we obtain ˆ l )uk2 2 + Cµ2 k(−Al )− 12 uk2 2 (1 − |µ|2 )kB2,l uk2L2 ≤ kMcos y B2,l uk2L2 + k(µ + Λ L L +

(3.165)

Cµ2 ˆ l )uk2 2 . kQl (µ + Λ L 1 − |µ|2

This proves (3.160) when |l| ≥ 2 and |µ| < 1 since Ql = I and B2,l is invertible in L2 (T) if |l| ≥ 2. ˆ l )uk2 2 = kQl (µ + Λ ˆ l )uk2 2 + k(I − Ql )(µ + Λ ˆ l )uk2 2 and then observe When |l| = 1 we use k(µ + Λ L L L R ˆ l )uk2 2 = | 1 2π sin y v dy|2 for u ∈ Yl and v = A−1 u. Thus (3.165) gives for that k(I − Ql )(µ + Λ l 2π 0 L |l| = 1, Z 2π 1 2 2 2 2 2 ˆ l )uk 2 + | sin yA−1 (1 − µ )kB2,l ukL2 ≤ kMcos y B2,l ukL2 + kQl (µ + Λ l u dy| L 2π 0 Cµ2 1 ˆ l )uk2 2 + Cµ2 k(−Al )− 2 uk2L2 + kQl (µ + Λ L 2 1−µ C 1 ˆ l )uk2 2 . ≤ kMcos y B2,l uk2L2 + Ck(−Al )− 2 uk2L2 + kQl (µ + Λ (3.166) L 1 − µ2 Then (3.161) follows from (3.166) and kuk2L2

= k(I −

Ql )uk2L2

+

Z kQl uk2L2

≤| 0

2π

u dy|2 + CkB2,l Ql uk2L2 .

The proof is complete. The main result of this subsection is stated as follows. Theorem 3.13 There exist C, M, α0 > 0 such that the following statement holds for all α ∈ R with |α| ≥ α0 . Let λ ∈ R and l ∈ Z \ {0} with |l| ≥ 2. Then  1 λ C    | > if 1 − | ,  2 1 1  λ 3  αl 3 (1 − | 2 |αl| |αl| |)   αl    C M λ 1  −1 if − ≤1−| |≤ , (3.167) k(iλ − Lα,l ) kL2 (T)→L2 (T) ≤  1 1 1   αl 2 2 2  |αl| |αl| |αl|    C λ M    if 1 − | | < − 1 .   |αl| (| λ | − 1) αl |αl| 2 αl The result is true also for the case |l| = 1 by replacing Lα,l and L2 (T) by Ql Lα,l and Yl , respectively. Proof. Set α˜ = αl, and it suffices to consider the case α˜ > 0 since the case α˜ < 0 is handled ˆ l . Let |l| ≥ 2. The proof is very similar to the in the same way. Note that Lα,l = Al − iα˜ Λ abstract one used in Theorem 2.9. Indeed, the only difference is that, instead of the interpolation 1 1 inequality kuk2L2 ≤ k(−Al ) 2 ukL2 k(−Al )− 2 ukL2 which is used in Theorem 2.9, we apply Lemma 3.12 in estimating kukL2 . Set µ = αλ˜ . Let us recall that (2.37) leads to ˆ l )uk2 2 ≤ C k(iλ − Lα,l )uk2 2 + CkB2,l ukL2 k(−Al )− 12 ukL2 + k(−Al ) 21 ukL2 kMcos y B2,l ukL2 αk(µ ˜ +Λ L L α˜ (3.168) 56

for any u ∈ L2 (T). Moreover, in virtue of Proposition 3.10, we also have the estimates of 1 Mcos y B2,l u and (−Al )− 2 u as in (2.38) and (2.39), that is, for m1 , m2 ≥ 102 , kMcos y B2,l uk2L2

Cm22 Cm22 1 2 ≤ 2 k(iλ − Lα,l )ukL2 + kB2,l ukL2 k(−Al )− 2 ukL2 α˜ α˜   m4 1 2 h (m , µ) k(−Al ) 2 uk2L2 + C 22 + m−2 2 2 2 α˜ 2 Cm4 Cm 1 ≤ 2 2 k(iλ − Lα,l )uk2L2 + 2 2 kB2,l uk2L2 + k(−Al )− 2 uk2L2 α˜ α˜   m4 1 2 + C 22 + m−2 h (m , µ) k(−Al ) 2 uk2L2 , 2 2 2 α˜

(3.169)

and 1

k(−Al )− 2 uk2L2 ≤

Cm21 Cm41 2 k(iλ − L )uk kB2,l uk2L2 + 2 α,l L α˜ 2 α˜ 2  m2 m2 m2 m−1 h (m , µ)  1 2 1 2 1 2 2 −4 2 +C + + m1 h1 (m1 , µ) k(−Al ) 2 uk2L2 . 2 α˜ α˜

(3.170)

We will take m1 ≤ m2 . Then 1

kMcos y B2,l uk2L2 + k(−Al )− 2 uk2L2 C(m21 + m22 ) C(m41 + m42 ) 2 k(iλ − L )uk + kB2,l uk2L2 α,l L2 α˜ 2 α˜ 2   m2 m2 + m4 m21 m−1 1 1 2 2 2 h2 (m2 , µ) −2 2 −4 2 + m2 h2 (m2 , µ) + + m1 h1 (m1 , µ) k(−Al ) 2 uk2L2 +C 2 α˜ α˜ 2 4 Cm Cm ≤ 2 2 k(iλ − Lα,l )uk2L2 + 2 2 kB2,l uk2L2 α˜ α˜ (3.171)  m4  1 2 −2 2 −4 2 2 + C 2 + m2 h2 (m2 , µ) + m1 h1 (m1 , µ) k(−Al ) 2 ukL2 . α˜ ≤

Combining (3.168), (3.170), and (3.171), we obtain from m1 ≤ m2 ,  21  m4 m2 m−1 h (m , µ) C 2 2 2 h (m , µ) k(iλ − Lα,l )uk2L2 + C 22 + 1 2 + m−4 kB2,l uk2L2 1 1 1 α˜ α˜ α˜  m4  12 1 2 −4 2 + C 22 + m−2 h (m , µ) + m h (m , µ) k(−Al ) 2 uk2L2 2 1 2 2 1 1 α˜  m4  21 C 2 2 −2 2 −4 2 ≤ k(iλ − Lα,l )ukL2 + C 2 + m2 h2 (m2 , µ) + m1 h1 (m1 , µ) kB2,l uk2L2 α˜ α˜  m4  12 1 2 −2 2 −4 2 + C 2 + m2 h2 (m2 , µ) + m1 h1 (m1 , µ) k(−Al ) 2 uk2L2 . α˜ (3.172)

ˆ l )uk2 2 ≤ αk(µ ˜ − lΛ L

Now we apply Lemma 3.12. If 1 − |µ| > 0 then (3.160) implies  C  C 1 ˆ l )uk2 2 , kuk2L2 ≤ kMcos y B2,l uk2L2 + k(−Al )− 2 uk2L2 + k(µ + Λ L 1 − |µ| (1 − |µ|)2 57

and thus, (3.171) and (3.172) yield kuk2L2

m42 C  m22 2 ≤ k(iλ − Lα,l )ukL2 + 2 kB2,l uk2L2 2 1 − |µ| α˜ α˜  m4   1 2 −2 2 −4 2 2 2 + 2 + m2 h2 (m2 , µ) + m1 h1 (m1 , µ) k(−Al ) ukL2 α˜ 1  m4  12 C 2 2 −2 2 −4 2 k(iλ − Lα,l )ukL2 + 2 + m2 h2 (m, µ) + m1 h1 (m1 , µ) kB2,l uk2L2 + α(1 ˜ − |µ|)2 α˜ α˜  12  m4  1 2 −2 2 −4 2 2 2 + 2 + m2 h2 (m, µ) + m1 h1 (m1 , µ) k(−Al ) ukL2 . α˜

Let us consider the case 3

1

1 − |µ| > κ 4 α˜ − 2 .

(3.173)

We take m1 and m2 as 1

m1 = m2 = α(1 ˜ − |µ|) 2 which satisfies 1 − |µ| >

κ , m2j


13

j = 1, 2, and gives the balance

,

m42 α˜ 2

(3.174) 1
2 2 −2 2 . = m−2 2 h2 (m2 , µ) = m2 (1 − |µ|)

Moreover, we have 2 −6 −1 −1 −2 2 m−4 < κ−1 m−2 1 h1 (m1 , µ) = m1 (1 − |µ|) 2 (1 − |µ|) = κ m2 h2 (m2 , µ) .

With this choice of m1 and m2 it follows that  m4  12 m42 1 2 −2 2 −4 2 + + m2 h2 (m2 , µ) + m1 h1 (m1 , µ) (1 − |µ|)α˜ 2 α(1 ˜ − |µ|)2 α˜ 2
C 1 1 + 4 ≤C 4 ≤ 1 1 2 5 α˜ 2 α˜ 3 (1 − |µ|) 3 α˜ 3 (1 − |µ|) 3 if α is large enough but independently of l, and then the terms with kB2,l uk2L2 in the right-hand side of the estimate for kuk2L2 above are absorbed by the left-hand side. Hence kuk2L2 is bounded from above as   m4  C  m22 1 2 2 −2 2 −4 2 2 uk2 (m , µ) k(−A ) + h (m , µ) + m h k(iλ − L )uk + m kuk2L2 ≤ α,l 2 1 l 2 2 1 1 L2 L2 1 − |µ| α˜ 2 α˜ 2 1  m4  12  C 1 2 2 −2 2 −4 2 2 2 + k(iλ − Lα,l )ukL2 + 2 + m2 h2 (m2 , µ) + m1 h1 (m1 , µ) k(−Al ) ukL2 . α(1 ˜ − |µ|)2 α˜ α˜ 2 m2 m42 m22

1 1 2 2 uk2 + C ≤C + k(iλ − L )uk + k(−A ) 2 α,l l L L2 (1 − |µ|)α˜ 2 α˜ 2 (1 − |µ|)2 α˜ 2 α˜ 2 (1 − |µ|)2 m22 m42 m22
1 2 ≤C k(iλ − Lα,l )ukL2 + C 2 + 2 k(−Al ) 2 uk2L2 . (3.175) 2 2 (1 − |µ|)α˜ α˜ α˜ (1 − |µ|) 1

The term k(−Al ) 2 uk2L2 is estimated as in (2.35), and thus, since Ql = I for |l| ≥ 2, 1

k(−Al ) 2 uk2L2 ≤ Ck(iλ − Lα,l )ukL2 kB2,l ukL2 . 58

(3.176)

Hence we have kuk2L2 ≤ C

m22 m42 m22
2 2 k(iλ − L )uk + C + k(iλ − Lα,l )uk2L2 , α,l L2 2 2 2 2 (1 − |µ|)α˜ α˜ α˜ (1 − |µ|)

that is, from (3.174), kukL2 ≤ C

2

1 α˜ (1 − |µ|) 2 3

1 3

+

(1 − |µ|) 3 α˜

2 3

+

1




α˜ (1 − |µ|) 4 3

5 3

k(iλ − Lα,l )ukL2 ≤

C α˜ (1 − |µ|) 3 2 3

1

k(iλ − Lα,l )ukL2 (3.177) 1

by taking into account the condition (3.173). Note that (3.177) provides the factor C α˜ − 2 in the 1 regime 1 − |µ| ∼ O(α˜ − 2 ). Next we consider the case 3

1

1

−M α˜ − 2 ≤ 1 − |µ| ≤ κ 4 α˜ − 2

(3.178)

with M ≥ 1 which will be taken large enough. In this case we simply apply Theorem 3.11 and thus, kukL2 ≤

C α˜ 2 1

k(iλ − Lα,l )ukL2 .

(3.179)

Note that the constant C is independent of M. This complete the proof of (3.167) for the case 1 1 − | αλ˜ | ≥ −M α˜ − 2 and |l| ≥ 2. Finally we consider the case 1

1 − |µ| < −M α˜ − 2 .

(3.180)

In this case Lemma 3.7 and the symmetry of the estimate for −µ and µ imply that C ˆ l )uk2 2 , k(µ + Λ L (1 − |µ|)2 C ˆ l )uk2 2 , k(µ + Λ ≤ L 1 − |µ| C ˆ l )uk2 2 . k(µ + Λ ≤ L |µ(1 − |µ|)|

kB2,l uk2L2 ≤ kMcos y B2,l uk2L2 1

k(−Al )− 2 uk2L2

Combining these estimates with (3.168) yield C C ˆ l )uk2 2 + C 1 k(µ + Λ ˆ l )ukL2 k(−Al ) 12 ukL2 k(iλ − Lα,l )uk2L2 + k(µ + Λ L 2 1 − |µ| 2 α˜ |µ|(1 − |µ|) C C ˆ l )uk2 2 + C k(−Al ) 21 uk2 2 ≤ k(iλ − Lα,l )uk2L2 + αk(µ ˜ +Λ L L 2 α˜ α|µ|(1 ˜ − |µ|) α˜ 1 − |µ| C C 1 k(−Al ) 2 uk2L2 , ≤ k(iλ − Lα,l )uk2L2 + (3.181) α˜ α˜ 1 − |µ|

ˆ l )uk2 2 ≤ αk(µ ˜ +Λ L

since (3.180) implies that |µ| > 1 and C C 1 ≤ ≤ α|µ|(1 ˜ − |µ|)2 M 2 2 59

if M is large enough but independently of l. Then (3.181) with the estimate of kB2,l uk2L2 gives kB2,l uk2L2 ≤


C 1 1 1 k(−Al ) 2 uk2L2 , k(iλ − Lα,l )uk2L2 + 2 α(1 ˜ − |µ|) α˜ α˜ 1 − |µ|

and thus, we have from (3.176), kB2,l uk2L2 ≤ C


1 1 C + k(iλ − Lα,l )uk2L2 ≤ 2 k(iλ − Lα,l )uk2L2 . 2 2 4 6 2 α˜ (1 − |µ|) α˜ (1 − |µ|) α˜ (1 − |µ|) (3.182)

In particular, (3.167) holds and the proof for the case |l| ≥ 2 is complete. The case |l| = 1 is handled in the same manner. As in the proof of Theorem 2.9, we devide into two cases |µ| < 21 and |µ| ≥ 12 . In the case |µ| < 12 we use (3.162) and apply the argument above for the case |l| ≥ 2 1 3 with 1 − |µ| > κ 4 α˜ − 2 . When |µ| ≥ 12 we divide into three cases as above, that is, 1 1 3 ≤ |µ| < 1 − κ 4 α˜ − 2 , 2 In the case

1 2

3

1

3

1

−M α˜ − 2 ≤ 1 − |µ| ≤ κ 4 α˜ − 2 ,

1

|µ| > 1 + M α˜ − 2 .

1

≤ |µ| < 1 − κ 4 α˜ − 2 we use (3.161) which in particular implies


C C 1 ˆ l )uk2 2 , k(−Al )− 2 uk2L2 + kMcos y B2,l uk2L2 + k(µ + Λ u ∈ L2 (T) , L 1 − |µ| (1 − |µ|)2 R 2π R 2π since 0 u dy = − 0 v dy for v = A−1 l u. Then one can discuss exactly in the same way as in the case |l| ≥ 2. The details are omitted here. The proof is complete. kuk2L2 ≤

3.4

Estimate for semigroup

The resolvent estimates in Theorem 3.13 provide a crucial information on the solution to the following nonstationary problem both in qualitative and quantitative point of views.  dw    − Lα w = 0 , t > 0,  dt (3.183)     w| = f ∈ L2 (T2 ) . t=0

0

ˆ is diagonalized in terms of the Fourier series with respect Note that the operator Lα = A − iαΛ to the x variable: Lα = ⊕l∈Z\{0} Lα,l ,

ˆl. Lα,l = Al − iαlΛ

(3.184)

Hence the estimates of the solution u to (3.201) are obtained from the estimates for each Fourier mode Pl u, which is given by the semigroup generated by Lα,l in L2 (T). For the estimate of the semigroup etLα,l it is convenient to use the representation in terms of the Dunford integral Z 1 tLα,l wl (t) = e fl = etζ (ζ − Lα,l )−1 fl dζ . (3.185) 2πi Γ

60

Here fl = (Pl f )e−ilx ∈ L2 (T), and Γ is first taken as  1 Γ = ζ ∈ C | Re ζ = − , | Im ζ| ≤ 4|αl| 2  1 1 ∪ ζ ∈ C | Im ζ = ∓(Re ζ + ) ± 4|αl| , Re ζ ≤ − 2 2 =: Γ0,− 12 + Γ±,− 12 ,

(3.186)

which is oriented counter-clockwisely. We note that k(ζ − Lα,l )−1 kL2 (T)→L2 (T) ≤

C , | Im ζ|

| Im ζ| ≥ 4α|l|

(3.187)

ˆ l )⊥ hold with a constant C independent of α. Set Pl = I − Ql , where Ql : L2 (T) → Yl = (Ker Λ is the orthogonal projection as used in the previous section. Note that Pl = 0 when |l| ≥ 2. Our aim is to establish the estimates for Ql etLα,l and Pl etLα,l . For the part Ql etLα,l the fast dissipation is expected for large α, while for the part Pl etLα,l the strong amplification is expected through the interaction term ˆ l Ql , iαlPl Λ

|l| = 1 ,

which does not vanish due to the lack of the invariance of the space Yl = Ql L2 (T) under the ˆ l , or in other words, due to the lack of the symmetry of Λ ˆ l , for this term automatically action of Λ tL α,l ˆ l is symmetric. To estimate Ql e we observe vanishes if Λ Z 1 tLα,l Ql e f = etζ Ql (ζ − Lα,l )−1 fl dζ 2πi Γ Z 1 etζ (ζ − Ql Lα,l )−1 Ql fl dζ . = 2πi Γ The last identity follows from (2.19). We observe that for each λ ∈ R the set 

ζ ∈ C | |ζ − iλ|
0   |αl|t2 |αl| 12 t kQl etLα,l fl kL2 ≤  (3.193)  1   − 12  Ce−c|αl| 2 t kQl fl k 2 , t ≥ |αl| . L On the other hand, the simple energy computation for

d hu (t), dt l

B2,l ul (t)iL2 (T) gives the identity

d hul , B2,l ul iL2 (T) = 2hAl ul , B2,l ul iL2 (T) . dt The term in the right-hand side is bounded from above by −(2l2 − 1))kQl ul k2L2 (T) ≤ −(2l2 − 1)hul , B2,l ul iL2 (T) , where we have used hul , B2,l ul iL2 (T) ≤ kQl uk2L2 (T) . Thus we have from the coercive estimate (3.11), 1

2

kQl etLα,l fl kL2 ≤ 2e− 2 (2l −1)t kQl fl kL2 , 62

t > 0.

(3.194)

This estimate is useful for a short time period. When |l| ≥ 2 we have obtained the desired semgroup bound since Ql = I in this case. For the estimate of Pl etLα,l in the case |l| = 1 we cannot shift the curve Γ as in Γα , and has to be computed in a different way. By the construction of the resolvent in the proof of Theorem 2.4, see (2.17), we observe that ˆ l (ζ − Ql Lα,l )−1 Ql fl + (ζ − Al )−1 Pl fl . Pl (ζ − Lα,l )−1 fl = −iαl(ζ − Al )−1 Pl Λ

(3.195)

Since k(ζ − Al )−1 kL2 (T)→L2 (T) ≤ |ζ|−1 , we have Z 1 tLα,l ˆ l (ζ − Ql Lα,l )−1 Ql fl dζkL2 etζ (ζ − Al )−1 Pl Λ kPl e fl kL2 ≤ α|l| k 2πi Γ Z 1 +k etζ (ζ − Al )−1 Pl fl dζkL2 2πi Γ = αI1 + I2 , and it is not difficult to see kI2 kX ≤ e−t kPl fl kL2 ,

t > 0,

|l| = 1

(3.196)

by the estimate for the semigroup generated by the self-adjoint operator Al in L2 (T). As for I1 , we replace Γ by Γ˜ α , where Γ˜ α = Γ˜ − 12 ,± + Γ˜ α,±,1 + Γα,±,2 + Γα,±,3 , where Γα,±. j with j = 2, 3 are the curves as in (3.189), while Γ˜ − 21 ,+ (Rep. Γ˜ − 12 ,− ) is the segment connecting ζ = − 21 and a point pα,+ of Γα,+,1 (Resp. pα,− of Γα,−,1 ), and Γ˜ α,±,1 is the part of Γα,±,1 which therefore connects pα,± and the end point of Γα,±,2 . We will take pα,± as | Im pα,± | = |αl| , 2 thus, they are away enough from the degenerate case such as | Im ζ| ∼ |αl| = |α|. On the curve Γ˜ − 21 ,± we have k(ζ − Ql Lα,l )−1 kYl →Yl ≤ C2 by the choice of pα,± . Thus we have |α| 3

1 k 2πi

Z

ˆ l (ζ − Ql Lα,l ) Ql fl dζkL2 ≤ e (ζ − Al ) Pl Λ tζ

Γ˜ − 1 ,±

−1

−1

2

≤

C

− 2t

Z

e 2

|α| 3

C| log α| |α|

2 3

ζ=pα,± ζ=− 12

| dζ| |ζ|

t

e− 2 kQl fl kL2 .

As for the integrals on the curve Γ˜ α,±,1 and Γα,±,2 we compute as in (3.190) and (3.191) respecˆ l , which is bounded tively, but the difference in this case is the presence of the factor (ζ −Al )−1 Pl Λ by ˆ l kL2 →L2 ≤ C ≤ C k(ζ − Al )−1 Pl Λ |ζ| |α| on Γ˜ α,±,1 and Γα,±,2 . Moreover, we need to compute the integral so that the singularity at t = 0 does not appear. Hence we have 1 k 2πi

Z Γ˜ α,±,1

ˆ l (ζ − Ql Lα,l )−1 Ql fl dζkL2 etζ (ζ − Al )−1 Pl Λ

63

Z −c|α| 21 C ≤ 2 |s|ets ds kQl fl kL2 |α| −c|α| 23 1 C ≤ 2 e−c|α| 2 t kQl fl kL2 , |α| 3

and similarly, 1

1 k 2πi

Z

ˆ l (ζ − Ql Lα,l ) Ql fl dζkL2 ≤ e (ζ − Al ) Pl Λ tζ

Γα,±,2

−1

−1

Z

C 3 2

|α|+M|α| 2 1

1

e−c|α| 2 t ds kQl fl kL2

|α| |α|−|α| 2 1 C ≤ e−c|α| 2 t kQl fl kL2 . |α|

Finally, the integrals over Γα,±,3 is computed as in (3.192), and we have Z 1 ˆ l (ζ − Ql Lα,l )−1 Ql fl dζkL2 k etζ (ζ − Al )−1 Pl Λ 2π Γα,+,3 √ Z 1 + c−2 C ts e ≤
ds kQl fl kL2 1 |αl| s≤−c|α| 21 (|s| + |α|) |αl| (− cs + |αl|) − 1 Z 1 ets ds kQl fl kL2 ≤C 1 (|s| + |α|)|s| s≤−c|α| 2 C| log α| −c|α| 12 t e ≤ kQl fl kL2 . |α| Collecting these above, we have 1

t

|α|I1 ≤ C|α| 3 | log α| e− 2 kQl fl kL2 .

(3.197)

Combining (3.196) and (3.197), we obtain 1

t

kPl etLα,l fl kL2 ≤ C|α| 3 | log α| e− 2 kQl fl kL2 + e−t kPl fl kL2 ,

t > 0,

|l| = 1 .

(3.198)

Thus we have arrived at the following theorem. Let us recall that Q : L02 (T2 ) → Y is the orthogonal projection on to Y. Theorem 3.14 For all sufficiently large |α| the following statement holds. The semigroup {etLα }t≥0 generated by Lα in L02 (T2 ) satisfies for any l ∈ Z \ {0},  − 21 (2l2 −1)t  2e kQPl f kL2 (T2 ) , t > 0,   1 kQPl etLα f kL2 (T2 ) ≤  (3.199) 1   Ce−c|αl| 2 t kQPl f kL2 (T2 ) , t ≥ |αl|− 2 , while 1

t

k(I − Q)Pl etLα f kL2 (T2 ) ≤ C|α| 3 | log α| e− 2 kQPl f kL2 (T2 ) + e−t k(I − Q)Pl f kL2 (T2 ) ,

t > 0,

|l| = 1 .

(3.200)

Here C and c are independent of α, l, and f . Theorem 3.14 immediately leads to the estimate of the solution to  dω    − Lν,a ω = 0 , t > 0,  dt     ω| = f ∈ L2 (T2 ) . t=0

(3.201)

0

ν,a

Here L is defined as in (1.7) with a ∈ R \ {0} and 0 < ν  1. Indeed, by introducing the rescaling ω(x, y, t) = w(x, y, νt) which gives α = aν in Theorem 3.14, we obtain 64

Corollary 3.15 For all sufficiently small aν > 0 the following statement holds. The semigroup ν,a {etL }t≥0 generated by Lν,a in L02 (T2 ) satisfies for any l ∈ Z \ {0},  1 2   t > 0, 2e− 2 (2l −1)νt kQPl f kL2 (T2 ) ,   ν,a  √ 1 kQPl etL f kL2 (T2 ) ≤  (3.202) −c a|l|ν t   kQPl f kL2 (T2 ) , t≥ √ ,   Ce a|l|ν while a 1 a νt ν,a k(I − Q)Pl etL f kL2 (T2 ) ≤ C( ) 3 | log | e− 2 kQPl f kL2 (T2 ) + e−νt k(I − Q)Pl f kL2 (T2 ) , ν ν t > 0 , |l| = 1 .

(3.203)

Here C and c are independent of ν, a, l, and f .

Acknowledgement The first author is partially supported by NSERC Discovery grant # 371637-2014, and also acknowledges the kind hospitality of the New York University in Abu Dhabi. The second author is partially supported by JSPS Program for Advancing Strategic International Networks to Accelerate the Circulation of Talented Researchers, ’Development of Concentrated Mathematical Center Linking to Wisdom of the Next Generation’, which is organized by Mathematical Institute of Tohoku University. The third author is partially supported by the NSF grant DMS1716466.

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