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Dec 12, 2022 •22 likes •392 views

Scattering in complex geometrical optics and the inverse problem for the conductivity equation Matti Lassas University of Helsinki In collaboration with Allan Greenleaf Matteo Santacesaria Samuli Siltanen Gunther Uhlmann fi fi Finnish

  1. Scattering in complex geometrical optics and the inverse problem for the conductivity equation Matti Lassas University of Helsinki In collaboration with Allan Greenleaf Matteo Santacesaria Samuli Siltanen Gunther Uhlmann fi fi Finnish Centre of Excellence in Inverse Problems Research

  2. Outline: ◮ Inverse conductivity problem ◮ Motivation of the edge detection problem ◮ Complex principal type operators ◮ Beltrami equation ◮ Single scattering and connection to Radon transform ◮ Analysis of higher order terms

  3. Inverse problem for the conductivity equation Conductivity equation on Ω ⊂ R d , d = 2 . ∇· σ ( x ) ∇ u ( x ) = 0 Inverse problem: Do the measurements made on the boundary determine the conductivity, that is, does the voltage-to-current map or the Dirichlet-to-Neumann operator Λ σ , Λ σ : u | ∂ Ω �→ ν · σ ∇ u | ∂ Ω determine the conductivity σ ( x ) in Ω ? Figure: EIT by Isaacson, Mueller, Newell and Siltanen.

  4. Some results on the Electrical Impedance Tomography (EIT) ◮ Calderón 1980: Linearized problem. ◮ Sylvester-Uhlmann 1987, Nachman 1988: Smooth conductivities in 3D. ◮ Nachman 1996: Smooth conductivities in 2D. ◮ Isaacson-Mueller-Newell-Siltanen 2004: Numerical reconstruction algorithm. ◮ Astala-Päivärinta 2006: Bounded conductivities in 2D, Astala-L.-Päivärinta 2016: Degenerated conductivities in 2D. ◮ Lee-Uhlmann 1989, L.-Uhlmann 2001, L.-Taylor-Uhlmann 2003, Dos Santos Ferreira-Kenig-Salo-Uhlmann 2009: Inverse problem for ∆ g on manifolds. ◮ Greenleaf-L.-Uhlmann 2003: Counterexaples related to invisibility cloaking. ◮ Daude-Kamran-Nicoleau 2016: New counterexamples with smooth conductivities.

  5. Exponentially decaying waves Let ξ = ( a , ib ) ∈ C 2 where a , b ∈ R , | a | = | b | , and b > 0. Then u ( x ) = e i ξ · x = e iax 1 · e − bx 2 , x = ( x 1 , x 2 ) ∈ R × R + are solutions of ∇ · ∇ u ( x ) = 0 in the half-space R × R + . These solutions decay as x 2 → ∞ and oscillate in the x 1 direction with the spatial frequency a . The vector ξ ∈ C 2 is called the complex wave number or the complex frequency and u a solution of Complex Geometrical Optics.

  6. Outline: ◮ Inverse conductivity problem ◮ Motivation of the edge detection problem ◮ Complex principal type operators ◮ Beltrami equation ◮ Single scattering and connection to Radon transform ◮ Analysis of higher order terms

  7. Edge detection in Electric Impedance Tomography Our aim in this talk is to determine jumps of the conductivity function. In particular, we want to determine locations of several jump surfaces in the presence of smooth, unknown background conductivity.

  8. Why a new edge detection method? Our main motivation: brain strokes imaging. • ischemic stroke: lower conductivity. Left: MRI image of ischemia (Hellerhoff 2010) • haemorrhagic stroke: higher conductivity. Challenges: • low conductive skull layer, • unknown background. Some existing work: • (Shi et al, 2009) experimental study on rhesus monkeys, • (Malone et al., 2014) simulated multi-frequency data.

  9. Outline: ◮ Inverse conductivity problem ◮ Motivation of the edge detection problem ◮ Complex principal type operators ◮ Beltrami equation ◮ Single scattering and connection to Radon transform ◮ Analysis of higher order terms { p R , p I } = 0

  10. Solutions in complex geometrical optics Let Ω = B ( 0 , 1 ) ⊂ R 2 and σ : R 2 → R , 0 < c 0 ≤ σ ( x ) ≤ c 1 , supp ( σ ) ⊂ Ω . Let us consider x = ( x 1 , x 2 ) ∈ R 2 . ∇ · ( σ ( x ) ∇ u ( x )) = 0 , (1) Let η = η R + i η I ∈ C 2 with | η R | = | η I | and τ ∈ R . We consider solutions of (1) of the form u ( x ) = e i τη · x v ( x , τ ) .

  11. Since u ( x ) = e i τη · x v ( x , τ ) satisfies the conductivity equation, 1 0 = σ ( x ) ∇ · ( σ ( x ) ∇ u ( x )) = (∆ + 1 σ ( ∇ σ ) · ∇ )( e i τη · x v ( x , τ )) � � ∆ v ( x , τ ) + 2 i τη · ∇ v ( x , τ ) + ( 1 e i τη · x = σ ∇ σ ) · ( ∇ + i τη ) v ( x , τ )

  12. Equation in time-domain Let � v ( x , t ) = F τ → t ( v ( x , τ )) be the Fourier transform of v ( x , τ ) in the τ variable, that is, � e − it τ v ( x , τ ) d τ. � v ( x , t ) = F τ → t v ( x , t ) = R We say that t is the pseudo-time corresponding to the complex frequency τ . The Fourier transform of the equation ∆ v ( x , τ ) + 2 i τη · ∇ v ( x , τ ) + ( 1 σ ∇ σ ) · ( ∇ + i τη ) v ( x , τ ) = 0 . is w ( x , t ) + 2 η ∂ v ( x , t ) + ( 1 σ ∇ σ ) · ( ∇ + η ∂ ∆ � ∂ t · ∇ � ∂ t ) � v ( x , t ) = 0 .

  13. The principal part of the equation v ( x , t ) + 1 v ( x , t ) + 2 η ∂ σ ( ∇ σ ) · ( ∇ + η ∂ ∆ � ∂ t · ∇ � ∂ t ) � v ( x , t ) = 0 is given by the operator � = P R + i P I = ∆ + 2 η ∂ � ∂ t · ∇ where η = η R + i η I and ∂ P R = ∆ + 2 η R ∂ t · ∇ ∂ P I = 2 η I ∂ t · ∇ They have symbols p R ( x , t , ξ, τ ) = ξ 2 1 + ξ 2 2 + 2 τη R · ξ p I ( x , t , ξ, τ ) = 2 τη I · ξ

  14. Complex principal type operator Let p ( x , t , ξ, τ ) = p R + ip I be the symbol of � � = P R + i P I . The characteristic variety of p is { ( x , t , ξ, τ ) ∈ T ∗ R 3 \ 0 ; p ( x , t , ξ, τ ) = 0 } . Σ = On Σ the Poisson brackets of p R ( x , t , ξ, τ ) and p I ( x , t , ξ, τ ) satisfy { p R , p I } = ( ∂ x p R · ∂ ξ p I + ∂ t p R · ∂ τ p I ) − ( ∂ ξ p R · ∂ x p I + ∂ τ p R · ∂ t p I ) = 0 and the differentials dp R and dp I are linearly independent on Σ . This implies that � � = P R + i P I is a complex principal type operator.

  15. Propagation of singularities By Hörmander-Duistermaat 1972, for a real principal type operator, e.g the wave operator � g , there are invertible Fourier Integral Operators A 1 and A 2 such that ∂ ( y 1 , y 2 , . . . , y 2 d ) ∈ R 2 d . � g = A 1 A 2 , ∂ y 1 For the wave equation the light-like singularities propagate along light rays. For the complex principal type operator � � = P R + i P I there are invertible Fourier integral operators A 1 and A 2 such that � = A 1 ( ∂ + i ∂ � ( y 1 , y 2 , . . . , y 2 d ) ∈ R 2 d . ) A 2 , ∂ y 1 ∂ y 2 For � � , singularities propagate along two dimensional surfaces, called strips. For example, if � � G ( x , t ) = δ ( x , t ) then G ( x , t ) is singular on planes t = 0 and 2 η R · x + t = 0.

  16. In this talk, our aim is to consider propagation and reflection of singularities in Complex Geometric Optics. In figure below the magenta plane wave hits to the blue surface and causes reflected light blue waves. Next we explain this in detail for equation ∇ · σ ∇ u = 0.

  17. Outline: ◮ Inverse conductivity problem ◮ Motivation of the edge detection problem ◮ Complex principal type operators ◮ Beltrami equation ◮ Single scattering and connection to Radon transform ◮ Analysis of higher order terms 2 ( ∂ ∂ x + i ∂ 2 ( ∂ ∂ x − i ∂ ∂ z = 1 ∂ z = 1 ∂ y ) , ∂ y ) , z = x + iy ∈ C

  18. Complex formulation of conductivity equation We denote z = x 1 + ix 2 ∈ C and identify R 2 with C . We use complex frequency k = τθ where τ ∈ R and θ ∈ C , | θ | = 1. In Astala-Päivärinta 2006, solutions for ∇ · σ ∇ u = 0 are written using the real-linear Beltrami equation, ∂ z f µ ( z , k ) = µ ( z ) ∂ z f µ ( z , k ) , z ∈ C , e ikz ( 1 + O ( | z | − 1 )) as | z | → ∞ . f µ ( z , k ) = Here, the Beltrami coefficient µ ( z ) is defined by µ ( z ) = 1 − σ ( z ) 1 + σ ( z ) . µ is a supported in Ω and � µ � L ∞ ( C ) < 1. The function u = Re ( f µ ) + i Im ( f − µ ) satisfies ∇ · σ ∇ u = 0. The map Λ σ determines f ± µ ( z , k ) for z ∈ C \ Ω .

  19. Notations Let us write the Complex Geometrical Optics solutions of the Beltrami equation ∂ z f µ = µ ∂ z f µ in the form of f = ‘incident wave’ + ‘scattered wave’, f µ ( z , k ) = e ikz v ( z , k ) , v ( z , k ) = 1 + v sc ( z , k ) , v sc ( z , k ) = O ( | z | − 1 ) , as z → ∞ . Let e k ( z ) = e i ( kz + kz ) = e i 2Re ( kz ) , so that | e k ( z ) | = 1 and e k ( z ) = e − k ( z ) .

  20. − 1 The solid Cauchy transform ∂ is z � f ( z ′ ) 1 − 1 z − z ′ d 2 z ′ , ∂ z f ( z ) = π C For any k ∈ C , the scattered wave v sc ( z , k ) satisfies v sc ( z , k ) = O ( | z | − 1 ) ∂ z v sc − µ e − k ( ∂ z + ik ) v sc = − ike − k µ, that yields the Lippmann-Schwinger type equation � � − 1 I − A v sc = − ik ∂ z ( e − k µ ) , where − 1 A v = ∂ z ( e k µρ ( ∂ z + ik ) v ) and ρ ( f ) := f denotes complex conjugation.

  21. Neumann series � � − 1 Using I − A v sc = − ik ∂ z ( e − k µ ) we can write � ∞ − 1 v sc ∼ v n , v 1 = − ik ∂ z ( e − k µ ) , v n + 1 = A v n . n = 1 More precisely, v n is the n:th Frechect derivative of the map V k : µ → v sc ( · , k ) , that is, v n = D n V k | 0 [ µ, µ, . . . , µ ] . Next we consider the term corresponding to single scattering, − 1 v 1 = − ik ∂ z ( e − k µ ) . The Dirichlet-to-Neumann map Λ σ determines v sc ( z , k ) for z ∈ ∂ Ω . The term v 1 ( z , k ) | z ∈ ∂ Ω determines singularities of µ . The terms v n ( z , k ) | z ∈ ∂ Ω , n ≥ 2 contribute ‘multiple scattering’, which explain artifacts in numerics.

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