Microlocal Analysis and Spectral Theory in honor of J. Sjöstrand Luminy, Sept 26, 2013 Two Partial Data Problems Gunther Uhlmann University of Washington and University of Helsinki
Outline: ◮ Calderón’s problem with partial data ◮ Travel time tomography with partial data
CALDER ´ ON’S PROBLEM Ω ⊂ R n ( n = 2 , 3) Can one determine the electrical conductivity of Ω , γ ( x ) , by making voltage and current measurements at the boundary? (Calder´ on; Geophysical prospection) Early breast cancer detection Normal breast tissue 0.3 mho Cancerous breast tumor 2.0 mho 1
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REMINISCENCIA DE MI VIDA MATEMATICA Speech at Universidad Aut´ onoma de Madrid accepting the ‘Doctor Honoris Causa’: My work at “Yacimientos Petroliferos Fiscales” (YPF) was very interesting, but I was not well treated, otherwise I would have stayed there. 3
ACT3 imaging blood as it leaves the heart (blue) and fills the lungs (red) during systole.
(Loading DBarPerfMovie1.avi) Thanks to D. Issacson
CALDER ´ ON’S PROBLEM (EIT) Consider a body Ω ⊂ R n . An electrical potential u ( x ) causes the current I ( x ) = γ ( x ) r u The conductivity γ ( x ) can be isotropic, that is, scalar, or anisotropic, that is, a matrix valued function. If the current has no sources or sinks, we have div( γ ( x ) r u ) = 0 in Ω 11
div( γ ( x ) r u ( x )) = 0 γ ( x ) = conductivity, � � u � @ Ω = f f = voltage potential at @ Ω � � Current flux at @ Ω = ( ν · γ r u ) � @ Ω were ν is the unit outer normal. Information is encoded in map � � Λ γ ( f ) = ν · γ r u � @ Ω EIT (Calder´ on’s inverse problem) Does Λ γ determine γ ? Λ γ = Dirichlet-to-Neumann map 12
Theorem n ≥ 3 (Sylvester-U, 1987) γ ∈ C 2 (Ω) , 0 < C 1 ≤ γ ( x ) ≤ C 2 on Ω Λ γ 1 = Λ γ 2 ⇒ γ 1 = γ 2 • Extended to γ ∈ C 3 / 2 (Ω) (Päivarinta-Panchenko-U, Brown-Torres, 2003) • γ ∈ C 1 + ǫ (Ω) , γ conormal (Greenleaf-Lassas-U, 2003) • γ ∈ C 1 (Ω) , (Haberman-Tataru, 2013). Complex-Geometrical Optics Solutions (CGO) • Reconstruction (A. Nachman, R. Novikov 1988) • Stability (G. Alessandrini 1988) • Numerical Methods (D. Issacson, J. Müller, S. Siltanen)
Reduction to Schr¨ odinger equation div( γ r w ) = 0 u = p γw Then the equation is transformed into: (∆ − q ) u = 0 , q = ∆ p γ � � n @ 2 X ∆ = p γ @x 2 i =1 i (∆ − q ) u = 0 � � u � @ Ω = f � Define Λ q ( f ) = @u � � @ Ω @ν ν = unit-outer normal to @ Ω. 16
IDENTITY Z Z � � � � � � Ω ( q 1 − q 2 ) u 1 u 2 = (Λ q 1 − Λ q 2 ) u 1 u 2 � @ Ω dS � @ Ω @ Ω (∆ − q i ) u i = 0 If Λ γ 1 = Λ γ 2 ) Λ q 1 = Λ q 2 and Z Ω ( q 1 − q 2 ) u 1 u 2 = 0 GOAL: Find MANY solutions of (∆ − q i ) u i = 0. 17
CGO SOLUTIONS on: Let ρ 2 C n , ρ · ρ = 0 Calder´ η, k 2 R n , | η | = | k | , η · k = 0 ρ = η + ik u = e x · ρ = e x · η e ix · k 8 > > exponentially decreasing , x · η < 0 > < ∆ u = 0, u = oscillating, x · η = 0 > > > : exponentially increasing , x · η > 0 18
COMPLEX GEOMETRICAL OPTICS (Sylvester-U) n ≥ 2 , q 2 L 1 (Ω) Let ρ 2 C n ( ρ = η + ik, η, k 2 R n ) such that ρ · ρ = 0 ( | η | = | k | , η · k = 0). Then for | ρ | sufficiently large we can find solutions of (∆ − q ) w ρ = 0 on Ω of the form w ρ = e x · ρ (1 + Ψ q ( x, ρ )) with Ψ q ! 0 in Ω as | ρ | ! 1 . 19
Proof Λ q 1 = Λ q 2 ) q 1 = q 2 Z Ω ( q 1 − q 2 ) u 1 u 2 = 0 u 1 = e x · ρ 1 (1 + Ψ q 1 ( x, ρ 1 )) , u 2 = e x · ρ 2 (1 + Ψ q 2 ( x, ρ 2 )) ρ 1 · ρ 1 = ρ 2 · ρ 2 = 0 , ρ 1 = η + i ( k + l ) ρ 2 = − η + i ( k − l ) | η | 2 = | k | 2 + | l | 2 η · k = η · l = l · k = 0 , Z Ω ( q 1 − q 2 ) e 2 ix · k (1 + Ψ q 1 + Ψ q 2 + Ψ q 1 Ψ q 2 ) = 0 Z Ω ( q 1 − q 2 ) e 2 ix · k = 0 Letting | l | ! 1 8 k = ) q 1 = q 2 20
PARTIAL DATA PROBLEM Suppose we measure supp f ⊆ Γ 0 Λ γ ( f ) | Γ , Γ, Γ 0 open subsets of @ Ω Can one recover γ ? Important case Γ = Γ 0 . 23
EXTENSION OF CGO SOLUTIONS u = e x · ρ (1 + Ψ q ( x, ρ )) ρ 2 C n , ρ · ρ = 0 (Not helpful for localizing) Kenig-Sj¨ ostrand-U (2007), u = e τ ( ' ( x )+ i ψ ( x )) ( a ( x ) + R ( x, τ )) τ 2 R , ' , ψ real-valued, R ( x, τ ) ! 0 as τ ! 1 . ' limiting Carleman weight, r ' · r ψ = 0 , |r ' | = |r ψ | Example: ' ( x ) = ln | x − x 0 | , x 0 / 2 ch (Ω) 24
CGO SOLUTIONS u = e τ ( ' ( x )+ i ψ ( x )) ( a 0 ( x ) + R ( x, τ )) R ( x, τ ) τ !1 ! 0 in Ω − ' ( x ) = ln | x − x 0 | Complex Spherical Waves Theorem (Kenig-Sj¨ ostrand-U) Ω strictly convex. � � � � Λ q 1 � Γ = Λ q 2 � Γ , Γ ⊆ @ Ω , Γ arbitrary ) q 1 = q 2 25
Complex Spherical Waves (Loading reconperfect1.mpg)
Theorem (Kenig-Sj¨ ostrand-U) Ω strictly convex. � � � � Λ q 1 � Γ = Λ q 2 � Γ , Γ ⊆ @ Ω , Γ arbitrary ) q 1 = q 2 u τ = e τ ( ' + i ψ ) a τ ' ( x ) = ln | x − x 0 | , x 0 2 \ ch (Ω) Eikonal: r ' · r ψ = 0 , |r ' | = |r ψ | ψ ( x ) = d ( x − x 0 | x − x 0 | , ! ) , ! 2 S n − 1 : smooth for x 2 ¯ Ω. Transport: ( r ' + i r ψ ) · r a τ = 0 (Cauchy-Riemann equation in plane generated by r ' , r ψ ) 27
' ( x ) = ln | x − x 0 | , x 0 2 \ ch (Ω) Carleman Estimates > u | @ Ω = @ u @ν | @ Ω − = 0 @ Ω ± = { x 2 @ Ω; r ' · ν < 0 } Z Z < r ' , ν > | e − τ' ( x ) @ u @ν | 2 ds ≤ C Ω | (∆ − q ) ue − τ' ( x ) | 2 ds @ Ω + τ This gives control of @ u @ν | @ Ω + , δ , @ Ω + , δ = { x 2 @ Ω , r ' · ν ≥ δ } 28
Outline: ◮ Calderón’s problem with partial data ◮ Travel time tomography with partial data
Travel Time Tomography (Transmission) Global Seismology Inverse Problem: Determine inner structure of Earth by measuring travel time of seismic waves. 1
Tsunami of 1960 Chilean Earthquake Black represents the largest waves, decreasing in height through purple, dark red, orange and on down to yellow. In 1960 a tongue of massive waves spread across the Pacific, with big ones through- out the region. 2
Human Body Seismology ULTRASOUND TRANSMISSION TOMOGRAPHY(UTT) Z 1 T = = Travel Time (Time of Flight) . c ( x ) ds γ 3
THIRD MOTIVATION OCEAN ACOUSTIC TOMOGRAPHY Ocean Acoustic Tomography Ocean Acoustic Tomography is a tool with which we can study average temperatures over large regions of the ocean. By measur- ing the time it takes sound to travel between known source and receiver locations, we can determine the soundspeed. Changes in soundspeed can then be related to changes in temperature. 4
REFLECTION TOMOGRAPHY Scattering Obstacle Points in medium 5
REFLECTION TOMOGRAPHY Oil Exploration Ultrasound 6
TRA VELTIME TOMOGRAPHY (Transmission) Motivation:Determine inner structure of Earth by measuring travel times of seismic waves Herglotz, Wiechert-Zoeppritz (1905) Sound speed c ( r ), r = | x | � � d r > 0 dr c ( r ) Reconstruction method of c ( r ) from lengths of geodesics 7
ds 2 = 1 c 2 ( r ) dx 2 More generally ds 2 = c 2 ( x ) dx 2 1 Velocity v ( x, ξ ) = c ( x ) , | ξ | = 1 (isotropic) Anisotropic case n X g = ( g ij ) is a positive defi- ds 2 = g ij ( x ) dx i dx j nite symmetric matrix i,j =1 qP n i,j =1 g ij ( x ) ξ i ξ j , Velocity v ( x, ξ ) = | ξ | = 1 g ij = ( g ij ) − 1 The information is encoded in the boundary distance function 8
More general set-up ( M, g ) a Riemannian manifold with boundary (compact) g = ( g ij ) x, y 2 @ M d g ( x, y ) = inf L ( σ ) σ (0)= x σ (1)= y L ( σ ) = length of curve σ r R 1 P n d σ j i,j =1 g ij ( σ ( t )) d σ i L ( σ ) = dt dt 0 dt Inverse problem Determine g knowing d g ( x, y ) x, y 2 @ M 9
dg ) g ? (Boundary rigidity problem) Answer NO ψ : M ! M di ff eomorphism � � � @ M = Identity ψ d ψ ∗ g = d g � D ψ ◦ g ◦ ( D ψ ) T � ψ ∗ g = ◦ ψ r R 1 P n d σ j i,j =1 g ij ( σ ( t )) d σ i L g ( σ ) = dt dt 0 dt σ = ψ ◦ σ L ψ ∗ g ( e σ ) = L g ( σ ) e 10
d ψ ∗ g = d g Only obstruction to determining g from d g ? No d g ( x 0 , @ M ) > sup x,y 2 @ M d g ( x, y ) Can change metric near SP 11
Def ( M, g ) is boundary rigid if ( M, e g ) satisfies d e g = d g . � � Then 9 ψ : M ! M di ff eomorphism, ψ � @ M = Identity, so that g = ψ ∗ g e Need an a-priori condition for ( M, g ) to be boundary rigid. One such condition is that ( M, g ) is simple 12
DEF ( M, g ) is simple if given two points x, y 2 @ M , 9 ! geodesic joining x and y and @ M is strictly convex CONJECTURE ( M, g ) is simple then ( M, g ) is boundary rigid ,that is d g determines g up to the natural obstruction. ( d ψ ∗ g = d g ) ( Conjecture posed by R. Michel, 1981 ) 13
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