A global stability estimate for the photo-acoustic inverse problem in layered media

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1 A global stability estimate for the photo-acoustic inverse problem in layered media Kui Ren Faouzi Triki May 11, 217 Abstract This paper is concerned with the stability issue in determining absorption and diffusion coefficients in photoacoustic imaging. Assuming that the medium is layered and the acoustic wave speed is known we derive global Hölder stability estimates of the photo-acoustic inversion. These results show that the reconstruction is stable in the region close to the optical illumination source, and deteriorate exponentially far away. Several experimental pointed out that the resolution depth of the photoacoustic modality is about tens of millimeters. Our stability estimates confirm these observations and give a rigorous quantification of this depth resolution. Key words. Inverse problems, wave equation, diffusion equation, Lipschitz stability, multiwave inverse problems. AMS subject classifications R3, 35J15, 35L5, 92C55 1 Introduction Photoacoustic imaging PAI [6, 8, 13, 3, 32, 42, 46] is a recent hybrid imaging modality that couples diffusive optical waves with ultrasound waves to achieve high-resolution imaging of optical properties of heterogeneous media such as biological tissues. In a typical PAI experiment, a short pulse of near infra-red photons is radiated into a medium of interest. A part of the photon energy is absorbed by the medium, which leads to the heating of the medium. The heating then results in a local temperature rise. The medium expanses due to this temperature rise. When the rest of the photons leave the Department of Mathematics and the Institute of Computational Engineering and Sciences ICES, University of Texas, Austin, TX 78712, USA; ren@math.utexas.edu Laboratoire Jean Kuntzmann, UMR CNRS 5224, Université Grenoble- Alpes, 7 Avenue Centrale, 3841 Saint-Martin-d Hères, France; faouzi.triki@univ-grenoble-alpes.fr 1

2 medium, the temperature of the medium drops accordingly, which leads to the contraction of the medium. The expansion and contraction of the medium induces pressure changes which then propagate in the form of ultrasound waves. Ultrasound transducers located on an observation surface, usually a part of the surface surrounding the object, measure the generated ultrasound waves over an interval of time, T with T large enough. The collected information is used to reconstruct the optical absorption and scattering properties of the medium. Assuming that the ultrasound speed in the medium is known, the inversion procedure in PAI proceeds in two steps. In the first step, we reconstruct the initial pressure field, a quantity that is proportional to the local absorbed energy inside the medium, from measured pressure data. Mathematically speaking, this is a linear inverse source problem for the acoustic wave equation [2, 3, 7, 19, 21, 22, 24, 25, 26, 28, 29, 31, 37, 38, 39, 43, 44]. In the second step, we reconstruct the optical absorption and diffusion coefficients using the result of the first inversion as available internal data [4, 5, 16, 17, 34, 36, 4, 41]. In theory, photoacoustic imaging provides both contrast and resolution. The contrast in PAI is mainly due to the sensitivity of the optical absorption and scattering properties of the media in the near infra-red regime. For instance, different biological tissues absorbs NIR photons differently. The resolution in PAI comes in when the acoustic properties of the underlying medium is independent of its optical properties, and therefore the wavelength of the ultrasound generated provides good resolution usually submillimeter. In practice, it has been observed in various experiments that the imaging depth, i.e. the maximal depth of the medium at which structures can be resolved at expected resolution, of PAI is still fairly limited, usually on the order of millimeters. This is mainly due to the limitation on the penetration ability of diffusive NIR photons: optical signals are attenuated significantly by absorption and scattering. The same issue that is faced in optical tomography [12]. Therefore, the ultrasound signal generated decays very fast in the depth direction. The objective of this work is to mathematically analyze the issue of imaging depth in PAI. To be more precise, assuming that the underlying medium is layered, we derive a stability estimate that shows that image reconstruction in PAI is stable in the region close to the optical illumination source, and deteriorates exponentially in the depth direction. This provides a rigorous explanation on the imaging depth issue of PAI. In the first section we introduce the PAI model and give the main global stability estimates in Theorem 2.1. Section 2 is devoted to the acoustic inversion, we derive observability inequalities corresponding to the internal data generated by well chosen laser illuminations. We also provide an observability inequality from one side for general initial states in Theorem 3.2. In section 3, we solve the optical inversion and show weighted stability estimates of the recovery of the optical coefficients from the knowledge of two internal data. Finally, the main global stability estimates are obtained by combining stability estimates from the acoustic and optical inversions. 2

3 2 The main results In our model we assume that the laser source and the ultrasound transducers are on the same side of the sample Γ m ; see Figure 1. This situation is quite realistic since in applications only a part of the boundary is accessible and in the exiting prototypes a laser source acts trough a small hole in the transducers. We also assume that the optical parameters D, µ a, similar to the acoustic speed c, only depend on the variable y following the normal direction to Γ m. We further consider the optical parameters Dy, µ a y within the set O M = {D, µ C 3 [, H] 2 ; D > D, µ > µ ; D C 3, µ C 3 M}, where D >, µ > and M > maxd, µ are fixed real constants. Γ m Γ Figure 1: The geometry of the sample. The propagation of the optical wave in the sample is modeled by the following diffusion equation 1 Dy ux + µ a yux = x, ux = gx x Γ m, ux = x Γ, u, y = ul, y y, H, where g is the laser illumination, D and µ a are respectively the diffusion and absorption coefficients. The part of the boundaries Γ j are given by Γ m =, L {y = H}, Γ =, L {y = }, and ν is the derivative along ν, the unit normal vector pointing outward of. We note that ν is everywhere defined except at the vertices of and we denote by Γ p the complementary of Γ Γ m in. We follow the approach taken in several papers [14, 15, 17] and consider two laser illuminations g j, j = 1, 2. Denote u j, j = 1, 2, the corresponding laser intensities. 3

4 Let V := {v H 1 ; v, y = vl, y, y, H; u = on Γ }. We further assume that g V Γm, where V Γm := {v Γm H 1 ; v, y = vl, y, y, H; u = on Γ }. Then, there exists a unique solution u V satisfying the system 1. The proof uses techniques developed in [23]. The first step is to show that the set V := {v H 1 ; v, y = vl, y, y, H; u = on Γ Γ m }, is a closed sub space of H 1, using a specific trace theorem for regular curvilinear polygons Theorem , page 5 in [23]. Then, applying the classical Lax-Milligram for elliptic operators in Lipschitz domain gives the existence and uniqueness of solution to the system 1. Remark 2.1. Using an explicit characterization of the trace theorem obtained in [23] one can derive the optimal local regularity for g H 1 2 Γ m that guarantees the existence and uniqueness of solutions to the system 1 see also [9]. For simplicity, we will further consider g j = ϕ kj, j = 1, 2, where k 1 < k 2, and ϕ k x, k N, is the Fourier orthonormal basis of L 2, L satisfying ϕ k y = λ2 k ϕ ky, with λ k = 2kπ, k Z. Direct calculation gives ϕ L kx = 1 L e iλkx, k Z. We assume that point-like ultrasound transducers, located on an observation surface Γ m, are used to detect the values of the pressure px, t, where x Γ m is a detector location and t is the time of the observation. We also assume that the speed of sound in the sample occupying =, L, H, is a smooth function and depends only on the vertical variable y, that is, c = cy >. Then, the following model is known to describe correctly the propagating pressure wave px, t generated by the photoacoustic effect tt px, t = c 2 y px, t x, t, ν px, t + β t px, t = x Γ m, t, 2 px, t = x Γ, t, p, y, t = pl, y, t y, H, t, px, = f x, t px, = f 1 x, x, Here β > is the damping coefficient, and f j x, j =, 1, are the initial values of the acoustic pressure, which one needs to find in order to determine the optical parameters of the sample. Remark 2.2. Notice that in most existing works in photoacoustic imaging the initial state f x is given by µ a xux, and is assumed to be compactly supported inside, while the 4

5 initial speed f 1 x is zero everywhere [29, 39, 43]. The compactly support assumption on f x simplifies the analysis of the inverse source initial-to-boundary problem and is necessary for almost all the existing uniqueness and stability results [1, 11, 27, 47]. Meanwhile the assumption is clearly in contradiction with the fact that f x coincides with µ a xux everywhere. We will show in section 4 that µ a xux is not only not compactly supported, it is also exponentially concentrated around the part of the boundary Γ m where we applied the laser illumination. In our model the initial speed t px, = f 1 x can be considered as the correction of the photoacoustic effect generated by the heat at Γ m. The following stability estimates are the main results of the paper, obtained by combining stability estimates from the acoustic and optical inversions. Theorem 2.1. Let D, µ a, D, µ a in O M, and k i, i = 1, 2 be two distinct integers. Let cy W 1,, H with < c m c 2 y and set θ = c 2 L. Denote u ki, i = 1, 2 and ũ ki, i = 1, 2 the solutions to the system 9 for g i = ϕ ki, i = 1, 2, with coefficients D, µ a and D, µ a respectively. Assume that DH = DH, D H = D H, µ ah = µ ah, k 1 < k 2, and k 1 is large enough. Then, for T > 2θH, there exists a constant C > that only depends on µ, D, k 1, k 2, M, L, and H, such that the following stability estimates hold. and where C C 2 i=1 2 i=1 u 2 mµ a µ a C 1 4 CM T 2θH + β t p i t p i 2L2Γm + xp i x p i 2L2Γm dt, u 2 md D C 1 4 CM T 2θH + β t p i t p i 2L2Γm + xp i x p i 2L2Γm dt, u m y = D 1 2 H D 1 2 y C M = He H c2 s yc 2 s ds c 2 H + β 2, sinhκ 1 2 m y D 1 2, κ sinhκ 1 m = min + µ a 2 y H m H D 1 2 D + λ2 k. Since the function u m y is exponentially decreasing between the value 1 on Γ m to the value on Γ, the stability estimates in Theorem 2.1 shows that the resolution deteriorate exponentially in the depth direction far from Γ m. 5

6 3 The acoustic inversion The data obtained by the point detectors located on the surface Γ m are represented by the function px, t = dx, t x Γ m, t. Thus, the first inversion in photoacoustic imaging is to find, using the data dx, t measured by transducers, the initial value f x at t = of the solution px, t of 2. We will also recover the initial speed f 1 x inside, but we will not use it in the second inversion. We first focus on the direct problem and prove existence and uniqueness of the acoustic problem 2. Denote by L 2 c the Sobolev space of square integrable functions with weight 1 c 2 y. Since the speed c2 is lower and upper bounded, the norm corresponding to this weight is equivalent to the classical norm of L 2. Let V = {p H 1 ; p, y = pl, y, y, H; p = on Γ }, and consider in V L 2 c the unbounded linear operator A defined by Ap, q = q, c 2 p, DA = {p, q V V ; p L 2 ; ν p + βq = on Γ m }. We have the following existence and uniqueness result. Proposition 3.1. For f, f 1 DA, the problem 2 has a unique solution px, t satisfying p, t p C [, +, DA C 1 [, +, V L 2 c. Proof. There are various methods for proving well-posedness of evolution problems: variational methods, the Laplace transform method and the semi-group method. Here, we will consider the semi-group method [45], and prove that the operator A is m-dissipative on the Hilbert space V L 2 c. Denote by, the scalar product in V L 2 c, that is, for p i, q i V L 2 c with i = 1, 2, dx p 1, q 1, p 2, q 2 = p 1 p 2 dx + q 1 q 2 c. 2 Now let p, q DA. We have Ap, q, p, q = q pdx + pqdx. Since p L 2 and ν p + βq = on Γ m, applying Green formula leads to Ap, q, p, q = q pdx q pdx β q 2 dσx. Γ m 6

7 Consequently R Ap, q, p, q = β q 2 dσx. Γ m Therefore the operator A is dissipative. The fact that is in the resolvent of A is straightforward. Then A is m-dissipative and hence, it is the generator of a strongly continuous semigroup of contractions [45]. Consequently, for f, f 1 DA there exists a unique strong solution to the problem 2. Now, back to the inverse problem of reconstructing the initial data f, f 1. We further assume that the initial data is generated by a finite number of Fourier modes, that is 3 f j x, y = f jk yϕ k x x, y j =, 1, k N with N being a fixed positive integer. As it was already remarked in many works, this linear initial-to-boundary inverse problem is strongly related to boundary observability of the source from the set Γ m see for instance [33, 43, 45, 48]. We will emphasize on the links between our findings and known results in this context later. Here we will use a different approach taking advantage of the fact that the wave speed cy only depends in the vertical variable y. Since px is L-periodic in the y variable, it has the following discrete Fourier decomposition px, y = p k y, tϕ k x x, y. k N One can check that p k y, tϕ k x is exactly the solution to the problem 2 with initial data f k yϕ k x, f 1k yϕ k x. Precisely, if λ k = 2kπ, the functions p L ky, t satisfy the following one dimensional wave equation 4 1 c 2 y ttpy, t = yy py, t λ 2 kpy, t, y, H, t, y ph, t + β t ph, t = t, p, t = t, py, = f k y, t py, = f 1k y, y, H, Next, we will focus on the boundary observability problem of the initial data f k at the extremity y = H. Taking advantage of the fact that the equation is one dimensional we will derive a boundary observability inequality with a sharp constant. Define Et the total energy of the system 4 by 5 Et = H c 2 y t py, t 2 + y py, t 2 + λ 2 k py, t 2 dy. 7

8 Multiplying the first equation in the system 4 by t py, t and integrating over, H leads to 6 E t = β t ph, t 2 for t. Consequently, Et is a non-increasing function, and the decay is clearly related to the magnitude of the dissipation on the boundary Γ m. It is well know that the system 4 has a unique solution. Here we establish an estimate of the continuity constant. Proposition 3.2. Assume that cy W 1,, H with < c m c 2 y. Then, for any T > we have for k N, where β 2 t p k H, t 2 dt Cm 1 + Cmλ 2 k T + CmE 3 k, H E k = c 2 y f 1k y 2 + f ky 2 + λ 2 k f k y 2 dy, Cm 1 = 1 + Hc 2 H H c 2 W 1,, c m Cm 3 = 1 + Hc 2 H H c 2 1/2 L, C 2 m = H1 + Hc 2 H 1. Proposition 3.3. Assume that cy W 1,, H with < c m c 2 y. Let θ = c 2 L and T > 2θH. Then the following inequalities hold H λ 2 k f k y 2 CM dy T 2θH + β t p k H, t 2 dt for k N, H for k N, with c 2 y f 1k y 2 + f ky 2 dy +λ 2 k CM T 2θH + β p k H, t 2 dt, +λ 2 k C M = He H c2 s yc 2 s ds c 2 H + β 2. t p k H, t 2 dt p k H, t 2 dt, The proofs of these results are given in the Appendix. The main result of this section is the following. 8

9 Theorem 3.1. Assume that cy W 1,, 1 with < c m c 2 y, and f, f 1 have a finite Fourier expansion 3. Let θ = c 2 L and T > 2θH. Then f x 2 CM dx T 2θH + β t px, t 2 L 2 Γ dt m and with c 2 y f 1 x 2 dx + CM T 2θH + β + x px, t 2 L 2 Γ m dt, C M = He H c2 s yc 2 s ds c 2 H + β 2. t p k x, t 2 L 2 Γ m dt x px, t 2 L 2 Γ m dt, Proof. The estimates are direct consequences of Proposition 3.2 and Proposition 3.3. The fact that the Fourier series of px, t has a finite number of terms justifies the regularity of the solution px, t, and allow interchanging the order between the Fourier series and the integral over, T. Using microlocal analysis techniques it is known that the boundary observability in a rectangle holds if the set of boundary observation necessarily contains at least two adjacent sides [18, 2]. Then, we expect that the the Lipschitz stability estimate in Theorem 3.1 will deteriorate when the number of modes N becomes larger. In fact the series on the right side does not converge because x px, t does not belong in general to L 2 Γ m, T. We here provide a hölder stability estimate that corresponds to the boundary observability on only one side of the rectangle. Theorem 3.2. Assume that cy W 1,, 1 with < c m c 2 y, and f, f 1 H 2 H 1 satisfying f H 1, f 1 H 2 M. Let θ = c 2 L and T > 2θH. Then f x 2 CM dx T 2θH + β t px, t 2 L 2 Γ dt m 2 px, t 2 3 dt, H 2 1 Γ m and c 2 y f 1 x 2 dx +C M CM T 2θH + β +C Mθ t p k x, t 2 L 2 Γ m dt px, t 2 H 1 2 Γ m dt 2 3,

10 with C M = He H c2 s yc 2 s ds c 2 H + β 2, C M = 2 M 2 3. Proof. The proof is again based on the results of Proposition 3.3. We first deduce from Proposition 3.1 that t px, t L 2 Γ m. Now, define fj N x = f jk yϕ k x x j =, 1, k N with f jk y are the Fourier coefficients of f j x, and N being a large positive integer. 7 8 Consequently f x 2 dx c 2 y f 1 x 2 dx f N x 2 dx + M 2 λ 2 N+1 c 2 y f1 N x 2 dx + θ2 2 M, λ 2 N+1, for N large. Applying now Proposition 3.3 to f N, f N 1, gives f x 2 dx c 2 y f 1 x 2 dx CM T 2θH + β t px, t 2 L 2 Γ dt m +λ N CM T 2θH + β px, t 2 2 M dt + H 1 2 Γ m λ 2 N+1 t px, t 2 L 2 Γ m dt +λ N px, t 2 dt + θ2 2 M. H 1 2 Γ m λ 2 N+1 By minimizing the right hand terms with respect to the value of λ N, we obtain the desired results., 4 The optical inversion Once the initial pressure f x, generated by the optical wave has been reconstructed, a second step consists of determining the optical properties in the sample. Although this second step has not been well studied in biomedical literature due its complexity, it is of importance in applications. In fact the optical parameters are very sensitive to the tissue condition and their values for healthy and unhealthy tissues are extremely different. 1

11 The second inversion is to determine the coefficients Dy, µ a y from the initial pressures recovered in the first inversion, that is, h j x = µ a yu j x, x, j = 1, 2. For simplicity, we will consider g j x = ϕ kj x, j = 1, 2 with k 1 and k 2 are two distinct Fourier eigenvalues that are large enough. We specify how large they should be later in the analysis. The main result of this section is the following. Theorem 4.1. Let D, µ a, D, µ a in O M, and k i, i = 1, 2 be two distinct integers. Denote u ki, i = 1, 2 and ũ ki, i = 1, 2 the solutions to the system 9 for g i = ϕ ki, i = 1, 2, with coefficients D, µ a and D, µ a respectively. Assume that DH = DH, D H = D H, µ ah = µ ah, k 1 < k 2, and k 1 is large enough. Then, there exists a constant C > that only depends on µ, D, k 1, k 2, M, L, H, such that the following stability estimates hold. u 2 md D C C u 2 mµ a µ a C C h 1 h 1 C 1 + h 2 h 2 C 1 h 1 h 1 C 1 + h 2 h 2 C 1 Classical elliptic operator theory implies the following result for the direct problem [35]. Proposition 4.1. Assume D, µ a be in O M solution u V to the system 1. It verifies where C = C µ, D, M, L, H >. u H 1 C g 2 H 1 2 Γ m,,. and g V Γm. Then, there exists a unique For gx = ϕ k x, the unique solution u has the following decomposition ux = u k yϕ k x x, where u k y satisfies the following one dimensional elliptic equation { Dyu y + µ a y + λ 2 kdyuy = y, H, 9 uh = 1, u =, Next we will derive some useful properties of the solution to the system 9. Lemma 4.1. Let uy be the unique solution to the system 9. Then uy C 2 [, H] and there exists a constant b = bµ, D, M, L, H > such that u C 2 b for all D, µ a O M. In addition the following inequalities hold for k large enough. for y H, where κ m = min y H u m y uy u M y, u m y = D 1 2 H sinhκ 1 2 m y, u D 1 2 y sinhκ 1 M y = D 1 2 H sinhκ 1 2 M y, 2 m H D 1 2 y sinhκ 1 2 M H D 1 2 D 1 2 D µ a D + λ2 k, κ M = max y H 11 D µ a D + λ2 k.

12 Proof. We first make the Liouville change of variables and introduce the function vy = D 1 2 y D 1 2 H uy. Forward calculations show that vy is the unique solution to the following system. { v y + κyvy = y, H, 1 vh = 1, v =, where κy = D D + µ a D + λ2 k. Assume now that k is large enough such that κ m >, and let v m y and v M y be the solutions to the system 1 when we replace κy by respectively the constants κ m and κ M. They are explicitly given by v m y = sinh κ m y sinh κ m H, v M y = sinh κ M y sinh κ M H. The maximum principle [35] implies that < vy, v m y, v M y < 1 for < y < H. By applying again the maximum principle on the differences v v m and v v M we deduce that v m y < vy < v M y for y H, which leads to the desired lower and upper bounds. We deduce from the regularity of the coefficients D and µ a and the classical elliptic regularity [35] that u H 3, H. Moreover there exist a constant b > that only depends on µ, D, M, L, H such that 11 u H 3 b. Consequently the uniform C 2 bound of u can be obtained using the continuous Sobolev embedding of H 3, H into C 2 [, H] [1]. Lemma 4.2. Let D, µ a O M, and uy be the unique solution to the system 9. Then, for k large enough there exists a constant ϱ = ϱd, µ, M, k > such that for y H. u y ϱ, 12

13 Proof. Since is the global minimum of uy, we have u >. enough, Lemma 4.1 implies that Moreover for k large uy DH Dy sinh κ m y sinh κ m H, for all y [, H]. Therefore u DH κm D L sinh κ m H Now integrating equation 9 over, y we obtain Dyu y = Du + y µ a s + λ 2 kdsusds Dyu y Du + y µ a s + λ 2 kds DH κm D DH D L sinh κ m H + µ + λ 2 kd D L DH sinh κ m s Ds sinh κ m H ds Taking into account the explicit expression of κ m finishes the proof. cosh κ m y 1 sinh. κ m H Since the illumination are chosen to coincide with the Fourier basis functions ϕ kj, j = 1, 2, the data h j x, j = 1, 2, can be rewritten as h j x = h j yϕ kj x, j = 1, 2, where h j y = µ a yu kj y. Therefore the optical inversion is reduced to the problem of identifying the optical pair D, µ a from the knowledge of the pair h 1 y, h 2 y over, H. Let D, µ a, D, µ a be two different pairs in O M, and denote u k and ũ k the solutions to the system 9, with coefficients D, µ a and D, µ a respectively. We deduce from Lemma 4.1 that 1 u k and 1 ũ k lie in L p, H for < p < 1. Unfortunately for or < p < 1 the usual L p is not anymore a norm on the vector space L p, H because it does not satisfy the triangle inequality see for instance [1]. In contrast with triangle inequality Hölder inequality holds for < p < 1, and we have 12 v u k L r 1 u k L p v L q, for all v L q, H with 1 r = 1 p + 1 q. 13

14 Consequently h = h 2 h 1 = u k 2 u k1 can be considered as a distribution that coincides with a C 2 function over, H. A forward calculation shows that h satisfies the equation 13 Du 2 k 1 h + Du 2 k1 hλ 2 2 λ 2 1 =, over, H. Since u kj, j = 1, 2 are in C 2 [, H], an asymptotic analysis of Dyu 2 k 1 yh y at and the results of Lemma 4.1, gives Similarly, we have lim y Du2 k 1 h =. lim h = 1. y 1 Integrating the equation 13 over, y, we get Dyu 2 k 1 yh y = λ 2 2 λ 2 1 y Dsu k1 su k2 sds. Dividing both sides by Dyu 2 k 1 y, and using again Lemma 4.1, imply which leads to 14 h y λ 2 2 λ 2 1D 1 yu 2 M y y Dsu 2 msds, y h y λ 2 2 λ 2 1 min y,h D 1 yu 2 M y Dsu 2 msds >. The right hand constant is strictly positive and only depends on D, µ, M, H, L and k. Now back to the optical inversion. The equation 13 can be written as Du 2 k 1 h + hλ 2 2 λ 2 1 h Du 2 k 1 =, over, H. Dividing both sides by Du 2 k 1 h, and integrating over, y, we obtain 15 Dyu 2 k 1 y = h hy + e λ2 2 λ2 1 y h h ds. This allows us to show the following result. Lemma 4.3. Under the assumptions of Theorem 4.1, there exists a constant C = Cµ, D, k 1, k 2, M, L, H such that the following inequality holds. Du 2 k 1 Dũ 2 k 1 C C h 1 h 1 C 1 + h 2 h 2 C 1. 14

15 Proof. Recall that the relation 15 is also valid for the pair D, µ a, that is Dyũ 2 k 1 y = h hy + e λ2 2 λ2 1 y h h ds, where h = ũk 2 ũ k1. Taking the difference between the last equation and the equation 15 we find Du 2 k 1 Dũ 2 k 1 C h h C + λ 2 2 λ 2 h 1H h h h C We then deduce the result from Lemma 4.1 and inequality 14. e λ2 2 λ2 1 H h h C + h C h. Now, we are ready to prove the main stability result of this section. We remark as in is a solution to the following equation. [14], that 1 u k1 Du 2 1 k 1 + λ 2 k1 Du 2 1 k1 = h 1, y, H. u k1 u k1 Since 1 ũ k1 solves the same type of equation, we obtain that w = 1 u k1 1 ũ k1, is the solution to the following system { Du 2 k 1 w + λ 2 k1 Du 2 k 1 w = e, y, H, 16 wh =, w H = 1 h µ ah 1H h 1H, ywy L 2, H, where e = Du 2 k 1 Dũ 2 k λ 2 k1 Du 2 k1 ũ Dũ 2 k1 1 + h 1 k1 ũ h 1. k1 We remark that to solve this system we have to deal with two main difficulties, the first is that the operator is elliptic degenerate, and the second is that the solution wy may be unbounded at y =. to Multiplying by signw, and integrating over s, H the first equation of the system leads Dsu 2 k 1 s w s = DH H H signw µ a H h 1H h 1H + eydy λ 2 k 1 Dyu 2 k 1 ywydy. s s Integrating again over t, H gives MH H h 1 µ h 1 C 1 + H s H eydy ds + λ2 k 1 H 15 t H Dsu 2 k 1 s w sds Dyu 2 k 1 y w ydy.

16 Since u k 1 > over, H Lemma 4.2, u k1 is increasing, and we have MH H h 1 µ h 1 C 1 + H s eydy ds + λ2 k 1 H H Now, we focus on the second term on the right hand side. 17 H Du 2 k 1 Dũ 2 k 1 ũ k 1 H ũ 2 k 1 ds + λ2 k 1 H Using the estimates in Lemma 4.3, we find Dtu 2 k 1 t w t Dyu 2 k 1 y w ydy. H H s eydy ds Du 2 k 1 Dũ 2 1 k 1 ũ k1 dy + H h 1 h 1 C. 18 H H s eydy h ds C 1 h 1 C 1 + h 2 h 2 C 1. Combining inequalities 17 and 18, leads to 19 Dtu 2 k 1 t w t C 1 h 1 h 1 C 1 + h 2 h H 2 C 1 + C 2 Dyu 2 k 1 y w ydy, for t H. t Using Gronwall s inequality we get Dtu 2 k 1 t w t C 1 e C 2 H Dyu2 k 1 ydy h 1 h 1 C 1 + h 2 h 2 C 1 for t H. Finally, we obtain the following estimate. 2 u 2 k 1 t w t D 1 C 1 e C H 2 Dyu2 k ydy 1 h 1 h 1 C 1 + h 2 h 2 C 1, The following Lemma is a direct consequence of the previous inequality. Lemma 4.4. Under the assumptions of Theorem 4.1, there exists a constant C = Cθ, µ, D, k 1, k 2, M, L, H such that the following inequality holds. u k1 u k1 ũ k1 C C h 1 h 1 C 1 + h 2 h 2 C 1., 16

17 Proof. Theorem 4.1 Recall that h 1 = µ a u k1 and h 1 = µ a ũ k1 over, H. Therefore Lemma 4.4 implies u 2 k 1 µ a µ a u k1 h 1 h 1 + µ a u k1 u k1 ũ k1. 21 u 2 k 1 µ a µ a C C h 1 h 1 C 1 + h 2 h 2 C 1. A simple calculation yields over, H. 22 Lemma 4.3 and 4.4 leads to u 2 k 1 D D D u 2 k 1 ũ 2 k 1 + Du 2 k 1 Dũ 2 k 1, u 2 k 1 D D C C h 1 h 1 C 1 + h 2 h 2 C 1. Applying the bounds in Lemma 4.1, we obtain the wanted results. 5 Proof of Theorem 2.1 The main idea here is to combine the stability results of the acoustic and optic inversions in a result that shows how the reconstruction of the optical coefficients is sensitive to the noise in the measurements of the acoustic waves. The principal difficulty is that the vector spaces used in both stability estimates are not the same due to the difference in the techniques used to derive them. We will use interpolation inequality between Sobolev spaces to overcome this difficulty. We deduce from the uniform bound on the solutions u i, i = 1, 2 see for instance 11 in the proof of Lemma 4.1 that 23 h i H 3, h i H 3 Mb, i = 1, 2, for all pairs D, µ a and D, µ a in O M. The Sobolev interpolation inequalities and embedding theorems [1] imply h i h i C 1 C h i h i H 2 C h i h i 1 2 H 1 h i h i 1 2 H 3, i = 1, 2, 17

18 which combined with 23 gives 24 h i h i C 1 C hi h i 1 2 H 1, i = 1, 2. Since the acoustic inversion is linear we obtain from Theorem 3.1 or Proposition 3.3 that, H λ 2 k i h i h i 2 CM dy T 2θH + β t p i H, t t p i H, t 2 dt +λ 2 k i p i H, t p i H, t 2 dt, for i = 1, 2, and H for i = 1, 2. c 2 y h i h i 2 dy CM T 2θH + β t p i H, t t p i H, t 2 dt +λ 2 k i p i H, t p i H, t 2 dt, Consequently, h i h i C 1 1 CM C T 2θH + β t p i H, t t p i H, t 2 + λ 2 k i p i H, t p i H, t 2 4 dt, for i = 1, 2. C and C Using the optical stability estimates in Theorem 4.1, we obtain 2 i=1 2 i=1 which ends the proof. u 2 mµ a µ a C 1 4 CM T 2θH + β t p i H, t t p i H, t 2 + λ 2 k i p i H, t p i H, t 2 dt, u 2 md D C 1 4 CM T 2θH + β t p i H, t t p i H, t 2 + λ 2 k i p i H, t p i H, t 2 dt, 18

19 6 Proof of Proposition 3.2 Multiplying the first equation of the system 4 by y y py, t and integrating by part one time over, T, we obtain H H y ph, t 2 dt = y py, t 2 dydt 2 c 2 tt py, ty y py, tdydt 2λ 2 k H py, ty y py, tdydt = A 1 + A 2 + A 3. In the rest of the proof we shall derive bounds of each of the constants A i, i = 1, 2, 3, in terms of the energy E. Due to the energy decay 6, we have A 1 T E. Integrating by part again over, T in the integral A 2, we get H A 2 = yc 2 y t py, t 2 dydt H H +2 yc 2 t py, T y py, T dy 2 yc 2 t py, y py, dy. Integrating by part now over, H, we find A 2 + Hc 2 H t ph, t 2 dt = H +2 yc 2 t py, T y py, T dy 2 H H y yc 2 t py, t 2 dydt yc 2 t py, y py, dy, which leads to the following inequality A 2 + Hc 2 H t ph, t 2 dt c 2 y yc 2 y L T E + H c 1 y ET + E. Using the energy decay 6, we finally obtain A 2 + Hc 2 H t ph, t 2 dt H c 2 W 1, T + 2H c 2 1/2 L E. c m Similar arguments dealing with the integral A 3 show that A 3 Hλ k T E. Combining all the previous estimates on the constants A i, i = 1, 2, 3, achieve the proof. 19

20 7 Proof of Proposition 3.3 Let θ = c 2 L and T > 2θH, and introduce the following function θy Φy= = θy θy θy c 2 H y t ph y, t 2 + y ph y, t 2 + λ 2 k ph y, t 2 dt, ϕy, tdt, for y H. We remark that 25 Φ = c 2 H + β 2 t ph, t 2 dt + λ 2 k ph, t 2 dt. On the other hand a forward calculation of the derivative of Φy gives Φ y = θy θy y ϕy, tdt θϕy, T θy θϕy, θy. Integrating by parts in the integral we deduce that θy Φ y = B θ y + y c 2 H y t ph y, t 2 dt, θy where B θ y = 2c 2 H y t ph y, t y ph y, t t=t θy θ c 2 H y t ph y, t 2 + y ph y, t 2 + λ 2 k ph y, t 2 t=t θy. The choice of θ implies B θ y < for y H. Hence, we obtain t=θy θy Φ y c 2 H y y c 2 H y c 2 H y t ph y, t 2 dt θy c 2 H y y c 2 H y Φy. Using Gronwall s inequality we get t=θy 26 Φy e H c2 s yc 2 s ds Φ, for y H. We deduce from the energy decay 6 that 27 T 2θHET T 2θHET θh θh θh Etdt. 2

21 Rewriting now the right hand side in terms of the function ϕ we found θh θh Etdt = H θh θh ϕy, tdtdy. Since θh, T θh θy, T θy for all y H, we have 28 θh θh Etdt H Φydy. Combining inequalities , we find 29 T 2θHET He H c2 s yc 2 s ds Φ. Back again to the energy derivative 6, and integrating the equality over, T we obtain E = ET + β The last equality and energy estimate 29 give t ph, t 2 dt. E T 2θH 1 He H c2 s yc 2 s ds Φ + β t ph, t 2 dt. Substituting Φ by its expression in 25 we finally find T 2θH 1 He H c2 s yc 2 s ds c 2 H + β 2 + β which combined with the fact that +λ 2 k E t ph, t 2 dt ph, t 2 dt, finishes the proof. E = H c 2 y f 1 y 2 + f y 2 + λ 2 k f y 2 dy, Acknowledgments The work of KR is partially supported by the US National Science Founda- tion through grant DMS The research of FT was supported in part by the LabEx PERSYVAL- Lab ANR-11-LABX FT would like to thank the Institute of Computational Engineering and Sciences ICES for the provided support during his visit. 21

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