Geometric optics expansion for weakly well-posed hyperbolic boundary value problem: The glancing degeneracy

Abstract

This article aims to finalize the classification of weakly well-posed hyperbolic boundary value problems in the half-space. Such problems with loss of derivatives are rather classical in the literature and appear for example in (Arch. Rational Mech. Anal. 101 (1988) 261–292) or (In Analyse Mathématique et Applications (1988) 319–356 Gauthier-Villars). It is known that depending on the kind of the area of the boundary of the frequency space on which the uniform Kreiss–Lopatinskii condition degenerates then the energy estimate can include different losses. The three first possible areas of degeneracy have been studied in (Annales de l’Institut Fourier 60 (2010) 2183–2233) and (Differential Integral Equations 27 (2014) 531–562) by the use of geometric optics expansions. In this article we use the same kind of tools in order to deal with the last remaining case, namely a degeneracy in the glancing area. In comparison to the first cases studied we will see that the equation giving the amplitude of the leading order term in the expansion, and thus initializing the whole construction of the expansion, is not a transport equation anymore but it is given by some Fourier multiplier. This multiplier needs to be invert in order to recover the first amplitude. As an application we discuss the existing estimates of (Discrete Contin. Dyn. Syst., Ser. B 23 (2018) 1347–1361; SIAM J. Math. Anal. 44 (2012) 1925–1949) for the wave equation with Neumann boundary condition.

Keywords

Hyperbolic boundary value problems geometric optics expansions weak well-posedness glancing modes

1. Introduction

This article deals with geometric optics expansion for hyperbolic boundary value problems and more precisely to the question of the loss of derivatives of such problems when the boundary condition leads to weak well-posedness. The considered problems read under the form: for fixed $T > 0$ $\begin{matrix} (1) & \{\begin{matrix} L (\partial) u : = \partial_{t} u + \sum_{j = 1}^{d} A_{j} \partial_{x_{j}} u = f & in Ω_{T}, \\ B u_{| x_{d} = 0} = g & on ω_{T}, \\ u_{| t ⩽ 0} = 0 & on R_{+}^{d}, \end{matrix} \end{matrix}$ where the space domain is the half-space $R_{+}^{d} : = {x = (x^{'}, x_{d}) \in R^{d}, x_{d} > 0}$ and where we defined the sets $Ω_{T} : =] - \infty, T] \times R_{+}^{d}$ and $ω_{T} : =] - \infty, T] \times R^{d - 1}$ .

In (1) the coefficient matrices $A_{1}, \dots, A_{d}$ belong to $M_{N} (R)$ with $N \in N^{*}$ fixed, the matrix B belongs to $M_{p, N} (R)$ (p is made more precise in Assumption 2.2), and the unknown u takes its values in $R^{N}$ .

The strong well-posedness of the problem (1) is well established from the seminal work of [14] in which the author characterizes all the boundary matrices B leading to strong well-posedness. By strong well-posedness we mean that for all choices of the sources $(f, g) \in L^{2} (Ω_{T}) \times L^{2} (ω_{T})$ the problem (1) admits a unique solution u (with trace on $ω_{T}$ in $L^{2} (ω_{T}))$ that satisfies the following energy estimate: there exists $C_{T} > 0$ such that we have the inequality $\begin{matrix} (2) & ‖ u ‖_{L^{2} (Ω_{T})}^{2} + ‖ u_{| x_{d} = 0} ‖_{L^{2} (ω_{T})}^{2} ⩽ C_{T} (‖ f ‖_{L^{2} (Ω_{T})}^{2} + ‖ g ‖_{L^{2} (ω_{T})}^{2}) . \end{matrix}$

In particular in equation (2) we have a control of the solution in the same functional space than the one of the data, this optimal control is referred as strong well-posedness. Without entering into technical details the full characterization is given in [14], the so-called Kreiss–Lopatinskii condition states that in the normal mode analysis of the problem (1) there is no stable modes solution to the homogeneous boundary condition. Indeed if we denote by $E^{s}$ the stable space, in the sense of dynamical systems, associated to the problem (1) in the frequency space then the existence of some non trivial element in $ker B \cap E^{s}$ gives rise to a non trivial solution to a homogeneous linear problem and thus leads to a contradiction.

However as firstly remarked in [17] and [1] on formal geometric optics expansions the uniform Kreiss–Lopatinskii condition can degenerate without that the problem generates a Hadamard instability and that it becomes ill-posed. Indeed when the uniform Kreiss–Lopatinskii condition fails then one can expect the problem to admit a unique solution u but in a less regular space than the one of the sources. Such phenomenon is referred as a loss of derivatives in the following and the associated concept of well-posedness is called the weak well-posedness.

Historically the first result establishing such weak well-posedness behaviour is due to [19] and is related to elastodynamics. In this paper the author shows the existence and the uniqueness of the solution u but with an energy estimate reading $\begin{matrix} (3) & ‖ u ‖_{L^{2} (Ω_{T})}^{2} + ‖ u_{| x_{d} = 0} ‖_{H^{- 1 / 2} (ω_{T})}^{2} ⩽ C_{T} (‖ f ‖_{L^{2} (Ω_{T})}^{2} + ‖ g ‖_{H^{1 / 2} (ω_{T})}^{2}), \end{matrix}$ that is to say that the solution exhibits a loss of one derivative on the boundary but no loss of derivatives in the interior. Then other kinds of estimates have also been demonstrated in [7] and [8] and the associated energy estimates show a loss of one derivative on the boundary coupled with a loss of one half derivative respectively one derivative in the interior. Without entering into technical details we know from the so-called block structure condition (see [18] and Theorem 2.1 below) that there are four kinds of degeneracy of the uniform Kreiss–Lopatinskii condition depending on the area of the frequency space where it fails. Namely the degeneracy can occur in the elliptic, mixed, hyperbolic or glancing area. Consequently there are four possible estimates, the ones described above correspond to the three first ones. The aim of the following article is to investigate the last case that is a degeneracy of the uniform Kreiss–Lopatinskii condition in the so-called glancing area.

Note that once an estimate of the form (3) has been established then a natural question is the one of the sharpness of the losses of derivatives. In order to investigate what can be the optimal losses then geometric optics expansions are commonly used. Indeed they are used for example in [9] and [3] in order to show that the estimates obtained in [7,19] and [8] are effectively optimal.

The whole idea of geometric optics (or WKB) expansions is to consider the highly oscillating problem $\begin{matrix} \{\begin{matrix} L (\partial) u^{ε} = f^{ε} & in Ω_{T}, \\ B u_{| x_{d} = 0}^{ε} = g^{ε} & on ω_{T}, \\ u_{| t ⩽ 0}^{ε} = 0 & on R_{+}^{d}, \end{matrix} \end{matrix}$ where the (small) parameter $0 < ε ≪ 1$ represents the typical wavelength of the highly oscillatory source terms $f^{ε}$ and $g^{ε}$ and to look for an approximate solution $u^{ε}$ as a sum of wave packets with amplitudes that are written as asymptotic expansion with respect to the small parameter ε. Of course such approximate solutions have an interest for their own but moreover if one is able to construct sufficiently enough terms in the geometric optics expansions then he/she can expect that this expansion is an approximation of $u^{ε}$ with high accuracy and to use the approximate solution to exhibit that some qualitative phenomenon (such that losses of derivatives) occur on the approximate solution and so they do on the exact one.

As already mentioned this method has been successfully used firstly in [9] and then in [3] to show the sharpness of the energy estimates with the different possible losses of derivatives for the degeneracies in the hyperbolic, the mixed and the elliptic area. In this paper we use the same kind of procedure to investigate what can be the loss of derivatives for the last remaining case namely the degeneracy in the so-called glancing area.

A main difference between the construction of geometric optics expansion for a failure of the uniform Kreiss–Lopatinskii condition in the glancing area compared to the other degeneracies is the nature of the propagation along the boundary which is the keystone that needs to be understood in order to start the resolution of the WKB cascade. To explain this, let us sketch the main ideas in the construction of the leading order amplitude as it is done in [9] and [3].

Let $u_{0}$ denotes the leading order amplitude in the geometric optics expansion and let $u_{1}$ be the first corrector. Then because some losses of derivatives are expected, we expect to have some amplification in the expansion compared with the source terms so that for a boundary source term of scale $ε^{α}$ , $α > 0$ , we expect $u_{0}$ to be of scale $ε^{0}$ .

The leading order term $u_{0}$ satisfies, in a classical setting, in geometric optics expansion some kind of polarization condition meaning essentially (and to simplify the exposition) that $u_{0} \in E^{s}$ , the stable subspace of the problem, and that to determine the whole $u_{0}$ then only the value of its trace on ${x_{d} = 0}$ is required.

However, because of the scale of the boundary source then this leading order should also satisfy the homogeneous boundary condition $B u_{0_{| x_{d} = 0}} = 0$ , equation that can not be used in order to determine $u_{0_{| x_{d} = 0}}$ but that implies that $u_{0_{| x_{d} = 0}} \in ker B \cap E^{s} : = span (e)$ because of the failure of the uniform Kreiss–Lopatinskii condition. So that we can write $\begin{matrix} u_{0_{| x_{d} = 0}} (t, x^{'}) : = α_{0} (t, x^{'}) e, \end{matrix}$ where e is precisely introduced in Assumptions 2.3 and 2.4 and thus the question is to determine $α_{0}$ . In order to do this following [9] and [3] we should have a look to the only boundary condition involving $u_{0_{| x_{d} = 0}}$ that is the boundary condition for $u_{1}$ in which $u_{0_{| x_{d} = 0}}$ appears via the unpolarized part of $u_{1}$ . This boundary condition essentially reads $\begin{matrix} (4) & B {\overline{u}}_{1_{| x_{d} = 0}} = g - B L u_{0_{| x_{d} = 0}}, \end{matrix}$ where ${\overline{u}}_{1}$ stands for the polarized part of $u_{1}$ , g is the amplitude of the boundary source and $L$ is some (explicit) operator. Then the common point in [9] and [3] is to show that the operator $L$ is in fact a transport operator (with respect to the boundary variables $(t, x^{'})$ ) so that (4) can be solved explicitly in order to determine $α_{0}$ .

For the failure of the uniform Kreiss–Lopatinskii condition in the glancing area we will follow the same approach but the operator $L$ will not be a transport operator anymore but some Fourier multiplier. However equation (4) will still be used in order to determine $α_{0}$ just by reversing the Fourier multiplier $L$ . Once that $α_{0}$ is determined we can initialize the resolution of the cascade of equations of the geometric optics expansion and then the order of resolution is rather classical and essentially follows [20] (see also [3]).

Let us point that in contrary to the other degeneracies then we do not have (in general) an energy estimate for such a degeneracy. Indeed it seems rather difficult to adapt the construction of the so-called Kreiss symetrizor (the classical tool used to show a priori energy estimates, see [14]) as it has been done for example in [19] in order to show an a priori estimate. That is why in our result about losses of derivatives we will only show that losses of one half a derivative in the interior or on the boundary occurs (we refer to Theorem 2.3 for a precise statement). However because we do not have an energy estimate with such losses we can not conclude that it is sharp.

But let us say that in some particular setting such estimates can be found in the literature (see [11] and [10]). More precisely in [11] the author obtains a weak well-posedness result when the so-called Kreiss–Sakamoto condition with power s holds. Even if a full characterization of the fulfilment of the Kreiss–Sakamoto condition in terms of the area of degeneracy of the uniform Kreiss–Loaptinskii condition has not been achieved yet we can use this estimate for the very interesting (in view of the applications) $2 d$ wave equation with Neumann boundary condition.

The paper is organized as follows Section 2 contains some classical definitions and notations used for the construction of geometric optics expansion for boundary value problems. The main results of the article namely Theorems 2.2 and 2.3 state respectively, the existence of a solution to the WKB expansion when the uniform Kreiss–Lopatinskii condition fails in the glancing area, and some results about what can be the losses of derivatives in such a framework can be found in Paragraph 2.2. Section 3 gives the construction of the WKB expansion and thus the proof of Theorem 2.2. It is the technical part of the article. Then the proof of Theorem 2.3 is given in Section 4. Finally some examples namely the $2 d$ -wave equation with Neumann boundary condition and a linearisation of Euler equation are discussed in Section 5.

2. Assumptions and main results

In all the article $C > 0$ stands for a constant which can change from one line to the other without changing of notation. When the constant C depends on some parameter A we write $C_{A}$ instead of C. For $a, b \in R$ the notation $a ≲ b$ will sometimes be used as a shorthand notation for: there exists $C > 0$ (independent of any paramater) such that $a ⩽ C b$ .

Let $z \in C$ we will write $z : = ℜ (z) + i ℑ (z)$ , $ℜ (z), ℑ (z)$ being real and denote respectively the real and imaginary part of z.

For two vectors of same size u and v we will denote $u \cdot v$ the euclidian scalar product of u and v. Finally for $p, q ⩾ 1$ and for a matrix $A \in M_{p, q}$ we will denote $A^{T} \in M_{q, p}$ for the transpose matrix of A.

We introduce the following frequency space Ξ and its boundary $Ξ_{0}$ : $\begin{array}{l} Ξ : = {ζ : = (γ + i τ, η) \in C \times R^{d - 1} ∖ {(0, 0)}, γ ⩾ 0}, \\ Ξ_{0} : = Ξ \cap {γ = 0} . \end{array}$

Finally for $s \in N$ we use the usual notation $H^{s}$ for the Sobolev space of order s. We denote1

¹
In fact the results of the article can also be stated in some space $H^{S}$ for S fixed large enough. But we choose $H^{\infty}$ instead for simplicity.

H^{\infty} : = ⋂_{s \in N} H^{s}

2.1. Assumptions and notations

In the following we consider the boundary value problem: $\begin{matrix} (5) & \{\begin{matrix} L (\partial) u = f & in Ω_{T}, \\ B u_{| x_{d} = 0} = g & on ω_{T}, \\ u_{| t ⩽ 0} = 0 & on R_{+}^{d} . \end{matrix} \end{matrix}$ We assume that the constant coefficients $A_{j} \in M_{N} (R)$ give rise to an operator $L (\partial)$ of hyperbolic type and more precisely a constantly hyperbolic operator. With more details we assume that:

Assumption 2.1.
There exists an integer $q ⩾ 1$ , q analytic, homogeneous of degree one functions $λ_{1}$ ,…, $λ_{q}$ on $R^{d} ∖ {0}$ and integers $μ_{1}$ ,…, $μ_{q}$ such that $\begin{matrix} \forall ξ : = (ξ_{1}, \dots, ξ_{d}) \in S^{d - 1}, det (τ I + \sum_{j = 1}^{d} ξ_{j} A_{j}) = \prod_{k = 1}^{q} {(τ - λ_{k} (ξ))}^{μ_{k}} . \end{matrix}$ The eigenvalues $λ_{1} (ξ), \dots, λ_{q} (ξ)$ are semi-simple and satisfy $λ_{1} (ξ) < \dots < λ_{q} (ξ)$ for all $ξ \in R^{d} ∖ {0}$ .

We restrict our analysis to a non characteristic boundary and therefore make the following assumption:
Assumption 2.2.
The matrix $A_{d}$ is invertible and the matrix B has maximal rank, its rank p being equal to the number of positive eigenvalues of $A_{d}$ (counted with their multiplicity). Moreover, the integer p satisfies $1 ⩽ p ⩽ N - 1$ .

Let us assume that the problem (5) is well-posed in the sense that it admits a unique solution $u \in L^{2} (Ω_{T})$ then we can perform a Laplace transform in the time variable t and a Fourier transform in the tangential space variable $x^{'}$ . Let $σ : = γ + i τ \in C_{+} : = {z \in C s.t. ℜ (z) > 0}$ and $η \in R^{d - 1}$ denote the dual variables of t and $x^{'}$ and let $\hat{\cdot}$ be the Fourier–Laplace transform. So, thanks to Assumption 2.2 the problem (5) reads in the resolvent form: $\begin{matrix} (6) & \{\begin{matrix} \frac{d}{d x_{d}} \hat{u} (ζ, x_{d}) = A (ζ) \hat{u} (ζ, x_{d}) + A_{d}^{- 1} \hat{f} for x_{d} \in R_{+}, \\ B \hat{u} (ζ, 0) = \hat{g} (ζ), \end{matrix} \end{matrix}$ in which $ζ \in Ξ$ acts like a parameter and where the so-called resolvent matrix is defined by: $\begin{matrix} A (ζ) : = - A_{d}^{- 1} (σ I + i \sum_{j = 1}^{d - 1} η_{j} A_{j}), where ζ : = (σ, η) \in C_{+} \times R^{d - 1} . \end{matrix}$

So, the spectrum of $A (ζ)$ encodes the behaviour of the solution to (6) and so the one to (5). For hyperbolic operators the spectrum of $A (ζ)$ is known and given by the following lemma due to Hersh (see [12]).
Lemma 2.1 (Hersh).

Under Assumptions 2.1 and 2.2 , for all frequency parameter $ζ \in Ξ ∖ Ξ_{0}$ , the resolvent matrix $A (ζ)$ only admits eigenvalues with non-zero real part, and thus does not have purely imaginary eigenvalues. We denote by $E^{s} (ζ)$ (resp. $E^{u} (ζ)$ ) the stable (resp. unstable) subspace of $A (ζ)$ that is the eigenspace associated with the negative (resp. positive) real part eigenvalues. Then, independently on $ζ \in Ξ ∖ Ξ_{0}$ ; $dim E^{s} (ζ) = p$ and $dim E^{u} (ζ) = N - p$ and we have the following decomposition: $\begin{matrix} (7) & C^{N} = E^{s} (ζ) \oplus E^{u} (ζ) . \end{matrix}$

However, this lemma only gives information as long as the frequency parameter lives away from the boundary $Ξ_{0}$ . The study of hyperbolic boundary value problems needs to have a look to the frequency parameters $ζ \in Ξ_{0}$ for which Lemma 2.1 does not apply anymore. Indeed in the limit $γ ↓ 0$ then the real parts of the eigenvalues may (and they do) vanish. For glancing modes in which we are especially interested in, such a degeneracy of the eigenvalue occurs at the order at least two meaning that in the limit $γ ↓ 0$ some eigenvalues associated to $E^{s}$ and some eigenvalues associated to $E^{u}$ coincide. More precisely the theorem describing the behaviour of the eigenvalues of $A (ζ)$ for $ζ \in Ξ_{0}$ is the so-called block structure theorem firstly shown in [14] for strictly hyperbolic systems (that is to say that Assumption 2.1 is satisfied with $μ_{1} = \dots = μ_{q} = 1$ ) and then extended by [18] for constantly hyperbolic operators.

Theorem 2.1 (Block structure [18]).

Under Assumptions 2.1 and 2.2 , for all $\underline{ζ} \in Ξ$ , there exists a neighbourhood, $V$ , of $\underline{ζ}$ in Ξ, an integer $L ⩾ 1$ , a partition $N = μ_{1} + \dots + μ_{L}$ , with $μ_{1}, \dots, μ_{L} ⩾ 1$ and an invertible matrix $T : = T (ζ)$ , regular on $V$ such that: $\begin{matrix} \forall ζ \in V, T^{- 1} (ζ) A (ζ) T (ζ) : = diag (A_{1} (ζ), \dots, A_{L} (ζ)) \end{matrix}$ where the blocks $A_{j} \in M_{μ_{j}} (C)$ satisfy one of the following alternatives:

all the elements in the spectrum of $A_{j} (\underline{ζ})$ have positive real part.

all the elements in the spectrum of $A_{j} (\underline{ζ})$ have negative real part.

$μ_{j} = 1$ , $A_{j} (\underline{ζ}) \in i R$ , $\partial_{γ} A_{j} (\underline{ζ}) \in R ∖ {0}$ and $A_{j} (\underline{ζ}) \in i R$ for all $ζ \in Ξ_{0} \cap V$ .

$μ_{j} > 1$ and there exists $k_{j} \in i R$ such that $\begin{matrix} A_{j} (\underline{ζ}) = (\begin{matrix} k_{j} & i & 0 \\ ⋱ & i \\ 0 & k_{j} \end{matrix}), \end{matrix}$ the coefficient in the lower left corner of $\partial_{γ} A_{j} (\underline{ζ}) \in R ∖ {0}$ and for all $ζ \in Ξ_{0} \cap V, A_{j} (\underline{ζ}) \in i M_{μ_{j}} (R)$ .

Such theorem motivates the following definition clarifying, in particular, the terminology of glancing modes used in the introduction.

Definition 2.1.
For $\underline{ζ} \in Ξ_{0}$ , we define:
$E$ the elliptic area which is the set of $\underline{ζ}$ such that Theorem 2.1 is satisfied with blocks of type 1 and 2 only.

$EH$ the mixed area which is the set of $\underline{ζ}$ such that Theorem 2.1 is satisfied with blocks of type 1, 2 and at least one block of type 3.

$H$ the hyperbolic area which is the set of $\underline{ζ}$ such that Theorem 2.1 is satisfied with blocks of type 3 only.

$G$ the glancing area which is the set of $\underline{ζ}$ such that Theorem 2.1 is satisfied with at least one block of type 4.

Thanks to Theorem 2.1, we thus have the following partition of $Ξ_{0}$ : $\begin{matrix} Ξ_{0} = E \cup EH \cup H \cup G, \end{matrix}$ and in the following we will be particularly interested in a boundary frequency $\underline{ζ} \in G$ .

However let us say that when $\underline{ζ} \in Ξ_{0} ∖ G$ , then the decomposition (7) still holds and we write: $\begin{matrix} (8) & C^{N} = E^{s} (\underline{ζ}) \oplus E^{u} (\underline{ζ}), \end{matrix}$ where $E^{s} (\underline{ζ})$ (resp. $E^{u} (\underline{ζ})$ ) stands for the extension by continuity of $E^{s} (ζ)$ (resp. $E^{u} (ζ)$ ) up to the boundary $Ξ_{0}$ .

These spaces admit the following decompositions: $\begin{matrix} E^{s} (\underline{ζ}) = E_{e}^{s} (\underline{ζ}) \oplus E_{h}^{s} (\underline{ζ}) and E^{u} (\underline{ζ}) = E_{e}^{u} (\underline{ζ}) \oplus E_{h}^{u} (\underline{ζ}), \end{matrix}$ where $E_{e}^{s} (\underline{ζ})$ (resp. $E_{e}^{u} (\underline{ζ})$ ) is the generalized eigenspace associated to eigenvalues of $A (\underline{ζ})$ with negative (resp. positive) real part and where the $E_{h}^{s} (\underline{ζ})$ , $E_{h}^{u} (\underline{ζ})$ are sums of eigenspaces associated to some purely imaginary eigenvalues of $A (\underline{ζ})$ .

But for $\underline{ζ} \in G$ , the decomposition (8) does not hold anymore because at a glancing frequency, we have $E^{s} (\underline{ζ}) \cap E^{u} (\underline{ζ}) \neq {0}$ . In this setting, we introduce the following decomposition of the stable and unstable spaces: $\begin{matrix} (9) & E^{s} (\underline{ζ}) = E_{e}^{s} (\underline{ζ}) \oplus E_{h}^{s} (\underline{ζ}) \oplus E_{g}^{s} (\underline{ζ}) and E^{u} (\underline{ζ}) = E_{e}^{u} (\underline{ζ}) \oplus E_{h}^{u} (\underline{ζ}) \oplus E_{g}^{u} (\underline{ζ}) \end{matrix}$ where $E_{g}^{s} (\underline{ζ})$ , $E_{g}^{u} (\underline{ζ})$ are sums of eigenspaces associated to the Jordan block of type 4 of $A (\underline{ζ})$ in Theorem 2.1 and consequently satisfying $E_{g}^{s} (\underline{ζ}) \cap E_{g}^{u} (\underline{ζ}) \neq {0}$ .

In this article we will make the rather classical assumption in geometric optics expansion construction for glancing modes that these modes are of order two (see [21]). Indeed without this restriction the construction of geometric optics expansions is a rather open question. Without this assumption it is shown on some explicit examples in [[20]-Part III] that the associated glancing boundary layer may blow up in $L^{\infty}$ -norm. We will also make the simplifying (and probably not necessary assumption) that there is only one block of type 4 in Theorem 2.1.
Assumption 2.3.
Let $\underline{ζ} \in G$ then Theorem 2.1 is satisfied with one block of type 4 only and moreover this block is of size two. In this setting, there exists $\tilde{e} \in C^{N} ∖ {0}$ such that: $\begin{matrix} E_{g}^{s} (\underline{ζ}) = E_{g}^{u} (\underline{ζ}) : = span (\tilde{e}) . \end{matrix}$

In order to construct the geometric optics expansion we need to be more precise about the hyperbolic and the glancing modes of $A (\underline{ζ})$ that is to say the ones associated with purely imaginary eigenvalues. Let $i {\underline{ξ}}_{m} \in i R$ be a purely imaginary eigenvalue of $A (\underline{ζ})$ then $\begin{matrix} det (\underline{τ} I + \sum_{j = 1}^{d - 1} {\underline{η}}_{j} A_{j} + {\underline{ξ}}_{m} A_{d}) = 0 . \end{matrix}$ From the hyperbolicity Assumption 2.1, there exists an index $k_{m}$ such that $\begin{matrix} \underline{τ} + λ_{k_{m}} (\underline{η}, {\underline{ξ}}_{m}) = 0 \end{matrix}$ where $λ_{k_{m}}$ is smooth in both variables so that we can introduce:
Definition 2.2.
The set of incoming (resp. outgoing) phases for the side $ω_{T}$ denoted by $I$ (resp. $O$ ) is the set of indices m such that the group velocity $v_{m} : = \nabla λ_{k_{m}} (\underline{η}, {\underline{ξ}}_{m})$ satisfies $v_{m, d} > 0$ (resp. $v_{m, d} < 0$ ).

The set of glancing phases for the side $ω_{T}$ denoted by $G$ is the set of indices m such that the group velocity $v_{m} : = \nabla λ_{k_{m}} (\underline{η}, {\underline{ξ}}_{m})$ satisfies $v_{m, d} = 0$ .

With this definition in hand, we can give the following description of the spaces $E_{h}^{s} (\underline{ζ})$ , $E_{h}^{u} (\underline{ζ})$ , $E_{g}^{s} (\underline{ζ})$ and $E_{g}^{u} (\underline{ζ})$ (see for example [4]).
Lemma 2.2.
Under Assumptions 2.1 and 2.2 ; for all $\underline{ζ} \in Ξ_{0}$ , we have: $\begin{array}{l} E_{h}^{s} (\underline{ζ}) = ⨁_{k \in I} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}), E_{h}^{u} (\underline{ζ}) = ⨁_{k \in O} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}), and \\ E_{g}^{s} (\underline{ζ}) = E_{g}^{u} (\underline{ζ}) = ⨁_{k \in G} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) \end{array}$ where $L$ stands for the symbol of $L (\partial)$ defined by $\begin{matrix} \forall ω : = (ω_{0}, \dots, ω_{d}) \in R^{d + 1}, L (ω) : = ω_{0} I + \sum_{j = 1}^{d} ω_{j} A_{j} . \end{matrix}$

Consequently, for $\underline{ζ} \in G$ , the decompositions ( 9 ) read: $\begin{array}{l} (10) & E^{s} (\underline{ζ}) = ⨁_{k \in I} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) ⨁_{k \in G} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) \oplus E_{e}^{s} (\underline{ζ}), \\ (11) & E^{u} (\underline{ζ}) = ⨁_{k \in O} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) ⨁_{k \in G} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) \oplus E_{e}^{u} (\underline{ζ}) . \end{array}$
Remark 2.1.
In fact, under Assumption 2.3, we can be more precise, the decompositions (10) and (11) are: $\begin{array}{l} E^{s} (\underline{ζ}) = ⨁_{k \in I} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) \oplus span (\tilde{e}) \oplus E_{e}^{s} (\underline{ζ}), \\ E^{u} (\underline{ζ}) = ⨁_{k \in O} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) \oplus span (\tilde{e}) \oplus E_{e}^{u} (\underline{ζ}), \end{array}$ where $\tilde{e}$ has been introduced in Assumption 2.3 can now be chosen as a generator of $ker L (\underline{τ}, \underline{η}, \underline{ξ_{k_{g}}})$ , where $k_{g} \in G$ stands for the only glancing index.

As mentioned in the introduction, the strong well-posedness of the boundary value problem (5) is totally characterized in [14] by the so-called uniform Kreiss–Lopatinskii condition (this condition is recalled below for the reader’s convenience). Because we are interested in weakly well-posed problems this condition will not be satisfied in our study and we describe in Assumption 2.4 how the uniform Kreiss–Lopatinskii condition degenerates.
Definition 2.3 (Uniform Kreiss–Lopatinskii condition).

Under Assumptions 2.1 and 2.2, let $ζ \in Ξ$ , and as previously we still denote by $E^{s} (ζ)$ the extension by continuity of $E^{s} (ζ)$ up to $Ξ_{0}$ of the well-defined (for $ζ \in Ξ ∖ Ξ_{0}$ ) stable subspace of $A (ζ)$ . Then the boundary condition B satisfies the uniform Kreiss–Lopatinskii condition (UKL) if: $\begin{matrix} \forall ζ \in Ξ, ker B \cap E^{s} (ζ) = {0} . \end{matrix}$ In other words, the restriction of B to $E^{s} (ζ)$ is invertible.

Assumption 2.4.
Under Assumptions 2.1, 2.2 we assume that the boundary value problem (5) satisfies:
For all $ζ \in Ξ ∖ G$ , $ker B \cap E^{s} (ζ) = {0}$ .

There exists $\underline{ζ} \in G$ such that $ker B \cap E^{s} (\underline{ζ}) = ker B \cap E_{g}^{s} (\underline{ζ}) \neq {0}$ . Moreover, we suppose that $ker B \cap E^{s} (\underline{ζ})$ is one-dimensional. So that there exists a vector $e \in C^{N} ∖ {0}, ker B \cap E^{s} (\underline{ζ}) = ker B \cap E_{g}^{s} (\underline{ζ}) = span (e)$ .

Remark 2.2.

The first point of Assumption 2.4 implies in particular the so-called (weak) Kreiss–Lopatinskii condition that is $\begin{matrix} \forall ζ \in Ξ ∖ Ξ_{0}, ker B \cap E^{s} (ζ) = {0} . \end{matrix}$ Indeed it is known (see for example [[6] Section 4.2]) that if this condition fails then the boundary value problem (5) develops a Hadamard instability meaning that we have an infinite number of losses of derivatives and thus we can not expected weak well-posedness. So that the uniform Kreiss–Lopatinskii condition can only degenerate for $\underline{ζ} \in Ξ_{0}$ .

The second point of Assumption 2.4 states that the uniform Kreiss–Lopatinskii condition is not satisfied at one frequency $\underline{ζ} \in G$ (so that it is a glancing frequency) and that the failure of the uniform Kreiss–Lopatinskii condition occurs on the component of the stable subspace which is associated to the glancing eigenvalue (that is the block of type 4 in Theorem 2.1). Moreover we assume in a classical setting (see [9] and [3]) that this non trivial intersection becomes one dimensional only.

We have $\tilde{e} = λ e$ for $λ \in C$ where $\tilde{e}$ and e are defined respectively in Assumption 2.3 and 2.4. Hence, we have $E_{g}^{s} (\underline{ζ}) = E_{g}^{u} (\underline{ζ}) = span (\tilde{e}) = span (e)$ .

Indeed, since $dim E_{g}^{s} = 1$ , we have $ker B \cap E_{g}^{s} (\underline{ζ}) = span (\tilde{e})$ and combining this with the fact that $ker B \cap E_{g}^{s} (\underline{ζ}) = span (e)$ , we obtain the result. So that in the following we will assume without loss of generality that $\tilde{e} = e$ .

We conclude this preliminary section with the introduction of some projectors that are commonly used in the construction of geometric optics expansions.
Definition 2.4.
Under Assumptions 2.1 and 2.2, for $\underline{ζ} = (i \underline{τ}, \underline{η}) \in Ξ_{0}$ , we define:
$Π_{e}^{s} : = Π_{e}^{s} (\underline{ζ})$ (resp. $Π_{e}^{u} : = Π_{e}^{u} (\underline{ζ})$ ) the spectral projector on $E_{e}^{s} (\underline{ζ})$ (resp. $E_{e}^{u} (\underline{ζ})$ ).

For $k \in I \cup O \cup G$ , $Π^{k} : = Π^{k} (\underline{ζ})$ the orthogonal projector on $ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}})$ .

For $k \in I \cup O \cup G$ , $Υ^{k} : = Υ^{k} (\underline{ζ})$ the partial inverse of $L (\underline{τ}, \underline{η}, \underline{ξ_{k}})$ characterized by the relations: $\begin{matrix} (12) & \{\begin{matrix} Υ^{k} L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) = L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) Υ^{k} = I - Π^{k}, \\ Im Π^{k} = ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) = ker Υ^{k}, \\ ker Π^{k} = Im L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) = Im Υ^{k} . \end{matrix} \end{matrix}$

Remark 2.3.
Since we have only one glancing phase, we denote $Π^{g} : = Π^{k}$ and $Υ^{g} : = Υ^{k}$ for the glancing index $k \in G$ .

We now introduce some material which are commonly used for geometric optics expansions without the uniform Kreiss–Lopatinskii condition (see [9]). From Assumption 2.4 the vector space $B E^{s} (\underline{ζ})$ is $(p - 1)$ -dimensional. We can therefore write it as the kernel of a complex linear form $\begin{matrix} (13) & B E^{s} (\underline{ζ}) : = {X \in C^{p}, b \cdot X = 0} \end{matrix}$ for a suitable vector $b \in C^{p} ∖ {0}$ .

Then, we can choose a supplementary vector space ${\tilde{E}}^{s} (\underline{ζ})$ of $span (e)$ in $E^{s} (\underline{ζ})$ : $\begin{matrix} (14) & E^{s} (\underline{ζ}) = span (e) \oplus {\tilde{E}}^{s} (\underline{ζ}) . \end{matrix}$ The matrix B induces an isomorphism from ${\tilde{E}}^{s} (\underline{ζ})$ to the hyperplane $B E^{s} (\underline{ζ})$ and we denote its inverse $Φ : = Φ (\underline{ζ}) : = B_{| {\tilde{E}}^{s} (\underline{ζ})}^{- 1}$ .
Remark 2.4.
In the following, we can take ${\tilde{E}}^{s} (\underline{ζ}) = E_{h}^{s} (\underline{ζ}) \oplus E_{e}^{s} (\underline{ζ})$ , due to Remark 2.2. Thus, we have the decomposition: $\begin{matrix} (15) & E^{s} (\underline{ζ}) = E_{h}^{s} (\underline{ζ}) \oplus span (e) \oplus E_{e}^{s} (\underline{ζ}) = ⨁_{k \in I} ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}}) \oplus span (e) \oplus E_{e}^{s} (\underline{ζ}) . \end{matrix}$

In the following, to determine the hyperbolic and elliptic amplitudes of high order (and more precisely their traces) we will need the following projectors2
²
Note in particular that we do not use or require the boundedness of the projector on $E_{g}^{s}$ in the decomposition (15). So for this special point we do not require the glancing mode to be of size two (see [20] and [13]).

.
Definition 2.5.
Under Assumptions 2.1, 2.2 and 2.3 for $\underline{ζ} \in Ξ_{0}$ , we define:
for $k \in I$ , $P^{k} : = P^{k} (\underline{ζ})$ the projector on $ker L (\underline{τ}, \underline{η}, \underline{ξ_{k}})$ with respect to the decomposition (15).

$P^{e} : = P^{e} (\underline{ζ})$ the projector on $E_{e}^{s} (\underline{ζ})$ with respect to the decomposition (15).

For technical reasons that will be made precise during the construction of the WKB expansion, we assume that the following holds. Assumption 2.5.
Under Assumptions 2.1, 2.2, 2.3 and 2.4, we assume that for the index $g \in G$ , the scalar quantity $λ : = b \cdot B Υ^{g} A_{d} e \neq 0$ .
Remark 2.5.
If we have $E^{s} (\underline{ζ}) = ker L (\underline{τ}, \underline{η}, ξ_{g}) = span (e)$ which is automatic for $2 \times 2$ systems, then we have $b \cdot B Υ^{g} A_{d} e \neq 0$ .

Indeed proceed by contradiction and assume that $Υ^{g} A_{d} e \in E^{s} (\underline{ζ}) = Im Π^{g}$ . However, $Υ^{g} A_{d} e \in Im Υ^{g} = ker Π^{g}$ and $ker Π^{g} \cap Im Π^{g} = {0}$ , because $Π^{g}$ is a projection.

So, we deduce that $Υ^{g} A_{d} e = 0$ and so, applying the symbol $L (\underline{τ}, \underline{η}, ξ_{g})$ , $\begin{matrix} Π^{g} A_{d} e = A_{d} e . \end{matrix}$ Next, since $e \in E^{s} (\underline{ζ}) = Im Π^{g}$ , we can write $e = Π^{g} f$ where $f \in C^{N} ∖ {0}$ . And, using Lemma 3.1, we obtain: $\begin{array}{l} A_{d} e & = Π^{g} A_{d} e = Π^{g} A_{d} Π^{g} f = 0 . \end{array}$ We deduce that $e = 0$ which is a contradiction and so $b \cdot B Υ^{g} A_{d} e \neq 0$ .

2.2. Main results

In this section we state the two main results of the article. The first one deals with the existence of a solution to the geometric optics expansion cascade of equations at any order. The amplitudes of the profiles are rather similar to the one introduced in [21] in order to treat glancing phases. In particular they involve two boundary layers one in $\frac{1}{ε}$ , it is associated to the elliptic modes, and one in $\frac{1}{\sqrt{ε}}$ which is proper to the glancing modes.

Theorem 2.2.
Under Assumptions 2.1 – 2.2 – 2.3 – 2.4 and 2.5 then for all $n \in N$ , the cascades of equations ( 22 )–( 23 ) and ( 24 ) admit a unique solution ${(u_{n, k}, u_{n}^{g}, U_{e v, n})}_{n \in N, k \in I \cup O, g \in G} \subset H^{\infty} (Ω_{T}) \times P_{g} \times P_{e v}$ where the evanescent and glancing profile sets, $P_{g}$ and $P_{e v}$ , are introduced in Definition 3.1 .

The construction of the geometric optics expansion is given in Section 3.

Note that in a general setting because there is no general well-posedness theory for hyperbolic boundary value problems when the uniform Kreiss–Lopatinskii condition fails in the glancing area then we can not justify that the approximated solution given by the geometric optics expansion is effectively an approximate solution to the boundary value problem (5) for highly oscillating source terms.

However if such a weak well-posedness theory is available then we can show (see Section 4) that the solution to the geometric optics expansion is effectively an approximate solution. Moreover for some examples like the interesting case of the wave equation for Neumann boundary condition such a weak well-posedness theory exists (see [11] and [10]) so, on some examples, we can also conclude that the geometric optics expansion is an approximate solution.

Our second result follows closely [9] and [3] by using the geometric optics expansion in order to understand what can be (we can not say “are” here because of the lack of a weak well-posedness theory) the losses of derivatives in the energy estimate. The result is the following and its proof is given in Section 4. Theorem 2.3.
Let Assumptions 2.1 , 2.2 , 2.3 , 2.4 and 2.5 be satisfied, let $s_{1}, s_{2} ⩾ 0$ and $T > 0$ . Assume that for all sources $f \in L^{2} (R_{x_{d}}^{+}; H_{t, x^{'}}^{s_{1}} (ω_{T}))$ and $g \in H^{s_{2}} (ω_{T})$ vanishing for negative times, there exists a unique $u \in L^{2} (Ω_{T})$ vanishing for negative times that is a weak solution to the problem ( 5 ), and that satisfies an energy estimate of the form: $\begin{matrix} (16) & ‖ u ‖_{L^{2} (Ω_{T})} ⩽ C_{T} (‖ f ‖_{L^{2} (R_{x_{d}}^{+}; H_{t, x^{'}}^{s_{1}} (ω_{T}))} + ‖ g ‖_{H^{s_{2}} (ω_{T})}) . \end{matrix}$ Then we have the following possible alternatives
If the extra Assumption 4.1 about the existence of some suitable for amplification outgoing mode holds then we have $s_{1} ⩾ \frac{1}{2}$ . So that a loss of at least one half derivative in the interior occurs.

If the system ( 5 ) admits an energy estimate with no loss in the interior $i . e$ . $s_{1} = 0$ then we have at least a lost of one half derivative on the boundary that is $s_{2} ⩾ \frac{1}{2}$ .

In the general framework we have a loss of derivatives in the interior or on the boundary. More precisely $s_{1} ⩾ \frac{1}{2}$ or $s_{2} ⩾ \frac{1}{2}$ .

3. Determination of the WKB expansion

This section is devoted to the construction of the geometric optics expansion that is the proof of Theorem 2.2. We consider the initial boundary value problem (5) with highly oscillating interior and boundary source terms. More precisely $\begin{array}{l} (17) & \{\begin{matrix} L (\partial) u^{ε} = f^{ε} & in Ω_{T}, \\ B u_{| x_{d} = 0}^{ε} = g^{ε} & on ω_{T}, \\ u_{| t ⩽ 0}^{ε} = 0 & on R_{+}^{d} . \end{matrix} \end{array}$

In all that follows we fix $\underline{ζ} : = (i \underline{τ}, \underline{η}) \in G$ such that Asumptions 2.3 and 2.4 are satisfied. In order to define the sources in (17), we introduce the phases functions: $\begin{matrix} ψ (t, x^{'}) : = \underline{τ} t + \underline{η} \cdot x^{'} and \{\begin{matrix} φ_{k} (t, x) : = ψ (t, x^{'}) + \underline{ξ_{k}} x_{d} & for k \in I \cup O, \\ φ^{g} (t, x) : = ψ (t, x^{'}) + \underline{ξ^{g}} x_{d} & for k \in G : = {g}, \end{matrix} \end{matrix}$ where the $\underline{ξ_{k}}$ stand for the roots in the ξ-variable of the dispersion relation $det L (\underline{τ}, \underline{η}, ξ) = 0$ .

We choose the sources $f^{ε}$ and $g^{ε}$ reading under the form, for $0 < ε ≪ 1$ $\begin{array}{l} (18) & f^{ε} : = ε^{\frac{1}{2}} (\sum_{k \in I \cup O} e^{\frac{i}{ε} φ_{k}} f_{k}) + e^{\frac{i}{ε} φ^{g}} f^{g} + e^{\frac{i}{ε} ψ} f_{e v}, \\ (19) & g^{ε} : = ε^{\frac{1}{2}} e^{\frac{i}{ε} ψ} g, \end{array}$ where for all $k \in I \cup O$ , the amplitudes $f_{k} \in H^{\infty} (Ω_{T})$ , $f^{g} \in P_{g}$ (the set of glancing profiles defined below) and $g \in H_{γ}^{\infty} (ω_{T})$ . All these terms vanish for negative times and for $X \subset Ω_{T}$ the space $H_{γ}^{\infty} (X)$ being defined by $\begin{matrix} (20) & H_{γ}^{\infty} (X) : = {u \in L^{2} (X) ∖ e^{- γ t} u \in H^{\infty} (X)} . \end{matrix}$

The elliptic source term $f_{e v}$ lies in $P_{e v}$ the space of elliptic profiles introduced in [16], note that this treatment of the elliptic modes differs a little from the one of [21] because it is done in a monoblock framework and in particular no diagonalization property is required on the elliptic block. Also remark that compared to [16] our elliptic profiles depend on $x_{d}$ only through the fast variable $\frac{x_{d}}{ε}$ . This will simplify some points of the proofs compared to [3] where some lifting of double traces on ${x_{d} = χ = 0}$ was required (see [16] or [3] for more details).

Definition 3.1.
[Boundary layer profiles] The set of elliptic (or evanescent) profiles $P_{e v}$ is defined by $\begin{matrix} P_{e v} : = {F : = F (t, x^{'}, X_{d}) \in H^{\infty} (ω_{T} \times R_{+}) ∖ \exists δ > 0, e^{δ X_{d}} F (t, x^{'}, X_{d}) \in H^{\infty} (ω_{T} \times R_{+})} . \end{matrix}$ We define similarly the set of glancing profiles $P_{g}$ : $\begin{matrix} P_{g} : = P_{e v} = {F : = F (t, x^{'}, χ) \in H^{\infty} (ω_{T} \times R_{+}) ∖ \exists δ > 0, e^{δ χ} F (t, x^{'}, χ) \in H^{\infty} (ω_{T} \times R_{+})} . \end{matrix}$

We postulate for ansatz $\begin{array}{l} u^{ε} (t, x) \sim & \sum_{n ⩾ 0} ε^{\frac{n}{2}} \sum_{k \in I \cup O} e^{\frac{i}{ε} φ_{k} (t, x)} u_{n, k} (t, x) + \sum_{n ⩾ 0} ε^{\frac{n}{2}} e^{\frac{i}{ε} φ^{g} (t, x)} u_{n}^{g} (t, x^{'}, \frac{x_{d}}{\sqrt{ε}}) \\ (21) & + \sum_{n ⩾ 0} ε^{\frac{n}{2}} e^{\frac{i}{ε} ψ (t, x^{'})} U_{e v, n} (t, x^{'}, \frac{x_{d}}{ε}), \end{array}$ where for all $n ⩾ 0$ the evanescent profile $U_{e v, n} \in P_{e v}$ , the hyperbolic and glancing amplitudes ${(u_{n, k})}_{k \in I \cup O} \in H^{\infty} (Ω_{T})$ and $u_{n}^{g} \in P_{g}$ .

We denote $χ : = \frac{x_{d}}{\sqrt{ε}}$ the fast variable for the glancing boundary layer and $X_{d} : = \frac{x_{d}}{ε}$ the one of the elliptic boundary layer.

Remark in particular that the amplitudes $u_{n}^{g}$ for glancing mode and $U_{e v, n}$ for the elliptic modes are functions of $(t, x^{'}, χ)$ and $(t, x^{'}, X_{d})$ so that their only dependency with respect to $x_{d}$ is made in the fast variable. It differs from the hyperbolic amplitudes $u_{n, k}$ .

Plugging the ansatz (21) in the evolution equation of (17) leads, after identification on the $ε^{\frac{n}{2}}$ and using the linear independence of the phases functions, to the following cascade of equations: $\begin{array}{l} (22) & \{\begin{matrix} I_{- 1} : \{\begin{matrix} L (d φ^{g}) u_{0}^{g} = 0, \\ L (d φ_{k}) u_{0, k} = 0 & \forall k \in I \cup O, \\ L (\partial_{X_{d}}) U_{e v, 0} = 0, \end{matrix} \\ I_{- \frac{1}{2}} : \{\begin{matrix} i L (d φ^{g}) u_{1}^{g} + A_{d} \partial_{χ} u_{0}^{g} = 0, \\ L (d φ_{k}) u_{1, k} = 0 & \forall k \in I \cup O, \\ L (\partial_{X_{d}}) U_{e v, 1} = 0, \end{matrix} \\ for n ⩾ 0, I_{\frac{n}{2}} : \{\begin{matrix} i L (d φ^{g}) u_{n + 2}^{g} + A_{d} \partial_{χ} u_{n + 1}^{g} + L^{'} (\partial) u_{n}^{g} = δ_{n, 0} f^{g}, \\ i L (d φ_{k}) u_{n + 2, k} + L (\partial) u_{n, k} = δ_{n, 1} f_{k} & \forall k \in I \cup O, \\ L (\partial_{X_{d}}) U_{e v, n + 2} + L^{'} (\partial) U_{e v, n} = δ_{n, 0} f_{e v}, \end{matrix} \end{matrix} \end{array}$ where $L (\partial_{X_{d}}) : = A_{d} (\partial_{X_{d}} - A (\underline{ζ}))$ is the operator of derivation with respect to the fast variable $X_{d}$ and $L^{'} (\partial) : = \partial_{t} + \sum_{j = 1}^{d - 1} A_{j} \partial_{x_{j}}$ is the operator of differentiation with respect to the tangential (slow) variables. In (22), $δ_{\cdot, \cdot}$ stands for the Kronecker symbol.

Plugging the ansatz (21) in the boundary condition of (17) gives: $\begin{matrix} (23) & \forall n ⩾ 0, B_{\frac{n}{2}} : B (\sum_{k \in I \cup O} u_{{n, k}_{| x_{d} = 0}} + u_{n_{| χ = 0}}^{g} + U_{{e v, n}_{| X_{d} = 0}}) = δ_{n, 1} g . \end{matrix}$

Finally, plugging the ansatz (21) in the initial condition (17) gives: $\begin{matrix} (24) & \forall n \in N, \forall k \in I \cup O, u_{n_{|} t ⩽ 0}^{g} = u_{{n, k}_{| t ⩽ 0}} = U_{{e v, n}_{| t ⩽ 0}} = 0 . \end{matrix}$

In the following, we describe how to construct the amplitudes $u_{n}^{g}$ , $u_{n, k}$ and $U_{e v, n}$ solving the cascades of equation (22), (23), (24). The construction is really classical for hyperbolic modes. Indeed because of polarization conditions and Lax lemma [15] the hyperbolic amplitudes solve transport equations. If the mode is incoming (resp. outgoing) there is (resp. there is not) a boundary condition on ${x_{d} = 0}$ to solve. However because the uniform Kreiss–Lopatinskii condition holds for hyperbolic modes this condition can be solved by inverting the boundary matrix B. The construction for such modes is performed in Paragraphs 3.1.1, 3.2.1 and 3.3.

The construction for elliptic modes is also rather classical. In the space $P_{e v}$ the solution of the evolution equation in the fast variable $X_{d}$ can be written via Duhamel formula as the evolution of the trace on ${X_{d} = 0}$ plus the contribution to the interior source (which are supposed to be known). Once again because the uniform Kreiss–Lopatinskii condition holds for elliptic modes we have an explicit formula for the trace on ${X_{d} = 0}$ . The determination of elliptic modes occupies Paragraphs 3.1.2, 3.2.2 and 3.3.

The main difficulty in the construction is the determination of the glancing mode. Indeed on the one hand Lax lemma applies so that it should solve the tangential (note that for a glancing mode the group velocity $v_{g} : = (v_{g}^{'}, 0)$ ) transport equation $\begin{matrix} (25) & \partial_{t} u_{0}^{g} + v_{g}^{'} \cdot \nabla_{x^{'}} u_{0}^{g} = 0, \end{matrix}$ equation that does not require any boundary condition on $ω_{T}$ . But on the other hand we require a boundary condition in (23) to have a good error term on the boundary. Boundary condition that overdetermined (25). This difficulty has been first encounter in [20] and then overcame in [21] by the introduction of the large boundary layer of size $\sqrt{ε}$ .

One extra difficulty compared to [20,21] is that to determine the value of the boundary layer trace on ${χ = 0}$ which is required for its determination we can not use the uniform Kreiss–Lopatinskii condition anymore. Indeed at some step we will obtain that $u_{0_{| χ = 0}}^{g} \in ker B \cap E_{g}^{s} (\underline{ζ})$ so that $u_{0_{| χ = 0}}^{g}$ reads $u_{0_{| χ = 0}}^{g} (t, x^{'}) : = α_{0} (t, x^{'}) e$ and the whole question is to find a way to determine the good amplitude value $α_{0}$ . To do so we will follow the method of [9] (see also [3]) consisting in considering the higher order boundary condition in order to derive some compatibility condition on the trace. Compared to [9] and [3] the equation determining $α_{0}$ will not be a simple transport equation anymore but it will involve some Fourier multiplier (see equation (49) for more details). The construction of the glancing boundary layer is made in Paragraph 3.1.3. For higher order terms, we refer to Paragraphs 3.2.3 and 3.3.
3.1. Construction of the leading order terms

In this paragraph, we first use the partial inverses introduced in Definition 2.4 in order to recover the usual so-called polarization condition for hyperbolic and glancing modes.

We next use Lax lemma [15] to obtain transport equation with respect to the group velocity for the hyperbolic and glancing leading amplitudes which are classical in geometric optics expansions. Finally, we will determine the hyperbolic and elliptic amplitudes first because they are mandatory to determine the glancing one.

From $I_{- 1}$ , we see that $u_{0}^{g} \in ker L (d φ^{g})$ and for all $k \in I \cup O$ we also have $u_{0, k} \in ker L (d φ_{k})$ . So that it turns out that we have the usual polarization conditions $\begin{array}{l} (26) & Π^{g} u_{0}^{g} = u_{0}^{g}, \\ (27) & Π^{k} u_{0, k} = u_{0, k}, \forall k \in I \cup O . \end{array}$ Next, applying the partial inverse $Υ^{g}$ to the equation for glancing mode of $I_{- \frac{1}{2}}$ gives: $\begin{array}{l} (28) & (I - Π^{g}) u_{1}^{g} = i Υ^{g} A_{d} \partial_{χ} u_{0}^{g} . \end{array}$

Now, we apply respectively the projector $Π^{g}$ and $Π^{k}$ to the equations for glancing and hyperbolic modes of $I_{0}$ to obtain: $\begin{array}{l} (29) & \{\begin{matrix} Π^{g} L^{'} (\partial) u_{0}^{g} + Π^{g} A_{d} \partial_{χ} u_{1}^{g} = Π^{g} f^{g}, \\ Π^{k} L (\partial) Π^{k} u_{0, k} = 0 k \in I \cup O . \end{matrix} \end{array}$ Decomposing $u_{1}^{g}$ in terms of polarized and unpolarized parts, gives: $\begin{matrix} (30) & Π^{g} L (\partial) u_{0}^{g} + Π^{g} A_{d} Π^{g} \partial_{χ} u_{1}^{g} + Π^{g} A_{d} \partial_{χ} (I - Π^{g}) u_{1}^{g} = Π^{g} f^{g} . \end{matrix}$

Using equations (26), (28) and (30), we thus obtain: $\begin{matrix} (31) & Π^{g} L (\partial) Π^{g} u_{0}^{g} + Π^{g} A_{d} Π^{g} \partial_{χ} u_{1}^{g} + i Π^{g} A_{d} Υ^{g} A_{d} Π^{g} \partial_{χ}^{2} u_{0}^{g} = Π^{g} f^{g} . \end{matrix}$

We now need some lemmas to recover a transport equation due to Williams [21] and Lax [15].

Lemma 3.1 ([21], Lemma 8.1).

Under Assumptions 2.1 , 2.2 and 2.3 we have: $\begin{matrix} Π^{g} A_{d} Υ^{g} A_{d} Π^{g} = \frac{1}{c} Π^{g} and Π^{g} A_{d} Π^{g} = 0, \end{matrix}$ where $c \in R ∖ {0}$ .

Lemma 3.2 (Lax lemma [15]).

Under Assumption 2.1 , we have: $\begin{matrix} \forall k \in I \cup O \cup G, Π^{k} L (\partial) Π^{k} = (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} \end{matrix}$ where the velocity $v_{k}$ is the so-called group velocity associated to k introduced in Definition 2.2 .

We recall that we have only one glancing frequency so that $G = {g}$ and the associated group velocity reads $v_{g} : = (v_{g}^{'}, 0)$ for some $v_{g}^{'} \in R^{d - 1}$ .

Using these two lemmas and equations (26), (27) permits to rewrite equations (31) and (29) as: $\begin{array}{l} (32) & - \partial_{χ}^{2} u_{0}^{g} + i c (\partial_{t} + v_{g}^{'} \cdot \nabla_{x^{'}}) u_{0}^{g} = i c Π^{g} f^{g}, \\ (33) & (\partial_{t} + v_{k} \cdot \nabla_{x}) u_{0, k} = 0, k \in I \cup O, \end{array}$ combined with the equation for the first elliptic mode $\begin{matrix} (34) & L (\partial_{X_{d}}) U_{e v, 0} = 0 . \end{matrix}$

Then we consider the first order boundary condition, namely $B_{0}$ . It reads: $\begin{matrix} B (\sum_{k \in I \cup O} u_{{0, k}_{| x_{d} = 0}} + u_{0_{| χ = 0}}^{g} + U_{{e v, 0}_{| X_{d} = 0}}) = 0, \end{matrix}$ so that we can decouple the boundary condition like: $\begin{array}{l} (35) & B u_{{0, k}_{| x_{d} = 0}} = 0 \forall k \in I \cup O, \\ (36) & B u_{0_{| χ = 0}}^{g} = 0, \\ (37) & B U_{{e v, 0}_{| X_{d} = 0}} = 0 . \end{array}$

Finally for the initial condition we will solve $\begin{matrix} \forall k \in I \cup O, u_{0_{| t ⩽ 0}}^{g} = u_{{0, k}_{| t ⩽ 0}} = U_{{e v, 0}_{| t ⩽ 0}} = 0 . \end{matrix}$

To determine the leading order amplitudes, we determine in the following the hyperbolic ones then the elliptic one and finally the glancing one. We can determine the hyperbolic and evanescent ones directly from the boundary condition but for the glancing one, since the uniform Kreiss–Lopatinskii condition does not hold, we follow the analysis of Coulombel–Guès [9] which requires to consider the boundary condition $B_{\frac{1}{2}}$ involving the hyperbolic and evanescent amplitudes of order one.

3.1.1. Determination of the hyperbolic leading order term

The transport for the outgoing phases goes from the interior of the domain $Ω_{T}$ to the boundary $ω_{T}$ and consequently in the resolution of (33) no boundary condition has to be imposed on the boundary $ω_{T}$ . Using (33) we thus determine the outgoing amplitudes by resolving the homogeneous transport equation: $\begin{matrix} \forall k \in O, \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) u_{0, k} = 0, \\ u_{{0, k}_{| t ⩽ 0}} = 0, \end{matrix} \end{matrix}$ and thus $u_{0, k} \equiv 0$ for all $k \in O$ .

To determine the glancing leading order term, we will also require to know the $u_{1, k} = Π^{k} u_{1, k}$ for $k \in O$ . Proceeding as it has been done for the leading order term, it is easy to see that these amplitudes satisfy the transport equations $\begin{matrix} \forall k \in O, \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{1, k} = Π^{k} f_{k}, \\ Π^{k} u_{{1, k}_{| t ⩽ 0}} = 0, \end{matrix} \end{matrix}$ so that integrating the equations along the characteristics, we determine the $u_{1, k}$ for all outgoing k in terms of the given sources $f_{k}$ . Because they are solution to a linear transport equation with a source in $H^{\infty} (Ω_{T})$ the $u_{1, k}$ are in $H^{\infty} (Ω_{T})$ .

We now turn to the construction of the incoming amplitudes. In that case the transport goes from the boundary $ω_{T}$ to the interior of $Ω_{T}$ and it is thus needed to know the value of $u_{{0, k}_{| x_{d} = 0}}$ in order to solve the transport equation. For these phases, due to Remark 2.4, we see that for $k \in I$ , $Π^{k} u_{0, k} \in {\tilde{E}}^{s} (\underline{ζ})$ and so we deduce that $Π^{k} u_{{0, k}_{| x_{d} = 0}} = Φ (0) = 0$ , for all $k \in I$ , where we recall that $Φ : = B_{| {\tilde{E}}^{s} (\underline{ζ})}^{- 1}$ . Hence, the amplitudes $u_{0, k}$ for incoming modes solve the transport equations $\begin{matrix} \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) u_{0, k} = 0, \\ u_{{0, k}_{| x_{d} = 0}} = 0, \\ u_{{0, k}_{| t ⩽ 0}} = 0, \end{matrix} \end{matrix}$ which are homogeneous linear transport equations so the $u_{0, k}$ for $k \in I$ are zero.

3.1.2. Determination of the evanescent leading order term

From the cascade of equations $I_{- 1}$ and the boundary condition (37), we have to solve for elliptic modes: $\begin{matrix} \{\begin{matrix} L (\partial_{X_{d}}) U_{e v, 0} = 0 & for (t, x) \in Ω_{T}, X_{d} \in R_{+}, \\ B U_{{e v, 0}_{| X_{d} = 0}} = 0, \\ U_{{e v, 0}_{| t ⩽ 0}} = 0, \end{matrix} \end{matrix}$ which is a standard ordinary differential equation with respect to the fast variable $X_{d}$ in which the variables $(t, x^{'})$ act as parameters. In order to solve this equation we will use the following lemma (see [16] or [3]):

Lemma 3.3 (Lescarret).

We define for $X_{d} ⩾ 0$ : $\begin{array}{l} (38) & (P_{e v} U) (X_{d}) : = e^{X_{d} A (\underline{ζ})} Π_{e}^{s} U (0), \\ (39) & (Q_{e v} F) (X_{d}) : = \int_{0}^{X_{d}} e^{(X_{d} - s) A (\underline{ζ})} Π_{e}^{s} A_{d}^{- 1} F (s) d s - \int_{X_{d}}^{\infty} e^{(X_{d} - s) A (\underline{ζ})} Π_{e}^{u} A_{d}^{- 1} F (s) d s . \end{array}$ Then, for all $F \in P_{e v}$ , the equation $L (\partial_{X_{d}}) U = F$ for $X_{d} ⩾ 0$ , admits a unique solution $U \in P_{e v}$ reading $U = P_{e v} U + Q_{e v} F$ .

Using this lemma, we have the polarization condition for elliptic modes $U_{e v, 0} = P_{e v} U_{e v, 0}$ . So it is sufficient to determine $U_{{e v, 0}_{| X_{d} = 0}}$ in order to determine $U_{e v, 0}$ .

Since $U_{e v, 0} = P_{e v} U_{e v, 0}$ , we deduce that $U_{e v, 0} \in E_{e}^{s} (\underline{ζ})$ . Hence, we can invert B and the boundary condition (37) gives $U_{{e v, 0}_{| X_{d} = 0}} = 0$ .

Hence, we deduce that $U_{e v, 0} \equiv 0$ because it is the free evolution of the trivial trace $U_{{e v, 0}_{| X_{d} = 0}}$ .

Next, we will have to be more precise about the form of $U_{e v, 1}$ to determine the glancing leading order term. Since $U_{e v, 0} = 0$ , we have from the cascade of equations $I_{- \frac{1}{2}}$ , $L (\partial_{X_{d}}) U_{e v, 1} = 0$ , so that from Lemma 3.3 we have, $U_{e v, 1} = P_{e v} U_{e v, 1}$ , polarization condition which will be sufficient for the determination of the leading order glancing term.

3.1.3. Determination of the glancing leading order term

For $γ > 0$ , let $v_{0}^{g} : = e^{- γ t} u_{0}^{g}$ . We observe that $\partial_{t} v_{0}^{g} = - γ v_{0}^{g} + e^{- γ t} \partial_{t} u_{0}^{g}$ . Hence, in the new unknown, the equation for the glancing leading order term (32) becomes: $\begin{matrix} (40) & - \partial_{χ}^{2} v_{0}^{g} + i c (γ + \partial_{t} + v_{g}^{'} \cdot \nabla_{x^{'}}) v_{0}^{g} = i c e^{- γ t} Π^{g} f^{g}, \end{matrix}$ where we recall that $c \in R ∖ {0}$ is defined in Lemma 3.1 and where $v_{g}^{'}$ stands for the (possibly) non vanishing part of the glancing group velocity.

We now perform a Fourier transform in the time variable t and in the tangential space variable $x^{'}$ to this equation. Let $τ \in R$ and $η \in R^{d - 1}$ denote the dual variables of t and $x^{'}$ and $\hat{\cdot}$ denotes the Fourier transform. We thus have: $\begin{matrix} - \partial_{χ}^{2} \hat{v_{0}^{g}} + i c (γ + i τ + i \sum_{j = 1}^{d - 1} v_{g, j}^{'} η_{j}) \hat{v_{0}^{g}} = i c Π^{g} \hat{e^{- γ t} f^{g}} . \end{matrix}$

Let $\begin{matrix} (41) & X = X (σ, η) : = i c (γ + i τ + i \sum_{j = 1}^{d - 1} v_{g, j}^{'} η_{j}) = c (- τ - \sum_{j = 1}^{d - 1} v_{g, j}^{'} η_{j} + i γ), \end{matrix}$ we obtain the following equation: $\begin{matrix} (42) & - \partial_{χ}^{2} \hat{v_{0}^{g}} + X \hat{v_{0}^{g}} = i c Π^{g} \hat{e^{- γ t} f^{g}} . \end{matrix}$

So, if $f^{g}$ has exponential decay with respect to χ for example if there exists $\tilde{f^{g}} \in L^{2} (R^{d})$ and $C, δ > 0$ such that $\begin{matrix} | f^{g} (t, x^{'}, χ) | ⩽ C | \tilde{f^{g}} (t, x^{'}) | e^{- δ χ}, \end{matrix}$ then we have a unique solution $\hat{v_{0}^{g}}$ which is in $L_{τ, η}^{2} (R^{d})$ with exponential decay with respect to χ (see [21], equation 8.40): $\begin{array}{l} \hat{v_{0}^{g}} (τ, η, χ) \\ (43) & = e^{\sqrt{X} χ} {\hat{v_{0}^{g}}}_{| χ = 0} (τ, η) + i c \int_{0}^{χ} e^{(χ - χ^{'}) \sqrt{X}} \int_{χ^{'}}^{\infty} e^{- (χ^{'} - χ^{″}) \sqrt{X}} Π^{g} \hat{e^{- γ t} f^{g}} (τ, η, χ^{″}) d χ^{″} d χ^{'} . \end{array}$ where $\sqrt{X}$ stands for the root of X with negative real part in order that $v_{0}^{g} \in L_{χ}^{2} (R_{+})$ .

Reversing the Fourier transform, we deduce that: $\begin{array}{l} v_{0}^{g} (t, x^{'}, χ) \\ = \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{\sqrt{X} χ} {\hat{v_{0}^{g}}}_{| χ = 0} (τ, η) d τ d η \\ (44) & + \frac{i c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{χ} e^{(χ - χ^{'}) \sqrt{X}} \int_{χ^{'}}^{\infty} e^{- (χ^{'} - χ^{″}) \sqrt{X}} Π^{g} \hat{e^{- γ t} f^{g}} (τ, η, χ^{″}) d χ^{″} d χ^{'} d τ d η . \end{array}$

So, to determine $v_{0}^{g}$ (and so $u_{0}^{g}$ ), we have to determine ${\hat{v_{0}^{g}}}_{| χ = 0}$ . Indeed the second term in (44) is a known function depending on the source $f^{g}$ . Moreover following [[20]-Proposition 9.6] we have that $v_{0}^{g}$ vanishes for negative times if $v_{0_{| χ = 0}}^{g}$ and $e^{- γ t} f^{g}$ do, so, if and only if $v_{0_{| χ = 0}}^{g}$ does.

If we first consider the boundary condition $B_{0}$ then (26) shows us that $u_{0}^{g}$ is polarized so $u_{0}^{g} \in E^{s} (\underline{ζ})$ and the boundary condition implies that the trace $u_{0_{| χ = 0}}^{g} \in ker B$ .

Thus, we deduce that $u_{0_{| χ = 0}}^{g} \in E_{g}^{s} \cap ker (B)$ so, using Assumption 2.4, there exists a scalar function $α_{0}$ defined on $ω_{T}$ such that $\begin{matrix} (45) & u_{0_{| χ = 0}}^{g} (t, x^{'}) : = α_{0} (t, x^{'}) e and thus v_{0_{| χ = 0}}^{g} (t, x^{'}) = e^{- γ t} α_{0} (t, x^{'}) e . \end{matrix}$ For compatibility reasons we should keep in mind that we should have $α_{0_{| t ⩽ 0}} = 0$ .

In order to determine $α_{0}$ , we look at the boundary condition $B_{\frac{1}{2}}$ which is the only other equation involving $u_{0_{| χ = 0}}^{g}$ (via the unpolarized part). More precisely, we have: $\begin{matrix} B (\sum_{k \in I \cup O} u_{{1, k}_{| x_{d} = 0}} + u_{1_{| χ = 0}}^{g} + U_{{e v, 1}_{| X_{d} = 0}}) = g . \end{matrix}$

We have already justified that for all $k \in I \cup O$ , we have $Π^{k} u_{1, k} = u_{1, k}$ so that we can replace $u_{1, k}$ by $Π^{k} u_{1, k}$ in the previous equation. Similarly we have that $U_{1, e v} = P_{e v} U_{1, e v}$ so that the boundary condition can be written as $\begin{matrix} B (\sum_{k \in I} Π^{k} u_{{1, k}_{| x_{d} = 0}} + u_{1_{| χ = 0}}^{g} + {(P_{e v} U_{e v, 1})}_{| X_{d} = 0}) = g - B \sum_{k \in O} Π^{k} u_{{1, k}_{| x_{d} = 0}} . \end{matrix}$

So, decomposing $u_{1}^{g}$ in terms of polarized and unpolarized parts and applying the vector b defined in (13), note that for all $k \in I$ , $Π^{k} u_{1, k} \in E_{h}^{s} (\underline{ζ})$ and that $P_{e v} U_{1, e v} \in E_{e}^{s} (\underline{ζ})$ , we obtain: $\begin{array}{l} b \cdot B (I - Π^{g}) u_{1_{| χ = 0}}^{g} & = b \cdot g - b \cdot B \sum_{k \in O} Π^{k} u_{{1, k}_{| x_{d} = 0}} . \end{array}$

Using (28), we can rewrite this equation as: $\begin{matrix} (46) & i b \cdot B Υ^{g} A_{d} {(\partial_{χ} u_{0}^{g})}_{| χ = 0} = b \cdot g - b \cdot B \sum_{k \in O} Π^{k} u_{{1, k}_{| x_{d} = 0}}, \end{matrix}$ where we recall that $Υ^{g}$ is the partial inverse of Definition 2.4. We multiply the above equation by $e^{- γ t}$ to obtain: $\begin{matrix} (47) & i b \cdot B Υ^{g} A_{d} {(\partial_{χ} v_{0}^{g})}_{| χ = 0} = e^{- γ t} (b \cdot g - b \cdot B \sum_{k \in O} Π^{k} u_{{1, k}_{| x_{d} = 0}}) . \end{matrix}$

Using (44) and (45), we see that $\begin{array}{l} {(\partial_{χ} v_{0}^{g})}_{| χ = 0} = & \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \sqrt{X} \hat{e^{- γ t} α_{0}} (τ, η) d τ d η e \\ (48) & + \frac{i c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} Π^{g} \hat{e^{- γ t} f^{g}} (τ, η, χ^{'}) d χ^{'} d τ d η . \end{array}$ This motivates the following definition: let T be the Fourier multiplier defined from $H_{γ}^{\infty} (R^{d})$ (we refer to (20)) into $H^{\infty} (R^{d})$ given by $\begin{matrix} (49) & (T f) (t, x^{'}) : = \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot ξ} \sqrt{X} \hat{e^{- γ t} f} (ξ) d ξ, \end{matrix}$ where ξ is a short hand notation for $(τ, η)$ . We also introduce $λ : = b \cdot B Υ^{g} A_{d} e \neq 0$ from Assumption 2.5. In this new notations using (48) in (47) thus gives the following condition on the unknown trace $α_{0}$ : $\begin{array}{l} (50) & i λ T (α_{0}) = & b \cdot e^{- γ t} g - b \cdot B \sum_{k \in O} e^{- γ t} Π^{k} u_{{1, k}_{| x_{d} = 0}} \\ + \frac{c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} b \cdot B Υ^{g} A_{d} Π^{g} \hat{e^{- γ t} f^{g}} (τ, η, χ^{'}) d χ^{'} d τ d η \\ (51) & : = & G_{0}, \end{array}$ where $G_{0}$ is explicit in terms of g, $f^{k}$ , $k \in O$ and $f^{g}$ . We also remark for later purposes that all the terms in the right hand side of (50) vanish for negative times if g, $f^{g}$ and the $f^{k}$ , $k \in O$ , do.

Proposition 3.1.
The operator T defined by $\begin{matrix} T (f) (t, x^{'}) : = \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot ξ} \sqrt{X} \hat{e^{- γ t} f} (ξ) d ξ \end{matrix}$ maps $H_{γ}^{\infty} (R^{d})$ into $H^{\infty} (R^{d})$ where $H_{γ}^{\infty} (R^{d})$ is defined in ( 20 ) and where X is defined in ( 41 ).

Moreover, T is invertible and $T^{- 1}$ is defined by $\begin{matrix} T^{- 1} (f) (t, x^{'}) : = \frac{e^{γ t}}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot ξ} \frac{1}{\sqrt{X}} \hat{f} (ξ) d ξ; \end{matrix}$ it maps $H^{\infty} (R^{d})$ into $H_{γ}^{\infty} (R^{d})$ .

Finally if the function f vanishes for negative times then so do $T (f)$ and $T^{- 1} (f)$ .
Proof.
In a first time, we show that T maps $H_{γ}^{\infty} (R^{d})$ into $H^{\infty} (R^{d})$ .

Let $f \in H_{γ}^{\infty} (R^{d})$ , we want to show that $T (f) \in H^{\infty} (R^{d})$ or equivalently that $\begin{matrix} \forall m \in N, \exists C_{m} > 0, \int_{R^{d}} {| \hat{T (f)} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ < C_{m} . \end{matrix}$

Then using the definition of X and Cauchy–Schwarz inequality, we have: $\begin{array}{l} {| \hat{T (f)} (ξ) |}^{2} & = | c | \sqrt{γ^{2} + {(τ + v_{g}^{'} \cdot η)}^{2}} {| \hat{e^{- γ t} f} (ξ) |}^{2}, \\ ⩽ \{\begin{matrix} | c | (γ + | (1, v_{g}^{'}) |) | \hat{e^{- γ t} f} (ξ) |^{2} & if | ξ | ⩽ 1, \\ | c | (γ + (1 + | ξ |^{2}) | (1, v_{g}^{'}) |) | \hat{e^{- γ t} f} (ξ) |^{2} & if | ξ | > 1 . \end{matrix} \end{array}$ Hence, denoting $v_{1}^{'} : = (1, v_{g}^{'})$ : $\begin{array}{l} \int_{R^{d}} {| \hat{T (f)} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ = & \int_{| ξ | ⩽ 1} {| \hat{T (f)} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ \\ + \int_{| ξ | > 1} {| \hat{T (f)} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ \\ ⩽ & | c | (γ + | v_{1}^{'} |) \int_{| ξ | ⩽ 1} {| \hat{e^{- γ t} f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ \\ + | c | γ \int_{| ξ | > 1} {| \hat{e^{- γ t} f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ \\ + | c | | v_{1}^{'} | \int_{| ξ | > 1} {| \hat{e^{- γ t} f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m + 1} d ξ, \\ ⩽ & | c | (γ + | v_{1}^{'} |) \int_{R^{d}} {| \hat{e^{- γ t} f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ \\ + | c | γ \int_{R^{d}} {| \hat{e^{- γ t} f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ \\ + | c | | v_{1}^{'} | \int_{R^{d}} {| \hat{e^{- γ t} f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m + 1} d ξ . \end{array}$ Because $f \in H_{γ}^{\infty} (R^{d})$ , we have that for all $m \in N$ $\int_{R^{d}} | \hat{T (f)} (ξ) |^{2} {(1 + | ξ |^{2})}^{m} d ξ ⩽ C_{m}$ , so that $T (f) \in H^{\infty} (R^{d})$ . We have shown that $‖ T (f) ‖_{H^{m} (R^{d})} ⩽ C ‖ e^{- γ t} f ‖_{H^{m + 1} (R^{d})}$ from which we deduce that if f vanishes for negative times then so do $T (f)$ .

Next, we show that the expression given for $T^{- 1}$ is really the inverse.

We rewrite $T (f) = F^{- 1} (\sqrt{X} \hat{e^{- γ t} f})$ and $T^{- 1} (f) = e^{γ t} F^{- 1} (\frac{1}{\sqrt{X}} \hat{f})$ , where $F^{- 1}$ stands for the inverse of the Fourier transform on $L^{2} (R^{d})$ .By easy computations, we can see that $T \circ T^{- 1} = T^{- 1} \circ T = I$ .

Finally, we show that $T^{- 1}$ maps $H^{\infty} (R^{d})$ into $H_{γ}^{\infty} (R^{d})$ . Let $f \in H^{\infty} (R^{d})$ . We have: $\begin{matrix} e^{- γ t} T^{- 1} (f) = F^{- 1} (\frac{1}{\sqrt{X}} \hat{f}) . \end{matrix}$ And so: $\begin{array}{l} {| \hat{e^{- γ t} T^{- 1} (f)} (ξ) |}^{2} = \frac{1}{| X (ξ) |} {| \hat{f} (ξ) |}^{2} . \end{array}$ But, $| X (ξ) | = | c | \sqrt{γ^{2} + {(τ + v_{g}^{'} \cdot η)}^{2}} ⩾ | c | γ$ and $c, γ \neq 0$ by definition of γ and Lemma 3.1. Hence $\begin{matrix} {| \hat{e^{- γ t} T^{- 1} (f)} (ξ) |}^{2} ⩽ \frac{1}{| c | γ} {| \hat{f} (ξ) |}^{2}, \end{matrix}$ and consequently $\begin{matrix} \int_{R^{d}} {| \hat{e^{- γ t} T^{- 1} (f)} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ ⩽ \frac{1}{| c | γ} \int_{R^{d}} {| \hat{f} (ξ) |}^{2} {(1 + | ξ |^{2})}^{m} d ξ, \end{matrix}$ that is to say that $‖ e^{- γ t} T^{- 1} (f) ‖_{H^{m} (R^{d})} ⩽ C ‖ f ‖_{H^{m} (R^{d})}$ , since $f \in H^{\infty} (R^{d})$ , we deduce that $T^{- 1} (f) \in H_{γ}^{\infty} (R^{d})$ and that it vanishes for negative times if f does. □

Then with Proposition 3.1 in hand we can easily determine the unknown trace $α_{0}$ from (50) by applying $T^{- 1}$ on each side to obtain $\begin{array}{l} α_{0} = \frac{1}{i λ} T^{- 1} G_{0} . \end{array}$ In order to do so we shall justify that $G_{0}$ defined in (51) is in $H^{\infty} (ω_{T})$ so that we can effectively apply $T^{- 1}$ . We recall that $\begin{array}{l} G_{0} : = & b \cdot e^{- γ t} g - b \cdot \sum_{k \in O} e^{- γ t} Π^{k} u_{{1, k}_{| x_{d} = 0}} \\ + \frac{c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} b \cdot B Υ^{g} A_{d} Π^{g} \hat{e^{- γ t} f^{g}} (τ, η, χ^{'}) d χ^{'} d τ d η, \end{array}$ the first term is clearly in $H^{\infty} (ω_{T})$ because $g \in H^{\infty} (ω_{T})$ . Similarly if we choose $f_{k} \in H^{\infty} (Ω_{T})$ then as a solution to a transport equation $Π^{k} u_{1, k} \in H^{\infty} (Ω_{T}) \subset H_{γ}^{\infty} (Ω_{T})$ and thus the second term is in $H_{γ}^{\infty} (ω_{T})$ . The last term reads under the form $\begin{array}{l} F^{- 1} \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} \hat{F} d χ^{'} \end{array}$ where $F \in H^{\infty} (ω_{T} \times R_{+})$ . And thus in order to show that this term is in $H^{\infty} (Ω_{T})$ we shall consider the integrals for a multi-index $δ \in N^{d}$ . $\begin{array}{l} \int_{R^{d}} ξ^{δ} {| \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} \hat{F} (ξ, χ^{'}) d χ^{'} |}^{2} d ξ ⩽ & \frac{C}{γ} \int_{R^{d}} ξ^{δ} sup_{χ^{'} \in R_{+}} {| \hat{F} (ξ, χ^{'}) |}^{2} d ξ, \end{array}$ which are finite for all δ because $F \in H^{\infty} (ω_{T} \times R_{+})$ .

To conclude, we have determined $u_{0}^{g}$ as $\begin{matrix} (52) & u_{0}^{g} (t, x^{'}, χ) = \frac{e^{γ t}}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{\sqrt{X} χ} \hat{e^{- γ t} α_{0}} (τ, η) d τ d η e, \end{matrix}$ where $α_{0} = \frac{1}{i λ} T^{- 1} (G_{0})$ vanishes for negative times and so do $u_{0}^{g}$ . In (52) we recall that X is defined in (41) and that e stands for a generator of $ker B \cap E_{g}^{s} (\underline{ζ})$ .

For later purpose let us remark that we can read on (52) that $u_{0}^{g}$ and its derivatives in the fast variable χ decay exponentially fast with respect to this variable. More precisely we give the following proposition establishing the regularity of $u_{0}^{g}$ .
Proposition 3.2.
With $u_{0}^{g}$ defined in ( 52 ) we have $u_{0}^{g} \in P_{g}$ and $u_{0}^{g}$ vanishes for negative times.
Proof.
Let us remark that in view of its expression $u_{0}^{g}$ has exponential decay with respect to χ.

In the following we consider $α \in N^{d - 1}$ a multi-index associated to $x^{'}$ and we use classical notations for multi-index.

From Leibniz formula to show that $u_{0}^{g} \in H_{t, x^{'}, χ}^{\infty} (Ω_{T} \times R_{+})$ it is sufficient to show that the $e^{γ t} \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{\sqrt{X} χ} τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} (τ, η) d τ d η$ are in $L_{t, x^{'}, χ}^{2} (Ω_{T} \times R_{+})$ for all $k, m \in N$ and $α \in N^{d - 1}$ .

Because we are in finite time the factor $e^{γ t}$ can be replaced by $e^{γ T}$ and we can integrate on the whole line for t.

Thus, using Plancherel theorem: $\begin{array}{l} {‖ F_{(t, x^{'}) \leftrightarrow (τ, η)}^{- 1} (e^{\sqrt{X} χ} τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}}) ‖}_{L_{t, x^{'}, χ}^{2} (R^{d} \times R_{+})} = {‖ e^{\sqrt{X} χ} τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} ‖}_{L_{τ, η, χ}^{2} (R^{d} \times R_{+})} . \end{array}$

Integrating with respect to χ first then gives: $\begin{array}{l} {‖ e^{\sqrt{X} χ} τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} ‖}_{L_{τ, η, χ}^{2} (R^{d} \times R_{+})}^{2} & = \int_{R^{d}} \int_{R_{+}} {| e^{\sqrt{X} χ} τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} |}^{2} d χ d τ d η, \\ = \int_{R^{d}} \int_{R_{+}} {| e^{\sqrt{X} χ} |}^{2} d χ {| τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} |}^{2} d τ d η, \\ = - \int_{R^{d}} \frac{1}{2 ℜ (\sqrt{X})} \times {| τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} |}^{2} d τ d η . \end{array}$ But if $ℜ (X) > 0$ we write $- ℜ (\sqrt{X}) = \sqrt{\frac{1}{2} (ℜ (X) + | X |)} ⩾ \frac{1}{\sqrt{2}} \sqrt{| c | γ}$ and if $ℜ (X) < 0$ we write $ℜ (\sqrt{X}) = - \frac{1}{\sqrt{2}} \frac{| ℑ (X) |}{\sqrt{| X | - ℜ (X)}}$ so that in both cases $- \frac{1}{2 ℜ (\sqrt{X})} ⩽ \frac{1}{\sqrt{2 | c | γ}}$ . Consequently $\begin{array}{l} {‖ e^{\sqrt{X} χ} τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} ‖}_{L_{τ, η, χ}^{2} (R^{d} \times R_{+})}^{2} ⩽ \frac{1}{\sqrt{2 | c | γ}} {‖ τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} ‖}_{L_{τ, η}^{2} (R^{d})}^{2} \end{array}$

We decompose like in Proposition 3.1 $\begin{array}{l} {‖ τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} ‖}_{L_{τ, η}^{2} (R^{d})}^{2} = & \int_{R^{d}} {| τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} |}^{2} d τ d η \\ ⩽ & | c |^{m} {(γ + | v_{1}^{'} |)}^{m} \int_{| ξ | ⩽ 1} {| \hat{e^{- γ t} α_{0}} |}^{2} d τ d η \\ + \int_{| ξ | > 1} {| τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} |}^{2} d τ d η \end{array}$ where we recall that $v_{1}^{'} = (1, v_{g}^{'})$ .

The first right hand side term is finite because $e^{- γ t} α_{0} \in H^{\infty} (ω_{T})$ and the second term can also be bounded using Holder inequality for ${\sqrt{X}}^{m}$ by $\begin{matrix} \int_{| ξ | > 1} {| τ^{k} η^{α} {\sqrt{X}}^{m} \hat{e^{- γ t} α_{0}} |}^{2} d τ d η ⩽ \int_{R^{d}} {(1 + | ξ |^{2})}^{k + | α | + m} {| \hat{e^{- γ t} α_{0}} |}^{2} d τ d η, \end{matrix}$ which is finite because $e^{- γ t} α_{0} \in H^{\infty} (ω_{T})$ . This proves that $u_{0}^{g} \in H^{\infty} (Ω_{T} \times R_{+})$ and so $u_{0}^{g} \in P_{g}$ because of the exponential decay with respect to χ. □

So we sum up the construction of the Paragraphs 3.1.1, 3.1.2 and 3.1.3 in the following proposition: Proposition 3.3.
Under Assumptions 2.1 , 2.2 , 2.3 , 2.4 and 2.5 , for all $k \in I \cup O$ , there exist $u_{0, k} \in H^{\infty} (Ω_{T})$ , $u_{0}^{g} \in P_{g}$ and $U_{e v, 0} \in P_{e v}$ satisfying the cascades of equations ( 22 ), ( 23 ) and ( 24 ) at order zero. In fact we have that for all $k \in I \cup O$ , $u_{0, k} = U_{e v, 0} = 0$ . In particular the leading term in the expansion is $u_{0}^{g}$ .

3.2. Construction of terms of order one

In this paragraph, we determine the amplitudes of order one. We follow the same steps than for the construction of the leading order term. But the determination of the amplitudes and more precisely the traces is a little more technical than for the leading amplitudes because the amplitudes are not polarized anymore.

Using (12), we apply the partial inverse $Υ^{g}$ (resp. $Υ^{k}$ ) to the equation for the glancing mode of $I_{0}$ (resp. to the equation for the hyperbolic modes of $I_{0}$ ) to derive: $\begin{array}{l} (53) & (I - Π^{g}) u_{2}^{g} = - i Υ^{g} f^{g} + i Υ^{g} A_{d} \partial_{χ} u_{1}^{g} + i Υ^{g} L (\partial) u_{0}^{g}, \\ (54) & (I - Π^{k}) u_{2, k} = i Υ^{k} L (\partial) u_{0, k}, k \in I \cup O . \end{array}$ We remark that (54) gives the unpolarized part of $u_{2, k}$ in terms of the constructed term $u_{0, k}$ . Similarly in the right hand side of (53) the last term is now a known function.

Now, we apply the projector $Π^{g}$ (resp. $Π^{k}$ ) to the glancing (resp. hyperbolic) equation of $I_{\frac{1}{2}}$ to obtain (recall that $u_{1, k}$ is polarized): $\begin{array}{l} Π^{g} L^{'} (\partial) u_{1}^{g} + Π^{g} A_{d} \partial_{χ} u_{2}^{g} = 0, \\ (55) & Π^{k} L (\partial) Π^{k} u_{1, k} = Π^{k} f_{k}, k \in I \cup O . \end{array}$ We have a look at the first equation. Decomposing $u_{1}^{g}$ and $u_{2}^{g}$ in terms of polarized and unpolarized parts, we have: $\begin{array}{l} (56) & Π^{g} L^{'} (\partial) Π^{g} u_{1}^{g} + Π^{g} A_{d} Π^{g} \partial_{χ} u_{2}^{g} + Π^{g} A_{d} \partial_{χ} (I - Π^{g}) u_{2}^{g} = - Π^{g} L^{'} (\partial) (I - Π^{g}) u_{1}^{g}, \end{array}$ where $(I - Π^{g}) u_{1}^{g}$ is known from equation (28) and thus the right hand side of (56) is known.

Using equations (53) and (56) and decomposing again $u_{1}^{g}$ in terms of polarized and unpolarized parts, we thus obtain: $\begin{array}{l} Π^{g} L^{'} (\partial) Π^{g} u_{1}^{g} + Π^{g} A_{d} Π^{g} \partial_{χ} u_{2}^{g} + i Π^{g} A_{d} Υ^{g} A_{d} \partial_{χ}^{2} Π^{g} u_{1}^{g} \\ = - Π^{g} L^{'} (\partial) (I - Π^{g}) u_{1}^{g} + i Π^{g} A_{d} Υ^{g} \partial_{χ} f^{g} \\ - i Π^{g} A_{d} Υ^{g} L^{'} (\partial) \partial_{χ} u_{0}^{g} - i Π^{g} A_{d} Υ^{g} A_{d} \partial_{χ}^{2} (I - Π^{g}) u_{1}^{g}, \\ (57) & : = F_{1} . \end{array}$

Using Lemmas 3.1 and 3.2, equations (57) and (55) can be rewritten as: $\begin{array}{l} (58) & - \partial_{χ}^{2} Π^{g} u_{1}^{g} + i c (\partial_{t} + v_{g}^{'} \cdot \nabla_{x^{'}}) Π^{g} u_{1}^{g} = i c F_{1}, \\ (59) & (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{1, k} = Π^{k} f_{k}, k \in I \cup O, \end{array}$ where we recall that c has been introduced in Lemma 3.1 and where $v_{k}$ , $v_{g}^{'}$ stand for the group velocities.

Then we consider the boundary condition: equation $B_{\frac{1}{2}}$ reads $\begin{matrix} (60) & B (\sum_{k \in I \cup O} u_{{1, k}_{| x_{d} = 0}} + u_{1_{| χ = 0}}^{g} + U_{{e v, 1}_{| X_{d} = 0}}) = g . \end{matrix}$

In the following, we need to determine the $Π^{k} u_{{1, k}_{| x_{d} = 0}}$ for $k \in I$ and $P_{e v} U_{{e v, 1}_{| X_{d} = 0}}$ so we will do this from this boundary condition. We recall that the $u_{1, k}$ for $k \in I \cup O$ are polarized and that we know $Π^{k} u_{1, k}, k \in O$ from Paragraph 3.1.1. And so: $\begin{matrix} (61) & B (\sum_{k \in I} Π^{k} u_{{1, k}_{| x_{d} = 0}} + u_{1_{| χ = 0}}^{g} + U_{{e v, 1}_{| X_{d} = 0}}) = g - B \sum_{k \in O} Π^{k} u_{{1, k}_{| x_{d} = 0}} . \end{matrix}$

Equation (28) gives the unpolarized part of $u_{1}^{g}$ . Remember that we have $Im Π^{g} = ker L (\underline{ζ}, \underline{ξ_{k}}) = E_{g}^{s} (\underline{ζ})$ , so we can write, using Assumption 2.3 and Remark 2.2: $\begin{matrix} (62) & Π^{g} u_{1_{| χ = 0}}^{g} (t, x^{'}) : = α_{1} (t, x^{'}) e where α_{1} is a scalar function defined on ω_{T}, \end{matrix}$ and where e stands for a generator of $ker B \cap E_{g}^{s} (\underline{ζ})$ .

Hence, $B Π^{g} u_{1_{| χ = 0}}^{g} = α_{1} B e = 0$ and we can rewrite equation (61) as: $\begin{matrix} B (\sum_{k \in I} Π^{k} u_{{1, k}_{| x_{d} = 0}} + U_{{e v, 1}_{| X_{d} = 0}}) = g - B \sum_{k \in O} Π^{k} u_{{1, k}_{|} x_{d} = 0} - B (I - Π^{g}) u_{1_{| χ = 0}}^{g} . \end{matrix}$ Finally using that $U_{e v, 1} = P_{e v} U_{e v, 1}$ , it turns out $\begin{array}{l} B (\sum_{k \in I} Π^{k} u_{{1, k}_{| x_{d} = 0}} + P_{e v} U_{{e v, 1}_{| X_{d} = 0}}) & = g - B \sum_{k \in O} Π^{k} u_{{1, k}_{| x_{d} = 0}} - B (I - Π^{g}) u_{1_{| χ = 0}}^{g}, \\ (63) & : = G_{1} . \end{array}$

Since $Π^{k} u_{{1, k}_{| x_{d} = 0}} \in E_{h}^{s} (\underline{ζ})$ and $P_{e v} U_{{e v, 1}_{| X_{d} = 0}} \in E_{e}^{s} (\underline{ζ})$ , from (14), we can invert B and so applying Φ, we deduce that: $\begin{matrix} \sum_{k \in I} Π^{k} u_{{1, k}_{| x_{d} = 0}} + P_{e v} U_{{e v, 1}_{| X_{d} = 0}} = Φ G_{1} . \end{matrix}$

Next, applying the projectors of Definition 2.5, we thus obtain: $\begin{array}{l} \forall k \in I, P^{k} Π^{k} u_{{1, k}_{| x_{d} = 0}} = P^{k} Φ G_{1}, \\ P^{e} P_{e v} U_{{e v, 1}_{| X_{d} = 0}} = P^{e} Φ G_{1} . \end{array}$

Because $Π^{k} u_{{1, k}_{| x_{d} = 0}} \in E_{h}^{s} (\underline{ζ})$ and $P_{e v} U_{{e v, 1}_{| X_{d} = 0}} \in E_{e}^{s} (\underline{ζ})$ , we see that $P^{k} Π^{k} u_{{1, k}_{| x_{d} = 0}} = Π^{k} u_{{1, k}_{| x_{d} = 0}}$ and $P^{e} P_{e v} U_{{e v, 1}_{| X_{d} = 0}} = P_{e v} U_{{e v, 1}_{| X_{d} = 0}}$ . Therefore, $\begin{array}{l} (64) & \forall k \in I, Π^{k} u_{{1, k}_{| x_{d} = 0}} = P^{k} Φ G_{1}, \\ (65) & P_{e v} U_{{e v, 1}_{| X_{d} = 0}} = P^{e} Φ G_{1} . \end{array}$ In the following paragraphs we describe the construction of the higher order hyperbolic and elliptic terms. We then use this construction for the determination of the glancing amplitude of order one in Paragraph 3.2.3.

3.2.1. Determination of the hyperbolic amplitudes of order one

The outgoing amplitudes of order one namely the $u_{1, k}$ for $k \in O$ have been determined in Paragraph 3.1.1. However as the leading order, the determination of the first order glancing amplitude will require the knowledge of one more order on the outgoing modes, more precisely we will need to know $u_{2, k}$ for $k \in O$ .

Reiterating the same analysis as the one performed in Paragraph 3.1.1, these terms can be determined independently of the other by solving an outgoing transport equation.

Indeed, applying the projector $Π^{k}$ on the hyperbolic equation of $I_{1}$ , gives: $\begin{matrix} Π^{k} L (\partial) u_{2, k} = 0 . \end{matrix}$ Equation (54) gives the unpolarized part of $u_{2, k}$ . So decomposing $u_{2, k}$ in terms of polarized and unpolarized parts and using Lemma 3.2, we deduce that: $\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{2, k} = - Π^{k} L (\partial) (I - Π^{k}) u_{2, k} : = f_{k, 2} . \end{matrix}$ Hence, we determine $Π^{k} u_{2, k}$ and so $u_{2, k}$ by solving the transport equation: $\begin{matrix} \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{2, k} = f_{k, 2}, \\ Π^{k} u_{{2, k}_{| t ⩽ 0}} = 0, \end{matrix} \end{matrix}$ by integration along the characteristics. The obtained solution is explicit in terms of $f_{k, 2}$ that is to say explicit in terms of $u_{0, k}$ from (54). Moreover the solution $u_{2, k}$ lies in the same functional space than the source $f_{2, k}$ .

For the incoming phases, we have to consider the boundary equation (64). For these phases, as in Paragraph 3.1.1 for the outgoing phases of order one, we have $(\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{1, k} = Π^{k} f_{k}$ . Hence, each of the incoming amplitude solves the transport equation: $\begin{matrix} \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{1, k} = Π^{k} f_{k}, \\ Π^{k} u_{{1, k}_{| x_{d} = 0}} = P^{k} Φ G_{1}, \\ Π^{k} u_{{1, k}_{| t ⩽ 0}} = 0, \end{matrix} \end{matrix}$ equation that can be explicitly solved in terms of $f_{k}$ , $k \in I$ and $G_{1}$ (which depends on g and on the $f_{k}$ for $k \in O$ , see (63)) so that $u_{1, k}$ depends on all the sources of the problem. We have that $u_{1, k}$ lies in $H^{\infty} (Ω_{T})$ vanishes for negative times if the sources do.

3.2.2. Determination of the evanescent amplitudes of order one

We recall that because of Paragraph 3.1.2 we have $U_{e v, 1} = P_{e v} U_{e v, 1}$ . So that in order to determine the whole elliptic amplitude it is sufficient, from the definition of $P_{e v}$ (see Lemma 3.3), to determine $U_{{e v, 1}_{| X_{d} = 0}}$ .

We use the boundary equation (65) to recover that $\begin{matrix} U_{e v, 1} (t, x^{'}, X_{d}) = (P_{e v} U_{e v, 1}) (t, x^{'}, X_{d}) = e^{X_{d} A (\underline{ζ})} P^{e} Φ G_{1} (t, x^{'}), \end{matrix}$ equation that determines $U_{e v, 1}$ in terms of $G_{1}$ . Moreover we readily see that $U_{e v, 1} \in P_{e v}$ the set of evanescent profiles.

Note that as mentioned in the introduction of Section 3 the fact that the elliptic profile is independent of the normal variable $x_{d}$ permits us to avoid the step of the lifting of some double trace as simple one as it is done in [4] or [16].

As for the leading order glancing term, the determination of the term of order one requires the knowledge of the elliptic amplitude of the same order plus one. More precisely we will need to know the form of $U_{e v, 2}$ to determine the glancing leading term of order one.

From the cascade of equations $I_{0}$ for elliptic modes we have $\begin{matrix} L (\partial_{X_{d}}) U_{e v, 2} = - \underset{= 0}{\underset{︸}{L^{'} (\partial) U_{e v, 0}}} + f_{e v}, \end{matrix}$ where we recall that $L (\partial_{X_{d}}) : = A_{d} (\partial_{X_{d}} - A (\underline{ζ}))$ and $L^{'} (\partial) : = \partial_{t} + \sum_{j = 1}^{d - 1} A_{j} \partial_{x_{j}}$ . Thanks to Lemma 3.3, we deduce that: $\begin{matrix} (66) & U_{e v, 2} = P_{e v} U_{e v, 2} + Q_{e v} f_{e v}, \end{matrix}$ equation that does not fully determine $U_{e v, 2}$ because the polarized part $P_{e v} U_{e v, 2}$ has not been determined yet (it is however easy to determine this term by determining the trace $U_{{e v, 2}_{| X_{d} = 0}}$ like it has been done for $U_{{e v, 1}_{| X_{d} = 0}}$ ) but which will be sufficient to determine the glancing amplitude of order one.

3.2.3. Determination of the glancing term of order one

For $γ > 0$ , let $v_{1}^{g} : = e^{- γ t} Π^{g} u_{1}^{g}$ . As for $u_{0}^{g}$ , equation (58) becomes: $\begin{matrix} (67) & - \partial_{χ}^{2} v_{1}^{g} + i c (γ + \partial_{t} + v_{g}^{'} \cdot \nabla_{x^{'}}) v_{1}^{g} = i c e^{- γ t} F_{1}, \end{matrix}$ where we recall that the right hand side $F_{1}$ depends on $u_{0}^{g}$ , the unpolarized part of $u_{1}^{g}$ and $f^{g}$ and eventually on their derivatives, it is defined in (57). In particular this term has exponential decay with respect to χ.

We now perform a Fourier transform with respect to t and $x^{'}$ , we thus have the same ordinary differential equation as the one determining the leading order glancing amplitude: $\begin{matrix} (68) & - \partial_{χ}^{2} \hat{v_{1}^{g}} + X \hat{v_{1}^{g}} = i c \hat{e^{- γ t} F_{1}}, \end{matrix}$ where we recall that $X = X (σ, η) = i c (γ + i τ + i \sum_{j = 1}^{d - 1} v_{g, j}^{'} η_{j})$ .

So, because $\hat{F_{1}}$ has exponential decay with respect to χ we have a unique solution with exponential decay with respect to χ (see [21], equation 8.40): $\begin{array}{l} \hat{v_{1}^{g}} (τ, η, χ) = & i c \int_{0}^{χ} e^{(χ - χ^{'}) \sqrt{X}} \int_{χ^{'}}^{\infty} e^{- (χ^{'} - χ^{″}) \sqrt{X}} \hat{e^{- γ t} F_{1}} (τ, η, χ^{″}) d χ^{″} d χ^{'} + e^{χ \sqrt{X}} {\hat{v_{1}^{g}}}_{| χ = 0} (τ, η), \\ (69) & : = & F_{2} (τ, η, χ) + e^{χ \sqrt{X}} {\hat{v_{1}^{g}}}_{| χ = 0} (τ, η) . \end{array}$ Reversing the Fourier transform, we deduce that: $\begin{array}{l} v_{1}^{g} (t, x^{'}, χ) = & \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} F_{2} (τ, η, χ) d τ d η \\ + \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{χ \sqrt{X}} {\hat{v_{1}^{g}}}_{| χ = 0} (τ, η) d τ d η \\ (70) & : = & F_{3} (t, x^{'}, χ) + \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{χ \sqrt{X}} {\hat{v_{1}^{g}}}_{| χ = 0} (τ, η) d τ d η \end{array}$

So, to determine $v_{1}^{g}$ (and so $Π^{g} u_{1}^{g}$ ), it is sufficient to determine ${\hat{v_{1}^{g}}}_{| χ = 0}$ because $F_{3}$ is known from (57) and (69).

Recall that from (62) we have: $\begin{matrix} (71) & v_{1_{| χ = 0}}^{g} = e^{- γ t} α_{1} e, \end{matrix}$ where $α_{1}$ is a scalar function defined on $ω_{T}$ and e a generator of $ker B \cap E_{g}^{s} (\underline{ζ})$ .

Thus, it remains to determine $α_{1}$ . For this, we look at the boundary condition $B_{1}$ that is: $\begin{matrix} B (\sum_{k \in I \cup O} u_{{2, k}_{| x_{d} = 0}} + u_{2_{| χ = 0}}^{g} + U_{{e v, 2}_{| X_{d} = 0}}) = 0 . \end{matrix}$

The $u_{2, k}$ , for $k \in O$ are known because the polarized part of them are determined from Paragraph 3.2.1 and the unpolarized part from (54). From (54) we also know $(I - Π^{k}) u_{2, k}, k \in I$

Hence, the precedent equation can be rewritten as $\begin{matrix} B (\sum_{k \in I} Π^{k} u_{{2, k}_{| x_{d} = 0}} + u_{2_{| χ = 0}}^{g} + U_{{e v, 2}_{| X_{d} = 0}}) = - B \sum_{k \in O} u_{{2, k}_{| x_{d} = 0}} - B \sum_{k \in I} (I - Π^{k}) u_{{2, k}_{| x_{d} = 0}} . \end{matrix}$

Furthermore, from (66), we see that: $\begin{matrix} U_{{e v, 2}_{| X_{d} = 0}} = Π_{e}^{s} U_{{e v, 2}_{| X_{d} = 0}} - \int_{0}^{+ \infty} e^{- s A (\underline{ζ})} Π_{e}^{u} A_{d}^{- 1} f_{e v} (t, x^{'}, s) d s . \end{matrix}$ The first term in the right hand side is in $E_{e}^{s} (\underline{ζ})$ so $b \cdot B Π_{e}^{s} U_{{e v, 2}_{| X_{d} = 0}} = 0$ and the second term is known.

So, decomposing $u_{2}^{g}$ in terms of polarized and unpolarized parts and testing against the vector b defined in (13), we obtain: $\begin{array}{l} b \cdot B (I - Π^{g}) u_{2_{| χ = 0}}^{g} = & - b \cdot B \sum_{k \in O} u_{{2, k}_{| x_{d} = 0}} - b \cdot B \sum_{k \in I} (I - Π^{k}) u_{{2, k}_{| x_{d} = 0}} \\ + b \cdot B \int_{0}^{+ \infty} e^{- s A (\underline{ζ})} Π_{e}^{u} A_{d}^{- 1} f_{e v} (t, x^{'}, s) d s, \\ (72) & : = & \tilde{G_{1}}; \end{array}$ clearly as the sum of solutions to transport equations and the free evolution of the regular source $f_{e v}$ then the new source $\tilde{G_{1}}$ lies in $H_{t, x^{'}}^{\infty} (ω_{T})$ .

Using (53), we can rewrite the left hand side of (72) as: $\begin{matrix} (73) & i b \cdot B Υ^{g} A_{d} {(\partial_{χ} u_{1}^{g})}_{| χ = 0} = \tilde{G_{1}} + i b \cdot B Υ^{g} f_{| χ = 0}^{g} - i b \cdot B Υ^{g} L^{'} (\partial) u_{0_{| χ = 0}}^{g}, \end{matrix}$ where we recall that $Υ^{g}$ stands for the partial inverse defined in (2.4) and where $L^{'} (\partial)$ is the tangential differentiation operator. We stress that the right hand side is known from (52). We multiply by $e^{- γ t}$ , to obtain: $\begin{array}{l} i b \cdot B Υ^{g} A_{d} {(\partial_{χ} v_{1}^{g})}_{| χ = 0} = & e^{- γ t} \tilde{G_{1}} + i e^{- γ t} b \cdot B Υ^{g} f_{| χ = 0}^{g} - i e^{- γ t} b \cdot B Υ^{g} L^{'} (\partial) u_{0_{| χ = 0}}^{g} \\ (74) & - i e^{- γ t} b \cdot B Υ^{g} A_{d} {(\partial_{χ} (1 - Π^{g}) u_{1}^{g})}_{| χ = 0} . \end{array}$

Using (70) and (71), we see that: $\begin{array}{l} {(\partial_{χ} v_{1}^{g})}_{| χ = 0} & = (\partial_{χ} F_{3}) (t, x^{'}, 0) + \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \sqrt{X} \hat{e^{- γ t} α_{1}} (τ, η) d τ d η e, \\ = \frac{i c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} \hat{e^{- γ t} F_{1}} (τ, η, χ^{'}) d χ^{'} d τ d η + T (α_{1}) e, \end{array}$ where T is the Fourier multiplier encounter in Paragraph 3.1.3. Combining (74) and the above equation, we deduce that: $\begin{array}{l} i λ T (α_{1}) = & e^{- γ t} \tilde{G_{1}} + i e^{- γ t} b \cdot B Υ^{g} f_{| χ = 0}^{g} - i e^{- γ t} b \cdot B Υ^{g} L^{'} (\partial) u_{0_{|} χ = 0}^{g} \\ - i e^{- γ t} b \cdot B Υ^{g} A_{d} {(\partial_{χ} (1 - Π^{g}) u_{1}^{g})}_{| χ = 0} \\ + \frac{c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{\infty} e^{χ^{'} \sqrt{X}} b \cdot B Υ^{g} A_{d} \hat{e^{- γ t} F_{1}} (τ, η, χ^{'}) d χ^{'} d τ d η, \\ : = & G_{1}, \end{array}$ where we recall that $λ : = b \cdot B Υ^{g} A_{d} e \neq 0$ .

The first terms of $G_{1}$ read $e^{- γ t} f$ where $f \in H_{t, x^{'}}^{\infty} (ω_{T})$ vanishes for negative times so that they are in $H_{t, x^{'}}^{\infty} (ω_{T})$ and hence in $H_{γ}^{\infty} (ω_{T})$ . The integral term in $G_{1}$ is shown to be in $H_{γ}^{\infty} (ω_{T})$ by the same arguments as for the integral term in the source term for the leading order glancing amplitude (see (51) and Proposition 3.1). Reversing the operator T, we find $α_{1}$ as $\begin{matrix} α_{1} (t, x^{'}) = \frac{1}{i λ} (T^{- 1} G_{1}) (t, x^{'}) . \end{matrix}$ Arguing like for the first boundary amplitude $α_{0}$ we can show that such $α_{1} \in H^{\infty} (ω_{T})$ vanishes for negative times. To conclude, we have determined $Π^{g} u_{1}^{g} \in P_{g}$ as $\begin{array}{l} Π^{g} u_{1}^{g} (t, x^{'}, χ) = & e^{γ t} F_{3} (t, x^{'}, χ) + e^{γ t} \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{\sqrt{X} χ} \hat{e^{- γ t} α_{1}} (τ, η) d τ d η e, \end{array}$ where $F_{3}$ is given by (57), (69) and (70). Reiterating the same kind of arguments as in the proof of Proposition 3.1 we have fully determined $u_{1}^{g} \in P_{g}$ such that $u_{1_{| t ⩽ 0}}^{g} = 0$ .

In this paragraph, we thus have shown:

Proposition 3.4.
Under Assumptions 2.1 , 2.2 , 2.3 , 2.4 and 2.5 , for all $k \in I \cup O$ , there exist $u_{1, k} \in H^{\infty} (Ω_{T})$ , $u_{1}^{g} \in P_{g}$ and $U_{e v, 1} \in P_{e v}$ satisfying the cascades of equations ( 22 ), ( 23 ) and ( 24 ) at order one.

3.3. Higher order terms

In this paragraph, we briefly describe how to construct terms of order n for $n ⩾ 2$ . The construction is very close to the construction of the terms of order one.

We suppose that we know the terms of order $n - 1, n - 2, \dots$ So, applying the partial inverses $Υ^{\cdot}$ on the equation for the glancing mode of $I_{\frac{n - 2}{2}}$ and the equation for the hyperbolic mode of $I_{\frac{n - 2}{2}}$ , we know the unpolarized parts of the glancing and hyperbolic amplitudes of order n as $\begin{array}{l} (I - Π^{g}) u_{n}^{g} = - i δ_{n - 2, 0} Υ^{g} f^{g} + i Υ^{g} A_{d} \partial_{χ} u_{n - 1}^{g} + i Υ^{g} L^{'} (\partial) u_{n - 2}^{g}, \\ (I - Π^{k}) u_{n, k} = - i δ_{n - 2, 1} Υ^{k} f_{k} + i Υ^{k} L (\partial) u_{n - 2, k}, \end{array}$ $δ_{\cdot, \cdot}$ being the Kronecker symbol.

Thus, we know the unpolarized parts of the glancing and the hyperbolic amplitudes of order n and it remains to determine the polarized parts. In order to do so, we will need to know the unpolarized parts of the terms of order $n + 1$ which are obtain applying the partial inverses to $I_{\frac{n - 1}{2}}$ : $\begin{array}{l} (75) & (I - Π^{g}) u_{n + 1}^{g} = i Υ^{g} A_{d} \partial_{χ} u_{n}^{g} + i Υ^{g} L^{'} (\partial) u_{n - 1}^{g}, \\ (76) & (I - Π^{k}) u_{n + 1, k} = - i δ_{n - 1, 1} Υ^{k} f_{k} + i Υ^{k} L (\partial) u_{n - 1, k} . \end{array}$

Then, applying the projectors $Π^{k}$ and $Π^{g}$ on $I_{\frac{n}{2}}$ and decomposing the amplitudes in terms of their polarized and unpolarized parts, we deduce that: $\begin{array}{l} Π^{g} A_{d} Π^{g} \partial_{χ} u_{n + 1}^{g} + i Π^{g} A_{d} Υ^{g} A_{d} Π^{g} \partial_{χ}^{2} u_{n}^{g} + Π^{g} L^{'} (\partial) Π^{g} u_{n}^{g} = F_{n, 1}, \\ Π^{k} L (\partial) Π^{k} u_{n, k} = f_{k, n}, \end{array}$ where $F_{n, 1} : = - Π^{g} L^{'} (\partial) (I - Π^{g}) u_{n}^{g} - i Π^{g} A_{d} Υ^{g} L^{'} (\partial) \partial_{χ} u_{n - 1}^{g} - i Π^{g} A_{d} Υ^{g} A_{d} \partial_{χ}^{2} (I - Π^{g}) u_{n}^{g}$ and $f_{k, n} : = - Π^{k} L (\partial) (I - Π^{k}) u_{n, k}$ are known.

So using Lemmas 3.1 and 3.2, we write these equations in a similar form to (58) and (59) that is: $\begin{array}{l} (77) & - \partial_{χ}^{2} Π^{g} u_{n}^{g} + i c (\partial_{t} + v_{g}^{'} \cdot \nabla_{x^{'}}) Π^{g} u_{n}^{g} = i c F_{n, 1}, \\ (78) & (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{n, k} = f_{k, n} . \end{array}$

Notice that from the equation of the evanescent mode of $I_{\frac{n - 2}{2}}$ , we have: $\begin{matrix} (79) & L (\partial_{X_{d}}) U_{e v, n} = δ_{n - 2, 0} f_{e v} - L^{'} (\partial) U_{e v, n - 2} : = F_{e v, n} . \end{matrix}$ So, using Lemma 3.3, we know the form of the evanescent profile: $\begin{matrix} U_{e v, n} = P_{e v} U_{e v, n} + Q_{e v} F_{e v, n} . \end{matrix}$

We now look at the boundary condition $B_{\frac{n}{2}}$ because we need to know $Π^{k} u_{{n, k}_{| x_{d} = 0}}$ for $k \in I$ and $P_{e v} U_{{e v, n}_{| X_{d} = 0}}$ to determine the incoming hyperbolic and the elliptic amplitude. We suppose that we know $Π^{k} u_{n, k}$ for $k \in O$ (it is not an issue because it is possible to determine the outgoing terms before the other amplitudes). We also recall that the unpolarized part of the incoming hyperbolic and $Q_{e v} F_{e v, n}$ are known. So the boundary condition gives: $\begin{array}{l} B (\sum_{k \in I} Π^{k} u_{{n, k}_{| x_{d} = 0}} + u_{n_{| χ = 0}}^{g} + P_{e v} U_{{e v, n}_{| X_{d} = 0}}) = & - B \sum_{k \in O} u_{{n, k}_{| x_{d} = 0}} - B \sum_{k \in I} (I - Π^{k}) u_{{n, k}_{| x_{d} = 0}} \\ - B Q_{e v} F_{{e v, n}_{| X_{d} = 0}} . \end{array}$

The unpolarized part of the glancing amplitude is known and from the fact that the polarized part can be write as (see equation (62) in Paragraph 3.2): $\begin{matrix} (80) & Π^{g} u_{n_{| χ = 0}}^{g} = α_{n} e where α_{n} is a scalar function defined on ω_{T}, \end{matrix}$ the above equation becomes: $\begin{matrix} B (\sum_{k \in I} Π^{k} u_{{n, k}_{| x_{d} = 0}} + P_{e v} U_{{e v, n}_{| X_{d} = 0}}) = G_{n}, \end{matrix}$ where $G_{n} : = - B \sum_{k \in O} u_{{n, k}_{| x_{d} = 0}} - B \sum_{k \in I} (I - Π^{k}) u_{{n, k}_{| x_{d} = 0}} - B Q_{e v} F_{{e v, n}_{| X_{d} = 0}} - B (I - Π^{g}) u_{n_{| χ = 0}}^{g}$ .

So, reversing B and using the boundary projectors (see Definition 2.5), we deduce the boundary conditions for the $Π^{k} u_{n, k}$ for $k \in I$ and $P_{e v} U_{e v, n}$ : $\begin{array}{l} \forall k \in I, Π^{k} u_{{n, k}_{| x_{d} = 0}} = P^{k} Φ G_{n}, \\ P_{e v} U_{{e v, n}_{| X_{d} = 0}} = P^{e} Φ G_{n}, \end{array}$ where we recall that $Φ : = B_{| {\tilde{E}}^{s} (\underline{ζ})}^{- 1}$ .

With this result in hand, we can construct the hyperbolic and elliptic amplitudes of order n:

The outgoing amplitudes of order n are determined by solving the transport equation without boundary condition $\begin{matrix} \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{n, k} = f_{k, n}, \\ Π^{k} u_{{n, k}_{| t ⩽ 0}} = 0 . \end{matrix} \end{matrix}$

Moreover, we will need to know the outgoing amplitudes of order $n + 1$ to determine the glancing one of order n. They are given by (76) and by solving the transport equation: $\begin{matrix} \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{n + 1, k} = f_{k, n + 1}, \\ Π^{k} u_{{n + 1, k}_{| t ⩽ 0}} = 0 . \end{matrix} \end{matrix}$

The incoming amplitudes of order n are determined by solving the transport equation with boundary condition: $\begin{matrix} \{\begin{matrix} (\partial_{t} + v_{k} \cdot \nabla_{x}) Π^{k} u_{n + 1, k} = f_{k, n + 1}, \\ Π^{k} u_{{n, k}_{| x_{d} = 0}} = P^{k} Φ G_{n}, \\ Π^{k} u_{{n, k}_{| t ⩽ 0}} = 0 . \end{matrix} \end{matrix}$

For the evanescent amplitude, we deduce from the boundary condition that $\begin{matrix} U_{e v, n} (t, x^{'}, X_{d}) = e^{X_{d} A (\underline{ζ})} P^{e} Φ G_{n} (t, x^{'}) + Q_{e v} F_{e v, n} (t, x^{'}, X_{d}) . \end{matrix}$ Furthermore, we will need to know the form of $U_{e v, n + 1}$ to determine the glancing amplitude. It solves $L (\partial_{X_{d}}) U_{e v, n + 1} = F_{e v, n + 1}$ , so (from Lemma 3.3), we have: $\begin{matrix} U_{e v, n + 1} = P_{e v} U_{e v, n + 1} + Q_{e v} F_{e v, n + 1}, \end{matrix}$ where the operators $P_{e v}$ and $Q_{e v}$ are defined in Lemma 3.3.

Now, we can construct the polarized part of the glancing amplitude. As for the term of order one, let $v_{n}^{g} : = e^{- γ t} Π^{g} u_{n}^{g}$ , where $γ > 0$ . Using this and performing a Fourier transform with respect to the tangential variables, (77) gives: $\begin{matrix} - \partial_{χ}^{2} \hat{v_{n}^{g}} + X \hat{v_{n}^{g}} = i c \hat{e^{- γ t} F_{n, 1}} . \end{matrix}$

So, following the same steps than for the glancing amplitude of order one (see Paragraph 3.2.3), we obtain: $\begin{matrix} (81) & v_{n}^{g} (t, x^{'}, χ) = F_{n, 3} (t, x^{'}, χ) + \frac{1}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} e^{χ \sqrt{X}} {\hat{v_{n}^{g}}}_{| χ = 0} (τ, η) d τ d η \end{matrix}$ where $\begin{matrix} F_{n, 3} : = \frac{i c}{{(2 π)}^{d}} \int_{R^{d}} e^{i (t, x^{'}) \cdot (τ, η)} \int_{0}^{χ} e^{(χ - χ^{'}) \sqrt{X}} \int_{χ^{'}}^{\infty} e^{- (χ^{'} - χ^{″}) \sqrt{X}} \hat{e^{- γ t} F_{n, 1}} (τ, η, χ^{″}) d χ^{″} d χ^{'} d τ d η . \end{matrix}$

So, to determine $v_{n}^{g}$ (and so $Π^{g} u_{n}^{g}$ ), it is sufficient to determine ${\hat{v_{n}^{g}}}_{| χ = 0}$ . Recall that from (80) we have: $\begin{matrix} (82) & v_{n_{| χ = 0}}^{g} = e^{- γ t} α_{n} e . \end{matrix}$ Thus, it remains to determine $α_{n}$ . For this, we look at the boundary condition $B_{\frac{n + 1}{2}}$ which can be rewritten as: $\begin{array}{l} B \sum_{k \in I} Π^{k} u_{{n + 1, k}_{| x_{d} = 0}} + B u_{n_{| χ = 0}}^{g} + B P_{e v} U_{{e v, n + 1}_{| X_{d} = 0}} \\ = - B \sum_{k \in O} u_{{n + 1, k}_{| x_{d} = 0}} - B \sum_{k \in I} (I - Π^{k}) u_{{n + 1, k}_{| x_{d} = 0}} - B Q_{e v} F_{{e v, n + 1}_{| X_{d} = 0}} \end{array}$ where the right hand side term is known.

Decomposing $u_{n + 1}^{g}$ in terms of polarized and unpolarized parts and testing against the vector b defined in (13), we obtain: $\begin{matrix} b \cdot B (I - Π^{g}) u_{{n + 1}_{| χ = 0}}^{g} = {\tilde{G}}_{n, 1} \end{matrix}$ where $\begin{matrix} {\tilde{G}}_{n, 1} : = - b \cdot B \sum_{k \in O} u_{{n + 1, k}_{| x_{d} = 0}} - b \cdot B \sum_{k \in I} (I - Π^{k}) u_{{n + 1, k}_{| x_{d} = 0}} - b \cdot B Q_{e v} F_{{e v, n + 1}_{| X_{d} = 0}} . \end{matrix}$

Now, using (75) and multiplying by $e^{- γ t}$ , it remains: $\begin{array}{l} i b \cdot B Υ^{g} A_{d} {(\partial_{χ} v_{n}^{g})}_{| χ = 0} = & e^{- γ t} {\tilde{G}}_{n, 1} - i e^{- γ t} b \cdot B Υ^{g} L^{'} (\partial) u_{{n - 1}_{| χ = 0}}^{g} \\ - i e^{- γ t} b \cdot B Υ^{g} A_{d} {(\partial_{χ} (I - Π^{g}) u_{n}^{g})}_{| χ = 0} . \end{array}$

Next, using (81), we deduce that: $\begin{matrix} i λ T (α_{n}) = G_{n, 1} \end{matrix}$ where $\begin{array}{l} G_{n, 1} : = & e^{- γ t} {\tilde{G}}_{n, 1} - i e^{- γ t} b \cdot B Υ^{g} A_{d} L^{'} (\partial) u_{{n - 1}_{| χ = 0}}^{g} - i b \cdot B Υ^{g} A_{d} {(\partial_{χ} F_{n, 2})}_{| χ = 0} \\ - i e^{- γ t} b \cdot B Υ^{g} A_{d} {(\partial_{χ} (I - Π^{g}) u_{n}^{g})}_{| χ = 0} . \end{array}$

Hence, we determine $α_{n} \in H^{\infty} (ω_{T})$ vanishing for negative times by the explicit formula $\begin{matrix} α_{n} (t, x^{'}) = \frac{1}{i λ} (T^{- 1} G_{n, 1}) (t, x^{'}), \end{matrix}$ where we recall that $λ : = b \cdot B Υ^{g} A_{d} e \neq 0$ . So, we have determined the polarized part of the glancing amplitude of order n.

To conclude this paragraph, we give the following proposition:

Proposition 3.5.
Under Assumptions 2.1 , 2.2 , 2.3 , 2.4 and 2.5 , for all $n \in N$ , for all $k \in I \cup O$ , there exist $u_{n, k} \in H^{\infty} (Ω_{T})$ , $u_{n}^{g} \in P_{g}$ and $U_{e v, n} \in P_{e v}$ satisfying the cascades of equations ( 22 ), ( 23 ) and ( 24 ) at the order n.

4. Discussion about the energy estimates

In this section, we consider the non-oscillatory initial boundary value problem $\begin{matrix} (83) & \{\begin{matrix} L (\partial) u = f & in Ω_{T}, \\ B u_{| x_{d} = 0} = g & on ω_{T}, \\ u_{| t ⩽ 0} = 0 & on R_{+}^{d} . \end{matrix} \end{matrix}$

The aim of the following is to use the geometric optics expansion to investigate what can be the losses of derivatives for the solution to (83). Indeed because the uniform Kreiss–Lopatinskii condition fails then such losses will occur. However for a failure of the uniform Kreiss–Lopatinskii condition on a glancing frequency then the weak well-posedness theory is not achieved yet and we do not have precise energy estimates. The existing estimates that can apply to this framework are the ones of [11] and [10]. In [11] the result is that if the uniform Kreiss–Lopatinskii condition fails at some order $δ > 0$ (or that it satisfies the Kreiss–Sakamoto condition of power δ) meaning that we have the estimate $\begin{matrix} \forall ζ \in Ξ, \forall v \in E^{s} (ζ), | B v | ≳ γ^{δ} | v | \end{matrix}$ then we have the following energy estimate, there exists $C_{T} > 0$ such that for large values of γ there holds $\begin{array}{l} γ ‖ u ‖_{L_{γ}^{2} (Ω_{T})}^{2} + ‖ u_{| x_{d} = 0} ‖_{L_{γ}^{2} (ω_{T})}^{2} ⩽ C_{T} (\frac{1}{γ^{1 + 2 δ}} {‖ L (\partial) u ‖}_{H_{γ}^{δ} (Ω_{T})}^{2} + \frac{1}{γ^{2 δ}} ‖ B u_{| x_{d} = 0} ‖_{H_{γ}^{δ} (ω_{T})}^{2}), \end{array}$ where for $X \subset R_{+}^{d}$ the space $L_{γ}^{2} (X)$ is defined via the norm $‖ \cdot ‖_{L_{γ}^{2} (X)} : = ‖ e^{- γ t} \cdot ‖_{L^{2} (X)}$ ; the Sobolev spaces $H_{γ}^{m} (X)$ being defined accordingly.

In [10] it is shown that if the boundary condition is in the particular so-called conservative form then one can recover an estimate without loss of derivatives in the interior up to the price of a larger loss along the boundary meaning that the estimate reads $\begin{array}{l} γ ‖ u ‖_{L_{γ}^{2} (Ω_{T})}^{2} + ‖ u_{| x_{d} = 0} ‖_{H_{γ}^{- 1 / 2} (ω_{T})}^{2} ⩽ \frac{C_{T}}{γ} ({‖ L (\partial) u ‖}_{L_{γ}^{2} (Ω_{T})}^{2} + ‖ B u_{| x_{d} = 0} ‖_{H_{γ}^{1 / 2} (ω_{T})}^{2}) . \end{array}$ For a sake of completeness we recall that conservative boundary conditions are defined in the following way:

Definition 4.1 (Conservative boundary condition).

We say that the boundary condition $B \in M_{p \times N} (R)$ is in conservative form if there exists a matrix $D \in M_{p \times N} (R)$ such that we can write $\begin{matrix} A_{d} = Re (D^{T} B) : = \frac{1}{2} (D^{T} B + B^{T} D) . \end{matrix}$

However let us point that the link between the failure of the Kreiss–Lopatinskii condition in the glancing area and the fulfilment of the Kreiss–Sakamoto condition and/or the fact that the boundary condition is conservative is not well-established. It seems that if the Kreiss–Lopatinskii condition fails on a glancing mode of order k then the Kreiss–Sakamoto condition of order $1 / k$ is satisfied but this conjecture is left for future studies.

Before to turn to a precise statement of the existing losses of derivatives first let us point that if one has an energy estimate for the problem (83) then it can be shown that the truncated geometric optics expansion is a good approximation of the exact solution $u^{ε}$ (we refer to the proof of Theorem 2.3 below for a precise proof of this fact (see equation (87)). This answers in particular the question of the uniqueness of the geometric optics expansion.

In order to investigate all the possible cases in the losses we will need the following extra assumption already met in [3] in order to ensure the existence of some suitable outgoing mode which can turn on the loss of derivatives in the interior. Note that the existence of such modes is automatic in the so-called $W R$ framework studied in [9] but that they are not in the general setting.

Assumption 4.1.
We assume that there exists $\underline{k} \in O$ such that $b \cdot B ker L (d φ_{\underline{k}}) \neq 0$ .

Let us also mention that the results of Theorem 2.3 agree with the conjecture about weakly well-posed problems and the failure of the uniform Kreiss–Lopatinskii condition for glancing modes of [[6]-Chapter 7.1].

We now turn to the proof of Theorem 2.3. Proof.
We first show that if Assumption 4.1 holds then the boundary value problem (5) has a loss of at least one half derivative in the interior. Then we show 3. of Theorem 2.3 that is the fact that in the general framework a loss of at least one half derivative in the interior or on the boundary occurs. In particular 2. of Theorem 2.3 (the case without loss of derivatives in the interior) is obtained as a particular case of 3. and we will omit the proof.

We assume that Assumption 4.1 holds and we argue by contradiction and assuming first that $s_{1} < \frac{1}{2}$ in order to show that $s_{1} ⩾ \frac{1}{2}$ .

Let $u^{ε}$ be the solution to the problem $\begin{matrix} (84) & \{\begin{matrix} L (\partial) u^{ε} = f^{ε} & in Ω_{T}, \\ B u_{| x_{d} = 0}^{ε} = 0 & on ω_{T}, \\ u_{| t ⩽ 0}^{ε} = 0 & on R_{+}^{d}, \end{matrix} \end{matrix}$ where $f^{ε} : = ε^{\frac{1}{2}} e^{\frac{i φ_{\underline{k}}}{ε}} f_{\underline{k}}$ with $\underline{k} \in O$ the index given by Assumption 4.1.

For $N_{0} ⩾ 0$ , let $u_{N_{0}, app}^{ε}$ be the truncated WKB expansion $\begin{matrix} (85) & u_{N_{0}, app}^{ε} : = \sum_{n = 0}^{N_{0}} ε^{\frac{n}{2}} \sum_{k \in I \cup O} e^{\frac{i}{ε} φ_{k}} u_{n, k} + \sum_{n = 0}^{N_{0}} ε^{\frac{n}{2}} e^{\frac{i}{ε} φ^{g}} u_{n}^{g} + \sum_{n = 0}^{N_{0}} ε^{\frac{n}{2}} e^{\frac{i}{ε} ψ} U_{e v, n}, \end{matrix}$ where the profiles ${(u_{n, k})}_{n \in N, k \in I \cup O} \in H^{\infty} (Ω_{T})$ , ${(u_{n}^{g})}_{n \in N} \in P_{g}$ and ${(U_{e v, n})}_{n \in N} \in P_{e v}$ are given by Theorem 2.2.

We have, using the triangle inequality: $\begin{array}{l} (86) & {‖ u_{N_{0}, app}^{ε} ‖}_{L^{2} (Ω_{T})} ⩽ {‖ u_{N_{0} + 2, app}^{ε} - u_{N_{0}, app}^{ε} ‖}_{L^{2} (Ω_{T})} + {‖ u_{N_{0} + 2, app}^{ε} - u^{ε} ‖}_{L^{2} (Ω_{T})} + {‖ u^{ε} ‖}_{L^{2} (Ω_{T})}, \end{array}$ and in the following we will estimate these terms separately. The term $u_{N_{0} + 2, app}^{ε} - u^{ε}$ satisfies the following problem: $\begin{matrix} \{\begin{matrix} L (\partial) (u_{N_{0} + 2, app}^{ε} - u^{ε}) = f_{1}^{ε} & in Ω_{T}, \\ B {(u_{N_{0} + 2, app}^{ε} - u^{ε})}_{|_{x_{d} = 0}} = 0 & on ω_{T}, \\ {(u_{N_{0} + 2, app}^{ε} - u^{ε})}_{|_{t ⩽ 0}} = 0 & on R_{+}^{d}, \end{matrix} \end{matrix}$ where $\begin{array}{l} f_{1}^{ε} : = & ε^{\frac{N_{0} + 1}{2}} (\sum_{k \in I \cup O} e^{\frac{i}{ε} φ_{k}} (L (\partial) u_{N_{0} + 1, k} + \sqrt{ε} L (\partial) u_{N_{0} + 2, k}) \\ + e^{\frac{i}{ε} φ^{g}} (A_{d} \partial_{χ} u_{N_{0} + 2}^{g} + L^{'} (\partial) u_{N_{0} + 1}^{g} + \sqrt{ε} L^{'} (\partial) u_{N_{0} + 2}^{g}) \\ + e^{\frac{i}{ε} ψ} (L^{'} (\partial) U_{e v, N_{0} + 1} + \sqrt{ε} L^{'} (\partial) U_{e v, N_{0} + 2})) . \end{array}$

We see that $‖ f_{1}^{ε} ‖_{L^{2} (Ω_{T})}$ is $O (ε^{\frac{N_{0} + 1}{2}})$ and $‖ f_{1}^{ε} ‖_{H^{1} (ω_{T})}$ is $O (ε^{\frac{N_{0} - 1}{2}})$ . So, using the interpolation inequalities, we deduce that $‖ f_{1}^{ε} ‖_{L^{2} (H^{s_{1}} (ω_{T}))} ⩽ C ε^{\frac{N_{0} + 1}{2} - s_{1}}$ where C is independent on ε.

Consequently, $\begin{array}{l} (87) & {‖ u_{N_{0} + 2, app}^{ε} - u^{ε} ‖}_{L^{2} (Ω_{T})} & ⩽ C {‖ f_{1}^{ε} ‖}_{L^{2} (R_{x_{d}}^{+}; H_{t, x^{'}}^{s_{1}} (ω_{T}))} ⩽ C ε^{\frac{N_{0} + 1}{2} - s_{1}} . \end{array}$

Moreover, $‖ f^{ε} ‖_{L^{2} (Ω_{T})}$ is $O (ε^{\frac{1}{2}})$ and $‖ f^{ε} ‖_{H^{1} (ω_{T})}$ is $O (ε^{- \frac{1}{2}})$ . So, using the interpolation inequalities, we deduce that $‖ f^{ε} ‖_{H^{s_{1}} (ω_{T})} ⩽ C ε^{\frac{1}{2} - s_{1}}$ . Thus, from the energy estimate (16) applied to problem (84) we obtain $\begin{matrix} {‖ u^{ε} ‖}_{L^{2} (Ω_{T})} ⩽ C ε^{\frac{1}{2} - s_{1}} . \end{matrix}$

Finally $‖ u_{N_{0} + 2, app}^{ε} - u_{N_{0}, app}^{ε} ‖_{L^{2} (Ω_{T})}$ is clearly $O (ε^{\frac{N_{0} + 1}{2}})$ ; and so, taking $N_{0} = 0$ it follows that $\begin{matrix} {‖ u_{0, app}^{ε} ‖}_{L^{2} (Ω_{T})} ⩽ C (ε^{\frac{1}{2}} + ε^{\frac{1}{2} - s_{1}} + ε^{\frac{1}{2} - s_{1}}) . \end{matrix}$

Because $u_{0, app}^{ε} = e^{\frac{i φ^{g}}{ε}} u_{0}^{g}$ , we have on the one hand $‖ u_{0, app}^{ε} ‖_{L^{2} (Ω_{T})} = ‖ u_{0}^{g} ‖_{L^{2} (Ω_{T})}$ . On the other hand because $s_{1} < \frac{1}{2}$ it turns out that $‖ u_{0}^{g} ‖_{L^{2} (Ω_{T})}$ tends to 0 as ε goes to 0, hence $u_{0}^{g} = 0$ .

However $u_{0}^{g}$ is given by (52) in which the source term $G_{0}$ defined in (51) reads $e^{- γ t} b \cdot B Π^{\underline{k}} u_{1, {\underline{k}}_{| x_{d} = 0}}$ where $Π^{\underline{k}} u_{1, {\underline{k}}_{| x_{d} = 0}}$ only depends on $f_{\underline{k}}$ as the solution of the associated transport equation. Let us point that from Assumption 4.1 the term $b \cdot B Π^{\underline{k}} u_{1, {\underline{k}}_{| x_{d} = 0}}$ is non-zero. We are free to choose the interior source term $f_{\underline{k}}$ in order that $Π^{\underline{k}} u_{1, {\underline{k}}_{| x_{d} = 0}}$ is not zero and so do $u_{g}^{0}$ . We have a contradiction and we get $s_{1} ⩾ \frac{1}{2}$ as desired.

Now we turn to the proof that $s_{1} ⩾ \frac{1}{2}$ or $s_{2} ⩾ \frac{1}{2}$ . Once again we argue by contradiction and assume that $s_{1} < \frac{1}{2}$ and $s_{2} < \frac{1}{2}$ .

Let $u^{ε}$ be the solution to the problem $\begin{matrix} (88) & \{\begin{matrix} L (\partial) u^{ε} = 0 & in Ω_{T}, \\ B u_{| x_{d} = 0}^{ε} = g^{ε} & on ω_{T}, \\ u_{| t ⩽ 0}^{ε} = 0 & on R_{+}^{d}, \end{matrix} \end{matrix}$ where $g^{ε} : = ε^{\frac{1}{2}} e^{\frac{i ψ}{ε}} g$ for some regular g, specified below, vanishing for negatives times. We consider again the approximated solution $u_{N_{0}, app}^{ε}$ given by (85) and once again we will use the triangle inequality (86) to estimate $u_{N_{0}, app}^{ε}$ .

The term $u_{N_{0} + 2, app}^{ε} - u^{ε}$ satisfies the same problem as before from which we infer that $\begin{matrix} {‖ u_{N_{0} + 2, app}^{ε} - u^{ε} ‖}_{L^{2} (Ω_{T})} ⩽ C ε^{\frac{N_{0} + 1}{2} - s_{1}} . \end{matrix}$

Now, because $‖ g^{ε} ‖_{L^{2} (ω_{T})}$ is $O (ε^{\frac{1}{2}})$ and $‖ g^{ε} ‖_{H^{1} (ω_{T})}$ is $O (ε^{- \frac{1}{2}})$ so we deduce that $‖ g^{ε} ‖_{H^{s_{2}} (ω_{T})} ⩽ C ε^{\frac{1}{2} - s_{2}}$ . Thus, from the energy estimate (16) applied to the problem (88) we obtain $\begin{matrix} {‖ u^{ε} ‖}_{L^{2} (Ω_{T})} ⩽ C ε^{\frac{1}{2} - s_{2}} . \end{matrix}$

Finally $‖ u_{N_{0} + 2, app}^{ε} - u_{N_{0}, app}^{ε} ‖_{L^{2} (Ω_{T})}$ is still $O (ε^{\frac{N_{0} + 1}{2}})$ . Consequently3
³
Taking $s_{1} = 0$ in estimate (89) gives the proof of point 2. of Theorem 2.3

for $N_{0} = 0$ : $\begin{array}{l} (89) & {‖ u_{0, app}^{ε} ‖}_{L^{2} (Ω_{T})} ⩽ C (ε^{\frac{1}{2}} + ε^{\frac{1}{2} - s_{1}} + ε^{\frac{1}{2} - s_{2}}) . \end{array}$

Since $u_{0, app}^{ε} = e^{\frac{i φ^{g}}{ε}} u_{0}^{g}$ , we have $‖ u_{0, app}^{ε} ‖_{L^{2} (Ω_{T})} = ‖ u_{0}^{g} ‖_{L^{2} (Ω_{T})}$ .

Because of our working assumption on $s_{1}$ and $s_{2}$ , equation (89) implies that $‖ u_{0, app}^{ε} ‖_{L^{2} (Ω_{T})}$ goes to zero with ε so that one should have $u_{0}^{g} = 0$ . But recall that $u_{0}^{g}$ is given by (52) in which the source $G_{0}$ reads $b \cdot g$ . In order to make sure that it gives a non zero leading order term we can for example set $g : = ϕ b$ where ϕ is some function in $D (ω_{T})$ . Once again we obtain the desired contradiction. □

5. Some examples

5.1. The $2 d$ -wave equation

We are considering the $2 d$ classical wave equation (where without loss of generality the velocity c equals one) with Neumann boundary condition: $\begin{matrix} (90) & \{\begin{matrix} \partial_{t}^{2} u - △ u = f & for (t, x_{1}, x_{2}) \in] - \infty, T] \times R_{+} \times R, \\ {(\partial_{1} u)}_{| x_{1} = 0} = g & for (t, x_{2}) \in] - \infty, T] \times R, \\ u_{| t ⩽ 0} = {(\partial_{t} u)}_{| t ⩽ 0} = 0 & for (x_{1}, x_{2}) \in R_{+} \times R, \end{matrix} \end{matrix}$ If we define $U : = {[\partial_{t} u - \partial_{1} u - \partial_{2} u, \partial_{t} u + \partial_{1} u - \partial_{2} u]}^{T}$ then equation (90) becomes $\begin{matrix} (91) & \{\begin{matrix} \partial_{t} U + A_{1} \partial_{1} U + A_{2} \partial_{2} U = F & for (t, x_{1}, x_{2}) \in] - \infty, T] \times R_{+} \times R, \\ B U_{| x_{1} = 0} = 2 g & for (t, x_{2}) \in] - \infty, T] \times R, \\ U_{| t ⩽ 0} = 0 & for (x_{1}, x_{2}) \in R_{+} \times R, \end{matrix} \end{matrix}$ where we defined $F : = {[f, f]}^{T}$ and $\begin{matrix} A_{1} : = [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}], A_{2} : = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] and B : = [\begin{matrix} - 1 & 1 \end{matrix}] . \end{matrix}$ We can easily check from the coefficients $A_{1}$ and $A_{2}$ that the boundary value problem (91) satisfies Assumptions 2.1 and 2.2. Moreover Assumption 2.3 is also satisfied because the problem is of size two.

Then in such a setting the boundary condition B breaks down the uniform Kreiss–Lopatinskii condition in the glancing area. Indeed the resolvent matrix associated to (91) is $\begin{matrix} A (ζ) = [\begin{matrix} - σ & - i η \\ i η & σ \end{matrix}], \end{matrix}$ whose eiganvalues $λ_{\pm}$ ( $λ_{-}$ being the one with negative real part) are the roots of the equation $\begin{matrix} λ^{2} = σ^{2} + η^{2}, \end{matrix}$ and we can thus parametrize the stable subspace $E^{s} (σ, η) : = span ({[- i η, λ_{-} + σ]}^{T})$ from which we can verify that $ker B \cap E^{s} (σ, η) = {0}$ for all $γ > 0$ . For $γ = 0$ we thus obtain that the eigenvalues are solution to $λ^{2} = τ^{2} - η^{2}$ so that the elliptic region is given by $\begin{matrix} E : = {(i τ, η) \in i R \times R ∖ | τ | > | η |}, \end{matrix}$ and that the hyperbolic region is $\begin{matrix} H : = {(i τ, η) \in i R \times R ∖ | τ | < | η |} . \end{matrix}$

For $| τ | = | η |$ we thus have zero as a root of multiplicity two from which we deduce that $\begin{matrix} G : = \partial H = {(i τ, η) \in i R \times R ∖ | τ | = | η |} . \end{matrix}$

Now fix $η < 0$ and choose $τ = - η$ for such parameters we have $\begin{matrix} A (- η, η) = i η [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}], \end{matrix}$ from which we can read that $E^{s} (- η, η) = span ({[1, 1]}^{T}) = ker B$ so that the uniform Kreiss–Lopatinskii condition fails in the glancing area. Moreover because $dim ker B = dim E^{s} (- η, η) = 1$ we are in the framework studied in this article and Assumption 2.5 holds. Indeed because $p = 1$ we can choose $b = 1$ and a simple computation gives $\begin{matrix} B Υ^{g} A_{1} e = [\begin{matrix} - 1 & 1 \end{matrix}] [\begin{matrix} 0 & 0 \\ 0 & - \frac{1}{2 η} \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] [\begin{matrix} 1 \\ 1 \end{matrix}] = \frac{1}{2 η} \neq 0 . \end{matrix}$

Consequently one can use Theorem 2.2 to perform a geometric optics expansion of the solution to (91). About the existence of such a solution we refer to [11] or [10]. Then if one applies Theorem 2.3 because the system is of size two the expansion only contains one glancing amplitude (and no hyperbolic outgoing modes) then Assumption 4.1 fails. So we are in cases 2. and 3. of Theorem 2.3. However in such a configuration we have energy estimates from the work of [11] and [10].

On the one hand it is shown in [11] that the so-called Kreiss–Sakamoto condition with power $1 / 2$ is satisfied meaning that $\begin{matrix} \forall ζ \in Ξ, \forall v \in E^{s} (ζ), | B v | ≳ γ^{1 / 2} | v | . \end{matrix}$

So that from [11] we have the energy estimate $\begin{array}{l} γ ‖ u ‖_{L_{γ}^{2} (Ω)}^{2} + ‖ u_{| x_{d} = 0} ‖_{L_{γ}^{2} (\partial Ω)}^{2} ⩽ C_{T} (\frac{1}{γ^{2}} ‖ f ‖_{L_{x_{d}}^{2} (H_{γ}^{1 / 2} (\partial Ω))}^{2} + \frac{1}{γ} ‖ g ‖_{H_{γ}^{1 / 2} (\partial Ω)}^{2}), \end{array}$ which is sharp because of Theorem 2.2 giving another justification of the fact that the power $1 / 2$ of the Kreiss–Sakamoto condition can not be lowered in [11].

On the other hand we can show that the boundary condition B is conservative in the sense of Definition 4.1 (take $D = B$ ) so that from [10] we also have the boundary estimate without loss of derivatives in the interior $\begin{array}{l} γ ‖ u ‖_{L_{γ}^{2} (Ω_{T})}^{2} + γ ‖ u_{| x_{d} = 0} ‖_{H_{γ}^{- 1 / 2} (ω_{T})}^{2} ⩽ \frac{C_{T}}{γ} (‖ f ‖_{L_{γ}^{2} (Ω_{T})}^{2} + ‖ g ‖_{H_{γ}^{1 / 2} (ω_{T})}^{2}) . \end{array}$ We can apply 2. of Theorem 2.3 which shows that the index $1 / 2$ in the norm of the right hand side is optimal.

5.2. Linearisation of Euler equation

We now consider the linearisation of isentropic $2 d$ -Euler equations around some subsonic outgoing fluid. This system of equations reads $\begin{matrix} (92) & \{\begin{matrix} \partial_{t} u + A_{1} \partial_{1} u + A_{2} \partial_{2} u = f, & for (t, x_{1}, x_{2}) \in] - \infty, T] \times R \times R_{+}, \\ B u_{| x_{2} = 0} = g, & for (t, x_{1}) \in] - \infty, T] \times R, \\ u_{| t ⩽ 0} = 0, & for (x_{1}, x_{2}) \in R \times R_{+}, \end{matrix} \end{matrix}$ where the coefficients $A_{1}$ , $A_{2}$ and B are given by $\begin{matrix} A_{1} : = [\begin{matrix} 0 & - 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}], A_{2} : = [\begin{matrix} M & 0 & - 1 \\ 0 & M & 0 \\ - 1 & 0 & M \end{matrix}] and B : = [\begin{matrix} 1 & b_{2} & b_{3} \end{matrix}] \end{matrix}$ in which $M \in] - 1, 0 [$ stands for the Mach number and where $(b_{2}, b_{3}) \in R^{2}$ are boundary parameters that can be chosen arbitrarily. It is rather easy to verifiy that the system (92) satisfies Assumptions 2.1 and 2.2. Let us mention by the way that the boundary condition B can not be conservative in the sense of Definition 4.1.

In [3] we can find a discussion about what the influence of the parameters on the strong (or weak) well-posedness of (92).

In particular the set of parameters $(b_{2}, b_{3}) \in R^{2}$ for which the Kreiss–Lopatinskii condition degenerates in the glancing area is given by $\begin{array}{l} Υ : = & {(b_{2}, b_{3}) \in R^{2} ∖ b_{2} = \pm \frac{1 + M b_{3}}{\sqrt{1 - M^{2}}}} : = Ω_{g}^{+} \cup Ω_{g}^{-}, \end{array}$ the union of these two lines constitutes the parameters $(b_{2}, b_{3}) \in R^{2}$ that were not covered by the analysis of [3]. Now Theorem 2.3 ends the picture.

More precisely in order to apply this theorem we should be more precise about the fulfilment of Assumption 4.1. In order to do so, we have a look to the stable and unstable subspaces for glancing frequencies. Let ω be an eigenvalue of $A (\underline{ζ})$ we thus have $det (σ I + i η A_{1} + ω A_{2}) = 0$ so that ω satisfies the dispersion relation $\begin{matrix} (93) & (M ω + σ) \cdot ({(M ω + σ)}^{2} + η^{2} - ω^{2}) = 0 . \end{matrix}$

For $\underline{ζ} \in Ξ_{0}$ we thus always have a hyperbolic eigenvalue $ω_{h} : = - \frac{i τ}{M}$ which differs from the roots of the second term of the left hand side of (93) for glancing frequencies $(i τ, η) \in G = {(i τ, η) \in i R \times R ∖ | τ | = \sqrt{1 - M^{2}} | η |}$ . In particular Assumption 2.3 holds.

Because $p = 1$ , the hyperbolic eigenvalue $ω_{h}$ contributes to the unstable subspace and some simple computations show that it is associated to the eigenspace $E^{u} (\underline{ζ}) : = span ({[0, \frac{τ}{M}, η]}^{T}) = span (v)$ . In order to study Assumption 4.1 we have to compute $B v$ for $τ = \pm \sqrt{1 - M^{2}} | η |$ and $(b_{2}, b_{3}) \in Ω_{g}^{+} \cup Ω_{g}^{-}$ to fix the ideas we set $τ = \pm \sqrt{1 - M^{2}} | η |$ and $(b_{2}, b_{3}) \in Ω_{g}^{+}$ the other case being treated similarly. We thus have $\begin{array}{l} B v = \{\begin{matrix} η (1 + b_{3} M + b_{3}) & if η > 0, \\ η (- 1 - b_{3} M + b_{3}) & if η < 0, \end{matrix} \end{array}$ from which we deduce that Assumption 4.1 is satisfied for all parameters $b_{3}$ except for $b_{3} \in {\frac{- 1}{M + 1}, \frac{1}{1 - M}}$ . So that when Assumption 4.1 holds we are in 1. of Theorem 2.3 and so equation (92) loses at least one half of derivative in the interior. This agrees with the fact that the boundary conditions of (92) can not be of conservative type so that the result of [10] does not apply.

On Fig. 1 is depicted the influence of the boundary parameters on the well-posedness of (92). In this figure $W R$ stands for the parameter giving a problem in the $W R$ class studied in [5], $S U$ stands for strongly unstable that is problems that do not satisfy the weak Kreiss–Lopatinskii condition and $S S$ stands for problem satisfying the uniform Kreiss–Lopatinskii condition.

Fig. 1.

The different boundary parameters for (92).

Footnotes

Acknowledgements

Research of the first author was supported by ANR project NABUCO, ANR 17-CE40-0025.

References

Artola and

Majda, Nonlinear development of instabilities in supersonic vortex sheets. I. The basic kink modes, Physica D: Nonlinear Phenomena 28(3) (1987), 253–281. doi:10.1016/0167-2789(87)90019-4.

Artola and

Majda, Nonlinear geometric optics for hyperbolic mixed problems, in: Analyse Mathématique et Applications, Gauthier-Villars, 1988, pp. 319–356.

Benoit, Geometric optics expansions for linear hyperbolic boundary value problems and optimality of energy estimates for surface waves, Differential Integral Equations 27(5–6) (2014), 531–562.

Benoit, WKB expansions for hyperbolic boundary value problems in a strip: Self interaction meets strong well posedness, Journal de l’ Institut de Mathématiques de Jussieu (2018).

Benzoni-Gavage,

Rousset,

Serre and

Zumbrun, Generic types and transitions in hyperbolic initial-boundary-value problems, Proc. Roy. Soc. Edinburgh Sect. A 132(5) (2002), 1073–1104. doi:10.1017/S030821050000202X.

Benzoni-Gavage and

Serre, Multidimensional Hyperbolic Partial Differential Equations, Oxford Mathematical Monographs, Oxford University Press, 2007.

J.-F.

Coulombel, Stabilité multidimensionnelle d’interfaces dynamiques. Application aux transitions de phase liquide-vapeur, PhD thesis, ENS, Lyon, 2002.

J.-F.

Coulombel, Well-posedness of hyperbolic initial boundary value problems, J. Math. Pures Appl. 84 (2005), 786–818. doi:10.1016/j.matpur.2004.10.005.

J.-F.

Coulombel and

Guès, Geometric optics expansions with amplification for hyperbolic boundary value problems: Linear problems, Annales de l’Institut Fourier 60(6) (2010), 2183–2233. doi:10.5802/aif.2581.

10.

Eller, On symmetric hyperbolic boundary problems with nonhomogeneous conservative boundary conditions, SIAM J. Math. Anal. 44(3) (2012), 1925–1949. doi:10.1137/110834652.

11.

Eller, Loss of derivatives for hyperbolic boundary problems with constant coefficients, Discrete Contin. Dyn. Syst., Ser. B 23(3) (2018), 1347–1361.

12.

Hersh, Mixed problems in several variables, Journal of Mathematics and Mechanics 12(3) (1963), 317–334.

13.

Kilque, Weakly nonlinear multiphase geometric optics for hyperbolic quasilinear boundary value problems: construction of a leading profile, 2021, working paper or preprint.

14.

H.-O.

Kreiss, Initial boundary value problems for hyperbolic systems, Communications on Pure and Applied Mathematics 23(3) (1970), 277–298. doi:10.1002/cpa.3160230304.

15.

P.D.

Lax, Asymptotic solutions of oscillatory initial value problems, Duke Mathematical Journal 24(4) (1957), 627–646. doi:10.1215/S0012-7094-57-02471-7.

16.

Lescarret, Diffractive wave transmission in dispersive media, Journal of Differential Equations 244(4) (2008), 972–1010. doi:10.1016/j.jde.2007.08.003.

17.

Majda and

Rosales, A theory for spontaneous Mach stem formation in reacting shock fronts. I. The basic perturbation analysis, SIAM J. Appl. Math. 43(6) (1983), 1310–1334. doi:10.1137/0143088.

18.

Métivier, The block structure condition for symmetric hyperbolic systems, Bulletin of the London Mathematical Society 32(6) (2000), 689–702. doi:10.1112/S0024609300007517.

19.

Sablé-Tougeron, Existence pour un problème de l’élastodynamique Neumann non linéaire en dimension 2, Arch. Rational Mech. Anal. 101(3) (1988), 261–292. doi:10.1007/BF00253123.

20.

Williams, Nonlinear geometric optics for hyperbolic boundary problems, Comm. Partial Differential Equations 21(11–12) (1996), 1829–1895. doi:10.1080/03605309608821247.

21.

Williams, Boundary layers and glancing blow-up in nonlinear geometric optics, Annales scientifiques de l’École Normale Supérieure 33(3) (2000), 383–432. doi:10.1016/S0012-9593(00)00116-6.

Geometric optics expansion for weakly well-posed hyperbolic boundary value problem: The glancing degeneracy

Abstract

Keywords

1. Introduction

2. Assumptions and main results

1 In fact the results of the article can also be stated in some space H S for S fixed large enough. But we choose H ∞ instead for simplicity.

Theorem 2.1 (Block structure [18]).

Lemma 3.1 ([21], Lemma 8.1).

Lemma 3.2 (Lax lemma [15]).

3.1.1. Determination of the hyperbolic leading order term

3.1.2. Determination of the evanescent leading order term

Lemma 3.3 (Lescarret).

3.1.3. Determination of the glancing leading order term

3.2.1. Determination of the hyperbolic amplitudes of order one

3.2.2. Determination of the evanescent amplitudes of order one

3.2.3. Determination of the glancing term of order one

Proposition 3.4. Under Assumptions 2.1 , 2.2 , 2.3 , 2.4 and 2.5 , for all k ∈ I ∪ O , there exist u 1 , k ∈ H ∞ ( Ω T ) , u 1 g ∈ P g and U e v , 1 ∈ P e v satisfying the cascades of equations ( 22 ), ( 23 ) and ( 24 ) at order one. 3.3. Higher order terms

Definition 4.1 (Conservative boundary condition).

5.1. The 2 d -wave equation

5.2. Linearisation of Euler equation

Footnotes

Acknowledgements

References

¹
In fact the results of the article can also be stated in some space $H^{S}$ for S fixed large enough. But we choose $H^{\infty}$ instead for simplicity.

Proposition 3.4.
Under Assumptions 2.1 , 2.2 , 2.3 , 2.4 and 2.5 , for all $k \in I \cup O$ , there exist $u_{1, k} \in H^{\infty} (Ω_{T})$ , $u_{1}^{g} \in P_{g}$ and $U_{e v, 1} \in P_{e v}$ satisfying the cascades of equations ( 22 ), ( 23 ) and ( 24 ) at order one.

3.3. Higher order terms

5.1. The $2 d$ -wave equation