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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

  1. 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

  2. 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

  3. 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

  4. ∇· 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. � � 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

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

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