The Asymptotic of Transmission Eigenvalues for a Domain with a Thin Coating

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Domain with a Thin Coating

Hanen Boujlida, Houssem Haddar, Moez Khenissi

To cite this version:

Hanen Boujlida, Houssem Haddar, Moez Khenissi. The Asymptotic of Transmission Eigenvalues for

a Domain with a Thin Coating. SIAM Journal on Applied Mathematics, Society for Industrial and

Applied Mathematics, 2018, 78 (5), pp.2348-2369. �hal-01646003v2�

WITH A THIN COATING

2

H. BOUJLIDA ^∗, H. HADDAR^†, AND M. KHENISSI ^‡ 3

Abstract. We consider the transmission eigenvalue problem for a medium surrounded by a thin layer of 4

inhomogeneous material with different refractive index. We derive explicit asymptotic expansion for the transmission 5

eigenvalues with respect to the thickness of the thin layer. We prove error estimate for the asymptotic expansion 6

up to order 1. This expansion can be used to obtain explicit expressions for constant index of refraction.

7

Key words. transmission eigenvalues, asymptotic expansions, thin layers, inverse scattering problems 8

AMS subject classifications. 35P30, 35P25 9

1. Introduction. This work is a contribution to the study of transmission eigenvalues [11, 4,

10

6] and their relation to the shape and material properties of scatterers. These special frequencies

11

are associated with the existence of an incident field that does not scatter. They can be equiva-

12

lently defined as the eigenvalues of a system of two coupled partial differential equations posed on

13

the inclusion domain. One of these equations refers to the equation satisfied by the total field and

14

the other one is satisfied by the incident field. The two equations are coupled on the boundary

15

by imposing that the Cauchy data coincide. This eigenvalue problem can then be formulated as

16

an eigenvalue problem for a non-selfadjoint compact operator. Although non intuitive, it can be

17

shown that this problem admits an infinite discrete set of real eigenvalues without finite accumu-

18

lation points [7, 26]. These special frequencies can be identified from far field data as proved in

19

[5, 19, 4]. Since they carry information on the material properties of the scatterer, transmission

20

eigenvalues would then be of interest for the inverse problem of retrieving qualitative information

21

on the material properties from measured multistatic data [14, 15]. In this perspective, it appears

22

important to study the dependence of these eigenfrequencies with respect to the material prop-

23

erties and the geometry. Several works in the literature have addressed this issue by considering

24

asymptotic regimes and quantifying the dependence of the first leading terms in the asymptotic

25

expansion of the transmission eigenvalue with respect to the small parameter [10, 8, 21, 16]. We

26

here consider the case of a scatterer made of a thin coating which corresponds to frequently en-

27

countered configurations in the stealth technology for instance. The goal is to characterize the

28

dependence of the first order term on the material properties and the thickness of the coating. A

29

first work on this topic was done in [10] where the case of coated perfect scatterer is considered.

30

One proves in particular for the latter case that the first order term depends only on the thickness.

31

We here address the more complicated configuration of a coated penetrable media. The analysis

32

indicates that the first order asymptotic resembles to the shape derivative for the buckling plate

33

equation [17] and contain non trivial dependence on the material properties. More importantly,

34

this expansion allows us to obtain explicit (approximate) expressions for the thin layer index of

35

refraction in terms of the thickness of the layer, the transmission eigenvalue for the coated medium

36

that can be extracted from the measurements and the transmission eigenvalues and eigenvectors

37

for the coated free medium that can be evaluated numerically. This indeed can be useful for the

38

solution of the inverse problem.

39

Although the formal derivation follows the systematic procedure using the classical scaled

40

expansion method (as in [3, 2, 13] for instance), the rigorous justification is much more involved.

41

For instance the arguments in [10] are hard to extend to the present case since special uniform

42

estimates have to be obtained for the transmission problem. We restrict ourselves here to the jus-

43

tification of the first two terms in the asymptotic expansion using the abstract theory developed

44

in [23, 21]. We follow the procedure developed in [8] for the case of small obstacles asymptotic.

45

The main technical point in the proof is to obtain the corrector for the main operator, which is

46

∗LAMMDA-ESST Hammam Sousse, University of Sousse,Tunisia (boujlida.hanen12@gmail.com).

†INRIA Saclay Ile de France/CMAP Ecole Polytechnique,France(houssem.haddar@inria.fr).

‡LAMMDA-ESST Hammam Sousse, University of Sousse, Tunisia (moez.khenissi@gmail.com).

1

here the biharmonic operator. Our main result provides explicit expansion for simple transmis-

47

sion eigenvalues and for multiple transmission eigenvalues that are associated with a generalized

48

eigenspace spanned only by eigenvectors.

49

We analyze the problem where the contrast in material properties affect only the lower order

50

term in the Helmholtz equation. We finally indicate that although the problem is considered

51

only in dimension 2, the results of the main theorem (including the expression of the first order

52

asymptotic term) remain true for three dimensions (up to more complicated technicalities in the

53

proof related to differential geometry).

54

The paper is organized as follows. We first introduce the transmission eigenvalues and write

55

them as the eigenvalues of a non selfadjoint operator. We then present the main result of our paper

56

and discuss applications to the inverse problem. We present next the outline of a classical formal

57

procedure to obtain the expression of the asymptotic expansion. We give the expression till the

58

second order term. We explain in particular why the expression of the second order term would

59

have less interest in practice. We then proceed with the main part of the paper that provides

60

explicit expressions and an error estimate for the first two terms in the asymptotic expansion.

61

2. Problem statement and main results. Let Ω ⊂ R

²

be a bounded domain with a

62

smooth boundary Γ. We denote by

63

Ω

⁰

= {x ∈ Ω, d(x, Γ) > }

64

and its boundary

65

Γ

= {x ∈ Ω, d(x, Γ) = } = ∂Ω

⁰

,

66

for > 0 a small enough parameter, where d(x, Γ) denotes the distance of a point x to the boundary

67

Γ. Let Ω = Ω\Ω

⁰

be the layer of thickness around Ω

⁰

(see Figure 1).

Fig. 1.Stretch of the geometry 68

We consider the following transmission eigenvalue problem:

69

(1)



 

 

 

 

∆w

+ k

²

n

(x)w

= 0 in Ω,

∆v

+ k

²

v

= 0 in Ω,

w = v on Γ,

∂w

∂ν = ∂v

∂ν on Γ,

70

where k denotes the unknown eigenfrequency and ν the unitary normal to Γ directed to the

71

interior of Ω. The index of refraction n is defined as follows:

72

n (x) =

( n

₀

(x) in Ω

⁰

, n

1

(x) in Ω ,

73

where n

0

and n

1

are non negative real valued functions ∈ L

^∞

( R

²

) that are independent from .

74

For the sake of simplicity, we assume that the restriction of n

0

and n

1

to Ω are constant functions

75

along the normal coordinate to Γ for sufficiently small. We finally assume that the function

76

1/(1 − n ) is either positive definite or negative definite on Ω. We remark that this assumption

77

also implies that 1/(1 − n

0

) is either positive definite or negative definite on Ω and that

78

(2) 1/|1 − n (x)| ≥ γ > 0 for a.e. x ∈ Ω

79

with γ being independent from (sufficiently small) .

The main goal of this paper is to find the asymptotic expansion of eigenfrequencies k with respect to . Assuming that 1

1 − n

∈ L

^∞

(Ω), the transmission eigenvalue problem (1) can be reformulated as the nonlinear eigenvalue problem for λ

:= k

²

∈ R and u

:= w

− v

∈ H

₀²

(Ω) such that

(∆ + λ

n

) 1 1 − n

(∆ + λ

)u

= 0 in Ω,

which in variational form, after integration by parts, is formulated as finding λ ∈ R and non-trivial

80

function u ∈ H

₀²

(Ω) such that

81

(3)

Z

Ω

1 1 − n

(∆u

+ λ

u

)(∆φ + λ

n

φ)dx = 0, ∀φ ∈ H

₀²

(Ω).

82

The space H

₀²

(Ω) denotes the closure in H

²

(Ω) of the set of regular compactly supported functions

83

in Ω. We shall work with the reformulation of (3) as a linear eigenvalue problem for a non

84

selfadjoint compact operator [4]. First observe that (3) can be written as

85

(4) A u + λ B u + λ

²

C u = 0 in H

₀²

(Ω)

86

where

87

A : H

₀²

(Ω) → H

₀²

(Ω), B : H

₀²

(Ω) → H

₀²

(Ω), C : H

₀²

(Ω) → H

₀²

(Ω)

88

are defined by the Riesz representation theorem as

89

(5) (A u , φ)

_H²

0(Ω)

:=

Z

Ω

1 1 − n

∆u ∆φdx,

90 91

(6) (B u , φ)

_H2

0(Ω)

:=

Z

Ω

1

1 − n

(u ∆φ + n ∆u φ)dx,

92

and

93

(7) (C u , φ)

_H²

0(Ω)

:=

Z

Ω

n

1 − n

u φdx.

94

Note that A : H

₀²

(Ω) → H

₀²

(Ω) is a bounded, self-adjoint and invertible operator (thanks to (2)),

95

B : H

₀²

(Ω) → H

₀²

(Ω) is a bounded, compact and self-adjoint operator and C : H

₀²

(Ω) → H

₀²

(Ω)

96

is a (non negative or non positive) bounded, compact and self-adjoint operator. Observe that

97

since A is invertible, λ 6= 0. In order to avoid distinguishing the cases of 1 − n being positive or

98

negative we shall abusively set C

1

2

≡ −(−C

¹²

) in the case where 1 − n non positive.

99

Setting U

= (u

, λ

C

¹²

u

), the transmission eigenvalue problem (4) can be transformed into

100

the linear eigenvalue problem, τ

∈ R , U

∈ H

₀²

(Ω) × H

₀²

(Ω) such that

101

(8) (T − τ

I)U

= 0, with τ

= 1

λ

,

102

for the compact non-selfadjoint operator T : H

₀²

(Ω) × H

₀²

(Ω) → H

₀²

(Ω) × H

₀²

(Ω) defined by

103

(9) T = −A

⁻¹

B

−A

⁻¹

C

¹²

C

¹²

0

! .

104

We set

105

(10) T

0

= −A

⁻¹₀

B

0

−A

⁻¹₀

C

1 2

0

C

1 2

0

0

!

106

where A

0

, B

0

and C

0

are defined by (5), (6) and (7) respectively for n = n

0

in Ω. We state here

107

the main result of this paper which will be proven in Section 4. In the following a transmission

108

eigenvalue λ

0

is called simple if the corresponding τ

0

= 1/λ

0

has an algebraic multiplicity equal

109

to 1. We refer to Theorem 4.11 for the case where λ

0

has an associated eigenspace formed

110

only by eigenvectors (and therefore an algebraic multiplicity that coincides with the geometrical

111

multiplicity).

112

Theorem 2.1. Assume that n

0

, n

1

∈ C

⁴

(Ω). Let λ

0

∈ R be a simple transmission eigenvalue of (3) with n = n

0

in Ω and let u

0

∈ H

₀²

(Ω) be an associated eigenfunction. This implies in particular that

β

₀

:=

Z

Ω

1 1 − n

0

λ

²₀

n

₀

|u

₀

|

²

− |∆u

₀

|

²

dx 6= 0.

If we suppose in addition that u

0

and A

⁻¹₀

u

0

are in C

⁶

(Ω), then, for sufficiently small > 0, there

113

exists a transmission eigenvalue λ of (3) such that

114

λ = λ

0

+ λ

1

+ O(

³²

)

115

where λ

1

is given by the following expression

116

λ

1

:= λ

0

β

₀

Z

Γ

n

0

− n

1

(1 − n

₀

)

²

|∆u

0

|

²

ds(x).

117

This theorem is an immediate consequence of Theorem 4.8 that is stated and proven in the

118

last section of this paper.

119

The formal calculations in Section 3 show that the formula for λ

1

is generically valid whenever

120

β

0

6= 0. However, we remark that in the case of transmission eigenvalues with multiplicity greater

121

than 1, this is not automatically ensured (See Theorem 4.11 for a rigorous expression of λ

1

that

122

involves all eigenvectors associated with λ

0

).

123

From the practical point of view, this theorem implies in particular that λ

1

gives a measure for

124

the contrast n

0

− n

1

. For instance, if n

1

is constant and n

0

is constant on Γ, one can approximate

125

the value of n

1

using the identity

126

(11) n

1

|

Γ

= n

0

|

Γ

− λ − λ

0

α

0

Z

Ω

1 1 − n

0

λ

0

n

0

|u

0

|

²

− 1 λ

0

|∆u

0

|

²

dx + O(

¹²

)

127

with

α

₀

:=

Z

Γ

|∆u

0

|

²

(1 − n

0

)

²

ds(x).

For the inverse problem where one would like to determine n

₁

from multistatic measurements of

128

scattered waves, the value of λ

can be approximated using sampling methods as in [5, 4] (see

129

also [19] for an alternative approach). The values of λ

₀

and u

₀

can be computed numerically if

130

one has a priori knowledge of n

0

and Ω (see for instance [12, 18, 20] for numerical methods to

131

approximate λ

0

and u

0

). We finally indicate that, although not carefully checked, we conjecture

132

that the expression for λ

1

remains true in three dimensions (corrections due to the curvature of Γ

133

only affect higher order terms).

134

3. Formal asymptotic expansion. In this section, we derive the formal asymptotic ex-

135

pansion for transmission eigenvalues and give explicit formulas for the terms up to order 2. The

136

idea here is to provide a systematic formal way to quickly obtain the explicit expression of λ

1

in

137

Theorem 2.1 and also higher order terms. The latter turn out to have complicated expressions

138

that would be of marginal interest for the solution of the inverse problem mentioned above. This

139

formal stage will also be helpful in establishing the rigorous proof based of Osborn’s theorem [23].

140

It allows one to have an intuition for the expression of the corrector in the asymptotic of the main

141

operator A

.

142

We assume the following expansions for the transmission eigenvalues :

143

(12) λ =

∞

X

j=0

^j

λ

j

,

144

and then follow a classical technique for thin layers asymptotics based on rescaling and asymptotic

145

expansion with respect to the thickness . We shall mainly follow the approach in [10].

146

3.1. Scaling. We assume that the boundary Γ is C

^∞

-smooth, although much less regularity

147

is needed if we restrict ourselves to only few terms in the expansion. The issue of optimal regularity

148

assumptions for Γ is not discussed here. However, one can check that at least a C

²

regularity is

149

needed to get the expression of λ

1

. We parametrize Γ as

150

Γ = {x

Γ

(s), s ∈ [0, L[},

151

with L being the length of Γ and s is the curvilinear abscissa. At the point x

_Γ

(s), the unit tangent

152

vector is τ(s) := dx

Γ

(s)

ds , the curvature κ(s) is defined by:

153

dτ (s)

ds = −κ(s)ν (s) or dν (s)

ds = κ(s)τ(s).

154

Within these notations, the boundary of Ω

⁰

is parametrized as

155

Γ = {x

Γ

(s) + ν(s), s ∈ [0, L[}.

156

This parametrization of the surface Γ is equivalent to the definition of Γ , for > 0 a small enough

157

parameter.

158

For a function u defined in Ω , we consider ˜ u defined on [0, L[×]0, [ by

159

(13) u(s, η) = ˜ u(ϕ(s, η)) where ϕ(s, η) := x

Γ

(s) + ην(s).

160

Then, the gradient and Laplace operators are expressed in the local coordinates as:

161

∇u = 1 (1 + ηκ(s))

∂

∂s τ(s) + ∂

∂η ν (s)

˜ u,

162

(14) ∆u = 1

(1 + ηκ)

∂

∂s 1 (1 + ηκ)

∂

∂s + κ (1 + ηκ)

∂

∂η + ∂

²

∂η

²

˜ u.

163

To make the formal calculations, we need to separate the thin layer and scaled it with respect

164

to the thickness so that the equation are posed on a domain independent from . We therefore

165

rewrite the transmission eigenvalue problem (1) in the following equivalent form

166

(15)



 

 

 

 

 

 

 

 

 

 

∆w

⁺

+ k

²

n

1

w

⁺

= 0 in Ω ,

∆w

⁻

+ k

²

n

₀

w

⁻

= 0 in Ω

⁰

,

∆v

+ k

²

v

= 0 in Ω,

w

⁺

= w

⁻

, ∂w

⁺

∂ν = ∂w

⁻

∂ν on Γ ,

w

⁺

= v on Γ,

∂w

⁺

∂ν = ∂v

∂ν on Γ.

167

We denote by ξ = η

the stretched normal variable inside Ω

and define

168

ϕ : G = [0, L[×]0, 1[ → Ω

(s, ξ) 7→ ϕ (s, ξ) = x

Γ

(s) + ξν(s).

169

Then the expression of the Laplace operator in the scaled layer is:

170

(16) ∆u = 1

(1 + ξκ)

∂

∂s 1 (1 + ξκ)

∂

∂s + κ (1 + ξκ)

∂

∂ξ + 1

²

∂

²

∂ξ

²

ˆ

u =: ∆

s,ξ

u ˆ

171

for ˆ u(s, ξ) := u(ϕ (s, ξ)).

172

The next step is to write the equation for w

⁺

in the scaled domain and solve for the asymptotic

173

expansion of w

⁺

in terms of the boundary values on Γ. These boundary values are given by the

174

asymptotic expansion of v

. More specifically, setting ˆ w

(s, ξ) := w

⁺

(ϕ

(s, ξ)), we have that

175

(17) ∆

s,ξ

w ˆ + λ n

1

w ˆ = 0 in G

176

together with the boundary conditions

177

(18)





 ˆ

w

(s, 0) = v

(x

_Γ

(s)) s ∈ [0, L[, 1

∂ w ˆ

∂ξ (s, 0) = ∂v

∂ν (x

_Γ

(s)) s ∈ [0, L[.

178

We assume that

179

(19) w ˆ (s, ξ) =

∞

X

j=0

^j

w ˆ

j

(s, ξ), (s, ξ) ∈ G and v (x) =

∞

X

j=0

^j

v

j

(x), x ∈ Ω

180

for some functions ˆ w

_j

defined on G and v

_j

defined on Ω that are independent from . Multiplying

181

(17) by

²

(1 + ξκ)

³

and using (12), we obtain

182

5

X

k=0

^k

A

_k

w ˆ

= 0,

183

where (A

k

)

k=0...5

are differential operators of order 2 at maximum with the following expressions

184

for the first fourth terms:

185

A

0

= ∂

²

∂ξ

²

,

186

A

1

=3ξκ ∂

²

∂ξ

²

+ κ ∂

∂ξ ,

187

A

₂

= ∂

²

∂s

²

+ 3ξ

²

κ

²

∂

²

∂ξ

²

+ 2ξκ

²

∂

∂ξ + λ

₀

n

₁

,

188

A

3

=ξ

³

κ

³

∂

²

∂ξ

²

+ ξ

²

κ

³

∂

∂ξ − ξ ∂κ

∂s

∂

∂s + ξκ ∂

²

∂s

²

+ 3λ

0

n

1

ξκ + λ

1

n

1

.

189 190

Inserting the ansatz (19) in (17) and (18) we obtain after equating the terms of same order in

191

and using the convention ˆ w

j

= v

j

= 0 for j < 0,

192

(20)



 

 

 



 

 

 



∂

²

∂ξ

²

w ˆ

j

= −

5

X

k=1

A

k

w ˆ

_j−k

in G, ˆ

w

j

(s, 0) = v

j

(x

Γ

(s)) s ∈ [0, L[,

∂ w ˆ

j

∂ξ (s, 0) = ∂v

_j−1

∂ν (x

Γ

(s)) s ∈ [0, L[.

193

These equations can be solved inductively to get the expressions of ˆ w

j

in terms of the boundary

194

values of v

_l

, l ≤ j. One gets for j = 0, 1, 2 and 3

195

ˆ

w

0

(s, ξ) = v

0

(x

Γ

(s)),

196

ˆ

w

₁

(s, ξ) = ∂v

₀

∂ν (x

_Γ

(s))ξ + v

₁

(x

_Γ

(s)),

197

ˆ

w

2

(s, ξ) = − ξ

²

2

κ ∂w

⁻₀

∂ν (x

Γ

(s)) + ∂

²

w

⁻₀

∂s

²

(x

Γ

(s)) + λ

0

n

1

w

₀⁻

(x

Γ

(s)) + ∂v

1

∂ν (x

Γ

(s)ξ + v

2

(x

Γ

(s)), (21)

198199 200

and

ˆ

w

₃

(s, ξ) = ξ

³

6

− 2κ

²

∂w

₀⁻

∂ν (x

_Γ

(s)) − 3κ ∂

²

w

⁻₀

∂s

²

(x

_Γ

(s)) − κλ

₀

n

₁

w

⁻₀

(x

_Γ

(s)) + λ

₀

n

₁

∂v

0

∂ν (x

_Γ

(s))

201

+ ξ

³

6

∂

³

v

₀

∂s

²

∂ν (x

_Γ

(s)) − κ ˙ ∂w

⁻₀

∂s (x

_Γ

(s))

202

+ ξ

²

2

κ ∂v

₁

∂ν (x

_Γ

(s)) + λ

₀

n

₁

v

₁

(x

_Γ

(s)) + λ

₁

n

₁

w

⁻₀

(x

_Γ

(s)) + ∂v

₂

∂ν (x

_Γ

(s))ξ + v

₃

(x

_Γ

(s)).

(22)

203 204

Now, we also postulate the following expansion for w

⁻

:

205

(23) w

⁻

(x) =

∞

X

j=0

^j

w

_j⁻

(x)

206

with w

⁻_j

: Ω → R are functions independent of . Then (w

⁻_j

, v

j

) solves

207

(24)



 

 



 

 



∆w

⁻_j

+ λ

0

n

0

w

⁻_j

= −

j

X

l=1

λ

l

n

0

w

⁻_j−l

in Ω,

∆v

j

+ λ

0

v

j

= −

j

X

l=1

λ

l

v

_j−l

in Ω.

208

Note that the functions w

⁻_j

are defined in all Ω and not only Ω

⁰

and therefore (23) gives a extension

209

of w

⁻

to all Ω. The continuity conditions at Γ can be written as

210

˜

w

⁻

(s, ) = ˆ w (s, 1) and ∂ w ˜

⁻

∂η (s, ) = 1

∂ w ˆ

∂ξ (s, 1)

211

where ˜ w

⁻

is defined from w

⁻

using the local change of variables (13) in a neighborhood of Γ.

212

Using Taylor’s expansion (up to the second order, which is sufficient to compute the first three

213

terms in the asymptotic expansion) we get

214

(25) w ˜

⁻

(s, ) = ˜ w

⁻

(s, 0) + ∂ w ˜

⁻

∂η (s, 0) +

²

2

∂

²

w ˜

⁻

∂η

²

(s, 0) + o(

²

) = ˆ w (s, 1)

215 216

and

(26) ∂ w ˜

⁻

∂η (s, ) = ∂ w ˜

⁻

∂η (s, 0) + ∂

²

w ˜

⁻

∂η

²

(s, 0) +

²

2

∂

³

w ˜

⁻

∂η

³

(s, 0) + o(

²

) = 1

∂ w ˆ

∂ξ (s, 1).

217

Injecting (19) and (23) into (25) and (26), we respectively obtain the following continuity conditions

218

on Γ,

219

w

₀⁻

(x

Γ

(s)) = ˆ w

0

(s, 1),

220

w

₁⁻

(x

Γ

(s)) + ∂w

₀⁻

∂ν (x

Γ

(s)) = ˆ w

1

(s, 1), (27)

221

w

₂⁻

(x

_Γ

(s)) + ∂w

₁⁻

∂ν (x

_Γ

(s)) + 1 2

∂

²

w

⁻₀

∂ν

²

(x

_Γ

(s)) = ˆ w

₂

(s, 1),

222 223

and

224

0 = ∂ w ˆ

0

∂ξ (s, 1),

225

∂w

₀⁻

∂ν (x

Γ

(s)) = ∂ w ˆ

1

∂ξ (s, 1), (28)

226

∂w

₁⁻

∂ν (x

Γ

(s)) + ∂

²

w

⁻₀

∂ν

²

(x

Γ

(s)) = ∂ w ˆ

2

∂ξ (s, 1).

227 228

System (24) coupled with the boundary conditions (28) and (27) provide an inductive way to

229

determine (w

_j⁻

, v

_j

). We obtain the set of equations satisfied by these terms after substituting the

230

expressions of ˆ w

j

(s, 1) given by (21),(22). We hereafter summarize the set of equations obtained

231

for (w

_j⁻

, v

j

) and how to use them to get the expressions of λ

j

, j = 0, 1, 2.

232

We first obtain that the couple (w

₀⁻

, v

0

) solves

233

(29)



 

 

 

 

∆w

⁻₀

+ λ

₀

n

₀

w

₀⁻

= 0 in Ω,

∆v

₀

+ λ

₀

v

₀

= 0 in Ω, w

₀⁻

− v

₀

= 0 on Γ,

∂w

⁻₀

∂ν − ∂v

0

∂ν = 0 on Γ.

234

This means in particular that λ

0

is a transmission eigenvalue for the limiting problem where the

235

thin layer is removed. We then obtain that the couple (w

₁⁻

, v

1

) satisfies

236

(30)



 

 

 

 

∆w

⁻₁

+ λ

₀

n

₀

w

₁⁻

= −λ

1

n

₀

w

⁻₀

in Ω,

∆v

₁

+ λ

₀

v

₁

= −λ

1

v

₀

in Ω,

w

₁⁻

− v

1

= 0 on Γ,

∂w

⁻₁

∂ν − ∂v

1

∂ν = λ

0

(n

0

− n

1

)w

⁻₀

on Γ.

237

Since λ

₀

is an eigenvalue of the associated homogeneous system, this problem is solvable only if a compatibility condition is satisfied by the right hand sides. This compatibility condition can be obtained by multiplying the first equation with w

₀⁻

and the second equation with v

₀

, taking the difference then integrating by parts and using (29). One ends up with

λ

₁

= Z

Γ

λ

0

(n

0

− n

1

)|w

⁻₀

|

²

ds(x) Z

Ω

n

0

|w

⁻₀

|

²

− |v

0

|

²

dx

which coincides with the expression of given in Theorem 2.1 expressed in terms of u

₀

= w

₀⁻

− v

₀

.

238

Although not covered by the analysis of convergence, we also provide the expression of the third

239

term in the asymptotic expression. One get that the couple (w

₂⁻

, v

2

) solves

240

(31)



 

 

 

 

∆w

₂⁻

+ λ

₀

n

₀

w

⁻₂

= −λ

1

n

₀

w

⁻₁

− λ

₂

n

₀

w

₀⁻

in Ω,

∆v

₂

+ λ

₀

v

₂

= −λ

1

v

₁

− λ

₂

v

₀

in Ω,

w

⁻₂

− v

2

= h

1

on Γ,

∂w

₂⁻

∂ν − ∂v

2

∂ν = h

2

on Γ,

241

where

242

h

1

= − 1 2

∂

²

w

₀⁻

∂ν

²

− 1

2 λ

0

(n

0

− n

1

)w

₀⁻

243

and

244

h

₂

= κ ∂

²

w

₀⁻

∂ν

²

− 7κ 2

∂

²

w

⁻₀

∂s

²

+

2κ

²

+ λ

₀

(n

₀

+ n

₁

2 ) ∂w

⁻₀

∂ν − 3κ 2

∂w

⁻₀

∂s + 3 2

∂

³

w

⁻₀

∂ν∂s

²

245

+

λ

1

(2n

1

− n

0

) + λ

0

(κ( n

₁

2 − n

0

))

w

⁻₀

− ∂

²

w

₁⁻

∂ν

²

+ κ ∂w

⁻₁

∂ν .

246247

Writing the compatibility condition for (31), we obtain the following formula for λ

2 248

λ

₂

Z

Ω

1 1 − n

₀

1

λ

₀

|∆u

₀

|

²

− λ

₀

n

₀

|u

₀

|

²

dx = −λ

²₁

Z

Ω

1

λ

₀

∆u

₀

u ¯

₀

+ 1

1 − n

₀

|u

₀

|

²

dx

249

− λ

₁

Z

Ω

1 1 − n

₀

u

₁

∆¯ u

₀

+ n

₀

∆u

₁

u ¯

₀

+ 2n

₀

λ

₀

u

₁

¯ u

₀

dx

250

+ Z

Γ

h

₁

∂

∂ν 1

1 − n

₀

(∆ + λ

₀

)¯ u

₀

ds(x) −

Z

Γ

h

₂

1

1 − n

₀

(∆ + λ

₀

)¯ u

₀

ds(x).

(32)

251 252

This complicated expression shows in particular a nonlinear dependence of λ

2

in terms of n

1

. It

253

suggests that the use of λ

2

for solutions to the inverse problem of determining n

1

may not be

254

appropriate.

255

4. Convergence analysis. The main goal of this section is to prove Theorem 2.1 that

256

provides a rigorous mathematical justification of the formal asymptotic expansion for simple real

257

transmission eigenvalues up to the first order. The proof is split into several steps. The first one is

258

to establish the convergence in norm of the operator T to T

0

. This ensures the convergence of λ

259

to λ

0

. In order to get to the term of order 1 in , we shall apply the Osborn theorem which requires

260

for instance a characterization of the pointwise asymptotic expansion of T (U ) up to order 1 in

261

(for some given function U ∈ H

₀²

(Ω) × H

₀²

(Ω)). The latter can be obtained from the asymptotic

262

expansions of A

⁻¹

u, B u and C u for some u ∈ H

₀²

(Ω). The difficult part to get the expansion of

263

A

⁻¹

u since for the two others, the first order terms are vanishing. This critical result is provided

264

by Lemma 4.5.

265

In all the following we use the notation

266

(f, g) := (f, g)

_H²

0(Ω)

= Z

Ω

∆f ∆gdx and kgk := (g, g)

1 2

H₀²(Ω)

.

267

For an operator A : V → V , kAk denotes the operator norm. To simplify the writing, C will

268

denote a generic constant whose value may change but remains independent from as → 0.

269

4.1. Pointwise convergence of the spectrum of T . In this first step, we prove pointwise

270

convergence of the spectrum of the operator T to the spectrum of T

0

. This is a direct consequence

271

of the following convergence in norm [23, 8].

272

Theorem 4.1. Assume that n

0

∈ C

²

(Ω). Let T and T

0

be defined by (9) and (10) respectively.

273

Then T converges to T

0

in the operator norm.

274

Proof. The proof follows from Lemma 4.2 and Lemma 4.4 below, using the definition of T

275

and T

0

.

276

In the first lemma we prove norm convergence for B and C .

277

Lemma 4.2. Let B

, C

, B

₀

and C

₀

be the operators defined by (6) and (7). Then, for suffi-

278

ciently small ,

279

(33) kB

− B

₀

k ≤ C

¹²

and kC

¹²

− C

₀¹²

k ≤ C.

280

Proof. From the definitions of B and B

0

, we have that for u, φ ∈ H

₀²

(Ω)

281

((B

− B

₀

)u, φ) = Z

Ω

1 1 − n

u∆φ + n

∆uφ dx −

Z

Ω

1 1 − n

0

u∆φ + n

₀

∆uφ dx

282

= Z

Ω

1

1 − n

₁

− 1 1 − n

₀

u∆φ + ∆uφ dx.

283 284

Therefore,

285

|((B − B

₀

)u, φ)| ≤C

kuk

L^∞(Ω)

k∆φk

L¹(Ω)

+ kφk

L^∞(Ω

k∆uk

L¹(Ω)

.

286287

Using the Sobolev embedding theorem and the Cauchy Schwartz inequality, we get

288

|((B − B

0

)u, φ)| ≤ C

¹²

(kuk

_H2

0(Ω)

kφk

_H2 0(Ω)

).

289

By choosing φ = (B − B

0

)u, we get

290

k(B − B

0

)uk

_H2

0(Ω)

≤ C

¹²

kuk

_H2 0(Ω)

.

291

The proof is similar for the second inequality. For u, φ ∈ H

₀²

(Ω), we have

292

((C − C

0

)u, φ) = Z

Ω

1 1 − n

1

− 1 1 − n

0

uφdx ≤ C

|Ω |kuk

L^∞(Ω)

kφk

L^∞(Ω)

293 294

From the Sobolev embedding theorem, we obtain

295

((C − C

0

)u, φ) ≤ C kuk

_H2

0(Ω)

kφk

_H2 0(Ω)

.

296

By choosing φ = (C

− C

₀

)u, we have

297

(34) k(C

− C

₀

)uk

_H2

0(Ω)

≤ Ckuk

_H2 0(Ω)

.

298

Using the square root Lemma in [24] and the fact that C

ⁿ

converges to C

₀ⁿ

at the same order

299

O(), we conclude that C

1

2

converges to C

1 2

0

at the same order O(). Hence we have

300

(35) k(C

¹²

− C

1 2

0

)uk

_H²

0(Ω)

≤ Ckuk

_H²

0(Ω)

.

301

Now we show convergence in the H

₀²

(Ω) norm for A

⁻¹

f assuming smoothness of f . This will be

302

useful in the proof of Lemma 4.4 since the operators B and C are regularizing.

303

Lemma 4.3. Let A and A

0

be defined by (5) for > 0 and = 0, respectively and f ∈ H

₀²

(Ω).

304

If A

⁻¹₀

f ∈ C

²

(Ω), then for sufficiently small ,

305

(36) kA

⁻¹

f − A

⁻¹₀

f k ≤ C

¹²

.

306

Proof. For a fixed f ∈ H

₀²

(Ω), define z and z

0

in H

₀²

(Ω) as z = A

⁻¹

f and z

0

= A

⁻¹₀

f . Since

307

A z = A

0

z

0

= f , we have that for φ ∈ H

₀²

(Ω)

308

(37) (A

(z

− z

₀

), φ) = (A

₀

z

₀

− A

z

₀

, φ) = Z

Ω

1 1 − n

0

− 1 1 − n

1

∆z

₀

∆φdx.

309

If z

0

∈ C

²

(Ω), we get

310

Z

Ω

1

1 − n

₀

− 1 1 − n

∆z

₀

∆φdx ≤ Ck∆z

₀

k

_∞

Z

Ω

∆φdx ≤C

¹²

kφk

_H2 0(Ω)

.

311 312

Thus, we have shown that

313

(A (z − z

0

), φ) ≤ C

¹²

kφk

_H2 0(Ω)

.

314

By plugging in φ = z − z

0

, we obtain the desired convergence using the coercivity of A .

315

Lemma 4.4. Assume that n

0

∈ C

²

(Ω). Let A , B , C , A

0

, B

0

and C

0

be defined by (5), (6)

316

and (7) for > 0 and = 0, respectively. Then for sufficiently small ,

317

kA

⁻¹

B − A

⁻¹₀

B

0

k −→

→0

0 and kA

⁻¹

C

1

2

− A

⁻¹₀

C

1 2

0

k −→

→0

0.

318

Proof. From (37), we have that for f, φ ∈ H

₀²

(Ω) and with z = A

⁻¹

f and z

0

= A

⁻¹₀

f

319

(A (z − z

0

), φ) ≤ Ck∆A

⁻¹₀

f k

L²(Ω)

kφk

_H2 0(Ω)

.

320321

Furthermore,

322

kA

⁻¹

B

f − A

⁻¹₀

B

₀

f k

_H2

0(Ω)

≤k(A

⁻¹

− A

⁻¹₀

)B

₀

f k

_H2

0(Ω)

+ kA

⁻¹

(B

− B

₀

)f k

_H2 0(Ω) 323

≤Ck∆A

⁻¹₀

B

0

fk

_L2(Ω)

+ kA

⁻¹

kk(B − B

0

)kkf k

_H2 0(Ω)

. (38)

324325

For estimating k∆A

⁻¹₀

B

0

f k

_L2(Ω)

, observe that B

0

u ∈ H

₀²

(Ω) is the weak solution

326

∆∆B

₀

u = ∆ n

₀

1 − n

0

u

+ 1

1 − n

0

∆u in Ω.

327

Classical regularity results [22, 25] and the fact that n

₀

∈ C

²

(Ω) imply that B

₀

u ∈ H

⁴

(Ω) ∩ H

₀²

(Ω)

328

and therefore

329

k∆A

⁻¹₀

B

₀

f k

_H1(Ω)

≤ Ckf k

_H2(Ω)

.

330

By the Sobolev embedding theorem, this implies that

331

k∆A

⁻¹₀

B

₀

f k

_Lp(Ω)

≤ Ckf k

_H2(Ω)

,

332

for p > 2 . Let ˜ p =

^p₂

> 1 and q such that 1

˜ p + 1

q = 1.

333

k∆A

⁻¹₀

B

0

f k

²_L2(Ω)

≤ k∆A

⁻¹₀

B

0

f k

²_Lp(Ω)

|Ω |

^1q

≤ C

^2q1

kf k

_H2(Ω)

. (39)

334335

From (33) we obtain

336

(40) kA

⁻¹

kk(B − B

₀

)kkf k

_H²

0(Ω)

≤ C

¹²

kf k

_H²

0(Ω)

.

337

Using (38), (39) and (40) we have that

338

kA

⁻¹

B

− A

⁻¹₀

B

₀

k −→

→0

0.

339

The second convergence result follows from similar arguments.

340

Now we would like to obtain explicit formula for the correction term in the asymptotic expansion

341

for the operator T . More precisely, we define explicit formula for the corrector term associated

342

with A

⁻¹

− A

⁻¹₀

.

343

4.2. Corrector term for A

⁻¹

− A

⁻¹₀

. In this subsection, we construct a corrector function

344

and use it to estimate the convergence rate of z = A

⁻¹

u for u ∈ H

₀²

(Ω). Let z

0

= A

⁻¹₀

u ∈ H

₀²

(Ω),

345

i.e z

0

∈ H

₀²

(Ω) solution of

346

(41) ∆ 1

1 − n

₀

∆z

0

= ∆∆u in Ω.

347

Inspired by the formal calculations on the previous section, namely problem (30), we define z

₁

348

solution of

349

(42)



 

 

 

 

∆ 1

1 − n

0

∆z

₁

= 0 in Ω,

z

₁

= 0 on Γ,

∂z

₁

∂ν = 1 − n

₁

1 − n

0

− 1

∆z

₀

on Γ.

350

We expect that z = z

0

+ z

1

+ O(

²

) in Ω

⁰

. We extend z

1

in Ω as ˜ z

₁

defined by

351

(43) z ˜

₁

=

( z

1

in Ω

⁰

, z

₁

− ψ in Ω

352

where ψ is a polynomial of order ≤ 3 and satisfying the boundary conditions:

353

(44)



 

 

ψ = 0 , ∂ψ

∂ν = 1 − n

₁

1 − n

0

− 1

∆z

₀

on Γ, ψ = ∂ψ

∂ν = 0 on Γ

.

354

This gives the following expression of ψ (that plays the role ˆ w

₂

in the formal calculations)

355

ψ(x) = ψ(ϕ(s, ξ)) = ˆ ψ(s, ξ) = 1 − n

₁

1 − n

0

− 1

∆z

₀

(ϕ(s, 0))ξ(1 − ξ)

²

.

356

The choice of ψ ensures in particular that ˜ z

₁

∈ H

₀²

(Ω). To simplify the notation we set m := 1

1 − n

0

− 1 1 − n

1

. Now we have the following Lemma.

357

Lemma 4.5. Assume that n

0

and n

1

are in C

⁴

(Ω). Let u ∈ H

₀²

(Ω) then set z = A

⁻¹

u and

358

z

0

= A

⁻¹₀

u. We define z ˜

₁

as in (43) and assume that z

0

∈ C

⁶

(Ω). Then we have, for sufficiently

359

small ,

360

kz − z

0

− ˜ z

₁

k

_H2

0(Ω)

≤ C

³²

.

361

Proof. For any φ ∈ H

₀²

(Ω) we have that

362

(45) (A (z − z

0

− ˜ z

₁

), φ) = (A (z − z

0

), φ) − (A z ˜

₁

, φ).

363

We recall that

364

(A (z − z

0

), φ) = Z

Ω

1 1 − n

0

− 1 1 − n

1

∆z

0

∆φdx.

365 366

Furthermore, we have that

367

(46) (A z ˜

₁

, φ) = Z

Ω⁰

1 1 − n

0

∆z

1

∆φdx + Z

Ω

1 1 − n

1

∆(z

1

− ψ)∆φdx.

368

Using the fact that ∆ 1 1 − n

0

∆z

₁

= 0 and the Green formula yield,

369

(A z ˜

₁

, φ) = Z

Γ

m∆z

1

∂φ

∂ν ds(x) + Z

Γ

∂

∂ν 1

1 − n

₁

∆z

1

− ∂

∂ν 1

1 − n

₀

∆z

1

φds(x)

370

+ Z

Γ

1 1 − n

1

∆ψ ∂φ

∂ν ds(x) − Z

Γ

∂

∂ν 1

1 − n

1

∆ψ

φds(x) − Z

Ω

∆ 1

1 − n

1

∆ψφdx.

371 372

Using the expression of ψ we have

373

1

1 − n

₁

∆ψ|

Γ

= 1

1 − n

₁

∆

s,ξ

ψ(s, ˜ 1) = 2

m∆z

0

(ϕ(s, 0)),

374

∂

∂ν ( 1 1 − n

1

∆ψ)|

Γ

= ∂

∂η ( 1 1 − n

1

∆ψ)|

Γ

= 1 1 − n

1

1

∂

∂ξ (∆

_s,ξ

ψ)(s, ˜ 1) = 6

²

m∆z

₀

(ϕ(s, 0)).

375

We then get after substitution of these expressions

376

(A z ˜

₁

, φ) = Z

Γ

m

∆z

1

(ϕ(s, )) + 2

∆z

0

(ϕ(s, 0)) ∂φ

∂ν ds(x)

377

− Z

Γ

m ∂

∂ν (∆z

₁

(ϕ(s, ) + 6

²

∆z

₀

(ϕ(s, 0)))φds(x) − Z

Ω

∆ 1

1 − n

1

∆ψφdx

378

= Z

Γ

φ

₁

∂φ

∂ν ds(x) − Z

Γ

φ

₂

φds(x) − Z

Ω

∆ 1

1 − n

₁

∆ψφdx

379 380

where we have set

381

φ

₁

(s) := m∆z

1

(ϕ(s, )) + 2

m∆z

0

(ϕ(s, 0)),

382

φ

₂

(s) := m ∂

∂ν (∆z

1

(ϕ(s, ))) + 6

²

m∆z

0

(ϕ(s, 0))

383

using the parametrization of the curve Γ , s 7−→ ϕ(s, ) with ϕ defined by (13). Using this

384

parametrization and setting ˜ φ(s, η) := φ(ϕ(s, η)) in Ω we have

385

Z

Γ

φ

₁

∂φ

∂ν ds(x) = Z

L

0

φ

₁

∂ φ ˜

∂η (s, )(1 + κ)ds = Z

L

0

Z

0

φ

₁

∂

²

φ ˜

∂η

²

(s, η)(1 + κ)dsdη.

386 387

From the definition of φ

₁

we then get for φ ∈ H

₀²

(Ω),

388

Z

Γ

φ

₁

∂φ

∂ν ds(x) = 2

Z

L 0

Z

0

m∆z

0

(ϕ(s, 0)) ∂

²

φ ˜

∂η

²

(s, η)dsdη + O(

¹²

)kφk

_H2(Ω)

.

389 390

Here and in all the following O(

^r

) denotes a function such that O(

^r

) ≤ C

^r

for a constant C

391

independent from the test function φ but that may depend on kz

0

k

_C6(Ω)

. Using Taylor’s expansion

392

we also get for φ ∈ H

₀²

(Ω),

393

Z

Γ

φ

₂

φds(x) = 2

Z

L 0

Z

0

φ

₂

∂

²

φ ˜

∂η

²

(s, η)(1 + κ)dsdη + O(

³²

)kφk

_H2(Ω) 394

= 3

Z

L 0

Z

0

m∆z

0

(ϕ(s, 0)) ∂

²

φ ˜

∂η

²

(s, η)dsdη + O(

¹²

)kφk

_H2(Ω) 395

396

where the last equality is obtained after substituting the expression of φ

₂

. One ends up with

Z

Γ

(φ

₁

∂φ

∂ν − φ

₂

φ)ds(x) = − Z

L

0

Z

0

m∆z

0

(ϕ(s, 0)) ∂

²

φ ˜

∂η

²

(s, η)dsdη + O(

³²

)kφk

H²(Ω)

. Equation (45) then gives

397

(A (z − z

0

− ˜ z

₁

), φ) = Z

Ω

m∆z

0

∆φdx − (A z ˜

₁

, φ)

398

= Z

L

0

Z

0

m∆z

0

(ϕ(s, η))∆φ(ϕ(s, η))(1 + ηκ)dsdη − Z

L

0

Z

0

m∆z

0

(ϕ(s, 0) ∂

²

φ

∂η

²

(ϕ(s, η))dsdη

399

− Z

Ω

∆ 1

1 − n

₁

∆ψφdx + O(

³²

)kφk

_H2(Ω)

.

400 401

We use the expression of the Laplacien in local coordinates

402

(1 + ηκ)∆φ(ϕ(s, η)) = ∂

∂s 1

1 + ηκ

∂ φ ˜

∂s (s, η) + κ ∂ φ ˜

∂η (s, η) + (1 + ηκ) ∂

²

φ ˜

∂η

²

(s, η)

403

to make the decomposition

404

Z

L 0

Z

0

m∆z

0

(ϕ(s, η))∆φ(ϕ(s, η))(1 + ηκ)dsdη = Z

L

0

Z

0

m∆z

0

(ϕ(s, η)) ∂

∂s 1

1 + ηκ

∂ φ ˜

∂s (s, η)

405

+ Z

L

0

Z

0

m∆z

0

(ϕ(s, η)) κ ∂ φ ˜

∂η (s, η) + ηκ ∂

²

φ ˜

∂η

²

(s, η) +

Z

L 0

Z

0

m∆z

0

(ϕ(s, η)) ∂

²

φ ˜

∂η

²

(s, η).

406 407

To estimate the first term, we integrate by parts on [0, L[, we obtain

408

Z

L 0

Z

0

m∆z

₀

(ϕ(s, η)) ∂

∂s 1

1 + ηκ

∂ φ ˜

∂s (s, η) dηds

409

= − Z

L

0

Z

0

1 1 + ηκ

∂

∂s (m∆z

0

(ϕ(s, η))) ∂ φ ˜

∂s (s, η)dsdη

410

= − Z

L

0

Z

1 0

m 1

1 + ξκ

∂

∂s ∆z

0

(ϕ(s, ξ)) ∂ φ ˜

∂s (s, ξ)dsdξ

411

= − Z

L

0

Z

1 0

m ∂

∂s ∆z

₀

(ϕ(s, 0)) Z

ξ 0

∂

²

φ ˜

∂η∂s (s, η)dη

dsdξ + O(

²

)kφk

_H2(Ω) 412

= O(

³²

)kφk

_H2(Ω)

.

413414

For the last term we proceed similarly to obtain

415

Z

L 0

Z

0

m∆z

₀

(ϕ(s, η)) ∂ φ ˜

∂η (s, η)dsdη = Z

L

0

Z

1 0

m∆z

₀

(ϕ(s, ξ)) ∂ φ ˜

∂η (s, ξ)dsdξ

416

= Z

L

0

Z

1 0

m∆z

0

(ϕ(s, 0)) Z

ξ 0

∂

²

φ ˜

∂η

²

(s, η)dη

dsdξ + O(

²

)kφk

_H2(Ω)

= O(

³²

)kφk

_H2(Ω) 417

418419

Observing in addition that

420

Z

L 0

Z

0

ηκm(∆z

0

(ϕ(s, η) ∂

²

φ ˜

∂η

²

(s, η))dsdη = O(

³²

)kφk

_H2(Ω)

,

421

one ends up with

422

(A

(z

− z

₀

− ˜ z

₁

), φ) = Z

L

0

Z

0

m

∆z

₀

(ϕ(s, η)) − ∆z

₀

(ϕ(s, 0)) ∂

²

φ ˜

∂η

²

(s, η)dsdη

423

− Z

Ω

∆ 1

1 − n

1

∆f ψ φdx + O(

³²

)kφk

_H2(Ω)

.

424 425

To conclude we just observe that the two remaining terms are also of the form O(

³²

)kφk

H²(Ω)

. For the first term, we simply use a Taylor expansion for ∆z

0

while for the second one we just use that, due to the regularity of n

0

and n

1

,

∆ 1

1 − n

₁

∆ψ ∈ L

^∞

(Ω).

In conclusion,

426

(A (z − z

0

− ˜ z

₁

), φ) ≤ C

³²

kφk

_H2(Ω)

.

427

Choosing φ = z − z

0

− ˜ z

₁

, since the coercivity constant associated with A is independent from

428

, we get

429

kz − z

0

− ˜ z

₁

k

_H2

0(Ω)

≤ C

³²

430

which ends the proof.

431

Lemma 4.6. Assume that n

0

and n

1

are in C

⁴

(Ω). If u ∈ C

⁶

(Ω) ∩ H

₀²

(Ω), then for sufficiently

432

small ,

433

(47) kB

0

(A

⁻¹

− A

⁻¹₀

)uk

_H2

0(Ω)

≤ C and kC

₀¹²

(A

⁻¹

− A

⁻¹₀

)uk

_H2

0(Ω)

≤ C

434

where C independent of .

435