30 YEARS OF CALDERÓN’S
PROBLEM
Gunther Uhlmann
University of Washington
Northwestern, November 18, 2009
Inverse Boundary Problems
Can one determine the internal properties of a medium
by making measurements outside the medium (noninvasive)?
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
R
− Lf
Idetector = e
Isource
Rf (s, θ) = g(s, θ) =
Z
hx,θi=s
1
f (x) =
p.v.
dθ
1
4π 2
S
Z
Z
f (x)dH =
Z
L
f
d g(s, θ)ds
ds
hx, θi − s
2
LIN EAR
(No Scattering)
X-ray tomography (CT)
PET
MRI
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N ON LIN EAR
(Scattering)
ULTRASOUND
ELECTRICAL
IMPEDANCE
TOMOGRAPHY (EIT)
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(Loading rawlong.mpg)
Mark Nelson, http://nelson.beckman.illinois.edu
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(Loading electro.mpg)
Mark Nelson, http://nelson.beckman.illinois.edu
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CALDERÓN’S PROBLEM
g(x)
Ω ⊂ Rn
(n = 2, 3)
W
Can one determine the electrical conductivity of Ω, γ(x),
by making voltage and current measurements at the
boundary?
(Calderón; Geophysical prospection)
Early breast cancer detection
Normal breast tissue
Cancerous breast tumor
0.3 mho
2.0 mho
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REMINISCENCIA DE MI VIDA MATEMATICA
Speech at Universidad Autónoma de Madrid accepting
the ‘Doctor Honoris Causa’:
My work at “Yacimientos Petroliferos Fiscales”
(YPF) was very interesting, but I was not well
treated, otherwise I would have stayed there.
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10
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CALDERÓN’S PROBLEM (EIT)
Consider a body Ω ⊂ Rn. An electrical potential u(x)
causes the current
I(x) = γ(x)∇u
The conductivity γ(x) can be isotropic, that is, scalar,
or anisotropic, that is, a matrix valued function. If the
current has no sources or sinks, we have
div(γ(x)∇u) = 0 in Ω
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div(γ(x)∇u(x)) = 0
u
∂Ω
=f
γ(x) = conductivity,
f = voltage potential at ∂Ω
Current flux at ∂Ω = (ν · γ∇u)
outer normal.
∂Ω
were ν is the unit
Information is encoded in
map
Λγ (f ) = ν · γ∇u
∂Ω
EIT (Calderón’s inverse problem)
Does Λγ determine γ ?
Λγ = Dirichlet-to-Neumann map
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Theorem n ≥ 3 (Sylvester-U, 1987)
γ ∈ C 2(Ω),
0 < C1 ≤ γ(x) ≤ C2
on Ω
Λγ1 = Λγ2 ⇒ γ1 = γ2
• Extended to γ ∈ C 3/2(Ω) (Päivarinta-Panchenko-U,
Brown-Torres)
• γ ∈ C 1+(Ω), γ conormal ( Greenleaf-Lassas-U)
Complex-Geometrical Optics Solutions (CGO)
• Reconstruction A. Nachman
• Stability G. Alessandrini
• Numerical Methods (yes in n=2, D. Issacson, J. Müller,
S. Siltanen
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Reduction to Schrödinger equation
div(γ∇w) = 0
√
u = γw
Then the equation is transformed into:
√
n
X
∆ γ
∂2
(∆ − q)u = 0, q = √
∆=
2
γ
∂x
i
i=1
(∆ − q)u = 0
u
∂Ω
=f
∂u Define Λq (f ) =
∂ν ∂Ω
ν = unit-outer normal to ∂Ω.
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IDENTITY
Z
Ω
(q1 − q2)u1u2 =
Z
∂Ω
(Λq1 − Λq2 )u1
u2
∂Ω
∂Ω
dS
(∆ − qi)ui = 0
If Λγ1 = Λγ2 ⇒ Λq1 = Λq2 and
Z
Ω
(q1 − q2)u1u2 = 0
GOAL: Find MANY solutions of (∆ − qi)ui = 0.
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CGO SOLUTIONS
Calderón: Let ρ ∈ Cn, ρ · ρ = 0
ρ = η + ik
η, k ∈ Rn, |η| = |k|, η · k = 0
u = ex·ρ = ex·η eix·k
∆u = 0,
u=




exponentially decreasing, x · η < 0
oscillating, x · η = 0



exponentially increasing, x · η > 0
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COMPLEX GEOMETRICAL OPTICS
(Sylvester-U) n ≥ 2, q ∈ L∞(Ω)
Let ρ ∈ Cn (ρ = η + ik, η, k ∈ Rn) such that ρ · ρ = 0
(|η| = |k|, η · k = 0).
Then for |ρ| sufficiently large we can find solutions of
(∆ − q)wρ = 0 on Ω
of the form
wρ = ex·ρ(1 + Ψq (x, ρ))
with Ψq → 0 in Ω as |ρ| → ∞.
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Proof Λq1 = Λq2 ⇒ q1 = q2
Z
Ω
(q1 − q2)u1u2 = 0
u1 = ex·ρ1 (1 + Ψq1 (x, ρ1)),
ρ1 · ρ1 = ρ2 · ρ2 = 0,
ρ1 = η + i(k + l)
ρ2 = −η + i(k − l)
η · k = η · l = l · k = 0,
Z
Ω
u2 = ex·ρ2 (1 + Ψq2 (x, ρ2))
|η|2 = |k|2 + |l|2
(q1 − q2)e2ix·k (1 + Ψq1 + Ψq2 + Ψq1 Ψq2 ) = 0
Letting |l| → ∞
Z
Ω
(q1 − q2)e2ix·k = 0
∀k =⇒ q1 = q2
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PARTIAL DATA PROBLEM
Suppose we measure
Λγ (f )|Γ,
suppf ⊆ Γ0
Γ, Γ0 open subsets of ∂Ω
Can one recover γ?
Important case Γ = Γ0.
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EXTENSION OF CGO SOLUTIONS
u = ex·ρ(1 + Ψq (x, ρ))
ρ ∈ Cn , ρ · ρ = 0
(Not helpful for localizing)
Kenig-Sjöstrand-U (2007),
u = eτ (ϕ(x)+iψ(x))(a(x) + ψ(x, τ ))
τ ∈ R, ϕ, ψ real-valued, ψ(x, τ ) → 0 as τ → ∞.
ϕ limiting Carleman weight,
∇ϕ · ∇ψ = 0,
Example: ϕ(x) = ln |x − x0|,
|∇ϕ| = |∇ψ|
x0 ∈
/ ch(Ω)
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CGO SOLUTIONS
u = eτ (ϕ(x)+iψ(x))(a0(x) + ψ(x, τ ))
τ →∞
ψ(x, τ ) −→ 0 in Ω
ϕ(x) = ln |x − x0|
Complex Spherical Waves
Theorem (Kenig-Sjöstrand-U) Ω strictly convex.
Λq1 = Λ q2 ,
Γ
Γ
Γ ⊆ ∂Ω,
Γ arbitrary
⇒ q1 = q2
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Theorem (Kenig-Sjöstrand-U) Ω strictly convex.
Λq1 = Λ q2 ,
Γ
Γ
Γ ⊆ ∂Ω,
Γ arbitrary
⇒ q1 = q2
uτ = eτ (ϕ+iψ)aτ
Eikonal:
ϕ(x) = ln |x − x0|, x0 ∈\ch(Ω)
∇ϕ · ∇ψ = 0, |∇ϕ| = |∇ψ|
x−x0
n−1 : smooth
,
ω),
ω
∈
S
ψ(x) = d( |x−x
0|
for x ∈ Ω̄.
Transport: (∇ϕ + i∇ψ) · ∇aτ = 0
(Cauchy-Riemann equation in plane generated by ∇ϕ, ∇ψ)
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ϕ(x) = ln |x − x0|,
x0 ∈\ch(Ω)
Carleman Estimates
u|∂Ω = ∂u
∂ν |∂Ω− = 0
Z
∂Ω+
>
∂Ω± = {x ∈ ∂Ω; ∇ϕ·ν < 0}
C
∂u
|(∆ − q)ue−τ ϕ(x)|2ds
< ∇ϕ, ν > |e−τ ϕ(x) |2ds ≤
∂ν
τ Ω
Z
This gives control of ∂u
∂ν |∂Ω+ ,δ ,
∂Ω+,δ = {x ∈ ∂Ω, ∇ϕ · ν ≥ δ}
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The Two Dimensional Case
Theorem (n = 2) Let γj ∈ C 2(Ω), j = 1, 2.
Assume Λγ1 = Λγ2 . Then γ1 = γ2 .
• Nachman (1996)
• Brown-U (1997) Improved to γj Lipschitz
• Astala-Päivärinta (2006) Improved to γj ∈ L∞(Ω)
Recall
div(γ∇u) = 0, γ ∈ L∞(Ω)
u|∂Ω = f
Qγ (f ) =
Z
Ω
γ|∇u|2dx = hΛγ f, f iL2(∂Ω).
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Partial data
Let Γ ⊆ ∂Ω, Γ open.
Let qj ∈ C 1+ε(Ω), ε > 0, j = 1, 2.
Theorem (Imanuvilov-U-Yamamoto) n=2. Assume
Λq1 (f )
Γ
= Λq2 (f )
Γ
∀ f , suppf ⊆ Γ. Then
q1 = q2 .
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Γ0 = ∂Ω − Γ
Partial data
Construct CGO solutions
∆uj + qj uj = 0 in Ω
uj |Γ0 = 0
In this case
Z
Ω
(q1 − q2)u1u2dx = 0
if Λq1 (f )|Γ = Λq2 (f )|Γ, suppf ⊆ Γ.
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Z
Ω
(q1 − q2)u1u2 = 0
uj |Γ0 = uj |∂Ω−Γ = 0
a (z)
a (z)
(1)
) + eτ Φ(z)(a(z) + 1
) + eτ ϕRτ
u1(x) = eτ Φ(z)(a(z) + 0
τ
τ
b (z)
b (z)
(2)
) + e−τ Φ̄(z)(ā(z) + 1
) + eτ ϕRτ
u2(x) = e−τ Φ̄(z)(ā(z) + 0
τ
τ
Φ = ϕ + iψ holomorphic
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u1 = Re eτ Φ(z)(a(z) + · · · ),
u2 = Re e−τ Φ̄(z)(ā(z) + · · · )
Φ(z) = ϕ + iψ
holomorphic
uj |∂Ω−Γ = 0
p ∈ Ω, Φ has non-degenerate critical point at p (Morse
function)
∂¯a = 0 Re a|∂Ω−Γ = 0
a = 0 at other critical points
Stationary phase in
Z
Ω
(q1 − q2)u1u2 = 0
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u1 = Re eτ Φ(z)(a(z) + · · · )
u2 = Re e−τ Φ̄(z)(ā(z) + · · · )
uj |∂Ω−Γ = 0
Φ(z) Morse function with non-degenerate critical point
at p.
Z
Ω
(q1 − q2)u1u2 = 0
Stationary phase
=⇒ (q1 − q2)(p) = 0
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Anisotropic case
Cardiac muscle
γ = (γ ij )
conductivity
6.3 mho
2.3 mho
(longitudinal)
(transversal)
positive-definite, symmetric
matrix
Ω ⊆ Rn, Ω bounded. Under assumptions of no sources
or sinks of current the potential u satisfies
div(γ∇u) = 0
n
X
∂
∂u
γ ij
=0 in Ω
∂x
∂x
i
i
i,j=1
u∂Ω =f
(*)
f = voltage potential at boundary
Isotropic
(
1,
γ ij (x) = α(x)δ ij ; δ ij =
0,
i=j
i 6= j
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n
X


∂  ij ∂u 
γ
=0 in Ω
∂xj
i,j=1 ∂xi
u
∂u
i
ij
Λγ (f ) =
νγ
∂x
j
i,j=1
∂Ω
(*)
=f
n
X
ν = ν 1, · · · , ν n
∂Ω
is the unit outer normal to ∂Ω
Λγ (f ) is the induced current flux at ∂Ω.
Λγ is the voltage to current map or Dirichlet - to Neumann map
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n
X


∂  ij ∂u 
γ
=0 in Ω
∂xj
i,j=1 ∂xi
u
∂Ω
(*)
=f
n
X
∂u
i
ij
Λγ (f ) =
νγ
∂x
j
i,j=1
∂Ω
EIT: Can we recover γ in Ω from Λγ ?
Reformulate it in geometric terms
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DIRICHLET-TO-NEUMANN MAP
(M, g) compact Riemannian manifold with boundary.
∆g Laplace-Beltrami operator g = (gij ) pos. def. symmetric matrix


n
X
∂ q
∂u 
1
det g g ij
(g ij ) = (gij )−1
∆g u = √
det g i,j=1 ∂xi
∂xj
∆g u = 0 on M
u
∂M
Conductivity:
√
ij
γ = det g g ij
=f
Λg (f ) =
n
X
ν j g ij
det g ∂u q
∂xi
∂M
i,j=1
ν = (ν 1, · · · , ν n) unit-outer normal
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∆g u = 0
u
Λg (f ) =
∂M
n
X
=f
∂u
=
ν j g ij
∂νg
∂xi
i,j=1
current flux at ∂M
det g ∂u q
∂M
Inverse-problem (EIT)
Can we recover g from Λg ?
Λg = Dirichlet-to-Neumann map or voltage to current
map
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ANOTHER MOTIVATION (STRING THEORY)
HOLOGRAPHY
Dirichlet-to-Neumann map is the “boundary-2pt function”
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
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∆g u = 0
u
∂M
=f
Answer: No
∂u Λg (f ) =
∂νg Λg ⇒ g
∂M
Λψ∗g = Λg where
ψ : M → M diffeomorphism, ψ ∂M
= Identity and
∗
T
ψ g = Dψ ◦ g ◦ (Dψ) ◦ ψ
38
?
Show Λψ∗g = Λg ; ψ : M → M diffeomorphism, ψ =
∂M
Identity
√
P R
∂u
∂u
ij
Qg (f ) = i,j M g ∂x ∂x det gdx
i
j
Z
Qg (f ) = −
∂M
Λg (f )f dS
Qg ⇔ Λg
v = u ◦ ψ, ∆ψ∗g v = 0
Qψ∗g = Qg ⇒ Λψ∗g = Λg
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Theorem (n ≥ 3) (Lassas-U, Lassas-Taylor-U) (M, gi), i =
1, 2, real-analytic, connected, compact, Riemannian manifolds with boundary. Let Γ ⊆ ∂M , Γ open. Assume
Λg1 (f )|Γ = Λg2 (f )|Γ,
∀f, f supported in Γ
Then ∃ψ : M → M diffeomorphism, ψ Γ
= Identity, so
that
g1 = ψ ∗ g2
In fact one can determine topology of M, as well (only
need to know Λg , ∂M ).
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Theorem (n = 2)(Lassas-U)
(M, gi), i = 1, 2, connected Riemannian manifold with
boundary. Let Γ ⊆ ∂M , Γ open. Assume
Λg1 (f )|Γ = Λg2 (f )|Γ,
∀f, f supported in Γ
Then ∃ψ : M → M diffeomorphism, ψ = Identity, and
Γ
β > 0, β = 1 so that
Γ
g1 = βψ ∗g2
In fact, one can determine topology of M as well.
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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 information 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.
42
Greenleaf-Lassas-U (2003 MRL)
Let
Ω = B(0, 2) ⊂ R3,
where B(0, r) = {x ∈ R3; |x| < r}
D = B(0, 1)
F : Ω \ {0} → Ω \ D
|x|
x
F (x) = (
+ 1)
2
|x|
F diffeomorphism, F ∂Ω
= Identity
43
g = identity metric in B(0, 2)
Let ĝ = (F −1)∗g on B(0, 2) \ B(0, 1)
σ̂ = conductivity associated to ĝ
In spherical coordinates (r, φ, θ) → (r sin θ cos φ, r sin θ sin φ, r cos θ)


2 sin θ
2(r
−
1)
0
0




σ̂ = 
0
2 sin θ
0

−1
0
0
2(sin θ)
Let ge be the metric in B(0, 2) (positive definite in B(0, 1))
s.t. ge = ĝ in B(0, 2) \ B(0, 1). Then
Theorem (Greenleaf-Lassas-U 2003)
Λge = Λg
44
Based on work of Greenleaf-Lassas-U, MRL 2003
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