Numerical studies of the reciprocity gap functional method in - PowerPoint PPT Presentation

The Reciprocity Gap Method An Analytic Case Numerical Examples Numerical studies of the reciprocity gap functional method in inverse scattering Jiguang Sun Delaware State University AIP2009, Vienna, Austria

The Reciprocity Gap Method An Analytic Case Numerical Examples Outline The Reciprocity Gap Method 1 An Analytic Case 2 Numerical Examples 3

The Reciprocity Gap Method An Analytic Case Numerical Examples Introduction x 0 B Γ D Ω The unknown scattering obstacle D is inside some domain B . Ω ⊂ B and the boundary of Ω is given by Γ . D ⊂ Ω . Point sources located at x 0 lie on a smooth curve ∂ B that is homotopic to Γ .

The Reciprocity Gap Method An Analytic Case Numerical Examples The direct scattering problem △ u + k 2 n ( x ) u = 0 D ∪ { x 0 } R \ � � in , (1) u = 0 ∂ D , on (2) u = u s + u i , (3) � ∂ u s √ � ∂ r − iku s r lim = 0 , (4) r →∞ where k is the wave number and r = | x | . Note that when n ( x ) is constant, u i is given by Φ( x ) defined as Φ( x , x 0 ) = i √ 4 H ( 1 ) 0 ( k n | x − x 0 | ) (5) where H ( 1 ) is the Hankel function of first kind and order zero. 0

The Reciprocity Gap Method An Analytic Case Numerical Examples The inhomogeneous background Green’s function When the background is inhomogeneous, u i can be written as u i ( x , x 0 ) = G ( x , x 0 ) (6) where G ( x , x 0 ) is the radiating solution of △ G ( x , x 0 ) + k 2 n ( x ) G ( x , x 0 ) = − δ ( x − x 0 ) in R 2 (7) with δ denoting the Dirac delta function. The inverse scattering problem we are interested in is to determine the support of the scattering obstacle D from the knowledge of the Cauchy data of u on ∂ Ω .

The Reciprocity Gap Method An Analytic Case Numerical Examples The Reciprocity Gap Functional H (Ω) := { v ∈ H 1 (Ω) : ∆ v + k 2 n ( x ) v = 0 in Ω } U is the set of solutions to (1-4) for all x 0 ∈ ∂ B . For v ∈ H (Ω) and u ∈ U we define the reciprocity gap functional by ∂ν − v ∂ u x 0 ∂ v � � � R ( u x 0 , v ) = u x 0 ds . (8) ∂ν ∂ Ω Define R : H (Ω) → L 2 ( ∂ B ) defined by R ( v )( x 0 ) = R ( u , v ) .

The Reciprocity Gap Method An Analytic Case Numerical Examples Herglotz wave functions H (Ω) contains the set of Herglotz wave functions defined by � e ik 1 ˆ d · x g (ˆ v g , k 1 ( x ) = d ) ds (ˆ d ) , (9) S where g ∈ L 2 ( S ) , with S = { ˆ d ∈ R 2 : | ˆ d | = 1 } and k 2 1 = k 2 n 0 for some constant n 0 . Let Φ k 1 , z = i 4 H ( 1 ) 0 ( k 1 | x − z | ) . For z ∈ Y , a sampling domain contained in Ω and containing D inside, we look for a solution g ∈ L 2 ( S 1 ) to R ( u , v g ) = R ( u , Φ k 1 , z ) for all u ∈ U , (10) where v g is a Herglotz wave function defined by (9).

The Reciprocity Gap Method An Analytic Case Numerical Examples Characterization of the target Theorem Assume that k 2 is not a Dirichlet eigenvalue for D. (a) If z ∈ D then there exist a sequence { g n } , such that n →∞ R ( u , v g n ) = R ( u , Φ z ) for all u ∈ U . lim Furthermore, v g n converges in H 1 ( D ) and v g n → Φ z in H 1 / 2 ( ∂ D ) . (b) If z ∈ Ω \ D then for every sequence { v g n } , such that n →∞ R ( u , v g n ) = R ( u , Φ z ) for all u ∈ U . lim we have n →∞ � v g n � H 1 ( D ) = ∞ . lim

The Reciprocity Gap Method An Analytic Case Numerical Examples Now equation (10) can be written in the form A g ( · , z ) = φ ( · , z ) (11) where A : L 2 ( S ) → L 2 ( C ) is the integral operator with kernel K : S × C ∈ C defined by K (ˆ d , x 0 ) = R ( u ( · , x 0 ) , v ( · , ˆ d )) with v ( x , ˆ d ) = exp ( ik ˆ d · x ) and where φ ( x 0 , z ) = R ( u ( · , x 0 ) , Φ z ) . We expect that for g ( · , z ) being the solution to (11) where A is replaced by a regularized operator, � g ( · , z ) � L 2 ( S ) should be large for z ∈ Ω \ D and bounded for z ∈ D .

The Reciprocity Gap Method An Analytic Case Numerical Examples Regularization Scheme To obtain regularized solution, Tikhonov regularization can be used. This means solving ( α + A ∗ A ) g α ( · , z ) = A ∗ φ ( · , z ) where α is the regularization parameter and A ∗ is the adjoint of A . If A is a compact operator with singular system ( σ n ; v n , u n ) , we have ∞ σ n g α = n + α < φ, u n > v n . � σ 2 n = 1 Hence it is very helpful to study the singular values of A in different cases.

The Reciprocity Gap Method An Analytic Case Numerical Examples Setting of the Problem Now we consider the simplified case when ∂ D , ∂ Ω and ∂ B are all circles whose centers are the origin of the coordinate system. Assume that the incident field u i is due to a point source at z = ( r s , φ ) and consider the scattering of u i by an infinite cylinder whose cross section is a circle with radius r D . The forward problem is to find a complex valued radiating function u s ( x ) such that △ u s + k 2 u s = 0 R 2 \ ¯ D in u s ( r , θ ) = − Φ( k | x − z | ) ∂ D on where Φ( k | x − z | ) = i 4 H ( 1 ) 0 ( k | x − z | ) .

The Reciprocity Gap Method An Analytic Case Numerical Examples Expansion of the solution Suppose D is a circle centered at the origin with radius r D . Due to symmetry, ∞ u ( r , θ ) = α n H ( 1 ) n ( kr ) cos n θ, r > r D . � n = 0 On ∂ D where r = r D , we have ∞ u ( r D , θ ) = α n H ( 1 ) n ( kr D ) cos n θ. � (12) n = 0 On the other hand, let x = ( ρ, ϕ ) , we have r 2 s − 2 r s ρ cos θ + ρ 2 where θ = φ − ϕ and � | z − x | = ∞ H 1 0 ( | z − x | ) = J 0 ( ρ ) H ( 1 ) 0 ( r s ) + 2 J n ( ρ ) H ( 1 ) n ( r s ) cos n θ � (13) n = 1 for r s > ρ .

The Reciprocity Gap Method An Analytic Case Numerical Examples Expansion coefficients Function J n is the Bessel’s function of the first kind of order n . Using (13), (12) and the boundary condition, we obtain − i J 0 ( kr D ) H ( 1 ) 0 ( kr s ) α 0 = , H ( 1 ) 0 ( kr D ) 4 − i 2 J n ( kr D ) H ( 1 ) n ( kr s ) n ≥ 1 . α n = 4 H ( 1 ) n ( kr D ) d n − 1 ( z ) − n n + 1 ( z )+ n dz H ( 1 ) n ( z ) = H ( 1 ) z H ( 1 ) n ( z ) = − H ( 1 ) z H ( 1 ) n ( z ) . (14)

The Reciprocity Gap Method An Analytic Case Numerical Examples Hence J 0 ( kr D ) H ( 1 ) 0 ( kr s ) u ( r , θ ) = − i H 1 0 ( kr ) 4 H ( 1 ) 0 ( kr D ) − i ∞ J n ( kr D ) H ( 1 ) n ( kr s ) H 1 n ( kr ) cos ( n θ ) , � H ( 1 ) n ( kr D ) 2 n = 1 J 0 ( kr D ) H ( 1 ) 0 ( kr s ) ∂ u ( r , θ ) = − ik {− H 1 1 ( kr ) } ∂ r 4 H ( 1 ) 0 ( kr 0 ) − ik ∞ J n ( kr D ) H ( 1 ) n ( kr s ) n − 1 ( kr ) − n { H 1 kr H 1 n ( kr ) cos ( n θ ) } . � H ( 1 ) n ( kr D ) 2 n = 1

The Reciprocity Gap Method An Analytic Case Numerical Examples In particular, if the point source is at z = ( r s , φ ) on a circle with radius r s , then the scattered field can be obtained easily from the above, J 0 ( kr D ) H ( 1 ) 0 ( kr s ) u ( r , θ ) = − i H 1 0 ( kr ) 4 H ( 1 ) 0 ( kr D ) − i ∞ J n ( kr D ) H ( 1 ) n ( kr s ) H 1 n ( kr ) cos ( n ( θ − φ )) , � H ( 1 ) n ( kr D ) 2 n = 1 J 0 ( kr D ) H ( 1 ) 0 ( kr s ) ∂ u ( r , θ ) = − ik {− H 1 1 ( kr ) } ∂ r 4 H ( 1 ) 0 ( kr D ) − ik ∞ J n ( kr D ) H ( 1 ) n ( kr s ) n − 1 ( kr ) − n � { H 1 kr H 1 n ( kr ) } cos ( n ( θ − φ )) . H ( 1 ) n ( kr D ) 2 n = 1

The Reciprocity Gap Method An Analytic Case Numerical Examples Now we consider the case when Γ = ∂ Ω := { x , | x | = r Γ > r D } . The solutions of the Helmholtz equation in Ω can be written as v n = J n ( kr Γ ) cos ( n θ ) or w n = J n ( kr Γ ) sin ( n θ ) . (15) ∂ v n J n − 1 ( kr ) − n � � ∂ r = k kr J n ( kr ) cos ( n θ ) , ∂ w n J n − 1 ( kr ) − n � � = k kr J n ( kr ) sin ( n θ ) . ∂ r

The Reciprocity Gap Method An Analytic Case Numerical Examples Now the reciprocity gap functional can be seen as an operator R : L 2 [ 0 , 2 π ] → L 2 [ 0 , 2 π ] (from L 2 ( ∂ Ω) → L 2 ( ∂ B ) ) defined by R ( v )( φ ) = R ( u , v ) . Thus combining the above equations we obtain, for n = 0, u ∂ v 0 ∂ u � R ( v 0 )( φ ) ∂ r − v 0 ∂ r ds = Γ J 0 ( kr D ) H ( 1 ) 0 ( kr s ) − i π H 1 0 ( kr Γ )( − kJ 1 ( kr Γ )) = H ( 1 ) 0 ( kr D ) 2 J 0 ( kr D ) H ( 1 ) 0 ( kr s ) + ik π {− H 1 1 ( kr Γ ) } J 0 ( kr Γ ) 2 H ( 1 ) 0 ( kr D ) λ 0 v 0 =

The Reciprocity Gap Method An Analytic Case Numerical Examples Case of n > 1 u ∂ v n ∂ u � R ( v n )( φ ) = ∂ r − v n ∂ r ds Γ = − ik π � J n ( kr D ) H ( 1 ) n ( kr s ) J n − 1 ( kr Γ ) − n � � H 1 n ( kr Γ ) J n ( kr Γ ) kr Γ 2 H ( 1 ) n ( kr D ) − J n ( kr D ) H ( 1 ) n ( kr s ) n − 1 ( kr Γ ) − n � { H 1 H 1 n ( kr Γ ) } J n ( kr Γ ) cos ( n φ ) kr Γ H ( 1 ) n ( kr D ) = λ n v n

The Reciprocity Gap Method An Analytic Case Numerical Examples Using the recurrence relation of Bessel’s functions, we obtain J n ( kr D ) H ( 1 ) λ n = − ik π c n ( kr s ) n ≥ 1 , (16) H ( 1 ) 2 n ( kr D ) where c = H 1 1 ( kr Γ ) J 0 ( kr Γ ) − H 1 0 ( kr Γ ) J 1 ( kr Γ ) (17) Similarly, we have R ( w n )( φ ) = λ n w n .