The inverse transmission eigenvalue problem

Transcription

1 The inverse transmission eigenvalue problem Drossos Gintides Department of Mathematics National Technical University of Athens Greece

2 Transmission Eigenvalue Problem Find (w, v) such that w + k 2 n(x)w = 0 in D It is a nonstandard eigenvalue problem v + k 2 v = 0 in D w = v on D w ν = v ν on D If n = 1 the interior transmission problem is degenerate If I(n) > 0 in D, there are no real transmission eigenvalues.

3 Outline Direct Transmission Eigenvalue Problem - Spherically Symmetric Medium Inverse Spectral Problem Results for the ISP for Spherically Symmetric n(x) Numerical Solution of the Direct Problem Numerical Solution of the Inverse Problem

4 Literature, Direct Interior Transmission Problem Kirsch (1986) and Colton-Monk (1988) Discreteness of transmission eigenvalues: Colton-Kirsch-Päivärinta (1989), Rynne-Sleeman (1991), Cakoni-Haddar (2007), Colton-Päivärinta-Sylvester (2007), Kirsch (2009), Cakoni-Haddar (2009), Hickmann (2011),etc. The first proof of existence of at least one transmission eigenvalues for large enough contrast is due to Päivärinta-Sylvester (2009). Existence of an infinite set of transmission eigenvalues is proven by Cakoni-Gintides-Haddar (2010) under only assumption that either n 1 > 0 or 1 n > 0. Other scattering problems by Kirsch (2009), Cakoni-Haddar (2010) Cakoni-Kirsch (2010), Bellis-Cakoni-Guzina (2011), Cossonniere (Ph.D. thesis) etc. Hitrik-Krupchyk-Ola-Päivärinta (2010), for general differential operators with constant coefficients.

5 Literature, cont. Finch thermo-acoustic imaging. Cakoni-Colton-Haddar (2010) and then Cossonniere-Haddar (2011), case when n = 1 in D 0 D and n 1 > α > 0 in D \ D 0. Sylvester (2012), the set of transmission eigenvalues is at most discrete if n 1 is positive (or negative) only in a neighborhood of D but otherwise could changes sign inside D. A similar result Bonnet Ben Dhia - Chesnel - Haddar (2011) and Lakshtanov-Vainberg (2011). Serov and Sylvester (2012), degenerate and singular cases. Numerical analysis of the Interior Transmission Problem, Colton, Monk, Sun (2010), Sun (2011), Ji and Turner (2012), Sun (2012). Inverse Problems 2013, Special Issue on Transmission eigenvalues

6 Literature, Inverse Interior Transmission Problem Inverse Interior Transmission Problem, McLaughlin, Polyakov, and Sacks, (1994), McLaughlin and Polyakov (1994). Cakoni, Colton and G. (2010), Aktosun, G., V. Papanicolaou (2011), Aktosun and V. Papanicolaou (2013), Leung and Colton (2012, 2013). Computational methods for the Inverse Transmission Problem, Cakoni, Colton, Monk and Sun (2010), Sun (2011),...

7 Spherically Symmetric Medium We consider the interior eigenvalue problem for a ball of radius b with index of refraction n(r) being a function of r := x,n C 2 [0, ) w + k 2 n(r)w = 0 v + k 2 v = 0 w r w = v = v r in B in B on B on B where B := { x R 3 : x < b }.

8 Spherically Symmetric Medium Separation of variables: v(r, θ) = a l j l (kr)p l (cos θ) and w(r, θ) = a l Y l (kr)p l (cos θ) j l is a spherical Bessel function and Y l is the solution of Y l + 2 ( r Y l + k 2 n(r) l(l + 1) r 2 ) Y l = 0 such that lim (Y l (r) j l (kr)) = 0. r 0 Values of k such that: Y l (b) l (k) := det Y l (b) j l (kb) kj l (kb) = 0. are the transmission eigenvalues. l (k) are entire function of k of finite type and bounded for k > 0.

9 Inverse Spectral Problems Determine properties of a system, from a set of spectral data (eigenvalues - eigenfunctions). Important problem in scattering theory: Determine the refractive index from a set of eigenvalues. Inverse spectral problems are not well posed. First inverse spectral problem (ISP): Determine n(x) from the zeros of 0 (λ), i.e. the λ n s (including multiplicities)

10 Equivalent Eigenvalue Problem for Spherically Symmetric Eigenfunctions The problem is equivalent to the following eigenvalue problem: v + λn(x)v = 0, 0 < x < b v(0) = 0, 0 (λ) := sin( λb) λ v (b; λ) cos( λb)v(b; λ) = 0 where λ = k 2 The zeros λ n of the entire function 0 (λ) are transmission eigenvalues corresponding to spherically symmetric eigenfunctions. If λ R then 0 (λ) R. The order of 0 (λ) is at most 1/2 0 (0) = 0 From Hadamard Factorization Theorem ) 0 (λ) = γλ (1 d λλn, γ R, d 1 n=1

11 Liouville transform Auxilliary initial value problem Let v(x) = v(x; λ) be the unique solution of the initial-value problem Liouville transformation x ξ := n(η)dη, 0 v (x) + λn(x)v(x) = 0, v(0) = 0, v (0) = 1. transforms the initial-value problem to: where q(ξ) = 1 n (x) 4 5 n (x) 2 n(x) 2 16 n(x) 3 w(ξ) = n(x) 1/4 v(x), w (ξ) q(ξ)w(ξ) + λw(ξ) = 0, w(0) = 0, w (0) = 1 n(0) 1/4,

12 Asymptotic estimates There exists a constant A > 0 such that [ v(x; λ) 1 λ x ] [n(0)n(x)] 1/4 λ sin n(η)dη 0 and A [ exp I{ x ] λ} n(η)dη λ 0 [ ] n(x) 1/4 [ λ x ] v (x; λ) cos n(η)dη n(0) 0 [ A exp I{ x ] λ} n(η)dη for all x [0, b] and all λ C 0

13 Inverse Spectral Problem Lemma 1. (a) Let f (λ) be an entire function such that f (λ) = exp(c I{ λ} ) λ O(1), λ where c > 0 ( π 2 n 2 If f = 0, for all n N := {1, 2,...} then there is a constant sin(c λ) C 1 such that f (λ) = C 1 λ = C 1 c ( ) n=1 1 c2 λ π 2 n 2 (b) Let g(λ) be an entire function such that g(λ) = exp(c I{ λ} )O(1), λ c 2 ) ( π 2 (2n 1) 2 If g = 0, for all n N then there is a constant C 2 such ( that g(λ) = C 2 cos c ) [ ] λ = C 2 n=1 1 4c2 λ π 2 (2n 1) 2 4c 2 )

14 Inverse Spectral Problem Lemma 2. (a) Assume that a := b 0 n(x)dx < b. If v(x; λ) satisfies the initial value problem, then v(b; λ) = γm(λ) and v (b; λ) = γn(λ), where M(λ) and N(λ) are entire functions uniquely determined from 0 (λ). (b) If a = b, then v(b; λ) = sin(b λ) [n(0)n(b)] 1/4 + γm(λ) and λ ] 1/4 ( v (b; λ) = cos b ) λ + γn(λ) where M(λ), N(λ) are as in case (a). [ n(b) n(0)

15 Inverse Spectral Problem Proof From the definition of 0 (λ) for λ = π 2 n 2 /b 2 for n N v (b; π2 n 2 ) ( π = ( 1) n 1 2 n 2 ) 0. b 2 Similarly, for λ = π 2 (2n 1) 2 /4b 2, for n N, and ( v b; π2 (2n 1) 2 ) 4b 2 = ( 1) and application of Lemma 1. n 1 π(2n 1) 2b b 2 ( π 2 (2n 1) 2 ) 0 4b 2

16 Inverse Spectral Problem Determine n(x) from the zeros of 0 (λ), i.e. the λ n s (including multiplicities) Theorem (Aktosun - G. - V. Papanicolaou, (Inverse Problems, 2011)) (a) If a < b, where a := b 0 n(r)dr, then the eigenvalues which are zeros of 0 (λ) determine n(r) uniquely. (b) If a = b, then the knowledge of all eigenvalues which are zeros of 0 (λ) together with the constant γ determine n(r) uniquely.

17 Inverse Spectral Problem Proof From the eigenvalues one can determine the zeros of both v(b; k 2 ) and of v (b; k 2 ). This means that the spectra of the two (standard) Sturm-Liouville problems: v + k 2 n(r)v = 0, 0 < r < b, v(0) = v(b) = 0 and v + k 2 n(r)v = 0, 0 < r < b, v(0) = v (b) = 0, are determined n(r) is uniquely determined.

18 Inverse Spectral Problem Determine n(x) from the zeros of 0 (λ), i.e. the λ n s (including multiplicities) If a > b then we the proof breaks down.

19 Inverse Spectral Problem Assume f 0 (λ) and g 0 (λ) are two entire functions, satisfying appropriate asymptotic estimates such that then so are sin( λb) g 0 (λ) cos( λb)f 0 (λ) = 0 (λ). λ f (λ) = f 0 (λ) + sin( λb) λ φ(λ) and g(λ) = g 0 (λ) + cos( λb)φ(λ), where φ(λ) is any entire function satisfying [ φ(λ) = exp (a b) I{ ] λ} O(1), λ. Therefore the above problem does not have a unique solution. Equivalently v(b; λ) and v (b; λ) cannot be determined uniquely, which suggests that our inverse problem may not have a unique solution.

20 Inverse Spectral Problem Determine n(x) from all transmission eigenvalues Uniqueness holds under a restriction. Theorem (Cakoni - G. - Colton) If n(0) is given then n(r) is uniquely determined from the knowledge of all transmission eigenvalues. l (k) is an entire function of k Hadamard factorization theorem From Müntz s theorem Partial results are given by McLaughlin-Polyakov (1994), McLaughlin-Polyakov-Sacks (1994)

21 Inverse Spectral Problem For general domains and refractive indices the determination of n(x) from all transmission eigenvalues is open.

22 Transmission Eigenvalues for general shapes and refractive indices n L (D), n = sup n(x) and 0 < n = inf n(x) and n > 1 (similar x D x D analysis if n < 1). Let u := w v H0 2(D). The transmission eigenvalue problem can be written for u as an eigenvalue problem for the fourth order equation: ( + k 2 1 n) n 1 ( + k 2 )u = 0 i.e. in the variational form 1 n 1 ( u + k 2 u)( ϕ + k 2 nϕ) dx = 0 D for all ϕ H 2 0 (D) Definition: k C is a transmission eigenvalue if there exists a nontrivial solution v L 2 (D), w L 2 (D), w v H0 2 (D) of the homogeneous interior transmission problem.

23 Operator representation - Discreteness Letting k 2 := τ, the transmission eigenvalue problem can be written as a quadratic pencil operator problem u τk 1 u + τ 2 K 2 u = 0, u H 2 0 (D) with selfadjoint compact operators K 1 = T 1/2 T 1 T 1/2 and K 2 = T 1/2 T 2 T 1/2 where (Tu, ϕ) H2 (D) = 1 u ϕ dx n 1 coercive D (T 1 u, ϕ) H2 (D) = (T 2 u, ϕ) H 2 (D) = D D 1 (n u ϕ + u ϕ) dx n 1 n n 1 u ϕ dx non-negative.

24 Operator representation - Linearization The transmission eigenvalue problem can be transformed to the eigenvalue problem ( ) u (K ξi)u = 0, U = τk 1/2, ξ := 1 2 u τ for the non-selfadjoint compact operator K: H0 2(D) H2 0 (D) H2 0 (D) H2 0 (D) given by ( K K := 1 K 1/2 2 K 1/2 2 0 ). However from here one can see that the transmission eigenvalues form a discrete set with + as the only possible accumulation point.

25 Existence of Transmission eigenvalues About existence of transmission eiganvalues we have the following theorem Theorem (Cakoni - G. - Haddar, 2010) Let n L (D) satisfy either one of the following assumptions: 1) 1 + α n n(x) n <, 2) 0 < n n(x) n < 1 β < 1 for some constants α, β. Then there exists an infinite set of transmission eigenvalues with + as the only accumulation point.

26 Numerical Methods for the Transmission Eigenvalue Problem-Literature Colton, Monk and Sun (2010) finite element methods (Argyris, continuous, mixed) square and disk both real and complex tr. eigen. Sun ( ) iterative methods square,disk and triangle only real tr. eigen. Ji,Sun and Turner (2012) mixed finite element method square and disk only real tr. eigen. Kleefeld, (2013)

27 Inverse Problem: Find a Piecewise Constant n(x) from a set of Transmission Eigenvalues, joint. work with N. Pallikarakis How many eigenvalues we need to compute a piecewise constant refractive index? Are complex eigenvalues useful in reconstructions? Can we use the method for a general domain?

28 Inverse Problem: Find a Piecewise Constant n(x) from a set of Transmission Eigenvalues joint. N. Pallikarakis Case 1: n(x) = n 0 constant. Then the first transmission eigenvalue completely defines n 0. Case 2: Spherically symmetric n(x) piecewise constant with different values on different subdomains of D. n(x) := n i, in D i, i = 1, 2,..., n where D = n i=1 D i, i = 1, 2,..., n. Case 3: General shape, n(x) piecewise constant with different values on different subdomains of D. n(x) := n i, in D i, i = 1, 2,..., n where D = n i=1 D i, i = 1, 2,..., n. (Case 3 is under investigation).

29 Discrete Inverse Eigenvalue Problem for n(x) Piecewise Constant Let {φ i } i=1 be a Hilbert basis in H2 0 (D). A good choice: the set of eigenfunctions of the problem: Lφ i = µ i φ i in D φ i = 0, φ i η L is a fourth order elliptic operator. = 0 on D If L = is the Bilaplacian operator, the eigenvalues and eigenfunctions (clamped plate problem) can be computed. For a circle we can compute them analytically. Other choices for L are also possible.

30 Discrete Representation of the Eigenvalue Problem Expand an eigenvalue u k in this system: u k = i=1 c iφ i. Using a Galerkin type procedure we approximate u k by := N i=1 c iφ i. Use as test functions the eigenfunctions φ i, i = 1,..., N. u (N) k The approximate non linear eigenvalue problem is: [ A (N) (k (N) ) 2 B (N) + (k (N) ) 4 C (N)] c = 0, where: ( B (N) := D and c = (c 1, c 2,..., c N ) T. ( ) A (N) 1 := D n(x) 1 φ i φ j dx, n(x) n(x) 1 φ 1 i φ j dx + D n(x) 1 φ i φ j dx ( ) C (N) n(x) := n(x) 1 φ i φ j dx D ),

31 Discrete Representation of the Eigenvalue Problem This is a typical quadratic eigenvalue problem (QEP). (see for example F. Tisseur and K. Meerbergen, SIAM Review 2000). After a linearization procedure the QEP can be written in the form: (K (N) ξ (N) I)U (N) = 0, U (N) = ( c τ (N) ( K (N)) 1/2 c existence: there exist 2N eigenvalues of the discrete eigenvalue problem (K (N) is a 2N 2N square matrix). ), ξ (N) := 1 τ (N)

32 Discrete Representation of the Eigenvalue Problem -Equivalent Operator Form we use the notation following Descloux, Nassif and Rapaz (1978): X := H 2 0 (D) and X N := H 2 0 (D),N, (N-dimensional subspace) orthogonal projections: Π N : X X N linear bounded operators: T (N) := Π N T XN, T (N) 1 := Π N T 1 XN, T (N) 2 := Π N T 2 XN : X N X N, which correspond to the N N Galerkin matrices A (N), B (N), C (N) resp. and K (N) 1 : = (T (N) ) 1/2 T (N) 1 (T (N) ) 1/2, K (N) 2 : = (T (N) ) 1/2 T (N) 2 (T (N) ) 1/2 K (N) := ( K (N) 1 (K (N) 2 ) 1/2 (K (N) 2 ) 1/2 0 )

33 Discrete Representation of the Eigenvalue Problem -Convergence Theorem (G.-Pallikarakis) Eigenvalues of the discrete problem converge to the corresponding operator problem as N. Proof convergence for Galerkin type eigenvalue problems (Babuska and Osborn 1991) operator norm for a block operator: K K (N) K 1 K (N) 1 +2 K 1/2 2 (K (N) 2 ) 1/2, and K 1 K (N) 1 = = T 1/2 T 1 T 1/2 (T (N) ) 1/2 T (N) 1 (T (N) ) 1/2 (T 1/2 (T (N) ) 1/2 )T 1 T 1/2 + + (T (N) ) 1/2 T 1 (T 1/2 (T (N) ) 1/2 ) + T 1 T (N) 1 (T (N) ) 1/2 2 compactness argument and pointwise convergence of bounded operators implies norm convergence

34 Complex Eigenvalues Recent results on complex transmission eigenvalues: eigenvalue free zones in the complex plane for media of general shape, Cakoni-Colton-Gintides ( 2010), Hitrik-Krupchyk-Ola-Päivärinta, (2011), almost all transmission eigenvalues are confined to a parabolic neighborhood of the positive real axis. existence results for spherically symmetric refractive index, Cakoni-Colton-Gintides ( 2010) Existence results of infinite complex eigenvalues for spherically symmetric refractive index, Leung - Colton ( 2012, 2013).

35 Reconstructions for the Constant Index Case We have reconstructions using as test data only the first eigenvalues for circles with constant index. The basic idea is to minimize the error f (n) := k 0 k (N) 0. original n first eigenvalue k 0 error k 0 k (N) 0 estimated n we used a basis of N = 12 eigenfunctions.

36 Reconstructions for the Constant Index Case-Error Plots k 0 k (N) 0 versus n for original n = 3, n = 6, n = 12, and n = 20 respectively error k 0 k 0 N error k 0 k 0 N refractive index n refractive index n error k 0 k 0 N error k 0 k 0 N refractive index n refractive index n

37 D 2, n 2 r 1 The Inverse Transmission eigenvalue Problem -Numerical Example R D 1, n 1 disc with two layers-piecewise constant index: { n 1, if 0 x < r 1 n(x) = n 2, if r 1 x R

38 Computation of Transmission Eigenvalues for Spherically Symmetric n(x) computation of transmission eigenvalues : Analytical scheme from the boundary conditions with sep. of variables, zeros of the det det J m (kr) 0 J m (k n 2 R) N m (k n 2 R) J m(kr) r=r 0 J m(k n 2 r) r=r N m(k n 2 r) r=r 0 J m (k n 1 r 1 ) J m (k n 2 r 1 ) N m (k n 2 r 1 ) 0 J m(k n 1 r) r=r1 J m(k n 2 r) r=r1 N m(k n 2 r) r=r1 using root finding software. = 0 Computational scheme Galerkin method with N eigenfunctions to compute the approximate tr. eig. k (N) i, using matlab function polyeig. (convergence of computed tr. eig. is ensured from the previous theorem).

39 Reconstructions for Piecewise Constant Index - Real Transmission Eigenvalues We have reconstructions using as test data the first L=4 eigenvalues for examples where 0.5 r1 1, 5 n 1, n The basic idea is to minimize the error f (n 1, n 2, r 1 ) := L k i k (N) 2. Original (n 1,n 2,r 1 ) AnalyticTE{k 0,k 1,k 2,k 3 } Reconstructions (ñ 1,ñ 2, r 1 ) (13,11,0.8) ,1.4884,1.8248, (13,11.2,0.8) (5,8,0.6) ,2.2483,2.6654, (5,8.4,0.6) (13,5,0.5) ,1.7336,2.1694, (10.7,5.8,0.6) (6,13,0.9) ,2.0239,2.5120, (6.3,9.6,0.8) (10,8,0.7) ,1.7304,2.1086, (10.1,7.9,0.7) we used a basis of N = 12 eigenfunctions. i=1 i

40 Reconstructions for Piecewise Constant Index - Non uniqueness example example where the algorithm does not converge: (13, 5, 0.9) is reconstructed as (11.4, 11.7, 0.5). (n 1, n 2, r 1 ) Analytic TE {k 0, k 1, k 2, k 3, k 4 } (13, 5, 0.9) , , , , (11.4, 11.7, 0.5) , , , , Solution: use more transmission eigenvalues as input data. using 5 instead of 4 tr. eigen. we have the reconstruction (13, 7, 0.9). Using more than 5 eigenvalues is much better.

41 Reconstructions for Piecewise Constant Index - Real and Complex Transmission Eigenvalues We have reconstructions using as test data the first L=4 eigenvalues for 0.5 r1 1, 2 n 1, n 2 6 including complex eigenvalues As n(x) is closer to 1 more complex eigenvalues exist. Original (n 1,n 2,r 1 ) Analytic TE{k 0,k 1,k 2,k 3 } Reconstr. (ñ 1,ñ 2, r 1 ) (5,3,0.6) i,2.885,3.183,3.293 (5.1,3,0.6) (3,6,0.8) i,3.025,3.669,3.796 (3,6,0.8) (2,4,0.5) i, i,3.908,4.11 (2,4.5,0.5) (6,2,0.9) 2.021,2.418, i,2.929 (6,2.9,0.9) (2,4,0.7) i, i, i, (3,5.2,0.7) we used a basis of N = 12 eigenfunctions.

42 Remarks Constant index case the first tr. eig. completely defines n Increasing the number of tr. eig. helps Real and complex eig. carry information about the index We can approximate a simple piecewise n(x) using few tr. eig.

43 The General Spherically Stratified Domain D is a spherically stratified domain with k layers Dk Dk 1 D2 D1 We can approximate any smooth function. n 1, x D 1 n(x) =. n k, x D k We can compute the tr. eig. from the boundary conditions analytically.

44 Newton s Method-The main idea The approximate non linear eigenvalue problem has the form: [ A (N) (k (N) ) 2 B (N) + (k (N) ) 4 C (N)] c = 0 A (N) = k l=1 a l A l, B (N) = k l=1 B(1) l + k l=1 a l (B (1) l +B (2) l ) and C (N) = k l=1 C l + k l=1 a l C l where a l := 1/(n l 1). Numerical Inverse Problem: Given: a set of tr. eig. S = {µ i } k i=1 estimate: {a l} k l=1.

45 Newton s Method-The main idea Compute constants a = (a 1, a 2,..., a k ) such that the nonlinear system holds f (a) = 0: where [ f i (a)=det µ 4 i ( k l=1 C l + k l=1 a l C l)+µ 2 i ( k l=1 B(1) l + ) k l=1 a l (B (1) l +B (2) l ) + ] k l=1 a l A l main advantage: we don t have to pair λ i (a) with µ i in each step. Remark: We have to pair eigenvalues (model and measured) if : f i (a) = λ i (a) µ i

46 The Algorithm Newton s method 1 starting value a (0) for m = 0, 1, 2 Jacobian J(a (m) ) and f (a (m) ) 3 solve the system: J(a (m) )ξ (m) = f (a (m) ) for ξ (m) 4 compute the new estimate of the coefficients vector a (m+1) = ξ (m) + a (m) 5 stop if ξ (m) is sufficiently small Jacobian: J(a (m) ) il = f i (a (m) )/ a l We can compute the Jacobian via a fast algorithm ( Elhay S and Ram Y M 2002: based on LU or QR decomposition)

47 Computing the Jacobian Input {A l } k l=1, {B l} k l=1, {C l} k l=1 of N N matrices. the k vector of a (m) to the m th iteration step a set S = {µ i } k i=1. Output the function f (a (m) ) the Jacobian matrix J(a (m) ) with il th component f i (a (m) )/ a l

48 Computing the Jacobian For each i = 1, 2,, k ( k 1 compute H = µ 4 i l=1 C l + ) k l=1 a lc l + ( k µ 2 i l=1 B(1) l + ) k l=1 a l(b (1) l + B (2) l ) + k l=1 a la l 2 use LU or QR factorization to H and compute f (a (m) ) =(product of the diagonal elements) 3 for each l = 1, 2,, k do the following loop

49 Computing the Jacobian Loop: 1 compute: ( D = k µ 4 i j=1 C j + ) k j=1,j l a jc j + µ 2 i ) B (2) j ) + k j=1,j l a ja j ( k j=1 B(1) j + k j=1,j l a j(b (1) j + 2 use QZ (Generalized Schur decomposition) algorithm to find Q and R with determinant unity which simultaneously triangularize A l, B (1) l, B (2) l, C l, D 3 denote as (α i, β i ) N i=1 the pairs of the diagonal elements of the QZ triangles 4 determine the number of nonzero α i 5 relabel (α i, β i ) so that α r+1 = α r+2 = = α N = 0 6 Then set: [ J(a (m) ) ] il = ( N i=r+1 β i ) r i=1 α i N j=1,j i (a(m) l α j + β j )

50 Numerical Example unit disc with 2 layers: n 1 =5, n 2 =8 and r 1 =0.6 (known a priori) and initial values (6.5, 6.5) is reconstructed as (4.99, 8.00) after 5 steps with tol= unit disc with 4 layers: n 1 =7, n 2 =3, n 3 =8, n 4 =11, width of each layer 0.25 (known a priori) and initial values (7, 7, 7, 7) is reconstructed as (6.99, 3.00, 8.00, 10.99) after 16 steps with tol=

51 The General Inverse Transmission Problem D = m i=1 D i, D i D j =, i j, D simply connected: D D1 D2 Dm n 1, x D 1 n(x) =. n m, x D m

52 The General Inverse Transmission Problem nonlinear system f (a) = 0: [ f i (a)=det µ 4 i ( m l=1 C l + m l=1 a l C l)+µ 2 i where: a l = 1/(1 n l ) C l = D l φ i φ j dx B (1) l B (2) l = D l φ i φ j dx = D l φ i φ j dx A l = D l φ i φ j dx for i, j = 1,..., N and l = 1,..., m ( m l=1 B(1) l + ) m l=1 a l (B (1) l +B (2) l ) + ] m l=1 a l A l

53 The Algorithm using similar arguments: Newton s method 1 starting value a (0) for m = 0, 1, 2 Jacobian J(a (m) ) and f (a (m) ) 3 solve the system: J(a (m) )ξ (m) = f (a (m) ) for ξ (m) 4 compute the new estimate of the coefficients vector a (m+1) = ξ (m) + a (m) 5 stop if ξ (m) is sufficiently small Jacobian: J(a (m) ) il = f i (a (m) )/ a l

54 Numerical Example unit disc with 2 inner elliptical layers: n 1 =10, n 2 =4 and n 3 = 7, geometry (known a priori) and initial values (7, 7, 7) is reconstructed as (10.00, 4.00, 6.99) after 7 steps with tol= more general geometry is under investigation

55 Open Problems Convergence and stability for the Newton s method Detection of n(x) for complicated domains Uniqueness of the inverse spectral problem for general domains