Local Inversions in Ultrasound Modulated Op5cal Tomography. Guillaume Bal Shari Moskow
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1 Local Inversions in Ultrasound Modulated Op5cal Tomography Guillaume Bal Shari Moskow
2 Ultrasound Modulated Op5cal Tomography (Acousto Op5cs Acous5c waves are emided which perturb the op5cal proper5es of the medium Light propaga5ng through the medium is used to recover the original op5cal parameters Model by G. Bal and J. C. Schotland Phys. Rev. LeDers, 2010.
3 Op5cal proper5es perturbed by acous5c waves σ ɛ = σ + ɛ(2β + 1 cos(k x + φ γ ɛ = γ + ɛ(2β 1 cos(k x + φ Lineariza5on wrt epsilon and some manipula5on yields boundary data [ Σ(k, φ = (2β 1γ( φ 2 + (2β + 1σφ 2] cos(k x + φ Ω Which is the Fourier transform of some internal data
4 Mathema5cal Problem Given internal data of the form H ij (x =γ u i u j + ησu i u j, η where is a known fixed constant and γ u j + σu j = 0 in Ω u j = f j on Ω γ Find and σ
5 Previous work Recovery of only, γ σ =0 Capdeboscq, Fehrenback, De Gournay, Kavian (n=2 Bal, Bonne5er, Monard, Triki (n=3 Bal, Monard (n>=4 Kuchment, Kunyansky Kuchment, Steinhauer pseudo differen5al calculus Ammari, Capdeboscq, Triki 2012 separa5on of terms
6 Assume we have some known γ 0 σ 0 background and. γ = γ 0 + δγ σ = σ 0 + δσ u j = u 0 j + δu j Where the background solu5ons sa5sfy γ 0 u 0 j + σ 0 u 0 j = 0 in Ω u 0 j = f j on Ω
7 L 0 := γ 0 + σ 0 δu j = L 1 0 ( δγ u0 j δσu 0 j
8 linearized problem dh ij = δγ u 0 i u 0 j + γ 0 δu i u 0 j + γ 0 u 0 i δu 0 j +ηδσu 0 i u 0 j + ησ 0 δu i u 0 j + ησ 0 u 0 i δu j Really have 3 unknowns here δu j, δγ, δσ But they are coupled L 0 δu j = δγ u 0 j δσu 0 j One approach: solve for and subs5tute back in δu j
9 Take Laplacian of data where T ij is compact, and dh ij (δγ, δσ =G ij (δγ, δσ+ T ij (δγ, δσ, G ij (δγ, δσ = u 0 i u 0 j δγ 2( u 0 i u 0 j s : D 2 δγ + ηu 0 i u 0 j δσ
10 Simplest case Case where n=2 and Take to get u 0 =1 Eliminate Take δσ u 0 i = x i γ 0 =1, σ 0 =0 dh 00 (δγ, δσ =ηδσ
11 then dh 11 =( 2 x 2 2 x 1 dh 12 = 2 x1 x 2 dh 22 =( 2 x 1 2 x 2 Separately get hyperbolic, not ellip5c Together ellip5c as a redundant system Hard to invert because redundant
12 but consider Γ T Γ = ij ( d H ij 2 Γ T Γ =2 4 x x 2 Which is ellip5c.
13 And for n 3 n i=1 d H ii =(n 2 Which we can invert
14 For general this doesn t work σ 0 So let us consider for general γ 0, σ 0 The highest order part G ij (δγ, δσ = u 0 i u 0 j δγ 2( u 0 i u 0 j s : D 2 δγ + ηu 0 i u 0 j δσ Define θ i = u0 i u 0 i
15 Then we are interested in the system A ij δγ + B ij δσ = F ij, where A ij = θ i θ j 2(θ i θ j s :, and B ij = ηd i d j, d i = u i u i.
16 Define the operator Γ = ( A ij B ij With a row for each pair (i,j We want to show this operator is ellip5c so that we can get a parametrix, or Inver5bility of the highest order part. Construc5on of parametrices for similar problems in Kuchment and Steinhauer for one coefficient.
17 Consider the 2x2 system Γ T Γ ( δγ δσ = Γ T F. Γ T Γ = ( ij AT ij A ij ij BT ij A ij ij AT ij B ij ij BT ij B ij.
18 Using symbols, the system is inver5ble when we always have at least one of the sub determinants not vanishing Det ( Aij B ij A kl B kl
19 These determinants are zero when (θ i θ j 2θ i ˆξθ j ˆξd p d q =(θ p θ q 2θ p ˆξθ q ˆξd i d j (i, j, p, q. θ i = u0 i u 0 i d i = u0 i u 0 i
20 Thanks to Gunther Uhlmann and CGOs u ρ = e ρ x = e ρr x (cos ρ I x + i sin ρ I x u ρ = e ρr x [(ρ r cos ρ I x ρ I sin ρ I x+i(ρ r sin ρ I x + ρ I cos ρ I x] Which gives, by taking real and imaginary parts θ 1 = ( cos ρi x sin ρ I x θ 2 = ( sin ρi x cos ρ I x d 1 = cos ρ I x ρ d 2 = sin ρ I x ρ
21 (1 2(θ 1 ξ 2 d 2 2 = (1 2(θ 2 ξ 2 d 2 1 2θ 1 ξθ 2 ξd 2 1 = (1 2(θ 1 ξ 2 d 1 d 2 2θ 1 ξθ 2 ξd 2 2 = (1 2(θ 2 ξ 2 d 1 d 2 Which is (s 2 c 2 d 2 2 = (c 2 s 2 d 2 1 2csd 2 1 = (s 2 c 2 d 1 d 2 2csd 2 2 = (c 2 s 2 d 1 d 2 (1 s 2 c 2 =0 (2 or (3 sc =0
22 But ellip5city doesn t guarantee injec5vity Need injec5vity for extensions to nonlinear problem
23 One approach: view as a differen5al operator with its natural square bilinear form (( v B w, ( v w := Ω Γ ( v w Γ ( v w on H 2 0(Ω H 2 0 (Ω
24 Varia5onal fomula5on: find (δγ, δσ H 2 0 (Ω H 2 0 (Ω Such that (( δγ B δσ, ( v w + L (( δγ δσ, ( v w = Ω F ΓT ( v w (v, w H 2 0 (Ω H 2 0 (Ω Where L is a lower order operator (generally nonlocal
25 B is clearly bounded above on Know ellip5c, can get coercivity bounds explicitly in some cases H 2 0(Ω H 2 0 (Ω
26 Case n=2, constant Have the two background solu5ons u 0 1 = e u 0 2 = e q σ0 γ x 1 0 q σ0 γ 0 x 2, σ 0, γ 0 Which give θ i = e i and d i = γ0 σ 0.
27 Γ =( A ij B ij Corresponding to (i,j=(1,1,(1,2,(2,2 where A 11 = yy xx A 12 = 2 xy A 22 = xx yy B := B 11 = B 12 = B 22 = η γ 0 σ 0.
28 (( v B w ( v, w = Ω 2(v xx 2 + 2(v yy 2 +3η 2 γ2 0 σ0 2 ( w 2 2η γ 0 v xy w σ 0 Use Cauchy s inequality v xy w ɛv 2 xy + ( w2 4ɛ and integra5on by parts Ω v2 xy = Ω v xxv yy
29 B (( v w ( v, w Ω ( 2 η γ 0 η γ 0 σ 0 ɛ 2 σ 0 ɛ ( vxx v yy ( vxx + v yy ( 3η 2 γ 2 0 σ 2 0 η γ 0 2ɛσ 0 ( w 2 choose ɛ = σ 0 γ 0 η. To get v xx 2 L 2 + v yy 2 L γ 2 0 σ 2 0 η 2 w 2 L 2.
30 If we have injec5vity, this means that the linearized solu5ons ˆ δγ H 2 0 (Ω, ˆ δσ H 2 0 (Ω C F L2 (Ω and we have explicit knowledge of C
31 System is ellip5c but don t yet know if injec5ve. But since problem is square: Ω Γ ( v w Γ ( v w =0 Γ ( v w = ( 0 0
32 Case where domain is small If the domain is small, and are u 0 u 0 i i close to constants dh ij (δγ, δσ = δγ u 0 i u0 j + γ 0 δu i u 0 j + γ 0 u 0 i δu j +ηδσu 0 i u0 j + ησ 0δu i u 0 j + ησ 0u 0 i δu j So when we take L_0 of data, lower order terms are differen5al operators. L 0 = γ 0 + σ
33 if is a differen5al operator and Γ ( ( v 0 Γ =. w 0 since = 0 on Ω v = v ν = w = w ν We can get that from Holmgren s v = w =0 theorem
34 Conclusions/Future Have ellip5city for linearized system Have injec5vity with boundary data if the domain is small enough (by varia5onal formula5on and Holmgren s theorem So for small domains, can extend to local nonlinear injec5vity/inversion S5ll to do: injec5vity for more general domains
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