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

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1 Nonlinear Fourier transform and the Beltrami equation Visibility and Invisibility in Impedance Tomography Kari Astala University of Helsinki

2 Beltrami equation: z f(z) = µ(z) z f(z)

3 non linear Fourier transform

4 What is the non linear Fourier transform?

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

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

7 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.

8 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 )

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

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

11 WHY the non linear Fourier transform? Dimension n = 2. Imaging in 2 dimensions (Electric Impedance Tomography)

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

13 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?

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

15 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,

16 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

17 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

18 Does NLFT exist, whenever σ, 1/σ L? - More generally? Can we invert the NLFT? τ σ (ξ) σ?

19 Inverting NLFT solves the Calderon problem: 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.

20 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

21 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 )),

22 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 )

23 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!

24 f ν (z) = e ikφ(z) with z f = ν z f z φ = k k ν( φ(z) ) z φ

25 4. Solution to conductivity equation: u(z) = u σ (z, ξ) = Ref ν (z, ξ) + i Imf ν (z, ξ) 5. By assumption: f ν (z, ξ) = e ikz ( 1 + β +(ξ) z + O( 1 z 2) ), f ν (z, ξ) = e ikz ( 1 + β (ξ) z + O( 1 z 2) ) u σ (z, ξ) has required asymptotics, is unique and τ σ (ξ) = 1 2 ( β+ (ξ) β (ξ) ), ξ R 2.

26 Inverting NLFT, τ σ σ??

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

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

29 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)!

30 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)!

31 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)

32 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?

33 A simple conductivity σ(z) with a discontinuity

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

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

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

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

39 Above σ(z) scalar. Howabout the case of σ(z) R 2 2? Obvious obstruction: Can change σ by F : Ω Ω, F Ω = Id Push forward F σ = H; DF (x)σ(x)df (x) t = H ( F (x) ) det(df (x))

40 Above σ(z) scalar. Howabout the case of σ(z) R 2 2? Theorem (Astala Lassas Päivärinta) Let σ 1, σ 2 : Ω R 2 2 be symmetric, with 1 K σ j K, K <. TFAE: (i) Λ σ1 = Λ σ2 (ii) τ σ1 τ σ2 (iii) σ 1 = F σ 2 ; F : Ω Ω homeo, F Ω = id, F W 1,2 (Ω)

41 Degenerate equations?

42 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?!

43 Artistic illustration by M. and J. Levin.

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

45 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.

46 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:

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48 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 <

49 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 <

50 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 ees 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.

51 Elliptic Partial Differential Equations and Quasiconformal Mappings in the Plane KARI ASTALA TADEUSZ IWANIEC GAVEN MARTIN