Lens Rigidity in Scattering by Unions of Strictly Convex Bodies in $\R^2$
aa r X i v : . [ m a t h . D S ] O c t LENS RIGIDITY IN SCATTERING BY UNIONS OF STRICTLYCONVEX BODIES IN R ∗ LYLE NOAKES † AND
LUCHEZAR STOYANOV ‡ Abstract.
It was proved in [NS1] that obstacles K in R d that are finite disjoint unions ofstrictly convex domains with C boundaries are uniquely determined by the travelling times ofbilliard trajectories in their exteriors and also by their so called scattering length spectra. Howeverthe case d = 2 is not covered in [NS1]. In the present paper we give a separate different proof of thisresult in the case d = 2. Key words. scattering by obstacles, billiard flow, scattering ray, travelling times spectrum,trapped trajectory
AMS subject classifications.
1. Introduction.
In scattering by an obstacle in R d ( d ≥
2) the obstacle K is acompact subset of R d with a C boundary ∂K such that Ω K = R d \ K is connected.A scattering ray in Ω K is an unbounded in both directions generalized geodesic (inthe sense of Melrose and Sj¨ostrand [MS1], [MS2]). Most of these scattering raysare billiard trajectories with finitely many reflection points at ∂K . In this paper weconsider the case when K has the form(1.1) K = K ∪ K ∪ . . . ∪ K k , where K i are strictly convex disjoint domains in R d with C smooth boundaries ∂K i .Then all scattering rays in Ω K are billiard trajectories, and the so called generalizedHamiltonian (or bicharacteristic) flow F ( K ) t : S ∗ (Ω K ) −→ S ∗ (Ω K )coincides with the billiard flow (see [CFS]).Given an obstacle K in R d , consider a large ball M containing K in its interior,and let S = ∂M be its boundary sphere. For any q ∈ ∂K let ν K ( q ) the outward unitnormal to ∂K . For q ∈ S we will denote by ν ( q ) the inward unit normal to S at q .Set S ∗ + ( S ) = { x = ( q, v ) : q ∈ S , v ∈ S d − , h v, ν ( q ) i ≥ } . Given x ∈ S ∗ + ( S ), define the travelling time t K ( x ) ≥ ∞ ) such that pr ( F ( K ) t ( x )) is in the interior of Ω K ∩ M for all 0 < t < t K ( x ), wherepr ( p, w ) = p (see Figure 1). For x = ( q, v ) ∈ S ∗ + ( S ) with h ν ( q ) , v i = 0 set t ( x ) = 0.The set { ( x ; t K ( x )) : x ∈ S ∗ + ( S ) } will be called the travelling times spectrum of K .It is natural to ask what information about the obstacle K can be derived fromits travelling times spectrum. For example: what is the relationship between twoobstacles K and L in R d if they have (almost) the same travelling times spectra? We ∗ Submitted to the editors DATE. † Department of Mathematics, University of Western Australia, Crawley WA 6009, Australia([email protected]. ‡ Department of Mathematics, University of Western Australia, Crawley WA 6009, Australia([email protected]). 1
L. NOAKES AND L. STOYANOV say that K and L have almost the same travelling times if there exists a subset R offull Lebesgue measure in S ∗ + ( S ) such that t K ( x ) = t L ( x ) for all x ∈ R .Similar questions can be asked about the so called scattering length spectrum(SLS) of an obstacle. Given a scattering ray γ in Ω K , if ω ∈ S d − is the incomingdirection of γ and θ ∈ S d − its outgoing direction, γ will be called an ( ω, θ ) -ray . Forany vector ξ ∈ S d − denote by Z ξ the hyperplane in R d orthogonal to ξ and tangentto S such that S is contained in the open half-space R ξ determined by Z ξ andhaving ξ as an inner normal. For an ( ω, θ )-ray γ in Ω, the sojourn time T γ of γ isdefined by T γ = T ′ γ − a , where T ′ γ is the length of the part of γ which is contained in R ω ∩ R − θ and a is the radius of S . It is known that this definition does not dependon the choice of the sphere S . The scattering length spectrum of K is defined to bethe family of sets of real numbers SL K = {SL K ( ω, θ ) } ( ω,θ ) where ( ω, θ ) runs over S d − × S d − and SL K ( ω, θ ) is the set of sojourn times T γ of all( ω, θ )-rays γ in Ω K . It is known (cf. [PS]) that for d ≥ d odd, and C ∞ boundary ∂K , we have SL K ( ω, θ ) = sing supp s K ( t, θ, ω )for almost all ( ω, θ ). Here s K is the scattering kernel related to the scattering operatorfor the wave equation in R × Ω K with Dirichlet boundary condition on R × ∂ Ω K (cf.[LP], [M], [PS]). Following [St3], we will say that two obstacles K and L have almostthe same SLS if there exists a subset R of full Lebesgue measure in S d − × S d − suchthat SL K ( ω, θ ) = SL L ( ω, θ ) for all ( ω, θ ) ∈ R . S q v Figure 1It is a natural and rather important problem in inverse scattering by obstacles toget information about the obstacle K from its SLS. It is known that various kinds ofinformation about K can be recovered from its SLS, and for some classes of obstacles K is completely recoverable (see [St3] for more information) – for example star-shapedobstacles are in this class. ENS RIGIDITY IN SCATTERING IN R K and L are finite disjoint unions of strictly convex bodies in R d with C boundaries and K and L have almost the same travelling times spectra (or almost the same SLS), then K = L . However the argument in [NS1] does not work in the case d = 2. We aregrateful to Antoine Gansemer who pointed this to us. As he showed in [Gan], when d = 2 and k > S ∗ + ( S ) \ Trap(Ω K ) is disconnected, and then the argumentin [NS1] does not work. Here Trap(Ω K ) is the set of all trapped points in S ∗ (Ω K ), i.e.points x = ( q, v ) ∈ S ∗ (Ω K ) such that either the forward billiard trajectory γ + K ( x ) = { pr ( F ( K ) t ( x )) : t ≥ } or the backward trajectory γ − K ( q, v ) = γ + K ( q, − v ) is infinitely long. That is, either thebilliard trajectory in the exterior of K issued from q in the direction of v is bounded(contained entirely in M ) or the one issued from q in the direction of − v is bounded.The obstacle K is called non-trapping if Trap(Ω K ) = ∅ .Here we prove the following. Theorem
Let K and L be obstacles in R such that each of them is a finitedisjoint union of strictly convex compact domains with C boundaries. If K and L have almost the same travelling times or almost the same scattering length spectra,then K = L . The argument we use is completely different from that in [NS1]. A result similarto that in [NS1] concerning non-trapping obstacles satisfying certain non-degeneracyconditions was proved recently in [St5].The set of trapped points plays a rather important role in various inverse problemsin scattering by obstacles, and also in problems on metric rigidity in Riemanniangeometry. It is known that Trap(Ω K ) ∩ S ∗ + ( S ) has Lebesgue measure zero in S ∗ + ( S )(see Sect. 4 for more information about this). However, as an example of M. Livshitsshows (see Ch. 5 in [M] or Figure 1 in [NS1]), in general the set of points x ∈ S ∗ (Ω K )for which γ K ( x ) = γ + K ( x ) ∪ γ − K ( x )is trapped in both directions may contain a non-trivial open set. In the latter case theobstacle cannot be recovered from travelling times (and also from the SLS). Similarexamples in higher dimensions are given in [NS3]. Definition
Let
K, L be two obstacles in R d . We will say that Ω K and Ω L have conjugate flows if there exists a homeomorphism Φ : S ∗ (Ω K ) \ Trap(Ω K ) −→ S ∗ (Ω L ) \ Trap(Ω L ) which is C on an open dense subset of S ∗ (Ω K ) \ Trap(Ω K ) and satisfies F ( L ) t ◦ Φ = Φ ◦ F ( K ) t , t ∈ R , and Φ = id on S ∗ ( R d \ M ) \ Trap(Ω K ) = S ∗ ( R d \ M ) \ Trap(Ω L ) . L. NOAKES AND L. STOYANOV
For
K, L in a generic class of obstacles in R d ( d ≥ K and L have almost the same SLS oralmost the same travelling times, then Ω K and Ω L have conjugate flows ([St3] and[NS2]). Thus, Theorem 1.1 is an immediate consequence of the following. Theorem
Let each of the obstacles K and L be a finite disjoint union ofstrictly convex compact domains in R with C boundaries. If Ω K and Ω L haveconjugate flows, then K = L . We prove Theorem 1.3 in Sect. 3 below. In Sect. 2 we state some useful resultsfrom [St2] and [St3]. It turns out that billiard trajectories with tangent points tothe boundary of the obstacle play an important role in the two-dimensional caseconsidered here. In Sect. 4 we prove that the set of trapped points Trap(Ω K ) hasLebesgue measure zero in S ∗ (Ω K ).
2. Preliminaries.
Next, we describe some propositions from [St2] and [St3] thatare needed in the proof of Theorem 1.3. We state them in the general case d ≥ d = 2. Standing Assumption. K and L are finite disjoint unions of strictly convex domainsin R d ( d ≥
2) with C boundaries and with conjugate flows F ( K ) t and F ( L ) t . Proposition ([St2]) (a) There exists a countable family { M i } = { M ( K ) i } ofcodimension submanifolds of S ∗ + ( S ) \ Trap(Ω K ) such that every σ ∈ S ∗ + ( S ) \ (Trap(Ω K ) ∪ i M i ) generates a simply reflecting ray in Ω K . Moreover the family { M i } is locally finite,that is any compact subset of S ∗ + ( S ) \ Trap(Ω K ) has common points with only finitelymany of the submanifolds M i .(b) There exists a countable family { R i } of codimension smooth submanifoldsof S ∗ + ( S ) such that for any σ ∈ S ∗ + ( S ) \ ( ∪ i R i ) the trajectory γ K ( σ ) has at most onetangency to ∂K .(c) There exists a countable family { Q i } of codimension smooth submanifoldsof S ∗ ∂K (Ω K ) such that for any σ ∈ S ∗ + ( ∂K ) \ ( ∪ i Q i ) the trajectory γ K ( σ ) has at mostone tangency to ∂K . It follows from the conjugacy of flows and Proposition 4.3 in [St3] that the sub-manifolds M i are the same for K and L , i.e. M ( K ) i = M ( L ) i for all i .The following is Lemma 5.2 in [St2]. In fact the lemma in [St2] assumes C ∞ smoothness for the submanifold X , however its proof only requires C smoothness. Proposition
Let X be a C smooth submanifold of codimension in R d ,and let x ∈ X and ξ ∈ T x X , k ξ k = 1 , be such that the normal curvature of X at x in the direction ξ is non-zero. Then for every ǫ > there exist an open neighbourhood V of x in X , a smooth map V ∋ x ξ ( x ) ∈ T x X and a smooth positive function t ( x ) ∈ [ δ, ǫ ] on V for some δ > such that Y = { y ( x ) = x + t ( x ) ξ ( x ) : x ∈ V } ENS RIGIDITY IN SCATTERING IN R is a smooth strictly convex surface with an unit normal field ν Y ( y ( x )) = ξ ( x ) , x ∈ V .That is, the normal field of Y consists of vectors tangent to X at the correspondingpoints of V . (See Figure 2.) As one would expect, the case d = 2 of the above proposition is rather easy toprove. X x Y ξ Figure 2An important consequence of the above is the following proposition which canbe proved using part of the argument in the proof of Proposition 5.5 in [St2]. Forcompleteness we sketch the proof in the Appendix.
Proposition
Let K be an obstacle in R which is a finite disjoint union ofstrictly convex compact domains with C boundaries. Then dim( S ∗ ( ∂K ) ∩ Trap(Ω K )) = 0 . In particular, S ∗ ( ∂K ) ∩ Trap(Ω K ) does not contain non-trivial open subsets of S ∗ ( ∂K ) . Here we denote by dim( X ) the topological dimension of a subset X of R .It turns out that for the type of obstacles considered in this paper the setTrap(Ω K ) of trapped points has Lebesgue measure zero in S ∗ (Ω K ). While formallythis fact is not necessary for the proof of Theorem 1.3, we mention it here since it is arather important feature of the billiard flow in the case considered in this paper (andalso in [NS1], [St3], etc.). This appears to be accepted as a ‘known fact’ althoughwe could not find a formal proof anywhere in the literature. However a simple prooffollows from known facts, e.g. using the ergodicity of the so called dispersive (Sinai)billiards (see [Si1], [Si1]). Proposition
Let K be an obstacle in R d of the form (1.1) . Then the set Trap(Ω K ) of all of trapped points of S ∗ (Ω K ) has Lebesgue measure zero in S ∗ (Ω K ) . We provide a proof of this proposition in Sect. 4 below.
3. Proof of Theorem 1.3.
Assume that the obstacles K and L in R satisfythe assumptions of Theorem 1.3.We claim that K ⊂ L . Assume this is not true and fix an arbitrary x ∈ ∂K suchthat x / ∈ L . Let ξ ∈ S be one of the unit vectors tangent to ∂K at x .It follows from Proposition 2.2 that there exists a small ǫ >
0, an open neigh-bourhood V of x in ∂K , a C map V ∋ x ξ ( x ) ∈ S ∗ x ( ∂K ) and a C positivefunction t ( x ) ∈ [ δ, ǫ ] on V for some δ ∈ (0 , ǫ ) such thatΣ = { y ( x ) = x + t ( x ) ξ ( x ) : x ∈ V } is a C strictly convex curve with unit normal field ν Σ ( y ( x )) = ξ ( x ), x ∈ V . So, forany x ∈ V the straight line through y ( x ) with direction ξ ( x ) is tangent to ∂K at x .Set y = x + ǫ ξ ∈ Σ. L. NOAKES AND L. STOYANOV
It follows from Proposition 2.3 that for the subsetΣ ′ = { y ∈ Σ : ( y, ν Σ ( y )) / ∈ Trap(Ω K ) } we have dim(Σ \ Σ ′ ) = 0. Thus, dim(Σ ′ ) = 1.Next, Proposition 2.1 implies that for all but countably many y ∈ Σ ′ the tra-jectories γ K ( y, ν Σ ( y )) and γ L ( y, ν Σ ( y )) have at most one tangency to ∂K and ∂L ,respectively. For such y , since γ K ( y, ν Σ ( y )) has a tangent point to ∂K , it must haveexactly one tangent point to ∂K . Since the flows F ( K ) t and F ( L ) t are conjugate byassumption, γ L ( y, ν Σ ( y )) also must have exactly one tangent point z ( y ) to ∂L . Moreprecisely, if ( y, ν Σ ( y )) = F ( K ) t ( σ ) for some σ ∈ S ∗ + ( S ) and some t >
0, then thetravelling time function t K has a singularity at σ . Since t K = t L on S ∗ + ( S ) near σ , the function t L also has a singularity at σ , so γ L ( y, ν Σ ( y )) = γ L ( σ ) must have atangent point to ∂L .Assume for a moment that for every z ∈ ∂L there exists an open neighbourhood W z of z in ∂L such that W z ∩ { z ( y ) : y ∈ Σ ′ } has topological dimension zero. Covering ∂L with a finite number of neighbourhoods W z , it follows that Σ ′ has topological dimension zero – a contradiction. Thus, thereexists z ∈ ∂L such that for every open neighbourhood W of z in ∂L the set W ∩ { z ( y ) : y ∈ Σ } has topological dimension one. Replacing y (and therefore x as well) by an appro-priate nearby point on Σ ′ , we may assume that z = pr ( F ( L ) t ( y , ν Σ ( y ))) for some t ∈ R , t = 0.We will assume that t >
0; otherwise we just have to replace ξ by − ξ and thecurve Σ by { x − t ( x ) ξ ( x ) : x ∈ V } . Let F ( L ) t ( y , ν Σ ( y )) = ( z , − ζ ) .∂K x Σ ξ ∂Lz X pζ Figure 3Using again Proposition 2.2, assuming ǫ > W of z in ∂L if necessary, there exist a C map W ∋ z ζ ( z ) ∈ S ∗ z ( ∂L )and a C positive function s ( z ) ∈ [ δ, ǫ ] on W for some δ ∈ (0 , ǫ ) such that ζ ( z ) = ζ and X = { p ( z ) = z + s ( z ) ζ ( z ) : z ∈ W } is a C strictly convex curve with unit normal field ν X ( p ( z )) = ζ ( z ), z ∈ W . So, forany z ∈ W the straight line through p ( z ) with direction ζ ( z ) is tangent to ∂L at z (see Figure 3). Set p = z + ǫ ζ ∈ X . ENS RIGIDITY IN SCATTERING IN R x = pr ( F ( L ) t ( x , ξ )) , . . . , x k = pr ( F ( L ) t k ( x , ξ ))be the common points of γ L ( x , ξ ) with ∂L (if any) with 0 < t < . . . < t k < t . Fixan arbitrary T ∈ ( t k , t ) close to t k . It then follows from a well-known result of Sinai([Si1]; see also [Si2]) that there exists an open neighbourhood Σ of y in Σ such that Y = { pr ( F ( L ) T ( y, ν Σ ( y ))) : y ∈ Σ } is a strictly convex curve in R with a unit normal field ν Y ( y, ν Σ ( y )) = pr ( F ( L ) T ( y, ν Σ ( y ))) .∂K x Σ ξ y ∂Lz Xp ζ x x Yq η x k Figure 4Set q = pr ( F ( L ) T ( y , ν Σ ( y ))) ∈ Y , η = ν Y ( q )(see Figure 4). It follows from the constructions of Σ, the point z ∈ ∂L , the neigh-bourhood W and the convex fronts X and Y that for y ∈ Σ ∩ Σ ′ the point q = pr ( F ( L ) T ( y, ν Σ ( y ))) ∈ Y is such that the straightline ray issued from q in direction ν Y ( q ) hits X perpendicularly.However, due to the strict convexity of X and Y , this is only possible when y = y ;a contradiction.This proves that we must have K ⊂ L .Using a similar argument we derive that L ⊂ K , as well. Therefore K = L .
4. On the set of trapped points.
Here we prove Proposition 2.4.Assume again that K is an obstacle in R d ( d ≥
2) of the form (1.1) where K i arestrictly convex disjoint domains in R d with C smooth boundaries ∂K i . Let λ be theLebesgue measure on S ∗ ( R d ). Let S be a large sphere in R d as in Sect. 1, and let µ be the Liouville measure on S ∗ + ( S ) defined by dµ = dρ ( q ) dω q | h ν ( q ) , v i | , where ρ is the measure on S determined by the Riemannian metric on S and ω q isthe Lebesgue measure on the ( d − S q ( S ) (see e.g. Sect. 6.1 in[CFS]).We will need the following generalisation of Santalo’s formula proved in [St4]. Infact, the latter deals with general billiard flows on Riemannian manifolds (under somenatural assumptions), however here we will restrict ourselves to the case consideredin Sect. 1. L. NOAKES AND L. STOYANOV
Theorem ([St4]) Let K be as above. Then for every λ -measurable function f : S ∗ (Ω K ) \ Trap(Ω K ) −→ C such that | f | is integrable we have Z S ∗ (Ω K ) \ Trap(Ω K ) f ( x ) dλ ( x ) = Z S ∗ + ( S ) \ Trap(Ω K ) Z t K ( x )0 f ( F ( K ) t ( x )) dt ! dµ ( x ) . As we mentioned earlier Trap(Ω K ) ∩ S ∗ + ( S ) has Lebesgue measure zero in S ∗ + ( S )(see Theorem 1.6.2 in [LP]; see also Proposition 2.3 in [St2] for a more rigorous proof).Using this and the above theorem with f = 1 gives the following. Corollary ([St4]) Let K be as above. Then λ ( S ∗ (Ω K ) \ Trap(Ω K )) = Z S ∗ + ( S ) \ Trap(Ω K ) t K ( x ) dµ ( x ) . That is, λ (Trap(Ω K )) = λ ( S ∗ (Ω K )) − Z S ∗ + ( S ) t K ( x ) dµ ( x ) . Proof of
Proposition 2.4. We can regard K as a subset of a domain Q in R d with apiecewise smooth boundary which is strictly convex inwards (see Figure 5). Considerthe billiard flow φ t on S ∗ ( Q ). It is well-known (see [CFS]) that φ t preserves theLebesgue measure λ (restricted to S ∗ ( Q )). Moreover φ t is ergodic with respect to λ ([Si1], [Si2]). ∂QK Figure 5Let T be the set of points x ∈ S ∗ (Ω K ) such that the billiard trajectory γ K ( x ) istrapped in both directions. Then Corollary 4.2 and the fact mentioned above thatTrap(Ω K ) ∩ S ∗ + ( S ) has Lebesgue measure zero in S ∗ + ( S ) imply that Trap(Ω K ) \ T has Lebesgue measure zero in S ∗ (Ω K ). So, it is enough to prove that λ ( T ) = 0. Or as a domain on the flat d -dimensional torus T d . Both embeddings will produce the requiredresult.ENS RIGIDITY IN SCATTERING IN R φ t coincides with the flow F ( K ) t on the set T , and T is aninvariant set with respect to F ( K ) t , and so with respect to φ t . Clearly T is a propersubset of S ∗ ( Q ) and S ∗ ( Q ) \ T has positive measure. Now the ergodicity of φ t impliesthat λ ( T ) = 0.
5. Appendix.
Here we prove Proposition 2.3 using part of the argument in theproof of Proposition 5.5 in [St2].It is enough to prove that every x ∈ ∂K has an open neighbourhood V in ∂K such that dim( S ∗ ( V ) ∩ Trap(Ω K )) = 0.Let x ∈ ∂K . As in the proof of Theorem 1.3, it follows from Proposition 2.2that there exists a small ǫ >
0, an open neighbourhood V of x in ∂K , a C map V ∋ x ξ ( x ) ∈ S ∗ x ( ∂K ) and a C positive function t ( x ) ∈ [ δ, ǫ ] on V for some δ ∈ (0 , ǫ ) such that Σ = { y ( x ) = x + t ( x ) ξ ( x ) : x ∈ V } is a C strictly convex curve with unit normal field ν Σ ( y ( x )) = ξ ( x ), x ∈ V . Set e Σ = { ( y, ν Σ ( y )) : y ∈ Σ } . It follows from Proposition 2.1(c) that there exists a countable subset X ′ = { Q i } of S ∗ ( ∂K ) such that for any σ ∈ S ∗ ( ∂K ) \ X ′ , the trajectory γ K ( σ ) has at most onetangency to ∂K , and therefore it has exactly one tangency to ∂K .Let X the set of those σ ∈ e Σ ∩ Trap(Ω K ) such that the trajectory γ + K ( σ ) has notangencies to ∂K . Set F = { , , . . . , k } , and consider e F = ∞ Y r =1 F with the product topology. It is well known that dim( e F ) = 0 and therefore everysubspace of e F has topological dimension zero (cf. e.g. [HW] or [E]). Consider themap f : X −→ e F , defined by f ( σ ) = ( i , i , . . . , i n , . . . ) , where the n th reflection point of γ + K ( σ ) belongs to ∂K i n for all n = 1 , , . . . . Clearly,the map f is continuous and it follows from [St1] that f is injective, so it defines ahomeomorphism f : X −→ f ( X ). Thus, X is homeomorphic to a subspace of e F and therefore dim( X ) = 0.Now the Sum Theorem for dim (cf. [HW] or [E]) shows that dim( X ′ ∪ X ) = 0 . Since S ∗ ( V ) ∩ Trap(Ω K ) is naturally homeomorphic to X ′ ∪ X , it follows thatdim( S ∗ ( V ) ∩ Trap(Ω K )) = 0 . This proves the proposition.0
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