Summer School on Mathematical Physics Inverse Problems ...smp2015/Gunther_Uhlmann_lecture_1.pdf · Summer School on Mathematical Physics Inverse Problems: Visibility and Invisibility

Summer School on Mathematical Physics

Inverse Problems Visibility and Invisibility

Lecture I

Gunther Uhlmann

University of Washington CMM (Chile)

HKUST (Hing Kong) amp University of Helsinki

Valparaiso Chile August 2015

Inverse Boundary Problems

Can one determine the internal properties of a medium by making

measurements outside the medium (non-invasive)

X-ray tomography (CAT-scans)

Problem Can we recover the density from attenuation of X-rays

1

Radon (1917) n = 2

f(x) = Unknown function

Idetector = eminusintL fIsource

Rf(s θ) = g(s θ) =int〈xθ〉=s

f(x)dH =intLf

f(x) =1

4π2pv

intS1dθint d

dsg(s θ)ds

〈x θ〉 minus s

2

LINEAR (No Scattering)

X-ray tomography (CT)

PET MRI

3

NONLINEAR (Scattering)

Ultrasound

Electrical

Impedance

Tomography

(EIT)

4

Hybrid Methods

Superposition of 2 images each obtained with a single wave

One single wave in sensitive only to a given contrast

Ultrasound to bulk compressibility

Photoacoustic

ImagingOptical wave to dielectric permittivity

Thermoacoustic

Imaging

LF Electromagnetic wave to electrical

impedance conductivity

5

Photoacoustic Tomography

Photoacoustic Effect The sound of light

Picture from Economist

(The sound of light)

Graham Bell When

rapid pulses of light are

incident on a sample of

matter they can be ab-

sorbed and the resulting

energy will then be radi-

ated as heat This heat

causes detectable sound

waves due to pressure

variation in the surround-

ing medium

6

Thermoacoustic Tomography

Wikipedia

7

(Loading Melanoma3DMovieavi)

Lihong Wang (Washington U)

8

Melanoma3DMovieavi
Media File (videoavi)

Mathematical Model

First Step in PAT and TAT is to reconstruct H(x) from u(x t)|partΩtimes(0T )

where u solves

(part2t minus c2(x)∆)u = 0 on Rn times R+

u|t=0 = βH(x)

parttu|t=0 = 0

Second Step in PAT and TAT is to reconstruct the optical or

electrical properties from H(x) (internal measurements)

9

CALDERONrsquoS PROBLEM and EIT

Ω sub Rn

(n = 23)

Can one determine the electrical conductivity of Ω γ(x) by making

voltage and current measurements at the boundary

(Calderon Geophysical prospection)

Early breast cancer detection

Normal breast tissue 03 mhoCancerous breast tumor 20 mho

10

11

REMINISCENCIA DE MI VIDA MATEMATICA

Speech at Universidad Autonoma de Madrid accepting the lsquoDoctor

Honoris Causarsquo

My work at ldquoYacimientos Petroliferos Fiscalesrdquo (YPF) was

very interesting but I was not well treated otherwise I would

have stayed there

12

(Loading imagesrawlongmpg)

Mark Nelson httpnelsonbeckmanillinoisedu

13

rawlongmpg
Media File (videompeg)

(Loading imageselectrompg)


14

electrompg

15

16

17

Combine with mammography for early detection

RPI Group (D Isaacson)

Sensitivity = predicted to have cancer

Total that have cancertimes 100

Specificity = predicted NOT to have cancer

Total that do NOT have cancertimes 100

Results for Equivocal Mammograms (N = 273)

Mamm T-Scanalone Adjunctive

Sensitivity 60 82(Biopsy ps=50)Specificity 41 57(Biopsy neg=223)

18

Other Applications

- Non-destructive testing (corrosion cracks)

- Seepage of groundwater pollutants

- Medical Imaging (EIT)

Tissue Conductivity (mho)Blood 67Liver 28

Cardiac muscle 63 (longitudinal)23 (transversal)

Grey matter 35White matter 15

Lung 10 (expiration)04 (inspiration)

19

ACT3 imaging blood as it leaves the heart (blue) and fills the lungs (red) during

systole

20

(Loading DBarPerfMovie1avi)

Thanks to D Issacson

21

DBarPerfMovie1avi

CALDERONrsquoS PROBLEM (EIT)

Consider a body Ω sub Rn An electrical potential u(x) causes the

current

I(x) = γ(x)nablau

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)nablau) = 0 in Ω

22

div(γ(x)nablau(x)) = 0

u∣∣∣partΩ

= f

γ(x) = conductivity

f = voltage potential at partΩ

Current flux at partΩ = (ν middot γnablau)∣∣∣partΩ

were ν is the unit outer normal

Information is encoded in

map

Λγ(f) = ν middot γnablau∣∣∣partΩ

EIT (Calderonrsquos inverse problem)

Does Λγ determine γ

Λγ = Dirichlet-to-Neumann map

23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int

Ωγ(x)|nablau(x)|2dx

Qγ(f g) =int

Ωγ(x)nablau middot nablavdx

div(γ(x)nablav(x)) = 0

v∣∣∣partΩ

= g

24

div(γnablau) = 0u|partΩ = f

div(γnablav) = 0

v|partΩ = gΛγ(f) = γ

partu

partν

∣∣∣∣partΩ

Qγ(f g) =int

Ωγnablau middot nablavdx

A P Calderon On an inverse boundary value problem in Seminar on Nu-

merical Analysis and its Applications to Continuum Physics RIo de Janeiro 1980

Qγ(f) =int

Ωγ|nablau(x)|2dx =

intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+

Qγ+εh(f g)minusQγ(f g)

ε=int

Ωhnablau middot nablavdx

Case γ = 1 div(γnablau) = ∆u = 0

Linearized Problem Suppose we knowintΩhnablau middot nablavdx forall∆u = ∆v = 0

Can we recover h

26

Linearized problem at γ = 1intΩhnablau middot nablavdx data forall∆u = ∆v = 0

Can we recover h

u = exmiddotρ

v = eminusxmiddotρ ρ isin Cn ρ middot ρ = 0

ρ =η minus iξ

2 ρ middot ρ = 0 hArr |η| = |ξ| η middot ξ = 0

|ξ|2int

Ωheminusixmiddotξdx known

we can recover χΩh(ξ) therefore h on Ω

27

Theorem (Kohn-Vogelius 1984)

Assume γ isin Cinfin(Ω) From Λγ we can determine partαγ∣∣∣partΩ

forallα

Proof (Sylvester-U Lee-U)

Λγ is a pseudodifferential operator of order 1 (Calderon)

partΩ = xn = 0 locally

Coordinates x = (xprime xn) xprime isin Rnminus1

Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime

28


λγ(xprime ξprime) = γ(0 xprime)|ξprime|+ a0(xprime ξprime) + middot middot middot+ aj(xprime ξprime) + middot middot middot

with aj(xprime ξprime) pos homogeneous of degree minusj in ξprime

aj(xprime λξprime) = λminusjaj(x

prime ξprime) λ gt 0

Result From aj we can determine partjγpartνj

∣∣∣∣xn=0

29

Theorem n ge 3 (Sylvester-U 1987)

γ isin C2(Ω) 0 lt C1 le γ(x) le C2 on Ω

Λγ1 = Λγ2 rArr γ1 = γ2

bull Extended to γ isin C32(Ω) (Paivarinta-Panchenko-U Brown-Torres

2003)

bull γ isin C1+ε(Ω) γ conormal (Greenleaf-Lassas-U 2003)

bull γ isin C1(Ω) (Haberman-Tataru 2012)

Complex-Geometrical Optics Solutions (CGO)

bull Reconstruction A Nachman (1988)

bull Stability G Alessandrini (1988)

bull Numerical Methods (D Issacson J Muller S Siltanen)

30

Reduction to Schrodinger equation

div(γnablaw) = 0

u =radicγw

Then the equation is transformed into

(∆minus q)u = 0 q =∆radicγ

radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f

Define Λq(f) =partu

partν

∣∣∣partΩ

ν = unit-outer normal to partΩ

31

IDENTITY

intΩ

(q1 minus q2)u1u2 =intpartΩ

((Λq1 minus Λq2)u1

∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0

If Λγ1 = Λγ2 rArr Λq1 = Λq2 andintΩ

(q1 minus q2)u1u2 = 0

GOAL Find MANY solutions of (∆minus qi)ui = 0

32

CGO SOLUTIONS

Calderon Let ρ isin Cn ρ middot ρ = 0

ρ = η + ik η k isin Rn |η| = |k| η middot k = 0

u = exmiddotρ = exmiddotηeixmiddotk

∆u = 0 u =

exponentially decreasing x middot η lt 0

oscillating x middot η = 0

exponentially increasing x middot η gt 0

33

COMPLEX GEOMETRICAL OPTICS

(Sylvester-U) n ge 2 q isin Linfin(Ω)

Let ρ isin Cn (ρ = η + ik η k isin Rn) such that ρ middot ρ = 0

(|η| = |k| η middot k = 0)

Then for |ρ| sufficiently large we can find solutions of

(∆minus q)wρ = 0 on Ω

of the form

wρ = exmiddotρ(1 + Ψq(x ρ))

with Ψq rarr 0 in Ω as |ρ| rarr infin

34

Proof Λq1 = Λq2 rArr q1 = q2intΩ


u1 = exmiddotρ1(1 + Ψq1(x ρ1)) u2 = exmiddotρ2(1 + Ψq2(x ρ2))

ρ1 middot ρ1 = ρ2 middot ρ2 = 0 ρ1 = η + i(k + l)ρ2 = minusη + i(k minus l)

η middot k = η middot l = l middot k = 0 |η|2 = |k|2 + |l|2

intΩ

(q1 minus q2)e2ixmiddotk(1 + Ψq1 + Ψq2 + Ψq1Ψq2) = 0

Letting |l| rarr infinint

Ω(q1 minus q2)e2ixmiddotk = 0 forallk =rArr q1 = q2

35

APPLICATIONS

n ge 3 (∆minus q) = 0 Λq determines q

bull EIT Λγ determines γ

bull Optical Tomography (Diffusion Approximation)

iωU minusnabla middotD(x)nablaU + σa(x)U = 0 in Ω

U= Density of photons D=Diffusion Coefficient σa(x)= optical

absorption

RESULT bull If ω 6= 0 we can recover both D(x) and σa(x)bull If ω = 0 we can recover either D(x) or σa(x)

36

OTHER APPLICATIONS (Fixed energy)

bull Optics (∆ minus k2n(x))u = 0 n(x) isotropic index of refraction

(q(x) = k2n(x))

bull Acoustic div( 1ρ(x)nablap) + ω2κ(x)p = 0 ρ density κ compress-

ibility (need two frequencies ω)

bull Inverse quantum scattering at fixed energy (∆minus qminusλ2)u = 0

q potential

bull Maxwellrsquos Equation (Isotropic)

(Ola-Somersalo) Reduction to (∆minusQ) Q an 8times 8 matrix

bull Quantitative Photoacoustic Tomography

(Bal-U)

37

PARTIAL DATA PROBLEM

Suppose we measure

Λγ(f)|Γ suppf sube Γprime

Γ Γprime open subsets of partΩ

Can one recover γ

Important case Γ = Γprime

38

EXTENSION OF CGO SOLUTIONS

u = exmiddotρ(1 + Ψq(x ρ))

ρ isin Cn ρ middot ρ = 0

(Not helpful for localizing)

Kenig-Sjostrand-U (2007)

u = eτ(ϕ(x)+iψ(x))(a(x) +R(x τ))

τ isin R ϕ ψ real-valued R(x τ)rarr 0 as τ rarrinfin

ϕ limiting Carleman weight

nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|

Example ϕ(x) = ln |xminus x0| x0 isin ch(Ω)

39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))

R(x τ)τrarrinfinminusrarr 0 in Ω

ϕ(x) = ln |xminus x0|

Complex Spherical Waves

Theorem (Kenig-Sjostrand-U) Ω strictly convex

Λq1

∣∣∣Γ

= Λq2

∣∣∣Γ Γ sube partΩ Γ arbitrary

rArr q1 = q2

40

Earlier result (Bukhgeim-U 2002)

n ge 3 Let ξ isin Snminus1 = x isin Rn |x| = 1 We define

partΩplusmn =

x isin partΩ 〈ν ξ〉 gt 0

lt 0

(ν is the unit outer normal) Let δ gt 0

partΩ+δ(ξ) = x isin partΩ 〈ν ξ〉 gt δpartΩminusδ(ξ) = x isin partΩ 〈ν ξ〉 lt δ

Theorem (Bukhgeim-U) Suppose we know

Λq(f)|partΩminusδ(ξ)supp f sube partΩ

Then we can recover q

41

Carleman estimate ξ isin Snminus1

Let q isin Linfin(Ω) u isin C2(Ω) u|partΩ = 0 For τ ge τ0

τ2int

Ω|eminusτ〈xξ〉u|2 dx+ τ

intpartΩ+

〈ξ ν(x)〉|eminusτ〈xξ〉partu

partν|2dS(x)

leC

intΩ|eminusτ〈xξ〉(∆minus q)u|2dxminus τ

intpartΩminus〈ξ ν(x)〉|eminusτ〈xξ〉

partu

partν|2dS

42

Remarks ∆ρu = eminusxmiddotρ∆(exmiddotρu)

bull Carleman estimate for domains with boundary for

∆ρ = ∆ + 2ρ middot nabla

bull Weight is linear 〈x ξ〉

Corollary u = 0 on partΩ partupartν |partΩminus = 0

τintpartΩ+

〈x ν(x)〉|eminusτ〈xξ〉partu

partν|2dS(x) le C

intΩ|eminusτ〈xξ〉(∆minus q)u|2dx

We need 〈ν(x) ξ〉 ge δ gt 0

43

Bukhgeim-U (n ge 3)

Λq1(f)|partΩminusδ(ξ)= Λq2(f)|partΩminusδ(ξ)

forallf =rArr q1 = q2

partΩminusδ(ξ) = x isin partΩ 〈ν ξ〉 lt δ

Sketch of proof

Choose u2 = exmiddotρ2(1 + Ψq2(x ρ2))

solution of (∆minus q2)u2 = 0

ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2

Let u1 be such that u1|partΩ = u2|partΩpartu1

partν|partΩminusδ(ξ)

=partu2


44

u2 = exmiddotρ2(1 + Ψq2(x ρ2))

(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)

u1|partΩ = u2|partΩpartu1

partν1|partΩminusδ(ξ)

=partu2


u = u1 minus u2 q = q1 minus q2

v1 = exmiddotρ1(1 + Ψq1(x ρ1)) ρ1 = minusτξ + i(k minus l)

solution of (∆minus q1)v1 = 0

intΩqu2v1dx =

intpartΩ

partu

partνv1dS

Note that u|partΩ = 0 partupartν |partΩminusδ(ξ)

= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS

u2 = exmiddotρ1(1 + Ψq1)v1 = exmiddotρ2(1 + Ψq2)

Fix k isin Rn ρ1 + ρ2 = 2ik

Carleman estimate

τintpartΩ+


partν|2dS(x) le C

intΩ|(∆minus q1)ueminusτ〈xξ〉|2dS(x)

Need 〈x ξ〉 ge δ gt 0 |RHS| le C as |ρ| rarr infin LHSrarrint

Ωqe2ixmiddotkdx

46

We getint

Ωe2ixmiddotkq(x)dx = 0

k perp ξ But we can move ξ a

little bit

ξ isin Snminus1

partΩminusδ(ξ) = x isin partΩ 〈ν(x) ξ〉 lt δ

We obtain χΩq(minus2k) = 0 in an open cone =rArr q = 0

Carleman estimate =rArr control of partupartν

∣∣∣partΩ+δ

(with appropriate linear

weights) (Stability estimates Heck-Wang)

47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

uτ = eτ(ϕ+iψ)aτ ϕ(x) = ln |xminus x0| x0 isinch(Ω)

Eikonal nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|ψ(x) = d( xminusx0

|xminusx0| ω) ω isin Snminus1 smooth

for x isin Ω

Transport (nablaϕ+ inablaψ) middot nablaaτ = 0

(Cauchy-Riemann equation in plane generated by nablaϕnablaψ)

48

ϕ(x) = ln |xminus x0| x0 isinch(Ω)

Carleman Estimates

u|partΩ = partupartν |partΩminus = 0 partΩplusmn = x isin partΩnablaϕ middot ν

gtlt 0

intpartΩ+

lt nablaϕ ν gt |eminusτϕ(x)partu

partν|2ds le

C

τ

intΩ|(∆minus q)ueminusτϕ(x)|2ds

This gives control of partupartν |partΩ+δ

partΩ+δ = x isin partΩnablaϕ middot ν ge δ

49

More general CGO solutions

uτ = eτ(ϕ+iψ)aτ

τ 0 τ = 1h (semicl) ϕψ real-valued

bull ϕ is a limiting Carleman weight

eϕhh2(minus∆ + q)eminus

ϕh

has semiclassical principal symbol

Pϕ(x ξ) = ξ2 minus (nablaϕ)2 + 2inablaϕ middot ξ

Hormanderrsquos condition

Re Pϕ Im Pϕ le 0 on Pϕ = 0

We need ϕminusϕ to be phase of solutions

LCW Re Pϕ Im Pϕ = 0

nablaϕ 6= 0 in an open neighborhood of Ω

50

CGO solutions uh = e1h(ϕ+iψ)ah

bull ϕ LCW ϕ real-valued

RePϕ ImPϕ = 0 on Pϕ = 0

nablaϕ 6= 0 on an open neighborhood of Ω

Examples (Dos Santos Ferreira-Kenig-Salo-U 2009)

(a) ϕ(x) = x middot ξ ξ isin Rn |ξ| = 1

(b) ϕ(x) = a ln |xminus x0|+ b (a b constants) x0 isinch(Ω)

(c) ϕ(x) =a〈xminus x0 ξ〉|xminus x0|2

+ b ξ isin Rn

(d) ϕ(x) = aarctan 2〈xminusx0ξ〉|xminusx0|2minus|ξ|2

+ b

(e) ϕ(x) = aarctanh 2〈xminusx0ξ〉|xminusx0|2minus|ξ|2

+ b

(f) n = 2 ϕ is a harmonic function

51

Instead of intΩe2ixmiddotkq(x)dx = 0

k perp ξ (ξ isin Snminus1) as in Bukhgeim-U argument we getintΩeiλf(x) q(x) a1a2dx = 0

λ any real number a1 a2 6= 0 f(x) real-analytic a1 a2 real analytic

Analytic microlocal analysis =rArr q = 0 (like inversion of real-analytic Radon transform)

52

Linearization (Analog of Calderon)

Theorem (Dos Santos Ferreira Kenig Sjostrand-U)

intΩhuv = 0

Γ sube partΩ Γ open

∆u = ∆v = 0 u v isin Cinfin(Ω)

supp u|partΩ supp v|partΩ sube Γ

rArr h = 0

53



ϕ(x) = ln |xminus x0| x0 isin ch(Ω)

Also used to determine inclusions obstacles etc

a) Conductivity Ide-Isozaki-Nakata-Siltanen-U

b) Helmholtz Nakamura-Yosida

c) Elasticity J-N Wang-U

d) 2D Systems J-N Wang-U

e) Maxwell T Zhou

54


(Loading reconperfect1mpg)

55

reconperfect1mpg

The Two Dimensional Case

Theorem (n = 2) Let γj isin C2(Ω) j = 12

Assume Λγ1 = Λγ2 Then γ1 = γ2

bull Nachman (1996)

bull Brown-U (1997) Improved to γj Lipschitz

bull Astala-Paivarinta (2006) Improved to γj isin Linfin(Ω)

Recall

div(γnablau) = 0 γ isin Linfin(Ω)u|partΩ = f

Qγ(f) =int

Ωγ|nablau|2dx = 〈Λγf f〉L2(partΩ)

56

This follows from more general result

Theorem (n = 2 Bukhgeim 2008) Let qj isin Linfin(Ω) j = 12

Assume Λq1 = Λq2 Then q1 = q2

Recall

(∆minus q)u = 0u|partΩ = f

Λq(f) =partu

partν

∣∣∣∣∣partΩ

with ν-unit outer normal

57

Λq1 = Λq2 rArr q1 = q2

Sketch of proof New class of CGO solutions

u1(z τ) = eτz2 (

1 + r1(z τ))

u2(z τ) = eminusτ z2 (

1 + r2(z τ)) τ 1

solve (∆minus qj)uj = 0 with rj(z τ)rarr 0 on Ω sufficiently fast

Notation z = x1 + ix2

Remark z2 = x21 minus x

22 + 2ix1x2 = ϕ+ iψ


ϕ harmonic ψ conjugate harmonic

59

Λq1 = Λq2 rArrint

Ω(q1 minus q2)u1u2dx = 0

(∆minus qj)uj = 0

u1 = eτz2 (

1 + r1(z τ)) u2 = eminusτ z

2 (1 + r2(z τ)

)Substitutingint

Ω(q1 minus q2)e4iτx1x2(1 + r1 + r2 + r1r2)dx = 0

Letting τ rarrinfin and using stationary phase

(q1 minus q2)(0) = 0

Changing z to z minus z0 we get

(q1 minus q2)(z0) = 0

60

Partial data

Let Γ sube partΩ Γ open

Let qj isin C1+ε(Ω) ε gt 0 j = 12

Theorem (Imanuvilov-U-Yamamoto 2010) n=2 Assume

Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ

forall f suppf sube Γ Then

q1 = q2

bull Riemann Surfaces Guillarmou-Tzou (2011)

61

Partial data Γ0 = partΩminus Γ

Construct CGO solutions

∆uj minus qjuj = 0 in Ωuj|Γ0

= 0

In this case intΩ

(q1 minus q2)u1u2dx = 0

if Λq1(f)|Γ = Λq2(f)|Γ suppf sube Γ

62

intΩ


uj|Γ0= uj|partΩminusΓ = 0

u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ

u2(x) = eminusτΦ(z)(a(z) +b0(z)

τ) + eminusτΦ(z)(a(z) +

b1(z)

τ) + eτϕR

(2)τ

Φ = ϕ+ iψ holomorphic

63

u1 = Re eτΦ(z)(a(z) + middot middot middot ) u2 = Re eminusτΦ(z)(a(z) + middot middot middot )

Φ(z) = ϕ+ iψ holomorphic

uj|partΩminusΓ = 0

p isin Ω Φ has non-degenerate critical point at p (Morse function)

parta = 0 Re a|partΩminusΓ = 0

a = 0 at other critical points

Stationary phase in intΩ


64

u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )


Φ(z) Morse function with non-degenerate critical point at pintΩ


Stationary phase

=rArr (q1 minus q2)(p) = 0

65

Corollary Obstacle Problem

Ω D sub R2 smooth boundary

such that D sub Ω

V sub partΩ open set

Let qj isin C2+α(ΩD) for some α gt 0 j = 12

Cqj =

(u|V partνu|V ) (∆minus qj)u = 0 in ΩD

supp u|partΩ sub V u isin H1(ΩD)

Then Cq1 = Cq2 =rArr q1 = q2

66

Carleman Estimate With Degenerate Weights

Lemma 1

Let partΩminus Γ = x isin partΩ ν middot nablaϕ = 0 Then for τ sufficiently large existsolution of

∆uminus qu = f in Ωu|partΩminusΓ = g

such that

ueminusτϕL2(Ω) le C(|τ |minus12feminusτϕL2(Ω) + geminusτϕL2(ΩminusΓ)

)

67

Phase Function

Lemma 2 (Vekua) Given points x1 xN in Ω and constants bi i =

12 C0j C1j C2j j = 1 N there exists an open and dense set

Θ sube C2(partΩminus Γ)times C2(partΩminus Γ)times C3N

solution of

φ+ iψ holomorphic in Ω

(φ ψ)|partΩminusΓ = (b1 b2)(φ+ iψ)(xj) = C0jpartpartz(φ+ iψ)(xj) = C1jpart2

partz2(φ+ iψ)(xj) = C2j

68

CGO Solutions (n = 2)

partminus1z g(z) = minus

1

π

intΩ

g(ξ1 ξ2)

ξ1 + iξ2 minus zdξ1dξ2 partminus1

z g = partminus1z g

RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))


H = non-degenerate critical points of Φ


∆u1 minus q1u1 = 0 in Ω

u1

∣∣∣partΩΓ

= 0

Let Φ be a holomorphic Morse function such that ImΦ = 0 on

partΩΓ Let Φ = ϕ+ iψ

u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)

+ eτϕu11 + eτϕu12

69



Choice of a a0 a1

a a0 a1 isin C2(Ω) partza = partza0 = partza1 equiv 0

Re a∣∣∣partΩΓ

= 0 a = partza = 0 on Hcap partΩ

(a0(z) + a1(z))∣∣∣partΩΓ

=partminus1z (aq1)minusM1(z)

4partzΦ

+partminus1z (a(z)q1)minusM3(z)

4partzΦ

70

The polynomials M1(z) and M3(z) satisfy

partjz(partminus1z (aq1)minusM3(z)

)= 0 z isin H j = 012

partjz

(partminus1z (aq1)minusM3(z)

)= 0 z isin H j = 012

Let ei i = 12 be smooth e1 + e2 = 1 on Ω with e1 = 0 in a

neighborhood of Hminus partΩ and e2 = 1 in a neighborhood of partΩ

71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1

4eminusiτψRΦminusτ

(e1

(partminus1z (a(z))q1 minusM3(z)

))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2

(partminus1z (a(z)q1)minusM3(z)

)4partzΦ

72

Other remainder term

Find u12 such that

∆(u12eτϕ)minus q1u12e

τϕ minus q1u11eτϕ + h1e

τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly

∆v minus q2v = 0 in Ω v∣∣∣partΩΓ

= 0

Construct solution v of the form

v(x) = eminusτΦ(z)(a(z) + b0(z)τ

)+ eminusτΦ(z)

(a(z) + b0(z)τ

)+eminusτϕv11 + eminusτϕv12

75

Main Term

R =int

Ω(q1 minus q2)(a(a0 + b0) + a(a1 + b1))dx

+1

4

intΩ

(q1 minus q2)

a partminus1z (aq2)minusM2(z)

partzΦ+ a

partminus1z (aq2)minusM4(z)

partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76

Proof of Uniqueness for Partial Data

bull Take geometiric optics solution u1 to

∆u1 minus q1u1 = 0 u1

∣∣∣partΩΓ

= 0

bull u2 ∆u2 minus q2u2 = 0 u2

∣∣∣partΩ

= u1

∣∣∣partΩ

DN maps are equal rArr nablau2 = nablau1 on Γ

u = u1 minus u2 rArr ∆uminus q2u = (q1 minus q2)u2

u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0

bull Take complex geometric optics solution v to

∆v minus q2v = 0 v∣∣∣partΩΓ

= 0

77

0 =int

Ωv(∆uminus q2u)dx = minus

intΩ

(q1 minus q2)vu1dx

Stationary phase + estimates for u12 rArr

2lsum

k=1

π((q1 minus q2)|a|2)(xk)Re ei2τ Im Φ(xk)

|(det Im Φprimeprime)(xk)|12

+R = o(1)

as τ rarrinfin

[left side] = almost perodic function in τ

Bohrrsquos theorem inplies [left side] = 0 for all τ

78

Phase function

We can choose Φ such that

Im Φ(xk) 6= Im Φ(xj) j 6= k

Let a(xk) 6= 0 Then stationary phase implies

q1(xk) = q2(xk)

79

Partial Data for Second Order Elliptic Equations (n = 2)

(ImanuvilovndashUndashYamamoto 2011)

∆g +A(z)part

partz+B(z)

part

partz+ q z = x1 + ix2

g = (gij) positive definite symmetric matrix

∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes

bull Anisotropic Calderonrsquos Problem

bull Magnetic Schrodinger Equation

bull Convection terms

80

Anisotropic case

Cardiac muscle 63 mho (longitudinal)23 mho (transversal)

γ = (γij)

conductivity

positive-definite symmetric

matrix

Ω sube RnΩ bounded Under assumptions of no sources or sinks of

current the potential u satisfies

div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()

f = voltage potential at boundary

Isotropic γij(x) = α(x)δij δij =

1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj

∣∣∣∣∣∣partΩ

ν =(ν1 middot middot middot νn

)is the unit outer normal to partΩ

Λγ(f) is the induced current flux at partΩ

Λγ is the voltage to current map or Dirichlet - to - Neumann map

82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj


EIT Can we recover γ in Ω from Λγ

83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ

Answer No Λψlowastγ = Λγ

where ψ Ωrarr Ω change of variables

ψ|partΩ = Identity

ψlowastγ =

(Dψ)T γ Dψ|detDψ|

ψminus1

v = u ψminus1

84

Theorem (ImanuvilovndashUndashYamamoto 2011) Ω sub R2 Γ sub partΩ Γ

open γk = (γijk ) isin Cinfin(Ω) k = 12 positive definite symmetric

Assume

Λγ1(f)|Γ = Λγ2(f)|Γ forallf suppf sub Γ

Then existF Ωrarr Ω Cinfin diffeomorphism F |Γ = Identity such that

Flowastγ1 = γ2

Full Data (Γ = partΩ)

bull γk isin C2(Ω) Nachman (1996)

bull γk Lipschitz SunndashU (2001)

bull γk isin Linfin(Ω) AstalandashLassasndashPavarinta (2006)

85

DIRICHLET-TO-NEUMANN MAP (Lee-U 1989)

(M g) compact Riemannian manifold with boundary∆g Laplace-Beltrami operator g = (gij) pos def symmetric matrix

∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g

∣∣∣∣∣partM

ν = (ν1 middot middot middot νn) unit-outer normal

86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g


current flux at partM

Inverse-problem (EIT)

Can we recover g from Λg

Λg = Dirichlet-to-Neumann map or voltage to current map

87

ANOTHER MOTIVATION (STRING THEORY)

HOLOGRAPHY

Dirichlet-to-Neumann map is the ldquoboundary-2pt functionrdquo

Inverse problem Can we recover (M g) (bulk) from boundary-2pt function

M Parrati and R Rabadan Boundary rigidity and holography JHEP

0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g

Answer No Λψlowastg = Λg where

ψ M rarrM diffeomorphism ψ∣∣∣partM

= Identity and

ψlowastg =(Dψ g (Dψ)T

) ψ

89

Show Λψlowastg = Λg ψ M rarrM diffeomorphism ψ∣∣∣partM

= Identity

Qg(f) =sumijintM gij partupartxi

partupartxj

radicdet gdx

Qg(f) = minusintpartM

Λg(f)fdS

Qg hArr Λgv = u ψ ∆ψlowastgv = 0

Qψlowastg = Qg rArr Λψlowastg = Λg

90

Theorem (n ge 3) (Lassas-U 2001 Lassas-Taylor-U 2003) (M gi) i =

12 real-analytic connected compact Riemannian manifolds with

boundary Let Γ sube partM Γ open Assume

Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ

Then existψ M rarrM diffeomorphism ψ∣∣∣Γ

= Identity so that

g1 = ψlowastg2

In fact one can determine topology of M as well (only need to know

Λg partM)

91

Theorem (Guillarmou-Sa Barreto 2009) (M gi) i = 12 are com-

pact Riemannian manifolds with boundary that are Einstein As-

sume

Λg1 = Λg2

Then existψ M rarrM diffeomorphism ψ|partM = Identity such that

g1 = ψlowastg2

Note Einstein manifolds with boundary are real analytic in the

interior

92

Theorem (n = 2)(Lassas-U 2001)

(M gi) i = 12 connected Riemannian manifold with boundary

Let Γ sube partM Γ open Assume



= Identity and

β gt 0 β∣∣∣Γ

= 1 so that

g1 = βψlowastg2

In fact one can determine topology of M as well

93

Moding Out the Diffeomorphism Group

Some conformal class Λβg = Λg β isin Cinfin(M)

=rArr β = 1

More general problem

(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f

Λg(f) = partupartνg|partM

Inverse Problem Does Λg determines q

94

(∆g minus q)u = 0 Λg(f) = partupartνg|partM Λg rarr q

Theorem (n=2) (Guillarmou-Tzou 2009)

YES

Earlier results

bull R2 q small (Sylvester-U 1986)

bull R2 q generic (Sun-U 2001)

bull R2 q = ∆radicλradicλ γ gt 0 (Nachmann 1996)

bull Riemannian surfaces q = ∆radicλradicλ γ gt 0 (Henkin-

Michel 2008)

bull q isin Linfin (Bukhgeim 2008)

95

MODING OUT GROUP OF DIFFEOMORPHISM

(n ge 3)



(lowast) g(x1 xprime) = c(x)

(1 00 g0(xprime)

) c gt 0

Theorem (Dos Santos-Kenig-Salo-U) Assume that there is a global

coordinate system so that () is true In addition g0 is simple

Then Λg determines uniquely q

Simple No conjugate points and strictly convex

96

g(x1 xprime) = c(x)

(1 00 g0(xprime)

) xprime isin Rnminus1

Examples

(a) g(x) conformal to Euclidean metric (Sylvester-U 1987)

(b) g(x) conformal to hyperbolic metric (Isozaki 2004)

(c) g(x) conformal to metric on sphere (minus a point)

97

Non-uniqueness for EIT (Invisibility)

Motivation (Greenleaf-Lassas-U MRL 2003)

When bridge connecting the two parts

of the manifold gets narrower the

boundary measurements give less infor-

mation about isolated area

When we realize the manifold in Euclidean space we should obtain

conductivities whose boundary measurements give no information

about certain parts of the domain

98

Greenleaf-Lassas-U (2003 MRL)

Let Ω = B(02) sub R3D = B(01)

where B(0 r) = x isin R3 |x| lt r

F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|

F diffeomorphism F∣∣∣partΩ

= Identity

99

Letg = identity metric in B(02)g = (Fminus1)lowastg on B(02) B(01)σ = conductivity associated to g

In spherical coordinates (r φ θ)rarr (r sin θ cosφ r sin θ sinφ r cos θ)

σ =

2(r minus 1)2 sin θ 0 00 2 sin θ 00 0 2(sin θ)minus1

Let g be the metric in B(02) (positive definite in B(01)) st g = g

in B(02) B(01) Then

Theorem (Greenleaf-Lassas-U 2003)

Λg = Λg

100

Based on work of Greenleaf-Lassas-U MRL 2003

101

Inverse Boundary Problems

Can one determine the internal properties of a medium by making

measurements outside the medium (non-invasive)

X-ray tomography (CAT-scans)

Problem Can we recover the density from attenuation of X-rays

1

Radon (1917) n = 2




f(x)dH =intLf

f(x) =1

4π2pv

intS1dθint d

dsg(s θ)ds

〈x θ〉 minus s

2



PET MRI

3


Ultrasound

Electrical

Impedance

Tomography

(EIT)

4

Hybrid Methods




Photoacoustic


Thermoacoustic

Imaging



5





Graham Bell When










ing medium

6


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Radon (1917) n = 2




f(x)dH =intLf

f(x) =1

4π2pv

intS1dθint d

dsg(s θ)ds

〈x θ〉 minus s

2



PET MRI

3


Ultrasound

Electrical

Impedance

Tomography

(EIT)

4

Hybrid Methods




Photoacoustic


Thermoacoustic

Imaging



5





Graham Bell When










ing medium

6


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



PET MRI

3


Ultrasound

Electrical

Impedance

Tomography

(EIT)

4

Hybrid Methods




Photoacoustic


Thermoacoustic

Imaging



5





Graham Bell When










ing medium

6


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Ultrasound

Electrical

Impedance

Tomography

(EIT)

4

Hybrid Methods




Photoacoustic


Thermoacoustic

Imaging



5





Graham Bell When










ing medium

6


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Hybrid Methods




Photoacoustic


Thermoacoustic

Imaging



5





Graham Bell When










ing medium

6


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101





Graham Bell When










ing medium

6


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Wikipedia

7



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



8

Melanoma3DMovieavi

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Mathematical Model


where u solves


u|t=0 = βH(x)

parttu|t=0 = 0



9


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Ω sub Rn

(n = 23)






10

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

11



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



Honoris Causarsquo



have stayed there

12



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



13

rawlongmpg



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



14

electrompg

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

15

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

16

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

17










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101










18

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Other Applications








19


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


systole

20



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



21

DBarPerfMovie1avi



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



current

I(x) = γ(x)nablau



sinks we have


22


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


u∣∣∣partΩ

= f






map





23


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


u∣∣∣partΩ

= f

Dirichlet Integral

Qγ(f) =int


Qγ(f g) =int



v∣∣∣partΩ

= g

24


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


div(γnablav) = 0


partu

partν

∣∣∣∣partΩ

Qγ(f g) =int




Qγ(f) =int


intpartΩ

Λγ(f)fdS

25

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Linearization

limεrarr0+


ε=int




Can we recover h

26


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Can we recover h

u = exmiddotρ


ρ =η minus iξ


|ξ|2int



27



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



forallα






28







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101







∣∣∣∣xn=0

29





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101





2003)







30


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


div(γnablaw) = 0

u =radicγw



radicγ

(∆ =

nsumi=1

part2

partx2i

)

(∆minus q)u = 0

u∣∣∣partΩ

= f


partν

∣∣∣partΩ


31

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

IDENTITY

intΩ



∣∣∣partΩ

)u2

∣∣∣partΩdS

(∆minus qi)ui = 0




32

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

CGO SOLUTIONS




∆u = 0 u =




33




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101




(|η| = |k| η middot k = 0)



of the form



34






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101






intΩ




35

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

APPLICATIONS






absorption


36



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



(q(x) = k2n(x))




q potential




(Bal-U)

37


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Suppose we measure



Can one recover γ


38











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101











39

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

CGO SOLUTIONS

u = eτ(ϕ(x)+iψ(x))(a0(x) +R(x τ))





Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2

40



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



partΩplusmn =


lt 0






41



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



τ2int


intpartΩ+


partν|2dS(x)

leC



partu

partν|2dS

42






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101






τintpartΩ+


partν|2dS(x) le C



43

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Bukhgeim-U (n ge 3)




Sketch of proof



ρ2 = τξ + i(k + l)

|k|2 + |l|2 = τ2



=partu2


44


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)



=partu2





intΩqu2v1dx =

intpartΩ

partu

partνv1dS


= 0

45

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

q = q1 minus q2

(lowast)int

Ωqu2v1dx =

intpartΩ+δ(ξ)

partu

partνv1dS



Carleman estimate

τintpartΩ+


partν|2dS(x) le C



Ωqe2ixmiddotkdx

46

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

We getint



little bit

ξ isin Snminus1




∣∣∣partΩ+δ



47


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Λq1

∣∣∣Γ

= Λq2


rArr q1 = q2




for x isin Ω



48


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Carleman Estimates


gtlt 0

intpartΩ+


partν|2ds le

C

τ




49






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101






ϕh








50









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101









+ b ξ isin Rn


+ b


+ b


51





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101





52



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



intΩhuv = 0




rArr h = 0

53









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101









e) Maxwell T Zhou

54



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



55

reconperfect1mpg




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101




bull Nachman (1996)



Recall


Qγ(f) =int


56




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101




Recall


Λq(f) =partu

partν



57



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



u1(z τ) = eτz2 (

1 + r1(z τ))


1 + r2(z τ)) τ 1




22 + 2ix1x2 = ϕ+ iψ



59

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Λq1 = Λq2 rArrint


(∆minus qj)uj = 0

u1 = eτz2 (


2 (1 + r2(z τ)

)Substitutingint



(q1 minus q2)(0) = 0


(q1 minus q2)(z0) = 0

60

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Partial data




Λq1(f)∣∣∣Γ

= Λq2(f)∣∣∣Γ


q1 = q2


61




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101




= 0

In this case intΩ



62

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

intΩ



u1(x) = eτΦ(z)(a(z) +a0(z)

τ) + eτΦ(z)(a(z) +

a1(z)

τ) + eτϕR

(1)τ



b1(z)

τ) + eτϕR

(2)τ


63









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101









64





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101





Stationary phase


65



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



such that D sub Ω



Cqj =




66


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Lemma 1



such that


)

67

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Phase Function




solution of




68



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



1

π

intΩ

g(ξ1 ξ2)


z g = partminus1z g






u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



u1

∣∣∣partΩΓ

= 0





69



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



Choice of a a0 a1






4partzΦ


4partzΦ

70



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



)= 0 z isin H j = 012

partjz


)= 0 z isin H j = 012



71

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Remainder term

Choice of u11

u11 = minus1

4eiτψRΦτ

(e1


))minus

1


(e1


))

minuseiτψ

τ

e2


)4partzΦ

minuseiτψ

τ

e2


)4partzΦ

72


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Find u12 such that



τϕ in Ω

u12

∣∣∣partΩΓ

=1

4RΦτ

(e1


))+

1

4RΦminusτ

(e1


))

u12L2(Ω) = o

(1

τ

) τ rarrinfin

73

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Here

h1 = eiτψ∆

e2


)4τpartzΦ

+eminusiτψ∆

e2


)4τpartzΦ

minusa0q1

τeiτψ minus

a1q1

τeminusiτψ

74

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Similarly


= 0



)+ eminusτΦ(z)

(a(z) + b0(z)τ


75

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Main Term

R =int


+1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx+

1

4

intΩ

(q1 minus q2)


partzΦ+ a


partzΦ

dx

76




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101




∣∣∣partΩΓ

= 0


∣∣∣partΩ

= u1

∣∣∣partΩ



u∣∣∣partΩ

= 0partu

partν

∣∣∣∣∣Γ

= 0



= 0

77

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

0 =int


intΩ

(q1 minus q2)vu1dx


2lsum

k=1



+R = o(1)

as τ rarrinfin



78

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Phase function




q1(xk) = q2(xk)

79



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



∆g +A(z)part

partz+B(z)

part



∆gu =1radic

det(g)

nsumij=1

part

partxi(radic

det(g)gijpartu

partxj) gij = (gij)

minus1

Includes




80

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

Anisotropic case


γ = (γij)

conductivity


matrix



div(γnablau) = 0

nsumij=1

part

partxi

(γij

partu

partxi

)=0 in Ω

u∣∣partΩ

=f

()



1 i = j

0 i 6= j

81

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj






82

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

nsumij=1

part

partxi

γij partupartxj

=0 in Ω

u∣∣∣partΩ

=f

()

Λγ(f) =nsum

ij=1

νiγijpartu

partxj



83

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

div(γnablau) = 0

u∣∣∣partΩ

= fΛγ(f) =

nsumij=1

γijνipartu

partxj

∣∣∣partΩ

Λγ rArr γ




ψlowastγ =


ψminus1

v = u ψminus1

84



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



Assume



Flowastγ1 = γ2





85



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



∆gu =1

radicdet g

nsumij=1

part

partxi

radicdet g gijpartu

partxj

(gij) = (gij)minus1

∆gu = 0 on M

u∣∣∣partM

= f

Conductivity

γij =radic

det g gij

Λg(f) =nsum

ij=1

νjgijpartu

partxi

radicdet g



86

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

∆gu = 0

u∣∣∣partM

= f

Λg(f) =partu

partνg=

nsumij=1

νjgijpartu

partxi

radicdet g






87


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


HOLOGRAPHY




0401 (2004) 034

88

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

∆gu = 0

u∣∣∣partM

= fΛg(f) =

partu

partνg


Λg rArr g



= Identity and


) ψ

89


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


= Identity


partupartxj

radicdet gdx


Λg(f)fdS



90






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101






= Identity so that

g1 = ψlowastg2


Λg partM)

91



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



sume

Λg1 = Λg2


g1 = ψlowastg2


interior

92






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101






= Identity and


= 1 so that

g1 = βψlowastg2


93



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



=rArr β = 1





94



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



YES

Earlier results





Michel 2008)


95


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


(n ge 3)




(1 00 g0(xprime)

) c gt 0





96

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101

g(x1 xprime) = c(x)

(1 00 g0(xprime)


Examples




97










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101










98


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101


Let Ω = B(02) sub R3D = B(01)


F Ω 0 rarr Ω D

F (x) = (|x|2

+ 1)x

|x|


= Identity

99



σ =



in B(02) B(01) Then


Λg = Λg

100


101



σ =



in B(02) B(01) Then


Λg = Λg

100


101


101

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