A mathematical framework for tunable metasurfaces. Part II

Abstract

This paper is concerned with wave propagation inside a cavity with a tunable boundary condition. It is a follow-up of [Asympt. Anal. (2019), to appear]. Cavities, because they trap waves for long times due to their reflecting walls, are used in a vast number of scientific domains. Indeed, in these closed media and due to interferences, the free space continuum of solutions becomes a discrete set of stationary eigenmodes. These enhanced stationary fields are commonly used in fundamental physics to increase wave-matter interactions. The eigenmodes and associated eigenfrequencies of a cavity are imposed by its geometrical properties through the boundary conditions. In this paper, we show the effect of a small change of boundary condition on the Green’s function of the cavity. This is achieved through the use of a tunable reflecting metasurface. The boundary condition can be switched from Dirichlet to Neumann at some specific resonant frequencies.

Keywords

Tunable metasurface cavity field enhancement

1. Introduction

Controlling waves in cavities, which are used in numerous domains of applied and fundamental physics, has become a major topic of interest [12,13,16,17]. The wave fields established in cavities are fixed by their geometry. They are usually modified by using mechanical parts. Nevertheless, tailoring the cavity boundaries permits one to design at will the wave fields they support. In [6], it is shown that we can locally modify the boundaries, switching them from Dirichlet to Neumann conditions, using tunable metasurfaces. The concept of metasurfaces is a powerful tool to shape waves by governing precisely the phase response of each constituting element through its subwavelength resonance properties [4,9,14,20]. Subwavelength resonators have been also used as the building block of super-resolution imaging [3,10,11].

In this paper, we propose an efficient approach to characterize the Green’s function of a cavity with mixed (Dirichlet and Neumann) boundary conditions. The use of tunable metasurfaces allows us to find a criterion ensuring that modifying parts of a cavity’s boundaries turn it into a completely different one. We provide a new and simple procedure for maximizing the Green’s function between two points at a chosen frequency in terms of the boundary conditions. Our algorithm is a one shot optimization algorithm and can then be used in real-time to focalize the wave on a given spot by maximizing the transmission between an emitter and a receiver through specific eigenmodes of the cavity or on the contrary, to minimize the field on a receiver.

The paper is organized as follows. Section 2 is devoted to the derivation of an asymptotic formula of the Green’s function of a cavity with mixed boundary conditions in terms of the size of the part of the cavity boundary where the boundary condition is switched from Dirichlet to Neumann. A closed form of the derivative of the Green’s function with respect to changes in the boundary condition is given in Theorem 2.4.

In Section 3, we consider the problem where we have a source in a bounded domain and we want to determine whether we activate a small part of the boundary to be reflecting or not in such a way the signal at a given receiving point is significantly enhanced. Based on Theorem 2.4, we propose a simple strategy aiming to maximize the norm of the Green’s function by nucleating Neumann boundary conditions. The basic idea follows the concept of topological derivative. The switching of parts of the boundary from Dirichlet to Neumann where the topological derivative of the norm of the Green’s function is positive allows for an increasing of the transmission between the point source and the receiver. Finally, we present some numerical experiments to show the applicability of the proposed methodology.

2. Changing a small part of the boundary from Dirichlet to Neumann

Let us consider a bounded domain, we can think of it as a cavity, where there is put up a source point, which emits a wave. On the boundary, we have mounted a device, which we can toggle to act like the other part of the boundary or to act in a reflecting manner. As discussed in the introduction, such a device can be constructed using a metasurface build upon of Helmholtz resonators. Given a receiving point in the domain, we want to be able to decide, which of the two options for the device give the higher signal at a given receiving point.

After establishing a mathematical set-up for the above described environment, we give the first order expansion term for the difference of the signal between the two option in terms of the size of the device. To this end, we establish the invertiblity of an operator, which emerges from Green’s formula, and compute then the inverse of that operator. Most of the results in this section are inspired by [7], where layer potential techniques were first introduced for solving the narrow escape problem of a Brownian particle through a small boundary absorbing part. It is worth emphasizing that, in the narrow escape problem, the small part of the boundary is absorbing while the remaining part is reflecting. Because of such a difference, the derivation of an asymptotic formula for the Green’s function here is technically more involved than in [7]. We refer the reader to [1,2,5,15,18] for the analysis of the mixed boundary value problem and the evaluation of the associated eigenvalues and eigenfunctions.

2.1. Preliminaries

2.1.1. Statement of the problem

Let Ω be an open, bounded, and simply connected subset of $R^{2}$ with a $C^{2}$ -boundary. Let $\partial Ω$ be partitioned in two open intervals $\partial Ω_{N}$ and $\partial Ω_{D}$ such that $\partial Ω_{N}$ is a line segment with length $2 ε$ , where $ε > 0$ , and with center ${(0, 0)}^{T} \in R^{2}$ . For simplicity, we assume that Ω is rotated so that $\partial Ω_{N}$ is parallel to the first coordinate axis, all points on $\partial Ω_{N}$ have height 0, and the normal on $\partial Ω_{N}$ is ${(0, 1)}^{T}$ . Then we fix two points, one is the source point $x_{S} \in Ω$ and the other the receiving point $x_{R} \in Ω$ . See Fig. 1. We are looking for an asymptotic expansion of the following function in terms of ε and an analytic expression for the first order term. This leading order term gives the topological derivative of the Green’s function of the cavity with respect to changes in the boundary conditions. In other terms, it describes the nucleation of a Neumann boundary condition.

Let $u_{x_{S}}^{k, ε} : Ω ∖ {x_{S}} \to C$ be the solution to $\begin{matrix} (2.1) & \{\begin{matrix} (△_{y} + k^{2}) u_{x_{S}}^{k, ε} (y) = δ_{0} (x_{S} - y) & in Ω, \\ u_{x_{S}}^{k, ε} (y) = 0 & on \partial Ω_{D}, \\ \partial_{ν_{y}} u_{x_{S}}^{k, ε} (y) = 0 & on \partial Ω_{N}, \end{matrix} \end{matrix}$ where we assume that $k \in (0, \infty)$ is not an eigenvalue to $- △$ with the above boundary conditions, and thus $u_{x_{S}}^{k, ε}$ is uniquely solvable.

Fig. 1.

This picture depicts $Ω \subset R^{2}$ with the two disjoint boundary components $\partial Ω_{D}$ and $\partial Ω_{N}$ . We have a source point at $x_{S}$ and a receiving point at $x_{R}$ .

Next, we need the function, which satisfies the above partial differential equation but has the Dirichlet condition on the whole boundary. It is often denoted as the Dirichlet function $G_{Ω}^{k} (z, \cdot) : Ω ∖ {z} \to C$ and satisfies $\begin{matrix} (2.2) & \{\begin{matrix} (△_{y} + k^{2}) G_{Ω}^{k} (z, y) = δ_{0} (z - y) & in Ω, \\ G_{Ω}^{k} (z, y) = 0 & on \partial Ω, \end{matrix} \end{matrix}$ for $k \in (0, \infty)$ not an eigenvalue to $- △$ with the above boundary condition.

We will see that we can express $u_{x_{S}}^{k, ε}$ as $\begin{matrix} u_{x_{S}}^{k, ε} (x_{R}) = G_{Ω}^{k} (x_{S}, x_{R}) + O (ε) . \end{matrix}$

2.1.2. The Dirichlet function

We have the following formula for the Dirichlet function:

Proposition 2.1.
Let $z \in Ω$ , $x \in Ω ∖ {z}$ and k not be an eigenvalue to $- △$ with the boundary condition given in PDE ( 2.2 ), then we have $\begin{matrix} G_{Ω}^{k} (z, x) = Γ^{k} (z, x) + R_{G, Ω}^{k} (z, x), \end{matrix}$ where $R_{G, Ω}^{k}$ is the solution to $\begin{matrix} \{\begin{matrix} (△_{x} + k^{2}) R_{G, Ω}^{k} (z, x) = 0 & in Ω, \\ R_{G, Ω}^{k} (z, x) = - Γ^{k} (z, x) & on \partial Ω . \end{matrix} \end{matrix}$

The uniqueness and existence of $R_{G, Ω}^{k}$ in Proposition 2.1 is a standard result. We are especially interested in a formula for $\partial_{ν_{x}} G_{Ω}^{k} (z, x)$ , for $z \in Ω$ , with a smooth enough remainder. To this end, we use $G_{Ω}^{k} (z, x) = Γ^{k} (z, x) + R_{G, Ω}^{k} (z, x)$ , then from [8, Chapter 2.3] we can extract the first two spatial, singular terms of $Γ^{k}$ , that is $\frac{1}{2 π} log (| z - x |)$ and $\frac{- 1}{8 π} k^{2} log (k | z - x |) | z - x |^{2}$ , and obtain a formula with a smooth enough remainder: Proposition 2.2.
Let $z \in Ω$ , $x \in \partial Ω$ and k not be an eigenvalue to $- △$ with the boundary condition given in PDE ( 2.2 ), then we have $\begin{matrix} \partial_{ν_{x}} G_{Ω}^{k} (z, x) = \frac{1}{2 π} \frac{ν_{x} \cdot (x - z)}{| z - x |^{2}} - \frac{1}{8 π} k^{2} ν_{x} \cdot (x - z) (2 log (k | z - x |) + 1) + R_{\partial G, Ω}^{k} (z, x), \end{matrix}$ where $R_{\partial G, Ω}^{k} (z, \cdot) \in H^{5 / 2} (Ω)$ .

2.2. Main results

We define $\begin{array}{l} X^{ε} : = {μ \in L^{2} ((- ε, ε)) | \int_{- ε}^{ε} \sqrt{ε^{2} - t^{2}} {| μ (t) |}^{2} d t < \infty}, \\ ‖ μ ‖_{X^{ε}} : = {(\int_{- ε}^{ε} \sqrt{ε^{2} - t^{2}} {| μ (t) |}^{2} d t)}^{1 / 2}, \\ Y^{ε} : = {μ \in C^{0} ([- ε, ε]) ∣ μ^{'} \in X^{ε}}, \\ ‖ μ ‖_{Y^{ε}} : = {(‖ μ ‖_{X^{ε}}^{2} + {‖ μ^{'} ‖}_{X^{ε}}^{2})}^{1 / 2}, \\ Y_{Π}^{ε} : = {μ \in Y^{ε} ∣ μ (- ε) = μ (ε) = 0} . \end{array}$ We define the operator $L^{ε} : X^{ε} \to Y^{ε}$ and the operator $J^{ε} : Y_{Π}^{ε} \to X^{ε}$ . Let $τ \in (- ε, ε)$ , then $\begin{array}{l} L^{ε} [μ] (τ) : = \int_{- ε}^{ε} μ (t) log (| τ - t |) d t, \\ J^{ε} [μ] (τ) : = H . f . p . \int_{- ε}^{ε} \frac{μ (t)}{{(τ - t)}^{2}} d t, \end{array}$ where the ‘H.f.p.’ denotes a Hadamard-finite-part integral, see also Remark 2.8. $L^{ε}$ is invertible and the inverse is given in Proposition 2.10. $J^{ε}$ is invertible and the inverse is given in Proposition 2.11. We then have the following result:

Theorem 2.3.
Let $ε > 0$ be small enough, $α, β > 0$ . The operator $- α J^{ε} + β L^{ε} : Y_{Π}^{ε} \to X^{ε}$ is linear and invertible and the inverse is given in Proposition 2.12 . The exact function ${(- α J^{ε} + β L^{ε})}^{- 1} [1]$ is given in Lemma 2.13 .
Theorem 2.4.
Let $ε > 0$ be small enough and let $k \in (0, \infty)$ not be an eigenvalue to $- △$ with the boundary condition given in PDE ( 2.1 ) as well as with the boundary condition given in PDE ( 2.2 ). The value $u_{x_{S}}^{k, ε} (x_{R})$ is determined through $\begin{matrix} (2.3) & u_{x_{S}}^{k, ε} (x_{R}) = G_{Ω}^{k} (x_{S}, x_{R}) + G_{(1)}^{k, ε} (x_{S}, x_{R}) + G_{(2)}^{k, ε} (x_{S}, x_{R}), \end{matrix}$ where $\begin{matrix} G_{(1)}^{k, ε} (x_{S}, x_{R}) : = - \int_{- ε}^{ε} \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} t \\ 0 \end{matrix})) {(\frac{- 1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [\partial_{ν_{x}} G_{Ω}^{k} (x_{S}, (\begin{matrix} τ \\ 0 \end{matrix}))] (t) d t, \end{matrix}$ where $G_{(1)}^{k, ε} (x_{S}, x_{R}) = O (\frac{ε}{| log (ε / 2) |})$ and $G_{(2)}^{k, ε} (x_{S}, x_{R}) = O (\frac{ε^{2}}{| log (ε / 2) |^{2}})$ .

2.3. Proof of the main results

The idea of the proof is inspired by [7] and is as follows. Using Green’s formula we readily establish that $u_{x_{S}}^{k, ε}$ is a small perturbation of $G_{Ω}^{k} (x_{S}, \cdot)$ . We see that the difference $v_{z}^{k, ε} : = u_{z}^{k, ε} - G_{Ω}^{k} (z, \cdot)$ satisfies the following two conditions: $\begin{array}{l} v_{z}^{k, ε} (x) = \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) v_{z}^{k, ε} (y) d σ_{y}, for x \in Ω, \\ \partial_{ν_{x}} v_{z}^{k, ε} (x) = - \partial_{ν_{x}} G_{Ω}^{k} (z, x), for x \in \partial Ω_{N}, \end{array}$ where the first one comes from Green’s formula and the second one from the partial differential equation for $u_{z}^{k, ε}$ and $G_{Ω}^{k} (z, \cdot)$ . Combining both leads us to the condition $\begin{array}{l} - \partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix})) & = - \frac{1}{2 π} J^{ε} [v_{z}^{k, ε}] + \frac{k^{2}}{4 π} L^{ε} [v_{z}^{k, ε}] \\ + \frac{k^{2}}{8 π} (2 log (k) + 1) \int_{- ε}^{ε} v_{z}^{k, ε} (t) d t + \int_{- ε}^{ε} v_{z}^{k, ε} (t) \partial_{x_{2}} R_{\partial G, Ω}^{k} ((\begin{matrix} τ \\ 0 \end{matrix}), (\begin{matrix} t \\ 0 \end{matrix})) d t . \end{array}$ The key now is to invert the operator $\frac{- 1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε} [v_{z}^{k, ε}]$ and proving that the integrals over $(- ε, ε)$ with integrand $v_{z}^{k, ε}$ are then of lower order. The proof for invertiblity uses a result given in [21, Chapter 11]. For finding the inverse, we use that $J^{ε}$ is of the form $\partial_{t} H^{ε}$ , where $H^{ε}$ is the finite Hilbert transform, and that ${(L^{ε})}^{- 1}$ is of the form ${(H^{ε})}^{†} [\partial_{t}] + C$ , where ${(H^{ε})}^{†}$ is the inverse of $H^{ε}$ on $ker {(H^{ε})}^{⊥}$ . This, together with $\begin{matrix} {(- α J^{ε} + β L^{ε})}^{- 1} = {(L^{ε})}^{- 1} {(β I - α J^{ε} {(L^{ε})}^{- 1})}^{- 1}, \end{matrix}$ leads us to the inverse. For the estimates we adapt the technique used in [8, Lemma 5.4]. To this end, we have to compute some integrals. To determine them, we use the mathematics tool Mathematica [22].

2.3.1. Condition on the gap

Proposition 2.5.
Let $z \in Ω$ and $x \in Ω ∖ {z}$ then $\begin{matrix} u_{z}^{k, ε} (x) = G_{Ω}^{k} (z, x) + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) (u_{z}^{k, ε} (y) - G_{Ω}^{k} (z, y)) d σ_{y} . \end{matrix}$
Proof.
With Green’s formula we have $\begin{array}{l} u_{z}^{k, ε} (x) & = \int_{Ω} (△ + k^{2}) G_{Ω}^{k} (x, y) u_{z}^{k, ε} (y) d y \\ = \int_{Ω} G_{Ω}^{k} (x, y) (△ + k^{2}) u_{z}^{k, ε} (y) d y \\ + \int_{\partial Ω} \partial_{ν_{y}} G_{Ω}^{k} (x, y) u_{z}^{k, ε} (y) d σ_{y} - \int_{\partial Ω} G_{Ω}^{k} (x, y) \partial_{ν_{y}} u_{z}^{k, ε} (y) d σ_{y} \\ = G_{Ω}^{k} (x, z) + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) u_{z}^{k, ε} (y) d σ_{y} - 0 . \end{array}$ We claim that for $z, x \in Ω$ , $z \neq x$ , we have that $G_{Ω}^{k} (z, x) = G_{Ω}^{k} (x, z)$ . With that claim we conclude then $\begin{array}{l} u_{z}^{k, ε} (x) & = G_{Ω}^{k} (z, x) + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) u_{z}^{k, ε} (y) d σ_{y} \\ = G_{Ω}^{k} (z, x) + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) u_{z}^{k, ε} (y) d σ_{y} - \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) G_{Ω}^{k} (z, y) d σ_{y} \\ = G_{Ω}^{k} (z, x) + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) (u_{z}^{k, ε} (y) - G_{Ω}^{k} (z, y)) d σ_{y} . \end{array}$ Thus the proof follows. To prove the claim, consider that with Green’s formula $\begin{array}{l} G_{Ω}^{k} (z, x) & = \int_{Ω} (△ + k^{2}) G_{Ω}^{k} (x, y) G_{Ω}^{k} (z, y) d y \\ = \int_{Ω} G_{Ω}^{k} (x, y) (△ + k^{2}) G_{Ω}^{k} (z, y) d y + 0 + 0 \\ = G_{Ω}^{k} (x, z), \end{array}$ which is exactly what we wanted. □

Let us define $v_{z}^{k, ε} (x) : = u_{z}^{k, ε} (x) - G_{Ω}^{k} (z, x)$ . Thus we see that $v_{z}^{k, ε}$ satisfies the partial differential equation $\begin{matrix} (2.4) & \{\begin{matrix} (△_{x} + k^{2}) v_{z}^{k, ε} (x) = 0 & in Ω, \\ v_{z}^{k, ε} (x) = 0 & on \partial Ω_{D}, \\ \partial_{ν_{x}} v_{z}^{k, ε} (x) = - \partial_{ν_{x}} G_{Ω}^{k} (z, x) & on \partial Ω_{N}, \end{matrix} \end{matrix}$ and from Proposition 2.5 we have for $x \in Ω ∖ {z}$ $\begin{matrix} (2.5) & v_{z}^{k, ε} (x) = \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) v_{z}^{k, ε} (y) d σ_{y} . \end{matrix}$ Combining (2.4) and (2.5) we obtain the following condition for $v_{z}^{k, ε}$ :
Lemma 2.6.
Let $z \in Ω$ and $x \in \partial Ω_{N}$ then $\begin{matrix} - \partial_{ν_{x}} G_{Ω}^{k} (z, x) = ν_{x} \cdot lim_{\begin{array}{c} \tilde{x} \to x \\ \tilde{x} \in Ω \end{array}} \nabla_{\tilde{x}} [\int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (\tilde{x}, y) v_{z}^{k, ε} (y) d σ_{y}], \end{matrix}$ where $ν_{x}$ is the outside normal at x.

Using Proposition 2.2, we have $\begin{array}{l} \nabla_{x} \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) v_{z}^{k, ε} (y) d σ_{y} \\ = \nabla_{x} \int_{\partial Ω_{N}} [\frac{1}{2 π} \frac{y_{2} - x_{2}}{| y - x |^{2}} - \frac{1}{8 π} k^{2} (y_{2} - x_{2}) (2 log (k | y - x |) + 1) \\ (2.6) & + R_{\partial G, Ω}^{k} (x, y)] v_{z}^{k, ε} (y) d σ_{y} . \end{array}$

Using that $\partial Ω_{N}$ is a line segment of length $2 ε$ with center ${(0, 0)}^{T}$ , we further compute that the right-hand-side in the last equation is $\begin{array}{l} \nabla_{x} \int_{- ε}^{ε} [\frac{1}{2 π} \frac{- x_{2}}{{(t - x_{1})}^{2} + x_{2}^{2}} - \frac{1}{8 π} k^{2} (- x_{2}) (2 log (k \sqrt{{(t - x_{1})}^{2} + x_{2}^{2}}) + 1) \\ (2.7) & + R_{\partial G, Ω}^{k} (x, {(t, 0)}^{T})] v_{z}^{k, ε} ({(t, 0)}^{T}) d t . \end{array}$ Pulling the ∇-operator inside the integral, then pulling the limes in Lemma 2.6 inside the integral, wherever possible, and considering that $\partial_{ν_{x}} = \partial_{x_{2}}$ , we obtain $\begin{array}{l} lim_{\begin{array}{c} h \to 0 \\ h > 0 \end{array}} \int_{- ε}^{ε} [\frac{- 1}{2 π} \frac{({(t - x_{1})}^{2} - h) v_{z}^{k, ε} (t)}{{({(t - x_{1})}^{2} + h)}^{2}} + \frac{k^{2} v_{z}^{k, ε} (t)}{8 π} (2 log (k | t - x_{1} |) + 1) \\ (2.8) & + v_{z}^{k, ε} (t) \partial_{x_{2}} R_{\partial G, Ω}^{k} (x, {(t, 0)}^{T})] d t, \end{array}$ where we identified $v_{z}^{k, ε} (t)$ with $v_{z}^{k, ε} ({(t, 0)}^{T})$ . This leads us to $\begin{array}{l} - \partial_{ν_{x}} G_{Ω}^{k} (z, x) & = \frac{- 1}{2 π} lim_{h \to 0} \int_{- ε}^{ε} \frac{({(t - x_{1})}^{2} - h) v_{z}^{k, ε} (t)}{{({(t - x_{1})}^{2} + h)}^{2}} d t + \frac{k^{2}}{4 π} \int_{- ε}^{ε} v_{z}^{k, ε} (t) log | t - x_{1} | d t \\ (2.9) & + \frac{k^{2}}{8 π} (2 log (k) + 1) \int_{- ε}^{ε} v_{z}^{k, ε} (t) d t + \int_{- ε}^{ε} v_{z}^{k, ε} (t) \partial_{x_{2}} R_{\partial G, Ω}^{k} (x, {(t, 0)}^{T}) d t . \end{array}$

Consider that for $h > 0$ $\begin{matrix} \int_{- ε}^{ε} \frac{({(t - x_{1})}^{2} - h) v_{z}^{k, ε} (t)}{{({(t - x_{1})}^{2} + h)}^{2}} d t = {[v_{z}^{k, ε} (t) \frac{x_{1} - t}{h + {(t - x_{1})}^{2}}]}_{t = - ε}^{ε} + \int_{- ε}^{ε} \frac{(t - x_{1}) \partial_{t} v_{z}^{k, ε} (t)}{{(t - x_{1})}^{2} + h} d t . \end{matrix}$

Using that $v_{z}^{k, ε} (- ε) = v_{z}^{k, ε} (ε) = 0$ , because of the Dirichlet boundary, we can readily compute that $\begin{matrix} lim_{h \to 0} \int_{- ε}^{ε} \frac{({(t - x_{1})}^{2} - h) v_{z}^{k, ε} (t)}{{({(t - x_{1})}^{2} + h)}^{2}} d t = - p . v . \int_{- ε}^{ε} \frac{\partial_{t} v_{z}^{k, ε} (t)}{x_{1} - t} d t = H . f . p . \int_{- ε}^{ε} \frac{v_{z}^{k, ε} (t)}{{(x_{1} - t)}^{2}} d t, \end{matrix}$ where the last integral is the Hadamard-finite-part integral.
Definition 2.7.
We define the operators $H^{ε} : X^{ε} \to X^{ε}$ and $J^{ε} : Y_{Π}^{ε} \to X^{ε}$ as $\begin{array}{l} H^{ε} [μ] (τ) : = p . v . \int_{- ε}^{ε} \frac{μ (t)}{τ - t} d t, \\ J^{ε} [μ] (τ) : = H . f . p . \int_{- ε}^{ε} \frac{μ (t)}{{(τ - t)}^{2}} d t . \end{array}$
Remark 2.8.
From the discussion above we especially obtain formulas for the operators $H^{ε}$ and $J^{ε}$ for $μ \in X^{ε}$ and $μ \in Y_{Π}^{ε}$ , respectively. These are: $\begin{array}{l} H^{ε} [μ] = p . v . \int_{- ε}^{ε} \frac{μ (t)}{τ - t} d t = lim_{\begin{array}{c} h \to 0 \\ h > 0 \end{array}} \int_{- ε}^{ε} \frac{(τ - t) μ (t)}{{(τ - t)}^{2} + h} d t, \\ J^{ε} [μ] = H . f . p . \int_{- ε}^{ε} \frac{μ (t)}{{(τ - t)}^{2}} d t = - H^{ε} [\partial_{τ} μ] = lim_{\begin{array}{c} h \to 0 \\ h > 0 \end{array}} \int_{- ε}^{ε} \frac{({(τ - t)}^{2} - h) μ (t)}{{({(τ - t)}^{2} + h)}^{2}} d t . \end{array}$

With Definition 2.7 and (2.9) we then obtain the following proposition: Proposition 2.9.
Let $z \in Ω$ and $τ \in (- ε, ε)$ , then $\begin{array}{l} - \partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix})) & = - \frac{1}{2 π} J^{ε} [v_{z}^{k, ε}] + \frac{k^{2}}{4 π} L^{ε} [v_{z}^{k, ε}] \\ + \frac{k^{2}}{8 π} (2 log (k) + 1) \int_{- ε}^{ε} v_{z}^{k, ε} (t) d t + \int_{- ε}^{ε} v_{z}^{k, ε} (t) \partial_{x_{2}} R_{\partial G, Ω}^{k} ((\begin{matrix} τ \\ 0 \end{matrix}), (\begin{matrix} t \\ 0 \end{matrix})) d t . \end{array}$

2.3.2. Hypersingular operator analysis

Let us show that $L^{ε}$ is bijective and let us consider its inverse.

Proposition 2.10.
Let $0 < ε < 2$ . The operator $L^{ε} : X^{ε} \to Y^{ε}$ is linear, bounded and invertible and has the inverse $\begin{matrix} {(L^{ε})}^{- 1} [η] (t) = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} η^{'} (τ)}{t - τ} d τ + \frac{C_{L} [η]}{π log (ε / 2) \sqrt{ε^{2} - t^{2}}}, \end{matrix}$ where $\begin{matrix} (2.10) & C_{L} [η] : = η (τ) - L^{ε} [- \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - s^{2}} η^{'} (s)}{t - s} d s] \end{matrix}$ is a constant depending on η and it is linear in η.
Proof.
The proof for invertiblity is given in [21, Chapter 11.5], the exact formula is derived in [8, Chapter 5.2.3]. □

Not only is $L^{ε}$ an isomorphism, $J^{ε} : Y_{Π}^{ε} \to X^{ε}$ is an isomorphism, too. However, the spaces are slightly different. Moreover, we have the following formula for the inverse:
Proposition 2.11.
Let $0 < ε < 2$ . The operator $J^{ε} : Y_{Π}^{ε} \to X^{ε}$ is linear and invertible and has the inverse $\begin{matrix} {(J^{ε})}^{- 1} [η] (t) = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \int_{- ε}^{τ} η (\tilde{τ}) d \tilde{τ}}{t - τ} d τ + t C_{J, 1} + C_{J, 2}), \end{matrix}$ where $\begin{array}{l} C_{J, 1} = - \int_{- ε}^{ε} \frac{1}{\sqrt{ε^{2} - τ^{2}}} (\int_{- ε}^{τ} η (\tilde{τ}) d \tilde{τ}) d τ, \\ C_{J, 2} = - \int_{- ε}^{ε} \frac{τ}{\sqrt{ε^{2} - τ^{2}}} (\int_{- ε}^{τ} η (\tilde{τ}) d \tilde{τ}) d τ, \end{array}$ are constants depending on η and they are linear in η.
Proof.
The proof for invertibility is given in [21, Chapter 11.5]. Thus for every $μ \in Y_{Π}^{ε}$ there exists exactly one $η \in X^{ε}$ such that $J^{ε} [μ] = η$ . Using the fact that the Hadamard-finite-part integral can be expressed as $J^{ε} [μ] = \partial_{τ} H^{ε} [μ]$ , and that $H^{ε}$ is isomorphic up to a one dimensional kernel of the form $ker (H^{ε}) = span {{(ε^{2} - t^{2})}^{- 1 / 2}}$ , and the inverse on $ker {(H^{ε})}^{⊥}$ , which we call ${(H^{ε})}^{†}$ , is of the following form (see for instance in [19]; it is also used in [8, Chapter 5.2.3]) $\begin{matrix} {(H^{ε})}^{†} [η] (t) = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} η (τ)}{t - τ} d τ, \end{matrix}$ we can rewrite $J^{ε} [μ] = η$ as $H^{ε} [μ] = \int_{- ε}^{τ} η + C_{\int}$ and then write $\begin{array}{l} μ (t) & = {(H^{ε})}^{†} [\int_{- ε}^{τ} η + C_{\int}] + \frac{C_{η}}{\sqrt{ε^{2} - t^{2}}} \\ (2.11) & = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \int_{- ε}^{τ} η}{t - τ} d τ + p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} C_{\int}}{t - τ} d τ - π^{2} C_{η}) \\ (2.12) & = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \int_{- ε}^{τ} η}{t - τ} d τ + π t C_{\int} - π^{2} C_{η}), \end{array}$ where we computed $p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{t - τ} d τ = π t$ , see [8, Chapter 5.2.3].

Let us find explicit expressions for the constants $C_{η}$ and $C_{\int}$ . Consider that the part in between the brackets in (2.12) has to be zero for the values $t = ε$ and $t = - ε$ , so that we can satisfy the condition $μ \in Y_{Π}^{ε}$ . This leads us to the system of equations $\begin{array}{l} \int_{- ε}^{ε} \frac{\sqrt{ε + τ}}{\sqrt{ε - τ}} (\int_{- ε}^{τ} η) d τ + π ε C_{\int} - π^{2} C_{η} = 0, \\ \int_{- ε}^{ε} \frac{\sqrt{ε - τ}}{- \sqrt{ε + τ}} (\int_{- ε}^{τ} η) d τ - π ε C_{\int} - π^{2} C_{η} = 0 . \end{array}$ After solving this, we obtain $\begin{array}{l} C_{\int} = - \frac{1}{2 ε π} p . v . \int_{- ε}^{ε} \frac{2 ε}{\sqrt{ε^{2} - τ^{2}}} (\int_{- ε}^{τ} η) d τ, \\ C_{η} = \frac{1}{2 π^{2}} p . v . \int_{- ε}^{ε} \frac{2 τ}{\sqrt{ε^{2} - τ^{2}}} (\int_{- ε}^{τ} η) d τ . \end{array}$ This proves Proposition 2.11. □

Let us consider the operator $- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε}$ .
Proposition 2.12.
Let ε be small enough, $α, β > 0$ . The operator $- α J^{ε} + β L^{ε} : Y_{Π}^{ε} \to X^{ε}$ is linear and invertible and for $η \in X^{ε}$ , the inverse is given by $\begin{matrix} (2.13) & {(- α J^{ε} + β L^{ε})}^{- 1} [η] (t) = {(L^{ε})}^{- 1} [μ_{ι}] (t), \end{matrix}$ where $\begin{array}{l} μ_{ι} (τ) & = C_{(2.14), 1} e^{\frac{\sqrt{β} τ}{\sqrt{α}}} + C_{(2.14), 2} e^{- \frac{\sqrt{β} τ}{\sqrt{α}}} \\ (2.14) & + \frac{1}{2 \sqrt{α β}} e^{- \frac{\sqrt{β} τ}{\sqrt{α}}} \int_{- ε}^{τ} η (s) e^{\frac{\sqrt{β} s}{\sqrt{α}}} d s - \frac{1}{2 \sqrt{α β}} e^{\frac{\sqrt{β} τ}{\sqrt{α}}} \int_{- ε}^{τ} η (s) e^{- \frac{\sqrt{β} s}{\sqrt{α}}} d s, \end{array}$ where $C_{(2.14), 1} \in C$ and $C_{(2.14), 2} \in C$ are given through solving the system of equations $\begin{array}{l} (2.15) & \int_{- ε}^{ε} \frac{\sqrt{ε + τ} \partial_{τ} μ_{ι} (τ)}{\sqrt{ε - τ}} d τ - \frac{π}{log (ε / 2)} C_{L^{ε} [μ_{ι}]} = 0, \\ (2.16) & \int_{- ε}^{ε} \frac{\sqrt{ε - τ} \partial_{τ} μ_{ι} (τ)}{- \sqrt{ε + τ}} d τ - \frac{π}{log (ε / 2)} C_{L^{ε} [μ_{ι}]} = 0 . \end{array}$
Proof.
From [21, Chapter 11.1] we have that $- α J^{ε} + β L^{ε}$ is a Fredholm operator with index 0. Thus we only have to show that it is injective.

Let us show that $- α J^{ε} + β L^{ε}$ is injective. To this end, consider that with Fubini’s theorem we have $\begin{matrix} (2.17) & L^{ε} [μ] (τ) = \int_{- ε}^{ε} log (| τ - t |) μ (t) d t = \int_{- ε}^{τ} H^{ε} [μ] (\tilde{τ}) d \tilde{τ} - C_{(2.17)}, \end{matrix}$ where $C_{(2.17)} = - \int_{- ε}^{ε} log (| ε + t |) μ (t) d t$ . Then we get that $\begin{matrix} (- α J^{ε} + β L^{ε}) [μ] (τ) = - α \partial_{τ} H^{ε} [μ] (τ) + β \int_{- ε}^{τ} H^{ε} [μ] (\tilde{τ}) d \tilde{τ} - β C_{(2.17)} . \end{matrix}$ Now, let $μ \in ker (- α J^{ε} + β L^{ε})$ , it follows that μ satisfies $\begin{matrix} (2.18) & - \frac{α}{β} \partial_{τ} H^{ε} [μ] (τ) + \int_{- ε}^{τ} H^{ε} [μ] (\tilde{τ}) d \tilde{τ} = C_{(2.17)} . \end{matrix}$ Deriving both sides, we obtain $\begin{matrix} - \frac{α}{β} \partial_{τ}^{2} H^{ε} [μ] (τ) + H^{ε} [μ] (τ) = 0 . \end{matrix}$ With the substitution $M : = H^{ε} [μ]$ we obtain a second-order linear ordinary differential equation with the solution $\begin{matrix} (2.19) & M (τ) = C_{(2.19), 1} exp (\frac{\sqrt{β} τ}{\sqrt{α}}) + C_{(2.19), 2} exp (- \frac{\sqrt{β} τ}{\sqrt{α}}) . \end{matrix}$ Inserting $M$ into (2.18), we obtain $\begin{matrix} (2.20) & - C_{(2.19), 1} e^{- ε \frac{\sqrt{α}}{\sqrt{β}}} + C_{(2.19), 2} e^{ε \frac{\sqrt{α}}{\sqrt{β}}} = \frac{\sqrt{β}}{\sqrt{α}} C_{(2.17)} . \end{matrix}$

Equation $M = H^{ε} [μ]$ has the general solution $\begin{matrix} (2.21) & μ (t) = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} M (τ)}{t - τ} d τ - π^{2} C_{M}), \end{matrix}$ compare the proof of Proposition 2.11. We insert this expression into $C_{(2.17)}$ and obtain $\begin{array}{l} C_{(2.17)} & = C_{(2.19), 1} \frac{1}{π^{2}} \int_{- ε}^{ε} log (| ε + t |) \frac{1}{\sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{t - τ} exp (\frac{\sqrt{β}}{\sqrt{α}} τ) d τ) d t \\ + C_{(2.19), 2} \frac{1}{π^{2}} \int_{- ε}^{ε} log (| ε + t |) \frac{1}{\sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{t - τ} exp (- \frac{\sqrt{β}}{\sqrt{α}} τ) d τ) d t \\ - C_{M} \int_{- ε}^{ε} log (| ε + t |) \frac{1}{\sqrt{ε^{2} - t^{2}}} d t . \end{array}$

Consider that $μ \in Y_{Π}^{ε}$ , this implies that the expression inside the brackets in (2.21) has to be zero for $t = ε$ and $t = - ε$ . This leads us to the system of equations $\begin{array}{l} (2.22) & C_{(2.19), 1} \int_{- ε}^{ε} \frac{\sqrt{ε + τ} e^{\frac{\sqrt{β} τ}{\sqrt{α}}}}{\sqrt{ε - τ}} d τ + C_{(2.19), 2} \int_{- ε}^{ε} \frac{\sqrt{ε + τ} e^{- \frac{\sqrt{β} τ}{\sqrt{α}}}}{\sqrt{ε - τ}} d τ - π^{2} C_{M} = 0, \\ (2.23) & C_{(2.19), 1} \int_{- ε}^{ε} \frac{\sqrt{ε - τ} e^{\frac{\sqrt{β} τ}{\sqrt{α}}}}{- \sqrt{ε + τ}} d τ + C_{(2.19), 2} \int_{- ε}^{ε} \frac{\sqrt{ε - τ} e^{- \frac{\sqrt{β} τ}{\sqrt{α}}}}{- \sqrt{ε + τ}} d τ - π^{2} C_{M} = 0 . \end{array}$ Using the power series of the exponential function, we have $\begin{matrix} \int_{- ε}^{ε} \frac{\sqrt{ε + τ} e^{\pm \frac{\sqrt{β} τ}{\sqrt{α}}}}{\sqrt{ε - τ}} d τ = ε π \pm \frac{\sqrt{β}}{\sqrt{α}} \frac{ε^{2} π}{2} + O (ε^{3}) . \end{matrix}$ We can compute that $\begin{matrix} \int_{- ε}^{ε} log (| ε + t |) \frac{1}{\sqrt{ε^{2} - t^{2}}} d t = π log (\frac{ε}{2}) . \end{matrix}$ With the mathematics tool Mathematica [22] we can further compute for $c \in R$ $\begin{matrix} \int_{- ε}^{ε} log (| ε + t |) \frac{1}{\sqrt{ε^{2} - t^{2}}} (p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{t - τ} exp (c τ) d τ) d t = ε π^{2} - c \frac{ε^{2} π^{2}}{4} + O (ε^{3}) . \end{matrix}$ Using (2.20), (2.22), (2.23), we get a $3 \times 3$ system of equations, whose only solution is $C_{(2.19), 1} = C_{(2.19), 2} = C_{M} = 0$ , for ε small enough. We conclude $μ = 0$ and that $- α J^{ε} + β L^{ε}$ is an injective Fredholm operator of index 0, hence it is invertible.

Let us find the inverse of $- α J^{ε} + β L^{ε}$ .

Now that we know that $- α J^{ε} + β L^{ε}$ is invertible, we can reformulate the inverse of the operator as $\begin{matrix} {(- α J^{ε} + β L^{ε})}^{- 1} = {((- α J^{ε} {(L^{ε})}^{- 1} + β I) L^{ε})}^{- 1} = {(L^{ε})}^{- 1} {(β I - α J^{ε} {(L^{ε})}^{- 1})}^{- 1}, \end{matrix}$ where $β I - α J^{ε} {(L^{ε})}^{- 1}$ is an operator from $L^{ε} (Y_{Π}^{ε})$ to $X^{ε}$ , and it is invertible, because $- α J^{ε} + β L^{ε}$ and $L^{ε}$ are, and where $I$ denotes the identity operator on $X^{ε}$ . Consider that $\begin{array}{l} J^{ε} {(L^{ε})}^{- 1} [η] & = \partial_{τ} H^{ε} [{(H^{ε})}^{†} [\partial_{τ} η] + \frac{C_{L} [η]}{π log (ε / 2) \sqrt{ε^{2} - t^{2}}}] \\ = \partial_{τ}^{2} η + 0, \end{array}$ where ${(H^{ε})}^{†}$ is discussed in the proof of Proposition 2.11. Now the general form of the solution to $(β I - α \partial_{τ}^{2}) [μ_{ι}] = η$ is $\begin{array}{l} μ_{ι} (t) & = C_{(2.24), 1} e^{\frac{\sqrt{β} t}{\sqrt{α}}} + C_{(2.24), 2} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} \\ (2.24) & + \frac{1}{2 \sqrt{α β}} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} \int_{- ε}^{t} η (s) e^{\frac{\sqrt{β} s}{\sqrt{α}}} d s - \frac{1}{2 \sqrt{α β}} e^{\frac{\sqrt{β} t}{\sqrt{α}}} \int_{- ε}^{t} η (s) e^{- \frac{\sqrt{β} s}{\sqrt{α}}} d s . \end{array}$ Then the solution of $(β L^{ε} - α J^{ε}) [μ] = η$ is given through $μ = {(L^{ε})}^{- 1} [μ_{ι}]$ , where the constant $C_{(2.24), 1}$ and $C_{(2.24), 2}$ are chosen such that ${(L^{ε})}^{- 1} [μ_{ι}] \in Y_{Π}^{ε}$ , which results in solving a $2 \times 2$ linear system. The uniqueness of the solution to this system follows from the invertibility of $L^{ε}$ . □
Lemma 2.13.
Let ε be small enough, and $α, β > 0$ . We have that $\begin{matrix} (2.25) & {(- α J^{ε} + β L^{ε})}^{- 1} [1] (t) = \frac{1}{β} {(L^{ε})}^{- 1} [1] - C_{(2.25)} {(L^{ε})}^{- 1} [cosh (\frac{\sqrt{β}}{\sqrt{α}} t)], \end{matrix}$ where $\begin{matrix} C_{(2.25)} = \frac{1}{β} \frac{1}{(1 - \frac{β}{2 α} ε^{2} (log (ε / 2) - \frac{1}{2}) + O (ε^{3}))}, \end{matrix}$

Using the power series of cosh, the difference between $C_{(2.25)} {(L^{ε})}^{- 1} [cosh (\frac{\sqrt{β}}{\sqrt{α}} t)]$ and $\frac{1}{β} {(L^{ε})}^{- 1} [1]$ yields a term in $O (ε / | log (ε) |)$ . Proof.
Using the notation in Proposition 2.12, we have $η = 1$ , thus $\begin{array}{l} μ_{ι} (t) & = C_{(2.26), 1} e^{\frac{\sqrt{β} t}{\sqrt{α}}} + C_{(2.26), 2} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} \\ (2.26) & + \frac{1}{2 \sqrt{α β}} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} \int_{- ε}^{t} e^{\frac{\sqrt{β} s}{\sqrt{α}}} d s - \frac{1}{2 \sqrt{α β}} e^{\frac{\sqrt{β} t}{\sqrt{α}}} \int_{- ε}^{t} e^{- \frac{\sqrt{β} s}{\sqrt{α}}} d s \\ (2.27) & = C_{(2.27), 1} e^{\frac{\sqrt{β} t}{\sqrt{α}}} + C_{(2.27), 2} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} + \frac{1}{β} . \end{array}$ Thus $\begin{matrix} \partial_{t} μ_{ι} (t) = \frac{\sqrt{β}}{\sqrt{α}} C_{(2.27), 1} e^{\frac{\sqrt{β} t}{\sqrt{α}}} - \frac{\sqrt{β}}{\sqrt{α}} C_{(2.27), 2} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} . \end{matrix}$ Let us solve the system of equations (2.15), (2.16). We readily see that $\begin{matrix} \int_{- ε}^{ε} \frac{2 ε \partial_{τ} μ_{ι} (τ)}{\sqrt{ε^{2} - τ^{2}}} d τ = 0 . \end{matrix}$ Using the mathematics tool Mathematica [22], we obtain that $\int_{- ε}^{ε} \frac{exp (c t)}{\sqrt{ε^{2} - t^{2}}} d t = π I_{0} (ε c)$ , for $c \in R$ , where $I_{n}$ is the modified Bessel function of the first kind. This leads us to $\begin{matrix} 0 = C_{(2.27), 1} I_{0} (ε \frac{\sqrt{β}}{\sqrt{α}}) - C_{(2.27), 2} I_{0} (- ε \frac{\sqrt{β}}{\sqrt{α}}) . \end{matrix}$ Since $I_{0}$ is even we have $C_{(2.27), 1} = C_{(2.27), 2}$ . Thus $\begin{matrix} (2.28) & \partial_{t} μ_{ι} (t) = \frac{\sqrt{β}}{\sqrt{α}} C_{(2.28)} sinh (\frac{\sqrt{β}}{\sqrt{α}} t) . \end{matrix}$ Now we solve $\begin{matrix} \int_{- ε}^{ε} \frac{\sqrt{ε + τ} \partial_{τ} μ_{ι} (τ)}{\sqrt{ε - τ}} d τ - \frac{π}{log (ε / 2)} C_{L^{ε} [μ_{ι}]} = 0 . \end{matrix}$ Then, we obtain $\begin{matrix} \int_{- ε}^{ε} \frac{\sqrt{ε + τ} \partial_{τ} μ_{ι} (τ)}{\sqrt{ε - τ}} d τ = \frac{\sqrt{β}}{\sqrt{α}} C_{(2.28)} ε π I_{1} (ε \frac{\sqrt{β}}{\sqrt{α}}) \end{matrix}$ and $\begin{array}{l} C_{L^{ε} [μ_{ι}]} & = μ_{ι} (0) - L^{ε} [- \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} μ_{ι} (τ)}{t - τ} d τ] (0) \\ = C_{(2.28)} + \frac{1}{β} + \frac{1}{π^{2}} \int_{- ε}^{ε} \frac{log | t |}{\sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} μ_{ι} (τ)}{t - τ} d τ d t \\ = C_{(2.28)} (1 + \frac{\sqrt{β}}{\sqrt{α}} \frac{1}{π^{2}} \int_{- ε}^{ε} \frac{log | t |}{\sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} sinh (\frac{\sqrt{β}}{\sqrt{α}} τ)}{t - τ} d τ d t) + \frac{1}{β} . \end{array}$ Using the power series for the sinus hyperbolicus, we have $\begin{matrix} \int_{- ε}^{ε} \frac{log | t |}{\sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} sinh (\frac{\sqrt{β}}{\sqrt{α}} τ)}{t - τ} d τ d t = \frac{\sqrt{β}}{\sqrt{α}} \frac{ε^{2} π^{2}}{4} + {(\frac{\sqrt{β}}{\sqrt{α}})}^{3} \frac{ε^{4} π^{2}}{64} + O (ε^{6}), \end{matrix}$ and $\begin{matrix} I_{1} (ε \frac{\sqrt{β}}{\sqrt{α}}) = \frac{\sqrt{β}}{\sqrt{α}} \frac{ε}{2} + {(\frac{\sqrt{β}}{\sqrt{α}})}^{3} \frac{ε^{3}}{16} + O (ε^{5}) . \end{matrix}$ We infer that $\begin{matrix} C_{(2.28)} \frac{\sqrt{β}}{\sqrt{α}} π ε (\frac{\sqrt{β}}{\sqrt{α}} \frac{ε}{2} + O (ε^{3})) - \frac{π}{log (ε / 2)} C_{(2.28)} (1 + \frac{β}{α} \frac{ε^{2}}{4} + O (ε^{4})) - \frac{π}{log (ε / 2)} \frac{1}{β} = 0 . \end{matrix}$ This leads us to $\begin{matrix} C_{(2.28)} (- \frac{π}{log (ε / 2)} + \frac{β}{α} \frac{π}{2} ε^{2} (1 - \frac{1}{2 log (ε / 2)}) + O (ε^{3})) = \frac{π}{log (ε / 2)} \frac{1}{β} . \end{matrix}$ Thus $\begin{matrix} C_{(2.28)} (1 + O (ε)) = - \frac{1}{β} . \end{matrix}$ Hence, we proved Lemma 2.13. □
Lemma 2.14.
Let ε be small enough, and $α, β > 0$ . Let $R$ be the integral operator defined from $Y_{Π}^{ε}$ into $X^{ε}$ by $\begin{matrix} R [μ] (τ) = \int_{- ε}^{ε} R (τ, t) μ (t) d t, \end{matrix}$ where $R (τ, t)$ is of class $C^{0, 1 / 2}$ in τ and t. For ε small enough, there exists a positive constant $C_{(2.29)}$ , independent of ε, such that for all $μ \in Y_{Π}^{ε}$ , we have $\begin{matrix} (2.29) & {‖ {(- α J^{ε} + β L^{ε})}^{- 1} R [μ] ‖}_{X^{ε}} ⩽ \frac{ε C_{(2.29)}}{| log (ε) |} ‖ R ‖_{C^{0, 1 / 2}} ‖ μ ‖_{X^{ε}} . \end{matrix}$
Proof.
We define $η : = R [μ]$ and use then the notation in Proposition 2.12. It follows with Fubini’s Theorem that $\begin{array}{l} μ_{ι} (t) & = C_{(2.30), 1} e^{\frac{\sqrt{β} t}{\sqrt{α}}} + C_{(2.30), 2} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} \\ + \frac{1}{2 \sqrt{α β}} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} \int_{- ε}^{t} η (s) e^{\frac{\sqrt{β} s}{\sqrt{α}}} d s - \frac{1}{2 \sqrt{α β}} e^{\frac{\sqrt{β} t}{\sqrt{α}}} \int_{- ε}^{t} η (s) e^{- \frac{\sqrt{β} s}{\sqrt{α}}} d s \\ (2.30) & = C_{(2.30), 1} e^{\frac{\sqrt{β} t}{\sqrt{α}}} + C_{(2.30), 2} e^{- \frac{\sqrt{β} t}{\sqrt{α}}} + \frac{1}{2 \sqrt{α β}} R_{⋆} [μ] (t), \end{array}$ where $\begin{array}{l} R_{⋆} [μ] (t) & : = \int_{- ε}^{ε} \int_{- ε}^{t} R (s, q) e^{\frac{\sqrt{β} (s - t)}{\sqrt{α}}} μ (q) d s d q - \int_{- ε}^{ε} \int_{- ε}^{t} R (s, q) e^{- \frac{\sqrt{β} (s - t)}{\sqrt{α}}} μ (q) d s d q \\ = \int_{- ε}^{ε} (\int_{- ε}^{t} 2 R (s, q) sinh (\frac{\sqrt{β} (s - t)}{\sqrt{α}}) d s) μ (q) d q \\ = : \int_{- ε}^{ε} R_{⋆} (t, q) μ (q) d q . \end{array}$ Let us examine those constants. They are given through solving the following system: $\begin{array}{l} (2.31) & \int_{- ε}^{ε} \frac{\sqrt{ε + τ} \partial_{τ} μ_{ι} (τ)}{\sqrt{ε - τ}} d τ - \frac{π}{log (ε / 2)} C_{L^{ε} [μ_{ι}]} = 0, \\ (2.32) & \int_{- ε}^{ε} \frac{2 ε \partial_{τ} μ_{ι} (τ)}{\sqrt{ε^{2} - τ^{2}}} d τ = 0 . \end{array}$ We compute using $L^{ε} [{(L^{ε})}^{- 1}] = I$ and $L^{ε} [1 / \sqrt{ε^{2} - t^{2}}] (0) = - π log (ε / 2)$ that $\begin{array}{l} C_{L^{ε} [μ_{ι}]} & = μ_{ι} (0) + L^{ε} [\frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} μ_{ι} (τ)}{t - τ} d τ] (0) \\ = C_{(2.30), 1} + C_{(2.30), 2} + \frac{1}{2 \sqrt{α β}} C_{L^{ε} [R_{⋆} [μ]]} \\ + \frac{1}{π^{2}} L^{ε} [p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{\sqrt{ε^{2} - t^{2}}} (\frac{\sqrt{β}}{\sqrt{α}} C_{(2.30), 1} \frac{e^{\frac{\sqrt{β}}{\sqrt{α}} τ}}{t - τ} - \frac{\sqrt{β}}{\sqrt{α}} C_{(2.30), 2} \frac{e^{- \frac{\sqrt{β}}{\sqrt{α}} τ}}{t - τ}) d τ] (0) \\ = C_{(2.30), 1} + C_{(2.30), 2} + \frac{1}{2 \sqrt{α β}} C_{L^{ε} [R_{⋆} [μ]]} \\ + \frac{\sqrt{β}}{\sqrt{α}} C_{(2.30), 1} (\frac{\sqrt{β}}{\sqrt{α}} \frac{ε^{2}}{4} + O (ε^{4})) - \frac{\sqrt{β}}{\sqrt{α}} C_{(2.30), 2} (- \frac{\sqrt{β}}{\sqrt{α}} \frac{ε^{2}}{4} + O (ε^{4})) \\ = C_{(2.30), 1} (1 + \frac{β}{4 α} ε^{2} + O (ε^{4})) + C_{(2.30), 2} (1 + \frac{β}{4 α} ε^{2} + O (ε^{4})) \\ + \frac{1}{2 \sqrt{α β}} C_{L^{ε} [R_{⋆} [μ]]} . \end{array}$ and for $c \in R$ $\begin{matrix} \int_{- ε}^{ε} \frac{\sqrt{ε + τ} exp (c τ)}{\sqrt{ε - τ}} d τ = π ε + π c \frac{ε^{2}}{2} + π c^{2} \frac{ε^{3}}{2} + O (ε^{4}), \end{matrix}$ and $\begin{matrix} \int_{- ε}^{ε} \frac{exp (c τ)}{\sqrt{ε^{2} - τ^{2}}} d τ = π + π c^{2} \frac{ε^{2}}{2} + O (ε^{4}) . \end{matrix}$ This leads us to the $2 \times 2$ system of equations $\begin{array}{l} (\begin{matrix} - \frac{π}{log (ε / 2)} + O (ε) & - \frac{π}{log (ε / 2)} + O (ε) \\ \frac{\sqrt{β}}{\sqrt{α}} π + O (ε^{2}) & - \frac{\sqrt{β}}{\sqrt{α}} π + O (ε^{2}) \end{matrix}) (\begin{matrix} C_{(2.30), 1} \\ C_{(2.30), 2} \end{matrix}) \\ (2.33) & = \frac{- 1}{2 \sqrt{α β}} (\begin{matrix} \int_{- ε}^{ε} \frac{\sqrt{ε + τ}}{\sqrt{ε - τ}} \partial_{τ} R_{⋆} [μ] (τ) d τ + C_{L^{ε} [R_{⋆} [μ]]} \\ \int_{- ε}^{ε} \frac{\partial_{τ} R_{⋆} [μ] (τ)}{\sqrt{ε^{2} - τ^{2}}} d τ \end{matrix}) . \end{array}$ Let us give an expansion in ε for the right-hand-side. Using the $L^{2}$ -Cauchy–Schwarz inequality, we have that $\int_{- ε}^{ε} μ (q) d q ⩽ \sqrt{π} ‖ μ ‖_{X^{ε}}$ . Hence, we establish that $\begin{array}{l} \partial_{t} R_{⋆} [μ] (t) & = - \int_{- ε}^{ε} (\int_{- ε}^{t} 2 R (s, q) cosh (\frac{\sqrt{β} (s - t)}{\sqrt{α}}) d s) μ (q) d q \\ ⩽ ‖ R ‖_{\infty} \frac{2 \sqrt{α}}{\sqrt{β}} | sinh (\frac{\sqrt{β}}{\sqrt{α}} (ε + t)) | \sqrt{π} ‖ μ ‖_{X^{ε}}, \end{array}$ Then we have $\begin{array}{l} \int_{- ε}^{ε} \frac{\sqrt{ε + τ}}{\sqrt{ε - τ}} \partial_{τ} R_{⋆} [μ] (τ) d τ & ⩽ ‖ R ‖_{\infty} \frac{2 \sqrt{α}}{\sqrt{β}} \int_{- ε}^{ε} | μ (q) | d q \int_{- ε}^{ε} \frac{\sqrt{ε + τ}}{\sqrt{ε - τ}} sinh (\frac{\sqrt{β}}{\sqrt{α}} (ε + τ)) d τ \\ ⩽ \frac{2 \sqrt{α}}{\sqrt{β}} ‖ R ‖_{\infty} π ‖ μ ‖_{X^{ε}} (\frac{\sqrt{β}}{\sqrt{α}} \frac{3 ε^{2} π}{2} + O (ε^{4})) \\ = π ‖ R ‖_{\infty} ‖ μ ‖_{X^{ε}} (3 ε^{2} π + O (ε^{4})) . \end{array}$ Next we have that $\begin{array}{l} \int_{- ε}^{ε} \frac{\partial_{τ} R_{⋆} [μ] (τ)}{\sqrt{ε^{2} - τ^{2}}} d τ & ⩽ π \frac{2 \sqrt{α}}{\sqrt{β}} ‖ R ‖_{\infty} ‖ μ ‖_{X^{ε}} \int_{- ε}^{ε} \frac{sinh (\frac{\sqrt{β}}{\sqrt{α}} (ε + t))}{\sqrt{ε^{2} - τ^{2}}} d τ \\ ⩽ π^{2} \frac{2 \sqrt{α}}{\sqrt{β}} ‖ R ‖_{\infty} ‖ μ ‖_{X^{ε}} (ε \frac{\sqrt{β}}{\sqrt{α}} + O (ε^{3})) . \end{array}$ Finally, $\begin{array}{l} C_{L^{ε} [R_{⋆} [μ]]} & = 0 + L^{ε} [\frac{1 / π^{2}}{\sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} R_{⋆} [μ] (τ)}{t - τ} d τ] (- ε) \\ = \int_{- ε}^{ε} \frac{log (ε + t) / π^{2}}{\sqrt{ε^{2} - t^{2}}} \int_{- ε}^{ε} \sqrt{ε^{2} - τ^{2}} \frac{\partial_{τ} R_{⋆} [μ] (τ) - \partial_{τ} R_{⋆} [μ] (t)}{t - τ} d τ d t \\ + \int_{- ε}^{ε} \frac{log (ε + t) / π^{2}}{\sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \sqrt{ε^{2} - τ^{2}} \frac{\partial_{τ} R_{⋆} [μ] (t)}{t - τ} d τ d t \\ (2.34) & ⩽ \int_{- ε}^{ε} \frac{log (ε + t) / π^{2}}{\sqrt{ε^{2} - t^{2}}} C_{(2.34)} π ε \sqrt{ε} d t {| \partial_{t} R_{⋆} [μ] |}_{C^{0, 1 / 2}} \\ + \int_{- ε}^{ε} \frac{log (ε + t) / π^{2}}{\sqrt{ε^{2} - t^{2}}} t π d t {‖ \partial_{τ} R_{⋆} [μ] ‖}_{C^{0}} \\ (2.35) & = C_{(2.34)} log (ε / 2) ε \sqrt{ε} {| \partial_{t} R_{⋆} [μ] |}_{C^{0, 1 / 2}} + ε {‖ \partial_{τ} R_{⋆} [μ] ‖}_{C^{0}}, \end{array}$ where we used that $\int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{\sqrt{| t - τ |}} d τ ⩽ C_{(2.34)} π ε \sqrt{ε}$ . We readily see that $\begin{array}{l} (2.36) & {| \partial_{t} R_{⋆} [μ] |}_{C^{0, 1 / 2}} ⩽ C_{(2.36)} | R |_{C^{0, 1 / 2}} ‖ μ ‖_{X^{ε}}, \\ (2.37) & {‖ \partial_{τ} R_{⋆} [μ] ‖}_{C^{0}} ⩽ ε C_{(2.37)} ‖ R ‖_{C^{0}} ‖ μ ‖_{X^{ε}}, \end{array}$ where $C_{(2.36)} = O (1)$ , and $C_{(2.37)} = O (1)$ , for $ε \to 0$ . Now we can solve (2.33), and obtain that $\begin{array}{l} (2.38) & C_{(2.30), 1} ⩽ C_{(2.38)} ε ‖ R ‖_{C^{0, 1 / 2}} ‖ μ ‖_{X^{ε}}, \\ (2.39) & C_{(2.30), 2} ⩽ C_{(2.39)} ε ‖ R ‖_{C^{0, 1 / 2}} ‖ μ ‖_{X^{ε}}, \end{array}$ for ε small enough, where $C_{(2.38)} = O (1)$ and $C_{(2.39)} = O (1)$ , for $ε \to 0$ .

We have examined the constant. Now we estimate $‖ {(L^{ε})}^{- 1} [μ_{ι}] ‖_{X^{ε}}$ . We have $\begin{array}{l} {(L^{ε})}^{- 1} [μ_{ι}] (t) & = C_{(2.30), 1} {‖ {(L^{ε})}^{- 1} [e^{\frac{\sqrt{β} t}{\sqrt{α}}}] ‖}_{X^{ε}} + C_{(2.30), 2} {‖ {(L^{ε})}^{- 1} [e^{- \frac{\sqrt{β} t}{\sqrt{α}}}] ‖}_{X^{ε}} \\ (2.40) & + \frac{1}{2 \sqrt{α β}} {‖ {(L^{ε})}^{- 1} [R_{⋆} [μ]] ‖}_{X^{ε}} . \end{array}$ Consider that for $c \in R$ , we have that $\begin{array}{l} (2.41) & {‖ {(L^{ε})}^{- 1} [e^{c t}] ‖}_{X^{ε}} & = {‖ {(L^{ε})}^{- 1} [\int_{- ε}^{ε} e^{c t} \frac{1}{π \sqrt{ε^{2} - s^{2}}} d s] ‖}_{X^{ε}} \\ (2.42) & ⩽ \frac{C_{(2.42)}}{| log (ε) |} {‖ \frac{1}{π \sqrt{ε^{2} - s^{2}}} ‖}_{X^{ε}} \\ (2.43) & = \frac{C_{(2.43)}}{| log (ε) |}, \end{array}$ where we used the estimate $‖ {(L^{ε})}^{- 1} R [μ] ‖_{X^{ε}} ⩽ \frac{C}{| log (ε) |} ‖ μ ‖_{X^{ε}}$ , given in [8, Lemma 5.4]. Then $\begin{matrix} {(L^{ε})}^{- 1} [R_{⋆} [μ]] = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} R_{⋆} [μ] (τ)}{t - τ} d τ + \frac{C_{L^{ε} [R_{⋆} [η]]}}{π log (ε / 2) \sqrt{ε^{2} - t^{2}}} . \end{matrix}$ In (2.35) we already found out that $C_{L^{ε} [R_{⋆} [η]]} = O (\frac{ε}{- log (ε / 2)})$ . For the other term, consider that $\begin{array}{l} - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} R_{⋆} [μ] (τ)}{t - τ} d τ \\ = \frac{- 1 / π^{2}}{\sqrt{ε^{2} - t^{2}}} \int_{- ε}^{ε} \sqrt{ε^{2} - τ^{2}} \frac{\partial_{τ} R_{⋆} [μ] (τ) - \partial_{τ} R_{⋆} [μ] (t)}{t - τ} d τ \\ + \frac{- 1 / π^{2}}{\sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \sqrt{ε^{2} - τ^{2}} \frac{\partial_{τ} R_{⋆} [μ] (t)}{t - τ} d τ, \end{array}$ thus $\begin{array}{l} {‖ \frac{- 1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} \partial_{τ} R_{⋆} [μ] (τ)}{t - τ} d τ ‖}_{X^{ε}}^{2} \\ ⩽ \int_{- ε}^{ε} \frac{1 / π^{4}}{\sqrt{ε^{2} - t^{2}}} {| \int_{- ε}^{ε} \sqrt{ε^{2} - τ^{2}} \frac{\partial_{τ} R_{⋆} [μ] (τ) - \partial_{τ} R_{⋆} [μ] (t)}{t - τ} d τ |}^{2} d t \\ + \int_{- ε}^{ε} \frac{1 / π^{4}}{\sqrt{ε^{2} - t^{2}}} {| p . v . \int_{- ε}^{ε} \sqrt{ε^{2} - τ^{2}} \frac{\partial_{τ} R_{⋆} [μ] (t)}{t - τ} d τ |}^{2} d t \\ ⩽ \int_{- ε}^{ε} \frac{1 / π^{4}}{\sqrt{ε^{2} - t^{2}}} {| \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{\sqrt{| t - τ |}} d τ |}^{2} d t {‖ \partial_{t} R_{⋆} [μ] ‖}_{C^{0, 1 / 2}}^{2} \\ + \int_{- ε}^{ε} \frac{1 / π^{4}}{\sqrt{ε^{2} - t^{2}}} {| p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}}}{t - τ} d τ |}^{2} d t {‖ \partial_{t} R_{⋆} [μ] ‖}_{C^{0}}^{2} \\ ⩽ C_{(2.34)}^{2} \frac{ε^{3}}{π} {| \partial_{t} R_{⋆} [μ] |}_{C^{0, 1 / 2}}^{2} + \frac{ε^{2}}{2 π} {‖ \partial_{t} R_{⋆} [μ] ‖}_{C^{0}}^{2} . \end{array}$ We conclude that $\begin{array}{l} {‖ {(L^{ε})}^{- 1} [μ_{ι}] ‖}_{X^{ε}} & = ‖ R ‖_{C^{0, 1 / 2}} (O (\frac{ε}{| log (ε) |}) + O (\frac{ε}{| log (ε) |}) + O (ε^{3 / 2}) + O (\frac{ε}{{(log (ε / 2))}^{2}})) \\ = ‖ R ‖_{C^{0, 1 / 2}} O (\frac{ε}{| log (ε) |}) . \end{array}$ This proves Lemma 2.14. □

2.3.3. Solution to the main problem

We know from Proposition 2.9, that $\begin{array}{l} - \partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix})) & = (- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε}) [v_{z}^{k, ε}] \\ + \int_{- ε}^{ε} v_{z}^{k, ε} (t) (\frac{k^{2}}{8 π} (2 log (k) + 1) + \partial_{x_{2}} R_{\partial G, Ω}^{k} ((\begin{matrix} τ \\ 0 \end{matrix}), (\begin{matrix} t \\ 0 \end{matrix}))) d t . \end{array}$ In the last subsection we saw that the operator $(- α J^{ε} + β L^{ε})$ , with $β = \frac{k^{2}}{4 π}$ and $α = \frac{1}{2 π}$ , is invertible, thus we have $\begin{array}{l} v_{z}^{k, ε} (t) & = {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [- \partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix}))] (t) - {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} \\ \times [\int_{- ε}^{ε} v_{z}^{k, ε} (t) (\frac{k^{2}}{8 π} (2 log (k) + 1) + \partial_{x_{2}} R_{\partial G, Ω}^{k} ((\begin{matrix} τ \\ 0 \end{matrix}), (\begin{matrix} t \\ 0 \end{matrix}))) d t] . \end{array}$

Then Proposition 2.5 yields $\begin{array}{l} u_{z}^{k, ε} (x) & = G_{Ω}^{k} (z, x) + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) v_{z}^{k, ε} (y) d σ_{y} \\ = G_{Ω}^{k} (z, x) - \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [\partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix}))] (y) d σ_{y} \\ + \int_{\partial Ω_{N}} \partial_{ν_{y}} G_{Ω}^{k} (x, y) {(\frac{- 1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [\int_{- ε}^{ε} v_{z}^{k, ε} (t) (\frac{k^{2}}{8 π} (2 log (k) + 1) \\ + \partial_{x_{2}} R_{\partial G, Ω}^{k} ((\begin{matrix} τ \\ 0 \end{matrix}), (\begin{matrix} t \\ 0 \end{matrix}))) d t] (y) d σ_{y} \end{array}$ from where we obtain $G_{(1)}^{k, ε}$ and $G_{(2)}^{k, ε}$ in Theorem 2.4.

With Lemma 2.14, that is $‖ {(- α J^{ε} + β L^{ε})}^{- 1} R [μ] ‖_{X^{ε}} ⩽ \frac{ε C_{(2.29)}}{| log (ε) |} ‖ R ‖_{C^{0, 1 / 2}} ‖ μ ‖_{X^{ε}}$ , and with the simple reformulation $\begin{matrix} - \partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix})) = \int_{- ε}^{ε} - \partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix})) \frac{1}{π \sqrt{ε^{2} - t^{2}}} d t, \end{matrix}$ we can infer $\begin{matrix} (2.44) & {‖ v_{z}^{k, ε} ‖}_{X^{ε}} ⩽ \frac{ε C_{(2.44), 1}}{| log (ε) |} {‖\partial_{ν_{x}} G_{Ω}^{k} (z, (\begin{matrix} τ \\ 0 \end{matrix}))‖}_{C^{0, 1 / 2}} + \frac{ε C_{(2.44), 2}}{| log (ε) |} {‖ v_{z}^{k, ε} ‖}_{X^{ε}} . \end{matrix}$ Consider that $\int_{- ε}^{ε} μ (t) d t ⩽ \sqrt{π} ‖ μ ‖_{X^{ε}}$ , due to the $L^{2}$ -Cauchy–Schwarz inequality. Hence we conclude for $z, x \in Ω$ , $z \neq x$ that $\begin{array}{l} (2.45) & | G_{(1)}^{k, ε} (z, x) | ⩽ \frac{ε C_{(2.45)}}{| log (ε) |}, \\ (2.46) & | G_{(2)}^{k, ε} (z, x) | ⩽ \frac{ε^{2} C_{(2.46)}}{| log (ε) |^{2}} . \end{array}$

3. Nucleation of the Neumann boundary condition

In this section, based on Theorem 2.4, we derive a simple procedure to maximize the norm of the Green’s function. The main idea is to nucleate the Neumann boundary conditions in order to increase the transmission between the point source and the receiver. By considering a disk shaped cavity, we illustrate by some numerical experiments the applicability of the proposed approach.

3.1. The disk case

Let Ω be the unit disk and let the source and the receiver be respectively $x_{S} = (x, y)$ and $x_{R} = (ξ, η)$ . Suppose that the opening $\partial Ω_{N}$ is an arc centered at $(1, 0)$ with length $2 ε$ .

Denote by $ρ = \sqrt{x^{2} + y^{2}}$ the distance between $x_{S}$ and the origin, and by $\tilde{ρ} = \sqrt{ξ^{2} + η^{2}}$ the distance between $x_{R}$ and the origin. Define $θ = arctan (y / x)$ , and $\tilde{θ} = arctan (η / ξ)$ . It is well known that the Green’s function in the unit disk is given by $\begin{matrix} (3.1) & G_{Ω}^{k} (x_{S}, x_{R}) = - \frac{i}{4} H_{0}^{(1)} (k | x_{S} - x_{R} |) + \frac{i}{4} \sum_{n = - \infty}^{+ \infty} A_{n} (k, ρ) J_{n} (k \tilde{ρ}) e^{i n (θ - \tilde{θ})} . \end{matrix}$ Recall that the cylindrical wave expansion of the free-boundary Green’s function is $\begin{matrix} (3.2) & - \frac{i}{4} H_{0}^{(1)} (k | x_{S} - x_{R} |) = - \frac{i}{4} \sum_{n = - \infty}^{+ \infty} J_{n} (k ρ_{<}) H_{n}^{(1)} (k ρ_{>}) e^{i n (θ - \tilde{θ})}, \end{matrix}$ where $ρ_{<} = min (ρ, \tilde{ρ})$ , $ρ_{>} = max (ρ, \tilde{ρ})$ . Substituting (3.2) into (3.1) yields $\begin{matrix} (3.3) & G_{Ω}^{k} (x_{S}, x_{R}) = \frac{i}{4} \sum_{n = - \infty}^{\infty} (A_{n} J_{n} (k ρ) - J_{n} (k ρ_{<}) H_{n}^{(1)} (k ρ_{>})) e^{i n (θ - \tilde{θ})} . \end{matrix}$ Imposing the Dirichlet boundary condition on (3.3) gives $\begin{matrix} A_{n} (k, ρ) = \frac{J_{n} (k ρ) H_{n}^{(1)} (k)}{J_{n} (k)}, \end{matrix}$ where $J_{n}$ is the Bessel function of first kind and order n.

Hence the Green’s function is $\begin{matrix} (3.4) & G_{Ω}^{k} (x_{S}, x_{R}) = - \frac{i}{4} H_{0}^{(1)} (k | x_{S} - x_{R} |) + \frac{i}{4} \sum_{n = - \infty}^{\infty} \frac{J_{n} (k ρ) H_{n}^{(1)} (k)}{J_{n} (k)} J_{n} (k \tilde{ρ}), \end{matrix}$ and its normal derivative on $\partial Ω_{N}$ is $\begin{array}{l} \frac{\partial G_{Ω}^{k}}{\partial \tilde{ρ}} (x_{S}, (\begin{matrix} t \\ 1 \end{matrix})) & = - \frac{i}{4} k H_{0}^{{(1)}^{'}} (k \sqrt{ρ^{2} + 1 - 2 ρ cos (θ - θ_{0})}) \frac{1 - ρ cos (θ - θ_{0})}{\sqrt{ρ^{2} + 1 - 2 ρ cos (θ - θ_{0})}} \\ (3.5) & + \frac{i}{4} \sum_{n = - \infty}^{\infty} \frac{J_{n} (k ρ) H_{n}^{(1)} (k)}{J_{n} (k)} k J_{n}^{'} (k), \end{array}$ with $arctan θ_{0} = 1 / t$ .

Define $\begin{matrix} I_{1} : = \int_{- ε}^{ε} \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} t \\ 1 \end{matrix})) {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [\partial_{ν_{x}} G_{Ω}^{k} (x_{S}, (\begin{matrix} s \\ 1 \end{matrix}))] (t) d t . \end{matrix}$

By Proposition 2.12, we have $\begin{matrix} {(- α J^{ε} + β L^{ε})}^{- 1} [f] (t) = {(L^{ε})}^{- 1} [u_{l}] (t), \end{matrix}$ where $\begin{matrix} u_{l} (t) = C_{1} e^{\sqrt{\frac{β}{α}} t} + C_{2} e^{- \sqrt{\frac{β}{α}} t} + \frac{1}{2 \sqrt{α β}} e^{- \sqrt{\frac{β}{α}} t} \int_{- ε}^{t} f (s) e^{\sqrt{\frac{β}{α}} s} d s - \frac{1}{2 \sqrt{α β}} e^{\sqrt{\frac{β}{α}} t} \int_{- ε}^{t} f (s) e^{- \sqrt{\frac{β}{α}} s} d s, \end{matrix}$ and $C_{1}$ , $C_{2}$ are constants determined by (2.15) and (2.16). Therefore, $\begin{matrix} u_{l}^{'} (τ) = \sqrt{\frac{β}{α}} (C_{1} e^{\sqrt{\frac{β}{α}} τ} - C_{2} e^{- \sqrt{\frac{β}{α}} τ} - \frac{1}{2 \sqrt{α β}} \int_{- ε}^{τ} f (s) 2 cosh (\sqrt{\frac{β}{α}} (s - τ)) d s) . \end{matrix}$ By Taylor expansion, $\begin{array}{l} u_{l}^{'} (τ) - u_{l}^{'} (0) \\ = \sqrt{\frac{β}{α}} (C_{1} e^{\sqrt{\frac{β}{α}} τ} - C_{2} e^{- \sqrt{\frac{β}{α}} τ} - (C_{1} - C_{2})) - \frac{1}{2 α} \int_{0}^{τ} f (s) 2 cosh (\sqrt{\frac{β}{α}} (s - τ)) d s \\ - \frac{1}{2 α} \int_{- ε}^{0} f (s) 2 (cosh (\sqrt{\frac{β}{α}} (s - τ)) - cosh (\sqrt{\frac{β}{α}} s)) d s \\ = \sqrt{\frac{β}{α}} (C_{1} \sqrt{\frac{β}{α}} + C_{2} \sqrt{\frac{β}{α}}) τ + \frac{1}{2} {(\frac{β}{α})}^{3 / 2} (C_{1} - C_{2}) τ^{2} \\ - \frac{1}{2 α} (2 f (0) τ + f^{'} (0) τ^{2} + \frac{β}{α} f (0) τ^{2} ε - \frac{β}{α} f (0) τ ε^{2} + O (τ^{3})) \\ = (\frac{β}{α} (C_{1} + C_{2}) τ - \frac{1}{α} f (0) + \frac{β}{2 α^{2}} f (0) ε^{2}) τ \\ (3.6) & + (\frac{1}{2} {(\frac{β}{α})}^{3 / 2} (C_{1} - C_{2}) - \frac{1}{2 α} f^{'} (0) - \frac{β}{2 α^{2}} f (0) ε) τ^{2} + O (ε, τ, 3), \end{array}$ where $O (ε, τ, 3)$ denotes the infinite sum of terms of the form $C ε^{n} τ^{m}$ , where $n + m ⩾ 3$ . Denote by $\begin{matrix} (3.7) & C_{u_{l}}^{1} : = \frac{β}{α} (C_{1} + C_{2}) - \frac{1}{α} f (0) + \frac{β}{2 α^{2}} f (0) ε^{2} \end{matrix}$ and by $\begin{matrix} (3.8) & C_{u_{l}}^{2} : = \frac{1}{2} {(\frac{β}{α})}^{3 / 2} (C_{1} - C_{2}) - \frac{1}{2 α} f^{'} (0) - \frac{β}{2 α^{2}} f (0) ε . \end{matrix}$

For $u_{l}^{'} (0)$ we have $\begin{array}{l} u_{l}^{'} (0) & = \sqrt{\frac{β}{α}} (C_{1} - C_{2} - \frac{1}{2 \sqrt{α β}} \int_{- ε}^{0} f (s) 2 cosh (\sqrt{\frac{β}{α}} s) d s) \\ = \sqrt{\frac{β}{α}} (C_{1} - C_{2}) \\ - \frac{1}{\sqrt{α β}} (f (0) \sqrt{\frac{α}{β}} sinh (\sqrt{\frac{β}{α}} ε) \\ + f^{'} (0) \frac{α}{β} (- 1 + cosh (\sqrt{\frac{β}{α}} ε) - ε \sqrt{\frac{β}{α}} sinh (\sqrt{\frac{β}{α}} ε)) + O (ε^{3})) \\ (3.9) & = \sqrt{\frac{β}{α}} (C_{1} - C_{2}) - \frac{1}{α} f (0) ε + \frac{1}{2 α} f^{'} (0) ε^{2} + O (ε^{3}), \end{array}$ while for $u_{l} (0)$ it holds that $\begin{array}{l} u_{l} (0) & = C_{1} + C_{2} + \frac{1}{2 \sqrt{α β}} \int_{- ε}^{0} f (s) (e^{\sqrt{\frac{β}{α}} s} - e^{- \sqrt{\frac{β}{α}} s}) d s \\ (3.10) & = C_{1} + C_{2} + \frac{1}{α} f (0) (- \frac{1}{2} ε^{2}) + O (ε^{3}) . \end{array}$ Recall the formula for ${(L^{ε})}^{- 1}$ and $C_{L^{ε}} [u_{l}]$ in Proposition 2.10. Since $C_{L^{ε}} [u_{l}]$ is a constant, we can simply evaluate it at $x = 0$ . We find that $\begin{array}{l} C_{L^{ε}} [u_{l}] & = u_{l} (0) - L^{ε} [- \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} u_{l}^{'} (τ)}{t - τ} d τ] (0) \\ = u_{l} (0) + \int_{- ε}^{ε} p . v . \int_{- ε}^{ε} \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} \frac{\sqrt{ε^{2} - τ^{2}} u_{l}^{'} (τ)}{t - τ} log | t | d τ d t . \end{array}$ From (3.6), we obtain $\begin{array}{l} \int_{- ε}^{ε} p . v . \int_{- ε}^{ε} \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} \frac{\sqrt{ε^{2} - τ^{2}} u_{l}^{'} (τ)}{t - τ} log | t | d τ d t \\ = \int_{- ε}^{ε} p . v . \int_{- ε}^{ε} \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} \frac{\sqrt{ε^{2} - τ^{2}} (u_{l}^{'} (0) + C_{u_{l}}^{1} τ + C_{u_{l}}^{2} τ^{2} + O (ε^{3}))}{t - τ} log | t | d τ d t \\ = u_{l}^{'} (0) \int_{- ε}^{ε} \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} π t log | t | d t + C_{u_{l}}^{1} \int_{- ε}^{ε} \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} (- \frac{π ε^{2}}{2} + π t^{2}) log | t | d t \\ + C_{u_{l}}^{2} \int_{- ε}^{ε} \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} (- \frac{1}{2} ε^{2} π t + π t^{3}) log | t | d t + O (ε^{4}) \\ = \frac{C_{u_{l}}^{1}}{π} (- \frac{ε^{2}}{2} (- π log (\frac{2}{ε})) - \frac{1}{4} ε^{2} π (- 1 + 2 log (\frac{2}{ε}))) + 0 + O (ε^{4}) \\ = \frac{C_{u_{l}}^{1}}{4} ε^{2} + O (ε^{4}) . \end{array}$ Here, we have used the fact that $\begin{array}{l} p . v . \int_{- ε}^{ε} \frac{τ \sqrt{ε^{2} - τ^{2}}}{t - τ} d τ = - \frac{π ε^{2}}{2} + π t^{2}, \\ \int_{- ε}^{ε} \frac{log | t |}{\sqrt{ε^{2} - t^{2}}} d t = - π log (\frac{2}{ε}), \end{array}$ and $\begin{matrix} \int_{- ε}^{ε} \frac{t^{2} log | t |}{\sqrt{ε^{2} - t^{2}}} d t = - \frac{1}{4} ε^{2} π (- 1 + 2 log (\frac{2}{ε})) . \end{matrix}$ Therefore, the following estimation holds: $\begin{matrix} (3.11) & C_{L^{ε}} [u_{l}] = u_{l} (0) + \frac{C_{u_{l}}^{1}}{4} ε^{2} + O (ε^{4}) . \end{matrix}$ We now compute $C_{1}$ and $C_{2}$ from (2.15) and (2.16). Applying (3.6) to (2.15), we get $\begin{matrix} \int_{- ε}^{ε} \sqrt{\frac{ε + τ}{ε - τ}} (u_{l}^{'} (0) + C_{u_{l}}^{1} τ + C_{u_{l}}^{2} τ^{2} + O (τ^{3})) d τ - \frac{π}{log (ε / 2)} C_{L^{ε}} [u_{l}] = 0, \end{matrix}$ which implies that $\begin{matrix} u_{l}^{'} (0) π ε + C_{u_{l}}^{1} \frac{π ε^{2}}{2} + C_{u_{l}}^{2} \frac{π ε^{3}}{2} - \frac{π}{log (ε / 2)} (u_{l} (0) + \frac{C_{u_{l}}^{1}}{4} ε^{2}) + O (ε^{4}) = 0 . \end{matrix}$ Combining the last equation together with (3.10) and (3.9), it follows that $\begin{array}{l} (\sqrt{\frac{β}{α}} (C_{1} - C_{2}) - \frac{1}{α} f (0) ε + O (ε^{2})) π ε + C_{u_{l}}^{1} \frac{π ε^{2}}{2} + C_{u_{l}}^{2} \frac{π ε^{3}}{2} + O (ε^{4}) \\ (3.12) & - \frac{π}{log (ε / 2)} (C_{1} + C_{2} - \frac{1}{2 α} f (0) ε^{2} + O (ε^{4}) + \frac{C_{u_{l}}^{1}}{4} ε^{2}) = 0 . \end{array}$ Thus, $\begin{array}{l} \sqrt{\frac{β}{α}} ε (C_{1} - C_{2}) - \frac{1}{log (ε / 2)} (C_{1} + C_{2}) \\ (3.13) & = \frac{1}{α} f (0) ε^{2} - \frac{C_{u_{l}}^{1}}{2} ε^{2} - \frac{C_{u_{l}}^{2}}{2} ε^{3} - \frac{1}{log (ε / 2)} (\frac{1}{2 α} f (0) ε^{2} - \frac{C_{u_{l}}^{1}}{4} ε^{2}) + O (ε^{4}) . \end{array}$ Similarly, from (2.16) we obtain $\begin{array}{l} - \sqrt{\frac{β}{α}} ε (C_{1} - C_{2}) - \frac{1}{log (ε / 2)} (C_{1} + C_{2}) \\ (3.14) & = - \frac{1}{α} f (0) ε^{2} - \frac{C_{u_{l}}^{1}}{2} ε^{2} + \frac{C_{u_{l}}^{2}}{2} ε^{3} - \frac{1}{log (ε / 2)} (\frac{1}{2 α} f (0) ε^{2} - \frac{C_{u_{l}}^{1}}{4} ε^{2}) + O (ε^{4}) . \end{array}$ Therefore, $\begin{matrix} (3.15) & C_{1} + C_{2} = (\frac{1}{2 α} f (0) - \frac{C_{u_{l}}^{1}}{4}) ε^{2} + \frac{C_{u_{l}}^{1}}{2} log (ε / 2) ε^{2} + O (ε^{3}), \end{matrix}$ and $\begin{matrix} (3.16) & C_{1} - C_{2} = \frac{1}{\sqrt{α β}} f (0) ε - \frac{1}{2} \sqrt{\frac{α}{β}} C_{u_{l}}^{2} ε^{2} + O (ε^{2}) . \end{matrix}$

Combining together (3.7) with (3.15) gives $\begin{array}{l} C_{1} + C_{2} & = \frac{\frac{- 1}{2 α} f (0) log (ε / 2) ε^{2} + \frac{3}{4 α} f (0) ε^{2}}{1 - \frac{β}{2 α} log (ε / 2) ε^{2} + \frac{β}{4 α} ε^{2}} + O (ε^{3} log ε) \\ (3.17) & = - \frac{1}{2 α} f (0) log (ε / 2) ε^{2} + \frac{3}{4 α} f (0) ε^{2} + O (ε^{2} log ε) \end{array}$ and $\begin{matrix} (3.18) & C_{u_{l}}^{1} = - \frac{1}{α} f (0) + \frac{β}{α} (- \frac{1}{2 α} f (0) log (ε / 2) ε^{2} + \frac{3}{4 α} f (0) ε^{2}) + O (ε^{2} log ε), \end{matrix}$ while combining (3.8) and (3.16) together leads to $\begin{matrix} (3.19) & C_{1} - C_{2} = \frac{1}{\sqrt{α β}} f (0) ε + \frac{1}{4 \sqrt{α β}} f^{'} (0) ε^{2} + O (ε^{3}), \end{matrix}$ and $\begin{matrix} (3.20) & C_{u_{l}}^{2} = - \frac{1}{2 α} f^{'} (0) + \frac{1}{8 α^{2}} f^{'} (0) ε^{2} + O (ε^{3}) . \end{matrix}$ Now, we are ready to estimate ${(L^{ε})}^{- 1} [u_{l}] (t)$ . From Proposition 2.10, we know that $\begin{array}{l} {(L^{ε})}^{- 1} [u_{l}] (t) & = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} p . v . \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} u_{l}^{'} (τ)}{t - τ} d τ + \frac{C_{L^{ε}} [u_{l}]}{π log (ε / 2) \sqrt{ε^{2} - t^{2}}} \\ = - \frac{1}{π^{2} \sqrt{ε^{2} - t^{2}}} \int_{- ε}^{ε} \frac{\sqrt{ε^{2} - τ^{2}} (u_{l}^{'} (0) + C_{u_{l}}^{1} τ + C_{u_{l}}^{2} τ^{2} + O (τ^{3}))}{t - τ} d τ \\ + \frac{C_{L^{ε}} [u_{l}]}{π log (ε / 2) \sqrt{ε^{2} - t^{2}}} \\ = - \frac{1}{π \sqrt{ε^{2} - t^{2}}} u_{l}^{'} (0) t - \frac{C_{u_{l}}^{1}}{π \sqrt{ε^{2} - t^{2}}} (- \frac{1}{2} ε^{2} + \frac{1}{2} t^{2}) \\ (3.21) & - \frac{C_{u_{l}}^{2}}{π \sqrt{ε^{2} - t^{2}}} (- \frac{1}{2} ε^{2} π t + π t^{3}) + O (ε^{4}) . \end{array}$ Plugging $α = 1 / (2 π)$ and $β = k^{2} / (4 π)$ into (3.9), we get $\begin{array}{l} {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [f] (t) \\ (3.22) & = - \frac{1}{\sqrt{ε^{2} - t^{2}}} (- f (0) t^{2} - f^{'} (0) π t^{3} + f (0) ε^{2} + \frac{1}{2} f^{'} (0) ε^{2} π t + \frac{3}{2} f^{'} (0) ε^{2} t + O (ε^{3})) . \end{array}$ Therefore, $\begin{array}{l} \int_{- ε}^{ε} \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} t \\ 1 \end{matrix})) {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [f] (t) d t \\ = (\partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} 0 \\ 1 \end{matrix})) + O (ε)), \\ (3.23) & \int_{- ε}^{ε} - \frac{1}{\sqrt{ε^{2} - t^{2}}} (- f (0) t^{2} - f^{'} (0) π t^{3} + f (0) ε^{2} + \frac{1}{2} f^{'} (0) ε^{2} π t + \frac{3}{2} f^{'} (0) ε^{2} t + O (ε^{3})) d t \\ = (\partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} 0 \\ 1 \end{matrix})) + O (ε^{1})) (f (0) \frac{ε^{2} π}{2} - f (0) ε^{2} π + O (ε^{3})) \\ = - f (0) \frac{π}{2} ε^{2} \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} 0 \\ 1 \end{matrix})) + O (ε^{3}) . \end{array}$ Recall in our setting that $f (s) : = \partial_{ν_{x}} G_{Ω}^{k} (x_{S}, (\begin{matrix} s \\ 1 \end{matrix}))$ . Thus, $\begin{array}{l} \int_{- ε}^{ε} \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} t \\ 1 \end{matrix})) {(- \frac{1}{2 π} J^{ε} + \frac{k^{2}}{4 π} L^{ε})}^{- 1} [\partial_{ν_{x}} G_{Ω}^{k} (x_{S}, (\begin{matrix} s \\ 1 \end{matrix}))] (t) d t \\ (3.24) & = - \frac{π}{2} ε^{2} \partial_{ν_{x}} G_{Ω}^{k} (x_{S}, (\begin{matrix} 0 \\ 1 \end{matrix})) \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} 0 \\ 1 \end{matrix})) + O (ε^{3}) . \end{array}$

3.2. Numerical illustrations

Now, two numerical experiments are presented in order to verify the applicability of the proposed methodology. In each one, the topological derivative is evaluated to detect the parts of the boundary where a Neumann boundary condition should be nucleated.

Denote by $\begin{matrix} (3.25) & F_{S} (θ) : = \frac{\partial G_{Ω}^{k}}{\partial \tilde{ρ}} (x_{S}, (\begin{matrix} 0 \\ 1 \end{matrix})), \end{matrix}$ and by $\begin{matrix} (3.26) & F_{R} (θ) : = \frac{\partial G_{Ω}^{k}}{\partial \tilde{ρ}} (x_{R}, (\begin{matrix} 0 \\ 1 \end{matrix})) . \end{matrix}$

Our goal is to maximize $| u_{x_{S}}^{k, ε} |^{2}$ , i.e., the norm of Green function at the receiver. We plot $y (θ) : = ℜ (π \partial_{ν_{x}} G_{Ω}^{k} (x_{S}, (\begin{matrix} 0 \\ 1 \end{matrix})) \partial_{ν_{y}} G_{Ω}^{k} (x_{R}, (\begin{matrix} 0 \\ 1 \end{matrix})))$ as a function of θ.

Fig. 2.

Plot of $y (θ)$ as a function of θ.

Fig. 3.

Nucleation of the Neumann boundary condition.

Set the wave number to be $k = 200$ , the distance between the source and the origin to be $ρ = 0.4$ , the distance between the receiver and the origin to be $\tilde{ρ} = 0.2$ , the angle difference from the receiver to the source to be $π / 3$ . We divide the whole boundary into $N = 10000$ parts, and set $N / 1000 = 10$ parts left and right of each local maximal point of $y (θ)$ to be a Neumann part. Figure 2 presents $y (θ)$ . Figure 3 gives the corresponding configuration of the boundary, where the red part corresponds to a Neumann boundary condition and the blue part corresponds to a Dirichlet boundary condition. The implementation shows that the norm of the Green function $G_{Ω}^{k}$ with Dirichlet boundary at $x_{R}$ is 0.067875, while the norm of the Green function $u_{x_{S}}^{k, ε}$ with modified boundary at $x_{R}$ is 32.403610.

4. Conclusion

In this paper, we have highlighted the mechanism of the Dirichlet function to exploit the intrinsic properties of the wave behaviour near and away from the gaps. We have established an asymptotic expansion of the Green’s function with a small Neumann boundary injection. With this knowledge we were able to answer both question; what are the zeroth-order and first-order terms in the asymptotic expansion and what is the vital decision to enhance the signal at a receiving point inside the cavity, when we are able to switch the boundary conditions from Dirichlet to Neumann on specific parts of the cavity boundary. We finally made a case study on the circle to illustrate the behaviour on a domain with well known Green’s functions.

Our approach opens many new avenues for mathematical imaging and focusing of waves in complex media. The results in Sections 2 and 3 can lead to enhanced communication between devices, like cell phones by improving the transmission between a source and a receiver through specific eigenmodes of the cavity. However, many challenging problems are still to be solved. For instance, how to optimize some specific cavity eigenmodes or how to design broadband metasurfaces which allow for broadband shaping and controlling of waves in complex media. These challenging problems would be the subject of forthcoming works.

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