Nonlinear Fourier transform for the conductivity equation. Visibility and Invisibility in Impedance Tomography

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1 Nonlinear Fourier transform for the conductivity equation Visibility and Invisibility in Impedance Tomography Kari Astala University of Helsinki CoE in Analysis and Dynamics Research

2 What is the non linear Fourier transform?

3 What is the non linear Fourier transform? Dimension n = 1.

4 What is the non linear Fourier transform? Dimension n = 1. For q L (R) with supp(q) [ M, M], consider ψ (x) + q(x)ψ(x) = k 2 ψ (k R) If ψ(x) = e ikx, x, then ψ(x) = a(k)e ikx + b(k)e ikx, x

5 What is the non linear Fourier transform? Dimension n = 1. For q L (R) with supp(q) [ M, M], consider ψ (x) + q(x)ψ(x) = k 2 ψ (k R) If ψ(x) = e ikx, x, then ψ(x) = a(k)e ikx + b(k)e ikx, x We call τ q (k) := ( a(k), b(k) ) the Non Linear Fourier Transform of q.

6 What is the non linear Fourier transform? Dimension n = 1. For q L (R) with supp(q) [ M, M], consider ψ (x) + q(x)ψ(x) = k 2 ψ (k R) If ψ(x) = e ikx, x, then ψ(x) = a(k)e ikx + b(k)e ikx, x We call τ q (k) := ( a(k), b(k) ) the Non Linear Fourier Transform of q. a(k) = 1+ i 2k q(y)dy+o( q 2 ), b(k) = i 2k F(q)( 2k)+O( q 2 )

7 What is the non linear Fourier transform? Dimension n = 2.

8 WHY the non linear Fourier transform? Dimension n = 2.

9 WHY the non linear Fourier transform? Dimension n = 2. Medical imaging in 2 dimensions (Electric Impedance Tomography)

10 WHY the non linear Fourier transform? Dimension n = 2. Medical imaging in 2 dimensions (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?

11 What is the non linear Fourier transform? Dimension n = 2. Conductivity equation: Div ( σ u ) = 0 ( σ(z) = I ) for z large

12 What is the non linear Fourier transform? Dimension n = 2. Conductivity equation: Div ( σ u ) = 0 ( σ(z) = I ) for z large A basic method in impedance tomography: Analysis of complex geometric optics solutions: u = u σ (z, ξ) = e iξz (1 + O ( )) 1 z as z,

13 What is the non linear Fourier transform? Dimension n = 2. Conductivity equation: Div ( σ u ) = 0 ( σ(z) = I ) for z large IF can find (unique) solutions of the form u = u σ (z, ξ) = e iξz (1 + O ( )) 1 z as z, then u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z

14 What is the non linear Fourier transform? Dimension n = 2. Conductivity equation: Div ( σ u ) = 0 ( σ(z) = I ) for z large IF can find (unique) solutions of the form u = u σ (z, ξ) = e iξz (1 + O ( )) 1 z as z, then u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z NLFT: τ σ (ξ) b(ξ), ξ R 2

15 Calderon s problem (1980) Mathematical model of Electric Impedance Tomography (EIT) Given the Dirichlet-to-Neumann map Λ σ, can we construct the conductivity σ inside Ω?

16 In two dimensions Calderon problem admits a 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. The proof yields, in principle, a method to construct σ from Λ σ.

17 Structure of Proof: Ω = D = { z < 1}, Set σ(z) 1 outside D; u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z NLFT: τ σ (ξ) b(ξ), ξ R 2, exists and is well defined. Λ σ determines τ σ (ξ), ξ R 2 (easy) τ σ determines the conductivity σ(z), z R 2 (difficult) τ σ σ Inverting NLFT

18 Finding exponentially growing u σ : Reduction to Complex Analysis. 1. If u W 1,2 loc (D) satisfies σ(x) u(x) = 0 Hodge * conjugate v(z) = J σ(z) u(z), J = 2. f = u + iv satisfies ( ) z f = ν z f ν = 1 σ 1+σ 3. Show: unique solution f = f ν such that f ν = e ikz ( 1 + O( 1 z )),

19 Finding exponentially growing u σ : Reduction to Complex Analysis. 1. If u W 1,2 loc (D) satisfies σ(x) u(x) = 0 Hodge * conjugate v(z) = J σ(z) u(z), J = 2. f = u + iv satisfies ( ) z f = ν z f ν = 1 σ 1+σ 3. Show: unique solution f = f ν such that f ν = e ikz ( 1 + O( 1 z )), f ν (z) = e ikφ(z), φ : C C homeo with φ(z) = z + O( 1 z )

20 Finding exponentially growing u σ : Reduction to Complex Analysis. 1. If u W 1,2 loc (D) satisfies σ(x) u(x) = 0 Hodge * conjugate v(z) = J σ(z) u(z), J = 2. f = u + iv satisfies ( ) z f = ν z f ν = 1 σ 1+σ 3. Show: unique solution f = f ν such that f ν = e ikz ( 1 + O( 1 z )), f ν (z) = e ikφ(z), φ : C C homeo a priori bounds!

21 4. Solution to conductivity equation: u(z) = u σ (z, ξ) = Ref ν (z, ξ) + i Imf ν (z, ξ)

22 Inverting NLFT, τ σ σ??

23 Inverting NLFT, τ σ σ?? Suppose: Div ( σ u ) = 0, (1) with u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z

24 Inverting NLFT, τ σ σ?? Suppose: Div ( σ u ) = 0, (1) with u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z Then: u σ (z, ξ) smooth in ξ

25 Inverting NLFT, τ σ σ?? Suppose: Div ( σ u ) = 0, (1) with u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z Then: u σ (z, ξ) smooth in ξ, ξ u σ satisfies (1)!

26 Inverting NLFT, τ σ σ?? Suppose: Div ( σ u ) = 0, (1) with u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z Then: u σ (z, ξ) smooth in ξ, ξ u σ satisfies (1)!

27 Inverting NLFT, τ σ σ?? Suppose: Div ( σ u ) = 0, (1) with u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z Then: u σ (z, ξ) smooth in ξ, ξ u σ satisfies (1)! ξ u σ (z, ξ) = iτ(ξ) u σ (z, ξ)

28 Inverting NLFT, τ σ σ?? Suppose: Div ( σ u ) = 0, (1) with u σ (z, ξ) = e iξz + a(ξ) z e i ξ z + b(ξ) z e i ξ z + O( 1 z 2) ei ξ z, z Then: u σ (z, ξ) smooth in ξ, ξ u σ satisfies (1)! ξ u σ (z, ξ) = iτ(ξ) u σ (z, ξ) σ smooth decay in ξ τ σ determines u σ (z, ξ), hence σ. For σ non smooth: asymptotics in ξ?

29 A simple conductivity σ(z) with a discontinuity

30 f µ (z, k) = e ikz (1 + ω(z, k)), here ω as a function k k=2 k=12 k= ! !

31

32 Theorem. Suppose σ, 1/σ L. Then τ σ (ξ) 1, ξ R 2.

33 Theorem. Suppose σ, 1/σ L. Then τ σ (ξ) 1, ξ R 2. Plancherel Formula? Riemann-Lebesgue lemma? etc.? etc.?

34 Above σ(z) scalar. Howabout the case of σ(z) R 2 2?

35 Above σ(z) scalar. Howabout the case of σ(z) R 2 2? Lemma. For σ L (Ω, R 2 2 ), symmetric and positive definite, τ σ = τ F σ, F σ det(σ F 1 ) I ( scalar!) where F : R 2 R 2 a quasiconformal mapping, F (z) z as z. Push forward F σ = H; DF (x)σ(x)df (x) t = H ( F (x) ) det(df (x))

36 Above σ(z) scalar. Howabout the case of σ(z) R 2 2? Lemma. For σ L (Ω, R 2 2 ), symmetric and positive definite, τ σ = τ F σ, F σ det(σ F 1 ) I ( scalar!) where F : R 2 R 2 a quasiconformal mapping, F (z) z as z. Push forward F σ = H; DF (x)σ(x)df (x) t = H ( F (x) ) det(df (x)) Corollary. Λ σ1 = Λ σ2 σ 1 = F σ 2 ; F : Ω Ω with F Ω = id

37 Degenerate equations?

38 Degenerate equations? Visibility in low fequences: Do the electric impedance tomography measurements on the boundary determine the inside of a body? Invisibility cloaking for low frequences: Can we coat a body with a special material so that it appears like homogeneous material in EIT-measurements? Limits of invisibility and visibility?!

39 Artistic illustration by M. and J. Levin.

40 Limits to Calderon s problem: Non-visible conductivities in 2D.

41 Limits to Calderon s problem: Non-visible conductivities in 2D. 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.

42 Limits to Calderon s problem: Non-visible conductivities in 2D. 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 (Greenleaf-Lassas-Uhlmann invisibility cloaking) σ = F 1 in B(2) \ B(1) with σ arbitrary in B(1). Then boundary measurements for σ and γ 1 coincide. Set:

43

44 Iwaniec Martin (2001): Whenever A(t) sublinear with exists a W 1,1 -homeo F : B(2) \ {0} B(2) \ B(1) with A(t) t 2 <, DF (x)σ 0 (x)df (x) t = det DF (x) Id, det(σ 0 ) = 1 Ω ea ( Trace(σ 0 ) ) dx <

45 Iwaniec Martin (2001): Whenever A(t) sublinear with exists a W 1,1 -homeo F : B(2) \ {0} B(2) \ B(1) with A(t) t 2 <, DF (x)σ 0 (x)df (x) t = det DF (x) Id, det(σ 0 ) = 1 ( Ω ea Trace(σ 0 ) ) dx < Set: γ(z) 1 in B(2) \ B(1) (homogeneous conductivity) with γ(z) 0, for z < 1 (perfect insulator). Then Λ σ0 = Λ γ Sharp limit to Calderon problem: Ω ea( Trace(σ) ) dx < ; A(t) t 2 <

46 Theorem. (A. Lassas Päivärinta) Let symmetric and positive definite σ 1, σ 2 : Ω R 2 2 satisfy Ω e A( Trace(σ) ) dx < ; A(t) t 2 = (+ mild growth control) and let (det σ j ) ±1 L. Then Λ σ1 = Λ σ2 σ 1 = F σ 2 for a W 1,1 loc -homeomorphism F : Ω Ω with F Ω = id. Moreover, for any E Ω, F (E) E = 0 (conditions N ±1 )

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