THE LINEAR SAMPLING METHOD FOR KIRCHHOFF-LOVE INFINITE PLATES

. This paper addresses the problem of identifying impenetrable obstacles in a Kirchhoﬀ-Love inﬁnite plate from multistatic near-ﬁeld data. The Linear Sampling Method is introduced in this context. We ﬁrstly prove a uniqueness result for such an inverse problem. We secondly provide the classi- cal theoretical foundation of the Linear Sampling Method. We lastly show the feasibility of the method with the help of numerical experiments.


1.
Introduction. In this contribution we consider the inverse problem of finding an impenetrable obstacle in an infinite elastic plate from multistatic scattering data in the frequency domain. Assuming that the thickness of the plate is small with respect to the wavelength, we consider that the behavior of the elastic plate is governed by the classical Kirchhoff-Love model in the purely bending case. The impenetrable obstacle D ⊂ R 2 is supposed to be a bounded open domain of class C 3 which is either characterized by a Dirichlet or a Neumann boundary condition. More precisely, by using the notations of [2], in particular Ω = R 2 \ D, the scattered field v s satisfies in the unbounded domain Ω the problem (1) Here k > 0 is the wave number, u i is an incident field which satisfies ∆ 2 u i −k 4 u i = 0 in a domain including D, B r is the open ball centered at 0 and of radius r, n is the outward normal to B r and s is the measure on ∂B r . The reader will refer to [2] for a short justification of how the system (1) is derived. Roughly speaking, the first line describes the motion of the plate in the frequency domain, the second one characterizes the boundary conditions on the boundary of the obstacle while the third one is the radiation condition, which specifies that only outgoing scattering waves are admissible. It is shown in [2] that the classical Sommerfeld condition for the Helmholtz equation is also valid for the bilaplacian case. In order to specify the surface differential operators B 1 and B 2 , we need to introduce some notations. A generic point x ∈ R 2 has Cartesian coordinates (x 1 , x 2 ). The outward unit normal to Ω is denoted n (note that n is oriented inside D). The unit tangent vector is denoted t and is such that the angle formed by the vectors (n, t) is π/2. The curvilinear abscissa associated with vector t is denoted s and coincides with the measure on ∂D. With the classical definitions either (B 1 , B 2 ) = (I, ∂ n ) (I is the identity), which corresponds to the Dirichlet boundary condition, or (B 1 , B 2 ) = (M, N ), which corresponds to the Neumann boundary condition, where the operators M and N are defined as follows: Here, ν ∈ [0, 1/2) is the Poisson's ratio and M 0 and N 0 are given by We mention that the Dirichlet boundary condition amounts to specify the out of plane displacement and the angle of rotation of the plate, while the Neumann boundary condition amounts to specify the bending moment and the shear force. In other words, the Dirichlet boundary condition corresponds to the clamped plate while the Neumann boundary condition corresponds to the free plate. Let us now consider the bounded domain Ω R = Ω ∩ B R , where R > 0 is such that B R contains the obstacle D. It is proved in [2] that the problem (1) is equivalent to the following problem set in the bounded domain Ω R : find u s ∈ H 2 (Ω R ) such that where T : is a Dirichletto-Neumann operator defined as follows. Assume that (f, We have . Here, H 1 m denotes the Hankel function of the first kind and of order m. Wellposedness of the forward problem (1) is the main purpose of [2]. The proof consists of a Fredholm analysis of the equivalent problem (2). Theorem 1.1. The problem (1) has a unique solution in H 2 loc (Ω) • for any k in the clamped case, • for k / ∈ K 0 in the free case, where the set K 0 is formed by some numbers k n > 0, n ∈ N, such that k n → +∞.
Note that we ignore if the restriction k / ∈ K 0 is purely technical or not. However, it is proved in [2] that for some particular obstacles D, for example a disc, we have The inverse problem we consider is the following. We assume that D is unknown but a priori contained in B R . Let us denote by G(·, y) the fundamental solution associated with the operator ∆ 2 − k 4 , that is the unique solution in H 2 loc (R 2 ) of the system It is well-known that G is given by For sake of self-containment, the well-posedness of problem (5) and the expression (6) are proved in Lemma 2.2 hereafter. For some point y ∈ ∂B R , we denote u s (·, y) the scattered field which is associated with the incident field u i = G(·, y) via (1). We also denoteũ s (·, y) the scattered field which is associated with the incident field u i = ∂ ny G(·, y) via (1), where n y is the outward normal to B R at point y ∈ ∂B R . The function G(·, y) can be seen as a point source located at y, while the function ∂ ny G(·, y) can be seen as a dipole located at y. For all points y ∈ ∂B R , we measure the scattered fields u s (·, y) andũ s (·, y) as well as their normal derivatives at all points x ∈ ∂B R . All these measurements constitute the so-called multistatic data. The goal of the inverse problem is to retrieve the obstacle D from those multistatic data. It arises in the framework of Non Destructive Testing, which is for example quite common in the aircraft industry. The bilaplacian model is interesting when the structure to inspect is thin and the frequency is low, which enables us to replace the 3D elastic model by such 2D approximate model. For example, Structural Health Monitoring would be a nice application. If we think of the SHM of the fuselage or the wings of an aircraft, even if some small hole in the skin would be visible to the naked eye when the aircraft is on the ground, it cannot be seen when the aircraft flies. Because structures may be complicated in real life, a possible extension of our present work would be to consider obstacles in a plate which has non homogeneous properties in a bounded region. Besides, in a view to consider defects such as corrosion or delamination, which could be modeled by a modified thickness of the plate or modified constants of elasticity, it would be useful to extend the present study to penetrable obstacles instead of impenetrable ones. All these extensions are non trivial, both concerning the uniqueness results and the effective reconstruction methods.
In order to address the inverse problem, we adapt the classical Linear Sampling Method introduced by Colton and Kirsch in [8] to the case of plates. Since this pioneering paper, the Linear Sampling Method has been applied in a large number of situations (see for example [9]), in particular in elasticity (see for example [1,10,12,3,4]). But the case of Kirchhoff-Love plates is new as far as we can judge. The Linear Sampling Method consists in testing, for all point of a sampling grid, if some test function depending on that point belongs to the range of an integral operator, the kernel of which only depends on the multistatic data. The LSM is both simple and efficient. In addition, the main feature of such method is that it works even if the nature of the obstacle is as priori unknown. To the best of our knowledge, the number of articles concerning inverse scattering problems in some infinite Kirchhoff-Love plate is very small. We highlight the very recent contribution [21] on that subject in the particular case of a bilaplacian operator with zero and first order perturbations. Concerning the static case in a bounded domain, we also mention recent contributions on uniqueness and stability issues for unknown boundaries in [17,18]. In order to build some synthetic data we need to solve the forward problem (1). To do so, we use a Finite Element Method based on the weak formulation associated with the problem (2) and a discretization of the Dirichlet-to-Neumann operator (3). The practical use of such a D-t-N operator to numerically compute a scattering solution in an infinite plate is new, as far as we know. Note that alternative approaches are the use of Perfectly Matched Layers, as in [13], or Boundary Integral Methods, as in [20].
The paper is organized as follows. The second section is devoted to the treatment of the inverse problem: we first derive an integral representation formula considered here as a preliminary tool, then prove the identifiability of the obstacle from the prescribed data, and lastly provide the justification of the Linear Sampling Method for the Dirichlet case and give some indications for the Neumann case. The third section presents some numerical results: we firstly describe the Finite Element Method we use to produce the artificial data with an example, then show the identification results obtained with the Linear Sampling Method, and lastly complete this numerical section by a short conclusion.
2.1. Integral representation. The integral representation formula is a basic tool in the proof of the identifiability of the obstacle and also in the justification of the Linear Sampling Method to retrieve it. Let us first recall some classical plateoriented Green formula (see [14]).
where we have introduced the bilinear form and n is the outward normal to O. Here the first part of the integral on ∂O has the meaning of duality pairing between H 1/2 (∂O) and H −1/2 (∂O) while the second part has the meaning of duality pairing between H 3/2 (∂O) and H −3/2 (∂O).
Given some functions (φ, ψ) ∈ H 3/2 (∂D) × H 1/2 (∂D), let us consider the following interior and exterior problems: find u − ∈ H 2 (D) and u + ∈ H 2 loc (Ω) such that where the unit normal vector n is oriented outside D in both problems (8) and (9). From Theorem 1.1, the exterior problem is well-posed. Concerning the interior problem, it is straightforward that well-posedness holds if and only if k / ∈ K D , where K D is the set formed by the fourth roots of the Dirichlet eigenvalues of operator ∆ 2 in domain D. Before proving our integral representation formula, let us derive the fundamental solution for operator ∆ 2 − k 4 . Lemma 2.2. Problem (5) has a unique solution which is given by (6).
Proof. Let u = G(·, y) be a solution to problem (5). We use the factorization By setting U = ∆u + k 2 u and V = ∆u − k 2 u, we hence obtain that In addition, by using a decomposition in the form of a series as in [2], it can be seen that if u satisfies the Sommerfeld radiation condition, the functions U and V both satisfy the radiation condition as well. This implies from the case of Helmholtz equation that Since u = (U − V )/2k 2 , we obtain the expression (6). It remains to prove that G(·, y) belongs to H 2 loc (R 2 ). Let us compare the regularity of G(·, y) with that of G 0 (·, y), that is the classical fundamental solution of the operator ∆ 2 (see [14]), the expression of which is It is clear that the function G(·, y) − G 0 (·, y) is infinitely smooth and it is easy to check by using polar coordinates that G 0 (·, y) is in H 2 loc (R 2 ). Then G(·, y) ∈ H 2 loc (R 2 ), which completes the proof.
Proof. Let us consider x ∈ Ω and r > 0 such that B(x, 2r) ∈ Ω. Next we define g x andg x such that where we have used the distribution brackets ·, · . It follows that By using Lemma 2.1, we obtain M y G(x, y) ∂ϕ ∂n y (y) + N y G(x, y)ϕ(y) ds(y) where n y is the unit normal oriented inside B(x, r). As a result, where we have used the notation If we now use Lemma 2.1 in the subdomain Ω r, where the normal n y involved in U(x, y) is oriented inside B(x, r) in the first integral, inside D in the second integral and outside B R in the third one. By using the Dirichlet-to-Neumann operator T : From (4) we observe that the operator T is symmetric, so that By using (12) and (13) we obtain that where n y is oriented inside D. Lastly we use again Lemma 2.1 in domain D, so that where n y is oriented inside D. From the two equations above, we obtain that for x ∈ Ω, where n y is oriented inside D. It could be similarly proved that the same formula is valid for x ∈ D. Given the boundary conditions satisfied by u + and u − on ∂D, we have that [u] = 0 and [∂ n u] = 0 on ∂D, so that the first integral vanishes, which completes the proof.

2.2.
Uniqueness. Before we detail the effective reconstruction, we prove uniqueness of the obstacle from the data. In the sequel, we simply denote Γ = ∂B R , which is the support of sources and measurements. More precisely, we have the following result. N )). Let us consider, for all y ∈ Γ and j = 1, 2, the scattered fields u s j (·, y) andũ s j (·, y) due to obstacle D j and associated via problem (1) with the incident fields u i = G(·, y) and u i = ∂ ny G(·, y), respectively. Assume that for all y ∈ Γ, the fields u s 1 (·, y) and u s 2 (·, y) coincide on Γ as well as their normal derivative and that the fieldsũ s 1 (·, y) andũ s 2 (·, y) coincide on Γ as well as their normal derivative. Then D 1 = D 2 . To prove such theorem, we need the following reciprocity relationships.

Lemma 2.4.
For all x, y ∈ Ω and z ∈ Γ, Proof. We detail the proof for the Dirichlet case, the Neumann case follows the same lines. From the proof of Proposition 1, for any x, z ∈ Ω, we have the integral representation where n y is oriented inside D. From the Green formula in D we obtain By introducing the total field u(·, z) = G(·, z) + u s (·, z), which satisfies the Dirichlet condition on ∂D, we obtain by adding the two above relationships that (15) u s (x, z) = ∂D ∂G(x, y) ∂n y M y u(y, z) + G(x, y)N y u(y, z) ds(y).
We now rewrite the first formula by inverting x and z, which yields From the Green formula in Ω and the radiation condition we obtain ∂u s (y, z) ∂n y + N y u s (y, x)u s (y, z) ds(y) By subtracting the two above relationships, we obtain by using the fact that G(y, z) = G(z, y), Subtracting (15) and (17) and using the fact that u(·, x) satisfies the Dirichlet condition on ∂D we obtain that u s (x, z) − u s (z, x) = 0 for all x, z ∈ Ω. Let us prove the second relationship. For any x, z ∈ Ω, we have the integral representatioñ where n y is oriented inside D. By computing the normal derivative of (14) with respect to z at point z ∈ Γ, we obtain that for all x ∈ Ω and all z ∈ Γ, 0 = − ∂D M y G(x, y) ∂ 2 G(y, z) ∂n y ∂n z + N y G(x, y) ∂G(y, z) ∂n z ds(y) + ∂D ∂G(x, y) ∂n y M y ∂G(y, z) ∂n z + G(x, y)N y ∂G(y, z) ∂n z ds(y).
By adding the above relationships, we get where we have used the fact that the total fieldũ(·, z) = ∂G(·, z)/∂n z +ũ s (·, z) satisfies the Dirichlet boundary condition on ∂D. Computing the normal derivative of (16) with respect to z at point z ∈ Γ, we obtain that for all x ∈ Ω and all z ∈ Γ, From the Green formula in Ω and the radiation condition we obtain 0 = − ∂D M yũ s (y, z) ∂u s (y, x) ∂n y + N yũ s (y, z)u s (y, x) ds(y) + ∂D ∂ũ s (y, z) ∂n y M y u s (y, x) +ũ s (y, z)N y u s (y, x) ds(y).
Adding the two above relationships implies that for all x ∈ Ω and all z ∈ Γ, Subtracting (18) and (19) and using the fact that u(·, x) satisfies the Dirichlet condition on ∂D we obtain thatũ s (x, z) − ∂u s (z, x)/∂n z = 0 for all x ∈ Ω and z ∈ Γ.
Proof of Theorem 2.3. Again, we detail the proof for the Dirichlet case, the Neumann case follows the same lines. Let us denoteΩ the unbounded connected component of R 2 \ D 1 ∪ D 2 . We have that for all x, y ∈ Γ, as well asũ From well-posedness of the forward diffraction problem when the obstacle is the ball B R with Dirichlet boundary condition, we have that for all x ∈ R 2 \ B R , for all y ∈ Γ, u s 1 (x, y) = u s 2 (x, y),ũ s 1 (x, y) =ũ s 2 (x, y). Unique continuation for the operator ∆ 2 − k 4 then implies that for all x ∈Ω, for all y ∈ Γ, u s 1 (x, y) = u s 2 (x, y),ũ s 1 (x, y) =ũ s 2 (x, y). We now use the reciprocity relationships of Lemma 2.4, which imply that for all x ∈Ω, for all y ∈ Γ, By repeating the same arguments as above we obtain that for all x, y ∈Ω, u s 1 (x, y) = u s 2 (x, y). Assume that D 1 ⊂ D 2 . Since R 2 \ D 2 is connected, there exists some non empty open set Γ * ⊂ (∂D 1 ∩ ∂Ω) \ D 2 . We now consider some point x * ∈ Γ * and the sequence where n 1 (x * ) denotes the unit normal to Γ * at point x * . For sufficiently large m, x m ∈Ω, so that for all x ∈Ω and all m, Using the boundary condition for u s 1 , we obtain Inverse Problems and Imaging Volume 14, No. 2 (2020), 363-384 Passing to the limit m → +∞, we get The function u s 2 is infinitely smooth in a vicinity of x * , while Γ * is a subset of the boundary of the C 3 domain Ω 1 := R 2 \ D 1 . This implies that which is also the regularity of (G(·, x * )| Γ * , ∂ n1 G(·, x * )| Γ * ). Hence the function G(·, x * ), which solves the equation ∆ 2 G(·, x * ) − k 4 G(·, x * ) = 0 inΩ, is H 3 in a vicinity of x * inΩ, from standard regularity results for elliptic problems [15]. But this is a contradiction in view of Remark 1. We conclude that D 1 ⊂ D 2 and we prove the same way that D 2 ⊂ D 1 . Eventually, D 1 = D 2 2.3. Justification of the Linear Sampling Method. We detail the classical theory of the Linear Sampling Method in the Dirichlet case, that is (B 1 , B 2 ) = (I, ∂ n ). The Neumann case follows the same lines and will be presented without justification in the next section. Let us start by introducing the mapping S D : We have the following property.
Proof. The proof is based on a comparison between the mapping S D and the analogue mapping S D,0 when the fundamental solution G of operator ∆ 2 −k 4 is replaced by the fundamental solution G 0 of operator ∆ 2 (see [14]) given by (10). The kernel G − G 0 is infinitely smooth, so that since the mapping S D,0 : is continuous (see for example [14]), this is also the case for S D . From [11, Lemma 3.1], the operator S D,0 is Fredlholm of index 0. Since S D −S D,0 is compact, this implies that the operator S D is Fredholm of index 0 as well. Let us prove that S D is surjective. We will then conclude that S D is an isomorphism. Assume that (φ, ψ) ∈ H 3/2 (∂D) × H 1/2 (∂D). We have to prove that there exits (τ, σ) ∈ H −3/2 (∂D) × H −1/2 (∂D) such that S D (τ, σ) = (φ, ψ). In this view, assuming that k / ∈ K D , let us consider the well-defined solutions u − in D and u + in Ω to the interior problem (8) and to the exterior problem (9) associated with the same Dirichlet data (φ, ψ). From proposition 1, the solution u = (u − , u + ) has expression (11). The trace and normal derivative of such solution, namely (u, ∂ n u), are continuous across ∂D and coincide with (φ, ψ), so that (φ, ψ) = S D (τ, σ), where we have set (τ, σ) = ([N u], [M u]) ∈ H −3/2 (∂D) × H −1/2 (∂D). The proof is complete.
We introduce the following operators: the obstacle-to-data operator FD : the data-to-obstacle operator H D : and the near-field operator N D : where u s (·, y) andũ s (·, y) are the solutions to problem (1)   Proof. Let us prove that G(x, y)τ (y) + ∂ ny G(x, y)σ(y) ds(y) t(x) ds(s) that is F D = H * D . In the computation above, we have used that G(x, y) = G(y, x) in view of (6) with (φ, ψ) ∈ H 3/2 (∂D) × H 1/2 (∂D). The corresponding solution to problem (24) is denoted v. Since the trace (v| Γ , ∂ n | Γ ) coincides with (G(·, z)| Γ , ∂ n G(·, z)| Γ ), we have that G(·, z) and v coincide outside the ball B R , and then in Ω\{z} by a unique continuation argument. The contradiction comes from a regularity comparison at point z: the function v is locally infinitely smooth while from Remark 1, G(·, z) is not locally H 3 .
From propositions 2, 3 and 4, we have the following Theorem. Since the proof mimics the one of [5] and is classical, it is left to the reader.
Theorem 2.5. We assume that k / ∈ K D .
• If z / ∈ D, then every solution (h ε (·, z), t ε (·, z)) of the inequality (25) satisfies lim ε→0 (h ε (·, z), t ε (·, z)) L 2 (Γ)×L 2 (Γ) = +∞ 2.4. The case of Neumann obstacle. In the Neumann case, that is (B 1 , B 2 ) = (M, N ), the important point is that the near-field operator N N is unchanged with respect to N D , provided in this case the scattered fields u s (·, y) andũ s (·, y) for y ∈ Γ are obtained for a Neumann obstacle instead of a Dirichlet obstacle. However the operators S N , F N , H N and B N which are involved in the theoretical justification of the Linear Sampling Method are modified as follows: (N y G(x, y)φ(y) + M y G(x, y)ψ(y)) ds(y) M ∂D (N y G(x, y)φ(y) + M y G(x, y)ψ(y)) ds(y) the obstacle-to-data operator is F N : the data-to-obstacle operator is H N : Lastly, the solution operator B N : It can be checked that the Propositions 2, 3, 4 and Theorem 2.5 are also valid for operators S N , B N , F N , H N and N N provided k / ∈ K N , where K N is the union of K 0 (see Theorem 1.1) and of the fourth roots of Neumann eigenvalues of operator ∆ 2 in domain D.

Numerical experiments.
3.1. Producing artificial data. In all the numerics, the Poisson's ratio is ν = 0.3. In order to obtain artificial data for the inverse problem, we need to solve the problem (2) numerically in Ω R . We hence derive a weak formulation which is equivalent to the strong problem (2). In the Neumann case, for example, such weak formulation is the following: find the scattered field u s ∈ H 2 (Ω R ) such that a is given by (7) with O = Ω R , t is defined by  [16,6]. This non-conforming finite element is probably the most simple element which enables one to solve bilaplacian problems. If Ω R is approximated by a polygonal domain and meshed with triangles, the finite element space is formed by functions u h such that their restriction on a given triangle is a second degree polynomial, so that each triangle has 6 degrees of freedom: the values of u h at the three vertices of the triangle and the values of the normal derivatives of u h at the middle of the three sides of the triangle. We use a very refined mesh, which is in particular consistent with the largest value of k (k = 30) that is used later on. The integer M has also to be sufficiently large with respect to k, we take M = kR +20.
The computation of all scattered fields that we need to produce the artificial data is implemented using the Matlab software and takes only a few minutes on a regular laptop. In the figure 1, we check the validity of the discretized Dirichlet-to-Neumann operator by computing the scattering solution either in Ω 1 or in Ω 2 , that is Ω R for R = 1 and R = 2. We verify that the two solutions coincide in the numerical sense in the intersection Ω 1 ∩ Ω 2 = Ω 1 . 3.2. Identification results. The procedure described in section 3.1 enables us to compute, for all y ∈ Γ, the solutions u s (·, y) andũ s (·, y) to the problem (1) with u i = G(·, y) and ∂ ny G(·, y), respectively. For each sampling point z ∈ Ω R we wish to solve in L 2 (Γ) × L 2 (Γ) the near-field equation for Dirichlet data. That the operator N D is compact implies that the previous equation is always ill-posed, this is why we solve it in the Tikhonov sense. In the realistic case when the data are perturbed by noise, the regularization parameter in the Tikhonov regularization is chosen as a function of the amplitude of noise by using the Morozov discrepancy principle exactly as in [7]. Of course the operator N D has to be replaced by N N for Neumann data. As it is done classically by LSM users, for each z ∈ Ω R we plot where (h ε , t ε ) is the Tikhonov regularized solution. Following Theorem 2.5, the function Ψ happens to be finite inside the unknown defect D and −∞ outside D, which means that imaging the defect D amounts to plotting the level sets of the function Ψ. In practice, the set of data is finite. In other words, we have to handle discretized versions of operators N D and N N corresponding to multistatic data u s (x i , y j ) andũ s (x i , y j ) where the points x i and y j for i, j = 1, . . . , I are equally distributed on the circle Γ (the locations of points x i and points y j are the same).
Here we have I = 500 in all the identification experiments. In the figure 2, we have represented the function Ψ for a Dirichlet obstacle D formed by the union of two discs and by using exact data, for various wave numbers k, that is k = 10, k = 20 and k = 30. In the figure 3, we show the identification results for an obstacle with  with k = 20. Note that such kite-shaped obstacle is not an obstacle of class C 3 as assumed in the introduction. However, the numerical resolution of the forward problem seems to work in that case, as well as the numerical resolution of the inverse one. Data are exact in all the previous cases. Now, let us analyze the impact of noise on the data u s (x i , y j ) andũ s (x i , y j ) for i, j = 1, . . . , I. This noise is such that for each j, the scattered fields u s (·, y j ) andũ s (·, y j ) as well as their normal derivatives are perturbed at each point x i by a Gaussian noise which is then rescaled in such a way that the artificial relative errors for the L 2 norm of all these fields have a prescribed value σ. We show in figure 5 the obtained results for the Dirichlet obstacle formed by two circles for k = 20 when the amplitude of noise is 5% and 10% (that is σ = 0.05 and σ = 0.1). These results have to be compared with the top right part of figure 2 obtained with unperturbed data. The identification is still efficient. However, it has to be noted that the contrast of the function Ψ between points inside and outside the obstacle D is much smaller in the noisy case, which stems from the Tikhonov/Morozov regularization procedure to solve the near-field equation.
We complete this numerical study by some identification attempts with much less data. Indeed, from a practical point of view, it could be considered as a complicated task to produce the fieldũ s (·, y), which for each y is the diffractive response to a  dipole, namely ∂ ny G(·, y). In addition, not only our sampling method needs the trace of fields u s (·, y) andũ s (·, y) on Γ but also their normal derivatives. Hence it is natural to wonder if one can use a sampling method using only u s (x, y) for (x, y) ∈ Γ × Γ, exactly as if the scattered field u s solved the Helmholtz equation ∆u s + k 2 u s = 0 instead of the true equation ∆ 2 u s − k 4 u s = 0. In [2], we observed that in Ω, any solution u to the equation ∆ 2 u − k 4 u = 0 is given by u = u pr + u ev , where u pr satisfies ∆u pr + k 2 u pr = 0 and u ev satisfies ∆u ev − k 2 u ev = 0. If we assume in addition that the function u is radiating, is was proved that outside some ball B R , the function u pr is an infinite linear combination of the functions H 1 n (kr)e inθ , which are oscillating and slowly decaying at infinity, while u ev is an infinite linear combination of the functions H 1 n (ikr)e inθ , which are exponentially decaying at infinity. This is why we call u pr the propagating part of the radiating solution u and u ev the evanescent part. As a conclusion, at a long distance of the obstacle, the scattered field u can be approximated by its propagating part u pr , which solves the Helmholtz equation (the evanescent part u ev of u is neglected).
Similarly, the fundamental solution G(·, y) given by (6) can be approximated at long distance of y by its propagating part i H 1 0 (k| · −y|)/8k 2 , which up to a constant coincides with the fundamental solution G (·, y) to the Helmholtz equation. In the Dirichlet case, it is then tempting, rather than solving the near-field equation (31), to solve the classical near-field equation corresponding to the Helmholtz equation. It consists, for each sampling point z ∈ Ω R , to solve in L 2 (Γ) the near-field equation where the operator N : L 2 (Γ) → L 2 (Γ) is defined, for h ∈ L 2 (Γ), by N h(x) = Γ u s (x, y)h(y) dy, x ∈ Γ.
We here use the fact that the Dirichlet boundary conditions for problem (1) implies the Dirichlet boundary condition for the Helmholtz equation. Comparing the operator N and the operator N D , we see that solving the equation (33) rather than the equation (31) leads us to use one block of data out of four. On the picture 6, we present the identification results for the Dirichlet obstacles formed by two or three circles, in the case of unperturbed data and k = 30, when we solve the equation (33) instead of (31). These results have to be compared with the bottom part of picture 2 and the left part of picture 4. We now consider the case when we have less data and when those data are noisy. The corresponding results are presented on picture 7 for the Dirichlet obstacle formed by two circles, for k = 20, when the amplitude of noise is 5% and 10%. These results have to be directly compared with the ones presented on picture 5.

Conclusion.
The numerical experiments of the previous section seem to show that the Linear Sampling Method is effective for Kirchhoff-Love plates, at least with data given on a circle Γ surrounding the unknown obstacle. It works both for a Dirichlet obstacle and for a Neumann obstacle, even in the presence of noisy data. As usual, the identification results improve when the wave number increases. Maybe we notice a slight degradation of the quality of the identification if we compare the results with those classically obtained for the Helmholtz equation in two dimensions.
If we compare to such case of Helmholtz equation in two dimensions, the LSM requires more data in the sense that we need the trace and the normal derivative on Γ of the scattered fields associated with both the point source G(·, y) and the dipole ∂ ny G(·, y) for all y ∈ Γ. However, we have shown experimentally in the Dirichlet case that using only the trace on Γ of the scattered field associated with the point source G(·, y) for all y ∈ Γ and hence proceeding exactly as if we solved the Dirichlet case for the Helmholtz equation, produces almost as good results as when using the complete data. This simplification would deserve a more quantitative justification.