Team organization may help swarms of flies to become invisible in closed waveguides

We are interested in a time harmonic acoustic problem in a waveguide containing flies. The flies are modelled by small sound soft obstacles. We explain how they should arrange to become invisible to an observer sending waves from $-\infty$ and measuring the resulting scattered field at the same position. We assume that the flies can control their position and/or their size. Both monomodal and multimodal regimes are considered. On the other hand, we show that any sound soft obstacle (non necessarily small) embedded in the waveguide always produces some non exponentially decaying scattered field at $+\infty$ for wavenumbers smaller than a constant that we explicit. As a consequence, for such wavenumbers, the flies cannot be made completely invisible to an observer equipped with a measurement device located at $+\infty$.


Introduction
Recently, questions of invisibility in scattering theory have drawn much attention. In particular, hiding objects is an activity in vogue. In this direction, the development of metamaterials with exotic physical parameters has played a fundamental role allowing the realization of cloaking devices. One of the most popular techniques which has been proposed consists of surrounding the object to hide by a well-chosen material so that waves go through as if there was no scatterer. In this approach, simple concepts of transformation optics allow one to determine the ad hoc material constituting the cloaking device (see e.g. [20,26,13]). It is important to emphasize that complex materials whose physical parameters exhibit singular values are required to build the device. From a practical point of view, constructing such materials is a challenging problem that people have not yet been able to solve.
For some applications, one may want to build invisible objects. But for others, it is better if they do not exist. In particular, for imaging methods, it is preferable that two different settings provide two different sets of measurements so that one can hope to recover features of the probed medium. In this field, invisible objects are interesting to study to understand the limits of a given technique. Indeed, it is important to have an idea of which objects can be reconstructed and which one cannot to assess how robust the existing algorithms are. Let us mention also that some techniques, like the Linear Sampling Method in inverse scattering theory, work only when invisible scatterers do not exist [18,19,8,2,9].
In the present article, we consider a scattering problem in a closed waveguide, that is in a waveguide which has a bounded transverse section. In such a geometry, at a given frequency, only a finite number of waves can propagate. We are interested in a situation where an observer wants to detect the presence of defects in some reference waveguide from far-field backscattering data. Practically, the observer sends waves, say from −∞, and measures the amplitude of the resulting scattered field at the same position. It is known that at −∞, the scattered field decomposes as the sum of a finite number of propagative waves plus some exponentially decaying remainder. We shall say that the defects are invisible if for all incident propagative waves, the resulting scattered field is exponentially decaying at −∞. In this setting, examples of invisible obstacles, obtained via numerical simulations, can be found in literature. We refer the reader to [22] for a water waves problem and to [1,21,47,48,25] for strategies using new "zero-index" and "epsilon near zero" metamaterials in electromagnetism (see also [23] for an application to acoustics). The technique that we propose in this article differs from the ones presented in the above mentioned works because it is exact in the sense that it is a rigorous proof of existence of invisible obstacles.
We will work with small sound soft obstacles of size ε (as in the so-called MUSIC algorithm [51,14,31]) that we call flies in the rest of the paper. To simplify the presentation, we assume here 1 that the frequency is such that the observer can send only one wave and measures one reflection coefficient s ε− (the amplitude of the scattered field at −∞). With this notation, our goal is to impose s ε− = 0. The approach we propose to help flies to become invisible is based on the following basic observation: when there is no obstacle in the waveguide, the scattered field is null so that s 0− = 0. When flies of size ε are located in the waveguide, we can prove that the reflection coefficient s ε− is of order ε. Our strategy, which is inspired from [40,6], consists in computing an asymptotic (Taylor) expansion of s ε− as ε tends to zero. In this expansion, the first terms have a relatively simple and explicit dependence with respect to the features of the flies (position and shape). This is interesting because it allows us to use theses parameters as control terms to cancel the whole expansion of s ε− (and not only the first term obtained with the Born approximation). More precisely, slightly perturbing the position or the size of one or several flies, it is possible to introduce some new degrees of freedom that we can tune to impose s ε− = 0. We underline that in principle, the sound soft obstacles have to be small compared to the wavelength. This explains the introduction of diptera terminology.
The technique described above mimics the proof of the implicit function theorem. It has been introduced in [40,41,44,45,11,42] with the concept of "enforced stability for embedded eigenvalues". In these works, the authors develop a method for constructing small regular and singular perturbations of the walls of a waveguide that preserve the multiplicity of the point spectrum on a given interval of the continuous spectrum. The approach has been adapted in [6,5] (see also [7,4,16] for applications to other problems) to prove the existence of regular perturbations of a waveguide, for which several waves at given frequencies pass through without any distortion or with only a phase shift. In the present article, the main novelty lies in the fact that we play with small obstacles and not with regular perturbations of the wall of the waveguide to achieve invisibility. This changes the asymptotic expansion of the reflection coefficient s ε− and we can not cancel it exactly as in [6]. In particular, as we will observe later (see Remark 5.1), the flies have to act as a team to become invisible: a single fly cannot be invisible. To some extent, our work shares similarities with the articles [39,10,43] where the authors use singular perturbations of the geometry to open gaps in the spectrum of operators considered in periodic waveguides.
In this article, we consider a scattering problem in a closed waveguide with a finite number of propagative waves. Note that the analysis we will develop can be easily adapted to construct sound soft obstacles in freespace which are invisible to an observer sending incident plane waves and measuring the far field pattern of the resulting scattered field in a finite number of directions (setting close to the one of [4]).
The text is organized as follows. In the next section, we describe the setting and introduce adapted notation. In Section 3 we compute an asymptotic expansion of the field u ε , the solution to the scattering problem in the waveguide containing small flies of size ε, as ε tends to zero. There is a huge amount of literature concerning scattering by small obstacles (see, among other references, [24,36,29,50,35,12,3]) and what we will do in this section is rather classical. Then in Section 4, from the expression of u ε , we derive an asymptotic expansion of the reflection coefficient s ε− appearing in the decomposition of u ε at −∞. In Section 5, slightly modifying the position of one fly and solving a fixed point problem we explain how to cancel all the terms in the asymptotic expansion of s ε− to impose s ε− = 0. Proposition 5.1 is the first main result of the paper. In Section 6, we study the question of invisibility assuming that the observer can send waves from −∞ and measure the resulting scattered field at +∞. More precisely, we show that it is impossible that the scattered field produced by the defect in the waveguide decays exponentially at +∞ for wavenumbers k smaller than a constant that can be explicitly computed. This result holds for all sound soft obstacles (not necessarily small) embedded in the waveguide. Proposition 6.1 is the second main result of the paper. In Section 7, we come back to backscattering invisibility for flies and instead of playing with their position, we modify slightly their size. With this degree of freedom, we explain how to cancel the reflection coefficient. In a first step, we consider the case where there is only one propagative wave. Then we work at higher frequency with several propagative waves. In this setting, the higher the frequency is, the more information the observer can get. Quite logically, in our approach, we shall need more and more flies to cancel the different reflection coefficients as the number of propagative waves increases. We provide a short conclusion in Section 8. Finally, Appendix 9 is dedicated to proving technical results needed in the justification of asymptotic expansions obtained formally in Section 5. Let Ω 0 := {x = (y, z) | y ∈ ω and z ∈ R} be a cylinder of R 3 . Here, ω ⊂ R 2 is a connected open set with smooth boundary. In the following, Ω 0 is called the reference or unperturbed waveguide (see Figure 1 on left). Let O ⊂ R 3 be an open set with Lipschitz boundary. Consider M 1 = (y 1 , z 1 ), M 2 = (y 2 , z 2 ) two points located in Ω 0 and define the sets, for n = 1, 2, ε > 0

Setting of the problem
Let ε 0 > 0 denote a positive parameter such that O ε n ⊂ Ω 0 for all ε ∈ (0; ε 0 ], n = 1, 2. We call perturbed waveguide (see Figure 1 on right) the set The sets O ε 1 and O ε 2 model the flies located in the waveguide. To begin with, and to simplify the exposition we assume that there are only two of them and not a "swarm". For the latter configuration, we refer the reader to §7.2. We are interested in the propagation of acoustic waves in time harmonic regime in Ω ε . Imposing soft wall boundary condition, it reduces to the study of the Dirichlet problem for the Helmholtz equation Here, u represents for example the pressure in the medium filling the waveguide, k corresponds to the wavenumber proportional to the frequency of harmonic oscillations, ∆ is the Laplace operator. Using separation of variables in the unperturbed waveguide Ω 0 , it is easy to compute the solutions of the problem To provide their expression, let us introduce λ j and ϕ j the eigenvalues and the corresponding eigenfunctions of the Dirichlet problem for the Laplace operator on the cross-section ω Here, δ j,j stands for the Kronecker symbol. In this paper, for any measurable set O ⊂ R r , r ≥ 1, we make no distinction between the complex inner products of the Lebesgue spaces L 2 (O) and L 2 (O) r , just using the notation (·, ·) O . The fact that the first eigenvalue λ 1 is simple is a consequence of the Krein-Rutman theorem. Moreover, we know that ϕ 1 has a constant sign on ω (see e.g. [27,Thm. 1.2.5]). Assume that k ∈ R is such that k 2 = λ j for all j ∈ N * . Then, the solutions of (3), the modes of the waveguide defined up to a multiplicative constant, are given by All through the paper, the complex square root is chosen so that if c = re iγ for r ≥ 0 and γ ∈ [0; 2π), According to the value of k 2 with respect to the λ j , the modes w ± j adopt different behaviours. For j ∈ N exp := {j ∈ N * | λ j > k 2 }, the function w + j (resp. w − j ) decays exponentially at +∞ (resp. −∞) and grows exponentially at −∞ (resp. +∞). For j ∈ N pro := {j ∈ N * | λ j < k 2 }, the functions w ± j are propagating waves in Ω 0 . In the present paper, except in §7.2, we shall assume that the wavenumber k verifies In this case, there are only two propagating waves w ± 1 and to simplify, we denote w ± := w ± 1 . In the perturbed waveguide Ω ε , the wave w + travels from −∞, in the positive direction of the (Oz) axis and is scattered by the obstacles. In order to model this phenomenon, classically, it is necessary to supplement equations (2) with proper radiation conditions at ±∞. Let us denote H 1 loc (Ω ε ) the set of measurable functions whose H 1 -norm is finite on each bounded subset of Ω ε . We will say that a function v ∈ H 1 loc (Ω ε ) which satisfies equations (2) is outgoing if it admits the decomposition for some constants s ± ∈ C and someṽ ∈ H 1 (Ω ε ). In (7), χ + ∈ C ∞ (Ω 0 ) (resp. χ − ∈ C ∞ (Ω 0 )) is a cut-off function equal to one for z ≥ (resp. z ≤ − ) and equal to zero for z ≤ /2 (resp. z ≥ − /2). The constant > 0 is chosen large enough so that Ω ε coincides with Ω 0 for x = (y, z) such that |z| ≥ /2. Using Fourier decomposition, we can show that the remainderṽ appearing in (7) is exponentially decaying at ±∞. Now, the scattering problem we consider states  (8) admits a unique solution u ε for ε small enough. In particular, there are no trapped modes for ε small enough. In the following, u ε − w + (resp. u ε ) will be referred to as the scattered (resp. total) field associated with the incident field w + . We emphasize that in (8), w + is the source term. The coefficients s ± appearing in (7) with v replaced by u ε − w + will be denoted s ε± , so that there holds whereũ ε ∈ H 1 (Ω ε ) is a term which is exponentially decaying at ±∞. With this notation, the usual reflection and transmission coefficients are respectively given by Our goal is to explain how the flies O ε n , n = 1, 2, should arrange so that there holds R ε = 0. In our analysis, we shall assume that the flies can play with their position or with their size. Note that when s ε− = 0 (or equivalently when R ε = 0), the scattered field u ε − w + defined from (8) is exponentially decaying at −∞. As a consequence, an observer located at z = −L, with L large, sending the wave w + and measuring the resulting scattered field is unable to detect the presence of the flies.

Asymptotic expansion of the total field
In this section, we compute an asymptotic expansion of the solution u ε to Problem (8) as ε tends to zero. The method to derive such an expansion is classical (see for example [36, §2.2]) but we detail it for the sake of clarity. In accordance with the general theory of asymptotic analysis, we make the ansatz where the dots stand for terms of high order unnecessary in the study. In this ansatz, the functions v k, n correspond to boundary layer terms. They depend on the rapid variables ε −1 (x − M n ) and compensate the residual of the principal asymptotic terms u k in a neighbourhood of M n , n = 1, 2. Moreover, for n = 1, 2, ζ n ∈ C ∞ 0 (R 3 , [0; 1]) denotes a cut-off function which is equal to one in a neighbourhood of M n and whose support is a compact set sufficiently small so that ζ n = 0 on Γ 0 , ζ n (M m ) = 0 in a neighbourhood of M m = M n . Now, we explain how to define each term in (11). For justification of the asymptotic expansion and error estimates, we refer the reader to Section 9.
At order ε 0 , the incident wave w + does not see the small obstacles and there is no scattered field. Therefore, we take u 0 = w + . For n = 1, 2, the function v 0, n allows us to impose the Dirichlet boundary condition on ∂O ε n at order ε 0 . For x ∈ ∂O ε n , computing a Taylor expansion, we find u 0 (x) = u 0 (M n ) + (x − M n ) · ∇u 0 (M n ) + . . . . Note that x − M n is of order ε. To simplify notations, for n = 1, 2, we introduce the fast variable ξ n = ε −1 (x − M n ). For the correction terms v k, n , in a neighbourhood of M n (remember that the cut-off function ζ n is equal to one in this region), we obtain Since there is no term of order ε −2 in the expansion (11), we impose ∆ ξn v 0, n = 0 in R 3 \ O. This analysis leads us to take v 0, n (ξ n ) = −u 0 (M n ) W (ξ n ).
Here, W is the capacity potential for O (i.e. W is harmonic in R 3 \ O, vanishes at infinity and verifies W = 1 on ∂O). In the sequel, the asymptotic behaviour of W at infinity will play a major role. As |ξ| → +∞, we have (see e.g. [33]) where Φ := ξ → −1/(4π|ξ|) is the fundamental solution of the Laplace operator in R 3 and q is some given vector in R 3 . The term cap(O) corresponds to the harmonic capacity [49] of the obstacle O. Let us translate the position of the origin, making the change of variable ξ → ξ θ := ξ + θ, for a given θ ∈ R 3 .
When |ξ| → +∞, we can write [49], this shows that there is exactly one value of θ ∈ R 3 such that W , in the new system of coordinates, admits the asymptotic expansion In the following, we shall always assume that W is defined in the system of coordinates centered at O + θ and to simplify, we shall denote ξ instead of ξ θ .

Remark 3.1. This trick to obtain a simple expansion for W at infinity is not a necessary step in our procedure. However, it allows one to shorten the calculus.
Now, we turn to the terms of order ε in the expansion of u ε . After inserting ). Remark that this discrepancy is defined only in Ω ε and not in Ω 0 . However, using (14), we replace it by its main contribution at infinity and we choose u 1 as the solution to the problem In (15), [∆ x , ζ n ] denotes the commutator such that [∆ x , ζ n ]ϕ := ∆ x (ζ n ϕ) − ζ n ∆ x ϕ = 2∇ϕ · ∇ζ n + ϕ∆ζ n . Since ζ n is equal to one in a neighbourhood of M n and compactly supported, note that the right hand side of (15) is an element of L 2 (Ω 0 ). As a consequence, by elliptic regularity results (see e.g. [34]), u 1 belongs to H 2 (K) for all bounded domains K ⊂ Ω 0 . A Taylor expansion at M n gives, for x ∈ ∂O ε n , In order to satisfy the Dirichlet boundary condition on the obstacles at order ε, we find that v 1, n must verify In (16), the vector valued function We remind the reader that Φ = ξ → −1/(4π|ξ|) is the fundamental solution of the Laplace operator in R 3 . The matrix M is called the polarization tensor [49,Appendix G]. It is symmetric and with our special choice of the origin of the system of coordinates (see (13)), the vector p is equal to zero. Therefore, for j = 1, 2, 3, we have Finally, we consider the expansion of u ε at order ε 2 . After inserting u 0 (x) (8), using formulas (14), (12) and (17), we get the discrepancy This leads us to define u 2 as the solution to the problem Find u 2 ∈ H 1 loc (Ω 0 ) such that u 2 is outgoing and The derivation of the problems satisfied by the terms v 2, n , n = 1, 2, will not be needed in the rest of the analysis. Therefore we do not describe it. In the next section, from the asymptotic expansion of u ε , solution to (8), we deduce an asymptotic expansion of the coefficients s ε± appearing in the decomposition (9).

Asymptotic expansion of the transmission/reflection coefficients
In accordance with the asymptotic expansion of u ε (11), for s ε± , we consider the ansatz Now, we wish to compute s 0± , s 1± , s 2± . First, we give an explicit formula for s ε± . For ±z > (we remind the reader that > 0 is chosen so that Ω ε coincides with Ω 0 for x = (y, z) such that |z| ≥ /2), using Fourier series we can decompose u ε − w + as where α ± n are some constants and where w ± n are defined in (5).
. A direct calculation using the orthonormality of the family (ϕ n ) n≥1 and the expression of the w ± (see (5)) yields the formulas where Observe that the correction terms ζ n v k, n appearing in the expansion of u ε (11) are compactly supported. As a consequence, they do not influence the coefficients s ε± . In particular, this implies s 0± = 0.
To compute s 1± , we plug (11) in (20) and identify with (19) the powers in ε. This gives Integrating by parts in ω × (− ; ) and using the equation ∆w ± + k 2 w ± = 0 as well as (15), we find vanishes in a neighbourhood of M n , n = 1, 2 (see the discussion after (15)) and using the estimate C > 0 being a constant independent of δ, we deduce from the last line of (21) that In (22), the set Ω 0δ is defined by Ω 0δ : in Ω 0δ . Using again that ζ n is equal to one in a neighbourhood of M n and integrating by parts in (22), we get In this expression, ν stands for the normal unit vector to ∂B δ 3 (M n ) directed to the interior of B δ 3 (M n ). Then, a direct computation using the relations u 0 ( Working analogously from the formulas we obtain

Perturbation of the position of one fly
In this section, we explain how to choose the position of the flies so that the reflection coefficient s ε− in the decomposition of u ε − w + vanishes. In the previous analysis, we obtained the formula First observe, that it is easy to cancel the term of order ε in the expansion (26). Indeed, remembering that w + (x) = (2β 1 ) −1/2 e iβ 1 z ϕ 1 (y), we obtain In order to satisfy (27), for example we choose y 1 , y 2 ∈ ω such that ϕ 1 (y 1 ) = ϕ 1 (y 2 ). Then we take z 1 = 0 and z 2 = (2m + 1)π/(2β 1 ) for some m ∈ N = {0, 1, 2, . . . }.
However, this is not sufficient since we want to impose s ε− = 0 at all orders in ε. The terms of orders ε 2 , ε 3 . . . in (26) have a less explicit dependence with respect to M 1 , M 2 . In particular, this dependence is non linear. Therefore, it is not obvious that we can find an explicit formula for the positions of the flies ensuring s ε− = 0. To cope with this problem, we will introduce new degrees of freedom slightly changing the position of one fly. To set ideas, we assume that the fly situated at M 1 moves from M 1 to M τ 1 = M 1 + ετ . Here, τ is an element of R 3 to determine which offers a priori three degrees of freedom. In the following, for a given ε > 0, we show how τ can be chosen as the solution of a fixed point problem to cancel the reflection coefficient.
has a unique solution u ε (τ ) for ε small enough. It admits the decomposition where s ε± (τ ) ∈ C and whereũ ε (τ ) ∈ H 1 (Ω ε (τ )) is a term which is exponentially decaying at ±∞. From the previous section (see formulas (24), (25)), we know that Formally, we also have (we will justify this expansion in Section 9) To remove the ε dependence hidden in the term w + (M τ 1 ), we use the Taylor expansion Plugging (31) and using the relation In (32),s ε− (τ ) denotes an abstract remainder (see §9.1 for its definition). With this new parameterization, for a given ε, our goal is to find Now, choose y 1 such that ∇ y ϕ 1 (y 1 ) = 0. Remark that it is possible since for the moment, we have only imposed that y 1 , y 2 are such that ϕ 1 (y 1 ) = ϕ 1 (y 2 ). Let us look for τ under the form where κ y , κ z are some real parameters to determine. Here, we emphasize that it is sufficient to parameterize the unknown τ ∈ R 3 with only two real degrees of freedom (and not three) because we want to cancel one complex coefficient. To impose s ε− (τ ) = 0, plugging (34) in (32), we see that we have to solve the problem Find κ = (κ y , κ z ) ∈ R 2 such that κ = F ε (κ), with Proposition 5.2 hereafter ensures that for any given parameter γ > 0, there is some ε 0 > 0 such that for all ε ∈ (0; ε 0 ], the map F ε is a contraction of B γ 2 (O) := {κ ∈ R 2 |κ| ≤ γ}. Therefore, the Banach fixed-point theorem guarantees the existence of some ε 0 > 0 such that for all ε ∈ (0; ε 0 ], Problem (35) has a unique solution in B γ 2 (O). Note that for a given γ > 0, ε 0 > 0 can be chosen small enough so that there holds O ε 1 (τ ) ⊂ Ω 0 (the first fly stays in the reference waveguide). From the previous analysis we deduce the following proposition.

Remark 5.3.
Let us denote κ sol ∈ B γ 2 (O) the unique solution to Problem (35). The previous proof ensures that there exists a constant c 0 > 0 independent of ε such that Introduce τ sol , τ 0 the vectors of R 3 respectively defined from κ sol , κ = 0 using formula (34). In particular, we have We know (see (34)) that |M τ sol As a consequence, we can say that M τ sol 1 is equal to M τ 0 1 = M 1 + ετ 0 at order ε.

Obstruction to transmission invisibility
Let us consider again u ε the solution to Problem (8). We have introduced the coefficients s ε± such that the scattered field u ε − w + admits the expansion whereũ ε ∈ H 1 (Ω ε ) is a term exponentially decaying at ±∞. With this notation, the usual reflection and transmission coefficients are respectively given by R ε = s ε− and T ε = 1 + s ε+ . Up to now, we have explained how to cancel s ε− so that the flies are invisible to an observer sending the incident waves w + and measuring the resulting scattered field at −∞. Now, assume that the observer can also measure the scattered field at +∞. Can we hide the flies in this setting? Equivalently, can we impose s ε+ = 0 (or T ε = 1) so that the scattered field is also exponentially decaying at +∞? First, remark that the approach of Section 5 can not be implemented to impose s ε+ = 0. And more generally, it is easy to see that the latter relation cannot be obtained when there are small flies in the reference waveguide. Indeed, from (19), (24), we know that s ε+ admits the asymptotic expansion Since cap(O) > 0, we have m s ε+ > 0 for ε small enough. Now, we prove another result showing that it is impossible to hide any (not necessarily small) sound soft obstacle for wavenumvers k smaller than a constant that we explicit. Since the result that we will prove holds in a more general setting than the one described at the beginning of the paper, we need to modify a bit the notation. Let O be a bounded domain with Lipschitz boundary verifying O ⊂ Ω 0 . We define the perturbed waveguide Ω := Ω 0 \ O and make the assumption that O is such that Ω is connected with a Lipschitz boundary (see Figure 2 for examples of geometries fulfilling these criteria). We consider the scattering problem Find u ∈ H 1 loc (Ω) such that u − w + is outgoing and −∆u = k 2 u in Ω u = 0 on Γ := ∂Ω.
Problem (43) always admits a solution u with the decomposition where R, T ∈ C are uniquely defined and whereũ ∈ H 1 (Ω) is exponentially decaying at ±∞. In (44), as in (7), χ ± ∈ C ∞ (Ω 0 ) are cut-off functions equal to one for ±z ≥ and equal to zero for ±z ≤ /2. Moreover, the constant > 0 is chosen large enough so that Ω coincides with Ω 0 for x = (y, z) verifying |z| ≥ /2. We wish to prove that for k < k with k 2 ∈ (λ 1 ; λ 2 ], we cannot have T = 1 so that perfect transmission invisibility cannot be imposed. We denote u i := w + and u s := u − u i . We also introduce the coefficients s ± such that s − = R and s + = T − 1. With this definition, according to (44), we have u s = χ + s + w + + χ − s − w − +û for someû ∈ H 1 (Ω) which is exponentially decaying at ±∞.

Lemma 6.1. Assume that the transmission coefficient T in the decomposition (44) satisfies T = 1.
Then there holds Proof. The conservation of energy ensures that Assume that T satisfies T = 1 ⇔ s + = 0. In this case, according to (46), we must have s − = 0 and u s is exponentially decaying at ±∞. Therefore, the integrals appearing in (45) are well-defined. Denote Ω := {x = (y, z) ∈ Ω | − < z < }. From the equation ∆u s + k 2 u s = 0, multiplying by u s and integrating by parts, we find Here, we set Σ = (ω × {− }) ∪ (ω × { }) (as in (20)) and ν stands for the normal unit vector to ∂Ω directed to the exterior of Ω . On ∂Ω, we have u = 0 which implies u s = −u i . This allows us to write The third equality of (48) is a direct result of integrations by parts. Now, we consider the last term of the right hand side of (48). On ∂Ω ∩ ∂O, remark that ν is directed to the interior of O. Therefore, integrating by parts, we obtain Finally, gathering (47), (48), (49) and using formula (20), we obtain identity (45).

Remark 6.1. Even though this result is stated in dimension d = 3, it holds in any dimension d ≥ 2.
And the proof for d = 3 is the same as the one presented below.

Remark 6.2. The authors do not know if the constant min(λ
Proof. Assume, by contradiction, that T = 1 so that (45) holds. Let us write Now, we estimate each of the two terms of the right hand side of (50).
Let us consider the first one. For ±z > L, using Fourier series, classically, we can decompose u s as u s (y, z) = s ± w ± (y, z) + +∞ n=2 α ± n w ± n (y, z).
Applied to γ ∈ X, estimate (54) gives Since γ = u s in Ω L and γ = −u i in O, this implies Using (52), (57) in (50), we obtain Therefore, for k 2 < µ 1 , since the interior of O is non empty, we are led to a contradiction and we must have T = 1. (44) verifies R = 0. From the conservation of energy |R| 2 + |T | 2 = 1, this implies |T | = 1. Proposition 6.1 shows that for k 2 < min(λ 1 + π 2 /(2L) 2 , λ 2 ), where L > 0 denotes the smallest number such that the flies are located in the region ω × (−L; L), we cannot have T = 1. This means that for such k 2 , the wave suffers a phase shift after passing the defects.

Remark 6.4.
For the problem considered in this article (see (43)), u is called a trapped mode if it satisfies −∆u = k 2 u in Ω, u = 0 on ∂Ω and u ∈ H 1 (Ω) (which implies that u is exponentially decaying both at ±∞). Following an approach similar to the proof of Proposition 6.1, we can show that that there are no trapped modes for k 2 ∈ (λ 1 ; min(λ 1 + π 2 /(2L) 2 , λ 2 )). As a consequence, for a given k 2 ∈ (λ 1 ; λ 2 ), an obstacle cannot be completely invisible (T = 1) and cannot trapped waves if it is too short in the (Ox) direction (L small).

Perturbation of the size of the flies
In Section 5, we played with the position of the flies to cancel the reflection coefficient. In this section, we change a bit the point of view considering that the flies can modify slightly their size. In §7.1, we explain how to cancel the reflection coefficient when there is only one incident wave (as in the previous sections). Then, in §7.2, we show how to impose reflection invisibility at higher frequency when there are several incident waves. To proceed, we will need more that two flies.
Remark 7.1. With two flies only, one can check that it is impossible in (59) to both cancel the term of order ε and to use the one of order ε 2 to compensate the whole expansion. This is the reason why we need to add at least one passive fly. Note that here, we disposed two passive flies in the waveguide (the ones located at M 3 , M 4 ) only to obtain a simple fixed point equation in (64). The scheme can also be implemented with one passive fly. For example, take y 1 , y 2 , y 3 ∈ ω such that ϕ 1 (y 1 ) = ϕ 1 (y 2 ) = ϕ 1 (y 3 ) and z 1 , z 2 , z 3 ∈ R such that e 2iβ 1 z 1 = 1, e 2iβ 1 z 2 = e 2iπ/3 , e 2iβ 1 z 3 = e 4iπ/3 . With such a choice, first we cancel the term of order ε in (61). Then proceeding like in (62), (63), we can derive a fixed point equation which admits a solution and from this solution we can find values for the parameters τ 1 , τ 2 in (61) to achieve s ε− (τ ) = 0.

Several incident waves
Assume that the wavenumber k verifies for some J ∈ N * . In this case, there are 2J propagating waves w ± 1 , . . . , w ± J in the waveguide. Let Ω ε be a waveguide obtained from Ω 0 adding a finite number of flies. For j = 1, . . . , J, denote u ε j the solution to the following scattering problem for ε small enough. In this paragraph, the sentence "u ε j − w + j ∈ H 1 loc (Ω ε ) is outgoing" means that there holds the decomposition where s ε± jj ∈ C and whereũ ε j ∈ H 1 (Ω ε ) is a term exponentially decaying at ±∞. Our goal is to find a domain Ω ε such that there holds s ε− jj = 0 for j, j = 1, . . . , J. In such a situation, for any combination of the incident plane waves w + 1 , . . . , w + J , the resulting scattered field is exponentially decaying at −∞ so that the flies are invisible to an observer measuring the scattered field at z = −R for large R.
It is known that the matrix made of the s ε− jj is symmetric. Therefore, there are P := J(J + 1)/2 degrees of freedom and we have to cancel 2P real coefficients. To proceed, we will play with the size of 2P flies. For n = 1, . . . , 2P , assume that the fly located at M n = (y n , z n ) coincides with the set where ε > 0 is small and where τ n ∈ R is a parameter to tune. In order to achieve reflection invisibility, as in the previous paragraph, we also need to adjust the position (but not the size) of additional flies.
The choice of the parameter N will be clarified later. Then, for τ := (τ 1 , . . . , τ 2P ) ∈ R 2P , define the perturbed waveguide Finally, we denote s ε− jj (τ ) the reflection coefficient for the scattering problem (65) considered in the geometry Ω ε (τ ). Working as in Sections 3-4 and §7.1, we obtain the asymptotic expansion wheres ε− jj (τ ) is a remainder and where u j, 1 is the solution to the problem Find u j, 1 ∈ H 1 loc (Ω 0 ) such that u j, 1 is outgoing and Now, we explain how to choose the positions and the sizes of the flies to obtain s ε− jj (τ ) = 0. To proceed, as in the previous paragraph, we need to find positions for the flies such that the term of order ε in (67) vanishes. But in the same time, we also wish to use the term of order ε 2 to cancel the complete expansion. Let us translate this into equations. We remind the reader that there holds w + j (M n ) = (2β j ) −1/2 e iβ j zn ϕ j (y n ) with β j = (k 2 − λ j ) 1/2 (see (5)). First, take y 1 , . . . , y N ∈ ω such that y 1 = · · · = y N . We want to impose To simplify, we assume that the wavenumber k ∈ (λ J ; λ J+1 ) is such that the numbers β j + β j = (k 2 − λ j ) 1/2 + (k 2 − λ j ) 1/2 , 1 ≤ j ≤ j ≤ J, are all distinct. Remark that, using the principle of isolated zeros, we can prove that wavenumbers such that this assumption is not verified form a set which is discrete or empty. Introduce γ 1 < · · · < γ P such that {γ p } 1≤p≤P = {β j + β j } 1≤j≤j ≤J . In order to use the parameters τ 1 , . . . , τ 2P to cancel the whole expansion in (67), we need to find z 1 , . . . , z 2P ∈ R (distinct) such that the matrix is invertible. Assume for a moment that we have constructed z 1 , . . . , z N ∈ R such that (68) holds and such that the matrix B is invertible. Then, let us look for τ under the form τ = (τ 1 , . . . , τ 2P ) with, for n = 1, . . . , P , Here, κ 1 , . . . κ 2P are some real parameters to determine and j ≤ j are the indices such that γ n = β j +β j (note that there is a one-to-one correspondence between the index n and the pair (j, j ) for j ≤ j ). To impose s ε− jj (τ ) = 0 for 1 ≤ j ≤ j ≤ J, plugging (70) in (67), we obtain that κ := (κ 1 , . . . , κ 2P ) must be a solution to the problem with In (72), U ∈ R 2P denotes the vector such that, for n = 1, . . . , P , U 2n−1 = es ε− (τ ) jj and U 2n = ms ε− (τ ) jj . Again, here j ≤ j are the indices such that γ n = β j + β j . Working as in Section 9, we can prove that for any given parameter γ > 0, there is some ε 0 > 0 such that for all ε ∈ (0; ε 0 ], the map F ε is a contraction of B γ 2P (O) := {κ ∈ R 2P |κ| ≤ γ}. Therefore, the Banach fixed-point theorem guarantees the existence of some ε 0 > 0 such that for all ε ∈ (0; ε 0 ], Problem (71) has a unique solution Now we explain how to construct z 1 , . . . , z 2P ∈ R such that the matrix B is invertible. We use a recursive approach. First, take z 1 , z 2 such that is invertible. Then, let us show that we can find z 3 ∈ R such that is analytic in C. Since γ 2 > γ 1 and since det(B 2 ) = 0, using Cramer's rule, we can prove that det(B 3 ) = 0 for z 3 = iL with L > 0 large enough. According to the principle of isolated zeros, we deduce that there is some z 3 ∈ R (different from z 1 , z 2 ) such that B 3 is invertible. Then, define The map z 4 → det(B 4 ) is analytic in C. Take z 4 = iL with L > 0 and pick the last column of B 4 to compute det(B 4 ) with Cramer's rule. Observe that sin(γ 2 iL) is purely imaginary whereas cos(γ 2 iL) is purely real. Using also that γ 2 > γ 1 and that det(B 3 ) is a non zero real number, we can prove that m (det(B 4 )) = 0 for L large enough. According to the principle of isolated zeros, we deduce that there is some z 4 ∈ R (different from z 1 , . . . , z 3 ) such that B 4 is invertible. Continuing the process, we can find z 1 , . . . , z 2P such that the matrix B defined in (69) is invertible.
With this approach, we can find N = 2 P +1 P numbers z 1 , . . . , z N such that (73) is satisfied. The technique is attractive because it is systematic and simple to implement. However, it requires a very high number of flies. For example, with J = 5 (in this case P = J(J + 1)/2 = 15), we find N = 983040. It would be interesting to find alternative algorithms which are less flies consuming.

Remark 7.2.
Above, we assumed that the wavenumber k ∈ (λ J ; λ J+1 ) is such that the numbers β j +β j = ( It is an open problem to impose reflection invisibility when this assumption is not satisfied.

Conclusion
We explained how flies (small Dirichlet obstacles) should arrange to become invisible to an observer sending waves from −∞ and measuring the resulting scattered field at the same position (see in particular Proposition 5.1). In other words, we constructed waveguides where the reflection coefficient R satisfies R = 0. We investigated a 3D setting. For 2D problems, the asymptotic calculus is a bit different, with the presence of a logarithm, but the analysis should be essentially the same. A possible direction to continue this work is to work with sound hard (Neumann) obstacles. We also considered the question of imposing T = 1 (T is the transmission coefficient). We observed that with small Dirichlet obstacles, it is impossible to have T = 1. Moreover, we showed that for any sound soft obstacle (non necessarily small) embedded in the waveguide, we cannot have T = 1 for wavenumbers smaller than an explicit value (see Proposition 6.1).
It is an open question to know whether or not this bound is optimal.

Appendix: justification of asymptotics
In this appendix, we explain how to justify the asymptotic expansion derived formally in Section 4. More precisely, we wish to show estimate (38) which was the key ingredient to obtain Proposition 5.1. The statement of this estimate is as follows: for all ϑ > 0, there is some ε 0 > 0 such that for all ε ∈ (0; ε 0 ], there holds where C > 0 is a constant independent of ε. The proof will be divided into several steps and will be the concern of the next three paragraphs. We shall use the same notation as in Sections 4, 5.

Explicit expression of the coefficients ε− (τ )
First, we provide an explicit formula for the coefficients ε− (τ ). According to (32),s ε− (τ ) is defined as the remainder appearing in the decomposition From (20), we know that Let us compute an asymptotic expansion of u ε (τ ), the solution to Problem (28), as ε tends to zero. We will work as in Section 3 where we obtained an asymptotic expansion of u ε (0) = u ε . Consider the decomposition In the above expression, the functions v 0, n , u 1 , v 1, 2 , u 2 , are respectively defined in (12), (15), (16), (18).