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 – p. 1/28

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? on’s inverse problem: Measure electric resistance Calder´ 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? – p. 2/28

2 Inverse conductivity problem Consider a body Ω ⊂ R n . 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 . – p. 3/28

∇· 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 Λ σ , Λ σ ( j ) = u | ∂ Ω determine the conductivity σ ( x ) in Ω ? Figure: University of Kuopio. – p. 4/28

3 Quantum mechanics Assume σ : Ω → R + . Let us continue σ ( x ) to R 3 \ Ω such that σ ( x ) = c for | x | > R . Using the conductivity equation in R 3 , ∇· σ ( x ) ∇ u ( x ) = 0 we define q ( x ) = − ∆ σ 1 / 2 ψ = σ 1 / 2 u, σ 1 / 2 . Then ψ is a solution of the Schrödinger equation in R 3 . (∆ + q ( x )) ψ = 0 – p. 5/28

Consider the Schrödinger equation in R 3 . (∆ + q ( x )) ψ = 0 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 R 3 \ Ω ? – p. 6/28

History of inverse conductivity problem: Calderón 1980: Solution of the linearized inverse problem. Sylvester-Uhlmann 1988: Nonlinear inverse problem in R n , n ≥ 3 Nachman 1996: Nonlinear inverse problem in R 2 Astala-Päivärinta 2003: Nonlinear inverse problem in R 2 with L ∞ -conductivity Sylvester 1990: Inverse problem for an anisotropic conductivity near constant in R 2 . Siltanen-Mueller-Isaacson 2000: Explicit numerical solution for the 2D-inverse problem. – p. 7/28

Theorem 1 (Sylvester-Uhlmann) Consider equation the equation (∆ + q ) u = 0 in Ω ⊂ R n , n ≥ 3 . The Neumann to Dirichlet map Λ q : ∂ n u | ∂ Ω → u | ∂ Ω determines q uniquely. q . For ρ ∈ C n , Idea of the proof. Assume Λ q = Λ e ρ = ( ρ 1 , ρ 2 , . . . , ρ n ) , ρ · ρ = ρ 2 1 + ρ 2 2 + · · · + ρ 2 n = 0 we have ∆ u ρ = 0 for u ρ ( x ) = exp( ρ · x ) . Let ξ ∈ R n . We can choose ρ + η = iξ , such that η · η = 0 . The equations ρ · ρ = 0 , (∆ + q ) w ρ = 0 in Ω , (∆ + � q ) � w η = 0 in Ω , have solutions such that for large ρ � w ρ − exp( ρ · x ) � L 2 (Ω , exp( Re ρ · x ) dx ) ≤ C | ρ | . – p. 8/28

q and η + ρ = iξ , ξ ∈ R n , we have Since Λ q = Λ e � O ( 1 | ρ | ) = ( w η ∂ n � w ρ − ∂ n w ρ � w η ) dS ∂ Ω � = ( w η ∆ � w ρ − ∆ w ρ � w η ) dx Ω � = ( − w η � w ρ + qw η � w ρ ) dx q � Ω � q ) exp(( η + ρ ) · x ) dx + O ( 1 = ( q − � | ρ | ) Ω q )( ξ ) + O ( 1 = F ( q − � | ρ | ) Thus the Fourier transform of q − � q vanishes. – p. 9/28

Theorem 2 (Astala-Päivärinta 2003) Let Ω ⊂ R 2 be a simply connected bounded domain and σ ∈ L ∞ (Ω; R + ) an isotropic conductivity function. Then the Neumann to Dirichlet map Λ σ for the equation ∇ · σ ∇ u = 0 determines uniquely the conductivity σ . – p. 10/28

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 in Ω ∇· γ ∇ v = 0 and Λ σ = Λ γ . – p. 11/28

5 Invariant formulation Assume n ≥ 3 . Consider Ω as a Riemannian manifold with g jk ( x ) = ( det σ ( x )) − 1 / ( n − 2) σ jk ( x ) . The conductivity equation is a Laplace-Beltrami equation ∆ g u = 0 in Ω , n � g − 1 / 2 ∂ ∂x j ( g 1 / 2 g jk ∂u ∆ g u = ∂x k ) , j,k =1 where g = det ( g ij ) , [ g ij ] = [ g jk ] − 1 . Inverse problem: Can we determine the Riemannian manifold ( M, g ) by knowing ∂M and Λ g : ∂ ν u | ∂ Ω �→ u | ∂ Ω . – p. 12/28

Theorem 3 (L.-Taylor-Uhlmann) Assume that ( M, g ) is a n -dimensional real-analytic Riemannian manifold and n ≥ 3 . Then ∂M and Λ g : ∂ ν u | ∂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 Λ g : ∂ ν u | ∂M �→ u | ∂M determine uniquely the conformal class α ∈ C ∞ ( M ) , α > 0 } . { ( M, αg ) : – p. 13/28

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 Λ g : ∂ ν u | ∂M �→ u | ∂M for equation ∆ g u = 0 . – p. 14/28

Theorem 6 (Astala-L.-Päivärinta 2005) Let Ω ⊂ R 2 be a simply connected bounded domain and σ 1 , σ 2 ∈ L ∞ (Ω; R 2 × 2 ) conductivity tensors. If Λ σ 1 = Λ σ 2 then there is a W 1 , 2 -diffeomorphism F : Ω → Ω , F | ∂ Ω = Id such that σ 1 = F ∗ σ 2 . Recall that if F : Ω → � Ω is a diffeomorphism, it transforms σ = F ∗ σ in � the conductivity σ in Ω to � Ω , � σ ( x ) = DF ( y ) σ ( y ) ( DF ( y )) t � � � � | det DF ( y ) | y = F − 1 ( x ) – p. 15/28

Proof. Identify R 2 = 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 1 x →∞ w ( x, k ) e − ikx = 1 , k log( w ( x, k ) e − ikx ) = 0 . lim lim k →∞ Let u ( x, k ) = w ( F − 1 ( x ) , k ) . The boundary data determines u ( x, k ) for x ∈ C \ Ω and log u ( x, k ) F − 1 ( x ) = lim x ∈ C \ Ω . , ik k →∞ – p. 16/28

� � 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 ) � � � � � � � � � � � � � � � � � � � � � � � � � σ jk ( x ) on Ω ⊂ R 2 Boundary measurements – p. 17/28

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 ⊂ R 2 is poorly known. Figure: Rensselaer Polytechnic Institute. – p. 18/28

Formulation of unknown boundary problem: 1. Assume that γ is an isotropic conductivity in Ω . 2. Assume that we are given a set Ω m that is our best guess for Ω . Let F m : Ω → Ω m be a map corresponding to the modeling error. 3. We are given the map � Λ = ( F m ) ∗ Λ γ that corresponds to the measurements done with electrodes on ∂ Ω . – p. 19/28

The boundary map � Λ on ∂ Ω m is equal to Λ γ 1 that corresponds to boundary measurements made with an anisotropic conductivity γ 1 = ( F m ) ∗ γ in Ω m . Assume we are given Ω m and � Λ . Our aim is to find a conductivity tensor in Ω m that is as close to an isotropic conductivity as possible. – p. 20/28

Definition 6.1 Let γ = γ jk ( x ) be a matrix valued conductivity. Let λ 1 ( x ) and λ 2 ( x ) , λ 1 ( x ) ≥ λ 2 ( x ) be its eigenvalues. Anisotropy of γ at x is � λ 1 ( x ) − λ 2 ( x ) � 1 / 2 K ( γ, x ) = . λ 1 ( x ) + λ 2 ( x ) The maximal anisotropy of γ in Ω is K ( γ ) = sup K ( γ, x ) . x ∈ Ω – p. 21/28

The anisotropy function K ( � γ, x ) is constant for � � λ 1 / 2 0 R − 1 γ ( x ) = η ( x ) R θ ( x ) � λ − 1 / 2 θ ( x ) 0 where λ ≥ 1 , η ( x ) ∈ R + , � � cos θ sin θ R θ = . − sin θ cos θ We say that � γ λ,θ,η is a uniformly anisotropic conductivity . γ = � – p. 22/28

Theorem 6.2 (Kolehmainen-L.-Ola 2005) Let Ω ⊂ R 2 be a bounded, simply connected C 1 ,α –domain and γ ∈ L ∞ (Ω , R ) be isotropic. Let Ω m be a model domain and F m : Ω → Ω m be a C 1 ,α –diffeomorphism. Assume that we know ∂ Ω m and � Λ = Λ ( F m ) ∗ γ . Then there is a unique anisotropic conductivity σ in Ω m such that Λ σ = � 1 . Λ . If σ 1 satisfies Λ σ 1 = � 2 . Λ then K ( σ 1 ) ≥ K ( σ ) . Moreover, the conductivity σ is uniformly anisotropic. – p. 23/28

Algorithm: The conductivity σ = � γ λ,η,θ can obtained by solving the minimization problem γ ( λ,θ,η ) = � min where S = { ( λ, θ, η ) : Λ b Λ } . ( λ,θ,η ) ∈ S λ, In implementation of the algorithm we approximate this by Λ || 2 + ǫ 1 | λ − 1 | 2 + ǫ 2 ( || θ || 2 + || η || 2 ) . γ ( λ,θ,η ) − � ( λ,θ,η ) || Λ b min – p. 24/28