Increasing stability in an inverse problem for the acoustic equation

Transcription

1 Increasing stability in an inverse problem for the acoustic equation Sei Nagayasu Gunther Uhlmann Jenn-Nan Wang Abstract In this work we study the inverse boundary value problem of determining the refractive index in the acoustic equation. It is known that this inverse problem is ill-posed. Nonetheless, we show that the ill-posedness decreases when we increase the frequency and the stability estimate changes from logarithmic type for low frequencies to a Lipschitz estimate for large frequencies. 1 Introduction In this paper we study the issue of stability for determining the refractive index in the acoustic equation by boundary measurements. It is well known that this inverse problem is ill-posed. However, one anticipates that the stability will increase if one increases the frequency. This phenomenon was observed numerically in the inverse obstacle scattering problem [5]. Several rigorous justifications of the increasing stability phenomena in different settings were obtained by Isakov et al [6, 7, 8, 10, 11]. Especially, in [8], Isakov considered the Helmholtz equation with a potential u k u + qu = 0 in Ω. 1.1 Department of Mathematical Science, Graduate School of Material Science, University of Hyogo, 167 Shosha, Himeji, Hyogo , Japan. sei@sci.u-hyogo.ac.jp Department of Mathematics, University of Washington, Box , Seattle, WA and Department of Mathematics, University of California, Irvine, CA , USA. gunther@math.washington.edu Department of Mathematics, NCTS Taipei, National Taiwan University, Taipei 106, Taiwan. jnwang@math.ntu.edu.tw 1

2 He obtained stability estimates of determining q by the Dirichlet-to-Neumann map for different ranges of k s. All of these results demonstrate the increasing stability phenomena in k. For the case of the inverse source problem for Helmholtz equation and an homogeneous background it was shown in [3] that the ill-posedness of the inverse problem decreases as the frequency increases. In this paper, we study the acoustic wave equation. Let Ω R n be a bounded domain, where n 3. Let Ω be smooth. We consider the equation + k qx ux = 0 in Ω, 1. where the real-valued qx is the refractive index. Assume that the kernel of the operator + k qx on H0Ω 1 is trivial. Associated with 1., we define the Dirichlet-to-Neumann map DN map Λ : H 1/ Ω H 1/ Ω by Λf = u ν, Ω where u is the solution to 1. with the Dirichlet condition u = f on Ω, and ν is the unit outer normal vector of Ω. The uniqueness of this inverse problem is well known [13]. This inverse problem is notoriously ill-posed. For this aspect, Alessandrini proved that the stability estimate for this problem is of log type [1] and Mandache showed that the log type stability is optimal [9]. In this paper, we would like to focus on how the stability behaves when the frequency k increases. Now we state the main result. Theorem 1.1. Assume that q 1 x and q x are two sound speeds with associated DN maps Λ 1 and Λ, respectively. Let s > n/ + 1, M > 0. Suppose q l H s Ω M l = 1, and suppq 1 q Ω. Denote q a zero extension of q 1 q. Then there exists a constant C 1, depending only on n, s, and Ω, such that if k 1/C 1 M and Λ 1 Λ 1/e then q H s R n C s n k expck Λ 1 Λ + C k 1 + log Λ 1 Λ 1.3 holds, where C > 0 depends only on n, s, Ω, M and suppq 1 q. Here is the operator norm from H 1/ Ω into H 1/ Ω. Remark The estimate 1.3 consists two parts Lipschitz and logarithmic estimates. As k increases, the logarithmic part decreases and the Lipschitz part

3 becomes dominated. In other words, the ill-posedness is alleviated when k is large.. We would like to remark on the constant C expck /k appearing in the Lipschitz part of /k comes from k q in the equation, which appears naturally, while, expck is due to the fact that we use the complex geometrical optics solutions in the proof. Even so, we expect that the there is an exponential growth of the constant with frequency since we do not assume any geometrical restriction on qx other than regularity. For the wave equation it has been shown by Burq for the obstacle problem [4] that the local energy decay is log-slow and this is due to the presence of trapped rays. Notice that in our case we can have trapped rays. For the case of simple sound speeds we expect that there is no exponential increase in the constant. In [1] a Hölder stability estimate was obtained for the hyperbolic DN map for generic simple metrics. For very general metrics there is not known modulus of continuity for the hyperbolic DN map, see [] for convergence results. However, in practice, k is fixed and so is the constant. Therefore, one should expect to obtain a better resolution of q from boundary measurements when the chosen k is large. 3. Unlike the result in [8, Theorem.1] for equation 1.1 where the stability estimates were derived in different ranges of k, estimate 1.3 is valid for all range of k provided k 1/C 1 M. The proof of Theorem 1.1 makes use of Alessandrini s arguments [1] and the CGO solutions constructed in [13]. The main task is to keep track of how k appears in the proof of the stability estimates. Complex geometrical optics solutions In this section, we construct CGO solutions to the equation 1. by using the idea in [13]. The main point is to express the dependence of constants on k explicitly. We first state two easy consequences from the results in [13]. Lemma.1 see [13, Proposition.1 and Corollary.]. Let s 0 be an integer. Let ε 0 > 0. Let ξ C n satisfy ξ ξ = 0 and ξ ε 0. Then for any f H s Ω there exists w H s Ω such that w is a solution to w + ξ w = f in Ω 3

4 and satisfies the estimate w H s Ω C 0 ξ f H s Ω, where a positive constant C 0 depends only on n, s, ε 0 and Ω. By using this lemma, we can obtain a solution to the equation ψ + ξ ψ + gψ = f.1 satisfying some decaying property as in the following lemma. Lemma. [13, Theorem.3 and Corollary.4]. Let s > n/ be an integer. Let ε 0 > 0. Let ξ C n satisfy ξ ξ = 0 and ξ ε 0. Let f, g H s Ω. Then there exists C 1 > 0 depending only on n, s, ε 0 and Ω such that if ξ C 1 g H s Ω then there exists a solution ψ H s Ω to the equation.1 satisfying the estimate ψ H s Ω C 0 ξ f H s Ω, where C 0 is the positive constant in Lemma.1. The needed CGO solutions are constructed as follows. Proposition.3. Let s > n/ be an integer. Let ε 0 > 0. Let ξ C n satisfy ξ ξ = 0 and ξ ε 0. Define the constants C 0 and C 1 as in Lemma.. Then if ξ C 1 k q H s Ω then there exists a solution u to the equation 1. with the form of ξ 1 ux = exp x + ψx,. where ψ has the estimate ψ H s Ω C 0k ξ Proof. Substituting. into 1., we have q H s Ω. ψ + ξ ψ + k qψ = k q. Then by Lemma., we obtain this proposition. 4

5 3 Proof of stability estimate This section is devoted to the proof of Theorem 1.1. We begin with Alessandrini s identity. Proposition 3.1. Let u l be a solution to 1. with q = q l, then we have k q q 1 u 1 u dx = Λ 1 Λ u 1 Ω, u Ω. Ω Now we would like to estimate the Fourier transform of the difference of two q s. We denote Ff the Fourier transformation of a function f. Lemma 3.. Let s > n/ + 1 be an integer and M > 0. Assume q l H s Ω M, suppq 1 q Ω and k 1/C 1 M, where C 1 is the constant defined in Lemma. corresponding to ε 0 = 1. Let q be a zero extension of q 1 q and a 0 C 1. Suppose that χ C 0 Ω satisfies χ 1 near suppq 1 q. Then for r 0 and η R n with η = 1 the following statements hold: if 0 r a 0 k M then F qrη C χ H s Ω a 0 q H s R n + C k expca 0k M Λ 1 Λ 3.1 holds; if r C 1 k M then F qrη CMk χ H s Ω q H r s R n + C k expcr Λ 1 Λ 3. holds, where C > 0 depends only on n, s and Ω. Proof. In the following proof, the letter C stands for a general constant depending only on n, s and Ω. By Proposition.3, we can construct CGO solutions u l x to the equation 1. with q = q l having the form of ξl 1 u l x = exp x + ψl x for l = 1,, and we have 1 q q 1 exp ξ 1 + ξ x 1 + ψ 1 + ψ + ψ 1 ψ dx Ω = 1 k Λ1 Λ u 1 Ω, u Ω 3.3 5

6 from Proposition 3.1, where ψ l satisfies ψ l H s Ω Ck ξ l q l H s Ω if ξ l C n satisfies ξ l ξ l = 0, ξ l 1 and ξ l C 1 k q l H s Ω. 3.4 We remark that ψ l H s Ω C also holds. Indeed, we have ψ l H s Ω Ck q l H s Ω ξ l Ck q l H s Ω C 1 k q l H s Ω = C C 1 = C. Now, let r 0, and η R n satisfy η = 1. We assume that α, ζ R n satisfy α η = α ζ = η ζ = 0 and ζ = α + r. 3.5 Define ξ 1 and ξ as ξ 1 = ζ + iα irη and ξ = ζ iα irη. Then we have ξ l ξ l = 0, ξ l = ζ + α + r = ζ l = 1, and 1 ξ 1 + ξ = irη. Hence by 3.3, we immediately obtain that F qrη = q q 1 exp irη xψ 1 + ψ + ψ 1 ψ dx Ω + 1 k Λ1 Λ u 1 Ω, u Ω 3.6 provided ξ l 1 and 3.4 are satisfied. We first estimate the first term on the 6

7 right hand side of 3.6 by q q 1 exp irη xψ 1 + ψ + ψ 1 ψ dx Ω = q q 1 exp irη xχψ 1 + ψ + ψ 1 ψ dx Ω q q 1 H s Ω χψ1 + ψ + ψ 1 ψ H s Ω q H s R n χ H Ω s ψ1 H s Ω + ψ H s Ω + ψ 1 H s Ω ψ H s Ω Ck q H s R n χ H s Ω + Ck + C Ck q l H ζ ζ ζ s Ω l=1 = Ck χ H s Ω q H ζ s R n q l H s Ω. l=1 since χψ 1 + ψ + ψ 1 ψ H0Ω s and s > n/. On the other hand, by taking R large enough such that Ω B R 0, we have ul L Re Ω Ω Ω 1/ u l C 0 Ω Ω 1/ ξl 1 exp R + ψl L Ω Re ξl 1 C exp R + ψl H s Ω Re ξl ζ C exp R 1 + C = C exp R. Likewise, we can get that u l L Ω = ξ l Ω u ξl l + exp ψ l ζ ζ C exp R ζ ζ C ζ exp + C exp C ζ exp C expc ζ ζ R L Ω ζ R ψ l C 0 Ω + Ω 1/ exp R R ψ l H s 1 Ω ζ + C exp R ψ l H s Ω 7

8 since s 1 > n/. Consequently, we have ul Ω H 1/ Ω C expc ζ. Therefore, we can estimate the second term of the right-hand side of 3.6 by Λ 1 Λ u 1 Ω, u Ω Λ1 Λ u1 H Ω u H 1/ Ω Ω 1/ Ω C expc ζ Λ 1 Λ. Summing up, we have shown that for r > 0 and for η R n with η = 1 if we take α and ζ satisfying the conditions 3.5, ζ 1/ and then ζ 1/ C 1 k q l H s Ω 3.7 F qrη Ck χ H s Ω q H ζ s R n q l H s Ω l=1 + C k expc ζ Λ 1 Λ 3.8 holds. Now assume that q l H s Ω M and k 1/C 1 M. Thus if ζ C 1 k M 3.9 holds, then 3.7 and ζ 1/ are satisfied. Pick a 0 C 1. We first consider the case where 0 r a 0 k M. By choosing α and ζ satisfying α η = α ζ = η ζ = 0, ζ = a 0 k M r and α = a 0 k M r both 3.5 and 3.9 are then satisfied since a 0 C 1. Hence we obtain 3.8, that is 3.1. On the other hand, when r C 1 k M, we can choose α = 0, η ζ = 0 and ζ = r. Then 3.5, 3.9 are satisfied and thus 3.8 holds and consequently 3. is valid. Now we prove our main result. 8

9 Proof. As above, C denotes a general constant depending only on n, s and Ω. Written in polar coordinates, we have q H s R n = C F qrη 1 + r s r n 1 dη dr 0 η =1 a0 k M = C T a 0 k M T η =1 η =1 F qrη 1 + r s r n 1 dη dr η =1 F qrη 1 + r s r n 1 dη dr F qrη 1 + r s r n 1 dη dr =: CI 1 + I + I 3, 3.10 where a 0 C 1 and T a 0 k M are parameters which will be chosen later. Here C 1 is the constant given in Lemma 3.. From now on, we take k 1/C 1 M. Our task now is to estimate each integral separately. We begin with I 3. Since F qrη C q 1 q L Ω, q 1 q H0Ω, s and s > n/, we have that I 3 C T q 1 q L Ω 1 + r s r n 1 dr CT m q 1 q L Ω CT ε q m 1 q H s Ω + C ε q 1 q H s Ω CT m ε q H s R n + M ε for ε > 0, where m := s n. On the other hand, by Lemma 3., we can estimate 3.11 I 1 C C a0 k M 1 + r s r n 1 dr 0 [ ] χ H s Ω q H s R n + expca 0k M Λ 1 Λ 0 a 0 a 0 k 4 [ C 1 + r s r n 1 χ dr q H s R n + expca ] 0k M Λ k 4 1 Λ = CC χ q a H s R n + C expca 0k M Λ 0 k 4 1 Λ, 3.1 9

10 where χ C0 Ω satisfies χ 1 near suppq q 1 and C χ := χ H s Ω. In view of T T 1 + r s r n 3 dr r s+n 3 dr Ca 0 k M s+n and T a 0 k M a 0 k M we have that a 0 k M Ca 0 k M C 1 k M m T expcr1 + r s r n 1 dr expct I CM k 4 χ H s Ω q H s R n + C k 4 Λ 1 Λ T expct a 0 k M 0 C expct, a 0 k M T a 0 k M C a 0k 4 M 1 + r s r n 1 dr 1 + r s r n 1 dr 1 + r s r n 3 dr expcr1 + r s r n 1 dr CC χ q a H s R n + C 0 k expct Λ 4 1 Λ Combining gives q H s R n CI 1 + I + I 3 CC χ a 0 q H s R n + C expca 0k M Λ k 4 1 Λ + CC χ a 0 C Cχ = q H s R n + C k expct Λ 4 1 Λ + CT m ε q H s R n + M a 0 ε + C 3 T m ε q H s R n + C expca0 k M + expct Λ k 4 1 Λ + CM T m, ε 10

11 where positive constants C and C 3 depend only on n, s and Ω. Now we pick a 0 and ε as a 0 = C C χ C 1 and ε = T m 4C 3 if needed, we take C large enough. We then obtain that q H s R n C k 4 [ expc CC χ k M + expct ] Λ 1 Λ + CT m M = C k 4 expcak A + CΦT 3.14 for T a 0 k M = C C χ k M = ak, where ΦT := 1 k 4 expc 4T A + M T m, A := Λ 1 Λ, a := C C χ M and C 4 > 0 depends only on n, s and Ω. To continue, we consider two cases: ak p log 1 A 3.15 and ak p log 1 A, 3.16 where p will be determined later see 3.4. For the first case 3.15, our aim is to show that there exists T ak such that ΦT C 5 k + log 1 A m Substituting 3.17 into 3.14 clearly implies 1.3. Now to derive 3.17, it is enough to prove that 1 k expc 4T A C 4 5 k + log A 1 m 3.18 and M T m C 5 k + log 1 A m

12 Remark that 3.19 in equivalent to T C 1/m 5 M 1/m k + log 1, A which holds if T C 1/m 5 M 1/m 1 + p log 1 a A 3.0 because of Setting T = p log1/a ak by 3.15, then 3.0 holds provided p C 1/m 5 M 1/m 1 + p. 3.1 a Now we turn to It is clear that 3.18 is equivalent to C 4 p log 1 A log C 5 + log k + log 1 A m log k + log 1 3. A since T = p log1/a. It follows from 3.15 that log k + log 1 p log A a log 1 A + log 1 A p = log a log log 1 A. Hence 3. is verified if we can show that C 4 p log 1 A log C 5 logmc 1 + log 1 p log A m a log log 1, A i.e. 1 C 4 p log 1 A m log log 1 p A + log C 5 logmc 1 m log a for log1/a 1. Now we choose Then 3.3 becomes 3.3 p = 1 C log 1 A 4m log log 1 p A + log C 5 4 logmc 1 4m log a

13 Notice that inf 0<A 1/e log 1 A 4m log log 1 A Hence if we choose C 5 such that = infz 4m log z z 1 e infz 4m log z = 4m log z>0 4m. C 5 MC 1 p a + 1 m 4m e m 3.6 then 3.5 follows. Finally, we take { } m 4m C 5 := max C1, p m M 1 + p m, e a which depends only on n, Ω, s, M and χ. With such choice of C 5, the conditions 3.6 and 3.1 hold, and thus estimate 3.17 is satisfied. Next we consider the second case By 3.14 with T = ak, we get that q H s R n C k 4 expcak A + C k 4 expc 4ak A + CM ak m Hence it remains to show that C k 4 expcak A + CM a m k 4m. k 4m C 6 k + log 1 A m, i.e. Since k C 1/m 6 k + log A k + log 1 A 1 + a k p by 3.16, we have 3.7 if we take C 6 large enough so that C a m. p The proof is completed. 13

14 Acknowledgements Nagayasu was partially supported by Grant-in-Aid for Young Scientists B. Uhlmann was partly supported by NSF and a Visiting Distinguished Rothschild Fellowship at the Isaac Newton Institute. Wang was partially supported by the National Science Council of Taiwan. We would also like to thank P. Stefanov for helpful discussions. References [1] G. Alessandrini, Stable determination of conductivity by boundary measurements, Appl. Anal., , [] M. Anderson, A. Katsuda, Y. Kurylev, M. Lassas, M. Taylor, Boundary regularity for the Ricci equation, Geometric Convergence, and Gel fand s Inverse Boundary Problem, Inventiones Mathematicae, , [3] G. Bao, J. Lin and F. Triki, A multi-frequency inverse source problem, J. Diff. Eq., 49, 010, [4] N. Burg, Decay of the local energy of the wave equation for the exterior problem and absence of resonance near the real axis, Acta Math., , 1-9. [5] D. Colton, H. Haddar, and M. Piana, The linear sampling method in inverse electromagnetic scattering theory, Inverse Problems, , S105-S137. [6] T. Hrycak and V. Isakov, Increased stability in the continuation of solutions to the Helmholtz equation, Inverse Problems, 0 004, [7] V. Isakov, Increased stability in the continuation for the Helmholtz equation with variable coefficient, Control methods in PDE-dynamical systems, 55V67, Contemp. Math., 46, AMS, Providence, RI, 007. [8] V. Isakov, Increasing stability for the Schrödinger potential from the Dirichlet-to-Neumann map, DCDS-S, 4 011, [9] N. Mandache, Exponential instability in an inverse problem for the Schrödinger equation, Inverse Problems, ,

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