L Halpern
email: halpern@math.univ-paris13.fr

S Petit-Bergez

J Rauch

INSTRUCTION FILE confluentesvf.hyper27237 Confluentes Mathematici

Keywords: PML. WKB, hyperbolic operators, weak well posedness, geometric optics, extrapolation, reflection, amplification. AMS Subject Classification: 65M12, 65M55, 30E10. Dedication

A systematic analysis of matched layers is undertaken with special attention to better understand the remarkable method of Bérenger. We prove that the Bérenger and closely related layers define well posed transmission problems in great generality. When the Bérenger method or one of its close relatives is well posed, perfect matching is proved. The proofs use the energy method, Fourier-Laplace transform, and real coordinate changes for Laplace transformed equations. It is proved that the loss of derivatives associated with the Bérenger method does not occur for elliptic generators. More generally, an essentially necessary and sufficient condition for loss of derivatives in Bérenger's method is proved. The sufficiency relies on the energy method with pseudodifferential multiplier. Amplifying and nonamplifying layers are identified by a geometric optics computation. Among the various flavors of Bérenger's algorithm for Maxwell's equations our favorite choice leads to a strongly well posed augmented system and is both perfect and nonamplifying in great generality. We construct by an extrapolation argument an alternative matched layer method which preserves the strong hyperbolicity of the original problem and though not perfectly matched has leading reflection coefficient equal to zero at all angles of incidence. Open problems are indicated throughout.

Introduction

This paper analyses absorbing layer methods for calculating approximations to the solution, U , of first order systems of hyperbolic partial differential equations,

L(∂ t , ∂ x ) U := ∂ t U + d l=1 A l ∂ l U = F , (t, x) ∈ R 1+d , U (t, x) ∈ C N . (1.1)
Approximate values are sought on a finite domain. The source term F and/or initial condition is compactly supported in the domain. The absorbing layer strategy surrounds the domain with a layer of finite thickness intended to be absorbing and weakly reflective. The simplest case is dimension d = 1 with computational domain x 1 < 0 and absorbing layer in x 1 > 0. For the first example consider inhomogeneous initial data and zero right hand side. The simplest absorbing layers add a lower order term σ1 x1>0 C U where 1 denotes the characteristic function, for example,

∂ t U + 1 0 0 -1 ∂ 1 U + σ 1 x1>0 1 0 c b U = 0.
If c = 0, then ∇ t,x1 U 2 is discontinuous across the ray {x 1 = -t}. From the perspective of a numerical method, such a reflected singularity is undesirable.

The reflected singularity from a discontinuous lower order term is weaker than the singularity of the incident wave. For the equation

∂ t U + A 1 ∂ 1 U + σ 1 x1>0 CU = 0,
if C is diagonal in a basis diagonalising A 1 , the reflections are avoided. The ease of eliminating reflections for this problem with d = 1 is deceptive. No such simple remedy exists in dimensions d > 1. For symmetric hyperbolic systems A 1 = A * 1 , it is wise to choose C = C * ≥ 0 so that the absorption term is dissipative in the L 2 (R d ) norm.

Consider next the wave equation with friction

∂ tt v -∂ 11 v + 2σ 1 x1>0 ∂ t v = 0 written in characteristic coordinates (U 1 , U 2 ) = (∂ t v -∂ 1 v, ∂ t v + ∂ 1 v) with absorption B = σC: ∂ t U + 1 0 0 -1 ∂ 1 U + σ 1 x1>0 C U = 0 , C = 1 1 1 1 .
The absorption matrix C is symmetric and nonnegative but does not commute with A 1 . It produces unacceptably strong reflections. The absorption from Israeli and Orszag [START_REF] Israeli | Approximation of radiation boundary conditions[END_REF], ∂ tt v-∂ 11 v+σ(∂ t v+∂ 1 v) = 0, absorbs only rightward waves and corresponds to

C = 1 0 0 0 = π + (A 1 ),
introducing the notation π + (A 1 ) for the spectral projector on the eigenspace corresponding to strictly positive eigenvalues of A 1 . The general nonnegative symmetric choice commuting with A 1 is a positive multiple of

C = π + (A 1 ) + ν π -(A 1 ), ν ≥ 0. (1.2)
We call these smart layers. They dissipate the L 2 norm. As observed by Israeli and Orszag, the numerical performance of the smart layers is not as good as one would hope. One reduces reflections by choosing σ(x) ≥ 0 vanishing to order k ≥ 0 at the origin. That reduces the rate of absorption and thereby increases the width of the layer required. The leading reflection by such smart layers of incoming wave packets of amplitude O(1) and wavelength ε is O(ε k+1 ). The leading reflection is linear in σ. In section 6, we introduce the method of Harmoniously Matched Layers which remove the leading order reflections (at all angles of incidence) by an extrapolation.

Open problem. Repeated extrapolation further reduces the order of reflection. It is easy to program and it is possible that an optimization could pay dividends.

Elaborate absorbing layer strategies, like Bérenger's PML introduce operators related to but often more complicated than the original operator L. The operators in the absorbing layer and in the domain of interest may not be the same. For the case of a layer in {x 1 > 0}, absorbing layer algorithms solve a transmission problem for an unknown (V, W ) where V is a C N valued function on x 1 < 0 and W is a function on x 1 > 0. The equations in x 1 > 0 are chosen to be absorbing and the transmission problem weakly reflective. The ingenious innovation of Bérenger was to realize that the operator R in the layer can differ substantially from L. He increased the number of unknown functions in the layer. So W is C M valued with M > N .

The pair (V, W ) is determined by a well posed transmission problem,

LV = F on R 1+d - := {(t, x) : x 1 < 0} R W = 0 on R 1+d + , (1.3)
with the homogeneous transmission condition (V, W ) ∈ N on {x 1 = 0}. (1.4) Here N ⊂ C N × C M is a linear subspace. a The choice of the hyperbolic operator R and transmission condition N is made with three goals,

• The transmission problem is well posed, and not hard to approximate numerically.

• Waves from the left are at most weakly reflected at x 1 = 0.

• Waves moving rightward decay rapidly in x 1 > 0 so that the layer can be chosen thin.

The criterion for perfection that we adopt is that of Appelo, Hagström and Kreiss [START_REF] Appelö | Perfectly matched layers for hyperbolic systems: General formulation, well-posedness and stability[END_REF] . In the case of one absorption, it is formulated as follows.

Definition 1.1. A well posed transmission problem is perfectly matched when for all F supported in x 1 < 0 , t ≥ 0, the solution supported in t ≥ 0 satisfies V = U x1<0 .

We prove in §4.1.4 that Bérenger's method with one discontinuous absorption σ 1 is perfect in this sense.

In practice one does not absorb in only one direction and the computational domain is rectangular. We give in §3.5 a definition with absorptions in more than one direction and a proof of perfection.

The strategy of Bérenger is quite ingenious. For an artificial boundary in two dimensions at {x 1 = r} and domain of interest {x 1 < r} it consists of two steps. The first is a doubling of the system and the second is insertion of an absorption term in {x 1 > r}. The doubled system involves the unknown U := (U 1 , U 2 ) ∈ C N × C N . When F = 0, the doubled equation without dissipation is

∂ t U j + A j ∂ j (U 1 + U 2 ) = 0 , j = 1, 2 .
The system with damping in x 1 changes the j = 1 equation to

∂ t U 1 + A 1 ∂ 1 (U 1 + U 2 ) + σ(x 1 ) U 1 = 0 , supp σ ⊂ {x 1 ≥ r} .
Then U := j U j satisfies L(∂)U = 0 in x 1 < r. In practice it is the restriction of U to x 1 < r that is of interest. There are three distinct ways to view this. One can think of the unknowns as U defined in x 1 < r and U in x 1 ≥ r with the transmission condition that A 1 U = A 1 (U 1 + U 2 ) on x 1 = r. One is given initial values of U and takes initial values of U vanishing. This is the most natural choice and the one presented by Bérenger.

From the computational point of view it is simpler to have the same unknowns throughout. The simplification is greater when one passes from the half space case to a computational domain equal to a rectangular domain in R d . One introduces U everywhere with transmission condition [A 1 (U 1 + U 2 )] = 0 where [ * ] denotes the jump at x 1 = r. The transmission condition is then equivalent to the validity of the differential equation satisfied by U in all of R d . When one uses U everywhere, the initial values of U are taken equal to zero outside the computational domain. The initial values are constrained to satisfy U = j U j within the computational domain. The choice is otherwise arbitrary. For the case of the doubling above the choice U j (0, x) = U (0, x)/2 for j = 1, 2 is common.

If the domain of interest is |x 1 | ≤ r one would choose σ > 0 on |x 1 | > r and vanishing for |x 1 | < r. The transmission condition is [A 1 (U 1 + U 2 )] = 0 with the jump at x 1 = r and also at x 1 = -r.

In a rectangular geometry in R d introduce U := (U 1 , . . . , U d ), where U l ∈ C N for 1 ≤ l ≤ d. Then U with values in C N d is required to satisfy (in the case F = 0),

( L(∂ t , ∂ x ) U ) l := ∂ t U l + A l ∂ l ( d j=1 U j ) + σ l (x l )U l = 0, 1 ≤ l ≤ d. (1.5)
Each absorption coefficient σ l (x l ) ≥ 0 depends on only one variable. It is strictly positive between the inside rectangle and a larger outside rectangle. In the layer between the rectangles the solution is expected to decay. If U solves (1.5), then U = d j=1 U j solves (1.1) on the set {x : σ l (x l ) = 0 for 1 ≤ l ≤ d} including the inner rectangle. In the case considered by Bérenger the σ were discontinuous and the a Transmission conditions which involve derivatives can also be treated. The algorithms of Bérenger and our HML do not require that generality.

Contents 5 equations (1.5) are equivalent to transmission problems where on the discontinuity surface of σ j one imposes the transmission condition of continuity of A j ℓ U ℓ .

Our first technique is the energy method. In §3.2 we show that if (ξ 1 , . . . , ξ d ) = 0 does not meet the characteristic variety of L then the Bérenger method is well posed without loss of derivatives. This applies in particular to linearized elasticity and suggests that in some ways the Bérenger method is better adapted to that situation than the Maxwell equations for which it was intended. In §3.3 we give a nontrivial extension of the method of Métral and Vacus to show that Bérenger's method for the Maxwell equations in dimension d = 2 (resp. d = 3) is well posed provided that σ j (x j ) ∈ W 1,∞ (R xj ) (resp. σ j (x j ) ∈ W 2,∞ (R xj )). The method introduces a norm that is the sum of L 2 (R d t,x ) norms of suitable differential operators P α (D) applied to U . It has the property that the norm at time t 1 is estimated in terms of the norm at time t 2 . If one introduces the vector of unknowns P α (D)U this shows that the Bérenger problem becomes strongly well posed without loss of derivatives. Such transformations are typical of weakly well posed problems. (see the Dominics' proof of Theorem 1.1 in §IV.1 of [START_REF] Taylor | Pseudodifferential operators[END_REF]).

When such an estimate is known, we prove sharp finite speed in §3. [START_REF] Bardos | Maximal positive boundary value problems as limits of singular perturbation problems[END_REF] and perfection in §3.5 and §3.6, the latter concerned with several variants of the Berenger strategy. The perfection proof passes by a study of the Laplace transform on {Im τ = 0}. The transformed problem is conjugated to the problem without absorption by a τ dependent change of independent variable x, an idea inspired by [START_REF] Diaz | A time domain analysis of PML models in acoustics[END_REF].

Our second method is the Fourier-Laplace method. Bérenger introduced his PML for Maxwell's equations with σ piecewise constant. Using a computation which resembles plane wave analysis of reflections for problems without lower order terms, Bérenger argued that the layers were perfectly matched for all wave numbers and all angles of incidence. Using variants of the same approach other closely related PML were constructed afterward. Performance is observed to be enhanced using σ which are not discontinuous. Twice differentiable cubic functions are the most common. The Bérenger method is a very good method for Maxwell's equations. The Fourier-Laplace method gives a framework for understanding the computations of Bérenger. In addition, it is the only method we know for proving well posedness of Bérenger's PML with discontinuous σ for Maxwell's equations.

Plane wave analysis is sufficient to study reflection and transmission for linear constant coefficient operators without lower order terms. Problems with lower order terms require other tools as it is no longer true that the plane waves generate all solutions. The first level of generalization is to use the Fourier-Laplace transform for problems where an absorbing layer occupies x 1 ≥ 0 and both L and R have constant coefficients. Hersh [START_REF] Hersh | Mixed problems in several variables[END_REF] found necessary and sufficient conditions for (weak) well posedness of transmission problems. We recall those ideas in §4.1.1 including the modifications needed for characteristic interfaces, and verify in §4.1.3 that the condition is satisfied for the Bérenger splitting of general systems with one discontinuous absorption coefficient. To our knowledge this is the first proof that the Bérenger split transmission problem with discontinuous σ(x 1 ) is well posed.

We give necessary and sufficient conditions for perfection at a planar interface. In §4.1.4 we verity that the condition is satisfied for the Bérenger splitting.

In §4.1.5 we prove that in the case of Maxwell's equations (and not in general) the perfection criterion follows by analytic continuation from the plane wave identities established by Bérenger.

In §4.2 we prove using the Fourier-Laplace method that Bérenger's method with one coefficient σ 1 (x 1 ) ∈ Lip(R x1 ) is well posed and perfectly matched. In our use of the Fourier-Laplace method, including this one, a central role is played by the Seidenberg-Tarski Theorem estimating the asymptotic behavior of functions defined by real polynomial equations and inequalities. The Fourier-Laplace method is limited to coefficients that depend only on x 1 .

Our third method of analysis is to study the behavior of short wavelength asymptotic solutions. For such solutions we examine in §5 the decay in the absorbing layers, and reflections at discontinuities of σ j (x j ) or its derivatives when smoother transitions are used. For problems other than Maxwell, Hu [START_REF] Hu | On absorbing boundary conditions of linearized Euler equations by a perfectly matched layer[END_REF] and Bécache, Fauqueux , and Joly [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF] have already shown that the supposedly absorbing layers may in fact lead to growth. The study of short wavelength solutions in the layer yields precise and clear criteria, also valid for variable coefficients, explaining the phenomenon.

The analysis of the reflection of short wavelength wave packets at the interface with the layer also leads us to propose in §6, a new absorbing layer strategy which we call Harmoniously Matched Layers. The method starts with a smart layer for a symmetric hyperbolic system. Then for wavelength ε asymptotic solutions of amplitude O(1) and discontinuous σ, the leading order reflected wave at nonnormal incidence typically has amplitude proportional to σε. Therefore an extrapolation using computations with two values of σ eliminates the reflections proportional to σ. This yields a method with leading order reflection O(ε 2 ) at all angles of incidence. The resulting method inherits the simple L 2 estimates of the symmetric systems. More generally if the first discontinuous derivative of the absorption coefficient is the k th then the reflection is O([D k σ]ε k+1 ) and the same extrapolation removes the leading order reflection. In §6. [START_REF] Bardos | Maximal positive boundary value problems as limits of singular perturbation problems[END_REF] we investigate several implementations of this idea and show that the method with cubic σ is competitive with that of Bérenger with the same σ. On short wavelengths or random data it performs better than the Bérenger method. On long wavelengths Bérenger performs better.

Though we provide satisfactory answers to a wide range of questions about absorbing layers, there is a notable gap.

Open problem. For the original strategy of Bérenger for Maxwell's equations with discontinuous absorptions in more than one direction we do not know if the resulting problem is well posed. Discussion. 1. In §3.6 we prove well posedness and perfection for a closely related method. 2. In practice discontinuous σ have been abandoned, but it is striking that this problem remains open. 3. Once well posedness is proved, perfection follows by the proof in §3.5.

Well posed first order Cauchy problems

Basic definitions

Consider a first order system of partial differential equations for C N valued functions on R 1+d ,

L(x, ∂ t , ∂ x ) U := ∂ t U + d l=1 A l ∂ l U + B(x) U = 0.
(2.1)

The principal part of L, denoted L 1 ,

L 1 (∂ t , ∂ x ) := ∂ t + d l=1 A l ∂ l ,
has constant matrix coefficients A l . In the Bérenger strategy, the operators L are the centerpieces and they differ from L. It is for this reason that we introduce L that can be L or L.

Definition 2.1. The characteristic variety Char(L) ⊂ C 1+d \ {0} of L is the set of (τ, ξ) such that det L 1 (τ, ξ) = 0. Definition 2.2. The smooth variety hypothesis is satisfied at (τ , ξ) ∈ Char(L) if there is a conic neighborhood Ω of ξ ∈ R d \ {0} and a C ∞ function ξ → τ (ξ)
on Ω so that on a neighbourhood of (τ , ξ), the characteristic variety has equation τ = τ (ξ). At such a point the associated group velocity is defined to be v := -∇ ξ τ (ξ).

Example 2.1. This hypothesis holds if an only if for ξ near ξ the spectrum of L(0, ξ) near -τ consists of a single point with multiplicity independent of ξ. For the polynomial (τ + ξ 1 )(τ 2 -|ξ| 2 ) with d > 1 the hypothesis fails at and only at τ + ξ 1 = 0 where two sheets of the variety are tangent. Replacing the first factor by τ + c ξ 1 with c > 1 the hypothesis fails where the two sheets cross transversally. For 0 ≤ c < 1 the hypothesis holds everywhere.

Contents 7

The Cauchy problem for L is to find a solution U defined on [0, ∞[×R d satisfying (2.1) with prescribed initial data U (0, •).

Definition 2.3. The Cauchy problem for L is weakly well posed if there exist q > 0, K > 0, and α ∈ R so that for any initial values in H q (R d ), there is a unique solution

U ∈ C 0 ([0, +∞[ ; L 2 (R d )) with ∀t ≥ 0, U (t, •) L 2 (R d ) ≤ Ke αt U (0, •) H q (R d ) . (2.2)
When the conclusion holds with q = 0, the Cauchy problem is called strongly well posed.

Theorem 2.1.

(i) The Cauchy problem for L 1 is weakly well posed if and only if for each ξ ∈ R d , the eigenvalues of L 1 (0, ξ) are real. (ii) The Cauchy problem for L 1 is strongly well posed if and only if for each ξ ∈ R d , the eigenvalues of L 1 (0, ξ) are real and L 1 (0, ξ) is uniformly diagonalisable, there is an invertible S(ξ) satisfying,

S(ξ) -1 L 1 (0, ξ) S(ξ) = diagonal, S , S -1 ∈ L ∞ (R d ξ ) . (iii) If B has
constant coefficients, then the Cauchy problem for L is weakly well posed if and only if there existes M ≥ 0 such that for any

ξ ∈ R d , det L(τ, ξ) = 0 =⇒ |Im τ | ≤ M . Remark 2.1.
1. The algebraic conditions in (i) and (iii) express weak hyperbolicity, in the sense of Gårding. The necessity of uniform diagonalisability in (ii) expressing strong hyperbolicity is due to Kreiss [START_REF] Chazarain | Introduction à la théorie des équations aux dérivées partielles linéaires[END_REF], [START_REF] Kreiss | Initial-Boundary Value Problems and the Navier-Stokes Equations[END_REF].

2. An application of Grönwall's inequality shows that if L 1 satisfies the condition of Theorem 2.1, (ii), then for all B(x) ∈ L ∞ (R d ; Hom(C N )), the Cauchy problem for L 1 + B is strongly well posed.

3.

By property (ii), if L is strongly hyperbolic, then every eigenvalue -τ of L 1 (0, ξ) is semi-simple. Equivalently, for any (τ, ξ) ∈ Char(L) the eigenvalue 0 of L 1 (τ, ξ) is semi-simple, i.e. its geometric multiplicity is equal to its algebraic multiplicity. It is equivalent to saying that Ker

L 1 (τ, ξ) = Ker (L 1 (τ, ξ)) 2 , or that C N = Ker L 1 (τ, ξ) ⊕ Range L 1 (τ, ξ).

Characteristic variety and projectors for Bérenger's L

To study the Cauchy problem for Bérenger's split operators L one starts with a study of the characteristic variety. The coefficients of Bérenger's operator L are the dN × dN matrices,

A l :=         0 . . . . . . . . . 0 . . . . . . A l . . . . . . . . . A l . . . . . . 0 . . . . . . . . . 0         , B(x) :=    σ 1 (x 1 )I N . . . 0 . . . . . . . . . 0 . . . σ d (x d )I N    . (2.3)
The principal symbol of L is

L 1 (τ, ξ) =      ξ 1 A 1 + τ I N ξ 1 A 1 . . . ξ 1 A 1 ξ 2 A 2 ξ 2 A 2 + τ I N . . . ξ 2 A 2 . . . . . . . . . . . . ξ d A d ξ d A d . . . ξ d A d + τ I N      . Theorem 2.2. (i) The characteristic polynomial of L is det L 1 (τ, ξ) = τ N (d-1) det L(τ, ξ) . (2.4)
The polynomial associated to the full symbol including the absorption is

det L(τ, ξ) = det L d j=1 (τ + σ j ) , ξ 1 j =1 (τ + σ j ) , ξ 2 j =2 (τ + σ j ) , . . . , ξ d j =d (τ + σ j ) . (2.5)
If (τ, ξ) ∈ Char L with τ = 0, the following properties hold.

(ii) The mapping

S : Φ = (Φ 1 , • • • , Φ d ) → - d j=1 Φ j
is a linear bijection from Ker L 1 (τ, ξ) onto Ker L 1 (τ, ξ). Its inverse is given by

Φ → ξ 1 τ A 1 Φ , . . . , ξ d τ A d Φ .
(iii) The kernel of the adjoint

L 1 (τ, ξ) * is equal to the set of vectors Φ = (Φ, • • • , Φ) such that Φ ∈ Ker L 1 (τ, ξ) * . The range of L 1 (τ, ξ) is equal to the set of vectors Ψ = Ψ 1 , • • • , Ψ d such that ( d j=1 Ψ j , Φ) = 0 for all Φ ∈ Ker L 1 (τ, ξ) * . (iv) If moreover the eigenvalue 0 of L 1 (τ, ξ) with τ = 0 is semi-simple, the eigenvalue 0 of L 1 (τ, ξ) is semi-simple. Equivalently, Ker L 1 (τ, ξ) ⊕ Range L 1 (τ, ξ) = C dN .
Proof. (i) Adding the sum of the other rows to the first row in the determinant of

L 1 (τ, ξ) yields, det L 1 (τ, ξ) = L(τ, ξ) . . . . . . L(τ, ξ) ξ 2 A 2 ξ 2 A 2 + τ I N . . . ξ 2 A 2 . . . . . . . . . ξ d A d . . . . . . ξ d A d + τ I N
Substracting the first column from the others yields,

det L 1 (τ, ξ) = L(τ, ξ) 0 . . . 0 ξ 2 A 2 τ I N . . . 0 . . . . . . . . . ξ d A d 0 . . . τ I N
The first result follows. For the second write, det L(τ, ξ) =

ξ 1 A 1 + (τ + σ 1 )I N ξ 1 A 1 . . . ξ 1 A 1 ξ 2 A 2 ξ 2 A 2 + (τ + σ 2 )I N . . . ξ 2 A 2 . . . . . . . . . ξ d A d . . . . . . ξ d A d + (τ + σ d )I N For each i divide the i th row by τ + σ i to find, det L(τ, ξ) = d j=1 (τ + σ j ) N det L 1 (1, ξ 1 τ + ξ 1 , • • • , ξ d τ + ξ d ) .
By formula (2.4)

this implies det L(τ, ξ) = d j=1 (τ + σ j ) N det L(1, ξ 1 τ + ξ 1 , • • • , ξ d τ + ξ d ),
which is equivalent to (2.5). (ii) Suppose that 0

= Φ = (Φ 1 , • • • , Φ d ) ∈ Ker L 1 (τ, ξ). Then, for any l, τ Φ l + ξ l A l d j=1 Φ j = 0, (2.6)
Add to find

L 1 (τ, ξ) d j=1 Φ j = 0.
Therefore the map Φ →j Φ j maps Ker L 1 (τ, ξ) to Ker L 1 (τ, ξ).

If d j=1 Φ j = 0, equation (2.6
) implies that all the Φ j vanish since τ = 0. Therefore the mapping is injective.

Let Φ ∈ Ker L 1 (0, ξ). Define

Φ j = ξ j τ A j Φ . (2.7) This defines an element Φ = (Φ 1 , • • • , Φ d ) in Ker L 1 (0, ξ) with S Φ = Φ
, so the mapping is surjective with inverse given by (2.7).

(iii) Since L 1 (τ, ξ) * Φ = (L 1 (τ, ξ) * Φ, . . . , L 1 (τ, ξ) * Φ) it follows that the set of Φ is included in the kernel. Since the matrices are square, Ker L 1 (τ, ξ) and Ker L 1 (τ, ξ) * have the same dimension. The set of Φ has dimension equal to this common dimension which by (ii) is equal to the dimension of Ker L 1 (τ, ξ) proving that they exhaust the kernel. The last property follows directly from the fact that Range L 1 (τ, ξ) is the orthogonal of Ker L 1 (τ, ξ) * .

(iv) It suffices to show that the intersection of these spaces consists of the zero vector. Equivalently, it suffices to show that there is no Φ = 0 in Ker L 1 (τ, ξ) such that

∀Ψ ∈ Ker L 1 (τ, ξ) * , ( d j=1 ξ j τ A j Φ, Ψ) = 0 .
The quantity above is equal to -(Φ, Ψ), and Φ would belong to (Ker L 1 (τ, ξ) * ) ⊥ = Range L 1 (τ, ξ). Since τ = 0 and Ker L 1 (τ, ξ) ∩ Range L 1 (τ, ξ) = 0, this would imply that Φ = 0, leading to a contradiction.

Denote by Π L (τ, ξ) (resp. Π L (τ, ξ)) the spectral projector onto the kernel of L 1 (τ, ξ) (resp. L 1 (τ, ξ)) along its range. For L it is given by

Π L (τ, ξ) = 1 2πi |z|=ρ z I -L 1 (τ, ξ) -1 dz
with ρ small. Like the characteristic variety, Π L depends only on the principal symbol L 1 . It is characterized by,

Π 2 L = Π L , Π L L 1 (τ, ξ) = 0, L 1 (τ, ξ) Π L = 0, rank Π L = dim Ker L 1 (τ, ξ), (2.8)
where the τ, ξ dependence of Π L is suppressed for ease of reading. The first three conditions assert that Π L (τ, ξ) is a projector annihilating Range L 1 (τ, ξ) and projecting onto a subspace of Ker L 1 (τ, ξ). That it maps onto the kernel is implied by the last equality.

Proposition 2.1. The matrix Π L (τ, ξ) is given by

Π L (τ, ξ) = -       ξ 1 A 1 τ Π L (τ, ξ) . . . ξ 1 A 1 τ Π L (τ, ξ) . . . . . . ξ d A d τ Π L (τ, ξ) . . . ξ d A d τ Π L (τ, ξ)       .
Proof. Call the matrix on the right M (τ, ξ). The properties of the projectors associated to L yield formulas for the (i, j) block of the products

(M (τ, ξ) L 1 (τ, ξ)) i,j = - ξ i A i τ Π L L 1 = 0, ( L 1 (τ, ξ)M (τ, ξ)) i,j = - ξ i A i τ L 1 Π L L 1 = 0, and, (M (τ, ξ)M (τ, ξ)) i,j = ξ i A i τ 2 Π L (L 1 -τ I)Π L = - ξ i A i τ Π 2 L = (M (τ, ξ)) i,j .
This proves the first three equalities of (2.8). Since M projects onto a subspace of Ker L 1 , rank M ≤ dim Ker L 1 . Apply M to a vector (Ψ, 0, . . . , 0) and compare with part (ii) of Theorem 2.2 to see that the range of M contains Ker L 1 (τ, ξ) so rank M ≥ dim Ker L 1 . This proves the last equality of (2.8).

Remark 2.2.

1. The characteristic varieties of L and L are identical in τ = 0. 2. In particular, the smooth variety hypothesis is satisfied at (τ, ξ) with τ = 0 for one if and only if it holds for both, and the varieties have the same equations and the same group velocities.

3. When the smooth variety hypothesis is satisfied, the spectral projection Π L (τ (ξ), ξ) is analytic in ξ, hence of constant rank. It follows that 0 is a semi-simple eigenvalue of L(τ (ξ), ξ) on a conic neighborhood of ξ.

If the eigenvalue 0 of L 1 (τ, ξ) is semi-simple, the kernel and the range of L 1 (τ, ξ) are complementary subspaces as mentioned in Remark 2.1 3., and the partial inverse

Q L (τ, ξ) of L 1 (τ, ξ) is uniquely determined by Q L (τ, ξ) Π L (τ, ξ) = 0, Q L (τ, ξ) L 1 (τ, ξ) = I -Π L (τ, ξ) . (2.9)
The partial inverse Q L(τ, ξ) is defined in the same way from L1 (τ, ξ).

The Cauchy problem for Bérenger's split operators

Part (i) of Theorem 2.2 proves the following.

Corollary 2.1. If the Cauchy problem for L is weakly well posed, then so is the Cauchy problem for the principal part L 1 .

An important observation is that though the Cauchy problem for L 1 is at least weakly well posed, the root τ = 0 is for all ξ a multiple root. When there are such multiple roots it is possible that order zero perturbations of L1 may lead to ill posed Cauchy problems. The next example shows that this phenomenon occurs for the Bérenger split operators with constant absorption σ j . Theorem 2.4 shows that when τ = 0 is a root of constant multiplicity of det L 1 (τ, ξ) = 0, the constant coefficient Bérenger operators have well posed Cauchy problems. Cases where the problem are strongly well posed are identified. In the latter cases, arbitrary bounded zero order perturbations do not destroy the strong well posedness.

Example 2.2. 1. For L := ∂ t + ∂ 1 + ∂ 2 , det L(τ, ξ) = τ + ξ 1 + ξ 2 . Therefore τ = 0 is a root if and only if ξ 1 + ξ 2 = 0. The doubled system with absorption σ = 1 in x 1 is L := ∂ t + 1 1 0 0 ∂ 1 + 0 0 1 1 ∂ 2 + 1 0 0 0 , and, det L(τ, ξ, -ξ) = det τ + ξ + 1 ξ -ξ τ -ξ = (τ + ξ + 1)(τ -ξ) + ξ 2 = τ 2 + τ -ξ .
The roots of det L(τ, ξ, -ξ) = 0 are τ = (-1 ± √ 1 + 4ξ)/2. Taking ξ → -∞ shows that the Cauchy problem for L is not weakly well posed by Theorem 2.1, (iii).

2. More generally if τ + ξ 1 + ξ 2 is a factor of det L 1 (τ, ξ) then for σ = 0 the operator L is not even weakly hyperbolic. In this case (2.5) implies that τ 2 + τξ is a factor of det L(τ, ξ, -ξ).

This generalizes to linear hyperbolic factors in arbitrary dimension.

A key tool is the following special case of Theorem A.2.5 in [START_REF] Hörmander | The analysis of linear partial differential operators. II. Differential operators with constant coefficients[END_REF]. Then either µ(ρ) = +∞ for ρ large, or there are a ∈ Q and A = 0 so that

µ(ρ) = A ρ a 1 + o(1) , ρ → ∞ .
Theorem 2.4. Suppose that τ = 0 is an isolated root of constant multiplicity m of det L 1 (τ, ξ) = 0.

(i) If the Cauchy problem for L 1 is strongly well posed, then for arbitrary constant absorptions σ j ∈ C, the Cauchy problem for L 1 + B is weakly well posed. (ii) If the Cauchy problem for L 1 is strongly well posed, and if there is a ξ = 0 such that Ker L(0, ξ) = ∩ ξj =0

Ker A j , then L 1 (0, ξ) is not diagonalizable. Therefore the Cauchy problem for L is not strongly well posed. (iii) If the Cauchy problem for L is strongly well posed and for all ξ, Ker L 1 (0, ξ) = ∩ ξj =0 Ker A j , then the Cauchy problem for L is strongly well posed. This condition holds if L 1 (0, ∂ x ) is elliptic, that is det L 1 (0, ξ) = 0 for all real ξ. Remark 2.3. 1. Part (i) is a generalisation of results in [START_REF] Hu | A stable, perfectly matched layer for linearized Euler equations in unsplit physical variables[END_REF] and Theorem 1 in [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF]. In the latter paper, Bécache et al. treated the case N = 2 assuming that the nonzero eigenvalues of L 1 (0, ξ) are of multiplicity one. They conjectured that the result was true more generally. Like them we treat the roots near zero differently from those that are far from zero. The treatment of each of these cases is different from theirs. The tricky part is the roots near zero. We replace their use of Puiseux series by the related Seidenberg-Tarski Theorem 2.3. [START_REF] Abarbanel | A mathematical analysis of the PML method[END_REF] proved (ii) in the special case of Maxwell's equations. The general argument below is simpler and yields a necessary and sufficient condition for loss of derivatives when the eigenvalue 0 of L(0, ξ) is of constant multiplicity. 3. Part (iii) is new, extending a result in the thesis of S. Petit-Bergez [START_REF] Petit-Bergez | Problèmes faiblement bien posés : discrétisation et applications[END_REF].

Arbarbanel and Gottlieb

Proof. (i) For ξ ∈ R d \ 0, define for ρ ∈ R + , E(ρ) := max Im (τ ) : det L(τ, ξ) = 0 , ξ ∈ R d , |ξ| 2 = ρ 2 .
Apply the Seidenberg-Tarski Theorem 2.3 with real variables ρ, ζ = (Re τ, Im τ, ξ) and polynomials

R(ρ, ζ) = | det L(τ, ξ)| 2 + (|ξ| 2 -ρ 2 ) 2 , S = 0 and Q(ρ, ζ) = Im τ .
Conclude that there is an α = 0 and a rational r so that

E(ρ) = α ρ r 1 + o(1) , ρ → ∞ .
To prove the result it suffices to prove that Im τ is bounded, i.e. to show that r ≤ 0. Suppose on the contrary that r > 0.

Given τ, ξ define k ∈ S d-1 , ρ ∈ R + , and, θ by

k := ξ |ξ| , ξ = ρ k, θ := τ ρ .
Choose sequences τ (n), and ξ(n) so that for n → ∞,

det L(τ (n), ξ(n)) = 0, Im τ (n) = α ρ(n) r (1 + o(1)) . (2.10) Write L(τ, ξ) = L 1 (τ, ξ) + B = ρ L 1 (θ, k) + 1 ρ B = ρ θI 2N ×2N + L 1 (0, k) + 1 ρ B .
The matrix L(τ, ξ) is singular if and only if -θ is an eigenvalue of

L 1 (0, k) + ρ -1 B.
For large ρ this is a small perturbation of L 1 (0, k). Choose µ > 0 so that for |k| = 1, the only eigenvalue of L 1 (0, k) in the disk |θ| ≤ 2µ is θ = 0.

Because of the strong well posedness of L, there is a uniformly independent basis of unit eigenvectors for the eigenvalues of L 1 (0, k) in |θ| ≥ µ. By part (iv) of Theorem 2.2 there is a uniformly independent basis of unit eigenvectors for the eigenvalues of L 1 (0, k) in |θ| ≥ µ.

It follows that there is a C 0 so that for ρ > C 0 the eigenvalues of L 1 (0, k) + ρ -1 B in |θ| > µ differ from the corresponding eigenvalues of L 1 (0, k) by no more than C 0 /ρ. In particular their imaginary parts are no larger than C 0 /ρ. Therefore, the corresponding eigenvalues τ = ρ θ have bounded imaginary parts. Thus for n large, E(ρ(n)) can be reached only for the eigenvalues -θ(n) which are perturbations of the eigenvalue 0 of L 1 (0, k(n)).

Perturbation by O(1/ρ) of the uniformly bounded family of dN × dN matrices, L 1 (0, k), can move the eigenvalues by no more than O(ρ -1 dN ). Since the unperturbed eigenvalue is 0,

|θ(n)| ≤ C ρ(n) -1/dN , so |τ (n)| ≤ Cρ(n) 1-1 dN , Im τ (n) = α ρ(n) r (1 + o(1)) , α = 0 . Therefore r ≤ 1 -1/dN < 1 and d j=1 (τ (n) + σ j ) = τ (n) d (1 + o(1)), ξ ℓ (n) j =ℓ (τ (n) + σ j ) = ξ ℓ (n) τ (n) d-1 (1 + o(1)) .
Insert in identity (2.5) to find

det L 1 τ (n) d (1 + o(1)) , ξ(n) τ (n) d-1 (1 + o(1)) = 0.
Divide the argument by ρ(n) τ (n) d-1 and use homogeneity to find

det L 1 τ (n) ρ(n) 1 + o(1) , k(n)(1 + o(1)) = 0.
The constant multiplicity hypothesis shows that

det L 1 (τ, ξ) = τ m F 1 (τ, ξ) , and ∀ξ ∈ R d , F 1 (0, ξ) = 0. (2.11) Since for n large (τ (n)/ρ(n))(1 + o(1)) = 0 we have F 1 τ (n) ρ(n) 1 + o(1) , k(n)(1 + o(1)) = 0.
Passing to a subsequence we may suppose that the bounded sequence k(n) → k. In addition, τ (n)/ρ(n) → 0 so passing to the limit yields F 1 0, k) = 0 contradicting (2.11). This contradiction proves (i).

Contents 13

(ii) Part (i) of Theorem 2.2 shows that 0 is an eigenvalue of L 1 (0, ξ) with algebraic multiplicity equal to N (d -1) + m. It remains to see that with the assumption, the dimension of Ker L 1 (0, ξ) is strictly smaller than N (d -1) + m. By definition

Ker L 1 (0, ξ) = Φ = (Φ 1 , • • • , Φ d ) : d j=1 Φ j ∈ ∩ p Ker (ξ p A p ) .
Define

E 1 := Φ = Φ 1 , • • • , Φ d : d j=1 Φ j = 0 . Then E 1 ⊂ Ker L 1 (0, ξ) and dim E 1 = N (d -1)
. Define

E 2 := Ker L 1 (0, ξ) ⊗ O d-1 , dim E 2 = m. If Φ ∈ Ker L 1 (0, ξ), d j=1 Φ j ∈ Ker L 1 (0, ξ) , write Φ = Φ j , 0 , . . . , 0 -W, Φ j , 0 , . . . , 0 ∈ E 2 , W ∈ E 1 . Thus, Ker L 1 (0, ξ) ⊂ E 1 ⊕ E 2 . Pick V in Ker L 1 (0, ξ), but not in ∩ ξj =0 Ker A j . Then V = (V, 0, • • • , 0) ∈ E 2 and V / ∈ Ker L 1 (0, ξ) .
This proves that Ker L 1 (0, ξ) is a proper subset of

E 1 ⊕ E 2 , so dim Ker L 1 (0, ξ) < dim E 1 + dim E 2 = N (d -1) + m .
Thus the geometric multiplicity of the eigenvalue 0 is strictly less than its algebraic multiplicity. Therefore, L 1 (0, ξ) is not diagonalizable. This proves (ii).

(iii) To prove that the split problem is strongly well posed it suffices to consider the principal part. Suppose L(0, ξ) is uniformly diagonalisable on a conic neighbourhood of ξ ∈ R N \0.

For U = (U 1 , • • • , U d ), introduce V = (V 1 , • • • , V d ) with V 1 := d j=1
U j , and V l := U l for 2 ≤ l ≤ d .

(2.12)

Then L1 (∂ t , ∂ x ) U = 0 ⇐⇒ ∂ t V + Q(∂ x ) V = 0, with, Q(ξ) :=      L 1 (0, ξ) 0 . . . 0 ξ 2 A 2 0 . . . 0 . . . . . . . . . . . . ξ d A d 0 . . . 0      .
The eigenvalues of Q(ξ) are those of L 1 (0, ξ), therefore real. It suffices to diagonalize uniformly Q(ξ) on the conic neighbourhood of ξ. By homogeneity it suffices to consider ξ with |ξ| = 1. By hypothesis there exist a real diagonal matrix D(ξ) and an invertible matrix S(ξ) so that

L 1 (0, ξ) = S(ξ)D(ξ)S -1 (ξ) , and, ∃K > 0, ∀ξ ∈ R d , S(ξ) + S -1 (ξ) ≤ K.
Seek a diagonalization of Q(ξ) on |ξ| = 1 in the form,

S(ξ) =      S(ξ) 0 . . . 0 ξ 2 A 2 Q(ξ)S(ξ) Id . . . 0 . . . . . . . . . . . . ξ d A d Q(ξ)S(ξ) 0 . . . Id      so, ( S(ξ)) -1 =      (S(ξ)) -1 0 . . . 0 -ξ 2 A 2 Q(ξ) Id . . . 0 . . . . . . . . . . . . -ξ d A d Q(ξ) 0 . . . Id      (2.13)
with Q(ξ) to be determined. Then,

S -1 (ξ) Q(ξ) S(ξ) =      D(ξ) 0 . . . 0 ξ 2 A 2 (I -Q(ξ)L 1 (0, ξ))S(ξ) 0 . . . 0 . . . . . . . . . . . . ξ d A d (I -Q(ξ)L 1 (0, ξ))S(ξ) 0 . . . 0      and S -1 (ξ) Q(ξ) S(ξ) is diagonal ⇐⇒ ξ j A j (I -Q(ξ)L 1 (0, ξ)) = 0, 2 ≤ j ≤ d . (2.14)
From the strong well posedness of L 1 it follows that uniformly in ξ one has Ker L 1 (0, ξ) ⊕ Range L 1 (0, ξ) = C N . Choose Q equal to the left inverse of L 1 (0, ξ) defined in (2.9). Since Ker L 1 (0, ξ) = ∩ Ker ξ j A j , the condition on the right in (2.14) holds so S(ξ) diagonalizes Q(ξ). Since S(ξ) and S(ξ) -1 are bounded on a conic neighborhood, it follows that S(ξ) and S(ξ) -1 are bounded on a neighborhood of ξ in |ξ| = 1. A finite cover of the sphere, completes the proof.

Remark 2.4. Denote by S(ξ) the function homogeneous of degree zero given by (2.13)

for |ξ| = 1 with Q constructed in the proof. Then S(D) V (t) L 2 (R d )) with V from (2.12) is a norm equivalent to U (t) L 2 (R d )
and is conserved for solutions of L 1 (∂) U = 0. Those solutions yield a unitary group with respect to the norm S(D) V L 2 (R d ) .

Analysis of the Bérenger's PML by energy methods

This section contains results proving that the initial value problems so defined are well posed. We begin with the case of Gevrey absorptions, then W 2,∞ , and finally the case of the Heaviside function.

In Section 3.1 we prove that when L is only weakly well posed, Gevrey regular σ l lead to well posed initial value problems in Gevrey classes. Commonly used σ are not this smooth.

The strongest result, from Section 3.2, applies when L(0, ∂ x ) is elliptic. Important cases are the wave equation and linearized elasticity. In these cases the operator L 1 is strongly hyperbolic so remains strongly hyperbolic even with general bounded zeroth order perturbations. Thus for bounded σ l (x l ) the initial value problem is strongly well posed.

In Section 3.3 we analyse the case of L associated to Maxwell's equations with finitely smooth σ. We follow the lead of [START_REF] Petit-Bergez | Problèmes faiblement bien posés : discrétisation et applications[END_REF] and extend the analysis of [START_REF]Vacus. Caractère bien posé du problème de Cauchy pour le système de Bérenger[END_REF] to several absorptions σ l and to higher dimensions. Related estimates for the linearized Euler equation have been studied by L. Métivier [START_REF] Métivier | Utilisation des équations Euler-PML en milieu hétérogène borné pour la résolution d'un problème inverse en géophysique[END_REF].

The results of this section do not treat the case of Bérenger's method for Maxwell's equations with discontinuous σ j . The case of one absorption is treated in §4. A closely related method is treated by an energy method in §3.6.

General operators and Gevrey absorption

The next result is implied by Bronstein's Theorem [6] [7] [23] [START_REF] Nishitani | sur les équations hyperboliques à coefficients hölderiens en t et de classe Gevrey en x[END_REF]. It shows that when L 1 (0, ξ) has only real eigenvalues and the σ j belong to the appropriate Gevrey class, then the Cauchy problem for L is solvable for Gevrey data.

Definition 3.1. For 1 ≤ s < ∞, f ∈ S ′ (R d ) belongs to the Gevrey class G s (R d ) when ∃C, M, ∀α ∈ N d , ∂ α f L 2 (R d ) ≤ M α! C |α| . Then G s ⊂ ∩ σ H σ (R d ) ⊂ C ∞ (R d ). For s > 1 the compactly supported elements of G s are dense. If | f (ξ)| ≤ C e -|ξ| a with 0 < a < 1, then u ∈ G 1/a . Theorem 3.1. If the principal part L 1 is weakly hyperbolic, and σ j ∈ G N/N +1 (R d ) then for arbitrary f ∈ G N/(N +1) (R d ) there is one and only one solution u ∈ C ∞ (R 1+d ) to L u = 0, u(0, •) = f .
The solution depends continuously on f .

Strong hyperbolicity when L(0, ∂) is elliptic

Theorem 3.2. If L is strongly well posed and L(0, ∂) is elliptic, then L is strongly well posed for any absorption

(σ 1 (x 1 ), . . . , σ d (x d )) in (L ∞ (R)) d .
Proof. Kreiss' theorem 2.1 asserts that an operator with constant coefficient principal part is uniformly well posed if and only if the principal part is uniformly diagonalisable on a conic neighborhood of each ξ = 0. Therefore the Corollary follows from the third part of Theorem 2.4.

Example 3.1. This result implies that the PML model for the elastodynamic system is strongly well posed. The system is written in the velocity-stress (v, Σ) formulation,

ρ ∂ t v -div Σ = 0 , ∂ t Σ -Cε(v) = 0 , ε ij (v) := (∂ i v j + ∂ j v i ) ,
with positive definite elasticity tensor C and Σ := Cε. See [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF], where the authors showed that such layers may be amplifying (see Section 5).

The method of Métral-Vacus extended to the 3d PML Maxwell system

Métral and Vacus proved in [START_REF]Vacus. Caractère bien posé du problème de Cauchy pour le système de Bérenger[END_REF] a stability estimate for Bérenger's two dimensional PML Maxwell system with one absorption

σ 1 (x 1 ) ∈ W 1,∞ (R) and x = (x 1 , x 2 ) ∈ R 2 .
There are two crucial elements in their method. First following Bérenger, they do not split all variables in all directions. This section begins by showing that the partially split model is equivalent to the fully split model restricted to functions U some of whose components vanish. The L evolution leaves this space invariant and its evolution on that subspace determines its behavior everywhere.

The second element is that on the partially split subspace there is an a priori estimate bounding the norm at time t by the same norm at time 0. This looks inconsistent with the fact that the Cauchy problem is only weakly well posed. However the norm is not homogeneous. Certain linear combinations of components have more derivatives estimated than others. The observation of [START_REF]Vacus. Caractère bien posé du problème de Cauchy pour le système de Bérenger[END_REF] is that the system satisfied by the fields and certain combinations of the fields and their derivatives, yields a large but symmetrizable first order system. These estimates have been obtained, and extended in Sabrina Petit's thesis [START_REF] Petit-Bergez | Problèmes faiblement bien posés : discrétisation et applications[END_REF] in the 2d case with two coefficients, and in the 3d case for an absorption in only one direction.

In this section, motivated in part by the clarification of the role of symmetrizers in the work of L. Métivier [START_REF] Métivier | Utilisation des équations Euler-PML en milieu hétérogène borné pour la résolution d'un problème inverse en géophysique[END_REF] for the 2d variable coefficient Euler equations in geophysics, we construct analogous more elaborate functionals which suffice for the general case of three absorptions in three dimensions. They require σ j ∈ W 2,∞ (R).

Maxwell's equations for ∂ t E 1 and ∂ t B 1 contain only partial derivatives with respect to x 2 , x 3 and not x 1 . In such a situation Bérenger splits the corresponding equations in directions x 2 , x 3 but not in direction x 1 . To see why this is a special case of the general splitting algorithm (1.5) reason as follows. If the equation for ∂ t U j from L does not contain any terms in ∂ k , that is the j th row of A k vanishes, then the equation for the j th component of the unknown U k corresponding to the splitting for the k th space variable is,

∂ t U k j + σ k (x k ) U k j = 0 , U k j = e -σ k (x k )t U k j (0, x). (3.1)
Plugging this into the other equations reduces the number of unknowns by one. The simplest strategy is to take initial data U k j (0, x) = 0 which yields the operator L restricted to the invariant subspace of functions so that U k j = 0. Conversely if one knows how to solve that restricted system then the full system can be reduced to the restricted system with an extra source term from (3.1).

Summary.

To study the fully split system it is sufficient to study the system restricted to {U k j = 0}. Performing this reduction for each missing spatial derivative, corresponds to splitting equations only along directions containing the corresponding spatial derivatives.

An extreme case of this reduction occurs if an equation contains no spatial derivatives, that unknown is eliminated entirely. For the Maxwell system which is the subject of this section this does not occur. The use of unsplit variables

• reduces the size of U reducing computational cost, • corresponds to Bérenger's original algorithm,

• is important for the method of Métral-Vacus which takes advantage of the vanishing components U k k . Consider the 3d Maxwell equations,

∂ t E -∇ × H = 0, ∂ t H + ∇ × E = 0.
Defining U = E + iH, they take the symmetric hyperbolic form (1.1) with hermitian matrices

A 1 =   0 0 0 0 0 -i 0 i 0   , A 2 =   0 0 i 0 0 0 -i 0 0   and A 3 =   0 -i 0 i 0 0 0 0 0   . (3.2)
Introduce the splitting (1.5) with some components unsplit. Define the subspace H with vanishing components corresponding to the unsplit components

H := U = (U 1 , U 2 , U 3 ) ∈ H 2 (R 3 ; C 3 ) 3 : U 1 1 = 0, U 2 2 = 0, U 3 3 = 0 . For U = (U 1 , U 2 , U 3 ) in H, define U := U 1 + U 2 + U 3 , V j := ∂ j U, V i,j := ∂ ij U, W := k σ k (x k )U k , W j := ∂ j W, Z := k ∂ k (W k + σ k (x k )U k ), Z j := ∂ j Z, V := U , V i , V i,j , W j , U j , W , Z j ∈ C 54 . (3.3)
The function Z and therefore Z j are C valued. The other slots in V are C 3 valued. The second derivatives

V i,j of U are ordered as V 1,1 , V 2,1 , V 3,1 , V 2,2 , V 3,2 , V 3,3
. This convention is important when the equations for V are written in matrix form. Computing in turn W j , Z, Z j requires two derivatives of σ j . The unknown in (1.5) is U = (U 1 , U 2 , U 3 ). The U j appear in the fifth slot of V. Therefore,

V(t, •) (L 2 (R 3 )) 54 ≥ U (t, •) (L 2 (R 3 )) 9 .
For the Cauchy problem the initial data is U 0 = (U 1 0 , U 2 0 , U 3 0 ), from which V 0 is deduced by the derivations above, and

V 0 (L 2 (R 3 )) 54 ≤ C U 0 (H 2 (R 3 )) 9 . Theorem 3.3. If σ j , for j = 1, 2, 3, belong to W 2,∞ (R), then for any U 0 = (U 1 0 , U 2 0 , U 3 0 ) in H there is a unique solution U in L 2 (0, T ; H) of the split Cauchy problem (1.5) with initial value U 0 . Furthermore there is a C 1 > 0 independent of U 0 so that for all positive time t, U (t, •) (L 2 (R 3 )) 9 ≤ C 1 e C1 t U 0 (H 2 (R 3 )) 9 .
(3.4)

Proof. The main step is to derive a system of equations satisfied by V(t, x) together with a symmetrizer S(D). These imply an estimate for t ≥ 0,

V(t) L 2 (R 3 ) ≤ C 2 e C2t V(0) L 2 (R 3 ) . (3.5)
From this estimate it easily follows that the Cauchy problem for the V-equations is uniquely solvable. It is true but not immediate that if the initial values of V are computed from those of U then the solution V comes from a solution U of the Bérenger system. The strategy has three steps,

• Discretize the Bérenger system in x only.

• Derive an estimate analogous to (3.5) for the semidiscrete problem. The estimate is uniform as the discretization parameter tends to zero. The proof is a semidiscrete analogue of (3.4).

• Solve the semidiscrete problem and pass to the limit to prove existence. This is done for the case d = 2 in [START_REF] Petit-Bergez | Problèmes faiblement bien posés : discrétisation et applications[END_REF] to which we refer for details. Uniqueness of the solutions to the V-system and therefore U is simpler and classical and is also in [START_REF] Petit-Bergez | Problèmes faiblement bien posés : discrétisation et applications[END_REF]. Equation (1.5) yields,

∂ t U j + k A k V j + σ j U j = 0. (3.6)
Summing on j yields

∂ t U + L(0, ∂)U + W = 0. (3.7)
Differentiate in direction x j to find,

∂ t V j + L(0, ∂)V j + W j = 0. (3.8)
Differentiate once more to get

∂ t V i,j + L(0, ∂)V i,j + ∂ i W j = 0. (3.9)
The quantity ∂ i W j on the left is replaced using the next lemma.

Lemma 3.1.

∂ j W = L(0, ∂)A j W + Ze j -k E jk (σ ′ k U + σ k V k ), (3.10)
where e j is the j th vector of the standard basis, and E ij is the 3 × 3 matrix all of whose entries vanish except the (i, j) element that is equal to 1.

Proof.

First evaluate L(0, ∂)A j W to find

L(0, ∂)A j W = k A k A j ∂ k W.
The matrices in Maxwell's equations satisfy A j A k = -E jk for j = k, and

A 2 j = k =j E kk = I -E jj . This yields L(0, ∂)A j W = - k =j E jk ∂ k W + (I -E jj )∂ j W = ∂ j W - k E jk ∂ k W = ∂ j W -div(W ) e j .
Introduce the definition of Z to find (3.10). The proof of the lemma is complete. Differentiate (3.10) in space to obtain

W j := ∂ j W = L(0, ∂)A j W + div(W )e j = L(0, ∂)A j W + Ze j -( k ∂ k (σ k U k )) e j . Compute k ∂ k (σ k U k ) e j = k E jk ∂ k (σ k U ) = k E jk σ ′ k U + k E jk σ k V k , which proves
∂ i W j = L(0, ∂)A j W i + ∂ i Ze j -k E jk ∂ i σ k U + σ ′ k V k . Inserting into (3.9) yields ∂ t V i,j + L(0, ∂)V i,j + L(0, ∂)A j W i + Z ij - k E jk ∂ i σ k U + σ ′ k V k = 0.
This is equivalent to

∂ t V i,j + L(0, ∂)V i,j + L(0, ∂)A j W i + Z i e j -σ ′ i E ji U -(σ i "E ji + k σ k E jk )V i - k σ ′ k E jk V i,k = 0. (3.11)
To close the system it remains to evaluate the time derivatives of W, W j and ∂ j Z.

∂ t W = k σ k ∂ t U k = - k σ k (A k U + σ k U k ) = - k σ k A k V k - k σ 2 k U k .
Using the particular form of the equations yields

k σ 2 k U k = ( k σ k )( k σ k U k ) -k σ k l =k σ l U l = ( k σ k )W -diag(σ 2 σ 3 , σ 1 σ 3 , σ 1 σ 2 )U -diag(σ 1 , σ 2 , σ 3 )W . Therefore ∂ t W + k σ k A k V k + ( k σ k )W -diag(σ 2 σ 3 , σ 1 σ 3 , σ 1 σ 2 )U -diag(σ 1 , σ 2 , σ 3 )W = 0 .
(3.12)

Differentiate in x i to find

∂ t W i + k (∂ i (σ k )A k V k + σ k A k V ki ) + ∂ i ( k σ k )W + k σ k W i -∂ i diag(σ 2 σ 3 , σ 1 σ 3 , σ 1 σ 2 ) U -diag(σ 2 σ 3 , σ 1 σ 3 , σ 1 σ 2 )V i -∂ i diag(σ 1 , σ 2 , σ 3 ) W -diag(σ 1 , σ 2 , σ 3 )W i = 0 (3.13)
Next compute

∂ t Z = i ∂ i ∂ t (W i + σ i U i ) .
Consider the pair of equations

∂ t U 2 1 + i∂ 2 U 3 + σ 2 U 2 1 = 0, and
∂ t U 3 1 -i∂ 3 U 2 + σ 3 U 3 1 = 0.
Add the two equations. Also add σ 2 times the first to σ 3 times the second. This yields two equations,

∂ t U 1 + i(∂ 2 U 3 -∂ 3 U 2 ) + W 1 = 0, and ∂ t W 1 + i(σ 2 ∂ 2 U 3 -σ 3 ∂ 3 U 2 ) + σ 2 2 U 2 1 + σ 2 3 U 3 1 = 0. Rewrite the last term as σ 2 2 U 2 1 + σ 2 3 U 3 1 = (σ 2 + σ 3 )W 1 -σ 2 σ 3 U 1 , to find ∂ t U 1 + i(∂ 2 U 3 -∂ 3 U 2 ) + W 1 = 0, and, ∂ t W 1 + i(σ 2 ∂ 2 U 3 -σ 3 ∂ 3 U 2 ) + (σ 2 + σ 3 )W 1 -σ 2 σ 3 U 1 = 0.
Multiply the first equation by σ 1 and add the second to obtain

∂ t (W 1 + σ 1 U 1 ) + i ((σ 1 + σ 2 )∂ 2 U 3 -(σ 1 + σ 3 )∂ 3 U 2 ) + ( k σ k )W 1 -σ 2 σ 3 U 1 = 0 .
The other indices follow by permutation. Differentiate in x k and add to find

k ∂ k ∂ t (W k + σ k U k ) + i k ∂ k ((σ k + σ k+1 )∂ k+1 U k+2 -(σ k + σ k+2 )∂ k+2 U k+1 ) + i ∂ i (( k σ k )W i ) -k ∂ k (σ k+1 σ k+2 U k ) = 0.
The terms with two spatial derivatives cancel. This leaves

∂ t Z + i k σ ′ k (∂ k+1 U k+2 -∂ k+2 U k+1 ) +( k σ k )( k W k k ) + k σ ′ k W k -k σ k+1 σ k+2 V k k = 0. Since Z = k ∂ k (W k + σ k (x k )U k ) = k (W k k + σ ′ k U k + σ k V k k ), we can replace ( k σ k )( k W k k ) in the previous equation by ( k σ k )(Z -k (σ ′ k U k + σ k V k k )) so ∂ t Z + ( k σ k )Z + i k σ ′ k (V k+1 k+2 -V k+2 k+1 ) -( k σ k )( k (σ ′ k U k + σ k V k k )) + k σ ′ k W k -k σ k+1 σ k+2 V k k = 0. Contents 19
Differentiating in x j yields

∂ t Z j + ( k σ k )Z j + σ ′ j Z + iσ j "(V j+1 j+2 -V j+2 j+1 ) + i k σ ′ k (V k+1,j k+2 -V k+2,j k+1 ) -σ ′ j ( k (σ ′ k U k + σ k V k k )) -( k σ k )(σ j "U j + σ ′ j V j j ) -( k σ k )( k (σ ′ k V j k + σ k V k j,k )) +σ ′ j W j + k σ ′ k W j k -k ∂ j (σ k+1 σ k+2 )V k k -k ∂ j (σ k+1 σ k+2 )V j,k k = 0. Replace Z by k (W k k + σ k V k k ) to end up with ∂ t Z j + ( k σ k )Z j + σ ′ j k (W k k + σ k V k k ) + iσ j "(V j+1 j+2 -V j+2 j+1 ) + i k σ ′ k (V k+1,j k+2 -V k+2,j k+1 ) -σ ′ j ( k (σ ′ k U k + σ k V k k )) -( k σ k )(σ j "U j + σ ′ j V j j ) -( k σ k )( k (σ ′ k V j k + σ k V k j,k )) + σ ′ j W j + k σ ′ k W j k -k ∂ j (σ k+1 σ k+2 )V k k -k ∂ j (σ k+1 σ k+2 )V j,k k = 0. (3.14)
Summarizing, V is solution of a first order system, ∂ t V+ P (∂ x )V+ B(x)V = 0, whose principal symbol is given by

P (∂) =            I 4 ⊗ L(0, ∂) 0 4,6 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 4,4 ⊗ 0 3,3 0 6,4 ⊗ 0 3,3 I 6 ⊗ L(0, ∂) (I 6 ⊗ L(0, ∂))M 0 6,3 ⊗ 0 3,3 0 6,4 ⊗ 0 3,3 0 3,4 ⊗ 0 3,3 0 3,6 ⊗ 0 3,3 0 3,3 ⊗ 0 3,3 0 3,3 ⊗ 0 3,3 0 3,4 ⊗ 0 3,3 0 3,4 ⊗ 0 3,3 0 3,6 ⊗ 0 3,3 0 3,3 ⊗ 0 3,3 0 3,3 ⊗ 0 3,3 0 3,4 ⊗ 0 3,3 0 4,4 ⊗ 0 3,3 0 4,6 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 4,4 ⊗ 0 3,3            .
Here the V i,j are ordered as indicated before the theorem and,

M :=          A 1 0 0 0 A 1 0 0 0 A 1 0 A 2 0 0 0 A 2 0 0 A 3          .
To symmetrize it suffices to construct a symmetrizer for the upper left hand block

Q(∂) :=     I 4 ⊗ L(0, ∂) 0 4,6 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 6,4 ⊗ 0 3,3 I 6 ⊗ L(0, ∂) (I 6 ⊗ L(0, ∂))M 0 3,4 ⊗ 0 3,3 0 3,6 ⊗ 0 3,3 0 3,3 ⊗ 0 3,3     .
We verify that

S :=   I 4 ⊗ I 3 0 4,6 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 6,4 ⊗ 0 3,3 I 6 ⊗ I 3 (I 6 ⊗ I 3 )M 0 3,4 ⊗ 0 3,3 0 3,6 ⊗ 0 3,3 I 3 ⊗ I 3   with S-1 =   I 4 ⊗ I 3 0 4,6 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 6,4 ⊗ 0 3,3 I 6 ⊗ I 3 -M (I 3 ⊗ I 3 ) 0 3,4 ⊗ 0 3,3 0 3,6 ⊗ 0 3,3 I 3 ⊗ I 3   is a symmetrizer for Q(iξ). Compute SQ S-1 =   I 4 ⊗ L(0, •) 0 4,6 ⊗ 0 3,3 0 4,3 ⊗ 0 3,3 0 6,4 ⊗ 0 3,3 I 6 ⊗ L(0, •) 0 6,4 ⊗ 0 3,3 0 3,4 ⊗ 0 3,3 0 3,6 ⊗ 0 3,3 0 3,3 ⊗ 0 3,3   which is symmetric since L(0, •) is.
Therefore P (ξ) is symmetrizable by a matrix independent of ξ. Hence, the Cauchy problem for (3.7), (3.8), (3.11), (3.12), (3.13), (3.14) is strongly well posed. The norm of the zero order terms depends on the coefficients σ j and their derivatives up to order 2. The estimate of the Theorem follows.

Remark 3.1. We recall a computation from [START_REF] Petit-Bergez | Problèmes faiblement bien posés : discrétisation et applications[END_REF], showing that when there are only 2 coefficients σ 1 and σ 2 only one derivative of σ j is needed. This is always the case in dimension d = 2. When σ 3 ≡ 0, split W as

W = E 33 W +   W 1 W 2 0   . Then   W 1 W 2 0   =   σ 2 U 2 1 σ 1 U 1 2 0   = diag(σ 2 , σ 1 , 0)U.
Rewrite (3.7) as

∂ t U + L(0, ∂)U + E 33 W + diag(σ 2 , σ 1 , 0)U = 0. (3.15)
Differentiate with respect to x 1 and x 2 to obtain

∂ t V j + L(0, ∂)V j + E 33 ∂ j W + ∂ j (diag(σ 2 , σ 1 , 0))U + diag(σ 2 , σ 1 , 0)W = 0. (3.16)
To find an equation on W , proceed as in the 3d proof to get,

∂ t W + k σ k A k V k + ( k σ k )W -σ 1 σ 2 U = 0 (3.17)
Therefore V is solution of a first order system, whose principal symbol is given by

P (∂) =     L(0, ∂) 0 0 0 0 L(0, ∂) 0 E 33 ∂ 1 0 0 L(0, ∂) E 33 ∂ 2 0 0 0 0     .
A symmetrizer is given by

S =     I 0 0 0 0 I 0 iE 23 0 0 I iE 13 0 0 0 I     , with S-1 =     I 0 0 0 0 I 0 -iE 23 0 0 I -iE 13 0 0 0 I 3     .

Sharp Finite Speed for Bérenger's PML

Recall some notions associated with estimates on the domains of influence and determinacy for a hyperbolic operator L (see [START_REF] Joly | Hyperbolic domains of determinacy and hamilton-jacobi equatiions[END_REF]). The timelike cones are the connected components of (1, 0, . . . , 0) in the complement of the characteristic variety.

τ = 0 as a root of det L 1 (τ, ξ) = 0 is independent of ξ ∈ R d \ 0.
The support of the solution of the Bérenger transmission problem is contained in the union of the propagation curves of L starting in the support of the source terms when either of the following conditions is satisfied.

(i) ∀ξ ∈ R d \ 0, Ker L 1 (0, ξ) = ∩ j Ker ξ j A j , and ∀j, σ j ∈ L ∞ . (ii) L 1 is Maxwell's equation and ∀j, σ j ∈ W 2,∞ . Contents 21
Proof. The characteristic varieties of L and L satisfy Char L = Char L ∪ {τ = 0}. When {τ = 0} has multiplicity as a root of det L 1 (τ, ξ) = 0 independent of ξ ∈ R d \ 0 the timelike cones of L and L coincide. Therefore the propagation cones and influence curves coincide too. Case (i). Part (ii) of Theorem 2.4 proves that L 1 defines a strongly well posed Cauchy problem. It follows that the sharp propagation conclusion of the Theorem is valid for L 1 + B(t, x) for any bounded B(t, x). This follows on remarking that the solution of ( L + B) U = 0 with initial data U 0 is the limit at ν → ∞ of Picard iterates U ν . The first, U 1 , is defined as the solution of the Cauchy problem without B. For ν > 1 the iterates are defined by,

L 1 U ν+1 + B(t, x) U ν = 0, U ν+1 (0, •) = U 0 .
Since L 1 has constant coefficients, sharp finite speed is classical for that operator. An induction proves that each iterate is supported in the union of influence curves starting in the support of U 0 . Case (ii). Reason as above constructing by Picard iteration approximations V ν converging to the solution V from (3.3). Since the equation satisfied by V is strongly well posed the iterates converge. An induction shows that they are supported in the set of influence curves starting in the support of U 0 .

Proof of perfection for Bérenger's PML by a change of variables

This section continues the analysis of Bérenger's method when the hypotheses of Theorem 3.4 are satisfied. In those cases well posedness is proved by an energy method. In addition suppose that ∀j, ∃L j > 0,

σ j = 0 when |x j | ≤ L j . (3.18) Denote by R := Π j ] -L j , L j [. Definition 3.2.
In this setting the method is perfectly matched when for arbitrary

F ∈ C ∞ 0 ]0, ∞[×R the unique solutions V and U of L V = F , V t≤0 = 0, L 1 U = F , U t≤0 = 0, (3.19)
with L as in (1.5) satisfies

V R×R = U R×R . (3.20)
Theorem 3.5. With the assumptions of Theorem 3.4 and σ j as above, Bérenger's method is perfectly matched.

Proof. Taking the Laplace transform of the V equation in (3.19) yields a transform holomorphic in Re τ > τ 0 with values in L 2 (R d ) satisfying for 1 ≤ j ≤ d,

V j + (τ + σ j (x j )) -1 A j ∂ j V = F j , with V := j V j , F := j F j . (3.21)
Multiply by τ and sum on j to obtain

τ V + j τ τ + σ j (x j ) A j ∂ j V = τ F . (3.22)
When τ is fixed real and positive this equation can be transformed to the corresponding equation without the σ j by a change of variables. The change of variables depends on τ . The resulting equation is exactly that determining U := j U j . In this way we find that V is obtained from U by this change of variables. This idea is inspired by Diaz and Joly in [START_REF] Diaz | A time domain analysis of PML models in acoustics[END_REF].

For real τ > 0 define d bilipschitzean homeomorphisms X j (x j ) of R to itself by

dX j (x j ) dx j = τ + σ j (x j ) τ , X j (0) = 0 .
Then,

∂ ∂x j = ∂X j ∂x j ∂ ∂X j = τ + σ j (x j ) τ ∂ ∂X j , τ τ + σ j (x j ) ∂ ∂x j = ∂ ∂X j . Therefore if U (X) is the solution of τ U (X) + j A j ∂ ∂X j U = F (X) , (3.23)
then the solution V of (3.22) is given by V (x) := Û (X(x)) since the latter function of x satisfies the equation determining V . Since X(x) = x for x ∈ R this proves that the transforms of U and V satisfy for real τ > τ 0 j

V j (τ, x) = j U j (τ, x) , x ∈ R . (3.24)
Since both sides of the identity in (3.24) are holomorphic in Re τ > τ 0 it follows that the identity extends to that domain by analytic continuation. Equation (3.21) and its analogue for U then imply that for all j, V j R = U j R . Uniqueness of the Laplace transform implies V j R = U j R for all t proving perfection.

Remark 3.2. The proof is very general. It shows that once the initial value problem defined by L is well posed there is perfect matching. The proof works more generally for at least weakly well posed methods for which the Laplace transform can be reduced to (3.22) for real τ . Our favorite version of the Bérenger algorithm is analysed this way in §3.6.

3.6.

Perfection for methods related to Bérenger's PML Consider (3.22) with F = 0. This equation is the starting point for many authors to construct well posed PML. It has been viewed as a complex stretching of coordinates (see [START_REF] Rappaport | Interpreting and improving the PML absorbing boundary condition using anisotropic lossy mapping of space[END_REF], [START_REF] Chew | A 3d perfectly matched medium from modified Maxwell's equations with stretched coordinates[END_REF], [START_REF] Petropoulos | Reflectionless sponge layers as absorbing boundary conditions for the numerical solution of Maxwell's equations in rectangular, cylindrical and spherical coordinates[END_REF], [START_REF] Hesthaven | On the analysis and construction of perfectly matched layers for the linearized Euler equations[END_REF]). This idea, for τ real, becomes an honest change of variables as in [START_REF] Diaz | A time domain analysis of PML models in acoustics[END_REF], that is at the heart of the proof in §3.5. In the case of Maxwell system, it can be viewed as a system with modified constitutive equations (a lossy medium [START_REF] Zhao | A reflectionless sponge layer absorbing boundary condition for the solution of Maxwell's equations with high-order staggered finite difference schemes[END_REF], [START_REF] Abarbanel | On the construction and analysis of absorbing layers in cem[END_REF]), or recovered as above from the Bérenger's system. The system (3.22) is not differential because of the division by τ + σ j (x j ). In order to recover a hyperbolic system, a change of unknowns is performed. We adopt the approach in [START_REF] Mazet | Interprétation dispersive du milieu PML de Bérenger[END_REF] for the Maxwell system.

Lemma 3.2. With matrices given in (3.2), define S j := (τ + σ j (x j ))/τ . There exists a pair of invertible matrices M, N , unique up to a multiplication by the same constant, such that

S -1 j N A j = A j M, j = 1, 2, 3. (3.25)
They are given by

M = γ   S 1 0 0 0 S 2 0 0 0 S 3   , N = γ   S 2 S 3 0 0 0 S 1 S 3 0 0 0 S 1 S 2   , γ ∈ C \ 0 . (3.26)
Proof. Since

A j e j = 0, A j e j+1 = -ie j+2 , A j e j+2 = ie j+1 , it is easy to see by applying This implies that N is also diagonal, equal to diag(m 2 S 3 , m 3 S 1 , m 1 S 2 ), and

m 1 S 3 = m 3 S 1 , m 2 S 1 = m 1 S 2 , m 3 S 2 = m 2 S 3
This leaves no choice but to choose (3.26).

In the rest of the analysis take γ = 1. In (3.22) with F = 0 replace V replace by U . Insert (3.25) to obtain

τ N U + j A j M ∂ j U = 0. (3.27)
The fact that σ j depends only on x j and the form of the matrices guarantees A j ∂ j M = 0. This yields

A j M ∂ j U ≡ A j ∂ j (M U ) .
Define a new unknown V := M U to find

N M -1 τ V + j A j ∂ j V = 0. (3.28) N M -1 = diag(S -1 1 S 2 S 3 , S -1 2 S 3 S 1 , S -1 3 S 1 S 2 )
. Next compute a rational fraction expansion of τ S -1 1 S 2 S 3 as

(τ + σ 2 )(τ + σ 3 ) τ + σ 1 = τ + (σ 2 + σ 3 -σ 1 ) + σ 2 1 + σ 2 σ 3 -σ 1 (σ 2 + σ 3 ) τ + σ 1 .
Introduce a new unknown W by

τ N M -1 V = τ V + Σ 1 V + Σ 2 W, equivalently W j = 1 τ + σ j (x j ) V j = 1 τ U j , with Σ := diag σ 1 , σ 2 , σ 3 , Σ (1)
:= diag σ 2 + σ 3 -σ 1 , σ 3 + σ 1 -σ 2 , σ 1 + σ 2 -σ 3 , Σ (2)
:= diag (σ 1 -σ 2 )(σ 1 -σ 3 ), (σ 2 -σ 1 )(σ 2 -σ 3 ), (σ 3 -σ 1 )(σ 3 -σ 2 ) .
This leads to a system in the unknowns V and W

L(∂ t , ∂ x )V + Σ (1) V + Σ (2) W = 0 , ∂ t W + ΣW -V = 0 . (3.29)
Finally, U is recovered from

U = ∂ t W = V -ΣW .
The system of equations for V, W is strongly well posed since L is symmetric hyperbolic. In the case of a single layer in the x 1 direction, there is only one coefficient σ and therefore a single complex valued suplementary variable. The equations for the magnetic and electric fields are

∂ t E 1 -(∇ ∧ H) 1 -σE 1 + σ 2 W 1 = 0, ∂ t E 2 -(∇ ∧ H) 2 + σE 2 = 0, ∂ t E 3 -(∇ ∧ H) 3 + σE 3 = 0, ∂ t W 1 + σW 1 -E 1 = 0, ∂ t H 1 + (∇ ∧ E) 1 -σH 1 + σ 2 W 2 = 0, ∂ t H 2 + (∇ ∧ E) 2 + σH 2 = 0, ∂ t H 3 + (∇ ∧ E) 3 + σH 3 = 0, ∂ t W 2 + σW 2 -H 1 = 0,
In 2d this is identical to the layers in [START_REF] Zhao | A reflectionless sponge layer absorbing boundary condition for the solution of Maxwell's equations with high-order staggered finite difference schemes[END_REF] and equivalent to those in [START_REF] Abarbanel | On the construction and analysis of absorbing layers in cem[END_REF].

The principal symbol and lower terms are

R 1 = L 0 0 I 3 ∂ t and B = Σ (1) Σ (2) -I 3 Σ .
Reversing the computation shows that (V, W ) ∈ Ker R(τ, ξ) if and only if V = M U , W = 1/τ U , and

L 1 (τ, ξ1τ τ +σ1 , • • • , ξ3τ τ +σ3 )U = 0.
The characteristic polynomial is therefore the same as for Bérenger's layer. Thus, by Theorem 2.

2 det R(τ, ξ) = τ 2 - ξ 2 j τ 2 (τ + σ j ) 2 .
Theorem 3.6. If σ j (x j ) ∈ L ∞ (R) and vanish for |x j | ≤ L j then the system (3.29) for V := M U and W j := V j /(τ + σ j (x j )) is strongly well posed in L 2 (R d ) and perfectly matched in the sense that for sources supported in R := Π j ]{|x j | ≤ L j } the function U computed from V, W agrees in R t × R with the solution of Maxwell's equation with corresponding sources.

Proof. The proof of Theorem 3.4 applies with only minor modifications. Remark 3.3. Note the ease with which strong well posedness is established and the lack of regularity required of the functions σ j .

Analysis of layers with only one absorption by Fourier-Laplace transform

There are cases where the energy method presented above does not prove well posedness. This is the case for the Bérenger algorithm when the ellipticity assumption is not satisfied and the absorptions are not regular. Notably for the Maxwell system and discontinuous absorptions. In this section we present a systematic analysis by Fourier-Laplace transformation of transmission problems with absorption in only one direction.

Fourier analysis of piecewise constant coefficient transmission problems

Return to the situation of (1.3) with operators L and R on the left and right half spaces and transmission condition (1.4). Suppose that both L and R are weakly hyperbolic in the sense of Gårding. An example is the classical method of Bérenger with one absorption. Among other things we will prove that the method is well posed and perfect. Note the open problem at the end of the introduction emphasizing that we do not know if the classic algorithm with two discontinuous absorptions is well posed. In addition, we show, by a non trivial analytic continuation argument in §4.1.5, that the perfection of Bérenger's method can be verified using the modified plane wave solutions from his original paper. It is our hope that the analysis may help in the construction of new perfectly matched layers.

Hersh's condition for transmission problems

This section takes up the analysis of mixed problems following Hersh in [START_REF] Hersh | Mixed problems in several variables[END_REF]. In the present context we treat transmission problems which are essentially equivalent. The analysis of Hersh supposed the interface is noncharacteristic which is never the case for Maxwell's equations. We address the changes that are needed to treat problems with characteristic interfaces.

First analyse the solution of the constant coefficient pure initial value problem LU = F on R 1+d by Laplace transform in time and Fourier transform in

x ′ = (x 2 , • • • , x d ). The transform U (τ, x 1 , η) := ∞ 0 e -τ t (2π) -d/2 e -ix ′ •η U (t, x ′ ) dt dx ′ decays as |x 1 | → ∞ and satisfies L(τ, d/dx 1 , iη) U = F in R .
When A 1 is invertible this is a standard ordinary differential equation in x 1 . When A 1 is singular, the analysis requires care. The homogeneous equation L(τ, d/dx 1 , iη) U = 0 has purely exponential solutions e ρx1 corresponding to the roots ρ of the equation det L(τ , ρ , iη) = 0 . be independent of τ, η. Since roots cannot cross the imaginary axis, the only way the integer can change is if roots escape to infinity. That can happen when the coefficient of the highest power of ρ vanishes. The next hypothesis rules that out. Definition 4.1. A hyperbolic operator L(∂ t , ∂ x ) is nondegenerate with respect to x 1 when there is a τ 1 > 0 so that the degree in ρ of the polynomial det L(τ, ρ, iη) is independent of (τ, η) for Re τ > τ 1 , η ∈ R d-1 .

Example 4.1. 1. In the noncharacteristic case, det A 1 = 0, the condition is satisfied and the degree with respect to ρ is equal to N . 2. For Maxwell's equations written in the real 6 × 6 form, the degree with respect to ρ is equal to 4. If written in the complex form (3.2), the degree is 2.

3. The formula for the characteristic polynomial in Theorem 2.2 shows that if L is nondegenerate then so is the Bérenger doubled operator L with one absorption σ 1 in x 1 > 0. The degree in ρ is the same for L and L. 4. If L = L 1 +B is nondegenerate with respect to x 1 then so is the operator

P := L 1 (∂)+a -1 B = a -1 L(a∂)
for any a > 0. If the degree for L is constant in Re τ > τ 1 , then the degree for P is constant for Re τ > a -1 τ 1 .

For the lemmas to follow it is useful to transform so that A 1 has block form.

Lemma 4.1. If L in (2.1) is nondegenerate with respect to x 1 then for Re τ > τ 1 , η ∈ R d-1 ,
(i) the degree in ρ of the polynomial det L(τ, ρ, iη) is equal to rank A 1 , (ii) the number of roots ρ with positive real part is equal to the number of negative eigenvalues of A 1 .

Proof.

Since L is nondegenerate, it suffices to study the case η = 0. (i) Choose invertible K so that

K -1 A 1 K =
A 0 0 0 , A an invertible square matrix of size rank A 1 .

Then

K -1 L(τ, ρ, 0) K = τ I + A ρ 0 0 τ I + matrix independent of τ, ρ .
It follows that the degree in ρ is no larger than rank A 1 .

The coefficient of ρ rank A1 in det L(τ, ρ, 0) is a polynomial in τ of degree ≤ Nrank A 1 . For large τ the coefficient is equal to det A τ N -rank A1 + lower order in τ .

Thus the degree in ρ is rank A 1 for such τ proving the result.

(ii) det L(τ, ρ, 0) = det τ I + ρA 0 0 τ I + B .
For fixed τ , sufficiently large, ρ = 0, and ρ/τ is a root of the polynomial p(x, 1/τ ) of degree rank A:

p(x, ε) = I + xA + εB 11 εB 12 εB 21 I + εB 22
This polynomial has exactly rank A roots. By Rouché's theorem,

ρ j τ ∼ - 1 λ j , τ ≫ 1,
where the λ j are the rank A eigenvalues of A repeated according to their algebraic multiplicity. Since the eigenvalues of A and the nonzero eigenvalues of A 1 are the same, this completes the proof.

Remark 4.1. For the transformed one dimensional hyperbolic operator L(∂ t , ∂ 1 , iη) the number of incoming characteristics at the boundary x 1 = 0 in the right half space is equal to the number of strictly positive eigenvalues of A 1 . The second part of the Lemma shows that this is equal to the number of roots with negative real part. The two natural ways to compute the number of necessary boundary conditions yield the same answer.

The next lemma shows that for nondegenerate operators, the characteristic case can be transformed to a standard ordinary differential equation. Lemma 4.2. Suppose that A, M ∈ Hom (C N ) and the equation det(Aρ + M ) = 0 has degree in ρ equal to rank A and no purely imaginary roots. Then, (i) The matrix M is invertible and all solutions of the homogeneous equation

A dU dx 1 + M U = 0 (4.2)
take values in the space G := M -1 (Range A) satisfying dim G = rank A. (ii) There is a M ∈ Hom G so that a function U satisfies (4.2) if and only if U is G valued and satisfies

dU dx 1 + M U = 0 . (4.3) (iii)
The vector space U of solutions of (4.2) is a linear subspace of C ∞ (R) with dimension equal to rank A. The Cauchy problem with data in G is well posed.

Proof.

(i) Since ρ = 0 is not a root, M is invertible. The equation U = M -1 A dU/dx 1 shows that continuously differentiable solutions U takes values in G. More generally, if U is a distribution solution and ψ ∈ C ∞ 0 (R) takes values in the annihilator, G ⊥ of G, then

U , ψ = M -1 A dU/dx 1 , ψ = dU/dx 1 , (M -1 A) * ψ
But G = range M -1 A so G ⊥ = ker (M -1 A) * . Therefore (M -1 A) * ψ = 0 so U, ψ = 0 which is the desired conclusion.

(ii) Multiplying the equation by an invertible P and making the change of variable U = KV transforms the equation to the equivalent form

P A K dV dx 1 + P M K V = 0 .
Choose invertible P, K so that P AK has block form The first part of the preceding lemma implies that the determinant on the left is a polynomical of degree rank A in ρ. It follows that H 22 is invertible.

P A K = I 0 0 0 , where I is the rank A × rank A identity matrix. With V = (V 1 , V 2 )
The solutions V satisfy

H 21 V 1 + H 22 V 2 = 0 so take values in V := V 2 = -H -1 22 H 21 V 1 .
The function V is a solution if and only if it takes values in V and

dV 1 dx 1 + RV 1 = 0, R := H 11 -H 12 H -1 22 H 21 . If N : V → V is the map, V 1 , V 2 → RV 1 , -H -1 22 H 21 RV 1 ,
then V is a solution if and only if it is V valued and satisfies dV /dx 1 = N V . Writing V = K -1 U and M = -K N implies (ii).

(iii) Follows from (ii).

Lemma 4.3. If L is hyperbolic and nondegenerate with respect to x 1 , then its principal part L 1 (∂ t , ∂ x ) is also nondegenerate with respect to x 1 . The degree in ρ of det L 1 (τ, ρ, iη) is constant for Re τ > 0 and η ∈ R d-1 .

Proof. With notation from the preceding proof,

L(∂ t , ∂ x1 , ∂ x ′ ) = P -1 I∂ x1 0 0 0 + H 11 (∂ t , ∂ x ′ ) H 12 (∂ t , ∂ x ′ ) H 21 (∂ t , ∂ x ′ ) H 22 (∂ t , ∂ x ′ ) K -1 . (4.4)
The proof of the last lemma showed that for Re τ > τ 1 and η

∈ R d-1 , H 22 (τ, η) is invertible.
The computation in Lemma 4.1 shows that for η = 0 and R ∋ τ → ∞ the coefficient of ρ rank A1 has modulus ≥ cτ N -rank A1 with c > 0. This implies that η = 0 is noncharacteristic for

H 22 . Therefore H 22 (∂ t , ∂ x ′ ) is hyperbolic.
Replacing L by its principal part L 1 has the effect of replacing each operator H ij (∂) by its principal part. This yields identity (4.4) with L and the H ij replaced by their principal parts.

Since the principal part of a hyperbolic operator is hyperbolic, it follows that (H 22 ) 1 (∂ t , ∂ x ′ ) is a homogeneous hyperbolic operator. Therefore (H 22 ) 1 (τ, iη) is invertible for η ∈ R d-1 and Re τ = 0. Thus, the coefficient of ρ rank A1 in det L 1 (τ, ρ, iη) is nonzero for η ∈ R d-1 and Re τ = 0. Lemma 4.4. Suppose that the ordinary differential equation (4.2) satisfies the hypotheses of Lemma 4.2.

Denote by E ± the linear space of solutions which tend exponentially to zero as x 1 → ±∞ and by Ė± their traces at x 1 = 0.

Then (i) Ė± ∩ ker A = {0} , (ii) dim AE ± = dim E ± , (iii) The map U → U (0) is an isomorphism from E ± to Ė± , (iv) A Ė+ ∩ A Ė-= {0}, (v) A Ė+ ⊕ A Ė-= Range A.
Proof. (i) The absence of purely imaginary roots shows that every solution is uniquely the sum of two solutions. One grows exponentially at +∞ and decays exponentially at x 1 = -∞. The second grows at -∞ and decays at +∞. In particular the only bounded solution is the zero solution.

If e + ∈ Ė+ ∩ ker A, denote by U (x 1 ) the solution with this Cauchy data. The function that is equal to U on x 1 > 0 and equal to 0 in x 1 ≤ 0 is a distribution solution of (4.2) on all of R since A[U ] x1=0 = 0. This solution is bounded hence identically equal to zero. Therefore e + = 0. The case for Ė-∩ ker A is analogous.

(ii) Follows from (i).

(iii) It is surjective by definition. If it were not injective for E + , there would be a nontrivial solution U (x) exponentially decaying as x 1 → +∞ with U (0) = 0 violating (i).

(iv) The set A Ė+ consists of the values AU + (0) with U + satisfying (4.2) and exponentially decreasing in x 1 > 0. If the intersection were nontrivial there would be a solutions U -decaying as x 1 → -∞ so that AU + (0) = AU -(0). The function V equal to U + in x 1 > 0 and U -in x 1 < 0 is then a distribution solution for all x 1 exponentially decaying in both directions. Hyperbolicity implies that V = 0 contradicting the nontriviality.

(v) Using (ii) and (iv), one sees that the direct sum on the left is a subspace of Range A of full dimension.

The next lemma is needed in in §4.1.2. Lemma 4.5. Assume that the hypotheses and notations of Lemma 4.2 are in force. Then for K ∈ Ė+ there is an F ∈ C ∞ 0 (] -∞, 0[) so that the unique solution of

A dU dx 1 + M U = F , lim |x1|→∞ U (x 1 ) = 0 , (4.5)
satifies U (0) = K.

Proof. Consider first the case of invertible A. A change of dependent variable yields the block form for the new variable still denoted

U dU dx 1 + M + 0 0 M - U = F, U = (U 1 , U 2 ), F = (F 1 , F 2 ) , spec M ± ⊂ {±Re z > 0}.
Then Ė+ = {U 2 = 0} so K = (K 1 , 0). Choose F = (F 1 , 0). Then U (0) = K if and only if,

K 1 = 0 -∞ e M+s F 1 (s) ds.
This is achieved with,

F 1 (s) = χ(s) e -M+s K 1 , χ ∈ C ∞ 0 (] -∞, 0[), χ(s) ds = 1 .
When A is not invertible change variable as in Lemma 4.2 to find the block form

I 0 0 0 dU dx 1 + H 11 H 12 H 21 H 22 U = F, with invertible H 22 . Part (i) of (4.2) implies that the map G ∋ G = (G 1 , G 2 ) → G 1 is an isomorphism. Write G ∋ K = (K 1 , K 2 ). Choose F = (F 1 , 0). Then choose a G valued solution U defined by dU 1 dx 1 + H 11 U 1 = F 1 , U 2 = -H -1 22 H 21 U 1 .
One has U (0) = K if and only if U 1 (0) = K 1 . The construction in the invertible case finishes the proof.

Suppose that

L = ∂ t + A 1 ∂ 1 + • • • and R = ∂ t + A 1 ∂ 1 + • • •
are nondegenerate with respect to x 1 . For Re τ > τ 0 and η ∈ R d-1 , define E ± L (τ, η) to be the set of solutions of

L(τ, d/dx 1 , iη)V = 0
Contents 29 tending to zero as x 1 → ±∞. Denote by Ė± L (τ, η) ⊂ C N the linear space of traces at x 1 = 0 of solutions in E ± L (τ, η).

Similarly with a possibly larger value still called τ 0 , there are Ė± R (τ, η) ⊂ C M so that the solutions of R(τ, d/dx 1 , iη)Z = 0 taking values in Ė± R (τ, η) are exactly those tending to zero exponentially as x 1 → ±∞. The subspaces E ± L (τ, η) and E ± R (τ, η) depend smoothly on τ, η for Re τ > τ 0 and η

∈ R d-1 . Lemma 4.1 implies that dim E - L (τ, η) = # positive eigenvalues of A 1 , dim E + R (τ, η) = # negative eigenvalues of A 1 . (4.6)
Consider the inhomogeneous transmission problem,

LV = 0 when x 1 < 0, RW = 0 when x 1 > 0 , (4.7) (V, W ) -g ∈ N when x 1 = 0 . (4.8)
The problem with inhomogeneous term F can be reduced to this form by subtracting on the left a solution of the hyperbolic Cauchy problem LU = F on R 1+d with U | t<0 = 0. Denote by V (τ, x 1 , η), W (τ, x 1 , η), ĝ(τ, η) the Fourier-Laplace transforms. The transform U is defined for x 1 ∈ R, while V (resp. W ) is defined for x 1 < 0 (resp. x 1 > 0). The transforms V , W decay as |x 1 | → ∞. V , W satisfy the ordinary differential transmission problem

L(τ, d/dx 1 , iη) V = 0 in x 1 < 0, R(τ, d/dx 1 , iη) W = 0 in x 1 > 0, (4.9)
( V (τ, 0, η) , W (τ, 0, η))ĝ(s, η) ∈ N . (4.10)

Hersh's necessary and sufficient condition for well posedness of the transmission problem is derived as follows. Uniqueness of solutions of (4.9), (4.10) for Re τ > τ 0 , η ∈ R d-1 is equivalent to the fact that there are no exponentially decaying solutions of the homogeneous transmission problem. That is,

N ∩ Ė- L (τ, η) × Ė+ R (τ, η) = {0}. (4.11)
In order to guarantee existence, one imposes the maximality condition,

N ⊕ Ė- L (τ, η) × Ė+ R (τ, η) = C N × C M . (4.12)
Using (4.6), this determines the dimension of N from the coefficients A 1 and A 1 of L and R respectively.

Definition 4.2. If the transmission problem (4.7), (4.8) satisfies (4.12) for all Re τ > τ 0 and η ∈ R d-1 it is said to satisfy Hersh's condition.

Theorem 4.1. Hersh's condition is satisfied if and only if there is an r and a λ 0 so that for all λ > λ 0 and g supported in t ≥ 0 with e -λt g ∈ H s+r (R d t,x ′ ) with values in

C N × C M there is a unique V, W supported in t ≥ 0 with e -λt V ∈ H s (] -∞, ∞[×{x 1 < 0}) and e -λt W ∈ H s (] -∞, ∞[×{x 1 > 0})
satisfying the transmission problem (4.7), (4.8).

Sketch of Proof.

We have shown that the Hersh condition permits one to compute a candidate Fourier-Laplace transform. We outline how the condition implies the desired estimate. The method is to use the Seidenberg-Tarski Theorem 2.3 to derive a lower bound on the real parts of the roots ν together with a contour integral representation. The same elements form the heart of [START_REF] Kasahara | On weak well-posedness of mixed problems for hyperbolic systems[END_REF], and §12.9 of [START_REF] Hörmander | The analysis of linear partial differential operators. II. Differential operators with constant coefficients[END_REF]. In the present context we treat a transmission problem rather than a boundary value problem. In addition, one needs to use the earlier lemmas to treat the case when x 1 = 0 is characteristic.

Choose Λ > max{τ 0 (L), τ 0 (R)}. The equations det L(τ, ν, iη) = 0, det R(τ, ν, iη) = 0 with Re τ ≥ Λ, η ∈ R d-1 have no purely imaginary roots. Define

ζ(R) := min |Re ν| : η ∈ R d-1 , Re τ ≥ Λ, |τ | 2 + |η| 2 ≤ R 2 , det L(τ, ν, iη) = 0 or det R(τ, ν, iη) = 0 .
The Seidenberg-Tarski Theorem 2.3 implies that there is a ρ ∈ Q and b = 0 so that

ζ(R) = R ρ (b + o(R)), as R → ∞ .
Thus, there are C, N so that, for any τ , η with Re τ ≥ Λ,

|Re ν| ≥ C 1 + |(τ, η)| N . (4.13)
The solutions in E + R (τ, η) are written using a contour integral representation of W in the block form of Lemma 4.2. Here the matrix H ij depend of (τ, η). Denote by D = D(τ, η) the finite union of squares with centers at the roots with Re ν < 0. The side of each square is the smaller of 1 and half the distance of the root to the imaginary axis. Then

W I = 1 2πi ∂D e τ x1 τ + H 11 + H 12 H -1 22 H 21 -1 dτ ˙ W I , W II = H -1 22 H 21 W I . (4.14)
The Seidenberg-Tarski Theorem 2.3 applied to max |w| 2 :

|z| 2 = 1, H 22 w = z, Re τ ≥ Λ, |τ | 2 + |η| 2 ≤ R 2
proves that

H 22 (τ, η) -1 = R β (a + o(1)), a = 0, β ∈ Q .
This estimate together with (4.13) yields with new C, N ,

∞ 0 | W (τ, x 1 , η)| 2 dx 1 ≤ C (1 + |(τ, η)| 2N )| W I (0)| 2 .
With the analogous expression for V the solution of (4.7) satisfies

∞ -∞ V (τ, x 1 , η) 2 dx 1 ≤ C (1 + |(τ, η)| 2N )| V I (0)| 2 .
The Hersh condition asserts that for each (τ, η), W I (0) and V I (0) are uniquely determined by ĝ(τ, η). Seidenberg-Tarski Theorem 2.3 yields an estimate

W I (0) , V I (0) ≤ C (1 + |(τ, η)|) a ĝ(τ, η) 2 .
The last three estimates together with Parseval's identity proves the desired estimate,

∃ C, N, ∀ g, ∀λ > Λ, e -λt U 2 L 2 (R 1+d ) ≤ C |α|≤N e -λt ∂ α t,x g L 2 (R 1+(d-1) ) .
This estimate proves the existence part of the Theorem.

Necessary and sufficient condition for perfection

The Fourier-Laplace method is used to derive a necessary and sufficient condition for perfection of an absorbing layer. Begin with a closer analysis of the transform, U (τ, x 1 , η), of the solution of the basic equation (1.1).

When A 1 is invertible, U is analysed as follows. Denote by Π ± (τ, η) the projectors associated with the direct sum decomposition Ė+

L (τ, η) ⊕ Ė- L (τ, η) = C N . Define S ± (τ, x 1 , η) as the Hom(C N ) valued solutions of L τ , d/dx 1 , iη S ± = 0, S ± x1=0 = A -1 1 Π ± .
Contents 31

Then S ± decays exponentially as

x 1 → ±∞ and χ ]-∞,0[ S -+ χ [0,∞[ S +
is the unique tempered fundamental solution of L(τ, d/dx 1 , iη).

Decompose F = F -+ F + , U = U + + U - according to E + L (τ, η) ⊕ E - L (τ, η) = C N . Then U -is the convolution of F-with χ ]-∞,0[ S -and U + is the convolution of F + with χ [0,∞[ S + . In particular U -(τ, 0, η) vanishes on a neighborhood of [0, ∞[ so U (τ, 0, η) = U + (τ, 0, η) ∈ Ė+ L (τ, η).
The value of U in x 1 ≥ 0 satisfy the homogeneous ordinary differential equation L(τ, d/dx 1 , iη) U = 0. with initial value U (0) ∈ Ė+ L (τ, η). To reach the same conclusion when A 1 is singular, apply the lemmas of the preceding section to the equation L(τ, d/dx 1 , iη)Z 0. Lemma 4.4 applied to A = A 1 and M = τ I + i d j=2 A j η j shows that both Ė± L (τ, η) are subspaces of G and that the space of solutions is a direct sum

E + L (τ, η) ⊕ E - L (τ, η). It follows that Ė- L (τ, η) ⊕ Ė+ L (τ, η) = G(τ, η) .
Repeating the analysis in the nonsingular case applied to (4.3) shows that U (τ, 0, η) ∈ Ė+ L (τ, η). (ii) There is a τ 0 ∈ R so that for all Re τ > τ 0 and η ∈ R d-1 , H(τ, η) = 0. (iii) There is a τ 0 ∈ R so that for all Re τ > τ 0 and η ∈ R d-1 ,

∈ Ė+ L (τ, η) there is a unique ( V , Ẇ ) ∈ Ė- L (τ, η) × Ė+ R (τ, η) so that (K, 0) ≡ ( V , Ẇ ) mod N . Define, H(τ, η) K := V .
∀K L ∈ Ė+ L (τ, η), ∃ ! K R ∈ Ė+ R (τ, η), such that (K L , K R ) ∈ N . (4.15)
Proof. Conditions (ii) and (iii) are clearly equivalent. For the equivalence with (i), compare the values of U and V in {x 1 < 0}. Since both satisfy LZ = F and decay as

x 1 → -∞ it follows that L( V -U ) = 0, so, V -U := Γ is an Ė- L valued solution of LΓ = 0. Since F = 0 in x 1 > 0, W ∈ E + R . The transmission condition requires that N ∋ ( V (0) , W (0)) = ( U (0) + Γ(0) , W (0)) = ( U (0) , 0) + (Γ(0) , W (0)) . (4.16) Since (Γ(0), W (0)) ∈ Ė- L (τ, η) × Ė+ R (τ, η), (4.16
) expresses ( U (0), 0) as a sum of an element in N and an element of Ė-L (τ, η) × Ė+ R (τ, η). The Hersh condition (4.12) asserts that such a decomposition is unique. Therefore ( V (0), W (0)) is uniquely determined from U (0).

The method is perfectly matched if and only if for all F supported in x 1 < 0, t ≥ 0

V = U x1<0 .
This occurs if and only Γ vanishes for x 1 < 0 which holds if and only if Γ(0) = 0. If the method is perfectly matched, then in the decomposition (4.16) one has Γ(0) = 0. Then ( U (0), W (0)) ∈ N . Lemma 4.5 asserts that for any K ∈ Ė+ L there is an F so that U (0) = K. This proves that (4.15) holds.

Conversely if (4.15) holds, then in the decomposition (4.16), Γ(0) = 0 so Γ = 0. It follows that

U | x1<0 = V .
Remark 4.2. 1. When (4.15) holds, the decomposition of (K, 0) ∈ C N × C M in the direct sum (4.12) is,

(K , 0) = (K , W (K)) -(0 , W (K)) ∈ N ⊕ (E - L × E + R ) .
2. With K = U (0) as above the solution (V, W ) of the ordinary differential equation transmission problem is given by V = U | x<0 and W is the solution of RZ = 0 with Z(0) = -W (K).

3. In the important case where N = M , invertible A 1 and A 1 and transmission condition N = {V = W }, the perfection criterion (iii) asserts that Ė+ L (τ, η) = Ė+ R (τ, η). We present a typical example showing that the natural absorbing layers are virtually never perfectly matched in dimension d ≥ 2.

A 1 = 1 0 0 -1 , A 2 = 0 1 1 0 , R = L + P, P = P * ≥ 0, N = {(V, W ) : V = W } . (i)
The transmission problem is perfectly matched if and only if P = 0.

(ii) The corresponding problem with d = 1 is perfectly matched if and only if P is diagonal.

Proof. Define

M L := A -1 1 [τ + iηA 2 ] , M R := A -1 1 [τ + iηA 2 ] + A -1 1 P , so that A -1 1 L(τ, ∂ 1 , iη) = ∂ 1 + M L (τ, η) and similarly for A -1 1 R(τ, ∂ 1 , iη).
For Re τ > 0 and η ∈ R, the matrices M L and M R have one eigenvalue with positive real part and one with negative real part. The eigenspace corresponding to positive (resp. negative) real part eigenvectors is equal to Ė+ L (τ, η) (resp. Ė+ R (τ, η)). Therefore the necessary and sufficient condition for perfection is that for Re τ > τ 0 and any η,

Ė+ L (τ, iη) = Ė+ R (τ, iη) . Since L(τ, ρ, iη) = τ + ρ iη iη τ -ρ , and, det L(τ, ρ, iη) = τ 2 -ρ 2 + η 2 , the eigenvalue of M L (τ, η) with positive real part is ρ = τ 2 + η 2 .
The eigenspace is the kernel of L(τ, ρ, iη). Therefore Ė+ L (τ, η) = C(-iη, τ + ρ) .

(4.17)

Since M R = M L + A -1 1 P , a necessary condition is that the family of vectors v(η, τ ) := (-iη, τ + ρ) be eigenvectors of the constant matrix A -1 1 P , which is possible only if A -1 1 P is a constant multiple of the identity. Therefore P = cA 1 . Since P ≥ 0 and A 1 has eigenvalues of both signs, it follows that c = 0 proving (i).

In the one dimensional case there is just one eigenvector (0, 1) which must be an eigenvector of A -1 1 P . Since (0, 1) is also an eigenvector of A 1 it follows that (0, 1) must be an eigenvector of P . Since P = P * , the orthogonal vector (1, 0) is also an eigenvector and P is diagonal. Conversely, if P is diagonal the condition is satisfied.

Remark 4.3. 1.
Examples verifying perfection for a family of absorbing layers related to but not including those of Bérenger are presented in [START_REF] Appelö | Perfectly matched layers for hyperbolic systems: General formulation, well-posedness and stability[END_REF]. To our knowlege Hersh's criterion for Bérenger's layers has not been verified before.

2. The perfection criterion is related to the plane wave criterion of Bérenger. We examine the relation in §4.1.6.

Hersh's condition for Bérenger's PML with piecewise constant σ 1

Of our earlier results, only those of Section 3.2 apply to discontinuous absorptions. So, if the generator is not elliptic, (for example. the PML Maxwell system of Bérenger), the preceding results do not prove that the initial value problem is well posed. In this section we prove that the doubled operators of Bérenger define a (weakly) well posed initial value problem provided that

σ j ≡ 0 for j ≥ 2 and σ 1 (x 1 ) ≡ σ ± in R d ± , (4.18)
and, the constant coefficient operators L on R d ± are both (weakly) hyperbolic. The unknown U satisfies (1.5). Denote by U ± = {U ± 1 , . . . , U ± d } the restriction of the unknown U to R d ± . They satisfy differential equations in the half spaces R d ± .

Lemma 4.6. For U locally square integrable on a neighborhood of (t, x) ∈ {x 1 = 0} the following are equivalent.

(i) L U ∈ L 2 on a neighborhood of (t, x) in R 1+d in the sense of distributions.

(ii) There is a neighborhood O of (t, x) so that

L U ± is square integrable on O ∩ R d ± and [ A 1 U ] = 0. Remark 4.4. 1.
The first hypothesis is often verified by combining

L U + B(x) U ∈ L 2 loc , U ∈ L 2 loc and B ∈ L ∞ loc . 2.
[ A 1 U ] makes sense since the differential equation implies

∂ 1 A 1 U + ∈ L 2 loc ]0, ε[ ; H -1 loc (R d t,x ′ ) . With U ∈ L 2 (]0, ε[ ; H -1 loc (R d ) this implies that A 1 U + ∈ C([0, ε[ ; H -1/2 loc (R d )
). An analogous result holds for A 1 U -. Therefore the traces from both sides and the jump are well defined elements of H -1/2 loc . 3. is clear on a formal level since if A 1 U were discontinuous there would be a δ(x 1 ) term from the differential operator L applied to U . 4. The standard proof based on these remarks is omitted.

We have supposed that the nonzero data are initial values f ± (x). By the usual subtraction one can convert the problem to one with homogeneous initial values and right hand side and inhomogeneous transmission condition. In this way, the determination of U ± is reduced to finding W ± satisfying the inhomogeneous transmission problem

L 1 (∂ t , ∂ x ) W ± + B ± W ± = 0 , A 1 [ W ] = g , (4.19)
where Proof. Drop the tildes on the Fourier-Laplace transforms of W , g for ease of reading. The transformed problem is

B ± :=      σ ± I N 0 . . .
A 1 d dx 1 + L(τ, 0, iη) W ± = 0 , A 1 [ W ] = ĝ. (4.21)
The condition of Hersh is that for an arbitrary right hand side ĝ in range A 1 this transmission problem has one and only one solution.

Denote by E ± L (τ, η, σ) the spaces associated to the Bérenger operator operator L with absorption σ. The uniqueness of solutions of (4.21) is equivalent to

A 1 Ė- L (τ, η, σ -) ∩ A 1 Ė+ L (τ, η, σ + ) = {0}. (4.22)
Existence is equivalent to (4.23). It remains to prove (4.22).

A 1 Ė- L (τ, η, σ -) + A 1 Ė+ L (τ, η, σ + ) = range A 1 . (4.23) Part (ii) of Lemma 4.4 implies that dim A 1 Ė- L (τ, η, σ -) + dim A 1 Ė- L (τ, η, σ + ) = dim range A 1 , so (4.22) implies
For σ > 0, the split Bérenger operator L is hyperbolic so for Re τ > τ 0 (σ) the solutions of

L(τ, d/dx 1 , iη) U = 0 (4.24)
are generated by exponentially growing and exponentially decaying solutions. The next lemma identifies these solutions in terms of the corresponding solutions of

L(τ, d/dx 1 , iη) V = 0 . (4.25)
The result shows that the traces at x 1 = 0, Ė± L , are independent of σ.

Lemma 4.7.

For σ > 0, Re τ > τ 0 (σ), η ∈ R d-1 , (i) The map V (x 1 ) → η 1 τ A 1 V (τ + σ)x 1 /τ , η 2 τ A 2 V (τ + σ)x 1 /τ , . . . , η 2 τ A d V (τ + σ)x 1 /τ
is an isomorphism from solutions of (4.25) onto the solutions of (4.24).

(ii) µ is a root of det L(τ, •, iη) = 0 if and only if ν = (τ + σ)µ/τ is a root of det L(τ, •, η) = 0.
(iii) For the roots in (ii), the real parts of µ and ν have the same sign. In particular, the map in

(i) is an isomorphism E ± L (τ, η) → E ± L (τ, η, σ). (iv) The map W = ( W 1 , . . . , W d ) → j W j is an isomorphism E ± L (τ, η, σ) → E ± L (τ, η).
Remark 4.5. In (i) it important to know that the solutions V (x 1 ) are entire analytic functions of x 1 so it makes sense to evaluate V at points off the x 1 -axis. In the literature this is sometimes called a complex change of variables. It is only reasonable for analytic solutions. A related idea is used in the Fourier-Laplace analysis for general σ 1 (x 1 ) presented in §4.2.

Proof of Lemma 4.7

(i) If U = ( U 1 , . . . , U d ) satisfies (4.24) then with W := j U j , A 1 d W dx 1 + (τ + σ) U 1 = 0, τ U j + iη j A j W = 0, j = 2, . . . , d . (4.26)
Multiply the first by τ and the last d -1 by (τ + σ). Sum and then divide by τ to find,

A 1 d W dx 1 + τ + σ τ L(τ, 0, iη) W ± = 0 . (4.27)
Conversely if W satisfies (4.27) and U j for j ≥ 2 is defined from W using the last equations in (4.26) and

U 1 := W -j≥2 U j then U satisfies (4.24).
The solutions W to (4.27) are exactly the V ((τ + σ)x 1 /τ ) with V satisfying (4.25). This proves that the mapping in (i) is surjective.

The set of solutions V of (4.25) has dimension rank A 1 . The set of solutions of (4.24) has dimension rank A 1 = rank A 1 (see (2.3)), so surjectivity implies injectivity.

(ii) and (iv) follow from (i). Proof. Verify condition (ii) of Theorem 4.2. For K ∈ Ė+ L (τ, η, σ + ) consider the unique decomposition guaranteed by the Hersh's condition,

(K, 0) = (W -, W + ) + (F -, F + ) , (4.30)
where,

W -, W + = (W - 1 , . . . , W - d ), (W + 1 , . . . , W + d ) ∈ N , (F -, F + ) ∈ Ė- L (τ, η, σ -) × Ė+ L (τ, η, σ + ).
Perfection is equivalent to F -= 0. By inspection, one such decomposition (4.30) is given by

(K, 0) = (K, K) + (0, -K)
where we use the fact from Lemma 4.7 that

Ė+ L (τ, η, σ -) = Ė+ L (τ, η, σ + ) .
As this decomposition satisfies F -= 0 the proof of the first assertion is complete.

For the partially split case, K ∈ Ė+ L (τ, η) has a unique decomposition from the Hersh's condition,

(K, 0) = (W -, W + ) + (F -, F + ), with (W -, W + ) = W -, (W + 1 , . . . , W + d ) ∈ N , (F -, F + ) ∈ Ė- L (τ, η) × Ė+ L (τ, η, σ + ). Define W + j := η j τ A j K .
Part (i) of Lemma 4.7 implies that W + ∈ Ė+ L (τ, η, σ + ). In addition, j W + j = K so (K, W + ) ∈ N . By inspection K , 0 = K , W + + 0 , -W + is the unique Hersh decomposition. Since F -vanishes for this one the proof is complete.

Analytic continuation for Maxwell like systems and Bérenger's plane waves

In this section we investigate Bérenger's method for operators, including the Maxwell system, whose characteristic polynomial is τ p (τ 2 -|ξ| 2 ) q . For ease of exposition we treat the case d = 2 and the explicit operator,

L = ∂ t + 1 0 0 -1 ∂ 1 + 0 1 1 0 ∂ 2 . (4.31)
Analogous results are valid for the Maxwell system with only slightly more complicated formulas. For Re τ > 0 and η ∈ R there is exactly one root of det L 1 (τ, ρ, iη) = 0 with Re ρ > 0 given by

ρ = τ 2 + η 2 , Re ρ > 0 . (4.32)
The corresponding eigenspace Ė+ L (τ, η) from (4.17) is spanned by Φ(τ, η) = (-iη, τ + ρ). If L 1 is the Bérenger operator doubled in the x 1 direction one has the same roots and Ė+ L1 is spanned by

ρA 1 Φ, iηA 2 Φ . Proposition 4.2.
Contents 37 (i) For each η, ρ(τ, η), Ė-L (τ, η), and Ė+ L (τ, η) are holomorphic in Re τ > 0 with continuous extension to Re τ ≥ 0. (ii) If σ 1 > 0 then for Re τ > 0 and η ∈ R the equation det L(τ, ν, iη) = 0 has exactly one root ν with positive real part. It is given by ν = (τ + σ 1 )ρ/τ . (iii) For σ 1 ≥ 0, the relation (4.12) with L 1 on the left and L on the right is satisfied on Re τ ≥ 0 , ρ = 0 . (iv) The mapping H(τ, η) is for each η holomorphic in Re τ > 0 with continuous extension to Re τ ≥ 0 , ρ = 0 .

Proof. (i) For Re τ > 0 there are two roots ±ρ with ρ from (4.32). One has strictly positive real part and the other strictly negative. Each is holomorphic in Re τ > 0. Holomorphy for Φ(τ, η) follows from its expression in terms of ρ. As Φ is a basis for Ė+ L (τ, η) holomorphy of the latter follows. So long as the eigenvalues ±ρ remain apart as Re τ → 0 they and their eigenspaces will be holomorphic. The delicate case is when τ + η 2 → 0. The limiting points are (±iη, η).

If η = 0, then ρ = τ and the eigenspace is (0, 1). Both are continuous up to the boundary.

When η = 0 one has ρ → 0 so ρ is continuous up to the boundary. Then Φ is continuous up to the boundary and nonvanishing from its expression in terms of ρ. Therefore Ė+ L is continuous up to the boundary.

(ii) It suffices to remark that this is an eigenvalue and then to show that the real part is positive. For the latter compute

∂ ∂σ τ + σ τ τ 2 + η 2 = τ 2 + η 2 τ = 1 + η 2 /τ 2 .
For Re τ > 0 this has positive real part so the real part of the eigenvalue is increasing as a function of σ so is positive for all σ ≥ 0.

(iii) It suffices to show that (4.11) is valid for such s, η.

Suppose (v, w) = (v 1 , v 2 , w 1 , w 2 ) ∈ Ė- L1 × Ė+ L . Must show that v 1 + v 2 = w 1 + w 2 . Since (w 1 , w 2 ) ∈ Ė+ L it follows that w 1 + w 2 ∈ Ė+ L . Similarly v 1 + v 2 ∈ Ė- L1 .
Thus it suffices to show that Ė-L1 and Ė+ L are uniformly transverse as Re τ → 0. It suffices to show that (iη, τ + ρ) and (iη, τρ) are uniformly independent. This follows from ρ = 0.

(iv) The holomorphy of H follows from (i). The continuous extension follows from (i) and (iii) .

Since the method is perfectly matched, H = 0 for Re τ > 0. By continuity the map vanishes for purely imaginary τ = 0. This shows that for {Re τ ≥ 0} \ 0, the function equal to e iτ t+ρ(τ,η)x1+iηx2 Φ for x 1 < 0, and e iτ t+ρ(τ,η)x1+iηx2 e -σρx1/τ Φ for x 1 > 0, satisfies the Bérenger transmission problem. For Re τ > 0 these solutions decay (resp. grow) exponentially as x 1 → ∞ (resp. x 1 → -∞). Though such solutions serve to verify perfection they don't look very physical in isolation.

On the other hand, when τ is purely imaginary and not equal to zero, the solution is a bounded plane wave in x 1 < 0 and is a plane wave modulated by an exponentially decaying factor in x 1 > 0. These are the solutions which Bérenger constructed to show that the method was perfectly matched.

In the language of the analytic objects constructed in the preceding lemma, Bérenger's plane wave solutions show that H(is, η) = 0 when s is real valued with s 2 > η 2 . For η fixed the function τ → H(τ, η) is holomorphic in the right half plane continuous up to the imaginary axis punctured at ±i|η|, and vanishes on the boundary interval τ = is ∈ iR with s 2 > η 2 . By Schwarz reflection and analytic continuation this implies that H vanishes in the right half plane.

In summary, the computation of Bérenger is actually sufficient to prove perfection for Maxwell's system given the structures provided in this paper.

Remark 4.6. The perfection argument based on plane waves is not valid in full generality where the objects like Ė and H are analytic in Re τ > τ 0 with τ 0 > 0. This is the case, for example, whenever the absorbing layer is amplifying.

Fourier-Laplace analysis with variable σ 1 (x 1 )

Consider the case of only one nonzero σ 1 (x 1 ). If L is hyperbolic for one constant value σ 1 = 0 the scaling (t, x) → (at, ax) shows that L is hyperbolic for σ 1 /a. Therefore L is hyperbolic for all constant values σ 1 .

The results of §4.1 will be extended to the case σ j = 0 for j ≥ 2 and variable coefficient σ 1 (x 1 ). The Fourier-Laplace transform U (τ, x 1 , η) of the Bérenger split operator satisfies

L(τ, d/dx 1 , η) U = F , < x 1 < ∞ , with variable coefficient σ 1 (x 1 ).
The first line of the proof of Lemma 4.7 yields (4.26) with σ = σ 1 (x 1 ). As in the proof of that lemma one derives (4.27) now with σ = σ 1 (x 1 ). The important observation is that the x 1 dependence of the coefficient appears only as a scalar prefactor in (4.27). Such equations will be analysed in the same way as the equations in Lemma 4.2.

4.2.1.

Well posedness by Fourier-Laplace with variable σ 1 (x 1 ) Theorem 4.5. Suppose σ j = 0 for j ≥ 2 and σ 1 (x 1 ) ∈ L ∞ (R) is real valued. Suppose in addition that L is nondegenerate with restect to x 1 , and for one value σ 1 = 0, L is hyperbolic. Then there is a τ 0 > 0 and m so that for all λ > τ 0 and F ∈ e λt L 2 R t : H m (R d t,x ′ ) there is a unique solution solution U ∈ e λt L 2 (R d+1 ) to the Bérenger split problem L U = F . In addition, there is a constant C independent of F, λ so that

e -λt U L 2 (R 1+d ) ≤ C e -λt F L 2 Rt : H m (R d t,x ′ ) . (4.33) Remark 4.7. 1. The condition U ∈ e λt L 2 implies that U tends to zero at t → -∞ as does F . 2. If F is supported in t ≥ t 0 it follows from (4.33) on sending λ → ∞ that U is supported in t ≥ t 0 .
Proof. The values of the Fourier Laplace Transform of W = U j are computed from the ordinary differential equation

A 1 d W dx 1 + τ + σ 1 (x 1 ) τ L 1 (τ, 0, iη) W = F . (4.34)
As in Lemmas 4.2 and 4.3, transform to the equivalent form,

I 0 0 0 d W dx 1 + τ + σ 1 (x 1 ) τ H 11 H 12 H 21 H 22 W = F , H 22 invertible.
Denote the decomposition as W = (W I , W II ) and similarly F . The invertibility of H 22 from Lemma 4.3 yields,

W II = H -1 22 F II -H 21 W I . (4.35)
It suffices to find W I which is determined from,

d W I dx 1 + τ + σ 1 (x 1 ) τ M (τ, η) W I = G , M (τ, η) := H 11 -H 12 H -1 22 H 21 , G := F I + H -1 22 F II .
The hyperbolicity of L implies that M has no purely imaginary eigenvalues. Correspondingly there is the decomposition, into the spectral parts with positive and negative imaginary parts,

W I = W + I + W - I , G = G + + G -, M = M + ⊕ M -.
For σ constant, part (iii) of Lemma 4.7 (using the hyperbolicity of L) implies that for Re τ sufficiently large (depending on σ), one has the spectral decomposition, corresponding to spectra with positive and negative real parts.

τ + σ τ M (τ, η) = τ + σ τ M (τ, η) + ⊕ τ + σ τ M (τ, η) -
Lemma 4.8. If g(x 1 ) satisfies dg(x 1 )/dx 1 = σ 1 (x 1 ) . Then

d dx 1 + M e g(x1)M/τ U I = e g(x1)M/τ d dx 1 + τ + σ 1 (x 1 ) τ M U I .
Proof of Lemma. Since d e gM/τ U I /dx 1 = e gM/τ g ′ M U I /τ + d U I /dx 1 one has

d dx 1 + M e g(x1)M/τ U I = e gM/τ d dx 1 U I + dg/dx 1 M τ + τ M τ U I ,
proving the desired identity.

Therefore

W I = e -g(x1)M/τ d dx 1 + M -1 e g(x1)M/τ G . The unique L 1 fundamental solution of ∂ 1 + M is equal to, e -x1M + χ [0,∞[ (x 1 ) + e -x1M -χ ]-∞,0] (x 1 ) .
Therefore,

e gM/τ W + I = e -x1M + χ [0,∞[ (x 1 ) * (e gM/τ G + , e gM/τ W - I = e -x1M -χ ]-∞,0] (x 1 ) * e gM/τ G -.
The kernel of the integral operator mapping

G + to W + I is equal to, exp -(x 1 -y 1 ) τ + (g(x 1 ) -g(y 1 )/(x 1 -y 1 ) τ M (τ, η) + χ x1≥y1 . (4.36)
Lemma 4.9.

∃ τ 0 = τ 0 (µ), ∀ Re τ ≥ τ 0 , ∀ η ∈ R d-1 , ∀ σ ∈ [-µ, µ], spec τ + σ τ M + (τ, η) ⊂ {Re z > 0} .
Proof of Lemma. Part (iii) of Lemma 4.7 allows one to choose τ 1 so that for σ = µ one has the desired conclusion for Re τ > τ 1 . Then for λ ∈ spec M (τ, η) + one has Re λ > 0, Re 1 + µ τ λ = Re τ + µ τ > 0 .

For 0 ≤ σ ≤ µ write σ = a + bµ with nonnegative a, b summing to 1. It follows that Re (1 + µ/τ )λ > 0. This proves that τ 1 suffices to treat the nonnegative values 0 ≤ σ ≤ µ.

Choosing τ 2 for σ = -µ, that value suffices for -µ ≤ σ ≤ 0. Set τ 0 equal to the maximum of τ 1 and τ 2 .

The Seidenberg-Tarski Theorem 2.3 shows that the absolute values of the real parts of the eigenvalues of M (τ, η) are bounded below by C(|τ | + |η|) -N for some N . And also that the spectral decomposition V → (V + , V -) and its inverse are both bounded polynomially in |τ, η|. More generally for τ, η, µ, σ as above,

spec τ + σ τ M + (τ, η) ⊂ Re z > C(|τ | + |η|) -N .
Taking µ := f L ∞ one finds that for all x 1 , y 1 , the matrix τ + (g(x 1 )g(y 1 )/(x This is proved using Schur's Theorem to reduce M ± to upper triangular form by orthogonal transformations of the spectral subspaces. Then solve the differential equation X ′ + M + X = 0 by back substitution to prove exp(ρM + ) ≤ C|τ, η| p e -cρ/|τ,η| N . The operator with kernel (4.38) is convolution by an element of L 1 (R) whose L 1 norm grows polynomially in |τ, η|. By Young's theorem one concludes that the operator with kernel (4.36) has norm in Hom(L 2 (R)) which grows at most polynomially in |τ, η|.

There is an entirely analogous estimate for the expression for the spectrum with negative real part. Therefore,

W I (τ, x 1 , η) L 2 (R) ≤ C 1 (1 + |τ | + |η|) N G(τ, x 1 , η) L 2 (R) ≤ C 2 (1 + |τ | + |η|) N F (τ, x 1 , η) L 2 (R) .
A

= f, g(0) = 0, so, g(x 1 ) = x1 0 f (s) ds .
Then for γ ∈ G the unique solution of the equivalent initial value problems for the G valued function U ,

A dU dx 1 + f (x 1 ) M U = 0, equivalently, dU dx 1 + f (x 1 ) M U = 0, U (0) = γ, is U (x 1 ) = e -g(x1) M γ .
Proof. Compute using the differential equation,

d dx 1 e g(x1) M U = e g(x1) M dg dx 1 M + dU dx 1 = e g(x1) M f M -f M = 0 .
The lemma follows.

The next result shows that when the Bérenger split problem with absorption σ(x 1 ) defines a stable time evolution, then the problem is perfectly matched. Either the split problem is ill posed, or it is well posed and perfect. Theorem 4.6. Suppose that σ 1 (x) ∈ L ∞ (R) has support in [0, ρ] for some ρ > 0, that σ j = 0 for j = 1, and that the operator L with these absorptions is nondegenerate with respect to x 1 and defines a weakly well posed time evolution. Then, the L evolution is perfectly matched in the sense that for F ∈ C ∞ 0 ({t > 0} ∩ {x 1 < 0}) the solutions U and U ′ with and without absorptions respectively, L 1 U = F , U t≤0 = 0, and,

L U ′ = F , U ′ t≤0 = 0 satisfy U x1<0 = U ′ x1<0 .
and extract the rapidly oscillating term e i(ξ•x+τ t)/ε to find, U ε (t, x) := e i(ξ•x+τt)/ε e i (τ (ξ+εζ)-τ (ξ))t/ε+ζx Π L (τ (ξ + εζ), ξ + εζ) G(ζ) dζ := e i(ξ•x+τt)/ε a(ε, t, x) .

(5.1)

Expanding in ε and keeping just the leading term yields the principal term in the geometric optics approximation

U ε ≈ e i(ξ•x+τ t)/ε e i(x•ζ-v(ξ)•ζt) Π L (τ , ξ) G(ζ) dζ , v(ξ) := -∂ ξ τ (ξ) .
One has

U ε ≈ e i(ξx+τ t)/ε a 0 (x -v(ξ)t) , a 0 (x) := e ix•ζ Π L (τ , ξ) G(ζ) dζ .
A complete Taylor expansion yields the corrected approximations which satisfy the equation with an error O(ε N ) for all N . We write O(ε ∞ ) for short. This yields infinitely accurate solutions,

U ε (t, x) := e i(ξx+τ t)/ε a(t, x, ε) , a(t, x, ε) ∼ a 0 (x -vt) + εa 1 (t, x) + • • • . (5.2)
If 0 is a semi-simple eigenvalue of L(τ, ξ), and Φ 0 ∈ Ker L(τ, ξ) \ {0} of dimension 1, then the leading amplitude a 0 in the case of (1.1) (resp. (1.5)) is of the form

α(t, x) Φ 0 , resp. α(t, x) ξ 1 τ A 1 Φ 0 , . . . , ξ d τ A d Φ 0
with scalar valued amplitude α satisfying,

∂ t + v.∂ x α = 0 .
This shows that α is constant on the rays which are lines with velocity equal to the group velocity v(ξ) := -∂ ξ τ (ξ). For g ∈ C ∞ 0 \ 0 the solutions do not have compact spatial support. This weakness is easily overcome. Choose χ ∈ C ∞ 0 (R d ) with χ = 1 on a neighborhood of the origin. For g ∈ S(R d ), define exact solutions by cutting off the integrand outside the domains of definition of τ (ξ) and Π L (τ (ξ), ξ), u ε (t, x) := e i(ξx+τ (ξ)t) Π L (τ (ξ), ξ) g(ξξ/ε) χ( √ ε(ξξ/ε)) dξ .

(5.

3)

The analysis above applies with the only change being the initial values. In the preceding case these values were equal to the transform of Π L (τ (ξ), ξ) g(ξξ/ε) and in the present case they are infinitely close to that quantity,

u ε (0, x) = e ix.ξ Π L (τ (ξ), ξ) g(ξ -ξ/ε) dξ + O(ε ∞ ) .
This yields infinitely accurate approximate solutions (2.5) which have support in the tube of rays with feet in the support of e ix.ξ Π L (τ (ξ), ξ) g(ξ) dξ.

Geometric optics with variable coefficients

The Fourier transform method of the preceding sections is limited to problems with constant coefficients. In this section the WKB method which works for variable coefficients is introduced. It will also serve for the analysis of reflected waves.

Let L be the general operator in (2.1). Fix (τ, ξ) ∈ Char L and seek asymptotic solutions

U ε ∼ e iS/ε +∞ j=0
ε j a j (t, x), with the phase S(t, x, ξ) = tτ + xξ.

(5.4)

Contents 43

More precisely we construct smooth functions a j (t, x) with supp a j ∩ ([0, T ] × R d ) compact so that if

a(t, x, ε) ∼ ∞ j=0 ε j a j (t, x) ,
in the sense of Taylor series at ε = 0, and supp a ∩ ([0, T ] × R d ×]0, ε]) is compact, then U ε := e iS/ε a(t, x, ε) satisfies for all s, N ,

L U ε H s ([0,T ]×R d ) = O(ε N ) .
In this case we say that (5.4) is a infinitely accurate approximate solution. The next result recalls some facts about such solutions.

Theorem 5.1. Suppose Problem (2.1) is hyperbolic, (τ, ξ) ∈ Char(L) satisfies the smooth variety hypothesis, and 0 is a semi-simple eigenvalue of L(τ, ξ).

(i) If the coefficients a j satisfy the recursion relation

a 0 (t, x) ∈ Ker L 1 (τ, ξ),
(5.5a)

∀j ≥ 0, iL 1 (τ, ξ) a j+1 (t, x) + L(∂ t , ∂ x ) a j (t, x) = 0, (5.5b)
then (5.4) is an infinitely accurate approximate solution of (2.1).

(ii) If g j = Π L (τ, ξ)g j ∈ C ∞ 0 (R d x )
are supported in a fixed compact K, then there is one and only one family of a j satisfying (5.5) together with the initial conditions, Π L (τ, ξ)a j (0, •) = g j and the polarization Π L (τ, ξ)a 0 = a 0 . They have support in the tube of rays with feet in K and speed of propagation v(ξ) = -∂ ξ τ (ξ). (iii) The principal term a 0 is a solution of the transport equation

∂ t a 0 + v(ξ) • ∂ x a 0 + Π L (τ, ξ)B(x)Π L (τ, ξ)a 0 = 0.
(

Proof. For simplicity note Π L := Π L (τ, ξ) when no ambiguity is to be feared. The equations (5.5) are obtained by injecting U ε in (2.1), to find an expression ∼ e iS/ε j≥0 ε j w j (t, x). In order that the w j vanish it is necessary and sufficient that the equations (5.5) are satisfied.

Next examine the leading order terms to find the relations determining a 0 . Projecting the case j = 0 of (5.5) onto Ker L 1 yields,

Π L (∂ t + d l=1 A l ∂ l + B(x)) a 0 = 0.
This yields a first order system satisfied by a 0 = Π L a 0 ,

∂ t a 0 + d l=1 Π L A l Π L ∂ l a 0 + Π L B(x) Π L a 0 = 0.
(5.7)

The leading order part of this equation is a scalar transport operator. To see this differentiate L 1 (τ (ξ), ξ)Π L (τ (ξ), ξ) = 0 with respect to ξ l to find

A l + ∂τ (ξ) ∂ξ l Id Π L (τ (ξ), ξ) + L 1 (τ (ξ), ξ) ∂ ∂ξ l (Π L (τ (ξ), ξ)) = 0.
Multiplying on the left by Π L (τ (ξ), ξ) eliminates the second term yielding,

Π L A l Π L + ∂τ (ξ) ∂ξ l Π L = 0.
Injecting this in (5.7) yields (5.6).

In order to compute the coefficients recursively, multiply (5.5b) on the left by the partial inverse Q L (τ, ξ), using the identity in (2.7), to obtain for j ≥ 1,

(I -Π L )a j = i Q L L(∂ t , ∂ x ) a j-1 .
(5.8) Projecting (5.5b) on the kernel yields,

Π L L(∂ t , ∂ x ) a j = 0 .
Writing a j as

a j = Π L a j + (I -Π L ) a j , yields Π L L(∂ t , ∂ x ) Π L a j = -Π L L(∂ t , ∂ x ) (I -Π L ) a j .
This is again a transport equation, but with a righthand side,

∂ t Π L a j + v • ∂ x Π L a j + Π L B Π L a j = -Π L L(∂ t , ∂ x ) (I -Π L ) a j .
(5.9)

(5.8) and (5.9) permit to calculate the coefficients recursively, knowing the initial values.

Next apply the above algorithm to the PML operator L. Fix (τ, ξ) ∈ Char L and seek asymptotic solutions

U ε ∼ e iS/ε +∞ j=0 ε j ãj (t, x),
with the phase S(t, x) = tτ + x • ξ.

(5.10)

Corollary 5.1. Suppose Problem (1.1) is strongly well posed, (τ, ξ) ∈ CharL satisfies the smooth variety hypothesis, and 0 is a semi-simple eigenvalue of L(τ, ξ).

(i) If the coefficients ãj satisfy the recursion relation

ã0 (t, x) ∈ Ker L 1 (τ, ξ), (5.11a
)

∀j ≥ 0, (I -Π L ) ãj (t, x) = iQ L L(∂ t , ∂ x ) ãj-1 (t, x), (5.11b)
∂ t Π L ãj + v • ∂ x Π L ãj + β(x) Π L ãj = -Π L L(∂ t , ∂ x )(I -Π L ) ãj .
(5.11c) then (5.10) is an infinitely accurate approximate solution of (1.5). (ii) If gj (x) = Π L gj ∈ C ∞ 0 (R d ) are supported in a fixed compact K, then there is one and only one family of ãj satisfying (5.11) together with the initial conditions, Π L (τ, ξ)ã j (0, x) = gj and the polarization Π L (τ, ξ)ã 0 = ã0 . They have support in the tube of rays with feet in K and speed of propagation v = -∂ ξ τ (ξ). (iii) The principal term ã0 is a solution of the transport equation

∂ t ã0 + v • ∂ x ã0 + β(x)ã 0 = 0, with β(x) = d l=1 σ l (x l )ξ l τ (ξ) ∂τ (ξ) ∂ξ l .
(5.12)

Proof. We need only identify the constant term in (5.6). Use the form of the projector given in Proposition 2.1, to obtain

Π L B(x) Π L = β(x)Π L .

Amplifying layers

The coefficient σ 1 (x 1 ) ≥ 0 is introduced with the idea that waves will be damped in the layer. In this section, we show that sometimes the anticipated decay is not achieved, and waves may be amplified. This was observed in [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF]. The authors analysed the phenomenon for σ constant in the layer. They showed that in an infinite layer solutions can in certain cases grow infinitely large. We present a related analysis using WKB solutions which has three advantages, 2. The growth is seen immediately and not expressed in terms of large time asymptotics.

3. The analysis in [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF] was in part restricted to d = 2 and eigenvectors of multiplicity one. We remove these restrictions.

It is because of 2 that we choose not to follow the authors of [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF] in calling this phenomenon instability.

Theorem 5.2. Suppose (τ, ξ) ∈ Char(L) satisfies the smooth variety hypothesis and β(x) is as in (5.12). Suppose in addition there is an interval on a ray

Γ := {(0, x) + t(1, -∂ ξ τ (ξ)), 0 ≤ t ≤ t 0 }, so that, t0 0 β(x -t∂ ξ τ (ξ))) dt < 0.
Then the corresponding WKB solution grows in the layer.

Proof. The solution of the transport equation (5.12) evaluated on Γ is

ã0 (t 0 , x) = exp - t0 0 β(x -s∂ ξ τ (ξ))))ds ã0 (0, x + t 0 ∂ ξ τ (ξ)).
The exponential is strictly greater than 1, so

|ã 0 (t 0 , x + t 0 ∂ ξ τ (ξ))| > |a 0 (0, x)|.
β = d j=1 σ j (x j ) ξ 2 j ξ 2 ≥ 0 .
Example 5.2 (Amplification is common). For the dispersion relation τ 2 = q(ξ) where q is a positive definite quadratic form so that the ξ axes are not major and minor axes of the ellipse q = 1, there are always τ > 0, ξ so that x 1 layers with σ 1 > 0 are amplifying ( [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF]). There are two lines on {τ = q(ξ) 1/2 } where ∂q/∂ξ 1 = 0. The half cone on which ∂q/∂ξ 1 < 0 corresponds to rays on which x 1 is increasing so they enter a layer x 1 > 0. The half cone {∂q/∂ξ 1 < 0} is divided into two sectors by the plane ξ 1 = 0. The sector on which ξ 1 > 0 (resp. ξ 1 < 0) corresponds to growing (resp. decaying) solutions (see Figure 1 on the left). This example shows that amplification is very common. Consequently for the dispersion ξ 1 = 0 outgoing, ∂q ∂ξ1 < 0 outgoing Fig. 1. Amplified outgoing wave numbers in bold relation τ 2 = q(ξ) it is wise to align coordinates along the major and minor axes of the ellipse to avoid amplification. However, if (τ 2q 1 )(τ 2q 2 ) divides the characteristic polynomial and the axes of q 1 and q 2 are distinct from each other then no linear change of coordinates can avoid amplification in the layer.

A second example from [START_REF] Diaz | A time domain analysis of PML models in acoustics[END_REF] is the linearized compressible Euler equation with nonzero background velocity (c, 0), c > 0 for which amplified wave numbers at a right hand boundary are indicated in bold in the right hand figure 1.

Denote by ρ ℓ the reflection coefficient of R µ . The leading amplitude of the reflected ℓ wave is then ρ ℓ ε. The preceding identity shows that ρ ℓ = ρ ℓ µ showing that the reflection coefficients are linear in µ. b

Reflection for Maxwell with smart layers

In this section L may denote one of two distinct operators. One option is the Maxwell operator L 1 from (3.2) for the C 3 valued field E + iB. The lower order term is B := µC from the smart layer (6.1). Alternatively L may denote the Bérenger operator operator L with lower order term B = µ C with

C =    σ(x 1 ) I N 0 . . . 0 . . . 0 . . . . . . 0 . . . 0 0    , supp σ ⊂ [0, ∞[ .
In both cases the absorption term is linear in µ. We compute the dependence of the reflection coefficient on µ. Lemma 4.6 shows that the Cauchy problem is equivalent to homogeneous problems in each half-space with a transmission condition on Γ := {x 1 = 0},

A 1 U Γ = 0 . (6.3)
In order to cover both cases the operator, coefficients, and unknown are indicated with round letters.

We study the reflection of high frequency waves in x 1 ≤ 0 which approach the boundary x 1 = 0. The input is an incident wave with phase S I (t, x) := τ t + ξ • x, where τ = 0 and τ (ξ) = ±|ξ|. The phase is chosen so that the group velocity v = -ξ/τ satisfies v 1 > 0. Denote x ′ := (x 2 , x 3 ), ξ ′ = (ξ 2 , ξ 3 ). Theorem 5.1 applies to the incident wave with B ≡ 0. In x 1 ≤ 0, the incident wave is

U ε := e iS I (t,x)/ε a I (t, x, ε) , a I (t, x, ε) ∼ ∞ j=0 ε j a I j (t, x), L 1 (∂ t , ∂ x ) U ε = O(ε ∞ ) . (6.4)
Suppose that the amplitudes a I j are supported in a tube, T , of rays with compact temporal crossections T ∩ {t = 0} ⊂⊂ {x 1 < 0}.

We construct a transmitted wave with the same phase, and a reflected wave with phase S R (t, x) := τ t+ ξ R x, with ξ R := (-ξ 1 , ξ ′ ). We first show that there are uniquely determined reflected and transmitted waves. Then we compute exactly the leading terms in their asymptotic expansions.

The reflected wave V ε is also supported in x 1 ≤ 0. The group velocity for the reflected wave is equal to v R := (-v 1 , v ′ ), and in x 1 ≤ 0,

V ε = e iS R (t,x)/ε a R (t, x, ε), a R (t, x, ε) ∼ ∞ j=0 ε j a R j (t, x), L 1 (∂ t , ∂ x )V ε = O(ε ∞ ) . (6.5)
The transmitted wave is supported in x 1 ≥ 0, (i) Given the incoming amplitudes a I j there are uniquely determined amplitudes a T j and a R j so that for any choice of the a I,R,T (t, x, ε) ∼ ε j a I,R,T (t, x), the U ε , V ε and W ε are infinitely accurate b This argument can be made rigorous under the following conditions. The incoming wave is a wave packet with oscillatory part e i(τ t+xξ)/ε with (τ, ξ) ∈ CharL. Denote (τ, ξ ′ ) the part determining the oscillations in x 1 = 0. Consider the roots ξ 1 of each of the equation, det L 1 (τ, ξ 1 , ξ ′ ) = 0. The nonreal roots are called elliptic. They lead to waves which have the structure of a boundary layer of thickness ∼ ε. The real roots are called hyperbolic. The favorable situation is when all the hyperbolic roots are at smooth points of the characteristic variety and the group velocities are transverse to the boundary. In that case one can construct infinitely accurate asymptotic solutions of the transmission problem consisiting of incoming, reflected, and transmitted wave packets corresponding to the hyperbolic roots, and, a finite number of boundary layers corresponding to elliptic roots. As this is a long story, we content ourselves with the Maxwell computation of the next subsection. solutions of the differential equations and the transmission condition is also satisfied to infinite order,

W ε := e iS I (t,x)/ε a T (t, x, ε), a T (t, x, ε) ∼ ∞ j=0 ε j a T j (t, x), L(∂ t , ∂ x )W ε = O(ε ∞ ) . ( 6
∀(t, x ′ ) ∈ R × R 2 , A 1 (U ε + V ε )(t, 0 -, x ′ ) = A 1 (W ε )(t, 0 + , x ′ ) + O(ε ∞ ) . (6.7)
(ii) In the case of Bérenger's PML, the coefficients ãR j vanish identically for j ≥ 0. (iii) For the smart layer (6.1, 6.2) with σ = 1 x1>0 , the coefficient a R 0 vanishes identically. The reflection coefficient of the layer is equal to

R(τ, ξ) = iµ (1 + ν) ξ 2 1 -τ 2 8τ ξ 2 1 = i µ(1 + ν) 8τ v 2 1 -1 v 2 1 . That is, if a I 0 (t, 0 -, x ′ ) = α(t, x ′ )Φ(τ, ξ) ∈ Ker L(τ, ξ), then a R (t, 0 -, x ′ ) = εR(τ, ξ) α(t, x ′ ) Φ(τ, ξ R ) + O(ε 2 ) .
Furthermore, the amplitudes a T , R are such that on the interface Γ, we have for all (i, j) ∈ N 2 , i = 0 and j ≤ 1, or i ≥ 1 and j ≥ 0, c

∂ j 1 a T i -∂ j 1 a I i ∈ µ C i+j-1 [µ] ⊗ C 3 , ∂ j 1 a R i ∈ µ C i+j-1 [µ] ⊗ C 3 . (6.8)
(iv) The smart layer with σ(x 1 ) satisfying σ(0

) = • • • = σ (k-1) (0) = 0, σ (k) (0) = 0 is nonreflecting at order k for any angle of incidence, i.e. if a I 0 (t, 0 -, x ′ ) = α(t, x ′ )Φ(τ, ξ), there exists R k (τ, ξ) such that a R (t, 0 -, x ′ ) = ε k σ (k) (0) R k (τ, ξ) α(t, x ′ ) Φ(τ, ξ R ) + O(ε k+1 ) .
Furthermore the amplitudes a T , R are linear functions of µ on the interface Γ. That is denoting

c R i (µ) = a R i |Γ and c T i (µ) = a T i |Γ -a I i |Γ , we have for all i ≥ k in N, c R, T i (µ) ∈ µ C i-1 [µ] ⊗ C 3 .
Remark 6.1. 1. There exist choices of a I,R,T so that U ε , V ε , and W ε is an exact solution. Since the transmission problem is well posed, there is a uniquely determined corrector c ε smooth and infinitely small on both sides so that adding c ε yields an exact solution. Adding c ε to the left corresponds to adding the infinitely small term c ε e iS I /ε to a I with a similar remark on the right. 2. Part (iv) of the theorem with k = 0 generalizes part (iii) to discontinuous and variable σ(x 1 ).

3. The basis elements, Φ R for a R 1 and Φ I for a I 0 are homogeneous of degree 2 in τ, ξ. Doubling τ, ξ and also ε leaves the incoming and reflected waves unchanged. Therefore εR(τ, ξ) must be equal to 2εR(2τ, 2ξ). This explains why R is homogeneous of degree -1. 4. The reflection coefficient vanishes when ξ ′ = 0. Since it is an even function of ξ, ∇ ξ R = 0 too.

Proof. The incoming solution is given. (i) Seek the leading amplitudes a T 0 and a R 0 . We will show that a R 0 = 0 so it is actually a R 1 that is the leading amplitude of the reflected wave. A jump discontinuity in a lower order coefficient does not lead to reflection at leading order. Denote , and a T 0 (resp. a R 0 ) satisfies a forward transport equation in x 1 ≥ 0 (resp. backward in x 1 ≤ 0) with zero initial values in time,

L T := ∂ t + A 2 ∂ 2 + A 3 ∂ 3 ; L 1 := L T + A 1 ∂ 1 ; L := L 1 + µC . T := ∂ t + v 2 ∂ 2 + v 3 ∂ 3 is
(v 1 ∂ 1 + T ) a I 0 = 0, x 1 ∈ R, (-v 1 ∂ 1 + T ) a R 0 = 0, x 1 ∈ R -, (v 1 ∂ 1 + T + µΠ L C Π L ) a T 0 = 0, x 1 ∈ R + .
(6.9) c C j [µ] denotes the space of polynomials of degree less than or equal to j with complex coefficients. C j [µ] ⊗ C 3 is the corresponding space of polynomials with coefficients in C 3 .

Using (I

-Π L )a R 1 = 0 in the transmission condition yields, A 1 Π R L a R 1 + Π L a I 1 -Π L a T 1 = iµA 1 Q L C 1 a I 0 . (6.17)
The eigenvalues of A 1 are 0 and ±1, with associated orthonormal set of eigenvectors Φ 0 = e 1 and Φ ± = 0, 1, ±i / √ 2. The projection operators on the positive and negative eigenspaces are π ± (A 1 ) = Φ ± Φ * ± , and

C = Φ + Φ * + + νΦ -Φ * -. The kernel of L(τ, ξ) is one-dimensional, it is spanned by Φ(τ, ξ) = ξ - τ 2 ξ 1 e 1 + i τ ξ 1 ξ ∧ e 1 = ξ 1 - τ 2 ξ 1 , i τ ξ 1 ξ 3 + ξ 2 , -i τ ξ 1 ξ 2 + ξ 3 , (6.18)
and the projection on Ker

L(τ, ξ) is Π L = ΦΦ * Φ * Φ . Compute Π L CΠ L = ΦΦ * Φ * Φ (Φ + Φ * + + νΦ -Φ * -) ΦΦ * Φ * Φ = 1 (Φ * Φ) 2 Φ(Φ * Φ + )(Φ * + Φ)Φ * + νΦ(Φ * Φ + )(Φ * + Φ)Φ * = |Φ * Φ + | 2 + ν|Φ * Φ -| 2 Φ * Φ Π L . Define γ := |Φ * Φ + | 2 + ν|Φ * Φ -| 2 Φ * Φ , so, Π L CΠ L = γΠ L . (6.19)
Since a I 0 is polarized,

C 1 a I 0 = (C - γ v 1 A 1 Π L ) a I 0 = (C - γ v 1 A 1 ) a I 0 := C 1 a I 0 .
To compute the righthand side of (6.17), use

C =       0 0 0 0 ν + 1 2 i ν -1 2 0 -i ν -1 2 ν + 1 2       , γ = (τ -ξ 1 ) 2 + ν(τ + ξ 1 ) 2 4τ 2 = 1 4 ((1 + v 1 ) 2 + ν(1 -v 1 ) 2 ) ,
to find

C 1 = ν + 1 2       0 0 0 0 1 i v 2 1 + 1 2v 1 0 -i v 2 1 + 1 2v 1 1       .
Write a I 0 = α I 0 Φ, and compute

C 1 Φ = (1 + ν) ξ 2 1 -τ 2 4τ ξ 2 1 Ψ, Ψ =   0 ξ 2 τ -iξ 1 ξ 3 , ξ 3 τ + iξ 2 ξ 3 ,   ,
to find a new version of (6.17), Since Q L is the left inverse of L, we have Q L ξ = 1 τ ξ, and Q L Φ 3 = 1 2τ Φ 3 , which gives

A 1 (Π R L a R 1 + Π L a I 1 -Π L a T 1 ) = iµ(1 + ν) ξ 2 1 -τ 2 4τ ξ 2 1 α I 0 A 1 Q L Ψ . ( 6
Q L Ψ = τ 2 -ξ 2 1 τ 2 ξ + ξ 2 1 2τ 2 Φ 3 .
Write the coefficients on Γ as

Π L a I, T 1 = α I, T 1 Φ, Π R L a R 1 = α R 1 Φ R
, and inject into the transmission condition to obtain

α R 1 A 1 Φ + (α I 1 -α T 1 )A 1 Φ R = iµ(1 + ν) ξ 2 1 -τ 2 4τ ξ 2 1 α I 0 A 1 ( τ 2 -ξ 2 1 τ 2 ξ + ξ 2 1 2τ 2 Φ 3 ).
Since the kernel of A 1 is e 1 , A 1 in the preceding identity may be replaced by the projection on (e 2 , e 3 ). The projection of Φ is ϕ = ξ ′i τ ξ1 ξ ∧ e 1 , and note that Φ 3 and Φ(τ, ξ R ) have the same projection, which is

ϕ 3 = ξ ′ + i τ ξ1 ξ ∧ e 1 . Write α R 1 ϕ 3 + (α I 1 -α T 1 )ϕ = iµ(1 + ν) ξ 2 1 -τ 2 4τ ξ 2 1 α I 0 ( τ 2 -ξ 2 1 τ 2 ξ ′ + ξ 2 1 2τ 2 ϕ 3 ) = iµ(1 + ν) ξ 2 1 -τ 2 4τ ξ 2 1 α I 0 ( τ 2 -ξ 2 1 2τ 2 (ϕ + ϕ 3 ) + ξ 2 1 2τ 2 ϕ 3 ) = iµ(1 + ν) ξ 2 1 -τ 2 4τ ξ 2 1 α I 0 ( τ 2 -ξ 2 1 2τ 2 ϕ + 1 2 ϕ 3 ).
The solutions are parameterized by α I 0 ,

α R 1 = iµ(1 + ν) ξ 2 1 -τ 2 8τ ξ 2 1 α I 0 , α I 1 -α T 1 = -iµ(1 + ν) (ξ 2 1 -τ 2 ) 2 8τ 3 ξ 2 1 α I 0 .
Now use the results in (i) and prove (6.8) by induction on i. Equation (6.12) asserts that a R 0 = 0 for x 1 < 0. Equations (6.9), and (6.10), imply that at x 1 = 0, a I, T 0 = Π L a I, T 0 , and v 1 (∂ 1 a T 0 -∂ 1 a I 0 ) + T (a T 0a I 0 ) + µγ(a T 0a I 0 ) + µγa I 0 = 0, x 1 ≥ 0.

Differentiation in x 1 several times yields ∂ j 1 a T 0 -∂ j 1 a I 0 ∈ µ C j-1 [µ] ⊗ C 3 , giving the results for i = 0 and j ≥ 1.

Assuming the inductive hypothesis is true for i we prove it for i + 1. Write (5.11) in the form

(I -Π R L ) a R i+1 (t, x) = iQ R L (L T (∂ t , ∂ x ′ ) + A 1 ∂ 1 ) a I i (t, x), (I -Π L ) a I i+1 (t, x) = iQ L (L T (∂ t , ∂ x ′ ) + A 1 ∂ 1 ) a I i (t, x), (I -Π L ) a T i+1 (t, x) = iQ L (L T (∂ t , ∂ x ′ ) + A 1 ∂ 1 + µC) a T i (t, x). By induction, ∂ j 1 (I -Π R L ) a R i (t, x) ∈ µ C i+j-1
[σ] ⊗ C 3 on the interface. Write for x 1 ≥ 0, (I -Π L ) (a T i+1a I i+1 )(t, x) = iQ L (L T + A 1 ∂ 1 + µC) (a T ia I i )(t, x) + iµ Q L C a I i (t, x). (6.21)

The inductive hypothesis, shows that on Γ, ∂ j 1 (L T +A 1 ∂ 1 +µC) (a T i -a I i ) ∈ µ C i+j [µ]⊗C 3 and ∂ j 1 (µC a I i ) ∈ µ C 0 [σ] ⊗ C 3 . The result follows for (I -Π L ) (a T i+1a I i+1 ), and for (I -Π R L ) a R i+1 in the same way. The transmission condition extends the assertion to the other parts Π L (a T i+1a I i+1 ) and Π R L a R i+1 .

Harmoniously matched layers

Based on Theorem 6.1 we construct an extrapolation method for symmetric hyperbolic operators with smart layers which eliminates the leading order reflection. The resulting method has desirable stability properties and is nearly as good as Bérenger's algorithm for the Maxwell equations where his method is at its best. We think that the new method provides a good alternative in situations where Bérenger's method is not so effective. Consider the computational domain x 1 ≤ b 1 . The domain of interest is the interval x 1 ≤ a 1 < b 1 . The absorbing layer is located in a 1 ≤ x 1 ≤ b 1 . The differential operator in the computational domain is symmetric hyperbolic L with smart layer L U + σ 1 (x 1 ) π + (A 1 ) + ν π -(A 1 ) U = 0 , σ 1 ≥ 0 , supp σ 1 ⊂ {x 1 ≥ a 1 } .

At the outer boundary x 1 = b 1 of the absorbing layer impose the simplest weakly reflecting boundary condition

π -(A 1 ) U = 0 when x 1 = b 1 .
This is a well posed problem provided that A 1 has constant rank on x 1 = b 1 . When L = L 1 (∂) has constant coefficients it generates a contraction group in L 2 ({x 1 ≤ b 1 }).

The hamoniously matched layer algorithms compute a smart layer with coefficient σ 1 and also with coefficient 2σ 1 . In view of Theorem 6.1, subtracting the second from twice the first, 2 U (σ 1 ) -U (2σ 1 ), yields a field with one more vanishing term in the reflected wave at the interface x 1 = a 1 . This extrapolation removes the leading reflection.

The harmonious matched layer algorithms in a rectangular domain R perform the same extrapolation with absorptions in all directions. With LU + d j=1 σ j (x j )(π + (A j ) + ν π -(A j ))U = 0, σ j ≥ 0, supp σ j ⊂ {|x j | ≥ a j } . Open Problem. For discontinuous σ j , the uniqueness of solutions to the initial boundary value problem on the rectangular computational domain is not known because of the discontinuity of the boundary space ker A j at the corner. Solutions are typically discontinuous. Uniqueness of strong solutions and existence of weak solutions is proved by the energy method. We do not know how to prove uniqueness of solutions with regularity not exceeding that of solutions known to exist. Similar problems plague virtually all methods on rectangular domains with absorbing boundary conditions imposed on the computational domain with corners. The present problem is one of the simplest of its kind. The fact that algorithms designed to compute solutions encounter no difficulties is reason for optimism.

Numerical experiments

Simulations are performed for the 2-D transverse electric Maxwell system in the (x, y) coordinates,

∂ t E x -∂ y H z = 0 , ∂ t E y + ∂ x H z = 0 , ∂ t H z + ∂ x E y -∂ y E x = 0 , (6.27)
d This can be proved by penalisation. Denote by Ω the rectangular computational domain. Add Λ 1 R d \Ω to L and solve on R 1+d t,x . The limit as Λ → ∞ provides a solution in L ∞ [0, T ] : L 2 (R) [START_REF] Bardos | Maximal positive boundary value problems as limits of singular perturbation problems[END_REF].

in a rectangle, with boundary conditions n ∧ E = 0 on the west, north and south boundaries. The layer will be imposed on the east boundary. Maxwell Bérenger is given by (0, 1, -1), the smart layers are:

∂ t E x -∂ y H z =
∂ t E x -∂ y H z = 0 , ∂ t E y + ∂ x H z + σ(x) 2 (E y + H z + ν(E y -H z )) = 0 , ∂ t H z -∂ y E x + ∂ x E y + σ(x)
2 (E y + H zν(E y -H z )) = 0 .

(6.30)

The boundary conditions (6.29) are imposed. The Yee scheme for Maxwell is The Yee scheme for Maxwell Bérenger using the notations σ i = σ(x i ) and σ i+ 1 2 = σ(x i+ 1 2 ) is,

(E x ) n i+ 1 2 ,j -(E x ) n-
(E x ) n i+ 1 2 ,j -(E x ) n-1 i+ 1 2 ,j ∆t - (H z ) n-1 2 i+ 1 2 ,j+ 1
In the first set of experiments, the absorption coefficient is constant in the layer, equal to 2. The initial magnetic field hits the layer at incidence 0 • ( v = (1, 0)) or 45 • ( v = (1, 1)).

In Table 1 we compare the performances on a high frequency wave (k = 10), while in Table 2 we consider a low frequency wave (k = 1).

In Table 1 we compare the performances on a high frequency wave (k = 10), while in Table 2 we consider a low frequency wave (k = 1). In Tables 3 and4, we perform the same set of experiments, but the absorption coefficient is now a third degree polynomial in the layer, equal to (x -6) 3 /8. The performance of the HMLV3 gives hope the method with its stronger well posedness, more robust absorption, and small reflection at all angles will be a good method where Bérenger has proven less good. For example, for non constant coefficients and nonlinear problems. We have taken pains to make the comparison where Bérenger is at its best. In 2D with a layer in a single direction the HML has an extra cost. Since there are 5 quantities to compute at each time step instead of 4 for Bérenger. This is no longer the case in three dimensions, since both strategies have to split 6 unknowns.

Open problems. 1. Our analysis does not explain the much better behavior with continuous absorption, nor the advantages of HMLV3. 2. A comparison with other methods where only supplementary ordinary differential equations are added should be made.

Theorem 2 . 3 (

 23 Seidenberg-Tarski Theorem). If Q(ρ, ζ), R(ρ, ζ), and S(ρ, ζ) are polynomials with real coefficients in the n + 1 real variables (ρ, ζ 1 , . . . , ζ n ) and the set M (ρ) := ζ : R(ρ, ζ) = 0, S(ρ, ζ) ≤ 0 is nonempty when ρ is sufficiently large, define µ(ρ) := sup ζ∈M(ρ) Q(ρ, ζ).

  (3.25) to e j that M is necessarily diagonal, M = diag(m 1 , m 2 , m 3 ). Applying(3.25) to e j+1 and e j+2 shows that for any j, N e j+1 = m j+2 S j e j+1 , N e j+2 = m j+1 S j e j+2 .

(4. 1 )

 1 Hyperbolicity of L guarantees that for Re τ > τ 0 and η ∈ R d-1 this equation has no purely imaginary roots.The number of boundary conditions at x 1 = 0 for the boundary value problem in the right half space is chosen equal to the number of roots with negative real part (see also Remark 4.1). That integer must May 6, 2011 19:12 WSPC/INSTRUCTION FILE confluentesvf.hyper27237 Contents 25

Definition 4 . 3 .

 43 For a transmission problem (L, R, N ) satisfying Hersh's condition (Definition 4.2), Re τ > τ 0 and η ∈ R d-1 , define the reflection operator, H(τ, η) : Ė+ L (τ, η) → Ė-L (τ, η) as follows. Hersh's condition implies that for each K

Theorem 4 . 2 .

 42 Suppose that the transmission problem (L, R, N ) satisfies the Hersh condition. The following are equivalent. (i) The transmission problem is perfectly matched in the sense of Definition 1.1.

Proposition 4 . 1 .

 41 Consider the dissipative symmetric hyperbolic example with d = N = M = 2,

4. 1 . 4 . 1 Theorem 4 . 4 .

 14144 Perfection for Bérenger's PML with piecewise constant σ With the hypotheses of Theorem 4.3, the Bérenger transmission problem is perfectly matched. The Bérenger transmission problem that is only split on the right is also perfectly matched.

May 6 ,

 6 2011 19:12 WSPC/INSTRUCTION FILE confluentesvf.hyper27237 Contents 39

Example 5 . 1 (

 51 No amplification for Maxwell/D'Alembert). If the dispersion relation is τ 2 = |ξ| 2 and σ ≥ 0 then there is no amplification since,

. 6 )

 6 Theorem 6.1.

  the tangential transport operator. By Theorem 5.1, the amplitudes are polarized, i.e. a I, R, T 0 = Π L a I, R, T 0

. 20 ) 1 (τ e 1 - 1 (τ e 1 +

 201111 Next compute Q L Ψ. First compute a basis of eigenvectors for L(τ, ξ). Φ 2 is such that L(τ, ξ)Φ 2 = τ Φ 2 , Φ 3 is such that L(τ, ξ)Φ 3 = 2τ Φ 3 . Choose Φ 2 = ξ, Φ 3 = Φ(-τ, ξ).Note thatΨ = τ ξξ 1 (τ e 1 + iξ ∧ e 1 ), iξ ∧ e 1 ), Φ 3 = ξ -τ ξ iξ ∧ e 1 ),which gives(τ e 1 + iξ ∧ e 1

  with π ∓ (A j ) U = 0 when x j = ±b j .This initial boundary value problem on a rectangle has weak solutions.d When L = L 1 (∂), the L 2 (R) norm is nonincreasing in time. The extrapolation is 2 U (σ 1 , • • • , σ d ) -U (2σ 1 , • • • , 2σ d ).

(6. 28 )

 28 For the computation, these equations are used in the whole rectangle (see the discussion in the introduction), with σ = 0 outside the layer. The boundary conditions are E y = H z and E x = 0 on the east , n ∧ E = 0 on the other boundaries. (6.29) Since Π + (A 1 ) = Ey+Hz 2 (0, 1, 1) and Π -(A 1 ) = Ey-Hz 2

  Theorem 3.4. Suppose that L defines a strongly well posed Cauchy problem and that the multiplicity of

The forward propagation cone is dual to the timelike cone. A lipschitzean curve [a, b] ∋ t → (t, γ(t) is an influence curve when γ ′ belongs to the propagation cone for Lebesgue almost all t ∈ [a, b].

  The elements of the matrix(4.37) are bounded above polynomially in |τ, η|. Therefore the kernel (4.36) is bounded above by |τ, η| N expc(x 1y 1 )/|τ, η| N χ x1≥y1 , c > 0 . (4.38)

	τ	1 -y 1 )	M (τ, η) +	(4.37)
	has spectrum in
	Re z > C(|τ | + |η|) -N ,	C > 0 .

  similar estimate for W II follows from(4.35). Estimates for U j follow from the second equation in (4.26). Plancherel's Theorem then implies (4.33), proving the existence part of well posedness.Uniqueness is proved by a duality argument of Hölmgren type using existence (backward in time) for the adjoint differential operator (details omitted).4.2.2.Perfection for Bérenger's PML with variable coefficient σ 1 (x 1 ) Lemma 4.10. Suppose that A, M satisfy the hypotheses of Lemma 4.2 with G and M ∈ Hom G are from that Lemma. Suppose in addition that f ∈ L ∞ loc (R ; C) and g is the unique solution of dg dx 1

  0 , ∂ t E y + ∂ x H z + σ(x)E y = 0 , ∂ t H zx + ∂ x E y + σ(x)H zx = 0 , ∂ t H zy -∂ y E x = 0 , H z = H zx + H zy .

Table 1 .

 1 Comparison of the L ∞ errors for high frequency, discontinuous absorption.

			normal incidence			45 • incidence
	refinement	0	1	2	0	1	2
	Bérenger 9.4e-02 3.9e-02 7.9e-03 1.3e-01 2.9e-02 5.6e-03
	Smart	5.2e-02 1.3e-02 5.1e-04	6.2e-02 1.1e-02 5.3e-03
	HMLV1	3.4e-02 3.1e-03 2.1e-05	4.5e-02 1.2e-03 5.5e-04
	HMLV2	2.5e-02 6.0e-03 1.2e-03	7.4e-02 1.1e-02 1.7e-03
	HMLV3 2.1e-02 4.2e-03 5.1e-04 4.5e-02 5.3e-03 5.7e-04

Table 2 .

 2 Comparison of the L ∞ errors for low frequency, discontinuous absorption.

			normal incidence		45 • incidence
	refinement	0	1	2	0	1	2
	Bérenger 1.5e-02 7.1e-03 3.5e-03 1.3e-02 6.1e-03 3.0e-03
	Smart	2.0e-02 2.0e-02 2.01e-02 4.3e-02	4.2e-02 4.2e-02
	HMLV1	1.7e-02 1.60e-02 1.6e-02	3.4e-02	3.3e-02 3.2e-02
	HMLV2	1.8e-02 1.1e-02 6.7e-03	3.1e-02	1.9e-02 1.1e-02
	HMLV3 4.3e-03 2.6e-03 1.4e-03 8.2e-03 4.8e-03 2.6e-03

Table 3 .

 3 Comparison of the L ∞ errors for high frequency, continuous absorption.

		normal incidence		45 • incidence
	refinement	0	1	2	0	1	2
	Bérenger	3.8e-05 1.9e-07 2.1e-09	2.0e-04 9.1e-07 1.6e-09
	Smart	2.7e-05	2.2e-07	1.7e-07	1.7e-04	9.0e-07	3.1e-08
	HMLV1	5.5e-07	6.0e-08	5.6e-08	5.6e-06	1.2e-08	4.7e-09
	HMLV2	6.8e-07	6.5e-08	3.1e-08	2.6e-06	8.1e-09	2.8e-09
	HMLV3	5.8e-08 2.4e-09 1.1e-09	1.5e-06 9.5e-10 9.0e-11

Table 4 .

 4 Comparison of the L ∞ errors for low frequency, continuous absorption.

			normal incidence			45 • incidence
	refinement	0	1	2	0	1	2
	Bérenger 6.2e-07 3.2e-08 7.8e-010 5.2e-07 2.9e-08 6.5e-010
	Smart	5.3e-04 5.3e-04 5.2e-04	3.9e-04 3.8e-04 3.7e-04
	HMLV1	1.6e-04 1.6e-04 1.5e-04	8.6e-05 8.3e-05 8.2e-05
	HMLV2	4.1e-04 2.0e-04 9.6e-05	2.0e-04 9.8e-05 4.8e-05
	HMLV3 1.1e-05 5.4e-06 2.7e-06 5.9e-06 2.9e-06 1.4e-06

Acknowledgments

This research project has spanned many years. It ows a great deal to the support of the University Paris 13 where J. Rauch was often invited for one month visits.

Research partially supported by the National Science Foundation under grant NSF DMS 0405899.

To get a feeling for the reflections, consider the solution U (t, x 1 ) so that, for t < 0 , U = δ(x 1t) , 0 .

Then

Contents 35 (iii) Denote by K the mapping from (i). When τ > 0, (τ + σ)/τ is also positive and real. Therefore K maps decaying (resp. increasing) solutions to decaying (resp. increasing) solutions. Thus for τ > τ 0 and real,

For all Re τ > τ 0 , K(E + L (τ, η)) is a subspace of solutions of (4. [START_REF] Nishitani | sur les équations hyperboliques à coefficients hölderiens en t et de classe Gevrey en x[END_REF] with dimension equal to dim E + L (τ, η). If (4.28) were violated, K(E + (τ, η)) would contain exponentially growing solutions. If this happened at τ , η with Re τ > τ 0 consider the values τ (r) = Re τ + rIm τ for 0 ≤ r ≤ 1. For r = 0 (4.28) is satisfied while for r = 1 it is violated. Let f (r) := max Re τ (r) + σ τ (r) µ : det L(τ (r) , µ , iη) = 0 , Re µ < 0 .

Then f (0) < 0, f (1) > 0 and f is continuous, so there is a 0 < r < 1 so that f (r) = 0. Then for τ = Re τ + r Im τ there is a purely imaginary root. This violates the hyperbolicity of L establishing (4.28) This proves (iii) completing the proof of the Lemma.

We now finish the proof of Theorem 4.3 by proving (4.22). Lemma 4.7 implies that the spaces of Cauchy data Ė± L are independent of σ. Therefore if (4.22) is violated then also

This contradicts part (iv) of Lemma 4.4 for the operator L with absorption σ + . The proof of Hersh's condition is complete.

In these problems with only one nonzero absorption coefficient σ 1 and σ 1 = 0 when x 1 < 0 one can consider a transmission problem which is only split in x 1 > 0. The next result shows that this partially split problem satisfies Hersh's condition if and only if the fully split problem does.

Introduce the partially split problem (L, R, N ) where

with B + given by (4.20) and the split variable on the right is W = (W 1 , . . . , W d ).

Corollary 4.1. Suppose that σ j = 0 for j ≥ 2, and σ + 1 > 0. Then the partially split Bérenger transmission problem (L 1 , L 1 + B + , N ) defined by (4.29) satisfies Hersh's condition if and only if the fully split problem does.

Proof. Denote by (V, W ) = (V, W 1 , . . . , W d ) the variables for the partially split problem and U , W = (U 1 , . . . , U d ), (W 1 , . . . , W d ) the split variables. If U (τ, x 1 , η), W (τ, x 1 , η) is an exponentially decaying solution of the split Laplace-Fourier transformed homogeneous transmission problem, then ( V , W ) = ( j U j , W ) is an exponentially decaying solution of the partially split homogeneous transmission problem.

) is a solution of the homogeneous partially split problem, the computation leading to (4.26) shows that

is an exponentially decaying solution of the fully split homogeneous transmission problem. Therefore, if either problem has decaying solutions for η real and Re τ arbitrarily large, then so does the other.

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Proof. Denote by U and U ′ the Fourier-Laplace transforms. The functions are characterized by

and, L(τ, d/dx 1 , η) U ′ = F , both required to decay exponentially as |x 1 | → ∞. The strategy is to construct a solution of the problem defining U ′ from the solution U . The equations for W = j U j and W ′ = j U ′ j in x 1 ≥ 0 have the form

Lemma 4.10 applies with f (x) := (τ + σ(x 1 ))/τ . Define g as in that lemma. Set V = W in x 1 ≤ 0. For x 1 ≥ 0 define

The resulting function satisfies the differential equation required of W ′ . In addition since e -g(x1)M is independent of x 1 for x 1 ≥ ρ, V decays as rapidly as W . Therefore V satisfies the conditions uniquely determining W ′ . Therefore V = W ′ , and

) is elliptic then Corollary 3.2 shows that the evolution of L is strongly well posed. This includes the case of anisotropic wave equations for which the layer is amplifying showing that perfection is not at all inconsistent with amplification.

2.

For the Maxwell equations and σ 1 (x 1 ) ∈ W 2,∞ (R) well posedness is proved in the remark following Theorem 3.3 and we deduce perfection.

Plane waves, geometric optics, and amplifying layers

This section includes a series of ideas all related to plane waves and short wavelength asymptotic solutions of WKB type. We first recall the derivation of such solutions from exact plane wave solutions by Fourier synthesis. Then we review the construction of short wavelength asymptotic expansions. These are then applied to examine the proposed absorption by the σ j . In many common cases the supposedly absorbing layers lead to asymptotic solutions which grow in the layer. Related phenomena are studied by Hu, and Becache, Fauqueux, Joly [START_REF] Hu | On absorbing boundary conditions of linearized Euler equations by a perfectly matched layer[END_REF], [START_REF] Bécache | Stability of perfectly matched layers, group velocities and anisotropic waves[END_REF]. For the Maxwell equations for which the PML were designed, the layers are not amplifying. At the end of Section 5.3, situations where the amplification does not occur are identified.

Geometric optics by Fourier synthesis

When the coefficient σ vanishes identically, both L and L are homogeneous constant coefficient systems. When (τ , ξ) is a smooth point of the characteristic variety, denote by τ = τ (ξ) the smooth parameterization, and Π L (τ, ξ) and Π L (τ, ξ) the associated spectral projections for ξ ≈ ξ, see (2.8). The function τ (ξ) is homogeneous of degree 1, while the projectors are homogeneous of degree 0. The next argument works equally well for L and L.

Make the change of variable

Summary. There is no amplification when the characteristic polynomial is a product of factors τ and τ 2 -q where q is a positive definite quadratic form with axes of inertia parallel to the coordinate axes. This includes the cases of Maxwell's equations in vacuum, for which the method was developed by Bérenger, the linearized Euler equations about the stationary state, and the linear isotrope elasticity equations. For these the quadratic forms q are multiples of |ξ| 2 .

Example 5.3 (Methods related to Bérenger, continued). For the model developed in Section 3.6 for the Maxwell equations, one can compute

Thus, this model has exactly the same good properties as Bérenger's, and is strongly well posed. For Maxwell equations, it is therefore an attractive alternative. The advantage is two fold. The system with the auxiliary variables is very compact. And it is strongly well posed, even for discontinuous σ.

On the surface this result sounds almost too good to be true. However the Bérenger system in the case of Maxwell's equations has almost exactly the same structure. The energy method proof when σ ′′ j ∈ L ∞ shows that there is a large vector V consisting of the components of U together with differential operators P α (D) applied to U and a strongly well posed equation for V. This means that if one were to introduce the additional variables in V one obtains a system with some of the desirable properties of SPML (strong PML). However, the SPML reduction is much more compact, and, has a good energy estimate even when σ is discontinuous. The extension of this strategy to other equations is not straightforward. For elastodynamic models, see [START_REF] Rahmouni | Un modèle PML bien posé pour l'élastodynamique anisotrope[END_REF].

Harmoniously matched layers

This section introduces a new absorbing layer method. It is based on the following strategy. Start with an operator L = L 1 (∂) on the left and consider a smart layer on the right

generalizing (1.2). This method is embedded in a family of absorbing layers parameterized by µ ≥ 0,

2

The method is nonreflective when µ = 0 and is both reflective and dissipative for µ > 0. When σ is discontinuous the leading order reflection coefficient for wave packets of amplitude 1 oscillating as e i(τ t+xξ)/ε is of the form ε µ r(τ, ξ). The leading order reflections can be removed by an extrapolation method using two values of µ. This simultaneously removes the leading reflections at all angles of incidence. We call the resulting method the harmoniously matched layer.

Reflection is linear in µ by scaling

In this subsection the linearity in µ of leading order reflections by the layer with R µ from (6.2) is demonstrated by a scaling argument when σ(x 1 ) = 1 x1>0 . In the next subsection the reflection is computed exactly for Maxwell's equations yielding additional information.

Suppose that U has an incoming wave of wavelength ε and reflected waves U ℓ with amplitudes ρ ℓ ε. Then U has an incoming wave with wavelength ε := ε/µ. The reflected waves have amplitudes

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Therefore, to determine a T 0 and a R 0 everywhere, it suffices to know a T 0 (t, 0 + , x ′ ) and a R 0 (t, 0 -, x ′ ). These values are determined from the transmission condition (6.3):

.10)

The matrix A 1 is singular. It is easy to see that (Ker L(τ, ξ) ⊕ Ker L(τ, ξ R )) ∩ Ker A 1 = 0. Therefore, A 1 Ker L(τ, ξ) and A 1 Ker L(τ, ξ R ) are complementary subspaces and generate Range A 1 . This proves that a R 0 (t, 0 -, x ′ ) = 0, a T 0 (t, 0 + , x ′ )a I 0 (t, 0 -, x ′ ) = 0. (6.11)

By the transport equation, we conclude that a R 0 ≡ 0, for x 1 < 0. (6.12)

The reflected zeroth order term vanishes identically when x ∈ R 3 -. We also deduce from the transport equation (6.9) that

Next determine inductively the correctors. For simplicity, throughout the proof we note Π L := Π L (τ, ξ) and Π R L := Π L (τ, ξ R ). Write the recursion relation (5.5) for j ≥ 1 for the incident, reflected and transmitted waves. Split the amplitudes as

x) are determined directly by (5.8). To determine the projection on the kernel, split the transmission condition (6.3) and insert (5.8) on the interface to get,

As for the terms of order 0, this determines Π L a T j (t, 0 + , x ′ ) and Π R L a R j (t, 0 -, x ′ ). By (5.9), the projections are solution of a transport equation, therefore uniquely determined by initial data and the values on the boundary . Borel's theorem allows one to construct a I (t, x, ε), a T (t, x, ε), and, a R (t, x, ε), so that the transmission condition is exactly satisfied. With this choice the approximate solution satisfies the transmission problem with infinitely small residual.

(ii) Theorem 4.4 implies that the exact solution in x 1 ≤ 0 is equal to U ε +O(ε ∞ ). The error of the approximation is O(ε ∞ ) so the exact solution is equal to

which is the desired conclusion.

(iii) For the smart layer (6.1, 6.2) with σ = 1 x1>0 , compute the first order term by (5.11) with j = 1. First deduce from (5.8) that

Replace in (6.15) the x 1 derivatives using (6.9),

.16)

(iv) Here σ vanishes to order k at x 1 = 0. (6.11) and (6.12) are still valid, and the transport equations (6.9) implies on the interface Γ that

From (6.21) for i = 0, (6.15) and (6.22), derive

Using the transmission conditions to obtain on the interface Γ,

Insert into the transport equations (5.9) to find a R 1 = 0 in R -, and

Now proceed iteratively, to see that, for i < k + 1,

Denote by s i the value of

1 a I i on Γ. Equation (6.26) yields the recursion relation

which can be solved as

The first nonzero reflected term is therefore a R k+1 = Π R L a R k+1 , and using the transmission condition yields

The incoming amplitude on Γ is

Using again the notation Φ ′ to denote the projection of a vector Φ on V ec(e 2 , e 3 ), this linear system is solved as

The proof of the linearity follows the same path as in (iii).

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The Yee scheme for the smart layers is

The schemes are implemented using time windows to save memory. The harmoniously matched layers can be implemented in several ways that we compare. The function σ(x) is as above.

HML Version 1. Global extrapolation. Compute the solution of (6.33) with an absorption of σ, (E 1 , H

We perform a series of experiments to illustrate the transmission properties of the layers. The coefficient ν is meant to achieve backward absorption and is taken equal to zero. The domain of interest is (0, 6) × (0, 10), the coefficient σ(x) is supported in 6 ≤ x ≤ 10. The time of computation is 4, the initial electric field is zero. The initial transverse magnetic field,

is compactly supported in the ball B(x c , r), with x c = (5, 5) and r = 0.8. The time of computation is fixed such that there is no reflection on the exterior walls. The initial mesh is taken to be ∆x = ∆y = 0.1, ∆t = 0.0702, and then divided by 2 twice.

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The Bérenger layer performs well on every frequency and every angle of incidence. Among the 3 versions for the HML, the third version is the best, which should be analyzed thoroughly.

Next compare the method on a gaussian initial value, supported in (0, 6) × (0, 10). Table 5 uses a constant absorption in the layer, while Table 6 uses the same smooth absorption as before. Finally, take unstructured random initial value, supported in the ball centered at (5, 5) and of radius 1. In Table 7, the absorption coefficient is constant in the layer, equal to 3. In Table 8, the absorption coefficient is a function of x in the layer, equal to (x -6) 3 /8. Summary. When comparing the reflection properties, the harmoniously matched layer, version 3, is competitive with the Bérenger layer. For very regular data, the Bérenger layers outperform everything.