Identification of generalized impedance boundary conditions in inverse scattering problems, Inverse Probl

In the context of scattering problems in the harmonic regime, we consider the problem of identification of some Generalized Impedance Boundary Conditions (GIBC) at the boundary of an object (which is supposed to be known) from far field measurements associated with a single incident plane wave at a fixed frequency. The GIBCs can be seen as approximate models for thin coatings, corrugated surfaces or highly absorbing media. After pointing out that uniqueness does not hold in the general case, we propose some additional assumptions for which uniqueness can be restored. We also consider the question of stability when uniqueness holds. We prove in particular Lipschitz stability when the impedance parameters belong to a compact subset of a finite dimensional space.


Introduction
We address in this work uniqueness and stability issues related to the identification of a medium impedance from the knowledge of far measurements of a scattered wave at a given frequency.We restrict ourselves in these first investigations to the scalar case (the Helmholtz equation), that models either acoustic waves or two dimensional settings of electromagnetic problems.Assuming that the unknown medium occupies a domain D, the medium impedance is understood as a "local" operator that links the Cauchy data of the field u on the medium boundary Γ := ∂D.More precisely we shall consider the cases where a boundary condition of the form ∂u/∂ν + Zu = 0 on Γ is satisfied, where Z is a boundary operator and ν denotes the outward normal field on Γ.The exact impedance operator Z corresponds to the so-called Dirichlet-to-Neumann (DtN) map, i.e. f → −∂u/∂ν| Γ where u solves the Hemholtz equation inside D and satisfies u = f on Γ. Consequently determining this map is "equivalent" to identify the physical properties inside D, which is in general a severely ill-posed problem that requires more than a finite number of measurements.
We are interested here in situations where the operator Z is an approximation of the exact DtN map.In general these approximations correspond to asymptotic models associated with configurations that involve a small parameter.These cases include small amplitude roughness, thin coatings, periodic gratings, highly absorbing media, . . .We refer to [18,8,2] for a review of these models.
The simplest form is the case where Z is a scalar function, which corresponds in general to the lowest order (non trivial) approximations, for instance in the case of very rough surfaces of highly absorbing media (the Leontovich condition [15]).However, for higher order approximations or in other cases the operator Z may involve boundary differential operators.For instance when the medium contains a perfect conductor coated with a thin layer of width δ then for TM polarization, the approximate boundary conditions of order 1 corresponds to Z = 1/δ while for the TE polarization it corresponds to Z = δ(∂ ss + k 2 n) where s denotes the curvilinear abscissa, k the wave number and n is the mean value of the thin coating index with respect to the normal coordinate.Higher order approximations would include curvature terms or even higher order derivatives [4,10,11].This type of conditions will be referred to as Generalized Impedance Boundary Conditions.
We shall address in the present work the question of unique identification and stability of the reconstruction of the operator Z from the knowledge of one scattered wave.One easily sees, from the given example, how the identification of the impedance would provide information on some effective properties of the medium (for instance, the thickness of the coating and the normal mean value of its index).Determining these effective properties would be less demanding in terms of measurements than solving the inverse problem with the exact DtN map (the unknown parameters have one dimension less) and we also expect that the inherent ill-posedness to be less severe.Motivated by the example above we shall consider generalized impedance boundary conditions of the form: (1) where µ is a complex constant and λ is a complex function and where ∆ Γ denotes the Laplace-Beltrami operator on Γ.The case of standard impedance problems (µ = 0) has been studied by several authors [14,19], where for instance optimal logarithmic stability is obtained.Our analysis here is different and is rather motivated by numerical considerations.It also applies to the case µ = 0. We shall prove injectivity of the Fréchet derivative (which is equivalent to local stability) of the inverse map by using the adjoint state technique.We also investigate the situations where Lipschitz stability can be obtained.In the same spirit as in [3] we shall prove a general result showing how Lipschitz stability holds when we restrict the impedance parameters to a compact subset of a finite dimensional space, assuming injectivity of the inverse map and of its derivative.Motivated by this result we investigate the uniqueness question from the knowledge of one scattered wave.We show that uniqueness when µ = 0 cannot be true in general, but holds under further assumptions on the parameters (typically assuming that part of the parameters are known) and on the geometry.
The outline of our article is the following.We introduce in the next section the forward scattering problem and briefly recall the mathematical property of the solution in the case of GIBC.We then formulate the inverse problem.Section 3 is dedicated to the question of uniqueness from the knowledge of the far field associated to one incident wave.Section 4 deals with the question of stability by analyzing the injectivity of the Fréchet derivative.It also includes a general result on Lipschitz stability.

The forward and inverse problems
Let D be an open bounded domain in R 3 , and Ω := R 3 \ D. The domain D will be referred to as the obstacle and a GIBC holds on its boundary Γ.An incident plane wave u i (x) := e ik d.x , with k > 0 denoting the wavenumber and ||d|| = 1, is scattered by the obstacle D and gives raise to a scattered field u s .The governing equations for u s in Ω are and where ν denotes the unit normal to Γ oriented to the exterior of D. If we set u := u s + u i as being the total field, then the second equation of ( 2) is equivalent to (3) ∂u/∂ν + µ∆ Γ u + λu = 0 on Γ.
The last equation in (2) is the classical Sommerfeld radiation condition where B R denotes a ball of radius R.
We assume that the boundary Γ is Lipschitz continuous, such that for a function v ∈ H 1 (Γ), the tangential gradient ∇ Γ v is defined in (L 2 (Γ)) 3 , and ∆ Γ v is defined in H −1 (Γ) by the identity The space of solutions to the forward problem differs between the cases µ = 0 and µ = 0. 1.In the case µ = 0, problem (2) coincides with the classical impedance problem.

It is uniquely solvable in
with Im(λ) ≥ 0 (see for instance [17]).This result remains true for f ∈ H −1/2 (Γ).In [7], it is proved that if moreover the boundary Γ is C 2 and λ ∈ C 0 (Γ), then the solution u s is continuous up to the boundary Γ. 2. In the case µ = 0, problem (2) will be referred to as the generalized impedance problem.It is uniquely solvable in with Im(λ) ≥ 0 as well as Re(µ) > 0 and Im(µ) ≤ 0. This result remains true for f ∈ H −1 (Γ).For sake of completeness, we shall sketch the proof in the case of the generalized impedance problem which can be seen as a slight adaptation of the proof in [17] for classical impedance problems.We restrict the problem to a bounded domain Ω R := Ω∩B R with the help of the Dirichlet-to-Neumann map S R : where u e is the solution to the Helmholtz equation in R 3 \ B R satisfying the Sommerfeld radiation condition and u e = g on ∂B R .This operator satisfies in particular Re S R g, g ≤ 0 and Im S R g, g ≥ 0 ∀ g ∈ H where •, • denotes the duality product between H − 1 2 (∂B R ) and Problem ( 4) is then proved to be equivalent to the weak formulation: find It is easily seen that V R equipped with the scalar product (•, is a Hilbert space.The weak formulation can then be written in the form: find u s in V R such that where the operators J R : and F R is uniquely defined by One then easily checks, by application of the Lax-Milgram theorem that J R is an isomorphism for Re(µ) > 0, and the application of the Rellich compact embedding theorem that K R is compact.Therefore, with the help of Fredholm alternative it is sufficient to prove the injectivity of the operator J R + K R , which is equivalent to prove uniqueness of solutions to problem (2).Following [7], a sufficient condition for uniqueness is that for all solution u ∈ V of (2) with f = 0, we have uniqueness in the case Im(µ) ≤ 0 and Im(λ) ≥ 0, which completes our sketch of proof.Let us finally notice that both operators J R and K R depend continuously on λ and µ.
Formulation of the inverse problem.Following [7], the solution u s of problem (2) has the asymptotic behavior uniformly in all directions x = x/r ∈ S 2 , where r := |x| and S 2 denotes the unit sphere in R 3 .The far field u ∞ ∈ L 2 (S 2 ) is given by where Φ ∞ (y, x) := 1 4π e −ikx.y and where the second integral has to be understood as a duality pairing between H −1/2 (Γ) and H 1/2 (Γ).
The inverse problem we are interested in is to determine both µ and the function λ on Γ from the knowledge of the far field u ∞ associated with one direction d of the incident field.We shall address the questions of uniqueness and stability of this problem.For both questions we shall first consider the classical impedance problem (µ = 0) and then the case of generalized impedance problem.Answering these questions amounts to study the properties of the non linear map T : λ −→ u ∞ for the first case and T : (λ, µ) −→ u ∞ for the second one.

Uniqueness
This section is devoted to the investigation of the uniqueness for the inverse problem formulated above which is equivalent to analyzing the injectivity of the operator T .
3.1.The classical impedance problem.Analyzing the injectivity of T as an operator acting on L ∞ (Γ) seems to be challenging.We shall restrict ourselves to the subspace of L ∞ (Γ) formed by piecewise-continuous functions.Let I be a given integer and Γ i , i = 1, ..., I, be open sets of Γ such that Γ i ∩ Γ j = ∅ for i = j, Γ = ∪ I i=1 Γ i , and the sets ∂Γ i are negligible in the sense of the Lebesgue surface measure supported by Γ.We define the subspace C I (Γ) as the set of functions u ∈ L ∞ (Γ) such that each restriction of u to Γ i belongs to C 0 (Γ i ).The space We shall also use the subsets where u ∞ is defined from u s by using (5) and u s is defined from λ by solving (2), is injective.
From the Rellich Lemma and the unique continuation principle (see [7]), the function u which are well defined in H 1/2 (Γ) and H −1/2 (Γ) respectively, also vanish.Considering now the boundary condition on Γ, We shall prove that λ 1 = λ 2 almost everywhere on Γ. Assume that there exists x 0 ∈ Γ, where x 0 does not belong to some ∂Γ i , such that λ 1 (x 0 ) = λ 2 (x 0 ).Since λ 1 and λ 2 are continuous in a neighborhood of x 0 , there exists η > 0 such that and unique continuation leads to u 1 = 0 in Ω.Since the incident wave u i does not satisfy the radiation condition, while the scattered field u s 1 does, we obtain a contradiction.
Remark 1.Let us notice that the proof remains valid even if I and the partition Γ i , i = 1, . . ., I are not a priori known.
The requirements of ( 6) will be referred to as assumption (H0) in the following.
It is constructed as follows: one first chooses a constant impedance λ 0 ∈ iR with Im(λ 0 ) > 0 such that the corresponding solution u 0 of the classical impedance problem, which is C ∞ up to the boundary Γ, does not vanish on Γ.This choice is possible as demonstrated in 2D by the case D = B(0, 1), d = (1, 0), k = 1, λ 0 = i where the solution u 0 (1, θ) on Γ is given (after straightforward calculations) by , where H n denotes the Hankel function of the first kind of order n.The modulus of the function u 0 (1, .) is depicted in Figure 1.Let us denote by u ∞ 0 the far field associated with u 0 .Since u 0 does not vanish on Γ, the function α = ∆ Γ u 0 /u 0 is a C ∞ function on Γ, and therefore bounded on Γ.Now let µ 1 = µ 2 be any complex numbers such that

Inverse Problems and Imaging
Volume X, No. X (200X), X-XX and let the associated C ∞ functions λ 1 and λ 2 be defined on Γ by We have on Γ As a result, u ∞ 0 is the far field associated to the solution u 0 of the generalized impedance problem with both (λ 1 , µ 1 ) and (λ 2 , µ 2 ).It should also be noted that even if µ is real (which would correspond to the thin coating case, as presented in the introduction, with µ = δ and λ = δk 2 n 2 ), our counterexample shows that unique determination of (λ, µ) does not hold in general.The counterexample above justifies the need to consider restricted cases where some a priori knowledge is introduced in addition to the assumption (H0) given by (6).We were able to recover injectivity in the following complementary cases: 1. Assumption (H1): λ ∈ C and µ ∈ C are two constants.
Furthermore, we assume there exist x 0 ∈ Γ and η > 0 such that Γ 0 = Γ ∩ B(x 0 , η) is either a portion of a plane, or a portion of a cylinder, or a portion of a sphere, and such that the set {x + γν(x), x ∈ Γ 0 , γ > 0} is included in Ω.
• (H2b) both Re(λ) and Im(µ) are fixed and known, the unknown being Im(λ) and Re(µ), • (H2c) both Im(λ) and Re(µ) are fixed and known, the unknown being Re(λ) and Im(µ), • (H2d) µ is fixed and known, the unknown being λ.Furthermore, in the three cases (H2a), (H2b), (H2c), we assume there is no constant C such that on Γ: Here S and T respectively denote the traces of the single and double layer potentials on Γ, while T * and −R denote the traces of the normal derivatives of the single and double layer potentials on Γ (see equation ( 16) below).This assumption will be referred to as (HC) in the following.It should be noted that the matrix P denotes the so-called Calderón projection for the interior problem (the operator P actually satisfies P 2 = P, [16]).In particular, the first of the two above equations can be specified as Assumption (HC) may be impossible to verify in practice, this is why we hereafter mention two particular situations where this condition is automatically satisfied (see Lemma   • the function λ is real and k 2 is not a Dirichlet eigenvalue of the operator −∆ in D, • the domain D is C 1 and both D and the function λ are assumed to be invariant by reflection against a plane which does not contain the direction d or by a rotation (different from identity) around an axis which is not directed by d.Let us remark that the proof under assumption (H2d) follows exactly the same lines as the proof in the case µ = 0 given in the previous section.We therefore shall detail the proofs only for assumptions (H1), (H2a) − (HC), (H2b) − (HC), (H2c) − (HC).The results are formulated and proved in Propositions 2, 3 and 4.
Proposition 2. Under assumption (H1) the farfield associated with one incident plane wave uniquely determines the coefficients λ and µ.
Proof.Using the same arguments and notation as in the first paragraph of the proof of Proposition 1, we arrive in the present case at: In order to simplify the notation we set in the following u := u 1 .If µ 1 = µ 2 , then we easily conclude that λ 1 = λ 2 using the same arguments as in the proof of Proposition 1.If We notice that α = −k 2 Γ , where k 2 Γ is an eigenvalue of the negative Laplace-Beltrami operator −∆ Γ .Let us consider first the case when Γ contains a portion of plane Γ 0 = Γ ∩ B(x 0 , η) of outward normal ν such that the set Q 0 = {x + γν, x ∈ Γ 0 , γ > 0} is included in Ω.Then there exists a system of coordinates (x 1 , x 2 , x 2 ) such that x(0, 0, 0) = x 0 and Inverse Problems and Imaging Volume X, No. X (200X), X-XX where the function c is uniquely defined by ( 10) In the domain On the surface Γ 0 we obtain The functions ũ and u are both solutions of the same Helmholtz equation in Q 0 , and they satisfy ũ = u and ∂ ν ũ = ∂ ν u on Γ 0 .Hence, unique continuation implies ũ = u in Q 0 .Since u s satisfies the radiation condition when ||x|| → +∞, we have in particular that lim x3→+∞ |u s (x 1 , x 2 , x 3 )| = 0. Recalling that u(x 1 , x 2 , x 3 ) = u s (x 1 , x 2 , x 3 ) + e ik(d1x1+d2x2+d3x3) , we obtain (11) u(x 1 , x 2 , x 3 ) ∼ e ik(d1x1+d2x2+d3x3) , x 3 → +∞, and in particular when To see that the asymptotic behavior ( 12) is impossible, we have to discuss separately the three cases k Γ < k, k Γ = k and k Γ > k.In the case k Γ < k, we obtain from ( 10) This implies that either In both cases, c(x 3 ) = e ikd3x3 and by using again that u(x 1 , x 2 , x 3 ) = u(x 1 , x 2 , 0)c(x 3 ) in Q 0 , we conclude from (11) that u(x 1 , x 2 , 0) = e ik(d1x1+d2x2) in Γ 0 , whence u = e ikd.x in Q 0 , and lastly u = e ikd.x in Ω.Now we use ∂u/∂ν = βu on Γ to see that d.ν is constant on Γ.This is forbidden by the fact there exist at least 4 points x ∈ Γ such that the 4 corresponding outward normals ν(x) are different.In the case k Γ = k, we have c(x 3 ) = 1 + βx 3 , while in the case k Γ > k, we have Inverse Problems and Imaging Volume X, No. X (200X), X-XX which in both cases contradicts the asymptotic behavior (12).We now briefly consider the cases when Γ 0 is a portion of cylinder or a portion of sphere.If Γ 0 is the portion of cylinder of radius R, then in Q 0 we obtain with appropriate cylindrical coordinates (r, θ, z) where the function c is uniquely defined, with Following the case κ = 0, κ 2 > 0 or κ 2 = (iκ) 2 < 0, the function c is respectively a linear combination of 1 and log r, a linear combination of H 1 0 (κr) and H 2 0 (κr) (Hankel functions of the first and second kind), or a linear combination of I 0 (κr) and K 0 (κr) (Modified Bessel functions).Whatever the case, the behavior of c(r) when r → +∞ is not consistent with u(r, θ, z) ∼ e ik(rdr+zdz) , r → +∞.
If Γ 0 is the portion of sphere of radius R, then in Q 0 we obtain with appropriate spherical coordinates (r, θ, φ) where the function c is uniquely defined, with Following the case κ = 0, κ 2 > 0 or κ 2 = (iκ) 2 < 0, the function c is respectively a linear combination of 1 and 1/r, a linear combination of e ikr /r and e −ikr /r, or a linear combination of e kr /r and e − kr /r .Whatever the case, the behavior of c(r) when r → +∞ is not consistent with u(r, θ, ψ) ∼ e ikrdr , r → +∞, which completes the proof.
Remark 2. The geometrical assumptions in (H1) are only technical and are dictated by the proof where an explicit solution, using separation of variables, is needed.It seems plausible that our technique can also be generalized to other geometries where explicit solutions that satisfy (8) can be obtained.However it does not apply to the general case.
Here Φ(x, y) := e ik||x−y|| /4π||x−y|| denotes the fundamental solution of the Helmholtz equation that satisfies the Sommerfeld radiation condition.Let γ − 0 (resp.γ + 0 ) be the interior (resp.exterior) trace application, γ − 1 (resp.γ + 1 ) be the interior (resp.exterior) trace of the normal derivative application.Following [16], we define ( 16) , and we recall the jump relations First, by passing to the limit x → Γ in ( 14) and ( 15), it follows that in the sense of trace on Γ, Using now u = u s + u i , the boundary conditions (13) and the equation ( 17), we obtain that 1 2 By subtracting (18) to the above equation we obtain so that by using once again Inverse Problems and Imaging Volume X, No. X (200X), X-XX on Γ.Secondly, by taking the trace on Γ of the normal derivative in ( 14) and ( 15) and by repeating the same calculations as above, we obtain (20) in the sense of H −1/2 (Γ).Equations ( 19) and ( 20) are equivalent to (7), which is forbidden by assumption (HC).
Remark 3. In view of lemma 3.1, we point out that since λ ∈ L ∞ (Γ), S(λ) reduces on Γ to and, if the boundary Γ is of class C 1 or is a polyhedral, T (1) reduces on Γ to Remark 4. One can easily show that if assumption (HC) is not satisfied, i.e. one can find a constant C and a field λ which satisfy (7), then one can reconstruct a total field u associated to the incident field u i that satisfies the generalized impedance boundary conditions for any value of µ.In other words, assumption (HC) is necessary to have uniqueness.Proof.The identity on Γ leads, by denoting again u 1 = u to simplify the notation, and if we assume By multiplying the above equation by u and by integrating by part over Γ, we obtain The assumption (H2b) implies that λ 2 − λ 1 ∈ iR and µ 2 − µ 1 ∈ R, while the assumption (H2c) implies that λ 2 − λ 1 ∈ R and µ 2 − µ 1 ∈ iR.In both cases (λ 2 − λ 1 )/(µ 2 − µ 1 ) ∈ iR.By taking the real part of the above equation we obtain whence there exists a constant C such that u = C on Γ.This implies that ∆ Γ u = 0 and by using the same arguments as in the proof of Proposition 1, one obtains in particular that λ 1 = λ 2 := λ.We are then in the same configuration as in ( 13), and therefore one can complete the proof as for Proposition 3. Now we specify two particular situations where the condition (HC) is automatically satisfied.Then, there exists no constant C such that Proof.Let us consider first the case when λ is real and k 2 is not a Dirichlet eigenvalue of the operator −∆ in D. Taking the imaginary part of (21), we obtain on Γ Let us define the two functions on D ∩ Ω, Considering the regularity of Let us now consider the second case where D and λ are assumed to be invariant by a transformation S which is either a reflection against a plane P with d / ∈ P or a rotation (different from identity) around an axis A with d / ∈ A. Without loss of generality, we assume that the plane P and the axis A contains O.We have We use the new variables x = Sx ′ and y = Sy ′ , and the new function φ ′ defined by φ(x) = φ ′ (x ′ ) = φ(Sx ′ ).We have for some x ∈ Γ, Using the chain rule .S and ν y = Sν y ′ , we obtain By using the change of variable y = Sy ′ in the integral, we obtain Finally, since Sx ′ = x ′ and given the form of φ, φ ′ (x ′ ) = φ(x ′ ), and thus T (1)(x) = T (1)(x ′ ).As a result T (1)(Sx) = T (1)(x) for all x ∈ Γ.We prove the same way that S(λ)(Sx) = S(λ)(x) for all x ∈ Γ, and if (21) holds for some constant C, then for all x ∈ Γ If x describes a sufficiently small part of Γ which contains the points x 0 such that Sx 0 = x 0 , then n = 0.In such small part of Γ, we hence have Sx − x ⊥ d.Since d does not belong to the plane or the axis of symmetry, we conclude that (Sx − x).d cannot vanish in the small part of Γ we consider, which completes the proof.
Remark 5.If the boundary Γ is C 1 , under either of assumptions (H2a), (H2b), (H2c), uniqueness of (λ, µ) from two far fields u ∞ 1 and u ∞ 2 generated by two plane waves u i 1 = e ikx•d1 and u i 2 = e ikx•d2 with d 1 = d 2 holds without assuming (HC).This follows from existence of two constants C 1 and C 2 such that (19) is satisfied for u i 1 and u i 2 respectively, and then differentiation of x) = 0 on Γ along the tangent planes at two points x 1 and x 2 of Γ which are perpendicular to d 1 and d 2 respectively.

4.1.
The classical impedance problem.The analysis of stability for the inverse problem here is based on the Fréchet derivative of operator T : ).We recall that this operator is differentiable in the sense of Fréchet if there exists a linear continuous operator dT λ : We have the following proposition.
Proposition 5.The operator T is differentiable and its Fréchet derivative is the operator where p(y, x) = Φ ∞ (y, x) + p s (y, x), Inverse Problems and Imaging Volume X, No. X (200X), X-XX p s (., x) is the solution of problem (2) in which u i is replaced by Φ ∞ (., x).Moreover, the operator dT λ : Proof.We denote u s and u s h , the solutions of (2) which are associated to λ and λ+h in L ∞ + (Γ).In order to prove differentiability, we first prove continuity.As we already remarked at the end of section 2, if we denote , where K h and F h satisfy K 0 = K, F 0 = F and depend continuously on h.By denoting A = I − K and A h = I − K h , we already know that A −1 and A −1 h exist and are bounded.Then the family of operators A h meet the assumptions of Theorem 10.1 in [13], and provided that |||A −1 (A h − A)||| < 1, which happens when ||h|| L ∞ (Γ) is sufficiently small, we obtain the error estimate The continuity of the operator T follows.Now let us denote e s h = u s h − u s .The function e s h satisfies the boundary condition on Γ ∂e s h ∂ν + λe s h = −h(u s h + u i ).This leads to the fact that there exists C > 0 (independent of h) such that (24)

Finally, let us define v ∞
h as the far field associated to the scattered field v s h , where v s h is the solution of problem (2) with data f = −h(u s + u i ).The mapping h → v ∞ h is clearly a linear continuous operator L ∞ (Γ) → L 2 (S 2 ).We obtain It follows that u s h − u s − v s h is the solution of problem (2) with data f = −he s h .By using (24), it is clear that T is differentiable in the sense of Fréchet with dT λ (h) = v ∞ h .We have from (5), Using the identity the definitions of p s and v s h lead to As a result, we obtain Inverse Problems and Imaging Volume X, No. X (200X), X-XX which is the result since u = u s + u i and p = p s + Φ ∞ .It remains to prove that dT λ : C I (Γ) → L 2 (S 2 ) is injective.Assume that for h ∈ C I (Γ), we have Γ p(y, x)u(y)h(y) ds(y) = 0, ∀x ∈ S 2 .
By using lemma 4.1, there exists a sequence xn , n ∈ N, with p(., xn )| Γ → u| Γ h in L 2 (Γ).It follows that u| Γ h = 0 on Γ, and by using the same arguments as in the proof of proposition 1, we conclude that h = 0.
The proof of the above proposition requires the following density lemma.Let u s be the solution of ( 2) which is associated to data f .We have Using now the boundary condition for p s , i.e. ∂p s /∂ν From (25) we have By using the two previous equalities and once again the boundary condition for u s , it follows that which is exactly u ∞ (x) = 0, for all x ∈ S 2 .We conclude from Rellich's lemma and unique continuation that u s = 0, and hence f = ∂u s /∂ν + λu s = 0.
We now introduce D I (Γ) a finite dimensional subspace of C I (Γ).We have the following result of local Lipschitz stability in , where u ∞ = T (λ) and u ∞ h = T (λ + h).
Lastly, K I (Γ) denotes a compact subset of D I (Γ) ∩ L ∞ + (Γ).In the same spirit as in [3], we obtain the following result of stability in K I (Γ), which is Lipschitz stability for the considered inverse impedance problem.
Remark 6.Let us remark that similar Lipschitz stability results were already established in [20] (see Theorem 2.4) for the Laplace equation and standard impedance problem (µ = 0) with a piecewise-constant impedance.In [20], the Lipschitz constant is proved to be independent of λ, with the help of a more complex technique based on a quantification of the unique continuation principle.One can also find in Inverse Problems and Imaging Volume X, No. X (200X), X-XX [20] a quantification of the exponential blowing up of the Lipschitz stability constant with respect to the space dimension of the parameters.For other results related to the impedance problem for the Laplace equation one can refer to [5,6,12,9, 1] and the references therein.
Proof.By reasoning exactly as in the proof of proposition 5, we obtain that T is differentiable and that dT λ,µ (h, l) coincide with the far field v ∞ h,l associated to the scattered field v s h,l , where v s h,l is the solution of problem (2) with data f = −l∆ Γ (u s + u i ) − h(u s + u i ) = −l∆ Γ u − hu.From (5), we have for all x ∈ S 2 v ∞ h,l (x) = v s h,l , ∂Φ ∞ (., x) ∂ν(y) = p s , l∆ Γ u + hu H 1 (Γ),H −1 (Γ) .
Finally we obtain v ∞ h,l (x) = p, l∆ Γ u + hu H 1 (Γ),H −1 (Γ) , which is the desired result.It follows from lemma 4.4 that l∆ Γ u+hu = 0 on Γ, and by using the same arguments as in the proof of proposition 4, we conclude that h = 0 and l = 0.
The proof of the above proposition requires the following lemma, which is the analogous of Lemma 4.1.Proof.Let u s be the solution of (2) which is associated to data f .We have = p s , f H 1 (Γ),H −1 (Γ) .
By using the same arguments as for the classical impedance problem, we obtain the following Lipschitz stability for the inverse problem when the parameters are restricted to the subset K(Γ) defined as follows.Let D(Γ) be a finite dimensional subspace of C I (Γ) × C, K(Γ) a compact subset of D(Γ) ∩ V (Γ).

Lemma 3 . 1 .
Let λ ∈ L ∞ (Γ) and d be a unitary vector of R 3 and assume that one of the following holds:1.The function λ is real and k 2 is not a Dirichlet eigenvalue of the operator −∆ in D. 2. The domain D is C 1 and both D and the function λ are assumed to be invariant by reflection against a plane which does not contain the direction d or by a rotation (different from identity) around an axis which is not directed by d.

Im
Φ(x, y)) ds(y), x ∈ D ∩ Ω and Im(SL)(λ)(x) = Γ Im(Φ(x, y))λ(y) ds(y), x ∈ D ∩ Ω,since Im(Φ(x, y)) = f (||x − y||) where f (r) = sin(kr)/4πr is a C ∞ function, it is readily shown by using differentiation under the integral sign that for general Lipschitz domain D, the function f 2 is in C 1 (R 3 ) .First, this implies that the traces on Γ of functions f 1 and f 2 coincide with the left-hand side and the right-hand side of equation (22), respectively.Secondly, f 1 and f 2 are functions in C 1 (R 3 ) which both satisfy the Helmholtz equation in D and Ω. Hence they satisfy the Helmholtz equation in R 3 .Since their traces on Γ coincide, and since k 2 is not a Dirichlet eigenvalue of the negative laplacian in D, f 1 and f 2 coincide in D, and then in R 3 by unique continuation.But f 2 tends to 0 when ||x|| → +∞, while f 1 does not, which completes the proof for the first case.