Unique determination of a transversely isotropic perturbation in a linearized inverse boundary value problem for elasticity

We consider a linearized inverse boundary value problem for the elasticity system. From the linearized Dirichlet-to-Neumann map at zero frequency, we show that a transversely isotropic perturbation of a homogeneous isotropic elastic tensor can be uniquely determined. From the linearized Dirichlet-to-Neumann map at two distinct positive frequencies, we show that a transversely isotropic perturbation of a homogeneous isotropic density can be identified at the same time.


Introduction and Main result
In this paper, we investigate the problem of determining interior material property of an elastic body from boundary measurements. For linear elasticity in equilibrium, the governing equations read (1) ∂ j C ijkl (x)∂ k u l (x) = 0, i = 1, 2, 3.
Here u is the displacement vector; C = C ijkl ∈ L ∞ (Ω) is the elastic tensor whose components obey the symmetry conditions (2) C ijkl = C jikl = C klij .
We have used Einstein's summation convention in (1) such that repeated indices are summed up over {1, 2, 3}. We note here that C with the above symmetry has a total number of 21 linearly independent components.
Let Ω be an open bounded domain in R 3 with C 1,1 boundary ∂Ω. If we assume further that the elasticity tensor C satisfies the following positivity condition: there exists δ > 0 such that for any 3 × 3 real-valued symmetric matrix (ε ij ), 3 i,j,k,l=1 then for any f ∈ H 1/2 (∂Ω), standard elliptic theory ensures a unique solution u f ∈ H 1 (Ω) to the boundary value problem We define the Dirichlet-to-Neumann map (DN map) Λ C by where ν = (ν 1 , ν 2 , ν 3 ) denotes the outer unit normal vector to ∂Ω. It follows that Λ C : H 1/2 (∂Ω) → H −1/2 (∂Ω) is a bounded linear operator, and the equivalent weak formulation is for any f, g ∈ H 1/2 (∂Ω). An important inverse boundary value problem in linear elasticity is whether one can recover C from Λ C . This is related to the invertibility of the non-linear map C → Λ C . The question is difficult for general C, so it is commonly studied under additional a-priori information. We say the elastic tensor C (or the medium) is homogeneous if it is a constant tensor (that is, independent of x); it is isotropic if it can be written as where the two functions λ(x) and µ(x) are known as Lamé parameters; and it is fully anisotropic if the components C ijkl are subject to no other relations other than (2). For isotropic C, a global uniqueness result can be found [11] in dimension two. The problem remains open in dimension three, yet some special cases have been tackled. Among them, Nakamura and Uhlmann [15] proved uniqueness when the Lamé parameters are smooth and µ(x) is close to a positive constant, see [5] for a similar result by Eskin and Ralston and [10] for a partial data version; uniqueness for recovering piecewise constant Lamé parameters was proved in [1,2]; and some boundary determination results were shown in [12,14,13] . For fully anisotropic C, uniqueness was proved in [4] for piecewise homogeneous medium.
In this article, we investigate the linearization of the map C → Λ C at a homogeneous isotropic elastic tensor. More specifically, suppose where C 0 = λ 0 δ ij δ kl + µ 0 (δ ik δ jl + δ il δ jk ) is a homogeneous, isotropic background tensor with Lamé parameters (λ 0 , µ 0 ) satisfying µ 0 > 0, 3λ 0 + 2µ 0 > 0, and δC(x) is viewed as a perturbation term with components δC ijkl (x). It is routine to verify that the map C → Λ C is Frechét differentiable at C 0 (we refer to [6] for more details), and the Frechét derivativeΛ The question we are interested in is whether the linearized mapΛ C 0 is injective. It was proved in [6] that the linearizationΛ C 0 is injective on isotropic perturbations. Our main theorem (see Theorem 1 below) gives an affirmative answer to this injectivity question on a class of anisotropic perturbations with certain symmetry. In linear elasticity, different types of anisotropy with extra symmetries in the internal structure of the material have been considered. With more symmetries, the elasticity tensor C would have fewer degrees of freedom. We list some commonly used anisotropy with symmetries in the table below. It is worth mentioning that these concepts of anisotropy are purely Cartesian (in a prescribed coordinate system (x 1 , x 2 , x 3 )).

Type of anisotropy
Symmetries Number of independent components cubic three mutually orthogonal planes of reflection symmetry plus π 2 rotation symmetry with respect to those planes 3 transversely isotropic three mutually orthogonal planes of reflection symmetry and one symmetry axis perpendicular to one symmetry plane 5 orthotropic (orthorhombic) three mutually orthogonal planes of reflection symmetry 9 monoclinic one plane of reflection symmetry 13 fully anisotropic no symmetry 21 We will focus on transversely isotropic elasticity in this article. Transverse isotropy means the elasticity have three mutually orthogonal planes of reflection symmetry and one symmetry axis perpendicular to one of the three symmetry planes. Assume the symmetry axis is x 3 , then δC obeys the invariance Q ip Q jq Q kr Q ls δC pqrs = δC ijkl , where Q can take any of the following reflection and rotation matrices.
The 5 linearly independent components of δC we will determine are δC 1111 , δC 1122 , δC 1133 , δC 1313 , and δC 3333 . The proof is based on construction of complex geometric optics (CGO) solutions for the system (4). CGO solutions were initiated by Sylvester and Uhlmann [16] in their solving Calderón's inverse conductivity problem [3]. Solutions of this type were introduced in [6] for the elasticity system with constant coefficients, and [5,15] with variable coefficients.
Injectivity of the linearized mapΛ C 0 has been studied in previous literature. In dimension two or higher, it is known thatΛ C 0 is injective on isotropic δC [6]. Our theorem can be viewed as generalization of such injectivity result from isotropic perturbations to transversely isotropic perturbations in dimension three. In dimension two, Ikehata [7,9,8] obtained a few characterizations of the injectivity allowing anisotropic C 0 .
The rest of the paper is devoted to the proof of Theorem 1.

Proof of the Theorem
In view of (3),Λ C 0 (δC) = 0 implies for any u, v satisfying (4). The key ingredient of our proof is constructing CGO solutions to (4) and inserting them into (6) to obtain sufficiently many linearly independent equations in the 5 independent components of δC. For the ease of notation, we abbreviate C for δC(x) and C ijkl for δC ijkl (x) from now on. We reserve the letter i for an index and write the bold face i for the imaginary unit.
Here χ Ω is the characteristic function of the domain Ω. Since s, t can be any real number, this Fourier transform vanishes on the punctured x 1 x 3 -plane. The axial symmetry with respect to x 3axis in the definition of transversal isotropy allows one to obtain similar vanishing result in any plane containing x 3 -axis. We conclude F[χ Ω (C 1212 + C 1313 )](ξ) = 0 for any ξ = 0. This forces C 1212 + C 1313 = 0 in Ω.
Using such u, v in (6), we have l e ζ (2) ·x dx =: We will analyze the asymptotic behavior as r → ∞. Direct calculation (though tedious) shows Equating the terms of order O(r 4 ) yields Let us put the three pieces of information (8)(9)(10) together We observe these combinations are linearly independent and thus can be used to eliminate 3 independent components of C. In fact, solving this linear system yields We are therefore left with only 2 independent components, say C 1111 and C 1133 .
The need for different solutions. The previous CGOs are not enough to determine the remaining independent components. To see this, we employ the known relations (11) (12) to simplify the integral identity (6), then All the solutions we have constructed have divergence zero, so they cannot give new information about the tensor C. Remark 1. With only CGO solutions of divergence zero, one cannot even determine an isotropic perturbation fromΛ C 0 (cf. [6]). To see this, suppose C ijkl = λδ ij δ kl + µ(δ ik δ jl + δ il δ jk ), then (6) reduces (with δC ijkl abbreviated as C ijkl ) to (14) Ω 2µSym(∇u) : Sym(∇v) + λ(∇ · u)(∇ · v)dx = 0.
Here Sym(∇u) := 1 2 (∇u + (∇u) T ) and A : B = 3 i,j=1 A ij B ij for any 3 × 3 matrices A, B. It is obvious that solutions with divergence zero cannot provide information about λ.
New type of solutions. This above analysis suggests the necessity to construct new solutions with non-vanishing divergence. We proceed to construct new CGO-type solutions with this property. They are of the form where ζ ∈ C 3 satisfies ζ · ζ = 0,ζ denotes ζ |ζ| , and b, c are constant vectors to be determined. This type of solutions can be constructed as in [6]. The divergence of u is On the other hand, the gradient of u is so ∆u can be computed: We then have Taking b = (λ 0 + µ 0 )ℜζ and c = − λ 0 +3µ 0 |ζ| ℜζ guarantees the right hand side is zero, making u a solution to (4). Notice that with such b, c, the divergence of u is ∇ · u = (b ·ζ + c · ζ)e ζ·x = −2µ 0 ℜζ ·ζ e ζ·x = −µ 0 e ζ·x , which is non-vanishing since µ 0 > 0.
Then we get C 1111 = 0. This completes the proof of the uniqueness of all parameters in C.