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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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 2
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 3
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 4
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
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Thanks to D Issacson
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DBarPerfMovie1avi
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 5
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 6
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 7
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 8
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 9
(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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 10
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 11
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 12
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 13
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)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 14
(Loading imagesrawlongmpg)
Mark Nelson httpnelsonbeckmanillinoisedu
13
rawlongmpg
Media File (videompeg)
(Loading imageselectrompg)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 15
(Loading imageselectrompg)
Mark Nelson httpnelsonbeckmanillinoisedu
14
electrompg
Media File (videompeg)
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 16
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 17
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 18
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 19
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 20
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 21
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
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 22
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Thanks to D Issacson
21
DBarPerfMovie1avi
Media File (videoavi)
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 23
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 24
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
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 25
div(γ(x)nablau(x)) = 0
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 26
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 27
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 28
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 29
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
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 30
Λγf(xprime) =inteixprimemiddotξprimeλγ(xprime ξprime)f(ξprime)dξprime
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 31
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 32
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 33
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 34
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 35
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
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 36
Proof Λq1 = Λq2 rArr q1 = q2intΩ
(q1 minus q2)u1u2 = 0
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 37
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 38
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 39
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 40
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 41
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 42
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 43
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 44
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 45
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
partν|partΩminusδ(ξ)
44
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 46
u2 = exmiddotρ2(1 + Ψq2(x ρ2))
(∆minus q2)u2 = 0 ρ2 = τξ + i(k + l)
u1|partΩ = u2|partΩpartu1
partν1|partΩminusδ(ξ)
=partu2
partν1|partΩminusδ(ξ)
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 47
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+
〈ξ ν(x)〉|eminusτ〈xξ〉partu
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 48
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
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 49
Theorem (Kenig-Sjostrand-U) Ω strictly convex
Λq1
∣∣∣Γ
= Λq2
∣∣∣Γ Γ sube partΩ Γ arbitrary
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 50
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 51
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 52
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 53
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 54
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
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 55
Complex Spherical Waves
uτ = eτ(ϕ+iψ)aτ
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
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 56
Complex Spherical Waves
(Loading reconperfect1mpg)
55
reconperfect1mpg
Media File (videompeg)
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 57
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 58
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 59
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
nablaϕ middot nablaψ = 0 |nablaϕ| = |nablaψ|
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 60
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 61
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 62
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
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 63
intΩ
(q1 minus q2)u1u2 = 0
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 64
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
(q1 minus q2)u1u2 = 0
64
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 65
u1 = Re eτΦ(z)(a(z) + middot middot middot )u2 = Re eminusτΦ(z)(a(z) + middot middot middot )
uj|partΩminusΓ = 0
Φ(z) Morse function with non-degenerate critical point at pintΩ
(q1 minus q2)u1u2 = 0
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 66
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 67
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 68
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 69
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)))
RΦτg = eτ(Φ(z)minusΦ(z))partminus1z (geτ(Φ(z)minusΦ(z)))
H = non-degenerate critical points of Φ
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 70
CGO Solutions (n = 2)
∆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
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 71
u1(x) = eτΦ(z)(a(z) + a0(z)τ) + eτΦz(a(z) + a0(z)τ)
+ eτϕu11 + eτϕu12
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 72
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
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 73
Remainder term
Choice of u11
u11 = minus1
4eiτψRΦτ
(e1
(partminus1z (aq1)minusM1(z)
))minus
1
4eminusiτψRΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
minuseiτψ
τ
e2
(partminus1z (aq1)minusM1(z)
)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
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 74
Other remainder term
Find u12 such that
∆(u12eτϕ)minus q1u12e
τϕ minus q1u11eτϕ + h1e
τϕ in Ω
u12
∣∣∣partΩΓ
=1
4RΦτ
(e1
(partminus1z (a(z)q1)minusM1(z)
))+
1
4RΦminusτ
(e1
(partminus1z (a(z))q1 minusM3(z)
))
u12L2(Ω) = o
(1
τ
) τ rarrinfin
73
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 75
Here
h1 = eiτψ∆
e2
(partminus1z (a(z)q1)minusM1(z)
)4τpartzΦ
+eminusiτψ∆
e2
(partminus1z (a(z)q1)minusM3(z)
)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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 76
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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 77
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)
a partminus1z (aq1)minusM1(z)
partzΦ+ a
partminus1z (aq1)minusM3(z)
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 78
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 79
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 80
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 81
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 82
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 83
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
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 84
nsumij=1
part
partxi
γij partupartxj
=0 in Ω
u∣∣∣partΩ
=f
()
Λγ(f) =nsum
ij=1
νiγijpartu
partxj
∣∣∣∣∣∣partΩ
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 85
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 86
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 87
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
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 88
∆gu = 0
u∣∣∣partM
= f
Λg(f) =partu
partνg=
nsumij=1
νjgijpartu
partxi
radicdet g
∣∣∣∣∣partM
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 89
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
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 90
∆gu = 0
u∣∣∣partM
= fΛg(f) =
partu
partνg
∣∣∣∣∣partM
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 91
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 92
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 93
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
Λg1(f)|Γ = Λg2(f)|Γ forallf f supported in Γ
Then existψ M rarrM diffeomorphism ψ∣∣∣Γ
= 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 94
Theorem (n = 2)(Lassas-U 2001)
(M gi) i = 12 connected Riemannian manifold 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 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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 95
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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 96
(∆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)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 97
MODING OUT GROUP OF DIFFEOMORPHISM
(n ge 3)
(∆g minus q)u = 0 q isin Cinfin(M)u|partM = f
Λg(f) = partupartνg|partM
(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
Page 98
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
Page 99
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
Page 100
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
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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
Page 101
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
Page 102
Based on work of Greenleaf-Lassas-U MRL 2003
101