A globally convergent numerical method and adaptivity for an - PowerPoint PPT Presentation

A globally convergent numerical method and adaptivity for an inverse problem via Carleman estimates Larisa Beilina ∗ , Michael V. Klibanov △ ∗ Chalmers University of Technology and Gothenburg University, Gothenburg, Sweden △ University of North Carolina at Charlotte, Charlotte, USA 1

Globally convergent method A numerical method X is globally convergent if: 1. A theorem is proved claiming convergence to a good approximation for the correct solution regardless on a priori availability of a good guess 2. This theorem is confirmed by numerical experiments for at least one applied problem 2

Challenges in solution of CIP • Solution of any PDE depends nonlinearly on its coefficients. y ′ − ay = 0 → y ( a, t ) = Ce at . • Any coefficient inverse problem is nonlinear. • Two major challenges in numerical solution of any coefficient inverse problem: NONLINEARITY and Ill-POSEDNESS. • Local minima of objective functionals. • Locally convergent methods: linearizaton, Newton-like and gradient-like methods. 3

A hyperbolic equation c ( x ) u tt = ∆ u − a ( x ) u in R n × (0 , ∞ ) , n = 2 , 3 , u ( x, 0) = 0 , u t ( x, 0) = δ ( x − x 0 ) . INVERSE PROBLEM . Let Ω ⊂ R n be a bounded domain. Let one of coefficients c ( x ) or a ( x ) be unknown in Ω but it is a given constant outside of Ω . Determine this coefficient in Ω , given the function g ( x, t ) , u ( x, t ) = g ( x, t ) , x ∈ ∂ Ω , t ∈ (0 , ∞ ) Similarly for the parabolic equation u in R n × (0 , ∞ ) , c ( x ) � u t = ∆ � u − a ( x ) � u ( x, 0) = δ ( x − x 0 ) . � 4

Applications 1. MEDICINE a. medical optical imaging; b. acoustic imaging. 2. MILITARY a. identification of hidden targets, like, e.g. landmines; improvised explosive devices via electric or acoustic sensing. b. detecting targets covered by smog or flames on the battlefield (via diffuse optics). 5

Laplace transform: � ∞ � ∞ u ( x, t ) e − s 2 t dt u ( x, t ) e − st dt = w ( x, s ) = � 0 0 � � s 2 c ( x ) + a ( x ) ∆ w − w = − δ ( x − x 0 ) , (1) ∀ s > s 0 = const. > 0 . | x |→∞ w ( x, s ) = 0 , ∀ s > s 0 = const. > 0 . lim (2) w ( x, s ) > 0 , ∀ s > s 0 . 6

THE TRANSFORMATION PROCEDURE. First, we eliminate the unknown coefficient from the equation: v = ln w. ∆ v + |∇ v | 2 = s 2 c ( x ) + a ( x ) in Ω , • Let, for example c ( x ) =? For simplicity let a ( x ) = 0 . It follows from works of V.G. Romanov that � � � 1 ��� − sl ( x, x 0 ) D α x D β s ( v ) = D α x D β 1 + O , s → ∞ . s g ( x, x 0 ) s • Introduce a new function v = v � s 2 . Then � 1 � v ( x, s ) = O , s → ∞ . � s 7

• Eliminate the unknown coefficient c ( x ) via the differentiation: ∂ s c ( x ) ≡ 0 q ( x, s ) = ∂ s � v ( x, s ) , � ∞ � s � v ( x, s ) = − q ( x, τ ) dτ ≈ − q ( x, τ ) dτ + V ( x, s ) . s s • V ( x, s ) is the tail function, V ( x, s ) ≈ 0 . But still we iterate with respect to the tail. • This truncation is similar to the truncation of high frequencies. 8

• Obtain Dirichlet boundary value problem for the nonlinear equation   2 s s � � ∆ q − 2 s 2 ∇ q ·   ∇ q ( x, τ ) dτ + 2 s ∇ q ( x, τ ) dτ (3) s s � s ∇ q ( x, τ ) dτ + 2 s ( ∇ V ) 2 = 0 , +2 s 2 ∇ q ∇ V − 2 s ∇ V · s q ( x, s ) = ψ ( x, s ) , ∀ ( x, s ) ∈ ∂ Ω × [ s, s ] . (4) • Backwards calculations v ) 2 , v + s 2 ( ∇ � c ( x ) = ∆ � 9

How To Solve the Problem (3), (4)? • Layer stripping with respect to the pseudo frequency s. • On each step the Dirichlet boundary value problem is solved for an elliptic equation. s = s N < s N − 1 < ... < s 1 < s 0 = s, s i − 1 − s i = h q ( x, s ) = q n ( x ) for s ∈ ( s n , s n − 1 ] . � s n − 1 � ∇ q ( x, τ ) dτ = ( s n − 1 − s ) ∇ q n ( x ) + h ∇ q j ( x ) , s ∈ ( s n , s n − 1 ] . j =1 s • Dirichlet boundary condition: q n ( x ) = ψ n ( x ) , x ∈ ∂ Ω , 10

s n − 1 � ψ n ( x ) = 1 ψ ( x, s ) ds. h s n 11

Hence,   n − 1 � � � s 2 − 2 s ( s n − 1 − s ) �  · ∇ q n  h L n ( q n ) := ∆ q n − 2 ∇ q j ( x ) j =1 � � s 2 − 2 s ( s n − 1 − s ) +2 ∇ q n · ∇ V ( x, s ) − εq n   2 n − 1 � � � ( ∇ q n ) 2 − 2 sh 2 s 2 − s ( s n − 1 − s )   = 2 ( s n − 1 − s ) ∇ q j ( x ) j =1   n − 1 �  − 2 s [ ∇ V ( x, s )] 2 , s ∈ ( s n − 1 , s n ]  h +4 s ∇ V ( x, s ) · ∇ q j ( x ) j =1 Introduce the s -dependent Carleman Weight Function C nµ ( s ) by C nµ ( s ) = exp [ µ ( s − s n − 1 )] , s ∈ ( s n , s n − 1 ] , where µ >> 1 is a parameter. 12

• Multiply the equation by C nµ ( s ) and integrate with respect to s ∈ [ s n , s n − 1 ] . � � n − 1 � L n ( q n ) := ∆ q n − A 1 n ( µ, h ) ∇ q i ( x ) · ∇ q n − εq n h i =1 � n − 1 � 2 � = 2 I 1 n ( µ, h ) I 0 ( µ, h ) ( ∇ q n ) 2 − A 2 n ( µ, h ) h 2 ∇ q i ( x ) i =1 � � n − 1 � +2 A 1 n ( µ, h ) ∇ V ( x, s ) · ∇ q i ( x ) h i =1 − A 2 n ( µ, h ) ∇ q n · ∇ V ( x, s ) − A 2 n ( µ, h ) [ ∇ V ( x, s )] 2 , where s n − 1 � C nµ ( s ) ds = 1 − e − µh I 0 ( µ, h ) = , µ s n 13

s n − 1 � � � s 2 − s ( s n − 1 − s ) I 1 n ( µ, h ) = ( s n − 1 − s ) C nµ ( s ) ds, s n s n − 1 � � � 2 s 2 − 2 s ( s n − 1 − s ) A 1 n ( µ, h ) = C nµ ( s ) ds, I 0 ( µ, h ) s n s n − 1 � 2 A 2 n ( µ, h ) = s C nµ ( s ) ds. I 0 ( µ, h ) s n • Important observation: s 2 | I 1 n ( µ, h ) | ≤ 4¯ µ , for µh > 1 . I 0 ( µ, h ) 14

• Iterative solution for every q n   n − 1 � ∆ q i  h  · ∇ q i nk − εq i nk + A 1 n ∇ q i nk · ∇ V i nk − A 1 n ∇ q j n = j =1   2 � � 2 n − 1 � 2 I 1 n ( µ, h ) ∇ q i − A 2 n h 2   ∇ q j ( x ) n ( k − 1) I 0 ( µ, h ) j =1   n − 1 � � � 2 , k ≥ 1 , +2 A 2 n ∇ V i  h  − A 2 n ∇ V i n · ∇ q j ( x ) n j =1 q i nk ( x ) = ψ n ( x ) , x ∈ ∂ Ω • Hence, we obtain the function nk , in C 2+ α � � q i k →∞ q i n = lim Ω . 15

CONVERGENCE THEOREM. • First, Schauder Theorem. Consider the Dirichlet boundary value problem 3 � ∆ u + b j ( x ) u x j − m ( x ) u = f ( x ) , x ∈ Ω , j =1 u | ∂ Ω = g ( x ) ∈ C 2+ α ( ∂ Ω) . Let b j , m, f ∈ C α � � � � Ω , d ( x ) ≥ 0; max | b j | α , | m | α ≤ 1 , Then � � | u | 2+ α ≤ K � g � C 2+ α ( ∂ Ω) + | f | α , where K = K (Ω) = const. ≥ 1 . 16

Global Convergence Theorem. Let Ω ⊂ R 3 be a convex bounded domain with the boundary ∂ Ω ∈ C 3 . Let the exact coefficient c ∗ ( x ) ∈ C 2 ( R 3 ) , c ∗ ∈ [2 d 1 , 2 d 2 ] and c ∗ ( x ) = 2 d 1 for x ∈ R 3 � Ω , where numbers d 1 , d 2 > 0 are given. For any function c ( x ) ∈ C α � R 3 � such that c ( x ) ≥ d 1 in Ω and c ( x ) = 2 d 1 in R 3 � Ω consider the solution u c ( x, t ) of the original Cauchy problem. Let C ∗ = const. ≥ 1 be a constant bounding certaon functions associated with the solution of this Cauchy problem . Let w c ( x, s ) ∈ C 3 � � R 3 � {| x − x 0 | < γ } , ∀ γ > 0 be the Laplace transform of u c ( x, t ) and V c ( x ) = s − 2 ln w c ( x, s ) ∈ C 2+ α � � Ω be the corresponding tail function. Suppose that the cut-off pseudo frequency s is so large that for any such function c ( x ) the following estimates hold | V ∗ | 2+ α ≤ ξ, | V c | 2+ α ≤ ξ, where ξ ∈ (0 , 1) is a sufficiently small number. 17

Let V 1 , 1 ( x, s ) ∈ C 2+ α � � Ω be the initial tail function and let | V 1 , 1 | 2+ α ≤ ξ. Denote η := 2 ( h + σ + ξ + ε ) . Let N ≤ N be the total number of functions q n calculated by the algorithm of section 5. Suppose that the number N = N ( h ) is connected with the step size h via N ( h ) h = β, where the constant β > 0 is independent on h . Let β be so small that 1 β ≤ 384 KC ∗ s 2 . In addition, let the number η and the parameter µ of the CWF satisfy the following estimates � � � � 16 KM ∗ , 3 1 256 KC ∗ s 2 , 3 1 η ≤ η 0 ( K, C ∗ , d 1 , s ) = min 8 d 1 = min 8 d 1 , � � ( C ∗ ) 2 , 48 KC ∗ s 2 , 1 µ ≥ µ 0 ( C ∗ , K, s, η ) = max . η 2 4 18

� � ∞ q k Then for each appropriate n the sequence k =1 converges in n, 1 C 2+ α � � Ω and the following estimates hold � 1 � � � | q n − q ∗ n | 2+ α ≤ 2 KM ∗ √ µ + 3 η , n ∈ 1 , N , � � | q n | 2+ α ≤ 2 C ∗ , n ∈ 1 , N , � � 2 · 9 n − 1 + 23 η | c n − c ∗ | α ≤ 8 η, n ∈ 2 , N . (5) In addition, functions c n,k ( x ) ≥ d 1 in Ω and c n,k ( x ) = 2 d 1 outside of Ω . 19