The MUSIC method and the factorization method in an inverse - PowerPoint PPT Presentation

The MUSIC method and the factorization method in an inverse scattering problem Pham Quy Muoi Slide form – p .1/24

Model of the problem Slide form – p .2/24

Model of the problem The func . n : Re n � 0 , Im n � 0 and n = 1 in R d \ Ω , d = 2 , 3 . Slide form – p .2/24

Model of the problem The incident wave u inc induce the scaterred wave u S , and the total wave u := u inc + u s : Slide form – p .3/24

Model of the problem The incident wave u inc induce the scaterred wave u S , and the total wave u := u inc + u s : ( I ) ∆ u + knu = 0 in R d \ ∂ Ω Slide form – p .3/24

Model of the problem The incident wave u inc induce the scaterred wave u S , and the total wave u := u inc + u s : ( I ) ∆ u + knu = 0 in R d \ ∂ Ω and u s satifies the Radiation Sommerfeld Condition Slide form – p .3/24

Model of the problem The incident wave u inc induce the scaterred wave u S , and the total wave u := u inc + u s : ( I ) ∆ u + knu = 0 in R d \ ∂ Ω and u s satifies the Radiation Sommerfeld Condition � � ( II ) ∂u 1 ∂n − iku = O , r = | x | → ∞ . r ( d +1) / 2 Slide form – p .3/24

Model of the problem The incident wave u inc induce the scaterred wave u S , and the total wave u := u inc + u s : ( I ) ∆ u + knu = 0 in R d \ ∂ Ω and u s satifies the Radiation Sommerfeld Condition � � ( II ) ∂u 1 ∂n − iku = O , r = | x | → ∞ . r ( d +1) / 2 Forward problem . Giving n, u inc , we find the solution of ( I ), ( II ). Slide form – p .3/24

Model of the problem The incident wave u inc induce the scaterred wave u S , and the total wave u := u inc + u s : ( I ) ∆ u + knu = 0 in R d \ ∂ Ω and u s satifies the Radiation Sommerfeld Condition � � ( II ) ∂u 1 ∂n − iku = O , r = | x | → ∞ . r ( d +1) / 2 Forward problem . Giving n, u inc , we find the solution of ( I ), ( II ). Inverse problem . Giving some information of the solu - tion u ( u ∞ ) , determine Ω . Slide form – p .3/24

Some well - known results The forward problem has unique solution and the solution of the problem is equivalent to the solution of the Lippmann - Schwinger integral equation : � u ( x ) − k 2 q ( y ) u ( y )Φ( x, y ) dy = u inc ( x ) , x ∈ Ω . Ω Slide form – p .4/24

Some well - known results The forward problem has unique solution and the solution of the problem is equivalent to the solution of the Lippmann - Schwinger integral equation : � u ( x ) − k 2 q ( y ) u ( y )Φ( x, y ) dy = u inc ( x ) , x ∈ Ω . Ω About inverse problem Slide form – p .4/24

Some well - known results The forward problem has unique solution and the solution of the problem is equivalent to the solution of the Lippmann - Schwinger integral equation : � u ( x ) − k 2 q ( y ) u ( y )Φ( x, y ) dy = u inc ( x ) , x ∈ Ω . Ω About inverse problem In R 3 , Giving u ∞ , Ω is determined uniquely . There are some algorithms to determine Ω such as iterative methods , the linear sampling method and the factorization method . Slide form – p .4/24

Some well - known results The factorization method ( FM ) In 1998, A . Kirsch introduce the FM to determine Ω in a scattering inverse problem . In 2002, Grinberg applied this method for some scattering inverse problems . Slide form – p .5/24

Overview Introduction Slide form – p .6/24

Overview Introduction The MUSIC method Slide form – p .6/24

Overview Introduction The MUSIC method The factorization method Slide form – p .6/24

The MUSIC method Let ’ s M point scatterers at locations y 1 , y 2 , . . . , y M ∈ R d ( d = 2 , 3) and θ ) = e ikx. ˆ u inc ( x, ˆ θ , x ∈ R d . Then the scattered wave u s is given by Slide form – p .7/24

The MUSIC method Let ’ s M point scatterers at locations y 1 , y 2 , . . . , y M ∈ R d ( d = 2 , 3) and θ ) = e ikx. ˆ u inc ( x, ˆ θ , x ∈ R d . Then the scattered wave u s is given by M u s ( x, ˆ � t i u inc ( y i , ˆ θ ) = θ )Φ( x, y i ) , i =1 Slide form – p .7/24

The MUSIC method Let ’ s M point scatterers at locations y 1 , y 2 , . . . , y M ∈ R d ( d = 2 , 3) and θ ) = e ikx. ˆ u inc ( x, ˆ θ , x ∈ R d . Then the scattered wave u s is given by M u s ( x, ˆ � t i u inc ( y i , ˆ θ ) = θ )Φ( x, y i ) , i =1 x.y + O ( | x | − ( d +1) / 2 ) , | x | → exp ( ikx ) | x | ( d − 1) / 2 e − ik ˆ Φ( x, y ) = γ d ∞ Slide form – p .7/24

The MUSIC method x, ˆ � M i =1 t i u inc ( y i , ˆ u ∞ (ˆ θ ) e − ik ˆ x.y i , θ ) = γ d Slide form – p .8/24

The MUSIC method x, ˆ � M i =1 t i u inc ( y i , ˆ u ∞ (ˆ θ ) e − ik ˆ x.y i , θ ) = γ d Inverse problem : to determine the locations of θ ∈ S d − 1 or x, ˆ x, ˆ scatterers y 1 , . . . , y M from u ∞ (ˆ θ ) , ∀ ˆ u ∞ (ˆ θ i , ˆ θ j ) , i, j = 1 . . . N. Slide form – p .8/24

The MUSIC method x, ˆ � M i =1 t i u inc ( y i , ˆ u ∞ (ˆ θ ) e − ik ˆ x.y i , θ ) = γ d Inverse problem : to determine the locations of θ ∈ S d − 1 or x, ˆ x, ˆ scatterers y 1 , . . . , y M from u ∞ (ˆ θ ) , ∀ ˆ u ∞ (ˆ θ i , ˆ θ j ) , i, j = 1 . . . N. In finite case , assuming N � M, we define the matrix F ∈ C N × N , S ∈ C N × M , and T ∈ C M × M by Slide form – p .8/24

The MUSIC method x, ˆ � M i =1 t i u inc ( y i , ˆ u ∞ (ˆ θ ) e − ik ˆ x.y i , θ ) = γ d Inverse problem : to determine the locations of θ ∈ S d − 1 or x, ˆ x, ˆ scatterers y 1 , . . . , y M from u ∞ (ˆ θ ) , ∀ ˆ u ∞ (ˆ θ i , ˆ θ j ) , i, j = 1 . . . N. In finite case , assuming N � M, we define the matrix F ∈ C N × N , S ∈ C N × M , and T ∈ C M × M by θ l ) , S jm = e − ik ˆ F jl = u ∞ ( ˆ θ j , ˆ θ j .y m , T = diag ( γ d t m ) . Slide form – p .8/24

The MUSIC method x, ˆ � M i =1 t i u inc ( y i , ˆ u ∞ (ˆ θ ) e − ik ˆ x.y i , θ ) = γ d Inverse problem : to determine the locations of θ ∈ S d − 1 or x, ˆ x, ˆ scatterers y 1 , . . . , y M from u ∞ (ˆ θ ) , ∀ ˆ u ∞ (ˆ θ i , ˆ θ j ) , i, j = 1 . . . N. In finite case , assuming N � M, we define the matrix F ∈ C N × N , S ∈ C N × M , and T ∈ C M × M by θ l ) , S jm = e − ik ˆ F jl = u ∞ ( ˆ θ j , ˆ θ j .y m , T = diag ( γ d t m ) . F = STS ∗ and R ( S ) = R ( F ) . (1 . 1) Slide form – p .8/24

The MUSIC method x, ˆ � M i =1 t i u inc ( y i , ˆ u ∞ (ˆ θ ) e − ik ˆ x.y i , θ ) = γ d Inverse problem : to determine the locations of θ ∈ S d − 1 or x, ˆ x, ˆ scatterers y 1 , . . . , y M from u ∞ (ˆ θ ) , ∀ ˆ u ∞ (ˆ θ i , ˆ θ j ) , i, j = 1 . . . N. In finite case , assuming N � M, we define the matrix F ∈ C N × N , S ∈ C N × M , and T ∈ C M × M by θ l ) , S jm = e − ik ˆ F jl = u ∞ ( ˆ θ j , ˆ θ j .y m , T = diag ( γ d t m ) . F = STS ∗ and R ( S ) = R ( F ) . (1 . 1) For z ∈ R d , we define the vector Φ z ∈ C N by Φ z = ( e − ik ˆ θ 1 .z , e − ik ˆ θ 2 .z , . . . , e − ik ˆ θ N .z ) Slide form – p .8/24

The MUSIC method θ n : n ∈ N } ⊂ S d − 1 with the Theorem 1.1. Let { ˆ property that any analytic function which vanishes in ˆ θ n , ∀ n ∈ N vanishes identically . Then there exists N 0 ∈ N such that for any N � N 0 the characterization holds z ∈ { y 1 , y 2 , . . . , y M } ⇔ Φ z ∈ R ( S ) . From (1.1) we have z ∈ { y 1 , y 2 , . . . , y M } ⇔ Φ z ∈ R ( F ) ⇔ P Φ z = 0 with P : C N → R ( F ) ⊥ is the orthogonal projection . Slide form – p .9/24

The MUSIC method Therefore , the plot of the function Slide form – p .10/24

The MUSIC method Therefore , the plot of the function 1 W ( z ) = | P Φ z | should result in sharp peaks at y 1 , . . . , y M . Slide form – p .10/24

The MUSIC method Therefore , the plot of the function 1 W ( z ) = | P Φ z | should result in sharp peaks at y 1 , . . . , y M . Example . d = 2 , M = 2 , N = 10 , k = 2 π and ˆ θ j , j = 1 , . . . , 10 , are equidistantly chosen directions . The values of t are 1 + i, 1 . 5 + i at ( − 1 , 1) , ( − 1 / 2 , − 1) , respectively . The plots of W ( z ) give by Slide form – p .10/24

The plots of W ( z ) Slide form – p .11/24

Main idea of two methods Firstly , we factorize operator F in the form F = SDS ∗ . Slide form – p .12/24

Main idea of two methods Firstly , we factorize operator F in the form F = SDS ∗ . Secondly , we define a function Φ z such that z ∈ Ω ⇔ Φ z ∈ R ( S ) . Slide form – p .12/24

Main idea of two methods Firstly , we factorize operator F in the form F = SDS ∗ . Secondly , we define a function Φ z such that z ∈ Ω ⇔ Φ z ∈ R ( S ) . Finally , we find an operator F ′ that only depend on F such that R ( F ′ ) = R ( S ) . Slide form – p .12/24

The factorization method Forward Problem . Let Ω ⊂ R d : bounded , open set and its complement is connected ; n = 1 + q, q ∈ L ∞ (Ω) , u inc = e ik ˆ θ.x , x ∈ R d . Slide form – p .13/24

The factorization method Forward Problem . Let Ω ⊂ R d : bounded , open set and its complement is connected ; n = 1 + q, q ∈ L ∞ (Ω) , u inc = e ik ˆ θ.x , x ∈ R d . The forward scattering problem is to detemine u = u s + u inc ∈ C 1 ( R d ) ∩ C 2 ( R d \ ∂ Ω) satisfies Slide form – p .13/24