Sharp stability for finite difference approximations of hyperbolic equations with boundary conditions
SSharp stability for finite difference approximations ofhyperbolic equations with boundary conditions
Jean-Fran¸cois
Coulombel & Gr´egory
Faye ∗ February 8, 2021
Abstract
In this article, we consider a class of finite rank perturbations of Toeplitz operators that have simpleeigenvalues on the unit circle. Under a suitable assumption on the behavior of the essential spectrum,we show that such operators are power bounded. The problem originates in the approximation ofhyperbolic partial differential equations with boundary conditions by means of finite difference schemes.Our result gives a positive answer to a conjecture by Trefethen, Kreiss and Wu that only a weak formof the so-called Uniform Kreiss-Lopatinskii Condition is sufficient to imply power boundedness.
AMS classification:
Keywords: hyperbolic equations, difference approximations, stability, boundary conditions, semigroup estimates,Toeplitz operators.
Throughout this article, we use the notation U := { ζ ∈ C , | ζ | > } , D := { ζ ∈ C , | ζ | < } , S := { ζ ∈ C , | ζ | = 1 } , U := U ∪ S , D := D ∪ S . If w is a complex number, the notation B r ( w ) stands for the open ball in C centered at w and with radius r >
0, that is B r ( w ) := { z ∈ C / | z − w | < r } . We let M n,k ( C ) denote the set of n × k matrices withcomplex entries. If n = k , we simply write M n ( C ).Eventually, we let C , resp. c , denote some (large, resp. small) positive constants that may varythroughout the text (sometimes within the same line). The dependance of the constants on the variousinvolved parameters is made precise throughout the article. ∗ Institut de Math´ematiques de Toulouse - UMR 5219, Universit´e de Toulouse ; CNRS, Universit´e Paul Sabatier,118 route de Narbonne, 31062 Toulouse Cedex 9 , France. Research of J.-F. C. was supported by ANRproject Nabuco, ANR-17-CE40-0025. G.F. acknowledges support from an ANITI (Artificial and Natural Intelli-gence Toulouse Institute) Research Chair and from Labex CIMI under grant agreement ANR-11-LABX-0040. Emails: [email protected] , [email protected] a r X i v : . [ m a t h . A P ] F e b Introduction
This article is devoted to the proof of power boundedness for a class of finite rank perturbations of someToeplitz operators. The problem originates in the discretization of initial boundary value problems forhyperbolic partial differential equations. From the standard approach in numerical analysis, convergenceof numerical schemes follows from stability and consistency. We focus here on stability. For discretizedhyperbolic problems with numerical boundary conditions, several possible definitions of stability havebeen explored. From a historic perspective, the first stability definition introduced for instance in [Kre68,Osh69b, Osh69a] is a power boundedness property and reads (here T denotes the discrete evolutionoperator which gives the solution at each time step, and the norm in (1) below corresponds to an operatornorm on (cid:96) ( N ) - the numerical boundary conditions are incorporated in the definition of the functionalspace): sup n ∈ N (cid:107) T n (cid:107) < + ∞ . (1)The notion of strong stability later introduced in the fundamental contribution [GKS72] amounts toproving a strengthened version of the resolvent condition :sup z ∈ U ( | z | − (cid:107) ( z I − T ) − (cid:107) < + ∞ . (2)We refer to [SW97] for a detailed exposition of the links between the conditions (1) and (2). Bothconditions (1) and (2) preclude the existence of unstable eigenvalues z ∈ U for the operator T , theso-called Godunov-Ryabenkii condition [GKO95].The notion of strong stability analyzed in [GKS72] has the major advantage of being stable withrespect to perturbations. It is an open condition, hence suitable for nonlinear analysis. However, it isrestricted to zero initial data and is therefore not so convenient in practical applications. A long line ofresearch has dealt with proving that strong stability implies power boundedness . As far as we know,the most complete answers in the discrete case are [Wu95] (for scalar 1D problems and one time stepschemes), [CG11] (for multidimensional systems and one time step schemes) and [Cou15] (for scalarmultidimensional problems and multistep schemes). In the continuous setting, that is for hyperbolicpartial differential equations, the reader is referred to [Rau72, Aud11, M´et17] and to references therein.All the above mentionned works are based on the fact that strong stability (or equivalently, the fulfillmentof the so-called Uniform Kreiss-Lopatinskii Condition) provides with a sharp trace estimate of the solutionin terms of the data. Summarizing the methodology in the strongly stable case, the goal is to control thetime derivative (the time difference in the discrete case) of the solution in terms of its trace. All thesetechniques thus break down if the considered problem is not strongly stable and a trace estimate is notavailable.However, it has been noted that several numerical boundary conditions do not yield strongly stableproblems, see for instance [Tre84]. As observed in [Tre84] and later made more formal in [KW93], eventhough the Uniform Kreiss-Lopatinskii Condition may not be fulfilled, it does seem that some numericalschemes remain stable in the sense that their associated (discrete) semigroup is bounded (property (1)).This is precisely such a result that we aim at proving here, in the case where the Uniform Kreiss-LopatinskiiCondition breaks down because of simple, isolated eigenvalues on the unit circle . Up to our knowledge, Recall that strong stability is actually stronger than just verifying (2). It is known that in general, (2) does not imply(1) in infinite dimension, see [SW97]. This is not the only possible breakdown for the Uniform Kreiss-Lopatinskii Condition, see [Tre84] or [BGS07] for theanalogous continuous problem. However, the case we deal here with is the simplest and therefore the first to tackle in viewof future generalizations. (cid:96) ( Z ) (or equivalently (cid:96) ∞ ( Z )) without any boundary condition. By the result in [Tho65], seemore recent developments in [Des08, DSC14], we thus base our analysis on the dissipation Assumption1 below. This does seem restrictive at first glance, but it is very likely that our methodology is flexibleenough to handle more general situations, up to refining some steps in the analysis. We shall explore suchextensions in the future. We consider the scalar transport equation ∂ t u + a ∂ x u = 0 , (3)in the half-line { x > } , and restrict from now on to the case of an incoming velocity, that is, a >
0. Thetransport equation (3) is supplemented with Dirichlet boundary conditions: u ( t,
0) = 0 , (4)and a Cauchy datum at t = 0. Our goal in this article is to explore the stability of finite differenceapproximations of the continuous problem (3), (4). We thus introduce a time step ∆ t > x >
0, assuming from now on that the ratio λ := ∆ t/ ∆ x is always kept fixed. The solution to(3), (4) is meant to be approximated by a sequence ( u nj ). We consider some fixed integers r, p withmin( r, p ) ≥
1. The interior cells are then the intervals [( j −
1) ∆ x, j ∆ x ) with j ∈ N ∗ , and the boundary cells are the intervals [( ν −
1) ∆ x, ν ∆ x ) with ν = 1 − r, . . . ,
0. The numerical scheme in the interiordomain N ∗ reads: u n +1 j = p (cid:88) (cid:96) = − r a (cid:96) u nj + (cid:96) , j ≥ , (5)where the coefficients a − r , . . . , a p are real and may depend only on λ and a , but not on ∆ t (or ∆ x ). Thenumerical boundary conditions that we consider in this article take the form: ∀ ν = 1 − r, . . . , , u n +1 ν = p b (cid:88) (cid:96) =1 b (cid:96),ν u n +1 (cid:96) , (6)where the coefficients b (cid:96),ν in (6) are real and may also depend on λ and a , but not on ∆ t (or ∆ x ). Weassume for simplicity that the (fixed) integer p b in (6) satisfies p b ≤ p . This is used below to simplify someminor technical details (when we rewrite high order scalar recurrences as first order vectorial recurrences).An appropriate vector space for the stability analysis of (5)-(6) is the Hilbert space H defined by: H := (cid:40) ( w j ) j ≥ − r ∈ (cid:96) / ∀ ν = 1 − r, . . . , , w ν = p b (cid:88) (cid:96) =1 b (cid:96),ν w (cid:96) (cid:41) . (7)Sequences in H are assumed to be complex valued (even though, in practice, the numerical scheme (5)-(6)applies to real sequences). Since any element w of H is uniquely determined by its interior values (those As usual, we identify the sequence ( u nj ) with its associated step function on the time-space grid. j ’s with j ≥ H : ∀ w ∈ H , (cid:107) w (cid:107) H := (cid:88) j ≥ | w j | . The numerical scheme (5)-(6) can be then rewritten as: ∀ n ∈ N , u n +1 = T u n , u ∈ H , where T is the bounded operator on H defined by: ∀ w ∈ H , ∀ j ≥ , ( T w ) j := p (cid:88) (cid:96) = − r a (cid:96) w j + (cid:96) . (8)Recall that a sequence in H is uniquely determined by its interior values so (8) determines T w ∈ H unambiguously. We introduce the following terminology. Definition 1 (Stability [Kre68, Osh69a]) . The numerical scheme (5) - (6) is said to be stable if there existsa constant C > such that, for any f ∈ H , the solution ( u n ) n ∈ N to (5) - (6) with initial condition u = f satisfies: sup n ∈ N (cid:107) u n (cid:107) H ≤ C (cid:107) f (cid:107) H . This means equivalently that the operator T in (8) is power bounded by the same constant C : sup n ∈ N (cid:107) T n (cid:107) H → H ≤ C .
Our goal in this article is to show that the scheme (5)-(6) is stable under some spectral assumptions onthe operator T . We make two major assumptions: one on the finite difference scheme (5), and one on the compatibilitybetween the scheme (5) and the numerical boundary conditions (6).
Assumption 1.
The finite difference approximation (5) is consistent with the transport equation (3) : p (cid:88) (cid:96) = − r a (cid:96) = 1 , p (cid:88) (cid:96) = − r (cid:96) a (cid:96) = − λ a < , (consistency). (9) Moreover, the coefficients a (cid:96) in (5) satisfy a − r a p (cid:54) = 0 and the dissipativity condition: ∀ θ ∈ [ − π , π ] \ { } , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < , (10) and for some nonzero integer µ and some positive real number β > , there holds: p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ = exp (cid:16) − i λ a θ − β θ µ + O (cid:16) θ µ +1 (cid:17)(cid:17) , (11) as θ tends to .
4n important consequence of Assumption 1 is the following Bernstein type inequality, which we prove inAppendix A.
Lemma 1.
Under Assumption 1, there holds λ a < r . The relevance of (11) for the (cid:96) stability of (5) on Z is the major result in [Tho65] (see [CF20, Des08,DSC14] for recent developments in this direction). This stability property will greatly simplify the finalsteps of the proof of our main result, which is Theorem 1 below. Relaxing (10) and (11) in order toencompass a wider class of finite difference schemes is postponed to some future works. We now state twoLemma whose proofs, which are relatively standard, can also be found in Appendix A. These two Lemmawill allow us to introduce our second spectral assumption on the operator T . Lemma 2.
There exists a constant c > such that, if we define the set: C := (cid:110) ρ e i ϕ ∈ C / ϕ ∈ [ − π , π ] and 0 ≤ ρ ≤ − c ϕ µ (cid:111) , then C is a compact star-shaped subset of D , and the curve: (cid:40) p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ / θ ∈ [ − π , π ] (cid:41) (12) is contained in C . The above Lemma 2 provides an estimate on the location of the essential spectrum of the operator T and shows that it is contained in C (see the reminder below on the spectrum of Toeplitz operators).Next, we introduce the following matrix: ∀ z ∈ C , M ( z ) := δ p − , z − a p − a p . . . . . . δ − r, z − a − r a p . . .
00 . . . . . . ...0 0 1 0 ∈ M p + r ( C ) . (13)Since min( r, p ) ≥
1, the upper right coefficient of M ( z ) is always nonzero (it equals − a − r /a p ), and M ( z )is invertible. We shall repeatedly use the inverse matrix M ( z ) − in what follows. Lemma 3 (Spectral splitting) . Let z ∈ C and let the matrix M ( z ) be defined as in (13) . Let the set C bedefined by Lemma 2. Then for z (cid:54)∈ C , M ( z ) has: • no eigenvalue on S , • r eigenvalues in D \ { } , • p eigenvalues in U (eigenvalues are counted with multiplicity).Furthermore, M (1) has as a simple eigenvalue, it has r − eigenvalues in D and p eigenvalues in U . We introduce some notation. For z (cid:54)∈ C , Lemma 3 shows that the so-called stable subspace, which isspanned by the generalized eigenvectors of M ( z ) associated with eigenvalues in D , has constant dimension r . We let E s ( z ) denote the stable subspace of M ( z ) for z (cid:54)∈ C . Because of the spectral splitting shown5n Lemma 3, E s ( z ) depends holomorphically on z in the complementary set of C . We can therefore find,near every point z (cid:54)∈ C , a basis e ( z ) , . . . , e r ( z ) of E s ( z ) that depends holomorphically on z . Similarly, theunstable subspace, which is spanned by the generalized eigenvectors of M ( z ) associated with eigenvaluesin U , has constant dimension p . We denote it by E u ( z ), and it also depends holomorphically on z in thecomplementary set of C . With obvious notation, the projectors associated with the decomposition: ∀ z (cid:54)∈ C , C p + r = E s ( z ) ⊕ E u ( z ) , are denoted π s ( z ) and π u ( z ).Let us now examine the situation close to z = 1. Since 1 is a simple eigenvalue of M (1), we canextend it holomorphically to a simple eigenvalue κ ( z ) of M ( z ) in a neighborhood of 1. This eigenvalue isassociated with the eigenvector: E ( z ) := κ ( z ) p + r − ... κ ( z )1 ∈ C p + r , which also depends holomorphically on z in a neighborhood of 1. Furthermore, the unstable subspace E u (1) associated with eigenvalues in U has dimension p . It can be extended holomorphically to a neigh-borhood of 1 thanks to the Dunford formula for spectral projectors. This holomorphic extension coincideswith the above definition for E u ( z ) if z is close to 1 and z (cid:54)∈ C . Eventually, the stable subspace of M (1)associated with eigenvalues in D has dimension r −
1. For the sake of clarity, we denote it by E ss (1)(the double s standing for strongly stable ). Using again the Dunford formula for spectral projectors, wecan extend this “strongly stable” subspace holomorphically with respect to z ; for z close to 1, E ss ( z )has dimension r − M ( z ). Namely, thesituation has no ambiguity: for z (cid:54)∈ C close to 1, the eigenvalue κ ( z ) necessarily belongs to D and thestable subspace E s ( z ) of M ( z ) (which has been defined above and has dimension r ) splits as: E s ( z ) = E ss ( z ) ⊕ Span E ( z ) . (14)Since the right hand side in (14) depends holomorphically on z in a whole neighborhood of 1 and not onlyin C c , the stable subspace E s ( z ) extends holomorphically to a whole neighborhood of 1 as an invariantsubspace of dimension r for M ( z ). In particular, we shall feel free to use below the notation E s (1) for the r -dimensional vector space: E s (1) := E ss (1) ⊕ Span , (15)which is, in our case, the direct sum of the stable and central subspaces of M (1).For future use, it is convenient to introduce the following matrix: B := · · · − b p b , · · · − b , · · · · · · − b p b , − r · · · − b , − r · · · ∈ M r,p + r ( R ) . (16)We can now state our final assumption. 6 ssumption 2. For any z ∈ U ∪ { } , there holds: C p + r = Ker B ⊕ E s ( z ) , or, in other words, B | E s ( z ) is an isomorphism from E s ( z ) to C r . Moreover, choosing a holomorphic basis e ( z ) , . . . , e r ( z ) of E s ( z ) near every point z ∈ S \ { } , the function: ∆ : z (cid:55)−→ det (cid:2) B e ( z ) · · · B e r ( z ) (cid:3) has finitely many simple zeroes in S \ { } . Let us recall that for z ∈ U \ { } , E s ( z ) denotes the stable subspace of the matrix M ( z ) in (13)since then z ∈ C c . At the point z = 1, E s (1) denotes the holomorphic extension of E s ( z ) at 1 and it isfurthermore given by (15).Of course, the function ∆ in Assumption 2 depends on the choice of the (holomorphic) basis e ( z ),. . . , e r ( z ) of E s ( z ). However, the location of its zeroes and their multiplicity does not depend on thatchoice, which means that Assumption 2 is an intrinsic property of the operator T . We shall refer lateron to the function ∆ as the Lopatinskii determinant associated with (5)-(6). It plays the role of acharacteristic polynomial for T which detects the eigenvalues in C c . This object already appears in[Kre68, Osh69b, Osh69a]. Its analogue in the study of discrete shock profiles is the so-called Evansfunction, see [God03]. Our main result is the following. Theorem 1.
Under Assumptions 1 and 2, the operator T in (8) is power bounded, that is, the numericalscheme (5) - (6) is stable. If the function ∆ in Assumption 2 does not vanish on U , the Uniform Kreiss-Lopatinskii Condition is saidto hold and the main result in [Wu95] implies that T is power bounded, see also [Kre68, Osh69b, Osh69a].The novelty here is to allow ∆ to vanish on S . The Uniform Kreiss-Lopatinskii Condition thus breaksdown. Power boundedness of T in this case was conjectured in [Tre84, KW93].The remainder of this article is organized as follows. The proof of Theorem 1 follows the same strategyas in [CF20]. In Section 2, we clarify the location of the spectrum of T and give accurate bounds on theso-called spatial Green’s function (that is, the Green’s function for the operator z I − T with z (cid:54)∈ σ ( T )).This preliminary analysis is used in Section 3 to give an accurate description of the so-called temporal Green’s function (that is, the Green’s function for the original problem (5)-(6)). Power boundedness of T easily follows by classical inequalities. An example of operator for which Theorem 1 applies is givenin Section 4. For later use, we let z , . . . , z K ∈ S \ { } denote the pairwise distinct roots of the Lopatinskii determinant∆ introduced in Assumption 2. We recall that these roots are simple. We first locate the spectrum ofthe operator T and then give an accurate description of the so-called spatial Green’s function. Precisedefinitions are provided below. 7 .1 A reminder on the spectrum of Toeplitz operators The operator T is a finite rank (hence compact) perturbation of the Toeplitz operator on (cid:96) ( N ) representedby the semi-infinite matrix: a · · · a p · · · · · · · · · ... . . . . . . . . . a − r · · · a · · · a p . . .0 . . . . . . . . . . . .... . . . . . . . . . . . . . Therefore T shares the same essential spectrum as the Toeplitz operator [Con90]. (The latter Toeplitzoperator corresponds to enforcing the Dirichlet boundary conditions u n +11 − r = · · · = u n +10 = 0 instead of themore general form (6)). The spectrum of Toeplitz operators is well-known, see for instance [Dur64] andfurther developments in [TE05]. The resolvent set of the above Toeplitz operator consists of all points z ∈ C that do not belong to the curve (12) and that have index 0 with respect to it. Moreover, any pointon the curve (12) is in the essential spectrum. In the particular case we are interested in, Assumption1 implies that the essential spectrum of T is located in the set C defined by Lemma 2 and that 1belongs to the essential spectrum of T . There remains to clarify the point spectrum of T . The situationwhich we consider here and that is encoded in Assumption 2 is that where the finite rank perturbationof the Toeplitz operator generates finitely many simple eigenvalues on the unit circle (there may alsobe eigenvalues within C but we are mainly concerned here with the eigenvalues of largest modulus). Aprecise statement is the following. Lemma 4 (The resolvant set) . Let the set C be defined by Lemma 2. Then there exists ε > such that ( C c \ { z , . . . , z K } ) ∩ { ζ ∈ C / | ζ | > − ε } is contained in the resolvant set of the operator T . Moreover,each zero z k of the Lopatinskii determinant is an eigenvalue of T .Proof. The proof of Lemma 4 is first useful to clarify the location of the spectrum of the operator T andit is also useful to introduce some of the tools used in the construction of the spatial Green’s functionwhich we shall perform below.Let therefore z ∈ C c \ { z , . . . , z K } and let f ∈ H . We are going to explain why we can uniquelysolve the equation: ( z I − T ) w = f , (17)with w ∈ H (up to assuming | z | > − ε for some sufficiently small ε > (cid:40) z w j − (cid:80) p(cid:96) = − r a (cid:96) w j + (cid:96) = f j , j ≥ ,w ν = (cid:80) p b (cid:96) =1 b (cid:96),ν w (cid:96) , ν = 1 − r, . . . , . We introduce, for any j ≥
1, the augmented vector: W j := w j + p − ... w j − r ∈ C p + r , : (cid:40) W j +1 = M ( z ) W j − a − p f j e , j ≥ , B W = 0 , (18)where we have used the notation e to denote the first vector of the canonical basis of C p + r , namely: e := ∈ C p + r . Our goal now is to solve the spatial dynamical problem (18). Since z ∈ C c , we know that M ( z )enjoys a hyperbolic dichotomy between its unstable and stable eigenvalues. We first solve for the unstablecomponents of the sequence ( W j ) j ≥ by integrating from + ∞ to any integer j ≥
1, which gives: ∀ j ≥ , π u ( z ) W j := a − p (cid:88) (cid:96) ≥ f j + (cid:96) M ( z ) − − (cid:96) π u ( z ) e . (19)In particular, we get the “initial value”: π u ( z ) W := a − p (cid:88) (cid:96) ≥ f (cid:96) M ( z ) − − (cid:96) π u ( z ) e . (20)The initial value for the stable components is obtained by using Assumption 2. Namely, if z ∈ U ,we know that the linear operator B | E s ( z ) is an isomorphism, and this property remains true near everypoint of S \ { } except at the z k ’s. We can thus find some ε > z ∈ C c \ { z , . . . , z K } verifying | z | > − ε , the linear operator B | E s ( z ) is an isomorphism. For such z ’s, we can therefore definethe vector π s ( z ) W ∈ E s ( z ) through the formula: π s ( z ) W := − a − p (cid:16) B | E s ( z ) (cid:17) − B (cid:88) (cid:96) ≥ f (cid:96) M ( z ) − − (cid:96) π u ( z ) e , (21)which is the only way to obtain both the linear constraint B W = 0 and the decomposition W = π s ( z ) W + π u ( z ) W in agreement with (20). Once we have determined the stable components π s ( z ) W of the initial value W , the only possible way to solve (18) for the stable components is to set: ∀ j ≥ , π s ( z ) W j := M ( z ) j − π s ( z ) W − a − p j − (cid:88) (cid:96) =1 f (cid:96) M ( z ) j − − (cid:96) π s ( z ) e . (22)Since the sequences ( M ( z ) − (cid:96) π u ( z )) (cid:96) ≥ and ( M ( z ) (cid:96) π s ( z )) (cid:96) ≥ are exponentially decreasing, we candefine a solution ( W j ) j ≥ ∈ (cid:96) to (18) by decomposing along the stable and unstable components andusing the defining equations (19) and (22). This provides us with a solution w ∈ H to the equation (17) This is the place where we use the assumption p b ≤ p in order to rewrite the numerical boundary conditions as a linearconstraint on the first element W of the sequence ( W j ) j ≥ . The case p b > p can be dealt with quite similarly but is justheavier in terms of notation.
9y going back to the scalar components of each vector W j . Such a solution is necessarily unique since if w ∈ H is a solution to (17) with f = 0, then the augmented vectorial sequence ( W j ) j ≥ ∈ (cid:96) satisfies: (cid:40) W j +1 = M ( z ) W j , j ≥ , B W = 0 . This means that the vector W belongs to E s ( z ) and to the kernel of the matrix B , and therefore vanishes.Hence the whole sequence ( W j ) j ≥ ∈ (cid:96) vanishes. We have thus shown that z belongs to the resolvant setof T .The fact that each z k is an eigenvalue of T follows from similar arguments. At a point z k , theintersection Ker B ∩ E s ( z k ) is not trivial, so we can find a nonzero vector W ∈ Ker B for which thesequence ( W j ) j ≥ defined by: ∀ j ≥ , W j +1 := M ( z k ) W j , is square integrable (it is even exponentially decreasing). Going back to scalar components, this provideswith a nonzero solution to the eigenvalue problem: (cid:40) z k w j − (cid:80) p(cid:96) = − r a (cid:96) w j + (cid:96) = 0 , j ≥ ,w ν = (cid:80) p b (cid:96) =1 b (cid:96),ν w (cid:96) , ν = 1 − r, . . . , . The proof of Lemma 4 is complete.We are now going to define and analyze the so-called spatial Green’s function. The main point, as in[ZH98, God03, CF20] and related works, is to be able to “pass through” the essential spectrum close to 1and extend the spatial Green’s function holomorphically to a whole neighborhood of 1. This was alreadyachieved with accurate bounds in [CF20] on the whole line Z (with no numerical boundary condition)and we apply similar arguments here, while adding the difficulty of the eigenvalues on S . Near all sucheigenvalues, we isolate the precise form of the singularity in the Green’s function and show that theremainder admits a holomorphic extension at the eigenvalue. All these arguments are made precise in thefollowing paragraph. For any j ≥
1, we let δ j denote the only element of the space H in (7) that satisfies: ∀ j ≥ , (cid:0) δ j (cid:1) j := (cid:40) , if j = j ,0 , otherwise.The boundary values of δ j are defined accordingly. Then as long as z belongs to the resolvant set of theoperator T , the spatial Green’s function, which we denote G z ( · , · ) is defined by the relation: ∀ j ≥ , (cid:0) z I − T (cid:1) G z ( · , j ) = δ j , (23)together with the numerical boundary conditions G z ( · , j ) ∈ H . We give below an accurate descriptionof G z in order to later obtain an accurate description of the temporal Green’s function, that is obtainedby applying the iteration (5)-(6) to the initial condition δ j . The analysis of the spatial Green’s functionsplits between three cases: • The behavior near regular points (away from the spectrum of T ),10 The behavior near the point 1 (the only point where the essential spectrum of T meets S ), • The behavior near the eigenvalues z , . . . , z K .Let us start with the easiest case. Lemma 5 (Bounds away from the spectrum) . Let z ∈ U \ { , z , . . . , z K } . Then there exists an openball B r ( z ) centered at z and there exist two constants C > , c > such that, for any couple of integers j, j ≥ , there holds: ∀ z ∈ B r ( z ) , (cid:12)(cid:12) G z ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c | j − j | (cid:1) . Proof.
Almost all ingredients have already been set in the proof of Lemma 4. Let therefore z ∈ U \{ , z , . . . , z K } , and let us first fix r > B r ( z ) is contained bothin C c and in the resolvant set of T . All complex numbers z below are assumed to lie within B r ( z ). Thenthe problem (23) can be recast under the vectorial form (18) with: ∀ j ≥ , f j := (cid:40) , if j = j ,0 , otherwise.Let us therefore consider the spatial dynamics problem (18) with the above Dirac mass type source term.The unstable components of the sequence ( W j ) j ≥ solution to (18) are given by (19), which gives here: ∀ j ≥ , π u ( z ) W j = (cid:40) , if j > j , a − p M ( z ) − ( j +1 − j ) π u ( z ) e , if 1 ≤ j ≤ j .In particular, we get the following uniform bounds with respect to z, j, j : ∀ z ∈ B r ( z ) , ∀ j ≥ , (cid:12)(cid:12)(cid:12) π u ( z ) W j (cid:12)(cid:12)(cid:12) ≤ (cid:40) , if j > j , C exp( − c ( j − j )) , if 1 ≤ j ≤ j . (24)The initial value π s ( z ) W of the stable components is then obtained by the relation (21), whichimmediately gives the bound : (cid:12)(cid:12)(cid:12) π s ( z ) W (cid:12)(cid:12)(cid:12) ≤ C exp( − c j ) . The stable components are then determined for any integer j ≥ ∀ j ≥ , π s ( z ) W j = (cid:40) M ( z ) j − π s ( z ) W , if 1 ≤ j ≤ j , M ( z ) j − π s ( z ) W − a − p M ( z ) j − j − π s ( z ) e , if j > j .By using the exponential decay of the sequence ( M ( z ) j π s ( z )) j ≥ , we get the following bounds for thestable components: ∀ j ≥ , (cid:12)(cid:12)(cid:12) π s ( z ) W j (cid:12)(cid:12)(cid:12) ≤ (cid:40) C exp( − c ( j + j )) , if 1 ≤ j ≤ j , C exp( − c ( j + j )) + C exp( − c ( j − j )) , if j > j . (25)Adding (24) with (25), and examining for each situation which among the terms is the largest, we getthe conclusion of Lemma 5 (recall that the scalar component G z ( j, j ) is just one among the coordinatesof the vector W j considered above). Here we use the fact that the linear map B | E s ( z ) is an isomorphism for all z ∈ B r ( z ).
11e are now going to examine the behavior of the spatial Green’s function G z close to 1. Let us firstrecall that the exterior U of the unit disk belongs to the resolvant set of T . Hence, for any j ≥
1, thesequence G z ( · , j ) is well-defined in H for z ∈ U . Lemma 6 below shows that each individual sequence G z ( j, j ) can be holomorphically extended to a whole neighborhood of 1. Lemma 6 (Bounds close to 1) . There exists an open ball B ε (1) centered at and there exist two constants C > and c > such that, for any couple of integers ( j, j ) , the component G z ( j, j ) defined on B ε (1) ∩ U extends holomorphically to the whole ball B ε (1) with respect to z , and the holomorphic extension satisfiesthe bound: ∀ z ∈ B ε (1) , (cid:12)(cid:12) G z ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c | j − j | (cid:1) , if 1 ≤ j ≤ j , C (cid:12)(cid:12)(cid:12) κ ( z ) (cid:12)(cid:12)(cid:12) | j − j | , if j > j , where κ ( z ) denotes the (unique) holomorphic eigenvalue of M ( z ) that satisfies κ (1) = 1 .Proof. Most ingredients of the proof are similar to what we have already done in the proof of Lemma5. The novelty is that there is one stable component which behaves more and more singularly as z ∈ U gets close to 1 since one stable eigenvalue, namely κ ( z ), gets close to S (its exponential decay is thusweaker and weaker). We thus recall that on some suitably small neighborhood B ε (1) of 1, we have the(holomorphic in z ) decomposition: C p + r = E u ( z ) ⊕ E ss ( z ) ⊕ Span E ( z ) , where all the above spaces are invariant by M ( z ), the spectrum of M ( z ) restricted to E u ( z ) lies in U , thespectrum of M ( z ) restricted to E ss ( z ) lies in D , and E ( z ) is an eigenvector for M ( z ) associated with theeigenvalue κ ( z ). With obvious notation, we use the corresponding decomposition: X = π u ( z ) X + π ss ( z ) X + µ E ( z ) . Let us from now on consider some complex number z ∈ B ε (1) ∩ U so that the Green’s function G z ( · , j ) is well-defined in H for any j ≥
1. As in the proof of Lemma 5, the Green’s function is definedby solving the spatial dynamics problem (18) with the Dirac mass datum: ∀ j ≥ , f j := (cid:40) , if j = j ,0 , otherwise.The unstable components are uniquely determined by: ∀ j ≥ , π u ( z ) W j = (cid:40) , if j > j , a − p M ( z ) − ( j +1 − j ) π u ( z ) e , if 1 ≤ j ≤ j ,and we readily observe that the latter right hand side depends holomorphically on z in the whole ball B ε (1) and not only in B ε (1) ∩ U . This already allows to extend the unstable components π u ( z ) W j to B ε (1), with the corresponding uniform bound similar to (24), that is: ∀ z ∈ B ε (1) , ∀ j ≥ , (cid:12)(cid:12)(cid:12) π u ( z ) W j (cid:12)(cid:12)(cid:12) ≤ (cid:40) , if j > j , C exp( − c ( j − j )) , if 1 ≤ j ≤ j . (26)12e can then use the fact that B | E s (1) is an isomorphism from E s (1) to C r , which implies that, up torestricting the radius ε , the matrix B restricted to the holomorphically extended stable subspace: E ss ( z ) ⊕ Span E ( z ) , is an isomorphism. We can thus uniquely determine some vector π ss ( z ) W ∈ E ss ( z ) and a scalar µ suchthat: B (cid:16) π ss ( z ) W + µ E ( z ) (cid:17) = − a − p B M ( z ) − j π u ( z ) e . In particular, we have the bound: ∀ z ∈ B ε (1) , (cid:12)(cid:12)(cid:12) π ss ( z ) W (cid:12)(cid:12)(cid:12) + | µ | ≤ C exp( − c j ) . For z ∈ B ε (1) ∩ U , the strongly stable components of ( W j ) j ≥ are then defined by the formula: ∀ j ≥ , π ss ( z ) W j = (cid:40) M ( z ) j − π ss ( z ) W , if 1 ≤ j ≤ j , M ( z ) j − π ss ( z ) W − a − p M ( z ) j − j − π ss ( z ) e , if j > j ,and the coordinate of ( W j ) j ≥ along the eigenvector E ( z ) is defined by the formula: ∀ j ≥ , µ j = (cid:40) κ ( z ) j − µ , if 1 ≤ j ≤ j , κ ( z ) j − µ − a − p κ ( z ) j − j − µ ( e ) , if j > j .As for the unstable components, we observe that for each couple of integers j, j , the above componentsof W j extend holomorphically to the whole ball B ε (1) since the spectral projectors of M ( z ) along E ss ( z )and Span E ( z ) do so. We thus consider from now on the holomorphic extension of W j for z ∈ B ε (1) andcollect the three pieces of the vector W j . For 1 ≤ j ≤ j , we have: W j = a − p M ( z ) − ( j +1 − j ) π u ( z ) e + M ( z ) j − π ss ( z ) W + κ ( z ) j − µ E ( z ) , which satisfies the bound: (cid:12)(cid:12)(cid:12) W j (cid:12)(cid:12)(cid:12) ≤ C exp( − c ( j − j )) + C exp( − c ( j + j )) + C | κ ( z ) | j exp( − c j ) , for some constants C > c > z . Since κ (1) = 1, we can alwaysassume that there holds | κ ( z ) | ≤ exp c on the ball B ε (1), and we are then left with the estimate: (cid:12)(cid:12)(cid:12) W j (cid:12)(cid:12)(cid:12) ≤ C exp( − c ( j − j )) , as claimed in the statement of Lemma 6.It remains to examine the case j > j for which we have the decomposition: W j = M ( z ) j − π ss ( z ) W − a − p M ( z ) j − j − π ss ( z ) e + κ ( z ) j − µ E ( z ) − a − p κ ( z ) j − j − µ ( e ) E ( z ) , and we can thus derive the bound: (cid:12)(cid:12)(cid:12) W j (cid:12)(cid:12)(cid:12) ≤ C exp( − c ( j + j )) + C exp( − c ( j − j )) + C | κ ( z ) | j exp( − c j ) + C | κ ( z ) | | j − j | . Since we can always assume that the ball B ε (1) is so small that | κ ( z ) | takes its values within the interval[exp( − c ) , exp c ], it appears that the largest term on the above right hand side is the last term, whichcompletes the proof of Lemma 6. 13et us observe that we can extend holomorphically each scalar component G z ( j, j ) but that does notmean that we can extend holomorphically G z ( · , j ) in H . As a matter of fact, the eigenvalue κ ( z ) startscontributing to the unstable subspace of M ( z ) as z (close to 1) crosses the curve (12). The holomorphicextension G z ( · , j ) then ceases to be in (cid:96) for it has an exponentially growing mode in j . The last case toexamine is that of the neighborhood of each eigenvalue z k . Lemma 7 (Bounds close to the eigenvalues) . For any eigenvalue z k ∈ S of T , there exists an open ball B ε ( z k ) centered at z k , there exists a sequence ( w k ( j, j )) j,j ≥ with w k ( · , j ) ∈ H for all j ≥ , and thereexist two constants C k > and c k > such that for any couple of integers ( j, j ) , the component G z ( j, j ) defined on B ε ( z k ) \ { z k } is such that: R z ( j, j ) := G z ( j, j ) − w k ( j, j ) z − z k , extends holomorphically to the whole ball B ε ( z k ) with respect to z , and the holomorphic extension satisfiesthe bound: ∀ z ∈ B ε ( z k ) , (cid:12)(cid:12) R z ( j, j ) (cid:12)(cid:12) ≤ C k exp (cid:0) − c k | j − j | (cid:1) . Moreover, the sequence ( w k ( j, j )) j,j ≥ satisfies the pointwise bound: ∀ j, j ≥ , (cid:12)(cid:12) w k ( j, j ) (cid:12)(cid:12) ≤ C k exp (cid:0) − c k ( j + j ) (cid:1) . Proof.
Many ingredients for the proof of Lemma 7 are already available in the proof of Lemma 5. Namely,let us consider an eigenvalue z k ∈ S of T . Since z k ∈ C c , the matrix M ( z ) enjoys the hyperbolic di-chotomy between its stable and unstable eigenvalues in the neighborhood of z k . Moreover, for a sufficientlysmall radius ε >
0, the pointed ball B ε ( z k ) \ { z k } lies in the resolvant set of T . In particular, for any z ∈ B ε ( z k ) \ { z k } , the spatial Green’s function is obtained by selecting the appropriate scalar componentof the vector sequence ( W j ) j ≥ defined by: ∀ j ≥ , π u ( z ) W j := (cid:40) , if j > j , a − p M ( z ) − ( j +1 − j ) π u ( z ) e , if 1 ≤ j ≤ j , (27)and: ∀ j ≥ , π s ( z ) W j := (cid:40) M ( z ) j − π s ( z ) W , if 1 ≤ j ≤ j , M ( z ) j − π s ( z ) W − a − p M ( z ) j − j − π s ( z ) e , if j > j , (28)where the vector π s ( z ) W ∈ E s ( z ) is defined by (see (21)): π s ( z ) W := − a − p (cid:16) B | E s ( z ) (cid:17) − B M ( z ) − j π u ( z ) e . (29)(Here we use the fact that for every z in the pointed ball B ε ( z k ) \ { z k } , the linear map B | E s ( z ) is anisomorphism.)The unstable component π u ( z ) W j in (27) obviously extends holomorphically to the whole ball B ε ( z k )and the estimate (24) shows that this contribution to the remainder term R z ( j, j ) satisfies the desireduniform exponential bound with respect to z . We thus focus from now on on the stable componentsdefined by (28), (29). We first observe that, as in the unstable component (27), the contribution: − a − p M ( z ) j − j − π s ( z ) e j > j also extends holomorphically to the ball B ε ( z k ) and contributesto the remainder term R z ( j, j ) with an O (exp( − c | j − j | )) term. We thus focus on the sequence: (cid:16) M ( z ) j − π s ( z ) W (cid:17) j ≥ , where the vector π s ( z ) W ∈ E s ( z ) is defined by (29) for z ∈ B ε ( z k ) \ { z k } . The singularity in the Green’sfunction comes from the fact that B | E s ( z k ) is no longer an isomorphism. We now make this singularityexplicit.We pick a basis e ( z ) , . . . , e r ( z ) of the stable subspace E s ( z ) that depends holomorphically on z near z k . Since the Lopatinskii determinant factorizes as:∆( z ) = ( z − z k ) ϑ ( z ) , where ϑ is a holomorphic function that does not vanish at z k , we can therefore write: (cid:0) B e ( z ) · · · B e r ( z ) (cid:1) − = 1 z − z k D ( z ) , where D ( z ) is a matrix in M r ( C ) that depends holomorphically on z near z k . We then define the vector: W ( j ) := (cid:0) e ( z k ) · · · e r ( z k ) (cid:1) D ( z k ) (cid:16) − a − p B M ( z k ) − j π u ( z k ) e (cid:17) , (30)which satisfies the bound: (cid:12)(cid:12) W ( j ) (cid:12)(cid:12) ≤ C exp( − c j ) , (31)for some positive constants C and c , uniformly with respect to j ≥
1. Moreover, since we have therelation: B (cid:0) e ( z k ) · · · e r ( z k ) (cid:1) D ( z k ) = 0 , the vector W ( j ) belongs to Ker B ∩ E s ( z k ). Hence, by selecting the appropriate coordinate, thegeometric sequence (which is valued in C p + r ): (cid:16) M ( z k ) j − W ( j ) (cid:17) j ≥ , provides with a scalar sequence ( w k ( j, j )) j,j ≥ with w k ( · , j ) ∈ H for all j ≥
1, and that satisfies thebound: ∀ j, j ≥ , (cid:12)(cid:12) w k ( j, j ) (cid:12)(cid:12) ≤ C k exp (cid:0) − c k ( j + j ) (cid:1) , as stated in Lemma 7. It thus only remains to show that the remainder term: R z ( j, j ) := M ( z ) j − π s ( z ) W − M ( z k ) j − W ( j ) z − z k (32)extends holomorphically to B ε ( z k ) and satisfies a suitable exponential bound.We decompose the vector π s ( z ) W in (29) along the basis e ( z ) , . . . , e r ( z ) of the stable subspace E s ( z )and write: π s ( z ) W = 1 z − z k (cid:0) e ( z ) · · · e r ( z ) (cid:1) D ( z ) (cid:16) − a − p B M ( z ) − j π u ( z ) e (cid:17) . R z ( j, j ) as follows: R z ( j, j ) = 1 z − z k (cid:16) ( M ( z ) π s ( z )) j − − ( M ( z k ) π s ( z k )) j − (cid:17) W ( j ) − a − p z − z k ( M ( z ) π s ( z )) j − (cid:16) (cid:0) e ( z ) · · · e r ( z ) (cid:1) D ( z ) B ( M ( z ) π u ( z )) − j e − (cid:0) e ( z k ) · · · e r ( z k ) (cid:1) D ( z k ) B ( M ( z k ) π u ( z k )) − j e (cid:17) . Both terms (the first line, and the difference between the second and third lines) in the above decompo-sition are dealt with by applying the following result combined with the hyperbolic dichotomy of M ( z )near z k . Lemma 8.
Let M be a holomorphic function on the open ball B δ (0) with values in M N ( C ) for some δ > and integer N , that satisfies: ∃ C > , ∃ r ∈ (0 , , ∀ j ∈ N , ∀ z ∈ B δ (0) , (cid:107) M ( z ) j (cid:107) ≤ C r j . Then up to diminishing δ and for some possibly new constants C > and r ∈ (0 , , there holds: ∀ j ∈ N , ∀ z , z ∈ B δ (0) , (cid:107) M ( z ) j − M ( z ) j (cid:107) ≤ C | z − z | r j . Applying Lemma 8 to the above decomposition of R z ( j, j ), and using the exponential decay of thesequences ( M ( z ) j π s ( z )) j ∈ N and ( M ( z ) − j π u ( z )) j ∈ N , we get the bound: (cid:12)(cid:12) R z ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c k ( j + j ) (cid:1) , which means that R z ( j, j ) remains bounded on B ε ( z k ) \ { z k } and can therefore be extended holomor-phically to the whole ball B ε ( z k ). The proof of Lemma 7 is complete. Proof of Lemma 8.
The argument is a mere application of the Taylor formula. Let us recall that thedifferential of the mapping: Ψ j : A ∈ M N ( C ) (cid:55)−→ A j , is given by: dΨ j ( A ) · B = j − (cid:88) (cid:96) =0 A (cid:96) B A j − − (cid:96) , so we have: M ( z ) j − M ( z ) j = ( z − z ) (cid:90) j − (cid:88) (cid:96) =0 M ( z + t ( z − z )) (cid:96) M (cid:48) ( z + t ( z − z )) M ( z + t ( z − z )) j − − (cid:96) d t . The result follows by using a uniform bound for the first derivative M (cid:48) , up to diminishing δ , and usingthe exponential decay of the sequence ( j r j ) j ∈ N . 16 .3 Summary Collecting the results of Lemma 5, Lemma 6 and Lemma 7, we can obtain the following bound for thespatial Green’s function away from the spectrum of T . Corollary 1.
There exist a radius ε > , some width η ε > and two constants C > , c > such that,for all z in the set: (cid:110) ζ ∈ C / e − η ε < | ζ | ≤ e π (cid:111) \ (cid:32) K (cid:91) k =1 B ε ( z k ) ∪ B ε (1) (cid:33) , and for all j ≥ , the Green’s function G z ( · , j ) ∈ H solution to (23) satisfies the pointwise bound: ∀ j ≥ , (cid:12)(cid:12) G z ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c | j − j | (cid:1) . Moreover, for z inside the ball B ε (1) , the Green’s function component G z ( j, j ) depends holomorphicallyon z and satisfies the bound given in Lemma 6, and for k = 1 , . . . , K and z in the pointed ball B ε ( z k ) \{ z k } , G z ( j, j ) has a simple pole at z k with the behavior stated in Lemma 7. The starting point of the analysis is to use inverse Laplace transform formula to express the temporalGreen’s function G n ( · , j ) := T n δ j as the following contour integral: ∀ n ∈ N ∗ , ∀ j ≥ , ∀ j ≥ , G n ( j, j ) = ( T n δ j ) j = 12 π i (cid:90) (cid:101) Γ z n G z ( j, j ) d z , (33)where (cid:101) Γ is a closed curve in the complex plane surrounding the unit disk D lying in the resolvent set of T and G z ( · , j ) ∈ H is the spatial Green’s function defined in (23).Following our recent work [CF20], the idea will be to deform (cid:101) Γ in order to obtain sharp pointwiseestimates on the temporal Green’s function using our pointwise estimates on the spatial Green’s functionsummarized in Corollary 1 above. To do so, we first change variable in (33), by setting z = exp( τ ), suchthat we get G n ( j, j ) = 12 π i (cid:90) Γ e n τ G τ ( j, j ) d τ , (34)where without loss of generality Γ = { s + i (cid:96) | (cid:96) ∈ [ − π, π ] } for some (and actually any) s >
0, and G τ ( · , j ) ∈ H is given by ∀ τ ∈ Γ , ∀ j ≥ , ∀ j ≥ , G τ ( j, j ) := G e τ ( j, j ) e τ . It is already important to remark that as T is a recurrence operator with finite stencil, for each n ≥ G n ( j, j ) = 0 , for j − j > r n or j − j < − p n . As a consequence, throughout this section, we assume that j , j and n satisfy − p n ≤ j − j ≤ r n . The very first step in the analysis of the temporal Green’s function defined in (34) is to translatethe pointwise estimates from Corollary 1 for the spatial Green’s function G z ( j, j ) to pointwise estimates17or G τ ( j, j ). We let be τ k = i θ k := log( z k ) for θ k ∈ [ − π, π ] \ { } for each k = 1 , · · · , K . Finally, wealso set α := λ a >
0. As in [CF20], the temporal Green’s function is expected to have a leading ordercontribution concentrated near j − j ∼ α n . An important feature of the situation we deal here with isthat the temporal Green’s function should also incorporate the contribution of the eigenvalues z k . Thesecontributions will not decay with respect to n since the z k ’s have modulus 1. Lemma 9.
There exist a radius ε > , some width η ε > and constants < β ∗ < β < β ∗ and C > , c > such that, for all z in the set: Ω ε := (cid:110) τ ∈ C | − η ε < Re( τ ) ≤ π (cid:111) \ (cid:32) K (cid:91) k =1 B ε ( i θ k ) ∪ B ε (0) (cid:33) , and for all j ≥ , the Green’s function G τ ( · , j ) ∈ H satisfies the pointwise bound: ∀ j, j ≥ , (cid:12)(cid:12) G τ ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c | j − j | (cid:1) . Moreover, for τ inside the ball B ε (0) , the Green’s function component G τ ( j, j ) depends holomorphicallyon τ and satisfies the bound ∀ τ ∈ B ε (0) , ∀ j, j ≥ , (cid:12)(cid:12) G τ ( j, j ) (cid:12)(cid:12) ≤ (cid:40) C exp (cid:0) − c | j − j | (cid:1) , if 1 ≤ j ≤ j , C exp (cid:0) | j − j | Re( (cid:36) ( τ )) (cid:1) , if j > j , with (cid:36) ( τ ) = − α τ + ( − µ +1 βα µ +1 τ µ + O (cid:0) | τ | µ +1 (cid:1) , ∀ τ ∈ B ε (0) , together with Re( (cid:36) ( τ )) ≤ − α Re( τ ) + β ∗ α µ +1 Re( τ ) µ − β ∗ α µ +1 Im( τ ) µ , ∀ τ ∈ B ε (0) . At last, for any k = 1 , . . . , K and τ in the pointed ball B ε ( i θ k ) \ { i θ k } , G τ ( j, j ) has a simple pole at i θ k with the following behavior. There exists a sequence ( w k ( j, j )) j,j ≥ with w k ( · , j ) ∈ H for all j ≥ ,such that: ∀ j, j ≥ , R τ ( j, j ) := G τ ( j, j ) − w k ( j, j ) τ − i θ k , extends holomorphically to the whole ball B ε ( i θ k ) with respect to τ , and the holomorphic extension satisfiesthe bound: ∀ τ ∈ B ε ( i θ k ) , (cid:12)(cid:12) R τ ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c | j − j | (cid:1) . Moreover, the sequence ( w k ( j, j )) j,j ≥ satisfies the pointwise bound: ∀ j, j ≥ , (cid:12)(cid:12) w k ( j, j ) (cid:12)(cid:12) ≤ C exp (cid:0) − c ( j + j ) (cid:1) . (35) Proof.
The proof simply relies on writing κ ( z ) = exp( ω ( z )) and using z = exp( τ ), such that after identi-fication we have (cid:36) ( τ ) := ω (exp( τ )). Next, using our assumption (11), we obtain the desired expansionfor (cid:36) ( τ ) near τ = 0. From this expansion, we getRe( (cid:36) ( τ )) = − α Re( τ ) − βα µ +1 Im( τ ) µ − βα µ +1 ( − µ Re( τ ) µ − βα µ +1 µ − (cid:88) m =1 ( − m (cid:18) µ m (cid:19) Re( τ ) m Im( τ ) µ − m ) + O ( | τ | µ +1 ) . j, n ) plane used in the analysis. In eachdomain, we use a different contour integral in (34). In domain I , that is for − n p ≤ j − j ≤ n α , wecan push the contour of integration Γ to Re( τ ) = − η for some well chosen η >
0. For values in domains II , III and IV , we can also push the contour of integration Γ to Re( τ ) = − η but this time we have touse a “parabolic” contour near the origin. We refer to Figure 2 for an illustration of such contours indomain III . Note that below the lines j − j = − n p and j − j = n r (blue) the Green’s function G n ( j, j )vanishes.for all τ ∈ B ε (0). We crucially note that the term Im( τ ) µ comes with a negative sign such that both O ( | τ | µ +1 ) and each term of the sum, using Young’s inequality, can be absorbed and we arrive at thedesired estimate for two uniform constants 0 < β ∗ < β < β ∗ . The remainder of the proof is a simpletransposition of Lemma 5, 6 and 7 in the new variable τ .With the notations introduced in the above Lemma, we can summarize in the following propositionthe results that we will prove in this section. Why Proposition 1 is sufficient to get the result of Theorem1 is explained at the end of this Section. Proposition 1.
There exist two constants
C > and ω > such that for any n ≥ the temporal Green’sfunction G n satisfies the pointwise estimate ∀ j, j ≥ , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) G n ( j, j ) − K (cid:88) k =1 w k ( j, j ) e n i θ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Cn µ exp (cid:32) − ω (cid:18) | j − j − α n | n µ (cid:19) µ µ − (cid:33) . From now on, we fix 0 < η < η ε such that the segment {− η + i (cid:96) | (cid:96) ∈ [ − π, π ] } intersects ∂B ε (0)outside the curve (12) of essential spectrum of T near the origin. We are going to distinguish severalcases depending on the relative position between j − j and n , as sketched in Figure 1. Formally, we willuse different contours of integration in (34) depending if j − j is near n α or away from n α . Indeed, when j − j ∼ n α , we expect to have Gaussian-like bounds coming from the contribution in B ε (0) near the origin19here the essential spectrum of T touches the imaginary axis. In that case, we will use contours similarto [God03, CF20] and that were already introduced in the continuous setting in [ZH98]. Let us note thatunlike in [CF20], we have isolated poles on the imaginary axis given by the τ k = i θ k , k = 1 , · · · , K , whosecontributions in (34) will be handled via Cauchy’s formula and the residue theorem. We thus divide theanalysis into a medium range, that is for those values of j − j away from n α , and short range when j − j is near n α . More specifically, we decompose our domain as • Medium range: − n p ≤ j − j < n α ; • Short range: n α ≤ j − j ≤ n r ;where we recall that α = λ a > In this section, we consider the medium range where − n p ≤ j − j < n α . In order to simplify thepresentation, we first treat the case where − n p ≤ j − j ≤ ≤ j − j < n α . Lemma 10.
There exist constants
C > and c > , such that for all integers j, j , n satisfying − n p ≤ j − j ≤ , the temporal Green’s function satisfies (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) G n ( j, j ) − K (cid:88) k =1 w k ( j, j ) e n i θ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η − c | j − j | . Proof.
We first recall that G n ( j, j ) = 12 π i (cid:90) Γ e n τ G τ ( j, j ) d τ , with Γ = { s + i (cid:96) | (cid:96) ∈ [ − π, π ] } for any 0 < s ≤ π . Next, we denote Γ − η = {− η + i (cid:96) | (cid:96) ∈ [ − π, π ] } . Usingthe residue theorem, we obtain that12 π i (cid:90) Γ e n τ G τ ( j, j ) d τ = 12 π i (cid:90) Γ − η e n τ G τ ( j, j ) d τ (cid:124) (cid:123)(cid:122) (cid:125) := (cid:101) G n ( j,j ) + K (cid:88) k =1 Res (cid:16) τ (cid:55)→ e n τ G τ ( j, j ) , τ k (cid:17) , where we readily have that K (cid:88) k =1 Res (cid:16) τ (cid:55)→ e n τ G τ ( j, j ) , τ k (cid:17) = K (cid:88) k =1 w k ( j, j ) e n i θ k , from Lemma 9. Here, and throughout, we use the fact that the integrals along {− v ± i π | v ∈ [ − η, s ] } compensate each other. Now, Γ − η intersects each ball B ε ( i θ k ) and we denote Γ k − η := Γ − η ∩ B ε ( i θ k ).Using once again Lemma 9, we have for each τ ∈ Γ k − η | G τ ( j, j ) | ≤ | R τ ( j, j ) | + (cid:12)(cid:12)(cid:12)(cid:12) w k ( j, j ) τ − i θ k (cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − c | j − j | + C e − c | j + j | ≤ C e − c | j − j | , for some positive constants C , >
0. Finally, we remark that for − n p ≤ j − j ≤
0, we have ∀ τ ∈ (cid:110) ω ∈ C | − η ε < Re( ω ) ≤ π (cid:111) \ (cid:32) K (cid:91) k =1 B ε ( i θ k ) (cid:33) , (cid:12)(cid:12) G τ ( j, j ) (cid:12)(cid:12) ≤ C e − c | j − j | , τ ∈ Γ − η and − n p ≤ j − j ≤ (cid:12)(cid:12) G τ ( j, j ) (cid:12)(cid:12) ≤ C e − c | j − j | . The estimate on (cid:101) G n ( j, j ) easily follows and concludes the proof.Next, we consider the range 1 ≤ j − j < n α . This time, the spatial Green’s function G τ ( j, j ) satisfiesa different bound in B ε (0). Nevertheless, we can still obtain some strong decaying estimates which aresummarized in the following lemma. Lemma 11.
There exists a constant
C > such that for all integers j, j , n satisfying ≤ j − j < n α ,the temporal Green’s function satisfies (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) G n ( j, j ) − K (cid:88) k =1 w k ( j, j ) e n i θ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η . Proof.
The beginning of the proof follows similar lines as the ones in the proof of Lemma 10. We deformthe initial contour Γ to Γ − η , and using the residue theorem we get G n ( j, j ) = K (cid:88) k =1 w k ( j, j ) e n i θ k + 12 π i (cid:90) Γ − η e n τ G τ ( j, j ) d τ . We denote by Γ in − η and Γ out − η the portions of Γ − η which lie either inside or outside B ε (0). Note that theanalysis along Γ out − η is similar as in Lemma 10, and we already get the estimate (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ out − η e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η − c | j − j | ≤ C e − n η . Along Γ in − η , we compute (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ in − η e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η (cid:90) Γ in − η e | j − j | Re( (cid:36) ( τ )) | d τ | . Next, for all τ ∈ Γ in − η we haveRe( (cid:36) ( τ )) ≤ − α Re( τ ) − β ∗ α µ +1 Im( τ ) µ (cid:124) (cid:123)(cid:122) (cid:125) ≤ + β ∗ α µ +1 Re( τ ) µ ≤ ηα + β ∗ α µ +1 η µ . As a consequence, − n η + | j − j | Re( (cid:36) ( τ )) ≤ n η (cid:18) − | j − j | n α + | j − j | n β ∗ α µ +1 η µ − (cid:19) ≤ n η (cid:18) −
12 + β ∗ α µ η µ − (cid:19) ≤ − n η , provided that η is chosen small enough (the choice only depends on β ∗ and α ).21 .2 Short range Throughout this section, we assume that n ≥ n α ≤ j − j ≤ n r . Following [ZH98, God03, CF20],we introduce a family of parametrized curves given byΓ p := (cid:26) Re( τ ) − β ∗ α µ Re( τ ) µ + β ∗ α µ Im( τ ) µ = Ψ( τ p ) | − η ≤ Re( τ ) ≤ τ p (cid:27) (36)with Ψ( τ p ) := τ p − β ∗ α µ τ µp . Note that these curves intersect the real axis at τ p . We also let ζ := j − j − n α µ n , and γ := j − j n β ∗ α µ > , and define ρ (cid:16) ζγ (cid:17) as the unique real root to the equation − ζ + γ x µ − = 0 , that is ρ (cid:18) ζγ (cid:19) := sgn( ζ ) (cid:18) | ζ | γ (cid:19) µ − . The specific value of τ p is now fixed depending on the ratio ζγ as follows τ p := − η ρ (cid:18) ζγ (cid:19) < − η ,ρ (cid:18) ζγ (cid:19) if − η ≤ ρ (cid:18) ζγ (cid:19) ≤ ε ,ε if ρ (cid:18) ζγ (cid:19) > ε , where 0 < ε < ε is chosen such that Γ p with τ p = ε intersects the segment {− η + i (cid:96) | (cid:96) ∈ [ − π, π ] } precisely on the boundary of B ε (0). Finally, let us note that as n α ≤ j − j ≤ n r , we have − α µ ≤ ζ ≤ r − α µ . As r > α = λ a (see Lemma 1), the region where − η ≤ ρ (cid:16) ζγ (cid:17) ≤ ε holds is not empty. From nowon, we will treat each subcase separately. Lemma 12.
There exist constants
C > and M > such that for n ≥ and − η ≤ ρ (cid:16) ζγ (cid:17) ≤ ε , thefollowing estimate holds: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) G n ( j, j ) − K (cid:88) k =1 w k ( j, j ) e n i θ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Cn µ exp (cid:32) − M (cid:18) | j − j − α n | n µ (cid:19) µ µ − (cid:33) . Proof.
We will consider a contour depicted in Figure 2 which consists of the parametrized curve Γ p nearthe origin and otherwise is the segment Γ − η . We will denote Γ in − η and Γ out − η , the portions of the segmentΓ − η which lie either inside or outside B ε (0) with | Im( τ ) | ≤ π . Using the residue theorem, we have that G n ( j, j ) = K (cid:88) k =1 w k ( j, j ) e n i θ k + 12 π i (cid:90) Γ in − η ∪ Γ out − η e n τ G τ ( j, j ) d τ + 12 π i (cid:90) Γ p e n τ G τ ( j, j ) d τ . This is possible because the curves Γ p are symmetric with respect to the real axis. − η ≤ ρ (cid:16) ζγ (cid:17) ≤ ε when n α ≤ j − j ≤ n r . Wedeform the initial contour Γ (dark red) into the contour which consists of the parametrize curve Γ p nearthe origin (blue) and a portion of the segment Γ − η = {− η + i (cid:96) | (cid:96) ∈ [ − π, π ] } (light blue). Note thatΓ − η ∩ B ε (0) (cid:54) = ∅ . Near each pole on the imaginary axis (pink cross), we use the residue theorem which issymbolized by the small oriented circles (dark blue) surrounding each of them.Computations along Γ out − η are similar to the previous cases, and we directly get (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ out − η e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η − c | j − j | . For all τ ∈ Γ in − η , we use that Im( τ ) ≥ Im( τ ∗ ) where τ ∗ = − η + i (cid:96) ∗ and (cid:96) ∗ > − η − β ∗ α µ η µ + β ∗ α µ (cid:96) µ ∗ = Ψ( τ p ) . That is, the point τ ∗ = − η + i (cid:96) ∗ lies at the intersection of Γ p and the segment {− η + i (cid:96) | (cid:96) ∈ [ − π, π ] } with τ ∗ ∈ B ε (0). As a consequence, for all τ ∈ Γ in − η we haveRe( (cid:36) ( τ )) ≤ ηα + β ∗ α µ +1 η µ − β ∗ α µ +1 Im( τ ) µ = − τ p α + β ∗ α µ +1 τ µp − β ∗ α µ +1 (cid:0) Im( τ ) µ − (cid:96) µ ∗ (cid:1)(cid:124) (cid:123)(cid:122) (cid:125) ≥ ≤ − τ p α + β ∗ α µ +1 τ µp . n Re( τ ) + ( j − j ) Re( (cid:36) ( τ )) ≤ − n η + ( j − j ) (cid:18) − τ p α + β ∗ α µ +1 τ µp (cid:19) = nα (cid:20) − η α + ( j − j ) n (cid:18) − τ p + β ∗ α µ τ µp (cid:19) (cid:21) = nα (cid:2) − ( η + τ p ) α − µ ζ τ p + γ τ µp (cid:3) = nα (cid:34) − ( η + τ p ) α + (1 − µ ) γ (cid:18) | ζ | γ (cid:19) µ µ − (cid:35) , for all τ ∈ Γ in − η . Finally, as − η ≤ ρ ( ζγ ) = τ p we have η + τ p ≥ η , and we obtain an estimate of the form (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ in − η e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η − nα (2 µ − γ (cid:16) | ζ | γ (cid:17) µ µ − ≤ C e − n η − c n | ζ | µ µ − , since γ is bounded from below and from above by positive constants.We now turn our attention to the integral along Γ p . We first notice that for all τ ∈ Γ p ⊂ B ε (0), wehave Re( τ ) ≤ τ p − c ∗ Im( τ ) µ , for some constant c ∗ >
0. As a consequence, we obtain the upper bound n Re( τ ) + ( j − j ) Re( (cid:36) ( τ )) ≤ n Re( τ ) − j − j α Re( τ ) + β ∗ ( j − j ) α µ +1 Re( τ ) µ − β ∗ ( j − j ) α µ +1 Im( τ ) µ ≤ n (Re( τ ) − τ p ) + nα (cid:104) − µ ζ τ p + γ τ µp (cid:105) ≤ − n c ∗ Im( τ ) µ − nα (2 µ − γ (cid:18) | ζ | γ (cid:19) µ µ − , for all τ ∈ Γ p ⊂ B ε (0). As a consequence, we can derive the following bound (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ p e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:90) Γ p e n Re( τ ) + ( j − j ) Re( (cid:36) ( τ )) | d τ |≤ C e − nα (2 µ − γ (cid:16) | ζ | γ (cid:17) µ µ − (cid:90) Γ p e − n c ∗ Im( τ ) µ | d τ |≤ C e − nα (2 µ − γ (cid:16) | ζ | γ (cid:17) µ µ − n µ ≤ C e − c n | ζ | µ µ − n µ , where we use again that γ is bounded from below and from above by positive constants. At the end ofthe day, we see that the leading contribution is the one coming from the integral along Γ p .Finally, we treat the last two cases altogether. 24 emma 13. There exist constants
C > and c (cid:63) > such that for n ≥ and ρ (cid:16) ζγ (cid:17) < − η or ρ (cid:16) ζγ (cid:17) > ε there holds: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) G n ( j, j ) − K (cid:88) k =1 w k ( j, j ) e n i θ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n c (cid:63) . Proof.
We only present the proof in case ρ ( ζ/γ ) > (cid:15) as the proof for ρ (cid:16) ζγ (cid:17) < − η follows similar lines.We deform the contour Γ into Γ p ∪ Γ out − η where Γ out − η are the portions of Γ − η which lie outside B ε (0) with | Im( τ ) | ≤ π . We recall that we choose τ p = ε here, so the curve Γ p intersects ∂B ε (0) precisely atRe( τ ) = − η . In that case, we have that for all τ ∈ Γ p n Re( τ ) + ( j − j ) Re( (cid:36) ( τ )) ≤ − n c ∗ Im( τ ) µ + nα (cid:16) − µ ζ ε + γ ε µ (cid:17) . But as ρ ( ζ/γ ) > ε we get that ζ > ζ > ε µ − γ , the last term in the previous inequality is estimatedvia − µ ζ ε + γ ε µ < (1 − µ ) γ ε µ . As a consequence, we can derive the following bound (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ p e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − nα (2 µ − γ ε µ n µ . With our careful choice of ε >
0, the remaining contribution along segments Γ out − η can be estimated asusual as (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) π i (cid:90) Γ out − η e n τ G τ ( j, j ) d τ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C e − n η − c ( j − j ) , as | τ | ≥ ε for τ ∈ Γ out − η . The conclusion of Lemma 13 follows.We can now combine Lemma 10, Lemma 11, Lemma 12 and Lemma 13 to obtain the result ofProposition 1. Indeed, we observe that in Lemma 10, Lemma 11 and Lemma 13, the obtained exponentialbounds can always be subsumed into Gaussian-like estimates. (Lemma 12 yields the worst estimate ofall.) For instance, in Lemma 11, the considered integers j, j , n satisfy 1 ≤ j − j ≤ n α/
2, which implies − n ≤ j − j − n αα ≤ − n , and therefore: − n ≤ − ω (cid:18) | j − j − n α | n µ (cid:19) µ µ − , for some sufficiently small constant ω >
0. It remains to explain why Proposition 1 implies Theorem 1.
We let w ∈ H , and first remark that for any integer n , the sequence T n w ∈ H is given by: ∀ j ≥ , ( T n w ) j = (cid:88) j ≥ G n ( j, j ) w j . G n ( j, j ) into two pieces G n ( j, j ) = K (cid:88) k =1 w k ( j, j ) e n i θ k + ˜ G n ( j, j ) , where the remainder term ˜ G n ( j, j ) satisfies the generalized Gaussian estimate of Proposition 1. Fromthe exponential bound (35) and Proposition 1, we have: (cid:88) j ≥ (cid:12)(cid:12)(cid:12) G n ( j, j ) w j (cid:12)(cid:12)(cid:12) ≤ C e − c j (cid:88) j ≥ e − c j | w j | + Cn µ (cid:88) j ≥ exp (cid:32) − ω (cid:18) | j − j − α n | n µ (cid:19) µ µ − (cid:33) | w j | . Noting that the sequence (cid:0) e − c j (cid:1) j ≥ is in (cid:96) , we get that (cid:88) j ≥ e − c j (cid:88) j ≥ e − c j | w j | ≤ C (cid:107) w (cid:107) H . Now for the second term, we observe that the sequence defined as ∀ j ∈ Z , g j := 1 n µ exp (cid:32) − ω (cid:18) | j − α n | n µ (cid:19) µ µ − (cid:33) , is bounded (with respect to n ∈ N ∗ ) in (cid:96) ( Z ). Using the Young’s convolution inequality (cid:96) ( Z ) (cid:63) (cid:96) ( Z ) → (cid:96) ( Z ), we thus obtain the uniform in time bound: (cid:88) j ≥ n µ (cid:88) j ≥ exp (cid:32) − ω (cid:18) | j − j − α n | n µ (cid:19) µ µ − (cid:33) | w j | ≤ C (cid:107) w (cid:107) H . This completes the proof that our operator T is power bounded on H . We illustrate our main result by considering the modified Lax-Friedrichs numerical scheme which reads u n +1 j = u nj + D (cid:0) u nj − − u nj + u nj +1 (cid:1) − λ a (cid:0) u nj +1 − u nj − (cid:1) , j ≥ , (37)where D > λa >
0, along with some specific boundary condition at j = 0 which we shall specifylater. Using our formalism from (5), we have p = r = 1 and a − = D + λ a , a = 1 − D , and a = D − λ a . We readily note that our consistency conditions (9) are satisfied. Next, if we denote F ( θ ) := (cid:88) (cid:96) = − a (cid:96) e i θ (cid:96) , θ ∈ [ − π, π ] , λa = 1 / D = 3 / b = − − √ . The blue curve is the essential spectrum of T and we have indicated by a cross thepresence of eigenvalue at z = −
1. Our Assumptions 1-2 are satisfied in that case.then we have F ( θ ) = 1 − D + D cos( θ ) − i λ a sin( θ ) . As a consequence, provided that 0 < λ a < λ a ) < D <
1, we get ∀ θ ∈ [ − π, π ] \ { } , | F ( θ ) | < , such that the dissipativity condition (10) is also verified. Next, we compute that F ( θ ) = 1 − i λ a θ − D θ + O ( θ ) , as θ tends to 0. We thus deduce that (11) is satisfied with µ := 1 , and β := D − ( λ a ) > . Assumption 1 is thus satisfied provided that we have 0 < λ a < λ a ) < D <
1. We also assumefrom now on D (cid:54) = λ a so that the coefficient a is nonzero.We now prescribe a boundary condition for (37) which will ensure that our Assumption 2 on theLopatinskii determinant is satisfied. That is, we want to find z ∈ S \{ } which is an eigenvalue for T . Thismeans that at this point z the boundary condition must be adjusted so as to have Ker B ∩ E s ( z ) (cid:54) = { } .We use a boundary condition of the form given in (6) with p b = p = 1: u n = b u n , n ≥ , where b ∈ R is a constant. In order to ensure that Ker B ∩ E s ( z ) (cid:54) = ∅ is satisfied, we impose that1 = b κ s ( z ) , λ a = 1 / D = 3 / b = − − √ . We started with an initial condition given bythe Dirac mass at j = 3. In the left figure, we represent the Green’s function at different time iterationsand compare with a fixed Gaussian profile centered at j − j = λ a n away from the boundary j = 1. Inthe right figure, we highlight the behavior of the Green’s function near the boundary. We represent thesolution (blue circules) after 500 time iterations and show that it corresponds to a so-called surface wavegiven by the eigenvalue at z = − T .where κ s ( z ) refers to the (unique) stable eigenvalue of M ( z ). Finally, we select z = −
1. This is the onlyvalue on the unit circle, apart from z = 1, which ensures that κ s ( z ) is real. Note that κ s ( −
1) has theexact expression κ s ( −
1) = − − a a + (cid:115)(cid:18) − − a a (cid:19) − a − a (cid:54) = 0 . Our actual boundary condition is thus u n = 1 κ s ( − u n , n ≥ . (38)With that specific choice, we easily see that Ker B ∩ E s ( z ) is nontrivial for z ∈ U \ { } if and onlyif z = −
1, for the Lopatinskii determinant equals 1 − κ s ( z ) /κ s ( − κ s ( z ) = κ s ( − z = z = −
1. Moreover, − ∂ t u + a ∂ x u = 0 , t > , x > ,u ( t,
0) = 0 , t > ,u (0 , x ) = u ( x ) , x > , for some given (smooth) initial condition u .We present in Figure 3 the spectrum of T associated to the modified Lax-Friedrichs numerical scheme(37)-(38) with λ a = 1 / D = 3 / b = − − √ . In Figure 4, we illustrate the decomposition given28n Proposition 1 where the temporal Green’s function decomposes into two parts: a boundary layer partgiven by w ( j, j ) ( − n which is exponentially localized in both j and j and a generalized Gaussian partwhich is advected away along j − j = λ a n . We start with an initial condition given by the Dirac massat j = 3. We remark that the Green’s function at different time iterations compares well with a fixedGaussian profile centered at j − j = λ a n away from the boundary j = 1. We also visualize the behaviorof the solution near the boundary for 1 ≤ j ≤
15 and shows that up to a constant, depending on j , theenvelope of the Green’s function is given by ± | κ s ( − | j − . A Proofs of intermediate results
This Appendix is devoted to the proof of several intermediate results, which are recalled here for thereader’s convenience.
A.1 The Bernstein type inequality
Lemma 14.
Under Assumption 1, there holds λ a < r .Proof.
We introduce the polynomial function: ∀ z ∈ C , P ( z ) := p (cid:88) (cid:96) = − r a (cid:96) z (cid:96) + r . Assumption 1 implies that P is a nonconstant holomorphic function on C and that the modulus of P isnot larger than 1 on S . By the maximum principle for holomorphic functions, P maps D onto D . Inparticular, since P has real coefficients, P achieves its maximum on [0 ,
1] at 1, and we thus have P (cid:48) (1) ≥ P (cid:48) (1) = r − λ a ≥
0. It remains to explain why λ a can not equal r .We assume from now on λ a = r and explain why this leads to a contradiction. Multiplying (11) byexp( i r θ ), we obtain: P (cid:0) e i θ (cid:1) = exp (cid:0) − β θ µ + O ( θ µ + 1 ) (cid:1) , for θ close to 0. By the unique continuation theorem for holomorphic functions, the latter expansion holdsfor either real or complex values of θ . We thus choose θ = ε exp( i π/ (2 µ )) for any sufficiently small ε > P (cid:0) e i ε exp( i π/ (2 µ )) (cid:1) = exp (cid:0) β ε µ + O ( ε µ + 1 ) (cid:1) , which is a contradiction since P maps D onto D and β >
0. We have thus proved λ a < r . A.2 Proof of Lemma 2
Lemma 15.
Under Assumption 1, there exists c > such that, if we define the set: C := (cid:110) ρ e i ϕ ∈ C / ϕ ∈ [ − π , π ] and 0 ≤ ρ ≤ − c ϕ µ (cid:111) , then C is a compact star-shaped subset of D , and the curve: (cid:40) p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ / θ ∈ [ − π , π ] (cid:41) is contained in C . roof. We first choose the constant c such that for any sufficiently small θ , the point: p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ lies in C . To do so, we use (11) from Assumption 1 and thus write for any sufficiently small θ : p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ = ρ ( θ ) e i ϕ ( θ ) , with: 0 ≤ ρ ( θ ) ≤ − β θ µ , and λ a | θ | ≤ | ϕ ( θ ) | ≤ λ a | θ | . Hence there exists c > θ > | θ | ≤ θ , there holds:0 ≤ ρ ( θ ) ≤ − c ϕ ( θ ) µ . Let us now examine the case θ ≤ | θ | ≤ π . By continuity and compactness, (10) yields:sup θ ≤| θ |≤ π (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 − δ , for some δ >
0. Up to choosing c smaller, we can always assume c π µ ≤ δ , so for any angle θ with θ ≤ | θ | ≤ π , the point: p (cid:88) (cid:96) = − r a (cid:96) e i (cid:96) θ lies in C . The proof is thus complete. A.3 Proof of Lemma 3 on the spectral splitting
Lemma 16.
Under Assumption 1, let z ∈ C and let the matrix M ( z ) be defined as in (13) . Let the set C be defined by Lemma 2. Then for z (cid:54)∈ C , M ( z ) has: • no eigenvalue on S , • r eigenvalues in D \ { } , • p eigenvalues in U (eigenvalues are counted with multiplicity).Furthermore, M (1) has as a simple eigenvalue, it has r − eigenvalues in D and p eigenvalues in U .Proof. We are first going to show that for z (cid:54)∈ C , M ( z ) has no eigenvalue on the unit circle S (this is aclassical observation that dates back to [Kre68]). From the definition (13), we first observe that for any z ∈ C , M ( z ) is invertible (its kernel is trivial since r ≥ a − r (cid:54) = 0 so the upper right coefficient of M ( z ) is nonzero). Therefore, for any z ∈ C , the eigenvalues of M ( z ) are those κ (cid:54) = 0 such that: z = p (cid:88) (cid:96) = − r a (cid:96) κ (cid:96) . (39)30n particular, Lemma 2 shows that for z (cid:54)∈ C , M ( z ) cannot have an eigenvalue κ on the unit circle forotherwise the right hand side of (39) would belong to C .Since C is closed and star-shaped, its complementary is pathwise-connected hence connected. There-fore, the number of eigenvalues of M ( z ) in D is independent of z (cid:54)∈ C (same for the number of eigenvaluesin U ). Following [Kre68] (see also [Cou13] for the complete details), this number is computed by letting z tend to infinity for in that case, the eigenvalues of M ( z ) in D tend to zero (the eigenvalues in D cannotremain uniformly away from the origin for otherwise the right hand side of (39) would remain boundedwhile the left hand side tends to infinity).The final argument is the following. For any z (cid:54)∈ C , the eigenvalues of M ( z ) are those κ (cid:54) = 0 suchthat: κ r = 1 z p (cid:88) (cid:96) = − r a (cid:96) κ r + (cid:96) , which is just an equivalent way of writing (39). Hence for z large, the small eigenvalues of M ( z ) behaveat the leading order like the roots of the reduced equation: κ r = a − r z , and there are exactly r distinct roots close to 0 of that equation. Hence M ( z ) has r eigenvalues in D forany z (cid:54)∈ C .There remains to examine the spectral situation for z = 1. Using (39) again, the eigenvalues of M (1)are exactly the roots κ (cid:54) = 0 to the equation: 1 = p (cid:88) (cid:96) = − r a (cid:96) κ (cid:96) . (40)Thanks to Assumption 1 (see (9) and (10)), the only root of (40) on the unit circle is κ = 1 and it isa simple root. This simple eigenvalue can therefore be extended holomorphically with respect to z as asimple eigenvalue of M ( z ) for z in a neighborhood of 1. Differentiating (39) with respect to z , we obtainthe Taylor expansion: κ ( z ) = 1 − λ a ( z −
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