Unique determination of the shape of a scattering screen from a passive measurement
aa r X i v : . [ m a t h . A P ] J u l Unique determination of the shape of a scatteringscreen from a passive measurement
Emilia Bl˚asten , Lassi P¨aiv¨arinta , and Sadia Sadique Division of Mathematics, Tallinn University of Technology, Department ofCybernetics, 19086 Tallinn, Estonia Department of Mathematics and Statistics, University of Helsinki, 00014 Helsinki,Finland
July 20, 2020
Abstract
We consider the problem of fixed frequency acoustic scattering froma sound-soft flat screen. More precisely the obstacle is restricted to atwo-dimensional plane and interacting with a arbitrary incident wave, itscatters acoustic waves to three-dimensional space. The model is partic-ularly relevant in the study and design of reflecting sonars and antennas,cases where one cannot assume that the incident wave is a plane wave.Our main result is that given the plane where the screen is located, thefar-field pattern produced by any single arbitrary incident wave deter-mines the exact shape of the screen, as long as it is not antisymmetricwith respect to the plane. This holds even for screens whose shape is anarbitrary simply connected smooth domain. This is in contrast to earlierwork where the incident wave had to be a plane wave, or more recent workwhere only polygonal scatterers are determined. keywords : inverse scattering; screen; uniqueness; single measurement;passive measurement
MSC : 35R30, 35P25, 35A02
The motivation for the study of wave scattering from thin and large ob-jects lies in the antenna theory. The starting point for this was when thePrussian Academy announced an open competition about who could bethe first to show the existence or non-existence of electromagnetic (EM) aves in 1879. The existence of these waves were predicted fifteen yearsearlier by the mathematical theory of James Clerk Maxwell [1]. The com-petition was won in 1882 by young Heinrich Hertz, in favour of Maxwell’stheory. He did this by constructing a dipole antenna radiating EM waveswhich he could measure. It is needless to mention the importance whichthis experiment together with Maxwell’s theory has had for modern soci-ety. Hertz’s antenna consisted of two identical perfectly conducting planarbodies, in his case squares, which create radiating EM waves. Since, byreciprocity, radiating antennas are identical to receiving antennas, thetheory of antennas is closely connected to EM scattering and inverse scat-tering theory.A key question in antenna design for scientific radio arrays is how tochoose the antenna topology so that its impedance and radiation patternare frequency independent (FI) over a wide range of frequencies and, si-multaneously, the radiation pattern supports beamforming. Well-knownexamples of FI antennas include log-periodic, log-spiral, and UHF frac-tal antennas on high-frequencies. While proven good for extremely wideband work, these are heavy and complicated structures and thus not cost-efficient for extremely large arrays.Instead of relying on traditional antenna forms, we aim to derive gen-eral principles for designing antennas with frequency independent charac-teristics. A major step in such a design strategy is to solve the inversescattering problem: given an input–output pair of waves, which antennashape produces it? The input is a given incident wave, and the output isthe far-field pattern produced by the antenna. The path to antenna designis a long one, so in this paper we study the technically easier accousticscattering problem.In acoustics, scattering surfaces or screens are not called antennas butsonars. Traditionally sonars are classified into active and passive sonars,depending of whether they act as a sound source or receiver. We consideracoustic scattering from screens, something which lies between these twoextremes. It is more correct to call these screens passive sonars as they donot have an energy source, however they are active in the sense that theireffect on the sound pattern is significant. In general the nomenclature“sonar” refers to probing using an active and passive sonar. Our researchis rather in the domain of acoustic design. The mathematical question offinding a screen that scatters a given incident wave into a particular far-field has applications like the following, for example: how to reduce echoin an office space? How to direct acoustic vibrations or reduce them?Of course, it also answers the probing question: can we determine theshape and location of a passive sonar by how it reflects sound? These arecomplex questions, only one part of which we are going to solve, namelythat a single input–output pair of sound waves uniquely determines theshape of a flat acoustic screen. The problem of inverse scattering with reduced measurement data hasgained a lot of interest lately. Traditionally determining a scatterer fromfar-field measurements requires sending all possible incident waves and ecording the corresponding far-field patterns. The method of using com-plex geometrical optics solutions and infinitely many far-field measure-ments in the fixed frequency setting was pioneered by Sylvester and Uhlmannin [2], and was the first method for uniquely determining an arbitrarysmooth enough scattering potential by far-field measurements. The fieldhas grown extremely fast since then, almost to the point of saturation,and we will only point the reader towards the surveys in [3] for referencesup to 2003, which gives a good picture of the situation except for scat-tering in two dimensions, which was solved by Bukhgeim [4] in 2007 andimproved by sevaral authors, e.g. [5–9].In many applications the scatterer is impenetrable, or we are only in-terested in its shape or location. The shape determination problem isknown as Schiffer’s problem in the literature [10]. M. Schiffer showedthat a sound-soft obstacle (with non-empty interior) can be uniquely de-termined by infinitely many far-field patterns. The proof appeared as aprivate communication in the monograph by Lax and Phillips [11]. Linearsampling [12] and factorization [13] methods were developped and they arevery well suited for shape determination, also from the numerical point ofview. These were applied in the context of curved screens in acoustic [14]and electromagnetic [15] scattering to determine the shape and location ofthe screen, also numerically. However these methods require the full useof infinitely many far-field patterns, except for a case of interest in [14] towhich we will return later on in more detail.There was still much to improve: counting dimensions shows that asingle far-field (a mapping S n − → C ) should be enough to determine theshape (a manifold of dimension n − cattering support [33, 34] by one far-field measurement. Again, none ofthe above are applicable to screens per se.Our work in this paper shows that given the far-field caused by anysingle given incident wave scattering off of a smooth flat screen, the latter’sshape is determined uniquely. Our methods are based on ideas which arepartly motivated by the study of certain integral operators in [35, 36]. Asin [14], we first show that the far-field is the restriction to a ball of radius k (the wavenumber) of the two-dimensional Fourier transform of a functionsupported on the screen. Next, since the incident wave might vanish onpart of the screen, we show that the shape of the screen is exactly thesupport of that function. This latter part involves a delicate analysis ofthe Taylor coefficients of the scattered wave at the screen, but it leadsto our main theorem: that Schiffer’s problem is uniquely solvable for flatscreens on a plane in three dimensions, for any incident wave that causesscattering.Let us discuss the significance of our result, with focus especially on ourimprovements over [14]: that any incident field is allowed. We will startwith the mathematical challenges. Unlike for infinite measurements in-verse problems such as [2,4], properties of the incident wave affect greatlythe solvability of single measurement inverse problems. Complex planewaves make things technically simpler in many scattering problems be-cause of their explicit form and non-vanishing everywhere. This oftenreduces the non-linear inverse scattering problem to the linear inversesource problem after a suitable interpretation, or avoids other challenges,as can be seen by comparing [26, 27, 31] to [28–30]. Futhermore, in situ-ations involving scattering from multiple objects, the total incident fieldimpinging on a given component is the sum of the original incident fieldand the fields scattered by the other components. This is relevant whenone wishes to uniquely determine a screen where space contains otherscatterers that are known. On the other hand, from the applied pointof view, solving the inverse problems for any given incident field enables passive measurements . This means that even if we do not have controlover the incident wave, or cannot afford to control it, the shape of thescatterer can be uniquely determined. This is both good and bad. Itmeans that the flat screen design problem of finding its shape such thatit scatters one given incident wave into a given far-field has no more thana unique solution. On the other hand it shows the impossibility of morecomplex input–output systems. One cannot require it to scatter two ormore incident waves into their corresponding far-fields in general. Thefirst incident-wave and far-field pair already determines the shape.Lastly, we remark that inverse scattering for screens has still manyopen problems. Current solutions require that the screen have at least adifferentiable boundary, something which arises from the way that the di-rect scattering problem has been shown solvable in [37] and other sources.To bring forward the range characterization condition from [14] to thesituations of let’s say Herglotz incident waves, one would need to solve adeconvolution problem. A more difficult and certainly more interestingquestion mathematically and from the point of view of applications, is theunique determination of the shape of a curved screen from one measure- ent, passive or fully controlled. The problem is solved for infinitely manymeasurements in [14], but counting dimensions suggests that it should besolvable with one measurement. Let us go forward to the mathematics. We start by defining what wemean by a screen and the scattering problem from screens. Then westate our three main theorems. They give representation formulas forthe scattered wave, the far-field pattern, and the unique solvability ofSchiffer’s problem for determining the shape of a scattering screen usinga single incident wave. In Section 2 we prove the representation formulas,and then in Section 3 we solve the inverse problem.We consider the scattering of a two dimensional sound-soft and flatobstacle Ω in three dimensional space. We will assume that Ω is an opensubset of R × { } . Definition 1.1.
We call a set Ω ⊂ R a screen , if Ω = Ω × { } for somesimply connected bounded domain Ω ⊂ R whose boundary is smooth,and which we call its shape .The scattering of acoustic waves by Ω leads to the study of the Helmholtzequation (∆ + k ) u = 0 where the wave number k is given by the positiveconstant k = ω/c where c is the constant speed of sound in the back-ground fluid (air, water, etc) and ω is the angular frequency of the wave.The pressure of the total wave vanishes on the boundary of a sound-softobstacle, and the total wave is a sum of the incident and scattered waves.This leads to the following set of partial differential equations. Definition 1.2.
We define the direct scattering problem for a screen
Ωas follows. Given an incident wave u i satisfying (∆ + k ) u i = 0 in R and a screen Ω, the direct scattering problem has a solution if there is u s ∈ H loc ( R \ Ω) that satisfies the following conditions(∆ + k ) u s = 0 , R \ Ω , (1.1) u i ( x ) + u s ( x ) = 0 , x ∈ Ω , (1.2) r (cid:16) ∂∂r − ik (cid:17) u s = 0 , r → ∞ , (1.3)where r = | x | and the limit is uniform over all directions ˆ x = x/r ∈ S as r → ∞ .There are a few things above that we should clarify. By H loc ( R \ Ω)we mean the set of distributions ψ on R \ Ω for which ψ | U ∈ H ( U )for any bounded convex open set U ⊂ R \ Ω. Secondly, since strictlyspeaking u s is not defined on Ω, by (1.2) we mean that the Sovolev traceof u s both from above ( x >
0) and below ( x <
0) coincides, and is equalto − u i on Ω.We shall start by showing a representation formula (1.4) for solutions u s of the direct scattering problem for the screen. This is mainly done sothat the reader would get a better intuition about this type of problemsand to fix notation and function spaces clearly. This formula is well known, nd it gives a unique solution to the direct problem [37]. After that wewill show that the far-field, defined below, corresponding to a single givennon-trivial incident wave uniquely determines the screen Ω. This type oftheorem was shown in [14] on the condition that the incident wave doesnot vanish on the plane R × { } . To get rid of this assumption, wehave to show Lemma 3.2. We remark that the far-field pattern exists andis unique for each u s satisfying the following assumptions. See [10] forreference. Definition 1.3.
Let u s satisfy the Sommerfeld radiation condition of(1.3) and the Helmholtz equation (∆ + k ) u s = 0 outside a ball B ⊂ R .We say that u ∞ s is the far-field of u s if u s ( x ) = e ik | x | | x | (cid:18) u ∞ s (ˆ x ) + O (cid:18) | x | (cid:19)(cid:19) uniformly over ˆ x as x → ∞ .We define some notation which will be useful throughout the wholetext. • x, y, . . . represent variables in R , and we associate to them variousprojections described below. • x ′ , y ′ , . . . mean variables in R or projections to R . For example if x = (1 , , ∈ R then in that context x ′ = (1 , ∈ R , but we couldhave dy ′ in an integral over a subset of R without having to definethe variable y separately. • x , y , . . . denote lifts to R , meaning x = ( x ′ , x ′ = ( − , −
2) then x = ( − , − , R → R × { } . So if x = (1 , ,
3) then x = (1 , , x ′ = ( x ′ ) = x and x ′ = ( x ) ′ = x ′ but we do not usethis combined notation explicitly. • Φ is reserved for the fundamental solution to (∆ + k ), defined inLemma 2.2. • u + , u − mean the function u restricted to R × R + and R × R − ,respectively. If their variable is in R × { } then they are the two-sided limits (traces) as x →
0. We often use ∂ u + and ∂ u − . Theseare simply the derivatives in the x -direction of u + and u − , respec-tively. Often this is evaluated on R × { } where it then denotes theone-sided derivative, i.e. the trace of ∂ u ± . • e H − / (Ω ): this is the set of H − / ( R ) distributions whose supportis contained in Ω , where we recall that Ω signifies the shape of ascreen Ω.Let us discuss the direct scattering problem (1.1)–(1.3) first. In Sec-tion 2, Proposition 2.4, we will show the well-known representation for-mula u s ( x ) = Z R Φ( x, y ) ρ ( y ′ ) dy ′ (1.4)for all x ∈ R \ Ω, where ρ ( y ′ ) = ∂ u + s ( y ) − ∂ u − s ( y ) (1.5) s an element of e H − / (Ω ) and the integral in (1.4) is interpreted as adistribution pairing between ρ and the smooth test function Φ restrictedto the screen. Taking the trace x → Ω in (1.4) and recalling that u s = − u i on Ω in the sense of traces, (1.2), we get u i ( x ) = − Z R Φ( x, y ) ρ ( y ′ ) dy ′ . (1.6)Now, for any candidate solution u s ∈ H loc ( R \ Ω), it solves the directproblem (1.1)–(1.3) if and only if ρ , as defined above, is in e H − / (Ω )and is the solution to (1.6). More precisely, given ρ solving the integralequation, we can define u s by (1.4), and it would solve the direct scatteringproblem. This was shown in Theorem 2.5 in [37]. Theorem 2.7 in the samesource proves that (1.6) has a unique solution ρ ∈ e H − / (Ω ) given any u i ∈ H / (Ω ).Our main contributions are the following. The first of which is thefamiliar far-field representation derived from (1.4) if ρ is a function. Wegeneralize is to distributions in H − / ( R ). This is required for consis-tency of the function spaces involved. This detail has not been statedexplicitely in earlier work involving scattering from screens. Theorem 1.4.
Let Ω ⊂ R be a screen and u s satisfy the direct scatteringproblem for some incident field u i and screen Ω . Then its far-field has therepresentation u ∞ s (ˆ x ) = 14 π D(cid:0) ∂ u + s − ∂ u − (cid:1) ( y ) , e − ik ˆ x · y E y ′ (1.7) for ˆ x ∈ S . If ∂ u + s − ∂ u − s is integrable on Ω , this formula is equivalentto u ∞ s (ˆ x ) = 14 π Z R e − ik ˆ x · y (cid:0) ∂ u + s − ∂ u − (cid:1) ( y ) dy ′ . Our main theorem shows that even with an unoptimal incident wave,the scattering caused by it from flat screens determines the shape uniquely.
Theorem 1.5.
Let Ω , ˜Ω ⊂ R be screens and k ∈ R + . Let u i be anincident wave and u s , ˜ u s be scattered waves that satisfy the direct scatteringproblem for screens Ω , ˜Ω , respectively.If u i is not antisymmetric with respect to R ×{ } and u ∞ s = ˜ u ∞ s , then Ω = ˜Ω . If it is antisymmetric then u ∞ s = ˜ u ∞ s = 0 for any screens Ω , ˜Ω . In this section we will prove that solutions to the direct scattering problemsatisfy (1.4). In essence we present the well-known but very condensedargument of [37] in more detail for the convenience of the readers. Wewill start with representation formulas for smooth functions and thenapproximate the H -smooth u s . At the end of the section we will proveTheorem 1.4. emma 2.1. Let D ⊂ R be a bounded domain whose boundary is piece-wise of class C and let ν denote the unit normal vector to the boundary ∂D directed to the exterior of D . Then, for u, v ∈ C ( D ) we have Green’ssecond formula Z D ( v ∆ u − u ∆ v ) dx = Z ∂D (cid:16) ∂u∂ν v − u ∂v∂ν (cid:17) ds (2.1) where ds is the surface measure of ∂D .Proof. Theorem 3 in Appendix C.2 of [38].
Lemma 2.2.
Let D ⊂ R be a bounded domain whose boundary is piece-wise of class C and k ∈ R + . Let Φ( x, y ) = e ik | x − y | π | x − y | for x, y ∈ R , x = y . Then for any ϕ ∈ C ( D ) and x ∈ R \ ∂D we have Z D Φ( x, y )(∆ + k ) ϕ ( y ) dy = Z ∂D (cid:0) Φ( x, y ) ∂ ν ϕ ( y ) − ϕ ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y )+ ( , x ∈ R \ D, − ϕ ( x ) , x ∈ D. (2.2) Proof.
We have (∆ + k ) ϕ bounded and y Φ( x, y ) integrable for any x , so Z D Φ( x, y )(∆ + k ) ϕ ( y ) dy = lim r → Z D \ B ( x,r ) Φ( x, y )(∆ + k ) ϕ ( y ) dy. Green’s second formula (2.1) applied to the integral on the right gives . . . = Z D \ B ( x,r ) (∆ + k )Φ( x, y ) ϕ ( y ) dy + Z S ( x,r ) ∩ D (cid:0) Φ( x, y ) ∂ ν ϕ ( y ) − ϕ ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y )+ Z ∂D \ B ( x,r ) (cid:0) Φ( x, y ) ∂ ν ϕ ( y ) − ϕ ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y ) . The first integral here vanishes because (∆ y + k )Φ( x, y ) = 0 when y = x .The integral over ∂D \ B ( x, r ) gives the second term in the claim when r → , ∂ Φ are integrable since x / ∈ ∂D . Let us estimate thefirst term in the first boundary integral. We have Z S ( x,r ) ∩ D Φ( x, y ) ∂ ν ϕ ( y ) ds ( y ) = Z S ( x,r ) ∩ D e ikr πr ∂ ν ( y ) ds ( y )and by the ML-inequality we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z S ( x,r ) ∩ D Φ( x, y ) ∂ ν ϕ ( y ) ds ( y ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ πr sup y ∈ S ( x,r ) ∩ D |∇ ϕ ( y ) | πr → s r → |∇ ϕ | has a uniform bound in D . In the last integral wehave ∂ n u Φ( x, y ) = − ∂ r (cid:0) e ikr / (4 πr ) (cid:1) = − ike ikr / (4 πr ) + e ikr / (4 πr ). Theintegral involving ike ikr / (4 πr ) can be estimated as above to conclude thatit vanishes when r →
0. The remaining integral is − e ikr πr Z S ( x,r ) ∩ D ϕ ( y ) ds ( y )= − e ikr πr Z S ( x,r ) ∩ D (cid:0) ϕ ( y ) − ϕ ( x ) (cid:1) ds ( y ) − e ikr πr ϕ ( x ) s (cid:0) S ( x, r ) ∩ D (cid:1) . We have | ϕ ( y ) − ϕ ( x ) | ≤ sup ξ ∈ D |∇ ϕ ( ξ ) || x − y | so the absolute value ofthe first integral above can be estimated as . . . ≤ sup |∇ ϕ | πr Z S ( x,r ) ∩ D | x − y | dy = sup |∇ ϕ | πr rs (cid:0) S ( x, r ) ∩ D (cid:1) → r →
0. The form of the remaining term implies the claim in each ofthe cases x ∈ D , x ∈ R \ D . Lemma 2.3.
Let D ⊂ R be a bounded domain with smooth boundaryand k ∈ R + . Let u s ∈ H ( D ) with (∆ + k ) u s ∈ L ( D ) . Then u s ( x ) = − Z D Φ( x, y )(∆ + k ) u s ( y ) dy + Z ∂D (cid:0) Φ( x, y ) ∂ ν u s ( y ) − u s ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y ) (2.3) for x ∈ D in the distribution sense. For x ∈ R \ D we have − Z D Φ( x, y )(∆ + k ) u s ( y ) dy + Z ∂D (cid:0) Φ( x, y ) ∂ ν u s ( y ) − u s ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y ) (2.4) in the distribution sense. Here the boundary integrals involving ∂ ν u s areto be interpreted as distribution pairings between a H − / ( ∂D ) functionand a test function.Proof. We will prove only the first case, namely x ∈ D . The second onefollows similarly. Let ( ϕ j ) ∞ j =0 be a sequence of smooth functions definedon D such that k u s − ϕ j k H ( D ) + (cid:13)(cid:13) (∆ + k )( u s − ϕ j ) (cid:13)(cid:13) L ( D ) → j → ∞ . Such a sequence exists, for example by convolving u s with amollifier ψ ε , as in ϕ j = ( u s ∗ ψ /j ) | D .We have Φ( x, y ) = Ψ( x − y ) for Ψ( z ) = exp( ik | z | ) / (4 π | z | ) which islocally integrable in R . Hence the first term in the right-hand side of(2.3), equal to Ψ ∗ (∆ + k ) u s , can be approximated by Ψ ∗ (∆ + k ) ϕ j inthe L ( D )-sense. or any x ∈ D the second integral in (2.3) is well defined because y Φ( x, y ) and y ∂ ν Φ( x, y ) are smooth on the smooth manifold ∂D .Moreover the x -dependence is smooth, so the mapping u s Z ∂D u s ( y ) ∂ ν Φ( x, y ) ds ( y )is bounded H ( D ) → H / ( ∂D ) → C ( D ) and similarly u s Z ∂D Φ( x, y ) ∂ ν u s ( y ) (cid:1) ds ( y )is bounded H ( D ) → H − / ( ∂D ) → C ( D ) when the integral is inter-preted as a distrubion pairing between a H − / ( ∂D )-function and a testfunction. The continuity does not necessarily hold up to the boundary.Because ϕ j → u s in H ( D ) and the trace operators map Tr : H ( D ) → H / /D ), ∂ ν : H ( D ) → H − / ( ∂D ), so the boundary integrals with u s replaced by ϕ j converge to the corresponding ones in C ( D ), namelyuniformly over compact subsets of D .In conclusion, for a test function ψ ∈ C ∞ ( D ) we have h u s , ψ i = lim j →∞ h ϕ j , ψ i = lim j →∞ * − Z D Φ( x, y )(∆ + k ) ϕ j ( y ) dy + Z ∂D (cid:0) Φ( x, y ) ∂ ν ϕ j ( y ) − ϕ j ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y ) , ψ ( x ) + x = * − Z D Φ( x, y )(∆ + k ) u s ( y ) dy + Z ∂D (cid:0) Φ( x, y ) ∂ ν u s ( y ) − u s ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y ) , ψ ( x ) + x so the equality holds in D ′ ( D ). Proposition 2.4.
Let Ω ⊂ R be a screen, k ∈ R + and Φ the fundamentalsolution from Lemma 2.2. Let u s ∈ H loc ( R \ Ω) . If (∆ + k ) u s = 0 in R \ Ω and it satisfies the Sommerfeld radiation condition, then u s ( x ) = Z R Φ( x, y )( ∂ u + s − ∂ u − s )( y ) dy ′ (2.5) for x ∈ R \ Ω . Also y ′ ( ∂ u + s − ∂ u − s )( y ) is in e H − / (Ω ) , andmore precisely the integral above represents the distribution pairing of a e H − / (Ω ) -function with the smooth test function Φ restricted to R ×{ } on the y -variable.Proof. Fix x ∈ R \ Ω. Let D ⊂ R be a bounded domain with smoothboundary for which x ∈ D and Ω ⊂ ∂D and furthermore we want this set o be on top of Ω, namely that its boundary normal pointing to the interiorat Ω is e and not − e . Let R > sup z ∈ D | x − z | . We will use the formulasof Lemma 2.3 on D , which has Ω on its boundary, and B ( x, R ) \ D .Firstly note that since (∆ + k ) u s = 0 only the boundary integralson the right-hand sides of (2.3) and (2.4) remain. We will see the firstintegral as is, namely u s ( x ) = Z ∂D (cid:0) Φ( x, y ) ∂ Dν u s ( y ) − u s ( y ) ∂ Dν Φ( x, y ) (cid:1) ds ( y ) , (2.6)where we denote by ∂ Dν the internal boundary normal derivative of D ,applied to functions on D . We will have the integrals in (2.4) to be overthe set B ( x, R ) \ D . The boundary of this set is S ( x, r ) ∪ ∂D , and theboundary normal pointing to its interior is − e on Ω ⊂ ∂ ( B ( x, R ) \ D ).We will split the boundary integral accordingly, and in the integral over ∂D we denote by ∂ D c ν the external boundary normal derivative applied tofunction on B ( x, R ) \ D . In conclusion (2.4) becomes0 = Z S ( x,R ) (cid:0) Φ( x, y ) ∂ ν u s ( y ) − u s ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y )+ Z ∂D (cid:0) Φ( x, y )( − ∂ D c ν ) u s ( y ) − u s ( y )( − ∂ D c ν )Φ( x, y ) (cid:1) ds ( y ) . (2.7)Finally, by interior elliptic regularity we see that u s is continuous (in factreal analytic) in some neighbourhood of x . Also, because x is outside of ∂D and S ( x, R ), the individual boundary integrals above are continuous.Hence the equality in the sense of distributions is in fact a pointwiseequality for continuous functions. In other words, both of (2.6) and (2.7)hold as continuous functions. We still remind that the integrals involving ∂ ν u s represent distribution pairings for an element of H − / ( ∂D ) withthat of a smooth Φ.Let us add (2.6) and (2.7). By smoothness, ∂ Dν Φ = ∂ D c ν Φ. Note thattwo-sided Sobolev traces of H -functions yield identical resuts, so theintegrals of u s ∂ Dν Φ and u s ∂ D c ν Φ in (2.6) and (2.7) cancel out. The sumthen gives u s ( x ) = Z S ( x,R ) (cid:0) Φ( x, y ) ∂ ν u s ( y ) − u s ( y ) ∂ ν Φ( x, y ) (cid:1) ds ( y )+ Z ∂D Φ( x, y ) (cid:0) ∂ Dν u s − ∂ D c ν u s (cid:1) ( y ) ds ( y ) . (2.8)Note that as R → ∞ the first integral in (2.8) vanishes because u s satisfiesthe Sommerfeld radiation condition. Also, u s is C outside of Ω by ellipticinterior regularity, so the second integral’s integrand is zero when y / ∈ Ω.Thus, letting R → ∞ gives u s ( x ) = Z Ω Φ( x, y ) (cid:0) ∂ Dν u s − ∂ D c ν u s )( y ) dy which implies the claim as ∂ Dν u s = ∂ u + s and ∂ D c ν u s = ∂ u − s on Ω ⊂ R × { } . Furthermore, as above, since u s is C outside of Ω, we see that ∂ u + s − ∂ u − s = 0 outside of Ω, so the integrand in the statement is in e H − / (Ω ), as claimed. ith the proposition above, we are almost ready to prove the formulafor the far-field of a wave scattered by a screen, Theorem 1.4. But firstlet us prove a lemma. Lemma 2.5.
Let k ∈ R + and K ⊂ R be a nonempty compact set. Then lim r →∞ sup | x | = r sup y ∈ K | x | (cid:12)(cid:12)(cid:12)(cid:12) ∂ αy (cid:18) e ik | x − y | | x − y | − e ik | x | | x | e − ik ˆ x · y (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = 0 for any multi-index α ∈ N with | α | ≤ . Recall that ˆ x = x/ | x | .Proof. The case of | α | = 0 is well known, see for example the proof ofTheorem 2.5 in [10]. For | α | = 1 we will instead show the equivalentstatement with ∂ αy replaced by ∇ y . Recall the following differentiationrules • ∇ y | x − y | s = − s x − y | x − y | | x − y | s − for all s ∈ R , • ∇ y e ik | x − y | = − ik x − y | x − y | e ik | x − y | , and • ∇ y e − ik ˆ x · y = − ik ˆ xe − ik ˆ x · y .These imply that ∇ y (cid:18) e ik | x − y | | x − y | − e ik | x | | x | e − ik ˆ x · y (cid:19) = − ik x − y | x − y | e ik | x − y | | x − y | + x − y | x − y | e ik | x − y | | x − y | + ik ˆ x e ik | x | | x | e − ik ˆ x · y = − ik (cid:18) x − y | x − y | − ˆ x (cid:19) e ik | x − y | | x − y | − ik ˆ x (cid:18) e ik | x − y | | x − y | − e ik | x | | x | e − ik ˆ x · y (cid:19) + x − y | x − y | e ik | x − y | | x − y | . Let us consider the three types of terms above. To prove the estimate, letus take the absolute value and multiply by | x | . The last one gives | x | (cid:12)(cid:12)(cid:12)(cid:12) x − y | x − y | e ik | x − y | | x − y | (cid:12)(cid:12)(cid:12)(cid:12) = | x || x − y | → y ∈ K , | x | = r and r → ∞ . The first term gives | x | (cid:12)(cid:12)(cid:12)(cid:12) − ik (cid:18) x − y | x − y | − ˆ x (cid:19) e ik | x − y | | x − y | (cid:12)(cid:12)(cid:12)(cid:12) = k | x || x − y | (cid:12)(cid:12)(cid:12)(cid:12) x − y | x − y | − x | x | (cid:12)(cid:12)(cid:12)(cid:12) where can still estimate (cid:12)(cid:12)(cid:12)(cid:12) x − y | x − y | − x | x | (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) x − y | x − y | | x | − | x − y || x | − y | x | (cid:12)(cid:12)(cid:12)(cid:12) ≤ || x | − | x − y ||| x | + | y || x | ≤ | y || x | because || x | − | x − y || ≤ | y | by the triangle inequality. Thus the first termalso tends to zero uniformly as r → ∞ . Lastly, the second one is estimatedas | x | (cid:12)(cid:12)(cid:12)(cid:12) − ik ˆ x (cid:18) e ik | x − y | | x − y | − e ik | x | | x | e − ik ˆ x · y (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = k | x | (cid:12)(cid:12)(cid:12)(cid:12) e ik | x − y | | x − y | − e ik | x | | x | e − ik ˆ x · y (cid:12)(cid:12)(cid:12)(cid:12) which tends to zero uniformly because this is the case | α | = 0 proven atthe beginning of this proof. roof of Theorem 1.4. By the definition of the far-field there is a finiteconstant
C > x such that (cid:12)(cid:12)(cid:12) u ∞ (ˆ x ) − | x | e − ik | x | u s ( x ) (cid:12)(cid:12)(cid:12) ≤ C | x | when | x | → ∞ . Let us denote ρ ( y ′ ) = ( ∂ u + s − ∂ u − s )( y ). Then (2.5)gives u ∞ s (ˆ x ) = lim | x |→∞ | x | e − ik | x | (cid:10) ρ ( y ′ ) , Φ( x, y ) (cid:11) y ′ should the limit exist. The distribution pairing is over y ′ ∈ R . We canrewrite | x | e − ik | x | (cid:10) ρ ( y ′ ) , Φ( x, y ) (cid:11) = D ρ ( y ′ ) , | x | e − ik | x | Φ( x, y ) − e − ik ˆ x · y / (4 π ) E y ′ + 14 π (cid:10) ρ ( y ′ ) , e − ik ˆ x · y (cid:11) y ′ . We can write the C -test function on the second line as | x | e − ik | x | Φ( x, y ) − e − ik ˆ x · y / (4 π )= e − ik | x | | x | π e ik | x − y || x − y | − e ik | x | | x | e − ik ˆ x · y ! which convergest to zero in the C topology over y ′ , and a fortiori y ,restricted to any compact set by Lemma 2.5. Note that the C -seminormsare taken with respect to the y ′ -variable, and the absolute value makes the e − ik | x | that doesn’t appear in the lemma disappear. Hence the applicationof the lemma is allowed. Elements of e H − / (Ω ) act well on C -functions,so the distribution pairing with ρ and the test function tends to zero. Thuslim | x |→∞ | x | e − ik | x | (cid:10) ρ ( y ′ ) , Φ( x, y ) (cid:11) y ′ = 14 π D ρ ( y ′ ) , e − ik ˆ x · y E y ′ as claimed. We are ready to tackle the inverse problem in this section.
Lemma 3.1.
Let k ∈ R + and ρ ∈ E ′ ( R ) be a distribution of compactsupport. Let u ∞ s (ˆ x ) = 14 π D ρ, e − ik ˆ x · y E (3.1) for ˆ x ∈ S and where the distribution pairing is over the variable y ′ =( y , y ) ∈ R . Then ρ is uniquely determined by u ∞ s . If ρ is integrable then u ∞ s (ˆ x ) = π R R e − ik ˆ x · y ρ ( y ′ ) dy ′ . roof. The operator mapping ρ u ∞ s is bounded and linear E ′ ( R ) → C ( S ). This is because ˆ x (cid:0) y ′ exp( − ik ˆ x · y ) (cid:1) is continuous S →E ( R ). So it is enough to show that ρ = 0 if u ∞ s = 0. Let us assume thelatter. For ξ ′ ∈ R we haveˆ ρ ( ξ ′ ) = 12 π D ρ, e − iξ ′ · y ′ E where the distribution pairing is over the variable y ′ ∈ R . This lookssimilar to the formula (3.1) in the statement. We can rewrite − ik ˆ x · y = − ik (ˆ x , ˆ x , ˆ x ) · ( y , y ,
0) = − i ( k ˆ x , k ˆ x ) · ( y , y ) . Thus u ∞ s (ˆ x ) = 12 ˆ ρ ( k ˆ x , k ˆ x ) . (3.2)The left-hand side is zero for all ˆ x ∈ S . When ˆ x goes through the wholeof S , the sum including only two of the squares, ˆ x + ˆ x , goes throughthe whole interval (0 , ρ ( ξ ′ ) = 2 u ∞ s (cid:18) ξ /k, ξ /k, q k − ξ + ξ /k (cid:19) = 0for all | ξ ′ | ≤ k . Since ρ has compact support, ˆ ρ can be extended to anentire function on C . Since it vanishes on an open subset of R it mustbe the zero function. Hence u ∞ s = 0 implies ρ = 0. Lemma 3.2.
Let (∆ + k ) u i = 0 in R . Let Ω ⊂ R be a screen and u s satisfy the direct scattering problem 1.2. Denote ρ ( x ′ ) = ∂ u + s ( x ) − ∂ u − s ( x ) for x ′ ∈ R and its properties are given in Proposition 2.4. If u i ( x ′ , x ) = − u i ( x ′ , − x ) for some x ∈ R then Ω = supp ρ (3.3) for the shape Ω of the screen Ω .Proof. The function ρ is a well-defined H − / (Ω )-function by Proposi-tion 2.4 so in particular supp ρ ⊂ Ω . It remains to prove that Ω ⊂ supp ρ .Assume the contrary, that Ω is not contained in the support of ρ .Then neither is Ω because if Ω ⊂ supp ρ then Ω ⊂ supp ρ = supp ρ .Because Ω is an open set and supp ρ is closed there is x ′ ∈ Ω and r > B ( x ′ , r ) ⊂ Ω \ supp ρ .Let us study the behaviour of u s in the tube B ( x ′ , r ) × R . We have ρ =0 on B ( x ′ , r ). Recall formula (1.4), which combined with the vanishingof ρ implies that (∆ + k ) u s = 0 in the whole tube, and interior ellipticregularity implies that u s is smooth there. In addition the formula impliesthat u s ( x , x , x ) = u s ( x , x , − x ) for all x in the tube. The vanishingof ρ gives ∂ u + s = ∂ u − s on the base of the tube. These two imply thatactually ∂ u s ( x ′ ,
0) = 0 for x ′ ∈ B ( x ′ , r ). e have the following u s = − u i , (3.4) ∂ u s = 0 (3.5)on B ( x ′ , r ) ×{ } . Let us calculate the higher order derivatives. Note that ∂ j and (∆ + k ) commute, and (∆ + k ) u s = 0 in the tube. Thus0 = ∂ j (∆ + k ) u s = (∆ + k ) ∂ j u s = (∆ ′ + k ) ∂ j u s + ∂ j +23 u s in the tube, and we denote ∆ ′ = ∂ + ∂ . This gives ∂ j +23 u s = − (∆ ′ + k ) ∂ j u s . Let us restrict ourselves to B ( x ′ , r ) × { } next. By inductionand (3.4)–(3.5) we see that ∂ j u s = ( ( − j +1 (∆ ′ + k ) j u i , j ∈ N , , j ∈ N + 1on B ( x ′ , r ) ×{ } . This can still be simplified! Recall that u i is an incidentwave, so (∆+ k ) u i = 0 everywhere. This means that (∆ ′ + k ) u i = − ∂ u i ,and a fortiori (∆ ′ + k ) j u j = ( − ∂ ) j u i everywhere by the commutatingof ∂ and (∆ ′ + k ). This implies ∂ j u s = ( − ∂ j u i , j ∈ N , , j ∈ N + 1 . (3.6)The other derivatives, ∂ and ∂ commute with each other and ∂ , sofinally we have ∂ α u s = ( − ∂ α u i , α ∈ N , , α ∈ N + 1 (3.7)on B ( x ′ , r ) × { } for all multi-indices α ∈ N .Let us define˜ u i ( x ) = 12 (cid:0) u i ( x , x , x ) + u i ( x , x , − x ) (cid:1) for all x ∈ R . This satisfies the Helmholtz equation everywhere, and isan incident wave because u i is one. We see that ∂ α ˜ u i ( x ) = 12 (cid:0) ∂ α u i ( x , x , x ) + ( − α ∂ α u i ( x , x , − x ) (cid:1) so ∂ α ˜ u i = ( ∂ α u i , α ∈ N , , α ∈ N + 1 (3.8)on B ( x ′ , r ) × { } . By (3.7) we see immediately that ∂ α u s = − ∂ α ˜ u i onthe base of the tube for all α ∈ N . Both functions u s and − ˜ u i satisfy theHelmholtz equation not only in the tube but also in R \ B (0 , R ), where R > ⊂ B (0 , R ). Solutions of the Helmholtzequation are real-analytic. Because their Taylor-expansions at ( x ′ ,
0) areequal, the functions are equal in the component of (cid:0) B ( x ′ , r ) × R (cid:1) ∪ (cid:0) R \ B (0 , R ) (cid:1) that contains ( x ′ , u s = − ˜ u i in all of R \ B (0 , R ). he function u s satisfies the Sommerfeld radiation condition, so sodoes ˜ u i . On the other hand (∆ + k )˜ u i = 0 in all of R , so ˜ u i is the zerofunction , which means that u i is antisymmetric with respect to R × { } ,a contradiction. Hence Ω ⊂ supp ρ .The solution to the inverse problem of determining a screen Ω fromthe knowledge of a single incident wave u i and the corresponding far-field u ∞ s scattered from the screen comes from a combination of determining ρ from the far-field, and then Ω from ρ . There is a slight suprise, namelythat the problem is only solvable for incident waves that are not too(anti)symmetric. However, one sees that antisymmetry is not the decidingfactor: what matters is whether u i is identically zero on the screen. Bya similar argument as that at the end of the proof of Lemma 3.2, we seethat if u i = 0 on a non-empty open subset of R × { } then u i ( x ′ , x ) = − u i ( x ′ , − x ) for all x ∈ R . It is interesting to see that partial invisibilityis achieved inside thickened screens as long as the incident plane wavecomes from a direction almost parallel to the screen’s normal [39]. Thedirection of incident waves seems very important in scattering from objectsthat are thin in one direction. Proof of Theorem 1.5.
Theorem 1.4 and Lemma 3.1 imply that ρ = ˜ ρ when u ∞ s = ˜ u ∞ s . If u i is not antisymmetric with respect to R × { } thenΩ = supp ρ = supp ˜ ρ = ˜Ω by Lemma 3.2. Because Ω is a smooth domain, we have Ω = int Ω ,and similarly for ˜Ω . Thus the equation above implies Ω = ˜Ω and bylifting, Ω = ˜Ω.If u i is antisymmetric then u i = 0 everywhere on R × { } and u s = 0satisfies all conditions of the direct scattering problem. Since solutionsto the direct scattering problem (1.2) are unique by [37, Thms 2.5–2.7],this is the only solution. Thus u s = ˜ u s = 0 and the same holds for theirfar-fields. This is irrespective of the shape of Ω , ˜Ω ⊂ R . Acknowledgements
The research of the authors is supported by Estonian Research Councilgrant PRG832. We also thank Markku Lehtinen for useful discussions.
References [1] Maxwell, J.C. On physical lines of force.
Philos. Mag. , ,11–23.[2] Sylvester, J.; Uhlmann, G. A global uniqueness theorem for an in-verse boundary value problem. Ann. Math. , , 153–169. Use e.g. (2.2) for a larbe ball whose radius grows to infinity. The boundary integraldecreases to zero as was seen for the first integral in (2.8).
3] Uhlmann, G.
Inside Out: Inverse Problems and Applications ; Cam-bridge University Press: Cambridge, UK, 2003.[4] Bukhgeim, A.L. Recovering a potential from Cauchy data in thetwo-dimensional case.
J. Inverse Ill-Posed Probl. , , 19–33.[5] Guillarmou, C.; Tzou, L. Calder´on inverse problem with partialdata on Riemann surfaces. Duke Math. J. , , 83–120.[6] Imanuvilov, O.Y.; Uhlmann, G.; Yamamoto,M. The Calder´on prob-lem with partial data in two dimensions. J. Am. Math. Soc. , , 655–691.[7] Dos Santos Ferreira, D.; Kenig, C.E. Salo, M. Determining anunbounded potential from Cauchy data in admissible geometries. Comm. Part. Differ. Eq. , , 50–68.[8] Bl˚asten, E.; Imanuvilov, O.Y.; Yamamoto, M. Stability and unique-ness for a two-dimensional inverse boundary value problem for lessregular potentials. Inverse Probl. Imaging , , 709–723.[9] Bl˚asten, E.; Tzou, L.; Wang, J. Uniqueness for the inverse boundaryvalue problem with singular potentials in 2D. Math. Z. publishedonline , doi:10.1007/s00209-019-02436-0.[10] Colton, D.; Kress, R. Inverse acoustic and electromagneticscattering theory. In
Applied Mathematical Sciences ; Springer:Berlin/Heidelberg, Germany, 1992; Volume 93.[11] Lax, P.; Phillips, R.
Scattering Theory ; Academic Press: New York,NY, USA; London, UK, 1967.[12] Colton, D.; Kirsch, A. A simple method for solving inverse scat-tering problems in the resonance region.
Inverse Probl. , ,383–393.[13] Kirsch A.; Grinberg, N. The Factorization Method for Inverse Prob-lems ; Oxford Lecture Series in Mathematics and Its Applications;Oxford University Press: Oxford, UK, 2008; Volume 36.[14] Alves, C.J.S.; Ha-Duong, T. On inverse scattering by screens.
In-verse Probl. , , 1161–1176.[15] Cakoni, F.; Colton, D.; Darrigrand, E, The inverse electromagneticscattering problem for screens. Inverse Probl. , , 627–642.[16] Colton, D.; Sleeman, B. Uniqueness theorems for the inverse prob-lem of acoustic scattering. IMA J. Appl. Math. , , 253–259.[17] Isakov, V. Inverse Problems for Partial Differential Equations , 2nded.; Springer: New York, NY, USA, 2006.[18] Alessandrini, G.; Rondi, L. Determining a sound-soft polyhedralscatterer by a single far-field measurement.
Proc. Am. Math. Soc. , , 1685–1691.[19] Cheng, J.; Yamamoto, M. Uniqueness in an inverse scattering prob-lem within non-trapping polygonal obstacles with at most two in-coming waves. Inverse Probl. ,
20] Elschner, J.; Yamamoto, M. Uniqueness in determining polyhedralsound-hard obstacles with a single incoming wave.
Inverse Probl. , , 035004.[21] Liu, H.; Petrini, M.; Rondi, L.; Xiao, J. Stable determination ofsound-hard polyhedral scatterers by a minimal number of scatteringmeasurements. J. Differ. Eq. , , 1631–1670.[22] Liu, H.; Rondi, L.; Xiao, J. Mosco convergence for H (curl) spaces,higher integrability for Maxwell’s equations, and stability in directand inverse EM scattering problems. J. Eur. Math. Soc. (JEMS) , , 2945–2993.[23] Liu, H.; Zou, J. Uniqueness in an inverse acoustic obstacle scatteringproblem for both sound-hard and sound-soft polyhedral scatterers. Inverse Probl. , , 515–524.[24] Rondi, L. Stable determination of sound-soft polyhedral scatterersby a single measurement. Indiana Univ. Math. J. , , 1377–1408.[25] Honda, N.; Nakamura, G.; Sini, M. Analytic extension and recon-struction of obstacles from few measurements for elliptic second or-der operators. Math. Ann. , , 401–427.[26] Bl˚asten, E.; P¨aiv¨arinta, L.; Sylvester, J. Corners always scatter. Commun. Math. Phys. , , 725–753.[27] P¨aiv¨arinta, L.; Salo, M.; Vesalainen, E.V. Strictly convex cornersscatter. Rev. Mat. Iberoam. , , 1369–1396.[28] E. Bl˚asten, Nonradiating sources and transmission eigenfunctionsvanish at corners and edges. SIAM J. Math. Anal. , , 6255–6270.[29] Bl˚asten, E.; Liu, H. On corners scattering stably, nearly non-scattering interrogating waves, and stable shape determination bya single far-field pattern. Indiana Univ. Math. J. , in press.[30] Bl˚asten, E.; Liu, H. Recovering piecewise constant refractive in-dices by a single far-field pattern.
Inverse Probl. , accepted,doi:10.1088/1361-6420/ab958f.[31] Hu, G.; Salo, M.; Vesalainen, E. Shape identification in inversemedium scattering problems with a single far-field pattern.
SIAMJ. Math. Anal. , , 152–165.[32] Ikehata, M. Reconstruction of a source domain from the Cauchydata. Inverse Probl. , , 637–645.[33] Kusiak, S.; Sylvester, J. The scattering support. Comm. Pure Appl.Math. , , 1525–1548.[34] Kusiak, S.; Sylvester, J. The convex scattering support in a back-ground medium. SIAM J. Math. Anal. , , 1142–1158.[35] P¨aiv¨arinta, L.; Rempel, S. A deconvolution problem with the kernel1 / | x | on the plane. Appl. Anal. , , 105–128.[36] P¨aiv¨arinta, L.; Rempel, S. Corner singularities of solutions to∆ ± / u = f in two dimensions. Asymptotic Anal. , , 429–460.
37] Stephan, E.P. Boundary integral equations for screen problems in R . Integr. Eq. Oper. Theory , , 236–257.[38] Evans, L.C. Partial Differential Equations , 2nd ed.; Graduate Stud-ies in Mathematics; American Mathematical Society: Providence,RI, USA, 2010; Volume 19.[39] Deng, Y.; Liu, H.; Uhlmann, G. On regularized full- and partial-cloaks in acoustic scattering.
Comm. Part. Differ. Eq. , ,821–851.,821–851.