Uniqueness and stability of Lamé parameters in elastography

Transcription

1 Uniqueness stability of Lamé parameters in elastography Ru-Yu Lai Abstract This paper concerns an hybrid inverse problem involving elastic measurements called Transient Elastography TE which enables detection characterization of tissue abnormalities. In this paper we assume that the displacements are modeled by linear isotropic elasticity system the tissue displacement has been obtained by the first step in hybrid methods. We reconstruct Lamé parameters of this system from knowledge of the tissue displacement. More precisely, we show that for a sufficiently large number of solutions of the elasticity system for an open set of the well-chosen boundary conditions, Lamé parameters can be uniquely stably reconstructed. The set of well-chosen boundary conditions is characterized in terms of appropriate complex geometrical optics solutions. 1 Introduction Medical imaging is the technique process used to create images of the human body for clinical purposes or medical science. Each available imagine method has its advantages disadvantages. Medical imaging modalities such as Computerized Tomography CT, Magnetic Resonance Imaging MRI Ultrasound Imaging UI are examples of modalities providing high resolution. In some situations, these modalities fail to exhibit a sufficient contrast between different types of tissues. For instance, in breast imaging ultrasound provides high resolution, while suffers from a low contrast. Other modalities, based on optical, elastic, or electrical properties of these tissues, display high contrast, such as Optical Tomography OT Electrical Impedance Tomography EIT. For example, some breast tumors on early stages might have no contrast with the healthy tissues with respect to ultrasound propagation, but a huge contrast in their optical electric properties. In order to obtain better image, the natural idea is to try to combine different imaging modalities. These are coupled-physics imaging methods, also called hybrid Department of Mathematics, University of Washington, Seattle, WA 98195, USA. rylai@uw.edu. 1 AMS subject classification codes: Primary: 35R3 ; Secondary: 35Q74. Key words: inverse problems, elasticity system, Lamé parameters. 1

2 methods. It is to combine the high resolution modality with another high contrast modality. Examples of possible physical couplings include: Photo-Acoustic Tomography PAT Thermo-Acoustic Tomography TAT, Ultrasound modulated Optical Tomography UMOT elasticity with ultrasound in Transient Elastography TE. Reconstructions in hybrid inverse problems involve two steps. The first step is to solve the inverse problem concerning the high-resolution-low-contrast modality. For instance, in PAT TAT, this corresponds to the reconstructing the initial condition of a wave equation from boundary measurements. In TE, this is to solving an inverse scattering problem in a time-dependent wave equation. In this paper, we assume that this first step has been performed. We will focus on the second step which is to reconstruct the coefficients that display high contrast from the mappings obtained during the first step. For more details about hybrid inverse problems, please refer to [1]. In this paper, the modality we consider is Transient Elastography. TE is a non-invasive tool for measuring liver stiffness. The device creases high-resolution shear stiffness images of human tissue for diagnostic purposes. Shear stiffness is targeted because shear wave speed is larger in abnormal tissue than in normal tissue. In the experiment tissue initially is excited with pulse at the boundary. This pulse creates the shear wave passing through the liver tissue. Then the tissue displacement is measured by using the ultrasound. The displacement is related to the tissue stiffness because the soft tissue has larger deformation than stiff tissue. When we have tissue displacement, we want to reconstruct shear modulus µ the first parameter λ. See [8], [9] references there for more details. In TE, the high resolution modality is also ultrasound. The tissue displacement data can be obtained by the ultrasound in the first step. The second step is to recover the Lamé parameters from the knowledge of the tissue displacement. In the following paper, we will assume the first step has been performed. We will formulate now the mathematical problem. Let Ω R n, n =, 3, be an open bounded domain with smooth boundary. Let u H 4 R n be the displacement satisfying the linear isotropic elasticity system { λ ui + S uµ + k u = in Ω, 1 u = g on Ω, where SA = A + A T / is the symmetric part of the matrix A. Here λ, µ are Lamé parameters k R is the frequency. For the forward problem of elasticity system, we refer the readers to [4]. Assume that k is not the eigenvalue of the elasticity system. The set of internal functions obtained by ultrasound in TE is given by : Hx = u j x 1 j J in Ω for some positive integer J. Hybrid inverse problems for the modality TE has less results than PAT TAT due to its complicated equations. The first result, to the author s best knowledge, is considered by Bal Uhlmann [] in a scalar model for the elastic displacement u, in which the parameters are uniquely stably determined by the

3 internal data. They also gave an explicit reconstruction procedure for those parameters. However, the scalar model is not accurate in many applications. Our main contribution in this paper is to consider a more general setting, that is, the linear isotropic elasticity system provide uniqueness stability results. To study the elasticity system we applied the CGO solutions due to [5] [1]. We used the strategies in [] to obtain a transport equation for µ or λ, then we can recover the parameters by following the methods of [3]. Denote that P = {λ, µ C 7 Ω C 9 Ω; < m λ C7 Ω, µ C 9 Ω M λ, µ > }. Let H H be two sets of internal data of the elasticity system with parameters λ, µ λ, µ, respectively. Below is our main result: Theorem 1.1. Let Ω be an open bounded domain of R n with smooth boundary. Suppose that the Lamé parameters λ, µ λ, µ P µ Ω = µ Ω. Let u j ũ j be the solutions of the elasticity system with boundary data g j for parameters λ, µ λ, µ, respectively. Let H = u j 1 j J H = ũ j 1 j J be the corresponding internal data for λ, µ λ, µ, respectively for some integer J 3n + 1. Then there is an open set of the boundary data g j 1 j J such that if H = H implies λ, µ = λ, µ in Ω. Moreover, we have the stability estimate µ µ CΩ + λ λ CΩ C H H C Ω. The remainder of this paper is organized as follows. In section, we introduce the CGO solutions of the elasticity system. Section 3 is devoted to constructing the Lamé parameters in the two-dimensional case. The reconstruction of λ, µ in 3D is presented in section 4. Complex geometric optics solutions of the elasticity system In this section, we will briefly introduce the complex geometric optics CGO solutions of the elasticity system. Based on [6]or see [5], we can derive the following reduced system. Let w, f H 5 R n w, f T satisfy w f w + V f 1 x + V w x =. f Here V contains the third derivative of µ µ 1/ µ 1 + µ 3/ k log µ V 1 x = Then the solution of the elasticity system 1 is u := µ 1/ w + µ 1 f f µ 1. λ+µ λ+µ µ1/. 3

4 Here g denotes the Hessian matrix g/ x i x j. Note that we will not need the explicit form of V in the construction of CGO solution. The construction of CGO solutions of with linear phase was first deduced by Nakamura Uhlmann in [1] [11], where they introduced the intertwining property in hling the first order terms. Eskin Ralston [5] also gave a similar result in. Later Uhlmann Wang [1] used Carleman estimate to deduce the CGO solutions in the two-dimensional case. In the following sections, we will use the CGO solutions constructed in Eskin Ralston s paper [5]. To construct CGO solutions of, it is convenient to work on R n instead of Ω. Since Ω is bounded, we pick a ball B R for R >, such that Ω B R extend λ µ to R n by preserving its smoothness also suppλ, suppµ B R. Let α β be two orthogonal unit vectors in R n. Denote that = τα+iβ C n θ = α + iβ with τ >. Eskin Ralston [5] showed the following important result in the three-dimensional case. For n =, it still holds see reference [1]. Lemma.1. Eskin-Ralston Consider the Schrödinger equation with external Yang-Mills potentials Lu = u iax u + Bxu =, x B R R n 3 where Ax = A 1 x,..., A n x C n B R, n 6 with A j x Bx are n + 1 n + 1 matrices. Then there are solutions of 3 of the form where C C n B R is the solution of u = e i x C x, θgθ x + Oτ 1 iθ x C x, θ = θ AxC x, θ with det C is never zero, gz is an arbitrary polynomial in complex variables z, Oτ 1 is bounded by C1/τ in H l B R, l n. By the lemma above, the CGO solutions of can be written as follows w = e i x r + Oτ 1, f s where r, s T is C x, θgθ x. We can write w = e i x r + Oτ 1 f = e i x s + Oτ 1. Plug w, f T into u = µ 1/ w + µ 1 f f µ 1. Then we have the CGO solutions of the elasticity system. Note that r, s T satisfies the equation r n n θ r θ = V s 1 x θ T. 4 s Remark 1. Since C is invertible at every point in Ω, we can conclude that r, s T is not zero everywhere in Ω. This does not imply that both r s never vanish in Ω. However, for any point y Ω, there is a small neighborhood B y of y in Ω a CGO solution of w, f T such that both r s do not vanish in B y. 4

5 3 Reconstruction of Lamé parameters in two dimensional case In the previous section, we already have the CGO solutions of the elasticity system. Now we want to use them to give a reconstruction of µ first. Let u = u 1, u T be the displacement which satisfies the elasticity system λ ui + S uµ + k u =. 5 We will recover µ λ separately in the following two subsections. 3.1 Reconstruction of µ in D From 5, we can deduce the following equation Here we denote u = 1 u u u u, F = u F + u G = k u. 6 λ + µ λ + µ 1 λ + µ λ + µ, u = a + b a b 1 a + b a b, G = 1 µ µ µ µ with a = u u, b = 1 u 1 u, u = u 1 + u. The component u u are known since they are vectors which only depend on the internal data u. In order to recover µ, we want to eliminate the first term of 6 so that µ satisfies the transport equation. Obverse that the vector u has three different entries, we only need to construct three linearly independent vectors on some subdomain in Ω. With these three vectors, we can remove the first term of the left h side of the equation 6. More precisely, suppose that u j, for j =, 1,, are three different solutions of 6 which satisfy 7 u j F + u j G = k u j. 8 The notations Rf If are defined to be the real imaginary part of f, respectively. Now for j = 1,, 3 =,,, we let u 1 = Rχxu, u = Iχxu u 1 = Rχ 3 xu 1, u = Iχ 3 xu 1, u 3 = R where χ j x is a nonzero function. Then we get u l F + u l χ 1 xu 1 + χ xu, G = k u l, l = 1, 9 5

6 u j F + u j G = k u j for j = 1,, 3. 1 Assume that { u 1, u, u 3 } are three linearly independent vectors in some subdomain of Ω, say Ω. Then there exist three functions Θ l 1, Θ l, Θ l 3 such that u l + 3 Θl j uj =. For l = 1,, multiplying 1 by Θ l j summing over j with equation 9, we have 3 v l G = k u l + Θ l ju j, where v l = u l + 3 Θl j uj. Let β l = v l e 1 ẽ 1 +v l e ẽ γ l = e 3 +e 4 v l. Here e j R 4, ẽ j R with the j th entry is 1 others are zero. Then the above equation can be rewritten as β l µ + γ l µ = k u l + 3 Θ l ju j. Suppose that β 1 x β x are linearly independent for every x Ω. Then we can recover µ in Ω Ω for each frequency k independently. Lemma 3.1. Let u j for j be C solutions of the elasticity system with boundary conditions u j = g j on Ω. Let us define assume that u = 1 u, u, u, u T 1. { u 1, u, u 3 } are three linearly independent vectors in Ω, the neighborhood of x in Ω.. {β 1 x, β x} are linearly independent in Ω. Then the reconstruction is stable in the sense that µ µ CΩ C µx + µx+ + H x H x C Ω 11 where x + Ω H x = u j j. Proof. Since {β 1 x, β x} are linearly independent in Ω, we can construct two vector-valued functions Γx, Φx CΩ such that µ + Γxµ = Φx in Ω. 1 Since µ µ are solutions of 1 with coefficients Γ, Φ Γ, Φ, respectively, we have µ µ + Γxµ µ = µ Γx Γx + Φx Φx. 6

7 Let x Ω, there exists a integral curve ψt with ψ = x + ψ1 = x. Thus µ µψt =µ µψe t Γψs ψ sds t + µ Γψs Γψs + Φ Φψs ψ sds, 13 for t [, 1]. Since Γ Φ only depend on Ω u j, j =, 1,, u j are in the class of C Ω, the value of Γ Γ Φ Φ are bounded by the sum of α u j α ũ j for α. There is a constant C such that µ µψt C µ µψ + C H x H x C Ω, t [, 1] for µ, µ P. Thus, for any x Ω, the value µ µx is controlled by the internal data µ at the boundary point x Ω Global reconstruction of µ in D Let λ, µ P, by Lemma.1, it implies that r s are in C 7 Oτ 1 H 5. Now we will show that how we can get three linear independent vectors of the form u. We plug the CGO solutions u = µ 1/ w + µ 1 f f µ 1 into u. Then we have the expression u = e i x µ λ + µ 1 r + O r + O i r i r + O1 by using the following equality which is part of the equation 4 14 s = λ + µ λ + µ µ1/ r. 15 Now we fix any point x Ω let = τ1, i = i C with τ >. Since, in Lemma.1, the matrix solution C x, θ is invertible, we can choose a constant vector g such that C x, θg = r, s T with the conditions s x = 1, s r in a neighborhood of x in B R, say U. Then we have the CGO solution of the elasticity system, that is, u = µ 1/ w + µ 1 f f µ 1, where w f = e i x r s + Oτ 1. 7

8 Let θ = /τ, θ = /τ. equations Let C 1 x, θ C x, θ satisfy the following two iθ x C 1x, θ = θ V 1 xc 1 x, θ, i θ x C x, θ = θ V 1 xc x, θ, respectively. Since = i, we can choose C x, θ = C 1 x, θ. By Lemma.1, the matrix solution C 1 x, θ is invertible, we can choose a suitable constant vector g such that C 1 x, θg = r 1, s 1 T, C x, θg = r, s T s l x =, r 1 x = 1, i = r x. Note that r = r 1 s = s 1. By continuity of r l, we have r l in a neighborhood U of x. Then the CGO solutions are u l = µ 1/ w l + µ 1 f l f l µ 1, where w 1 f 1 w f = e i x r 1 s 1 = e i x r s + Oτ 1, + Oτ 1. So far we have three CGO solutions, that is, u, u 1, u. We consider e i x u 1 = 1 λ+µ r1 1 + ir 1 λ+µ ir1 1 r 1 + Oτ 1, e i x 1 u 1 = 1 λ+µ τr1 1 + ir 1 + O1 λ+µ τir1 1 r 1 + O1 λ+µ r1 1 + ir 1 λ+µ r1 1 + ir 1 + Oτ 1 e i x 1 u = 1 λ+µ τ r 1 ir + O1 λ+µ τ ir 1 + r + O1 λ+µ ir 1 r λ+µ ir 1 r + Oτ 1. 8

9 Since r = r 1 s = s 1, we have We define { Then u 1 e i x 1 u 1 = 1 + e i x 1 u O1 O1 λ+µ 1 + ir ir 1 λ+µ 1 + ir ir 1 + Oτ 1. = Re i x u ; u = Ie i x u ; =,, u 1 u 3 = Re i x u 1 ; u = Ie i x u 1 ; = R e i x 1 u 1 } + e i x 1 u. u 1, u, u 3 are linearly independent in a small neighborhood U of x when τ is sufficiently large. Therefore, for l = 1,, there exist functions Θ l j, j = 1,, 3 such that u l + 3 Θ l ju j =, l = 1,. 16 Since u, u 1, u are solutions of the equation 6, we have the following equations: where u j u l F + u l F + u j G = k u l, l = 1, ; G = k u j, j = 1,, 3. Summing over the equations above using 16, we have two equations 3 β,l µ + γ,l µ = k u l + Θ l ju j, for l = 1,, 17 β,l = v l e 1 ẽ 1 + v l e ẽ γ,l = v l e 3 + v l e 4, Here we define v l = u l + 3 Θl j uj. 9

10 Remark. By choosing suitable g, Θ l 3x can be as small as we want. To show that, we choose a new constant vector ĝ, instead of the original g, such that ˆr, ŝ T = C x, θĝ θ x where ˆr x = r x /M, M >, ŝ x = s x ˆr, ŝ T satisfies the original assumption, that is, ŝ x = 1 ŝ ˆr in a neighborhood of x. Then Θ l 3x = P 3j 1 R λ+µ 1, i r x λ+µ i1, i r x + Oτ 1 p j ˆΘ l 3x = P 3j 1 R M λ+µ 1, i r x λ+µ i1, i r x + Oτ 1 p j, where P = p ij P 1 = P ij with p ij = p i p j. Note that P is a boundedly invertible symmetric matrix. Here p 1 = u 1, p = u, p 3 = u 3. From above, we obtain that ˆΘ l 3 is close to Θ l 3/M as τ is large. Therefore, the new ˆΘ l 3x is small when M τ is sufficiently large. Lemma 3.. Given any point x Ω, there exists a small neighborhood V of x such that β,j is not zero in V for j = 1,. Note that we denote by equalities up to terms which are asymptotically negligible as τ goes to infinity. Proof. It is sufficiently to prove that β,1 does not vanish in some neighborhood of x. Recall that where β,1 = v 1 e 1 ẽ 1 + v 1 e ẽ, 18 v 1 = u Θ 1 ju j. By 16, we have that Θ 1 j τ for j = 1,, 3. Since s 1 x =, we get that u 1 e 1 ẽ 1 + u 1 e ẽ x τ 1 19 u e 1 ẽ 1 + u e ẽ x τ 1. 1

11 Since s k x =, r = r 1, s = s 1, 1, i r 1 x =, we obtain that u 3 e 1 ẽ 1 + u 3 e ẽ x = 4µ 1 R 1 i 1 s 1 + s 1 1 s 1 + i s 1 + Oτ 1 x. 1 Combining from 18 to 1, since Θ 1 3x can be taken as small as we want See Remark in the construction of CGO solutions above, it follows that 4 β,1 x µ 1 + Θ 1 i 1 s 1 + s 1 3R 4 1 s 1 + i s 1 x. Similarly, β, x µ Θ i 1 s 1 + s 1 3R 4 1 s 1 + i s 1 x. By continuity of β,j, we complete the proof. Let g j = u j Ω be the given boundary data. By Lemma.1, since λ, µ P, we knew that u j H 4 B R. Let g j C 1,α Ω be the boundary data close to, that is, g j g j g j C 1,α Ω < ε for some ε >, then we can find solutions u j of the elasticity system with boundary data g j the existence of such solutions, see e.g. Ch.4 of [7]. By elliptic regularity theorem, we have u j u j C Ω < Cε 3 for some constant C which is independent of λ, µ. Then we obtain that u j u j C Ω < Cε. Here the notation u j is constructed in the same way as u j with the CGO solutions u j replaced by the solutions u j. Therefore, { u 1, u, u 3 } are also linearly independent when ε is sufficiently small. We construct β j as in the equation 17 with u j replaced by u j, j =, 1,. Therefore, by the definitions of β j β,j 3, it follows that β j β,j C1 Ω is small when ε is small. Since β,j is not zero in V by Lemma 3., we can deduce that β j is also not zero in V if ε is small enough τ is sufficiently large. Moreover, with the suitable chosen CGO solutions u, {β,1, β, } are linearly 11

12 independent in V as τ is sufficiently large See the proof of Lemma 3.. When ε is sufficiently small, it implies that {β 1, β } are also linearly independent in V. Then we have the following equations: β l µ + γ l µ = k u l + 3 Θ l ju j, l = 1, 4 with {β 1, β } a basis in R for every point x Ω. Here we denote Ω = U V Ω. 5 Thus, there exists an invertible matrix A = a ij such that β l = a lk ẽ k with inverse of class C Ω. Thus, we have constructed two vector-valued functions Γx, Φx CΩ such that 4 can be rewritten as Then we obtain the following result: µ + Γxµ = Φx in Ω. 6 Theorem 3.3. Suppose that λ, µ λ, µ P. For any fixed x Ω, let u j be the corresponding CGO solutions for λ, µ u j constructed as above with internal data H x = u j j with ε as in sufficiently small. Let H x be constructed similarly with the parameters λ, µ. Assume that µ Ω = µ Ω. Then H x = H x implies that µ = µ in Ω which is defined in 5, the neighborhood of x in Ω. Proof. Based on the construction above, the domain Ω can be taken as a small open ball with center x Ω B R. Since H x = H x, we have that µ µ solve the same equation µ + Γxµ = Φx in Ω where the functions Γ Φ depend on u j. Let x Ω denote ψt = 1 tx + tx, t [, 1]. Restricted to this curve, we have { ψ t µ + Γψt ψ tµ = Φψt ψ t in Ω 7 µx = µx, The solution of 7 is given by µψt = µψe t Γψs ψ sds + t Φψs ψ sds, t [, 1]. The solution µx is given by the same formula since µ Ω = µ Ω so that µ = µ in Ω. We have constructed { u 1, u, u 3 } are linearly independent {β 1, β } forms a basis in R for every point x Ω when ε is sufficiently small τ is large. Applying Lemma 3.1, we have 1

13 Theorem 3.4. Suppose that λ, µ λ, µ P. For any fixed x Ω, let u j be the corresponding CGO solutions for λ, µ u j constructed as above with internal data H x = u j j with ε as in sufficiently small. Let H x = ũ j j be constructed similarly for the parameters λ, µ with u j Ω = ũ j Ω. Then there exists an open neighborhood Ω which is defined in 5 of x in Ω such that µ µ CΩ C µx µx + H x H x C Ω, x Ω. 8 µ µ CΩ C µx + µx+ + H x H x C Ω, x + Ω, x Ω. The global uniqueness stability result are stated as follows. Theorem 3.5. Global reconstruction of µ Let Ω be an open bounded domain of R with smooth boundary. Suppose that the Lamé parameters λ, µ λ, µ P µ Ω = µ Ω. Let u j ũ j be the solutions of the elasticity system with boundary data g j for parameters λ, µ λ, µ, respectively. Let H = u j 1 j J H = ũ j 1 j J be the corresponding internal data for λ, µ λ, µ, respectively for some integer J 3. Then there is an open set of the boundary data g j 1 j J such that if H = H implies µ = µ in Ω. Moreover, we have the stability estimate µ µ CΩ C H H C Ω. Note that for the uniqueness of µ, we suppose that the two set of internal data are the same, that is, H = H. Since µ is uniquely reconstructed near a fixed point x Ω under the condition µ Ω = µ Ω, from the stability of µ in Ω, we can obtain that µ = µ in Ω. Proof. In section, we constructed CGO solutions in a ball B R which contains Ω. First, we consider any point x in Ω. Then we can find an open neighborhood B x B R of x. By Theorem 3.4, we have the estimate 9 µ µ CBx Ω C H x H x C Ω 3 since µ Ω = µ Ω. Second, for any point y Ω, by Theorem 3.4, there exists an open neighborhood B y Ω of y with B y Ω = such that µ µ CBy C µy + µy + + H y H y C Ω for some y + B y. Therefore, the compact set Ω is covered by x Ω B x. Then there exists finitely many B x, say, B x1,..., B xn, such that Ω N l=1 B x l

14 Now for arbitrary point z Ω, there is B xj such that z B xj. Suppose that B xj Ω, this means that x j Ω. Then, by 3, we have µz µz C H xj H xj C Ω. 3 Otherwise, if B xj Ω is empty, then, by 31, we get that µz µz C µx + j µx+ j + H x j H xj C Ω for x + j B x j. 33 For the point x + j, since Ω is covered by finitely many subdomain B x l, after at most N 1 steps, we have µx + j µx+ j C Combining Then we have µz µz C N l j,l=1 H xl H xl C Ω. 34 N H xl H xl C Ω. 35 l=1 With 3 35, we have the global stability µ µ CΩ C N H xl H xl C Ω. 36 l=1 3. Reconstruction of λ in D The elasticity system can also be written in this form where u = u u a + b a b, F = u F + u G = k u, 37 1 λ + µ λ + µ 1 µ µ, u = 1 u u 1 a + b a b, G = λ + µ λ + µ µ µ As in the reconstruction of µ, we will construct there linear independent vectors such that the first term of the equation 37 can be eliminated. Suppose that u j, for j =, 1,, 3, are three different solutions of 6 which satisfy. u j F + u j G = k u j

15 Now for j = 1,, 3 =,,, we let u = Rχxu, u 1 = Rχ 3 xu 1, u = Iχ 3 xu 1, u 3 = R where χ j x is a nonzero function. Then we get χ 1 xu + χ xu 3, u F + u G = k u 39 u j F + u j G = k u j for j = 1,, 3. 4 Assume that { u 1, u, u 3 } are three linearly independent vectors in some subdomain of Ω, say Ω. Then there exist three functions Θ 1, Θ, Θ 3 such that u + 3 Θ ju j =. Multiplying 4 by Θ j summing over j with equation 39, we have 3 v G = k u + Θ j u j, where v = u + 3 Θ ju j. Let κ = 1, 1,, T v, σ = 1, 1, 1, 1 T v. Then the above equation can be rewritten as κλ = σµ k u + Suppose that κx does not vanish in Ω. lemma. 3 Θ j u j. Then we can deduce the following Lemma 3.6. Let u j for j 3 be C solutions of the elasticity system with boundary conditions u j = g j on Ω. Let u = u 1, u, a = u u b = 1 u 1 u. We define u = u, u, a + b, a b assume that 1. { u 1, u, u 3 } are three linearly independent vectors in Ω, the neighborhood of x in Ω.. κx does not vanish in Ω. Then the reconstruction is stable in the sense that λ λ CΩ C µ µ CΩ + H x H x C Ω where H x = u j j

16 3..1 Global reconstruction of λ We will find three linearly independent vectors { u 1, u, u 3 } first. Then we can deduce stability of λ by using Lemma 3.6 Theorem 3.5. Plugging the CGO solution u = µ 1/ w + µ 1 f f µ 1 into u, we get u = ie i x µ λ+µ r µ λ+µ r µ 1 is ω 1 s + µ 1/ ω 1 r µ 1 is ω s + µ 1/ ω r + O1, where ω 1 = 1 +, 1 ω = 1, 1 +. For the same fixed point x Ω. Denote that = τe 1 + ie = i. We choose a constant vector g such that C x, θg = r, s with s, r, r x = 1 in a neighborhood of x, U. Then we get the CGO solution of the elasticity system u = µ 1/ w + µ 1 f f µ 1 with w f = e i x r s + Oτ 1. We choose another constant vector g 1 such that C 1 x, θg 1 = r 1, s 1 with s 1 in U r 1 x =. Then we get the CGO solution of the elasticity system u 1 = µ 1/ w 1 + µ 1 f 1 f 1 µ 1 with w 1 f 1 = e i x r 1 s 1 + Oτ 1. For l =, 3, we choose a constant vector g l such that C l x, θg l = r l, s l with r l in U. Here we can choose r = r 3, s = s 3 16

17 by taking g = g 3 C x, θ = C 3 x, θ. Then we get the CGO solution of the elasticity system u l = µ 1/ w l + µ 1 f l f l µ 1 with w f w 3 f 3 = e i x r s = e i x r 3 s 3 + Oτ 1, + Oτ 1. We let e i x i 1 u 1 = iµ 1 s 1 + i i 1 + Oτ 1, u,1 := e i x i 1 1 u = 1 µ λ+µ µ λ+µ 1, i r 1, i r µ 1 τ1 + iis + ν µ 1 τi 1is + iν + Oτ 1 u, := e i x i 1 1 u 3 = 1 µ λ+µ µ λ+µ i, 1 r3 i, 1 r3 µ 1 τ 1 iis 3 + ν 3 µ 1 τ1 iis 3 + iν 3 + Oτ 1, where ν = µ i, 1 i s + µ 1/ 1 + i, 1 i r ν 3 = µ i, i + 1 s 3 + µ 1/ 1 + i, i + 1 r 3. To eliminate the higher order term of u,j, we consider the summation of two vectors : e i x i 1 1 u + e i x i 1 1 u 3 µ = 1 λ+µ 1 + i, i 1 r3 µ λ+µ 1 + i, i 1 r3 µ 1 i, s + µ 1/ i, r 3 µ 1, i s 3 + µ 1/, i r 3 We define u 1 u = Re i x i 1 u ; =,, + Oτ 1. = Re i x i 1 u 1 ; u = Ie i x i 1 u 1 ; 17

18 u 3 = R e i x i 1 1 u + e i x i 1 1 u 3. { Thus we have constructed three linear independent vectors u 1, u }, u 3 in U as τ is sufficiently large. Therefore, there are three functions Θ j, j = 1,, 3, such that u + 3 They also satisfy the following equations: u j F + u j Θ j u j =. 4 G = k u j, j 3. Summing over j using 4, we get the following equation 3 v G = k u + Θ j u j, 43 where Then we obtain that v = u + κλ = σµ k 3 u + Θ j u j. 3 Θ j u j, 44 where κ = 1, 1,, T v, σ = 1, 1, 1, 1 T v. Lemma 3.7. κ does not vanish in some neighborhood of x. Proof. Since Θ j τ r 1 x =, by using similar argument as in the proof of Lemma 3., we have that Observe that 1, 1,, T Θ 1 u 1 τ 1, 1, 1,, T Θ u τ κ 4 := 1, 1,, T u 3 1 e i x 1 u =R e i x 1 i u + i 1 u 3 u 3. 18

19 The Oτ term of κ 4 is 1 µ i τ λ + µ 1 + r µ λ + µ r 3 = since r = r 3. Thus the leading order term of κ 4 is O1, that is, where Since ϕ l = 1 i τ ϕ = ϕ + ϕ 3, µ 1/ i l 1r l 1 + il 1 r l + il 1 r l 1 + il 1r l + i r l + µ 1 i 1 s l + s l + i 1 + s l + i 1 µ 1 l s l 11 µ 1 i l 1 + l sl + i µ 1 l s l µ 1 i l 1 + l sl + 1 µ 1/ i l rl 1 + il rl + il 1 rl 1 + µ 1/ i l 1 rl 1 + il 1 rl + il rl. κ 1 :=1, 1,, T u x 1 ie i x 1 u = R 1 u 45, we get that x = µ λ + µ + Oτ 1 κx =1, 1,, T vx 46 =κ 1 x + Θ 3 κ 4 x + Oτ 1 47 µ λ + µ + Θ 3x ϕx. 48 Since µ P, ϕx is some fixed number. Moreover, we can take s small such that Θ 3 x to be sufficiently small It can be done by following similar argument in Remark, then we can obtain that µ λ+µ + Θ 3x ϕx. By continuity of κ, there exists a neighborhood V such that κ never vanishes in V. Let Ω = U V Ω. 49 { } We have u 1, u, u 3 are linearly independent in Ω as τ is sufficiently large also κ does not vanish in Ω. Then it follows that λ = σµ 3 κ k u + Θ j u j in Ω. 5 κ 19

20 As in section 3.1, we also can find the solutions u j of the elasticity system such that u j u j C Ω < Cε, j =, 1,, Now we let the internal data H x contains the three solutions we constructed in Theorem 3.3 the four solutions u j, j =, 1,, 3 in this section. Then we have the following result. Theorem 3.8. Suppose that λ, µ λ, µ P. For any fixed x Ω, let u j be the corresponding CGO solutions for λ, µ u j constructed in sections 3.1, 3. with internal data H x with ε as in sufficiently small. Let H x be constructed similarly with the parameters λ, µ. Assume that µ Ω = µ Ω. Then H x = H x implies that λ = λ in Ω which is defined in 49. Proof. Applying Theorem 3.3 equation 5, we have the uniqueness of λ near the point x. We deduce the following result by applying Lemma 3.6 Theorem 3.4. Theorem 3.9. Suppose that λ, µ λ, µ P. For any fixed x Ω, let u j be the corresponding CGO solutions for λ, µ u j constructed in sections 3.1, 3. with internal data H x with ε as in sufficiently small. Let Hx be constructed similarly for the parameters λ, µ with u j Ω = ũ j Ω. Then we have the estimates λ λ CΩ C µx µx + H H C Ω, x Ω. 5 λ λ CΩ C Here Ω is defined in 49. µx + µx+ + H H C Ω, x + Ω, x Ω. With 5 Theorem 3.9, we follow the same proof as in Theorem 3.5, then we can get the following global reconstruction of λ. Theorem 3.1. Global reconstruction of λ Let Ω be an open bounded domain of R with smooth boundary. Suppose that the Lamé parameters λ, µ λ, µ P µ Ω = µ Ω. Let u j ũ j be the solutions of the elasticity system with boundary data g j for parameters λ, µ λ, µ, respectively. Let H = u j 1 j J H = ũ j 1 j J be the corresponding internal data for λ, µ λ, µ, respectively for some integer J 7. Then there is an open set of the boundary data g j 1 j J such that if H = H implies λ = λ in Ω. Moreover, we have the stability estimate λ λ CΩ C H H C Ω. 53

21 4 Reconstruction of Lamé parameter in three dimensional case The reconstruction of λ µ in R 3 basically follows the similar argument as in Section 3. The major difference in proving the stability of Lamé parameters between dimensions two three is that more CGO solutions are needed to get linearly independent vectors locally in R 3 than in R. 4.1 Global reconstruction of µ in 3D Let u = u 1, u, u 3 T be the displacement which satisfies the elasticity system Denote u = 1 u u 3 u u u u λ ui + S uµ + k u =. 54, F = λ + µ λ + µ λ + µ 1 λ + µ λ + µ 3 λ + µ, u = b 3 b 13 b 1 1 b 3 b 13 3 b 1, G = 1 µ µ 3 µ µ µ µ where b ij = l u l i u i j u j + i u l + l u i + j u l + l u j with l, i, j {1,, 3} are distinct numbers. From 54, we can deduce the following equation: u F + u G = k u. Here u = u 1 + u + u 3. In the following we will show that how we can get four linearly independent vectors of the form u on some subdomain of Ω. The key thing is to observe the behavior of u. We plug the CGO solutions u = µ 1/ w + µ 1 f f µ 1 into u. Then we get u = e i x λ+µ 1 r + O λ+µ r + O λ+µ 3 r + O λ+µ r λ+µ r λ+µ r, + O1. 55 Note that r = r 1, r, r 3 T = 1,, 3 T. Now we fix any point x Ω. Let = τ1, i, = τ1,, i with τ >. Let θ = /τ, θ = /τ. Since, in Lemma.1, the matrix solutions C x, θ C x, θ are invertible, we can choose two constant vectors g g such that C x, θg = r, s T C x, θ g = r, s T with 1

22 s x = 1 = s x s, s r, r in a neighborhood of x, say U. Then we have the CGO solutions of the elasticity system, that is, u = µ 1/ w + µ 1 f f µ 1, ũ = µ 1/ w + µ 1 f f µ 1 with w f w f = e i x r s = e i x r s + Oτ 1, + Oτ 1. Let = τi, 1, θ = /τ. Let C 1 x, θ C x, θ satisfy that iθ x C 1x, θ = θ V 1 xc 1 x, θ, i θ x C x, θ = θ V 1 xc x, θ, respectively. Since = i, we can choose C x, θ = C 1 x, θ. Moreover, r = r 1 s = s 1. With suitable constant vector g, we can get that s l is zero at point x r 1 x = 1, i, = r x. By continuity of r l, we have r l in a neighborhood U 1 of x. Then the CGO solution is u l = µ 1/ w l + µ 1 f l f l µ 1 with w 1 f 1 w f = e i x r 1 s 1 = e i x r s + Oτ 1, + Oτ 1. For, with a suitable constant vector g, we can get that s 3 is zero at point x r 3 x = 1,, i. By continuity of r 3, we have r 3 in a neighborhood U of x. Then the CGO solution is u 3 = µ 1/ w 3 + µ 1 f 3 f 3 µ 1 with w 3 f 3 = e i x r 3 s 3 + Oτ 1. Let U = l= U l. So far we have five CGO solutions, that is, u, ũ, u 1, u, u 3.

23 Let r 1 = r 1 1, r1 u,1 u,, r1 3. We define u 1 := e i x u 1 = 1 := e i x 1 u 1 = 1 := e i x 1 u = 1 λ+µ r1 1 + ir 1 λ+µ ir1 1 r 1 λ+µ τr1 1 + ir 1 + O1 λ+µ τir1 1 r 1 + O1 λ+µ r1 1 + ir 1 λ+µ r1 1 + ir 1 λ+µ r1 1 + ir 1 + Oτ 1 λ+µ τ r 1 ir + O1 λ+µ τ ir 1 + r + O1 λ+µ ir 1 r λ+µ ir 1 r λ+µ ir 1 r Let u = u,1 + u,. Since r = r 1 s = s 1, we have u = 1 O1 O1 λ+µ 1 + ir ir 1 λ+µ λ+µ 1 + ir ir ir ir 1 + Oτ 1 + Oτ 1. + Oτ 1. We also define u 3 := e i x u 3 = 1 λ+µ r3 1 + ir 3 3 λ+µ ir3 1 r Oτ 1 3

24 We denote u 1 = Re i x u ; u = Ie i x u ; u 3 = Re i x ũ ; =,, u 3 u 1 = R e i x 1 u 1 = Re i x u 1 ; u + e i x 1 u = Ie i x u 1 ; ; u 4 = Re i x u 3. { } Then u j : 1 j 4 are linearly independent in the neighborhood U of x as τ is sufficiently large. Therefore, for fixed l = 1,, 3 there exist functions Θ l j, j = 1,, 3, 4, such that u l + 4 Θ l ju j =. As in Section 3.1, we summing over equations, then we have 4 β,l µ + γ,l µ = k u l + Θ l ju j for l = 1,, 3, where β,l γ,l are functions which depend on, Ω CGO solutions ũ, u j for j =,..., 3. Lemma 4.1. Given any point x Ω, there exists an open neighborhood V of x such that β,l is not zero in V, for l = 1,, 3. Proof. Following the similar proof as in Lemma 3., we can prove this Lemma. Let Ω = U V Ω. Based on the lemma above, we may suppose that β,j are linearly independent in Ω as τ sufficiently large. Let g j = u j Ω g = ũ Ω be the given boundary data for j =, 1,, 3. Let g, g j C 1,α Ω be the boundary data close to g, g j, respectively, that is, g j g j C 1,α Ω < ε, g g C 1,α Ω < ε. Then we can find solutions ũ, u j of the elasticity system with boundary data g, g j, respectively. By regularity theorem, it follows that From 56, we have u j u j C Ω < Cε, ũ ũ C Ω < Cε. 56 u j u j CΩ Cε, 1 j 4. 4

25 Therefore, { u j : 1 j 4 } are also linearly independent when ε is sufficiently small. We construct β l by replacing u j by u j. Then from 56, we can deduce that β l β,l C1 Ω is small when ε is sufficiently small. Then we have the following equation 4 β l µ + γ l µ = k u l + Θ l ju j, l = 1,, 3, 57 with {β l x} l=1,,3 a basis in R 3 for every point x Ω. There exists an invertible matrix A = a ij such that β l = a lk e k with inverse of class CΩ. Thus, we have constructed two vector-valued functions Γx, Φx CΩ such that 57 can be rewritten as µ + Γxµ = Φx in Ω. 58 Then we have the following uniqueness stability theorem. Theorem 4.. Global reconstruction of µ Let Ω be an open bounded domain of R 3 with smooth boundary. Suppose that the Lamé parameters λ, µ λ, µ P µ Ω = µ Ω. Let u j ũ j be the solutions of the elasticity system with boundary data g j for parameters λ, µ λ, µ, respectively. Let H = u j 1 j J H = ũ j 1 j J be the corresponding internal data for λ, µ λ, µ, respectively for some integer J 5. Then there is an open set of the boundary data g j 1 j J such that if H = H implies µ = µ in Ω. Moreover, we have the stability estimate µ µ CΩ C H H C Ω. Proof. The proof is similar to Theorem Global reconstruction of λ in 3D The elasticity system can also be written in this form where u = u u u b 3 b 13 b 1, F = u F + u G = k u, 59 1 λ + µ λ + µ 3 λ + µ 1 µ µ 3 µ, u = 1 u u 3 u 1 b 3 b 13 3 b 1, G = λ + µ λ + µ λ + µ µ µ µ. 5

26 As in the reconstruction of µ, we will construct CGO solutions such that the first term of 59 can be eliminated. Plug the CGO solution u = µ 1/ w + µ 1 f f µ 1 into u. Then we get u = e i x λ+µ r λ+µ r λ+µ r µ s + O µ s + O µ s + O + O1. For the same fixed point x Ω. We choose a constant vector g such that C x, θg = r, s with s r in a neighborhood of x, say U, r x = 1. Then we get the CGO solution of the elasticity system with u = µ 1/ w + µ 1 f f w f = e i x r s µ 1 + Oτ 1. We choose another constant vector g 1 such that C 1 x, θg 1 = r 1, s 1 with s 1 in a neighborhood of x, say U 1, r 1 x =. Then we get the CGO solution of the elasticity system with u 1 = µ 1/ w 1 + µ 1 f 1 f 1 w 1 f 1 = e i x r 1 s 1 µ 1 + Oτ 1. For l =, 3, we choose a constant vector g l such that C l x, θg l = r l, s l with r l in a neighborhood of x, say U. Here we can choose r = r 3, s = s 3 by taking g = g 3 C x, θ = C 3 x, θ. Then we get the CGO solution of the elasticity system with u l = µ 1/ w l w f w 3 f 3 + µ 1 f l = e i x r s = e i x r 3 s 3 f l µ 1 + Oτ 1, + Oτ 1. 6

27 For, we choose another constant vector g 4 such that C 4 x, θg 4 = r 4, s 4 with s 4 in a neighborhood of x, say U 3, r 4 x =. Then we get the CGO solution of the elasticity system with We define u 4 = µ 1/ w 4 + µ 1 f 4 f 4 w 4 f 4 u 1 := e i x u 1 u,1 u, :=e i x 1 u = 1 :=e i x 1 u 3 = 1 u = 1 = e i x r 4 s 4 = µ 1 s µ 1 + Oτ 1. λ+µ r λ+µ r 1 + i i 1 λ+µ r µ 1 τ1 + is + O1 µ 1 τi 1s + O1 O1 λ+µ r3 r3 λ+µ r3 µ 1 τ 1 is 3 + O1 µ 1 τ1 is 3 + O1 O1 λ+µ + Oτ 1 + Oτ 1 + Oτ 1. Let u = u,1 + u,, then the higher order is eliminated. Thus we have λ+µ + r3 + r3 λ+µ λ+µ + r3 O1 O1 O1 + Oτ 1. 7

28 Also, we define that We denote u 3 := e i x u 4 = µ 1 s 1 + i i 1 + Oτ 1. u = Re i x u ; =,, u 3 Then u 1 = R e i x 1 u = Re i x u 1 ; u + e i x 1 u 3 = Ie i x u 1 ; ; u 4 = Re i x u 4. { } u j : 1 j 4 are linearly independent in the neighborhood U = 3 l= U l of x as τ is sufficiently large. Therefore, there exist functions Θ j, j = 1,, 3, 4, such that 4 u + Θ j u j =. Summing over, we get the following equation as in Section 3.: 4 v G = k u + Θ j u j, 6 with v = u + 4 Θ ju j. We obtain that 4 κλ = σµ k u + Θ j u j, 61 where κ = 1, 1, 1,,, T v, σ = 1, 1, 1, 1, 1, 1 T v. Lemma 4.3. κ does not vanish in some neighborhood of x. Proof. Similar argument as Lemma 3.7. Since 1 u λ+µ u = e i x 1 r + O λ+µ r + O 3 u λ+µ r + O 8 + O1

29 r 1 x = = r 4 x, we have 1, 1, 1,,, T Hence, we obtain that κx = 1, 1, 1,,, T µ Θ 1 u 1 + Θ u u λ + µ + Θ 3u 3 x. + Θ 4 u 4 x τ 1. + Θ 3 u 3 x + Oτ 1 We can take s small enough such that Θ 3 x is small. Thus, κx. By continuity of κ, κ does not vanish in some neighborhood V of x. Let Ω = U V Ω. Since κ does not vanish in Ω, we have λ = σµ 4 κ k u + Θ j u j in Ω. 6 κ Applying the similar proof as in Theorem 3.5, we can deduce the following result. Theorem 4.4. Global reconstruction of λ Let Ω be an open bounded domain of R 3 with smooth boundary. Suppose that the Lamé parameters λ, µ λ, µ P µ Ω = µ Ω. Let u j ũ j be the solutions of the elasticity system with boundary data g j for parameters λ, µ λ, µ, respectively. Let H = u j 1 j J H = ũ j 1 j J be the corresponding internal data for λ, µ λ, µ, respectively for some integer J 1. Then there is an open set of the boundary data g j 1 j J such that if H = H implies λ = λ in Ω. Moreover, we have the stability estimate Acknowlegments λ λ CΩ C H H C Ω. The author are grateful to professor Gunther Uhlmann for his encouragements helpful discussions. The author also would like to thank Professor Jenn-Nan Wang for taking the time to discuss some properties of the elasticity system with her. The author is partially supported by the NSF. References [1] G. Bal, Hybrid inverse problems internal functionals, Inside Out II, Cambridge University Press, Cambridge 1. [] G. Bal G. Uhlmann, Reconstruction of coefficients in scalar second-order elliptic equations from knowledge of their solutions, Comm. on Pure Applied Math, ,

30 [3] G. Bal G. Uhlmann, Inverse diffusion theory for photoacoustics, Inverse Problems, 6 1, 511. [4] P. G. Ciarlet, Mathematical Elasticity, Elsevier Science Publishers B.V., Amsterdam [5] G. Eskin J. Ralston, On the inverse boundary value problem for linear isotropic elasticity, Inverse Problems, 18, [6] M. Ikehata, A remark on an inverse boundary value problem arising in elasticity, preprint. [7] W. McLean, Strongly Elliptic Systems Boundary Integral Equations, Cambridge University Press, Cambridge. [8] J. R. McLaughlin D. Renzi, Shear wave speed recovery in transient elastography supersonic imaging using propagating fronts, Inverse Problems, 6, [9] J. R. McLaughlin, N. Zhang, A. Muca, Calculating tissue shear modulus pressure by D log-elastographic methods, Inverse Problems, 6 1, 857. [1] G. Nakamura G. Uhlmann, Global uniequness for an inverse boundary problem arising in elasticity, Invent. Math., , [11] G. Nakamura G. Uhlmann, Erratum: Global uniqueness for an inverse boundary problem arising in elasticity, Invent. Math., 15 3, 5 7. [1] G. Uhlmann J.-N. Wang, Complex geometrical optics solutions reconstruction of discontinuities, SIAM J. Appl. Math., 68 8,