Local Stability on Hybrid Inverse Problems

Transcription

1 Purdue University January 5, 2013 Joint work with P. Stefanov and A. Nachman

2 Table of contents Mathematical model 1 Mathematical model 2 3 4

3 Mathematical Model Mathematical model Let Ω be a bounded simply connected open set of R n with C k boundary, k 1. Consider the elliptic boundary value problem where σ is in C k (Ω), such that σ u = 0 in Ω, u Ω = f, (1) 0 < λ σ(x), x Ω. (2) We address the question of whether we can determine σ, in a stable way, from the functional where p is fixed. F (σ) = σ u p, (3)

4 : Kutchment and Steinhauer 12. : Nachmann, Tamasan and Timonov 07, 09. Seo, Kwon and Woo 05. MREIT or CDII. Hyperbolic case p = 2: Bal, Bonnetier, Mondard, Triki 11, 12. Capdeboscq et al 09, Bal, Schotland 10, Kutchment and Kuyanski 11, and more. UMEIT or EAT

5 Mathematical model Linearized. Analyze the linearization in the different cases, 0 p < 1, p = 1 and p > 1. Use SU paper Linearizing non-linear inverse problems and an application to inverse backscattering JFA 09 to get conditional stability for the non-linear problem.

6 Mathematical model The derivative of F at some fixed σ 0 is given by ( df σ0 (ρ) = σ 0 u 0 p ρ + p u ) 0 v(ρ) u 0 2, ρ := δσ/σ 0, (4) where v solves σ 0 v = σ 0 u 0 ρ in Ω, v Ω = 0, (5) and u 0 is the solution of (1) for σ = σ 0. Notice that (4) makes sense as long as u 0 0 in Ω.

7 Solve (4) for the free ρ term and plug that into (5) to get ( ) ( ) u 0 v σ 0 v + p σ 0 u 0 2 u dfσ0 (ρ) 0 = u 0 p u 0 ( ) dfσ0 (ρ) = σ 0 u 0 σ 0 u 0 p. in Ω, with v Ω = 0.

8 Representation of df σ0 Proposition Let u 0 be σ 0 -harmonic, with u 0 0 in Ω. Then σ 0 T 0 df σ0 (ρ) σ 0 u 0 p = L 1 σ 0,D T 0ρ, where T 0 = u 0 is a transport operator along the gradient field of u 0, σ,d is the Dirichlet realization of σ := σ and ( ) u 0 v Lv := σ 0 v + p σ 0 u 0 2 u 0. Remark: The operator L is the only interesting object to study.

9 Example: Consider the particular case when σ 0 = 1, f = x n. Then u 0 = x n and L = x (1 p) 2 x n, where x = (x, x n ). Notice that: For 0 p < 1, L is an elliptic operator. For p = 1, L becomes the restriction of the Laplacion over x n = 0. For p > 1, L is a hyperbolic operator. This characterization remains true in the general case.

10 Proposition Let u 0 be σ 0 -harmonic, with u 0 0 in Ω. There exist local coordinates (y, y n ) such that dx 2 = c 2 (dy n ) 2 + g αβ dy α dy β, where c = u 0 1. In this coordinates g αβ := i x i x i y α y β (6) L = Q + 1 p det g y n c 2 σ 0 det g y n, where Q is a second order elliptic positively defined differential operator in the variables y smoothly dependent on y n. In dimension two this representation is global.

11 Proof. Recall Lv = σ 0 v + p ( ) u 0 v σ 0 u 0 2 u 0 Let - Π 0 ω the the orthogonal projection of the covector ω onto u 0. - Π := Id Π 0. Take a test function φ C 0 (Ω), and compute (Lv, φ) = (σ 0 v, φ) p(σ 0 Π 0 v, φ), = (σ 0 Π v, Π φ) + (1 p) (σ 0 Π 0 v, Π 0 φ). (7)

12 Hence L = (Π ) σ 0 (Π ) (1 p)(π 0 ) σ 0 (Π 0 ), (8) where the prime stands for transpose. Notice that trivially c 2 u 0 2 = 1 for c = u 0 1. Near x 0 choose boundary local coordinates to the level surface u 0 (x) = u 0 (x 0 ), with y n = u 0 then and then c 2 dx 2 = (dy n ) 2 + c 2 g αβ dy α dy β dx 2 = c 2 (dy n ) 2 + g αβ dy α dy β, g αβ := i x i x i y α y β

13 In this coordinates, Π 0 v = (0,..., 0, v/ y n ) Π v = ( v/ y 1,..., v/ y n 1, 0) (Lv, φ) = Hence ( αβ v σ 0 g φ v y α + (1 φ ) det p)c 2 y β g dy y n y n L = 1 ( det g y β σ 0g αβ det g y ) + (1 p) α y n c 2 σ 0 det g y n

14 In the case n = 2 we have a global coordinate system. Take y 2 = u 0 and y 1 = v 0 with v 0 H 1 (Ω) being the σ 0 -conjugate harmonic of u 0 (u 0 + iv 0 is analitic on Ω). The level curves to u 0 and v 0 are perpendicular and, under the non-vanishing gradient assumption, they can not intersect in more than one point. Remark In the 2D case we can ensure u 0 0 in Ω by imposing condition in the boundary. For instance if u Ω = f is two-to-one then u 0 0 in Ω.

15 Using the representation of df σ0 and decomposition of L we get ( v, df ) σ(ρ) u 0 p u 0 = (Lv, v) = σ 1/2 0 Π v 2 + (1 p) σ 1/2 0 Π 0 v 2 min(1, (1 p)) σ 1/2 0 v 2. If df σ (ρ) = 0 = v = 0. We then have σ 0 u 0 ρ = 0. This implies ρ = 0. Hence the linearization is injective. Together with the ellipticity, this has the usual consequences.

16 In this case we have the following stability of L. Proposition (p=1) Let u 0 be σ 0 harmonic, with u 0 0 in Ω, then (Lv, v) C v 2 L 2 (Ω), for v Ω = 0, (9) for C > 0.

17 Sketch proof: For each x 0 Ω there exist ɛ 0 such that for all 0 < ɛ < ɛ 0 a+ɛ a ɛ Q 0 Lvvdy dy n = a+ɛ a ɛ C C a+ɛ a ɛ a+ɛ a ɛ Q 0 σ 0 g αβ v y α v y β det(dx/dy) dy dy n Q 0 y v(y, y n ) 2 dy dy n Q 0 v(y, y n ) 2 dy dy n C v 2 L 2 (Ω 0 ɛ ) where a = u 0 (x 0 ) and Ω 0 ɛ = (a ɛ, a + ɛ) Q 0. Here C depends on λ,n and the upper bound or the level curves of u 0 intersecting the domain of integration.

18 Once we have established stability for the operator L we can get stability for the linearization using the decomposition σ 0 T 0 df σ0 (ρ) σ 0 u 0 p = L 1 σ 0,D T 0ρ. Hence for some s 1, s 2. ρ H s 1 df σ0 (ρ) H s 2 (10)

19 Finally using SU technique to transfer stability from the linearization to the non-linear problem we get Hölder type of stability. For L > 0 there exist ɛ > 0, so that for any f with σ σ 0 H s ɛ, σ H s L one has the conditional stability estimate for some 0 < µ < 1. σ σ 0 H s C F (σ) F (σ 0 ) µ H s

20 Final Remarks Critical points (easier in n=2). General approach to get stability for a non-linear problem. Case p = 2.

21 Thank you!