Reconstructing conductivities with boundary corrected D-bar method Janne Tamminen June 24, 2011
Short introduction to EIT The Boundary correction procedure The D-bar method Simulation of measurement data, numerical D-bar
Short introduction to EIT The aim of electrical impedance tomography (eit) is to form an image of the conductivity distribution inside an unknown body using electric boundary measurements. Applications in medical imaging, nondestructive testing, subsurface monitoring: monitoring heart and lungs of unconscious patients, detecting pulmonary edema, breast cancer detection, detecting cracks in concrete structures, environmental applications...
Short introduction to EIT The inverse conductivity problem introduced by Calderón is to find the conductivity σ when the Dirichlet-to-Neumann -map (dn-map) Λ σ is known. { (σ u) = 0 in Ω1, (1) u = f on Ω 1. Λ σ f = σ u ν Ω 1 (2) The problem of reconstructing σ from Λ σ is nonlinear and ill-posed.
Short introduction to EIT In the two-dimensional case, unique reconstruction can be obtained: For σ W 2,p (Ω 1 ),p > 1, Nachman 1996 (the D-bar method), Knudsen-Lassas-Mueller-Siltanen 2009 For σ W 1,p (Ω 1 ),p > 2, Brown-Uhlmann 1997 and for σ L (Ω 1 ) Astala-Päivärinta 2003. Nachman s method reduces the reconstruction problem to the case σ = 1 near the boundary. This procedure is tested numerically in this joint work with S. Siltanen.
The Boundary correction procedure Short introduction to EIT The Boundary correction procedure The D-bar method Simulation of measurement data, numerical D-bar
The Boundary correction procedure We transform the original conductivity equation to Schrödinger equation: by writing u = σ 1/2 ũ we get ( q)ũ = 0 in Ω 1, (3) where q = σ 1/2 (σ 1/2 ). This means we have to have σ c 0 > 0. The dn-map becomes Λ q = σ 1/2 (Λ σ + 1 σ 2 ν )σ 1/2. (4) In order for this transformation to be useful we need to have σ 1 near Ω 1 so that Λ q = Λ σ.
The Boundary correction procedure Ω 2 Ω 1 Λ σ Λ γ? We extend σ: { σ(x), x Ω 1, γ(x) = σ(x), x Ω 2 \Ω 1, σ σ W 2,p σ 1 σ Ω1 = σ Ω1 (5) σ ν Ω 1 = σ ν Ω 1 (6) giving us γ W 2,p (Ω 2 ) and Λ q = Λ γ.
The Boundary correction procedure In this work, Ω 1 = D(0,r 1 ), r 1 = 1.0 Ω 2 = D(0,r 2 ), r 2 = 1.2 σ L (Ω 1 ) σ W 2,p (Ω 2 \Ω 1 ) γ L (Ω 2 ),
The Boundary correction procedure Define two Dirichlet problems, for j = 1,2, ( σ u j ) = 0 in Ω 2 \Ω 1 u j = f j on Ω j u j = 0 on Ω i, i = 1,2, i j. (7) Four new dn maps in Ω 2 \Ω 1 can be characterized by so λ ij : H 1/2 ( Ω j ) H 1/2 ( Ω i ). Λ ij f j = σ u j ν Ω i, i,j = 1,2 (8)
The Boundary correction procedure Proposition (6.1 in Nachman 1996) Let γ W 2,p (Ω 2 ),p > 1, then Λ γ = Λ 22 +Λ 21 (Λ σ Λ 11 ) 1 Λ 12. (9) The proof consists of showing that the operator Λ σ Λ 11 is invertible, and the identities (Λ γ Λ 22 )f 2 = Λ 21 (u Ω1 ) (10) (Λ σ Λ 11 )(u Ω1 ) = Λ 12 f 2 (11) for any u that solves the conductivity equation in Ω 2 with u Ω2 = f 2.
The D-bar method Short introduction to EIT The Boundary correction procedure The D-bar method Simulation of measurement data, numerical D-bar
The D-bar method The D-bar method: based on Nachman 1996 robust algorithm was given by Siltanen-Mueller-Isaacson The method has been successfully tested on a chest phantom and on in vivo human chest data (Isaacson-Mueller-Newell-Siltanen). Λ γ t(k) γ
The D-bar method The CGO-solution (Complex Geometric Optics) on Ω 2 can be solved from ψ(,k) Ω2 = e ikx S k (Λ γ Λ 1 )ψ(,k) Ω2, (12) in the Sobolev space H 1/2 ( Ω 2 ) for all k C\{0}. Here S k is a single-layer operator (S k φ)(x) := G k (x y)φ(y)ds, Ω 2 and G k is Faddeev s Green function defined by G k (x) := e ikx g k (x), g k (x) := 1 R2 e ix ξ (2π) 2 ξ 2 +2k(ξ 1 +iξ 2 ) dξ.
The D-bar method Using the CGO-solution we can define the scattering transform: t(k) = e i k x (Λ γ Λ 1 )ψ(,k)ds. (13) Ω 2 For each fixed x Ω 2, we would solve the following integral formulation of the D-bar equation: µ(x,k) = 1+ 1 t(k ) x+k x) (2π) 2 µ(x,k )dk R 2 (k k ) k ei(k 1 dk 2, (14) where µ(x,k) = exp( ik(x 1 +ix 2 ))ψ(x,k). Then the conductivity would be perfectly reconstructed as γ(x) = µ(x,0) 2.
Simulation of measurement data, numerical D-bar Short introduction to EIT The Boundary correction procedure The D-bar method Simulation of measurement data, numerical D-bar
Simulation of measurement data, numerical D-bar In this work we omit the requirement of continuity of the derivative γ ν and use the method of Nakamura-Tanuma-Siltanen-Wang to calculate g(θ) σ Ω1. We extend σ to γ using the following extension in polar coordinates: σ(ρ,θ), ρ r 1, γ(ρ,θ) = (g(θ) 1)f m (ρ)+1, r 1 < ρ r e, 1, r e < ρ r 2, (15) where r e = 1.175 and f m (ρ) 0 is a suitable third-degree polynomial satisfying f m (r 1 ) = 1 and f m (r e ) = 0.
Simulation of measurement data, numerical D-bar Any linear operator Λ : H 1/2 ( Ω i ) H 1/2 ( Ω j ) can be represented by a matrix in the following way. Define a truncated orthonormal basis at the boundary Ω k : φ (n) k (θ) = 1 2πrk e inθ, n = N,...,N, k = 1,2. (16) Write any function f : Ω i C as a vector f = [ˆf( N),ˆf( N+1),...,ˆf(N 1),ˆf(N)] T, ˆf(n) = Ω i fφ (n) i ds. Then the operator Λ is approximated by the matrix L = [û(l,n)], where û(l,n) = (Λφ (n) i )φ (l) j ds. (17) Ω j
Simulation of measurement data, numerical D-bar In practice in eit we measure the Neumann-to-Dirichlet -map approximated by the matrix R σ. We simulate measurement noise by using the matrix R ε σ := R σ +ce, (18) where E is a matrix with random entries independently distributed according to the Gaussian normal density. Then, the noisy DN-matrix L ε σ is roughly speaking the inverse of Rε σ, and the boundary correction procedure gives us L ε γ = L22 +L 21 (L ε σ L11 ) 1 L 12, (19) provided that the matrix L ε σ L 11 is invertible.
Simulation of measurement data, numerical D-bar The D-bar method numerically: expand e ikx Ω2 as a vector g in our finite trigonometric basis, solve the CGO-solution ψ k := [I +S k (L ε γ L 1 )] 1 g. (20) for k ranging in a grid inside the disc k < R ( the truncation radius R > 0). Define the truncated scattering transform by { t R (k) = Ω 2 e i k x F 1 ((L ε γ L 1) ψ k )(x)ds for k < R, (21) 0, otherwise, Finally solve the equation (14) with the numerical algorithm of Knudsen-Mueller-Siltanen using t R and denote the solution by µ R (x,k). Then γ(x) µ R (x,0) 2.
Short introduction to EIT The Boundary correction procedure The D-bar method Simulation of measurement data, numerical D-bar
Solutions to the Dirichlet problems are calculated using FEM, with 1048576 triangles in the disk Ω 1 and 425984 triangles in the annulus Ω 2 \Ω 1. R ǫ 1 R 1 / R 1 = 0.0001 (the ACT3 impedance tomography imager of Rensselaer Polytechnic Institute has SNR of 95.5 db noise of 0.0017%) ǫ fem = R th 1 R 1 / R th 1 = 0.0000173, the error R ǫ σ R σ / R σ ranges between 0.00011 and 0.00076, The condition number of the matrix L ε σ L 11 was less than 27 in all our test cases. The error L ǫ γ L 2 γ / L 2 γ, where L 2 γ is the dn map calculated directly on the boundary Ω 2, was less than 2.2% in all cases.
% 50 40 30 20 10 L ǫ γ=1 L2 γ=1 / L2 γ=1 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 Lǫ γ=1 L γ=1 / L γ=1 0 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 0 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 r 2
Example 1 σ Ω1 4 3.5 3 2.5 2 1.5 1 0.5 0 π 2π θ over boundary
Example 2 6 σ Ω1 5 4 3 2 1 0 0 π 2π θ over boundary
Example 3 7 σ Ω1 6 5 4 3 2 1 0 0 π 2π θ over boundary
Example 4 σ Ω1 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0 π 2π θ over boundary
Re(t(k)) Im(t(k)) non-noisy L σ 10 noisy L ε σ 5 0 5 Scattering transform, real and imaginary parts, of Example 4. 10 noisy and corrected L ε γ
R = 3.0 3.4 3.8 4.2 4.6 5.0 5.4 Original Uncorrected Corrected
% 50 Example 1 50 Example 2 40 40 30 30 20 20 10 % 75 65 Example 3 Example 4 10 35 30 55 25 45 20 35 3 3.4 3.8 4.2 4.6 5 5.4 5.8 6 R 15 3 3.4 3.8 4.2 4.6 5 5.4 5.8 6 R
R = 3.0 Uncorrected R = 5.0 Uncorrected R = 6.0 Uncorrected Example 1 Original conductivity 30% 25% 37% R = 3.0 Corrected R = 5.0 Corrected R = 6.0 Corrected 26% 18% 49%
R = 3.0 Uncorrected R = 5.4 Uncorrected R = 6.0 Uncorrected Example 2 Original conductivity 42% 29% 35% R = 3.0 Corrected R = 5.4 Corrected R = 6.0 Corrected 35% 15% 39%
R = 3.0 Uncorrected R = 5.0 Uncorrected R = 6.0 Uncorrected Example 3 Original conductivity 66% 60% 67% R = 3.0 Corrected R = 5.0 Corrected R = 6.0 Corrected 59% 39% 75%
R = 3.0 Uncorrected R = 4.8 Uncorrected R = 6.0 Uncorrected Example 4 Original conductivity 25% 22% 49% R = 3.0 Corrected R = 4.8 Corrected R = 6.0 Corrected 25% 21% 63%
Thank you!
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