Some issues on Electrical Impedance Tomography with complex coefficient Elisa Francini (Università di Firenze) in collaboration with Elena Beretta and Sergio Vessella E. Francini (Università di Firenze) EIT 1 / 42
Outline 1 Electrical Impedance Tomography with complex coefficient 2 Reconstruction from the DN map: a stability result. 3 Size estimate from a single boundary measurement. E. Francini (Università di Firenze) EIT 2 / 42
Electrical Impedance Tomography Electrical Impedance Tomography: Determine electrical conductivity and permittivity in the interior of a body by measurements of electric currents and voltages at the boundary of the body. Medical diagnostics: detection of pulmonary emboli, breast tumours, screening of organs in transplant surgery Nondestructive testing of materials: detection of corrosion or cracks Geophysical prospection: detection of underground mineral deposits E. Francini (Università di Firenze) EIT 3 / 42
Mathematical model γ : Ω R n C complex admittivity function γ = σ + iωɛ electrical conductivity frequency electrical permittivity σ λ 1 > 0 (dissipation of energy), γ λ The potential u H 1 (Ω, C) is a weak solution to equation div (γ u) = 0 in Ω. Boundary data: u Ω and γ u ν Ω. E. Francini (Università di Firenze) EIT 4 / 42
Mathematical model γ : Ω R n C complex admittivity function γ = σ + iωɛ electrical conductivity frequency electrical permittivity σ λ 1 > 0 (dissipation of energy), γ λ The potential u H 1 (Ω, C) is a weak solution to equation div (γ u) = 0 in Ω. Boundary data: u Ω and γ u ν Ω. E. Francini (Università di Firenze) EIT 4 / 42
Mathematical model γ : Ω R n C complex admittivity function γ = σ + iωɛ electrical conductivity frequency electrical permittivity σ λ 1 > 0 (dissipation of energy), γ λ The potential u H 1 (Ω, C) is a weak solution to equation div (γ u) = 0 in Ω. Boundary data: u Ω and γ u ν Ω. E. Francini (Università di Firenze) EIT 4 / 42
Part I Reconstruction from infinite boundary measurements E. Francini (Università di Firenze) EIT 5 / 42
DN map Given f H 1/2 ( Ω) there exists a unique weak solution u H 1 (Ω) of the problem { div (γ u) = 0 in Ω u = f on Ω. Consider the Dirichlet to Neumann map Λ γ : H 1/2 ( Ω) H 1/2 ( Ω) given by Λ γ f = γ u ν Ω, where ν is the exterior unit normal vector to Ω. EIT Inverse Problem: Determine γ from the knowledge of the Dirichlet-to Neumann map Λ γ E. Francini (Università di Firenze) EIT 6 / 42
DN map Given f H 1/2 ( Ω) there exists a unique weak solution u H 1 (Ω) of the problem { div (γ u) = 0 in Ω u = f on Ω. Consider the Dirichlet to Neumann map Λ γ : H 1/2 ( Ω) H 1/2 ( Ω) given by Λ γ f = γ u ν Ω, where ν is the exterior unit normal vector to Ω. EIT Inverse Problem: Determine γ from the knowledge of the Dirichlet-to Neumann map Λ γ E. Francini (Università di Firenze) EIT 6 / 42
DN map Given f H 1/2 ( Ω) there exists a unique weak solution u H 1 (Ω) of the problem { div (γ u) = 0 in Ω u = f on Ω. Consider the Dirichlet to Neumann map Λ γ : H 1/2 ( Ω) H 1/2 ( Ω) given by Λ γ f = γ u ν Ω, where ν is the exterior unit normal vector to Ω. EIT Inverse Problem: Determine γ from the knowledge of the Dirichlet-to Neumann map Λ γ E. Francini (Università di Firenze) EIT 6 / 42
Known results Uniqueness n 3 If γ is sufficiently smooth the results obtained by Sylvester and Uhlmann (1987, Ann. of Math.) for the conductivity case can be extended to the complex case. (Λ γ1 = Λ γ2 implies γ 1 = γ 2 ) n = 2 F.(2000, Inverse Problems) for small frequencies ω, Bukhgeim for arbitrary frequencies and smooth coefficients (2008, Inv. Ill Posed Problems). n 3, γ L (Ω) completely open n = 2 Astala and Päivärinta for real conductivities σ L (Ω) ( 2006, Ann. of Math.) E. Francini (Università di Firenze) EIT 7 / 42
Known results Stability n 3 Alessandrini (1988, Appl. Anal.) has proved for the conductivity case that if σ j W 2, (Ω) E for j = 1, 2, then σ 1 σ 2 L (Ω) ω( Λ σ1 Λ σ2 ) where ω(t) = C log t η. n = 2 Barcelo, Faraco and Ruiz (2007, Jour. Math. Pures Appl.) have proved for the conductivity case an analogous logarithmic estimate under the a priori assumption σ C α (Ω) E. Clop, Faraco, Ruiz (2009, IPI) for σ j W α,p, α > 0, j = 1, 2, and 1 < p < then where ω(t) = C log t η. σ 1 σ 2 L 2 (Ω) ω( Λ σ1 Λ σ2 ) E. Francini (Università di Firenze) EIT 8 / 42
A priori assumptions Mandache (2001, Inverse Problems) has proved that logarithmic stability is the best possible stability also using as a-priori assumption σ C k (Ω) E, k = 0, 1, 2... E. Francini (Università di Firenze) EIT 9 / 42
Strategy: Look for a-priori assumptions on γ that are physically relevant give rise to a better type of stability Assume: γ is piecewise constant on known domains. In the real case: Alessandrini and Vessella (2005, Adv. in Appl. Math.) E. Francini (Università di Firenze) EIT 10 / 42
Strategy: Look for a-priori assumptions on γ that are physically relevant give rise to a better type of stability Assume: γ is piecewise constant on known domains. In the real case: Alessandrini and Vessella (2005, Adv. in Appl. Math.) E. Francini (Università di Firenze) EIT 10 / 42
Strategy: Look for a-priori assumptions on γ that are physically relevant give rise to a better type of stability Assume: γ is piecewise constant on known domains. In the real case: Alessandrini and Vessella (2005, Adv. in Appl. Math.) E. Francini (Università di Firenze) EIT 10 / 42
Main assumptions (H1) Ω R n, bounded, Ω A, Lipschitz boundary (H2) γ(x) = N j=1 γ j1 Dj (x) unknown complex numbers known open sets (H3) D i D j =, i j, Lipschitz boundaries and N j=1 D j = Ω D 1 Ω contains an open flat portion Σ 1. For every j {2,..., N} there exists a chain of subdomains connecting D 1 to D j so that consecutive domains share a flat portion of boundary E. Francini (Università di Firenze) EIT 11 / 42
A possible configuration D j Σ 1 D 1 E. Francini (Università di Firenze) EIT 12 / 42
Chain D j Σ 1 D 1 E. Francini (Università di Firenze) EIT 13 / 42
[Beretta, F., Comm. PDE, 2011] E. Francini (Università di Firenze) EIT 14 / 42 Main result Theorem Let Ω satisfy assumption (H1). Let γ (k), k = 1, 2 be two complex piecewise constant functions of the form γ (k) (x) = N j=1 γ (k) j 1 Dj (x), where γ (k) satisfy for k = 1, 2 assumption (H2) and D j, j = 1,..., N satisfy assumption (H3). Then there exists a positive constant C such that γ (1) γ (2) L (Ω) C Λ γ (1) Λ γ (2) L(H 1/2 ( Ω),H 1/2 ( Ω)). The constant C depends on the Lipschitz constant of the boundaries, on the volume of Ω, on n, N and λ.
Main idea of the proof (uniqueness) Assume Λ γ (1) = Λ γ (2) ) ) div (γ (1) u 1 = 0 div (γ (2) u 2 = 0 in Ω 0 =< (Λ γ (1) Λ γ (2))u 1, u 2 >= We want to use this identity for singular solutions Ω ( γ (1) γ (2)) u 1 u 2 dx E. Francini (Università di Firenze) EIT 15 / 42
Main idea of the proof (uniqueness) Assume Λ γ (1) = Λ γ (2) ) ) div (γ (1) u 1 = 0 div (γ (2) u 2 = 0 in Ω 0 =< (Λ γ (1) Λ γ (2))u 1, u 2 >= We want to use this identity for singular solutions Ω ( γ (1) γ (2)) u 1 u 2 dx E. Francini (Università di Firenze) EIT 15 / 42
Main idea of the proof (uniqueness) Assume Λ γ (1) = Λ γ (2) ) ) div (γ (1) u 1 = 0 div (γ (2) u 2 = 0 in Ω 0 =< (Λ γ (1) Λ γ (2))u 1, u 2 >= We want to use this identity for singular solutions Ω ( γ (1) γ (2)) u 1 u 2 dx E. Francini (Università di Firenze) EIT 15 / 42
Main idea of the proof (uniqueness) Assume Λ γ (1) = Λ γ (2) ) ) div (γ (1) u 1 = 0 div (γ (2) u 2 = 0 in Ω 0 =< (Λ γ (1) Λ γ (2))u 1, u 2 >= We want to use this identity for singular solutions Ω ( γ (1) γ (2)) u 1 u 2 dx E. Francini (Università di Firenze) EIT 15 / 42
Main idea of the proof (uniqueness) D M γ (1) γ (2) L (D M ) = γ (1) γ (2) L (Ω) Extend γ (k) = 1 in D 0, an open cylinder connected to Σ 1. D 0 Set G y Σ k (x, y) the Green s function related to γ 1 z D and Ω D 0 1 For y, z D 0 ( ) D 1 γ (1) 1 γ (2) 1 G 1 (x, y) G 2 (x, z) dx ( ) + γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0. Ω\D 1 by letting y and z tend to Σ 1 γ (1) 1 = γ (2) 1. E. Francini (Università di Firenze) EIT 16 / 42
Main idea of the proof (uniqueness) D M γ (1) γ (2) L (D M ) = γ (1) γ (2) L (Ω) Extend γ (k) = 1 in D 0, an open cylinder connected to Σ 1. D 0 Set G y Σ k (x, y) the Green s function related to γ 1 z D and Ω D 0 1 For y, z D 0 ( ) D 1 γ (1) 1 γ (2) 1 G 1 (x, y) G 2 (x, z) dx ( ) + γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0. Ω\D 1 by letting y and z tend to Σ 1 γ (1) 1 = γ (2) 1. E. Francini (Università di Firenze) EIT 16 / 42
Main idea of the proof (uniqueness) D M γ (1) γ (2) L (D M ) = γ (1) γ (2) L (Ω) Extend γ (k) = 1 in D 0, an open cylinder connected to Σ 1. D 0 Set G y Σ k (x, y) the Green s function related to γ 1 z D and Ω D 0 1 For y, z D 0 ( ) D 1 γ (1) 1 γ (2) 1 G 1 (x, y) G 2 (x, z) dx ( ) + γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0. Ω\D 1 by letting y and z tend to Σ 1 γ (1) 1 = γ (2) 1. E. Francini (Università di Firenze) EIT 16 / 42
Main idea of the proof (uniqueness) Ω\D 1 ( ) γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0, y, z D 0. Can we let y, z into D 1? Yes because ( ) S 1 (y, z) := γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx Ω\D 1 solves in D 0 D 1 D M and div y (γ (1) (y) y S 1 (y, z))) = 0 D 0 D 1 Σ 1 div z (γ (2) (z) z S 1 (y, z))) = 0 By letting y and z tend to D 2 γ (1) 2 = γ (2) 2. E. Francini (Università di Firenze) EIT 17 / 42
Main idea of the proof (uniqueness) Ω\D 1 ( ) γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0, y, z D 0. Can we let y, z into D 1? Yes because ( ) S 1 (y, z) := γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx Ω\D 1 solves in D 0 D 1 D M and div y (γ (1) (y) y S 1 (y, z))) = 0 D 0 D 1 Σ 1 div z (γ (2) (z) z S 1 (y, z))) = 0 By letting y and z tend to D 2 γ (1) 2 = γ (2) 2. E. Francini (Università di Firenze) EIT 17 / 42
Main idea of the proof (uniqueness) Ω\D 1 ( ) γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0, y, z D 0. Can we let y, z into D 1? Yes because ( ) S 1 (y, z) := γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx Ω\D 1 solves in D 0 D 1 D M and div y (γ (1) (y) y S 1 (y, z))) = 0 D 0 D 1 Σ 1 div z (γ (2) (z) z S 1 (y, z))) = 0 By letting y and z tend to D 2 γ (1) 2 = γ (2) 2. E. Francini (Università di Firenze) EIT 17 / 42
Main idea of the proof (uniqueness) Ω\D 1 ( ) γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx = 0, y, z D 0. Can we let y, z into D 1? Yes because ( ) S 1 (y, z) := γ (1) (x) γ (2) (x) G 1 (x, y) G 2 (x, z) dx Ω\D 1 solves in D 0 D 1 D M and div y (γ (1) (y) y S 1 (y, z))) = 0 D 0 D 1 Σ 1 div z (γ (2) (z) z S 1 (y, z))) = 0 By letting y and z tend to D 2 γ (1) 2 = γ (2) 2. E. Francini (Università di Firenze) EIT 17 / 42
Main technical ingredients Construction of singular solutions and study of their asymptotic behavior near the discontinuity interfaces. Quantitative estimates of unique continuation. E. Francini (Università di Firenze) EIT 18 / 42
Main technical ingredients Construction of singular solutions and study of their asymptotic behavior near the discontinuity interfaces. Quantitative estimates of unique continuation. E. Francini (Università di Firenze) EIT 18 / 42
Singular solutions Difficulty: Lack of a result of existence of the Green s function in Ω for n 3 n 3 Hofmann and Kim (2007, Manuscripta Math. ) existence of Green s function for weak solutions satisfying interior Hölder continuity estimates. n = 2 Dong and Kim (2009, Trans. Amer. Math. Soc. ) existence of Green s function E. Francini (Università di Firenze) EIT 19 / 42
Singular solutions Difficulty: Lack of a result of existence of the Green s function in Ω for n 3 n 3 Hofmann and Kim (2007, Manuscripta Math. ) existence of Green s function for weak solutions satisfying interior Hölder continuity estimates. n = 2 Dong and Kim (2009, Trans. Amer. Math. Soc. ) existence of Green s function E. Francini (Università di Firenze) EIT 19 / 42
Walkway Define a walkway K M 1 j=0 D j that crosses only flat interfaces D M D 0 D 1 E. Francini (Università di Firenze) EIT 20 / 42
Tools Fundamental solutions at the interfaces Let γ, δ C. Then, if Γ(x, y) is the fondamental solution to the Laplace equation 1 γ Γ(x, y) + sγ(x, y ) if x n > 0, y n > 0, ( ) Γ γ,δ (x, y) = 1 γ + s Γ(x, y) if x n y n < 0, 1 δ Γ(x, y) + tγ(x, y ) if x n < 0, y n < 0, γ δ where y = (y 1,..., y n 1, y n ), s = γ(γ+δ) and t = δ γ γ(γ+δ), is a fundamental solution for the differential operator ) ) div ((δχ R + γχ n R in R n+ n. Li,Nirenberg, Comm. Pure Appl. Math., 2003 Hölder gradient estimates up to the interfaces for solutions to elliptic systems. E. Francini (Università di Firenze) EIT 21 / 42
Singular solutions Theorem Let γ satisfy assumptions (H1)-(H3) in Ω 0 = Ω D 0. For y K there exists a unique function G(, y), continuous in Ω 0 \ {y} Ω 0 γ G(, y) φ = φ(y), φ C 0 (Ω 0 ), G(, y) H 1 (Ω 0 \B r (y)) Cr 1 n/2, r < d(y, Ω)/2, G(x, y) = G(y, x) for every x, y K. Moreover for x D k, x 0 Σ k and y r = x 0 + rν, x G(x, y r ) = x Γ k (x, y r ) + O(1) E. Francini (Università di Firenze) EIT 22 / 42
Estimates of unique continuation Theorem Let v H 1 (K) be a solution to div (γ v) = 0 in K, such that v L (K 0 ) ɛ 0, v(x) (ɛ 0 + E 0 )dist(x, Σ M ) 1 n 2 for every x K. Then v( x r ) C ( ɛ0 E 0 + ɛ 0 ) τ (M)N 1 δ M 1 τ r (E 0 + ɛ 0 )r (1 n 2)(1 τ r ), where x r = P M 2rν (P M ), P M Σ M, r (0, r 0 ) and τ, τ r, δ 1 (0, 1), N 1 = A r0 n and C depends on r 0, L, A, n, λ, N. E. Francini (Università di Firenze) EIT 23 / 42
D M Σ M x r D 0 D 1 E. Francini (Università di Firenze) EIT 24 / 42
Some remarks We expect that Lipschitz continuous dependence of the admittivities on the data still holds replacing the flatness condition on the interfaces with C 1,α regularity. E. Francini (Università di Firenze) EIT 25 / 42
Some remarks Local Dirichlet to Neumann map γ 1 γ 2 L (Ω) C Λ Σ 1 1 Λ Σ 1 2 L(H 1/2 co,h 1/2 co ). { } Hco 1/2 = φ H 1/2 ( Ω) : suppφ Σ 1 Ω E. Francini (Università di Firenze) EIT 26 / 42
An important application V de Hoop, Qiu, Scherzer, Inverse problems, 2012 Hölder stability implies local convergence of Landweber iteration. The constant in Hölder stability determines the ray of convergence of iterations. Lipschitz continuity implies linear convergence. E. Francini (Università di Firenze) EIT 27 / 42
Part II Size estimates E. Francini (Università di Firenze) EIT 28 / 42
The problem D γ = γ 0 1 Ω\D + γ 1 1 D Ω Size estimate Determine D from FINITELY many boundary measurements. Motivation: Necrosis in liver tissue E. Francini (Università di Firenze) EIT 29 / 42
The solutions Let u 0 H 1 (Ω) be the background solution { div(γ0 u 0 ) = 0 in Ω, u γ 0 0 ν = h on Ω, γ 0 = σ 0 + iɛ 0, σ 0 c 0 > 0 Let u 1 H 1 (Ω) solution to { div(γ u1 ) = 0 in Ω, γ u 1 ν = h su Ω. γ = γ 0 1 Ω\D + γ 1 1 D γ 1 = σ 1 + iɛ 1, σ 1 c 0 > 0 Goal We want to bound D in terms of δw := Ω h(u 1 u 0 ) E. Francini (Università di Firenze) EIT 30 / 42
The solutions Let u 0 H 1 (Ω) be the background solution { div(γ0 u 0 ) = 0 in Ω, u γ 0 0 ν = h on Ω, γ 0 = σ 0 + iɛ 0, σ 0 c 0 > 0 Let u 1 H 1 (Ω) solution to { div(γ u1 ) = 0 in Ω, γ u 1 ν = h su Ω. γ = γ 0 1 Ω\D + γ 1 1 D γ 1 = σ 1 + iɛ 1, σ 1 c 0 > 0 Goal We want to bound D in terms of δw := Ω h(u 1 u 0 ) E. Francini (Università di Firenze) EIT 30 / 42
The solutions Let u 0 H 1 (Ω) be the background solution { div(γ0 u 0 ) = 0 in Ω, u γ 0 0 ν = h on Ω, γ 0 = σ 0 + iɛ 0, σ 0 c 0 > 0 Let u 1 H 1 (Ω) solution to { div(γ u1 ) = 0 in Ω, γ u 1 ν = h su Ω. γ = γ 0 1 Ω\D + γ 1 1 D γ 1 = σ 1 + iɛ 1, σ 1 c 0 > 0 Goal We want to bound D in terms of δw := Ω h(u 1 u 0 ) E. Francini (Università di Firenze) EIT 30 / 42
The conductivity case D γ 0 = σ 0 γ 1 = σ 1 Ω dist(d, Ω) d 0 > 0 Kang, Seo and Sheen, Alessandrini and Rosset, Capdeboscq and Vogelius (low volume fraction) Kang, Kim and Milton ( Different approach for the two dimensional case. Two boundary measurement. No need of background. Explicit constants) E. Francini (Università di Firenze) EIT 31 / 42
Strategy (Alessandrini, Seo, Rosset) Energy estimates δw := Ω h(u 1 u 0 ) C 1 δw u 0 2 C 2 δw D Lower and upper bounds for D in terms of D u 0 2 using respectively regularity estimates and quantitative estimates of unique continuation K 2 W 0 D p u 0 2 K 1 W 0 D D E. Francini (Università di Firenze) EIT 32 / 42
Strategy (Alessandrini, Seo, Rosset) Energy estimates δw := Ω h(u 1 u 0 ) C 1 δw u 0 2 C 2 δw D Lower and upper bounds for D in terms of D u 0 2 using respectively regularity estimates and quantitative estimates of unique continuation K 2 W 0 D p u 0 2 K 1 W 0 D D E. Francini (Università di Firenze) EIT 32 / 42
Strategy (Alessandrini, Seo, Rosset) Energy estimates δw := Ω h(u 1 u 0 ) C 1 δw u 0 2 C 2 δw D Lower and upper bounds for D in terms of D u 0 2 using respectively regularity estimates and quantitative estimates of unique continuation K 2 W 0 D p u 0 2 K 1 W 0 D D E. Francini (Università di Firenze) EIT 32 / 42
General assumptions (H1) Assumptions on Ω: Bounded with Lipschitz boundary. (H2) Assumptions on D: Lebesgue measurable subset of Ω; (H3) Assumptions on the coefficients: we assume the reference medium and the inclusion have admittivities γ 0 = σ 0 + iɛ 0 and γ 1 = σ 1 + iɛ 1 satisfying σ j c 0, γ j c0 1 in Ω, for j = 0, 1. (H4) Assumptions on the boundary data: h H 1/2 ( Ω), h 0 such that h = 0 Ω E. Francini (Università di Firenze) EIT 33 / 42
Assumptions for energy estimates (H3a) γ 0 and γ 1 are different constants with positive real part. (H3b) ɛ 0 (x) 0 in Ω, and there exists a positive constant µ 0 such that ɛ 1 (x) µ 0 or σ 1 (x) σ 0 (x) µ 0 in Ω. E. Francini (Università di Firenze) EIT 34 / 42
Energy estimates Proposition Assume (H3a) then 1 (1 + γ 1 γ 0 c 0 ) γ 1 γ 0 δw Proposition D u 0 2 Assume γ 0 and γ 1 satisfy assumption (H3b) then F 1 δw u 0 2 F 2 δw, where. F 1 = D c0 3 [ 1 2(2 + c0 2) and F 2 = 2 µ 0 c0 2 ( ) 1 2 + δw c 0 γ 1 γ 0 + 1 µ 0 + 1 c 0 ] E. Francini (Università di Firenze) EIT 35 / 42
The one dimensional case Let Ω = ( 1, 1) and let D = (a, b) Ω. Consider { (γ 0 u 0 ) = 0 in ( 1, 1) (γ 0 u 0 )( 1) = (γ 0u 0 )(1) = K C, u 0( 1) + u 0 (1) = 0 { (γu 1 ) = 0 in ( 1, 1) (γu 1 )( 1) = (γu 1 )(1) = K C, u 1( 1) + u 1 (1) = 0 E. Francini (Università di Firenze) EIT 36 / 42
The one dimensional case δw = K 2 2 b a ( 1 γ 1 1 γ 0 If ( 1 R 1 ) = σ 1 γ 1 γ 0 σ1 2 + σ 0 ɛ2 1 σ0 2 + ɛ2 0 or ( 1 I 1 ) = ɛ 1 γ 1 γ 0 σ1 2 + ɛ 0 ɛ2 1 σ0 2 + ɛ2 0 we can estimate b a from δw. ɛ 0 = 0 ) dx < (>)0 in ( 1, 1) < (>)0 in ( 1, 1) σ 1 (σ 1 σ 0 ) ɛ 2 1 σ 0 (σ 2 1 + ɛ2 1 ) < (>)0, ɛ 1 σ 2 1 + ɛ2 1 < (> 0) E. Francini (Università di Firenze) EIT 37 / 42
The one dimensional case δw = K 2 2 b a ( 1 γ 1 1 γ 0 If ( 1 R 1 ) = σ 1 γ 1 γ 0 σ1 2 + σ 0 ɛ2 1 σ0 2 + ɛ2 0 or ( 1 I 1 ) = ɛ 1 γ 1 γ 0 σ1 2 + ɛ 0 ɛ2 1 σ0 2 + ɛ2 0 we can estimate b a from δw. ɛ 0 = 0 ) dx < (>)0 in ( 1, 1) < (>)0 in ( 1, 1) σ 1 (σ 1 σ 0 ) ɛ 2 1 σ 0 (σ 2 1 + ɛ2 1 ) < (>)0, ɛ 1 σ 2 1 + ɛ2 1 < (> 0) E. Francini (Università di Firenze) EIT 37 / 42
The one dimensional case δw = K 2 2 b a ( 1 γ 1 1 γ 0 If ( 1 R 1 ) = σ 1 γ 1 γ 0 σ1 2 + σ 0 ɛ2 1 σ0 2 + ɛ2 0 or ( 1 I 1 ) = ɛ 1 γ 1 γ 0 σ1 2 + ɛ 0 ɛ2 1 σ0 2 + ɛ2 0 we can estimate b a from δw. ɛ 0 = 0 ) dx < (>)0 in ( 1, 1) < (>)0 in ( 1, 1) σ 1 (σ 1 σ 0 ) ɛ 2 1 σ 0 (σ 2 1 + ɛ2 1 ) < (>)0, ɛ 1 σ 2 1 + ɛ2 1 < (> 0) E. Francini (Università di Firenze) EIT 37 / 42
Estimate of D u 0 2 in terms of D (far from the boundary) Proposition Assume that σ 0 is real valued and Lipschitz continuous dist(d, Ω) d 0. There is a constant K 1 depending on the a priori constants and on d 0 and two constants K 2 and p > 1 depending on the a priori constants, on d 0 and also on F (h) = h H 1/2 ( Ω) h, such that H 1 ( Ω) K 2 W 0 D p u 0 2 K 1 W 0 D. D E. Francini (Università di Firenze) EIT 38 / 42
Ingredients Three Spheres Inequality Lipschitz Propagation of Smallness Doubling Inequality E. Francini (Università di Firenze) EIT 39 / 42
Size estimates Theorem Let Ω satisfy (H1) and let D be a measurable subset of Ω such that dist(d, Ω) d 0. Let γ 0 and γ 1 be coefficients satisfying assumptions (H3a) or (H3b) and let h with zero mean value. Then, δw C 1 W 0 D C δw 1/p 2 W 0 where C 1 depend on the a priori constants and on d 0 and the numbers p > 1 and C 2 depend on the a priori constants, on d 0 and, in addition, on F (h). [Beretta, Francini, Vessella, submitted] E. Francini (Università di Firenze) EIT 40 / 42
Inclusions touching the boundary D Ω P D E. Francini (Università di Firenze) EIT 41 / 42
Estimate of D u 0 2 in terms of D (touching the boundary) Assume D Ω, but there exist r 1 > 0 and P Ω such that D Ω\B r1 (P). h H 1/2 ( Ω) be a complex valued nontrivial current density at Ω such that Ω h = 0, and supp h Γ 0 := Ω B r1 /2(P). There is a constant K 1 depending on the a priori constants and r 1 and two constants K 2 and p > 1 depending on the same a priori constants, on r 1 and also on F (h) = h H 1/2 ( Ω) h, such that H 1 ( Ω) K 2 W 0 D p u 0 2 K 1 W 0 D. D E. Francini (Università di Firenze) EIT 42 / 42