geometric inverse problems for linear and non linear wave
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Geometric inverse problems for linear and non-linear wave equations - PowerPoint PPT Presentation

Geometric inverse problems for linear and non-linear wave equations Matti Lassas University of Helsinki Finnish Centre of Excellence in Inverse Modelling and Imaging 2018-2025 2018-2025 Outline: Inverse problem for linear wave equation


  1. Geometric inverse problems for linear and non-linear wave equations Matti Lassas University of Helsinki Finnish Centre of Excellence in Inverse Modelling and Imaging 2018-2025 2018-2025

  2. Outline: ◮ Inverse problem for linear wave equation ◮ Generalised time reversal methods and point sources ◮ Inverse problem for non-linear wave equation

  3. Boundary measurements for anisotropic wave operator Let n ≥ 2, M ⊂ R n have a smooth boundary and g ( x ) be a positive definite matrix depending on x ∈ M . Let ν be the unit normal of the boundary ∂ M and ( x , t ) ∈ M × R + . Let u ( x , t ) = u f ( x , t ) solve the wave equation ( ∂ 2 t − ∇ · g ( x ) ∇ ) u ( x , t ) = 0 on ( x , t ) ∈ M × R + , ν · g ∇ u ( x , t ) | ∂ M × R + = f ( x , t ) , u | t = 0 = 0 , ∂ t u | t = 0 = 0 , with the Neumann boundary value f . Then the Neumann-to-Dirichlet map is � � Λ g f = u f ( x , t ) � � ( x , t ) ∈ ∂ M × R + Observation: There are different matrix valued functions g ( 1 ) ( x ) and g ( 2 ) ( x ) such that Λ g ( 1 ) = Λ g ( 2 ) .

  4. Next we consider M ⊂ R n as Riemannian manifold with a boundary. Then the travel time metric is given by ds 2 = � n j , k = 1 g jk ( x ) dx j dx k , where ( g jk ) = ( g jk ) − 1 . The distance of points x , y ∈ M is dist M ( x , y ) = travel time between x and y . Manifold is given by a collection of local coordinate charts.

  5. Inverse problem on manifold Let M be a Riemannian manifold with boundary, ( g jk ( x )) n j , k = 1 be the Riemannian metric and ∆ g be the Laplace operator n � | g | − 1 / 2 ∂ ∂ x j ( | g | 1 / 2 g jk ∂ ∆ g u = ∂ x k u ) , j , k = 1 ( g jk ) = ( g jk ) − 1 . | g | = det ( g jk ) , Let u f = u f ( x , t ) be the solution of ∂ 2 t u − ∆ g u = 0 on M × R + , ∂ ν u | ∂ M × R + = f , u | t = 0 = 0 , ∂ t u | t = 0 = 0 , where ν is unit interior normal of ∂ M . Define Λ g f = u f | ∂ M × R + . Assume that we are given the boundary data ( ∂ M , Λ g ) .

  6. Inverse problem on a manifold Inverse problem: Let ∂ M and the map Λ g : ∂ ν u | ∂ M × R + �→ u | ∂ M × R + be given. Can we determine the manifold ( M , g ) by representing it as a collection of images of g on local coordinate charts? Near the boundary ∂ M , we can represent the metric tensor in the time migration coordinates, i.e., the boundary normal coordinates. (Figure by Cameron, et al, Berkeley.)

  7. Results of Belishev-Kurylev 1992 and Tataru 1994 yield Theorem Let M ⊂ R n and g ( 1 ) ( x ) and g ( 2 ) ( x ) be smooth matrix-valued functions such that Λ g ( 1 ) = Λ g ( 2 ) for the equation ∂ 2 t u − ∆ g u = 0 . Then ( M , g ( 1 ) ) and ( M , g ( 2 ) ) isometric, that is, there is a diffeomorphism F : M → M , F | ∂ M = Id such that � � g ( 2 ) ( x ) = DF ( y ) t · g ( 1 ) ( y ) · DF ( y ) � (1) � y = F − 1 ( x ) If Λ g ( 1 ) = Λ g ( 2 ) for the equation ( ∂ 2 t − ∇ · g ( x ) ∇ ) u = 0 then the map F in (1) satisfies also det ( DF ( x )) = 1, that is, F preserves the Euclidean volume [Katchalov-Kurylev-L. 2001].

  8. Results on the hyperbolic inverse problem: ◮ Uniqueness for inverse problem ( ∂ 2 t − c ( x ) 2 ∆) u = 0 in Ω ⊂ R n by combining the Boundary Control method by Belishev ’87, Belishev-Kurylev ’87 and the controllability results by Tataru ’95. ◮ Spectral problem for ∆ g on manifold, Belishev-Kurylev 1992. ◮ Bingham-Kurylev-L.-Siltanen: Solution by modified time reversal and focusing of waves 2008. ◮ de Hoop-Kepley-Oksanen: Numerical methods for focusing of waves 2016. ◮ Partial data: L.-Oksanen 2014, Mansouri-Milne 2016.

  9. Outline: ◮ Inverse problem for linear wave equation ◮ Generalised time reversal methods and point sources ◮ Inverse problem for non-linear wave equation Figure on the left: Inverse problems group of Kuopio, J. Kaipio et al. Figure on the right: University of Karlsruhe, R. Riedlinger et al.

  10. Time reversal methods: M. Fink; G. Bal, L. Borcea, G. Papanicolaou. Basic steps of the iterated time reversal are ◮ Send a wave that hits to a point scatterer ◮ Record the scattered signals s j ( t ) ◮ Send the time reversed signals s j ( T − t ) Next we apply a modified time reversal iteration to solve an inverse problem. In particular, we do not assume the existence of point scatterers.

  11. Time reversal methods: M. Fink; G. Bal, L. Borcea, G. Papanicolaou. Basic steps of the iterated time reversal are ◮ Send a wave that hits to a point scatterer ◮ Record the scattered signals s j ( t ) ◮ Send the time reversed signals s j ( T − t ) Next we apply a modified time reversal iteration to solve an inverse problem. In particular, we do not assume the existence of point scatterers.

  12. Time reversal methods: M. Fink; G. Bal, L. Borcea, G. Papanicolaou. Basic steps of the iterated time reversal are ◮ Send a wave that hits to a point scatterer ◮ Record the scattered signals s j ( t ) ◮ Send the time reversed signals s j ( T − t ) Next we apply a modified time reversal iteration to solve an inverse problem. In particular, we do not assume the existence of point scatterers.

  13. Recall that u ( x , t ) = u f ( x , t ) solves ( ∂ 2 t − ∆ g ) u ( x , t ) = 0 on ( x , t ) ∈ M × R + , ∂ ν u ( x , t ) | ∂ M × R + = f ( x , t ) , u | t = 0 = 0 , ∂ t u | t = 0 = 0 . Let T > 0 and denote � � Λ 2 T f = u f ( x , t ) � . � ( x , t ) ∈ ∂ M × [ 0 , 2 T ] Moreover, let � � u f ( T ) , u h ( T ) � L 2 ( M ) = u f ( x , T ) u h ( x , T ) dV g ( x ) M and � u f ( T ) � L 2 ( M ) = � u f ( T ) , u f ( T ) � 1 2 .

  14. Inner products of waves By Blagovestchenskii identity, � � u f ( T ) , u h ( T ) � L 2 ( M ) = ( Kf )( x , t ) h ( x , t ) dS ( x ) dt , ∂ M × [ 0 , 2 T ] where K = J Λ 2 T − R Λ 2 T RJ , Rf ( x , t ) = f ( x , 2 T − t ) “time reversal operator” , � 2 T − t Jf ( x , t ) = 1 f ( x , s ) ds “low-pass filter” . 2 t Moreover, � u f ( T ) , 1 � L 2 ( M ) = � f , φ T � L 2 ( ∂ M × [ 0 , 2 T ]) , φ T ( t ) = ( T − t ) +

  15. Domains of influence Let dist M ( x , y ) denote the travel time from x to y . Definition Let s > 0 and Γ ⊂ ∂ M . The set M (Γ , s ) = { x ∈ M : dist M ( x , Γ) ≤ s } is the domain of influence of Γ at time s . t = s M (Γ , s ) ❈ ✄ ❈ ✄ ❈ ✄ ❈ ✄ t = 0 s ❈ ✄ Γ ❈ ✄ Γ

  16. Lemma Let f ∈ L 2 ( ∂ M × [ 0 , s ]) be such that f ( x , t ) = 0 for x �∈ Γ . Then u f ( x , s ) = 0 , for x �∈ M (Γ , s ) . Proof. The result follows the finite velocity of the wave propagation. � t = s M (Γ , s ) ❈ ✄ ❈ ✄ ❈ ✄ ❈ ✄ t = 0 ❈ ✄ ❈ ✄ Γ ❈ ✄

  17. We use following Tataru’s approximate controllability result. Theorem Let s > 0 and Γ ⊂ ∂ M be an open subset. Let v ( x ) ∈ L 2 ( M ) be zero outside M (Γ , s ) . Then for any ε > 0 there is f ∈ L 2 ( ∂ M × [ 0 , s ]) , that is zero outside Γ × [ 0 , s ] , such that �� � 1 / 2 � u f ( s ) − v � L 2 ( M ) = | u f ( x , s ) − v ( x ) | 2 dV ( x ) < ε. M Stability estimates of the equivalent unique continuation result by Bosi-Kurylev-L. 2016, Laurent-Léautaud 2017 t = s M (Γ , s ) ❈ ✄ ❈ ✄ ❈ ✄ ❈ ✄ t = 0 ❈ ✄ Γ ❈ ✄

  18. Blind control problem: Find a boundary source f such that � 1 , when x ∈ M (Γ , s ) , u f ( x , s ) ≈ χ M (Γ , s ) ( x ) = 0 , otherwise . The above control problem is equivalent to the minimisation � u f ( T ) − 1 � 2 min L 2 ( M ) = � Kf , f � − 2 � f , φ T � + const. f where f ∈ L 2 (Γ × [ T − s , T ]) , that is, f is zero outside Γ × [ T − s , T ] . t = T M (Γ , s ) ❈ ✄ ❈ ✄ ❈ ✄ t = T − s ❈ ✄ ❈ ✄ ❈ ✄

  19. The minimization problem has usually no solution and is ill-posed. We consider the regularized minimization problem f ∈ L 2 (Γ × [ T − s , T ]) G α ( f ) min where α ∈ ( 0 , 1 ) , � Kf , f � L 2 − 2 � f , φ T � L 2 + α � f � 2 G α ( f ) = L 2 . This leads to a linear equation ( K + α ) f = φ T . As K = J Λ 2 T − R Λ 2 T RJ , this equation can be solved using an iteration of operators R , Λ 2 T and J .

  20. The modified time reversal iteration is the following algorithm: Let f 0 = 0 and define ( 1 − α 2 ) f n + α ( R Λ 2 T RJf n − J Λ 2 T f n ) + αφ T , f n + 1 := where f n ∈ L 2 (Γ × [ T − s , T ]) and α > 0.

  21. Theorem Let Γ ⊂ ∂ M , 0 ≤ s ≤ T . For α > 0 , the modified time reversal iteration gives sources f n = f n ( α ) . Then in L 2 ( ∂ M × [ 0 , 2 T ]) , f ( α ) = lim n →∞ f n ( α ) , and α → 0 u f ( α ) ( x , T ) = χ M (Γ , s ) ( x ) , in L 2 ( M ) . lim s Γ Bingham-Kurylev-L.-Siltanen 2008, de Hoop-Kepley-Oksanen 2016

  22. The iteration to focus waves and numerical methods have been further developed by M. de Hoop, P. Kepley, and L. Oksanen (a) (c) (d) Above, the wave u ( x , s ) is non-zero near γ z ,ν ( s ) , where z ∈ ∂ M , s > 0. Figures by M. de Hoop, P. Kepley, and L. Oksanen, SIAP 2016.

  23. Theorem Let z ∈ ∂ M and s > 0 . Using an iteration of Λ 2 T , R and J , we find boundary sources � f ( α ) , α > 0 , such that � f ( α ) ( x , T ) = C δ y ( x ) α → 0 u lim in D ′ ( M ) , y = γ z ,ν ( s ) . y s s z Bingham-Kurylev-L.-Siltanen 2008, de Hoop-Kepley-Oksanen 2016

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