An inverse problem for a waveguide Virginia Selgas Peter Monk - PowerPoint PPT Presentation

An inverse problem for a waveguide Virginia Selgas Peter Monk University of Delaware University of Oviedo Spain USA PICOF 2012

Goal: non-destructive testing of pipes Problem. We want to detect and locate bounded inhomogeneous obstacles in an infinite tubular waveguide, when we are given measurements of pressure waves due to point sources inside the waveguide. We have in mind a potential application of acoustic techniques for the inspection of underground pipes such as sewers. thanks to S.N. Chandler-Wilde for bringing this to our attention! We assume that an acoustic source and array of microphones could be lowered into a sewer and then used to collect multistatic scattering data from the mouth of the pipe.

Previous work Engineers have suggested using acoustic means to inspect sewers (but have not considered multistatic data). [F . Podd, M. Ali, K. Horoshenkov, et al., Water Science and Technology , 56 (2007), pp. 131–9] [M.T.Bin Ali, K.V.Horoshenkov, S.J.Tait, NOVATECH 2010] Linear Sampling Method (LSM) [D. Colton and A. Kirsch, Inv. Prob. , 12 (1996), pp. 383–93] Reciprocity Gap Method (RGM) [D. Colton and H. Haddar, Inv. Prob. , 21 (2005), pp. 383–98] LSM for infinite sound hard acoustic waveguides (i.e. pipes) and sound soft obstacle. [L. Bourgeois and E. Lunéville, Inv. Prob. , 24 (2008), pp. 15–8] Several authors have considered the use of time-reversal methods for small inclusions in waveguides. e.g. [P . Roux and M. Fink, J. Acoust. Soc. Am. , 107 (2000), pp. 2418–29]

Forward problem: geometry Geometry. Cross-section of the waveguide: Σ ⊂ R 2 , bounded, smooth, simply connected later a disc Infinite tubular waveguide: R × Σ ⊂ R 3 Penetrable obstacle: D ⊂ R × Σ , bounded, with connected complement R × Σ \ D ∂ D denotes the boundary of D We identify each point x ∈ R 3 with ( x 1 , ˆ x ) ∈ R × R 2 . ✓ ✏ { 0 }× Σ ✲ ν 0 D � ✒ ✄ � � ν D ✒ ✑ ✂ ✁ ν ❄

Forward problem: equations A known incident field u i is excited in the pipe (e.g. by a loudspeaker). Forward problem. Find the fields u and v such that  △ u + k 2 u = 0 in ( R × Σ) \ D ,    △ v + k 2 n v  = 0 in D ,       = u i v − u across ∂ D ∩ ( R × Σ) ,   = ∂ ν D u i ∂ ν D v − ∂ ν D u across ∂ D ∩ ( R × Σ) ,      on ( R × ∂ Σ) \ ∂ D , ∂ ν u = 0       ∂ ν v = 0 on ( R × ∂ Σ) ∩ ∂ D ,  The inhomogeneity is represented by the coefficient n ∈ C ( D ) with Re ( n ) ≥ C > 0 and Im ( n ) ≥ 0 in D . The total field is u t = u i + u , being u the scattered field (as above). A further radiation condition required for uniqueness!

Forward problem: radiation condition The radiation condition requires the scattered wave to propagate (or decay) away from the inhomogeneity D . To formalize it, we need: cross-sectional modes k n ∈ [ 0 , + ∞ ) , where k 2 n is an eigenvalue of the Neumann problem for the negative Laplacian on Σ , sorted in such way that k n ր + ∞ ; θ n an eigenfunction associated to the eigenvalue k 2 n and with { θ n } n ∈ N forming an orthonormal basis of L 2 (Σ) ; waveguide modes g ± x ) e ± ıβ n x 1 , where n ( x 1 , ˆ x ) := θ n (ˆ k 2 − k 2 � β n := n ∈ C and we choose Re ( β n ) ≥ 0 and Im ( β n ) ≥ 0. Note: obviously ∆ g ± n + k 2 g ± n = 0 in R × Σ , and ∂ ν g ± n = 0 on R × ∂ Σ ; there are a finite number of propagating modes, and the remainder are evanescent modes.

Forward problem: radiation condition formalization The Dirichlet–to–Neumann operator is T ± R : H 1 / 2 (Σ ± R ) → ˜ H − 1 / 2 (Σ ± R ) given by � T ± R g = ± ı � β n g θ n dS θ n , Σ ± R n ∈ N where Σ ± R := {± R } × Σ . The radiation condition can be written as T ± R u s = ∂ ν 0 u s on Σ ± R . In addition we assume that the incident field u i is a smooth solution of △ u i + k 2 u i = 0 � in ( − R S , R S ) × Σ , ∂ ν u i = 0 on ( − R S , R S ) × ∂ Σ , for some R S ∈ ( 0 , R ) with D ⊂ ( − R S , R S ) × Σ .

Forward problem: summary Forward problem. We consider R > 0 large enough so that D ⊂ ( − R , R ) × Σ := B R . The forward problem is to find u ∈ H 1 ( B R \ D ) and v ∈ H 1 ( D ) such that △ u + k 2 u  in B R \ D , = 0    △ v + k 2 n v  = 0 in D ,       u i v − u = across ∂ D ∩ B R ,   ∂ ν D u i ∂ ν D v − ∂ ν D u = across ∂ D ∩ B R ,      ∂ ν u = 0 and ∂ ν v = 0 on (( − R , R ) × ∂ Σ) ∩ ∂ D ,       T ± R u ∂ ν 0 u = on Σ ± R .  Result. Existence and uniqueness except for a discrete set of frequencies (now assumed to be avoided).

Inverse problem: geometry We consider two surfaces Σ S := Σ a S and Σ M := Σ a M where we place sources and take measurements, respectively. We assume that − R < a S < a M < R and D ⊂ ( a M , R ) × Σ , We denote B S R := ( a S , R ) × Σ and B M R := ( a M , R ) × Σ , so that D ⊂ B M R ⊂ B S R ⊂ B R . ✓ ✏ ✓ ✏ ✓ ✏ ✓ ✏ Σ S Σ M Σ − R Σ R ✄ D � ✒ ✑ ✒ ✑ ✒ ✑ ✂ ✁ ✒ ✑

Inverse problem: point sources For each source point x 0 ∈ Σ S , the incident field u i x 0 is u i with x 0 ∈ Σ S , x ∈ R × Σ , x 0 ( x ) := Φ( x 0 , x ) where Φ is the fundamental solution for the waveguide e ıβ n | x 1 − y 1 | � θ n (ˆ x ) θ n (ˆ Φ( x , y ) := Φ x ( y ) := − y ) 2 ıβ n n ∈ N for x = ( x 1 , ˆ x ) , y = ( y 1 , ˆ y ) ∈ R × Σ with x � = y . For this incident field u i x 0 , the corresponding total field is u t x 0 = u s x 0 + u i x 0 in B R \ D and v x 0 in D .

Inverse problem Given u t x 0 ( x ) and ∂ ν 0 u t x 0 ( x ) for all source points x 0 ∈ Σ S and all measurement points x ∈ Σ M , determine the boundary of D . Remark. Of course in practice we would only have a finite number of sources and a finite number of measurements.

LSM and RGM algorithms Two possible methods considered in our paper: Linear Sampling Method (LSM) It uses only measurements of u x 0 , but requires the fundamental solution of the pipe and entire domain (i.e. pipe and manhole). Reciprocity Gap Method (RGM) It uses all measurements, of u t x 0 ( x ) and ∂ ν 0 u t x 0 ( x ) , but isolates the pipe and only requires the fundamental solution of the pipe. For the remainder of the talk we shall concentrate on the RGM.

Reciprocity Gap operator: function spaces For B ⊆ B R , define the function space � u ∈ H 1 ( B ) ; △ u + k 2 u = 0 in B , H ( B ) = ∂ ν u = 0 on ∂ B ∩ (( − R , R ) × ∂ Σ) } , as well as its subspace H R ( B ) := { u ∈ H ( B ) ; ∂ ν 0 u = T R u on Σ R } if Σ R ⊂ ∂ B . We also define U := { u t x 0 = u s x 0 + u i x 0 ∈ Σ S } . x 0 ;

Reciprocity Gap operator For sources placed on Σ S and measurements made on Σ M , we consider the RG operator R : H R ( B S R ) → L 2 (Σ S ) given by R v ( x 0 ) := R ( u t for x 0 ∈ Σ S , v ∈ H R ( B S x 0 , v ) R ) , where the Reciprocity Gap (RG) bilinear form is � R ( u t u t x 0 ∂ ν 0 v − ∂ ν 0 u t for x 0 ∈ Σ S , v ∈ H R ( B S � � x 0 , v ) := x 0 v dS R ) . Σ M

Interior Transmission Problem To analyze the RGM we first need to consider the following ITP . Interior Transmission Problem. Find v 1 ∈ H 1 ( D ) , v 2 ∈ H 1 ( D ) such that △ v 1 + k 2 v 1  = 0 in D ,    △ v 2 + k 2 n v 2  = 0 in D ,      ITP : v 1 − v 2 = G across ∂ D ∩ B R ,   ∂ ν D v 1 − ∂ ν D v 2 = g across ∂ D ∩ B R ,       ∂ ν v 1 = ∂ ν v 2 = 0 on (( − R , R ) × ∂ Σ) ∩ ∂ D .  for suitable data g and G . Remark. This is a generalized interior transmission problem because of the mixed boundary conditions.

Interior Transmission Problem: analysis Using the arguments of [D.Colton, L.Paeivaerinta and J.Sylvester, IPI , 1 (2007) 13-28] we can rewrite ITP as a mixed 4th order boundary valued problem for w = v 1 − v 2 . Noticing that w satisfies an additional natural boundary condition on the boundary (( − R , R ) × ∂ Σ) ∩ ∂ D 1 , we can show: Lemma If Re ( n ) > C > 1 in D, the interior transmission problem is well posed for any k except for, at most, a discrete set of k values called transmission eigenvalues . If Im ( n ) > 0 there are no transmission eigenvalues. 1 Personal communication of F.Cakoni

First properties of R We use this lemma and follow .Cakoni, M.Cayoeren and D.Colton, Inv. Prob. , 24 (2008), 065016] . [F Lemma Provided k is not a transmission eigenvalue and D � = ∅ , the RG operator R : H R ( B S R ) → L 2 (Σ S ) is injective.

Single Layer operator We need a suitable parametrization for the RG operator argument, v . This is provided by a single layer operator using densities on yet another open surface Σ SL := Σ a SL with a SL ∈ ( − R , a S ) . We define S Σ SL : ˜ S Σ SL : ˜ H − 1 / 2 (Σ SL ) → H ± R ( B R \ Σ SL ) H − 1 / 2 (Σ SL ) → H 1 / 2 (Σ SL ) and by � for a.e. x ∈ B R \ Σ SL , ( S Σ SL g )( x ) = Σ SL Φ( x , y ) g ( y ) dS y � for a.e. x ∈ Σ SL . ( S Σ SL g )( x ) = Σ SL Φ( x , y ) g ( y ) dS y