The Complex-Scaled Half-Space Matching Method

Thecompl ex scal edhal f spacemat chi ng met hod Ar t i cl e Accept edVer si on Bonnet BenDhi a,A. S. ,Chandl er Wi l de,S.N. ,Fl i ss,S. , Hazar d,C. ,Per f ekt ,K. M.andTj andr awi dj aj a,Y.( 2022)The compl ex scal edhal f spacemat chi ngmet hod.SI AM Jour nalon Mat hemat i calAnal ysi s( SI MA) ,54( 1) .pp.512557.I SSN 00361410doi :ht t ps: / / doi . or g/ 10. 1137/ 20M1387122Avai l abl e atht t ps: / / cent aur . r eadi ng. ac. uk/ 100188/ I ti sadvi sabl et or ef ert ot hepubl i sher ’ sver si oni fyoui nt endt oci t ef r om t he wor k.SeeGui danceonci t i ng. Tol i nkt ot hi sar t i cl eDOI :ht t p: / / dx. doi . or g/ 10. 1137/ 20M1387122 Publ i sher :Soci et yf orI ndust r i alandAppl i edMat hemat i cs Al lout put si nCent AUR ar epr ot ect edbyI nt el l ect ualPr oper t yRi ght sl aw, i ncl udi ngcopyr i ghtl aw.Copyr i ghtandI PR i sr et ai nedbyt hecr eat or sorot her copyr i ghthol der s.Ter msandcondi t i onsf oruseoft hi smat er i alar edefinedi n eement . t heEndUserAgr www. r eadi ng. ac. uk/ cent aur Cent AUR Cent r alAr chi veatt heUni ver si t yofReadi ng Readi ng’ sr esear chout put sonl i ne THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD∗ ANNE-SOPHIE BONNET-BEN DHIA† , SIMON N. CHANDLER-WILDE‡ , SONIA FLISS† , CHRISTOPHE HAZARD† , KARL-MIKAEL PERFEKT§ , AND YOHANES TJANDRAWIDJAJA¶ Abstract. The Half-Space Matching (HSM) method has recently been developed as a new method for the solution of 2D scattering problems with complex backgrounds, providing an alternative to Perfectly Matched Layers (PML) or other artificial boundary conditions. Based on half-plane representations for the solution, the scattering problem is rewritten as a system of integral equations in which the unknowns are restrictions of the solution to the boundaries of a finite number of overlapping half-planes contained in the domain: this integral equation system is coupled to a standard finite element discretisation localised around the scatterer. While satisfactory numerical results have been obtained for real wavenumbers, wellposedness and equivalence to the original scattering problem have been established only for complex wavenumbers. In the present paper, by combining the HSM framework with a complex-scaling technique, we provide a new formulation for real wavenumbers which is provably well-posed and has the attraction for computation that the complex-scaled solutions of the integral equation system decay exponentially at infinity. The analysis requires the study of double-layer potential integral operators on intersecting infinite lines, and their analytic continuations. The effectiveness of the method is validated by preliminary numerical results. Key words. Helmholtz equation, scattering, integral equation, artificial radiation condition AMS subject classifications. 35J05, 35J25, 35P25, 45B05, 45F15, 65N30, 65N38, 78A45 1. Introduction and the scattering problem. The mathematical and numerical analysis of scattering by bounded obstacles and/or inhomogeneities in a homogeneous background is a mature research area, and there are many effective numerical schemes, at least for low to moderately high frequencies. However, when the background is heterogeneous (stratified, periodic,...) and/or anisotropic, especially when electromagnetic or elastic waves are considered, many theoretical questions are still open and the design of efficient numerical methods remains a significant challenge. It is well known that, for homogeneous backgrounds, the Sommerfeld radiation condition ensures well-posedness of the scattering problem [25]. The extension of this standard radiation condition to the aforementioned more complex backgrounds is really intricate (see, e.g., [10, 13, 29, 33, 36]). Moreover, a Green’s function or tensor, which could be used to derive an integral equation formulation of the problem, is in general not available or hard to compute. Finally, Perfectly Matched Layer (PML) techniques, which are popular in homogeneous backgrounds because they are easy to implement, can produce spurious effects for complex backgrounds, as is well-known for anisotropic backgrounds in relation to instabilities in the time domain [8]. ∗ Submitted to the editors DATE. Funding: The research of K.-M. Perfekt was partially supported by UK Engineering and Physical Sciences Research Council (EPSRC) grant EP/S029486/1. The work of Y. Tjandrawidjaja was supported by the French Commissariat à l’Energie Atomique et aux Energies Alternatives (CEA), through funding for PhD studies. † POEMS (CNRS-INRIA-ENSTA Paris), Institut Polytechnique de Paris, Palaiseau, France (annesophie.bonnet-bendhia@ensta-paris.fr, sonia.fliss@ensta-paris.fr, christophe.hazard@ensta-paris.fr) ‡ Department of Mathematics and Statistics, University of Reading, Whiteknights PO Box 220, Reading RG6 6AX, UK (s.n.chandler-wilde@reading.ac.uk, k.perfekt@reading.ac.uk) § Department of Mathematical Sciences, Norwegian University of Science and Technology (NTNU), NO-7491 Trondheim, Norway karl-mikael.perfekt@ntnu.no) ¶ Lehrstuhl I, Fakultät für Mathematik, Technische Universität Dortmund, Vogelpothsweg 87, 44227 Dortmund, Germany (Yohanes.Tjandrawidjaja@math.tu-dortmund.de) 1 2 A.-S. BONNET-BEN DHIA ET AL. To overcome these difficulties a new method, called the Half-Space Matching (HSM) method, has been developed as an (exact) artificial boundary condition for two-dimensional scattering problems. This method is based on explicit or semi-explicit expressions for the outgoing solutions of radiation problems in half-planes, these expressions established by using Fourier, generalized Fourier, or Floquet transforms when the background is, respectively, homogeneous [12, 11] (and possibly anisotropic [46, 7, 45]), stratified [42], or periodic [32]. The domain exterior to a bounded region enclosing the scatterers is covered by a finite number of half-planes (at least three). The unknowns of the formulation are the traces of the solution on the boundaries of these half-planes and the restriction of the solution to the bounded region. The system of equations which couples these unknowns is derived by writing compatibility conditions between the different representations of the solution. This coupled system includes second-kind integral equations on the infinite boundaries of the half-planes. This new formulation is attractive and versatile as a method to truncate computational domains in problems of scattering by localised inhomogeneities in complex backgrounds (including backgrounds that may be different at infinity in different directions). It has been employed successfully in numerical implementations for a range of problems, namely periodic [34] and stratified media (including cases with different stratifications in different parts of the background domain) [42], and anisotropic scalar and elastic problems [46, 45]. Up to now the theoretical and numerical analysis of the method has remained an open question in the challenging, and more practically relevant, non-dissipative case when waves radiate out to infinity. But a rather complete analysis has been carried out in the simpler dissipative case, when the solution (and its traces) decay exponentially at infinity. In that case the analysis can be done using an L2 framework for the traces, and the associated formulation has been shown to be of Fredholm type and well-posed in a number of cases [12, 11], with the sesquilinear form of the weak formulation coercive plus compact, enabling standard numerical analysis arguments [11]. One of the main difficulties in the non-dissipative case is the slow decay at infinity of the solution which results in non-L2 traces. The possibility, to address this, of working in the framework initially introduced in [14, 15] was investigated by the authors, but it seems to be inappropriate for the present analysis. The objective of this paper is to propose a new formulation of the HSM method which is well-suited for theoretical and numerical analysis (and practical computation) in the non-dissipative case. For the sake of clarity and as a first step, we restrict ourselves in the present paper to a relatively simple model problem for which the justification of the method is based on the simple form of the associated Green’s function. (Let us mention that the extension of this formulation to anisotropic backgrounds has been already validated and will be the subject of a forthcoming paper. See Section 8 for a brief discussion of extensions to other more complex configurations.) This new formulation exploits a fundamental property of the solution in the spirit of the ideas behind complex-scaling methods (e.g. the pioneering works of [6],[2]): in any given direction, the solution, as a function of the associated curvilinear abscissa, has an analytic continuation into the complex plane which is exponentially decaying in the upper complex plane. This enables us to replace the system of equations for the traces by similar equations for exponentially-decaying analytical continuations of these traces. This recovers well-posedness in an L2 framework, with coercive plus compact sesquilinear forms; moreover, attractive for computation, the rate of exponential decay of these analytically-continued traces increases with increasing wavenumber. Let us mention that in [39] a similar integral-equation-based complex-scaling idea, namely THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 3 a boundary integral equation formulation of PML, is used to compute 2D scattering by localised perturbations in a straight interface between different media. In the present paper we consider the rather simple model case of a scalar equation, the isotropic Helmholtz equation (1) − ∆u − k 2 ρu = f in Ω, deduced from the wave equation assuming a time-dependence e−iωt for a given angular frequency ω > 0. Here ρ is a function in L∞ (Ω), bounded from below by a positive constant, and such that ρ − 1 is compactly supported, and the constant k = ω/c is the wavenumber, where c is the wave speed outside the support of ρ − 1, so that (1) models propagation in a domain with a local perturbation in wave speed. The propagation domain Ω is R2 , or R2 minus a set of obstacles which are included in a bounded region. We assume that, for a positive constant a, ∂Ω ⊂ Ωa where Ωa := (−a, a)2 . In the presence of obstacles, boundary conditions have to be added to the model. The source term f is supposed to be a function in L2 (Ω) with compact support included in Ωa , and we assume that supp(ρ − 1) ⊂ Ωa . As already mentioned, in order to get well-posedness, one has to prescribe in addition the Sommerfeld radiation condition, that, for x := (x1 , x2 ) ∈ R2 ,   ∂u (x) − iku(x) = o r−1/2 as r := |x| → +∞, (2) ∂r b := x/r. uniformly with respect to x In the sequel, we will consider two configurations in order to focus first on the HalfSpace Matching formulation and then on its coupling with a variational formulation in a bounded region. In Sections 2 to 5 we consider a Dirichlet problem set in the exterior of the square Ωa . The application of the analysis in Sections 2-5 to general configurations, with source terms, inhomogeneities, and/or obstacles contained in Ωa , is the object of Section 6. The outline of the paper is as follows. In Section 2 we recall the main results concerning the HSM formulation in the dissipative case (that is with a complex wavenumber k). In previous papers the HSM formulation has been derived using Fourier representations for the solution in half-planes. Here we introduce a new formulation using double-layer potential integral representations to derive the so-called complex-scaled version of the method, valid for a real wavenumber k. The derivation and the analysis of this new formulation is the object of Section 3, which finishes with a statement of our main well-posedness result for the new method. In Section 4 we establish properties of the solution that can be reconstructed a posteriori, from knowledge of the complex-scaled traces, notably elements of the farfield pattern. These properties are used in Section 5 to establish the uniqueness result for the complex-scaled HSM problem. The HSM method approach is extended to general configurations in Section 6; the analysis in this section depends throughout on the well-posedness and other results obtained in Sections 3-5. In Section 7 the implementation of a finite element discretization of the complex-scaled HSM formulation is described and numerical results 4 A.-S. BONNET-BEN DHIA ET AL. are presented, for both the model Dirichlet problem and more general configurations. The paper finishes with three appendices to which we defer certain technical details of the analysis. 2. The HSM method for complex wavenumber. In this section, as preparation for studying the HSM method for real wavenumber, we first recall what is known about the method in the dissipative case. It is enough for this purpose to consider the Dirichlet problem for complex wavenumber (ℑ(k) > 0, ℜ(k) > 0) in the exterior of the square Ωa , i.e. ( −∆u − k 2 u = 0 in Ω := R2 \ Ωa , (3) u = g on Σa := ∂Ωa , for a given g ∈ H 1/2 (Σa ). It is well-known that Problem (3) has a unique solution u ∈ H 1 (Ω). Let us first recall the main results of [12]. The domain Ω is the union of 4 overlapping half-planes Ωj that abut the 4 edges of the square Ωa . We introduce the following local coordinates for all j ∈ J0, 3K := {0, 1, 2, 3}:   j   cos(jπ/2) sin(jπ/2) x1 x1 := (4) . j −sin(jπ/2) cos(jπ/2) x2 x2 The half-planes and their boundaries are defined as follows for all j ∈ J0, 3K: (5) Ωj := {(xj1 , xj2 ) : xj1 > a, xj2 ∈ R}, Σj := {(xj1 , xj2 ) : xj1 = a, xj2 ∈ R}. Finally, we denote Σja := Σa ∩ Σj . (6) These notations are summarised in Figure 1. As explained in the introduction, the formulation uses the representation of the solution in each half-plane Ωj in terms of its trace on Σj . More precisely, let us denote ϕj := u (7) Σj for j ∈ J0, 3K so that (8) u Ωj = U j (ϕj ) for j ∈ J0, 3K where, for any ψ ∈ H 1/2 (Σj ), U j (ψ) ∈ H 1 (Ωj ) is the unique solution of −∆U j − k 2 U j = 0 (9) Uj = ψ in Ωj , on Σj . In the sequel, we identify any function defined on Σj , in particular the function ϕj , with a function of the real variable xj2 . We can express U j (ψ) explicitly in terms of its trace ψ in two manners: using the Fourier transform or using a Green’s function representation. First, using the Fourier transform in the xj2 −direction, it is easy to see that the solution of (9) is given by (10) 1 U j (ψ)(xj ) = √ 2π Z R − b ψ(ξ)e √ ξ 2 −k2 (xj1 −a) iξxj2 e dξ, xj := (xj1 , xj2 ) ∈ Ωj , THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD Ω1 5 Σ0 Σ1a Σ1 x2 Ω Σ2a 2 Ωa x1 Σ0a Ω0 Σ3a Σ3 Σ2 Ω3 Fig. 1. The notations defined in (4-5-6). √ where the square root is defined with the convention ℜ( z) ≥ 0, for z ∈ C \ R− (with R− := (−∞, 0]) and ψb is the Fourier transformation of ψ using the convention Z j b := √1 ψ(xj2 ) e−iξx2 dxj2 , ξ ∈ R. (11) ψ(ξ) 2π R Secondly, using a Green’s function representation, we can show that Z ∂Gj (xj , y j ) ψ(y j ) ds(y j ), xj ∈ Ωj , (12) U j (ψ)(xj ) = ∂n(y j ) Σj where Gj (xj , y j ) is the Dirichlet Green’s function for Ωj and n(y j ) is the unit normal to Σj that points into Ωj . Explicitly, Gj (xj , y j ) = Φ(xj , y j ) − Φ(x̃j , y j ), with x̃j the image of xj in Σj , where Φ(x, y) is the standard fundamental solution of the Helmholtz equation defined by (13) Φ(x, y) := i (1) H (k|x − y|), 4 0 x, y ∈ R2 , x 6= y, so that, equivalently, (14) j j U (ψ)(x ) = 2 Z ∂Φ(xj , y j ) ψ(y j ) ds(y j ), ∂n(y j ) xj ∈ Ωj . h(xj1 − a, xj2 − y2j ) ψ(y2j ) dy2j , xj ∈ Ω j , Σj This leads to (15) j j U (ψ)(x ) = Z R where (1) (16) h(x1 , x2 ) := ikx1 H1 (kR(x1 , x2 )) , 2 R(x1 , x2 ) 6 A.-S. BONNET-BEN DHIA ET AL. and (17) R(x1 , x2 ) := (x21 + x22 )1/2 , x1 , x2 ∈ R. Let us remark that the two representations (10) and (15) of U j (ψ) can be derived the one from the other by using simply a Plancherel formula (e.g. [19, p. 821]). To derive the system of equations whose unknowns are the traces ϕj of the solution, it suffices to write that the half-plane representations must coincide where they coexist. For instance, in the quarter plane Ω0 ∩ Ω1 we have Ω 0 ∩ Ω1 Ω1 Σ1 Ωa Ω0 Σ0 Fig. 2. Construction of the compatibility condition. (18) u = U 0 (ϕ0 ) = U 1 (ϕ1 ) in Ω0 ∩ Ω1 , and in particular (19) ϕ1 = U 0 (ϕ0 ) on Ω0 ∩ Σ1 , which leads to a first integral equation linking ϕ0 and ϕ1 . Indeed, for any point of Σ1 ∩ Ω0 , represented by (x01 , x02 ) and (x11 , x12 ) in local coordinates systems, we have x12 = −x01 < −a and x11 = x02 = a (indeed, the point is on Σ1 ∩ Ω0 if and only if these equations hold). Thus the compatibility relation (19) can be rewritten (identifying Σ1 with R in the way noted above) as (20) ϕ1 (x12 ) = U 0 (ϕ0 )(−x12 , a), x12 < −a, where we can use either of the two integral representations (10) and (15) for the half-plane solution U 0 (ϕ0 ). From (18) we have also ϕ0 = U 1 (ϕ1 ) on Σ0 ∩ Ω1 , which leads to another integral equation linking ϕ0 and ϕ1 : ϕ0 (x02 ) = U 1 (ϕ1 )(x02 , −a), x02 > a. Repeating this for each quarter plane we get 8 equations linking the 4 traces. In order to write the system of equations in a condensed manner, we shall use the same THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 7 notation t instead of the various variables xj2 so that the 8 equations become (21) ϕj (t) = S D ϕj−1 (t), ϕj (t) = D S ϕj+1 (t), ∀j ∈ J0, 3K, t < −a, t > a, where we have set ϕ−1 := ϕ3 and ϕ4 := ϕ0 , and where the operators S and D are defined as follows. For any ψ ∈ L2 (R), (22) Sψ(t) := ψ(−t), t ∈ R, and the integral operator D is defined by Dψ(t) := U 0 (ψ)(t, a), (23) t > a. D can be given explicitly by either of the following two expressions: Z √ 1 i k2 −ξ 2 (t−a) iξa b ψ(ξ)e e dξ, t > a, (24) Dψ(t) = √ 2π R or (25) Dψ(t) = Z R h(t − a, s − a) ψ(s) ds, t > a, where the kernel h is defined in (16)1 . The system of equations has to be completed with the Dirichlet boundary condition rewritten as (26) ϕj (t) = g|Σja (t), −a < t < a, j ∈ J0, 3K. One can easily check that (21)-(26) is equivalent to the original problem (3). More precisely, if {ϕ0 , ϕ1 , ϕ2 , ϕ3 } ∈ (H 1/2 (R))4 is a solution to (21)-(26) then one can recover the solution u to (3), from the knowledge of the ϕj ’s, thanks to the half-plane representations (10) or (15). Indeed, by uniqueness of Dirichlet quarterplane problems, two half-plane representations U j (ϕj ) and U j+1 (ϕj+1 ) coincide on the quarter-plane Ωj ∩ Ωj+1 since the compatibility conditions (21) imply that they coincide on its boundary. For the analysis and the computation it is convenient to consider the formulation in an L2 -framework: Find {ϕ0 , ϕ1 , ϕ2 , ϕ3 } ∈ (L2 (R))4 such that (27) ϕj (t) = S D ϕj−1 (t), ϕj (t) = g|Σja (t), j j+1 ϕ (t) = D S ϕ t < −a, −a < t < a, (t), t > a. j ∈ J0, 3K One attraction of this L2 -framework is that it allows the use of elementary operations of restriction and extension. More precisely, for any function of L2 (R), its restriction to an open interval I ⊂ R is in L2 (I). More significantly, any function of L2 (I) extended by 0 belongs to L2 (R). For simplicity, any function defined on a part of R is identified hereafter with its extension by 0. With this convention we can write (28) 1 Note L2 (R) = L2 (−∞, −a) ⊕ L2 (−a, a) ⊕ L2 (a, +∞), that, as is clear from (14), D is precisely a double-layer potential operator (in the sense, e.g., of [26] or [17]) from {(a, t) : t ∈ R} to {(t, a) : t ≥ a}. 8 A.-S. BONNET-BEN DHIA ET AL. which will be extensively used hereafter. In line with this convention, we define Dψ(t) for all t ∈ R by setting (29) Dψ(t) := 0, t ≤ a. With these various conventions (27) can be rewritten shortly as ϕj = SDϕj−1 + DSϕj+1 + g|Σja , (30) j ∈ J0, 3K. Also, with the above conventions, results proved in [17, 12] can be stated as follows: Proposition 2.1. (i) D is a continuous operator on L2 (R), with range in L2 (a, +∞) ⊂ L2 (R). √ (ii) As an operator on L2 (a, +∞), D is the sum of an operator of norm ≤ 1/ 2 and a compact operator. (iii) As an operator from L2 (−∞, −a) to L2 (a, +∞), D is compact. Proof. Note first that (i) follows immediately from (ii) and (29), since (ii), together with a symmetry argument with respect to a, implies that D : L2 (−∞, a) → √ 2 L (a, +∞) is also the sum of an operator of norm ≤ 1/ 2 and a compact operator. Consider the expression (25) for D. Because of the dissipation (ℑ(k) > 0), the kernel h is exponentially decaying at infinity (i.e. as t or s tends to ∞). Further, the mapping (t, s) 7→ h(t − a, s − a) is continuous except at t = s = a. Thus (iii) is clear since the kernel of D is Hilbert-Schmidt, i.e. (t, s) 7→ h(t − a, s − a) ∈ L2 ((−∞, −a) × (a, +∞)). To show (ii), the only difficulty comes from the singularity of the kernel h at t = s = a. As in Proposition C.2 and Remark C.3 in Appendix C, let h0 and D0 denote h and D, respectively, when k = 0. For b > a let χ(a,b) denote the characteristic function of (a, b). Then it is straightforward to see that, for every b > a, D − χ(a,b) D0 is an integral operator with kernel h(t − a, s − a) − χ(a,b) (t)h0 (t − a, s − a) that is Hilbert-Schmidt, so D − χ(a,b) D0 is compact. Further (Remark C.3), as an operator √ √ on L2 (a, +∞), kD0 k = 1/ 2, so also kχ(a,b) D0 k ≤ kχ(a,b) k kD0 k = 1/ 2. Now the system (27) can be formulated in an operator form. Let us introduce (31) L20 (R) := {ψ ∈ L2 (R) : ψ(t) = 0 for − a < t < a} and Φ := {ϕ0 , ϕ1 , ϕ2 , ϕ3 } ∈ (L2 (R))4 , (32) Φg := {g|Σ0a , g|Σ1a , g|Σ2a , g|Σ3a } ∈ (L2 (−a, a))4 ⊂ (L2 (R))4 . Then, noting (30), the system (27) can be rewritten as: (33) where (34) Find Φ ∈ (L2 (R))4 such that Φ − Φg ∈ (L20 (R))4 and (I − D)(Φ − Φg ) = D Φg ,  0 S D D :=   0 DS In [12], the following result is proven: DS 0 SD 0 0 DS 0 SD  SD 0  . D S 0 THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 9 Theorem 2.2. (i) D is a continuous operator on (L2 (R))4 and D((L2 (R))4 ) ⊂ (L20 (R))4 . √ (ii) As an operator on (L20 (R))4 , D is the sum of an operator of norm ≤ 1/ 2 and a compact operator. (iii) Problem (33) is well-posed. Let us give some ideas of the proof which will be relevant for the following sections. The property (iii) is largely a consequence of (ii) since (ii) gives that (I − D), as an operator acting on (L20 (R))4 , is the sum of a coercive operator2 and a compact one. By Fredholm theory, it suffices then to show uniqueness (which is not straightforward in the L2 framework, see [12] for more details). The properties (i) and (ii) are consequences (see Appendix A) of Proposition 2.1. 3. The complex-scaled HSM method for real wavenumber. In this section we consider the Dirichlet problem of the previous section, but now with real 1 1 1 wavenumber (k > 0). Where Hloc (Ω) := {v|Ω : v ∈ Hloc (R2 )}, we seek u ∈ Hloc (Ω) such that ( −∆u − k 2 u = 0 in Ω := R2 \ Ωa , (35) u = g on Σa , for a given g ∈ H 1/2 (Σa ), and such that the radiation condition (2) holds. It is well known that this problem has a unique solution. The half-plane representations (8), with U j (ϕj ) given by (12) (equivalently (14) or (15)), still hold for k real, and can be derived via Green’s theorem using the radiation condition (2) (cf. [18, Theorem 2.1]). As a consequence, the traces ϕj , j ∈ J0, 3K, of the solution u on Σj , still satisfy the system of equations (27) when k > 0. We note that, although the solution of (35)-(2) decays only slowly, like r−1/2 as r → +∞, the integrals (15) still make sense. The bound (110), that follows from asymptotics of the (1) Hankel function H1 , implies that, for some constant C > 0 depending only on k, |h(x1 , x2 )| ≤ Cx1 (R−3/2 + R−2 ), x1 > 0, x2 ∈ R, where R := (x21 + x22 )1/2 , so that (15) is well-defined, even when k is real, for every xj ∈ Ωj and ψ ∈ L2loc (R) with ψ(t) = O(1) as |t| → +∞. Moreover, though we shall not need this, it is still possible to rewrite (15) equivalently as (10), provided that care is taken in interpreting the right hand side of (10); see the discussion in [4, 20] and [14, 15]. From a numerical point of view, the HSM method for real k works well [12]. However, from a theoretical point of view, the formulation does not make sense in an L2 setting. Indeed, as the solution of (35)-(2) decays only like r−1/2 as r → +∞, we cannot expect that its traces on Σj , j ∈ J0, 3K, are in L2 (R). In parallel work [9] we have shown that the HSM formulation for real k is equivalent with the original problem (35) if we supplement it with radiation conditions analogous to the Sommerfeld condition (2) (see (74) below). However, there are still significant gaps in our understanding of this formulation when k is real. In particular, while (27) can be written formally in operator form as (33), just as in the dissipative case, in the case 2 Recall that, given a Hilbert space H with inner product (·, ·), we call a bounded linear operator A on H coercive if the corresponding sesquilinear form a(·, ·), defined by a(φ, ψ) = (Aφ, ψ), ∀φ, ψ ∈ H, is coercive, i.e., if, for some constant γ > 0, ℜ(a(φ, φ)) ≥ γkφk2 , ∀φ ∈ H. 10 A.-S. BONNET-BEN DHIA ET AL. when k is real there is no obvious function space setting (e.g., Lp , or a weighted Lp space) for which this formulation makes sense, with D a well-defined bounded linear operator. Consequently, we are not able to justify the numerical method and neither provide a priori error estimates. These difficulties with the standard formulation for real k are part of the motivation for the method proposed in this paper that we term the complex-scaled HSM method. The idea behind this method is to “complexify”. Since the pioneering works of Aguilar, Balslev and Combes [2, 6], complex-scaling methods have been used intensively to construct the analytic continuation of resolvents in mathematical physics, see for instance [31] and the references therein. The complex-scaling method in [2, 6] is closely related to the idea behind PML (see for instance [24]). Our use of complexscaling is somewhat different since we consider only analytic continuation of traces of the solution on particular infinite half-lines and apply that complex scaling in an integral equation context. This is similar to manipulations made to understand analyticity of boundary traces in high frequency scattering problems in [22, §4.1]. Precisely our plan is as follows: 1. From properties of the solution u of (35)-(2) we deduce that the traces ϕj , j ∈ J0, 3K, have analytic continuations into the complex plane from (−∞, −a) and from (a, +∞). Further, we introduce paths in the complex plane on which the ϕj ’s are L2 (in fact, decay exponentially), see Proposition 3.3. The objective of the next steps is to derive an equivalent of the HSM formulation for these “complex-scaled” traces. 2. For real wavenumbers equations (8) and (15) provide half-plane representations of the solution u in terms of the traces ϕj , j ∈ J0, 3K. The magic result is that the solution u can also be represented in terms of the complex-scaled traces, see Theorem 3.4. The price to pay is that the jth representation, for j ∈ J0, 3K, is valid only in a part, which depends on the path chosen in step 1, of the corresponding half-plane Ωj . These new representations are deduced from the initial ones (8) by applying Cauchy’s integral theorem. 3. Fortunately, the part of Ωj where the new representation holds contains the half-lines Σj±1 ∩ Ωj . By complexifying – by which we mean analytically continuing – this new representation into the complex plane from the halflines Σj±1 ∩Ωj , we can derive compatibility conditions for the complex-scaled traces which constitute a complex-scaled version of the HSM formulation in an L2 setting. Fredholmness of this formulation can be proven using similar arguments as for the standard HSM for complex wavenumbers, see Theorem 3.7. 4. Once the complex-scaled traces are computed the solution can be reconstructed using the new representations in terms of the complex-scaled traces established in step 2. Let us mention that, while the initial motivation for the complexification was a theoretical one, it turns out that the new formulation is very attractive computationally, because of the fast decay at infinity of the complex-scaled traces. Let us note also that this idea of complexification is potentially valuable for computation in the dissipative case too, and it is likely that the formulation in the non-dissipative case could be derived from a complex-scaled formulation in the dissipative case by a limiting absorption argument. (Something similar has been done in the context of scattering by wedges in [28, 37].) An attraction of such a derivation would be that traces are in L2 at each step of the derivation. THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 11 3.1. The complex-scaled traces. The construction of the so-called complex1 scaled traces is based on an analyticity property of any solution u ∈ Hloc (Ω) to (2)-(35), which can be derived as follows. Following [16] (and see [21, Theorem 2.27, Corollary 2.28]), u can be expressed as a combined single- and double-layer potential on Σa , i.e. as (36) u(x) = Dφ(x) − ikSφ(x), x ∈ Ω, for some φ ∈ H 1/2 (Σa ) (specified below). Here Sφ and Dφ are the (acoustic) singleand double-layer potentials, respectively, with density φ, defined for φ ∈ L2 (Σa ) by Z Z ∂Φ(x, y) φ(y) ds(y), x ∈ Ω, Sφ(x) := Φ(x, y)φ(y) ds(y), Dφ(x) := Σa Σa ∂n(y) where the normal n is directed into Ω and Φ is the outgoing Green’s function of the Helmholtz equation given in (13). The function u defined in (36) satisfies the Helmholtz equation (35) and the Sommerfeld radiation condition (2) for any choice of φ ∈ H 1/2 (Σa ) (in fact, any φ ∈ L2 (Σa )), and (see [21, §2.6]), satisfies the boundary condition u = g on Σa provided (37) Aφ = g, where Aφ is defined for φ ∈ L2 (Σa ) and almost all x ∈ Σa by  Z  φ(x) ∂Φ(x, y) Aφ(x) := + − ikΦ(x, y) φ(y)ds(y), 2 ∂n(y) Σa with the integral understood as a Cauchy principal value. Since A : H s (Σa ) → H s (Σa ) is invertible for 0 ≤ s ≤ 1 [23, Corollary 2.8], in particular for s = 1/2, (37) has a unique solution φ ∈ H 1/2 (Σa ). For a given j ∈ J0, 3K, we apply (36) for x ∈ Σj \ Σja , and we use the coordinate system (xj1 , xj2 ) defined in (4). Defining xj (t) := (a, t) with |t| > a, this yields, by definition of ϕj , ϕj (t) = Dj φ(t) − ikS j φ(t), (38) for real t with |t| > a, where (39) Z Z ∂Φ(xj (t), y j ) j j j φ(y )ds(y ), S φ(t) := Φ(xj (t), y j )φ(y j ) ds(y j ), Dj φ(t) := ∂n(y j ) Σa Σa and we recall from (13) that Φ(xj (t), y j ) = i (1) kR(a − y1j , t − y2j )), H 4 0 where R is defined in (17). Let us use (38) to prove that the function ϕj , defined by (7), can be continued analytically into the complex plane from (−∞, −a) and from (a, +∞). Consider a fixed y j ∈ Σa . The function z 7→ R(a − y1j , z − y2j ) has an analytic continuation from (a, +∞) (respectively, (−∞, −a)) to the complex half-plane ℜ(z) > a (respectively, ℜ(z) < −a). Indeed, to obtain this analytic continuation we simply have to use, in 12 A.-S. BONNET-BEN DHIA ET AL. the definition (17) of R(a−y1j , z −y2j ) for real √ z, the principal square root of a complex number, which we will denote by z 1/2 or z, defined as √ z := |z|1/2 eiArg(z)/2 with Arg(z) ∈ (−π, +π], which is analytic in C\R− , where R− := (−∞, 0]. The analyticity of z 7→ R(a−y1j , z − y2j ) follows by noticing that (y1j − a)2 + (y2j − z)2 ∈ C \ R− if ℜ(z) > a (respectively, if (1) ℜ(z) < −a), since y2j ∈ [−a, +a]. Since also z 7→ H0 (z) is analytic in ℜ(z) > 0, we j j conclude that the function z 7→ Φ(x (z), y ) is analytic in |ℜ(z)| > a. And the same arguments and conclusion apply also for z 7→ ∂Φ(xj (z), y j )/∂n(y j ). Finally, using standard results about analyticity of functions defined as integrals (e.g. [3, Corollary X.3.19]), we conclude that z 7→ S j φ(z), z 7→ Dj φ(z), and so also z 7→ ϕj (z), have analytic continuations from (a, +∞) (respectively, (−∞, −a)) to the complex halfplane ℜ(z) > a (respectively, ℜ(z) < −a). The behavior of these analytic continuations as |z| → +∞ depends on ℑ(z). Indeed, for m ∈ N := {0, 1, ...}, we have [1, Equation (9.2.30)] r
 2 i(z−mπ/2−π/4)  (1) (40) Hm (z) = e 1 + O |z|−1 as |z| → +∞, πz uniformly in Arg(z) for |Arg(z)| < ζ, for every ζ < π. Further, as a consequence of Lemma B.2 and Remark B.3, (41) (42) R(a − y1j , z − y2j ) = z − y2j + O(|z|−1 ) R(a − y1j , z − y2j ) = y2j − z + O(|z| −1 ) in ℜ(z) > a, and in ℜ(z) < −a, as |z| → +∞, uniformly in Arg(z) and y j , for y j ∈ Σa . Using (38) it follows from the above asymptotics that   ϕj (z) = O e−kℑ(z) |z|−1/2 , (43) in ℜ(z) > a, and   ϕj (z) = O ekℑ(z) |z|−1/2 , (44) in ℜ(z) < −a, as |z| → +∞, uniformly with respect to Arg(z), and we note that ℑ(R(a − y1j , z − y2j )) and ℑ(z) have the same sign if ℜ(z) > a (opposite  signs if ℜ(z) < −a). Thus j ϕ (z) is exponentially decreasing in the quadrants ℜ(z) > a and ℑ(z) > 0 and  ℜ(z) < −a and ℑ(z) < 0 . Our idea is to choose half-lines in these quadrants and consider the analytic continuations of the ϕj on these half-lines as new unknowns instead of the initial traces. In other words, choosing some θ ∈ (0, π/2), we introduce the complex path (Figure 3) parameterized by   −a + (s + a) eiθ if s < −a, s if − a ≤ s ≤ +a, (45) z = τθ (s) :=  a + (s − a) eiθ if s > a, and we define the complex-scaled traces by
(46) ϕjθ (s) := ϕj τθ (s) for s ∈ R and j ∈ J0, 3K. THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 13  Remark 3.1. Wehave chosen particular complex paths in the quadrants ℜ(z) > a and ℑ(z) > 0 and ℜ(z) < −a and ℑ(z) < 0 , given by (45), that move into the complex plane already from ±a, the corners of Σa . Note that it is possible, alternatively, to start to complexify at a positive distance from the corners, i.e. from ±b, for some b > a. It is also possible to choose a smoother complex change of variable as usually done in PML methods. ℑ(z) θ −a θ a ℜ(z) Fig. 3. The complex path s → τθ (s) We note that, by definition, ϕjθ |(−a,a) ∈ L2 (−a, a), since ϕjθ (t) = ϕj (t) = u(xj (t)) = g(xj (t)), for −a < t < a, and g ∈ H 1/2 (Σa ) ⊂ L2 (Σa ). Moreover, thanks to the analyticity of ϕj , it is clear that the restrictions of ϕjθ to (a, +∞) and (−∞, −a) are continuous. Further, it follows from (43-44) that   (47) ϕjθ (s) = O e−k|s| sin θ |s|−1/2 , as |s| → +∞. Thus whether or not ϕjθ ∈ L2 (R) depends on the behaviour of ϕjθ (t) as t → a+ and t → −a− . The following propositions bound ϕjθ (t) on |t| > a, in particular near ±a, and show that ϕjθ ∈ L2 (R) for 0 < θ < π/2. We relegate some of the technical details to Appendix C. Applying the bounds from Proposition C.1 to (the analytic continuation of) (38) we obtain the following proposition, on observing, from (37), that φ = A−1 g and that [23, Corollary 2.8] A−1 is bounded as an operator on L2 (Σa ). Proposition 3.2. For every θ ∈ (0, π/2) and j ∈ J0, 3K there exists a constant C > 0, that depends only on a, k, and θ, such that  C|z − a|−1/2 e−kℑ(z) kgkL2 (Σa ) , if ℜ(z) > a with |Arg(z − a)| ≤ θ, j |ϕ (z)| ≤ C|z + a|−1/2 ekℑ(z) kgkL2 (Σa ) , if ℜ(z) < −a with |Arg(−z − a)| ≤ θ. The bound in the above proposition implies that ϕjθ ∈ L1 (R), for 0 < θ < π/2, and is sufficient for our Cauchy’s integral formula arguments below in §3.2. But it is not quite strong enough to establish ϕjθ ∈ L2 (R), as discussed in the proof of Proposition C.4 in Appendix C. It follows from (the analytic continuation of) (38) that, in the notation of Proposition C.4, ϕjθ (s) = Dθj φ(s) − ikSθj φ(s), for |s| > a, where φ = A−1 g. Thus, and arguing as above Proposition 3.2, we deduce the following result from the above proposition and Proposition C.4. Proposition 3.3. For 0 < θ < π/2 and j ∈ J0, 3K, ϕjθ ∈ L2 (R). Further, for some constant C > 0 depending only on θ, a, and k, |ϕjθ (s)| ≤ C(|s| − a)−1/2 e−k sin(θ)(|s|−a) kgkL2 (Σa ) , and kϕjθ kL2 (R) ≤ CkgkL2 (Σa ) . |s| > a, 14 A.-S. BONNET-BEN DHIA ET AL. 3.2. The deformed half-plane representations. We have introduced in the previous section the complex-scaled traces ϕjθ and proved that they belong to L2 (R). The objective now is to derive a HSM formulation for these new unknowns. The first step is to establish new representation formulas for the solution u of (1)-(2) in the half-planes using these complex-scaled traces instead of the original traces ϕj . We recall that the solution u of (1-2) can be represented in each half-plane in terms of its traces as Z j h(xj1 − a, xj2 − y2j ) ϕj (y2j ) dy2j , xj ∈ Ωj , (48) u(x ) = R where the kernel h is defined by (16). Our objective is to derive a similar formula using the complex-scaled trace ϕjθ instead of the trace ϕj . This can be done by deforming the path of integration into the complex plane. This leads to the following crucial result. Theorem 3.4. Let u be the solution of (1)-(2) and ϕjθ be defined as in (46). For 0 < θ < π/2 we have Z h(xj1 − a, xj2 − τθ (s)) ϕjθ (s) τθ′ (s) ds, xj ∈ Ωjθ , j ∈ J0, 3K, (49) u(xj ) = R where Ωjθ := {xj = (xj1 , xj2 ) ∈ Ωj : xj1 − a > (|xj2 | − a) tan(θ)}. (50) Remark 3.5. Let us point out that the new representation formula (49) is valid only in a subdomain Ωjθ of the half-plane Ωj (Figure 4). The larger the angle θ, the faster the decay of the complex-scaled traces ϕjθ (Proposition 3.3) but the smaller the domain of validity Ωjθ of the new representation (49). xj2 xj2 Σj Σj xj xj xj1 xj1 2a Ωj θ Ωjθ Fig. 4. The domains of validity of the half-plane representations: left is without complex-scaling (Ωj ); right is with complex-scaling (Ωjθ ). Proof. To derive (49) from (48) it suffices to show (note the symmetry τθ (−s) = −τθ (s)) that, for all xj ∈ Ωjθ , Z +∞ Z +∞ h(xj1 − a, xj2 − y2j ) ϕj (y2j ) dy2j = h(xj1 − a, xj2 − τθ (s)) ϕjθ (s) τθ′ (s) ds a and Z +∞ a a h(xj1 − a, xj2 + y2j ) ϕj (−y2j ) dy2j = Z +∞ a h(xj1 − a, xj2 + τθ (s)) ϕjθ (−s) τθ′ (s) ds. THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 15 For 0 < δ < M and 0 < θ < π/2 we introduce the complex domains Tθ δ,M Tθ := {z = a + reiα : 0 < α < θ, r > 0} and := {z = a + reiα : 0 < α < θ, δ < r < M } (see Figure 5). First we show that, if xj = (xj1 , xj2 ) ∈ Ωj , then  z 7→ h(xj1 − a, xj2 ± z) is analytic in Tθ j j (51) x ∈ Ωθ ⇔ and continuous in Tθ . Indeed, for each xj , the function z 7→ h(xj1 − a, xj2 − z) has two branch points z± , the points where (xj1 − a)2 + (xj2 − z)2 vanishes, given by z± = xj2 ± i(xj1 − a). If xj = (xj1 , xj2 ) ∈ Ωj these branch points are outside Tθ if and only if xj1 − a > (xj2 − a) tan(θ); similarly, the branch points of the mapping z 7→ h(xj1 − a, xj2 + z) are outside Tθ if and only if xj1 − a > (−xj2 − a) tan(θ). Thus, and since also z 7→ ϕj (±z) is analytic in Tθδ,M and continuous in Tθδ,M , for 0 < δ < M , applying Cauchy’s integral theorem we have Z h(xj1 − a, x2 ∓ z) ϕj (±z) dz = 0, xj ∈ Ωjθ . ∂Tθδ,M To complete the proof, we have to show that lim δ→0 Z θ 0 h(xj1 − a, xj2 ∓ (a + δeiα ))ϕj (±(a + δeiα ))iδeiα dα = 0 and lim M →+∞ Z θ 0 h(xj1 − a, xj2 ∓ (a + M eiα ))ϕj (±(a + M eiα ))iM eiα dα = 0. These two limits are a consequence of Proposition 3.2, since the constraint xj ∈ Ωjθ ensures, by (51), that h(xj1 − a, xj2 ∓ z) is a continuous function of z in Tθ , and the
bound (110) and the asymptotics (41)-(42) imply that h(xj1 − a, xj2 ∓ z) = O |z|−1/2 as z → +∞ in Tθ , uniformly in Arg(z). 3.3. Derivation and analysis of the complex-scaled HSM method. We derive now an analogue of (21) for the complex-scaled traces ϕjθ . We know, thanks to Theorem 3.4, that Z 0 h(x01 − a, x02 − τθ (s)) ϕ0θ (s) τθ′ (s) ds, x0 ∈ Ω0θ . u(x ) = R As Ω0 ∩ Σ1 ⊂ Ω0θ , for all θ ∈ (0, π/2), this holds, in particular, when x0 = (x01 , x02 ) ∈ Ω0 ∩ Σ1 , i.e. for x01 = −x12 > a and x02 = x11 = a, so that (cf. (20)) Z 1 1 h(−x12 − a, a − τθ (s)) ϕ0θ (s) τθ′ (s) ds, x12 < −a. (52) ϕ (x2 ) = R Remember that ϕ1 (x12 ) is analytic in ℜ(x12 ) < −a and, by the definition (46), ϕ1θ (t) for t < −a is the value of the analytic continuation of ϕ1 at x12 = τθ (t). 16 A.-S. BONNET-BEN DHIA ET AL. ℑm z xj1 − a θ a xj2 a+M ℜe z −(xj1 − a) Fig. 5. The contour ∂Tθδ,M and the branch points z± = xj2 ± i(xj1 − a) of h(xj1 − a, xj2 − z). To obtain the new complex-scaled compatibility relation, the idea is to complexify x12 in (52), i.e. simply to substitute x12 = τθ (t). By the uniqueness of analytic continuation, this is valid provided the right-hand side is analytic as a function of x12 in a connected domain containing the half-line x12 < −a and the half-line x12 = τθ (t), t < −a. We check this in the next lemma. Lemma 3.6. Let ψ ∈ L2 (R). For 0 < θ < π/2 the function Z (53) z→ h(z − a, a − τθ (s))ψ(s)τθ′ (s)ds R is analytic in the domain Gθ := {z ∈ C : z 6= a, −π/2 + θ < Arg(z − a) < π/2}. Proof. We will prove that this function is analytic using standard results about analyticity of functions defined as integrals based on the dominated convergence theorem (e.g., [3, Corollary X.3.18]). By the definition (16), for each s ∈ R the kernel h(z − a, a − τθ (s)) is locally an analytic function of z wherever the quantity R(z − a, τθ (s) − a) = [(z − a)2 + (τθ (s) − a)2 ]1/2 does not vanish. Lemma B.5 shows that, for every θ0 ∈ (θ, π/2), (54) |R(z − a, τθ (s) − a)|2 ≥ cos(θ0 )|z − a|2 if −θ0 + θ ≤ Arg(z − a) ≤ θ0 . Consequently, R(z − a, τθ (s) − a) does not vanish in Gθ . Further, (54) implies that, for every bounded subdomain U of Gθ that is bounded away from a, there exists mU > 0 such that |R(z − a, τθ (s) − a)| ≥ mU , z ∈ U, s ∈ R. Moreover, for any such subdomain it follows from Lemma B.4 that, for some constant C > 0 independent of s ∈ R and z ∈ U , ℑ(R(z − a, τθ (s) − a)) ≥ |s| sin(θ) − C. From the above bounds, and the bound (110), it follows that |h(z − a, a − τθ (s))| ≤ C ′ e−k sin(θ)|s| , THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 17 for some constant C ′ > 0 independent of s ∈ R and z ∈ U . Thus, for every z ∈ U and s ∈ R, the integrand in (53) has modulus ≤ H(s), where H ∈ L1 (R) is defined by H(s) := C ′ e−k sin(θ)|s| |ψ(s)|, s ∈ R. This domination property, and the analyticity of z 7→ h(z−a, a−τθ (s)) in U ⊂ Gθ , imply (e.g., [3, Corollary X.3.18]) that the function (53) is analytic in U , for every U , and so analytic in Gθ . Noting that, with Gθ as defined in the above lemma, −x12 ∈ Gθ for x12 < −a and −τθ (t) ∈ Gθ for t < −a, we see that we have justified the analytic continuation of (52) from x12 < −a to the path x12 = τθ (t), t < −a. Thus we obtain finally the new complex-scaled compatibility relation Z ϕ1θ (t) = h(−τθ (t) − a, a − τθ (s)) ϕ0θ (s) τθ′ (s) ds, t < −a. R By applying similar reasoning, and noting that τθ (−t) = −τθ (t), we get 8 equations linking the four complex-scaled traces (cf. (21)), namely (55) ∀j ∈ J0, 3K, ϕjθ (t) = S Dθ ϕj−1 (t), θ ϕjθ (t) = Dθ S ϕj+1 (t), θ t < −a, t > a, 3 4 0 where we have set ϕ−1 θ := ϕθ and ϕθ := ϕθ . In this system the operator S is defined as in (22), and, where h is as given in (16), Dθ is defined, for 0 < θ < π/2, by Z h(τθ (t) − a, a − τθ (s)) ψ(s) τθ′ (s) ds, t > a, ψ ∈ L2 (R), (56) Dθ ψ(t) := R and, similarly to (29), by Dθ ψ(t) := 0, t ≤ a, ψ ∈ L2 (R). The system (55) has to be completed with the Dirichlet boundary condition (57) ϕjθ (t) = g|Σja (t), −a < t < a, j ∈ J0, 3K. As in Section 2, these equations can be formulated as a single operator equation. Introducing Φθ := {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ } ∈ (L2 (R))4 , and recalling the definition (32) of Φg , the systems of equations (55) and (57) can be rewritten as (58) Find Φθ ∈ (L2 (R))4 such that Φθ − Φg ∈ (L20 (R))4 and (I − Dθ )(Φθ − Φg ) = Dθ Φg , where Dθ is obtained by replacing D by Dθ in (34), i.e.   0 Dθ S 0 S Dθ  S Dθ 0 Dθ S 0  . (59) Dθ :=   0 S Dθ 0 Dθ S  Dθ S 0 S Dθ 0 Our first main result is that, as proved for D in the dissipative case (Theorem 2.2), the operator I − Dθ , as an operator on (L20 (R))4 , is Fredholm of index zero, indeed (importantly for numerical analysis of Galerkin methods) is a compact perturbation of a coercive operator. 18 A.-S. BONNET-BEN DHIA ET AL. Theorem 3.7. For 0 < θ < π/2: (i) Dθ is a continuous operator on (L2 (R))4 , and Dθ ((L2 (R))4 ) ⊂ (L20 (R))4 ; √ (ii) as an operator on (L20 (R))4 , Dθ is the sum of an operator of norm ≤ 1/ 2 and a compact operator. To show well-posedness of (58), it remains to prove a uniqueness result which is the subject of section 5 (Theorem 5.1). At the end of this section we will combine the above theorem with Theorem 5.1 to write down a result expressing this well-posedness and the equivalence of (58) with the original scattering problem (35)-(2). The proof of Theorem 3.7, which we defer to Appendix A, mirrors the proof of Theorem 2.2, once we establish properties of the operator Dθ to mirror those proved for D in Proposition 2.1. Establishing these properties of Dθ , in Propositions 3.8, 3.9, and 3.10, is the focus of most of the rest of this section: these propositions give the properties of Dθ when it acts on functions whose support is, respectively, in (a, +∞), (−∞, −a), and the whole of R. (This splitting is necessary because of the piecewise definition of the complex-scaling function τθ .) Let us point out a useful fact (see the proof of the following proposition): when Dθ acts on functions whose support is in (a, +∞), it is equal to the operator D defined in (23) for the dissipative case with wavenumber keiθ . Proposition 3.8. Suppose that θ ∈ (0, π/2). For all ψ ∈ L2 (a, +∞) we have (60) Dθ ψ(t) = i(t − a)keiθ 2 Z +∞ a (1) H1 (keiθ R(t − a, s − a)) ψ(s) ds, R(t − a, s − a) t > a, with R defined in (17). As a consequence, as an operator on L2 (a, +∞), Dθ is the √ sum of an operator of norm ≤ 1/ 2 and a compact operator. Proof. Using the definition (56) of Dθ and the expression (16) for the kernel h, we easily see (60). This implies that Dθ , when it acts on functions whose support is in (a, +∞), is exactly the operator D defined in (23) for the dissipative case if we set the wavenumber in the dissipative case to be keiθ . The result is therefore a direct consequence of item (ii) of Proposition 2.1. Proposition 3.9. Dθ is compact as an operator from L2 (−∞, −a) to L2 (a, +∞), for θ ∈ (0, π/2). Proof. Using the definition (56) of Dθ and the expression (16) for the kernel h, we can show easily that, for all ψ ∈ L2 (−∞, −a), we have ik(t − a)e2iθ Dθ ψ(t) = 2 Z −a −∞ (1) H1 (kRθ (t, s)) ψ(s) ds, Rθ (t, s) t > a, with Rθ (t, s) := R(eiθ (t − a), 2a − (s + a)eiθ ) and R defined in (17). We prove the compactness of Dθ , as we prove the compactness of D in (iii) of Proposition 2.1, by showing that Dθ is a Hilbert-Schmidt operator from L2 (−∞, −a) to L2 (a, +∞). By a simple change of variable t 7→ t − a and s 7→ −(s + a), it suffices to show that Z +∞ Z +∞ |K(t, s)|2 dt ds < +∞, (61) 0 0 where (1) K(t, s) := t eθ (t, s)) H1 (k R eθ (t, s) R and eθ (t, s) := R(eiθ t, 2a + eiθ s). R THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 19 Where R+ := [0, +∞), K is continuous on R+ × R+ , since ℜ(R̃θ ) > 0 on R+ × R+ . We want to use now the bound on the Hankel function given in (110) which implies that, for some c > 0, (1) (62) H1 (z) ≤ c|z|−3/2 e−ℑ(z) , z ℜ(z) > 0, |z| ≥ 1. eθ can be rewritten as R eθ (t, s) = R̂(ẑ, z), where ẑ := 4aseiθ +4a2 , z := eiθ (t2 +s2 )1/2 , R and R̂ is defined in (98). Thus we can use Lemma B.4 and deduce that, for some C > 0 and all s, t ∈ R+ with s2 + t2 sufficiently large, it holds that and eθ (t, s) − eiθ (t2 + s2 )1/2 | ≤ C |R 1 + |s| + t2 )1/2 (s2 eθ (t, s)) ≥ sin(θ)(t2 + s2 )1/2 − C ℑ(R 1 + |s| . + t2 )1/2 (s2 eθ (t, s)| ≥ 1 (t2 + s2 )1/2 and Thus, for some C ′ > 0 and all s2 + t2 large enough, |R 2 2 2 1/2 ′ eθ (t, s)) ≥ sin(θ)(t + s ) − C , so that, by (62), ℑ(R ! √ 2 2 p e−k sin(θ) t +s |K(t, s)| = O as t2 + s2 → +∞, 2 2 1/4 (t + s ) uniformly in t and s. Thus (61) is clear. Proposition 3.10. For 0 < θ < π/2, Dθ is a continuous operator from L2 (R) to L (a, +∞). 2 Proof. From Propositions 3.8 and 3.9, it suffices to show that Dθ is a continuous operator from L2 (−a, a) to L2 (a, +∞). But this is immediate from the continuity of Dθj from L2 (Σa ) to L2 (a, +∞), established in Proposition C.4. We finish this section with the promised statement of well-posedness of (58), and of its equivalence with the original scattering problem (35)-(2). Theorem 3.11. For every θ ∈ (0, π/2), the operator I − Dθ is invertible on (L20 (R))4 . Thus, for every g ∈ L2 (Σa ), (58) has exactly one solution Φθ ∈ (L2 (R))4 such that Φθ − Φg ∈ (L20 (R))4 . Moreover, for some constant c > 0 depending on θ, (63) kΦθ k(L2 (R))4 ≤ ckΦg k(L2 (R))4 = ckgkL2 (Σa ) , for every g ∈ L2 (Σa ). Further, if g ∈ H 1/2 (Σa ) and Φθ = {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ } ∈ (L2 (R))4 is the solution of (58), then, for j ∈ J0, 3K: (i) ϕjθ satisfies (46), i.e. ϕjθ is the analytic continuation to the path τθ of the restriction to Σj of the solution u of (35)-(2); (ii) the solution u of (35)-(2) is given in terms of ϕjθ in Ωjθ by (49); (iii) for some constant C > 0 that depends only on a, k, and θ, ϕjθ (s) satisfies the bounds of Proposition 3.3, for |s| > a. Proof. Theorem 3.7 implies that, as an operator on (L20 (R))4 , I − Dθ is Fredholm of index zero, and Theorem 5.2 implies that it is injective, so that I − Dθ is invertible with a bounded inverse. This implies, since Dθ is a bounded operator from (L2 (R))4 20 A.-S. BONNET-BEN DHIA ET AL. to (L20 (R))4 by Theorem 3.7(i), that (58) has exactly one solution, and this solution satisfies the bound (63). In the case that g ∈ H 1/2 (Σa ), that (i) holds follows from the derivation of (58) from (35)-(2) in Section 3.3, and since (58) has only one solution; that (ii) holds follows from (i) and Theorem 3.4; that (iii) holds follows from (i) and Proposition 3.3. 4. Reconstruction of the solution and far-field formula. Suppose that we have computed the solution Φθ to (58). Then the solution u of the problem (35-2) can be recovered a posteriori through the representation formulas (49). More precisely, as we have observed in Theorem 3.11(ii), it can be reconstructed in the union for j ∈ J0, 3K of the domains Ωjθ defined by (50). Let us point out that, if θ < π/4, the union of the Ωjθ covers the whole domain Ω, so that the whole solution u can be reconstructed a posteriori. It is well known (e.g., [21, Lemma 2.5]) that the solution of (35-2) satisfies
eikr F (b x) + O(r−1 ) , as r → +∞, r1/2 b := x/r, where F ∈ C ∞ (S 1 ), with S 1 the unit circle, is the far-field uniformly in x pattern. By analogy with classical boundary integral methods, one can wonder if this far-field pattern can also be recovered from properties of the ϕjθ . A partial answer will be given in this section, by deriving far-field formulas in the four directions orthogonal to the edges of the square Ωa . The proof of this result requires first that we establish some properties of the solution u that can be deduced from the representation formulas (49). The far-field behaviour that we will establish, indeed all of the results of this section, will be ingredients in the proof of uniqueness for problem (58) that will be the focus of the next section. From (49), let us consider the representation formula for any ψ ∈ L2 (R) Z 0 h(x01 − a, x02 − τθ (s)) ψ(s) τθ′ (s) ds, x0 ∈ Ω0θ , (65) Uθ (ψ)(x ) := (64) u(x) = R and the associated integral operator defined by e θ ψ(t) := Uθ (ψ)(t, a), D (66) t > a. Let us note that in Lemma 3.6 we have shown that, for any data ψ ∈ L2 (R), the e θ ψ(t) can be continued analytically from (a, +∞) into the domain of function t 7→ D the complex plane {z ∈ C : z 6= a, −π/2 + θ < Arg(z − a) < π/2} just replacing t e θ ψ(z), in particular near a and for by z in (66). The following proposition bounds D large values of z. Lemma 4.1. There exists a constant C > 0, depending only on θ, a and k, such that, for all ψ ∈ L2 (R) and z ∈ C with ℜ(z) > a and Arg(z − a) ∈ [0, θ], e θ ψ(z)| ≤ C(|z − a|−1/2 + 1) exp(−kℑ(z)) kψkL2 (R) . |D Proof. Throughout this proof C will denote any positive constant, depending only on a, k, and θ, not necessarily the same at each occurrence. Using the expressions (65-66), the definition (16) of the kernel h, and the definition e θ ψ(z)| ≤ I(z − a), where of the complex-scaling function (45), we see that |D k|w| I(w) := 2 Z (1) R H1 (k R(w, τθ (s) − a)) |ψ(s)| ds, R(w, τθ (s) − a) ℜ(w) > 0, Arg(w) ∈ [0, θ], 21 THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD with R defined in (17). The bound on the Hankel function given in (110) implies that   H1 (z) ≤ C |z|−2 + |z|−3/2 e−ℑz , z (1) ℜ(z) > 0. Further, Lemma B.5 (applied with θ0 = θ) gives, for Arg(w) ∈ [0, θ], that |R(w, τθ (s) − a)|2 ≥ cos(θ)(|w|2 + |τθ (s) − a|2 ) and ℑR(w, τθ (s) − a) ≥ cos(θ − Arg(w))ℑ(w) − C. Letting t := |w| and γ := Arg(w), we deduce from the above bounds that  Z  t t k cos(θ−γ)ℑ(w) + |ψ(s)| ds. |I(w)| e ≤C 3/4 t2 + |τθ (s) − a|2 R (t2 + |τθ (s) − a|2 ) Applying the Cauchy-Schwarz inequality, and noticing that |τθ (s) − a|2 = (s − a)2 , for s ≥ −a, while |τθ (s) − a|2 = (s + a)2 + 4a2 (1 − cos(θ)) − 4as cos(θ) ≥ (s + a)2 , for s < −a, yields |I(w)| e k cos(θ−γ)ℑ(w) ≤C Z +∞ 0 t2 ds + (t2 + s2 )2 Z +∞ 0 t2 ds (t2 + s2 )3/2 1/2 kψkL2 (R) . We see, by substituting s = tp, that the first and second integrals on the right hand side of this last inequality are ≤ Ct−1 and ≤ C, respectively. Thus we have shown that e θ ψ(z)| ≤ C(|z − a|−1/2 + 1) exp(−k cos(θ − Arg(z − a))ℑ(z − a)) kψkL2 (R) , |D for all ψ ∈ L2 (R) and z ∈ C with ℜ(z) > a and Arg(z − a) ∈ [0, θ]. Now, defining e θ ψ(z), F(z) := (1 + (z − a)−1/2 )−1 exp(−ikz) D this last bound implies that, for ℜ(z) > a with 0 ≤ Arg(z − a) ≤ θ, F is analytic and |F(z)| ≤ CkψkL2 (R) B(z), where B(z) := exp(kℑ(z)(1 − cos(θ − Arg(z − a)))). Now B(z) = 1 when Arg(z − a) = 0 or θ, and |B(z)| ≤ exp(k|z|), for all z with ℜ(z) > a and 0 ≤ Arg(z − a) ≤ θ. Thus, by a standard Phrágmen-Lindelöf principle (e.g., [27, Chapter VI, Cor. 4.2]), |F(z)| ≤ CkψkL2 (R) , for all z with ℜ(z) > a and e θ ψ(z)| follows. 0 ≤ Arg(z − a) ≤ θ, and the required bound on |D e θ ψ(t), for t > a, Lemma 4.1 implies that φ ∈ L1 (a, b), for every Defining φ(t) := D 2 b > a, if ψ ∈ L (R). In the uniqueness proof in the next section we will need also the following stronger result. e θ ψ(t), for t > a, then φ ∈ L2 (a, b), for Lemma 4.2. If ψ ∈ L2 (R) and φ(t) := D every b > a. Proof. Arguing as in the proof of Lemma 4.1, we see that ! Z Z (t − a) |ψ(s)| (t − a) |ψ(s)| |φ(t)| ≤ C ds + ds , 2 2 3/4 R (t − a) + |τθ (s) − a| R ((t − a)2 + |τθ (s) − a|2 ) for t > a. Further, arguing as at the end of the proof of Lemma 4.1, using that |τθ (s) − a|2 = (s − a)2 , for s ≥ −a, while |τθ (s) − a|2 ≥ (s + a)2 , for s < −a, we 22 A.-S. BONNET-BEN DHIA ET AL. see that the second integral in the above sum is bounded on (a, +∞), and so is in L2 (a, b), for every b > a, while the first integral is Z +∞ Z +∞ t−a t−a |ψ(s)| ds + |ψ(−s)| ds. ≤ 2 + (s − a)2 2 + (s − a)2 (t − a) (t − a) a −a The right hand side of this last inequality is in L2 (a, +∞) by Remark C.3. Let us remark that by definition (56) we have, for any ψ ∈ L2 (R), (67) Dθ ψ(t) = D̃θ ψ(τθ (t)) = Uθ (ψ)(τθ (t), a), t > a. We deduce thus from Lemma 4.1 the following result. Corollary 4.3. There exists a constant C > 0, depending only on θ, a and k, such that, for all ψ ∈ L2 (R), |Dθ ψ(t)| ≤ C[(t − a)−1/2 + 1] e−k sin(θ)t kψkL2 (R) , t > a. Now we are able to prove the main result of this section, which provides far field formulas in the direction orthogonal to the edges of the square Ωa . Proposition 4.4. Let ψ ∈ L2 (R) be such that (t2 + 1)ψ(t) ∈ L1 (R). Then the function Uθ (ψ) defined by (65) has the following behaviour at infinity:

eikx1 1 + O x−1 , Uθ (ψ)(x1 , x2 ) = C∞ √ 1 x1 where as x1 → +∞, for each x2 ∈ [−a, a] fixed, r Z k 1−i ψ(s)τθ′ (s) ds. π 2 R Proof. Throughout this proof, C will denote a positive constant, depending only on a, k, and θ, not necessarily the same at each occurrence. For any fixed x2 ∈ [−a, a], let φx2 (t) := Uθ (ψ)(t + a, x2 ), for t > 0. By the definition (16) of h, we have Z (1) ikt H1 (kR(t, x2 − τθ (s))) φx2 (t) = ψ(s) τθ′ (s) ds. 2 R R(t, x2 − τθ (s)) C∞ := In order to use (40) that says that (68)
 1 + i eiz  (1) H1 (z) = − √ √ 1 + O |z|−1 π z as |z| → +∞, uniformly in Arg(z) for |Arg(z)| < ζ, for every ζ < π, we write r Z k 1 − i eikt √ (69) φx2 (t) − ψ(s) τθ′ (s) ds = φ1x2 (t) + φ2x2 (t) π 2 t R where φ1x2 (t) := φ2x2 (t) := r  Z  k 1−i eikR(t,x2 −τθ (s)) eikt t − ψ(s) τθ′ (s) ds, 3/2 3/2 π 2 [R(t, x − τ (s))] t 2 θ R Z  (1) H1 (kR(t, x2 − τθ (s))) ikt 2 R R(t, x2 − τθ (s))  eikR(t,x2 −τθ (s)) 1+i ψ(s) τθ′ (s) ds. + √ kπ [R(t, x2 − τθ (s))]3/2 THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 23 To estimate these quantities we note first that, if t > 0 and z ∈ Zθ := {z = reiα : 0 ≤ α ≤ θ, r ≥ 0} (or −z ∈ Zθ ), then 0 ≤ Arg(R(t, z)) ≤ π/2, | cos(Arg(z))| ≥ cos(θ), and, by Lemma B.1, (70) |R(t, z)| ≥ cos(θ) t. Since, from the definition (45), x2 − τθ (s) or τθ (s) − x2 is in Zθ for all x2 ∈ [−a, a] and all s ∈ R, these observations hold in particular if z = x2 − τθ (s), so that (68) applies and implies that Z C 2 (71) |φx2 (t)| ≤ 3/2 |ψ(s)| ds, t R for all sufficiently large t > 0. To estimate φ1x2 (t) we observe that, for t > 0 and z ∈ Zθ ,  ikR(t,w)  ∂ eikt e eikR(t,z) ≤ sup − |z| 3/2 3/2 3/2 [R(t, z)] t w∈U (z) ∂w [R(t, w)] where U (z) := {w ∈ Zθ : |w| ≤ |z|} and where we have used that R(t, 0) = t. Using again (70), this yields that |z|2 eikt eikR(t,z) ≤ C − , [R(t, z)]3/2 t3/2 t5/2 z ∈ Zθ , t > 0. Since |x2 − τθ (s)| ≤ C(1 + |s|), it follows that Z C (72) |φ1x2 (t)| ≤ 3/2 (1 + |s|)2 |ψ(s)| ds, t R t > 0. Combining (69), (71) and (72), we get that φx2 (t) − r k 1 − i eikt √ π 2 t Z ψ(s) τθ′ (s) ds R ≤ C t3/2 Z (1 + s2 )|ψ(s)| ds, R for all sufficiently large t > 0, which ends the proof. This proposition proves in particular that (73) F (cos(jπ/2), sin(jπ/2)) = r k 1−i π 2 Z R ϕjθ (s)τθ′ (s) ds, j ∈ J0, 3K, where the far-field F is defined by (64) and ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ are the complex-scaled traces of u, which are exponentially decaying at infinity thanks to Proposition 3.3, or thanks to (55) and Corollary 4.3. 5. Uniqueness. In this section we will prove uniqueness of solution for the complex-scaled HSM method (58). This result, important in its own right, is also key to the proof of uniqueness for the complex-scaled HSM method for more general configurations; see Proposition 6.1 below. Our proof depends on the following uniqueness result for the standard HSM (27) that we prove, using completely different arguments, in [9]. 24 A.-S. BONNET-BEN DHIA ET AL. Theorem 5.1. If k > 0 and {ϕ0 , ϕ1 , ϕ2 , ϕ3 } ∈ (L2loc (R))4 satisfies (27) with g = 0, and satisfies the radiation condition that
 j ik|t| −1/2 1 + O(|t|−1 )
, as t → +∞, c+ e |t| j (74) ϕ (t) = cj− eik|t| |t|−1/2 1 + O(|t|−1 ) , as t → −∞, for some constants cj± ∈ C and every j ∈ J0, 3K, then ϕj = 0, for j ∈ J0, 3K. To see how this result is relevant to uniqueness for the complex-scaled HSM method (58), recall that to derive (55) and (57) we started from (27), satisfied for real k by ϕj , the trace on Σj of the solution u of (35)-(2), for j ∈ J0, 3K. We showed that ϕjθ , defined by (46) as the analytic continuation of ϕj from the real line to the path τθ of Figure 3, satisfies (55) and (57), equivalently (58). A key component in this argument was to deform paths of integration from the real line to the path τθ (Theorem 3.4). In the following uniqueness proof we reverse this derivation. We show that if the ϕjθ satisfy (55) and (57), with g = 0, then, in the sense that (46) holds, they are the analytic continuations onto the path τθ of functions ϕj that satisfy the system (27) with g = 0. (A key component in this argument is a deformation of paths of integration from τθ back to the real line, justified as in the proof of Theorem 3.4.) Moreover, by an application of Proposition 4.4 (justified by Corollary 4.3), the functions ϕj satisfy the radiation conditions (74), so that ϕj = 0 by Theorem 5.1. Thus ϕjθ , which is the analytic continuation of ϕj , is also zero. Theorem 5.2. Suppose that {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ } ∈ (L2 (R))4 is a solution of (55) such that (75) ϕθj (t) = 0, −a < t < a, j ∈ J0, 3K. Then ϕθj = 0 for j ∈ J0, 3K. Proof. Suppose that {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ } ∈ (L2 (R))4 is a solution of (55) that satisfies (75), and define the functions ϕ0 , ϕ1 , ϕ2 , ϕ3 by (76) e θ ϕj−1 (t), ϕj (t) := S D θ j ϕ (t) := 0, e θ S ϕj+1 (t), ϕj (t) := D θ t < −a, −a ≤ t ≤ a, t > a, e θ is defined by (66). As discussed below (66), since the ϕj ’s for j ∈ J0, 3K, where D θ 2 are in L (R), it follows from this definition, the definition (22) of S, and Lemma 3.6, that the ϕj ’s have analytic continuations from (a, +∞) (respectively (−∞, −a)) to the part of the half-plane ℜ(z) > a with −π/2 + θ < Arg(z − a) < π/2 (respectively ℜ(z) < −a with −π/2 + θ < Arg(−z − a) < π/2). In particular, by (67), and recalling (22) and that τθ (−t) = τθ (t) for t < −a, ϕj (τθ (t)) = S Dθ ϕj−1 (t), θ j ϕ (τθ (t)) = 0, j ϕ (τθ (t)) = Dθ S ϕj+1 (t), θ t < −a, −a ≤ t ≤ a, t > a, for j ∈ J0, 3K. Comparing this equation with (55) we see that
(77) ϕjθ (s) = ϕj τθ (s) for s ∈ R and j ∈ J0, 3K. THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 25 Thus, for s > a and for s < −a, the ϕjθ ’s are the analytic continuations of the ϕj ’s to the complex path parametrized by τθ . Therefore, to complete the proof of the theorem it is enough to show, for j ∈ J0, 3K, that ϕj (t) = 0 for t ∈ R with |t| > a, for this will imply, by the uniqueness of analytic continuation, that ϕj (τθ (s)) = 0 for s > a and s < −a, which will imply, using (77), that each ϕjθ = 0. To establish, for j ∈ J0, 3K, that ϕj (t) = 0 for t > a and t < −a, we show, via an application of Cauchy’s integral theorem as in the proof of Theorem 3.4, that uses the analyticity of the ϕj ’s noted above, Lemma 4.1, and (77), that (76) implies that the ϕj satisfy (27) with g = 0, i.e. ϕj (t) = S D ϕj−1 (t), j ϕ (t) = 0, j t < −a, −a < t < a, ϕ (t) = D S ϕj+1 (t), t > a, for j ∈ J0, 3K, where D is defined in (25), or equivalently by (56) with θ = 0. (The key step in the argument is to show, as we have done in the proof of Theorem 3.4, that Z +∞ Z +∞ h(t − a, a ∓ s)ϕ(s) ds h(t − a, a ∓ τθ (s))ϕ(s)τθ′ (s) ds = a a if, for every 0 < δ < M , ϕ is analytic in the domain Tθδ,M introduced in the proof of Theorem 3.4, and continuous in its closure, and if the behaviour of ϕ(z) as z → 0 and as z → +∞ is suitably constrained, satisfying the bounds of Lemma 4.1.) We note also that it follows from (76) and Lemma 4.2 that ϕj ∈ L2loc (R), for j ∈ J0, 3K. Moreover, the ϕjθ are in L2 (R) and it follows, from (55) and Corollary 4.3, that each ϕjθ decays exponentially at infinity, so that (t2 + 1)ϕjθ (t) ∈ L1 (R), j ∈ J0, 3K. Thus we can apply Proposition 4.4 to (76), noting the second equality in (67), to see that the radiation conditions (74) hold. But this is enough to conclude that ϕj = 0, for j ∈ J0, 3K, by Theorem 5.1. 6. The complex-scaled HSM method for the general case. Let us now explain how to extend the complex-scaled HSM method of section 3 to solve the general problem presented in the introduction. More precisely, for a real wavenumber k > 0 and for a function ρ and a subdomain Ω of R2 satisfying the hypotheses described in section 1, the objective is to derive a complex-scaled HSM formulation 1 to compute the solution u ∈ Hloc (Ω) of (1) and (2). For the sake of simplicity, we will restrict attention to the case Ω = R2 , but adding bounded obstacles contained in Ωa is completely straightforward (see Section 7.3). As in section 1, the source term f is in L2 (Ω), with compact support that is a subset of Ωa . The idea is to introduce, in addition to the lines Σj , j ∈ J0, 3K, a square Ωb := (−b, b)2 for some b > a. (As in section 2 we set Σb := ∂Ωb and denote the sides of Σb by Σjb , j ∈ J0, 3K; see Figure 6.) We will show how to derive a formulation of problem (1,2) whose unknowns are the complex-scaled traces ϕjθ , j ∈ J0, 3K, associated to the infinite lines Σj , and the restriction ub := u|Ωb of the solution u to the square Ωb . To do that, we need to make the following assumption on the parameter θ: (78) θ< π . 4 Let us derive the equations linking the ϕjθ , j ∈ J0, 3K, and ub . On the one hand, the ϕjθ still satisfy the system of compatibility relations (55). But, instead of (57), we 26 A.-S. BONNET-BEN DHIA ET AL. Ω1 Σ0 Σ1b Σ1 2 Ω2 Σb Ωb Σ0b Ω0 Σ0b Ω0θ Ωb Σ3 Σ3b Σ2 θ Ω3 Fig. 6. The notations for the general case. have to impose equality between ϕjθ and ub on Σja : ϕjθ (t) = ub |Σja (t), −a < t < a, j ∈ J0, 3K. On the other hand, we can derive a variational formulation for ub in Ωb . Since −∆ub − k 2 ρub = f in Ωb and f is supported in Ωa , the following Green’s identity holds for all vb ∈ H 1 (Ωb ), where n is the normal unit vector pointing out of Ωb : Z Z Z
∂ub f vb . vb = ∇ub · ∇vb − k 2 ρub vb − (79) Σb ∂n Ωa Ωb The last idea is to replace in the previous identity the normal derivative of ub on the jth side of the square by an integral representation as a function of ϕjθ . Indeed, we must have, for j ∈ J0, 3K, ∂ub ∂U j − ikub = − ikU j on Σjb , ∂n ∂n where U j denotes the restriction of the solution u to the half-plane Ωj . (Our choice of Robin traces instead of normal derivatives is so that later we have uniqueness for the boundary value problem (85) for all k > 0.) We have proved in Theorem 3.4 that U j (xj ) has an integral representation in terms of ϕjθ , as soon as xj belongs to the domain Ωjθ defined by (50). Notice that under condition (78) one has Σjb ⊂ Ωjθ (see Figure 6). Consequently, one can use the formula (49) to rewrite the above Robin compatibility condition on Σjb . Precisely, we have   ∂ub − ikub (t) = Σjb Z ∂n (∂1 h(b − a, t − τθ (s)) − ikh(b − a, t − τθ (s))) ϕjθ (s)τθ′ (s) ds, −b < t < b, R where ∂1 h denotes the derivative of h with respect to its first variable. This leads us to define the following Dirichlet-to-Robin operator Λθ . For ψ ∈ L2 (R) and 0 < θ < π/4, define Λθ ψ ∈ L2 (−b, b) by Z λ(b − a, t − τθ (s))ψ(s)τθ′ (s) ds, −b < t < b, (80) Λθ ψ(t) := R THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 27 where we have set λ(x1 , z) := ∂1 h(x1 , z) − ikh(x1 , z), x1 > 0, z ∈ C, which one can easily check, using [30, (10.6.2)] and the definition (16) of h, takes the explicit form   ik kx21 (1) (1) (81) λ(x1 , z) = [1 − ikx1 ] H1 (kR) − H (kR) , R = [x21 + z 2 ]1/2 . 2R R 2 With this notation, the previous equations linking ub and the ϕjθ can be written as   ∂ub − ikub (t) = Λθ ϕjθ (t), −b < t < b, j ∈ J0, 3K. (82) ∂n Σjb Our complete formulation reads as follows: Find {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ } ∈ (L2 (R))4 and ub ∈ H 1 (Ωb ) such that ϕjθ (t) = S Dθ ϕj−1 (t) for t < −a, θ ϕjθ (t) = ub |Σja (t) for − a < t < a, ϕjθ (t) (83) and such Z = Dθ S ϕj+1 (t) for θ 1 that, ∀vb ∈ H (Ωb ), 2 t > a.
∇ub · ∇vb − k ρub vb − ik Ωb 3 Z X − j=0 b −b j ∈ J0, 3K, Λθ ϕjθ (t)vb |Σj (t)dt b 3 Z X j=0 = Z Σjb ub v b f vb , Ωa where S, Dθ , and Λθ are defined by (22), (56), and (80). e θ the following elements of (L2 (R))4 : Let us denote by Φθ , Φ(ub ), and Φ Φθ := {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ }, Φ(ub ) := {ub |Σ0a , ub |Σ1a , ub |Σ2a , ub |Σ3a }, e θ = {ϕ Φ e0θ , ϕ e1θ , ϕ e2θ , ϕ e3θ } := Φθ − Φ(ub ). e θ ∈ (L2 (R))4 . Note that, using the second equation of (83) and recalling (31), we have Φ 0 With these notations the first block of equations in (83) can be rewritten (cf. (58)) e θ = Dθ Φ(ub ), where Dθ is defined by (59), so that problem (83) can be as (I − Dθ )Φ rewritten as e θ ∈ (L2 (R))4 and ub ∈ H 1 (Ωb ) such that Find Φ 0 e ∈ (L2 (R))4 and ∀vb ∈ H 1 (Ωb ), ∀Ψ 0 e (L2 (R))4 e θ , Ψ) e (L2 (R))4 − (Dθ Φ(ub ), Ψ) ((I − Dθ )Φ 0 0 Z Z 3 X
ub v b ∇ub · ∇vb − k 2 ρub vb − ik + (84) Ωb − 3 Z X j=0 b −b j=0 Σjb Z   Λθ ϕ ejθ + ub |Σja (t) vb |Σj (t) dt = b Let us first prove a uniqueness result for this problem: f vb . Ωa 28 A.-S. BONNET-BEN DHIA ET AL. Proposition 6.1. If f = 0 then the only solution of problem (84) is the trivial e θ = 0. solution ub = 0, Φ e θ ∈ (L2 (R))4 and ub ∈ H 1 (Ωb ) are such that (84) holds Proof. Suppose that Φ 0 e θ = Dθ Φ(ub ), and the second of equations (83) holds with with f = 0. Then (I − Dθ )Φ f = 0 and ϕjθ := ϕ ejθ + ub |Σja , j ∈ J0, 3K, so that −∆ub − k 2 ρub = 0 in Ωb and (82) holds for j ∈ J0, 3K. 1 Let us denote by u∞ ∈ Hloc (R2 \Ωa ) the unique solution of (35)-(2) with g = 1/2 ub |Σa ∈ H (Σa ). Then, as we have shown in section 3.3, the vector of complexscaled traces of u∞ satisfies (58) with Φg = Φ(ub ), so that, by Theorem 5.2, it coincides with the vector Φθ := {ϕ0θ , ϕ1θ , ϕ2θ , ϕ3θ }. Thus, applying Theorem 3.4 (as we did above to derive (82) from (1)-(2)) we see that   ∂u∞ j − iku∞ , j ∈ J0, 3K. Λ θ ϕθ = ∂n Σjb Consequently, v := ub − u∞ belongs to H 1 (Ωb \Ωa ) and satisfies ∆v + k 2 v = 0 in Ωb \Ωa , v=0 on Σa , ∂v − ikv = 0 on Σb . ∂n But, for every k > 0, this homogeneous problem has no solutionR except v = 0. (To see this apply Green’s identity (cf. (79)) in Ωb \Ωa to deduce that Σb |v|2 = 0, so that v = ∂v/∂n = 0 on Σb , and apply Holmgren’s uniqueness theorem; [21, p. 104].) Thus ub = u∞ in Ωb \Ωa so that the function ( ub in Ωb , w := 2 u∞ in R \Ωa , (85) is well-defined, and is a solution of the homogeneous Helmholtz equation in R2 which satisfies the radiation condition (2). As a consequence (e.g., [25, Theorem 8.7]), w = 0 in R2 , so that ub = 0 in Ωb and u∞ = 0. Further, each ϕjθ is zero, since it is a complex trace of u∞ . This completes the proof. The well-posedness of problem (84) follows from classical arguments combined with the results of the previous sections and the following lemma: Lemma 6.2. The operator Λθ , defined by (80), is a compact operator from L2 (R) to L2 (−b, b). Proof. To establish this lemma, we will prove that the kernel of Λθ is HilbertSchmidt, i.e. that (86) (t, s) 7→ λ(b − a, t − τθ (s)) ∈ L2 ((−b, b) × R), by using similar arguments as in the proof of Proposition 3.9. On the one hand, since b > a and θ < π/4, the function R which appears in expression (81) never vanishes for −b ≤ t ≤ b and s ∈ R. As a consequence, the kernel is continuous on [−b, b] × R. On the other hand we have, using (41), (42), and (40), the asymptotic estimate  −k sin θ|s|  e |λ(b − a, t − τθ (s))| = O as |s| → +∞, |s|1/2 uniformly in t, for |t| ≤ b. Together, these properties prove (86). THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 29 Recall that we call a sesquilinear form a(·, ·) on a Hilbert space H compact if the associated linear operator A on H, defined by (Aφ, ψ) = a(φ, ψ), for all φ, ψ ∈ H, is compact. (Here (·, ·) denotes the inner product on H.) Equivalently, a(·, ·) is compact is, whenever φn ⇀ 0 and ψn ⇀ 0 (weak convergence in H), it holds that a(φn , ψn ) → 0. In our final theorem we show that the sesquilinear form on the Hilbert space (L20 (R))4 × H 1 (Ωb ) which appears on the left-hand side of (84) is the sum of coercive plus compact sesquilinear forms, and that, as a consequence of this and of the above uniqueness result, problem (84) is well-posed. Regarding the last sentence of the theorem, note that, given the constraint 0 < θ < π/4, Ω ⊂ Ωb ∪ 3 [ Ωjθ , j=0 so that the solution of the original scattering problem can be recovered in the whole e θ , ub ) of (84). of Ω from the solution (Φ Theorem 6.3. For every θ ∈ (0, π/4) and every f ∈ L2 (Ω) with support in Ωa , e θ , ub ) ∈ (L2 (R))4 × H 1 (Ωb ). Further, for Problem (84) has exactly one solution (Φ 0 some constant c > 0 depending on θ, (87) e θ k(L2 (R))4 + kub kH 1 (Ω ) ≤ ckf kL2 (Ω ) , kΦ a b 0 e θ , ub ) ∈ (L2 (R))4 × H 1 (Ωb ) is for all f ∈ L2 (Ω) with support in Ωa . Moreover, if (Φ 0 1 the solution of (84) and u ∈ Hloc (Ω) is the solution of (1)-(2), then u = ub in Ωb , while u is given in terms of ϕjθ in Ωjθ by (49), for j ∈ J0, 3K. Proof. As an operator on (L20 (R))4 we have, by Theorem 3.7, that Dθ = D1θ + D2θ , √ 1 where kDθ k ≤ 1/ 2 and D2θ is compact. Thus we can decompose the sesquilinear form which appears on the left-hand side of (84) as the sum of a first sesquilinear form e θ , Ψ) e (L2 (R))4 + ((I − D1θ )Φ 0 Z Ωb 3 X
∇ub · ∇vb + ub vb − ik j=0 Z Σjb ub v b , which is coercive on (L20 (R))4 × H 1 (Ωb ), and a second sesquilinear form Z 2 2e e e ub v b −(Dθ Φθ , Ψ)(L20 (R))4 − (Dθ Φ(ub ), Ψ)(L20 (R))4 − (1 + k ) − 3 Z X j=0 Ωb b −b   Λθ ϕ ejθ + ub |Σj (t) vb |Σj (t) dt b b which is compact on the same space. The proofs of compactness of the four terms of this second sesquilinear form rely on different arguments. The first is compact because D2θ is compact. For the second term, we notice that the operator ub 7−→ Dθ Φ(ub ) is compact from H 1 (Ωb ) to (L20 (R))4 , as it is the composition of the bounded operator Dθ on (L20 (R))4 and the compact map ub 7−→ Φ(ub ). (To see the compactness of this last map, note that it can be thought of as a composition of the bounded trace map from H 1 (Ωb ) to H 1/2 (Σb ) and the compact embedding H 1/2 (Σb ) ⊂ L2 (Σb ).) The compactness of the third term is a consequence of the Rellich embedding theorem, that 30 A.-S. BONNET-BEN DHIA ET AL. the embedding H 1 (Ωb ) ⊂ L2 (Ωb ) is compact. To see that the last term is compact one can use Lemma 6.2. Indeed, note that continuity of Λθ , combined with compactness of the map vb 7−→ vb |Σj from H 1 (Ωb ) to L2 (Σjb ), suffices to conclude. b Since the sesquilinear form is coercive plus compact, the Fredholm alternative holds (e.g. [40, Theorem 2.33]), so that unique solvability of (84) and the stability bound (87) are a consequence of Proposition 6.1. We have shown the last sentence of the theorem in our derivation, earlier in this section, of (84) from (1)-(2), using Theorem 3.4. 7. Numerical implementation and results. In this section we demonstrate, through some illustrative numerical experiments implemented in XLiFE++ [38], that the complex-scaled HSM formulations (58) and (84) can be solved numerically to compute solutions to the scattering problems (35)-(2) and (1)-(2), respectively. 7.1. Numerical implementation of the deformed half-space representation. Before considering the discretization of the HSM systems, we just want to provide an illustration of Theorem 3.4. More precisely, let (88) u(x) := i (1) H (k R(x1 , x2 )), 4 0 x ∈ R2 \ {0}, where R is defined by (17), so that u satisfies (35)-(2) in the case that g := u|Σa . Then ϕ0 := u|Σ0 is given by ϕ0 (s) = i (1) H (k R(s, a)), 4 0 s ∈ R. The corresponding complex-scaled trace ϕ0θ , defined by (45) and (46), is an even function given, for |s| < a, by ϕ0θ (s) := ϕ0 (s), and, for s > a, by ϕ0θ (s) (89)  q  i (1) 2 iθ 2 k a + (a + (s − a)e ) = H0 4 0 := ϕ (τθ (s)) ∼ ei(ka+π/4−θ/2) eik(s−a)e √ 2 2πks iθ as s → +∞, by (40) and (41). The asymptotic behaviour (89) agrees with (47) and Proposition 3.3, indeed demonstrates that (47) and Proposition 3.3 are sharp. We represent, on Figure 7, ϕ0θ for four different values of θ, with a = 1 and k = 2π. We see that ϕ0θ is more and more rapidly decaying at infinity as θ increases, in line with (89), (47), and Proposition 3.3. Then we represent in Figure 8 the function Z (90) Uθ (ϕ0θ )(x0 ) = h(x01 − a, x02 − τθ (s)) ϕ0θ (s) τθ′ (s) ds R (cf. (65)) in the half-space Ω0 , for θ = π/6, π/4, π/3, evaluating this integral accurately by standard numerical quadrature methods (namely a 5th order composite Gauss quadrature rule on a fine mesh of step length 0.1). From Theorem 3.4 we know that (91) Uθ (ϕ0θ )(x0 ) = u(x0 ) := i (1) H (k R(x01 , x02 )), 4 0 x0 ∈ Ω0θ , 31 THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD ·10−2 ·10−2 θ=0 θ = π/6 θ = π/4 θ = π/3 5 5 0 0 −5 −5 −4 −2 0 s 2 4 −4 −2 0 s 2 4 Fig. 7. Left: representation of the exact complex-scaled traces s 7→ ϕ0θ (s) for θ = 0, π/6, π/4, π/3. Right: comparison of the exact (red line) and computed (blue dots) complex-scaled trace s 7→ ϕ0θ (s) for θ = π/6. In both plots a = 1 and k = 2π. where Ω0θ := {x0 = (x01 , x02 ) : x01 − a > (|x02 | − a) tan θ}. The boundary of Ω0θ is indicated on the figures by dashed lines and, as predicted in (91), Uθ (ϕ0θ ) coincides with u in Ω0θ . Uθ (ϕ0θ ) is not equal to u outside Ω0θ ; in particular, it is easy to see from the definition that Uθ (ϕ0θ )(a, x02 ) = 0 for |x02 | > a. It appears at first glance that Uθ (ϕ0θ ) is continuous across the dashed lines in Figure 8, but an application of the residue theorem, modifying the argument of Theorem 3.4, shows a jump in the value of Uθ (ϕ0θ ) across the dashed lines of ϕ0 (x02 ± i(x01 − a)) = ϕ0θ (s) at the point x0 = (x01 , x02 ) where x02 = ±(a + (x01 − a) cot(θ)) = ±a + (s ∓ a) cos(θ). This jump across the dashed lines is just about visible very close to x0 = (a, ±a), but not visible elsewhere because ϕ0θ (s) is exponentially decaying as s → ±∞; see the zoom for θ = π/3. θ = π/6 θ = π/4 Fig. 8. Representation of Uθ (ϕ0θ ) in Ω0 (see (90)), where ϕ0θ (s) = a = 1 and k = 2π. The dotted lines are part of the boundary of Ω0θ . θ = π/3 i (1) H (k R(τθ (s), a)), with 4 0 7.2. Discretization of the complex-scaled HSM formulation and valie θ,h ∈ V0,h , where V0,h dations. We approximate the solution Φθ − Φg of (58) by Φ 32 A.-S. BONNET-BEN DHIA ET AL. e θ,h is the is a finite dimensional space V0,h ⊂ (L20 (R))4 that we specify below and Φ solution of the following Galerkin approximation: e θ,h ∈ V0,h such that Find Φ (92) e θ,h , Ψ e h )(L2 (R))4 = (Dθ Φg,h , Ψ e h )(L2 (R))4 , ((I − Dθ )Φ 0 0 e h ∈ V0,h , ∀Ψ where Dθ is defined by (59). To define the approximation space V0,h , where h := (h, q, T ), let us first introduce Vh ⊂ L2 (R). To construct Vh we truncate the infinite line at some distance T > 0 and build Vh with 1D Lagrange finite elements of degree q ≥ 1 and maximum element length h supported on [−T, T ]. The space V0,h is nothing else but Vh ∩ (L20 (R))4 e θ,h = {ϕ̃0 , ϕ̃1 , ϕ̃2 , ϕ̃3 } ∈ V0,h , each ϕ̃j where Vh := (Vh )4 . Thus, for Φ θ,h θ,h θ,h θ,h θ,h is a continuous piecewise polynomial function supported in [−T, T ] which vanishes on [−a, a]. In (92), Φg,h ∈ Vh ∩ (L2 (−a, a))4 is an interpolate of Φg . Finally, we e θ,h +Φg,h . approximate Φθ by Φθ,h = {ϕ0θ,h , ϕ1θ,h , ϕ2θ,h , ϕ3θ,h } ∈ Vh , given by Φθ,h := Φ It is clear that the approximation space V0,h that we have constructed has the approximation property that, for all Φ ∈ (L20 (R))4 , inf e h ∈V0,h Ψ e h k(L2 (R))4 → 0 kΦ − Ψ 0 as h → 0 and T → +∞. Thus, and since the sesquilinear form in (92) is coercive plus compact on (L20 (R))4 by Theorem 3.7(ii) (cf. Theorem 6.3), standard convergence results for Galerkin methods apply (e.g., [44, Theorem 4.2.9]). These give that, for e θ,h of (92) is well-defined for all 0 < h ≤ h0 some h0 > 0 and T0 > 0, the solution Φ and T ≥ T0 , and a quasi-optimality error estimate holds, that, for some constant C > 0 and all 0 < h ≤ h0 and T ≥ T0 , kΦθ − Φθ,h k(L2 (R))4 ≤ C inf Ψh ∈Φg,h +V0,h  kΦθ − Ψh k(L2 (R))4 ≤ C kΦθ k(L2 (R\(−T,T )))4 (93) + inf Ψh ∈Φg,h +V0,h  kΦθ − Ψh k(L2 (−T,T ))4 . This right hand side tends to zero as h → 0 and T → +∞, i.e., our Galerkin method is convergent, as long as kΦg − Φg,h k(L2 (−a,a))4 → 0 as h → 0. To implement the Galerkin method (92), the integrals ! Z Z T a T −T h(τθ (t) − a, a − τθ (s)) ϕjθ,h (s) τθ′ (s) ds ψ̃hj±1 (t)dt, ψ̃hj±1 ∈ Vh , which appear in the variational formulation, need to be approximated. In the results below we use a standard quadrature formula, without any specific treatment of the singularity. To validate the method, we consider u given by (88) which is the solution of (58)(2) with g := u|Σa . In the case that θ = π/6, a = 1, and k = 2π, we draw in the right hand side of Figure 7 the exact complex-scaled trace ϕ0θ of u (by symmetry, the four traces are equal in this case) and the computed complex-scaled trace ϕ0θ,h , obtained by solving (92) with h = 0.1, q = 1, and T = 5. We observe a very good agreement. 33 THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD More quantitatively, to explore the dependence of the error on T , we plot in Figure 9, for a fixed small value of h, the error kϕ0θ − ϕ0θ,h kL2 (R) (94) as a function of T for θ = π/6, π/4, π/3. As ϕ0θ,h is zero outside (−T, T ) the error cannot be smaller than the L2 norm of ϕ0θ on R \ (−T, T ), which decreases like e−k sin(θ)T /T 1/2 as T tends to +∞ by (89) and Proposition 3.3. On the other hand, when h is small enough so that the second term on the right hand side of (93) is negligible, the quasi-optimality bound (93) and Proposition 3.3 imply that e−k sin(θ)T /T 1/2 is also an upper bound for the error, precisely that the error (94) is ≤ Ce−k sin(θ)T /T 1/2 , for some constant C > 0 and all sufficiently large T . And, indeed, we observe in Figure 9 this rate of exponential behaviour as T increases, until the other sources of error become significant. log kϕ0θ − ϕ0θ,h kL2 (R) −1 −1 1 θ = π/6 θ = π/4 θ = π/3 k sin(π/6) k sin(π/4) k sin(π/3) −2 −2 −3 −3 1 1.5 2 2.5 3 1 1.5 2 T 2.5 3 T Fig. 9. Absolute error in ϕ0θ,h for three different values of θ computed with P 3 elements (q = 3), a = 1, and h = 0.002 and k = π (left), h = 0.001 and k = 2π (right). Using a standard numerical quadrature applied to formula (49), with ϕjθ approximated by ϕ0θ,h , we represent finally in Figure 10 the numerical solution in Ω0θ ∪ Ω1θ (left) and the whole of Ω (right) for θ = π/6. To reconstruct the solution in the whole of Ω several choices are possible, since θ < π/4 is such that the reconstruction domains overlap (Ωjθ overlaps with Ωj±1 , for j ∈ J0, 3K). Here we have reconstructed θ the solution using the identity u(xj ) = Uθ (ϕjθ )(xj ), xj ∈ Ωjπ/4 , j ∈ J0, 3K, where Uθ is defined in (65), and where we have used that Ωjπ/4 ⊂ Ωjθ . We notice in Figure 10 that these different representations are compatible, up to a small discretization error not visible in the plots. Finally, we validate formula (73) for the far-field pattern. In the present case, the far-field pattern is the same in all directions and one has for j ∈ J0, 3K: r Z k 1−i 1−i (95) F (cos(jπ/2), sin(jπ/2)) = √ = ϕjθ (s)τθ′ (s) ds. π 2 kπ R In Figure 11 we plot the real and imaginary part of the right hand side of (95) when 34 A.-S. BONNET-BEN DHIA ET AL. u u Ω0θ ∪Ω1θ Ω Fig. 10. Reconstruction of the computed solution in Ω0θ ∪ Ω1θ (left) and in the whole domain (right), with θ = π/6, a = 1, k = 2π. j = 0, a = 1, and k = 2π, with ϕ0θ approximated by ϕ0θ,h , that is we plot (96) Fθ,h := r k 1−i π 2 Z T −T ϕjθ,h (s)τθ′ (s) ds as a function of T for different values of θ. We again observe a rapid convergence towards the exact value as T increases. 0.3 0.28 −0.18 0.26 −0.2 0.24 −0.22 0.22 0.2 θ = π/6 θ = π/4 θ = π/3 −0.16 −0.24 1 1.5 2 T 2.5 1 1.5 2 2.5 T Fig. 11. Real (left figure) and imaginary (right figure) parts of the far-field coefficient for three different θ with P 3 elements (q = 3), h = 0.001, a = 1, and k = 2π, computed using the formula (96). The black lines indicate the exact values. 7.3. Numerical results for the general case. Finally, to discretize problem (84), we combine the previous tools with a classical Lagrange finite element approximation of the 2D unknown ub . Precisely, we use the HSM method to solve the problem THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 35 of diffraction of the incident plane wave ui (x) = exp(ik(x1 cos(π/6) + x2 sin(π/6))) by a perfectly reflecting scatterer which is the union of a disk and a triangle, this union contained in the square Ωa with a = 0.8. The scattering problem is (1)-(2) in Ω, the domain exterior to the scatterer, with f = 0, ρ = 1, and the Dirichlet boundary condition u = g := −ui on ∂Ω. The HSM problem we solve is a variation e b := Ω ∩ Ωb ; ii) we replace H 1 (Ωb ) by an affine on (84) in which: i) we replace Ωb by Ω e b ) that respects the Dirichlet boundary condition; precisely, we seek subspace of H 1 (Ω 1 e ub ∈ {v ∈ H (Ωb ) : v = g on ∂Ω}. It is easy to see, by a straightforward variation of the arguments of Proposition 6.1 and Theorem 6.3, that this modification of (84) remains well-posed: uniqueness holds and the modified sesquilinear form is coercive 1 e 1 e e b ) : v = 0 on ∂Ω}. plus compact on (L20 (R))4 × HD (Ωb ), where HD (Ωb ) := {v ∈ H 1 (Ω As in Section 7.2, this implies convergence of Galerkin methods of numerical solution, provided the sequence of approximation spaces used is asymptotically dense in 1 e (Ωb ). (L20 (R))4 × HD In our numerical implementation of a Galerkin method for this HSM formulation we fix a = 0.8 and b = 1.2, we use P2-elements for both the 2D unknown ub and the 1D unknowns ϕjθ , j ∈ J0, 3K, with a maximum element diameter for both meshes of h = 0.05, choose the truncation parameter T = 5, and choose θ = π/6. The additional integrals, involving the operator Λθ , which appear in the variational formulation are approximated, like the other ones, by a standard quadrature formula. To reconstruct the solution everywhere several choices are possible, since θ < π/4 e b with Ω0 , or Ω0 with and the different reconstruction domains overlap (for instance Ω θ θ 1 Ωθ , ...). We have reconstructed the solution as u(x) = ub (x), e b, x∈Ω u(xj ) = Uθ (ϕjθ )(xj ), e b , j ∈ J0, 3K, xj ∈ Ωjπ/4 \ Ω where Uθ is defined in (65) and where we have used that Ωjπ/4 ⊂ Ωjπ/6 . See Figure e b ∪ Ω0 ∪ Ω1 (left), the reconstruction in the whole 12 for the reconstruction in Ω π/4 π/4 domain Ω (middle), and the corresponding total field u + ui (right). We notice that e b, the different deformed half-space representations (the representations in Ωjπ/4 \ Ω j ∈ J0, 3K) are compatible between themselves, and are also compatible with the 2D e b. solution ub in Ω 8. Perspectives. Our new complex-scaled HSM method has been presented in this paper for a relatively simple configuration. We expect that it can be extended easily to more complex problems for which the relevant half-space Green’s functions are known sufficiently explicitly, such as acoustic scattering in stratified media, including cases where the stratification is different in different half-spaces (see e.g. [42] for a presentation of the method in the dissipative case). The method is also expected to work well in at least some cases with infinite boundaries, for instance scattering by an infinite wedge with Dirichlet, Neumann or Robin boundary conditions. In all these cases, the complex-scaled HSM should be a convenient way to take into account possible surface/guided waves that propagate towards infinity. Elastic scattering in isotropic media can also be considered. More challenging extensions are to the cases where PMLs are observed to fail, such as anisotropic media. A potential advantage 36 A.-S. BONNET-BEN DHIA ET AL. Diffracted field Total field Fig. 12. The diffracted field u reconstructed in Ωb ∪ Ω0π/4 ∪ Ω1π/4 (left) and in the whole of Ω (middle), and the corresponding total field u + ui (right) for the case of scattering by a Dirichlet obstacle. over PML of the complex-scaled HSM method in such cases is that it requires the existence of exponentially decaying analytical continuation of the traces of the solution only in a few directions (on the boundaries of a few half-planes). Appendix A. Properties of the operators D and defined by (34) can be rewritten as  0 1  0 0 D = D S ⊗ J + S D ⊗ J∗ with J :=  0 0 1 0 Dθ . 0 1 0 0 The operator D  0 0 , 1 0 where the operator S ∈ L(L2 (R)) (the space of continuous linear operators on L2 (R)) is defined in (22) and the operator D ∈ L(L2 (R)) is defined in (23) and (29). Here A ⊗ M denotes the tensor product of an operator A ∈ L(L2 (R)) with a 4 × 4 scalar matrix M (see, e.g., [5, §12.4]), which yields an operator of L((L2 (R))4 ). But it is actually enough to see this as a simple notation which makes the writing of the proof below easier: A ⊗ M is the block operator matrix obtained by multiplying each scalar component of M by the operator A. One can easily verify that it satisfies the basic property kA ⊗ Mk ≤ kAk kMk. The operator Dθ , given by (59), has the same definition as D just replacing D by Dθ . The operators D and Dθ satisfy similar properties, given respectively in Proposition 2.1 and in Propositions 3.8, 3.9, and 3.10. We show in this appendix properties (i) and (ii) of Theorem 2.2 and Theorem 3.7. These results are properties of D and Dθ , respectively, that are based on the above properties of D and Dθ . Let us give now the proof for (i) and (ii) of Theorem 2.2 for D; the same proof holds for Theorem 3.7 for Dθ . As D and S are continuous operators on L2 (R), the continuity of D on (L2 (R))4 (part (i) of Theorem 2.2) is obvious. To show part (ii) of Theorem 2.2 we consider D as an operator on (L20 (R))4 . Let us denote by χ+ (respectively χ− ) the characteristic function of (a, +∞) (respectively (−∞, −a)). We have, by using (28), that, for ϕ ∈ L20 (R) ϕ = χ+ ϕ + χ− ϕ, THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 37 and kϕk2L2 (R) = kχ+ ϕk2L2 (R) + kχ− ϕk2L2 (R) . (97) We can reformulate items (ii) and (iii) of Proposition 2.1 using these characteristic functions as follows: (ii) Dχ+ = Lχ+ + Kχ+ ∈√ L(L2 (R), L2 (a, +∞)), where L, K ∈ L(L2 (a, +∞)) are such that kLk ≤ 1/ 2 and K is compact. (iii) Dχ− ∈ L(L2 (R), L2 (a, +∞)) is compact. These properties, and that S χ± = χ∓ S, lead us to decompose D as an operator on (L20 (R))4 as D=L+K where K := (D Sχ+ + K S χ− ) ⊗ J + (S Dχ− + S Kχ+ ) ⊗ J∗ and L := (L Sχ− ) ⊗ J + (S Lχ+ ) ⊗ J∗ . It follows from (ii) and (iii) that K is compact. Moreover, noting that χ+ L = L, so that χ− S L = S L, we have L = χ+ (L S ⊗ J) χ− + χ− (S L ⊗ J∗ ) χ+ . We deduce then by (97) that, for all Φ ∈ (L20 (R))4 , kLΦk2L2 (R)4 = k (L S ⊗ J) χ− Φk2L2 (R)4 + k (S L ⊗ J∗ ) χ+ Φk2L2 (R)4 1 1 1 ≤ kχ− Φk2L2 (R)4 + kχ+ Φk2L2 (R)4 = kΦk2L2 (R)4 , 2 2 2 √ where we have used that kL S ⊗ Jk ≤ kL Sk kJk ≤ 1/ 2, and the same bound for S L ⊗ J∗ . Appendix B. Technical lemmas. The lemmas in this annex (cf. [22, Lemma 4.4]) concern the complex functions R(ẑ, z) := (ẑ 2 + z 2 )1/2 , (98) R̂(ẑ, z) := (ẑ + z 2 )1/2 , z, ẑ ∈ C. Note that, as throughout the rest of the paper, all square roots in this appendix are principal square roots, i.e. square roots with argument in the range (−π/2, π, 2]. Lemma B.1. We have
|R(t, z)|2 ≥ | cos(Arg(z))| t2 + |z|2 , t ∈ R, z ∈ C, z 6= 0. Proof. Let γ := Arg(z). We have |R(t, z)|4 = |t2 + |z|2 e2iγ |2 = t4 + |z|4 + 2t2 |z|2 cos(2γ) = t2 − |z|2 which yields |R(t, z)|4 ≥ cos2 (γ) h t2 − |z|2
2
2 + 4t2 |z|2 cos2 (γ) i
2 + 4t2 |z|2 = cos2 (γ) t2 + |z|2 . 38 A.-S. BONNET-BEN DHIA ET AL. Lemma B.2. There exists a constant C > 0 such that, for all z ∈ C and t ∈ R, |R(t, z) − z| ≤ C t2 , |z| ℜ(z) > 0, |z| ≥ |t|; in particular ℑ(R(t, z)) ≥ ℑ(z) − C t2 , |z| ℜ(z) > 0, |z| ≥ |t|. Moreover, for all A > 0, there exists a constant C ′ > 0 such that, for −A ≤ t ≤ A, ℑ(R(t, z)) ≥ ℑ(z) − C ′ , ℜ(z) > 0. Proof. Since the function z 7→ (1 + z)1/2 − 1 is an analytic function of z in the open unit disk that vanishes at the origin and is bounded in the closed disk, for some constant C > 0, |(1 + z)1/2 − 1| ≤ C|z|, (99) |z| ≤ 1. Further, the function 2 2 1/2 z 7→ (t + z )  t2 −z 1+ 2 z 1/2 is analytic in the domain {z ∈ C : ℜ(z) > 0} and it vanishes for z ∈ R with z > 0, so that it vanishes everywhere in {z ∈ C : ℜ(z) > 0}. This implies that R(t, z) − z = z " t2 1+ 2 z 1/2 # −1 , ℜ(z) > 0. Combined with (99), this proves the first inequality. The second inequality is a direct consequence of the first since |R(t, z) − z| ≥ ℑ(z − R(t, z)). Finally, the second inequality yields that, for any constant A > 0 and for all |t| ≤ A ≤ |z|, ℑ(R(t, z)) ≥ ℑ(z) − CA. But a similar √ inequality also holds for |z| ≤ A, possibly with another constant, since |R(t, z)| ≤ 2A for |t| ≤ A and |z| ≤ A. Remark B.3. Note that, replacing z by −z, one directly deduces from the previous lemma similar results valid for ℜ(z) < 0. In particular, |R(t, z) + z| ≤ C t2 , |z| ℜ(z) < 0, |z| ≥ |t|, and ℑ(R(t, z)) ≥ −ℑ(z) − C t2 , |z| ℜ(z) < 0, |z| ≥ |t|. Analogous results can be proved in the more general case where the positive real number t2 is replaced by any complex number ẑ. THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 39 Lemma B.4. There exists a constant C > 0 such that, for all z, ẑ ∈ C, z 6= 0, ẑ , z ẑ , ℑ(R̂(ẑ, z)) ≥ ℑ(z) − C z |R̂(ẑ, z) − z| ≤ C (100) ℜ(z) ≥ |ẑ|1/2 , ℜ(z) ≥ |ẑ|1/2 . Moroever, if U is a bounded subset of C and 0 < γ < π/2, there exists a constant C ′ > 0 such that ℑ(R̂(ẑ, z)) ≥ ℑ(z) − C ′ , ẑ ∈ U, |Arg(z)| < γ. Proof. To proceed as in the proof of Lemma B.2, we just have to show that the function 1/2  ẑ z 7→ (ẑ + z 2 )1/2 − z 1 + 2 z is analytic in the domain {ℜ(z) > |ẑ|1/2 }. For that, we have to check that the branch
1/2 cuts of the functions z 7→ (ẑ + z 2 )1/2 and z 7→ 1 + ẑ/z 2 do not intersect the domain ℜ(z) > |ẑ|1/2 .
1/2 This is clear for the function z 7→ 1 + ẑ/z 2 . The branch cut of the function z 7→ (ẑ + z 2 )1/2 is the subset of the hyperbola {z ∈ C : ℑ(z 2 + ẑ) = 0} where ℜ(z 2 + ẑ) ≤ 0. The intersection of this hyperbola with the domain ℜ(z) > |ẑ|1/2 is the connected set Ĥ = {z ∈ C : ℑ(z 2 + ẑ) = 0, ℜ(z) > |ẑ|1/2 }, which describes a curve which is asymptotic to the real axis when ℜ(z) → +∞. To conclude, we have to prove that for all z ∈ Ĥ, ℜ(z 2 + ẑ) > 0. This is clearly true for large values of ℜ(z). If it were not true for all z ∈ Ĥ, there would exist some z0 ∈ Ĥ such that ℜ(z02 + ẑ) = 0. But then z02 + ẑ = 0, so that |z0 | = |ẑ|1/2 , which is impossible since ℜ(z0 ) > |ẑ|1/2 . An important application of the previous lemmas for our purpose is the following: Lemma B.5. For θ ∈ (0, π/2), let τθ be defined by (45), and suppose that θ0 ∈ [θ, π/2). Then, for w ∈ C such that w 6= 0 and −θ0 + θ ≤ Arg(w) ≤ θ0 , and for all s ∈ R, it holds that (101) |R(w, τθ (s) − a)|2 ≥ cos(θ0 )(|w|2 + |τθ (s) − a|2 ) ≥ cos2 (θ0 )(|w|2 + |τθ (s) − a|2 ). Further, there exists a constant C > 0, depending only on a and θ, such that, for all s ∈ R, (102) ℑ(R(w, τθ (s)−a)) ≥ [cos(θ−Arg(w))]1/2 ℑ(w)−C ≥ cos(θ−Arg(w))ℑ(w)−C, provided w 6= 0 and 0 ≤ Arg(w) ≤ θ. Proof. Let γ := Arg(w) ∈ [θ − θ0 , θ0 ]. We have, for all s ∈ R, |R(w, τθ (s) − a)| = |R(|w|eiγ , τθ (s) − a)| = |R(|w|, (τθ (s) − a)e−iγ )|. Applying Lemma B.1 we obtain the inequality (103) |R(w, τθ (s) − a)|2 ≥ | cos(γ̂(s))|(|w|2 + |τθ (s) − a|2 ) 40 A.-S. BONNET-BEN DHIA ET AL.
where γ̂(s) := Arg (τθ (s) − a)e−iγ . We now consider separately the three cases |s| ≤ a, s > a, and s < −a to derive the two inequalities of the lemma. Case 1. For |s| ≤ a, since τθ (s) = s, we have | cos(γ̂(s))| = cos(γ) ≥ cos(θ0 ) so that (101) holds. Further, applying the third inequality of Lemma B.2 we get that there exists a constant c > 0, dependent only on a, such that (104) ℑ(R(w, τθ (s) − a)) = ℑ(R(s − a, w)) ≥ ℑ(w) − c. Case 2. For s > a, since τθ (s) − a = s̃eiθ , where s̃ := s − a > 0, and γ̂(s) = θ − γ so that | cos(γ̂(s))| = cos(θ − γ) ≥ cos(θ0 ), we have from (103) that (105) |R(w, τθ (s) − a)|2 = |R(w, s̃eiθ )|2 ≥ cos(θ − γ)(|w|2 + s̃2 ), so that (101) holds. To see that (102) holds, for γ ∈ [0, θ] as required in the lemma, note that (106) Arg(R(w, τθ (s) − a)) = Arg(R(|w|eiγ , s̃eiθ )) ∈ [γ, θ]. Combining this with the bound (105) we see that ℑ(R(w, τθ (s) − a)) = ℑ(R(w, s̃eiθ )) = |R(w, s̃eiθ )| sin Arg(R(w, s̃eiθ )) p p ≥ cos(θ − γ) |w| sin(γ) = cos(θ − γ) ℑ(w). (107)
Case 3. For s < −a, setting s̃ := −(s + a) > 0, we have τθ (s) − a = −(s̃eiθ + 2a) and | cos(γ̂(s))| = | cos(θ̃(s) − γ)|, where θ̃(s̃) := Arg(s̃eiθ + 2a). Since 0 ≤ θ̃(s̃) ≤ θ for s̃ > 0, and γ ∈ [θ − θ0 , θ0 ], we have | cos(γ̂(s))| ≥ min | cos(γ − γ̃)| ≥ cos(θ0 ), 0≤γ̃≤θ so that (101) follows from (103). To show (102), for 0 ≤ γ ≤ θ as required, we rewrite R(w, s̃eiθ + 2a) = [R(w, s̃eiθ )2 + 4a2 + 4as̃eiθ ]1/2 (108) = R̂(4a2 + 4as̃eiθ , R(w, s̃eiθ )), where R̂ is defined by (98). Using (105) and (106) we see that |R(w, s̃eiθ )|2 ≥ cos(θ − γ)s̃2 ≥ cos(θ)s̃2 and Arg(R(w, s̃eiθ )) ∈ [γ, θ]. Thus, by (100) applied to (108), there exist constants C > 0 and s̃0 > 0, depending only on θ and a, such that ℑ(R(w, s̃eiθ + 2a)) ≥ ℑ(R(w, s̃eiθ )) − C 1 + s̃ , s̃ for s̃ ≥ s̃0 . Hence, and using (107), it follows that, for s̃ ≥ max(1, s̃0 ), p ℑ(R(w, s̃eiθ + 2a)) ≥ cos(θ − γ)ℑ(w) − 2C. On the other hand, for s̃ ≤ max(1, s̃0 ) the last inequality of Lemma B.4 gives that there exists a constant C ′ > 0, depending only on a, such that ℑ(R(w, s̃eiθ + 2a)) ≥ ℑ(w) − C ′ . THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 41 Gathering the two estimates we have that, for s < −a, p (109) ℑ(R(w, τθ (s) − a))) ≥ cos(θ − γ)ℑ(w) − max(C ′ , 2C). We have shown that (101) holds in each case. That (102) also holds follows from (104), (107), and (109), trivially noting that (cos(t))1/2 ≥ cos(t) for t ∈ R. Appendix C. Mapping properties of complex-scaled integral operators. In this appendix we prove mapping properties of the analytic continuations into the complex plane of the single- and double-layer potential operators S j and Dj , defined by (39) for j ∈ J0, 3K, φ ∈ L2 (Σa ), and |t| > a. The proofs use the bounds established in Appendix B and bounds on the relevant Hankel functions. Specifically [35, Lemma 3.4], for some constant c1 > 0,   (1) (110) e−iz H1 (z) ≤ c1 |z|−1 + (1 + |z|)−1/2 , ℜ(z) > 0. Similarly, it follows from (40) and [30, §10.2(ii)] that, for some constant c0 > 0, (1) e−iz H0 (z) ≤ c0 M(|z|), (111) ℜ(z) > 0, where M(t) := log(1 + t−1 ) + (1 + t)−1/2 , for t > 0. Proposition C.1. For every θ ∈ (0, π/2) there exists some constant C > 0, that depends only on a, k, and θ, such that, for j ∈ J0, 3K and every φ ∈ L2 (Σa ), |S j φ(z)| ≤ C M(|z − a|) exp(−kℑ(z)) kφkL2 (Σa ) , (112) |Dj φ(z)| ≤ C |z − a|−1/2 exp(−kℑ(z)) kφkL2 (Σa ) , (113) for ℜ(z) > a with |Arg(z − a)| ≤ θ, while |S j φ(z)| ≤ C M(|z + a|) exp(kℑ(z)) kφkL2 (Σa ) , (114) |Dj φ(z)| ≤ C |z + a|−1/2 exp(kℑ(z)) kφkL2 (Σa ) , (115) for ℜ(z) < −a with |Arg(−z − a)| ≤ θ. Proof. We prove only the bounds (112-113); the proofs of (114-115) are identical. Throughout this proof C will denote any positive constant depending only on a, k, and θ, not necessarily the same at each occurrence, and we assume that y j ∈ Σa and that ℜ(z) > a with |Arg(z − a)| ≤ θ. It follows from (the analytic continuation of) (39) and the Cauchy-Schwarz inequality that 1/2 |S j φ(z)| ≤ CIS kφkL2 (Σa ) , 1/2 |Dj φ(z)| ≤ CID kφkL2 (Σa ) , where IS := Z j Σa j 2 j |Φ(x (z), y )| ds(y ), ID := Z Σa ∂Φ(xj (z), y j ) ∂n(y j ) 2 ds(y j ). Now, from (111), and since xj (z) := (a, z) and recalling the definition (17),   |Φ(xj (z), y j )| ≤ C M k|R(a − y1j , z − y2j )| exp(ikR(a − y1j , z − y2j )) . 42 A.-S. BONNET-BEN DHIA ET AL. Further, it follows from Lemma B.2 that, for some constant Ca > 0 depending only on a, (116) | exp(ikR(a − y1j , z − y2j ))| ≤ exp(−k(ℑ(z) − Ca )) ≤ C exp(−kℑ(z)), and from Lemma B.1 that (117) |R(a − y1j , z − y2j )| ≥ cos(θ)((a − y1j )2 + |z − a|2 )1/2 ≥ cos(θ)|z − a|. Thus, noting that M (t) is decreasing as t increases and that, for every c > 0, t 7→ M (ct)/M (t) is a bounded function on t > 0, it follows that |Φ(xj (z), y j )| ≤ CM (|z − a|) exp(−kℑ(z)), 1/2 so that IS ≤ CM (|z − a|) exp(−kℑ(z)) and (112) follows. Writing n(y j ) = (n1 (y j ), n2 (y j )) in the (xj1 , xj2 ) coordinate system, we see that ∂Φ(xj (z), y j ) = ∂n(y j ) (118) n1 (y j )(y1j − a) + n2 (y j )(y2j − z) k (1) j j H (kR(a − y1 , z − y2 )) . 4 1 R(a − y j , z − y j ) 1 2 How we bound the right hand side of this equation depends on which side of Σa the point y j is located. When y j ∈ Σja ⊂ Σa the right hand side of (118) vanishes, since j then y1j = a and n2 (y j ) = 0. When y j ∈ ∂Ω \ (Σja ∪ Σj+1 a ) it holds that y1 = −a or j y2 = −a, so that |R(xj (z), y j )| ≥ C by Lemma B.1, from which it follows, applying (110) and (116), that the right hand side of (118) is bounded by C exp(−kℑ(z)). Thus Z 2 ∂Φ(xj (z), y j ) ∗ ∗ ds(y j ). ID ≤ C exp(−2kℑ(z)) + ID , where ID := ∂n(y j ) Σj+1 a Finally, when y j ∈ Σj+1 it holds that n1 (y j ) = 0 and y2j = a, so that, where ρ := a |z − a|, (118) says that (119) ∂Φ(xj (z), y j ) k ρ (1) = H (kR(a − y1j , z − y2j )) ∂n(y j ) 4 1 R(a − y1j , z − y2j ) ! ρ −1/2 ≤C + (1 + ρ) exp(−kℑ(z)), ρ2 + (a − y1j )2 using (110), (116), and (117). Thus ∗ e2kℑ(z) ID ≤ C +C 1+ρ Z ρ2 dy1j a −a (ρ2 + (y1j − a)2 )2 ≤ C , ρ 1/2 where we see the final bound by substituting s := (y1j − a)/ρ. Thus ID a|−1/2 e−kℑ(z) and the bound (113) follows. ≤ C|z − The final result of this appendix, and arguments we make elsewhere in the paper, depend on a mapping property of the classical double-layer potential operator on the boundary of a quadrant. For the convenience of the reader we state this key and well-known result and sketch its proof in the following proposition. THE COMPLEX-SCALED HALF-SPACE MATCHING METHOD 43 Proposition C.2. Let h0 denote the kernel function h defined in (16) in the case that k = 0, so that h0 (x1 , x2 ) = x1 /(π(x21 + x22 )), for x1 > 0 and x2 ∈ R. For φ ∈ L2 (0, +∞) define Dφ : (0, +∞) → C by Dφ(s) := Z +∞ h0 (s, t) φ(t) dt = 0 1 π Z +∞ 0 s φ(t) dt, s 2 + t2 s > 0. 2 2 Then Dφ ∈ L2 (0, +∞) √ and the mapping D : L (0, +∞) → L (0, +∞) is bounded, with norm kDk = 1/ 2. Proof. This result can be proved by making use of the equivalence of (24) and (25) in the static case k = 0 as in [12], or directly via Mellin transform methods (cf. [41, 43]). Equivalently, we observe as in [17] that the mapping I : L2 (0, +∞) → L2 (R), given by Iφ(t) = φ(e−t )e−t/2 , t ∈ R, is unitary, as is the Fourier transform b given by (11). Further [17], for φ ∈ L2 (R), operator F : L2 (R) → L2 (R), φ 7→ φ, Z −1 IDI φ(s) = κ(s − t)φ(t) dt, s ∈ R, R where κ(τ ) := eτ /2 /(π(e2τ + 1)), for τ ∈ R, and [17] Thus, for φ ∈ L2 (R), 1 sinh(π(ξ − i/2)/2) , κ b(ξ) = √ 2π sinh(π(ξ − i/2)) FIDI−1 F−1 φb = √ ξ ∈ R. b 2π κ b φ, b ∈ L∞ (R), D is bounded on L2 (0, +∞) with so that [17], since κ √ √ √ kDk = kFIDI−1 F−1 k = 2π kb κkL∞ (R) = 2π |b κ(0)| = 1/ 2. Remark C.3. Let D0 denote the operator D, given by (25) and (29), in the case that k = 0, so that, for φ ∈ L2 (R), D0 φ(t) = 0, for t ≤ a, while Z Z t−a 1 D0 φ(t) = φ(t) dt, t > a, h0 (t − a, s − a) φ(s) ds = 2 + (s − a)2 π (t − a) R R where h0 is as defined in Proposition C.2. Then it is clear from the above proposition √ that, as an operator on L2 (a, +∞), D0 is bounded with norm kD0 k = kDk = 1/√2, and that D0 is also bounded as an operator on L2 (R), with norm kD0 k = 2kDk = 2. Proposition C.4. For 0 < θ < π/2 and j ∈ J0, 3K, Sθj and Dθj are continuous operators from L2 (Σa ) to L2 (−∞, −a) ⊕ L2 (a, +∞), where, for φ ∈ L2 (Σa ), Sθj φ(s) := S j φ(τθ (s)), Dθj φ(s) := Dj φ(τθ (s)), |s| > a. Proof. It is clear from the bounds (112) and (114) that Sθj maps L2 (Σa ) continuously to L2 (−∞, −a) ⊕ L2 (a, +∞). The analogous bounds (113) and (115) do not quite imply that Dθj φ ∈ L2 (−∞, −a) ⊕ L2 (a, +∞) for each φ ∈ L2 (Σa ), since (s − a)−1/2 e−k sin(θ)(s−a) is in L1 (a, +∞) but not in L2 (a, +∞). To see that Dθj : L2 (Σa ) → L2 (a, +∞) and is continuous we argue as in the proof of Proposition C.1, in particular using (118) and (119), and recalling that, except when y j ∈ Σj+1 a , the 44 A.-S. BONNET-BEN DHIA ET AL. right hand side of (118) is ≤ C exp(−kℑ(z)). These bounds imply that, for φ ∈ L2 (Σa ) and s > a, Z Z |s − a| |φ(y j )| j −k sin(θ)(s−a) j j ds(y j ), |Dθ φ(s)| ≤ Ce |φ(y )| ds(y ) + C (s − a)2 + (y1j − a)2 Σa Σj+1 a where, throughout the proof, C > 0 denotes a constant that depends only on θ, a, and k. Thus |Dθj φ(s)| ≤ Ce−k sin(θ)(s−a) kφkL2 (Σa ) + CD0 ψ(s), where D0 is as in Remark C.3 above, while ψ ∈ L2 (−a, a) denotes the restriction of j j j j |φ| to Σj+1 a , precisely ψ(y1 ) := |φ((y1 , a))|, for −a < y1 < a, while ψ(y1 ) := 0, for j |y1 | ≥ a. Since (Remark C.3) D0 is a bounded operator on L2 (R), it follows that kDθj φkL2 (a,+∞) ≤ CkφkL2 (Σa ) + CkD0 k kψkL2 (−a,a) ≤ CkφkL2 (Σa ) . Arguing in the same way, we see that Dθj is also continuous as a mapping from L2 (Σa ) to L2 (−∞, −a), and the proof is complete. REFERENCES [1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, Dover, 1964. [2] J. Aguilar and J.-M. 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