Characterization of the acoustic fields scattered by a cluster of small holes

Abstract

We deal with the time-harmonic acoustic waves scattered by a large number of small holes, with radius a, $a ≪ 1$ , arbitrarily distributed in a bounded part of the homogeneous background $R^{3}$ . We assume no periodicity in distributing these holes. Using the asymptotic expansions of the scattered field by this cluster, we show that as their number M grows following the law $M \sim a^{- s}$ , $a \to 0$ , the collection of these holes has one of the following behaviors:

(1) if $s < 1$ , then the scattered fields tend to vanish as a tends to zero, i.e., the cluster is a soft one.

(2) if $s = 1$ , then the cluster behaves as an equivalent medium modeled by a refraction index, supported in a given bounded domain Ω, which is described by certain geometry properties of the holes and their local distribution. The cluster is a moderate (or intermediate) one.

(3) if $s > 1$ , with additional conditions, then the cluster behaves as a totally reflecting extended body, modeled by a bounded domain Ω, i.e., the incident waves are totally reflected by the surface of this extended body. The cluster is a rigid one.

Explicit errors estimates between the scattered fields due to the cluster of small holes and the ones due to equivalent media (i.e., the refraction index) or the extended body are provided.

Keywords

Scattering by small particles Foldy–Lax approximation effective medium theory

1. Introduction and statement of the results

1.1. The acoustic scattered fields generated by a cluster of small holes

We set $D_{m} : = ϵ B_{m} + z_{m}$ to be the small bodies characterized by the parameter $ϵ > 0$ and the locations $z_{m} \in R^{3}$ , $m = 1, \dots, M$ , where $B_{1}, B_{2}, \dots, B_{M}$ are M open, bounded and simply connected sets in $R^{3}$ with Lipschitz boundaries containing the origin. We assume that the Lipschitz constants of $B_{j}$ , $j = 1, \dots, M$ are uniformly bounded. We denote by $U^{s}$ the acoustic field scattered by the M small and rigid bodies $D_{m} \subset R^{3}$ due to the incident plane wave $U^{i} (x, θ) : = e^{i k x \cdot θ}$ , with the incident direction $θ \in S^{2}$ , with $S^{2}$ being the unit sphere. Hence the total field $U^{t} : = U^{i} + U^{s}$ satisfies the following exterior Dirichlet problem of the acoustic waves $\begin{array}{l} (1.1) & (Δ + κ^{2}) U^{t} = 0 in R^{3} ∖ (⋃_{m = 1}^{M} {\bar{D}}_{m}), \\ (1.2) & U^{t} |_{\partial D_{m}} = 0, 1 ⩽ m ⩽ M, \\ (1.3) & \frac{\partial U^{s}}{\partial | x |} - i κ U^{s} = o (\frac{1}{| x |}), | x | \to \infty, (S.R.C) \end{array}$ where $κ > 0$ is the wave number, $κ = 2 π / λ$ , λ is the wave length and S.R.C stands for the Sommerfeld radiation condition. The scattering problem (1.1)–(1.3) is well-posed in appropriate spaces, see [10,15] for instance, and the scattered field $U^{s} (x, θ)$ has the following asymptotic expansion: $\begin{matrix} (1.4) & U^{s} (x, θ) = \frac{e^{i κ | x |}}{| x |} U^{\infty} (\hat{x}, θ) + O (| x |^{- 2}), | x | \to \infty, \end{matrix}$ with $\hat{x} : = \frac{x}{| x |}$ , where the function $U^{\infty} (\hat{x}, θ)$ for $(\hat{x}, θ) \in S^{2} \times S^{2}$ is called the far-field pattern. We recall that the fundamental solution, $Φ_{κ} (x, y)$ , of the Helmholtz equation in $R^{3}$ with the fixed wave number κ is given by $\begin{array}{rcl} (1.5) & Φ_{κ} (x, y) & : = & \frac{e^{i κ | x - y |}}{4 π | x - y |}, for all x, y \in R^{3} . \end{array}$

Definition 1.1.
We define $\begin{matrix} a : = max_{1 ⩽ m ⩽ M} diam (D_{m}) [= ϵ max_{1 ⩽ m ⩽ M} diam (B_{m})], d : = min_{\begin{array}{c} m \neq j \\ 1 ⩽ m, j ⩽ M \end{array}} d_{m j}, \end{matrix}$ where $d_{m j} : = dist (D_{m}, D_{j})$ . We assume that $0 < d ⩽ d_{max}$ , and $d_{max}$ is given. Finally, we assume that we have $κ_{max}$ as the upper bound of the used wave numbers, i.e., $κ \in [0, κ_{max}]$ .

Let us now focus on the case where we have a large number of holes of the form $M \sim a^{- s}$ with $s > 0$ and the minimum distance $d \sim a^{t}$ with $t > 0$ . In [7], we have shown using the single layer representation of the solution that if $s ⩽ 1 - 2 t$ , then we have the asymptotic expansion (1.7)–(1.10), see below, of the far-field pattern $U^{\infty} (\hat{x}, θ)$ . Using also the single layer representation of the solution, this condition is weakened to (1.6) in the following proposition. In particular, (1.6) allows to take $s > 1$ and study the scattering by an extremely large number of holes, which is the main topic of this work.
Proposition 1.2.
Assume that the cluster of the holes is modeled by their number M following the law $M \sim a^{- s}$ and their minimum distance $d \sim a^{t}$ with s and t satisfying the following conditions $\begin{array}{l} (1.6) & 0 ⩽ t ⩽ 1 and 0 ⩽ s ⩽ min {3 t, 3 - 3 t, \frac{1}{2} (3 - t)} \end{array}$ then $\begin{array}{rcl} (1.7) & U^{\infty} (\hat{x}, θ) & = & \sum_{m = 1}^{M} e^{- i κ \hat{x} \cdot z_{m}} Q_{m} + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) \end{array}$ uniformly in $\hat{x}$ and θ in $S^{2}$ . The coefficients $Q_{m}$ , $m = 1, \dots, M$ , are the solutions of the following linear algebraic system $\begin{array}{rcl} (1.8) & Q_{m} + \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} C_{m} Φ_{κ} (z_{m}, z_{j}) Q_{j} & = & - C_{m} U^{i} (z_{m}, θ), \end{array}$ for $m = 1, \dots, M$ , with $\begin{matrix} (1.9) & C_{m} : = \int_{\partial D_{m}} σ_{m} (s) d s \end{matrix}$ as the capacitance of $D_{m}$ defined through the density $σ_{m}$ solution of the integral equation of the first kind $\begin{array}{l} (1.10) & \int_{\partial D_{m}} \frac{σ_{m} (s)}{4 π | t - s |} d s = 1, t \in \partial D_{m} . \end{array}$ The algebraic system ( 1.8 ) is invertible under the conditions: $\begin{array}{l} (1.11) & \frac{a}{d} ⩽ c_{1} and min_{j \neq m} cos (κ | z_{j} - z_{m} |) ⩾ 0, \end{array}$ where $c_{1}$ depends only on the Lipschitz character of the obstacles $B_{j}$ , $j = 1, \dots, M$ .

By a simple scale, we have $C_{m} = ϵ \int_{B_{m}} {\tilde{σ}}_{m} (s) d s$ where ${\tilde{σ}}_{m}$ solves the integral equation $\int_{\partial B_{m}} \frac{{\tilde{σ}}_{m} (s)}{4 π | t - s |} d s = 1$ , $t \in \partial B_{m}$ . Hence $C_{m} \sim ϵ$ .
1.2. The distribution of the small holes

Let Ω be a bounded domain, say of unit volume, containing the obstacles $D_{m}, m = 1, \dots, M$ . We shall divide Ω into $[a^{- s}]$ subdomains $Ω_{m}, m = 1, \dots, [a^{- s}]$ , such that each $Ω_{m}$ contains $D_{m}$ , i.e., $z_{m} \in Ω_{m}$ , and some of the other $D_{j}$ ’s. As an example, taking $a : = N^{- \frac{1}{s}}$ , with N an integer and $N ≫ 1$ , we have $a ≪ 1$ and $[a^{- s}] = N$ . It is natural then to assume that the number of obstacles in each $Ω_{m}$ , for $m = 1, \dots, [a^{- s}]$ , to be uniformly bounded in terms of m. To describe correctly this number of obstacles, let $K : R^{3} \to R$ be a given function which is non-negative, continuous and bounded potential. Let each $Ω_{m}$ , $m \in N$ , be a cube of volume $a^{s} \frac{[K (z_{m}) + 1]}{K (z_{m}) + 1}$ and contains $[K (z_{m}) + 1] (= [K (z_{m})] + 1)$ obstacles, see Fig. 1. For a given real and positive number x, we denote here by $[x]$ , the unique integer n such that $n ⩽ x < n + 1$ , i.e., n is the floor number. We set $K_{max} : = {sup}_{z_{m}} [K (z_{m}) + 1]$ , hence $M = \sum_{m = 1}^{[a^{- s}]} [K (z_{m}) + 1] ⩽ K_{max} [a^{- s}] = O (a^{- s})$ . The function K describes then the local distribution of the holes, i.e., the number of holes in each $Ω_{m}$ is fixed as $[K (z_{m}) + 1]$ . A few remarks are in order:

If we distribute the obstacles periodically, then $K \equiv 0$ , the $Ω_{m}$ ’s are identical (modulo a translation) and $| Ω_{m} | = a^{s}$ .

K can be identically zero but the $Ω_{m}$ ’s can be distributed non-periodically. In general, if $K ∣_{Ω}$ is an integer, then also $| Ω_{m} | = a^{s} \frac{[K (z_{m}) + 1]}{K (z_{m}) + 1} = a^{s}$ and the $Ω_{m}$ ’s can be distributed non-periodically.

Assume now that K is, eventually, a variable function and not an integer. Hence $| Ω_{m} | = a^{s} \frac{[K (z_{m}) + 1]}{K (z_{m}) + 1} < a^{s}$ and $Vol ({lim}_{a \to 0} ⋃_{m = 1}^{[a^{- s}]} Ω_{m}) = \int_{Ω} \frac{[K (z) + 1]}{K (z) + 1} d z < | Ω |$ . Hence ${lim}_{a \to 0} ⋃_{m = 1}^{[a^{- s}]} Ω_{m} ⊊ Ω$ . In this case, to Ω we cut a layer of volume $| Ω | - \int_{Ω} \frac{[K (z) + 1]}{K (z) + 1} d z$ and a constant depth starting from $\partial Ω$ . We denote the resulting domain by Ω too. Note that this last domain has the same shape (and hence the same regularity) as the former.

In every case, we see that Ω can be filled with the $Ω_{m}$ ’s. One way to do it, for a general function K, is to put the center $z_{1}$ of the first obstacle $D_{1}$ in the ‘center’ of Ω and then surround it with the cube $Ω_{1}$ of volume $a^{s} \frac{[K (z_{1}) + 1]}{K (z_{1}) + 1}$ . Inside $Ω_{1}$ , add the other $[K (z_{1})]$ obstacles. Starting from $Ω_{1}$ , build up the other $Ω_{m}$ ’s in a Rubik style respecting their volumes and the number of obstacles included inside them using the function K.

Observe that for the cubes $Ω_{m}$ ’s intersecting with $\partial Ω$ , the sets $Ω \cap Ω_{m}$ will have a volume of the order $a^{s}$ but the exact volume is not easy to estimate as it depends strongly on the shape of Ω (unless if Ω has a simple shape itself). We do not put obstacles in those cubes touching $\partial Ω$ .

As Ω can have an arbitrary shape, the set of the cubes intersecting $\partial Ω$ is not empty (unless if Ω has a simple shape as a cube). Later in our analysis, we will need the estimate of the volume of this set. Since each $Ω_{j}$ has volume of the order $a^{s}$ , its maximum radius is of the order $a^{\frac{1}{3} s}$ , and then the intersecting surfaces with $\partial Ω$ has an area of the order $a^{\frac{2}{3} s}$ . As the area of $\partial Ω$ is of the order one, we conclude that the number of such cubes will not exceed the order $a^{- \frac{2}{3} s}$ . Hence the volume of this set will not exceed the order $a^{- \frac{2}{3} s} a^{s} = a^{\frac{1}{3} s}$ , as $a \to 0$ .

Fig. 1.

A schematic example on how the obstacles are distributed in Ω.

1.3. The equivalent media

By applying the estimate (2.11) of Lemma 2.3 to the system (1.8) and knowing that the capacitances $C_{m}$ behave as a, $a ≪ 1$ , we have the following asymptotic expansion: $\begin{matrix} (1.12) & \sum_{m = 1}^{M} e^{- i κ \hat{x} \cdot z_{m}} Q_{m} = O (a^{1 - s}) . \end{matrix}$ We distinguish the following cases:

1.3.1. Case $s < 1$

If the number of obstacles is $M : = M (a) : = a^{- s}$ , $s < 1$ and t satisfies (1.6), $a \to 0$ , then from (1.7), we deduce that $\begin{matrix} (1.13) & U^{\infty} (\hat{x}, θ) \to 0, as a \to 0, uniformly in terms of θ and \hat{x} in S^{2} . \end{matrix}$ This means that this collection of obstacles has no effect on the homogeneous medium as $a \to 0$ .

1.3.2. Case $s = 1$

In this case we divide the bounded domain Ω in the way we explained in Section 1.2 above. In [1], we proved the following result. Let the small obstacles be distributed in a bounded domain Ω, say of unit volume, with their number $M : = M (a) : = O (a^{- 1})$ and their minimum distance $d : = d (a) : = a^{t}$ , $\frac{1}{3} ⩽ t ⩽ \frac{5}{12}$ , as $a \to 0$ , as described above. Then

If the obstacles are distributed arbitrarily in Ω, i.e., with different capacitances, then there exists a potential $C_{0} \in ⋂_{p ⩾ 1} L^{p} (R^{3})$ with support in Ω such that $\begin{matrix} (1.14) & lim_{a \to 0} U^{\infty} (\hat{x}, θ) = U_{0}^{\infty} (\hat{x}, θ) uniformly in terms of θ and \hat{x} in S^{2}, \end{matrix}$ where $U_{0}^{\infty} (\hat{x}, θ)$ is the far-field corresponding to the scattering problem $\begin{array}{l} (1.15) & (Δ + κ^{2} - (K + 1) C_{0}) U_{0}^{t} = 0 in R^{3}, \\ (1.16) & U_{0}^{t} = U_{0}^{s} + e^{i κ x \cdot θ}, \\ (1.17) & \frac{\partial U_{0}^{s}}{\partial | x |} - i κ U_{0}^{s} = o (\frac{1}{| x |}), | x | \to \infty . \end{array}$

If in addition $K |_{Ω}$ is in $C^{0, γ} (Ω)$ , $γ \in (0, 1]$ and the obstacles have the same capacitances, i.e., which we denote by $\bar{C}$ , then $\begin{matrix} (1.18) & U^{\infty} (\hat{x}, θ) = U_{0}^{\infty} (\hat{x}, θ) + O (a^{min {γ, \frac{1}{3} - \frac{4}{5} t}}) uniformly in terms of θ and \hat{x} in S^{2}, \end{matrix}$ where $C_{0} = \bar{C}$ in Ω and $C_{0} = 0$ in $R^{3} ∖ \overline{Ω}$ .

This result shows the ‘equivalent’ behaviour between a cluster of, appropriately dense, small holes and an extended penetrable obstacle modeled by an additive potential. Such an observation goes back at least to the works by Cioranescu and Murat [8,9] and also the reference therein. Their analysis, made for the Poisson problem, is based on homogenization via energy methods and, in particular, they assume that the obstacles are distributed periodically. In [1], we have confirmed this result without the periodicity assumption and provided the error of the approximation. In this work, compared to [1], the error of the approximation $a^{min {γ, \frac{1}{3} - \frac{4}{5} t}}$ and the interval to which t belongs, i.e., $\frac{1}{3} ⩽ t ⩽ \frac{5}{12}$ , are improved to $O (a^{min {\frac{1}{3}, 2 - 3 t}})$ , and $\frac{1}{3} ⩽ t < \frac{2}{3}$ respectively, at least for the case K is a constant function. For instance, for $t = \frac{1}{3}$ , the former error is of the order $a^{\frac{1}{12}}$ while the latter is of the order $a^{\frac{1}{3}}$ .

1.3.3. Case $s > 1$

Here we assume, for simplicity,1

¹
See the footnote after Lemma 2.3 for more explanation on these assumptions.

that the density K is a constant function (i.e., in the distribution of the holes, the subdomains

Ω_{m}

’s contain the same number of hole, i.e.,

K + 1

, see Section 1.2), and the holes have the same capacitance, i.e., they have the same shape for instance. The main contribution of this work is to prove the following result.

For β, a real and positive number, we use the notation $β_{-}$ to describe the property $α < β_{-}$ if $α ⩽ β - r$ for some small $r > 0$ .

Theorem 1.3.

Let the cluster of the holes be modeled by their number M following the law $M \sim a^{- s}$ and their minimum distance $d \sim a^{t}$ . Let also Ω be bounded domain of class $C^{1, 1}$ . We divide it in terms of the $Ω_{m}$ ’s as described in Section 1.2 above where the function K is assumed to be constant (but arbitrary) and the capacitances $C_{m}$ are taken to be equal (ex. the holes have the same shapes). There exists a constant $κ_{0} (Ω)$ depending only on Ω,2

For more details on this condition, see (2.63).

such that if

\begin{matrix} (1.19) & κ ⩽ κ_{0} (Ω) \end{matrix}

and

\begin{array}{l} (1.20) & 0 ⩽ t < 1 and 0 ⩽ s < min {{(\frac{33}{29})}_{-}, 3 t, 3 - 3 t, \frac{4 - t}{3}, \frac{4 - 3 t}{2}} \end{array}

then we have the following expansion:

\begin{array}{l} U^{\infty} (\hat{x}, θ) - U_{D}^{\infty} (\hat{x}, θ) \\ (1.21) & = O (a^{\frac{s - 1}{4}} + a^{\frac{{(33 - 29 s)}_{-}}{12}} + a^{4 - 3 s - t} + a^{4 - 2 s - 3 t}) uniformly in terms of θ and \hat{x} in S^{2}, \end{array}

where

U_{D}^{\infty} (\hat{x}, θ)

is the far-field corresponding to the Dirichlet scattering problem

\begin{array}{l} (1.22) & (Δ + κ^{2}) U_{D}^{t} = 0 in R^{3} ∖ \overline{Ω}, \\ (1.23) & U_{D}^{t} : = U_{D}^{s} + e^{i κ x \cdot θ} = 0 on \partial Ω, \\ (1.24) & \frac{\partial U_{D}^{s}}{\partial | x |} - i κ U_{D}^{s} = o (\frac{1}{| x |}), | x | \to \infty . \end{array}

To have an idea on the order of convergence, we take $t = \frac{s}{3}$ , then if $s < {(\frac{33}{29})}_{-}$ the conditions in (1.20) are satisfied. Now, choosing $s : = 1.1$ , then the error is approximately of the order $a^{\frac{1}{10}}$ , as $a ≪ 1$ .

To our best knowledge, this kind of result has never been published before even if in a few references, as [8,9] and also the cited reference therein, it is claimed that with such a dense cluster of holes the scattered fields should behave as one of the exterior problem. Let us also emphasize that the arguments we developed to prove Theorem 1.3 can also be used for general settings in acoustic, as the gas bubbles model [2–5], electromagnetic or elastic metamaterials.

1.4. A brief description of the arguments

Let us now give our arguments why the claimed results make sense. Here we describe the whole steps of the proof and the formal computations including the function K. However, in the detailed proof, we take K constant to simplify the exposition (in particular for the case $s > 1$ ). But certainly the proof should work without this assumption.

We recall that $C_{j} = {\bar{C}}_{j} a$ where ${\bar{C}}_{j}$ is the capacitance of the reference obstacle $B_{j}$ and we rewrite (1.8) as: $\begin{array}{rcl} (1.25) & \frac{Q_{m}}{C_{m}} + a^{1 - s} \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} {\bar{C}}_{j} a^{s} Φ_{κ} (z_{m}, z_{j}) \frac{Q_{j}}{C_{j}} & = & - U^{i} (z_{m}, θ), \end{array}$ for $m = 1, \dots, M$ , which we can rewrite, recalling that $M = \sum_{m = 1}^{[a^{- 1}]} [K (z_{m}) + 1]$ , see Section 1.2, as $\begin{array}{rcl} (1.26) & \frac{Q_{m}}{C_{m}} + a^{1 - s} \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{[a^{- 1}]} {\bar{C}}_{j} [K (z_{j}) + 1] a^{s} Φ_{κ} (z_{m}, z_{j}) \frac{Q_{j}}{C_{j}} & = & - U^{i} (z_{m}, θ), \end{array}$ Similarly, we rewrite the representation (1.7) as $\begin{array}{rcl} (1.27) & U^{\infty} (\hat{x}, θ) = a^{1 - s} \sum_{m = 1}^{[a^{- 1}]} e^{- i κ \hat{x} \cdot z_{m}} \bar{C_{m}} [K (z_{m}) + 1] a^{s} \frac{Q_{m}}{C_{m}} + o (1), a \to 0 \end{array}$ We assume, for simplicity, that the reference obstacles are all the same, precisely they have the same capacitance ${\bar{C}}_{m} = {\bar{C}}_{j}$ , $j \neq m$ and we set $\bar{C}$ equal to the constant. Let us introduce the Lippmann–Schwinger equation $\begin{matrix} (1.28) & U_{a} + a^{1 - s} \int_{Ω} Φ_{κ} (\cdot, z) \bar{C} (K (z) + 1) U_{a} (z) d z = u^{i} (\cdot, θ) . \end{matrix}$ The equation (1.28) is the integral equation modeling the unique solution of the acoustic problem $\begin{matrix} (1.29) & - Δ U_{a} - κ^{2} U_{a} + a^{1 - s} \bar{C} (K (z) + 1) U_{a} = 0 \end{matrix}$ where $U_{a} (\cdot, θ) : = U_{a}^{s} (\cdot, θ) + u^{i} (\cdot, θ)$ and $U_{a}^{s} (\cdot, θ)$ satisfies the Sommerfeld radiation condition. The far-field corresponding to the solution of (1.28) [or (1.29)] has the form $\begin{matrix} (1.30) & U_{a}^{\infty} (\hat{x}, θ) : = - a^{1 - s} \int_{Ω} e^{- i κ \hat{x} \cdot z} \bar{C} (K (z) + 1) U_{a} (z) d z . \end{matrix}$ Based on (1.26)–(1.28), noticing that $M = \sum_{m = 1}^{[a^{- 1}]} [K (z_{m}) + 1]$ and $| Ω_{m} | = a^{s} \frac{[K (z_{m}) + 1]}{K (z_{m}) + 1}$ , $m = 1, \dots, M$ , we derive the following error estimate: $\begin{matrix} (1.31) & U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) = o (1), a \to 0 \end{matrix}$ with an explicit expression of the error term $o (1)$ .

Our final step is to estimate the term $U_{a}^{\infty} (\hat{x}, θ)$ , as $a \to 0$ . For $s < 1$ , it is clear from (1.30) that $U_{a}^{\infty}$ tends to zero, as $a \to 0$ uniformly in terms of $\hat{x}$ and θ. For $s = 1$ , we see that $U_{a}^{\infty}$ is the far-field corresponding to the scattering problem by the potential $(K + 1) C_{0}$ described in (1.15)–(1.17). The most delicate case is when $s > 1$ . In this case, we need to study the scattering problem (1.29) modeled by the Schrodinger equation with the potential $a^{1 - s} \bar{C} (K + 1) 1_{Ω}$ . We set $h : = a^{\frac{s - 1}{2}}$ and $V_{0} : = \bar{C} (K + 1) 1_{Ω}$ , hence (1.29) becomes $\begin{matrix} (1.32) & - Δ u^{t} - κ^{2} u^{t} + h^{- 2} V_{0} u^{t} = 0 in R^{3} \end{matrix}$ with $u^{t} : = u^{t} (\cdot, θ) : = u^{s} (\cdot, θ) + u^{i} (\cdot, θ)$ where $u^{s}$ ( $: = U_{a}^{s} (\cdot, θ)$ ) satisfies the Sommerfeld radiation condition.

When $h \to 0$ , the tunneling effect through the potential barrier $h^{- 2} V_{0} 1_{Ω}$ becomes negligible, hence in this limit our model corresponds to a hard obstacle supported in Ω and does not allow any source coming from outside to penetrate inside Ω, as $h \to 0$ . We find it quite interesting to link the scattering by a collection of impenetrable obstacles $D_{m}$ , $m = 1, \dots, M$ , to the scattering by potential barriers.

To derive our result with the error estimates in terms of h and hence justify our claim, we prove the following trace estimate: $\begin{matrix} (1.33) & {‖ u^{t} (\cdot, θ) ‖}_{H^{t} (\partial Ω)} = O (h^{\frac{1}{2} - t}), t \in [0, 1 / 2], as h ≪ 1 . \end{matrix}$

Taking $t = 0$ in (1.33), we have the estimate $‖ u^{t} (\cdot, θ) |_{\partial Ω} ‖_{L^{2} (\partial Ω)} = O (h^{\frac{1}{2}}) = O (a^{\frac{s - 1}{4}})$ , as $a ≪ 1$ . As $u^{s} (\cdot, θ)$ satisfies $(Δ + k^{2}) u^{s} (\cdot, θ) = 0$ in $R^{3} ∖ \bar{Ω}$ with the radiation conditions and the boundary estimate $‖ u^{s} (\cdot, θ) + e^{i κ x \cdot θ} |_{\partial Ω} ‖_{L^{2} (\partial Ω)} = O (a^{\frac{s - 1}{4}})$ as $a ≪ 1$ , then the well-posedness of the forward scattering problem in the exterior domain $R^{3} ∖ \bar{Ω}$ implies that the corresponding far-fields satisfy the estimate, recalling that $u^{s} (\hat{x}, θ) = U_{a}^{s} (\hat{x}, θ)$ , $\begin{matrix} (1.34) & | U_{a}^{\infty} (\hat{x}, θ) - U_{D}^{\infty} (\hat{x}, θ) | = O (a^{\frac{s - 1}{4}}) as a ≪ 1 . \end{matrix}$ The claim follows by combining (1.31) and (1.34).

Let us now describe the main idea to derive the estimate (1.33). For this, we write (1.28) in the form $\begin{matrix} (1.35) & (h^{2} + R (κ)) u^{t} = h^{2} u^{i}, in Ω, \end{matrix}$ where $R (κ)$ , defined through $R (κ) u^{t} (x) : = \int_{Ω} V_{0} (y) Φ_{κ} (x, y) u^{t} (y) d y$ , is the Newtonian potential operator restricted to Ω. We rewrite (1.35) as $\begin{matrix} (1.36) & (h^{2} + R (0)) u^{t} = h^{2} u^{i} - (R (κ) - R (0)) u^{t} in Ω . \end{matrix}$ For simplicity of exposition, we set $V_{0} \equiv 1$ . The key observation is that $(λ, e)$ is an eigenvalue and eigenfunction, respectively, of $R (0)$ if and only if $(λ^{- 1}, e)$ is an eigenvalue and eigenfunction of the operator defined, on Ω, by $- Δ$ with the boundary condition $\partial_{ν} e - S^{- 1} (- \frac{1}{2} I + K) e = 0$ on $\partial Ω$ , where S and K are the single and double layer potential, respectively. Using variational formulations with boundary integral equations, we show that this second operator is self-adjoint with a compact inverse and allows the following characterization of $H^{1} (Ω)$ , i.e., $H^{1} (Ω) = {u \in L^{2} (Ω) : \sum_{n} λ_{n}^{- 1} {(u, e_{n})}^{2} < \infty}$ , with the spectral family $(λ_{n}, e_{n}), n \in N$ [of $R (0)$ ]. Using this characterization and the spectral decomposition (Galerkin method) in (1.36), we show that $‖ u^{t} ‖_{H^{1} (Ω)} = O (1)$ , $h ≪ 1$ . With this estimate and an integration by parts in (1.32), in Ω, we deduce that $‖ u^{t} ‖_{L^{2} (Ω)} = O (h)$ , $h ≪ 1$ . We end up with interpolation and trace estimates.

We finish this introduction by making a link between the estimates we derived and the semiclassical resolvent estimates. Recall that the solution $u^{t} = u^{i} + u^{s}$ of the equation (1.35) is nothing but the solution of the equation $(- Δ - κ^{2} + h^{- 2} V_{0} 1_{Ω}) u^{t} = 0$ , and then the scattered field $u^{s}$ is the solution of the equation $(- Δ - κ^{2} + h^{- 2} V_{0} 1_{Ω}) u^{s} = - h^{- 2} V_{0} 1_{Ω} u^{i}$ satisfying the Sommerfeld radiation condition. This last equation can be written, in the semiclassical setting, as $(- h^{2} Δ - κ^{2} h^{2} + V_{0} 1_{Ω}) u^{s} = - V_{0} 1_{Ω} u^{i}$ . We set $Res (- h^{2} Δ + V_{0} 1_{Ω}; k^{2})$ to be the semiclasscial resolvent of $- h^{2} Δ + V_{0} 1_{Ω}$ at the frequency $k^{2}$ . The estimates discussed above imply, in particular, that $‖ 1_{Ω} Res (- h^{2} Δ + V_{0} 1_{Ω}; k^{2}) 1_{Ω} ‖_{L^{2} (R^{3})} = O (h^{- 1})$ at low frequencies, precisely for $k = O (h)$ , $h ≪ 1$ . Hence, with the method described above, we can derive the semiclassical resolvent estimates at low frequencies, i.e., $κ = O (h)$ , $h ≪ 1$ , for positive, compactly supported but not necessarily smooth potentials. The semiclassical resolvent estimates for frequencies k away from zero, and under non trapping conditions, are well known for general, but smooth enough, potentials, see for instance [6,16,18] and the reference therein.

The rest of the paper is organized as follows. In Section 2, we prove Theorem 1.3 by using the expansions of Proposition 1.2. The section is divided into two subsections, Section 2.1 and Section 2.2 corresponding to the estimates of $U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ)$ and $U_{a}^{\infty} (\hat{x}, θ) - U_{D}^{\infty} (\hat{x}, θ)$ respectively. The justification of estimate (1.33) is provided in Section 2.2. The proof of Proposition 1.2 is postponed to Section 3.

2. Proof of Thereom 1.3

The existence and uniqueness of the solution of the following scattering problem is guaranteed based on the Fredholm alternative, as it is discussed in Section 2.2. Let $U_{a}^{t}$ be its unique solution. $\begin{array}{l} (2.1) & (Δ + κ^{2} - a^{1 - s} C_{0}) U_{a}^{t} = 0 in R^{3}, \\ (2.2) & U_{a}^{t} = U_{a}^{s} + e^{i κ x \cdot θ}, \\ (2.3) & \frac{\partial U_{a}^{s}}{\partial | x |} - i κ U_{a}^{s} = o (\frac{1}{| x |}), | x | \to \infty . \end{array}$

Theorem 1.3 can split into the following two propositions;

Proposition 2.1.
Let the cluster of the holes be modeled by their number M following the law $M \sim a^{- s}$ and their minimum distance $d \sim a^{t}$ . Let also Ω be bounded domain of class $C^{1, 1}$ . We divide it in terms of the $Ω_{m}$ ’s as described in Section 1.2 where the function K is assumed to be constant (but arbitrary) and the capacitances $C_{m}$ are taken to be equal. There exists a constant $κ_{0} (Ω)$ depending only on Ω, such that if $\begin{matrix} (2.4) & κ ⩽ κ_{0} (Ω) \end{matrix}$ and $\begin{array}{l} (2.5) & \frac{s}{3} ⩽ t < 1 and 0 ⩽ s < min {{(\frac{33}{29})}_{-}, 2 - t, 3 - 3 t, \frac{1}{2} (3 - t), \frac{4 - t}{3}, \frac{4 - 3 t}{2}} \end{array}$ we have the following expansion: $\begin{matrix} (2.6) & U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) = O (a^{\frac{{(33 - 29 s)}_{-}}{12}} + a^{4 - 3 s - t} + a^{4 - 2 s - 3 t}) as a \to 0, \end{matrix}$ uniformly in terms of θ and $\hat{x}$ in $S^{2}$ .
Proposition 2.2.
Assume that Ω is a bounded domain with $C^{1, 1}$ regularity. There exists a constant $κ_{0} (Ω)$ depending only on Ω, such that if $\begin{matrix} (2.7) & κ ⩽ κ_{0} (Ω) \end{matrix}$ and $s > 1$ , then we have the following expansion $\begin{matrix} (2.8) & U_{a}^{\infty} (\hat{x}, θ) - U_{D}^{\infty} (\hat{x}, θ) = O (a^{\frac{s - 1}{4}}), as a \to 0, \end{matrix}$ uniformly in terms of θ and $\hat{x} in S^{2}$ .

2.1. Proof of Proposition 2.1

We rewrite the algebraic system (1.8) in form $\begin{array}{rcl} (2.9) & Y_{m} + \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} Φ_{κ} (z_{m}, z_{j}) {\bar{C}}_{j} Y_{j} a & = & U^{i} (z_{m}, θ), \end{array}$ with $Y_{m} : = - \frac{Q_{m}}{C_{m}}$ and $C_{m} : = {\bar{C}}_{m} a$ , for $m = 1, \dots, M$ and ${\bar{C}}_{m}$ are related to the capacitances of $B_{m}$ ’s, i.e., they are independent of a. We set $\hat{C} : = {({\bar{C}}_{1}, {\bar{C}}_{2}, \dots, {\bar{C}}_{M})}^{⊤}$ and define $\hat{Y} : = {(Y_{1}, Y_{2}, \dots, Y_{M})}^{⊤} and U^{I} : = {(U^{i} (z_{1}), U^{i} (z_{2}), \dots, U^{i} (z_{M}))}^{⊤}$ . The following lemma ensures the invertibility of the algebraic system (2.12), see its proof in [7]:

Lemma 2.3.
If $a < \frac{5 π}{3} \frac{d}{‖ \hat{C} ‖}$ and $t : = {min}_{j \neq m, 1 ⩽ j, m ⩽ M} cos (κ | z_{m} - z_{j} |) ⩾ 0$ , then the matrix B is invertible and the solution vector $\hat{Y}$ of ( 2.9 ) satisfies the estimate $\begin{matrix} (2.10) & \sum_{m = 1}^{M} | Y_{m} |^{2} ⩽ 4 {(1 - \frac{3 t a}{5 π d} ‖ \hat{C} ‖)}^{- 2} \sum_{m = 1}^{M} {| U^{i} (z_{m}) |}^{2}, \end{matrix}$ and hence the estimate $\begin{matrix} (2.11) & \begin{matrix} \sum_{m = 1}^{M} | Y_{m} | ⩽ 2 {(1 - \frac{3 t a}{5 π d} ‖ \hat{C} ‖)}^{- 1} M max_{1 ⩽ m ⩽ M} | U^{i} (z_{m}) | . \end{matrix} \end{matrix}$

Recalling that $M = \sum_{m = 1}^{[a^{- 1}]} [K + 1]$ , see Section 1.2, then we rewrite (2.9) as $\begin{array}{rcl} (2.12) & Y_{m} + a^{1 - s} \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{[a^{- 1}]} Φ_{κ} (z_{m}, z_{j}) {\bar{C}}_{j} [K + 1] Y_{j} a^{s} & = & - U^{i} (z_{m}, θ) . \end{array}$

Consider the Lippmann–Schwinger equation $\begin{array}{l} (2.13) & Y (z) + a^{1 - s} \int_{Ω} Φ_{κ} (z, y) \bar{C_{0}} (K + 1) (y) Y (y) d y = U^{i} (z, θ), z \in Ω, \end{array}$ where $\bar{C_{0}}$ is a piecewise constant function such that $\bar{C_{0}} |_{Ω_{m}} = {\bar{C}}_{m}$ for all $m = 1, \dots, M$ and vanishes outside Ω. As we have assumed that obstacles have the same capacitances, ${\bar{C}}_{m}$ is the same for every m and we denote it by $\bar{C}$ .

For $m = 1, \dots, M$ , the equation (2.13) can be rewritten as3
³
As the functions K and $\bar{C}$ are assumed to be constants, we need only to differentiate $Φ_{κ} Y$ to estimate A. If K (or $\bar{C}$ ) is a variable function, then we need to add intermediate steps. To avoid such complications, in the presentation, we assumed K and $\bar{C}$ to be constant functions.

$\begin{array}{l} Y (z_{m}) + a^{1 - s} \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} Φ_{κ} (z_{m}, z_{j}) \bar{C} Y (z_{j}) a^{s} \\ = U^{i} (z_{m}, θ) + a^{1 - s} [\sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} Φ_{κ} (z_{m}, z_{j}) \bar{C} Y (z_{j}) a^{s} - \int_{Ω} Φ_{κ} (z_{m}, y) \bar{C_{0}} (y) (K + 1) Y (y) d y] \\ = U^{i} (z_{m}, θ) + a^{1 - s} [\sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{[a^{- s}]} Φ_{κ} (z_{m}, z_{j}) \bar{C} [K + 1] Y (z_{j}) a^{s} - \int_{Ω} Φ_{κ} (z_{m}, y) \bar{C_{0}} (y) (K + 1) Y (y) d y] \\ = U^{i} (z_{m}, θ) + a^{1 - s} \bar{C} (K + 1) [\sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{[a^{- s}]} Φ_{κ} (z_{m}, z_{j}) Y (z_{j}) a^{s} \frac{[K + 1]}{K + 1} - \sum_{j = 1}^{[a^{- s}]} \int_{Ω_{j}} Φ_{κ} (z_{m}, y) Y (y) d y] \\ - a^{1 - s} \int_{Ω ∖ ⋃_{j = 1}^{[a^{- s}]} Ω_{j}} Φ_{κ} (z_{m}, y) {\bar{C}}_{0} (y) (K + 1) Y (y) d y \\ = U^{i} (z_{m}, θ) + a^{1 - s} \bar{C} (K + 1) \\ \times [\underset{= : A}{\underset{︸}{\sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{[a^{- s}]} \int_{Ω_{j}} [Φ_{κ} (z_{m}, z_{j}) Y (z_{j}) - Φ_{κ} (z_{m}, y) Y (y)] d y}}] (as | Ω_{j} | = a^{s} \frac{[K + 1]}{K + 1}) \\ - a^{1 - s} \bar{C} (K + 1) \underset{= : B}{\underset{︸}{\int_{Ω_{m}} Φ_{κ} (z_{m}, y) Y (y) d y}} \\ (2.14) & - a^{1 - s} \underset{= : D}{\underset{︸}{\int_{Ω ∖ ⋃_{j = 1}^{[a^{- s}]} Ω_{j}} Φ_{κ} (z_{m}, y) {\bar{C}}_{0} (y) (K + 1) Y (y) d y}} . \end{array}$

First, we prove the following lemma.
Lemma 2.4.
There exists a constant $κ_{0} (Ω)$ depending only on Ω, such that if $\begin{matrix} (2.15) & κ ⩽ κ_{0} (Ω) \end{matrix}$ then the function Y satisfying the Lippmann–Schwinger equation ( 2.13 ) satisfies the following estimates. $\begin{array}{l} (2.16) & ‖ Y ‖_{L^{\infty} (Ω)} = O (a^{\frac{1 - s}{2}}), ‖ \nabla Y ‖_{L^{\infty} (Ω)} = O (a^{\frac{3 - η}{4} (1 - s)}), \end{array}$ where η is arbitrary positive quantity.
Proof.
In Proposition 2.6, under the above condition on the frequency κ, we have the estimate $‖ Y ‖_{H^{α} (Ω)} = O (a^{(s - 1) \frac{1 - α}{2}})$ , $α \in [0, 1]$ , $a ≪ 1$ . For $α = 0$ , we have $‖ Y ‖_{H^{α} (Ω)} = O (a^{\frac{s - 1}{2}})$ , and (2.13) implies that $\begin{matrix} (2.17) & | Y (z) | = O (1) + O (a^{1 - s}) O (a^{\frac{s - 1}{2}}) = O (a^{\frac{1 - s}{2}}), \end{matrix}$ for $s ⩾ 1$ . Again, from (2.13), we deduce that $\begin{matrix} | \nabla Y (z) | = O (1) + a^{1 - s} {‖ C (y) (K + 1) \nabla_{z} Φ_{κ} (z, y) ‖}_{L^{p} (Ω)} ‖ Y ‖_{L^{p^{'}} (Ω)} \end{matrix}$ where $\frac{1}{p} + \frac{1}{p^{'}} = 1$ . As for $p < \frac{3}{2}$ , we have $‖ \nabla_{z} Φ_{κ} (z, y) ‖_{L^{p} (Ω)} < \infty$ , we need only to estimate $‖ Y ‖_{L^{p^{'}} (Ω)}$ for $p^{'} > 3$ . We know also, by Sobolev embeddings, that $‖ Y ‖_{L^{p^{'}} (Ω)} ⩽ C ‖ Y ‖_{H^{α} (Ω)}$ for $\frac{1}{p^{'}} = \frac{1}{2} - \frac{α}{3}$ , i.e., $α > \frac{1}{2}$ . Then $‖ Y ‖_{L^{p^{'}} (Ω)} = O (a^{(s - 1) \frac{1 - α}{2}})$ , $α > \frac{1}{2}$ and then $\begin{array}{l} (2.18) & \begin{aligned} | \nabla Y (z) | = O (1) + O (a^{1 - s + (s - 1) \frac{1 - α}{2}}) = O (a^{(1 - s) \frac{1 + α}{2}}), α > \frac{1}{2}, or \\ ‖ \nabla Y ‖_{L^{\infty} (Ω)} = O (a^{\frac{3 - η}{4} (1 - s)}) \end{aligned} \end{array}$ where η is an arbitrary positive quantity. □

Based on Lemma 2.4, we derive the following estimates of A, B and D [as defined in (2.14) above].
Lemma 2.5.
The quantities A, B and D enjoy the following estimates $\begin{array}{rcl} (2.19) & | A | = O (a^{\frac{3 - s}{6}} + a^{\frac{(9 - 3 η) - (5 - 3 η) s}{12}}), | B | = O (a^{\frac{3 + s}{6}}) and | D | = O (a^{\frac{3 - s}{6}}) \end{array}$ where η is an arbitrary positive quantity.
Proof.
To evaluate A and D for every $z_{m}$ , we need to perform the sum of the integrals when the $Ω_{j}$ ’s are located inside Ω (for A) and when they are intersecting $\partial Ω$ (for D). We proceed as follows in counting these $Ω_{j}$ ’s:
Evaluation of A. We start by distinguishing between the near and far-by obstacles to each obstacle. Let us suppose that these cubes are arranged in a cuboid, for example Rubik’s cube, in different layers such that the total cubes up to the nth layer consists ${(2 n + 1)}^{3}$ cubes for $n = 0, \dots, [a^{- \frac{s}{3}}]$ , and $Ω_{m}$ is located on the center. Hence the number of obstacles located in the nth, $n \neq 0$ layer will have almost $[{(2 n + 1)}^{3} - {(2 n - 1)}^{3}]$ elements and their distance from $D_{m}$ is more than $n (a^{\frac{s}{3}} - \frac{a}{2})$ .

Evaluation of D. The corresponding $Ω_{j}$ ’s are touching the surface $\partial Ω$ . We distinguish two situations.

In the first situation, the point $z_{m}$ is away from $\partial Ω$ so that $Φ_{κ} (z_{m}, y)$ is bounded in y on $\partial Ω$ .

In the second situation, the point $z_{m}$ is close to $\partial Ω$ , i.e., when $z_{m}$ is located near one of the $Ω_{j}$ ’s touching $\partial Ω$ . Since the radius a is small, and hence the $Ω_{j}$ ’s, the point $z_{m}$ is close to $\partial Ω$ . As the function $Φ_{κ} (z_{m}, y)$ is singular only for y near $z_{m}$ , we split the estimate of D into two parts. One part involves the $Ω_{j}$ ’s close to $z_{m}$ , that we denote by $N_{m}$ , and the other part involves the reminder, that we denote by $F_{m}$ . The latter part can be estimated as in the first situation (i.e., 1. above). To estimate the former part (which is the worst part), we observe that, as a is small enough, the $Ω_{j}$ ’s close to $z_{m}$ are located near a small part of $\partial Ω$ that we can assume to be flat, recalling that $\partial Ω$ is smooth. The point $z_{m}$ being close to this flat part, we divide this part into concentric layers as in the case when we evaluated A. But now, up to the nth layer at most ${(2 n + 1)}^{2}$ cubes are intersecting the surface, for $n = 0, \dots, [a^{- \frac{s}{3}}]$ . As a consequence, the number of obstacles located in the nth, $n \neq 0$ layer will have at most $[{(2 n + 1)}^{2} - {(2 n - 1)}^{2}]$ elements and their distance from $D_{m}$ is more than $n (a^{\frac{s}{3}} - \frac{a}{2})$ .

Let us set $f (z_{m}, y) : = Φ_{κ} (z_{m}, y) Y (y)$ . Using Taylor series, we can write $\begin{matrix} f (z_{m}, y) - f (z_{m}, z_{l}) = (y - z_{l}) R_{l} (z_{m}, y), \end{matrix}$ with $\begin{array}{rcl} R_{l} (z_{m}, y) & = & \int_{0}^{1} \nabla_{y} f (z_{m}, y - β (y - z_{l})) d β \\ = & \int_{0}^{1} [\nabla_{y} Φ_{κ} (z_{m}, y - β (y - z_{l}))] Y (y - β (y - z_{l})) d β \\ (2.20) & + \int_{0}^{1} Φ_{κ} (z_{m}, y - β (y - z_{l})) [\nabla_{y} Y (y - β (y - z_{l}))] d β . \end{array}$ From the explicit form of $Φ_{κ}$ , we have $\nabla_{y} Φ_{κ} (x, y) = Φ_{κ} (x, y) [\frac{1}{| x - y |} - i κ] \frac{x - y}{| x - y |}$ , $x \neq y$ . Hence, we obtain that
for $l \neq m$ , we have $\begin{array}{l} | Φ_{κ} (z_{m}, y - β (y - z_{l})) | ⩽ \frac{1}{4 π n (a^{\frac{s}{3}} - \frac{a}{2})}, and \\ | \nabla_{y} Φ_{κ} (z_{m}, y - β (y - z_{l})) | ⩽ \frac{1}{4 π n (a^{\frac{s}{3}} - \frac{a}{2})} [\frac{1}{n (a^{\frac{s}{3}} - \frac{a}{2})} + κ] . \end{array}$ These values give us $\begin{array}{rcl} | R_{l} (z_{m}, y) | & ⩽ & \frac{1}{n (a^{\frac{s}{3}} - \frac{a}{2})} ([\frac{1}{n (a^{\frac{s}{3}} - \frac{a}{2})} + κ] \int_{0}^{1} | Y (y - β (y - z_{l})) | d β \\ + \int_{0}^{1} | \nabla_{y} Y (y - β (y - z_{l})) | d β) \\ (2.21) & ⩽ & \frac{c_{1}}{n (a^{\frac{s}{3}} - \frac{a}{2})} ([\frac{1}{n (a^{\frac{s}{3}} - \frac{a}{2})} + κ] ‖ Y ‖_{L^{\infty} (Ω)} + c_{5} ‖ \nabla Y ‖_{L^{\infty} (Ω)}) . \end{array}$

Hence, for $l \neq m$ using (2.21) we get the following estimate for A; $\begin{array}{rcl} | A | & = & | \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{[a^{- s}]} \int_{Ω_{j}} [Φ_{κ} (z_{m}, z_{j}) Y (z_{j}) - Φ_{κ} (z_{m}, y) Y (y)] d y | \\ ⩽ & \sum_{n = 1}^{[a^{- \frac{s}{3}}]} [{(2 n + 1)}^{3} - {(2 n - 1)}^{3}] a^{\frac{s}{3}} a^{s} \frac{c_{1}}{n (a^{\frac{s}{3}} - \frac{a}{2})} \\ \times ([\frac{1}{n (a^{\frac{s}{3}} - \frac{a}{2})} + κ] ‖ Y ‖_{L^{\infty} (Ω)} + c_{5} ‖ \nabla Y ‖_{L^{\infty} (Ω)}) \\ = & O (\sum_{n = 1}^{[a^{- \frac{s}{3}}]} [24 n^{2} + 2] a^{\frac{4}{3} s} [\frac{1}{n^{2}} a^{- \frac{2}{3} s} ‖ Y ‖_{L^{\infty} (Ω)} + \frac{1}{n} a^{- \frac{1}{3} s} ‖ \nabla Y ‖_{L^{\infty} (Ω)}]) \\ (2.22) & = & O (a^{\frac{3 - s}{6}} + a^{\frac{(9 - 3 η) - (5 - 3 η) s}{12}}) . \end{array}$

Let us estimate the integral value $\int_{Ω_{m}} Φ_{κ} (z_{m}, y) Y (y) d y$ . We have the following estimates: $\begin{array}{rcl} | \int_{Ω_{m}} Φ_{κ} (z_{m}, y) Y (y) d y | & ⩽ & ‖ Y ‖_{L^{\infty} (Ω)} | \int_{Ω_{m}} Φ_{κ} (z_{m}, y) d y | \\ ⩽ & c_{1} a^{\frac{1 - s}{2}} | \int_{Ω_{m}} Φ_{κ} (z_{m}, y) d y | (taking r < \frac{1}{2} a^{\frac{s}{3}}) \\ ⩽ & \frac{1}{4 π} c_{1} a^{\frac{1 - s}{2}} (\int_{B (z_{m}, r)} \frac{1}{| z_{m} - y |} d y + \int_{Ω_{m} ∖ B (z_{m}, r)} \frac{1}{| z_{m} - y |} d y) \\ (2.23) & = & O (a^{\frac{3 + s}{6}}) . \end{array}$

Let us estimate D. Recall that $| Ω ∖ ⋃_{j = 1}^{[a^{- s}]} Ω_{j} |$ , and hence $| F_{m} |$ , is of the order $a^{\frac{1}{3} s}$ as $a \to 0$ . $\begin{array}{rcl} | D | & = & | \int_{Ω ∖ ⋃_{j = 1}^{[a^{- s}]} Ω_{j}} Φ_{κ} (z_{m}, y) {\bar{C}}_{0} (y) Y (y) d y | \\ = & | \int_{N_{m}} Φ_{κ} (z_{m}, y) {\bar{C}}_{0} (y) Y (y) d y | + | \int_{F_{m}} Φ_{κ} (z_{m}, y) {\bar{C}}_{0} (y) Y (y) d y | \\ ⩽ & \sum_{l = 1}^{[a^{- \frac{2}{3} s}]} ‖ Y ‖_{L^{\infty} (Ω)} ‖ {\bar{C}}_{0} ‖_{L^{\infty} (Ω)} \frac{1}{d_{m l}^{'}} | Ω_{l} | + {‖ Φ_{κ} (z_{m}, \cdot) ‖}_{L^{\infty} (F_{m})} ‖ {\bar{C}}_{0} ‖_{L^{\infty} (Ω)} ‖ Y ‖_{L^{\infty} (Ω)} | F_{m} | \\ ⩽ & ‖ Y ‖_{L^{\infty} (Ω)} ‖ {\bar{C}}_{0} ‖_{L^{\infty} (Ω)} a^{s} \sum_{l = 1}^{[a^{- \frac{2}{3} s}]} \frac{1}{d_{m l}^{'}} \\ + C ‖ Y ‖_{L^{\infty} (Ω)} | F_{m} | (as Φ_{κ} (z_{m}, \cdot) is not singular in F_{m}) \\ ⩽ & ‖ Y ‖_{L^{\infty} (Ω)} ‖ {\bar{C}}_{0} ‖_{L^{\infty} (Ω)} a^{s} \sum_{l = 1}^{[a^{- \frac{1}{3} s}]} [{(2 n + 1)}^{2} - {(2 n - 1)}^{2}] (\frac{1}{n (2^{- \frac{1}{3}} a^{\frac{s}{3}} - \frac{a}{2})}) + C a^{\frac{1 - s}{2}} a^{\frac{1}{3} s} \\ (2.24) & = & O (a^{\frac{3 - s}{6}}) . \end{array}$
□

From these estimates of A, B and D, we deduce that: $\begin{array}{rcl} (2.25) & Y (z_{m}) + \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} Φ_{κ} (z_{m}, z_{j}) {\bar{C}}_{j} Y (z_{j}) a = U^{i} (z_{m}, θ) + O (a^{\frac{9 - 7 s}{6}} + a^{\frac{(21 - 3 η) - (17 - 3 η) s}{12}}) . \end{array}$

Taking the difference between (2.12) and (2.25) produces the algebraic system $\begin{array}{rcl} (Y_{m} - Y (z_{m})) + \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} Φ_{κ} (z_{m}, z_{j}) {\bar{C}}_{j} (Y_{j} - Y (z_{j})) a & = & O (a^{\frac{9 - 7 s}{6}} + a^{\frac{(21 - 3 η) - (17 - 3 η) s}{12}}) . \end{array}$

Comparing this system with (2.12) and by using Lemma 2.3, we obtain the estimate $\begin{array}{rcl} (2.26) & \sum_{m = 1}^{M} | Y_{m} - Y (z_{m}) | & = & O (M a^{\frac{9 - 7 s}{6}} + M a^{\frac{(21 - 3 η) - (17 - 3 η) s}{12}}) . \end{array}$

Recalling that $d = a^{t}$ , $M = O (a^{- s})$ with $t, s > 0$ , we have the following approximation of the far-field from the Foldy–Lax asymptotic expansion (1.7) and from the definitions $Y_{m} : = - \frac{Q_{m}}{C_{m}}$ and $C_{m} : = {\bar{C}}_{m} a$ , for $m = 1, \dots, M$ : $U^{\infty} (\hat{x}, θ) = - \bar{C} \sum_{j = 1}^{M} e^{- i κ \hat{x} \cdot z_{j}} Y_{j} a + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t})$ which we rewrite as $\begin{array}{rcl} U^{\infty} (\hat{x}, θ) & = & - \sum_{j = 1}^{[a^{- s}]} e^{- i κ \hat{x} \cdot z_{j}} \bar{C} (K + 1) Y_{j} a \\ (2.27) & + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) . \end{array}$ Consider the far-field $\begin{array}{rcl} (2.28) & U_{a}^{\infty} (\hat{x}, θ) & = & - a^{1 - s} \int_{Ω} e^{- i κ \hat{x} \cdot y} {\bar{C}}_{0} (y) (K + 1) Y (y) d y . \end{array}$ Taking the difference between (2.28) and (2.27) gives us $\begin{array}{l} U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) \\ = a^{1 - s} [\int_{Ω ∖ ⋃_{j = 1}^{[a^{- 1}]} Ω_{j}} e^{- i κ \hat{x} \cdot y} {\bar{C}}_{0} (y) Y (y) d y + \sum_{j = 1}^{[a^{- 1}]} \int_{Ω_{j}} e^{- i κ \hat{x} \cdot y} {\bar{C}}_{0} (y) (K + 1) Y (y) d y \\ - \sum_{j = 1}^{[a^{- s}]} e^{- i κ \hat{x} \cdot z_{j}} \bar{C} (K + 1) Y_{j} a^{s}] \\ + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) \\ = \bar{C} (K + 1) a^{1 - s} [\sum_{j = 1}^{[a^{- s}]} \int_{Ω_{j}} [e^{- i κ \hat{x} \cdot y} Y (y) - e^{- i κ \hat{x} \cdot z_{j}} Y (z_{j})] d y] \\ + \bar{C} (K + 1) a [\sum_{j = 1}^{[a^{- s}]} e^{- i κ \hat{x} \cdot z_{j}} [Y (z_{j}) - Y_{j}]] + a^{1 - s} \int_{Ω ∖ ⋃_{j = 1}^{[a^{- s}]} Ω_{j}} e^{- i κ \hat{x} \cdot y} {\bar{C}}_{0} (y) Y (y) d y \\ + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) \\ \underset{(2.26)}{=} \bar{C} (K + 1) a^{1 - s} [\sum_{j = 1}^{[a^{- s}]} \int_{Ω_{j}} [e^{- i κ \hat{x} \cdot y} Y (y) - e^{- i κ \hat{x} \cdot z_{j}} Y (z_{j})] d y] \\ + \bar{C} (K + 1) a O (M a^{\frac{9 - 7 s}{6}} + M a^{\frac{(21 - 3 η) - (17 - 3 η) s}{12}}) \\ + O (a^{1 - s} {‖ {\bar{C}}_{0} (K + 1) ‖}_{L^{\infty} (Ω)} ‖ Y ‖_{L^{\infty} (Ω)} | Ω ∖ ⋃_{j = 1}^{[a^{- s}]} Ω_{j} |) \\ + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) \\ \underset{(2.16)}{=} \bar{C} (K + 1) a^{1 - s} [\sum_{j = 1}^{M} \int_{Ω_{j}} [e^{- i κ \hat{x} \cdot y} Y (y) - e^{- i κ \hat{x} \cdot z_{j}} Y (z_{j})] d y] \\ + \bar{C} a O (M a^{\frac{9 - 7 s}{6}} + M a^{\frac{(21 - 3 η) - (17 - 3 η) s}{12}}) + O (a^{\frac{9 - 7 s}{6}}) \\ (2.29) & + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) . \end{array}$

By following the similar computations as it was done in (2.20)–(2.22), we can estimate the quantity ‘ $\sum_{j = 1}^{[a^{- s}]} \int_{Ω_{j}} [e^{- i κ \hat{x} \cdot y} Y (y) - e^{- i κ \hat{x} \cdot z_{j}} Y (z_{j})] d y$ ’ by $O (a^{\frac{3 - s}{6}} + a^{\frac{(9 - 3 η) - (5 - 3 η) s}{12}})$ . Hence, we obtain $\begin{array}{l} U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) \\ = O (a^{\frac{9 - 7 s}{6}} + a^{\frac{(21 - 3 η) - (17 - 3 η) s}{12}} + a^{\frac{15 - 13 s}{6}} + a^{\frac{(33 - 3 η) - (29 - 3 η) s}{12}} \\ (2.30) & + a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) \end{array}$

Due to the fact that $d = a^{t}$ is the minimum distance between the small bodies, we will have $t ⩾ \frac{s}{3}$ . Otherwise, we will have contradiction as the volume of the collection of obstacles, which is of order $a^{- s} {(\frac{a}{2} + \frac{d}{2})}^{3}$ , explodes. From (1.6), we also have that $\begin{array}{l} (2.31) & \frac{s}{3} ⩽ t < 1 and 0 ⩽ s < min {3 - 3 t, \frac{1}{2} (3 - t)} . \end{array}$ We observe that for the particular case where $s = 1$ , we have $\begin{matrix} U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) = O (a^{\frac{1}{3}} + a^{1 - t} + a^{2 - 3 t}) = O (a^{\frac{1}{3}} + a^{2 - 3 t}), \frac{1}{3} ⩽ t < \frac{2}{3} . \end{matrix}$ Under the general conditions $\begin{array}{l} (2.32) & \frac{s}{3} ⩽ t < 1 and 0 ⩽ s < min {\frac{33 - 3 η}{29 - 3 η}, 2 - t, 3 - 3 t, \frac{1}{2} (3 - t), \frac{4 - t}{3}, \frac{4 - 3 t}{2}} \end{array}$ we have $\begin{array}{rcl} U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) & = & O (a^{\frac{(33 - 3 η) - (29 - 3 η) s}{12}} + a^{4 - 3 s - t} + a^{4 - 2 s - 3 t}) \\ (2.33) & = & \{\begin{matrix} O (a^{\frac{(33 - 3 η) - (29 - 3 η) s}{12}} + a^{4 - 2 s - 3 t}) & if s < 2 t; \\ O (a^{\frac{(33 - 3 η) - (29 - 3 η) s}{12}} + a^{\frac{8 - 7 s}{2}}) & if s = 2 t; \\ O (a^{\frac{(33 - 3 η) - (29 - 3 η) s}{12}} + a^{4 - 3 s - t}) & if s > 2 t . \end{matrix} \end{array}$ As η can be taken arbitrary, we can rewrite this result as follows. Under the general conditions $\begin{array}{l} (2.34) & \frac{s}{3} ⩽ t < 1 and 0 ⩽ s < min {{(\frac{33}{29})}_{-}, 3 - 3 t, \frac{4 - t}{3}, \frac{4 - 3 t}{2}} \end{array}$ we have $\begin{array}{l} (2.35) & U^{\infty} (\hat{x}, θ) - U_{a}^{\infty} (\hat{x}, θ) = O (a^{\frac{{(33 - 29 s)}_{-}}{12}} + a^{4 - 3 s - t} + a^{4 - 2 s - 3 t}) . \end{array}$
2.2. Proof of Proposition 2.2

2.2.1. Reduction to a semiclassical type estimate

Let us set $h : = a^{\frac{s - 1}{2}}$ , $V_{0} : = C_{0} (K + 1)$ , $u^{i} : = u^{i} (x, κ) : = e^{i κ (x \cdot d)}$ and $u^{t} : = U_{a}^{t}$ , then the scattering problem (2.1)–(2.2)–(2.3) reduces to $\begin{array}{l} (2.36) & (Δ + κ^{2} - h^{- 2} V_{0}) u^{t} = 0 in R^{3}, \\ (2.37) & u^{t} = u^{s} + e^{i κ x \cdot θ} \\ (2.38) & \frac{\partial u_{s}^{t}}{\partial | x |} - i κ u^{s} = o (\frac{1}{| x |}), | x | \to \infty . \end{array}$ The solution of this scattering problem satisfies the following Lippmann–Schwinger integral equation $\begin{matrix} (2.39) & u^{t} (\cdot, κ) = u^{i} (\cdot, κ) - \frac{1}{h^{2}} \int_{Ω} Φ_{κ} (\cdot, y) V_{0} u^{t} (y, κ) d y . \end{matrix}$ Using (2.39) and the large-x behavior of $Φ_{κ} (x, y)$ , the radiation condition $\begin{matrix} (2.40) & | x | (\hat{x} \cdot \nabla - i κ) u^{s} = o (\frac{1}{| x |}) \end{matrix}$ follows for the scattered field $u^{s} = u^{t} - u^{i}$ . Conversely, any solution of (2.39) is a solution of (2.36)–(2.38). As the potential $h^{- 2} V_{0}$ is real valued, then standard arguments based on the Fredholm alternative show that (2.36)–(2.38) has one and only one solution which is in $H_{loc}^{2} (R^{3})$ , see [10] for instance.

The proof of Proposition 2.2 is reduced to the proof of the following property:

Proposition 2.6.
Let $\partial Ω$ be of class $C^{1, 1}$ . There exists a constant $κ_{0} (Ω)$ depending only on Ω, such that if $\begin{matrix} (2.41) & κ ⩽ κ_{0} (Ω) \end{matrix}$ we have the following estimate $\begin{matrix} (2.42) & ‖ {u^{t} |_{\partial Ω}‖}_{H^{t} (\partial Ω)} ⩽ C h^{1 / 2 - t}, t \in [0, 1 / 2], \end{matrix}$ with $C > 0$ .

Indeed, assuming that Proposition 2.6 is valid and taking $t = 0$ in (2.42), we have the estimate $‖ u^{t} |_{\partial Ω} ‖_{L^{2} (\partial Ω)} = O (h^{\frac{1}{2}}) = O (a^{\frac{s - 1}{4}})$ , as $a ≪ 1$ . As $u^{s}$ satisfies $(Δ + κ^{2}) u^{s} = 0$ in $R^{3} ∖ \bar{Ω}$ , the radiation conditions with $‖ u^{s} (x) + e^{i κ x \cdot θ} |_{\partial Ω} ‖_{L^{2} (\partial Ω)} = O (a^{\frac{s - 1}{4}})$ as $a ≪ 1$ , then the well-posedness of the forward scattering problem in the exterior domain $R^{3} ∖ \bar{Ω}$ implies that the corresponding far-fields satisfy the estimate $\begin{matrix} (2.43) & | U^{\infty} (\hat{x}, θ) - U_{D}^{\infty} (\hat{x}, θ) | = O (a^{\frac{s - 1}{4}}) as a ≪ 1 . \end{matrix}$
2.2.2. Proof of Proposition 2.6

The starting point is the following Lippmann–Schwinger equation $\begin{matrix} (2.44) & (h^{2} + R (κ)) u^{t} = h^{2} u^{i} in Ω \end{matrix}$ where $\begin{matrix} (2.45) & R (κ) u^{t} (x) : = \int_{Ω} Φ_{κ} (x, y) V_{0} u^{t} (y) d y . \end{matrix}$ We write $R (κ) : = R (0) + P (κ)$ , where now $\begin{array}{l} (2.46) & \begin{aligned} R (0) u^{t} : = \int_{Ω} Φ_{0} (x, y) V_{0} u^{t} (y) d y and \\ P (κ) u^{t} (y) : = \int_{Ω} (Φ_{κ} (x, y) - Φ_{0} (x, y)) V_{0} u^{t} (y) d y \end{aligned} \end{array}$ with $Φ_{0} (x, y) : = {(4 π | x - y |)}^{- 1}$ . We rewrite the equation (2.44) as: $\begin{matrix} (2.47) & (h^{2} + R (0)) u^{t} = h^{2} u^{i} - P (κ) u^{t} . \end{matrix}$ Let $u_{1}^{t}$ be defined as the solution of the equation $\begin{matrix} (2.48) & (h^{2} + R (0)) u_{1}^{t} = h^{2} u^{i} \end{matrix}$ and $u_{2}^{t}$ be the one of the equation $\begin{matrix} (2.49) & (h^{2} + R (0)) u_{2}^{t} = - P (κ) u^{t} . \end{matrix}$ It is clear that $u^{t} = u_{1}^{t} + u_{2}^{t}$ . The estimates of both $u_{1}^{t}$ and $u_{2}^{t}$ are based on the following properties of the operator $R (0)$ .

Proposition 2.7.
Let $(λ_{n}, e_{n})$ be the sequence of the eigenelements of the self-adjoint and compact operator $R (0) : L^{2} (Ω) ⟶ L^{2} (Ω)$ . We have the characterization $\begin{matrix} (2.50) & u \in H^{1} (Ω) ⟺ \sum_{n} λ_{n}^{- 1} {(u, e_{n})}_{L^{2} (Ω)}^{2} < \infty \end{matrix}$ and there exist two positive constants $c (Ω)$ and $C (Ω)$ depending only on Ω such that for every $u \in H^{1} (Ω)$ $\begin{matrix} (2.51) & c (Ω) ‖ u ‖_{H^{1} (Ω)} ⩽ {(\sum_{n} λ_{n}^{- 1} V_{0} {(u, e_{n})}_{L^{2} (Ω)}^{2})}^{\frac{1}{2}} ⩽ C (Ω) ‖ u ‖_{H^{1} (Ω)} \end{matrix}$ and $\begin{matrix} (2.52) & {(C (Ω))}^{- 1} ‖ u ‖_{{(H^{1} (Ω))}^{'}} ⩽ {(\sum_{n} λ_{n} V_{0}^{- 1} {(u, e_{n})}_{L^{2} (Ω)}^{2})}^{\frac{1}{2}} ⩽ {(c (Ω))}^{- 1} ‖ u ‖_{{(H^{1} (Ω))}^{'}} \end{matrix}$ where we used the scalar product ${(u, v)}_{L^{2}} : = \int_{Ω} u (x) v (x) d x$ .
Proof.
We observe that if ${(λ_{n}, e_{n})}_{n \in N}$ are the eigenvalues and eigenfunctions of the (Newtonian potential) operator $R (0)$ then ${(λ_{n}^{- 1} V_{0}, e_{n})}_{n \in N}$ are exactly the ones of the problem $\begin{matrix} (2.53) & - Δ e_{n} = λ_{n}^{- 1} V_{0} e_{n}, in Ω and \partial_{ν} e_{n} = S^{- 1} (- \frac{1}{2} I + K) e_{n}, on \partial Ω \end{matrix}$ where S is the single layer operator $S f (x) : = \int_{\partial Ω} Φ_{0} (x, y) f (y) d s (y)$ and K the double layer operator $K f (x) : = \int_{\partial Ω} \partial_{ν (y)} Φ_{0} (x, y) f (y) d s (y)$ with ν as the exterior unit normal to Ω. We recall that the operators S and K are well defined, linear and bounded from $H^{- \frac{1}{2}} (\partial Ω)$ to $H^{\frac{1}{2}} (\partial Ω)$ and from $H^{\frac{1}{2}} (\partial Ω)$ to itself respectively. In addition, S is invertible. This observation has been made since at least Kac [13]. More details are provided in [14,17]. The reader can get the boundary condition above by simply writing the Green’s formula for $Φ_{0}$ and $e_{n}$ inside Ω and then taking the trace on $\partial Ω$ using the jumps of the double layer operator.

We set $B : = S^{- 1} (- \frac{1}{2} I + K)$ . We define the operator $(A_{0}, D (A_{0}))$ as $D (A_{0}) : = {u \in H^{1} (Ω), Δ u \in L^{2} (Ω) and \partial_{ν} u = B u}$ and $A_{0} u = - Δ u$ for $u \in D (A_{0})$ . We set the corresponding quadratic form $\begin{matrix} (2.54) & a_{0} (u, v) : = \int_{Ω} \nabla u (x) \cdot \nabla v (x) d x - \int_{\partial Ω} Bu (x) v (x) d s (x) . \end{matrix}$ This form is well defined in $H^{1} (Ω) \times H^{1} (Ω)$ . In addition, it is symmetric. Indeed, we have $\begin{matrix} \int_{\partial Ω} Bu (x) v (x) d s (x) & = \int_{\partial Ω} S^{- 1} (- \frac{1}{2} I + K) u (x) v (x) d s (x) \\ = \int_{\partial Ω} (- \frac{1}{2} I + K) u (x) S^{- 1} v (x) d s (x) . \end{matrix}$ We write $u = S S^{- 1} u$ , then as $K S = S K^{}$ , see [11] for instance, where $K^{}$ is the dual of double layer potential K, we derive $\begin{matrix} \int_{\partial Ω} (- \frac{1}{2} I + K) u (x) S^{- 1} v (x) d s (x) & = \int_{\partial Ω} S (- \frac{1}{2} I + K^{}) S^{- 1} u (x) S^{- 1} v (x) d s (x) \\ = \int_{\partial Ω} u (x) S^{- 1} (- \frac{1}{2} I + K) v (x) d s (x) . \end{matrix}$ Hence $\int_{\partial Ω} Bu (x) v (x) d s (x) = \int_{\partial Ω} u (x) Bv (x) d s (x)$ and then $a_{0} (u, v) = a_{0} (v, u)$ , $for u, v \in H^{1} (Ω)$ . Now, we show that $a_{0} (u, u) ⩾ \int_{Ω} | \nabla u |^{2}$ . Indeed, $\begin{matrix} - \int_{\partial Ω} Bu (x) u (x) d s (x) & = \int_{\partial Ω} S^{- 1} (\frac{1}{2} I - K) u (x) u (x) d s (x) \\ = \int_{\partial Ω} u (\frac{1}{2} I - K^{}) S^{- 1} u (x) d s (x) . \end{matrix}$ As the operator $\frac{1}{2} I - K^{}$ is positive definite on $H^{- \frac{1}{2}} (\partial Ω)$ equipped with scalar product $< S u, v >$ , see [11] for instance, we obtain $\begin{matrix} - \int_{\partial Ω} Bu (x) u (x) d s (x) = \int_{\partial Ω} S (S^{- 1} u) (x) (\frac{1}{2} I - K^{}) S^{- 1} u (x) d s (x) ⩾ c_{0} {‖ S^{- 1} u ‖}_{H^{- \frac{1}{2}} (\partial Ω)} \end{matrix}$ with a positive constant $c_{0}$ . Hence $a_{0} (u, u) ⩾ \int_{Ω} | \nabla u |^{2}$ .

Let us define the bilinear form $a_{α} (u, v) : = a_{0} (u, v) + α {(u, v)}_{L^{2} (Ω)}$ for $u, v \in H^{1} (Ω)$ , with a positive constant α. This form is defined on $H^{1} (Ω)$ , symmetric, continuous and coercive. To $a_{α}$ , corresponds a self-adjoint operator with a compact inverse $A_{α} : L^{2} (Ω) ⟶ L^{2} (Ω)$ . From the spectral theory, see [12], we know that the eigenvalues and eigenfunctions ${(λ_{n}^{α}, e_{n}^{α})}_{n \in N}$ of $A_{α}$ define a basis in $L^{2} (Ω)$ and, in addition, we have $\begin{matrix} H^{1} (Ω) = D ({(A_{α})}^{\frac{1}{2}}) . \end{matrix}$ This means that $\begin{matrix} (2.55) & u \in H^{1} (Ω) ⟺ \sum_{n} λ_{n}^{α} {(u, e_{n}^{α})}_{L^{2} (Ω)}^{2} < \infty and ‖ u ‖_{H^{1} (Ω)}^{2} \sim \sum_{n} λ_{n}^{α} {(u, e_{n}^{α})}_{L^{2} (Ω)}^{2} . \end{matrix}$ The constants appearing in the two inequalities of this equivalence might depend on Ω but they are independent on $V_{0}$ . As $A_{0}$ is the operator corresponding to $a_{0}$ , then we have $\begin{matrix} λ_{n}^{α} = λ_{n}^{- 1} V_{0} + α and e_{n}^{α} = e_{n}, for every n . \end{matrix}$ Finally, as $0 < λ_{n}$ for every n, $λ_{n}^{- 1} ⟶ \infty$ , as $n ⟶ \infty$ , and α is arbitrary small, say $α < 1$ , we deduce from (2.55) that $\begin{matrix} u \in H^{1} (Ω) ⟺ \sum_{n} λ_{n}^{- 1} V_{0} {(u, e_{n})}_{L^{2} (Ω)}^{2} < \infty \end{matrix}$ and $\begin{matrix} c (Ω) ‖ u ‖_{H^{1} (Ω)} ⩽ {(\sum_{n} λ_{n}^{- 1} V_{0} {(u, e_{n})}_{L^{2} (Ω)}^{2})}^{\frac{1}{2}} ⩽ C (Ω) ‖ u ‖_{H^{1} (Ω)} . \end{matrix}$ Using the definition of the ${(H^{1} (Ω))}^{'}$ norm, we derive the estimate (2.52). As mentioned above, the constants appearing in these inequalities are independent on $V_{0}$ . □

Estimates of $u_{1}^{t}$ . We recall that $(h^{2} + R (0)) u_{1}^{t} = h^{2} u^{i}$ . By the spectral decomposition, we write $u_{1}^{t} = \sum_{n} (u_{1}^{t}, e_{n}) e_{n}$ , then $\begin{matrix} (2.56) & (u_{1}^{t}, e_{n}) = \frac{h^{2}}{h^{2} + λ_{n}} (u^{i}, e_{n}) . \end{matrix}$ Multiplying both sides by $λ_{n}^{- \frac{1}{2}}$ , and as $λ_{n} > 0$ , for every n, we get $\begin{matrix} \sum_{n} λ_{n}^{- 1} V_{0} {| (u_{1}^{t}, e_{n}) |}^{2} ⩽ \sum_{n} λ_{n}^{- 1} V_{0} {| (u^{i}, e_{n}) |}^{2} \end{matrix}$ By Proposition 2.7, we deduce that $\begin{matrix} (2.57) & {‖ u_{1}^{t} ‖}_{H^{1} (Ω)} ⩽ {(c (Ω))}^{- 1} C (Ω) {‖ u^{i} ‖}_{H^{1} (Ω)} . \end{matrix}$ In addition, from (2.56), we have $| (u_{1}^{t}, e_{n}) | ⩽ h^{2} λ_{n}^{- 1} | (u^{i}, e_{n}) |$ and then $\begin{matrix} V_{0}^{2} \sum_{n} λ_{n} V_{0}^{- 1} {| (u_{1}^{t}, e_{n}) |}^{2} ⩽ h^{4} \sum_{n} λ_{n}^{- 1} V_{0} {| (u^{i}, e_{n}) |}^{2} . \end{matrix}$ As $‖ u_{1}^{t} ‖_{{(H^{1} (Ω))}^{'}}^{2} ⩽ C^{2} (Ω) \sum_{n} λ_{n} V_{0}^{- 1} | (u_{1}^{t}, e_{n}) |^{2}$ , this means that $\begin{matrix} (2.58) & {‖ u_{1}^{t} ‖}_{{(H^{1} (Ω))}^{'}} ⩽ C {(Ω)}^{2} V_{0}^{- 1} h^{2} {‖ u^{i} ‖}_{H^{1} (Ω)} = O (V_{0}^{- 1} h^{2}), h ≪ 1 . \end{matrix}$

Estimates of $u_{2}^{t}$ . We start with the equation $(h^{2} + R (0)) u_{2}^{t} = - P (κ) u^{t}$ . By the spectral decomposition, we get $\begin{matrix} (2.59) & (u_{2}^{t}, e_{n}) = - \frac{(P (κ) u^{t}, e_{n})}{h^{2} + λ_{n}} . \end{matrix}$ Hence, $| (u_{2}^{t}, e_{n}) | ⩽ λ_{n}^{- 1} | (P (κ) u^{t}, e_{n}) |$ or $λ_{n}^{\frac{1}{2}} | (u_{2}^{t}, e_{n}) | ⩽ λ_{n}^{- \frac{1}{2}} | (P (κ) u^{t}, e_{n}) |$ . But $P (κ) u^{t} = \sum_{m} (u^{t}, e_{m}) P (κ) e_{m}$ as $P (κ)$ is linear and bounded. Hence, $\begin{matrix} λ_{n}^{\frac{1}{2}} | (u_{2}^{t}, e_{n}) | & ⩽ λ_{n}^{- \frac{1}{2}} | \sum_{m} (u^{t}, e_{m}) (P (κ) e_{m}, e_{n}) | \\ ⩽ {(\sum_{m} λ_{m} {(u^{t}, e_{m})}^{2})}^{\frac{1}{2}} {(\sum_{m} λ_{m}^{- 1} λ_{n}^{- 1} {(P (κ) e_{m}, e_{n})}^{2})}^{\frac{1}{2}} \end{matrix}$ and then $\begin{matrix} (2.60) & V_{0}^{2} \sum_{n} λ_{n} V_{0}^{- 1} {| (u_{2}^{t}, e_{n}) |}^{2} ⩽ (\sum_{m} λ_{m} V_{0}^{- 1} {(u^{t}, e_{m})}^{2}) \sum_{n} \sum_{m} λ_{m}^{- 1} V_{0} λ_{n}^{- 1} V_{0} {(P (κ) e_{m}, e_{n})}^{2} . \end{matrix}$ Let us estimate the term $\sum_{n} \sum_{m} λ_{m}^{- 1} V_{0} λ_{n}^{- 1} V_{0} {(P (κ) e_{m}, e_{n})}^{2}$ . First, as $P (k)$ is symmetric, we observe that $\begin{matrix} \sum_{m} λ_{m}^{- 1} V_{0} {(P (κ) e_{m}, e_{n})}^{2} = \sum_{m} λ_{m}^{- 1} V_{0} {(P (κ) e_{n}, e_{m})}^{2} ⩽ C^{2} (Ω) {‖ P (κ) e_{n} ‖}_{H^{1} (Ω)}^{2} . \end{matrix}$ Then, we have $\begin{matrix} \sum_{n} \sum_{m} λ_{m}^{- 1} V_{0} λ_{n}^{- 1} V_{0} {(P (κ) e_{m}, e_{n})}^{2} ⩽ C^{2} (Ω) \sum_{n} λ_{n}^{- 1} V_{0} {‖ P (κ) e_{n} ‖}_{H^{1} (Ω)}^{2} . \end{matrix}$ But $\begin{array}{rcl} \sum_{n} λ_{n}^{- 1} V_{0} {‖ P (κ) e_{n} ‖}_{H^{1} (Ω)}^{2} & = & \sum_{n} λ_{n}^{- 1} V_{0} \int_{Ω} {(\int_{Ω} p_{κ} (x, y) e_{n} (y) d y)}^{2} d x \\ + \sum_{n} λ_{n}^{- 1} V_{0} \int_{Ω} {(\int_{Ω} \nabla_{x} p_{κ} (x, y) e_{n} (y) d y)}^{2} d x \end{array}$ recalling that $P (κ) e_{n} (x) : = \int_{Ω} p_{κ} (x, y) e_{n} (y) V_{0} d y$ where $p_{κ} (x, y) : = Φ_{κ} (x, y) - Φ_{0} (x, y)$ . Then by Lebesgue dominated convergence Theorem, we get $\begin{array}{rcl} \sum_{n} λ_{n}^{- 1} V_{0} {‖ P (κ) e_{n} ‖}_{H^{1} (Ω)}^{2} & = & \int_{Ω} \sum_{n} λ_{n}^{- 1} V_{0} {(\int_{Ω} p_{κ} (x, y) e_{n} (y) V_{0} d y)}^{2} d x \\ + \int_{Ω} \sum_{n} λ_{n}^{- 1} V_{0} {(\int_{Ω} \nabla_{x} p_{κ} (x, y) e_{n} (y) V_{0} d y)}^{2} d x \\ ⩽ & C^{2} (Ω) V_{0}^{2} [\int_{Ω} {‖ p_{κ} (x, \cdot) ‖}_{H^{1} (Ω)}^{2} d x + \int_{Ω} {‖ \nabla_{x} p_{κ} (x, \cdot) ‖}_{H^{1} (Ω)}^{2} d x] \\ = : & C^{2} (Ω) C^{2} (κ) V_{0}^{2}, \end{array}$ where we have set $\begin{matrix} (2.61) & C (κ) : = \int_{Ω} {‖ [Φ_{κ} - Φ_{0}] (x, \cdot) ‖}_{H^{1} (Ω)}^{2} d x + \int_{Ω} {‖ \nabla_{x} [Φ_{κ} - Φ_{0}] (x, \cdot) ‖}_{H^{1} (Ω)}^{2} d x \end{matrix}$ as $p_{κ} (x, y) = Φ_{κ} (x, y) - Φ_{0} (x, y)$ . Observe that $p_{κ}$ satisfies $Δ_{y} p_{κ} (x, y) = - κ^{2} Φ (x, y)$ and as $Φ (x, \cdot)$ is in $L_{loc}^{2} (R^{3})$ , then by interior estimates, we deduce that $p_{κ} (x, \cdot)$ is in $H_{loc}^{2} (R^{3})$ and $‖ p_{κ} (x, \cdot) ‖_{H^{2} (Ω)}$ is uniformly bounded with respect to $x \in Ω$ . Finally as $\nabla_{x} p_{κ} (x, y) = - \nabla_{y} p_{κ} (x, y)$ , then $C (κ)$ makes sense, i.e., it is finite. Then from (2.60), we deduce that $\begin{matrix} (2.62) & {‖ u_{2}^{t} ‖}_{{(H^{1} (Ω))}^{'}} ⩽ {(c (Ω))}^{- 1} C^{2} (Ω) C (κ) {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} . \end{matrix}$ Let us now go back to the relation $(u_{2}^{t}, e_{n}) = - \frac{(P (κ) u^{t}, e_{n})}{h^{2} + λ_{n}}$ and derive the estimate $| (u_{2}^{t}, e_{n}) | ⩽ \frac{| (P (κ) u^{t}, e_{n}) |}{h^{2}}$ or $λ_{n}^{- \frac{1}{2}} | (u_{2}^{t}, e_{n}) | ⩽ λ_{n}^{- \frac{1}{2}} \frac{| (P (κ) u^{t}, e_{n}) |}{h^{2}}$ and then $\begin{matrix} \sum_{n} λ_{n}^{- 1} V_{0}^{- 1} | (u_{2}^{t}, e_{n}) | ⩽ h^{- 4} \sum_{n} λ_{n}^{- 1} V_{0}^{- 1} | (P (κ) u^{t}, e_{n}) | \end{matrix}$ which means that $\begin{matrix} {‖ u_{2}^{t} ‖}_{H^{1} (Ω)} ⩽ {(c (Ω))}^{- 1} C (Ω) h^{- 2} {‖ P (κ) u^{t} ‖}_{H^{1} (Ω)} . \end{matrix}$ Let us now estimate $‖ P (κ) u^{t} ‖_{H^{1} (Ω)}$ . From the equality $\begin{matrix} {‖ P (κ) u^{t} ‖}_{H^{1} (Ω)}^{2} = \int_{Ω} {(\int_{Ω} p_{κ} (x, y) u^{t} (y) d y)}^{2} d x + \int_{Ω} {(\int_{Ω} \nabla_{x} p_{κ} (x, y) u^{t} (y) d y)}^{2} d x \end{matrix}$ and the fact that $\begin{matrix} | \int_{Ω} p_{κ} (x, y) u^{t} (y) d y | ⩽ {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} {‖ p_{κ} (x, \cdot) ‖}_{H^{1} (Ω)} \end{matrix}$ and $\begin{matrix} | \int_{Ω} \nabla_{x} p_{κ} (x, y) u^{t} (y) d y | ⩽ {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} {‖ \nabla_{x} p_{κ} (x, \cdot) ‖}_{H^{1} (Ω)} \end{matrix}$ we have $\begin{matrix} {‖ P (κ) u^{t} ‖}_{H^{1} (Ω)}^{2} & ⩽ [\int_{Ω} {‖ p_{κ} (x, \cdot) ‖}_{H^{1} (Ω)}^{2} d x + \int_{Ω} {‖ \nabla_{x} p_{κ} (x, \cdot) ‖}_{H^{1} (Ω)}^{2} d x] {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}}^{2} \\ = C (κ) {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}}^{2} . \end{matrix}$ From the singularities of the function $p_{κ} (\cdot, \cdot)$ , as described above, we know that the two quantities $\int_{Ω} ‖ p_{κ} (x, \cdot) ‖_{H^{1} (Ω)}^{2} d x$ and $\int_{Ω} ‖ \nabla_{x} p_{κ} (x, \cdot) ‖_{H^{1} (Ω)}^{2} d x$ are finite.

Finally, we have $\begin{matrix} {‖ u_{2}^{t} ‖}_{H^{1} (Ω)} ⩽ {(c (Ω))}^{- 1} C (Ω) C (κ) h^{- 2} {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} . \end{matrix}$

Estimates of $u^{t}$ . We have shown so far that $\begin{array}{l} {‖ u_{1}^{t} ‖}_{H^{1} (Ω)} ⩽ {(c (Ω))}^{- 1} C (Ω) {‖ u^{i} ‖}_{H^{1} (Ω)} = O (1) and \\ {‖ u_{1}^{t} ‖}_{{(H^{1} (Ω))}^{'}} ⩽ C^{2} (Ω) V_{0}^{- 1} h^{2} {‖ u^{i} ‖}_{H^{1} (Ω)} = O (V_{0}^{- 1} h^{2}), \\ {‖ u_{2}^{t} ‖}_{H^{1} (Ω)} ⩽ {(c (Ω))}^{- 1} C (Ω) C (κ) h^{- 2} {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} and \\ {‖ u_{2}^{t} ‖}_{{(H^{1} (Ω))}^{'}} ⩽ {(c (Ω))}^{- 1} C^{2} (Ω) C (κ) {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} \end{array}$ From these estimates, we deduce that $\begin{matrix} {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} & ⩽ {‖ u_{1}^{t} ‖}_{{(H^{1} (Ω))}^{'}} + {‖ u_{2}^{t} ‖}_{{(H^{1} (Ω))}^{'}} \\ ⩽ C^{2} (Ω) V_{0}^{- 1} h^{2} {‖ u^{i} ‖}_{H^{1} (Ω)} + {(c (Ω))}^{- 1} C^{2} (Ω) C (κ) {‖ u^{t} ‖}_{{(H^{1} (Ω))}^{'}} . \end{matrix}$ Hence under the condition on κ, recalling (2.61), $\begin{matrix} (2.63) & C (κ) < \frac{c (Ω)}{C^{2} (Ω)} \end{matrix}$ we derive the estimate $‖ u^{t} ‖_{{(H^{1} (Ω))}^{'}} ⩽ \frac{C^{2} (Ω) h^{2} ‖ u^{i} ‖_{H^{1} (Ω)}}{1 - \frac{C^{2} (Ω) C (κ)}{c (Ω)}}$ . Using this estimate, we derive the following one $\begin{matrix} {‖ u^{t} ‖}_{H^{1} (Ω)} & ⩽ {‖ u_{1}^{t} ‖}_{H^{1} (Ω)} + {‖ u_{2}^{t} ‖}_{H^{1} (Ω)} \\ ⩽ {(c (Ω))}^{- 1} C (Ω) {‖ u^{i} ‖}_{H^{1} (Ω)} + {(c (Ω))}^{- 1} C (Ω) C (κ) h^{- 2} \frac{h^{2} C^{2} (Ω) ‖ u^{i} ‖_{H^{1} (Ω)}}{1 - \frac{C^{2} (Ω) C (κ)}{c (Ω)}} \end{matrix}$ and then $\begin{matrix} {‖ u^{t} ‖}_{H^{1} (Ω)} ⩽ {(c (Ω))}^{- 1} C (Ω) [1 + \frac{C^{2} (Ω) C (κ) c (Ω)}{c (Ω) - C^{2} (Ω) C (κ)}] {‖ u^{i} ‖}_{H^{1} (Ω)} . \end{matrix}$

With this estimate, we have also the ones of the traces $‖ u^{t} ‖_{H^{\frac{1}{2}} (\partial Ω)} = O (1)$ . In addition, observe that $(Δ + κ^{2}) u^{t} = 0$ in $R^{3} ∖ \overline{Ω}$ , recalling that $u^{t} = u^{s} + u^{i}$ and $u^{s}$ satisfies the radiation condition. From the well-posedness of this exterior problem stated with Dirichlet boundary condition on the boundary $\partial Ω$ , and the previous estimate, i.e., $‖ u^{t} ‖_{H^{\frac{1}{2}} (\partial Ω)} = O (1)$ , we deduce that $‖ \partial_{ν} u^{t} ‖_{H^{- \frac{1}{2}} (\partial Ω)} = O (1)$ , as $h ≪ 1$ . By an integration by parts, we have $\begin{matrix} \int_{Ω} {| \nabla u^{t} (x) |}^{2} + (h^{- 2} V_{0} - κ^{2}) {| u^{t} (x) |}^{2} d x ⩽ {‖ \partial_{ν} u^{t} ‖}_{H^{- \frac{1}{2}} (\partial Ω)} {‖ u^{t} ‖}_{H^{\frac{1}{2}} (\partial Ω)} = O (1), h ≪ 1 . \end{matrix}$ Hence $‖ u^{t} ‖_{L^{2} (Ω)} = O (h)$ , $h ≪ 1$ . By interpolation, we deduce that $\begin{matrix} {‖ u^{t} ‖}_{H^{t} (Ω)} = O (h^{1 - t}), h ≪ 1, for t \in [0, 1] \end{matrix}$ and, by the trace operator estimates, that $\begin{matrix} {‖ u^{t} ‖}_{H^{t} (\partial Ω)} = O (h^{\frac{1}{2} - t}), h ≪ 1, for t ⩽ \frac{1}{2} . \end{matrix}$
3. Proof of Proposition 1.2

Using a single layer representation of the solution $\begin{matrix} u^{s} (x, d) : = \sum_{j = 1}^{M} \int_{\partial D_{j}} Φ_{κ} (x, s) σ_{j} (s) d s, \end{matrix}$ one can write the Dirichlet boundary condition in a compact form as $\begin{array}{rcl} (3.1) & (L + K) σ = - U^{I} \end{array}$ where $L : = {(L_{m j})}_{m, j = 1}^{M}$ and $K : = {(K_{m j})}_{m, j = 1}^{M}$ , where $\begin{array}{l} (3.2) & L_{m j} = \{\begin{matrix} S_{m j} & m = j, \\ 0 & else, \end{matrix} K_{m j} = \{\begin{matrix} S_{m j} & m \neq j, \\ 0 & else, \end{matrix} \\ (3.3) & U^{I} = U^{I} (s_{1}, \dots, s_{M}) : = {(U^{i} (s_{1}), \dots, U^{i} (s_{M}))}^{T}, and \\ (3.4) & σ = σ (s_{1}, \dots, s_{M}) : = {(σ_{1} (s_{1}), \dots, σ_{M} (s_{M}))}^{T} . \end{array}$ Here, for the indices m and j fixed, $S_{m j}$ is the integral operator acting as $\begin{array}{rcl} (3.5) & S_{m j} (σ_{j}) (t) : = \int_{\partial D_{j}} Φ_{κ} (t, s) σ_{j} (s) d s, t \in \partial D_{m} . \end{array}$

3.1. A priori estimate of σ

We can rewrite (3.1) as $\begin{array}{rcl} (3.6) & (L + \overset{´}{K}) σ = - U^{I} - Φ^{c} σ \end{array}$ where $\begin{array}{l} (3.7) & \begin{aligned} {\overset{´}{K}}_{m j} = \{\begin{matrix} S_{m j}^{'} & m = j, \\ 0 & else, \end{matrix} \\ Φ_{m j}^{c} = \{\begin{matrix} S_{m j}^{c} & m \neq j, \\ 0 & else, \end{matrix} \end{aligned} \end{array}$ with $\begin{array}{l} (3.8) & S_{m j}^{'} (σ_{j}) (t) : = \int_{\partial D_{j}} [Φ_{κ} (t, s) - Φ_{κ} (z_{m}, z_{j})] σ_{j} (s) d s, t \in \partial D_{m} \\ S_{m j}^{c} (σ_{j}) (t) : = \int_{\partial D_{j}} Φ_{κ} (z_{m}, z_{j}) σ_{j} (s) d s, t \in \partial D_{m} \\ (3.9) & = Φ_{κ} (z_{m}, z_{j}) Q_{j}, t \in \partial D_{m} . \end{array}$

Here $Q_{j}$ are called total charge on each surface $\partial D_{j}$ , associated to the surface charge distributions $σ_{j}$ and are defined as $\begin{array}{rcl} (3.10) & Q_{j} : = \int_{\partial D_{j}} σ_{j} (s) d s and Q : = {(Q_{1}, Q_{2}, \dots, Q_{M})}^{T} . \end{array}$

Further, (3.6) can be written as, $\begin{array}{rcl} (3.11) & L_{0} σ = - U^{I} - Φ^{c} Q - \overset{´}{K} σ - (L - L_{0}) σ \end{array}$

Here, $L_{0}$ is defined same as the matrix operator $L$ but for the zero frequency by denoting the corresponding single layer operators by $S_{m j}^{0}$ . By making use of invertibility of $L_{0}$ , we can rewrite (3.11) as below: $\begin{array}{rcl} (3.12) & σ = - L_{0}^{- 1} U^{I} - L_{0}^{- 1} Φ^{c} Q - L_{0}^{- 1} \overset{´}{K} σ + L_{0}^{- 1} (L - L_{0}) σ . \end{array}$ Observe that $‖ S_{m m} - S_{m m}^{0} ‖_{L (L^{2} (\partial D_{m}), H^{1} (\partial D_{m}))}$ behaves as $O (a^{2})$ , and hence the norm $‖ L - L_{0} ‖_{L (\prod_{m = 1}^{M} L^{2} (\partial D_{m}), \prod_{m = 1}^{M} H^{1} (\partial D_{m}))}$ also behaves as $O (a^{2})$ . Next, we show that $‖ {(S_{m m}^{0})}^{- 1} S_{m j}^{'} σ_{j} ‖_{L^{2} (\partial D_{m})}$ , for $m \neq j$ behaves as $O (\frac{a^{2}}{d_{m j}^{2}}) = C^{'} \frac{a^{2}}{d_{m j}^{2}}$ (a uniform constant). Indeed, using the analyticity (or the Taylor series at the order $l + 1$ if $t ⩽ \frac{l}{l + 1}$ , see Remark 3.1 below) of $Φ_{κ}$ in ${\bar{D}}_{m} \times {\bar{D}}_{j}$ for $m \neq j$ around $z_{m}$ and then the Taylor series of first order for the first term around $z_{l}$ , we can write $\begin{array}{rcl} Φ_{κ} (t, s) & = & Φ_{κ} (z_{m}, s) + (t - z_{m}) \cdot \nabla_{t} Φ_{κ} (z_{m}, s) + \sum_{| η | = 1}^{\infty} \frac{D_{x}^{η} Φ_{κ} (z_{m}, s)}{η!} {(t - z_{m})}^{η} \\ = & Φ_{κ} (z_{m}, z_{j}) + (s - z_{j}) \cdot \int_{0}^{1} \nabla_{s} Φ_{κ} (z_{m}, s - α (s - z_{j})) d α + (t - z_{m}) \cdot \nabla_{t} Φ_{κ} (z_{m}, s) \\ (3.13) & + \sum_{| η | = 2}^{\infty} \frac{D_{x}^{η} Φ_{κ} (z_{m}, s)}{η!} {(t - z_{m})}^{η} . \end{array}$ Observe that, the above expansion of $Φ_{κ}$ is due to expansion of infinite order around $z_{m}$ . However, if we just want to use the $(l + 1)$ th order expansion of $Φ_{κ}$ , then the last two terms of the equation (3.13) will be replace by $\sum_{| η | = 1}^{l + 1} \frac{D_{x}^{η} Φ_{κ} (z_{m}, s)}{η!} {(t - z_{m})}^{η} + \sum_{| η | = l + 2} {(t - z_{m})}^{η} \frac{| η |}{η!} \int_{0}^{1} D_{t}^{η} Φ_{κ} (z_{m} + α (t - z_{m}), s) d α$ .

Now write $S_{m m}^{0} ψ = {(t - z_{m})}^{n}$ . Then using scaling, we can have $‖ ψ ‖_{L^{2} (\partial D_{m})} = ϵ ‖ \hat{ψ} ‖_{L^{2} (\partial B_{m})}$ which gives us $‖ {(S_{m m}^{0})}^{- 1} ({(t - z_{m})}^{n}) ‖_{L^{2} (\partial D_{m})} = O (ϵ^{n})$ and it sufficient to prove the claim $‖ {(S_{m m}^{0})}^{- 1} S_{m j}^{'} σ_{j} ‖_{L^{2} (\partial D_{m})} = O (\frac{a^{2}}{d_{m j}^{2}})$ . Indeed, $\begin{array}{l} {‖ {(S_{m m}^{0})}^{- 1} S_{m j}^{'} σ_{j} ‖}_{L^{2} (\partial D_{m})} \\ \equiv {‖ {(S_{m m}^{0})}^{- 1} (\int_{\partial D_{j}} [Φ_{κ} (\cdot, s) - Φ_{κ} (z_{m}, z_{j})] σ_{j} (s) d s) ‖}_{L^{2} (\partial D_{m})} \\ ⩽ {‖ {(S_{m m}^{0})}^{- 1} (\int_{\partial D_{j}} (s - z_{j}) \cdot [\int_{0}^{1} \nabla_{s} Φ_{κ} (z_{m}, s - α (s - z_{j})) d α] σ_{j} (s) d s) ‖}_{L^{2} (\partial D_{m})} \\ + {‖ {(S_{m m}^{0})}^{- 1} (\int_{\partial D_{j}} [(\cdot - z_{m}) \cdot \nabla_{x} Φ_{κ} (z_{m}, s)] σ_{j} (s) d s) ‖}_{L^{2} (\partial D_{m})} \\ + {‖ {(S_{m m}^{0})}^{- 1} (\int_{\partial D_{j}} [\sum_{| η | = 2}^{\infty} \frac{D_{x}^{η} Φ_{κ} (z_{m}, s)}{η!} {(\cdot - z_{m})}^{η}] σ_{j} (s) d s) ‖}_{L^{2} (\partial D_{m})} \\ (3.14) & = O (\frac{a^{2}}{d_{m j}^{2}}) . \end{array}$

Let us now estimate $‖ L_{0}^{- 1} \overset{´}{K} ‖$ . $\begin{array}{rcl} ‖ L_{0}^{- 1} \overset{´}{K} ‖ & \equiv & max_{1 ⩽ m ⩽ M} \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} {‖ {(S_{m m}^{0})}^{- 1} S_{m j}^{'} ‖}_{L (L^{2} (\partial D_{j}), H^{1} (\partial D_{m}))} \\ ⩽ & max_{1 ⩽ m ⩽ M} \sum_{\begin{array}{c} j = 1 \\ j \neq m \end{array}}^{M} C^{'} \frac{a^{2}}{d_{m j}^{2}} \\ ⩽ & C^{'} M_{max} [\frac{a^{2}}{d^{2}} + \sum_{n = 1}^{[a^{- \frac{s}{3}}]} [{(2 n + 1)}^{3} - {(2 n - 1)}^{3}] {(\frac{a}{n (2^{- \frac{1}{3}} a^{\frac{s}{3}} - \frac{a}{2})})}^{2}] \\ = & C^{'} M_{max} [\frac{a^{2}}{d^{2}} + {(\frac{a}{(2^{- \frac{1}{3}} a^{\frac{s}{3}} - \frac{a}{2})})}^{2} \sum_{n = 1}^{[a^{- \frac{s}{3}}]} [24 + \frac{2}{n^{2}}]] \\ (3.15) & ⩽ & C^{'} M_{max} [a^{2} d^{- 2} + 26 a^{2} a^{- s}] . \end{array}$

From the above discussions and due to the fact that $‖ L_{0}^{- 1} ‖$ behaves as $O (a^{- 1})$ , we can observe that $‖ L_{0}^{- 1} (L - L_{0}) ‖_{L (\prod_{m = 1}^{M} L^{2} (\partial D_{m}), \prod_{m = 1}^{M} H^{1} (\partial D_{m}))}$ and $‖ L_{0}^{- 1} \overset{´}{K} ‖_{L (\prod_{m = 1}^{M} L^{2} (\partial D_{m}), \prod_{m = 1}^{M} L^{2} (\partial D_{m}))}$ behaves as $O (a)$ and $O (a^{2 - s} + a^{2 - 2 t})$ respectively, which enable us to write (3.12) as below; $\begin{array}{rcl} (3.16) & σ = - L_{0}^{- 1} U^{I} - L_{0}^{- 1} Φ^{c} Q + O (a + a^{2 - s} + a^{2 - 2 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} in L^{2} . \end{array}$ Now, for $m = 1, \dots, M$ , by integrating the mth row of the above equation over $\partial D_{m}$ , we obtain: $\begin{array}{rcl} (3.17) & Q_{m} = - \int_{\partial D_{m}} {(L_{0}^{- 1} U^{I} + L_{0}^{- 1} Φ^{c} Q)}_{m} + O (a^{2} + a^{3 - s} + a^{3 - 2 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} . \end{array}$ To make the notation simple, introduce the diagonal matrix integral operator ∫ such that $\int = Diag (\int_{\partial D_{1}}, \dots, \int_{\partial D_{m}})$ . Then, from the above equations we can have: $\begin{array}{rcl} (3.18) & [I + \int L_{0}^{- 1} Φ^{c}] Q = - \int L_{0}^{- 1} U^{I} + O (a^{2} + a^{3 - s} + a^{3 - 2 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} . \end{array}$ Observe that, $\int L_{0}^{- 1} Φ^{c} = Diag (\int_{\partial D_{1}} {(S_{11}^{0})}^{- 1}, \dots, \int_{\partial D_{m}} {(S_{m m}^{0})}^{- 1}) Φ^{c}$ and hence, $\int L_{0}^{- 1} Φ^{c}$ is a off-diagonal matrix operator of size $M \times M$ such that $\begin{matrix} {[\int L_{0}^{- 1} Φ^{c}]}_{p q} = \{\begin{matrix} \int_{\partial D_{p}} {(S_{p p}^{0})}^{- 1} Φ_{κ} (z_{p}, z_{q}), & for p \neq q; \\ 0, & for p = q . \end{matrix} \end{matrix}$ Observe also that, $\int L_{0}^{- 1} Φ^{c} Q = Diag (\int_{\partial D_{1}} {(S_{11})}^{- 1}, \dots, \int_{\partial D_{m}} {(S_{m m})}^{- 1}) Φ^{c} Q$ and hence, $\int L_{0}^{- 1} Φ^{c} Q$ is a vector of length M such that $\begin{array}{rcl} {[\int L_{0}^{- 1} Φ^{c} Q]}_{m} & = & \int_{\partial D_{m}} {(S_{m m}^{0})}^{- 1} (\sum_{\begin{array}{c} q = 1 \\ q \neq m \end{array}}^{M} Φ_{κ} (z_{m}, z_{q}) Q_{q}) \\ = & \sum_{q = 1, q \neq m}^{M} Φ_{κ} (z_{m}, z_{q}) Q_{q} \int_{\partial D_{m}} {(S_{m m}^{0})}^{- 1} (1) (s) d s \\ (3.19) & = & C B Q . \end{array}$ Here, $B$ and C are the off-diagonal and diagonal matrices, respectively. These are defined as $\begin{matrix} B_{n} (p, q) = \{\begin{matrix} Φ_{κ} (z_{p}, z_{q}), & p \neq q; \\ 0, & p = q; \end{matrix} \end{matrix}$ and $C = Diag (C_{1}, \dots, C_{m})$ , with $C_{m} : = \int_{\partial D_{m}} {(S_{m m}^{0})}^{- 1} (1) (s) d s$ . We call these $C_{m}$ ’s as acoustic capacitances. From (3.19) and the fact that $[I + C B] Q = C [C^{- 1} + B] Q$ , we can rewrite (3.18) as: $\begin{array}{rcl} [C^{- 1} + B] Q & = & - \underset{= : Y_{1}}{\underset{︸}{C^{- 1} \int L_{0}^{- 1} U^{I}}} - \underset{= : Y_{2}}{\underset{︸}{C^{- 1} \int [L_{0}^{- 1} \overset{´}{K} σ + L_{0}^{- 1} (L - L_{0}) σ]}} \\ (3.20) & = & O (1) + O (a + a^{2 - s} + a^{2 - 2 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} . \end{array}$ Now, by applying [7, Lemma 2.22] to the system (3.20), we can prove the invertibility of the matrix $[C^{- 1} + B]$ for the conditions “ ${max}_{1 ⩽ m ⩽ M} | C_{m} | < \frac{5 π}{3 γ} d$ and $γ : = {min}_{j \neq m} cos (κ | z_{m} - z_{j} |) ⩾ 0$ ” and derive the estimate $\begin{matrix} (3.21) & \begin{matrix} \sum_{m = 1}^{M} | Q_{m} |^{2} | C_{m} |^{- 1} ⩽ 4 {(1 - \frac{3 γ}{5 π d} max_{1 ⩽ m ⩽ M} | C_{m} |)}^{- 2} \sum_{m = 1}^{M} {| {(Y_{1} + Y_{2})}_{m} |}^{2} | C_{m} | \end{matrix} \end{matrix}$ which gives us: $\begin{matrix} (3.22) & \begin{matrix} \sum_{m = 1}^{M} | Q_{m} | ⩽ 2 {(1 - \frac{3 γ}{5 π d} max_{1 ⩽ m ⩽ M} | C_{m} |)}^{- 1} M max_{1 ⩽ m ⩽ M} | C_{m} | max_{1 ⩽ m ⩽ M} | {(Y_{1} + Y_{2})}_{m} | . \end{matrix} \end{matrix}$

We have the estimates $\begin{array}{l} {‖ L_{0}^{- 1} Φ^{c} Q ‖}_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} \\ = {Max}_{m} {‖ {(S_{m m}^{0})}^{- 1} (\sum_{j = 1, j \neq m}^{M} Φ_{κ} (z_{m}, z_{j}) \int_{\partial D_{j}} σ_{j}) ‖}_{L^{2} (\partial D_{m})} \\ ⩽ {Max}_{m} {‖ {(S_{m m}^{0})}^{- 1} ‖}_{L (H^{1} (\partial D_{m}), L^{2} \partial D_{m})} \sum_{j = 1, j \neq m}^{M} {‖ Φ_{κ} (z_{m}, z_{j}) \int_{\partial D_{j}} σ_{j} ‖}_{H^{1} (\partial D_{m})} \\ ⩽ {Max}_{m} ϵ^{- 1} {‖ {(S_{m m}^{0})}^{- 1} ‖}_{L (H^{1} (\partial B_{m}), L^{2} \partial B_{m})} \sum_{j = 1, j \neq m}^{M} | Φ_{κ} (z_{m}, z_{j}) | | Q_{j} | ‖ 1 ‖_{H^{1} (\partial D_{m})} \\ \underset{(3.22)}{⩽} ({Max}_{m} {‖ {(S_{m m}^{0})}^{- 1} ‖}_{L (H^{1} (\partial B_{m}), L^{2} \partial B_{m})} | \partial B_{m} |^{\frac{1}{2}}) 2 {(1 - \frac{3 γ}{5 π d} max_{1 ⩽ m ⩽ M} | C_{m} |)}^{- 1} \\ \times \frac{M}{d} max_{1 ⩽ m ⩽ M} | C_{m} | max_{1 ⩽ m ⩽ M} | {(Y_{1} + Y_{2})}_{m} | \\ ⩽ ({Max}_{m} {‖ {(S_{m m}^{0})}^{- 1} ‖}_{L (H^{1} (\partial B_{m}), L^{2} \partial B_{m})} | \partial B_{m} |^{\frac{1}{2}}) 2 {(1 - \frac{3 γ}{5 π d} max_{1 ⩽ m ⩽ M} | C_{m} |)}^{- 1} \\ \times M \frac{a}{d} max_{1 ⩽ m ⩽ M} | {(Y_{1} + Y_{2})}_{m} | \\ (3.23) & \underset{(3.20)}{=} O (M \frac{a}{d}) + O (a^{2 - s - t} + a^{3 - 2 s - t} + a^{3 - s - 3 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} . \end{array}$

From, (3.16), we deduce that $\begin{array}{l} ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} \\ ⩽ {‖ L_{0}^{- 1} U^{I} ‖}_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} + {‖ L_{0}^{- 1} Φ^{c} Q ‖}_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} + O (a + a^{2 - s} + a^{2 - 2 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} \\ \underset{(3.23)}{⩽} O (1) + O (M \frac{a}{d}) + O (a^{2 - s - t} + a^{3 - 2 s - t} + a^{3 - s - 3 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} \\ (3.24) & + O (a + a^{2 - s} + a^{2 - 2 t}) ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} . \end{array}$

Hence $‖ σ_{m} ‖_{L^{2} (\partial D_{m})}$ behaves as $O (1 + M \frac{a}{d})$ , if we have a, $a^{2 - s - t}$ , $a^{2 - s}$ , $a^{3 - 2 s - t}$ , $a^{2 - 2 t}$ , $a^{3 - s - 3 t}$ are small enough. Hence, we should have $2 - s - t$ , $2 - s$ , $3 - t - 2 s$ , $2 - 2 t$ and $3 - s - 3 t$ are positive. Due to the fact that $t ⩾ \frac{s}{3}$ , the above inequalities are equivalent to the following conditions on t and s: $\begin{array}{l} (3.25) & \frac{s}{3} ⩽ t ⩽ 1 and 0 ⩽ s ⩽ min {2, 2 - t, 3 - 3 t, \frac{1}{2} (3 - t)} . \end{array}$

3.2. Derivation of the asymptotic expansion

For simplicity, let $C_{σ}$ be the uniform constant such that $‖ σ_{m} ‖_{L^{2} (\partial D_{m})} ⩽ ‖ σ ‖_{\prod_{m = 1}^{M} L^{2} (\partial D_{m})} ⩽ C_{σ} (1 + M \frac{a}{d})$ . Using this, we have the following estimates for the total charge $Q_{m}$ and the far-field $U^{\infty}$ ; $\begin{array}{l} (3.26) & | Q_{m} | ⩽ C_{σ} a (1 + M \frac{a}{d}), for m = 1, \dots, M, \\ (3.27) & U^{\infty} (\hat{x}) = \sum_{m = 1}^{M} [e^{- i κ \hat{x} \cdot z_{m}} Q_{m} + O (κ a^{2} (1 + M \frac{a}{d}))], \end{array}$ with $Q_{m}$ given by (3.10), if $κ a < 1$ where $O (κ a^{2} (1 + M \frac{a}{d})) ⩽ C_{σ} κ (1 + M \frac{a}{d}) a^{2}$ .

Again by making use of $L^{2}$ estimate of σ, we can rewrite (3.20) as below, for $m = 1, \dots, M$ , $\begin{array}{l} \frac{Q_{m}}{C_{m}} + \sum_{j \neq m} C_{j} Φ_{κ} (z_{m}, z_{j}) \frac{Q_{j}}{C_{j}} \\ = - \frac{1}{C_{m}} \int_{\partial D_{m}} {(L_{0}^{- 1} U^{I})}_{m} + O (a + a^{2 - s - t} + a^{2 - s} + a^{3 - 2 s - t} + a^{2 - 2 t} + a^{3 - s - 3 t}) \\ = - \frac{1}{C_{m}} \int_{\partial D_{m}} ({(S_{m m}^{0})}^{- 1} U^{I}) (s_{m}) d s_{m} + O (a + a^{2 - s - t} + a^{2 - s} + a^{3 - 2 s - t} + a^{2 - 2 t} + a^{3 - s - 3 t}) \\ = - \frac{1}{C_{m}} \int_{\partial D_{m}} {(S_{m m}^{0})}^{- 1} (U^{i} (z_{m})) (s_{m}) d s_{m} - \frac{1}{C_{m}} \int_{\partial D_{m}} {(S_{m m}^{0})}^{- 1} [U^{i} (s_{m}) - U^{i} (z_{m})] d s_{m} \\ + O (a + a^{2 - s - t} + a^{2 - s} + a^{3 - 2 s - t} + a^{2 - 2 t} + a^{3 - s - 3 t}) \\ (3.28) & = - U^{i} (z_{m}) + O (a + a^{2 - s - t} + a^{2 - s} + a^{3 - 2 s - t} + a^{2 - 2 t} + a^{3 - s - 3 t}) . \end{array}$

For $m = 1, \dots, M$ , let ${\bar{Q}}_{m}$ be the potentials such that, $\begin{array}{rcl} (3.29) & \frac{{\bar{Q}}_{m}}{C_{m}} + \sum_{j \neq m} C_{j} Φ_{κ} (z_{m}, z_{j}) \frac{{\bar{Q}}_{j}}{C_{j}} & = & - U^{i} (z_{m}) . \end{array}$

Now the difference between (3.28) and (3.29), for $m = 1, \dots, M$ , provides us: $\begin{array}{l} \frac{Q_{m} - {\bar{Q}}_{m}}{C_{m}} + \sum_{j \neq m} C_{j} Φ_{κ} (z_{m}, z_{j}) \frac{Q_{j} - {\bar{Q}}_{j}}{C_{j}} \\ (3.30) & = O (a + a^{2 - s - t} + a^{2 - s} + a^{3 - 2 s - t} + a^{2 - 2 t} + a^{3 - s - 3 t}) . \end{array}$

Comparing this system of equations with (3.20), we obtain: $\begin{array}{rcl} \sum_{m = 1}^{M} | Q_{m} - {\bar{Q}}_{m} | & ⩽ & 2 {(1 - \frac{3 γ}{5 π d} max_{1 ⩽ m ⩽ M} | C_{m} |)}^{- 1} \\ \times M max_{1 ⩽ m ⩽ M} | C_{m} | | O (a + a^{2 - s - t} + a^{2 - s} + a^{3 - 2 s - t} + a^{2 - 2 t} + a^{3 - s - 3 t}) | \\ (3.31) & = & O (a^{2 - s} + a^{3 - 2 s - t} + a^{3 - 2 s} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) . \end{array}$

We can rewrite the far-field (3.27) as, $\begin{array}{rcl} U^{\infty} (\hat{x}) & = & \sum_{m = 1}^{M} [e^{- i κ \hat{x} \cdot z_{m}} Q_{m} + O (κ a^{2} (1 + M \frac{a}{d}))] \\ = & \sum_{m = 1}^{M} e^{- i κ \hat{x} \cdot z_{m}} Q_{m} + O (a^{2 - s} + a^{3 - 2 s - t}) \\ = & \sum_{m = 1}^{M} e^{- i κ \hat{x} \cdot z_{m}} {\bar{Q}}_{m} + \sum_{m = 1}^{M} e^{- i κ \hat{x} \cdot z_{m}} (Q_{m} - {\bar{Q}}_{m}) + O (a^{2 - s} + a^{3 - 2 s - t}) \\ (3.32) & \underset{(3.30)}{=} & \sum_{m = 1}^{M} e^{- i κ \hat{x} \cdot z_{m}} {\bar{Q}}_{m} + O (a^{2 - s} + a^{3 - 2 s - t} + a^{4 - 3 s - t} + a^{3 - s - 2 t} + a^{4 - 2 s - 3 t}) . \end{array}$

Remark 3.1.

Instead of using the analyticity of $Φ_{κ}$ around $z_{m}$ in (3.15), we use the Taylor series expansion till the order $(l + 1)$ , $l = 0, 1, 2, \dots$ , then following similar computations as in (3.14), we can prove that $‖ {(S_{m m}^{0})}^{- 1} S_{m j}^{'} σ_{j} ‖_{L^{2} (\partial D_{m})}$ , for $m \neq j$ , behaves as $O (\frac{a^{2}}{d_{m j}^{2}} + \frac{a^{2 + l}}{d_{m j}^{3 + l}})$ and then deduce that $‖ L_{0}^{- 1} \overset{´}{K} ‖_{L (\prod_{m = 1}^{M} L^{2} (\partial D_{m}), \prod_{m = 1}^{M} L^{2} (\partial D_{m}))}$ behaves as $O (a^{(2 + l) - (4 + l) \frac{s}{3}} + a^{(2 + l) - (3 + l) t} + a^{3 - s} + a^{(2 - 2 t})$ , which eventually leads to the estimate $‖ σ_{m} ‖_{L^{2} (\partial D_{m})} = O (1 + M \frac{a}{d})$ under the following assumptions $\begin{array}{l} (3.33) & \begin{aligned} \frac{s}{3} ⩽ t ⩽ \frac{2 + l}{3 + l} and \\ 0 ⩽ s ⩽ min {3 \frac{(2 + l)}{(4 + l)}, 2 - t, (3 + l) - (4 + l) t, \frac{3}{7 + l} (3 + l - t), 2, 3 - 3 t, \frac{1}{2} (3 - t)} . \end{aligned} \end{array}$

Footnotes

Acknowledgement

The authors would like to express their warm gratitude to the referee for the comments and suggestions that improved much the quality of the presentation. The three authors were partially supported by the Austrian Science Fund (FWF): P28971-N32. The first author was also supported by DST SERB MATRICS: MTR/2017/000539.

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