Enhanced resolution in structured media

Transcription

1 Enhanced resolution in structured media Eric Bonnetier w/ H. Ammari (Ecole Polytechnique), and Yves Capdeboscq (Oxford) Outline : 1. An experiment of super resolution 2. Small volume asymptotics 3. Periodicity defect in a composite 4. The main ingredients for the proofs 5. Back to super resolution

2 1. Super (enhanced) resolution : Experiment conducted at LOA by M. Fink and co-workers Focusing beyond the diffraction limit with far-field time reversal (G. Lerosey et al, Science, 315, 1120, 2007) λ λ/30 Microwave sources surrounded by scatterers made of thin copper wires Inter-scatterer distance = λ/100. Sources are λ/30 appart

3 The time reversal experiment : - One of the sources emits a signal at time 0, at a frequency ω = 2π/λ - The signal is recorded far away from the source by a series of transducers, during some time interval (0, T) - The signal is then time reversed φ(t) = φ(t t) and sent back into the medium by the transducers - As the wave equation is symmetric, the signal back propagates towards the orginating source In the absence of scatterers, the signal gets time-reversed into a spot of diameter λ/2. The originating source cannot be distinguished In the presence of scatterers, the time-reversed signal sharply refocuses on the originating antenna

4 2. Small volume asymptotics Ω Ω a(x) a(x) k Reference medium conductivity a(x) Perturbed medium a d (x) = a(x) + (k j a(x)) 1 ωj (x) 8 < : div(a u) = 0 a u / Ω n = g R Ω u = 0 8 < : div(a d u d ) = 0 R a u d/ Ω n = g Ω u d = 0

5 Asymptotics of the voltage difference In the presence of N inclusions centered at the points z j (Fengya-Moskow-Vogelius) u d (x) u(x) = ε n X N M j : u(z j ) N(x, z j ) + o(ε n ) j=1 x Ω where M j is a polarization tensor that contains some information on the geometry of the j-th inclusion + conductivity contrast N(x, z) is the Neumann s function of the reference medium

6 Such asymptotics have been extended to the Helmholtz equations, the Maxwell system, elasticity, to strip like inclusions (Ammari Kang, Ammari Moskow, Beretta Francini Vogelius, Movchan Serkov, Volkov Vogelius,...) Compensated compactness approach and bounds on the polarization tensors (Capdeboscq Vogelius) The singularity in the Neumann or the Green function in the RHS is very interesting for detection purposes : MUSIC type algorithms prove quite efficient to determine the location of inclusions (Brühl Hanke Vogelius, Ammari Iakovleva, see also factorisation method) Echoscan : EIT measurements perturbed by localized ultrasound waves. Small volume asymptotics show that asymptotically the data determine the pointwise values of the electrostatic energy density solve the 0 Laplacian (Ammari B Capdeboscq Tanter Fink) j(x) div( u(x) 2 u(x)) = 0

7 3. Misplaced inclusions in a periodic composite A situation where the size of the defect compares to the size of the variations of the background conductivity Ω ε R 3 perfectly periodic medium, with conductivity a ε (x) = a(x/ε) Y periodicity cell, contains an inclusion D a(y) = k in D a(y) = 1 in Y \ D Ω ε,d perturbed medium : the p-th cell ε(p + Y ) contains a misplaced inclusion. The pth cell lies in ω ε,2 = ε(p + d + D), d < 1 instead of ω ε,1 = ε(p + D). ω ε = ω ε,1 ω ε,2

8 The perturbed conductivity is 8 < : a ε,d (x) = a ε (x) x Ω \ ω ε ω ε = (ω ε,1 ω ε,2 ) a ε,d (x) = 1 x ω ε,1 a ε,d (x) = k x ω ε,2 State equations : Given g on Ω, smooth, 8 < : 8 < : L ε u ε = div(a ε u ε ) = 0 in Ω a ε u ε ν = g on Ω R Ω u ε = 0 L ε,d u ε,d = div(a ε,d u ε,d ) = 0 in Ω a ε,d u ε,d ν = g on Ω R Ω u ε,d = 0 Let z Ω, dist(z, ω ε ) >> ε. Assume the defect is centered at x = 0. Asymptotics of u ε,d (z) to u ε (z) as ε 0

9 Representing both u ε and u ε,d with the Green function G ε, obtain (u ε,d u ε )(z) + term on Ω = [a ε a ε,d ] u ε,d (x) G ε (x, z) dσ x. ωε As ε 0, u ε, u ε,d converge to the homogenized potential u 8 >< >: div(a u ) = 0 in Ω A u ν = g on Ω u = 0, Ω where χ is the vector the components of which solve the cell problems 8 < div(a (χ j (y) + y j )) = 0 in Y : χ j H 1 #(Y ), Y χ j = 0 on Ω

10 Ansatz : u ε,d (x) u ε (x) + εv d (x/ε) + r ε,d (x) where v d solves the rescaled problem (y = x/ε) 8 < : div(a d (y) [(I + y χ) x u (0) + v d (y)]) = 0 in R 3 v d (y) 0 as y v d is the correction to the oscillatory potential with linear growth that would be solution if there were no perturbation Inject in the representation formula u ε,d (z) u ε (z) + term on Ω = (a ε a ε,d )[ x u ε (x) + y v d (x/ε)] x G ε (x, z) ωε + ωε (a ε a ε,d ) r ε,d (x) x G ε (x, z) = I 1 + I 2.

11 Approximate I 1 = ωε (a ε a ε,d )[ x u ε (x) + y v d (x/ε)] x G ε (x, z) u ε (x) u (x) + εχ(x/ε) u (x) u ε (x) [I + y χ(x/ε)] u (0) G ε (x, z) [I + y χ(x/ε)] G (0, z) to obtain I 1 ε n ω 1 (a a d )[(I + y χ(y)) u (0) + y v d (y)] (I + y χ(y)) x G (0, z) while I 2 = o(ε n )

12 Rewritting the auxiliary function v d (y) = ϕ(y) u (0), with ϕ defined by j div [ad (y) ϕ j ] = div[(a a d )(y) (y j + χ j )] in R n, ϕ j (y) 0 as y 0 to get the first order asymptotics : (u ε,d u ε )(z) + Ω a ε ν xg ε (x, z) (u ε,d u ε )(x) = ε n M : x u (0) x G (0, z) + o(ε n ) where the polarization tensor is given by M ij = a ω a d! 1 (y i + χ i ) ha + ν ϕ + j + a (y j + χ j ) i Same structure as in the case of a homogeneous medium, with homogenized potential and Green s function

13 4. Ingredients for the proofs The expansions are based on W 1, estimates on u ε and on G ε, wich are uniform with respect to ε. In the case n = 3 : x u ε L (ω ε) C x u ε (.) (I + y χ(./ε)) x u (.) L (ω ε) Cε x G ε (., z) (I + y χ(./ε)) x G (., z) L (ω ε) Cε 1/4 y r ε,d L 2 (ε 1 Ω) Cε 3/2 y v d (y) = O( y 2 ), where the constant C is uniform w.r.t. ε (and w.r.t z away from the position of the defect).

14 Uniform estimates on the gradients u ε of the potential for a perfectly periodic medium, were first established by M. Avellaneda and Fang Hua Lin under the assumption that the coefficients are regular : Theorem 1. Let f L n+δ (Ω), g C 1,µ ( Ω), a C 0,γ (Y ) M, + uniform ellipticity (with constants λ, Λ) Let u ε solve j div(a(x/ε) uε ) = f in Ω u ε = g on Ω, Then for C = C(n, Ω, λ, Λ, δ, µ, γ, M), one has u ε C 0,1 (Ω) C g C 1,µ(Ω) + f L n+δ(ω).

15 Extension to the case of piecewise Hölder coefficients : - Y = L l=1 D l, matrix phase is D L = Y \ l 1 l=1 D l - Each D l has regularity C 1,α - Each point of Y belongs to at most 2 D l s Theorem 2. (YanYan Li and M. Vogelius, YanYan Li and L. Nirenberg) Assume that a C µ (D l ) 1 l L + uniform ellipticity. Let f L (Ω), g C µ (D l ) 1 l L, and let u solve div(a u) = f + div(g) in Ω. Then, ρ > 0, there is a constant C such that LX l=1 u C 1,α (D l Ωρ) " C u L 2 (Ω) + f L (Ω) + X l g C 0,α (D l ) # for all 0 < α min(µ, α 2(α + 1) ).

16 One can combine both previous theorems to get uniform Lipschitz estimates on u ε Theorem 3. Let B = B(0, 1) and a, elliptic, Y periodic, let b Y periodic, such that a, b C 0,µ (Y \ D) and a, b C 0,µ (D) Let f L (B), h C 0,µ (B) and u ε solves div(a ε u ε ) = f + εdiv(b ε h) in B then, for some C independent on ε, u ε C 0,µ (B1/2 ) + u ε L (B1/2 ) i C h u ε L 2(B) + f L (B) + h C 0,µ(B)

17 Consequences : Corrector result : Assume that u ε u 0 L 2 (Ω) Cε σ, with 0 < σ < 1, then, for some C independent on ε, u ε u 0 L (ω) Cε σ, u ε (.) [I + y χ(./ε)] u 0 (.) L (ω) Cε σ. Estimation of G ε (., z) : Let ω Ω, then for some C independent on ε G ε (., z) G 0 (., z) L (ω) Cε 1/4, G ε (., z) [I + y χ(./ε)] G 0 (., z) L (ω) Cε 1/4.

18 5. Back to super resolution : λ λ/30 Time reversal operator in the case of a small inclusion centered at z, embedded in a homogeneous medium (uδ u)(x, t) where T = y z, ωδ M Uy (z, T ) [Uz (x, t0 t) Uz (x, t t0 )] Uy (x, t) = δt= x y 4π x y In Fourier space F (uδ u)(x, k) ωδ p sin(k x z ) x z «

19 Harmonic regime Û z (x, k) = eik x z 4π x z δû(x) = (û δ û)(x, k) ω δ Û y (z, k)m Û x (z, k) The response function corresponds to ŵ(x) = Û x (x, k) ν δû(x ) δû(x ) ν Û x (x, k) dσ(x ) S «sin(k x z ) is proportionnal to Im(Û z (x, k)) x z In the case of a homogeneous background medium, the width of the first lobe of the Green s function is roughly equal to k 1 = (ω ε 0 µ) 1

20 In O R 2, consider inclusions distributed periodically in a region Ω, which also contains a defect a δ,d = 8 < : (ε s + iσ s /ω) 1 in the inclusions ε 1 d in the defect ε 1 0 elsewhere and the Helmholtz equation j div(aδ,d (x) u δ,d ) + ω 2 µu δ,d = 0 u δ,d/ O = φ We assume that j ε0, ε d, ε s > 0, σ s > 0 µ 0 = µ d = µ s, k = ω 2 µ 0 As above, consider corresponding defect-free field u δ and homogenized field u Assume that ω is not an eigenvalue of div(a v) + ω 2 µv

21 Ingredients : - The smoothness of the homogenized limit u - Pointwise interior estimates on u δ u - A convergence estimate u δ,d u δ 1,2,O = O( D δ 1/2 ), u δ,d u δ 0,2,O = O( D δ 1/2+η ) to prove the asymptotic expansion : (u δ,d u δ )(x) = ω δ (a δ,d a δ ) / Ω M : u (0) G (0, x) + o( ω δ ) The response function involves the homogenized Green function. When the homogenized matrix is isotropic (and real), the resolution is proportional to the effective wavelength k 1 = `ω ε, µ 1

22 Conclusions/perspectives - Small volume asymptotic in a periodic background medium : The asymptotics of the fields have a structure similar to the asymptotics obtained in the case of a smooth (slowly varying) background - Super-resolution experiment : Choosing the proper dielectric parameters for the scatterers may indeed improve resolution : In the case when σ s = 0, ε s > ε 0 then k 1 < k What are the distributions/geometries of the scatterers that achieve optimal resolution (generically, homogenized media are anisotropic)? - When σ s > 0 energy is absorbed in the scatterers and the diameter of the scatterers, the size of Ω might play a role

23 The 3 step compactness method to obtain uniform bounds on u ε Theorem 4. Assume f h L (B 1 ) and u ε solves div(a ε u ε ) = f, in B 1 Then there exists C = C(B 1, A) independent of h such that u ε W 1, (B1/2 ) C u ε L 2 (B1 ) + f L (B 1 )

24 Sketch of proof : for a uniform Hölder estimate step 1 : Let 0 < µ = α/2(α + 1). Show that 0 < θ < 1, 0 < ε 0 < 1 such that, if 8 >< div(a ε u ε ) = f in B 1 R B1 u ε 2 1 then for 0 < ε < ε 0, >: f L (B1 ) ε 0 B θ u ε (u ε ) 0,θ 2 θ 2µ proof : Let µ < µ < 1. By elliptic regularity, the solutions to the limiting problem div(a u ) = 0 in B 1 are smooth : 0 < θ < 1 such that u u θ 2 θ 2µ u B θ B1 2

25 Assume that a sequence ε n 0, u ε n, f ε n does not satisfy the estimate We can extract a subsequence such that u ε n u in H 1 loc (B 1), a solution to the limiting equation Passing to the limit one gets θ 2µ liminf u ε n (u ε n) 0,θ 2 B 1 = u (u ) θ 2 = θ 2µ B 1 a contradiction

26 step 2 : Iteration Show that for all k 1, such that ε/θ k ε 0 u ε (u ε ) 0,θ 2 θ 2kµ ( u ε L 2 (B1 ) + ε 1 0 f L (B 1 ) ) 2 B θ k proof : by induction, applying step 1 to the rescaled functions w ε (x) = J 1 u ε (θ k x) u ε0,θ k J = ( u ε L 2 (B1 ) + ε 1 0 f L (B 1 ) ) 2 where we use the fact that the discretization is uniform

27 step 3 : Blow up Show the Hölder estimate sup 0<r<1/2 sup x <1/2 u ε (u ε ) x,r 2 C r 2µ u ε L 2 (B1 ) + f L (B 1 ) Bx,r When r ε/ε 0, the estimate holds by step 2 Let w ε (y) = ε µ ( u ε (εy) (u ε ) 0,2ε/ε0 ) Then L 1 w ε = f ε in B 2/ε0 i.e., w ε solves an equation the coefficients of which are independent of ε, and on a domain independent of ε. Regularity results for piecewise Hölder coefficients Hölder estimate on w ε, then on u ε