Stability Estimates in the Inverse Transmission Scattering Problem

Transcription

1 Stability Estimates in the Inverse Transmission Scattering Problem Michele Di Cristo Dipartimento di Matematica Politecnico di Milano Abstract We consider the inverse transmission scattering problem with piecewise constant refractive index. Under mild a priori assumptions on the obstacle we establish logarithmic stability estimates. 1 Introduction In this paper we consider the scattering of acoustic time-harmonic waves in an inhomogeneous medium. More precisely we shall consider a penetrable obstacle D and we want to recover information on its location from a knowledge of Cauchy data on the boundary of a region Ω containing the obstacle D. Given a spherical incident wave u i (, x 0 ) = Φ(, x 0 ), where the point source x 0 is located on the boundary of a ball B of radius R, B such that Ω B, and Φ denotes the fundamental solution to the Helmholtz equation Φ(x, x 0 ) = 1 e ik x x0 4π x x 0, x R, x x 0, we denote by G(x, x 0 ) = u i (x, x 0 ) + u s (x, x 0 ) the Green s function of the equation (1.1a) div (γ(x) G(x, x 0 )) + k 2 n(x)g(x, x 0 ) = δ(x x 0 ), in R, where the scattered field u s satisfies the Sommerfeld radiation condition ( ) u s (1.1b) lim x x r (x) ikus (x) = 0. Here k > 0 is the wave number and r = x. We shall study equation (1.1a) with piecewise constant coefficients, in particular we shall consider γ and n to be of the following form γ(x) = 1 + (a 1)χ D (x) n(x) = 1 + (b 1)χ D (x) a λ > 0, b λ > 0, (a 1) 2 + (b 1) 2 δ 2 > 0, 1

2 where λ and δ are given constants. We refer to [Co-Kr, Is06] for basic information on scattering problem of this type. The unique determination of D from a knowledge of the far field data has been established by Isakov [Is90]. The purpose of the present paper is to establish a stability result. Under reasonable mild assumptions on the regularity of D we show that there is a continuous dependance of D on the Cauchy data on Ω with a modulus of continuity of logarithmic type. This rate of continuity appears optimal in view of the recent paper [DC-Ro] indicating the strong ill-posedness of the inverse problem. The main ideas employed to obtain stability rely on the study of the behavior of G(x, x 0 ) when x and x 0 get close and the use of unique continuation. These ideas go bach to [Is88] where a uniqueness result for the inverse inclusion problem is proved and it has also been used in inverse scattering theory in [Is90]. In order to apply these ideas to stability some further properties on singular solutions and quantitative estimates of unique continuation are needed. We refer to [Al-DC] where similar ideas are developed for studying the stability of the inverse inclusion problem. The stability issue in inverse scattering theory has been considered by Isakov [Is92, Is9] for the determination of a sound-soft obstacle. Hähner and Hohage [Ha-Ho] considered equation (1.1a) with a = 1 and n(x) smooth. They showed that n depends on G(x, x 0 ), x, x 0 B, with a logarithmic rate of continuity. They considered both far field data and near field data. They improve and simplify a previous result of Stefanov [St]. We finally mention a result obtained by Potthast [Po] for impenetrable obstacles which is also based on the use of singular solutions. The plan of the paper is the following. In Section 2 we give the a priori assumptions we need and we state the stability theorem. In Section the proof of the stability theorem is given based on some auxiliary results whose proofs are collected in Section 4 and Section 5. In particular, in Section 4 we establish some results on singular solutions of equation (1.1a) and in Section 5 we study quantitative estimates of unique continuation. 2 The Main Result In this section we state the stability theorem. Before doing this we shall give some definitions we need and introduce the a priori assumptions on the regularity of the obstacle. For any x = (x 1, x 2, x ) R and any r > 0 we denote by B r (x) the open ball in R of radius r centered in the point x, B r (0) = B r and for x = (x 1, x 2 ) R 2 we denote by B r(x ) the open ball in R 2 of radius r centered in the point x. In places, we shall denote a point x R by x = (x, x ) where x R 2, x R. Definition 2.1. Let Ω be a bounded domain in R. Given α, 0 < α 1, we shall say that a portion S of Ω is of class C 1,α with constants r 0, L > 0 if for any P S, there exists a rigid transformation of coordinates under which we have P = 0 and Ω B = {x B : x > ϕ(x )}, where ϕ is a C 1,α function on B r 0 R 2 satisfying ϕ(0) = ϕ(0) = 0 and ϕ C 1,α (B r ) Lr

3 Definition 2.2. We shall say that a portion S of Ω is of Lipschitz class with constants r 0, L > 0 if for any P S, there exists a rigid transformation of coordinates under which we have P = 0 and Ω B = {x B : x > ϕ(x )}, where ϕ is a Lipschitz continuous function on B r 0 R 2 satisfying ϕ(0) = 0 and ϕ C 0,1 (B r ) Lr 0. 0 Remark 2.1. We use the convention to scale all norms in such a way that they are dimensionally equivalent to their argument. For instance, for any ψ C 1,α (B r 0 ) we set ψ C 1,α (B r 0 ) = ψ L (B r 0 ) + r 0 ψ L (B r 0 ) + r 1+α 0 ψ α,b. Assumptions on the obstacle D For given numbers r 0, L > 0, 0 < α < 1, we shall assume there exists a bounded domain Ω such that the obstacle D satisfies the following conditions: (2.2a) (2.2b) (2.2c) D Ω; Ω D is connected; D is of class C 1,α with constants r 0, L. In the sequel we shall refer to numbers r 0, L, α, R, a, b and k as the a priori data. The inverse problem we are concerned with is the determination of the obstacle D from the knowledge of the Cauchy data of the singular solutions G(, x 0 ) on Ω for all points source x 0 located on B. For two possible obstacles D 1, D 2 satisfying (2.2) we shall denote by G i, i = 1, 2, the corresponding solutions to (1.1a) satisfying the Sommerfeld radiation condition (1.1b). Theorem 2.2. Let D 1, D 2 be two obstacles satisfying (2.2). If, given ε > 0, we have ( G 1 (, x) (2.) sup G 2(, x) x B ν ν + L 2 ( Ω) ) G 1 (, x) G 2 (, x) L2 ( Ω) ε, then d H ( D 1, D 2 ) ω(ε), where ω is an increasing function on [0, + ), which satisfies ω(t) C log t η, for every 0 < t < 1 and C, η, C > 0, 0 < η 1, are constants only depending on the a priori data. Remark 2.. We stress the fact that we don t need any assumption on k.

4 Proof of the Stability Theorem We denote by G the connected component of Ω \ (D 1 D 2 ) such that Ω G and Ω D = Ω \ G. Theorem 2.2 evaluates how close the two inclusions are in term of the Hausdorff distance d H. We recall a definition of this metric. { } d H (D 1, D 2 ) = max sup dist(x, D 2 ), sup dist(x, D 1 ). x D 1 x D 2 In order to deal with the Hausdorff distance we introduce a simplified variation of it which we call modified distance. Definition.1. We shall call modified distance between D 1 and D 2 the number { } (.4) d µ (D 1, D 2 ) = max sup dist(x, D 2 ), sup dist(x, D 1 ). x D 1 Ω D x D 2 Ω D We wish to remark here that such modified distance does not satisfy the axioms of a metric and in general does not dominate the Hausdorff distance (see [Al-Be-Ro-Ve, ] for related arguments). Proposition.1. Let D 1, D 2 be two obstacles satisfying (2.2). Then (.5) d H ( D 1, D 2 ) cd µ (D 1, D 2 ), where c depends only on the a priori assumptions. Proof. See [Al-DC, Proposition.1] With no loss of generality, we can assume that there exists a point O of D 1 Ω D, where the maximum in the Definition.1 is attained, that is (.6) d µ = d µ (D 1, D 2 ) = dist(o, D 2 ). We remark that G is solution to div (γ(x) G(x, y)) + k 2 n(x)g(x, y) = δ(x, y). We shall denote by G 1 and G 2 Green s functions when D = D 1 and D 2 respectively and γ i, n i, i = 1, 2, the corresponding coefficients. Integrating by parts we have (.7) = = { } (a 1) G 1 (, y) G 2 (, w) G 1 (, y) G 2 (, w) D 1 D 2 { } +k 2 (b 1) G 2 (, w)g 1 (, y) G 1 (, y)g 2 (, w) D 1 D 2 Ω Ω + ( G1 (, y) ν G 2 (, w) G 1 (, y) G 2(, w) ν G 1 (, y) (G 2 (, w) G 1 (, w)) ν ( G1 (, w) G 1 (, y) G ) 2(, w) ν ν Ω 4 ) y, w CB.

5 (.8) Let us define for y, w CB S 1 (y, w) = (a 1) G 1 (, y) G 2 (, w) + k 2 (b 1) D 1 G 1 (, y)g 2 (, w), D 1 S 2 (y, w) = (a 1) G 1 (, y) G 2 (, w) + k 2 (b 1) D 2 G 1 (, y)g 2 (, w), D 2 f(y, w) = S 1 (y, w) S 2 (y, w). Thus (.7) can be rewritten as f(y, w) = Ω + G 1 (, y) (G 2 (, w) G 1 (, w)) ν ( G1 (, w) Ω G 1 (, y) ν G ) 2(, w) ν y, w CB. Let us fix P D. We can assume P 0. We denote by ν(p ) the outer unit normal vector to Ω D in P and we rotate the coordinate system in such a way that ν(p ) = (0, 0, 1). Let us denote by χ + (x) the characteristic function of the half-space and by G + the Green s function of div((1 + (a 1)χ + ) ) + k 2 (1 + (b 1)χ + ). Proposition.2. Let D Ω be a bounded open set whose boundary is of class C 1,α with constants r 0, L. Then there exist constants c 1, c 2 depending on a, α, k and L such that (.9) (.10) for every x, y R. x G(x, y) c 1 x y 2, x G + (x, y) c 2 x y 2 Proof. (.9) and (.10) can be obtained following [Al-DC, Proposition.1]. In [Al-DC] the key point is the piecewise regularity of the transmission problem. For a proof of that we refer to [DB-El-Fr] and [Li-Vo]. We shall state now two propositions that describe the behavior of f(y) and S 1 (y) when we move the singularity y toward the boundary of the inclusion. We postpone their proofs in the last Section 5. Proposition.. Let D 1, D 2 two obstacles verifying (2.2) and let y = hν(o), with O defined in (.6). If, given ε > 0 we have ( G 1 (, x) sup G 2(, x) x B ν ν + L2 ( Ω) ) G 1 (, x) G 2 (, x) L2 ( Ω) ε, then for every h, 0 < h < cr 0, with c (0, 1) depending on L, f(y, y) c εbhf h A, where 0 < A < 1 and c, B, F > 0 are constants that depend only on the a priori data. 5

6 Proposition.4. Let D 1, D 2 two obstacles verifying (2.2) and let y = hν(o), with O defined in (.6). Then for every h, 0 < h < min{r 2, d µ } (.11) S 1 (y, y) c 1 h 2 c 2 (d µ h) 2 + c where c 1, c 2, c and r 2 are positive constants only depending on the a priori data. Proof of Theorem 2.2. Let O D 1 as defined (.6), that is d µ (D 1, D 2 ) = dist(o, D 2 ) = d µ. Then, for y = hν(o), with 0 < h < h 1, where h 1 = min{d µ, cr 0, r 2 /2}, using (.9), we have 1 (.12) S 2 (y, y) c (d µ h) (d µ h) dx = c 1 (d µ h) 2 D 2. Using Proposition., we have D 2 1 S 1 (y, y) S 2 (y, y) S 1 (y, y) S 2 (y, y) = f(y, y) c εbhf h A. On the other hand, by Proposition.4 and (.12), there exists h 0 > 0, only depending on the a priori data, such that for h, 0 < h < h 0 Thus we have S 1 (y, y) S 2 (y, y) c 1 h 2 c 4 (d µ h) 2. c 1 h 2 c 4 (d µ h) 2 εbhf h A. Let h = h(ε) where h(ε) = min{ ln ε 1 2F, d µ }, for 0 < ε ε 1, with ε 1 (0, 1) such that exp( B ln ε 1 1/2 ) = 1/2. If d µ ln ε 1 2F the theorem follows using Proposition.1. In the other case we have c 4 (d µ h) 2 c h 2 εbhf h A where à = 2 A, à > 0. Since c 5h 2( 1 ε Bh hã) F, for any ε, 0 < ε < ε 1, ε Bh(ε)F h(ε)ã ε B ln ε 1/2 exp ( B ln ε 1/2), (d µ h(ε)) 2 c 6 h(ε) 2, that is, solving for d µ, and recalling that, in this case, h(ε) = ln ε 1 2F where δ = 1/(2F ). When ε ε 1, then and, in particular when ε 1 ε < 1 d µ c 7 ln ε δ 2 d µ diamω d µ diamω ln ε 1 2F ln ε 1 1 2F Finally, using Proposition.1, the theorem follows.. 6

7 4 Remarks on Singular Solutions Proposition 4.1. Let D R be an open set with C 1,α boundary with constants r 0, L, let P be a point in D and let us denote with ν(p ) the outer normal vector to D in P. There exist positive constants c, c 4 depending on a, k, α and L such that (4.1) (4.14) G(x, y) G + (x, y) c α x y 1+α, x G(x, y) x G + (x, y) c 4 x y 2+α2, α2 for every x D B r (P ) and y = hν(p ), with 0 < r < (min{ 1 2 (8L) 1/α, 1 2 })r 0 = r 0, 0 < h < (min{ 1 2 (8L) 1/α, 1 2 }) 2. Proof. Let us fix r 1 = min{ 1 2 (8L) 1/α r 0, 2 }. In the ball B r 0 (P ) the boundary of D can be represented as the graph of a C 1,α function ϕ. Let us introduce now the following change of variable that transform in B (P ) D in the x -axis. For every r > 0, let Q r (P ) be the cube centered at P, with sides of length 2r and parallel to the coordinates axes. We have that the ball B r (P ) is inscribed into Q r (P ). We define Ψ : Q 2r1 (P ) Q 2r1 (P ) ( ) ( x ξ = x x n ξ n = x n ϕ(x )θ ( x ) ( r 1 θ xn where θ C (R) be such that 0 θ 1, θ(t) = 1, for t < 1, θ(t) = 0, for t > 2 and dθ dt 2. Since the C1,α regularity of ϕ, it is possible to verify that the following inequalities hold: (4.15a) (4.15b) (4.15c) r1 ) c 1 x 1 x 2 Ψ(x 1 ) Ψ(x 2 ) c x 1 x 2, Ψ(x) x c α x 1+α x R, DΨ(x) I c α x α x R where c 1 depends on L and α only. Ψ is a C 1,α diffeomorphism from R into itself. Let us define the cylinder C r1 as C r1 = {x R : x < r 1, x n < r 1 }. For x, y C r1, we shall denote (4.16) G(x, y) = G(Ψ 1 (x), Ψ 1 (y)). G(x, y) is solution of (4.17) div((1 + (a 1)χ + )B G(x, y)) + k 2 ζ(1 (b 1)χ + (x))b G(x, y) = δ(x y), where B = JJT ξ det J, with J = x (Ψ 1 (ξ)), is of class C α, B(0) = I and ζ = det J. Since G + is solution to (4.18) div((1 + (a 1)χ + )G + (x, y))+ k 2 (1 (b 1)χ + (x))g + (x, y) = δ(x, y), ), 7

8 subtracting (4.18) to (4.17) we obtain that R(x, y) = G(x, y) G + (x, y) is solution to (4.19) div((1 + (a 1)χ + ) R(x, y)) + k 2 (1 + (b 1)χ + ) R(x, y) = div((1 + (a 1)χ + )[B(x) I] G(x, y)) + k 2 (1 ζ)(1 + (b 1)χ + ) G(x, y). Let L, depending on the a priori data, be such that Ω B L(0), then using the fundamental solution G + we get R(x, y) = B L(0) [ + [B(z) I] B L(0) + k 2 (1 ζ) (1 + (a 1)χ + )[B(z) I] x G(z, y) x G + (z, x)dz R(x, z) G + ν (z, y) + G +(z, y) R ] (x, z) dσ(z) ν B L(0) [ k 2 (1 ζ) (1+(a 1)χ + ) B L(0) (1 + (b 1)χ + ) G(z, x)g + (z, y)dz+ R(x, z) G + ν (z, y) + G +(z, y) R ] (x, z) dσ(z) ν Integrals over B L(0) are bounded by a constant. Let us split B L(0) = (B L(0) C r1 ) (B L(0) C r1 ). For x, y r 1 /2, in B L(0) C r1 we are away from the singularity thus the integrals over B L(0) C r1 are bounded. Let us evaluate integrals over B L(0) C r1. We have (1 + (a 1)χ + )[B(z) I] x G(z, y) x G + (z, x)dz B L(0) C r1 c z α z y 2 z x 2 dz = I B L(0) C r1 where c depends on L, α, a and n. We can split I = I 1 + I 2 where I 1 = z α x z 2 y z 2 dz, { z <4h} C r1 I 2 = z α x z 2 y z 2 dz. { z >4h} C r1 8

9 Now I 1 w <4 = h α 1 h α w α h 2 x h w 2 h 2 y h w 2 h dw w <4 w α x h w 2 y h w 2 dw where h = x y and h α 1 F (ξ, η), F (ξ, η) = 4 α w <4 ξ w 2 η w 2 dw and ξ = x/h and η = y/h. From standard bounds (see, for instance, [Mi, Ch. 2, 11]), it is not difficult to see that for all ξ, η R, ξ η = 1. Thus F (ξ, η) const. <, I 1 c x y α 1. Let us consider now I 2. Since y = y n x y = h, we can deduce z 4 y z and z 2 x z and thus obtain that I 2 c z α+1 n+1 n dz ch α 1. Then we conclude z >4h (4.20) R(x, y) c x y 1+α, for every x, y r 1 /2, where c depends on L, α, k and a only. We observe that if x Ψ 1 (B + r 1/2 (0)) and y = e y, with y ( r 1 /2, 0) then (4.21) c 1 x Ψ(x) Ψ(x) y c x y. From (4.20) and (4.21) we can conclude (4.22) R(x, y) c x y 1+α. Now, since G(x, y) G + (x, y) =G(x, y) G + (x, y) + G + (Ψ(x), Ψ(y)) G + (Ψ(x), Ψ(y)) = R(Ψ(x), Ψ(y)) + G + (Ψ(x), y) G + (x, y), using Theorem 4.1 of [Li-Vo], the properties of Ψ and (4.22) we obtain G(x, y) G + (x, y) c α x y α 1 + c α G + (, y) L (Q r1 ) x Ψ(x) c α x y α 1 + c α x y 1+α h 2 c α x y α 1, 9

10 where c depends on k, α and L only. We estimate now the first derivative of R. To estimate the first derivative of R let us consider a cube Q B + r 1/4 (x) of side cr 1/4, with 0 < c < 1, such that x Q. The following interpolation inequality holds: R(, y) L (Q) c R(, y) 1 δ L (Q) R(, y) δ α,q, where δ = 1 1+α, c depends on L only and R α,q = sup R(x, y) R(x, y) x,x Q,x x x x α. Since, from the piecewise Hölder continuity of G and of G +, we have that R(x, y) α,q G(x, y) α,q + G + (x, y) α,q ch α 2, where c depends on L only, thus we conclude where η = α2 1+α. Thus c x R(x, y) r η h (α 1)(1 δ) h ( α 2)δ = c 0 r η h 2+η, 0 c (4.2) x R(x, y) r η x y η 2, 0 where η = α2 1+α and c depends on L only. Concerning G + we have x G + (Ψ(x), y) x G + (x, y) = DΨ(x) T G + (, y) Ψ(x) x G + (x, y) (DΨ(x) T I) G + (, y) Ψ(x) + G + (, y) Ψ(x) x G + (x, y) c G + (, y) L (Q r1 ) x Ψ(x) + G + (, y) α,q Ψ(x) x α r α 0 c α h 1+α h 2 + c r α2 0 h 2+α2, c r α2 0 h α 2 h (1+α)α where c depends on k, α and L only. Let us denote by G 0 + the Green s function of the operator div((1 + (a 1)χ + ) ). Proposition 4.2. Let G + and G 0 + as above, then there exist positive constants c 5, c 6 depending on the a priori data such that for every x, y R we have (4.24) G + (x, y) G 0 +(x, y) c 5 x y (4.25) x G + (x, y) x G 0 +(x, y) c 6 x y 1 10

11 Proof. Defining R(x, y) = G + (x, y) G 0 +(x, y), we have that (4.26) div((1 + (b 1)χ + ) R(x, y)) = k 2 (1 + ((b 1)χ + )G + (x, y). Thus R(x, y) = k 2 Ω (1 + (b 1)χ + )G + (z, y)g 0 +(x, z)dz. Hence for [Li-St-We] we have R(x, y) C Let decompose Ω = B x y For z B x y Thus B x y (x) (x) we have that Ω (x) B x y (y) G. x z 1 y z 1 dz. y z y z y z y x x y x y = 2 x y. x z 1 y z 1 dz 2 x y x y 1 0 ρ dρ c x y 2. Similarly it can be evaluated the integral over B x y (y). Let us consider now the integral over G. For z G we have that z y x z, then we obtain G c x z 1 y z 1 dz c x z 1 x z 1 dz G Ω B x y (x) x z 1 x z 1 dz 2 L c ρdρ c 1 x y 2 + c 2. x y Let us prove now (4.25). We use the interpolation inequality As in Proposition 4.1, since R( ) L (Q) R( ) 1 δ L (Q) R(, y) δ α,q. R(, y) α,q h α 2, we obtain R(x, y) ch 2+η ch 1. 11

12 5 Proof of Proposition. and.4 Proof of Proposition.. Let us consider f(y, w), where w is a fixed point in CB. Since f, as a function of y, is a radiating solution of L y f = y f + k 2 f = 0 in CΩ D, then by [Co-Kr, Theorem 2.14], for y CB we have f(y, w) = n n=0 m= n a m n h (1) n (k y )Yn m (ŷ), where ŷ = y/ y, Yn m is a spherical harmonic of order n and h (1) n is a spherical Hankel function of the first kind of order n. Let us consider y such that R < R 1 < y < R 2. For an integer N, using Schwarz inequality and the asymptotic behavior of Hankel function (see [Co-Kr, (2.8) pg. 28]) we have [ N n n=0 m= n N n=0 c N (k y ) h (1) n (kr) h (1) n n n=0 m= n a m n h (1) n (k y )Yn m (ŷ) 2 N n n=0 m= n ] 2 a m n 2 h (1) n (kr) 2 Y m n (ŷ) 2, a m n 2 h (1) n (kr) 2 Y m n (ŷ) 2. for some constant c depending on R, R 1 and R 2. N +, we can conclude that Thus, taking the limit as f(y, w) 2 c f(, w) B 2, y B R2 B R1, where c depends on R, R 1 and R 2. Analogous considerations can be carried on fixing y and varying w. Thus, we can conclude that for all (y, w) [ BR2 B R1 ] 2 f(y, w) f B B cε. For y G h, where G h = {x G : dist(x, Ω D ) h}, S 1 (y, w) c x y 2 ch 2, D 1 where c = c(l, R). Similarly S 2 (y, w) ch 2. Then we conclude that (5.27) f(y, w) ch 2 in G h. At this stage we shall make use iteratively of the three spheres inequality (see [La, Ku]). Let u be a solution of Lu = 0 in G, let x G. There exist r 1, r, r 2, 0 < r 1 < r < r 2 < R and τ (0, 1) such that (5.28) u L (B r(x)) c u τ L (B r1 (x)) u 1 τ L (B r2 (x)), 12

13 where c and τ depend on R, r/r 2, r 1 /r 2 and L. Applying (5.28) to u( ) = f(, w), with x = x B 4R B R, r 1 = r 0 /2, r = r 0 /2 and r 2 = 2r 0 we obtain f L (B /2(x)) c f τ L (B /2(x)) f 1 τ L (B 2 (x)), For every y G h, we denote by γ a simple arc in G joining x to y. Let us define {x i }, i = 1,..., s as follows x 1 = x, x i+1 = γ(t i ), where t i = max{t : γ(t) x i = r 0 } if x i y > r 0, otherwise let i = s and stop the process. By construction, the balls B /2(x i ) are pairwise disjoint, x i+1 x i = r 0 for i = 1,..., s 1, x s y r 0. There exists β such that s β. An iterated application of the three spheres inequality (5.28) for f (see for instance [Al-Be-Ro-Ve, pg. 780], [Al-DB, Appendix E]) gives that for any r, 0 < r < r 0 (5.29) f L (B r/2 (y)) c f τ s L (B r/2 (x)) f 1 τ s L (G). We can estimate the right hand side of (5.29) by (5.27) and obtain for any r, 0 < r < r 0 (5.0) f L (B r/2 (y)) c(h 2 ) 1 τ s ε τ s ch A ε β, where β = τ β and A = 2(1 β). Let O D 1 as defined in (.6), that is d(o, D 2 ) = d µ (D 1, D 2 ). There exists a C 1,α neighborhood U of O in Ω D with constants r 0 and L. Thus there exists a non-tangential vector field ν, defined on U such that the truncated cone { } (5.1) C(O, ν(o), θ, r 0 ) = x R (x O) ν(o) : > cos θ, x O < r 0 x O satisfies C(O, ν(o), θ, r 0 ) G, where θ = arctan(1/l). Let us define { } λ 1 = min 1 + sin θ, r 0, θ 1 = arcsin sin θ ( sin θ G 1 = O + λ 1 ν, ρ 1 = λ 1 sin θ 1. We have that B ρ1 (G 1 ) C(O, ν(o), θ 1, r 0 ), B 4ρ1 (G 1 ) C(O, ν(o), θ, r 0 ). Let G = G 1, since ρ 1 r 0 /2, we can use (5.0) in the ball B ρ1 (G) and we can approach O D 1 by constructing a sequence of balls contained in the cone C(O, ν(o), θ 1, r 0 ). We define, for k 2 G k = O + λ k ν, λ k = χλ k 1, ρ k = χρ k 1, with χ = 1 sin θ sin θ 1. Hence ρ k = χ k 1 ρ 1, λ k = χ k 1 λ 1 and B ρk+1 (G k+1 ) B ρk (G k ) B ρ4k (G k ) C(O, ν, θ, r 0 ). 4 ), 1

14 Denoting d(k) = G k O ρ k = λ k ρ k, we have d(k) = χ k 1 d(1), with d(1) = λ 1 (1 sin θ). For any r, 0 < r d(1), let k(r) be the smallest integer such that d(k) r, that is log r d(1) log r d(1) log χ k(r) 1 log χ + 1. By an iterated application of the three spheres inequality over the chain of balls B ρ1 (G 1 ),..., B ρk(r) (G k(r) ), we have (5.2) f(, w) L (B ρk(r) (G k(r) )) ch A(1 τ k(r) 1) ε βτ k(r) 1 ch A ε βτ k(r) 1, for 0 < r < cr 0, where c, 0 < c < 1, depends on L. Let us consider now f(y, w) as a function of w. First we observe that L w f = 0 in CΩ D, for all y CΩ D. For y, w G h, y w, using (.9) S 1 (y, w) c x y 2 x w 2 dx ch 4. D 1 Similarly for S 2. Therefore f(y, w) ch 4 with y, w G h. For w B 4R B R and y G h, using (5.2), we have f(y, w) ch A ε βτ k(r) 1. Proceeding as before, let us fix y G such that dist(y, Ω D ) = h and w B 4R B R such that dist( w, B R ) = R/2. Taking r = R/2, r 1 = r, r 2 = 4r, w 1 = O + λ 1 ν and using iteratively the three spheres inequality, we have f(y, w) L (B R/2 (w 1) f(y, w) τ s s L (B R/2 ( w) f(y, w) 1 τ L (G), where τ and s are as above. Therefore f(y, w) L (B R/2 (w 1) c(h 4 ) 1 τ s h Aτ s (ε βτ k(h) 1 ) τ s c(h 4 ) 1 γ h Aτ s (ε βτ k(h) 1 ) γ ch A (ε βτ k(h) 1 ) γ, where γ = τ β, with β as above, so 0 < γ < 1 and A = Aτ s 4 + γ. Once again, let us apply the three spheres inequality over a chain of balls contained in a cone with vertex in O, choosing y = w = hν(o) we obtain (5.) f(y, y) ch A (ε βτ k(h) 1 ) γτ k(h) 1. We observe that, for 0 < h < cr 0, where 0 < c < 1 depends on L, k(h) c log h = c log h, so we can write τ k(h) = e c log h log τ = h c log τ = h c log τ = h F, 14

15 with F = c log τ. Therefore f(y, y) h A ε Bτ k(h) = e A log h e Bτ k(h) log ε = e A log h+b h F log ε Then in (5.) we obtain f(y, y) e A log h+b h F log ε h F = εb. h A Proof of Proposition.4. Let us define r 2 = min{r 0, r 2 }, where r 0 is the one of Proposition 4.1 and r 2 will be fixed later. For every x, y such that x y < r, with 0 < r < r 2, the following asymptotic formula holds (cf. Proposition 4.1) We now distinguish two situations: 1) x B r (D 1 D 2 ); 2) x B r (D 1 D 2 ). G 1 (x, y) G + (x, y) c x y 1+α. If case 1) occurs then the asymptotic formula (4.14) holds also for G 2 since the hypothesis of Proposition 4.1 are met. From [Al, Lemma.1] there exists r 2, depending on the a priori data, such that (5.4) G 1 (x, y) G 2 (x, y) c x y 2. Let us consider case 2). In B r (D 1 D 2 ) we consider a smaller ball B ρ (0) with radius ρ where 0 < ρ < min{d µ, r 2 }. Since the definition of d µ we have B ρ D 2 =. If x and y are in B ρ and denoting by L = + k 2 we have L (G 2 (x, y) Φ(x, y)) = 0 in B ρ where Φ is the fundamental solution of the Helmholtz equation, with the boundary condition [G 2 (x, y) Φ(x, y)] Bρ cρ 1. Thus by maximum principle and by interior gradient bound G 2 (x, y) Φ(x, y) c 1 ρ 1 x, y B ρ G 2 (x, y) Φ(x, y) c 2 ρ 2 x B ρ/2, y B ρ. Thus using Lemma.1 of [Al], in B ρ/2 (O) we obtain the formula formula (5.5) G 1 (x, y) G 2 (x, y) c x y 2 c 4 ρ 2. 15

16 Let us consider h < r 2 /2 and 0 < r < r 2. Then we have G 1 (x, y) G 2 (x, y)dx D 1 = G 1 (x, y) G 2 (x, y) + G 1 (x, y) G 2 (x, y) D 1 B r(o) D 1 B r(o) G 1 (x, y) G 2 (x, y) D 1 B r(o) D 1 B r(o) The first integral can be estimated as follows G 1 (x, y) G 2 (x, y)dx = + D 1 B r(o) (D 1 D 2) B r(o) (D 1 D 2) B r(o) G 1 (x, y) G 2 (x, y)dx [(D 1 D 2) B ρ(o)] [(D 1 D 2) B ρ] [(D 1 D 2) B r(o)] B ρ G 1 (x, y) G 2 (x, y)dx G 1 (x, y) G 2 (x, y) G 1 (x, y) G 2 (x, y)dx G 1 (x, y) G 2 (x, y)dx In conclusion, choosing ρ = h and using (5.4), (5.5) and (.9) we obtain S 1 (y) c 1 x y 2 dx c 2 [(D 1 D 2) B ρ(o)] [(D 1 D 2) B ρ] [(D 1 D 2) B r(o)] B ρ x y 1 x y 1 dx c D 1 B r(o) x y 1 x y 1 dx c 4 h 2 c 5 d 2 µ c 6. References [Al] G. Alessandrini, Singular solutions of elliptic equations and the determination of conductivity by boundary measurement, J. Differential Equations 84 (1990), pp [Al-Be-Ro-Ve] G. Alessandrini, E. Beretta, E. Rosset, S. Vessella, Optimal stability for inverse elliptic boundary value problems with unknown boundaries, Ann. Scuola Norm. Sup. Pisa 4 (2000), pp

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