Determination of thin elastic inclusions from boundary measurements.

Transcription

1 Determination of thin elastic inclusions from boundary measurements. Elena Beretta in collaboration with E. Francini, S. Vessella, E. Kim and J. Lee September 7, 2010 E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 1 / 37

2 Setting Ω Ω is a bounded smooth plane domain, σ is a line segment in Ω, for some small ɛ ω ɛ = {x Ω : d(x, σ) < ɛ} E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 2 / 37

3 Setting σ Ω is a bounded smooth plane domain, σ is a line segment in Ω, for some small ɛ ω ɛ = {x Ω : d(x, σ) < ɛ} E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 2 / 37

4 Setting Ω is a bounded smooth plane domain, σ is a line segment in Ω, for some small ɛ ω ɛ = {x Ω : d(x, σ) < ɛ} E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 2 / 37

5 Thin inclusions in a homogeneous conductor E. Beretta, E. Francini, M. Vogelius, Asymptotic formula for the steady state voltage potentials in the presence of thin inhomogeneities. A rigorous analysis, J. Math. Pures Appl, (2003). H. Ammari, E. Beretta, E. Francini, Reconstruction of thin conductivity imperfections, Appl. Anal., (2004). H. Ammari, E. Beretta, E. Francini, Reconstruction of thin conductivity imperfections, II. The case of multiple segments, Appl. Anal., (2006). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 3 / 37

6 Problems Given a traction field g on Ω, study how the displacement field generated in Ω by this traction depends on the presence of the thin inclusion. Recover the position of the segment σ from boundary measurements of the displacement field. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 4 / 37

7 Problems Given a traction field g on Ω, study how the displacement field generated in Ω by this traction depends on the presence of the thin inclusion. Recover the position of the segment σ from boundary measurements of the displacement field. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 4 / 37

8 Problems Given a traction field g on Ω, study how the displacement field generated in Ω by this traction depends on the presence of the thin inclusion. Recover the position of the segment σ from boundary measurements of the displacement field. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 4 / 37

9 Outline The direct problem 1 Thin inclusions in an elastic body 2 Asymptotic expansion The inverse problem 1 The correction term: a crack model 2 Uniqueness 3 Stability 4 Reconstruction E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 5 / 37

10 Outline The direct problem 1 Thin inclusions in an elastic body 2 Asymptotic expansion The inverse problem 1 The correction term: a crack model 2 Uniqueness 3 Stability 4 Reconstruction E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 5 / 37

11 References H. Ammari, H. Kang, G. Nakamura, and K. Tanuma,Complete asymptotic expansions of solutions of the system of elastostatics in the presence of an inclusion of small diameter and detection of an inclusion, J. Elasticity, (2002). H. Kang, E. Kim an J.Y. Lee,Identification of elastic inclusions and elastic moment tensors by boundary measurements, Inverse Problems (2003). H. Ammari and H. Kang, Reconstruction of Small Inhomogeneities from Boundary Measurements, Lecture Notes in Math. 1846, Springer-Verlag, Berlin, E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 6 / 37

12 References H. Ammari, H. Kang, G. Nakamura, and K. Tanuma,Complete asymptotic expansions of solutions of the system of elastostatics in the presence of an inclusion of small diameter and detection of an inclusion, J. Elasticity, (2002). H. Kang, E. Kim an J.Y. Lee,Identification of elastic inclusions and elastic moment tensors by boundary measurements, Inverse Problems (2003). H. Ammari and H. Kang, Reconstruction of Small Inhomogeneities from Boundary Measurements, Lecture Notes in Math. 1846, Springer-Verlag, Berlin, E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 6 / 37

13 References H. Ammari, H. Kang and M. Lim, Effective parameters of elastic composites, Indiana Univ. Math. J. (2006). H. Ammari, H. Kang and H. Lee, Asymptotic expansions for eigenvalues of the Lamï 1 2 system in the presence of small inclusions, Comm. PDE (2007). H. Ammari, H. Kang and H. Lee, A boundary integral method for computing elastic moment tensors for ellipses and ellipsoids, J. Comput. Math. (2007). H. Ammari, P. Calmon and E. Iakovleva, Direct elastic imaging of a small inclusion, SIAM J. Imaging Sci. (2008). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 7 / 37

14 References H. Ammari, H. Kang and M. Lim, Effective parameters of elastic composites, Indiana Univ. Math. J. (2006). H. Ammari, H. Kang and H. Lee, Asymptotic expansions for eigenvalues of the Lamï 1 2 system in the presence of small inclusions, Comm. PDE (2007). H. Ammari, H. Kang and H. Lee, A boundary integral method for computing elastic moment tensors for ellipses and ellipsoids, J. Comput. Math. (2007). H. Ammari, P. Calmon and E. Iakovleva, Direct elastic imaging of a small inclusion, SIAM J. Imaging Sci. (2008). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 7 / 37

15 Part I The direct problem E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 8 / 37

16 Thin inclusions in an isotropic elastic body Ω R 2 is a plane region occupied by a homogeneous elastic material containing an inclusion of the form ω ɛ = {x Ω : d(x, σ) < ɛ} where σ is a line segment. σ ω ε Let C 0 and C 1 be the elastic tensor fields in Ω \ ω ɛ and ω ɛ respectively. We assume that both C 0 and C 1 are isotropic: {C l } 2 ijhk=1 = λ lδ ij δ hk + µ l (δ hi δ kj + δ hj δ ki ) for l = 0, 1. λ l and µ l are the Lamé coefficients. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 9 / 37

17 Thin inclusions in an isotropic elastic body Ω R 2 is a plane region occupied by a homogeneous elastic material containing an inclusion of the form ω ɛ = {x Ω : d(x, σ) < ɛ} where σ is a line segment. σ ω ε Let C 0 and C 1 be the elastic tensor fields in Ω \ ω ɛ and ω ɛ respectively. We assume that both C 0 and C 1 are isotropic: {C l } 2 ijhk=1 = λ lδ ij δ hk + µ l (δ hi δ kj + δ hj δ ki ) for l = 0, 1. λ l and µ l are the Lamé coefficients. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 9 / 37

18 Thin inclusions in an isotropic elastic body Ω R 2 is a plane region occupied by a homogeneous elastic material containing an inclusion of the form ω ɛ = {x Ω : d(x, σ) < ɛ} where σ is a line segment. σ ω ε Let C 0 and C 1 be the elastic tensor fields in Ω \ ω ɛ and ω ɛ respectively. We assume that both C 0 and C 1 are isotropic: {C l } 2 ijhk=1 = λ lδ ij δ hk + µ l (δ hi δ kj + δ hj δ ki ) for l = 0, 1. λ l and µ l are the Lamé coefficients. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 9 / 37

19 The direct problem A traction g : Ω R 2 on Ω induces a displacement u ɛ : Ω R 2 that solves the system { ) div (C ɛ ˆ u ɛ = 0 in Ω (C ɛ ˆ u ɛ ) ν = g on Ω, where C ɛ = C 0 χ Ω\ωɛ + C 1 χ ωɛ, ˆ u ɛ = 1 ( 2 uɛ + ( u ɛ ) T ) is the symmetric deformation tensor, and ν is the unit outer normal to Ω. The solution is unique if we impose some normalization conditions, for example u ɛ = 0, ( u ɛ uɛ T ) = 0. Ω Ω E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 10 / 37

20 The direct problem A traction g : Ω R 2 on Ω induces a displacement u ɛ : Ω R 2 that solves the system { ) div (C ɛ ˆ u ɛ = 0 in Ω (C ɛ ˆ u ɛ ) ν = g on Ω, where C ɛ = C 0 χ Ω\ωɛ + C 1 χ ωɛ, ˆ u ɛ = 1 ( 2 uɛ + ( u ɛ ) T ) is the symmetric deformation tensor, and ν is the unit outer normal to Ω. The solution is unique if we impose some normalization conditions, for example u ɛ = 0, ( u ɛ uɛ T ) = 0. Ω Ω E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 10 / 37

21 Assumptions d(σ, Ω) δ. Lamé coefficients satisfy µ l α 0 > 0, λ l + µ l β 0 > 0 for l = 0, 1. ( C ɛ is strongly convex in Ω.) (λ 0 λ 1 ) 2 + (µ 0 µ 1 ) 2 > 0 and (λ 0 λ 1 )(µ 0 µ 1 ) > 0. g H 1/2 ( Ω) satisfies the compatibility condition g r = 0, Ω for every infinitesimal rigid displacement r(x) = c + Mx where c is constant and M is a skew symmetric matrix. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 11 / 37

22 Assumptions d(σ, Ω) δ. Lamé coefficients satisfy µ l α 0 > 0, λ l + µ l β 0 > 0 for l = 0, 1. ( C ɛ is strongly convex in Ω.) (λ 0 λ 1 ) 2 + (µ 0 µ 1 ) 2 > 0 and (λ 0 λ 1 )(µ 0 µ 1 ) > 0. g H 1/2 ( Ω) satisfies the compatibility condition g r = 0, Ω for every infinitesimal rigid displacement r(x) = c + Mx where c is constant and M is a skew symmetric matrix. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 11 / 37

23 Assumptions d(σ, Ω) δ. Lamé coefficients satisfy µ l α 0 > 0, λ l + µ l β 0 > 0 for l = 0, 1. ( C ɛ is strongly convex in Ω.) (λ 0 λ 1 ) 2 + (µ 0 µ 1 ) 2 > 0 and (λ 0 λ 1 )(µ 0 µ 1 ) > 0. g H 1/2 ( Ω) satisfies the compatibility condition g r = 0, Ω for every infinitesimal rigid displacement r(x) = c + Mx where c is constant and M is a skew symmetric matrix.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 11 / 37

24 Assumptions d(σ, Ω) δ. Lamé coefficients satisfy µ l α 0 > 0, λ l + µ l β 0 > 0 for l = 0, 1. ( C ɛ is strongly convex in Ω.) (λ 0 λ 1 ) 2 + (µ 0 µ 1 ) 2 > 0 and (λ 0 λ 1 )(µ 0 µ 1 ) > 0. g H 1/2 ( Ω) satisfies the compatibility condition g r = 0, Ω for every infinitesimal rigid displacement r(x) = c + Mx where c is constant and M is a skew symmetric matrix.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 11 / 37

25 Assumptions d(σ, Ω) δ. Lamé coefficients satisfy µ l α 0 > 0, λ l + µ l β 0 > 0 for l = 0, 1. ( C ɛ is strongly convex in Ω.) (λ 0 λ 1 ) 2 + (µ 0 µ 1 ) 2 > 0 and (λ 0 λ 1 )(µ 0 µ 1 ) > 0. g H 1/2 ( Ω) satisfies the compatibility condition g r = 0, Ω for every infinitesimal rigid displacement r(x) = c + Mx where c is constant and M is a skew symmetric matrix.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 11 / 37

26 Main point: ɛ is small Strategy: expand u ɛ with respect to ɛ The zero order term the background displacement field u 0 Energy estimate ) div (C 0 u0 = 0 in Ω (C 0 u0 ) ν = g on Ω, Ω u 0 = 0, Ω ( u 0 u0 T ) = 0. u ɛ u 0 H 1 (Ω) C ɛ g H 1/2 ( Ω). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 12 / 37

27 Main point: ɛ is small Strategy: expand u ɛ with respect to ɛ The zero order term the background displacement field u 0 Energy estimate ) div (C 0 u0 = 0 in Ω (C 0 u0 ) ν = g on Ω, Ω u 0 = 0, Ω ( u 0 u0 T ) = 0. u ɛ u 0 H 1 (Ω) C ɛ g H 1/2 ( Ω). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 12 / 37

28 Main point: ɛ is small Strategy: expand u ɛ with respect to ɛ The zero order term the background displacement field u 0 Energy estimate ) div (C 0 u0 = 0 in Ω (C 0 u0 ) ν = g on Ω, Ω u 0 = 0, Ω ( u 0 u0 T ) = 0. u ɛ u 0 H 1 (Ω) C ɛ g H 1/2 ( Ω). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 12 / 37

29 Main point: ɛ is small Strategy: expand u ɛ with respect to ɛ The zero order term the background displacement field u 0 Energy estimate ) div (C 0 u0 = 0 in Ω (C 0 u0 ) ν = g on Ω, Ω u 0 = 0, Ω ( u 0 u0 T ) = 0. u ɛ u 0 H 1 (Ω) C ɛ g H 1/2 ( Ω). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 12 / 37

30 Asymptotic expansion: ingredients Neumann function for the background problem ) div y (C 0 ˆ N(x, y) = δ y (x)i d in Ω (C 0 ˆ N) ν = 1 Ω I d su Ω, Ω N = 0, Ω ( N NT ) = 0. For x = y the Neumann function has the same singularities of the fundamental solution Γ ij (x, y) := A 2π δ ij ln x y B 2π (x y) i (x y) j x y 2, ( ) ( ) where A = µ λ 0 +2µ 0 and B = µ 1 0 λ 0 +2µ 0 E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 13 / 37

31 Asymptotic expansion: ingredients Neumann function for the background problem ) div y (C 0 ˆ N(x, y) = δ y (x)i d in Ω (C 0 ˆ N) ν = 1 Ω I d su Ω, Ω N = 0, Ω ( N NT ) = 0. For x = y the Neumann function has the same singularities of the fundamental solution Γ ij (x, y) := A 2π δ ij ln x y B 2π (x y) i (x y) j x y 2, ( ) ( ) where A = µ λ 0 +2µ 0 and B = µ 1 0 λ 0 +2µ 0 E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 13 / 37

32 Asymptotic expansion: ingredients Neumann function for the background problem ) div y (C 0 ˆ N(x, y) = δ y (x)i d in Ω (C 0 ˆ N) ν = 1 Ω I d su Ω, Ω N = 0, Ω ( N NT ) = 0. For x = y the Neumann function has the same singularities of the fundamental solution Γ ij (x, y) := A 2π δ ij ln x y B 2π (x y) i (x y) j x y 2, ( ) ( ) where A = µ λ 0 +2µ 0 and B = µ 1 0 λ 0 +2µ 0 E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 13 / 37

33 Asymptotic expansion Theorem For every y Ω and for ɛ 0 (u ɛ u 0 )(y) = 2ɛ M(x) u 0 (x) N(x, y) dσ(x) + o(ɛ), where M u 0 = a div u 0 I d + b u 0 + c (u 0 τ) τ σ τ τ + d (u 0 n) n n, n where τ and n are the tangential and normal direction to segment σ and a, b, c, d depend only on the Lamé coefficients of C 0 and C 1. The term o(ɛ) is bounded by Cɛ 1+θ g H 1/2 ( Ω), with 0 < θ < 1 and C depending only on Ω, α 0, β 0 and K. [Beretta, Francini, SIAM J. Math. Anal., 2006] E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 14 / 37

34 Asymptotic expansion Theorem For every y Ω and for ɛ 0 (u ɛ u 0 )(y) = 2ɛ M(x) u 0 (x) N(x, y) dσ(x) + o(ɛ), where M u 0 = a div u 0 I d + b u 0 + c (u 0 τ) τ σ τ τ + d (u 0 n) n n, n where τ and n are the tangential and normal direction to segment σ and a, b, c, d depend only on the Lamé coefficients of C 0 and C 1. The term o(ɛ) is bounded by Cɛ 1+θ g H 1/2 ( Ω), with 0 < θ < 1 and C depending only on Ω, α 0, β 0 and K. [Beretta, Francini, SIAM J. Math. Anal., 2006] E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 14 / 37

35 The solution u ɛ By our assumptions u ɛ is uniquely determined in H 1 (Ω). Set u e ɛ := u ɛ Ω\ωɛ and u i ɛ := u ɛ ωɛ. Then div(c 0 u e ɛ ) = 0 in Ω \ ω ɛ, div(c 1 u i ɛ ) = 0 in ω ɛ. and u e ɛ = u i ɛ su ω ɛ. and, if n is the outer normal to ω ɛ, (C 0 u e ɛ ) n = (C 1 u i ɛ ) n su ω ɛ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 15 / 37

36 The solution u ɛ By our assumptions u ɛ is uniquely determined in H 1 (Ω). Set u e ɛ := u ɛ Ω\ωɛ and u i ɛ := u ɛ ωɛ. Then div(c 0 u e ɛ ) = 0 in Ω \ ω ɛ, div(c 1 u i ɛ ) = 0 in ω ɛ. and u e ɛ = u i ɛ su ω ɛ. and, if n is the outer normal to ω ɛ, (C 0 u e ɛ ) n = (C 1 u i ɛ ) n su ω ɛ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 15 / 37

37 The solution u ɛ By our assumptions u ɛ is uniquely determined in H 1 (Ω). Set u e ɛ := u ɛ Ω\ωɛ and u i ɛ := u ɛ ωɛ. Then div(c 0 u e ɛ ) = 0 in Ω \ ω ɛ, div(c 1 u i ɛ ) = 0 in ω ɛ. and u e ɛ = u i ɛ su ω ɛ. and, if n is the outer normal to ω ɛ, (C 0 u e ɛ ) n = (C 1 u i ɛ ) n su ω ɛ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 15 / 37

38 The solution u ɛ By our assumptions u ɛ is uniquely determined in H 1 (Ω). Set u e ɛ := u ɛ Ω\ωɛ and u i ɛ := u ɛ ωɛ. Then div(c 0 u e ɛ ) = 0 in Ω \ ω ɛ, div(c 1 u i ɛ ) = 0 in ω ɛ. and u e ɛ = u i ɛ su ω ɛ. and, if n is the outer normal to ω ɛ, (C 0 u e ɛ ) n = (C 1 u i ɛ ) n su ω ɛ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 15 / 37

39 The solution u ɛ By our assumptions u ɛ is uniquely determined in H 1 (Ω). Set u e ɛ := u ɛ Ω\ωɛ and u i ɛ := u ɛ ωɛ. Then div(c 0 u e ɛ ) = 0 in Ω \ ω ɛ, div(c 1 u i ɛ ) = 0 in ω ɛ. and u e ɛ = u i ɛ su ω ɛ. and, if n is the outer normal to ω ɛ, (C 0 u e ɛ ) n = (C 1 u i ɛ ) n su ω ɛ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 15 / 37

40 Proof First, we show that y Ω, (u ɛ u 0 )(y) = (C 1 C 0 ) u ɛ(x) i N(x, y) dx. ω ɛ The yellow part of ω ɛ gives a contribution of order ɛ 2, so we ignore it. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 16 / 37

41 Proof First, we show that y Ω, (u ɛ u 0 )(y) = (C 1 C 0 ) u ɛ(x) i N(x, y) dx. ω ɛ The yellow part of ω ɛ gives a contribution of order ɛ 2, so we ignore it. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 16 / 37

42 (u ɛ u 0 )(y) = ω ɛ (C 1 C 0 ) u i ɛ(x) N(x, y) + o(ɛ) u i ɛ is a Hölder continuous function in ω ɛ. xε x σ ε Then, we can approximate u i ɛ (x) σ by u ɛ(x i ɛ ), and get (u ɛ u 0 )(y) = 2ɛ (C 1 C 0 ) u ɛ(x i ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ By transmission condition, we pass from uɛ i to uɛ e, and get (u ɛ u 0 )(y) = 2ɛ M u ɛ e (x ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ Finally we show that we can approximate u e ɛ by u 0 and get the final form. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 17 / 37

43 (u ɛ u 0 )(y) = ω ɛ (C 1 C 0 ) u i ɛ(x) N(x, y) + o(ɛ) u i ɛ is a Hölder continuous function in ω ɛ. xε x σ ε Then, we can approximate u i ɛ (x) σ by u ɛ(x i ɛ ), and get (u ɛ u 0 )(y) = 2ɛ (C 1 C 0 ) u ɛ(x i ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ By transmission condition, we pass from uɛ i to uɛ e, and get (u ɛ u 0 )(y) = 2ɛ M u ɛ e (x ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ Finally we show that we can approximate u e ɛ by u 0 and get the final form. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 17 / 37

44 (u ɛ u 0 )(y) = ω ɛ (C 1 C 0 ) u i ɛ(x) N(x, y) + o(ɛ) u i ɛ is a Hölder continuous function in ω ɛ. xε x σ ε Then, we can approximate u i ɛ (x) σ by u ɛ(x i ɛ ), and get (u ɛ u 0 )(y) = 2ɛ (C 1 C 0 ) u ɛ(x i ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ By transmission condition, we pass from uɛ i to uɛ e, and get (u ɛ u 0 )(y) = 2ɛ M u ɛ e (x ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ Finally we show that we can approximate u e ɛ by u 0 and get the final form. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 17 / 37

45 (u ɛ u 0 )(y) = ω ɛ (C 1 C 0 ) u i ɛ(x) N(x, y) + o(ɛ) u i ɛ is a Hölder continuous function in ω ɛ. xε x σ ε Then, we can approximate u i ɛ (x) σ by u ɛ(x i ɛ ), and get (u ɛ u 0 )(y) = 2ɛ (C 1 C 0 ) u ɛ(x i ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ By transmission condition, we pass from uɛ i to uɛ e, and get (u ɛ u 0 )(y) = 2ɛ M u ɛ e (x ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ Finally we show that we can approximate u e ɛ by u 0 and get the final form. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 17 / 37

46 (u ɛ u 0 )(y) = ω ɛ (C 1 C 0 ) u i ɛ(x) N(x, y) + o(ɛ) u i ɛ is a Hölder continuous function in ω ɛ. xε x σ ε Then, we can approximate u i ɛ (x) σ by u ɛ(x i ɛ ), and get (u ɛ u 0 )(y) = 2ɛ (C 1 C 0 ) u ɛ(x i ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ By transmission condition, we pass from uɛ i to uɛ e, and get (u ɛ u 0 )(y) = 2ɛ M u ɛ e (x ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ Finally we show that we can approximate u e ɛ by u 0 and get the final form. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 17 / 37

47 (u ɛ u 0 )(y) = ω ɛ (C 1 C 0 ) u i ɛ(x) N(x, y) + o(ɛ) u i ɛ is a Hölder continuous function in ω ɛ. xε x σ ε Then, we can approximate u i ɛ (x) σ by u ɛ(x i ɛ ), and get (u ɛ u 0 )(y) = 2ɛ (C 1 C 0 ) u ɛ(x i ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ By transmission condition, we pass from uɛ i to uɛ e, and get (u ɛ u 0 )(y) = 2ɛ M u ɛ e (x ɛ ) N(x ɛ, y) dσ ɛ + o(ɛ). σ ɛ Finally we show that we can approximate u e ɛ by u 0 and get the final form. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 17 / 37

48 Remark An analogue asymptotic expansion holds true if σ is a regular curve, σ ω ε Ω or a collection of multiple disjoint curves. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 18 / 37

49 Part II The inverse problem E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 19 / 37

50 The correction term The asymptotic expansion can be read this way: (u ɛ u 0 )(y) = ɛu σ (y) + o(ɛ). where u σ (y) = 2 M(x) u 0 (x) N(x, y) dσ(x) σ is the first order approximation of u ɛ that we will call correction term. From now on we will consider consider a linear background u 0 (x) = Wx + c, where W is a symmetric matrix and c is a constant determined by the normalization conditions E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 20 / 37

51 The correction term The function u σ con be expressed in the following way u σ (y) = C 0 N(x, y)n ϕ dσ(x) + (N(Q, y) τ)f (Q) (N(P, y) τ)f (P). Here σ P and Q are the endopints of segment σ, τ = PQ PQ and n = τ, ϕ is a vector valued function, f is a scalar function and they both depend only on the background u 0, on τ and n and on the Lamé coefficients. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 21 / 37

52 The crack model The correction term u σ is the trace of a function that has the following properties: In Ω \ σ it solves the background system ) div (C 0 ˆ u σ = 0. At the endpoints {P, Q} of the segment σ it has singularities proportional to N(Q, y) τ and N(P, y) τ. It jumps across the segment σ. The correction term u σ has a zero conormal derivative on Ω. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 22 / 37

53 The crack model The correction term u σ is the trace of a function that has the following properties: In Ω \ σ it solves the background system ) div (C 0 ˆ u σ = 0. At the endpoints {P, Q} of the segment σ it has singularities proportional to N(Q, y) τ and N(P, y) τ. It jumps across the segment σ. The correction term u σ has a zero conormal derivative on Ω.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 22 / 37

54 The crack model The correction term u σ is the trace of a function that has the following properties: In Ω \ σ it solves the background system ) div (C 0 ˆ u σ = 0. At the endpoints {P, Q} of the segment σ it has singularities proportional to N(Q, y) τ and N(P, y) τ. It jumps across the segment σ. The correction term u σ has a zero conormal derivative on Ω.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 22 / 37

55 The crack model The correction term u σ is the trace of a function that has the following properties: In Ω \ σ it solves the background system ) div (C 0 ˆ u σ = 0. At the endpoints {P, Q} of the segment σ it has singularities proportional to N(Q, y) τ and N(P, y) τ. It jumps across the segment σ. The correction term u σ has a zero conormal derivative on Ω.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 22 / 37

56 The crack model The correction term u σ is the trace of a function that has the following properties: In Ω \ σ it solves the background system ) div (C 0 ˆ u σ = 0. At the endpoints {P, Q} of the segment σ it has singularities proportional to N(Q, y) τ and N(P, y) τ. It jumps across the segment σ. The correction term u σ has a zero conormal derivative on Ω.. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 22 / 37

57 Inverse problem Problem: Given the trace of the correction term on an open subset of Ω, determine the segment σ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 23 / 37

58 Inverse problem Problem: Given the trace of the correction term on an open subset of Ω, determine the segment σ. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 23 / 37

59 Uniqueness Proposition Let σ 0 and σ 1 be two segment contained in the interior of the open set Ω. Let u σ0 and u σ1 be the correction terms corresponding to the same background displacement field u 0 = Wx + c. Let Γ be an open subset of Ω. If u σ0 = u σ1 on Γ, then σ 0 = σ 1. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 24 / 37

60 Sketch of proof By unique continuation, since their Cauchy data coincide on Γ, u σ0 = u σ1 on Ω \ (σ 0 σ 1.) If σ 0 σ 1, there is a part γ of σ 0 not contained in σ 1. σ 0 σ 1 γ This implies that u σ0 is bounded and has no jumps on γ. Going back to the properties of u σ0 we see that this is not possible unless u 0 0. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 25 / 37

61 Sketch of proof By unique continuation, since their Cauchy data coincide on Γ, u σ0 = u σ1 on Ω \ (σ 0 σ 1.) If σ 0 σ 1, there is a part γ of σ 0 not contained in σ 1. σ 0 σ 1 γ This implies that u σ0 is bounded and has no jumps on γ. Going back to the properties of u σ0 we see that this is not possible unless u Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 25 / 37

62 Stability Theorem Let σ 0 and σ 1 be two segments contained in Ω, far from the boundary and of positive length. Let u σ0 and u σ1 be the correction terms corresponding to the same non zero background displacement field u 0 = Wx + c, and let Γ be an open subset of Ω. There exists a constant C depending only on the a priori data such that d H (σ 0, σ 1 ) C u σ0 u σ1 L 2 (Γ). where d H denotes the Hausdorff distance. [Beretta, Francini, Vessella, SIAM J. Math. Anal.,2008] E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 26 / 37

63 Idea of the proof We first derive a rough stability estimate in the form d H (σ 0, σ 1 ) w ( ) u σ0 u σ1 L 2 (Γ) where w(δ) is a log log type modulus of continuity. For t [0, 1], let σ t = tσ 0 + (1 t)σ 1 and let u σt be the correction term corresponding to the background displacement field u 0 = Wx + c and to the crack σ t. The trace on Γ of function u σt is Frechet differentiable with respect to t and we can explicitly write, for x Ω u (x, t) := t u σ t (x). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 27 / 37

64 Idea of the proof We first derive a rough stability estimate in the form d H (σ 0, σ 1 ) w ( ) u σ0 u σ1 L 2 (Γ) where w(δ) is a log log type modulus of continuity. For t [0, 1], let σ t = tσ 0 + (1 t)σ 1 and let u σt be the correction term corresponding to the background displacement field u 0 = Wx + c and to the crack σ t. The trace on Γ of function u σt is Frechet differentiable with respect to t and we can explicitly write, for x Ω u (x, t) := t u σ t (x).. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 27 / 37

65 Idea of the proof We first derive a rough stability estimate in the form d H (σ 0, σ 1 ) w ( ) u σ0 u σ1 L 2 (Γ) where w(δ) is a log log type modulus of continuity. For t [0, 1], let σ t = tσ 0 + (1 t)σ 1 and let u σt be the correction term corresponding to the background displacement field u 0 = Wx + c and to the crack σ t. The trace on Γ of function u σt is Frechet differentiable with respect to t and we can explicitly write, for x Ω u (x, t) := t u σ t (x).. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 27 / 37

66 Idea of the proof We first derive a rough stability estimate in the form d H (σ 0, σ 1 ) w ( ) u σ0 u σ1 L 2 (Γ) where w(δ) is a log log type modulus of continuity. For t [0, 1], let σ t = tσ 0 + (1 t)σ 1 and let u σt be the correction term corresponding to the background displacement field u 0 = Wx + c and to the crack σ t. The trace on Γ of function u σt is Frechet differentiable with respect to t and we can explicitly write, for x Ω u (x, t) := t u σ t (x).. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 27 / 37

67 Idea of the proof, continued By using the explicit expression of u (x, t) we show that there exist constants C 0, m 0 > 0 depending only on the a priori data, such that u (, 0) L 2 (Γ) m 0 (d H (σ 0, σ 1 )) and u (, 0) u (, t) L 2 (Γ) C 0 (d H (σ 0, σ 1 )) 2 E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 28 / 37

68 Idea of the proof, continued By using the explicit expression of u (x, t) we show that there exist constants C 0, m 0 > 0 depending only on the a priori data, such that u (, 0) L 2 (Γ) m 0 (d H (σ 0, σ 1 )) and u (, 0) u (, t) L 2 (Γ) C 0 (d H (σ 0, σ 1 )) 2 E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 28 / 37

69 Idea of the proof, continued Then, for x Ω, u σ1 (x) u σ0 (x) = and t 0 u (x, η) dη = u (x, 0) t 0 (u (x, η) u (x, 0)) dη u σ1 u σ0 L 2 (Γ) u (, 0) L 2 (Γ) u (, 0) u (, t) L 2 (Γ) (m 0 C 0 d H (σ 0, σ 1 ))d H (σ 0, σ 1 ) Since, by the rough estimate, d H (σ 0, σ 1 ) is smaller than m 0 2C 0 u σ1 u σ0 L 2 (Γ) is small enough, then if d H (σ 0, σ 1 ) C u σ0 u σ1 L 2 (Γ) E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 29 / 37

70 Idea of the proof, continued Then, for x Ω, u σ1 (x) u σ0 (x) = and t 0 u (x, η) dη = u (x, 0) t 0 (u (x, η) u (x, 0)) dη u σ1 u σ0 L 2 (Γ) u (, 0) L 2 (Γ) u (, 0) u (, t) L 2 (Γ) (m 0 C 0 d H (σ 0, σ 1 ))d H (σ 0, σ 1 ) Since, by the rough estimate, d H (σ 0, σ 1 ) is smaller than m 0 2C 0 u σ1 u σ0 L 2 (Γ) is small enough, then if d H (σ 0, σ 1 ) C u σ0 u σ1 L 2 (Γ) E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 29 / 37

71 Idea of the proof, continued Then, for x Ω, u σ1 (x) u σ0 (x) = and t 0 u (x, η) dη = u (x, 0) t 0 (u (x, η) u (x, 0)) dη u σ1 u σ0 L 2 (Γ) u (, 0) L 2 (Γ) u (, 0) u (, t) L 2 (Γ) (m 0 C 0 d H (σ 0, σ 1 ))d H (σ 0, σ 1 ) Since, by the rough estimate, d H (σ 0, σ 1 ) is smaller than m 0 2C 0 u σ1 u σ0 L 2 (Γ) is small enough, then if d H (σ 0, σ 1 ) C u σ0 u σ1 L 2 (Γ) E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 29 / 37

72 Idea of the proof, continued Then, for x Ω, u σ1 (x) u σ0 (x) = and t 0 u (x, η) dη = u (x, 0) t 0 (u (x, η) u (x, 0)) dη u σ1 u σ0 L 2 (Γ) u (, 0) L 2 (Γ) u (, 0) u (, t) L 2 (Γ) (m 0 C 0 d H (σ 0, σ 1 ))d H (σ 0, σ 1 ) Since, by the rough estimate, d H (σ 0, σ 1 ) is smaller than m 0 2C 0 u σ1 u σ0 L 2 (Γ) is small enough, then if d H (σ 0, σ 1 ) C u σ0 u σ1 L 2 (Γ) E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 29 / 37

73 Rescaled stability estimate Corollary Let σ 0 and σ 1 be two segments contained in Ω, far from the boundary and of positive length and let Γ be an open subset of Ω. For j = 0, 1, let uɛ j be the solution of { ( ) div C ɛ ˆ u ɛ j = 0 in Ω (C ɛ ˆ u j ɛ) ν = g on Ω, corresponding to ωɛ j = {x Ω : d(x, σ j ) < ɛ}. Then, there exist a positive constant C and θ (0, 1), depending only on the a priori data, such that, ( d H (σ 0, σ 1 ) C ɛ 1 uɛ 0 uɛ 1 L 2 (Γ) + ɛ θ). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 30 / 37

74 Rescaled stability estimate Corollary Let σ 0 and σ 1 be two segments contained in Ω, far from the boundary and of positive length and let Γ be an open subset of Ω. For j = 0, 1, let uɛ j be the solution of { ( ) div C ɛ ˆ u ɛ j = 0 in Ω (C ɛ ˆ u j ɛ) ν = g on Ω, corresponding to ωɛ j = {x Ω : d(x, σ j ) < ɛ}. Then, there exist a positive constant C and θ (0, 1), depending only on the a priori data, such that, ( d H (σ 0, σ 1 ) C ɛ 1 uɛ 0 uɛ 1 L 2 (Γ) + ɛ θ). E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 30 / 37

75 Algorithm The MUSIC-type algorithm first computes the end-points P, Q of an inclusion using three measured data, (Λ ɛ Λ 0 )(g j ) Ω for j = 1, 2, 3. Once endpoints are detected, the thickness ɛ can be easily recovered using the ratio between (Λ ɛ Λ 0 )(g j ) L 2 ( Ω) and w g j σ L 2 ( Ω). We take Ω to be the unit disk centered at zero and choose N equi-spaced points y i along the boundary Ω, say {(cos θ i, sin θ i ) : θ i = 2π(i 1)/N, i = 0, 1,, N 1}. Suppose that the 2N 3 matrix A := (w g j σ (y i )) has the spectral decomposition A = 3 σ p u p v p, (1) p=1 where σ p are the singular values of A, and u p, v p are the corresponding singular vectors. Let P : R N span{u 1, u 2, u 3 } be the orthogonal projector P = 3 p=1 up u p. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 31 / 37

76 Algorithm The most important observation to derive a MUSIC-type algorithm is that lies in the space spanned by columns of A or equivalently, (N Q N P ) (Q P) Q P ( ) (NQ N P )(y i ) (Q P) (I P) = 0. Q P E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 32 / 37

77 Algorithm Proof Let g = ( (C 0 W ) )ν be the Neumann data corresponding a symmetric a b matrix W =. Since σ := [P, Q] is a straight line segment, ϕ is b c a constant vector The first order expansion term w σ (x) has the form: w σ (x) = C 0 Nz (x)ν ϕdσ z + f (N Q N P )(x) τ. σ E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 33 / 37

78 Algorithm It is easy to see that we can find W 0 so that ϕ = 0 and f 0 For such a Neumann data g = (C 0 W )ν, w σ (x) becomes f (N Q N P )(x) τ. Then which completes the proof. (N Q N P )(y i ) τ span{u 1, u 2, u 3 }, (2) E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 34 / 37

79 Algorithm Let A ɛ := ((Λ ɛ Λ 0 )(g j )(y i )). The 2N 3 matrix A ɛ has a spectral decomposition 3 A ɛ = σɛ p uɛ p vɛ p, p=1 where σɛ p are the singular values of A ɛ, and uɛ p, vɛ p are the corresponding singular vectors. Let P ɛ : R N span{uɛ 1, uɛ 2, uɛ 3 } be the orthogonal projector P ɛ = 3 p=1 up ɛ uɛ p. we seek to find P and Q minimizing (I P ɛ ) L(P, Q) := ( (NQ N P ) (Q P) Q P P ɛ ( (NQ N P ) (Q P) Q P ) L 2 ( Ω) ) L 2 ( Ω) E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 35 / 37

80 Numerical experiments Table: Computational domain and Reconstruction error. Case P Q ɛ P P c P Q Q Q c P Q ɛ c ɛ ɛ 1 (0.5,0.5) (0.6,0.45) 5.0e (0.0,0.2) (0.7,0.6) 5.0e (0.0,0.2) (0.1,0.3) 5.0e (0.7,0.6) (0.8,0.4) 5.0e Err E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 36 / 37

81 Figure: Reconstruction results of the MELIR algorithm. The solid line represent the actual domains and the dashed lines for the computed results. Bottom figures are 6.67 times zoom up of the corresponding results. E. Beretta (Università di Roma La Sapienza ) Linear elastic cracks Msri,Berkeley 37 / 37