fi Finnish Centre of Excellence in Inverse Problems Research p. - - PowerPoint PPT Presentation

Geometric methods for inverse problems

Matti Lassas

AB HELSINKI UNIVERSITY OF TECHNOLOGY

Institute of Mathematics

fi

Finnish Centre of Excellence in Inverse Problems Research

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1 Inverse problem in applications

Travel time problem: Measure travel times of sound waves

between all boundary points of a body. Can the wave speed be determined in the body?

Calder´

• n’s inverse problem: Measure electric resistance

between all boundary points of a body. Can the conductivity be determined in the body?

Inverse problem for heat equation: Let us warm the boundary of

a body and measure the temperature at the boundary. Can the heat conductivity be determined in the body?

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2 Inverse conductivity problem

Consider a body Ω ⊂ Rn. An electric potential u(x) causes the current

J(x) = σ(x)∇u(x).

Here the conductivity σ(x) can be a scalar or matrix valued function. If the current has no sources inside the body, we have

∇· σ(x)∇u(x) = 0.

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∇· σ(x)∇u(x) = 0

in Ω,

n · σ∇u|∂Ω = j.

Imaging problem: Do the measurements made on the

boundary determine the conductivity, that is, does the map

Λσ, Λσ(j) = u|∂Ω

determine the conductivity σ(x) in Ω?

Figure: University of Kuopio.

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3 Quantum mechanics

Assume σ : Ω → R+. Let us continue σ(x) to R3 \ Ω such that σ(x) = c for |x| > R. Using the conductivity equation

∇· σ(x)∇u(x) = 0

in R3, we define

ψ = σ1/2u, q(x) = −∆σ1/2 σ1/2 .

Then ψ is a solution of the Schrödinger equation

(∆ + q(x))ψ = 0

in R3.

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Consider the Schrödinger equation

(∆ + q(x))ψ = 0

in R3.

Inverse problems:

Can we determine q(x) when we are given scattering data? Can we determine q(x) if we know all possible solutions ψ in the domain R3 \ Ω?

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History of inverse conductivity problem: Calderón 1980: Solution of the linearized inverse problem. Sylvester-Uhlmann 1988: Nonlinear inverse problem in

Rn, n ≥ 3

Nachman 1996: Nonlinear inverse problem in R2 Astala-Päivärinta 2003: Nonlinear inverse problem in

R2 with L∞-conductivity

Sylvester 1990: Inverse problem for an anisotropic conductivity near constant in R2. Siltanen-Mueller-Isaacson 2000: Explicit numerical solution for the 2D-inverse problem.

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Theorem 1 (Sylvester-Uhlmann) Consider equation the equation (∆ + q)u = 0 in Ω ⊂ Rn, n ≥ 3. The Neumann to Dirichlet map Λq : ∂nu|∂Ω → u|∂Ω determines q uniquely.

Idea of the proof. Assume Λq = Λe

• q. For ρ ∈ Cn,

ρ = (ρ1, ρ2, . . . , ρn), ρ · ρ = ρ2

1 + ρ2 2 + · · · + ρ2 n = 0 we have

∆uρ = 0

for uρ(x) = exp(ρ · x). Let ξ ∈ Rn. We can choose ρ + η = iξ, such that

ρ · ρ = 0, η · η = 0. The equations (∆ + q)wρ = 0 in Ω, (∆ + q) wη = 0 in Ω,

have solutions such that for large ρ

wρ − exp(ρ · x)L2(Ω,exp(Re ρ·x)dx) ≤ C |ρ|.

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Since Λq = Λe

q and η + ρ = iξ, ξ ∈ Rn, we have

O( 1 |ρ|) =

• ∂Ω

(wη∂n wρ − ∂nwρ wη)dS =

• Ω

(wη∆ wρ − ∆wρ wη)dx =

• Ω

(−wη q wρ + qwη wρ)dx =

• Ω

(q − q) exp((η + ρ)· x)dx + O( 1 |ρ|) = F(q − q)(ξ) + O( 1 |ρ|)

Thus the Fourier transform of q −

q vanishes.

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Theorem 2 (Astala-Päivärinta 2003) Let Ω ⊂ R2 be a simply connected bounded domain and σ ∈ L∞(Ω; R+) an isotropic conductivity function. Then the Neumann to Dirichlet map Λσ for the equation

∇ · σ∇u = 0

determines uniquely the conductivity σ.

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4 Anisotropic inverse problems

Assume that ∇· σ∇u = 0 in Ω, where σ is matrix valued. Let

F be diffeomorphism F : Ω → Ω, F|∂Ω = id.

Let v(x) = u(F(x)) and

γ(y) = F∗σ(y) = (DF)(x)· σ(x)· (DF(x))t

det(DF(x))

• x=F −1(y)

Then

∇· γ∇v = 0

in Ω and Λσ = Λγ.

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5 Invariant formulation

Assume n ≥ 3. Consider Ω as a Riemannian manifold with

gjk(x) = (det σ(x))−1/(n−2)σjk(x).

The conductivity equation is a Laplace-Beltrami equation

∆gu = 0

in Ω,

∆gu =

n

• j,k=1

g−1/2 ∂ ∂xj (g1/2gjk ∂u ∂xk ),

where g = det(gij), [gij] = [gjk]−1.

Inverse problem: Can we determine the Riemannian manifold

(M, g) by knowing ∂M and Λg : ∂νu|∂Ω → u|∂Ω.

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Theorem 3 (L.-Taylor-Uhlmann) Assume that (M, g) is a

n-dimensional real-analytic Riemannian manifold and n ≥ 3.

Then ∂M and

Λg : ∂νu|∂M → u|∂M

determine uniquely (M, g). Theorem 4 (L.-Uhlmann) Assume that (M, g) is a compact

2-dimensional C∞ smooth Riemannian manifold. Then ∂M

and

Λg : ∂νu|∂M → u|∂M

determine uniquely the conformal class

{(M, αg) : α ∈ C∞(M), α > 0}.

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Theorem 5 (Pestov-Uhlmann) Assume that (M, g) is a compact 2-dimensional Riemannian manifold such that any

x, y ∈ M can be connected with a unique geodesic in M.

Then ∂M and

d(x, y), x, y ∈ ∂M

determine (M, g). The reason for this is that the distances of the boundary points determine the boundary map

Λg : ∂νu|∂M → u|∂M

for equation ∆gu = 0.

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Theorem 6 (Astala-L.-Päivärinta 2005) Let Ω ⊂ R2 be a simply connected bounded domain and

σ1, σ2 ∈ L∞(Ω; R2×2) conductivity tensors. If Λσ1 = Λσ2 then

there is a W 1,2-diffeomorphism

F : Ω → Ω, F|∂Ω = Id

such that

σ1 = F∗σ2.

Recall that if F : Ω →

Ω is a diffeomorphism, it transforms

the conductivity σ in Ω to

σ = F∗σ in Ω,

• σ(x) = DF(y) σ(y) (DF(y))t

|detDF(y)|

• y=F −1(x)

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• Proof. Identify R2 = C. Let σ be an anisotropic conductivity,

σ(x) = I for x ∈ C \ Ω. There is F : C → C such that γ = F∗σ

is isotropic. There are solutions w(x, k) such that

∇· γ∇w = 0 in C

and

lim

x→∞ w(x, k)e−ikx = 1,

lim

k→∞

1 k log(w(x, k)e−ikx) = 0.

Let u(x, k) = w(F −1(x), k). The boundary data determines

u(x, k) for x ∈ C \ Ω and F −1(x) = lim

k→∞

log u(x, k) ik , x ∈ C \ Ω.

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Generally, solutions of anisotropic inverse problem are not

• unique. However, if we have enough a priori knowledge of

the form of the conductivity, we can sometimes solve the inverse problem uniquely. Manifold (M, g)

• Boundary measurements
• σjk(x) on Ω ⊂ R2

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6 Electrical Impedance Tomography with an unknown boundary

Practical task: In medical imaging one often wants to find an

image of the conductivity, when the domain Ω ⊂ R2 is poorly known.

Figure: Rensselaer Polytechnic Institute.

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Formulation of unknown boundary problem:

• 1. Assume that γ is an isotropic conductivity in Ω.
• 2. Assume that we are given a set Ωm that is our best

guess for Ω. Let Fm : Ω → Ωm be a map corresponding to the modeling error.

• 3. We are given the map

Λ = (Fm)∗Λγ that corresponds to

the measurements done with electrodes on ∂Ω.

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The boundary map

Λ on ∂Ωm is equal to Λγ1 that

corresponds to boundary measurements made with an

anisotropic conductivity γ1 = (Fm)∗γ in Ωm.

Assume we are given Ωm and

Λ. Our aim is to find a

conductivity tensor in Ωm that is as close to an isotropic conductivity as possible.

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Definition 6.1 Let γ = γjk(x) be a matrix valued

• conductivity. Let λ1(x) and λ2(x), λ1(x) ≥ λ2(x) be its
• eigenvalues. Anisotropy of γ at x is

K(γ, x) = λ1(x) − λ2(x) λ1(x) + λ2(x) 1/2 .

The maximal anisotropy of γ in Ω is

K(γ) = sup

x∈Ω

K(γ, x).

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The anisotropy function K(

γ, x) is constant for

• γ(x) = η(x)Rθ(x)
• λ1/2

λ−1/2

• R−1

θ(x)

where

λ ≥ 1, η(x) ∈ R+, Rθ =

• cos θ

sin θ − sin θ cos θ

• .

We say that

γ = γλ,θ,η is a uniformly anisotropic conductivity.

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Theorem 6.2 (Kolehmainen-L.-Ola 2005) Let Ω ⊂ R2 be a bounded, simply connected C1,α–domain and γ ∈ L∞(Ω, R) be isotropic. Let Ωm be a model domain and Fm : Ω → Ωm be a C1,α–diffeomorphism. Assume that we know ∂Ωm and

Λ = Λ(Fm)∗γ. Then there is a

unique anisotropic conductivity σ in Ωm such that

1. Λσ = Λ. 2.

If σ1 satisfies Λσ1 =

Λ then K(σ1) ≥ K(σ).

Moreover, the conductivity σ is uniformly anisotropic.

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Algorithm:

The conductivity σ =

γλ,η,θ can obtained by solving the

minimization problem

min

(λ,θ,η)∈S λ,

where S = {(λ, θ, η) : Λb

γ(λ,θ,η) =

Λ}.

In implementation of the algorithm we approximate this by

min

(λ,θ,η) ||Λb γ(λ,θ,η) −

Λ||2 + ǫ1|λ − 1|2 + ǫ2(||θ||2 + ||η||2).

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Let Fm : Ω → Ωm be the boundary modeling map and σ be the conductivity with the smallest possible anisotropy such that Λσ =

Λ. Then

Corollary 6.3 Then there is a unique map

Fe : Ω → Ωm,

depending only on Fm|∂Ω : ∂Ω → ∂Ωm such that det(σ(x))1/2 = γ(F −1

e

(x)).

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7 Idea of the proof

If F : Ω → Ω is a diffeomorphism and γ1 is isotropic conductivity, then

K(F∗γ1, x) = |µF(x)|

where

µF = ∂F ∂F , ∂ = 1 2(∂x1 − i∂x2).

To find the minimally anisotropic conductivity we need to find a quasiconformal map with the smallest possible dilatation and the given boundary values. This is called the Teichmüller map.

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1 2 0.78 1.41 0.03 1.3 0.77 1.41 – p. 27/28

0.05 8.08 0.2 16.06 0.55 8.23 – p. 28/28