Linearized internal functionals for anisotropic conductivities - PowerPoint PPT Presentation

Linearized internal functionals for anisotropic conductivities Chenxi Guo Chenxi Guo Chenxi Guo Chenxi Guo joint work with Guillaume Bal Guillaume Bal Guillaume Bal and Fran Guillaume Bal Fran Fran Franç ç ç çois Monard ois Monard ois Monard ois Monard Dept. of Applied Physics and Applied Mathematics, Columbia University. June 20th, 2012 UC Irvine Conference in honor of Gunther Uhlmann 1

1. Background n Ω = ∑ u | g −∇⋅ γ ∇ ≡ − ∂ γ ∂ = ij ( u ) ( u ) 0 Ω ( ) ∂ i j = i j , 1 γ is a real-valued symmetric positive definite tensor with bounded coefficients, satisfying a uniform ellipticity condition for some κ ≥ 1 ξ ∈ n 2 − R x ∈Ω κ 1 ξ ≤ ξ γ ⋅ ξ ≤ κ ξ ( ) x • Conductivity equation: rules the equilibrium distribution of the electrostatic potential u inside the domain , in response to a prescribed boundary voltage g. Ω Electrical Impedance Tomography(EIT). → γ ∇ ⋅ u v | Calderon’s problem ∂Ω • Internal measurement = ∇ ⋅ γ ∇ H [ ]( ): g x u x ( ) ( ) x u x ( ) power density of a solution u γ Application: Hybrid imaging How to construct power densities: Ammari et al. (2008), Kuchment-Kunyansky (2010) 2

−∇ ⋅ γ ∇ = Ω ∂Ω = = ∇ ⋅ γ ∇ ( u ) 0 ( ) u | g H [ ]( ) : g x u x ( ) ( ) x u x ( ) γ History (non-linear case): 2 = σ ∇ γ = σ H u I • Isotropic case from only one measurement n H Ω = Ω ∇⋅ ∇ = u | g ( u ) 0 ( ) ∂ 2 ∇ u • Newton-based method see (Gebauer and Scherzer (2009)) • Theoretically, by Bal (2012). 3

The conductivity equation u | ∇ ⋅ γ ∇ = = ( ' u ) 0 g ( 1 ) ( Ω ), � � ∂ Ω known unknown g ≡ g 1 , , ⋯ g ' g u solving (1) with Fixing a few boundary condtions with m i i γ → γ = γ = γ ∇ ⋅ ∇ ' ' H : ' H ( ' ) H ( ' ) ' u u ≤ , ≤ Measurement operator : 1 i j m ij i j γ from ' Problem: recover H • Isotropic, 2 dimensional setting, see (Capdeboscq et al. (2009)) • Explicit reconstruction using a large number of functionals in an isotropic case in dimension 3. see (Bal,Monard, Bonnetier and Triki) • Generalized to dimension n, isotropic tensor with more general type of 2 measurements σ α ∇ 2 u 1 where α not necessary 2 see Monard and Bal (2012). • reconstruction formulas for the anisotropic two-dimensional problem see Monard and Bal (2012). 4

Linearization of the problem The partial differential equation u | = ∇ ⋅ γ ∇ = g ( 1 ) ( ' u ) 0 ( Ω ), � � ∂ Ω known unknown g ≡ g 1 , , ⋯ g u solving (1) with ' g Fix a few boundary conditions with m i i γ → γ = γ = γ ∇ ' ⋅ ∇ ' H : ' H ( ' ) H ( ' ) ' u u ≤ , ≤ Measurement operator : 1 i j m ij i j γ ' Non-linear problem: recover from H γ = γ + εγ + ο ε 2 Fréchet derivative: ' ( ) � 0 known = + ε + ο ε ' 2 u u v ( ) i i i ( ε o ( 1 ) o ) PDE (1) of order and = − ∇ ⋅ γ ∇ = u | g ( u ) 0 ( Ω ), ∂ Ω � i i � 0 i known known = − ∇ ⋅ γ ∇ = ∇ ⋅ γ ∇ ( Ω v | 0 ( v ) ( u ) ), � ∂ Ω i 0 i i unknown 5

= γ ∇ ⋅ ∇ + ε γ ∇ ⋅ ∇ + γ ∇ ⋅ ∇ + γ ∇ ⋅ ∇ + ο ε 2 The measurements look like H u u ( u u u v u v ) ( ) ij 0 i j i j 0 i j 0 j i ↓ = γ ∇ ⋅ ∇ + γ ∇ ⋅ ∇ + γ ∇ ⋅ ∇ Linearized measurements dH u u u v u v � ij i j 0 i j 0 j i known Linearized problem: recover γ from dH ij References on the linearized problem • isotropic case in dimension 2 and 3 with numerical implementation see (Kuchment and Kunyansky (2011)) • isotropic case, studied using pseudo-differential calculus, inversion modulo a compact operator. see (Kuchment and Steinhauer (2011)) 6

2.Microlocal inversion of Study of the principal symbol ξ dH M ij ( x , ) ij Recall the equations and measurements: − ∇ ⋅ γ ∇ = = ( Ω ( u ) 0 u | g ), � � ∂ Ω 0 i i i known known ( Ω = ), v | 0 − ∇ ⋅ γ ∇ = ∇ ⋅ γ ∇ ( v ) ( u ) ∂ Ω i � 0 i i unknown = γ ∇ ⋅ ∇ + γ ∇ ⋅ ∇ + γ ∇ ⋅ ∇ dH u u u v u v � ij i j 0 i j 0 j i known = −∇ ⋅ γ ∇ = − ∇ ⋅ γ ∇ Ω Denote L : ( ) 1 v L ( ( u )) 0 0 i 0 i γ supp suppose γ compactly supported inside Ω = − ∇ ⋅ γ ∇ 1 v L ( ( u )) into Insert and express as a pseudo-DO dH i 0 i ij ∫∫ γ = π − ξ ⋅ − ξ + ξ γ ξ n i ( x y ) dH ( , x ) ( 2 ) e ( M ( x , ) M | ( x , )) : ( y ) d d y − ij ij ij 1 n R n × R − 0 1 ξ = ο ξ ξ = ο ξ M ij ( x , ) ( ) M ij | ( x , ) ( ) − 1 ↓ ↓ pricipal symbol the symbol of order -1 7

∫∫ γ = π − ξ ⋅ − ξ + ξ γ ξ n i ( x y ) dH ( , x ) ( 2 ) e ( M ( x , ) M | ( x , )) : ( y ) d d y − ij ij ij 1 n R × n R − 0 1 ξ = ο ξ ξ = ο ξ M ij ( x , ) ( ) M ij | ( x , ) ( ) − 1 ↓ ↓ the symbol of order -1 pricipal symbol = Goal: determine under which conditions the operator dH { dH } ≤ ≤ ij 1 i , j m is an elliptic pseudo-differential operator ξ = Conclusion: with only principal symbols , M ij ( x , ) dH { dH } will never ≤ ≤ ij 1 i , j m be elliptic, no matter how large m ∧ 1 = ∇ = γ ξ = ξ ξ Define A ( ) and V : A u 0 : A rewrite the principal symbol M ij ( x , ) 2 i 0 i 0 0 0 ~ ξ = ξ M ij ( x , ) A M ( x , ) A 0 ij 0 = − ξ ⋅ ξ − ξ ⋅ ξ V ⊙ V ( V ) ⊙ V ( V ) ⊙ V i j 0 i 0 j 0 j 0 i 1( Notation: = ⊗ + ⊗ U ⊙ V U V V U ) 2 8

Lemma: i , and vector fields = ∇ For any j V , i V defined as above V : A u the symbol j i 0 i ~ ξ satisfies M ij ( x , ) ~ η ⊥ ξ − η ∈ n 1 x ξ ξ η = S for all and M ( , ): ⊙ 0 ij 0 0 M can never control a subspace of − Conclusion: the S n ( R ) of dimension n 1 ij Basic Hypothesis: the gradients ∇ n { u } form a frame in n R = i i 1 Lemma: n { V } Suppose that the vector fields = form a basis of If a matrix ∈ P S ( R ) n R i i 1 n ~ ξ = ≤ ≤ ≤ M x ( , ): P 0 1 i j n ij Then P is of the form ⊙ for some vector η satisfying P ξ = η η ⊥ ξ 0 0 Conclusion: the only only only only directions that are not controlled by the principal symbols are P ξ = η η ⊥ ξ ⊙ 0 0 9

dH The second term of ij ∫∫ − ξ ⋅ − γ = π ξ + ξ γ ξ n i ( x y ) dH ( , x ) ( 2 ) e ( M ( x , ) M | ( x , )) : ( y ) d d y − ij ij ij 1 n R n × R = ξ + ξ dH { dH } M ( x , ) M | ( x , ) Goal: invert microlocally from ≤ ≤ ij 1 i , j m − ij ij 1 ~ γ is constant • case ξ = ξ M | 1 ( x , ) A M | ( x , ) A ij − 0 ij − 0 1 0 − 1 = ξ − ξ ⋅ − ξ ⊗ ξ + ⊗ ξ A 1[ H (( V )( I 2 ) V ) 0 i 0 j n 0 0 j 0 + ξ ⋅ − ξ ⊗ ξ + ⊗ ξ ] sym H (( V )( I 2 ) V j 0 i n 0 0 i 0 = ∇ 2 H A u A where i 0 i 0 Lemma γ is constant, pick Suppose u = ≤ i ≤ and add an additional solution 1 n x 0 i i denoted by with full-rank Hessian If there exist such ∇ 2 ∈ u u P Q , S ( ) R + + n 1 n 1 n that for all ≤ ≤ 1 i j , n ~ ~ + ξ + − = ( M M | )( , ) :( x P 1 ) Q 0 ij − ij 1 ~ ~ + + ξ + − = ( M M | )( , ):( x P 1 ) Q 0 + i n , 1 − i n , 1 1 then = = P Q 0 10

γ is not constant • case 0 Basic Hypotheses − ∇ ⋅ γ ∇ = ∇ ∇ = ( u ) 0 u | g ( u , ⋯ u ) form a frame • � � ∂ Ω 0 i i i 1 n known known n ∑ u + µ ∇ +∇ = • Add one addition such that u u 0 + n 1 j j n 1 j we construct the matrix assume Z to be invertible = ∇ µ ∇ µ Z [ , ⋯ , ] 1 n Remark: in the case constant, the above hypothesis is automatically γ 0 satisfied, by choosing ≤ ≤ = 1 i n u x i i 1 + = = t invertible u x Qx tr ( Q ) 0 Q n 1 2 ∫∫ γ = π − ξ ⋅ − ξ + ξ γ ξ n i ( x y ) dH ( , x ) ( 2 ) e ( M ( x , ) M | ( x , )) : ( y ) d d y Recall − ij ij ij 1 n R n × R ~ Goal: under the above hypotheses, M | can control all bad directions ij − 1 ξ η ⊙ 0 11