Nonlinear Fourier transform for the conductivity equation Visibility and Invisibility in Impedance Tomography Kari Astala University of Helsinki CoE in Analysis and Dynamics Research
What is the non linear Fourier transform?
What is the non linear Fourier transform? Dimension n = 1.
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
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.
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 )
What is the non linear Fourier transform? Dimension n = 2.
WHY the non linear Fourier transform? Dimension n = 2.
WHY the non linear Fourier transform? Dimension n = 2. Medical imaging in 2 dimensions (Electric Impedance Tomography)
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?
What is the non linear Fourier transform? Dimension n = 2. Conductivity equation: Div ( σ u ) = 0 ( σ(z) = I ) for z large
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,
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
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
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?
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.
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
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 ( 0 1 1 0 ) 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 )),
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 ( 0 1 1 0 ) 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 )
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 ( 0 1 1 0 ) 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!
4. Solution to conductivity equation: u(z) = u σ (z, ξ) = Ref ν (z, ξ) + i Imf ν (z, ξ)
Inverting NLFT, τ σ σ??
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
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
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)!
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)!
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)
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?
A simple conductivity σ(z) with a discontinuity
f µ (z, k) = e ikz (1 + ω(z, k)), here ω as a function k k=2 k=12 k=22 0.6 0!0.6 0.6 0!0.6 0 0.6
Theorem. Suppose σ, 1/σ L. Then τ σ (ξ) 1, ξ R 2.
Theorem. Suppose σ, 1/σ L. Then τ σ (ξ) 1, ξ R 2. Plancherel Formula? Riemann-Lebesgue lemma? etc.? etc.?
Above σ(z) scalar. Howabout the case of σ(z) R 2 2?
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))
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
Degenerate equations?
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?!
Artistic illustration by M. and J. Levin.
Limits to Calderon s problem: Non-visible conductivities in 2D.
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.
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:
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 <
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 <
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 )