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Submitted on 17 Feb 2021
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SEMICLASSICAL PARAMETRIX FOR THE
MAXWELL EQUATION AND APPLICATIONS TO
THE ELECTROMAGNETIC TRANSMISSION
EIGENVALUES
Georgi Vodev
To cite this version:
Georgi Vodev. SEMICLASSICAL PARAMETRIX FOR THE MAXWELL EQUATION AND AP-PLICATIONS TO THE ELECTROMAGNETIC TRANSMISSION EIGENVALUES. Research in the Mathematical Sciences , Springer, In press. �hal-03143832�
APPLICATIONS TO THE ELECTROMAGNETIC TRANSMISSION EIGENVALUES
GEORGI VODEV
Abstract. We introduce an analog of the Dirichlet-to-Neumann map for the Maxwell equation in a bounded domain. We show that it can be approximated by a pseudodifferential operator on the boundary with a matrix-valued symbol and we compute the principal symbol. As a consequence, we obtain a parabolic region free of the transmission eigenvalues associated to the Maxwell equation.
Key words: Maxwell equation, semiclassical parametrix, transmission eigenvalues.
1. Introduction
Let Ω ⊂ R3 be a bounded, connected domain with a C∞ smooth boundary Γ = ∂Ω, and
consider the Maxwell equation
(1.1) ∇ × E = iλµ(x)H in Ω, ∇ × H = −iλε(x)E in Ω, ν× E = f on Γ,
where λ ∈ C, |λ| ≫ 1, ν = (ν1, ν2, ν3) denotes the Euclidean unit normal to Γ, µ, ε ∈ C∞(Ω)
are scalar-valued strictly positive functions. The functions E = (E1, E2, E3) ∈ C3 and B =
(B1, B2, B3) ∈ C3 denote the electric and magnetic fields, respectively. The equation (1.1)
describes the propagation of electromagnetic waves in Ω with a frequency λ moving with a speed (εµ)−1/2. Recall that given two vectors a = (a1, a2, a3) and b = (b1, b2, b3), a× b denotes
the vector (a2b3− a3b2, a3b1− a1b3, a1b2− a2b1) and it is perpendicular to both a and b. Thus
we have
∇ × E = (∂x2E3− ∂x3E2, ∂x3E1− ∂x1E3, ∂x1E2− ∂x2E1)
and similarly for ∇ × H. Throughout this paper, given s ∈ R we will denote by Hs(Γ) the
Sobolev space Hs(Γ; C3). Introduce the spaces
Hts(Γ) :={f ∈ Hs(Γ) :hν(x), f(x)i = 0}, s = 0, 1,
where hν, fi := ν1f1+ ν2f2+ ν3f3. In view of Theorem 3.1 we can introduce the operator
N (λ) : Ht1(Γ)→ Ht0(Γ)
defined by
N (λ)f = ν × H|Γ,
which can be considered as an analog of the Dirichlet-to-Neumann map. Set h = |Re λ|−1 if |Re λ| ≥ |Im λ| and h = |Im λ|−1 if |Im λ| ≥ |Re λ|, z = hλ and θ = |Im z| ≤ 1. Clearly, in
the first case we have z = 1 + iθ, while in the second case we have θ = 1. We would like to approximate the operator N (λ) by a matrix-valued h − ΨDO. It is proved in [8], [10] that the
Dirichlet-to-Neumann operator associated to the Helmholtz equation with refraction index εµ can be approximated by Oph(ρ), where
ρ(x′, ξ′, z) =p−r0(x′, ξ′) + z2(ε0µ0)(x′), Im ρ > 0, (x′, ξ′)∈ T∗Γ,
where ε0 = ε|Γ, µ0 = µ|Γ, and r0 ≥ 0 is the principal symbol of the operator −∆Γ. Here ∆Γ
denotes the negative Laplace-Beltrami operator on Γ with Riemannian metric induced by the Euclidean one. It is well-known (see Section 2) that r0 =hβ, βi, where β = β(x′, ξ′)∈ R3 is a
vector-valued homogeneos polynomial of order one in ξ′, which is perpendicular to the normal
ν(x′), that is,hβ, νi = 0. Set
m = (zµ0)−1 ρI + ρ−1B
,
where I is the identity 3× 3 matrix, while the matrix B is defined by Bg = hβ, giβ, g∈ R3.
Our main result is the following
Theorem 1.1. Let θ ≥ h2/5−ǫ, where 0 < ǫ≪ 1 is arbitrary. Then for every f ∈ Ht
1 we have
the estimate
(1.2) kN (λ)f − Oph(m + h em)(ν× f)kH0 .hθ−5/2kfkH−1
where em ∈ C∞(T∗Γ) is a matrix-valued function independent of h, belonging to the space S0 0,1
uniformly in z and such that µ0m is independent of ε and µ.e
Hereafter the Sobolev spaces are equipped with the h-semiclassical norm. Clearly, the estimate (1.2) provides a good approximation of the operator N (λ) as long as θ ≥ h2/5−ǫ. It also implies
the following improvement upon the estimate (3.4). Corollary 1.2. Let θ≥ h2/5−ǫ. Then for every f ∈ Ht
1 we have the estimate
(1.3) kN (λ)fkH0 .θ−1/2kfkH1.
Note that analog estimates for the Dirichlet-to-Neumann operator associated to the Helmholtz equation are proved in [8], [10] for θ ≥ h1/2−ǫ, in [12] for θ ≥ h2/3−ǫ and in [9] for θ ≥ h1−ǫ,
0 < ǫ ≪ 1 being arbitrary. In the last case it is assumed that the boundary is strictly concave. In all these papers the approximation of the Dirichlet-to-Neumann map is used to get parabolic regions free of transmission eigenvalues.
To prove Theorem 1.1 we build in Section 4 a semiclassical parametrix near the boundary for the solutions to the equation (1.1). It takes the form of oscilatory integrals with a complex-valued phase function ϕ satisfying the eikonal equation modO(xN
1 ) (see (4.5)), where N ≫ 1 is
arbitrary and 0 < x1 ≪ 1 denotes the normal variable near the boundary, that is, the distance
to Γ. The amplitudes satisfy some kind of transport equations mod O(xN
1 ) (see (4.2)). Thus
the parametrix satisfies the Maxwell equation modulo an error term which is given by oscilatory integrals with amplitudes of the form O(xN
1 ) +O(hN). To estimate the difference between the
exact solution to equation (1.1) and its parametrix we use the a priori estimate (3.5). Note that there exists a different approach suggested in [2] which could probably lead to (1.2) as well. It consists of using the results in [8], [10] to approximate the normal derivatives −ih∂νE|Γ and
−ih∂νH|Γ by Oph(ρ)E|Γ and Oph(ρ)H|Γ. Thus the equation (1.1) can be reduced to a system
of h− ΨDOs on Γ by restricting the equations in (1.1) on the boundary.
In analogy with the Helmholtz equation, Theorem 1.1 can be used to study the location on the complex plane of the transmission eigenvalues associated to the Maxwell equation (see Section 5). It can also be used to study the complex eigenvalues associated to the Maxwell equation with dissipative boundary conditions like that one considered in [2].
2. Preliminaries
We will first introduce the spaces of symbols which will play an important role in our analysis and will recall some basic properties of the h− ΨDOs. Given k ∈ R, δ1, δ2 ≥ 0, we denote by
Sδk1,δ2 the space of all functions a∈ C∞(T∗Γ), which may depend on the semiclassical parameter
h, satisfying ∂xα′∂β ξ′a(x′, ξ′, h) ≤ Cα,βhξ′ik−δ1|α|−δ2|β|
for all multi-indices α and β, with constants Cα,β independent of h. More generally, given a
function ω > 0 on T∗Γ, we denote by Sδk1,δ2(ω) the space of all functions a∈ C∞(T∗Γ), which
may depend on the semiclassical parameter h, satisfying ∂xα′∂β ξ′a(x′, ξ′, h) ≤ Cα,βωk−δ1|α|−δ2|β|
for all multi-indices α and β, with constants Cα,β independent of h and ω. Thus Sδk1,δ2 =
Skδ1,δ2(hξ′i). Given a matrix-valued symbol a, we will say that a ∈ Sk
δ1,δ2 if all entries of a
belong to Sδk1,δ2. Also, given k ∈ R, 0 ≤ δ < 1/2, we denote by Sk
δ the space of all functions
a∈ C∞(T∗Γ), which may depend on the semiclassical parameter h, satisfying
∂xα′∂β ξ′a(x′, ξ′, h) ≤ Cα,βh−δ(|α|+|β|)hξ′ik−|β|
for all multi-indices α and β, with constants Cα,β independent of h. Again, given a
matrix-valued symbol a, we will say that a∈ Sk
δ if all entries of a belong to Sδk. The h− ΨDO with a
symbol a is defined by (Oph(a)f ) (x′) = (2πh)−2 Z Z e−hihx ′ −y′ ,ξ′ ia(x′, ξ′, h)f (y′)dξ′dy′. If a∈ Sk
0,1, then the operator Oph(a) : Hhk(Γ)→ L2(Γ) is bounded uniformly in h, where
kukHk h(Γ) := Oph(hξ′ik)u L2(Γ).
It is also well-known (e.g. see Section 7 of [3]) that, if a ∈ Sδ0, 0 ≤ δ < 1/2, then Oph(a) : Hs
h(Γ)→ Hhs(Γ) is bounded uniformly in h. More generally, we have the following (see Section
2 of [8]):
Proposition 2.1. Let hℓ±a±∈ S±k
δ , 0≤ δ < 1/2, where ℓ±≥ 0 are some numbers. Assume in
addition that the functions a± satisfy
(2.1) ∂α1 x′ ∂ β1 ξ′ a+(x′, ξ′)∂α 2 x′∂ β2 ξ′ a−(x′, ξ′) ≤ κCα1,β1,α2,β2h −(|α1|+|β1|+|α2|+|β2|)/2
for all multi-indices α1, β1, α2, β2 such that|αj|+|βj| ≥ 1, j = 1, 2, with constants Cα1,β1,α2,β2 >
0 independent of h and κ. Then we have
(2.2) Oph(a+)Oph(a−)− Oph(a+a−) L2(Γ)→L2(Γ).h + κ.
Let η ∈ C∞(T∗Γ) be such that η = 1 for r
0 ≤ C0, η = 0 for r0 ≥ 2C0, where C0 > 0 does
not depend on h. It is easy to see (e.g. see Lemma 3.1 of [8]) that taking C0 big enough we can
arrange
C1θ1/2≤ |ρ| ≤ C2, Im ρ≥ C3|θ||ρ|−1≥ C4|θ|
on supp η, and
on supp(1− η) with some constants Cj > 0. We will say that a function a∈ C∞(T∗Γ) belongs to Sk1 δ1,δ2(ω1) + S k2 δ3,δ4(ω2) if ηa∈ S k1 δ1,δ2(ω1) and (1− η)a ∈ S k2 δ3,δ4(ω2). It is shown in Section 3 of
[8] (see Lemma 3.2 of [8]) that
(2.3) ρk∈ S2,2k (|ρ|) + S0,1k (|ρ|) ⊂ S1,1−ek/2(θ) + S0,1k ⊂ θ−ek/2S1/2−ǫ−N + S k
0,1⊂ θ−ek/2S1/2−ǫk
as long as θ ≥ h1/2−ǫ, uniformly in θ and h, where ek = 0 if k ≥ 0, ek = −k if k ≤ 0 and N ≫ 1
is arbitrary. Proposition 2.1 implies the following Proposition 2.2. Let h1/2−ǫ≤ θ±≤ 1, ℓ± ≥ 0, and let
a± ∈ S−ℓ± 1,1 (θ±) + S0,1k±⊂ θ −ℓ± ± S k± 1/2−ǫ. Then we have (2.4) Oph(a+)Oph(a−)− Oph(a+a−) Hk h(Γ)→L2(Γ) . hθ−1−ℓ+ + θ −1−ℓ− − , where k = k++ k−− 1.
Proof. Let η0, η1, η2 ∈ C0∞(T∗Γ) be such that η1 = 1 on supp η, η2 = 1 on supp η1, η = 1 on
supp η0. Then we have
Oph(a+a−)− Oph(ηa+η1a−)Oph(η2)− Oph((1− η)a+(1− η0)a−)
= Oph(ηa+η1a−)Oph(1− η2) =O(h∞) : Hhk(Γ)→ L2(Γ),
Oph(a+)Oph(a−)− Oph(ηa+)Oph(η1a−)Oph(η2)− Oph((1− η)a+)Oph((1− η0)a−)
= Oph(ηa+)Oph((1− η1)a−) + Oph((1− η)a+)Oph(η0a−)
+Oph(ηa+)Oph(η1a−)Oph(1− η2) =O(h∞) : Hhk(Γ)→ L2(Γ).
By assumption, ηa+ ∈ S−ℓ+
1,1 (θ+), η1a− ∈ S1,1−ℓ−(θ−), which implies that the functions ηa+ and
η1a− satisfy the condition (2.1) with κ = hθ+−1−ℓ+θ −1−ℓ−
− . Therefore, by (2.2) we have
Oph(ηa+η1a−)− Oph(ηa+)Oph(η1a−)Oph(η2)f
L2 .hθ−1−ℓ+ + θ−1−ℓ − − kOph(η2)fkL2 .hθ−1−ℓ + + θ−1−ℓ − − kfkHk h.
On the other hand, (1− η)a+∈ Sk+
0,1, (1− η0)a− ∈ S0,1k−. The standard pseudodifferential calculas
gives that, modO(h∞), the operator
Oph((1− η)a+(1− η0)a−)− Oph((1− η)a+)Oph((1− η0)a−)
is an h− ΨDO with symbol hω, ω ∈ S0,1k uniformly in h, where k = k++ k−− 1. Therefore,
Oph((1− η)a+(1− η0)a−)f − Oph((1− η)a+)Oph((1− η0)a−)f
L2 .hkfkHk h.
Clearly, (2.4) follows from the above estimates. ✷
We also have
Proposition 2.3. Let h1/2−ǫ≤ θ ≤ 1, ℓ ≥ 0, and let
a∈ S1,1−ℓ(θ) + S0,1k ⊂ θ−ℓS1/2−ǫk .
Then we have
(2.5) kOph(a)kHk
h(Γ)→L2(Γ).θ
Note that these propositions remain valid for matrix-valued symbols.
We will next write the gradient ∇ in the local normal geodesic coordinates near the boundary (see also Section 2 of [2]). Fix a point y0 ∈ Γ and let U ⊂ R3 be a small neighbourhood of y0.
Let U0 be a small neighbourhood of x′ = 0 in R2 and let x′ = (x2, x3) be local coordinates in
U0. Then there exists a diffeomorphism s : U0 → U ∩ Γ. Let y = (y1, y2, y3) ∈ U ∩ Ω, denote
by y′ ∈ Γ the closest point from y to Γ and let ν′(y′) be the unit inner normal to Γ at y′. Set x1= dist(y, Γ), x′ = s−1(y′) and ν(x′) = ν′(s(x′)) = (ν1(x′), ν2(x′), ν3(x′)). We have
y = s(x′) + x1ν(x′) and hence ∂ ∂yj = νj(x′) ∂ ∂x1 + 3 X k=2 αj,k(x) ∂ ∂xk , where αj,k = ∂x∂ykj, provided x1 is small enough. Note that the matrix
∂xk ∂yj , 1≤ k, j ≤ 3, is the inverse of ∂yk ∂xj
, 1≤ k, j ≤ 3. In particular, this implies the identities
3
X
j=1
νj(x′)αj,k(x) = 0, k = 2, 3.
Set ζ1 = (1, 0, 0), ζ2 = (0, 1, 0), ζ3 = (0, 0, 1). Clearly, we can write the Euclidean gradient
∇ = (∂y1, ∂y2, ∂y3) in the coordinates x = (x1, x ′) as ∇ = γ(x)∇x= ν(x′) ∂ ∂x1 + 3 X k=2 γ(x)ζk ∂ ∂xk ,
where γ is a smooth matrix-valued function such that γ(x)ζ1 = ν(x′), γ(x)ζk = (α1,k, α2,k, α3,k),
k = 2, 3. Notice that the above identities can be rewritten in the form
(2.6) hν(x′), γ(x)ζki = 0, k = 2, 3.
Let (ξ1, ξ′), ξ′ = (ξ2, ξ3), be the dual variable of (x1, x′). Then the symbol of the operator
−i∇|x1=0 in the coordinates (x, ξ) takes the form ξ1ν(x
′) + β(x′, ξ′), where β(x′, ξ′) = 3 X k=2 ξkγ(0, x′)ζk.
Thus we get that the principal symbol of −∆|x1=0 is equal to
ξ12+hβ(x′, ξ′), β(x′, ξ′)i.
This implies that the principal symbol, r0(x′, ξ′), of the positive Laplace-Beltrami operator on
Γ is equal to
hβ(x′, ξ′), β(x′, ξ′)i. Note also that (2.6) implies the identity
(2.7) hν(x′), β(x′, ξ′)i = 0
for all (x′, ξ′).
In what follows in this section we will solve the linear system
(2.8) ψ0× a − zµ0b = a♯, ψ0× b + zε0a = b♯, ν× a = g,
where ψ0 = ρν− β and hg, νi = 0. To this end, we rewrite it in the form (2.9) β× a + zµ0b = ρg− a♯, ρν× b − β × b + zε0a = b♯, ν× a = g.
Using the identity −β × (β × a) = hβ, βia − hβ, aiβ, we obtain zρµ0ν× b = zµ0β× b − z2ε0µ0a + zµ0b♯
=−β × (β × a) − z2ε0µ0a + β× (ρg − a♯) + zµ0b♯
= (hβ, βi − z2ε0µ0)a− hβ, aiβ + β × (ρg − a♯) + zµ0b♯
= (r0− z2ε0µ0)a− hβ, aiβ + β × (ρg − a♯) + zµ0b♯
=−ρ2a− hβ, aiβ + β × (ρg − a♯) + zµ0b♯.
Taking the scalar product of this identity with ν and using that hν, βi = 0 and hν, ν × bi = 0, we get
hν, ai = ρ−1hν, β × gi − ρ−2hβ × a♯, νi + zµ0ρ−2hb♯, νi.
On the other hand, at= a− hν, aiν satisfies ν × at= ν× a = g. Hence,
ν× g = ν × (ν × at) =−hν, νiat+hν, atiν = −at.
Thus we find
a =−ν × g + ρ−1hν, β × giν − ρ−2hβ × a♯, νiν + zµ0ρ−2hb♯, νiν,
zµ0b = ρg + β× (ν × g) − ρ−1hν, β × giβ × ν
−a♯+ ρ−2hβ × a♯, νiβ × ν − zµ0ρ−2hb♯, νiβ × ν,
zµ0ν× b = −ρa + β × g + ρ−1hβ, ν × giβ − ρ−1β× a♯+ zρ−1µ0b♯
= ρν× g + β × g − hν, β × giν + ρ−1hβ, ν × giβ −ρ−1β× a♯+ ρ−1hβ × a♯, νiν + zρ−1µ0b♯− zρ−1µ0hb♯, νiν.
Since hν, gi = 0 and hν, βi = 0, we have
β× g − hν, β × giν = 0. Thus we obtain
zµ0ν× b = ρν × g + ρ−1hβ, ν × giβ
3. A priori estimates
Let ef ∈ H1t and let the functions U1, U2 ∈ L2(Ω; C3) be such that div U1, div U2 ∈ L2(Ω),
u1 :=hν, U1|Γi ∈ L2(Γ). In this section we will prove a priori estimates for the restrictions on
the boundary of the solutions E and H to the Maxwell equation
(3.1) h∇ × E = izµ(x)H + U1 in Ω, h∇ × H = −izε(x)E + U2 in Ω, ν× E = ef on Γ.
Since h∇, ∇ × Ei = 0, the solutions to (3.1) must satisfy the equation (3.2) ( h∇, Ei = (izε)−1h∇, U 2i − ε−1h∇ε, Ei in Ω, h∇, Hi = −(izµ)−1h∇, U 1i − µ−1h∇µ, Hi in Ω.
To simplify the notations, in what follows we will denote by k · k (resp. k · k0) the norm on
L2(Ω; C3) (resp. L2(Γ; C3)) or on L2(Ω) (resp. L2(Γ)). We also set Y = (E, H), U = (U 1, U2),
and define the norms kY k, kUk and kdiv Uk by
kY k2 =kEk2+kHk2, kUk2=kU1k2+kU2k2, kdiv Uk2 =kdiv U1k2+kdiv U2k2.
By the Gauss divergence theorem we have the identity (3.3) Z ΩhE, ∇ × Hi − Z ΩhH, ∇ × Ei = Z ΓhH × E, νi.
We will use (3.3) to prove the following
Theorem 3.1. Let θ > 0 and 0 < h ≪ 1. Suppose that E and H satisfy equation (3.1) with U1= U2= 0. Then the functions f = E|Γ, g = H|Γ satisfy the estimate
(3.4) kfkH0+kgkH0 .θ
−1k efk H1.
Suppose that E and H satisfy equation (3.1) with ef = 0. Then the functions f = E|Γ, g = H|Γ
satisfy the estimate
(3.5) kfkH0+kgkH0 .ku1k0+ h
−1/2θ−1kUk + h1/2kdiv Uk.
Proof. We decompose the vector-valued functions f and g as f = ft+ fn, g = gt+ gn, where
fn=hν, fiν, gn =hν, giν. Clearly, we have the idenities hft, fni = hgt, gni = 0 and ν ×f = ν ×ft,
ν× g = ν × gt, ft=−ν × (ν × f), gt=−ν × (ν × g). Applying (3.3) to the solutions of equation
(3.1) leads to the idenity iz Z Ω ε|E|2− iz Z Ω µ|H|2 = Z ΩhH, U 1i − Z ΩhE, U 2i + h Z Γhgt× f t, νi.
Taking the real part yields the estimate
(3.6) kY k2.θ−2kUk2+ hθ−1kgtk0kftk0.
By equation (3.2) we also have
(3.7) |h∇, Ei| + |h∇, Hi| . |div U| + |Y |.
Restricting the first equation of (3.1) on Γ and taking the scalar product with ν leads to the estimate
In the normal coordinates (x1, x′), x′ ∈ Γ, the gradient takes the form ∇ = γeν∂x1+ γ e∇x′, where
e
ν = (1, 0, 0) and e∇x′ = (0,∇x′). So, we have
∇|x1=0= γ0ν∂e x1 + γ0∇ex′ = ν∂x1 + γ0∇ex′, γ0(x
′) = γ(0, x′).
Hence
hν, h∇ × Ei|Γ = hhν, ν × ∂x1E|x1=0i + hν, hγ0∇ex′ × E|x1=0i
=hν, hγ0∇ex′× fi = hν, hγ0∇ex′ × fti + hhν, γ0∇ex′ × fni.
On the other hand,
hν, γ0∇ex′ × fni = hν, fihν, γ0∇ex′× νi + hν, γ0∇ex′(hν, fi) × νi
=hν, fihν, γ0∇ex′× νi.
Therefore (3.8) gives
(3.9) kgnk0 .k efkH1+ku1k0+ hkfk0.
We will now bound the norms of fnand gt. Let the function φ0 ∈ C0∞(R) be such that φ0(σ) = 1
for|σ| ≤ 1, φ0(σ) = 0 for|σ| ≥ 2, and set φ(σ) = φ0(σ/δ), where 0 < δ≪ 1. Then the functions
Y♭:= (E♭, H♭) = (φ(x
1)E, φ(x1)H) satisfy equation
(3.10) ( h(γeν∂x1+ γ e∇x′)× E ♭= izµH♭+ U♭ 1 in Ω, h(γeν∂x1+ γ e∇x′)× H ♭=−izεE♭+ U♭ 2 in Ω, where U♭:= (U♭
1, U2♭) satisfy kU♭k0.kUk0+ hkY k0. By (3.7) the functions
p =hγeν, ∂x1E ♭i + hγ e∇ x′, E♭i, q = hγeν, ∂x 1H ♭i + hγ e∇ x′, H♭i, satisfy (3.11) |p| + |q| . |div U| + |Y |.
Denote by h·, ·i0 the scalar product in L2(Γ; C3) or in L2(Γ), that is,
ha, bi0= Z Γha, bi if a, b ∈ L 2(Γ; C3), ha, bi0 = Z Γ ab if a, b∈ L2(Γ). Introduce the functions
F1(x1) = γeν× E♭ 2 0− hγeν, E♭i 2 0, F2(x1) = γeν× H♭ 2 0− hγeν, H♭i 2 0.
Since ν = γ0eν = γeν|x1=0, we have
F1(0) =kftk20− kfnk20, F2(0) =kgtk20− kgnk20.
Using equation (3.10) we will calculate the first derivatives Fj′(x1) = dFdxj1. In view of (3.11), we
get F1′(x1) = 2Re D γeν× ∂x1E ♭ , γeν× E♭E 0+ 2Re D γ′eν× E♭, γeν× E♭E 0 −2Re Dhγeν, ∂x1E ♭ i, hγeν, E♭iE 0− 2Re D hγ′ν, Ee ♭i, hγeν, E♭iE 0 =−2Re Dγ e∇x′ × E♭, γeν× E♭ E 0+ 2h −1Re D(izµH♭+ U♭ 1), γeν× E♭ E 0
+2Re Dhγ e∇x′, E♭i, hγeν, E♭i E 0− 2Re D p,hγeν, E♭iE 0+O kE♭k20 =−2ReDγ e∇x′× E♭, γeν× E♭ E 0+ 2Re D hγ e∇x′, E♭i, hγeν, E♭i E 0+R
with a remainder term R satisfying the estimate
|R| . h−1kY k20+ h−1kUk02+kEk0kdiv U1k0.
Clearly, we have a similar expression for F2′(x1) as well. Let us see now that
(3.12) Re Dγ e∇x′× E♭, γeν× E♭ E 0− Re D hγ e∇x′, E♭i, hγeν, E♭i E 0 =O kE♭k20 .
It suffices to check (3.12) at a symbol level. Let eξ′ = (0, ξ′) denote the symbol of −i e∇x′. We
have the identity D
γ eξ′× E♭, γeν× E♭E=Dγ eξ′, γeνE DE♭, E♭E−DE♭, γeνE Dγ eξ′, E♭E=−DE♭, γeνE Dγ eξ′, E♭E where we have used that Dγ eξ′, γeνE= 0 (see (2.6)). Hence
ImDγ eξ′× E♭, γeν× E♭E− Im Dhγ eξ′, E♭i, hγeν, E♭iE= 0 which clearly implies (3.12). Thus we conclude
(3.13) F1′(x1) + F′ 2(x1) . h−1kY k2 0+ h−1kUk20+ hkdiv Uk20. Since Fj(0) =− Z 2δ 0 Fj′(x1)dx1, we deduce from (3.13), (3.14) |F1(0)| + |F2(0)| . h−1kY k2+ h−1kUk2+ hkdiv Uk2. By (3.6) and (3.14), kfnk20+kgtk02.kftk20+kgnk20+ θ−1kftk0kgtk0+ h−1θ−2kUk2+ hkdiv Uk2, which implies (3.15) kfnk20+kgtk20 .θ−2kftk20+kgnk20+ h−1θ−2kUk2+ hkdiv Uk2.
Clearly, the estimates (3.4) and (3.5) follow from (3.9) and (3.15) by taking h small enough. ✷ 4. Parametrix construction
We keep the notations from the previous sections and will suppose that θ≥ h2/5−ǫ, 0 < ǫ≪ 1.
Let (x1, x′) be the local normal geodesic coordinates introduced in Section 2. Clearly, it suffices
to build the parametrix locally. Then the global parametrix is obtained by using a suitable partition of the unity on Γ and summing up the corresponding local parametrices. We will be looking for a local parametrix of the solution to equation (1.1) in the form
e E = (2πh)−2 Z Z ehi(hy ′ ,ξ′ i+ϕ(x,ξ′ ,z))χ(x, ξ′)a(x, y′, ξ′, z, h)dξ′dy′, e H = (2πh)−2 Z Z ehi(hy ′ ,ξ′ i+ϕ(x,ξ′ ,z))χ(x, ξ′)b(x, y′, ξ′, z, h)dξ′dy′, where χ = φ0(x1/δ)φ0(x1/|ρ|3δ),
the function φ0 being as in the previous section and 0 < δ ≪ 1 is a parameter independent
of h and θ, which is fixed in Lemma 4.1. We require that eE satisfies the boundary condition ν× eE = f on x1= 0, where f ∈ Ht1. The phase function is of the form
ϕ =
N −1X k=0
xk1ϕk, ϕ0 =−hx′, ξ′i, ϕ1 = ρ,
where N ≫ 1 is an arbitrary integer and the functions ϕk, k ≥ 2, are determined from the
eikonal equation (4.5). The amplitudes are of the form a = N −1X j=0 hjaj, b = N −1X j=0 hjbj.
In what follows we will determine the functions aj and bj in terms of f so that ( eE, eH) satisfy
the Maxwell equation modulo an error term. We have
e−iϕ/hh∇ × (eiϕ/ha)− izµeiϕ/hb= i(γ∇xϕ)× a − izµb + h(γ∇x)× a
=
N −1X j=0
hj(i(γ∇xϕ)× aj− izµbj+ (γ∇x)× aj−1) + hN(γ∇x)× aN −1,
e−iϕ/hh∇ × (eiϕ/hb) + izεeiϕ/ha= i(γ∇xϕ)× b + izεa + h(γ∇x)× b
=
N −1X j=0
hj(i(γ∇xϕ)× bj+ izεaj+ (γ∇x)× bj−1) + hN(γ∇x)× bN −1,
where a−1 = b−1 = 0. We let now the functions aj and bj satisfy the equations
(4.1) (γ∇xϕ)× a0− zµb0 = xN1 Ψ0, (γ∇xϕ)× b0+ zεa0 = xN1 Ψe0, ν× a0= g on x1 = 0, where ν = ν(x′) = (ν
1(x′), ν2(x′), ν3(x′)) is the unit normal vector at x′ ∈ Γ,
g =−ν(x′)× (ν(y′)× f(y′)) = f (y′)− (ν(x′)− ν(y′))× (ν(y′)× f(y′)), and (4.2) (γ∇xϕ)× aj − zµbj = i(γ∇x)× aj−1+ xN1 Ψj, (γ∇xϕ)× bj+ zεaj = i(γ∇x)× bj−1+ xN1 Ψej, ν× aj = 0 on x1 = 0,
for 1≤ j ≤ N − 1. We will be looking for solutions of the form aj = N −1X k=0 xk1aj,k, bj = N −1X k=0 xk1bj,k.
Let us expand the functions µ, ε and γ as µ(x) =
N −1X k=0
ε(x) = N −1X k=0 xk1εk(x′) + xN1 E(x), γ(x) = N −1X k=0 xk1γk(x′) + xN1 Θ(x). Observe that ∇xϕ = N −1X k=0 xk1ek with e0 = (ρ,−ξ2,−ξ3), ek = (ϕk+1, ∂x2ϕk, ∂x3ϕk), k≥ 1. Hence γ∇xϕ = N −1X k=0 xk1ψk+ xN1 Θ,e where ψ0 = γ0e0= ρν− β, ψk= k X ℓ=0 γℓek−ℓ, k≥ 1, e Θ = 2N −2X k=N xk−N1 ψk+ Θ(∇xϕ).
We will first solve equation (4.1). We let the functions a0,0 and b0,0 satisfy the equation
(4.3) ψ0× a0,0− zµ0b0,0 = 0, ψ0× b0,0+ zε0a0,0= 0, ν× a0,0 = g.
The equation (4.3) is solved in Section 2 and we have
a0,0=−ν × g + ρ−1hν, β × giν,
b0,0 = ρ(zµ0)−1g + (zµ0)−1β× (ν × g) − (zµ0)−1ρ−1hν, β × giβ × ν,
(4.4) zµ0ν× b0,0 = ρν× g + ρ−1hβ, ν × giβ.
Next we let ϕ satisfy the eikonal equation mod O(xN 1 ):
(4.5) hγ∇xϕ, γ∇xϕi − z2ε0µ0 = xN1 Φ.
This equation is solved in Section 4 of [8]. The functions ϕk, k ≥ 2, are determined uniquely
and have the following properties (see Lemma 4.1 of [8]): Lemma 4.1. We have
(4.6) ϕk∈ S2,24−3k(|ρ|) + S0,11 (|ρ|), k≥ 1,
(4.7) ∂xk1Φ∈ S2,22−3N −3k(|ρ|) + S0,12 (|ρ|), k≥ 0,
uniformly in z and 0 ≤ x1 ≤ 2δ min{1, |ρ|3}. Moreover, if δ > 0 is small enough, independent
of ρ, we have
(4.8) Im ϕ≥ x1Im ρ/2 for 0≤ x1 ≤ 2δ min{1, |ρ|3}.
Furthermore, there are functions ϕ♭
k ∈ S0,11 , independent of ε and µ, such that
Set e ϕ = N −1X k=1 xk1ϕk.
Using the above lemma we will prove the following
Lemma 4.2. There exists a constant C > 0 such that we have the estimates (4.9) ∂xα′∂β ξ′ ei eϕ/h ≤ ( Cα,βθ−|α|−|β|e−Cx1θ/h on supp η, Cα,β|ξ′|−|β|e−Cx1|ξ ′|/h on supp(1− η),
for 0≤ x1 ≤ 2δ min{1, |ρ|3} and all multi-indices α and β with constants Cα,β > 0 independent
of x1, θ, z and h.
Proof. Let us see that the functions cα,β = e−i eϕ/h∂xα′∂β
ξ′
ei eϕ/h, |α| + |β| ≥ 1, satisfy the bounds
(4.10) ∂xα′′∂β ′ ξ′cα,β . |α|+|β|+|α′ |+|β′ | X j=1 x1 h|ρ| j |ρ|−2(|α|+|β|+|α′ |+|β′ |−j) on supp η, and (4.11) ∂xα′′∂β ′ ξ′cα,β . |α|+|β|+|α′ |+|β′ | X j=1 x1 h j |ξ′|−(|β|+|β′|−j)
on supp(1− η), for all multi-indices α′ and β′. We will proceed by induction in |α| + |β|. Let α1 and β1 be multi-indices such that |α1| + |β1| = 1 and observe that
cα+α1,β+β1 = ∂
α1
x′ ∂ξβ′1cα,β+ ih−1cα,β∂xα′1∂ξβ′1ϕ.e
More generally, we have (4.12) ∂xα′′∂β ′ ξ′cα+α1,β+β1 = ∂α1+α ′ x′ ∂ β1+β′ ξ′ cα,β+ ih−1∂α ′ x′∂β ′ ξ′ cα,β∂xα′1∂ β1 ξ′ ϕe . By Lemma 4.1 we have (4.13) x−11 ϕe∈ S2,21 (|ρ|) + S0,11 (|ρ|)
for 0≤ x1 ≤ 2δ min{1, |ρ|3}. By (4.12) and (4.13), it is easy to see that if (4.10) and (4.11) hold
for cα,β, they hold for cα+α1,β+β1 as well. Using (4.10) together with (4.8) we obtain
ei eϕ/hcα,β . |α|+|β| X j=1 x1 h|ρ| j |ρ|−2(|α|+|β|−j)e−2Cθx1(h|ρ|)−1 . |α|+|β|X j=1 θ−j|ρ|−2(|α|+|β|−j)e−Cx1θ/h .θ−|α|−|β|e−Cx1θ/h. Similarly, by (4.11) we obtain ei eϕ/hcα,β . |α|+|β| X j=1 x1 h j |ξ′|−|β|+je−2Cx1|ξ′|/h.|ξ′|−|β|e−Cx1|ξ′|/h. ✷
We take a0,k= ea0,kν for k ≥ 1, where ea0,k are scalar functions to be determined such that
(4.14) hγ∇xϕ, a0i = xN1 Φ.e
Using that hψ0, a0,0i = 0 we can expand the left-hand side as 2N −2X k=1 xk1 k−1 X ℓ=0 hψℓ, νiea0,k−ℓ+hψk, a0,0i ! + xN1 heΘ, a0i. Therefore, if hψk, a0,0i + k−1 X ℓ=0 hψℓ, νiea0,k−ℓ= 0, 1≤ k ≤ N − 1,
then (4.14) is satisfied with e Φ = 2N −2X k=N xk−N1 k−1 X ℓ=0 hψℓ, νiea0,k−ℓ+heΘ, a0i.
Since hψ0, νi = ρ, we arrive at the relations
(4.15) ea0,k=−ρ−1hψk, a0,0i − ρ−1 k−1
X
ℓ=1
hψℓ, νiea0,k−ℓ
which allow us to find all ea0,k, and hence to find a0. To find b0 we will use the expansion
(γ∇xϕ)× aj = N −1X k=0 xk1ψk× N −1X k=0 xk1aj,k+ xN1 Θe× aj, = N −1X k=0 xk1 k X ℓ=0 ψk−ℓ× aj,ℓ+ xN1 2N −2X k=N xk−N1 k X ℓ=0 ψk−ℓ× aj,ℓ+ xN1 Θe× aj with j = 0. We take (4.16) b0,k= (zµ0)−1 k X ℓ=0 ψk−ℓ× a0,ℓ, 0≤ k ≤ N − 1.
Then the first equation of (4.1) is satisfied with Ψ0 = 2N −2X k=N xk−N1 k X ℓ=0 ψk−ℓ× a0,ℓ+ eΘ× a0.
On the other hand, we have the identity
(γ∇xϕ)× ((γ∇xϕ)× a0) =−hγ∇xϕ, γ∇xϕia0+hγ∇xϕ, a0iγ∇xϕ.
Therefore, in view of (4.5) and (4.14), the second equation of (4.1) is satisfied with e Ψ0 = (zµ)−1 −Φa0+ eΦγ∇xϕ . To solve equation (4.2) we will use the expansion
(γ∇x)× aj = N −1X k=0 xk1(γk∇x)× N −1X k=0 xk1aj,k+ xN1 (Θ∇x)× aj
= N −1X k=0 xk1 k X ℓ=0 (γk−ℓ∇ex′)× aj,ℓ+ (ℓ + 1)γk−ℓνe× aj,ℓ+1 +xN1 2N −2X k=N xk−N1 k X ℓ=0 (γk−ℓ∇ex′)× aj,ℓ+ (ℓ + 1)γk−ℓeν× aj,ℓ+1 + xN1 (Θ∇x)× aj,
where eν = (1, 0, 0) and e∇x′ = (0,∇x′). Clearly, we have similar expansions with aj replaced by
bj. We let the functions aj,k satisfy the equations
ψ0× aj,k− zµ0bj,k =− k−1 X ℓ=0 (ψk−ℓ× aj,ℓ− zµk−ℓbj,ℓ) + k X ℓ=0 i(γk−ℓ∇ex′)× aj−1,ℓ+ (ℓ + 1)γk−ℓνe× aj−1,ℓ+1 =: a♯j,k, ψ0× bj,k+ zε0aj,k =− k−1 X ℓ=0 (ψk−ℓ× bj,ℓ+ zεk−ℓaj,ℓ) + k X ℓ=0 i(γk−ℓ∇ex′)× bj−1,ℓ+ (ℓ + 1)γk−ℓνe× bj−1,ℓ+1 =: b♯j,k,
ν× aj,k = 0, for 1≤ j ≤ N − 1 and 0 ≤ k ≤ N − 1. Then the equation (4.2) is satisfied with
Ψj = 2N −2X k=N xk−N1 k X ℓ=0 (ψk−ℓ× aj,ℓ− zµk−ℓbj,ℓ) + eΘ× aj− zMbj + 2N −2X k=N xk−N1 k X ℓ=0 (γk−ℓ∇ex′)× aj−1,ℓ+ (ℓ + 1)γk−ℓνe× aj−1,ℓ+1 + (Θ∇x)× aj−1, e Ψj = 2N −2X k=N xk−N1 k X ℓ=0 (ψk−ℓ× bj,ℓ+ zεk−ℓaj,ℓ) + eΘ× bj+ zEaj + 2N −2X k=N xk−N1 k X ℓ=0 (γk−ℓ∇ex′)× bj−1,ℓ+ (ℓ + 1)γk−ℓνe× bj−1,ℓ+1 + (Θ∇x)× bj−1,
where a−1,ℓ = b−1,ℓ= 0. The above equations are solved in Section 2 and we have the formulas
aj,k= ρ−2hβ × a♯j,k, νiν + zµ0ρ−2hb♯j,k, νiν,
bj,k = (zµ0)−1a♯j,k− (zµ0)−1ρ−2hβ × a♯j,k, νiβ × ν − ρ−2hb♯j,k, νiβ × ν,
(4.17) zµ0ρν× bj,k = β× a♯j,k− hβ × a ♯
j,k, νiν + zµ0b♯j,k− zµ0hb♯j,k, νiν.
Thus we can express all functions aj,k, bj,k in terms of g. More precisely, they are of the form
aj,k= Aj,k(x′, ξ′) ef (y′), bj,k = Bj,k(x′, ξ′) ef (y′),
where ef (y′) = ν(y′)× f(y′) = ι
ν(y′)f (y′), ιν being a 3× 3 matrix, and Aj,k, Bj,k are smooth
matrix-valued functions whose main properties are given in Lemma 4.3 below. In what follows, given a vector-valued function a of the form A(x′, ξ′) ef (y′), we will write a∈ Sℓk1,ℓ2f if all entriese of A belong to Sℓk1,ℓ2.
Lemma 4.3. We have (4.18) Aj,k ∈ S2,2−1−3k−5j(|ρ|) + S −j 0,1(|ρ|), j≥ 0, k ≥ 0, (4.19) Bj,k∈ S2,2−1−3k−5j(|ρ|) + S1−j0,1 (|ρ|), j ≥ 0, k ≥ 0, (4.20) ιν(x′)Bj,k ∈ S2,2−3k−5j(|ρ|) + S 1−j 0,1 (|ρ|), j ≥ 1, k ≥ 0, (4.21) ∂xk1Ψj ∈ S−1−3(N +k)−5j2,2 (|ρ|) ef + S0,11−j(|ρ|) ef , j≥ 0, (4.22) ∂xk1Ψej ∈ S −1−3(N +k)−5j 2,2 (|ρ|) ef + S 2−j 0,1 (|ρ|) ef , j≥ 0,
uniformly in z and 0≤ x1≤ 2δ min{1, |ρ|3}.
Proof. By Lemma 4.1,
(4.23) ψk∈ S2,21−3k(|ρ|) + S0,11 (|ρ|), Θe ∈ S1−3N2,2 (|ρ|) + S0,11 (|ρ|).
It is easy to see from (4.15) and (4.16) by induction in k that (4.23) implies (4.18) and (4.19) for j = 0 and all k ≥ 0. To prove the assertion for all j ≥ 1 and k ≥ 0 we will proceed by induction in j + k. Suppose it is fulfilled for all 0 ≤ j ≤ J, k ≥ 0, as well as for j = J + 1 and k ≤ K, where J ≥ 0, K ≥ −1 are integers. This implies
(4.24) a♯J+1,K+1∈ S2,2−7−3K−5J(|ρ|) ef + S0,1−J(|ρ|) ef ,
(4.25) b♯J+1,K+1∈ S2,2−7−3K−5J(|ρ|) ef + S0,11−J(|ρ|) ef .
Recall that a♯j,0 = b♯j,0 = 0. Using (2.3) with k =−2 and the formulas for aj,k and bj,k in terms
of a♯j,k and b♯j,k, we get from (4.24) and (4.25) that (4.18) and (4.19) hold with j = J + 1 and k = K + 1, as desired. It is also clear that (4.20) follows from (4.24) and (4.25)(used with K = k− 1, J = j − 1) and (4.17) together with (2.3) with k = −1. Since the functions Ψj
and eΨj are expressed in terms of Aj,k, Bj,k, ψk and eΘ, one can derive (4.21) and (4.22) from
(4.18),(4.19) and (4.23). One just needs the following simple observation: if a∈ Sℓ1 2,2(|ρ|) + S0,1ℓ2 (|ρ|), then xk1a∈ Sℓ1+3k 2,2 (|ρ|) + Sℓ 2 0,1(|ρ|), k ≥ 0. ✷ Clearly, we have ν× eE|x1=0 = f and
ν× eH|x1=0= ιν(x ′) eH| x1=0 = N −1X j=0 hjOph(ινBj,0) ef = Oph(ινB0,0+ h(1− η)ινB1,0) ef +K1f ,e where K1= hOph(ηινB1,0) + N −1X j=2 hjOph(ινBj,0) .
Lemma 4.4. There exists a matrix-valued function B1,0♭ ∈ S0,10 such that (4.26) (1− η)ινB1,0− B1,0♭ ∈ S0,1−1
and µ0B1,0♭ is independent of ε and µ.
Proof. In view of (4.15) and (4.16) we have
−ia♯1,0 = γ0∇ex′× a0,0+ ν× a0,1= γ0∇ex′× a0,0, −ib♯1,0= γ0∇ex′× b0,0+ ν × b0,1 = γ0∇ex′× b0,0+ (zµ0)−1(ψ1× a0,0+ ψ0× a0,1) = γ0∇ex′× b0,0+ (zµ0)−1(ψ1× a0,0− ρ−1hψ1, a0,0iψ0× ν) = γ0∇ex′ × b0,0+ (zµ0)−1(ψ1× a0,0+ ρ−1hψ1, a0,0iβ × ν). Thus by (4.17) we obtain −izµ0ινB1,0f =e −izµ0ν× b1,0 = ρ−1β× (γ0∇ex′ × a0,0)− ρ−1hβ × (γ0∇ex′× a0,0), νiν +zµ0ρ−1γ0∇ex′× b0,0− zµ0ρ−1hγ0∇ex′ × b0,0, νiν +ρ−1ψ1× a0,0− ρ−1hψ1× a0,0, νiν + ρ−2hψ1, a0,0iβ × ν.
Observe now that
ρ = i√r0(1 +O(r0−1)) = i√r0+O 1 √r 0 as r0 → ∞.
More generally, we have
(1− η)(ρ − i√r0)∈ S0,1−1,
(1− η)ρ−k− (i√r0)−k
∈ S0,1−k−2, k = 1, 2.
Define a♭0,0 and b♭0,0 by replacing in the formulas for a0,0 and b0,0 above the function ρ by i√r0.
Clearly, a♭0,0 and µ0b♭0,0 are independent of ε and µ. Moreover, we have
(1− η)(a0,0− a♭0,0)∈ S0,1−1f ,e (1− η)(b0,0− b♭0,0)∈ S0,10 f .e
Define ψ1♭ ∈ S1
0,1 by replacing in the definition of ψ1 the function ϕ2 by ϕ♭2 and ρ by i√r0. We
also define eB1,0♭ by replacing in the formula for ινB1,0 above the function ρ by i√r0, ψ1 by ψ1♭,
a0,0 and b0,0 by a♭0,0 and b0,0♭ . Set B♭1,0= (1− η) eB1,0♭ . With this choice one can easily check that
the conclusions of the lemma hold. ✷
Clearly, we can write the matrix ιν in the formP3j=1νjIj, where Ij are constant matrices. In
view of (4.4) we have ινB0,0f = νe × b0,0= m(ν× g) = m ef + mιν 3 X j=1 (νj(y′)− νj(x′))Ijfe
where m = (zµ0)−1(ρI + ρ−1B). Set m0= i(zµ0)−1√r0(I − r−10 B). We have
Oph(ινB0,0) ef = Oph(m) ef + 3
X
j=1
= Oph(m) ef + 3 X j=1 [Oph((1− η)m0ινIj), νj] ef + 3 X j=1 [Oph((ηm + (1− η)(m − m0))ινIj), νj] ef = Oph(m + hn) ef + 3 X j=1 ([Oph((1− η)m0ινIj), νj]− Oph(hnj)) ef + 3 X j=1 [Oph((ηm + (1− η)(m − m0))ινIj), νj] ef , where n =P3j=1nj with nj =−i X |α|=1 ∂xα′νj∂ξα′((1− η)m0)ινIj. Thus we obtain (4.27) ν× eH|x1=0 = Oph(m + h em) ef +K ef ,
where we have put em = n + B1,0♭ andK = K1+K2+K3 with
K2 = hOph (1− η)ινB1,0− B1,0♭ , K3= 3 X j=1 ([Oph((1− η)m0ινIj), νj]− Oph(hnj)) + 3 X j=1 [Oph((ηm + (1− η)(m − m0))ινIj), νj] .
Furthermore, it is easy to see that
∇ × (χa) = χ∇ × a + eχa,
where eχ is a smooth matrix-valued function, which is a linear combinations of ∂xjχ. Therefore
e χ is supported in δ min{1, |ρ|3} ≤ x 1 ≤ 2δ min{1, |ρ|3}. We have h∇ × eE− izµφ eH = (2πh)−2 Z Z ehi(hy ′ ,ξ′ i+ϕ)V 1(x, y′, ξ′, h, z)dξ′dy′ =: U1, h∇ × eH + izεφ eE = (2πh)−2 Z Z ehi(hy ′ ,ξ′ i+ϕ)V 2(x, y′, ξ′, h, z)dξ′dy′=: U2, where V1 = heχa + hNχ(γ∇x)× aN −1+ xN1 N −1X j=0 hjχΨj, V2 = heχb + hNχ(γ∇x)× bN −1+ xN1 N −1X j=0 hjχ eΨj.
Let α be a multi-index such that |α| ≤ 1. Then we can write ((h∂x)αUj)(x1,·) = Oph
where Vj(0) = Vj and
Vj(α)= i∂xαϕVj + (h∂x)αVj
if |α| = 1. Since (E − eE, H− eH) satisfy equation (3.1) with ef = 0, by (3.5) together with (4.27) we get the estimate
kN (λ)f − Oph(m + h em)(ν× f)kH0
(4.28) .h−1/2θ−1kUk + h1/2kdiv Uk + ku1k0+kK efk0.
We need now the following
Lemma 4.5. We have the estimates
(4.29) kK efk0 .hθ−5/2kfkH−1,
(4.30) ku1k0+kUk + kdiv Uk . h5ǫN/2−ℓkfkH−1,
with some constant ℓ > 0. Proof. By (4.20),
ηινBj,0ιν ∈ S2,2−5j(|ρ|) ⊂ S −5j/2
1,1 (θ), j≥ 1,
(1− η)ινBj,kιν ∈ S0,11−j(|ρ|) ⊂ S−10,1, j≥ 2.
Therefore Proposition 2.3 yields kK1fek0≤ N −1X j=1 hjkOph(ηινBj,0ιν)fk0+ N −1X j=2 hjkOph((1− η)ινBj,0ιν)fk0 . N −1X j=1 hjθ−5j/2kfkH−1 + N −1X j=2 hjkfkH−1 .hθ −5/2kfk H−1.
Furthemore (4.26) clearly implies K2 =O(h) : H−1 → H0. To bound the norm of K3 we will
use Proposition 2.2 twice – with
a+= (ηm + (1− η)(m − m0))ινIj, θ+= θ, a−= νj, θ−= 1, and with a+= νj, θ+= 1, a−= (ηm + (1− η)(m − m0))ινIj, θ−= θ. Since (ηm + (1− η)(m − m0))ινIj ∈ S2,2−1(|ρ|) + S0,1−1(|ρ|) ⊂ S −1/2 1,1 (θ) + S0,1−1, by Proposition 2.2, k[Oph((ηm + (1− η)(m − m0))ινIj), νj]kH−1→H0 .hθ−3/2.
On the other hand, the standard pseudodifferential calculas gives that, modO(h∞), the operator
[Oph((1− η)m0ινIj), νj] is an h− ΨDO with a principal symbol hnj, nj ∈ S00,1 being as above.
This implies that
[Oph((1− η)m0ινIj), νj]− Oph(hnj)
is an h− ΨDO with a symbol h2ω, with ω∈ S−1
0,1. Hence
k[Oph((1− η)m0ινIj), νj]− Oph(hnj)kH−1→H0 .h
2,
which completes the proof of (4.29). Furthermore, since xN1 e−Cx1θ/h
.hNθ−N, xN1 e−Cx1|ξ′|/h
we deduce from Lemma 4.2 that
(4.31) h−NxN1 ei eϕ/h∈ S1,1−N(θ) + S0,1−N
uniformly in x1 and h. On supp eχ we have the bounds
e−Cx1θ/h≤ e−Cδ|ρ|3θ/h≤ e− eCθ5/2/h.hNθ−5N/2,
e−Cx1|ξ′|/h
≤ e−Cδ|ξ′|/h.hN|ξ′|−N. Therefore, by Lemma 4.2 we have
(4.32) h−Nχee i eϕ/h ∈ S1,1−5N/2(θ) + S0,1−N.
Notice that hjθ−5j/2≤ 1 for j ≥ 1 as long as θ ≥ h2/5−ǫ. Taking this into account one can easily
check that (4.31) and (4.32) together with Lemma 4.3 imply (4.33) h−Nei eϕ/hVj(α)∈ S−5N/2−ℓα
1,1 (θ) ef + S0,1−N +eℓαfe
with some ℓα, eℓα > 0 independent of N , whose exact values are not important in the analysis
that follows. Let N > eℓα+ 1. By (4.33) and Proposition 2.3 we get
(4.34) k((h∂x)αUj)(x1,·)kH0 .h Nθ−5N/2−ℓα ef H−1 .h5ǫN/2−2ℓα/5kfk H−1
as long as θ ≥ h2/5−ǫ, uniformly in x1. Observe also that
h−NV1|x1=0 = (γ∇x)× aN −1|x1=0= (γ0∇ex′)× aN −1,0+ ν× aN −1,1
= (γ0∇ex′)× (AN −1,0f ) + νe × (AN −1,1f ) =: ω ee f .
By Lemma 4.3,
ω ∈ S1,1−5N/2(θ) + S0,1−N +1,
which together with Proposition 2.3 yield
Oph(ω) =Oθ−5N/2:H−1→ H0. Since U1|x1=0 = h NOp h(ω) ef , we get (4.35) kU1|x1=0kH0 .h Nθ−5N/2 ef H−1 .h5ǫN/2kfkH−1.
Clearly, (4.30) follows from (4.34) and (4.35). ✷
Taking N big enough depending on ǫ, it is easy to see that the estimate (1.2) follows from (4.28) and Lemma 4.5.
5. Electromagnetic transmission eigenvalues
A complex number λ is said to be an electromagnetic transmission eigenvalue if the following boundary-value problem has a nontrivial solution:
(5.1) ∇ × E1= iλµ1(x)H1 in Ω, ∇ × H1 =−iλε1(x)E1 in Ω, ∇ × E2= iλµ2(x)H2 in Ω, ∇ × H2 =−iλε2(x)E2 in Ω, ν× (E1− E2) = 0 on Γ, ν× (c1H1− c2H2) = 0 on Γ,
where µj, εj ∈ C∞(Ω), cj ∈ C∞(Γ), j = 1, 2, are scalar-valued strictly positive functions.
The most important question that arrises in the theory of the transmission eigenvalues is to know the conditions on the coefficients under which they form a discreet set on the complex plane. This question has been largely investigated in the context of the acoustic transmission eigenvalues, that is, those associated to the Helmholtz equation. Several sufficient condition have been found that guarantee not only the discreteness, but also Weyl asymptotics for the counting function of the acoustic transmission eigenvalues (see [5], [6], [7]). In particular, it was proved in [6] that the existence of parabolic eigenvalue-free regions implies the Weyl asymptotics. On the other hand, such regions were obtained in [8], [9], [10], [11] and [12] under various conditions, by approximating approprietly the Dirichlet-to-Neumann operator associated to the Helmholtz equation with smooth refraction index. It was proved in [11] that, under quite general conditions on the coefficients on the boundary, all transmission eigenvalues are located in a strip |Im λ| ≤ C, which turns out to be optimal. The situation, however, is very different as far as the electromagnetic transmission eigenvalues are concerned. In this context there are few results and they are mainly concerned with the question of discreteness (e.g see [1], [4]). The most general one is in [1], where the authors considered the case c1 ≡ c2 ≡ 1 and proved the discreteness
under the condition
(5.2) ε1 6= ε2, µ1 6= µ2, ε1 µ1 6= ε2 µ2 on Γ.
They also proved that given any γ > 0 there is Cγ > 0 such that there are no electromagnetic
transmission eigenvalues in the region |Im λ| ≥ γ|Re λ|, |λ| ≥ Cγ.
Our goal is to obtain a parabolic eigenvalue-free region under the condition
(5.3) c1
µ1
= c2 µ2
, ε1µ16= ε2µ2 on Γ.
Indeed, using Theorem 1.1 we will prove the following
Theorem 5.1. Under the condition (5.3), there exists a constant C > 0 such that there are no electromagnetic transmission eigenvalues in the region
(5.4) |Im λ| ≥ C(|Re λ| + 1)57.
Proof. Denote by Nj(λ), j = 1, 2, the operator introduced in Section 1 corresponding to
(εj, µj), and set T (λ) = c1N1(λ)− c2N2(λ). We define the functions ρj by replacing in the
definition of ρ the function εµ|Γ by εjµj|Γ. Set f = ν× E1 = ν × E2 ∈ Ht1. Then λ is an
electromagnetic transmission eigenvalue if f 6= 0 and T (λ)f = 0. Therefore, to get the free region (5.4) we need to show that the operator T (λ) is invertible there. By Theorem 1.1 we have
(5.5) kOph(T )(ν× f)kH0 =kT (λ)f − Oph(T )(ν× f)kH0 .hθ
−5/2kfk H−1
for θ≥ h2/5−ǫ, where T = c1ρ1 µ1 I + c1 ρ1µ1B − c2ρ2 µ2 I− c2 ρ2µ2B = c1 µ1 (ρ1− ρ2) I− (ρ1ρ2)−1B. Since (ρ1− ρ2)(ρ1+ ρ2) = ρ21− ρ22 = z2ε1µ1− z2ε2µ2, we have T = w eT , where w = z2c1 µ1 (ε1µ1− ε2µ2)6= 0, e T = (ρ1+ ρ2)−1 I − (ρ1ρ2)−1B.
Using that B2 = r0B, one can easily check the identity
(5.6) (I + (ρ1ρ2− r0)−1B) I − (ρ1ρ2)−1B= I.
Lemma 5.2. For all integers k≥ 1 and all multi-indices α and β we have the estimates (5.7) ∂xα′∂β ξ′(r0− ρ1ρ2)−k ≤ ( Ck,α,βθ−k−|α|−|β| on supp η, Ck,α,β|ξ′|−2k−|β| on supp(1− η), (5.8) ∂xα′∂β ξ′(ρ1+ ρ2)−k ≤ ( Ck,α,βθ−|α|−|β| on supp η, Ck,α,β|ξ′|−k−|β| on supp(1− η), (5.9) ∂xα′∂β ξ′ (ρ1ρ2)−1(ρ1+ ρ2)−1 ≤ ( Cα,βθ−1/2−|α|−|β| on supp η, Cα,β|ξ′|−3−|β| on supp(1− η).
Proof. We will first prove the estimates on supp(1− η). Since ρj = i√r0 1 +O(r0−1)
as r0 → ∞, we have r0− ρ1ρ2 = 2r0 1 +O(r0−1) , ρ1+ ρ2= 2i√r0 1 +O(r0−1) .
Therefore, |r0− ρ1ρ2| ≥ r0 and |ρ1+ ρ2| ≥√r0 on supp(1− η), provided the constant C0 in the
definition of η is taken large enough (what we can do without loss of generality). To prove (5.7) for all α and β we will proceed by induction in|α| + |β|. Suppose that (5.7) holds on supp(1− η) for α, β such that |α| + |β| ≤ K and all integers k ≥ 1. We will show that it holds for all α, β such that |α| + |β| = K + 1 and all integers k ≥ 1. Let α1 and β1 be multi-indices such that
|α1| + |β1| = 1. We have ∂α1 x′ ∂ β1 ξ′ (r0− ρ1ρ2)−k=−k(r0− ρ1ρ2)−k−1∂xα′1∂ β1 ξ′ (r0− ρ1ρ2)
and more generally, if α, β are such that |α| + |β| = K, we have ∂α+α1 x′ ∂ξβ+β′ 1(r0− ρ1ρ2)−k=−k∂xα′∂β ξ′ (r0− ρ1ρ2)−k−1∂xα′1∂ξβ′1(r0− ρ1ρ2) . Recall now that r0is a homogeneous polynomial of order two in ξ′. Hence ∂xα′∂β
ξ′r0=O(hξ′i2−|β|).
Furthermore, by (2.3) we have ∂xα′∂ξβ′(ρ1ρ2) =O(hξ′i2−|β|) on supp(1− η). Uisng this, one can
easily deduce from the above identity that (5.7) holds on supp(1− η) for α + α1, β + β1 and
all integers k ≥ 1. Clearly, the same argument also works for (5.8). The estimate (5.9) on supp(1− η) follows from (5.8).
To prove (5.7) on supp η, we will use the identity
which we rewrite in the form
(r0− ρ1ρ2)−k = w−k1 (r0+ ρ1ρ2)k(w2r0− z2)−k.
By induction, in the same way as above, one can easily prove the estimates ∂xα′∂β ξ′(w2r0− z2)−k ≤ Ck,α,βθ−k−|α|−|β|
on supp η. On the other hand, by (2.3) we have ∂α
x′∂ξβ′(r0 + ρ1ρ2)k = O(θ−|α|−|β|) on supp η.
Therefore (5.7) on supp η follows from the above estimates. The estimates (5.8) and (5.9) on supp η can be obtained in the same way, using (2.3) and the identities
(ρ1+ ρ2)−k= w−k3 (ρ1− ρ2)k, w3 := z2(ε1µ1− ε2µ2),
(ρ1ρ2)−1(ρ1+ ρ2)−1 = w−13 ρ−12 − ρ−11
.
✷ We rewrite the identity (5.6) in the form
(5.10) T1T =e hξ′i−1I,
where
T1 =hξ′i−1(ρ1+ ρ2)(I + (ρ1ρ2− r0)−1B).
It follows from Lemma 5.2 together with (2.3) that
T1 ∈ S1,1−1(θ) + S0,10 ⊂ θ−1S1/2−ǫ0 ,
e
T ∈ S1,1−1/2(θ) + S0,1−1⊂ θ−1/2S1/2−ǫ−1 , as long as θ ≥ h1/2−ǫ. Therefore, by Proposition 2.3 we get
(5.11) kOph(T1)kH0→H0 .θ
−1,
while Proposition 2.2 yields
(5.12) kOph(T1T )e − Oph(T1)Oph( eT )kH−1→H0 .hθ
−7/2.
Combining (5.10), (5.11) and (5.12) leads to kOph(hξ′i−1) efkH0 .hθ −7/2k efk H−1+kOph(T1)Oph( eT ) efkH0 (5.13) .hθ−7/2k efkH−1+ θ −1kOp h( eT ) efkH0
where ef = ν× f. Since the norms kOph(hξ′i−1) efkH0, k efkH−1 and kfkH−1 are equivalent, by
(5.5) and (5.13) we obtain
(5.14) kfkH−1 .hθ
−7/2kfk H−1.
Thus, if hθ−7/2 ≪ 1 we deduce from (5.14) that f = 0. In other words, the region hθ−7/2≪ 1
is free of transmission eigenvalues. It is easy to see that this region is equivalent to (5.4) on the
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