Clair Poignard

RIGOROUS ASYMPTOTICS FOR THE ELECTRIC FIELD IN TM MODE AT MID-FREQUENCY IN A BIDIMENSIONAL MEDIUM WITH A THIN LAYER

Keywords: 1991 Mathematics Subject Classification. 34E05, 34E10, 35J05 Asymptotics, Helmholtz equation, Thin Layer, Transmission conditions. 1

Consider an ambient medium and a heterogeneous entity composed of a bidimensional material surrounded by a thin membrane. The electromagnetic constants of these three materials are different. By analogy with biological cells, we call this entity a cell. We study the asymptotic behavior of the electric field in the transverse magnetic (TM) mode, when the thickness h of the membrane tends to zero. We provide a rigorous derivation of the first two terms of the asymptotic expansion for h tending to zero. In the membrane, these terms are given explicitly in local coordinates in terms of the boundary data and of the function f , while outside the membrane they are the solutions of a scalar Helmholtz equation with appropriate boundary and transmissions conditions given explicitly in terms of the boundary data. We prove that the remainder terms are of order O(h 3/2 ). In addition, if the complex dielectric permittivity in the membrane, denoted by zm, tends to zero faster than h, we give the difference between the exact solution and the above asymptotic with zm = 0; it is of order O(h 3/2 + |zm|).

Introduction

We study in this paper the behavior of the solution of Helmholtz equation in a bidimensional medium in transverse magnetic (TM) mode (see Balanis and Constantine [START_REF] Balanis | Advanced Engineering Electromagnetics[END_REF]). The medium is made out of three materials: a central region surrounded by a thin membrane of thickness h, with θ a curvilinear coordinate, and a third material, which is not assumed to be thin; see Fig. 1. This assemblage is submitted to a field of pulsation ω; after proper scalings, ω is included in the complex dielectric permittivity, which may be different in the three materials. By analogy with the biological cell, we call this entity a cell in an environment. In this article, we show that as the thickness of the membrane tends to zero, i.e as h tends to zero, the electric field tends to the solution of a Helmholtz equation with an appropriate transmission condition at the boundary between the cell and the ambient medium. This work is a sequel to the author's former article on the static case [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF].

Let us give now precise notations. Let Ω be a bidimensional bounded domain composed of three subdomains: a bounded domain O c surrounded by a thin membrane O h with small thickness h, and an exterior domain O e,h : We suppose that the cell is strictly embbeded in the ambient domain, that is:

Ω = O c ∪ O h ∪ O e,h . q c µ c q m µ m q e µ e h O c O h O e,h
(O c ∪ O h ) ∩ ∂Ω = ∅.
The closed curves Γ 0 and Γ h are supposed to be of class C ∞ . We denote by Γ 0 and Γ h respectively the boundaries of ∂O c and of ∂O h ∩ ∂O e,h :

Γ 0 = ∂O c , Γ h = ∂O h ∩ ∂O e,h .
Let µ e , µ m and µ c be the magnetic permittivities: they are constant and strictly positive. Let q e , q m and q c be three complex numbers with strictly negative imaginary part and strictly positive real part; they are non dimensionalized complex permittivities ( see [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF] or [START_REF] Poignard | Méthodes asymptotiques pour le calcul de champs électromagnétiques[END_REF] for a description of the non dimensionalization). We define piecewise constant functions µ and q on Ω as follows:

µ =      µ e , in O e,h , µ m , in O h , µ c , in O c , q =      q e , in O e,h , q m , in O h , q c , in O c .
To simplify, we denote by z the product µq, and z e , z m and z c designate the restrictions of z respectively to the domains O e,h , O h and O c . We summarize our hypotheses in Fig 1 .  Let φ be a given function on ∂Ω. Its regularity will be chosen later on. We consider the electric field u, which solves the following Helmholtz equation with Neumann boundary condition:

∇.
1 µ ∇u + qu = 0, in Ω, (1a)

∂ n u| ∂Ω = φ, on ∂Ω. (1b)
Here ∂ n denotes the derivative in the direction of the vector n: n is the exterior normal to Γ 0 , and is oriented by continuity on Γ h and also on ∂Ω (see Fig. 1). Let us denote by u e , u h and u c the restrictions of u respectively to the domains O e,h , O h and O c . These restrictions satisfy the following transmission conditions:

1 µ c ∂ n u c | Γ0 = 1 µ m ∂ n u h | Γ0 , (2a) 1 µ e ∂ n u e | Γ h = 1 µ m ∂ n u h | Γ h , (2b)
u c | Γ0 = u h | Γ0 , (2c)
u e | Γ h = u h | Γ h . (2d)
We would like to understand the behavior for h tending to zero of the solution u of Problem [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF].

In our proof, we assume that µ m and q m are given constants; µ c , µ e , q c and q e could be continuous functions of the spatial coordinates with the imaginary part of q c and q e bounded away from zero, without changing the argument.

Beretta and Francini have worked on a similar problem in [START_REF] Beretta | Asymptotic formulas for perturbations in the electromagnetic fields due to the presence of thin inhomogeneities[END_REF]. They considered a thin dielectric material O h in an ambient medium, and they studied a Helmholtz equation with Dirichlet boundary condition. They compared on the boundary of the domain ∂Ω the exact solution to the so-called background solution defined by replacing the material of the membrane by the interior material. The difference between these two solutions restricted to the boundary ∂Ω is then given through an integral involving the polarization tensor, plus some remainder terms. This polarization tensor is defined for instance in [START_REF] Ammari | Properties of the generalized polarization tensors[END_REF], [START_REF] Ammari | Reconstruction of conductivity inhomogeneities of small diameter via boundary measurements[END_REF], [START_REF] Beretta | Asymptotic formulas for steady state voltage potentials in the presence of thin inhomogeneities. A rigorous error analysis[END_REF], [START_REF] Beretta | Asymptotic formulas for steady state voltage potentials in the presence of conductivity imperfections of small area[END_REF], [START_REF] Capdeboscq | A general representation formula for boundary voltage perturbations caused by internal conductivity inhomogeneities of low volume fraction[END_REF]. In this paper, we do not use this approach since we are interested in the transmembranar potential (see Fear and Stuchly [START_REF] Fear | Modeling assemblies of biological cells exposed to electric fields[END_REF]), and in the behavior of the field in the whole domain. We work with bidimensional domains and we expect that the same analysis could be performed in higher dimensions.

The heuristics of this work are the same as in [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF]. We parameterize O h by local coordinates (η, θ) varying in the h-independent domain [0, 1] × R/LZ. Here L is simply the length of the curve ∂O c A change of coordinates in the membrane O h is performed, so as to parameterize it by local coordinates (η, θ), which vary in a domain independently of h; in particular, if we denote by L the length of ∂O c , the variables (η, θ) belong to [0, 1] × R/LZ. This change of coordinates leads to h-independent expression of the Laplacian in the membrane. Once the transmission conditions of the new problem are derived, we perform a formal asymptotic expansion of the solution of (1) in terms of h. It remains to validate this expansion.

This paper is structured as follows. First, we suppose that the parameters q c , q m , q e , µ c , µ m and µ e are constant with respect to space and to h and do not vanish.

Moreover we assume that the imaginary parts of z c , z m and z e , which always have the same sign for physical reasons are negative and bounded away from 0.

In Section 1, we define our geometric conventions. We perform the above described change of variables in the membrane. We refer the reader to [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF] for more information on the local coordinates. In Section 3, we derive formally the first two terms of the asymptotic expansion of the solution of our problem in terms of h. Section 4 contains regularity result, which is necessary for estimating the error, and Section 5 is devoted to estimating the error.

In addition, in Remark 8, we give the first two terms of the asymptotic expansion of the electric field for a thin membrane on the boundary of O c , and in Remark 10, we consider the case z m = 0. The proofs of these asymptotics are very close to the proof performed in Section 5 and therefore, they are omitted.

In the case of a biological cell, µ is identically equal to 1, z c and z e are constants as above and z m is very small. In Section 6, we show that if |z m | is small compared to h, we just have to replace z m by 0 in the asymptotics found in Section 3 to obtain the electric field in all of the domain Ω with an error in O(h

3/2 + |z m |).
Let us present now our main result.

Main result. We give the first two terms of the asymptotic expansion of the above function u for h tending to zero, and we estimate rigourously the error made by this aproximation.

• The 0 th order terms. The electric fields u e 0 and u c 0 are solution of the following problem in Ω:

∆u e 0 + z e u e 0 = 0, in Ω \ O c , ∆u c 0 + z c u c 0 = 0, in O c , (3a)
with transmission conditions

u c 0 | Γ0 = u e 0 | Γ0 , (3b) 1 µ c ∂ n u c 0 | Γ0 = 1 µ e ∂ n u e 0 | Γ0 , (3c)
and with Neumann boundary condition

∂ n u e 0 | ∂Ω = φ. (3d)
• The first order terms. The fields u e 1 and u c 1 are solution of the following problem in Ω:

     ∆u e 1 + z e u e 1 = 0, in Ω \ O c , ∆u c 1 + z c u c 1 = 0, in O c , ∂ n u e 1 | ∂Ω = 0, (4a)
with the transmission conditions

1 µ c ∂ n u c 1 | Γ0 - 1 µ e ∂ n u e 1 | Γ0 = 1 µ m - 1 µ e ∂ 2 t u c 0 | Γ0 + (q m -q e ) u c 0 | Γ0 , (4b) u c 1 | Γ0 -u e 1 | Γ0 = µ e -µ m µ c ∂ n u c 0 | Γ0 . (4c)
We have the following theorem.

Theorem 1. Let O c be a bouded domain with smooth boundary, and denote by κ the curvature of ∂O in local coordinates. Let h 0 be such that:

0 < h 0 < 1 κ ∞ .
Let h belong to (0, h 0 ) and φ to H s (∂Ω), s > 7/2. We denote by u the solution of Problem [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF]. Define (v e , v c ) in Ω by:

v e = u e 0 + hu e 1 , in Ω \ O c , v c = u c 0 + hu c 1 , in O c . Then there exists an h-independent constant C > 0 such that u -v c H 1 (Oc) ≤ Ch 3/2 φ H s (∂Ω) , u -v e L 2 (O h ) + 1 µ m ∇u - 1 µ e ∇v e L 2 (O h ) ≤ Ch 3/2 φ H s (∂Ω) , u -v e H 1 (O e,h ) ≤ Ch 3/2 φ H s (∂Ω) .

Geometry of the problem

The boundary of the domain O c is assumed to be smooth. The boundary Γ 0 is counterclockwise oriented, and we denote by ∂ t the tangential derivative along O c . Thanks to a change of units of length, we may suppose that the length of Γ 0 is equal to 2π. We denote by T the flat torus:

T = R/2πZ.
Since Γ 0 is of class C ∞ , we can parameterize it by a smooth function Ψ from T to R 2 satisfying:

∀θ ∈ T, |Ψ ′ (θ)| = 1.
The following identities hold: Γ 0 = {Ψ(θ), θ ∈ T}, and Γ h = {Ψ(θ) + hn(θ), θ ∈ T}.

Here n(θ) is the unitary exterior normal at Ψ(θ) to Γ 0 . We parameterize the membrane O h as follows:

O h = {Ψ(θ) + hηn(θ), (η, θ) ∈]0, 1[×T}.
We define now:

Φ(η, θ) = Ψ(θ) + hηn(θ).
We denote by κ the curvature of the Γ 0 in curvilinear coordinate, and by K the curvature in Euclidean coordinates:

∀x ∈ ∂O, K = κ o Φ -1 0 (x). ( 5
)
Let h 0 satisfy:

0 < h 0 < 1 κ ∞ . (6)
Then, for all h in [0, h 0 ], there exists an open interval I containing (0, 1) such that Φ is a diffeomorphism of class C ∞ from I × R/2πZ to its image, which is a neighborhood of the membrane. The metric in O h is given by:

h 2 dη 2 + (1 + hηκ(θ)) 2 dθ 2 . ( 7
)
We use two systems of coordinates, depending on the domains O e,h , O c and O h : in the interior and exterior domains O e,h and O c , we use Euclidean coordinates (x, y) and in the membrane O h , we use local coordinates with metric (7).

Statement of the problem

In this section, we express our Problem (1) in local coordinates. It is convenient to write:

∀θ ∈ T, Φ 0 (θ) = Φ (0, θ) , Φ 1 (θ) = Φ (1, θ) .
Let us denote by u e and u c respectively the electric field in O e,h and in O c , written in Euclidean coordinates, and by u m the electric field in O h in the local coordinates:

u e = u, in O e,h , u c = u, in O c , u m = u o Φ, in [0, 1] × T.
We have shown in [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF] that the Laplacian in the local coordinates is given by:

∆| Φ(η,θ) = 1 h(1 + hηκ) ∂ η 1 + hηκ h ∂ η + ∂ θ h 1 + hηκ ∂ θ . (8)
Therefore, we rewrite Problem (1) as follows:

∆u e + z e u e = 0, in O e,h , (9a)
∆u c + z c u c = 0, in O c , (9b) ∀ (η, θ) ∈ [0, 1] × T, ∂ η 1 + hηκ h ∂ η u m + ∂ θ h 1 + hηκ ∂ θ u m + z m hf (1 + hηκ)u m = 0, (9c)
with transmission conditions (2) expressed in local coordinates at η = 0:

1 µ c ∂ n u c o Φ 0 = 1 hµ m ∂ η u m η=0 , (9d)
u c o Φ 0 = u m | η=0 , (9e) at η = 1: 1 µ e ∂ n u e o Φ 1 = 1 hµ m ∂ η u m η=1 , (9f) u e o Φ 1 = u m | η=1 , (9g)
and with boundary condition

∂ n u e | ∂Ω = φ. (9h)

Formal asymptotic expansion

In this section, we derive asymptotic expansions of the electric field (u e , u c , u m ) solution of ( 9) in terms of the parameter h. In the limit, we want to be able to replace the membrane by transmission conditions.

We multiply (9c) by h(1 + hηκ) 2 and we order the result in powers of h, in order to obtain the partial differential equation (PDE) satisfied by u m :

∀(η, θ) ∈ [0, 1] × T, ∂ 2 η u m + hκ 3η∂ 2 η u m + ∂ η u m + h 2 3η 2 κ 2 ∂ 2 η u m + 2ηκ 2 ∂ η u m + ∂ 2 θ u m + z m u m + h 3 η 3 κ 3 ∂ 2 η u m + η 2 κ 3 ∂ η u m + ηκ∂ 2 θ u m -ηκ ′ ∂ θ u m + 3z m ηκu m + 3h 4 z m η 2 κ 2 u m + h 5 η 3 κ 3 z m u m = 0 (10)
We assume the following ansatz:

u e = u e 0 + hu e 1 + • • • , (11a)
u c = u c 0 + hu c 1 + • • • , (11b)
u m = u m 0 + hu m 1 + • • • . (11c)
We will to derive the first two terms of the asymptotic expansions of u e , u c and u m by identifying the terms of coefficients of a given power of h.

We extend formally u e to Ω \ O c , by extending a finite number of coefficients of the powers of h. Moreover, we suppose that φ is as regular as needed. We will also need the first two terms of

u e o Φ(η, θ) = u e o (Ψ(θ) + hηn(θ))
and ∂ n u e o Φ(η, θ). This amounts to composing two asymptotics series. We remember that we introduced Φ 0 = Ψ and Φ 1 = Φ(1, •) to homogeneize our notations. A simple calculation gives:

u e o Φ = u e o Φ 0 + h (u e 1 o Φ 0 + η∂ n u e 0 o Φ 0 ) + • • • , and similarly ∂ n u e o Φ = ∂ n u e o Φ 0 + h ∂ n u e 1 o Φ 0 + η∂ 2 n u e 0 o Φ 0 + • • • .

These expansions enable us to rewrite transmission condition (9f) as:

hµ m µ e ∂ n u e 0 o Φ 0 + h ∂ n u e 1 o Φ 0 + ∂ 2 n u e 0 o Φ 0 + • • • = ∂ η u m 0 | η=1 + h ∂ η u m 1 | η=1 + h 2 ∂ η u m 2 | η=1 + • • • , (12a)
and transmission condition (9g) as

u e 0 o Φ 0 + h (u e 1 o Φ 0 + ∂ n u e 0 o Φ 0 ) + • • • = u m 0 | η=1 + hu m 1 | η=1 + • • • . (12b)
Observe that we have chosen to limit the order of explicit asymptotic expansions to what will be needed below.

We systematically substitute the fields u e , u c and u m by their asymptotic expansion [START_REF] Fear | Modeling assemblies of biological cells exposed to electric fields[END_REF] in [START_REF] Capdeboscq | A general representation formula for boundary voltage perturbations caused by internal conductivity inhomogeneities of low volume fraction[END_REF]. For transmission condition, at η = 1, it is more convenient to use transmission conditions [START_REF] Yan | Gradient estimates for solutions to divergence form elliptic equations with discontinuous coefficients[END_REF] instead of (9f)-(9g).

We are going to select all terms of an appropriate order in these expanded equations in order to get the conditions satisfied by u m 0 , u e i , u c i and u m i+1 (i = 0, 1).

First step : identification of 0 th order terms. Substituting into [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF] the field u m by its expansion (11c) we obtain:

∂ 2 η u m 0 = 0, ∀(η, θ) ∈ (0, 1) × T. (13)
Moreover, we obtain easily:

∆u e 0 + z e u e 0 = 0, in Ω \ O c , (14a) ∆u c 0 + z c u c 0 = 0, in O c , (14b)
and the boundary condition

∂ n u e 0 | ∂Ω = φ. (14c) Equality (9d) implies: ∂ η u m 0 | η=0 = 0,

and equality (12a) implies:

∂ η u m 0 | η=1 = 0. Therefore, u m 0 depends only on θ. By identifying 0 th order term in (9d)-(12a), we infer:

u c 0 o Φ 0 = u m 0 = u e 0 o Φ 0 , (15)
thus we obtain the following transmission condidtion:

u c 0 o Φ 0 = u e 0 o Φ 0 . ( 16
)
We will determine u m 0 later on. Second step : identification of first order terms. Substituting into equality [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF] the field u m by its expansion (11c), and using that in (0, 1) × T, we have

∂ 2 η u m 0 = ∂ η u m 0 = 0, we obtain: ∂ 2 η u m 1 = 0. (17)
Moreover, we obtain easily:

∆u e 1 + z e u e 1 = 0, in Ω \ O c , (18a) ∆u c 1 + z c u c 1 = 0, in O c , (18b)
and the boundary condition

∂ n u e 1 | ∂Ω = 0. (18c)
Equality (9d) implies:

∂ η u m 1 | η=0 = µ m µ c ∂ n u c 0 o Φ 0 , (19a)
and equality (12a) implies:

∂ η u m 1 | η=1 = µ m µ e ∂ n u e 0 o Φ 0 . (19b)
We infer the following transmission condition between ∂ n u e 0 and ∂ n u c 0 :

1 µ c u c 0 o Φ 0 = 1 µ e u e 0 o Φ 0 . (20)
Therefore, with ( 14), ( 16) and (20) we infer that (u e 0 , u c 0 ) satisfies the following PDE in Ω:

∆u e 0 + z e u e 0 = 0, in O e,h , ∆u c 0 + z c u c 0 = 0, in O c , (21a)
with the transmission conditions

u c 0 | Γ0 = u e 0 | Γ0 , (21b) 1 µ c ∂ n u c 0 | Γ0 = 1 µ e ∂ n u e 0 | Γ0 , (21c)
and with Neumann boundary condition:

∂ n u e 1 | ∂Ω = 0. (21d)
According to [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF], u m 0 is equal to:

∀ (η, θ) ∈ [0, L] × T, u m 0 (η, θ) = u c 0 o Φ 0 (θ). ( 22
)
We have determined u e 0 , u c 0 and u m 0 . Observe that the identification of the first order term in (9e) implies:

u m 1 | η=0 = u c 1 o Φ 0 , (23)
and u c 1 will be determine later.

Third step : identification of second order terms. According to (17) and ( 22), we have

∂ η u m 0 ≡ ∂ 2 η u m 1 ≡ 0.
Therefore, by identifying the second order term in h of (10), we obtain:

∂ 2 η u m 2 + m 1 = 0, (24)
with

m 1 = κ∂ η u m 1 + ∂ 2 θ u m 0 + z m u m 0 . ( 25
)
Observe that m 1 depends only on θ.

The identification of second order terms of (9d) implies

∂ η u m 2 | η=0 = µ m µ c ∂ n u c 1 o Φ 0 ,
and those of (12a) implies

∂ η u m 2 | η=1 = µ m µ e ∂ n u e 1 o Φ 0 + ∂ 2 n u e 0 o Φ 0 . (26)
Observe that ∂ η u m 1 depends only on θ thanks to (17). Thereby integrating (24) with respect to η we obtain:

∂ η u m 2 (η, •) = -ηm 1 + µ m µ c ∂ n u c 1 o Φ 0 . (27)
From ( 26) and (27) we will obtain a transmission condition for (u e 1 , u c 1 ). More precisely, taking η = 1 in (27), with the help of (26) we obtain:

µ m µ c ∂ n u c 1 o Φ 0 - µ m µ e ∂ n u e 1 o Φ 0 = m 1 + µ m 1 µ e ∂ 2 n u e 0 o Φ 0 , (28)
and the right-hand side of (28) is entirely determined.

By identifying the terms of order 1 in (12b) we obtain:

u c 1 o Φ 0 -u e 1 o Φ 0 = 1 - µ m µ c ∂ n u e 0 o Φ 0 . ( 29
)
For convenience we write equations satisfied by (u e 1 , u c 1 ) in Euclidean coordinates. Thanks to ( 14), ( 28) and (29) we infer that (u e 1 , u c 1 ) solves:

     ∆u e 1 + z e u e 1 = 0, in Ω \ O c , ∆u c 1 + z c u c 1 = 0, in O c , ∂ n u e 1 | ∂Ω = 0, (30a)
with transmission conditions

µ m µ c ∂ n u c 1 | Γ0 - µ m µ e ∂ n u e 1 | Γ0 = ∂ 2 t u c 0 | Γ0 + z m u c 0 | Γ0 + µ m µ e ∂ 2 n u e 0 | Γ0 + µ m µ c K∂ n u c 0 | Γ0 , (30b)
u c 1 | Γ0 -u e 1 | Γ0 = 1 - µ m µ e ∂ n u e 0 | Γ0 . (30c)
In Section 4, we prove the existence and uniqueness of (u e 1 , u c 1 ) defined by (30). Remark that u m 1 given by equality (23) is entirely determined and ∂ η u m 2 is entirely determined by (27).

Observe that (30b) contains a second normal derivative; this is a feature of the asymptotics of a cell in an ambient medium; no second derivative appeared in [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF], where there is a cell with boundary condition on the exterior of the membrane. Let us summarize the first two terms of the asymptotics we obtained formally.

• The 0 th order terms. The electric fields u e 0 and u c 0 are solution of the following problem in Ω:

∆u e 0 + z e u e 0 = 0, in Ω \ O c , ∆u c 0 + z c u c 0 = 0, in O c , (31a)
with transmission conditions

u c 0 | Γ0 = u e 0 | Γ0 , (31b) 1 µ c ∂ n u c 0 | Γ0 = 1 µ e ∂ n u e 0 | Γ0 , (31c)
and with Neumann boundary condition

∂ n u e 0 | ∂Ω = φ. (31d)
In the membrane, the field u m 0 is equal to:

∀(η, θ) ∈ [0, 1] × T, u m 0 = u c 0 o Φ 0 (θ). (32)
• The first order terms. The fields u e 1 and u c 1 are solution of the following problem in Ω:

     ∆u e 1 + z e u e 1 = 0, in Ω \ O c , ∆u c 1 + z c u c 1 = 0, in O c , ∂ n u e 1 | ∂Ω = 0, (33a)
with the transmission conditions

1 µ c ∂ n u c 1 | Γ0 - 1 µ e ∂ n u e 1 | Γ0 = 1 µ m ∂ 2 t u c 0 | Γ0 + z m u c 0 | Γ0 + 1 µ e ∂ 2 n u e 0 | Γ0 + 1 µ c K∂ n u c 0 | Γ0 , (33b)
u c 1 | Γ0 -u e 1 | Γ0 = µ e -µ m µ c ∂ n u c 0 | Γ0 . (33c)
In the membrane, we have:

∀(η, θ) ∈ [0, 1] × T, u m 1 = η µ m µ c ∂ n u c 0 o Φ 0 + u c 1 o Φ 0 . (34) Remark 2. We may write ∂ 2 n u e 0 | Γ0 in terms of ∂ n u e 0 | Γ0 , of u e 0 | Γ0
and of its tangential derivatives. Actually, we perform the change in local coordinates in a neighborhood of ∂O c . According to (31), the following identity holds along Γ 0 :

∂ 2 n u e 0 | Γ0 = -K∂ n u e 0 | Γ0 -∂ 2 t u e 0 | Γ0 -z e u e 0 | Γ0 ,
thus we may rewrite transmission conditions (33b)-(31c) as follows: We have given the first two terms of the asymptotic expansion of u e , u c and u m . It remains to prove that the remainder terms are small. First we need to study the regularity of u e 0 and u e 1 in a neighborhood of Γ 0 .

1 µ c ∂ n u c 1 | Γ0 - 1 µ e ∂ n u e 1 | Γ0 = 1 µ m - 1 µ e ∂ 2 t u c 0 | Γ0 + (q m -q e ) u c 0 | Γ0 , hence (31)

Regularity Result

In this section, we study the regularity of the solution of Helmhotz equation with our transmission condition, which is non usual. This result is required to prove Theorem 1 of Section 5, which estimates the errors between the asymptotics and the exact solution. The following result is natural and expected; it is very close to a result of [START_REF] Yan | Gradient estimates for solutions to divergence form elliptic equations with discontinuous coefficients[END_REF] (Appendix, page 147) by Li and Vogelius, but different enough to require a proof. We thank very warmly Michael Vogelius for his suggestions on the reflection principle.

Theorem 3. Let G belong to H s (Γ 0 ), s ≥ -1/2. Let (U e , U c ) be the solution of the following problem:

∇. 1 µ c ∇U c + q c U c = 0, in O c , ∇. 1 µ e ∇U e + q e U e = 0, in Ω \ O c ,
with the following transmission condition:

U e | Γ0 = U c | Γ0 , 1 µ e ∂ n U e | Γ0 - 1 µ c ∂ n U c | Γ0 = G,
and with the Neumann boundary condition on ∂Ω

∂ n U e | ∂Ω = 0.
Then we have:

U e ∈ H s+3/2 (Ω \ O c ), U c ∈ H s+3/2 (O c ).
Moreover let m be a non negative integer, and s > m + 1/2. Then,

U e ∈ C m (Ω \ O c ), U c ∈ C m (O c ).
Proof. Since Γ 0 is smooth, we use local coordinates in a neighborhood of Γ 0 . Actually, as in Section 1, there exists h 1 such that:

V 1 = {Ψ(θ) + h 1 ηn(θ), (η, θ) ∈ (-1, 1) × T} ,
is an open neighborhood of Γ 0 and

(η, θ) → Ψ(θ) + h 1 ηn(θ)
is a diffeomorphism from (-1, 1) × T to V 1 . We denote by g the function G written in local coordinates:

∀θ ∈ T, g(θ) = G o Ψ(θ).
We denote by C the cylinder [0, 1] × T and by H 1 m (C) the space of the functions α defined on C such that:

α H 1 m (C) = 1 0 2π 0 h 1 (1 + h 1 ηκ)|α(η, θ)| 2 dη dθ + 1 0 2π 0 1 + h 1 ηκ h 1 |∂ η α(η, θ)| 2 + h 1 1 + h 1 ηκ |∂ θ α(η, θ)| 2 dη dθ 1/2
, is finite. We equip H 1 m (C) with such a norm, which is equivalent to the ordianry norm

1 0 2π 0 |α| 2 + |∂ η α| 2 + |∂ θ α| 2 dθdη 1/2
, because we have the following equality:

v o Φ -1 H 1 (V1) = v H 1 m (C) .
We use a partition of unity and classical elliptic regularity to reduce our problem to establishing the regularity of the solutions (V e , V c ) of the following problem:

∀(η, θ) ∈ [-1, 0] × T, ∂ η 1 + h 1 ηκ h 1 µ c ∂ η V c + ∂ θ h 1 (1 + h 1 ηκ) µ c ∂ θ V c + q c h 1 (1 + h 1 ηκ)V c = 0, ∀(η, θ) ∈ [0, 1] × T, ∂ η 1 + h 1 ηκ h 1 µ e ∂ η V e + ∂ θ h 1 (1 + h 1 ηκ) µ e ∂ θ V e + q e h 1 (1 + h 1 ηκ)V e = 0,
with Dirichlet boundary conditions

V c | η=-1 = 0, V e | η=1 = 0
and with transmission conditions

V c | η=0 = V e | η=0 , 1 µ ∂ η V e | η=0 - 1 µ c ∂ η V c | η=0 = g,
We use the reflection principle, suggested by Vogelius and coming from an idea of Nirenberg (see [START_REF] Yan | Gradient estimates for solutions to divergence form elliptic equations with discontinuous coefficients[END_REF], page 147 or [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF] and [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF]). With the help of this principle, we transform transmission conditions into boundary conditions. We define V r on [0, 1] × T by:

∀(η, θ) ∈ [0, 1] × T, V r (η, θ) = V c (-η, θ).
The functions V e , V r satisfy the following problem in (0, 1) × T:

∀(η, θ) ∈ (0, 1) × T, ∂ η 1 + h 1 ηκ h 1 µ e ∂ η V e + ∂ θ h 1 (1 + h 1 ηκ)µ e ∂ θ V e + q e h 1 (1 + h 1 ηκ)V e = 0, (35a)
∂ η 1 -h 1 ηκ h 1 µ c ∂ η V r + ∂ θ h 1 (1 -h 1 ηκ)µ c ∂ θ V r + q c h 1 (1 -h 1 ηκ)V r = 0, (35b) with Dirichlet boundary conditions in η = 1 V r | η=1 = 0, V e | η=1 = 0, (35c)
with boundary conditions in η = 0:

V r | η=0 -V e | η=0 = 0, (35d) 1 µ c ∂ η V r | η=0 + 1 µ e ∂ η V e | η=0 = g, (35e)
Multiplying (35a) by V e and (35b) by V r , integrating by parts and summing, we obtain:

1 0 2π 0 1 + h 1 ηκ h 1 µ e |∂ η V e | 2 + 1 -h 1 ηκ h 1 µ c |∂ η V r | 2 + h 1 (1 + h 1 ηκ)µ e |∂ θ V e | 2 + h 1 (1 -h 1 ηκ)µ c |∂ θ V r | 2 -q c h 1 (1 -h 1 ηκ)|V r | 2 -q e h 1 (1 + h 1 ηκ)|V e | 2 dη dθ = 2π 0 1 h 1 µ e ∂ η V e | η=0 V e | η=0 + 1 h 1 µ c ∂ η V r | η=0 V r | η=0 (36a)
Using boundary conditions (35d)-(35e), we obtain:

2π 0 1 µ e ∂ η V e | η=0 V e | η=0 + 1 µ c ∂ η V r | η=0 V r | η=0 = 2π 0 gV e dθ. ( 36b
)
We argue as in [START_REF] Nečas | Les méthodes directes en théorie des équations elliptiques[END_REF] or in [START_REF] Ammari | Reconstruction of conductivity inhomogeneities of small diameter via boundary measurements[END_REF], and the reader will verify that (36a) and (36b) suffice to give existence and uniqueness of solutions of (35) in H 1 m (C). To obtain the regularity result, we just have to apply the method of frozen coefficients. Let θ 0 ∈ T, and denote by κ 0 the value of κ at θ 0 . A classical argument (see for instance [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF], [START_REF] Agmon | Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions[END_REF], [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF], [START_REF] Lions | Problèmes aux limites non homogènes et applications[END_REF] or [START_REF] Nečas | Les méthodes directes en théorie des équations elliptiques[END_REF] ) shows that (V e , V r ) have the same respective regularity as (V ′ , V ′′ ) solution of:

∀(η, θ) ∈ [0, 1] × T, 1 h 1 ∂ 2 η V ′ + κ 0 1 + h 1 ηκ 0 ∂ η V ′ + h 1 ∂ 2 θ V ′ (1 + h 1 ηκ 0 ) 2 = 0, 1 h 1 ∂ 2 η V ′′ - κ 0 1 -h 1 ηκ 0 ∂ η V ′′ + h 1 ∂ 2 θ V ′′ (1 -h 1 ηκ 0 ) 2 = 0,
with Dirichlet boundary conditions

V ′ | η=1 = 0, V ′′ | η=1 = 0, with transmission conditions V ′ | η=0 = V ′′ | η=0 , 1 µ e ∂ η V ′ | η=0 + 1 µ c ∂ η V ′′ | η=0 = g.
The regularity results of this last problem is obtained directly by working in Fourier coefficients, hence the regularity result, in V 1 . The end of the proof follows by classical regularity theorems (see [START_REF] Nečas | Les méthodes directes en théorie des équations elliptiques[END_REF] for instance).

Error Estimates

We give an error estimate, which proves that the first two terms obtained in Section 3 through a formal argument are indeed the first terms, i.e. the remainder is smaller.

Remark 4. Recall that the L 2 norm of a 0-form α in C with the metric [START_REF] Beretta | Asymptotic formulas for steady state voltage potentials in the presence of thin inhomogeneities. A rigorous error analysis[END_REF], denoted by α Λ 0 L 2 m (C) , is equal to:

α 2 Λ 0 L 2 m (C) = 1 0 2π 0 h(1 + hηκ)|α(η, θ)| 2 dη dθ, = α o Φ -1 2 L 2 (O h ) ,
and that the L 2 norm of the exterior derivative dα of α, denoted by dα

Λ 1 L 2 m (C)
is equal to

dα 2 Λ 1 L 2 m (C) = 1 0 2π 0 1 + hηκ h |∂ η α(η, θ)| 2 + h 1 + hηκ |∂ θ α(η, θ)| 2 dη dθ, = ∇ α o Φ -1 2 L 2 (O h ) .
Let us prove now Theorem 1. We remember that u is the solution to Problem (1) and that (v e , v c ) are defined in Ω by:

v e = u e 0 + hu e 1 , in Ω \ O c , (37a) v c = u c 0 + hu c 1 , in O c . (37b)
We have to prove that there exists an h-independent constant C > 0 such that

u -v c H 1 (Oc) ≤ Ch 3/2 φ H s (∂Ω) , (38a) u -v e Λ 0 L 2 m (C) + 1 µ m du - 1 µ e dv e Λ 1 L 2 m (C) ≤ Ch 3/2 φ H s (∂Ω) , (38b) u -v e H 1 (O e,h ) ≤ Ch 3/2 φ H s (∂Ω) . (38c)
Remark 5. The estimates of Theorem 1 are piecewise H 1 estimates since estimate (38b) involves

1 µ m du - 1 µ e dv e Λ 1 L 2 m (C)
, which is not the norm of a difference of gradients. However, we could have global estimate with an appropriate norm involving the permeabilities µ c , µ m and µ e and by defining an appropriately modified µ; details are left to the reader.

Since φ belongs to H s (∂Ω), s > 7/2 and Theorem 3 holds, u e 0 belongs to C 3 (O h ) and u e 1 to C 2 (O h ). To prove Theorem 1, we need the following lemma.

Lemma 6. Let h belong to (0, h 0 ).

Let u e 0 , u e 1 , u m 0 and u m 1 be defined by (31) and (33). We denote by v the following function:

∀(η, θ) ∈ [0, 1] × T, v(η, θ) = u m 0 + hu m 1 . (39)
Then, there exists an h-independent constant C > 0 such that

v e o Φ -v Λ 0 L 2 (C) ≤ Ch 3/2 φ H s (∂Ω) , (40) 1 µ e d (v e o Φ) - 1 µ m d v Λ 1 L 2 (C) ≤ Ch 3/2 φ H s (∂Ω) (41) and (42)      v e o Φ 1 -v| η=1 Λ 0 L 2 (T) ≤ Ch 2 φ H s (∂Ω) , ∂ θ v e o Φ 1 -∂ θ v| η=1 Λ 0 L 2 (T) ≤ Ch 2 φ H s (∂Ω) , ∂ 2 θ v e o Φ 1 -∂ 2 θ v| η=1 Λ 0 L 2 (T) ≤ Ch 2 φ H s (∂Ω) .
Remark 7. A priori, one would have expected that the statement of Theorem 1 would have given a comparison of the exact solution with its asymptotics in the three regions O e,h , O h and O c . Actually, Lemma 6 shows that we may dispense with the asymptotics in O h , provided that v e defined by (37) has been extended up to the inner boundary of the membrane, and this is precisely how v e has been constructed.

Proof of Lemma 6. Since u e 0 o Φ belongs to C 3 ([0, 1] × T) and since u e 1 o Φ belongs to C 2 ([0, 1] × T), using Taylor formula with integral remainder, we obtain for all (η, θ) ∈ [0, 1] × T:

v e o Φ| (η,θ) = u e 0 o Φ| (0,θ) + hη∂ n u e 0 o Φ| (0,θ) + hu e 1 o Φ| (0,θ) + h 2 η 2 1 0 (1 -t) ∂ n u e 1 o Φ(tη, θ) + (1 -t) 2 ∂ 2 n u e 0 o Φ(tη, θ) dt,
and

∂ η v e o Φ (η,θ) = h∂ n u e 0 o Φ| (0,θ) + h 2 ∂ n u e 1 o Φ| (η,θ) + η 1 0 (1 -t)∂ 2 n u e 0 o Φ(t, θ) dt .
Since we have:

v(η, θ) = u c 0 o Φ 0 (θ) + hη µ m µ c ∂ n u c 0 o Φ 0 (θ) + hu c 1 o Φ 0 (θ), ∂ η v(η, θ) = h µ m µ c ∂ n u c 0 o Φ 0 (θ),
using transmission condition (31b), we obtain for all (η, θ) ∈ [0, 1] × T:

(v e o Φ -v)| η,θ = h η 1 - µ m µ e ∂ n u e 0 o Φ 0 + u e 1 o Φ 0 -u c 1 o Φ 0 + hη 1 0 (1 -t) ∂ n u e 1 o Φ(tη, θ) + η (1 -t) 2 ∂ 2 n u e 0 o Φ(tη, θ) dt .
This equality implies directly estimate (40). Moreover, using transmission conditions (31b) and (33c), we obtain for all θ ∈ T:

v e o Φ 1 (θ) -v(1, θ) = h 2 1 0 (1 -t) ∂ n u e 1 o Φ(t, θ) + (1 -t) 2 ∂ 2 n u e 0 o Φ(t, θ) dt,
which implies the first estimate of (42). Applying the same reasoning to ∂ θ v e o Φ and to ∂ 2 θ v e o Φ, we obtain the two last estimates of (42).

Observe that:

1 µ e ∂ η v e o Φ (η, θ) - 1 µ m ∂ η v(η, θ) = h 2 f µ e (θ) ηf 1 0 (1 -t)∂ 2 n u e 0 o Φ(tη, θ) dt + ∂ n u e 1 o Φ(η, θ) ,
hence estimate (41). This ends the proof of Lemma 6

Let us prove now Theorem 1.

Proof of Theorem 1. Define m 0 by

m 0 = µ m µ c ∂ n u c 0 o Φ 0 ,
and ūm 2 by

ūm 2 (η, •) = - η 2 2 m 1 + ηm 0 ,
where m 1 is defined by (25).

Let

W e = u e -(u e 0 + hu e 1 ) , in O e,h , (43a)
W c = u c -(u c 0 + hu c 1 ) -h 2 B c , in O c , (43b)
W m = u m -(u m 0 + hu m 1 ) -h 2 B m , in [0, 1] × T, (43c)
where

B m (η, θ) = ūm 2 + a m (θ) + ηb m (η), (44)
and B c , a m and b m are allowed to depend on h and will be chosen later, so that yield the easiest estimates of W e , W c and W m .

Let us write the problem satisfied by (W e , W c , W m ). In order to simplify the notations, we introduce L , the Helmholtz operator written in the local coordinates (η, θ) given by

L =∂ η 1 + hηκ h ∂ η + ∂ θ h 1 + hηκ ∂ θ + z m h(1 + hηκ).
We obtain

∆W e + z e W e = 0, in O e,h , (45a)
∆W c + z c W c = -h 2 (∆B c + z c B c ) , in O c , (45b) ∀(η, θ) ∈ [0, 1] × T, L W m = -L u m 0 + hu m 1 + h 2 B m ,
with transmission conditions coming from (2)

1 µ c ∂ n W c o Φ 0 = 1 hµ m ∂ η W m | η=0 + h 1 µ m ∂ η B m | η=0 - 1 µ c ∂ n u c 1 o Φ 0 -h 2 ∂ n B c o Φ 0 , (45c)
W c o Φ 0 = W m | η=0 + h 2 (B m | η=0 -B c o Φ 0 ) , (45d) 1 µ e ∂ n W e o Φ 1 = 1 hµ m ∂ η W m | η=1 + h∂ η u m 1 + h 2 ∂ η B m | η=1 (45e) - 1 µ e ∂ n u e 0 o Φ 1 - h µ e ∂ n u e 1 o Φ 1 , (45f)
W e o Φ 1 = W m | η=1 + h 2 B m | η=1 + u m 0 | η=1 + hu m 1 | η=1 -u e 0 o Φ 1 -hu e 1 o Φ 1 , (45g)
and the boundary condition

∂ n W e | ∂Ω =0. (45h)
We calculate L W m , knowing that ∂ η u m 0 and ∂ 2 η u m 1 vanish and we obtain:

L W m = -hκ∂ η u m 1 -h∂ η ((1 + hηκ)∂ η B m ) -h∂ θ 1 1 + hηκ ∂ θ u m 0 + hu m 1 + h 2 B m -z m h (1 + hηκ) u m 0 + hu m 1 + h 2 B m , (46)
and we find that in the above expression, the coefficient of terms of order 1 in h is:

-∂ 2 η B m -κ∂ η u m 1 -∂ 2 θ u m 0 -z m u m 0 . (47)
By definition (44) of B m , the first term of (47) is m 1 , and according to the definition (25) of m 1 , the expression (47) vanishes.

We will determine a m and b m so as to have nice transmission conditions. Observe that if y satisfies, in the weak sense:

∇. 1 µ ∇y + qy = 0, in O e,h ∪ O h ,
with discontinuous µ on the outer boundary of the membrane O h , then the transmission conditions on this boundary are:

y e o Φ 1 = y m | η=1 , µ m µ e ∂ n y e o Φ 1 = 1 h ∂ η y m | η=1 .
Therefore, it is natural to write transmission conditions of this form on the outer boundary of O h . The continuity condition (45g) may be rewritten :

W e o Φ 1 -W m | η=1 = h 2 B m | η=1 -v| η=1 -v e o Φ 1 ,
where v and v e are respectively defined by ( 39) and (37). We choose B m so that the right-hand side of the above equality vanishes:

a m + b m = v| η=1 -v e o Φ 1 h 2 -ūm 2 | η=1 , (48)
and thanks to Lemma 6 estimate (42), the right-hand side of (48) is bounded in H 2 (T).

The condition (45f) is rewritten into:

1

µ e ∂ n W e o Φ 1 - 1 hµ m ∂ η W m | η=1 = 1 µ m (∂ η u m 1 + h∂ η B m | η=1 ) - 1 µ e ∂ n u e 0 o Φ 1 - h µ e ∂ n u e 1 o Φ 1 , . (49)
Observe that:

∂ n u e 0 o Φ 1 = ∂ n u e 0 o Φ 0 + h∂ 2 n u e 0 o Φ 0 + h 2 1 0 (1 -t) 2 2 ∂ 3 n u e 0 o Φ dt, ∂ n u e 1 o Φ 1 = ∂ n u e 1 o Φ 0 + h 1 0 (1 -t)∂ 2 n u e 0 o Φ dt,
and recall that

∂ η u m 1 = µ m µ e ∂ n u e 0 o Φ 0 .
Therefore, in order for the right-hand side of (49) to be of order 2, we impose:

1 µ m ∂ η B m | η=1 - 1 µ e ∂ 2 n u e 0 o Φ 0 - 1 µ e ∂ n u e 1 o Φ 0 = 0, which implies b m = m 1 -m 0 + µ m µ e ∂ 2 n u e 0 o Φ 0 + µ m µ e ∂ n u e 1 o Φ 0 , (50)
thanks to (44). We infer from (48) and (50) that a m and b m are bounded in H 2 (T) independently of h, and therefore, since B m is polynomial in η, it belongs to C ∞ [0, 1]; H 2 (T) , the space of functions, which are C ∞ in η ∈ [0, 1] with values in H 2 (T). Particularly, there exists C > 0 independent of h such that

∀η ∈ [0, 1], B m (η, •) H s-1 (T) ≤ C φ H s (∂Ω) . (51)
Observe that with such B m , we have:

∂ η B m | η=0 = b m + m 0 , = m 1 + µ m µ e ∂ 2 n u e 0 o Φ 0 + µ m µ e ∂ n u e 1 o Φ 0 ,
thanks to (50). Transmission condition (28) with definition (25) of m 1 imply:

∂ η B m | η=0 = µ m µ c ∂ n u c 1 o Φ 0 .
Therefore, transmission condition (45c) is simplified into:

1 µ c ∂ n W c o Φ 0 = 1 hµ m ∂ η W m | η=0 -h 2 ∂ n B c o Φ 0 , (52)
It remains to define B c . It is simply define by

∆B c + z c B c = 0, in O c , B c | ∂Oc = B m o Φ -1 0 . Since B m ∈ H 2 (
T), a classical argument and estimate (51) imply that there exists C > 0 independent on h such that:

∂ n B c | ∂Oc H 1 (∂Oc) ≤ C φ H s (∂Ω) , (53a) B c H 2+1/2 (Oc) ≤ C φ H s (∂Ω) . (53b)
To simplify our notations, we define

g(η, θ) = 1 h 2 L u m 0 + hu m 1 + h 2 B m , g 1 (θ) = 1 h 2 1 µ m (∂ η u m 1 + h∂ η B m | η=1 ) - 1 µ e ∂ n u e 0 o Φ 1 - h µ e ∂ n u e 1 o Φ 1 .
We equip L 2 (C) with the ordinary norm

α L 2 (C) = 1 0 2π 0 |α| 2 dθ dη 1/2
, and L 2 (T) with the ordinary norm

γ L 2 (T) = 2π 0 |γ| 2 dθ 1/2
.

We have chosen B m and B c such that there exists an h-independent constant C > 0 such that

g L 2 (C) ≤ C φ H s (∂Ω) , (54a)
and

g 1 L 2 (T) ≤ C φ H s (∂Ω) . (54b)
We rewrite Problem (45) as follows:

∆W e + z e W e = 0, in O e,h , (55a)
∆W c + z c W c = 0, in O c , (55b)
∀(η, θ) ∈ [0, 1] × T, L W m = -h 2 g, (55c) with transmission conditions 1 µ c ∂ n W c o Φ 0 = 1 hµ m ∂ η W m | η=0 -h 2 ∂ n B c o Φ 0 , (55d) W c o Φ 0 = W m | η=0 , (55e) 1 µ e ∂ n W e o Φ 1 = 1 hµ m ∂ η W m | η=1 + h 2 g 1 , (55f) W e o Φ 1 = W m | η=1 (55g)
and the boundary condition

∂ n W e | ∂Ω =0. (55h)
Now we are ready to perform L 2 estimates as it has been performed in [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF]. In O c parameterized by Euclidean coordinates, the L 2 norm of a 0-form β, denoted by β L 2 (Oc) , is equal to:

β Λ 0 L 2 (Oc) = β L 2 (Oc) ,
and the L 2 norm of its exterior derivative dβ, denoted by du

Λ 1 L 2 (Oc) is equal to dβ Λ 1 L 2 (Oc) = ∇β L 2 (Oc) .
In O e,h parameterized by Euclidean coordinates, the L 2 norm of a 0-form γ, denoted by γ L 2 (O e,h ) , is equal to:

γ Λ 0 L 2 (O e,h ) = γ L 2 (O e,h ) ,
and the L 2 norm of its exterior derivative dv, denoted by du

Λ 1 L 2 (O e,h ) is equal to dγ Λ 1 L 2 (O e,h ) = ∇γ L 2 (O e,h ) .
We multiply equalities (55a), (55b) and (55c) respectively by the conjugates of W e , W c and W m . Using transmission conditions (55d)-(55g), we integrate by parts and we take the imaginary part of the result. To simplify the notations, we define W 2 Λ 0 L 2 (Ω) and dW 2 Λ 1 L 2 (Ω) as follows:

W 2 Λ 0 L 2 (Ω) = W e 2 Λ 0 L 2 (O e,h ) + W m 2 Λ 0 L 2 m (C) + W c 2 Λ 0 L 2 (Oc) , dW 2 Λ 1 L 2 (Ω) = dW e 2 Λ 1 L 2 (O e,h ) + dW m 2 Λ 1 L 2 m (C) + dW c 2 Λ 1 L 2 (Oc) , Defining σ = min ℑ(z e ), ℑ(z m ), ℑ(z c ) ,
we obtain:

σ W 2 Λ 0 L 2 (Ω) ≤ h 2 µ c Γ0 ∂ n B c W c dvol Γ0 + h 2 µ m 2π 0 (1 + hκ) g 1 W e o Φ 1 dθ + h 2 1 0 2π 0 gW m dη dθ .
Therefore, there exists an h-independent constant C such that:

W 2 Λ 0 L 2 (Ω) ≤ Ch 3/2 √ h ∂ n B c L 2 (Γ0) + g 1 L 2 (T) + 1 0 2π 0 |g| 2 dη dθ 1/2 W Λ 0 L 2 (Ω) + dW Λ 1 L 2 (Ω) . (56)
Observe that C depends on the dielectric parameters and on the geometry of the domains. One more time, we multiply equalities (55a)-(55b)-(55c) respectively by the conjugates of W e , W c and W m . Using transmission conditions we integrate by parts and we take the real part of the result. Defining

a = max {ℜ(z e ), ℜ(z m ), ℜ(z c )} ,
We infer:

dW 2 Λ 1 L 2 (Ω) ≤ a W 2 Λ 0 L 2 (Ω) + h 2 µ c Γ0 ∂ n B c W c dvol Γ0 + h 2 µ m 2π 0 g 1 W e o Φ 1 dθ + h 2 1 0 2π 0 gW m dη dθ .
The field u c 1 is the solution of the following problem in O c :

∆u c 1 + z c u c 1 = 0, in O c , (µ m /µ c ) ∂ n u c 1 | Γ0 = Kg + ∂ 2 t u c 0 | Γ0 + z m u c 0 | Γ0 , on Γ 0 .
In the membrane, we have:

∀(η, θ) ∈ [0, 1] × T, u m 1 = ηγ + u c 1 o Φ 0 .
Let W be the function defined on Ω h by:

W = u -(u c 0 + hu c 1 ) , in O c , u -u m 0 o Φ -1 + hu m 1 o Φ -1 , in O h . Then, there exists an h-independent constant C > 0 such that W H 1 (Ω h ) ≤ Ch 3/2 g H s (∂Oc) .
Remark 10 (The case z m = 0 in O h ). In Theorem 1, we can replace z m by zero. The proof is then very similar, except that we need the following inequality.

Proposition 11. Let h be as in Theorem 1. Let u be a function of class C 1 ([0, 1] × T). In the cylinder [0, 1]×T, we use Euclidean metric [START_REF] Beretta | Asymptotic formulas for steady state voltage potentials in the presence of thin inhomogeneities. A rigorous error analysis[END_REF] written in local coordinates defined at Section 1, that is

h 2 dη + (1 + hηκ)dθ.
Then, there exists an h-independent constant C such that We deduce directly inequality (60).

Application to the biological cell

In biological cells, the membrane is insulating (see Fear and Stuchly [START_REF] Fear | Modeling assemblies of biological cells exposed to electric fields[END_REF] or Sebastián al. [17]). This means that at mid frequencies, the ratio |z m |/|z c | and |z m |/|z e | are small compared to h. Actually, the thickness is of order 10 -3 , while |z m |/|z c | is about 10 -5 (see [START_REF] Poignard | Full asymptotics for steady state voltage potentials in a bidimensionnal highly contrasted medium[END_REF]). We say that we work at mid frequency since we suppose that z c and z e are of order 1. Moreover, the relative permeabiltiy is constant equal to 1, thus in the following, we suppose:

µ c = µ m = µ e = 1.
The following results show that the asymptotics obtained by replacing z m by zero in the expansions of Theorem 1 give a good approximation of the electric field in the biological cell. We have the following proposition.

Proposition 12. Let z c and z e be complex constants with strictly negative imaginary part.

We suppose that |z m | tends to zero and that there exists a constant c > 0 such that:

0 < - |z m | ℑ(z m ) < c. ( 62
)
Let z and z be such that:

z =      z e , in O e,h , z m , in O h , z c , in O c , z =      z e , in O e,h , 0, in O h , z c , in O c .
Let φ in H 1/2 (∂Ω). Let u the solution of the following problem: ∆u + zu = 0, in Ω, (63a) ∂ n u| ∂Ω = φ, in Ω, (63b) and let v be such that ∆v + zv = 0, in Ω, (63c)

∂ n v| ∂Ω = φ, in Ω. (63d)
Then, there exists a constant C such that:

u -v H 1 (Ω) ≤ C|z m | φ H 1/2 (∂Ω) .
Proof. First, using hypothesis (62), we prove by classical argument that there exists an h-independent constant C such that:

u H 1 (Ω) ≤ C φ H 1/2 (∂Ω) .
Then, we just have to write the problem satisfies by uv in local coordinates in the membrane. As usual, we multiply in by uv and we integrate by parts. Then, according to inequality (60) the following inequality holds:

u -v H 1 (Ω) ≤ C|z m | u H 1 (Ω) ,
which ends the proof of the theorem.

Using Proposition 12 and Remark 10, we infer the following theorem.

[17] J.L. Sebastián, S. Muñoz, M. Sancho, and J.M. Miranda. Analysis of the influence of the cell geometry and cell proximity effects on the electric field distribution from direct rf exposure. Phys. Med. Biol., 46:213-225 (electronic), 2001.

Institut Camille Jordan and CEGELY, Université Claude Bernard Lyon1 poignard@math.univ-lyon1.fr

Figure 1 .

 1 Figure 1. Geometric and dielectric data.

  -(33) are equivalent to (3)-(4).

u 2 Λ 0 L 2 m 2 Λ 1 L 2 m 2 Λ 0 L 2 m 2 h 2 +

 222122222 (C) ≤ C du θ)| 2 dθ .(60)Proof. Actually, according to the definition of h 0 in (6) there exists two constants C 1 and C 2 depending on the domain O such that the following inequalities hold:u + h |∂ θ u| 2 dθ dη . (61b)Let us denote by ( u) k for k ∈ Z the k th -Fourier coefficient (with respect to θ) of u:u k = π 0 u(θ) e -2iπk/L dθ. Since ∂ θ u k = 2iπk u k , itis easy to see that: 2| u 0 (0)| 2 dη.

Acknowledgments

The author thanks very warmly Michelle Schatzman for reading carefully the manuscript and Michael Vogelius for his advice on the regularity result of Section 4.

Using (56) we infer that there exists an h-independent constant C > 0 such that:

Therefore, estimating the right-hand side of (57) with estimates (53) and (54), we infer that there exists an h-independent constant C > 0 such that:

Since we have respectively in the cylinder C

and in O e u eu e 0hu e 1 = W c , we have proved Theorem 1.

Remark 8 (Neumann boundary condition imposed on the cell). Consider the domain Ω h defined by:

Let γ be in H s (∂Ω h ), s > 7/2, and we denote by g and g the following function defined on the torus:

Let u be the solution of the following problem:

Then, we have the following theorem:

Theorem 9. We remember that h 0 is defined in Theorem 1. Let γ be in H s (∂Ω h ) We denote by u c 0 , u m 0 , u c 1 , and u m 1 the functions defined as follows:

In the membrane, the field u m 0 is equal to:

Theorem 13. Let h 0 be in (0, 1) such that

Let h be in (0, h 0 ). Let z c and z e be complex constants with strictly negative imaginary part. We suppose that |z m | = o(h) and that there exists a constant c > 0 such that:

Let φ be in H s (∂Ω), s > 7/2. We denote by u the solution of the following problem:

We define (u e 0 , u c 0 ), u m 0 , (u e 1 , u c 1 ), and u m 1 are defined as follows. • The 0 th order terms. The electric fields u e 0 and u c 0 are solution of the following problem in O e,h ∪ O c :

with transmission conditions

with Neumann boundary condition:

In the membrane, the field u m 0 is equal to:

• The first order terms. The fields u e 1 and u c 1 are solution of the following problem in O e,h ∪ O c :

with the following transmission conditions

Recall that K is the curvature of ∂O c defined by (5). In the membrane, we have:


Beatrice Pradarelli
email: beatrice.pradarelli@lirmm.fr

Pascal Nouet

Laurent Latorre

Problem-Based Learning Approach to Teach Printed Circuit Boards Test

Keywords: problem based learning, collaborative, education, evaluation, skills, competencies, PCB test, risk analysis, adfa, p. 8, 2011

This article presents a return of experience of a collaborative educational approach put in place to build a new course about the test of Printed Circuits Boards (PCB). This course was dedicated to 5 th year engineer apprentice students from Montpellier Polytech School. The teaching method followed the "problem-based learning" (PBL) approach, which is a student-centered pedagogy: the students learn ingroup about a topic (knowledge and know-how) through the experience of solving an open-ended problem proposed by the teacher. This approach targets two types of objectives: technical with the learning of PCB test methods and educational with the learning of thinking strategies, decision-making, negotiation and collaborative work. A course evaluation form has been created, based on the F2A Louvain University one, to help the students providing classifiable feedback about this "problem based learning" approach.

Introduction

In 2014-2015, a new course relative to printed circuit boards (PCB) testing has been added to the syllabus of the 5 th year Embedded Systems curricula of Polytech Montpellier engineering school. 5 th year Embedded Systems students are engineer apprentices students who, since the 3 rd year at Polytech school, spend half of a month attending courses and studying and, the remaining half working for a company. In order to position them in a teaching context close to their real world working, the teacher freshly trained to problem-based learning approach from Louvain University teachers [1] decided to apply this student-centered pedagogy to the PCB testing course. Students would learn in-group about a topic (knowledge and know-how) through the experience of solving an open-ended problem proposed by the teacher. The choice of the problem-based active teaching approach was appropriated to motivate the students elaborating a learning strategy where they were protagonists of their learning and the teacher's role was only to guide them to be skilled before the visit of a PCB manufacturing and test plant organized in order to compensate the lack of lab material for practice.

This paper presents a return of experience of the "problem based learning" (PBL) approach developed to teach PCB testing and is composed of several sections. Section II is dedicated to the description of the context of PCB manufacturing and test. Section III briefly introduces the Electrical Engineering department of Polytech Montpellier engineering school. Section IV presents the problem-based learning approach concepts, its process phases, its organization and the impacts on the teacher's role. Then section IV describes how the PBL concepts have been applied to the PCB testing course and the technical and pedagogical objectives. Section V presents the outcome of this new student-centered teaching method while the section VI states about the required preparation to implement this collaborative approach and the improvements that could be applied. Section VII is the conclusion of this return experience of collaborative learning and the perspectives.

Printed Circuits Boards manufacturing and test

A printed circuit board (PCB) is a thin board made by combining different sheets of non-conductive material like fiberglass, composite epoxy, plastic or other laminate material that easily holds copper circuitry. The conductive pathways are etched or "printed" onto board, connecting different surfacemounted and socketed components on the PCB, such as transistors, resistors, and integrated circuits (IC).

PCBs manufacturing first requires designing the board with CAD tools and then fabricating it through several chemical operations.

PCBs are used in both desktop and laptop computers. They serve as the foundation for many internal computer components. Even if PCBs are often associated with computers, they are used in many other consumers' electronic devices like TVs, radios, digital cameras, cellular phones, and tablets and all of them include one or more printed circuit boards. Figure 1 shows the photograph of a PCB with soldered components. PCB testing targets several objectives and happens at different stages of the board manufacturing. Only the main tests are briefly described here. The first test consists in verifying that the naked printed circuits (before soldering any component) are free of any damaged conductive pathways and missing printed note-pad. Verifying the soldering process is part of the second test as well as checking the soldered components presence, value, polarization and alignment. This step of verifications is performed either with an AOI (Automated Optical Inspection) machine or an AXI (Automated X-ray Inspection) one, which both scan every PCB going of the soldering process. Then, an electrical test is executed using a bed of nails or a flyer probe tester to check voltages between nodes. Each of these tests requires dedicated method and test equipment.

Embedded System Engineer Students @ POLYTECH MONTPELLIER

Polytech is an engineering school inside the University of Montpellier, which offers to students starting their 3 rd year of study in the Electrical Engineering department, the choice between 2 curricula: "MEA" (Microelectronics Electronics & Automation) and "ES" (Embedded Systems). The students who will select the "ES" curricula will become apprentice students during 3 years and will spend half of a month attending courses and studying and, the remaining half working for a company. The students who will choice "MEA" curricula will be full time studying at the school.

Both future engineers will be able to work in the same fields like the consumers' electronics, automotive, air space and military ones but at different levels: "MEA" engineers will deal with integrated circuits and "ES" engineers will manage systems (from PCB to satellite).

4

Problem Based Learning Approach

Choice of the teaching approach

In 2014-2015, a new course relative to the test of printed circuit boards has been added to the syllabus of the 5 th year Embedded Systems curricula. To get students fully skilled in this discipline, it was necessary to provide them knowledge from lecture and know-how from labs. Unfortunately, no test equipment (bed of nails, AOI or AXI) that would have allowed students acquiring practise in PCB testing technics was available at the school. To compensate this lack of lab material, the teacher organised the visit of the Company OMICRON [2] located nearby Montpellier, which was a PCBs manufacturing and test plant. To benefit from the learning of this visit, the teacher wanted the students to know the basics about PCB testing in order to be active and ask questions during the company tour.

Because "ES" students are apprentices working for a company half of their time, the teacher chose Problem-Based Learning (PBL) approach to set up the course into a teaching context close to their real world working. Students would learn ingroup about PCB testing through the experience of solving an open-ended problem proposed by the teacher. The challenge was to motivate the students collaborating to define in-group a learning strategy that would have promoted individual and teamwork and achieve the learning objectives. The teacher's role was only to guide them to get skilled.

Educational approach organization

Dedicated course syllabus and training material have been elaborated to match the PBL approach (based on the FA2L Louvain teachers 'manual), which basis consists of two sessions of collaborative work and time in between (2-3 weeks) during which students study individually. The 12hrs of the course were distributed as follow: 4hrs for the PBL approach itself, 4hrs for the visit of a PCB manufacturing and test plant, 3 additional hours to allow the students finishing solving the open-ended problem and working on their final report, and 1hr during which the students evaluated their learning outcome in this particular teaching approach. The course ended with a cake party. The exact course schedule is described in table I. The beginning of the first session (2hrs) was dedicated to the presentation of the PLB approach and the learning objectives to the students: "Go" and "Return" sessions agenda, the students' role inside the group, the problem they had to solve ingroup and the learning strategy they had to define. It was crucial that they understood the teacher's expectations about their collaborative and individual learning in order to achieve the course goals. The classroom organization was modified to look like a meeting room facilitating the group discussions. Paper and white boards were also provided.

Teacher' s role

In the PBL approach the teacher's role is to guide the students to solve in-group the open-ended problem [1, 3, 4]: he had to diagnose any obstacle (conflict inside the group, misunderstanding of the problem) that would alter each group's progress according to the procedure. He also had to question indirectly the students about their previous know-how of the problem context to orient the discussion and he had to facilitate the exchange of ideas, the negotiations and communication inside the groups.

Course progress

Problem Based Learning approach presentation to students First, the teacher explained to the 12 students (only men) the motivation of the teaching approach choice. Then, she presented the teaching method it-self: an open-ended problem will be proposed to the students and they will have to solve it in-group, without having any preliminary course. Students will have to identify what they already knew about the topic, what they needed to learn, and how and where to access the information that would lead to the resolution of the problem. It was clearly mentioned to the students that the allocated time to work in-group on the problem in the classroom was limited to 2 sessions (Go and Return ones) of 2hrs. So, each group had to identify the 3 or 4 members that will animate both sessions: the leader (who lead the group's discussion during the session), the moderator (who checked the schedule and took care of any logistics aspect), the scribe (who was the instant memory of the group's ideas, concepts and questions) and the secretary (who took notes during the session and sent the minutes back to all group members). The number and the title of these roles can slightly change from one PBL usage to another.

Between the 2 sessions, each student was expected to perform individual work. The teacher concluded presenting the technical and educational objectives. Getting skilled in PCB test concepts and risks analysis methodology and practice were the technical goals. Cooperate actively within the group to the definition of the learning strategy, individually acquire technical knowledge and share it to the group to build collaboratively the competencies allowing the team to solve the initial problem were the educational targets.

Collaborative work during the "Go session"

During the "Go" session, two groups of 6 students were created based on students' friendship. One group was composed of 5 good classmate students. The 6 th member had his friend in the second group, which gathered the remaining students, some of them also being good classmates.

Each group received from the teacher the same note-pad describing the problem to solve, the step by step procedure and the timing to follow to collaboratively understand the problem, identify their prior-knowledge and ignorance to end it having established a learning strategy plan that each student would performed before the "Return session".

The first step of the procedure was to assign a role to at least 4 group members: the leader, the moderator, the scribe and the secretary. The group composed of the good classmates who approved the learning approach, shown a high motivation and quickly performed this task. The second group, composed of only 3 motivated students over 6, met the first signs of internal conflicts at this step as none of the 3 remaining students wanted to take a role. The teacher neither forced the missing assignment nor questioned the reluctant students about their motivation but rather reminded the timing and the objectives of the "Go session" to the group. Facing the fact that their lack of collaboration would penalize the whole group, the unenthusiastic students started cooperating.

In the second step, the students of each group had to read the open-ended problem statement, to highlight the important words and to list the raised questions, the misunderstood or unclear items. The leader had to collect all inputs from the team members, which were written on a white board by the scribe. The moderator had to take care that this task was performed within 10 minutes.

In the third step, the group had to provide a summarized reformulation of the problem to be sure that the mission was understood and a list of deliverables: to deliver a risk analysis template relative to PCB tests. The secretary had to consign this information. The allocated time was 15 minutes. The next step was to make a status of the group members' current knowledge about the topics of PCB manufacturing and test, and risk analysis methodology. It was the first time when the cooperation of each group member presenting his knowledge contributed to build the group knowledge. So, this stage was important because the learning strategy of a group depended of its individual's erudition on the topic. The teacher had to carefully watch the communication inside each group to check that each student gave his statement and opinion. If not, he had to question the leader to make him realizing his mistake. 30 minutes were allocated to this task.

The 5 th step consisted in 15 minutes to list the learning objectives to solve the problem with the assessment conditions for each goal.

The 6 th and last step of the "Go session" was the creation of a Gantt chart describing each task with its duration, allocated resource (documentation) and owner(s), and the list of deliverables to produce. It was important that the students identify the tasks directly linked to the learning objectives, to keep them focused.

The group with the 6 motivated students went through all the 5 steps after the roles assignment without any major problem. The behavior of the 3 unwilling students of the second group, the authority issue that faced the leader and the lack of interaction from the teacher often conduced the group to unorganized discussion, unstructured animation, no teamwork collaboration and a communication restricted to some members. Laboriously, the group followed each procedure step and established a learning strategy plan at the end of the "Go session".

Individual work

During the 3 weeks separating the "Go" and "Return" sessions, the students had to read the documents provided in the note-pad according to the learning strategy both groups had defined. Each student read all the documentation in order to acquire alone the entire requested knowledge to solve the problem, focused on a particular topic and produced a summary he will share with the group during the "Return session".

Even if it was not requested, each group sent to the teacher a report of the "Go session" which detailed its Gantt chart.

Collaborative work during the "Return session"

During the "Return session" (2hrs) 3 weeks later, both groups discussed their findings and started elaborating together a solution based on what they learned individually. Once again, they followed a step-by-step procedure. First, each group organized, in 10 minutes, the assignment of the leader, moderator, scribe and secretary roles. They decided to keep their previous ones even if the teacher suggested them changing. The second and main step (50 minutes) was the synthesis of the learning. It required the cooperation of each group member who had to share the outcome of his individual work. Any student who would not have performed his assigned task would impact the technical learning objectives of the group. The individual work was crucial for the group collaborative success. Once again, one group achieved almost straight forwards this synthesis task, whereas the other one faced team conflicts. The passive behaviour of the 3 students influenced a lot team spirit and its outcome as the group ran out of the time allocated to this procedure step. One again, both groups produced a report of the "Return session".

Visit of OMICRON Company and outcome

The director spent 3 hours presenting the different PCB test steps and equipment to the students. Thanks to the students' involvement in the active learning approach, most of them were able to link the test concepts and technics they learnt to the real usage of the equipment during the demonstrations. All students were happy of the visit and admitted that the experi-ence would not have been so fruitful with a classic lecture teaching method in which they would not have been the protagonists of their learning.

Students 'skills evaluation

The students had to deliver one risk analysis report per group about the test technics applied to verify PCB based on the knowledge they acquired reading documentation selected by the teacher and the outcomes of the visit.

Students' learning self-evaluation

Individually, the students had to evaluate their learning in the context of the PBL approach. They had to complete a dedicated form developed based on an evaluation form from FA2L Louvain teachers [1].

Three main sections composed the document. They were organized as follow:

Student's evaluation of the group performances. Using a spider diagram (see figure 2), they had to rank the following criteria:

• Group learning outcome: the group produced something satisfactory, this production was really the result of a collective effort, the meetings were effective, the exchanges allowed bringing to the foreground different points of view to handle the problem. • Group members' investment and communication: each of the participants contributed significantly to the efficiency of the group, the group gave the opportunity to each of his members to express its point of view even to the passive participants, all the members of the group proposed their individual work done between both sessions • Relationship with the teacher: in a general way, the group took advantage of the teacher's presence and guidelines to progress. • Relationship with the learning context: the group became quickly motivated to work the theme; the group found that the problem was well adapted to the members that it was enough complex and rich enough for a meaningful learning. • Group organization: the group succeeded in coordinating its activities, the group remained centred on the task to be carried out, the group made a good use of the white board, the secretary kept tracks of the exchanges, the leader played its role and the timing was respected • Atmosphere in the group and in the working context: the agreement between the members of the group was good, the participants encouraged each other's, the group managed to surmount its differences of view, and nobody managed to impose his point of view.

The ranking scale was from 1 to 4 with the following meaning: 1= not satisfying; 2= not really satisfying; 3= satisfying; 4=very satisfying.

Cake party

The teacher proposed a cake party to celebrate the end of the course. All students brought food and beverages and shared them as shown on figure 3. Students took this opportunity to give a subjective feedback about the educational approach. The motivated students were very satisfied as they considered they learnt in a collaborative context close their real work. They liked being the protagonists of their learning and it increased their self-esteem to contribute to the group success. The unenthusiastic students explained they did not like the approach because preferred the classic top-down teaching method rather than this student-centered one. Educational Approach outcomes

Analysis of students 'skill evaluation

As mentioned earlier, each group produced a summary report of the "Go" and "Return" sessions in addition to the final risks analysis one. Each report was marked. The final mark was the average mark of the 3 reports.

As each member of one group fairly collaborated to solve the initial open-ended problem being all very invested in their own learning as well as in the success of the group, they all obtained the same mark. Concerning the second group, only the 3 motivated students received the highest mark produced by the evaluation of the 3 reports. The 3 others were penalized and lost one point.

This experience shown that evaluating the students' skills using the PBL approach was not an easy task. It was almost not possible to objectively mark the learning a single student acquired during the course. So, when selecting a teaching approach, it is not only important to define the learning objectives but also to identify the "best way" to verify the students have achieved them. The report was an appropriate choice to represent the work produced by each group. But, may be, an additional quiz would have allowed to individually check the students' learning acquisition. As it was the time this course was put in place, there is no reference to compare the students' marks with an estimate the relevance of the teaching approach.

Analysis of the students' learning self-evaluation

This paragraph presents the synthesis and analysis of the students' answers about three sections of the evaluation form.

Student's evaluation of the group performances.

First, the 12 spider diagrams were studied all together and the average values to the 6 items varied between 3 and 3.38, which were high scores as the maximum value is 4. So, as a preliminary conclusion, it appeared that all students recognized the interest of the collaborative work.

As 3 students out of 12 clearly shown a lack of motivation to learn in this collaborative context, a second analysis was Atmosphere in the group and in t h e w o r k i n g context performed on the spider diagrams, this time separating the ones from the motivated students to the unwilling ones. The compilation of numbers obviously pointed out a major difference of appreciation of the collaborative teamwork between the motivated and not motivated students. All the compiled numbers are presented in table III and reported on the spider diagram in figure 4 using a specific color that code is mentioned in the table. The students enumerated the actions that went fine and the ones that did not concerning learning in-group. It appeared that several actions obtained opposite opinions like the assignment of the roles, the atmosphere inside the group, the cooperation of the students, the organization and the distribution of the tasks inside a group that conduced to different workloads between students, the communication inside the group, the animation of the group and the collaboration of each student during the "Return session" to share his outcomes of the whole group.

The students insisted on the fact that the time allocated to perform the procedures during the "Go" and "Return" sessions was too short, in particular to clearly understand the problem and its context. Nevertheless, most of the students were satisfied of the overall process: defining in-group a learning strategy to solve the open-ended problem, studying individually all the documents but focusing on a dedicated topic and sharing their outcome to the whole group to collaboratively allow all members to achieve the learning objectives. All students liked the cake party.

Concerning the actions the students could personally do to improve the work in-group, they mentioned: to be more involve to provide a better collaboration, communicate and helped reaching the learning objectives in less time and in a better team spirit.

Students' learning self-evaluation.

Even if each student was free to list any relevant learning skill, three main topics appeared from the synthesis of all the lists: knowledge about PCB manufacturing, PCB testing and risk analysis methodology. Each student ranked from 0 to 5 his level of knowledge for each skill before and after the course. The compiled results of all students, of the 9 motivated and 3 unwilling ones are presented in table IV.

It was surprising to discover, from the analysis of table IV, that the 3 not motivated students achieved a better level of knowledge on 2 out of 3 skills compared to the 9 motivated students. Were they over-estimating themselves? The number of hours each students had spent learning was studied and it appeared that the 3 students spent an average time of 6.2hrs per student studying whereas the 9 motivated ones declared an average of 5.4hrs per student. This shown that the 3 students who demonstrated a none collaborative behavior wanted to learn to be skilled in these topics but disagreed on the teaching approach.

The analysis of table IV also put on evidence that even if the students improved this initial knowledge on the 3 topics, they were far away to be fully skilled (level 5) at the end of the course.

Statement

Problem-Based Learning approach required a long preparation from the teacher in terms of research on papers, documents, web sites relative to the topics. The teacher has to read all documents to identify the most relevant ones to be proposed to the students in the note-pad.

Despite the fact that the students who studied the documents became skilled on the new topics and able to solve the given problem, the real challenge was to motivate each group Atmosphere in the group and in t h e w o r k i n g context members to fairly collaborate and to acquire these competencies. The teacher who became a tutor meaning that her role was to guide the students, had to build their confidence to take on the problem, and to encourage them cooperating in this new learning approach, which represented a paradigm shift from traditional teaching and learning philosophy, which was more often lecture-based. This new tutor role was not easy to play.

Improvements

Changes have to be applied to this collaborative educational approach to improve the technical and educational outcomes:

• In order to build a real teamwork spirit and to facilitate the collaboration and communication within the group, groups should not be created based on students' friendship.

• The teacher has to interact more often, in particular when diagnosing any conflict between group members and normalize it, or when the group shows a lack of dynamism du to none fairly collaboration of each member in order to minimize the impacts on the group success. • The teacher should add regular milestones to check students' knowledge progresses. • A solution has to be found to allow students practicing and acquiring know-how on these technical topics. • The teacher has to spend more time explaining the different roles and has to ask the leader and the moderator to respect the timing. May be a game that rules would fit to the PBL ones could be a good way to initiate the students to the new collaborative approach. • The teacher has to identify the appropriated exam mode to verify each student achieved the technical learning objectives.

Conclusion

From pedagogical point of view, the "problem-based learning" approach was an interesting experience for the 5 th year apprentice engineer students as they discovered an innovative learning method where they were the protagonists of their own learning and which allowed them to study in conditions close to real world work when collaboration is in purpose. For the students, the gains of this approach were the acquisition of technical knowledge in PCB test and risks analysis methods and they got trained in thinking strategies, decision-making, negotiation and collaborative work.

For the teacher, this approach requested a long preparation, as it was necessary to develop dedicated course syllabus as well as course material to fit the PBL approach procedure and the learning objectives. The teacher had to get familiar with his new role of instructor or tutor, which was not a natural thing. He had to facilitate learning by supporting, guiding, and moni-toring the learning process of the students in-group without ordering them.

Based on this first positive experience, the PBL approach will be repeated in 2015-2016 taking into account all the enumerated improvements.

ACKNOWLEDGMENT Authors acknowledge the support of the French Agence Nationale de la Recherche (ANR), under grant ANR-11-IDFI-0017 (project IDEFI-FINMINA).

Fig. 1 .

 1 Fig. 1. Photograph of a mounted PCB.

Fig. 2 .

 2 Fig. 2. Spider diagram for the "Problem-Based" approach evaluation by the groups of students

Fig. 3 .

 3 Fig. 3. Photograph of the ES students at the cake party

Fig. 4 .

 4 Fig. 4. Graphical representation of the group performances compilation

Table 1 .

 1 PCB TESTING COURSE ORGANIZATION

	Date	Title of the session	Hrs
	15/3/4	PBL Go session: collaborative work	2h
		Individual work at home
	15/4/1	PBL Return session: collaborative work	2h
	15/4/8	OMICRON's visit	4h
	15/5/6	Collaborative work on risk analysis & report	3h
	15/5/25 Student's learning self-evaluation & Cake party	1h

Table 2 .

 2 RANKING SCALE MEANING

	0	1	2	3		4		5
	I don't	I	I can say	I	can	I can list the	I can argue,
	anything	have	one or two	describe	relevant		justify my
	about	heart	things	the main	items of the	statement,
	the topic	about	about the	items of	topic	and	choices and
		the	topic	the topic	the relation	decisions
		topic				between		about the
						them		topic

Table 3 .

 3 RESULTS COMPILATION

	Group Per-	12 students	9 moti-	3 not motivated
	formances	(Blue dots)	vated stud.	stud.
	/ # Student		(Black dots)	(Red dots)
	a) Outcome	3.32	3.53	2.67
	b) Investment	3.17	3.34	2.67
	c) Teacher	3.38	3.17	4
	d) Problem	3.13	3.17	3
	e) Organization	3	3.28	2.34
	f) Atmosphere	3.5	3.8	2.67

Table 4 .

 4 RESULTS SYNTHESIS

			12 students	9 motivated stu-	3 not moti-
					dents		vated students
	Knowledge /	Be	A	Before	Af	Be	Af
		# students	fore	fter		ter	fore	ter
	PCB manuf.	1.92	3.1	1.89	2.89	2	3.67
		PCB test	1.43	3.4	1.61	3.61	0.89	2.75
	Risk analysis	0.75	2.42	0.89	2.11	0.33	3.33
	6	Statement & Improvements