aa r X i v : . [ m a t h - ph ] J a n An inverse problem without the phase information
Michael V. Klibanov ∗† Abstract
We prove a new uniqueness theorem for an inverse scattering problem withoutthe phase information for the 3-D Helmholtz equation. The spatially distributeddielectric constant is the subject of the interest in this problem. We consider thecase when the modulus of the scattered wave field | u sc | is measured. The phase isnot measured. Key Words : phaseless data, inverse scattering problem, uniqueness theorem
Phaseless Inverse Scattering Problem (PISP) for a wave-like Partial Differential Equation(PDE) with a complex valued solution is the problem of the reconstruction of an unknowncoefficient of this PDE from measurements of the modulus of its solution on a certainset. The phase is not measured. On the other hand, in conventional inverse scatteringproblems both the modulus and the phase of the complex valued wave field are measuredon certain sets, see, e.g. [4, 7, 9, 10, 25, 26, 27].Let u = u + u sc be the total wave field, where u is the incident wave field and u sc is thewave field scattered by a scatterer. The main result of this paper is a uniqueness theoremfor the PISP for the 3-D Helmholtz equation in the case when the modulus | u sc | of thescattered wave field is measured on a certain surface. The closest previous publication is[19]. In [19] uniqueness was proven for the case when the modulus | u | of the total wavefield is measured. Compared with [19], the main difficulty here is due to the interferenceof two wave fields: u and u . To handle this difficulty, we develop here some significantlynew ideas , which are presented in section 5 as well as in (6.10)-(6.15).The first uniqueness result for a PISP was proven in [13] for the 1-D case. In [16, 17]uniqueness theorems were proven for the case when the Schr¨odinger equation∆ v + k v − q ( x ) v = − δ ( x − y ) , x ∈ R (1.1)was the underlying one. However, equation (1.1) is easier to investigate since the unknowncoefficient q ( x ) is not multiplied by k here, unlike the Helmholtz equation. Hence, unlike ∗ Department of Mathematics and Statistics, University of North Carolina at Charlotte, Charlotte, NC28223, USA; [email protected] . † This work was supported by US Army Research Laboratory and US Army Research Office grantW911NF-15-1-0233 and by the Office of Naval Research grant N00014-15-1-2330. g ( x ) rather than the δ − function. This is because the last step of the proof of each of those theorems is theapplication of the method of [6], which is based on Carleman estimates, see, e.g. [15] forthe recent survey of this method. However, the method of [6] does not work for the casewhen the δ − function stands in the right hand side of a PDE.PISPs were also considered in [28, 29]. Statements of problems of these works aredifferent from ours, so as the results. These results include both uniqueness theorems andreconstruction procedures.Recall that a PISP is about the reconstruction of an unknown coefficient from phaselessmeasurements. Along with PISPs, inverse problems of the reconstruction of unknownsurfaces of scatterers from phaseless data are also of an obvious interest. In this regard, werefer to [1, 2, 3, 11, 12] for numerical solutions of some inverse scattering problems withoutthe phase information in the case when the surface of a scatterer was reconstructed.As to the applications of PISPs, they are in the lenseless imaging of nanostructures.In the case when the size of a nanostructure is of the order of 100 nanometers, whichis 0.1 micron, the wavelength of the probing radiation must be also about 0.1 micron.This corresponds to the frequency of 2,997,924.58 Gigahertz, see, e.g. [32]. It is wellknown that it is impossible to measure the phase of an electromagnetic radiation at suchhuge frequencies [8, 30, 36]. Therefore, to image such a nanostructure, one needs tocompute its unknown spatially distributed dielectric constant using measurements of onlythe intensity of the scattered wave field (intensity is the square of the modulus). Thesecond application is in optical imaging of biological cells, since their sizes are between 1and 10 microns [31].In section 2 we formulate our PISP as well as the main result of our paper. The restof the paper is devoted to the proof of this result. In section 3 we explore a connectionvia the Fourier transform between our forward problem for the Helmholtz equation andthe Cauchy problem for a certain hyperbolic equation. In section 4 we formulate threelemmata. In two lemmata and one corollary of section 5 we present the above mentionednew ideas, which handle the interference between the wave fields u and u . In section 6we finalize the proof of the main result. 2
Problem statement
Below x = ( x , x , x ) ∈ R . Consider a non-magnetic and non-conductive medium, whichoccupies the whole space R . Let c ( x ) be the spatially varying dielectric constant of thismedium. It was established in chapter 13 of the classical textbook of Born and Wolf[5] that if the function c ( x ) varies slowly enough on the scales of the wavelength, thenthe scattering problem for Maxwell’s equations can be approximated by the scatteringproblem for the Helmholtz equation for a certain component of the electric field. Thisjustifies, from the Physics standpoint, our work with the Helmholtz equation.Let Ω , Ψ , G ⊂ R be three bounded domains and Ω ⊂ Ψ ⊂ G. Let ∂ Ψ = S ∈ C . Denote 2 ρ = min ( dist ( S, ∂ Ω) , dist ( S, ∂G )) , where “ dist ” denotes the Hausdorffdistance. We assume that ρ > . For any number ω > y ∈ R denote P ω ( y ) = { x ∈ R : | x − y | < ω } . We impose throughout the paper the followingconditions on the function c ( x ): c ∈ C ( R ) , (2.1) c ( x ) ≥ R , (2.2) c ( x ) ≥ β in Ψ , β = const. > , (2.3) c ( x ) = 1 for x ∈ R \ G. (2.4)Inequality (2.2) means that the dielectric constant of the medium is not less than thedielectric constant of the vacuum, which is 1. So, (2.4) means that the domain G isembedded in the vacuum. In section 3 we use the fundamental solution of a hyperbolicequation with the coefficient c ( x ) in the principal part of its operator. The constructionof this solution works only if c ∈ C ( R ) [20, 34]. In addition, the constructions of[20, 34] require the regularity of geodesic lines, see Condition below. We also note thatthe minimal smoothness requirements for unknown coefficients are rarely a significantconcern in uniqueness theorems for multidimensional coefficient inverse problems, see,e.g. [26, 27], theorem 4.1 in Chapter 4 of [33] and [15].The function c ( x ) generates the conformal Riemannian metric, dτ = p c ( x ) | dx | , | dx | = p ( dx ) + ( dx ) + ( dx ) . (2.5)We assume throughout the paper that the following condition holds: Condition . Geodesic lines generated by the metric (2.5) are regular. In other words,each pair of points x, y ∈ R can be connected by a single geodesic line Γ ( x, y ) . A sufficient condition for the regularity of geodesic lines was derived in [35]. For anarbitrary pair of points x, y ∈ R consider the travel time τ ( x, y ) between them due tothe Riemannian metric (2.5). Then [20, 33] |∇ x τ ( x, y ) | = c ( x ) , (2.6) τ ( x, y ) = O ( | x − y | ) as x → y. (2.7)The solution of the problem (2.6), (2.7) is [20, 33] τ ( x, y ) = Z Γ( x,y ) p c ( ξ ) dσ, (2.8)3here dσ is the euclidean arc length. Using the above Condition, we conclude that τ ( x, y )is a single-valued function in R × R .Let y ∈ R be the position of the point source, r = | x − y | and k > u + k c ( x ) u = − δ ( x − y ) , x ∈ R , (2.9) ∂ r u − iku = o (1 /r ) , r → ∞ . (2.10)Theorem 8.7 of [7] implies that for each k > u ∈ C ( | x − y | ≥ ε ) , ∀ ε > . Consider the incident spherical wave u and the scatteredwave u sc , u ( x, y, k ) = exp ( ik | x − y | )4 π | x − y | , (2.11) u sc ( x, y, k ) = u ( x, y, k ) − u ( x, y, k ) . (2.12)In this paper we consider the following PISP: Phaseless Inverse Scattering Problem (PISP) . Assume that the function c ( x ) is given for x ∈ R (cid:31) Ω and it is unknown for x ∈ Ω . Suppose that the following function F ( x, y, k ) is known F ( x, y, k ) = | u sc ( x, y, k ) | , ∀ y ∈ ∂S, ∀ x ∈ P ρ ( y ) , x = y, ∀ k ∈ ( a, b ) , (2.13) where ( a, b ) ⊂ R + = { k : k > } is a certain interval. Determine the function c ( x ) for x ∈ Ω . The main result of this paper is Theorem 1:
Theorem 1 . Assume that (2.1)- (2.4), (2.12), (2.13) and Condition hold. Then thePISP has at most one solution.
The rest of this paper is devoted to the proof of this theorem. We assume below thatits conditions hold.
As in [19, 20], consider the following Cauchy problem c ( x ) U tt = ∆ U + δ ( x − y, t ) , x ∈ R , t > , (3.1) U | t< ≡ . (3.2)For an arbitrary T > K ( y, T ) and K ∗ ( y, T ) as K ( y, T ) = { ( x, t ) : 0 < t ≤ T − τ ( x, y ) } ,K ∗ ( y, T ) = { ( x, t ) : τ ( x, y ) ≤ t ≤ T − τ ( x, y ) } . Let H ( t ) be the Heaviside function, H ( t ) = (cid:26) , t > , , t < . emma 3.1 [20]. For any fixed source position y ∈ R and for any T > , thereexists unique solution U ( x, y, t ) of the problem (3.1), (3.2), which can be represented inthe domain K ( y, T ) in the form U ( x, y, t ) = A ( x, y ) δ ( t − τ ( x, y )) + e U ( x, y, t ) H ( t − τ ( x, y )) , (3.3) where the function e U ( x, y, t ) ∈ C ( K ∗ ( y, T )) and A ( x, y ) is a certain function suchthat A ( x, y ) > , ∀ x ∈ R , x = y and A ( x, y ) is continuous with respect to x, y for x = y. Furthermore, for any bounded domain D ⊂ R the function U ( x, y, t ) decays exponentiallywith respect to t together with its x, t derivatives up to the second order. In other words,there exist numbers γ = γ ( D, c, y ) > , Y = Y ( D, c, y ) > , t = t ( D, c, y ) > depending only on listed parameters such that (cid:12)(cid:12) D αx,t U ( x, y, t ) (cid:12)(cid:12) ≤ Y e − γ t , ∀ t ≥ t , ∀ x ∈ D. (3.4) In (3.4) α = ( α , α , α , α ) is the multi-index with non-negative integer coordinates and | α | = α + α + α + α ≤ . Consider the Fourier transform F ( U ) with respect to t ofthe function U , F ( U ) ( x, y, k ) = ∞ Z U ( x, y, t ) e ikt dt := V ( x, y, k ) . (3.5) Then V ( x, y, k ) = u ( x, y, k ) , (3.6) where the function u ( x, y, k ) is the solution of the problem (2.9), (2.10). We note that the assertions of this lemma about both the exponential decay and (3.6)were derived in [20] from the results of [37, 38].For an arbitrary number θ > C θ = { z ∈ C : Im z > − θ } , C + = { z ∈ C : Im z > } . Lemma 3.2 follows immediately from (3.4)-(3.6).
Lemma 3.2 [19].
For every x ∈ G the function u ( x, y, k ) is analytic with respect to thereal variable k ∈ R + . Furthermore, the function u ( x, y, k ) can be analytically continuedwith respect to k from R + in the half complex plane C γ , where γ = γ ( G, c, y ) > is thenumber of Lemma 3.1. Lemma 3.3 follows immediately from (3.3)-(3.6).
Lemma 3.3 [19].
Let A ( x, y ) be the function in (3.3). The asymptotic behavior ofthe function u ( x, y, k ) is u ( x, y, k ) = A ( x, y ) e ikτ ( x,y ) (cid:18) O (cid:18) k (cid:19)(cid:19) , | k | → ∞ , k ∈ C γ/ , x ∈ G, (3.7) where (cid:12)(cid:12)(cid:12)(cid:12) O (cid:18) k (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ M | k | + 1 , k ∈ C γ/ , x ∈ G, (3.8) where the number M = M ( G, c, y ) > depends only on listed parameters .5 Three lemmata
Lemma 4.1 [19].
Let the function f ( k ) be analytic for all k ∈ R . Then the function | f ( k ) | can be uniquely determined for all k ∈ R by the values of | f ( k ) | for k ∈ ( a, b ) . Lemma 4.2 [19].
Let { p j } N j =1 and { q j } N j =1 be two sets of integers, all of which arenon-negative. In addition, consider two sets of complex numbers { λ j } N j =1 , { θ j } N j =1 ⊂ C + . Assume that there exist two sets of non-zero numbers { l j } N j =1 , { s j } N j =1 ⊂ C suchthat N X j =1 l j t p j exp (cid:0) − iλ j t (cid:1) = N X j =1 s j t q j exp (cid:0) − iθ j t (cid:1) , ∀ t > . (4.1) Then N = N = N and numbers involved in (4.1) can be re-numbered in such a way that l j = s j , p j = q j , λ j = θ j , ∀ j = 1 , ..., N. Lemma 4.3 follows immediately from Proposition 4.3 of [14].
Lemma 4.3.
Let f ( k ) be an analytic function in the half plane C γ . Assume that thefunction f ( k ) has no zeros in C + ∪ R . Also, let the asymptotic behavior of the function f ( k ) be: f ( k ) = Ck n [1 + o (1) + p exp ( ikL )] exp ( ikL ) , | k | → ∞ , k ∈ C + ∪ R , where C, p ∈ C , L ∈ R are some numbers, L > , n ≥ and | p | < . (4.2) Then the values of | f ( k ) | for k ∈ R uniquely determine the function f ( k ) for k ∈ C + ∪ R . u and u main difficulties in the proofs of this section are caused by the above mentionedinterference of wave fields u and u . Lemma 5.1 . Fix the point y ∈ S. Then there exists a sufficiently small number ω = ω ( c, β ) ∈ (0 , ρ ) depending only on the function c and the number β , such that theasymptotic behavior of the function u sc ( x, y, k ) for | k | → ∞ , k ∈ C + ∪ R , x ∈ P ω ( y ) is u sc ( x, y, k ) = − exp ( ik | x − y | )4 π | x − y | (cid:20) − B ( x, y, k ) + O (cid:18) k (cid:19)(cid:21) , (5.1) where the function B ( x, y, k ) is B ( x, y, k ) = B ( x, y ) exp [ ik ( τ ( x, y ) − | x − y | )] , (5.2) B ( x, y ) = 4 π | x − y | A ( x, y ) . (5.3) The functions B ( x, y ) and | − B ( x, y, k ) | can be estimated as B ( x, y ) ≤ β/
21 + β , ∀ x ∈ P ω ( y ) , (5.4)6 − B ( x, y, k ) | ≥ β β ) , ∀ k ∈ C + ∪ R , ∀ x ∈ P ω ( y ) . (5.5) Furthermore, there exists a sufficiently large number K = K ( c, β ) > depending onlyon c and β such that the following estimate holds | u sc ( x, y, k ) | ≥ π | x − y | · β β ) , ∀ k ∈ ( C + ∪ R ) ∩ {| k | ≥ K } , ∀ x ∈ P ω ( y ) . (5.6) Proof . Formulas (5.1)-(5.3) follow immediately from formulas (2.12), (3.7) and (3.8).However, an essentially new element here are estimates (5.4)-(5.6).By the formula (3.9) of [20] the function A ( x, y ) has the form A ( x, y ) = p J ( x, y )4 π p c ( x ) τ ( x, y ) , (5.7)where J ( x, y ) > x, y ∈ R . Also, formula (3.7) of [20] implies that J ( y, y ) = 1 . Hence, there exists a sufficiently smallnumber ω ∈ (0 , ρ ) such that p J ( x, y ) < β , ∀ x ∈ P ω ( y ) . (5.8)Since S = ∂ Ψ , then by (2.3) there exists a sufficiently small number ω ∈ (0 , ρ ) such that c ( x ) ≥ β, ∀ x ∈ P ω ( y ) . (5.9)Let ω = min ( ω , ω ) . Then (2.8) and (5.9) imply that τ ( x, y ) ≥ p β | x − y | , ∀ x ∈ P ω ( y ) . (5.10)Hence, by (5.7)-(5.10) A ( x, y ) ≤ β/ β ) 4 π | x − y | , ∀ x ∈ P ω ( y ) . (5.11)Hence, using (5.3), we obtain (5.4).We now prove (5.5). By (5.10) τ ( x, y ) − | x − y | ≥ (cid:16)p β − (cid:17) | x − y | ≥ , ∀ x ∈ P ω ( y ) . Hence, | exp [ ik ( τ ( x, y ) − | x − y | )] | ≤ , ∀ k ∈ C + ∪ R , ∀ x ∈ P ω ( y ) . Hence, using (5.2) and (5.4), we obtain | B ( x, y, k ) | ≤ β/
21 + β , ∀ k ∈ C + ∪ R , ∀ x ∈ P ω ( y ) . Hence, | − B ( x, y, k ) | ≥ − | B ( x, y, k ) | ≥ β β ) , K = K ( c, β ) > (cid:12)(cid:12)(cid:12)(cid:12) O (cid:18) k (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ β β ) , ∀ k ∈ ( C + ∪ R ) ∩ {| k | ≥ K } , ∀ x ∈ P ω ( y ) . This, (5.1) and (5.5) prove (5.6). (cid:3)
Everywhere below each complex/real valued zero of the function u sc ( x, y, k ) := ϕ x,y ( k )as the function of the variable k is counted as many times as its multiplicity is. For anynumber z ∈ C its complex conjugate is denoted as z. Lemmata 3.2 and 5.1 imply Corollary 5.1.
Corollary 5.1 . Fix the point y ∈ S. Then there exists a sufficiently small number ω = ω ( c, β ) ∈ (0 , ρ ) depending only on the function c and the number β such that forevery fixed point x ∈ P ω ( y ) the function ϕ x,y ( k ) has at most a finite number of zeros inthe set C + ∪ R . Lemma 5.2 . Fix a point y ∈ S and a point x ∈ P ω ( y ) , x = y, where ω is thenumber of Lemma 5.1. Let { a j } m j =1 ⊂ R be the set of all real zeros of the function ϕ x,y ( k ) and { b j } m j =1 ⊂ C + be the set of all those complex zeros of ϕ x,y ( k ), which are located inthe upper half complex plane C + (Corollary 5.1). Consider the function e ϕ x,y ( k ) definedas e ϕ x,y ( k ) = ϕ x,y ( k ) m Y j =1 k − a j ! m Y j =1 k − b j k − b j ! . (5.12) Then the values of the modulus (cid:12)(cid:12) ϕ x,y ( k ) (cid:12)(cid:12) for k ∈ R together with the values of all realzeros uniquely determine the function e ϕ x,y ( k ) for k ∈ R . Proof . Note that (cid:12)(cid:12)(cid:12)(cid:12) k − zk − z (cid:12)(cid:12)(cid:12)(cid:12) = 1 , ∀ k ∈ R , ∀ z ∈ C . Hence, (cid:12)(cid:12)e ϕ x,y ( k ) (cid:12)(cid:12) = (cid:12)(cid:12) ϕ x,y ( k ) (cid:12)(cid:12) m Y j =1 | k − a j | , ∀ k ∈ R . (5.13)We now need to apply Lemma 4.3. To do this, we use Lemma 5.1. In notations ofLemma 4.3 C = − π | x − y | , L = | x − y | ,p = B ( x, y ) > , L = τ ( x, y ) − | x − y | > . It follows from (5.4) that p < , which means that condition (4.2) holds. Next, (5.12)implies that the function e ϕ x,y ( k ) does not have zeros in C + ∪ R . To finish this proof, werefer to Lemma 4.3 and (5.13). (cid:3)
In this proof we use results of sections 3-5. The part of the proof which handles theinterference of wave fields u and u is from (6.10) to (6.15).8ssume that there exist two functions c , c which correspond to the same function F ( x, y, k ) in (2.13). Then c ( x ) = c ( x ) , x ∈ R (cid:31) Ω . (6.1)Our goal is to prove that c ( x ) = c ( x ) , x ∈ Ω . (6.2)Let u ( x, y, k ) and u ( x, y, k ) be two functions u ( x, y, k ) which correspond to coeffi-cients c and c respectively and let u ,sc ( x, y, k ) and u ,sc ( x, y, k ) be two correspondingfunctions u sc ( x, y, k ). Fix a point y ∈ S. Let ω ∗ = min ( ω ( c , β ) , ω ( c , β )) , where thenumber ω = ω ( c, β ) ∈ (0 , ρ ) was defined in Lemma 5.1. Fix a point x ∈ P ω ∗ ( y ) , x = y. Denote ϕ ( k ) = u ,sc ( x, y, k ) , ϕ ( k ) = u ,sc ( x, y, k ) . (6.3)It follows from (2.13) and Lemma 4.1 that | ϕ ( k ) | = | ϕ ( k ) | , ∀ k ∈ R . (6.4)Using (6.4), we obtain, similarly with [19], that real zeros of functions ϕ ( k ) and ϕ ( k )coincide.Consider now zeros of functions ϕ ( k ) and ϕ ( k ) in the upper half complex plane C + . By Corollary 5.1 each of these functions has at most a finite number of such zeros.Let { d j } n j =1 ⊂ C + and { e j } n j =1 ⊂ C + be zeros of functions ϕ ( k ) and ϕ ( k ) respectively.Then (6.4) and Lemma 5.2 imply that ϕ ( k ) n Y j =1 k − d j k − d j = ϕ ( k ) n Y j =1 k − e j k − e j , ∀ k ∈ R . (6.5)Hence, ϕ ( k ) n Y j =1 k − e j k − e j = ϕ ( k ) n Y j =1 k − d j k − d j , ∀ k ∈ R . Hence, ϕ ( k ) + ϕ ( k ) n Y j =1 k − e j k − e j − ! = ϕ ( k ) + ϕ ( k ) n Y j =1 k − d j k − d j − ! , ∀ k ∈ R . (6.6)Denote g ( k ) = n Y j =1 k − d j k − d j − , g ( k ) = n Y j =1 k − e j k − e j − . Let the multiplicity of the zero e j be p j and the multiplicity of the zero d j be q j . Thenthe partial fraction expansion implies that there exist numbers X j and Y j such that g ( k ) = n ′ X j =1 X j (cid:0) k − d j (cid:1) p j , g ( k ) = n ′ X j =1 Y j ( k − e j ) q j , (6.7)In (6.7) n ′ , n ′ are some positive integers such that n ′ ≤ n , n ′ ≤ n . Applying the inverseFourier transform F − to functions g ( k ) and g ( k ) in (6.6), we obtain [19] F − ( g ) = Q ( t ) = H ( t ) n ′ X j =1 X j ( − p j − i p j ( p j − · t p j − exp (cid:0) − id j t (cid:1) . (6.8)9 − ( g ) = Q ( t ) = H ( t ) n ′ X j =1 Y j ( − q j − i q j ( q j − · t q j − exp ( − ie j t ) , (6.9)By Lemma 3.1 and (6.3) the inverse Fourier transform of each of functions ϕ ( k ) and ϕ ( k ) exists and F − ( ϕ ) = U ( x, y, t ) − δ ( t − | x − y | )4 π | x − y | , (6.10) F − ( ϕ ) = U ( x, y, t ) − δ ( t − | x − y | )4 π | x − y | , (6.11)where functions U ( x, y, t ) and U ( x, y, t ) are solutions of the Cauchy problem (3.1), (3.2)with c ( x ) = c ( x ) and c ( x ) = c ( x ) respectively. In (6.10) and (6.11) we have used thefact that F − ( u ) = δ ( t − | x − y | )4 π | x − y | . Therefore, using the convolution theorem and (6.6)- (6.11), we obtain U ( x, y, t ) + t Z (cid:18) U ( x, y, t − ξ ) − δ ( t − ξ − | x − y | )4 π | x − y | (cid:19) Q ( ξ ) dξ (6.12)= U ( x, y, t ) + t Z (cid:18) U ( x, y, t − ξ ) − δ ( t − ξ − | x − y | )4 π | x − y | (cid:19) Q ( ξ ) dξ. Since x ∈ P ω ∗ ( y ) , x = y and by (5.10) τ ( x, y ) ≥ √ β | x − y | , then (3.3) implies that U ( x, y, t ) = U ( x, y, t ) = 0, ∀ t ∈ (cid:16) | x − y | , p β/ | x − y | (cid:17) . (6.13)Hence, using (6.12) and (6.13), we obtain Q ( t − | x − y | ) = Q ( t − | x − y | ) , ∀ t ∈ (cid:16) | x − y | , p β/ | x − y | (cid:17) . (6.14)Since by (6.8) and (6.9) each of functions Q ( t ) and Q ( t ) is analytic as the function ofthe real variable t > , then (6.14) implies that Q ( t ) = Q ( t ) , ∀ t > . (6.15)Thus, using Lemma 4.2, (6.8), (6.9) and (6.15), we obtain that zeros of functions ϕ ( k )and ϕ ( k ) in C + ∪ R coincide.Hence, by (6.5) ϕ ( k ) = ϕ ( k ) , ∀ k ∈ R . Hence, (2.11), (2.12) and (6.3) imply that u ( x, y, k ) = u ( x, y, k ) , ∀ k ∈ R . (6.16)Since y is an arbitrary point of the surface S ⊂ R (cid:31) Ω, x = y is an arbitrary point of theball P ω ∗ ( y ) and P ω ∗ ( y ) ∩ Ω = ∅ , then, using (6.1), (6.16) and the well known theoremabout the uniqueness of the continuation of the solution of the elliptic equation of thesecond order (see, e.g. [24]), we obtain u ( x, y, k ) = u ( x, y, k ) , ∀ k ∈ R , ∀ y ∈ S, ∀ x ∈ R (cid:31) Ψ . U ( x, y, t ) = U ( x, y, t ) , ∀ t > , ∀ y ∈ S, ∀ x ∈ R (cid:31) Ψ . (6.17)Thus, (3.3) and (6.17) imply that τ ( x, y ) = τ ( x, y ) , ∀ x, y ∈ S, (6.18)where functions τ ( x, y ) and τ ( x, y ) correspond to the function τ ( x, y ) for c = c and c = c respectively.As the last step of the proof, we now apply to (6.18) theorem 3.4 of Chapter 3 of thebook [33]. We follow notations of that theorem. Let n ( x ) = p c ( x ) and n ( x ) = p c ( x ) . (6.19)By (2.2) n ( x ) , n ( x ) ≥ . (6.20)Also, using (2.1) and (2.4) we obtain that there exists a number n > k n ( x ) k C ( Ψ ) , k n ( x ) k C ( Ψ ) ≤ n . (6.21)Denote Λ (1 , n ) the class of functions n ( x ) such that the following two conditions holdfor every function n ( x ) ∈ Λ (1 , n ) :1. The function c ( x ) = n ( x ) satisfies conditions (2.1)-(2.4) as well as Condition ofsection 2.2. k n ( x ) k C ( Ψ ) ≤ n . By (6.19)-(6.21) both functions n ( x ) , n ( x ) ∈ Λ (1 , n ) . Therefore, (6.1), (6.18) andthe estimate (3.66) of theorem 3.4 of Chapter 3 of the book [33] imply (6.2). (cid:3)
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