Calderón s inverse problem in 2D Electrical Impedance Tomography

Transcription

1 Calderón s inverse problem in 2D Electrical Impedance Tomography Kari Astala (University of Helsinki) Joint work with: Matti Lassas, Lassi Päivärinta, Samuli Siltanen, Jennifer Mueller and Alan Perämäki Isaac Newton Institute, August 2011

2 Matti Lassas, University of Helsinki Jennifer Mueller, Colorado State University Lassi Päivärinta, University of Helsinki Allan Perämäki, Aalto University Samuli Siltanen, University of Helsinki

3 Electric Impedance Tomography. Inverse conductivity problem: Measure electric resistance between all boundary points of a body. Can one determine from this data the conductivity inside the body?

4 Calderon s problem (1980) Mathematical model of Electric Impedance Tomography (EIT) Ω Λ σ f = σ u n Ω, σ, σ 1 L (Ω). σ(x) σ u = 0, in Ω, u = f on Ω Given the Dirichlet-to-Neumann map Λ σ, can we construct the conductivity σ(x) inside Ω?

5 In two dimensions Calderon problem admits a complete solution: Theorem 1 (Astala Päivärinta) Let Ω R 2 be a bounded, simply connected domain. Assume that Then σ 1, σ 2, σ 1 1, σ 1 2 L. Λ σ1 = Λ σ2 σ 1 = σ 2 a.e. Here conductivity σ(x) isotropic, i.e. a scalar function. May take Ω = D, the unit disk.

6 Basis of the EIT-uniqueness proof: Analysis of the complex geometric optics solutions u k, u(x) = e ik x (1 + O( 1 x )), where σ(x) u(x) = 0 with k k = 0, and their dependence on the complex wave-number k. (For σ smooth, analysis reduces to the Schrödinger equation: Sylvester, Uhlmann, Nachmann, Brown, Päivärinta, Kenig, Sjöstrand, Bukhgeim,... )

7 Finding exponentially growing u σ : Reduction to Complex Analysis. 1. If u W 1,2 loc (C) satisfies σ(x) u(x) = 0 Hodge * conjugate v(z) = J σ(z) u(z), J = ( f = u + iv quasiregular (= locally quasiconformal), i.e. satisfies ) z f = ν z f ν = 1 σ 1+σ [ν := 0 for z > 1] 3. Show: unique solution f = f ν (z, k) such that f ν = e ikz ( 1 + O( 1 z )), u(z) = u σ (z, k) = Ref ν (z, k) + i Imf ν (z, k)

8 The method of solution to Calderón problem leads to computerized imaging algorithm. (A., Mueller, Siltanen, Perämäki and Päivärinta) Remarkable contrast resolution and good spatial correspondence. Applicable to all non-smooth and L -discontinuous conductivities. Non-iterative, Based on solving a PDE on Fourier side. Regularization by simple low-pass filtering on the Fourier side. Stability (Clop-Faraco-Ruiz): σ W α,p, some α > 0, 1 < p <.

9 11.6% 12.7% Conductivity σ 1 Reconstruction from ideal data, R = 6 Reconstruction with 0.01% noise, R = 5.5 Practical EIT data modelled by trig. basis functions (N=32)

10 The k -equation and Non-linear Fourier transform. Suppose: Div ( σ u ) = 0, (1) where u σ (z, k) = e ikz ( 1 + O( 1 z )), z. Comparing coefficients gives, from uniqueness of CGO solutions, k u σ (z, k) = iτ σ (k) u σ (z, k) (Beals-Coiffman, Nachman, A. - Päivärinta,...) NLFT: τ σ (k), k C, Determined by DN-map Λ σ

11 Nachman s (1996) reconstruction for smooth coefficients: For σ C 2 ψ(z, k) := σ(z) u σ (z, k) solves ψ + qψ = 0; q = σ 1/2 σ 1/2 ; ψ = e ikz ( 1 + O( 1 z )) Let µ(z, k) := e ikz ψ(z, k) (i) k µ(z, k) = iτ σ (k)e z (k) µ(z, k), e k (z) := e i (kz+ k z) (ii) µ(z, k) 1 0 as k 0

12 Nachman s (1996) reconstruction for smooth coefficients: For σ C 2 ψ(z, k) := σ(z) u σ (z, k) solves ψ + qψ = 0; q = σ 1/2 σ 1/2 ; ψ = e ikz ( 1 + O( 1 z )) Let µ(z, k) := e ikz ψ(z, k) (i) k µ(z, k) = iτ σ (k)e z (k) µ(z, k), e k (z) := e i (kz+ k z) (ii) µ(z, k) 1 0 as k 0 Conversely, solving (i) - (ii) gives σ(z) = lim k 0 µ(z, k). (Computer implementation by Isaacson-Mueller-Siltanen, et.al.)

13 Nachman s (1996) reconstruction for smooth coefficients: For σ C 2 ψ(z, k) := σ(z) u σ (z, k) solves ψ + qψ = 0; q = σ 1/2 σ 1/2 ; ψ = e ikz ( 1 + O( 1 z )) Let µ(z, k) := e ikz ψ(z, k) (i) k µ(z, k) = iτ σ (k)e z (k) µ(z, k), e k (z) := e i (kz+ k z) (ii) µ(z, k) 1 0 as k 0 Conversely, solving (i) - (ii) gives σ(z) = lim k 0 µ(z, k). Can (i) - (ii) work for rough, nonsmooth conductivities?

14 Nachman s (1996) reconstruction for smooth coefficients: For σ C 2 ψ(z, k) := σ(z) u σ (z, k) solves ψ + qψ = 0; q = σ 1/2 σ 1/2 ; ψ = e ikz ( 1 + O( 1 z )) Let µ(z, k) := e ikz ψ(z, k) (i) k µ(z, k) = iτ σ (k)e z (k) µ(z, k), e k (z) := e i (kz+ k z) (ii) µ(z, k) 1 0 as k 0 Conversely, solving (i) - (ii) gives σ(z) = lim k 0 µ(z, k). Q: For L -conductivities, the k-asymptotics of a CGO-sol n =?

15 A simple conductivity σ(z) with a discontinuity

16 f ν (z, k) = e ikz (1 + ω(z, k)) ; here ω(, k) as a function k 0.9 Sup norm of!(.,k) as function of k

17 What to do??

18 Reconstruction strategy for rough coefficients: 1. From measured EIT data determine traces of CGO-solutions. This gives u σ (z, k) for all z 1, k C. 2. Compute u σ (z, k), for z inside Ω, using Transport matrix and Low-pass filtering. 3. Reconstruct the conductivity from: f ν (z) = Re u σ (z, k)+i Im u 1/σ (z, k), ν(z) = zf(z) z f(z), σ = 1 ν 1+ν [σ := 1 for z > 1]

19 1. Traces of CGO solutions from EIT -data via explicit Boundary integral equation: For CGO-solutions f ν = e ikz M ν, where M ν (, k) D + 1 = (P k ν + P 0 )M ν (, k) D Λ σ = T H ν ν-hilbert transform 2P ν g = g + ih ν g + D g; Pk ν = e ikz P ν e ikz Here Id (P k ν + P 0 ) invertible; P ν Riesz-projection For practical data with noise, solve BIE only for k < R. Ill-posedness arises only here!

20 2. Transport matrix. Fix w 0 outside Ω = D, z D. Then: (i) u σ (z, k) = a(z, w 0 ; k) u σ (w 0, k) + b(z, w 0 ; k) i u 1/σ (w 0, k), where a( ), b( ) R. The k -equation φ := a + ib satisfies (ii) k φ = ν w0 (k) k φ, ν w0 (k) = u 1/σ (w 0,k) u σ (w 0,k) u 1/σ (w 0,k)+u σ (w 0,k), ν w 0 (k) < 1 Subexponential growth : u σ (z, k) = e ikz+kε z(k), where ε z (k) = o(k). For practical data with noise, solve (ii) with ν R w 0 (k) = ν w0 (k) χ D(R),

21 2. Transport matrix. Fix w 0 outside Ω = D, z D. Then: (i) u σ (z, k) = a(z, w 0 ; k) u σ (w 0, k) + b(z, w 0 ; k) i u 1/σ (w 0, k), where a( ), b( ) R. The k -equation φ := a + ib satisfies (ii) k φ = ν w0 (k) k φ, ν w0 (k) = u 1/σ (w 0,k) u σ (w 0,k) u 1/σ (w 0,k)+u σ (w 0,k), ν w 0 (k) < 1 Subexponential growth : u σ (z, k) = e ikz+kε z(k), where ε z (k) = o(k). φ(z, w 0 ; k) := e ik(z w 0)+k ε z (k), φ(0) = 1 (ii) determines u σ (z, k).

22 Actual conductivity Reconstruction σ % σ 3 Subtract 16.3% σ 2 σ 3

23 Conductivity σ 4 Reconstruction x < % x < %

24 Degenerate or infinite conductivities? Above algorithm required only Existence and uniqueness of CGO-solutions, u σ (z, k), z C, k C. Subexponential growth, u σ (z, k) = e ikz+kε z(k), where ε z (k) 0 as k.

25 Degenerate or infinite conductivities? Non-visible conductivities: Let B(ρ) be a 2-dimensional disc of radius ρ. Consider the map F : B(2) \ {0} B(2) \ B(1), F (x) = ( x 2 + 1) x x. Theorem 2 (Greenleaf-Lassas-Uhlmann invisibility cloaking) Set σ = F 1 in B(2) \ B(1) with σ arbitrary in B(1). Then boundary measurements for σ and σ 1 coincide.

26 Here push forward σ = F σ defined by: DF (x)σ(x)df (x) t = σ ( F (x) ) det(df (x)) for an W 1,1 loc -homeomorphism F : Ω Ω. Note that for σ = 1, we have det σ 1.

27 Theorem 4. (Astala Lassas Päivärinta) Let symmetric and positive definite σ 1, σ 2 : Ω R 2 2 satisfy Ω [ Tr(σ es j )+Tr(σ 1 j ) ] dx <, for some s > 0, and det(σj ) ±1 L. Then Λ σ1 = Λ σ2 σ 1 = σ 2 up to a coordinate change, i.e. σ 1 = F σ 2 for a W 1,1 loc -homeomorphism F : Ω Ω with F Ω = I.

28 Can one do better?

29 Limit to Calderon problem: Whenever A(t) sublinear with A(t) t 2 <, there is a W 1,1 -homeo F : B(2) \ {0} B(2) \ B(1) with DF (x)σ(x)df (x) t = det DF (x) Id, det(σ) = 1 Ω ea ( Tr(σ)+Tr(σ 1 ) ) dx < [Iwaniec-Martin (2001)]

30 Limit to Calderon problem: Whenever A(t) sublinear with A(t) t 2 <, there is a W 1,1 -homeo F : B(2) \ {0} B(2) \ B(1) with DF (x)σ(x)df (x) t = det DF (x) Id, det(σ) = 1 Ω ea ( Tr(σ)+Tr(σ 1 ) ) dx < Then Λ σ = Λ σ where σ = F σ 1 in B(2) \ B(1) and σ(z) 0 for z < 1.

31 Theorem (Astala Lassas Päivärinta) Let symmetric and positive definite σ 1, σ 2 : Ω R 2 2 satisfy Ω ea ( Tr(σ)+Tr(σ 1 ) ) < where A(t) t 2 =, ta (t), and Ω eees(det σ j +1/ det σ j ) <. Then Λ σ1 = Λ σ2 σ 1 = F σ 2 for a W 1,1 loc -homeomorphism F : Ω Ω with F Ω = id. Makes tunneling of electrons possible in classical electrostatistics!

32 Finding exponentially growing u σ : Reduction to Quasiconformal maps.

33 Finding exponentially growing u σ : Reduction to Quasiconformal maps. If σ(x) symmetric, positive definite matrix function, and σ(x) u(x) = 0, then f = u + iv satisfies: z f = µ z f + ν z f Here µ = σ 22 σ 11 2iσ 12 1+Trace(σ)+det σ, ν = 1 det σ 1+Trace(σ)+det σ

34 Finding exponentially growing u σ : Reduction to Quasiconformal maps. If σ(x) symmetric, positive definite matrix function, and σ(x) u(x) = 0, then f = u + iv satisfies: z f = µ z f + ν z f Mappings of finite distortion ( early 90 s ) f W 1,1 loc, max α =1 αf K(z, f) min α =1 α f, K(z) < a.e.

35 Finding exponentially growing u σ : Reduction to Quasiconformal maps. If σ(x) symmetric, positive definite matrix function, and σ(x) u(x) = 0, then f = u + iv satisfies: z f = µ z f + ν z f Ellipticity: 1 4 [ Tr σ(z) + Tr σ 1 (z) ] K(z, f) 1 1 ( µ(z) + ν(z) ) Tr σ(z)+tr σ 1 (z) Uniform ellipticity µ(z) + ν(z) k < 1.

36 Finding exponentially growing u σ : Reduction to Quasiconformal maps. If σ(x) symmetric, positive definite matrix function, and σ(x) u(x) = 0, then f = u + iv satisfies: z f = µ z f + ν z f Now, look for solutions f = f µ,ν with f = e ikz (1+O( 1 z )) as z. Ansatz: f ν (z) = e ikφ(z), φ : C C homeo with φ(z) = z + O( 1 z ).

37 z φ = µ(z) z φ k k ν(z) e k ( φ(z, k) ) z φ, e k (z) := e i (kz+ k z) z φ ( µ(z) + ν(z) ) z φ

38 z φ = µ(z) z φ k k ν(z) e k ( φ(z, k) ) z φ, e k (z) := e i (kz+ k z) z φ ( µ(z) + ν(z) ) z φ If Ω esk(z,φ) <, get Energy estimates: Since Dφ 2 K(z, φ)j(z, φ) Dφ 2 log(e+ Dφ 2 ) K(z,φ)J(z,φ) log(e+k(z,φ)j(z,φ)) [ a b log(e+a) a + eb] 1 p ( J(z,φ) log(e+j(z,φ) ) pk(z, φ) 1 p ( J(z, φ) + e pk(z,φ) ),

39 z φ = µ(z) z φ k k ν(z) e k ( φ(z, k) ) z φ, e k (z) := e i (kz+ k z) z φ ( µ(z) + ν(z) ) z φ If Ω esk(z,φ) <, get Energy estimates: Since Dφ 2 K(z, φ)j(z, φ) Dφ 2 log(e+ Dφ 2 ) 1 p ( J(z,φ) log(e+j(z,φ) and integrate Ω K(z,φ)J(z,φ) log(e+k(z,φ)j(z,φ)) ) pk(z, φ) 1 p ( J(z, φ) + e pk(z,φ) ), Dφ 2 log(e+ Dφ 2 ) < C 0 [ Ω J(z, φ) φ(ω) < ]

40 For homeomorphisms the regularity Ω Dφ 2 log(e + Dφ 2 ) C 0 < gives a uniform modulus of continuity, depending only on the Orlicz-norm. Thus solutions to approximate (uniformly elliptic) systems converge! [ e.g. σ n (x) σ(x), T r(σ n ) + T r(σ 1 n ) min{n, T r(σ) + T r(σ 1 )} ]

41 Complex geometric optics solutions: Infinite conductivities Let σ(x) be symmetric, positive matrix function, σ = Id outside Ω. Theorem. Suppose Ω e s( Trace σ+trace σ 1) dx <, where s > 0. Then for every ξ C there is a unique solution to σ(x) u(x) = 0, x C, such that 1. u(z, ξ) = e iξz (1+O( 1 z )) as z. [(ξ, iξ) (x, y) = (x+iy)ξ] 2. Ω u 2 log(e+ u ) dx <.

42 Why the Orlicz-norms like Ω u 2 log(e+ u ) dx <?

43 Why the Orlicz-norms like Ω u 2 log(e+ u ) dx <? 1. Ω u 2 dx < not true 2. u 1<q<2 W 1,q loc (C) not enough for uniqueness

44 Subexponential growth requires energy bounds for φ(z, k), as well as for ψ = φ 1. This requires Ω eee s (T r(σ)+t r(σ 1 )) <!

45 THANK YOU!