Stéphanie Delair

René Guyonnet

Alexandre Govin

Bernard Guilhot

STUDY OF EFFLORESCENCE FORMING PROCESS ON CEMENTITIOUS MATERIALS ETUDE DU PROCESSUS DE FORMATION DES EFFLORESCENCES SUR DES MATERIAUX CIMENTAIRES

Efflorescence is a white deposit of calcium carbonate on the surface of cementitious materials. It appears under specific climatic conditions. Besides the aesthetic problems, an important economical problem also exists, especially for colored materials. Indeed, this phenomenon is a cause for rejection of products by customers. The aims of the study presented herein are to understand the mechanisms involved and identify the main parameters which control efflorescence formation. A procedure was previously performed to produce primary efflorescence in a climatic chamber. As pozzolans are frequently used in order to decrease the intensity of efflorescence, substitution of cement by different pozzolans was studied. This latest research allowed the Authors to underline which parameters were modified by the substitutions and to link decreases in the intensity of efflorescence with those modifications. RÉSUMÉ -Les efflorescences forment un dépôt blanc de carbonate de calcium à la surface de matériaux cimentaires. Elles apparaissent pour des conditions climatiques spécifiques. Au delà du problème esthétique, un important problème économique intervient car ce phénomène peut être la cause du rejet d'un produit par les clients. Les objectifs de l'étude sont la compréhension des mécanismes et l'identification des paramètres principaux qui contrôlent la formation d'efflorescences. Au préalable, un test a été mis au point dans une chambre climatique pour reproduire des efflorescences. Comme les pouzzolanes sont souvent utilisées pour atténuer les efflorescences, la substitution du ciment par différentes pouzzolanes a été étudiée. Cette étude a permis d'identifier les paramètres affectés par ces substitutions et de relier ces modifications à l'atténuation des efflorescences.

Introduction

Cementitious constructions are subjected to natural conditions of temperature and relative humidity. Some of these climatic conditions are in favor of efflorescence phenomenon. Efflorescence is a white deposit of CaCO 3 on the surface of cementitious materials. This white deposit on dark constructions can be a cause of rejection of products by customers and thus sometimes becomes a significant economic problem. Understanding the mechanisms involved and identifying the main parameters which control efflorescence formation are the first steps for reducing the formation of efflorescence.

Two kinds of efflorescence can be noticed: primary and secondary. The distinction comes from at witch time in the curing process the precipitation of CaCO 3 occurs [START_REF] Bolte | Efflorescence on concrete products -causes and strategies for avoidance[END_REF]. Secondary efflorescence which occurs in cured concretes, is not the subject of this program and is not detailed in this article. Primary efflorescence forms during the curing process. Calcium ions from cement phases and carbonate ions from carbon dioxide in air dissolve in the interstitial solution. Excess water in the cementitious matrix diffuses to the surface with Ca 2+ (Figure 1.a). When water evaporates, the concentrations of Ca 2+ and CO 3 2-increase and reach the saturation limit of CaCO 3 . Then CaCO 3 precipitates (Figure 1

.b-d). (a) (b) (c) (d)
Figure 1. Mechanism of appearance of efflorescence (Dossier Ciment 5, 1995).

The main parameters identified that influence primary efflorescence on cement are [START_REF] Dow | Calcium carbonate efflorescence on Portland cement and building materials[END_REF]: capillarity and permeability, ionic concentration in pores and climatic conditions (temperature and relative humidity) (Figure 1

.c-d).

Concerning the climatic conditions, it is known that for a fast drying (wind, sunshine) precipitation of CaCO 3 occurs on the inside of pores (Figure 1.c), while for a longer drying (rainy weather, low temperature) the precipitation occurs on the surface of sample (Figure 1.d) [START_REF] Dow | Calcium carbonate efflorescence on Portland cement and building materials[END_REF]. In order to study the efflorescence forming process, a procedure was performed to produce reproducible primary efflorescence under laboratory conditions in a climatic chamber.

Pozzolanic reactivity is known to decrease the intensity of efflorescence. Metakaolin is used in industry in substitution of cement for its beneficial action on efflorescence. This action is realized thanks to the pozzolanic reaction of metakaolin when it is mixed with cement. According to Garcia-Diaz, contact between metakaolin and basic solution supersaturated in calcium is necessary for the pozzolanic reaction [START_REF] Garcia-Diaz | Réactivité pouzzolanique des métakaolinites : corrélations avec les caractéristiques minéralo-gitologiques des kaolinites[END_REF]. In this case, it was the process of the tricalcium silicate hydration, the majority compound of the anhydrous cement, which gave birth to this basic solution according to reaction (1).

Ca 3 SiO 5 + 3 H 2 O → 3 Ca 2+ + 4 OH -+ H 2 SiO 4 2- (1)
Metakaolin (Al 2 O 3 (SiO 2 ) 2 ), after the hydroxylic dissolution, reacts with Ca 2+ and OH -ions in the solution. The straetlingite ((CaO) 2 Al 2 O 3 SiO 2 (H 2 O) 8 ) and silico-calciques hydrated gels (C-S-H) form in detriment of portlandite (Ca(OH) 2 hydrate of cement) according to reaction (2):

Al 2 O 3 (SiO 2 ) 2 + (2+C/S) Ca 2+ + 2.(2+C/S) OH -+ (y+8-2-C/S) H 2 O → C C/S -S-H y + (CaO) 2 Al 2 O 3 SiO 2 (H 2 O) 8 (2)
An acceleration of cement hydration is observed in case of a mixture of cement and pozzolans [START_REF] Takemoto | Hydratation des ciments pouzzolaniques[END_REF].

Materials and method

Materials

The investigated cement was a grey Portland cement CPA CEM I 52.5, according to the NF EN 197-1 standard. Three metakaolins (MK) were used. These differed by physical parameters such as BET surface, given in Table I, andparticle size distribution, given in Figure 2 Average diameters of the three MKs are specified in the Table II. Specific surface and particle size distribution are indicators of the reactivity of powder. Thus, we can firstly suppose that MK2 should have a higher reactivity than MK0 or MK1.

Sample preparation

Four types of cementitious paste were prepared and cast in moulds of cubic shape (2.10 - 6 m 3 ) with a water-to-solid ratio (W/S) of 0.4. The first paste was a pure cement paste. The three others paste were mixes of cement and MK0, MK1 or MK2 in ratio of substitution of cement between 3 to 30 %. The hydration period before demoulding the sample is called the curing period. During the curing period, samples were stored under nitrogen and in a water-vapor-saturated atmosphere (100% relative humidity) to avoid carbonation.

Efflorescence test

According to the literature in this area, a layer of water and a sufficient duration are necessary to the formation of efflorescence on concrete [START_REF] Dow | Calcium carbonate efflorescence on Portland cement and building materials[END_REF]. Indeed, the film water allows ions to be propagated on the surface of the samples.

The test developed simulated these climatic conditions and it allowed for producing efflorescence in a reproducible way. This test could be divided into two parts: -Curing period -Sample demoulding and placing into the climatic chamber:

-Stage of saturation of water (100% of relative humidity) -Drying by decrease in the relative humidity (100% to 50%). The temperature of the climatic chamber was kept constant during the test (16°C).

Measurement of the intensity of efflorescence on cement: Spectrocolorimetry

To validate and exploit the test, variations in the intensity of efflorescence had to be quantified. As efflorescence is a white deposit on the grey surface of cement, the clarity of the sample surface varied: there is a correlation between intensity of efflorescence and the clarity of the cement. This was why the intensity of efflorescence was measured with a spectrocolorimeter. This apparatus gives, in the L*.a*.b* coordinates, the numerical color of a surface. L* corresponds to clarity and goes from 0 for black to 100 for white. a* and b* correspond to chromaticity. Only L* was studied in this work. Indeed, few variations in a* and b* were noticed with grey cement. Each measurement was conducted in triplicate. As the clarity of the cement could vary with introduction of MKs (nearly white powders), we chose to study the difference of clarity between a sample reference and a sample after the efflorescence test. The reference was kept under nitrogen and water-vapor-saturated at 16°C so that no CaCO 3 could be formed and interfere with the measurements.

Mercury porosimetry

Intrusion porosimetry deals with analysing the penetration of a liquid in the porosity of the specimen material studied, via the intrusion at various pressures of mercury. As pressure increases, mercury moves into the sample pores. Intrusion of different size pores occurs at different pressures. The greater pressure, the smallest is the pore diameter into which mercury can be forced. Pore diameters are calculated with the pressure at which mercury is intruded, assuming that the pores have a cylindrical shape.

As the pores of cement paste are not of cylindrical shape, the diameter calculated is the diameter of entry of the mercury in the pore. This implies that one must be careful in interpretating the pore diameter distribution obtained by this method.

Electrical conductivity

Electrical conductivity measurements were performed in dilute suspensions with a water-tosolid ratio of 20 and 50. For experiments with cement, a water-to-cement ratio of 50 was used. Conductivity equipment consisted of a 25°C thermostated reactor, two stainless steel annular electrodes, and a conductimeter. The constant cell was first calibrated with a 0.1 M KCl solution. For experiments with cement, MK (0.1g), with a MK-to-cement ratio of 0.05, was mixed with 100 ml of deionized water (Millipore, mQ, USA). Then, after 30s, cement (2g) was introduced.

Results and discussion

Firstly the influence of MK on the intensity of efflorescence was considered. Then the pozzolanic reactivity of MK was studied, as well as the influence of MK on cement hydration and on porosity of cement past. Finally, links between the evolution of intensity of efflorescence and the parameters studied were highlighted.

Efflorescence test

After the efflorescence test, clarity of samples was measured. The evolution of the difference of clarity v.s. MK content in the cement is given in Figure 3. We observed for low metakaolin content that there was an increase of efflorescence. Then for content over 10%, we noticed that MK2 is better for attenuation of efflorescence.

Pozzolanic reactivity of MK

MKs were introduced in a lime solution to evaluate their pozzolanic reactivity. The conductivity of the solution decreased, meaning that ions were more consumed than they were dissolved (Figure 4). MK2 showed a greater decrease of conductivity than other MKs. This order is kept for two various water-to-solid ratios (W/S) 20 and 50. The pozzolanic reaction (2) indicates that the decrease of conductivity should be due to calcium consumption. This hypothesis was verified by ionic chromatography measurements (Figure 5) in the lime solution where MKs were introduced with a water-to-solid ratio of 50. It could be noticed that MK2 was more efficient in consuming Ca 2+ and OH -so MK2 had a pozzolanic reactivity higher than MK1 and MK0.

MK and cement

Influence on cement hydration

The influence of MK on cement hydration was studied by conductimetry. The precipitation of portlandite (CH) was taken as a benchmark to identify acceleration or delay in the cement hydration (Figure 6). The time differences are given in Table III. Acceleration by MK was observed as it was expected from the literature [START_REF] Takemoto | Hydratation des ciments pouzzolaniques[END_REF]. We noticed that MK2 was the one that accelerated more the cement hydration.

Table III. Acceleration in time of hydration.

Sample name

Cement -MK0 Cement -MK1 Cement -MK2 Δ T (min) 9 16 24 3.3.2. Influence on porosity of cement paste For three rates (5, 10 and 20%) of substitution of cement by MK, we studied the modifications induced in the distribution of the pore size of the paste after 6 h of hydration. This time corresponds to the curing period before the efflorescence test. By porosimetry, whatever the MK and the ratio, two groups of pore diameters were revealed (Figure 7). One was around 1 µm and the other one was under 0.1 µm. The porous distribution of a cement matrix presents inter residual grain pores which are commonly named capillary pores, for which the size is around 0.003-10 µm [START_REF] Moon | Relationship between average pore diameter and chloride diffusivity in various concretes[END_REF]. Both families can be ascribed to capillary pores and their sizes are in accordance with the literature. We noticed that there was a decrease in the size of the pore of the first family with the increase of MK, in particular for MK2 (Figure 7.d). This result can be linked with the decrease in efflorescence, because capillary pores that permit the transport of ions are reduced.

Discussion

Thanks to the pozzolanic reactions, MKs consume calcium dissolved from anhydrous cement, leaving less calcium available for efflorescence formation. MK2 is the metakaolin which consumed more calcium. This fact can explain the acceleration of the precipitation of CH in this case. Acceleration of cement hydration explains that MKs reduce pore size entry, making transport of ions more difficult. MK2 reduced pore sizes more than MK1 and MK0.

Considering that MK2 reduced the efflorescence intensity by acting in a favorable way on two parameters, physical and chemical actions couldn't be distinguished.

Conclusion

In this work, several parameters that could influence the formation of efflorescence on cementitious construction have been studied.

A test of wet/dry has been established for the formation of efflorescence, the variation of intensity of efflorescence was digitized and quantified by spectrocolorimeter.

It was determined that the formation of the efflorescence was reduced by substitution of cement by MK. Moreover, by studying the impact of substitution of cement by MK on the intensity of efflorescence, the correlation between efflorescence, calcium consumption, and porosity was highlighted. It was determined that a MK could be chosen by the way it consumes calcium and reduces pore size.

Figure 2 .

 2 Figure 2. Particle size distribution of MK obtained by laser granulometry measurement.

Figure 3 .

 3 Figure 3. Influence of the MK content in cement on intensity of efflorescence.

Figure 4 .

 4 Figure 4. Evolution of conductivity of a lime solution in which MK was introduced.

Figure 5 .

 5 Figure 5. Evolution of calcium concentration.

Figure 6 .

 6 Figure 6. Influence of MK on cement hydration.

Table II .

 II Average diameter of MK.

	Sample name	MK0	MK1	MK2
	D a (µm)	10	13.6	4.7

diameter (nm) dV/dD (mL/g/nm)

		0.0014			Cement			0.0014		Cement
	dV/dD (mL/g/nm)	0.0002 0.0004 0.0006 0.0008 0.001 0.0012			Cement -5% MK0 Cement -5% MK1 Cement -5% MK2	dV/dD (mL/g/nm)	0.0002 0.0004 0.0006 0.0008 0.001 0.0012		Cement -10% MK0 Cement -10% MK1 Cement -10% MK2
		0							0
			1	10	100	1000	10000			1	10	100	1000	10000
				Pore diameter (nm)					Pore diameter (nm)
				(a)							(b)
		0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.0016 0.0018 0.002	1	10 Pore Cement 100 1000 Cement -20% MK0 10000 Cement -20% MK1 Cement -20% MK2		0 0.0002 0.0004 0.0006 0.0008 0.001 0.0012 0.0014 0.0016 0.0018 0.002	1	10 Pore	100	1000	10000

diameter (nm) dV/dD (mL/g/nm)

		Cement
		Cement -5% MK2
		Cement -10% MK2
		Cement -20% MK2
	(c)	(d)
	Figure 7. Influence of MK content on porosity.

Acknowledgements

The authors would like to acknowledge SOCAREL for materials and financial supports.


Giorgi -Houssem Haddar

Giovanni Giorgi

Houssem Haddar

Computing estimates on material properties from transmission eigenvalues

Keywords: direct and inverse scattering problem, linear sampling method, transmission eigenvalues

This work is motivated by inverse scattering problems, those problems where one is interested in reconstructing the shape and the material properties of an inclusion from electromagnetic farfields measurements. More precisely we are interested in complementing the so called sampling methods, (those methods that enables one to reconstruct just the geometry of the scatterer), by providing estimates on the material properties. We shall use for that purpose the so-called transmission eigenvalues. Our method is based on reformulating the so-called interior transmission eigenvalue problem into an eigenvalue problem for the material coefficients. We shall restrict ourselves to the two dimensional setting of the problem and treat the cases of both TE and TM polarizations. We present a number of numerical experiments that validate our methodology for homogeneous and inhomogeneous inclusions and backgrounds. We also treat the case of a background with absorption and the case of scatterers with multiple connected components of different refractive indexes.

Rapport de Recherche Inria

Résumé : Ce travail est motivé par les problèmes de diffraction inverses en lectromagnétisme o l'on est intéressé par la reconstruction de la forme et les propriétés physiques d'une l'inclusion a partir de mesures de champs lointains. Plus précisément, nous nous intéressons à complémenter les méthodes d'échantillonnage (permettant la reconstruction de la géométrie de l'inclusion), en fournissant des estimations sur les propriétés physiques du matériau. Nous utiliserons à cette fin les fréquences de transmission qui sont les valeurs propres du problème de transmission intérieur.

Notre méthode est basée sur la reformulation de ce problème en un problème de valeur propre pour l'indice de réfraction. Nous nous limitons dans ce travail au cas bidimensionnel et traitons les deux cas de polarisations TE et TM. Nous présentons un certain nombre d'expériences numériques qui valident notre méthodologie pour les inclusions homogènes et inhomogènes. Nous traitons également le cas d'un milieu de référence avec ou sans absorption ainsi que le cas d'inclusions avec un ou plusieurs composantes connexes possédant des indices de réfraction différents.

Mots-clés : problème de diffraction direct et inverse, linear sampling method, transmission eigenvalues

Introduction

This work is motivated by inverse scattering problems, those problems where one is interested in reconstructing the shape and the material properties of an inclusion from electromagnetic farfields measurements. More precisely we are interested in complementing the so called sampling methods [START_REF] Cakoni | Qualitatives methods in inverse scattering theory[END_REF][START_REF] Kirsch | The factorization method for inverse problems[END_REF], (those methods that enables one to reconstruct just the geometry of the scatterer), by providing estimates on the material properties. We shall use for that purpose the so-called transmission eigenvalues [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF]. These special frequencies can be determined from the knowledge of the farfields for a range of frequencies. They also correspond to values of the wavenumbers for which the so called homogeneous interior transmission problem has a non trivial solution. Our work is in the same spirit of [START_REF] Colton | Analytical and computational methods for transmission eigenvalues[END_REF] but we shall employ a different and (in principle) a simpler method.

The proposed method is based on reformulating the interior transmission eigenvalue problem into a (standard, generalized) eigenvalue problem for the material coefficients. In the case of constant coefficients the value of the material parameter corresponds with the smallest eigenvalue [START_REF] Cakoni | Houssem Haddar The existence of an infinite descrete set of transmission eigenvalues[END_REF]. The obtained eigenvalue problem corresponds to a fourth order operator that we shall solve using an appropriate mixed second order formulation of the problem. To obtain the desired approximation of the refractive index, one needs to determine the first transmission eigenvalue from the farfield data. We shall review here the method proposed in [START_REF] Cakoni | On the use of transmission eigenvalues to estimate the index of refraction from farfield data[END_REF][START_REF] Cakoni | Houssem Haddar The computation of lower bounds for the norm of the index of refraction in an anisotropic media from farfield data[END_REF] based on a frequency sampling of the farfield solution. We then extend this method to the case of scatterers with multiply connected components so that one is able to detect the transmission eigenvalues associated with each connected components.

In these first investigations we restrict ourselves to a two dimensional setting of the problem that models electromagnetic scattering from infinite cylinders and treat both transverse electric (TE) and transverse magnetic (TM) polarizations. We present a number of numerical experiments that validate our methodology for homogeneous and non homogeneous inclusions and backgrounds. We also treat the case of a background with absorption and the case where the geometry is also unknown (but reconstructed using the linear sampling algorithm).

The outline of this report is the following. The second section is dedicated to introducing the forward and the inverse scattering problem for TM polarizations. We introduce our method for constructing constant approximation of the refractive index in the third section. The fourth section corresponds with the extension of our results to the case of TE polarizations. The last section is dedicated to numerical experiments that validate our approach and illustrate its effectiveness. We finally provide in the appendix a fast implementation procedure of the linear sampling method in the case of backgrounds with piecewise constant and axisymmetric refractive indexes. The latter has been used for instance to perform some of the numerical experimentations in the last section of this report.

Forward and inverse scattering problem

We consider the scattering of a time harmonic electromagnetic plane wave by an inhomogeneous infinite cylinder with cross section D such that the electric field E = (0, 0, u exp -iωt ) is polarized parallel to the axis of the cylinder (TM polarization). Supposing the index of refraction n dependent only on the two coordinates orthogonal to the cylinder axis and factoring out the time harmonic component we obtain that the total field u satisfies [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF]:

∆u(x) + k 2 n(x)u(x) = 0 in R 2 (1)
u(x) = u i (x) + u s (x), with u i (x) = exp ikx•d , (2)
lim r→∞ √ r( ∂u s (x) ∂r -iku s (x)) = 0 uniformly with respect to x := x |x| , (3)
where x ∈ R 2 , k > 0 is the wave number, r = |x| and d ∈ Ω = { x |x| : x ∈ R 2 }. The index of refraction n is assumed to be a piecewise continuously differentiable function with discontinuities across non intersecting C 2 curves. Moreover we suppose that Im(n) ≥ 0 and that, with m := 1-n and D being the support of m, D is compact with the complement of D connected and boundary ∂D smooth.

Problem (1)-( 3) has a unique solution u ∈ H 2 loc (R 2 ) and the corresponding scattered field u s has the asymptotic behavior [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF]:

u s (x) = exp ikr √ r u ∞ (x, d) + O(r -3/2 ) (4)
as r → ∞ uniformly with respect to x ∈ Ω. The knowledge of the so-called farfield pattern u ∞ (x, d) for all observation directions x ∈ Ω, all incident directions d ∈ Ω and for a range of frequencies k (that will be specified later) forms the data of the inverse problem of determining D and n. In fact, although the shape of D will be determined at a fixed frequency k, for the reconstruction of n, the knowledge of u ∞ (x, d) for a range of k will be required. As it will be better specified in the following, even in this last case the method will operate without combining data of different frequencies; for this reason we suppressed the mention to k in the notation.

Let Φ be the radiating fundamental solution of the Helmholtz equation

Φ(x, y) = i 4 H (1) 0 (k|x -y|), (5)
where H

(1) 0

is the Hankel function of the first kind of order 0 and consider

Φ ∞ (x, z) := exp iπ/4 √ 8πk exp -ikx•z (6)
the farfield pattern of Φ(•, z). Central to the Linear Sampling Method is the farfield equation

(F g z )(x) = Φ ∞ (x, z) (7)
where the farfield operator F : L 2 (Ω) → L 2 (Ω) is defined by:

(F g)(x) = Ω u ∞ (x, d)g(d)ds(d) (8)
We recall that F g is the farfield associated to the incident wave

v g (x) := Ω exp ikx•d g(d)ds(d) (9)
called Herglotz wave function with density g. We recall hereafter the main theorem associated with the LSM (see [START_REF] Colton | The linear sampling method in inverse electromagnetic scattering theory[END_REF] for instance). We shall assume in the sequel that Re(n) -1 or 1 -Re(n) is positive definite on D.

Theorem 2.1. Except possibly for a discrete set of values of k when Im(n) = 0, it holds:

1. if z ∈ D, for every ǫ > 0 there is a solution g ǫ z ∈ L 2 (Ω) of the inequality F g ǫ z -Φ ∞ (•, z) L 2 (Ω) < ǫ (10) such that lim ǫ→0 v g ǫ z L 2 (D) < ∞ ( 11
)
and that lim

z→∂D v g ǫ z L 2 (D) = ∞. ( 12
)
2. if z is not in D, then for every ǫ > 0 and g ǫ z satisfying

F g ǫ z -Φ ∞ (•, z) L 2 (Ω) < ǫ ( 13
)
we have that lim

ǫ→0 v g ǫ z L 2 (D) = ∞. (14)
The linear sampling method consists in solving the regularized (F is compact between infinite dimensional spaces) farfield equation for each z in a grid containing D and visualizing the indicator function z → g z L 2 (Ω) . This method therefore solves the inverse scattering problem of determining the shape of an unknown object D from the knowledge of its farfield pattern u ∞ for all observation directions and all incident directions at a given frequency.n. However its fails to do so if this frequency coincides with so called transmission eigenvalues that will be defined in the sequel.

We recall that for z ∈ R 2 , the equation F g z = Φ ∞ (•, z) has a solution g z ∈ L 2 (Ω) if and only if there exists w ∈ H 2 (D) and a Herglotz wave function v gz such that w and v gz are solution of

∆w + k 2 nw = 0, ∆v gz + k 2 v gz = 0 in D, (15)
w -v gz = Φ(•, z), ∂w ∂ν - ∂v gz ∂ν = ∂Φ ∂ν (•, z) on ∂D. (16)
Equations ( 15)-( 16) form the so called interior transmission problem. As it will be better clarified in the next section, transmission eigenvalues identify wave numbers for which the farfield operator may not be complete in L 2 (Ω). In particular, when k is such that the homogeneous problem associated to equations ( 15)-( 16) admits a non trivial solution (w, v) with v a Herglotz wave function, then farfield operator F is not complete in L 2 (Ω) [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF]. In [START_REF] Cakoni | Houssem Haddar The existence of an infinite descrete set of transmission eigenvalues[END_REF] the set of transmission eigenvalues has been proved to be infinite and countable.

The case of inhomogeneous background. We conclude this section by noting that an adaptation of the linear sampling method to the problem of detecting an object embedded in an inhomogeneous background has been well established for example in [START_REF] Colton | The linear sampling method in inverse electromagnetic scattering theory[END_REF] (for TM polarized plane waves and impenetrable objects) and in [START_REF] Colton | A linear sampling method for the detection of leukemia using microwaves[END_REF] (for near field measurements and penetrable objects). In the inhomogeneous cases, the kernel of the farfield operator has to be modified by subtracting its background analogous and the right hand side should be replaced by the farfield of the background Green's function. The index n b of the inhomogeneous background is assumed to verify similar regularity assumptions as n and to be equal to 1 outside a ball.

Following [START_REF] Colton | The linear sampling method in inverse electromagnetic scattering theory[END_REF] if u s b,∞ denotes the farfield pattern of the background medium (i.e. without the inclusion D) and if G ∞ (•, z) is the farfield of the Green function corresponding to the inhomogeneous background and to a point source at z, then Theorem 2.1 holds with

(F g)(x) := Ω [u ∞ (x, d) -u s b,∞ (x, d)]g z (d)ds(d) (17)
and Φ ∞ replaced by G ∞ . If u b (•, d) denotes the total field associated with the background, i.e. solution to (1)-(3) with n = n b , then the Herglotz wave is defined in the present case as

v g (x) := Ω u b (x, d)g(d)ds(d) (18)
and forms the incident wave that generates, for the inclusion D, a farfield equals to F g.

Derivation of a constant approximation of the refractive index

We consider the case of a non-absorbing (i.e. Im(n) = 0) inhomogeneous medium embedded in a (possibly inhomogeneous) background and we assume that LSM or other qualitative methods [START_REF] Cakoni | Qualitatives methods in inverse scattering theory[END_REF][START_REF] Kirsch | The factorization method for inverse problems[END_REF] have already been used to determine the support D of the target. The aim of this section is to determine an estimate of the index of refraction n(x) under the assumption that the support D is known.

To achieve this goal we shall make use of the first transmission eigenvalue and consider the interior transmission eigenvalue problem as an eigenvalue problem where the eigenvalue is the index of refraction. This is why we first need to provide a clearer definition of transmission eigenvalues and comment on their identification using farfield data.

Transmission eigenvalues and their identification using farfield data

We start analyzing the case of a TM polarized scattering problem of an inhomogeneous medium in the vacuum. Using the notation and assumptions of the second section, we define the interior transmission problem [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF][START_REF] Colton | John Sylvester The interior transmission problem[END_REF] as the problem of finding w, v

∈ L 2 (D), with w -v ∈ H 2 (D), such that ∆w + k 2 nw = 0 in D (19) ∆v + k 2 v = 0 in D (20) w -v = f on ∂D (21) ∂w ∂ν - ∂v ∂ν = g on ∂D (22)
for some boundary data f ∈ H 1 2 (∂D) and g ∈ H 3 2 (∂D). We have that the farfield operator F is injective with dense range provided that k is not a transmission eigenvalue, i.e. a value of k for which the homogeneous (i.e. f = 0 and g = 0 in (21) and ( 22)) interior transmission problem has a non trivial solution. We also recall that for z ∈ D, the farfield equation has a nearby solution g ǫ z ∈ L 2 (Ω) that satisfies (10)- [START_REF] Colton | Analytical and computational methods for transmission eigenvalues[END_REF] if there exists a couple (w z , v z ) solution of (19)-( 22) with f = Φ(•, z) and g = ∂ ∂ν Φ(•, z). Since the linear sampling method is expected to fail when k is a transmission eigenvalue and, in particular, for almost all z ∈ D the norm of a regularized solution g z,α satisfying

lim α→0 F g z,α -Φ ∞ (•, z) L 2 (Ω) = 0 (23)
is expected to be large for such values of k [5, 3], a possible method for finding transmission eigenvalues would be to solve (23) for few fixed z ∈ D and for values of k in an interval: the transmission eigenvalues would correspond with large values of k → g z L 2 (Ω) .

The following theorem formalizes previous considerations (the formulation is an adaptation of the results in [START_REF] Cakoni | Houssem Haddar On the determination of Dirichlet or transmission eigenvalues from farfield data[END_REF]). Theorem 3.1. Let g z,α satisfy (23) and z be a point of R 2 ; let v gz,α be the Herglotz wave function associated to g z,α and let us assume that k is a transmission eigenvalue. Then for almost every z ∈ D, v gz,α L 2 (D) cannot be bounded as α → 0. Otherwise, if k is not a transmission eigenvalue, then there exists g z,α satisfying (23) such that v gz,α L 2 (D) is bounded as α → 0 for each z ∈ D.

Let us denote by P (D, k, n) the interior transmission problem (19)-( 22) for a domain D of refractive index n and with boundary data f = Φ(•, z) and g = ∂ ∂ν Φ(•, z). We give in following a corollary of the previous theorem that indicates how, for multiply connected inclusions D, one can identify the transmission eigenvalues associated with each connected component. More precisely, assume that D = ĵ∈{j0,...,jm} D ĵ with m ∈ N and D ĵ connected for each ĵ ∈ {j 0 , ..., j m } and refer to P (D ĵ , k, n ĵ ) as the interior transmission problem associated to INRIA D ĵ , with n ĵ restriction of n to D ĵ . We remark that, obviously, the set of transmission eigenvalues for D is the union of transmission eigenvalues for D ĵ for ĵ ∈ {j 0 , ..., j m }.

Let us consider the farfield equation associated to the whole D and let g z,α satisfy (23).

Corollary 3.1. Let j p , j q ∈ {j 0 , ..., j m } and let k * be a transmission eigenvalue for P (D jp , k, n jp ) and not for P (D jq , k, n jq ). Then

• v gz,α L 2 (D) is unbounded for almost every z ∈ D jp .
• For each z ∈ D jq there exists a g z,α satisfying ( 23) such that v gz,α L 2 (D) is bounded.

Proof. First, we point out that if (w, v) is a solution of P (D, k, n), then (wj, vj) (restriction of (w, v) to Dj) would be a solution of P (Dj, k, nj) and that

v 2 L 2 (D) = j∈{j0,...,jm} vj 2 L 2 (Dj ) . (24)
• Let k * be a transmission eigenvalue for P jp . Then, from Theorem 3.1, it follows that v gz,α,jp L 2 (Dj p ) has to be unbounded for almost every z ∈ D jp . Then, by (24

), v gz,α L 2 (D) is unbounded for almost every z ∈ D jp . • If z ∈ D jq , (0, Φ(•, z)) is a (bounded) solution of P jp . Hence, (w, v) such that w = 0 on D, v = 0 on Dj for each j = j p and v = Φ(•, z) on D jp is a bounded solution of P (D, k * , n).
Denseness of the set of Herglotz wave functions implies the existence of a g z,α such that

v gz,α -v L 2 (D) → 0 for α → 0. This implies that F g z,α -Φ(•, z) L 2 (Ω) → 0 for α → 0.

Identification of the refractive index

As explained before, solving the interior transmission eigenvalue problem means, knowing the shape of D and the pointwise values of n, find k * such that there exists a non trivial solution u of the problem P (D, k, n).

Let us change the point of view: since transmission eigenvalues are detectable without knowing n(x), it is possible to consider directly the problem P (D, k * , n) where D and k * are known and n is unknown. Moreover, supposing n constant, P (D, k * , n) could be treated as a classical generalized eigenvalue problem. Now, we are ready to explain the idea of the whole method. Given the inverse scattering problem of detecting an inhomogeneous medium of index n, one can first reconstruct the shape of the unknown object D using the LSM (or other similar qualitative methods). Then, fixing a z in D and plotting the norm of g (solution of ( 23)) for every k in an interval, transmission eigenvalues k * of P (D, k, n) can be detected. With their knowledge, it will be possible to solve the problem of finding n * so that P (D, k * , n * ) has a non trivial solution; i.e. finding n * for which k * is a transmission eigenvalue of P (D, k, n * ).

Proposition 3.1. Assume that n is constant. Then, the homogeneous interior transmission problem

P (D, k, n) is equivalent to the problem of finding u ∈ H 2 0 (D) satisfying (∆ + k 2 )∆u = -k 2 n(∆ + k 2 )u in D. ( 25
)
Further, the eigenvalue problem: find (u, n) ∈ H 2 0 (D) × R satisfying (25), admits a countable number of eigenvalues n ∈ R.

Proof. The equivalence between (20)-(22) (with n constant) and (25) can be easily proved by setting

u = w -v ([2, 15]). Problem (25) is equivalent to (∆ + k 2 ) 2 u = λ(∆ + k 2 )u with λ = k 2 (1 -n). The operator (∆+k 2 ) 2 : H 2 0 (D) → H 2 0 (D)
is bijective and self adjoint while operator (∆+k 2 ) : H 2 0 (D) → H 2 0 (D) is compact, injective and self adjoint. Then, the existence of a countable number of real eigenvalues λ can be proved applying classical spectral theory for compact self adjoint operators.

The main difficulty related to numerical solutions of the eigenvalue problem (25) comes from the fourth order operator which requires the use of C 1 finite elements. We shall in the following provide mixed formulations of the problem that allows the use of C 0 finite elements. These formulations, inspired by classical mixed formulation for the biharmonic problem, are those used in the numerical simulations (done with standard P 1 Lagrange finite elements).

Proposition 3.2. Assume that n is constant. Then if u ∈ H 2 0 (D) ∩ H 3 (D) is such that (25) is valid, then u and u = (∆ + k 2 )u are solution of the variational problem: "Find u ∈ H 1 0 (D) and u ∈ H 1 (D) such that D (uv + ∇u∇v -k 2 uv -∇u∇v)dx = -k 2 n D (k 2 uv -∇u∇v)dx ( 26
)
for all v ∈ H 1 0 (D), v ∈ H 1 (D)". Conversely, if u ∈ H 1 0 (D) ∩ H 3 (D) and u ∈ H 2 (D)
are solutions of ( 26), then u ∈ H 2 0 (D) and satisfies (25).

Proof. Substituting u = (∆ + k 2 )u, equation ( 25) becomes ∆u = -k 2 nu. Multiplying both equations respectively against test functions v ∈ H 1 (D) and v ∈ H 1 0 (D) and applying Green's identity we obtain

D (∇u∇v -k 2 uv)dx = - D uvdx (27)
and

- D ∇u∇vdx = -k 2 n D (k 2 uv -∇u∇v)dx. (28)
Summing the two equations we obtain (26). Reciprocally, starting with (26) we can go back to equations (25) by assuming that u ∈ H 1 0 (D)∩ H 3 (D) and u ∈ H 2 (D). Using Green's identity we obtain that (27) is equivalent to

D (u -∆u -k 2 nu)vdx = ∂D ∂u ∂ν v ds(x) (29) for all v ∈ H 1 (D). Since C ∞ c (D) ⊂ H 1 (D) we get D (u -∆u -k 2 nu)vdx = 0 for all v ∈ C ∞ c (D)
. Hence, u -∆u -k 2 nu = 0 almost everywhere in D and consequently ∂u ∂ν = 0 almost everywhere on ∂D. The latter implies in particular that u ∈ H 2 0 (D). Similar arguments applies to (28) and prove that ∆u = -k 2 nu almost everywhere in D, which implies (25).

The case of an inhomogeneous background

The approach described in the previous subsection can be extended to the case in which the medium is embedded in an inhomogeneous background. Let us consider the case where the inclusion D is embedded in a background with index n b . Then the homogeneous interior transmission problem is formulated as follows:

∆w + k 2 n D w = 0 in D, ( 30
) ∆v + k 2 n b v = 0 in D, (31) w = v on ∂D, ( 32
) ∂w ∂ν = ∂v ∂ν on ∂D. ( 33
)
where n D denotes the index of D. Taking u = w -v in (30)-(33) and applying the operator ∆ + k 2 n D to the difference (30)-(31) we obtain

(∆ + k 2 n D ) 1 k 2 (n b -n D ) (∆ + k 2 n b )u = 0 in D (34) INRIA u = 0, ∂u ∂ν = 0 on ∂D. (35)
Substituting

u := 1 k 2 (n b -n D ) (∆ + k 2 n b )u (36) we obtain ∆u = -k 2 n D u in D. (37)
Multiplying (37) by a test function v ∈ H 1 0 (D) and (36) by

k 2 (n b -n D )v with v ∈ H 1 (D)
, applying the Green's formula, taking into account the boundary conditions satisfied by u and adding the two equations, one ends up with the following variational formulation

D (k 2 n b (uv -uv) -∇u∇v -∇u∇v)dx = λ D (uv + uv)dx, (38)
which is valid for all v ∈ H 1 0 (D) and v ∈ H 1 (D) and with λ := -k 2 n D . Using this procedure and similar arguments as in the proof of Proposition 3.2, one can prove the following.

Proposition 3.3. If n D is assumed to be constant and ∃w, v ∈ L 2 (D) and w -v =: u ∈ H 2 0 (D) ∩ H 3 (D) such that (30)-(33) is valid, then u and u = 1 k 2 (n b -n D ) (∆ + k 2 n b )u are solution of the variational problem: "Find u ∈ H 1 0 (D) and u ∈ H 1 (D) such that D (k 2 n b (uv -uv) -∇u∇v -∇u∇v)dx = λ D (uv + uv)dx (39) for all v ∈ H 1 0 (D), v ∈ H 1 (D)". Conversely, if u ∈ H 1 0 (D) ∩ H 3 (D)
and u ∈ H 2 (D) are solutions of (39), then u ∈ H 2 0 (D) and satisfies (34). Remark 3.1. In the case of inhomogeneous background, [START_REF] Colton | The linear sampling method in inverse electromagnetic scattering theory[END_REF] takes the form of

[F g z ](•) = G ∞ (•, z), ( 40
)
where G ∞ and F are defined as ( 17), taking into account the presence of the inhomogeneous background. As in the homogeneous case, k → g z L 2 (Ω) is expected to have peaks for k being a transmission eigenvalue for a.e. z ∈ D. This procedure allows us to compute the transmission eigenvalues from farfield data.

A remark on the case of absorbing background

It is well known that, when a non-absorbing object of support D is embedded in a background such that Im(n b ) > 0, then there are no real transmission eigenvalues associated with (30)-(33) [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF].

Since complex transmission eigenvalues are not detectable by the procedure previously indicated in (Remark 3.1), this means that our method providing estimates on the index of refraction is not applicable. However, in this preliminary remark something is not sufficiently precise: the definition of the background index inside D has no physical meaning since the "physical index" of the medium inside D is n. For backgrounds with constant indexes or for layered backgrounds, a "natural extension" of the definition of the background index inside D can be defined. But this is mainly only for a mathematical or numerical convenience.

This simple consideration suggests that there can be some degrees of freedom in the definition of n b inside D, which is the part used in the interior transmission problem (30)-(33). As far as the inverse problem of determining n is concerned, this is feasible since at this stage we assume that D is known. In particular if we choose n b to be a real constant (different from n D ) in D, then we are back again to the case where the interior transmission problem has real eigenvalues and therefore one would be able to apply the procedure previously explained to get estimates on n D . This procedure will be validated by some numerical experiment in the numerical section.

Of course there is an extra price to pay: it is hidden in the step of determining the transmission eigenvalue from farfield data. Indeed when one complexifies the background index, this makes the computation of u s b,∞ in (17) more costly. We finally observe that this procedure can also be applied in the case without absorption if one would like to use the (simpler) formulation of the interior transmission problem with constant index.

The case of TE polarization

Our results for TM polarized waves can be extended to the case of transverse electric (TE) polarization, i.e. the case for which the magnetic component of the field (originated by the incidence of a time harmonic plane wave) vibrates parallel to the axis of the infinite cylinder that is assumed to have cross section D. The main steps of the inversion algorithm are the same as in the TM case and we shall hereafter only give a sketch of the main differences.

Let us first quickly introduce the direct scattering problem for a scatterer D of refractive index n in the case of TE incident plane waves. The total field u ∈ H 1 loc (R 2 ) satisfies

∇ • ( 1 n(x) ∇u(x)) + k 2 u(x) = 0 in R 2 , (41) u
(x) = u i (x) + u s (x) with u i (x) = exp ikx•d , (42)
lim r→∞ √ r( ∂u s (x) ∂r -iku s (x)) = 0. ( 43
)
The associated farfield pattern as well as the farfield operator can be deduced as in the TM case. Theorem 2.1 is also still valid in the present case under the same assumption on the refractive index.

Transmission eigenvalues

In this case, the homogeneous interior transmission problem is defined as follows [START_REF] Cakoni | Houssem Haddar The computation of lower bounds for the norm of the index of refraction in an anisotropic media from farfield data[END_REF]:

find w, v ∈ L 2 (D) such that w -v =: u ∈ H 1 (D), ∇ • ( 1 n ∇ w -∇v) ∈ L 2 (D) and ∇ • ( 1 n ∇ w) + k 2 w = 0 in D, ( 44
) ∆v + k 2 v = 0 in D, (45) w
= v on ∂D, (46) 1 n
∂ w ∂ν = ∂v ∂ν on ∂D. ( 47
)
Since the Linear Sampling Method works independently from wave polarization, results obtained in the case of transverse magnetic waves as well as Theorem 3.1 and Corollary 3.1 could be extended to the case of transverse electric polarization. Hence, again, transmission eigenvalues are detectable from the knowledge of farfield patterns and the resolution of the eigenvalue problem P (D, k * 0 , n) can provide us with a constant approximation n * of n.

Mixed formulations of the refractive index eigenvalues problems

We shall explain here how problem (44)-(47) can be formulated as an eigenvalue problem for a constant n, assuming that k and D are known.

(∆ + k 2 )∆u = -k 2 n(∆ + k 2 )u in D (48) ∆u = 0, ∂u ∂ν = 0 on ∂D. ( 49
)
Proof. Substituting w = 1 n w and using ( 44) and (45) it is easy to obtain

∆w + k 2 nw = 0 in D, ( 50
) ∆v + k 2 v = 0 in D, (51) ∆w
= -k 2 w = -k 2 v = ∆v on ∂D (52) ∂w ∂ν = ∂v ∂ν on ∂D. (53)
The equivalence between (50)-( 53) and ( 48)-( 49) can be shown as is the TM case. 49) is valid, then u and u := ∆u are solution of the problem: "Find u ∈ H 1 (D) and u ∈ H 1 0 (D) such that

Proposition 4.2. Assuming n constant then: if u ∈ H 2 (D) ∩ H 3 (D) is such that (48)-(
D (uv + ∇u∇v + k 2 uv -∇u∇v)dx = -k 2 n D (k 2 uv -∇u∇v)dx. ( 54
)
for all v ∈ H 1 (D), v ∈ H 1 0 (D)". Conversely, if u ∈ H 3 (D) and u ∈ H 1 0 (D) ∩ H 2 (D)
are solutions of (54), then ( 48)-( 49) is satisfied by u.

Proof. Substituting u = ∆u, equation (48) becomes (∆+k 2 )u = -k 2 n(∆+k 2 )u. Multiplying both equations respectively against test functions v ∈ H 1 (D) and v ∈ H 1 0 (D) and applying Green's identity we obtain

D uvdx = - D ∇u∇vdx (55)
and

D (-∇u∇v + k 2 uv)dx = -k 2 n D (k 2 uv -∇u∇v)dx. (56)
Summing the two equations we obtain (54). The reverse implication is a classical exercise similar to the one done at the end of the proof of Proposition 3.2.

The case of an inhomogeneous background

The case of a scatterer embedded in an inhomogeneous background illuminated by TE polarized plane waves is more complex. Nevertheless, following the approach of [START_REF] Cakoni | Houssem Haddar The computation of lower bounds for the norm of the index of refraction in an anisotropic media from farfield data[END_REF], it is possible to write the homogeneous interior transmission problem as a generalized eigenvalue problem with λ = -k 2 n D as eigenvalues.

Let us again consider an inclusion D with index n D embedded in a background with index n b ; the homogeneous interior transmission problem is formulated as follows:

∇ • ( 1 n D ∇w) + k 2 w = 0 in D, (57)
∇ • ( 1 n b ∇v) + k 2 v = 0 in D, (58)
w = v on ∂D, (59) 1 n D ∂ ∂ν w = 1 n b ∂ ∂ν v on ∂D. (60)
Taking the gradient of (57) and (58) and substituting w = 1 n D ∇w and v = 1 n b ∇v in (57)-(60) we obtain

∇(∇ • w) + k 2 n D w = 0 in D, (61) ∇(∇ • v) + k 2 n b v = 0 in D, (62)
with boundary conditions

ν • w = ν • w and ∇ • w = ∇ • v on ∂D. (63)
Setting u = wv and taking the difference between ( 61) and ( 62) we obtain

1 k 2 (n b -n D ) (∇∇ • +k 2 n b )u = w. (64)
Then, applying (∇∇ • +k 2 n D ) to both sides,

(∇∇ • +k 2 n D ) 1 k 2 (n b -n D ) (∇∇ • +k 2 n b )u = 0 in D. ( 65
) Substituting U = 1 k 2 (n b -n D ) (∇∇ • +k 2 n b )u we obtain (∇∇ • +k 2 n D )U = 0 in D, which gives rise to the system (∇∇ • +k 2 n b )u -k 2 n b U = -k 2 n D U in D, (66)
∇∇ • U = -k 2 n D U in D, (67)
ν • u = 0 and ∇ • u = 0 on ∂D. (68)
The latter eigenvalue problem for eigenvalues λ = -k 2 n D can be formulated in a variational form as in the inhomogeneous TM case by respectively replacing the variational spaces H 1 (D) and

H 1 0 (D) with H div (D) := {V ∈ L 2 (D) 2 ; ∇ • V ∈ L 2 (D)} ∋ U,
and {v ∈ H div (D) ; ν • u = 0 on ∂D} ∋ u.
The numerical implementation of the resulting variational formulation requires the use of H div elements. We did not validate our procedure in the present case.

5 Numerical Experiments

The case where the shape is known

Before presenting some numerical results concerning the reconstruction of the shape and the constant approximation of the refractive index of unknown objects, let us show how -solving P (D, k * 0 , n) as a generalized eigenvalue problem -we obtain a good approximation of n when both D and its first transmission eigenvalue are known.

We focus on the case of a circular geometry and we shall investigate two configurations: in the first one, the refractive index is constant (Figure 1 In both cases the first transmission eigenvalue is analytically computable using the following result (that can be obtained using separation of variables).

Proposition 5.1. For a circle of radius R and refractive index n, if k * 0,TM and k * 0,TE denote the first transmission eigenvalues respectively in the case of transverse magnetic and transverse electric polarized waves, then

k * 0,TM = min{k|∃p ∈ N s.t. det(A TM p,k ) = 0}, k * 0,TE = min{k|∃p ∈ N s.t. det(A TE p,k ) = 0}, with A TM p,k := J p (kR) J p (k √ nR) kJ p-1 (kR) -p R J p (kR) k √ nJ p-1 (k √ nR) -p R J p (k √ nR) , A TE p,k := J p (kR) J p (k √ nR) kJ p-1 (kR) -p R J p (kR) 1 √ n (k √ nJ p-1 (k √ nR) -p R J p (k √ nR)) .
Proposition 5.2. For a circle made of two concentric layers of external radius R e and internal radius R i , if n e and n i respectively denote the refractive index of the exterior and interior layers and if k * 0,TM and k * 0,TE are the first transmission eigenvalues respectively in the case of transverse magnetic and transverse electric polarized waves, then Problem P (D, k * 0 , n) is solved with P 1 -finite elements based on the variational formulations ( 26) and (54). The discretization step is ∼ 1 10 of the wavelength inside the medium. The matrices are assembled using FreeFem++ [START_REF]FreeFem++[END_REF] and the associate eigenvalue problem is solved in Matlab with the preset function eigs, in modality 'sm' (i.e. look for smaller magnitude eigenvalues). Since the eigenvalue problem formulation correspond with a generalized eigenvalue problem for non self-adjoint matrices and with degenerate mass matrices, there is no guarantee for convergence of the algorithm. Moreover, the mixed formulation is equivalent to the original one only under regularity assumptions on the solution. This is why parasite values are observed in the numerical computations of the eigenvalues. These parasite values correspond to solutions u that do not satisfy the boundary condition ∂u/∂ν = 0 on ∂D. From the practical point of view we observed that these bad values can simply be cut off by respecting the lower bound n * = min{-λ/k * 2 0 > 1} [START_REF] Cakoni | On the use of transmission eigenvalues to estimate the index of refraction from farfield data[END_REF][START_REF] Cakoni | Houssem Haddar The computation of lower bounds for the norm of the index of refraction in an anisotropic media from farfield data[END_REF].

k * 0,TM = min{k|∃p ∈ N s.t. det(B TM p,k ) = 0}, k * 0,TE = min{k|∃p ∈ N s.t. det(B TE p,k ) = 0}, with B TM p,k :=      J p (kR e ) -J p (k √ n e R e ) -H (1)
p (k √ n e R e ) 0 J ′ p (kR e ) -J ′ p (k √ n e R e ) -H (1) ′ p (k √ n e R e ) 0 0 J p (k √ n e R i ) H (1) p (k √ n e R i ) -J p (k √ n i R i ) 0 J ′ p (k √ n e R i ) H (1) ′ p (k √ n e R i ) -J ′ p (k √ n i R i )      , B TE p,k :=      J p (kR e ) -J p (k √ n e R e ) -H (1)
p (k √ n e R e ) 0 J ′ p (kR e ) -1 ne J ′ p (k √ n e R e ) -1 ne H (1) ′ p (k √ n e R e ) 0 0 J p (k √ n e R i ) H (1) p (k √ n e R i ) -J p (k √ n i R i ) 0 1 ne J ′ p (k √ n e R i ) 1 ne H (1) ′ p (k √ n e R i ) -1 ni J ′ p (k √ n i R i )      .
Figure 2 shows results achievable when D is a circle of radius 0.5 and has a constant refractive index. The value of n is well recovered, with a better precision occurring for small n. As it is clear from Figure 2, with the increasing of the value of n, k * 0 decreases, clustering around k = 1. Since the computation of transmission eigenvalues is precise up to two digits, the previous consideration suggests that the difference in precision between reconstructions for small and large values of n is due to a not enough precise approximation of k * 0 for large n.

Understanding how transmission eigenvalues distribute depending on shape and dimension of the scatterer, on its index of refraction and on polarization of the wave, is, at present, still an open problem. The next few experiments are an overview of this complex relation between n * and different configurations of the material inside D which is still a circle of radius 0.5. With Figures 5 and6, (respectively for TM and TE polarization), we want to investigate how different structures of n would lead to different transmission eigenvalues (and consequently to different n * ) even if min D n and max D n are not modified. In both cases we consider a circular scatterer of radius 0.5 containing a concentric circular layer of varying radius. For fixed values of the refractive index n i = 6 and n e = 12 we show how k * 0 and n * behave with respect to the radius of the internal circle R i . As expected, since the first transmission eigenvalue decreases with respect to the index of refraction [START_REF] Cakoni | Houssem Haddar The existence of an infinite descrete set of transmission eigenvalues[END_REF], it should be increasing with respect to R i (since n e > n i ) and therefore decreasing with respect to the area of the external layer A e . However, the dependence is completely nonlinear. The values of n * have reversed monotonicity property with respect to the first transmission eigenvalue and therefore increases with A e . The behavior of n * is also nonlinear as demonstrated by the comparison with the weighted average of the exact n.

Numerical results for cases where the geometry is not known a priori

In this last section we present some numerical experiments and results that are related to a more realistic inverse scattering experiment: i.e. a situation where both the geometry D of the scatterer and its refractive index n are unknown and the only data at our disposal are the farfields associated with many incident directions. Since in the previous section it has been shown that the method performance for TM polarized waves and for TE polarized waves is similar, we show only simulations for TM waves. The inversion algorithm is the following:

• Apply the LSM algorithm (as described in Section 2) to find an approximation D of the shape of D.

• Determine the smallest transmission eigenvalue k * 0 (for each connected component of D) using the method described in Section 3.

• Solve the eigenvalue problem P ( D, k * 0 , n) to obtain a constant approximation n * of the exact index n inside D.

Following this scheme, for each example, we shall present both the LSM reconstruction and the graph of k → g z L 2 (Ω) for several z ∈ D. Simulations are done using 20 emitting antennas, 20 measuring antennas and by corrupting the data with 5% random noise.

Figures are collected in panels (Figures 78910111213) and the values of the corresponding constant approximation of n are indicated in Tables 1234567.

The case of circular geometries.

In this case comparisons can be made with the exact calculations made in previous section for transmission eigenvalues. It has also the advantage of allowing fast computation of the Green function for inhomogeneous (layered) backgrounds and fast implementation of LSM (see Appendix).

The first example corresponds with the two layered configuration of D (shown in Figure 1right) with an external radius 0.5 and an internal radius 0.354, embedded in the vacuum where we considered four configurations for the indexes of refraction n = n e for the exterior layer and n = n i for the interior layer. Figure 7-(a1-a2) shows the LSM reconstruction for n e = 22 and n i = 19; for four different configurations of the index of refraction (n e = 8, n i = 8), (n e = 11, n i = 5), (n e = 22, n i = 19) and (n e = 67, n i = 61), we present k → g z L 2 (Ω) that allows us to determine the first transmission eigenvalue. The corresponding values of n * are collected in Table 1. 2,3, describe the results achievable when the two layered inclusion of previous experiment is now embedded in an inhomogeneous background. The inhomogeneous background consists into a circular domain of radius 0.75 and with refractive index n b embedded in the vacuum.

Figure 8 and Table 2 show how the method perform for objects embedded in a non-absorbing background. As expected, we observe that the transmission eigenvalues are different from those corresponding to the homogeneous background but the estimates n * are roughly the same. 8.

Figure 9 and Table 3 correspond with the case of an absorbing background. We develop here the idea explained in Section 3.3.1 and show that, when an artificial non-absorbing background of index ñb is used inside the reconstructed domain, real transmission eigenvalues can be computed, allowing to find roughly the same values of n * as in the non absorbing case. We remark that in this configuration the accuracy of D (approximation of the shape of D) has a stronger influence on the accuracy of n * since it affects the computation of the incident field for the background. The latter is used in computing g z L 2 (Ω) and therefore affect the precision of the first transmission eigenvalue.

Cases of non circular geometries

In Figure 10 and Figure 11 the case of scatterers that are not circular is investigated. In both figures a two layered inclusion is considered. The exact profile of D, the LSM reconstruction for n e = 22, n i = 19 and transmission eigenvalues for (n e , n i ) = (8, 8), [START_REF] Colton | Analytical and computational methods for transmission eigenvalues[END_REF][START_REF] Cakoni | Houssem Haddar On the determination of Dirichlet or transmission eigenvalues from farfield data[END_REF], (22,19) are respectively shown for a layered ellipse and for an ellipse embedded into a rectangle. Tables 4 and5 present the values of n * achievable using the values of k * 0 and the extracted profile D. One can draw roughly similar conclusions as in the case of circular geometries.

Cases of multiply connected inclusions

Finally, inspired by Corollary 3.1, we present a simulation for a scatterer consisting in two distinct objects of different refractive indexes. Figures 12 and13 jointly with Tables 6 and7 prove that a reconstruction of both indexes of refraction is effectively achievable by considering k → g z L 2 (Ω) first for several z in one connected component and second for several z in the other connected component. As one cas observe in part (b 1 ) of Figure 13, small peaks are present in k → g z L 2 (Ω) preceding the first transmission eigenvalue. A closer look reveals that those peaks correspond with the transmission eigenvalues of the other connected component (see part (b 3 )). We think that this is due to the fact that the two connected components are close to each others. For instance, such perturbation is not observed in the case of Figure 12 where the distance between the two connected components is larger. A Fast implementation of LSM in the case of a background with axisymmetric index

We here explain a procedure that accelerates the LSM algorithm for inhomogeneous backgrounds that have piecewise constant and axisymmetric indexes. In this configuration, by exploiting the analytic expression of a solution to the Helmholtz equation (obtained by separation of variables) and by imposing a single regularization to the family of equations ( 7), we shall indicate how one can avoid a computation of the right hand sides for all sampling points, and therefore speed up the inversion of the data. Such a procedure can be useful in situations in which, for reasons related to the resolution power of the output or related to the material properties, we need to reconstruct the shape of the scatterer on a grid made of a large amount of points. We explain the procedure in the case of TM polarized wave (the case of TE case can be treated in a similar way). m = 0 and β 0 m = 0 correspond to the expansion of the incident field in the neighborhood of the outer boundary. More precisely, using the Jacobi-Anger expansion [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF], the incident field is expanded as

u i (x) = +∞ m=-∞ β 0 m J m (kr x ) exp imθx . (71)
The remaining coefficients in expansion (70) are determined by imposing the continuity of the total field and of its normal derivative on the interfaces. One obtains for each harmonic m a system of 2L equations for 2L unknowns. The matrix of the obtained system is independent of the right hand side. This enables for instance a quick evaluation of the background farfields (see ( 17)) since

u s ∞,b (x, d) = +∞ m=-∞ α 0 m (d) √ 2 π exp -iπ 2 (2m+1) exp imθ x . (72)
One can also use this technique to evaluate the background Green tensor. In that case, G(•, z) has similar expression as (70) in the layers that does not contain z with the additional condition that β 0 m = 0. In the layer that contain z we add to that expression i 4 H

(1) 0 (k n(x)|x -z|). Using the addition formula in order to expand x → i 4 H

(1) 0 (k n(x)|x -z|) as a Fourier series one ends up, as in the previous case, with a 2L × 2L system to solve for each harmonic or order m, with a matrix independent of the point z. We therefore infer an expression of G ∞ (see (17)) of the form

G ∞ (x, z) = +∞ m=-∞ ξ 0 m (z) √ 2 π exp -iπ 2 (2m+1) exp imθ x . (73)
The linear sampling method requires the resolution of

F g z = G ∞,z (74)
for each z point of R 2 belonging to a grid containing the scatterer and where F is defined in (17). If F * denotes the adjoint of F , the Tikhonov regularization of ( 74) is

(F * F + α z I)g z = F * G ∞,z (75)
which has to be solved for each z belonging to a grid containing the unknown object. Supposing α z independent from z, i.e. supposing to choose the regularization parameter only once during all the reconstruction process (for instance, in [START_REF] Aramini | The linear sampling method without sampling[END_REF] a no sampling approach to the linear sampling method is proposed and the problem of properly choosing α is analyzed), some simplification inside the far field equation is possible. In fact, defining

assuming that we truncate the infinite series at M ∈ N. The solution g z can be expanded as

g z = M m=-M ξ 0 m (z)g m , (78)
where the functions g m are independent from z and are solutions to:

(F * F + α z I)g m = F * ψ m . (79)
Practical experiments suggest that generally M does not need to be large (due to the behavior of Bessel functions) and, in particular the needed M is much smaller than the dimension of the grid points for z.

INRIA

Proposition 4 . 1 .

 41 Considering n constant, problem (44)-(47) is equivalent to the problem of findingu := 1 n w -v ∈ H 2 (D) such that

Figure 1 :

 1 Figure 1: Configuration of the refractive index in a circular domain D: homogeneous case (a), piecewise constant case (b).

Remark 5 . 1 .

 51 Studying or plotting the behavior of det(A TM p,k ), det(A TE p,k ), det(B TM p,k ), det(B TE p,k ) for k ∈ R, one can observe that few p ∈ N (the firsts) are enough for determining k * 0 .

Figure 2 :

 2 Figure 2: Reconstructed index for a homogeneous circle of radius 0.5. TM case (left) and TE case (right).

Figures 3 and 4

 4 Figures 3 and 4 are associated with the two layered case where the areas of the two layers are equal (R e = 0.5 and R i = 1 2 √ 2 ). The values of n e and n i are varied by maintaining the difference n e -n i equals to 6, (for example (n e , n i ) = (8, 2), (9, 3), ...) for Figure 3. As in previous figures, we plot the value of the reconstructed value n * of n (y-axis) against the first transmission eigenvalue (x-axis); blue lines corresponding to the value of n i and n e . They are over plotted in order to clarify wether or not n * fall in [n i , n e ]. In Figure 4 the role of n e and n i are reversed to emphasize how a different structure (and not only different values of n or different layer areas) could affect the reconstruction.

Figure 3 :

 3 Figure 3: Reconstruction of the index in the case of a two layered circle (R e = 0.5 and R i = 1 2 √ 2 ). The exact refractive index is such that n e -n i = 6. The reconstructed value is shown by * and the solid line indicate the exact values of n e and n i . TM case (left) and TE case (right).

Figure 4 :

 4 Figure 4: Same settings as in Figure 3 but with n i -n e = 6. TM case (left) and TE case (right). In the previous examples one observes that it is not simple to relate the value of the reconstructed constant index n * with non-constant refractive indexes. This is confirmed by next examples where we have not been able to observe a relation more accurate than min x∈D n(x) ≤ n * ≤ max x∈D n(x).(69)

Figure 5 :

 5 Figure 5: Results for TM polarized incident waves in the case of a two layered circle with R e = 0.5 and a varying R i . The exact index values are n i = 6 and n e = 12. The x-axis represents the values of A e /A tot , ratio of external layer area to the total circle area. Left: first transmission eigenvalue. Right: reconstructed n * .

Figure 6 :

 6 Figure 6: Results for TE polarized incident waves in the case of a two layered circle with R e = 0.5 and a varying R i . The exact index values are n i = 6 and n e = 12. The x-axis represents the values of A e /A tot , ratio of external layer area to the total circle area. Left: first transmission eigenvalue. Right: reconstructed n * .

Figure 7 :

 7 Figure 7: Two layered circular inclusion embedded in the vacuum (Figure 1-right). (a 1 ) LSM reconstruction at k = 4.2, n e = 22 and n i = 19. (a 2 ) reconstructed shape (red) versus exact shape (blue). (a 3 ), (b 1 ), (b 2 ), (b 3 ) plots of k → g z L 2 (Ω) for several choices of z, respectively for (n e , n i ) = (8, 8), (11, 5), (22, 19), (67, 61).

Figure 8 :

 8 Figure 8: Two layered circular inclusion (a 1 ) embedded in an inhomogeneous background (a 2 ). (a 3 ): reconstructed shape (red) versus exact shape (blue). (b 1 ), (b 2 ), (b 3 ): plots of k → g z L 2 (Ω) for several choices of z, respectively for (n e , n i , n b ) = (8, 8, 2), (11, 5, 2), (67, 61, 10).

Figure 9 :

 9 Figure 9: Two layered circular inclusion (n e , n i ) = (8, 8), embedded in an inhomogeneous absorbing background (a 3 ) with n b = 2 + ı. (a 1 ): reconstructed shape (red and green) versus exact shape (blue). (a 2 ): considered artificial background with ñb = 1 inside D. (b 1 ), (b 2 ), (b 3 ): plots of k → g z L 2 (Ω) for several choices of z, respectively for the exact shape and artificial background, reconstructed shape (red) and artificial background, exact shape and true background. (c 1 ), (c 2 ), (c 3 ): same as (b 1 ), (b 2 ), (b 3 ) but for the case (n e , n i ) = (11, 5) and for the green reconstructed shape.

Figure 10 :

 10 Figure 10: Two layered ellipsoidal inclusion (a 1 ) embedded in the vacuum. (a 2 ): LSM reconstruction for k = 4.16, n e = 22 and n i = 19. (a 3 ): reconstructed shape (red) versus exact shape (blue). (b 1 ), (b 2 ), (b 3 ): plots of k → g z L 2 (Ω) for several choices of z, respectively for (n e , n i ) = (8, 8), (11, 5), (22, 19).

Figure 11 :

 11 Figure 11: Rectangular inclusion containing an ellipsoidal layer (a 1 ) embedded in the vacuum. (a 2 ): LSM reconstruction for k = 5.64, n e = 22 and n i = 19. (a 3 ): reconstructed shape (red) versus exact shape (blue). (b 1 ), (b 2 ), (b 3 ): plots of k → g z L 2 (Ω) for several choices of z, respectively for (n e , n i ) = (8, 8), (11, 5), (22, 19).

Figure 12 :

 12 Figure 12: Two homogeneous circular inclusions (a 1 ): a circle c 1 centered at (0, 0) of refractive index n = 8 and a circle c 2 centered at (5, 5) of refractive index n = 25. (a 2 ): LSM reconstruction for k = 4.2. (a 3 ) and (b 2 ): reconstructed shape (red) versus exact shape (blue) respectively for c 1 and c 2 . (b 1 ) and (b 3 ): plots of k → g z L 2 (Ω) for several choices of z, respectively for c 1 and c 2 .

Figure 13 :

 13 Figure 13: Two homogeneous circular inclusions (a 1 ): a circle c 1 centered at (0, 0) of refractive index n = 8 and a circle c 2 centered at (5, 5) of refractive index n = 25. (a 2 ): LSM reconstruction for k = 4.2. (a 3 ) and (b 2 ): reconstructed shape (red) versus exact shape (blue) respectively for c 1 and c 2 . (b 1 ) and (b 3 ): plots of k → g z L 2 (Ω) for several choices of z, respectively for c 1 and c 2 .

Table 7 :

 7 Transmission eigenvalues and reconstructed index associated with Figure13.In the case of incident plane waves, the solution to the Helmholtz equation in the case of piecewise constant and axisymmetric indexes is of the formu(x) = +∞ m=-∞ [α ℓ m H (1) m (kr x n(x)) + β ℓ m J m (kr x n(x))] exp imθx(70)in the region ℓ where n(x) is constant, where (r x , θ x ) are the polar coordinates of x and where H m and J m are respectively Hankel and Bessel functions. The coefficients α ℓ m and β ℓ m only depend on the incident field and on the geometry. If ℓ = 0 denotes the outer domain (vacuum) and ℓ = L denotes the inner homogeneous domain then α L

2 (

 2 ψ m (x) := √ 2 π exp -iπ 2 (2m+1) exp imθx (76) we have that ∀x ∈ Ω G ∞ (x, z) = 2m+1) exp imθ x = M m=-M ξ 0 m (z)ψ m (r, θ x),

Table 1 :

 1 Transmission eigenvalues and reconstructed index associated with Figure7.

	n e	n i	k * 0	n * -exact shape	n * -reconstructed shape
	8	8	2.98	8.07	7.61
	11	5	3.27	7.05	6.69
	22	19	1.76	20.28	18.86
	67	61	0.97	64.11	59.42
	Figures 8, 9 and corresponding Tables

), (22, 19), (67, 61).

Table 2 :

 2 ). Transmission eigenvalues and reconstructed index associated with Figure

	n e	n i	n b	k * 0	n * -exact shape	n * -reconstructed shape
	8	8	2	4.11	8.09	7.70
	11	5	2	4.89	6.92	5.04
	67	61	10	1.11	64.35	61.44

Table 3 :

 3 Transmission eigenvalues and reconstructed index associated with Figure9.

	n e	n i	n b	ñb	(k * 0 , n * ) -exact shape (k * 0 , n * ) -reconst. shape
	8	8	2+ı	1	(2.98, 8.07)	(2.96, 7.69)
	11	5	2+ı	1	(3.27, 7.05)	(3.28, 7.43)

Table 4 :

 4 ). Transmission eigenvalues and reconstructed index associated with Figure10.

	n e	n i	k * 0	n * -exact shape	n * -reconstructed shape
	8	8	3.42	8.08	7.74
	11	5	2.81	11.10	10.50
	22	19	1.92	22.16	20.73

Table 5 :

 5 ). Transmission eigenvalues and reconstructed index associated with Figure11.

	n e	n i	k * 0	n * -exact shape	n * -reconstructed shape
	8	8	3.05	8.94	9.37
	11	5	2.48	12.62	13.27
	22	19	1.69	25.53	26.90

Table 6 :

 6 Transmission eigenvalues and reconstructed index associated with Figure12.

	circle c 1	n 8	k * 0 2.99	n * -exact shape 8.03	n * -reconstructed shape 7.17
	c 2	25	1.57	25.17	24.60

2 .

RR n°7729