AN INVERSE PROBLEM FOR THE WAVE EQUATION WITH A TIME DEPENDENT COEFFICIENT

Size: px

Start display at page:

Download "AN INVERSE PROBLEM FOR THE WAVE EQUATION WITH A TIME DEPENDENT COEFFICIENT"

Alban Hardy
6 years ago
Views:

1 AN INVERSE PROBLEM FOR THE WAVE EQUATION WITH A TIME DEPENDENT COEFFICIENT Rakesh Department of Mathematics University of Delaware Newark, DE A.G.Ramm Department of Mathematics Kansas State University Manhattan, KS 6656

2 Abstract Let Ω R n, n 2, be a bounded domain with smooth boundary. Consider u tt x u + q(x, t)u = in Ω [, T ] u(x, ) =, u t (x, ) = if x Ω u(x, t) = f(x, t) on Ω [, T ] We show that if u and f are known on Ω [, T ], for all f C ( Ω [, T ]), then q(x, t) may be reconstructed on C = { (x, t) : x Ω, < t < T, x tω & x + (T t)ω Ω ω R n, ω = 1 } provided q is known at all points not in C. If u, f are known on Ω [, T ], and u is known on t = T, for all f C ( Ω [, T ]) then q(x, t) may be reconstructed on D = { (x, t) : x Ω, < t < T, x tω Ω ω R n, ω = 1} provided q is known at all points not in D.

3 Let Ω R n be a bounded domain with smooth boundary, and q(x, t) a function on Ω. Suppose u(x, t) satisfies u + q(x, t)u u tt x u + q(x, t)u = in Ω [, T ] (1) u(x, ) =, u t (x, ) = if x Ω u(x, t) = f(x, t) on Ω [, T ] For a fixed q(x, t), (1) is a well posed initial boundary value problem, hence one may define the Dirichlet to Neumann map and the boundary to final data map We prove the following two results - Λ q : H 1 ( Ω [, T ]) L 2 ( Ω [, T ]) f(x, t) u n Γ q : H 1 ( Ω [, T ]) H 1 (Ω) L 2 (Ω) f(x, t) ( u(x, T ), u t (x, T ) ) THEOREM A If n > 1, and q(x, t) is in C 1 (Ω [, T ]), then knowing Λ q we can reconstruct q(x, t) on the region C = { (x, t) : x Ω, < t < T, x tω & x + (T t)ω Ω ω R n, ω = 1 } provided q(x, t) is known at all points not in C. C is region III in the figure, and it is a subset of Ω [, T ] consisting of those points through which every line making an angle of 45 o with the vertical meets the planes t = and t = T outside Ω. THEOREM B If n > 1, and q(x, t) is in C 1 (Ω [, T ]), then knowing Λ q and Γ q we can reconstruct q(x, t) on the region D = { (x, t) : x Ω, < t < T, x tω Ω ω R n, ω = 1 } provided q(x, t) is known at all points not in D. D is region III IV V in the figure, and is a subset of Ω [, T ] consisting of those points through which every line making an angle of 45 o with the vertical meets the planes t = outside Ω. 1

4 V IV IV III II II I REMARKS In [5] Rakesh and Symes showed that Λ q uniquely determines q provided q is independent of t. Romanov in [7] also proved a uniqueness result for a related problem in the half plane when the (known) wave speed is of special form, again when q is independent of t. [4] contains a reconstruction formula for recovering q from Λ q, again, provided q is independent of t. We refer the reader to [5] for other articles related to the time independent case. In [3] Isakov proved a theorem related to Theorem B. He proved uniqueness for q(x, t) over the region Ω [, T ], except his data was the response of the medium not just for zero initial data (and all possible Dirichlet data), as in Theorem B, but all possible initial data (and all possible Dirichlet data). So he proved uniqueness over a larger region but he needed much more information. The proof of Theorem B may be easily modified to prove Isakov s result. Given Λ q and Γ q, from domain of dependence arguments it is clear that with zero 2

5 initial data one cannot hope to recover q(x, t) over the region I given by { (x, t) : x Ω, < t < T, dist(x, Ω) < t } But our inability to determine q on region II, we feel, is just a shortcoming of our technique. If we are given only Λ q, the response measured on the boundary, then from domain of dependence arguments one can see that we cannot hope to recover q(x, t) on the subregion I V given by { (x, t) : x Ω, < t < T, dist(x, Ω) < t, or dist(x, Ω) < T t } However, we feel that one should be able to recover q(x, t) on regions II and IV, but we have not succeeded in doing so. The derivations of the two results start from an identity which was motivated by one in [6] for elliptic differential equations. Alessandrini, Sylvester and Uhlmann (for elliptic problems), and Ziqi Sun (for hyperbolic problems) amongst others, have used similar identities. We also draw upon ideas of Sylvester and Uhlmann in [1] and upon the work of the first author with Bill Symes in [5]. NOTE For the rest of this article we will assume that q(x, t) is in C 1 (Ω [, T ]), and q(x, t) is defined to be zero outside Ω [, T ]. DERIVATION OF IDENTITY Suppose (2) u + q(x, t)u = in Ω [, T ] u(x, ), u t (x, ) = if x Ω and (3) Then using an integration by parts Ω [,T ] dx dt q(x) u v = Ω [,T ] dx dt u v v = in Ω [, T ] = Ω [,T ] dxdt u v + Ω [,T ] 3 ds (v u n u v n )

6 (4) + Ω {t=t } dx (uv t vu t ) = Ω [,T ] ds (v u n u v n ) + Ω {t=t } dx (uv t vu t ) We construct special solutions of (2) and (3) as described in the following Lemma which is proved later. LEMMA 1 Given a unit vector ω R n, σ >, and χ C (R n ) with (5) supp χ Ω = φ we can construct special solutions of (2) and (3) of the form u = χ(x + tω)e +iσ(x ω+t) + R 1 v = χ(x + tω)e iσ(x ω+t) + R 2 with R 1, R 2 zero on Ω [, T ], (R 1, R 1t ) zero on t =, (R 2, R 2t ) zero on t = T, and (6) R i (x, t) L 2 (Ω [,T ]) C i = 1, 2 σ where C depends only on Ω, T, q C 1 (Ω [,T ]), and χ C 3 (Ω). With such u and v we have Ω [,T ] dx dt q(x, t) u v = Ω [,T ] χ2 (x + tω)q(x, t) + Ω [,T ] χ(x + tω)e iσ(x ω+t) q(x, t)r 1 (7) + Ω [,T ] q(x, t)r 1R 2 + Ω [,T ] χ(x + tω)e+iσ(x ω+t) q(x, t)r 2 So for q(x, t) bounded on Ω [, T ] we have from (4), (6), and (7) Ω [,T ] χ2 (x + tω)q(x, t) { u = lim σ Ω [,T ] ds (v n u v n ) + } (8) Ω {t=t } dx (uv t vu t ) for any ω R n, ω = 1, and any χ C (R n ) satisfying (5). Given any point a Ω, we can find a sequence of χ ɛ C (R n ) satisfying (5) and so that, as ɛ approches zero, χ 2 ɛ approaches δ(x a) as distributions on R n. So if q(x, t) is defined to be zero outside Ω [, T ] then (8) implies T dt q(a tω, t) = lim ɛ lim σ { Ω [,T ] 4 ds (v u n u v n ) + Ω {t=t } dx (uv t vu t ) }

7 But q(x, t) is zero for t < or t > T. So if we define f(x, t) u = χ ɛ (x + tω)e iσ(x ω+t) (x, t) Ω [, T ] then (9) dt q(a tω, t) = lim ɛ lim σ { Ω [,T ] ds (vλ q(f) f v n ) + Ω {t=t } dx (uv t vu t ) } for any ω R n, ω = 1, and any a Ω. We shall need the following lemma proved later in this article. LEMMA 2 If q(x, t) is a bounded, measurable function on R n R, with compact support, then knowing the light ray transform Rq(ω, a) = we can reconstruct q. PROOF OF THEOREM A dt q(a tω, t) ω R n, ω = 1, a R n Let { (a tω, t) : < t < } be a 45 o line through a point in C. Then from the definition of C we have that this line meets t = and t = T outside Ω i.e. a Ω and a T ω Ω. So we can construct a sequence of functions χ ɛ in C (R n ) such that suppχ ɛ Ω = φ, suppχ ɛ T ω Ω = φ and χ 2 ɛ approaches δ(x a) as distributions in R n. So the v constructed by Lemma 1 and used in (9) may be chosen so that v, v t are zero on t = T. So if Λ q is known then (9) implies that Rq(ω, a) is known for all 45 o lines which go through a point of C. For lines not going through any point of C, Rq(ω, a) is known anyway because q is known at all points outside C (from the hypothesis of Theorem A). Hence Rq(ω, a) is known for all a R n and all unit vectors ω R n. Hence q is determined on C by Lemma 2. PROOF OF THEOREM B Since v in (3) is independent of q, if Λ q and Γ q are known then from (9), Rq(ω, a) is known a Ω and ω R n, ω = 1. For a Ω the line { (a tω, t) : < t < } 5

8 lies in the complement of D. From the hypothesis of Theorem B, q is known at all points not in D. So in fact Rq(ω, a) is known is known ω R n, ω = 1, and a R n. So Lemma 2 allows us to recover q on D. PROOF OF LEMMA 1 Our proof of Lemma 1 is a modification of the proof of a similar result in [5] which dealt with the situation where q did not depend on t. To prove our Lemma it would be enough to show that if R 1 + q(x, t)r 1 = ( + q) ( χ(x + tω) e +iσ(x ω+t)) in Ω [, T ] (1) and R 1 (x, ) =, t R 1 (x, ) = if x Ω R 1 = on Ω [, T ] R 2 = ( χ(x + tω) e iσ(x ω+t)) in Ω [, T ] R 2 (x, T ) =, t R 2 (x, T ) = if x Ω R 2 = on Ω [, T ] then the estimates (6) hold. We shall prove the estimate for R 1, and the R 2 case may be handled in a similar fashion. Let us define (11) φ(x, t) = ( + q) ( χ(x + tω) e +iσ(x ω+t)) = e +iσ(x ω+t) ( + q)χ(x + tω) From estimates for hyperbolic initial boundary value problems as in [2], for any w H 1 (Ω [, T ]) with (w, w t ) zero on t =, w zero on Ω [, T ], and large τ τ Ω [,T ] dx dt e 2τt w t (x, t) 2 C Ω [,T ] dx dt e 2τt w(x, t) 2 with C depending only on Ω and T. So taking w(x, t) = t ds R 1(x, s) in the previous equation and integrating equation (1) with respect to t one obtains τ Ω [,T ] dx dt e 2τt R 1 (x, t) 2 C { t t } Ω [,T ] dx dt e 2τt ds φ(x, s) 2 + ds q(x, s) R 1 (x, s) 2 C { t t } Ω [,T ] dx dt e 2τt ds φ(x, s) 2 + ds e τs q(x, s) R 1 (x, s) 2 6

9 { t C 1 Ω [,T ] dx dt e 2τt C 2 Ω [,T ] dx dt { e 2τt t ds φ(x, s) 2 + T ds φ(x, s) 2 + e 2τt R 1 (x, t) 2 } ds e 2τs q(x, s) 2 R 1 (x, s) 2 } using the boundedness of q. So by taking τ large enough one obtains (12) Ω [,T ] dx dt R 1(x, t) 2 C t Ω [,T ] dx dt ds φ(x, s) 2 Looking at the form of φ(x, t) in (11), we define χ 1 (x, t) Then t dsφ(x, s) = Using this in (12) we have χ 1 (x, t) = ( + q(x, t))χ(x, t) t = 1 iσ = 1 iσ ds χ 1 (x, s) e +iσ(x ω+s) t t ds χ 1 (x, s) d ds (e+iσ(x ω+s) ds χ 1 (x, s)e+iσ(x ω+s) s + 1 iσ χ 1(x, t)e +iσ(x ω+t) 1 iσ χ 1(x, )e iσx ω Ω [,T ] dx dt R 1(x, t) 2 C σ with C depending only on Ω, T, χ C 3 (R n ), and q C 1 (Ω [,T ]). PROOF OF LEMMA 2 Since q has compact support and is bounded, from Fubini s theorem R R n n da Rq(ω, a)e ia ξ = da dt q(a tω, t) e ia ξ = dt da q(a tω, t) e ia ξ R n = dt da q(a, t) e ia ξ e itω ξ = ˆq(ξ, ω ξ) for all ξ R n and all unit vectors ω R n. Here ˆq is the fourier transform of q. So ˆq(ξ, τ) is known in { (ξ, τ) : ξ R n, τ R, τ < ξ }, which contains an open set 7 R n

10 in R n R. Since ˆq(ξ, τ) is analytic in the variables ξ and τ, the above information is enough to recover ˆq everywhere, hence q may be recovered. ACKNOWLEDGEMENTS: The second author thanks NSF and ONR for support for this work, and Professor J. Sjöstrand and Dr. P Stefanov for discussions related to this problem. Professor Sjöstrand and Dr. Stefanov showed the authors different proofs of uniqueness results similar to Theorem A for the case when q(x, t) C. References [1] G. Alessandrini. Stable determination of conductivity by boundary measurements, Appl. Anal., 27, (1988). [2] L. Hörmander. The Analysis of Linear Partial Differential Operators III, Springer- Verlag, New York, (1985). [3] Victor Isakov. Completeness of products of solutions and some inverse problems for PDE, preprint, (1989). [4] Rakesh. Reconstruction for an inverse problem for the wave equation, preprint, (1989). [5] Rakesh and William W. Symes. Uniqueness for an inverse problem for the wave equation, Comm. in PDE, 13(1), (1988). [6] A. G. Ramm. A uniqueness theorem for a boundary inverse problem, Inverse Problem, 4, L1-5 (1988). [7] V. G. Romanov. Integral Geometry and Inverse Problems for Hyperbolic Equations. Springer Verlag, New York, (1974). [8] J. Sjöstrand. private communication, (1989). [9] P. Stefanov. Uniqueness of the multidimensional inverse scattering problem for time dependent potentials, Math. Z., 21, (1989). [1] J. Sylvester and G. Uhlmann. A global uniqueness theorem for an inverse boundary value problem, Annals of Mathematics, 125 (1987), [11] Ziqi Sun. On continuous dependence of an inverse initial boundary value problem of wave equation, preprint, (1988). 8

On uniqueness in the inverse conductivity problem with local data

On uniqueness in the inverse conductivity problem with local data Victor Isakov June 21, 2006 1 Introduction The inverse condictivity problem with many boundary measurements consists of recovery of conductivity