# Design of an acoustic energy distributor using thin resonant slits

We consider the propagation of time harmonic acoustic waves in a device with three channels. The wave number is chosen such that only the piston mode can propagate. The main goal of this work is to present a geometry which can serve as an energy distributor. More precisely, the geometry is first designed so that for an incident wave coming from one channel, the energy is almost completely transmitted in the two other channels. Additionally, tuning a bit two geometrical parameters, we can control the ratio of energy transmitted in the two channels. The approach is based on asymptotic analysis for thin slits around resonance lengths. We also provide numerical results to illustrate the theory.

## Authors

• 5 publications
• 3 publications
• ### Design of a mode converter using thin resonant ligaments

The goal of this work is to design an acoustic mode converter. More prec...
02/15/2021 ∙ by Lucas Chesnel, et al. ∙ 0

• ### Acoustic passive cloaking using thin outer resonators

We consider the propagation of acoustic waves in a 2D waveguide unbounde...
05/03/2021 ∙ by Lucas Chesnel, et al. ∙ 0

• ### Modelling wave dispersion in fluid saturating periodic scaffolds

Acoustic waves in a slightly compressible fluid saturating porous period...
01/06/2021 ∙ by Eduard Rohan, et al. ∙ 0

• ### A high-order discontinuous Galerkin method for nonlinear sound waves

We propose a high-order discontinuous Galerkin scheme for nonlinear acou...
12/04/2019 ∙ by Paola F. Antonietti, et al. ∙ 0

• ### Damping of Propagating Kink Waves in the Solar Corona

Alfvénic waves have gained renewed interest since the existence of ubiqu...
04/18/2019 ∙ by Ajay K. Tiwari, et al. ∙ 0

• ### Acoustic Structure Inverse Design and Optimization Using Deep Learning

From ancient to modern times, acoustic structures have been used to cont...
01/29/2021 ∙ by Xuecong Sun, et al. ∙ 16

• ### Investigating Wave Energy Potential in Southern Coasts of the Caspian Sea and Evaluating the Application of Gray Wolf Optimizer Algorithm

There is a significantly accelerating trend in the application of the wa...
12/31/2019 ∙ by Erfan Amini, et al. ∙ 4

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

In this article, we are interested in the design of an acoustic energy distributor. More precisely, we study the propagation of time harmonic waves at a given wave number in a structure with one input/output and two outputs (see Figure 1). Our goal is to find a geometry where the energy of an incoming wave is almost completely transmitted and additionally where we can control the ratio of energy transmitted in the two other output channels. The main difficulty of this problem lies in the fact that the dependence of the acoustic field with respect to the geometry is non linear and implicit. A device closed to the one that we wish to create is the acoustic power divider [LaLI05, AnEl05, NgCa07, EsBA20]. The difference is that in our case, we want to be able to control the ratio of energy transmitted in the two output channels. We mention that such devices are very interesting for applications not only in acoustics but also for example in optics [PLLPBV15, FESJY13] or in radio electronics [ChWu11] (see the literature concerning radio diplexers).

To exhibit such particular waveguides, contrary to for example [EsBA20], we will not use zero-index material or similar metamaterials which are, from our understanding, still hard to handle in practice. Instead we will work with thin slits as illustrated in Figure 1. In general, due to the geometrical features, almost no energy passes through the slits and it may seem a bit paradoxical to use them to have almost complete transmission. However working around the resonance lengths, it has been shown that we can observe this phenomenon. This has been studied for example in [Krieg, BoTr10, LiZh17, LiZh18, LiSZ19] in the context of the scattering of an incident wave by a periodic array of subwavelength slits. The approach that we will consider is based on matched asymptotic expansions. For related techniques, we refer the reader to [Beal73, Gady93, KoMM94, Naza96, Gady05, Naza05, JoTo06, BaNa15, BoCN18]. We emphasize that an important feature of our work distinguishing it from the previous references is that the lengths, and not only the widths, of the slits depend on (see (1)). This way of considering the problem is an essential ingredient of the analysis.

Let us mention that techniques of optimization (see e.g. [LDOHG19, LGHDO19, Lebb19]) have been applied to exhibit energy acoustic distributors. However they involve non convex functionals and unsatisfying local minima exist. Moreover, they offer no control on the obtained shape compare to the approach we propose here. In particular, with our geometry, a small change of the geometry allows us to transmit the energy in one channel instead of the other. In the context of propagation of acoustic waves, another device which is interesting in practice is the modal converter [HoHu05, CXJD06, KTSM09]. At higher frequency, when several modes can propagate, the aim is to have a structure where the energy of an incident mode is transferred onto another mode. We do not know if thin slits can be useful to obtain such effect.

The outline is as follows. In the next section, we present the geometry and the notation. Then in Section 3, we introduce two auxiliary problems which will be used in the analysis. The Section 4 constitutes the heart of the article: here we compute an asymptotic expansion of the acoustic field and of the scattering coefficients with respect to , the width of the thin slits. Then we exploit the results in Section 5 to exhibit situations where the device acts as an energy distributor. We illustrate the theory in Section 6 with numerical experiments before proving two technical lemmas needed in the study in a short appendix.

## 2 Setting

First, we describe in detail the geometry (see Figure 1). Set . Pick two different points . For , define the lengths

 Lε±:=L±+εL′± (1)

where the values , will be fixed later on to observe interesting phenomena. Define the thin strips

 Sε±:=(p±−ε/2;p±+ε/2)×[1;1+Lε±].

Define the points such that . Set

 Πε±:=(p±−1/2;p±+1/2)×(1+Lε±;+∞).

And finally, we define the geometry

 Ωε:=Π0∪Sε+∪Sε−∪Πε+∪Πε−.

Interpreting the domain as an acoustic waveguide, we are led to consider the following problem with Neumann boundary condition

 (2)

Here, is the Laplace operator while corresponds to the derivative along the exterior normal. Furthermore, is the acoustic pressure of the medium while is the wave number of the piston modes (resp. ) propagating in (resp. ). We fix so that no other mode can propagate. We are interested in the solution to the diffraction problem (2) generated by the incoming wave in the trunk . This solution admits the decomposition

 uε(x,y)=w+h(x)+Rεw−h(x)+… in Π0w+h(x)Tε+w+v(y−Lε)+… in Πε+w+h(x)Tε−w+v(y−Lε)+… in Πε− (3)

where is a reflection coefficient and are transmission coefficients. In this decomposition, the ellipsis stand for a remainder which decays at infinity with the rate in and in . Due to conservation of energy, one has

 |Rε|2+|Tε+|2+|Tε−|2=1.

In general, almost no energy of the incident wave passes through the thin strips and one observes almost complete reflection. More precisely, one finds that there holds

 Rε=1+~Rε,Tε±=~Tε±, (4)

where , tend to zero as goes to zero. The main goal of this work is to show that choosing carefully the lengths of the thin strips as well as their positions, the energy of the wave can be almost completely transmitted. Moreover we can control the energy transmitted respectively in and . More precisely, we will prove that choosing carefully , as tends to zero we can have

 Rε=~Rε,Tε±=T0±+~Tε,

where , tend to zero as goes to zero, and can be any number in (see formulas (31) below). Thus we can select the energy ratio transmitted in and the device acts as an energy distributor.

## 3 Auxiliary objects

In this section, we discuss a couple of boundary value problems whose solutions will appear in the construction of the asymptotic expansions of the acoustic field .

Considering the limit in the equation (2) restricted to the strips , we are led to study the one-dimensional Helmholtz equations

 ∂2yv+ω2v=0 in ℓ±:=(1;1+L±) (5)

supplied with the artificially imposed Dirichlet conditions

 v(1)=v(1+L±)=0. (6)

Eigenvalues and eigenfunctions (up to a multiplicative constant) of the boundary value problem (5)–(6) are given by

 μm:=(πm/L±)2,vm(y)=sin(πm(y−1)/L±),

with .

Now we present a second problem which is involved in the construction of asymptotics and which will be used to describe the boundary layer phenomenon near the points , . To capture rapid variations of the field for example in the vicinity of , we introduce the stretched coordinates . Observing that

 (Δz+ω2)uε(ε−1(z−A±))=ε−2Δξ±uε(ξ±)+…, (7)

we are led to consider the Neumann problem

 −ΔξY=0 in Ξ,∂νY=0 % on ∂Ξ, (8)

where (see Figure 2) is the union of the half-plane and the semi-strip such that

 Ξ−:=R2−={ξ=(ξx,ξy):ξy<0},Ξ+:={ξ:ξy≥0,|ξx|<1/2}.

In the method of matched asymptotic expansions (see the monographs [VD, Ilin], [MaNaPl, Chpt. 2] and others) that we will use, we will work with solutions of (8) which are bounded or which have polynomial growth in the semi-strip as . One of such solutions is evident and is given by . Another solution, which is linearly independent with , is the unique function satisfying (8) and which has the representation

 Y1(ξ)=⎧⎪⎨⎪⎩ξy+CΞ+O(e−πξy)% as ξy→+∞,ξ∈Ξ+1πln1|ξ|+O(1|ξ|) as |ξ|→+∞,ξ∈Ξ−. (9)

Here, is a universal constant. Note that the coefficients in front of the growing terms in (9) are related due to the fact that a harmonic function has zero total flux at infinity.

## 4 Asymptotic analysis

In this section, we compute the asymptotic expansion of the field in (3) as tends to zero. The final results are summarized in (26). We assume that the limit lengths of the thin strips (see (1)) are such that

 L±=πm±ωfor somem±∈N∗. (10)

In other words, we assume that is an eigenvalue of the problems (5)–(6). We emphasize that these problems are posed in the fixed lines but the true lengths of the strips depend on the parameter . In the channels, we work with the ansatz

 uε=u00+εu′0+…\rm in Π0,uε=u0±+εu′±+…\rm in Πε±, (11)

while in the thin strips, we deal with the expansion

 uε(x,y)=ε−1v−1±(y)+v0±(y)+…\rm in Sε±.

Taking the formal limit , we find that must solve the homogeneous problem (5)–(6). Under the assumption (10) for the lengths , non zero solutions exist for this problem and we look for in the form

 v−1±(y)=a±v±(y) with a±∈C, v±(y)=sin(ω(y−1)).

Let us stress that the values of are unknown and will be fixed during the construction of the asymptotics of . At , the Taylor formula gives

 ε−1v−1±(y)+v0±(y)=0+(CA±ξ±y+v0±(1))+… with CA±:=a±∂yv(1)=a±ω. (12)

Here is the stretched variable introduced just before (7). At , we have

 ε−1v−1±(y)+v0±(y)=0+(CB±ζ±y+v0±(1+L±))+… (13)

with

 CB±:=−a±∂yv(1+L±)=−a±ωcos(ωL±)=(−1)1+m±a±ω. (14)

Here, we use the stretched coordinates (mind the sign of ).

We look for an inner expansion of in the vicinity of of the form

 uε(x)=CA±Y1(ξ±)+cA±+…

where is introduced in (9), are defined in (12) and are constants to determine. In a vicinity of , we look for an inner expansion of of the form

 uε(x)=CB±Y1(ζ±x,ζ±y+L′±)+cB±+…

where are defined in (14) and are constants to determine.

Let us continue the matching procedure. Taking the limit , we find that the main term in (11) must solve the problem

 Δu00+ω2u00=0  in Π0,∂νu00=0 on ∂Π0∖{A+,A−},

with the expansion

 u00=w+h+R0w−h+~u00.

Here and decay exponentially at infinity. Moreover, we find that the term in (11) must solve the problem

 Δu0±+ω2u0±=0  in Πε±,∂νu0±=0 on ∂Πε±∖Bε±,

with the expansion

 u0±(x,y)=T0±w+v(y−Lε)+~u0±(x,y),

Here and decay exponentially at infinity. The coefficients , will provide the first terms in the asymptotics of , :

 Rε=R0+… and Tε±=T0±+….

Matching the behaviours of the inner and outer expansions of in , we find that at the points , the function must expand as

 u00(x,y)=CA±1πln1rA±+U00±+O(rA±) as rA±:=((x−p±)2+(y−1)2)1/2→0+,

where is a constant. Observe that is singular both at and . Integrating by parts in

 0=∫Π0,ρ(Δu00+ω2u00)(e+iωx+e−iωx)−u00(Δ(e+iωx+e−iωx)+ω2(e+iωx+e−iωx)dxdy,

with , and taking the limit , we get . From the expressions of (see (12)), this gives

 R0=1+i(a+cos(ωp+)+a−cos(ωp−)). (15)

Then matching the behaviours of the inner and outer expansions of in , we find that at the points , the function must expand as

 u0±(x,y)=CB±1πln1rB±+U0±+O(rB±) as rB±:=((x−p±)2+(y−1−Lε)2)1/2→0+,

where are constants. Note that is singular at . Integrating by parts in

 0=∫Πε±,ρ(Δu0±+ω2u0±)(e+iω(y−1−Lε)+e−iω(y−1−Lε))dxdy−∫Πε±,ρu0±(Δ(e+iω(y−1−Lε)+e−iω(y−1−Lε))+ω2(e+iω(y−1−Lε)+e−iω(y−1−Lε))dxdy,

with , and taking the limit , we get . From the expressions of (see (14)), this gives

 T0±=i(−1)1+m±a±cos(ωp±). (16)

Matching the constant behaviour inside , we get

 U00±=CA±π−1lnε+cA±=−CA±π−1|lnε|+cA±.

This sets the value of . However depends on and we have to explicit this dependence. For , we have the decomposition

 u00=w+h+w−h+CA+γ++CA−γ− (17)

where are the outgoing functions such that

 Δγ±+ω2γ±=0 in % Π0∂νγ±=δA± on ∂Π0. (18)

Here stands for the Dirac delta function at . Denote by the constant behaviour of at , that is the constant such that behaves as

 γ±(x,y)=1πln1rA±+Γ±+O(rA±) when rA±=((x−p±)2+(y−1)2)1/2→0+.

In Lemma 6.2 below, we will prove that the constant behaviours of at are equal. We denote by the value of this coupling constant. Then from (17), we derive

 U00±=2cos(ωp±)+ω(a±Γ±+a∓~Γ).

Matching the constant behaviour at inside the thin strips, we obtain

 v0±(1)=CA±CΞ+cA±=U00±+CA±(π−1|lnε|+CΞ)=2cos(ωp±)+a∓ω~Γ+a±ω(π−1|lnε|+CΞ+Γ±). (19)

Now, matching the constant behaviour inside , we get

 U0±=CB±π−1lnε+cB±=−CB±π−1|lnε|+cB±.

This sets the value of . However depends on and we have to explicit this dependence. For , we have the decomposition

 u0±(x,y)=CB±g(x−p±,y+1+Lε)

where is the outgoing function such that

 Δg+ω2g=0 in Π:=(−1/2;1/2)×(0;+∞)∂νg=δO on ∂Π. (20)

Here stands for the Dirac delta function at . Denote by the constant behaviour of at , that is the constant such that behaves as

 g(x,y)=1πln1r+G+O(r) when r=(x2+y2)1/2→0+.

Then we have

 U0±=CB±G=(−1)1+m±a±ωG.

Matching the constant behaviour at inside the thin strips, we obtain

 v0±(1+L±)=CB±(L′±+CΞ)+cB±=U0±+CB±(π−1|lnε|+L′±+CΞ)=(−1)1+m±a±ω(π−1|lnε|+L′±+CΞ+G). (21)

Writing the compatibility condition so that the problem (5) supplemented with the boundary conditions (19)–(21) admits a non zero solution, we get

 v0±∂yv|1−v0±∂yv|1+L±−(v∂yv0±|1−v∂yv0±|1+L±)=0.

Since , we obtain

 v0±(1)+(−1)1+m±v0±(1+L±)=0.

This gives the relations

 2cos(ωp±)+a∓ω~Γ+a±ω(2π−1|lnε|+L′±+2CΞ+Γ±+G)=0. (22)

Below, we will prove that , (Lemma 6.2) and (Lemma 6.1). Therefore (22) writes equivalently

 a±(β±+i(1+|cos(ωp±)|2))+a∓ω~Γ=−2cos(ωp±) (23)

where are the real valued quantities such that

 β±:=ω(2π−1|lnε|+L′±+2CΞ+ReΓ±+ReG). (24)

Identities (23) form a system of two equations whose unknowns are . Solving it, we get

 a±=2cos(ωp∓)ω~Γ−2(β∓+i(1+|cos(ωp∓)|2))cos(ωp±)(β++i(1+|cos(ωp+)|2))(β−+i(1+|cos(ωp−)|2))−ω2~Γ2. (25)

And from (15), (16), we obtain explicit expressions for , . This ends the asymptotic analysis. To sum up, when tends to zero, we have obtained the following expansions

 \framebox$uε(x,y)=w+h(x,y)+w−h(x,y)+a+ωγ+(x,y)+a−ωγ−(x,y)+… in Π0,uε(x,y)=(−1)1+m±a±ωg(x−p±,y+1+Lε)+… in Πε±,uε(x,y)=ε−1a±sin(ω(y−1))+… in% Sε±,\omit\span\omit\span\omitRε=1+i(a+cos(ωp+)+a−cos(ωp−))+…,Tε±=i(−1)1+m±a±cos(ωp±)+…,$ (26)

where are given by (25). Here, the functions , are respectively introduced in (18), (20). Note in particular that when , the amplitude of the field blows up in the thin slits as tends to zero.

## 5 Analysis of the results

In this section, we explain how to use the asymptotic results (26) to exhibit settings where the waveguide

acts as an energy distributor. The degrees of freedom we can play with are

and , that is the lengths and the abscissa of the thin strips.

When are chosen such that , from (25) we get . This implies and . In this case, the energy brought to the system is almost completely backscattered and the thin strips have almost no influence on the incident field. Definitely, this is not an acoustic distributor. Roughly speaking, in this situation what happens is that the resonant eigenfunctions associated with complex resonances existing due to the presence of the thin slits are not excited.

When are chosen such that , we find

 a±=2ω~Γ−2(β∓+2i)(β++2i)(β−+2i)−ω2~Γ2.

Then, we have

 R0=1+i(a++a−)=β+β−+4−ω2~Γ2+4iω~Γ(β++2i)(β−+2i)−ω2~Γ2.

and

 T0±=i(−1)1+m±2ω~Γ−2(β∓+2i)(β++2i)(β−+2i)−ω2~Γ2.

According to Lemma 6.2 below, we have for a certain which characterizes the coupling between the two strips. With this notation, we find

 R0=β+β−+1−η2+2iηβ+β−+2i(β++β−)−3−η2−2iη,T0±=2i(−1)m±(β∓+i−η)β+β−+2i(β++β−)−3−η2−2iη. (27)

Notice that the denominator in the expressions for , cannot vanish. Indeed it vanishes if and only if and . One can verify that this cannot occur for .

If was zero, then we would have , with

 R0=β+β−+1β+β−+2i(β++β−)−3,T0±=2i(−1)m±(β∓+i)β+β−+2i(β++β−)−3. (28)

In particular, we would have for all pairs such that

 β+β−=−1, (29)

and then (conservation of energy) as well as

 |T0+(β+,β−)||T0−(β+,β−)|= ⎷β2−+1β2++1= ⎷1/β2++1β2++1. (30)

Observing in (24) that varies in when runs in , we see that the ratio (30) could take any value in .

However, the coupling constant cannot be chosen as we wish because we have already set to impose . In Lemma 6.2 below, we will prove that the function , with , is real and exponentially decaying at infinity. As a consequence, we infer that for with large, we have

 ω~Γ=ωγ+(p−,1)≈ωs+w−h(p−)=icos(ωp+)e−iωp−.

With the above choice for as well as the relation (29) for , which translates into a condition relating to , we have . From the previous analysis, we deduce that

 R0≈0 together with |T0+(β+,β−)||T0−(β+,β−)|≈ ⎷