Thermoacoustic tomography with an arbitrary elliptic operator

Abstract

Abstract. Thermoacoustic tomography is a term for the inverse problem of determining of one of initial conditions of a hyperbolic equation from boundary measurements. In the past publications both stability estimates and convergent numerical methods for this problem were obtained only under some restrictive conditions imposed on the principal part of the elliptic operator. In this paper logarithmic stability estimates are obatined for an arbitrary variable principal part of that operator. Convergence of the Quasi-Reversibility Method to the exact solution is also established for this case. Both complete and incomplete data collection cases are considered.

Carleman estimates for the regularization of ill-posed Cauchy problems

“…Although results of this section actually follow from the earlier work of Klibanov and Malinsky [48] (1991), this problem did not have that name at that time. More details can be found in the paper of the author [55]. Numerical studies by the method of this section were performed in [29,54].…”

“…It is shown in section 7 that the original technique of [48] allows one to obtain the Lipschitz stability estimate, to construct the Tikhonov functional and to obtain the Lipschitzlike convergence rate of its minimizers for the problem of determining an initial condition of a hyperbolic PDE from boundary measurements. This problem is called nowadays "the problem of thermoacoustic tomography"; also, see more details in the paper of the author [55].…”

Uniqueness of two phaseless non-overdetermined inverse acoustics problems in 3-d

“…Next, we apply the operator F −1 to both sides of (32). Using (19), (22), (32), (35), (36) and the convolution theorem, we obtain…”

Reconstruction Procedures for Two Inverse Scattering Problems Without the Phase Information

“…Using theorems 3.2, 4.1, Corollary 4.2 and energy estimates of Chapter 4 of [19], one can easily prove that w s ∈ C l−1 (D(y, T )) , see, e.g. Theorem 2.2 of [12] for a similar result. Choosing m = 15 and s = 3, we obtain l = 3.…”