Recovering discontinuous conductivity from internal current : case - PowerPoint PPT Presentation

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography Pierre Millien DMA, Ecole Normale Sup´ erieure de Paris October 6, 2014 Joint work with H. Ammari, P. G Grasland-Mongrain, and L. Seppecher projet ERC : MULTIMOD

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography Position of the problem Electrical impedance tomography Cheap Low side effects Good differentiation of soft tissues Good differentiation of pathological state Poor resolution (ill posed inverse problem) Ultrasound Imaging Cheap Low side effects Good resolution Poor differention of soft tissues Poor differentiation of pathological state

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography Goal Image conductivity map, especially the conductivity jumps in a medium with the resolution of ultrasound imaging.

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant 1 magnetic field ? The ultrasonically induced Lorenz force tomography Ionic description of the conductivity in aqueous tissues Boundary measurements From boundary measurements to meaningful internal data 2 Introduction of a virtual potential Deconvolution Geometric integral transform or asymptotic formula Recovering the conductivity from an internal current 3 By optimization By solving a transport equation

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography The experiment transducer (500 kHz) sample with electrodes magnet (300 mT) oil tank degassed water absorber

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? The ultrasonically induced Lorenz force tomography Be 3 Support du faisceau acoustique Γ 1 Ω y Γ 0 I ( t ) τ Γ 0 e 2 ξ e 1 Γ 2 e 3 Assumptions Ω mechanically homogeneous and is a conductive medium. Γ 1 and Γ 2 are perfect conductors. Γ 0 is a perfect isolator. B is constant.

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? The ultrasonically induced Lorenz force tomography Be 3 Support du faisceau acoustique Γ 1 Ω y Γ 0 I ( t ) τ Γ 0 e 2 ξ e 1 Γ 2 e 3 Velocity field For any x ∈ Ω, written x = y + z ξ + r with z > 0, r ∈ ξ ⊥ , v y ,ξ ( y + z ξ + r , t ) = A ( z , | r | ) w ( z − ct ) ξ

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? The ultrasonically induced Lorenz force tomography Be 3 Support du faisceau acoustique Γ 1 Ω y Γ 0 I ( t ) τ Γ 0 e 2 ξ e 1 Γ 2 e 3 As Ω is electrically neutral, can we explain the origin of the current measured at the electrodes ?

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? Ionic description of the conductivity in aqueous tissues Assume that Ω is an electrolyte medium (saline gel, living tissues,. . . ) the conductivity phenomenon is due to the presence of ions. Assume that we have N types of ions of charge q i and volume density n i ( x ), i ∈ { 1 , . . . , N } . We have, for any x ∈ Ω Neutrality � q i n i ( x ) = 0 i Kolhrausch’s law σ ( x ) = e + � µ i q i n i ( x ) i with µ i ∈ R , satisfying µ i q i > 0 is called the ionic mobility and e + is the elementary charge.

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? Ionic description of the conductivity in aqueous tissues We can understand now understand the source of current as the deviation of the ions by the magnetic field B . Consider an ion i at position x at time t . The acoustic beam imposes to it a velocity in the direction ξ : v ( x , t ) ξ. The Lorentz force applied to i is F i = q i v ξ × Be 3 and the ion get almost immediately an additional drift speed v d , i = µ i F i = B µ i v τ q i where τ = ξ × e 3 . At first order in the displacement length, its total velocity is v i = v ξ + B µ i v τ. Defining the current as the total amount of charges displacement, i n i µ i q i ) v τ = B j S = � i n i q i v i = ( � i n i q i ) v ξ + B ( � e + σ v τ .

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? Boundary measurements The interaction between the velocity field v ( x , t ) ξ and the magnetic field Be 3 create a source of current j S ( x , t ) = B e + σ ( x ) v ( x , t ) τ Our measure is the indirect effect of j S on the boundary. Assume that the electromagnetic propagation is much faster than the acoustic propagation, we adopt the electrostatic approximation. j = j S + σ ∇ u satisfying ∇ · j = 0 then the potential satisfies at a fixed time t , −∇ · ( σ ∇ u ) = ∇ · j S in Ω

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography How to create currents with an acoustic beam and a constant magnetic field ? Boundary measurements Be 3 support of j S support of the acoustic beam Γ 1 Ω y Γ 0 I ( t ) τ Γ 0 e 2 ξ e 1 Γ 2 e 3  −∇ · ( σ ∇ u ) = ∇ · j S in Ω   u : u = 0 on ∂ Γ1 ∪ Γ 2  ∂ ν u = 0 on Γ 0  The intensity that we measure is � I = σ∂ ν u Γ 2

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography From boundary measurements to meaningful internal data Introduction of a virtual potential In order to understand the measurements, we multiply the potential equation by a well chosen test function U called virtual potential defined by  −∇ · ( σ ∇ U ) = 0 in Ω    U = 0 on Γ1  U = 1 on Γ2     ∂ ν U = 0 on Γ 0 and through integration by part it comes � � j S · ∇ U = B I = v ( x , t ) σ ( x ) ∇ U ( x ) dx · τ e + Ω Ω and we define the measurments function as � x , z � � M y ,ξ ( z ) = v y ,ξ σ ( x ) ∇ U ( x ) dx · τ ξ c Ω

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography From boundary measurements to meaningful internal data Introduction of a virtual potential The inverse problem posed by this hybrid method is Inverse problem Find σ : Ω → R from the knowledge of � x , z � � M y ,ξ : z → v y ,ξ σ ( x ) ∇ U ( x ) dx · τ ξ c Ω known for any y ∈ Y ⊂ R d and ξ ∈ Θ ⊂ S d − 1 In general, Y is supposed to be a bounded smooth surface of R d . Idea If Y and Θ are well chosen, we show that the virtual current J ( x ) = ( σ ∇ U )( x ) can be recovered.

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography From boundary measurements to meaningful internal data Deconvolution Step 1 : Deconvolution = w ( z ′ − z ) A ( z ′ , | r | ) we rewrite the y + z ′ ξ + r , z � � As v y ,ξ c measurments M y , ξ as M y ,ξ ( z ) = ( w ∗ Φ y ,ξ )( z ) where � Φ y ,ξ ( z ) = ξ ⊥ ( σ ∇ U )( y + z ξ + r ) A ( z , | r | ) dr · τ ξ To recover Φ y ,ξ with stability, we need short pulses and/or changes of the frequency. To recover the largest spectral band in the Fourrier domain.

Recovering discontinuous conductivity from internal current : case of the ultrasonically-induced Lorentz force electrical impedance tomography From boundary measurements to meaningful internal data Geometric integral transform or asymptotic formula Step 2 : Getting the current Once we know � Φ y ,ξ ( z ) = ξ ⊥ ( σ ∇ U )( y + z ξ + r ) A ( z , | r | ) dr · τ ξ we can notice that it looks like a weighted Radon transform of the current density. If we assume that the support of A is thin, � Φ y ,ξ ( z ) = ( σ ∇ U )( y + z ξ ) ξ ⊥ A ( z , | r | ) dr · τ ξ + O ( R ) where R is such that supp( ρ �→ A ( z , ρ )) ⊂ [0 , R ] and with a remainder depending on | σ ∇ U | TV (Ω) . Finally, choosing x ∈ Ω and consider Φ y ,ξ ( z ) for any ( y , ξ, z ) such that x = y + z ξ we reconstruct J ( x ) = ( σ ∇ U )( x )