Complex geometrical optics solutions for Lipschitz conductivities

Transcription

1 Rev. Mat. Iberoamericana 19 (2003), Complex geometrical optics solutions for Lipschitz conductivities Lassi Päivärinta, Alexander Panchenko and Gunther Uhlmann Abstract We prove the existence of complex geometrical optics solutions for Lipschitz conductivities. Moreover we show that, in dimensions n 3 that one can uniquely recover a W 3/2, conductivity from its associated Dirichlet-to-Neumann map or voltage to current map. 1. Introduction Let Ω R n be a bounded domain with Lipschitz boundary. Let γ L (Ω) be the electrical conductivity of Ω. We assume throughout the paper that the conductivity is strictly positive on Ω. Given a voltage potential f on Ω, under the assumption of no sources or sinks of current on Ω, the induced potential u on Ω satisfies the Dirichlet problem (1.1) div(γ u) =0 on Ω u Ω = f. The Dirichlet-to-Neumann map (DN), or voltage to current map, is defined by ( (1.2) Λ γ (f) = γ u ) ν where u is a solution to (1.1) and ν denotes the unit outer normal to Ω Mathematics Subject Classification: 35R30 35Q60. Keywords: Electrical Impedance Tomography, Complex Geometrical Optics, Lipschitz Conductivities. Ω,

2 58 L. Päivärinta, A. Panchenko and G. Uhlmann The inverse problem of determining γ from Λ γ has been extensively studied since Calderón s pioneer paper [4]. The essential tool for the solution is the construction of complex geometrical optics solutions for the underlying conductivity equation. See [16] for a recent survey. Kohn and Vogelius proved that one can uniquely determine piecewise real-analytic conductivities from the DN map [7]. Sylvester and Uhlmann proved in dimension n 3 a global identifiability result for smooth conductivities [13]. Brown extended this result to conductivities in C 3/2+ε (Ω) for any ε>0, [2]. In two dimensions Nachman proved a global identifiability result for conductivities having two derivatives [8]. This was improved to Lipschitz conductivities in [3]. We point out that the above inverse conductivity problem makes sense for conductivities that are only in L. There are neither proofs nor counter-examples for this case in any dimension. As mentioned above a crucial ingredient in the proof of the results in [2], [3], [8], and [13] is the construction of complex geometrical optics solutions for the equation (1.1), [13, 12]. Let C n with =0. Moreover,letγ C 2 (R n ), n 2, with γ =1 for x R and R sufficiently large. Then for sufficiently large there exist solutions of div(γ u) =0inR n of the form (1.3) u = e x γ 1/2 (1 + ψ γ (x, )). In two dimensions these solutions are constructed for all C 2 \{0} with =0,[8]. The function ψ γ satisfies (1.4) ( + 2 ) ψ γ = q(1 + ψ γ ), where (1.5) q = γ γ. An important property of ψ γ is that ψ γ 0 in an appropriate sense so that the solutions of (1.4) behave like e x γ 1/2 for large. In this paper we construct complex geometrical optics solutions for conductivities γ W 1, (R n ). We sketch some of the steps in the construction.

3 Complex geometrical optics solutions for Lipschitz conductivities 59 We first rewrite (1.1) in the form (1.6) u + A u =0, where A = log γ L (R n ) has compact support. Let A ε = A Φ ε where Φ ε C0 (R n ), ε>0, is an approximation to the δ-function. We consider instead of (1.6) the Schrödinger equation with a smooth first order term (1.7) u ε + A ε u ε =0. In [9] complex geometrical optics solutions to (1.7) are constructed of the form u ε = e x e ϕ ε(x) 2 r ε (x, ), with r ε satisfying appropriate decay conditions as. The functions ϕ ε are defined by (1.8) ϕ ε = ϕ Φ ε. We remark that in the case that A ε = log α with α smooth, then e ϕ ε(x) 2 = α 1/2. Our complex geometrical optics solutions are of the form (1.9) u = e x (ω 0 (x, ε)+ω 1 (x, ε, )), with (1.10) ω 0 (x, ε) =e ϕ ε(x) 2. In Section 2 we see that ω 0 γ 1/2 as ε 0, in appropriate sense. Since ω 0 is smooth we have avoided the problem of taking two derivatives of γ. Now ω 1 solves the equation (1.11) (( + 2 )+A ( + ))ω 1 = (( + 2 )+A ( + ))ω 0. We show that ω 1 0asε 0, in appropriate sense. In this paper we show the existence of the complex geometrical solutions for all Lipschitz conductivities and give a global identifiability result for W 3/2, conductivities in dimension three and higher. This latter result improves on the result of [2].

4 60 L. Päivärinta, A. Panchenko and G. Uhlmann More precisely, we will prove: Theorem 1.1 Let γ W 1, (R n ), γ strictly positive and γ =1outside a large ball. Let 1 <δ<0. Then for sufficiently large there is a unique solution of div(γ u) = 0 in R n of the form (1.12) u = e x (γ 1/2 + ψ γ (x, )) with ψ γ L 2 δ (Rn ). Moreover, ψ γ has the form (1.13) ψ γ (x, ) =(ω 0 (x, ) γ 1/2 )+ω 1 (x, ) where ω 0 and ω 1 satisfy (1.14) lim ω 0(x, ) γ 1/2 H 1 δ =0, and (1.15) lim ω 1 (x, ) L 2 δ =0. Theorem 1.2 Let n 3. Let γ i W 3/2, (Ω), be strictly positive on Ω, i =1, 2. Assume Λ γ1 =Λ γ2.thenγ 1 = γ 2 on Ω. Note that if one could replace L 2 δ norm in (1.15) with H1 δ norm then Theorem 1.2 would easily follow for any W 1, (Ω) conductivity. In Section 2 we construct solutions as in Theorem 1.1. The key point of the proof is the observation that the operator on the left-hand side of (1.11) has an explicit approximative inverse that involves only first order derivatives of γ (c.f. Theorem 2.1 below). In Section 3 we prove Theorem 1.2, using a new identity that is satisfied if the DN maps of two conductivities are the same and plugging the solutions (1.9) into this identity. The standard identity used in previous results doesn t seem to give the desired result. As for the case of W 2, (Ω) conductivities, we expect that the methods of this paper will lead to stability estimates and a reconstruction method. ThecaseofC 2/3+ε (Ω), ε>0, first order perturbations of the Laplacian was considered in [15].

5 Complex geometrical optics solutions for Lipschitz conductivities Construction of the complex geometrical optics solutions In this section we construct ω 0 and ω 1 as in Theorem 1.1 and we derive their asymptotic properties as. We denote by Φ ε (x) = ε n Φ(x/ε), ε > 0, an approximation to the δ-function. More specifically we assume Φ(x) = n Φ 0 (x j ) with Φ 0 C0 (R), Φ 0 0, Φ 0 = 1 near 0 and Φ 0 (x)dx =1. We define the weighted L 2 -space, L 2 α,by { } L 2 α(r n )= f : (1 + x 2 ) α/2 f(x) 2 dx <, j=1 with norm given by f 2 L 2 α = (1 + x 2 ) α f(x) 2 dx. We denote by Hα(R s n ) the corresponding Sobolev space defined by interpolation for non integer s. Throughout this section we assume that γ W 1, (R n ), with γ =1 outside a large ball. Let (2.1) (2.2) (2.3) ϕ ε =Φ ε log γ, A = log γ, A ε =Φ ε A. Clearly (2.4) ϕ ε = A ε. Next we define (2.5) ω 0 (x, ε) =e ϕ ε(x)/2.

6 62 L. Päivärinta, A. Panchenko and G. Uhlmann The proof of the following proposition is standard. Proposition 2.1 If A W 1/2, (Ω) we have as ε 0 and A ε A L 2 = ε 1/2 o(ε) xj A ε L 2 = ε 1/2 o(ε). From the above proposition we conclude Lemma 2.1 Let ω 0 be as in (2.5). Then for any δ>0 lim ω 0 γ 1/2 H 1 ε 0 δ =0. Let C n with = 0. We define the operators (2.6) u = e x (e x u)= u +2 u, (2.7) v = e x (e x u)= v + v. Let f C0 (R n ). We define (2.8) 1 f = 1 (2π) n and recall e ix ξ ˆf(ξ) ξ 2 +2i ξ dξ Proposition 2.2 ([13, 12]). Let n 2. Let 1 <δ<0 and s 0. Then extends to a bounded operator 1 with 1 : H s δ+1(r n ) H s δ (R n ) (2.9) 1 H s δ+1 H s δ for some C>0. Moreover is bounded and C(s, δ, n), 1 : L 2 δ+1(r n ) H k δ (R n ), k =1, 2 (2.10) 1 L 2 δ+1 H k δ C(δ, k, n) k 1, k =1, 2, for some C>0.

7 Complex geometrical optics solutions for Lipschitz conductivities 63 Next we construct ω 1. To have a solution of the form (1.9), ω 1 must satisfy (2.11) ( + A )ω 1 = ( + A )ω 0 =: f ε. We write then ω must satisfy ω 1 = 1 ω, (2.12) (I + A 1 ) ω = f ε. A straightforward calculation gives that (2.13) f ε = e ϕ ε 2 ( 1 2 ϕ ε ( ϕ ε) 2 1 ) 2 A ϕ ε ϕ ε + A Since ϕ ε = A ε,weget (2.14) f ε = e ϕ ε 2 ( 1 2 ϕ ε+ 1 4 ( ϕ ε) 2 1 ) 2 A ϕ ε +(A A ε ) In order to solve (2.12) we define the operators T (γ) ands (γ) by (2.15) T (γ) :=(I + A 1 ) and (2.16) S (γ) :=γ 1/2 (I A 1 )γ 1/2. Note that (2.17) S (γ) =γ 1/2 T (γ 1 )γ 1/2. The main result of this section is that T (γ) ands (γ) are approximate inverses to each other. This result is used to prove that (2.12) has a unique solution in an appropriate space. Theorem 2.1 Let 1 <δ<0 and γ W 1, (R n ) with γ strictly positive and γ =1outside a large ball. Then the operator (2.18) T (γ) :L 2 δ+1(r n ) L 2 δ+1(r n ),.. is bounded and it has a bounded inverse T 1 for large. Moreover (2.19) T 1 (γ) S (γ) L 2 δ+1 L 2 0 as. δ+1

8 64 L. Päivärinta, A. Panchenko and G. Uhlmann We first prove a stronger version of (2.19) for C 2 conductivities. More precisely we show Lemma 2.2 Let γ C 2 (R 2 ), γ =1outside a large ball and strictly positive. Then for largewehavetheestimate (2.20) T 1 (γ) S (γ) L 2 δ+1 L 2 C γ C 2, δ+1 for some C>0. Proof. We have the following identity which was implicitly used in [13, 12] to reduce the construction complex geometrical optics solutions for the conductivity equation to the construction of complex geometrical optics solutions for the Schrödinger equation. The identity was explicitly stated in [5] and used to construct complex geometrical optics solutions for Maxwell s equations. (2.21) ( + A )γ 1/2 = γ 1/2 ( q), with q = γ γ.then and therefore ( + A ) 1 (2.22) I +(A ) 1 By Proposition 2.2 q 1 = γ 1/2 ( q)γ 1/2 1 = γ 1/2 γ 1/2 1 L 2 δ+1 L 2 C q, δ+1 q 1. for some C>0. Therefore ( ) I +(A ) 1 1 ( ) = γ 1/2 γ 1/ B, for large with B L 2 δ+1 L 2 C q. δ+1 Since ( ) γ 1/2 γ 1/2 1 1 = γ 1/2 1 γ 1/2, we conclude that (2.23) ( ) I +(A ) 1 1 γ 1/2 1 γ 1/2 L 2 δ+1 L 2 C γ C 2. δ+1

9 Complex geometrical optics solutions for Lipschitz conductivities 65 Now we use formula (2.22) again by changing γ to 1/γ, to conclude (2.24) γ 1/2 1 γ 1/2 = γ ( ) 1/2 I (A ) 1 γ 1/2 +γ 1/2 (γ 1/2 ) 1. From (2.23) and (2.24) we conclude the proof of the Lemma since by Proposition 2.2 again γ 1/2 (γ 1/2 ) 1 L 2 δ+1 L 2 C γ C 2. δ+1 Lemma 2.3 Let γ W 1,, γ strictly positive and γ =1outside a large ball. Let γ ε = γ Φ ε.then (2.25) T (γ ε ) T (γ) L 2 δ+1 L 2 δ+1 Proof. We have that ε 0 T (γ ε ) T (γ) =I +( log γ ε log γ) 1. Using Proposition 2.2 we conclude that 1 L 2 δ+1 L 2 δ+1 C, for some C>0 independent of. Therefore T (γ ε ) T (γ) L 2 δ L 2 δ+1 C log γ ε log γ L (R n ) proving the Lemma. Proof of Theorem 2.1. Theorem 2.1 follows directly from Lemma 2.2 and Lemma 2.3 by standard approximation procedure. Namely, for γ W 1, (R n )asbeforewehave S T I = (S (γ ε )T (γ ε ) I) (2.26) (S (γ ε ) S (γ))t (γ ε ) S (γ)(t (γ ε ) T (γ)). Since γ ε C 2 = O(ε 2 ) we obtain from Lemma 2.2 (2.27) S (γ ε )T (γ ε ) I L 2 δ+1 L 2 δ+1 C γ ε C 2 ( ) 1 = O. ε 2

10 66 L. Päivärinta, A. Panchenko and G. Uhlmann Now by Proposition 2.2 S ( γ) L 2 δ+1 L 2 δ+1 + T ( γ) L 2 δ+1 L 2 δ+1 C log γ L (R n ) holds for any conductivity γ W 1, (R n ). Therefore by taking ε = 1/4 and large we obtain from Lemma 2.3 and (2.27) the desired conclusion (2.19). Next we study the behavior of ω 1 as ε 0and. By Theorem 2.1 and (2.12) we can define (2.28) ω 1 (x, ε, ) = 1 S f ε + h where (2.29) h = 1 C f ε with C = T 1 S. Lemma 2.4 lim h H 1 δ (R n ) =0 Proof. By Theorem 2.1 we have that T 1 L 2 δ+1 L 2 C δ+1 with C independent of. Bydefinition (2.30) T ω 1 = f ε, and ω 1 = S f ε + C f ε. Therefore by Theorem 2.1 ω 1 + S f ε L 2 δ+1 L 2 δ+1 0as. This implies that (2.31) A := ω S f ε L 2 δ L 2 δ since 0as A = 1 ( ω 1 + S f ε ) L 2 δ L 2 δ C ω 1 S f ε L 2 δ+1 L 2 δ+1

11 Complex geometrical optics solutions for Lipschitz conductivities 67 Proof of Theorem 1.1. The choice ω 0 (x, ) =ω 0 (x, ε) andω 1 (x, ) = ω 1 (x, ε, ), with say ε = α for any 0 <α 1 will lead to the desired result. Indeed, by Proposition 2.2 and by (2.16) On the other hand by Lemma 2.3 f ε C(ε 1 + )o(ε). ω 1 L 2 δ 1 S f ε L 2 δ + h L 2 δ C f ε which proves that ω 1 L 2 δ This together with Lemma 2.1 prove Theorem The uniqueness proof In this section we prove Theorem 1.2. We shall use the following identity Lemma 3.1 Let γ W 1, (Ω), γ strictly positive on Ω and a = γ. Let u H 1 (Ω) be a solution of div(γ u) = 0. Letv H 1 (Ω). Then (3.1) ( a (uv) (au) v)dx = av u Ω Ω ν ds, where ds denotes surface measure on Ω. Proof. By the divergence theorem we have ( v (3.2) a 1 div(a 2 u)vdx = a 2 u dx a Ω Ω a) u Ω ν vds. Therefore ( v (3.3) a 2 u dx = a Ω a) u Ω ν vds. Now setting w = au and using that a 1 = a 2 a we obtain a 2 (a 1 w) (a 1 v)dx Ω ( ( = w v a w (a 1 v)+v (a 1 w) )) dx Ω = w v a (a 1 wv)dx Ω = (au) v a (uv)dx, proving the Lemma. Ω

12 68 L. Päivärinta, A. Panchenko and G. Uhlmann The right-hand side of (3.1) is determined by the DN map if we know v Ω. This is a consequence of the following two Lemmas. Lemma 3.2 [1,6,14]Letγ i W 1, (Ω) conductivities i =1, 2 with Λ γ1 = Λ γ2.then ( ) ( ) γ1 γ 1 Ω = γ 2 Ω and γ2 ν = Ω ν. Ω Lemma 3.3 [13] Assume γ i W 1, (Ω) conductivities i =1, 2 with Λ γ1 = Λ γ2. Extend γ i W 1, (R n ) with γ 1 = γ 2 in Ω c and γ i =1outside a large ball. Let u i, i =1, 2 be the complex geometrical solutions as in Theorem 1.1. Then u 1 = u 2 in Ω c. We include the proof here for the sake of completeness. Proof. Let z H 1 (Ω) be the solution of div(γ 1 z) =0 on Ω z Ω = u 2 Ω. By the assumption Λ γ1 =Λ γ2,wehave ( ) z (3.4) γ 1 ν = Ω ( γ 2 u 2 ν ), Ω and therefore, since γ 1 equals γ 2 on the boundary, ( ) ( ) z u2 ν = Ω ν. Ω Then z solves div(γ 1 z) =0inR n if we extend z = u 2 on Ω c. Since z has the same asymptotic behavior as u 1 we have from the uniqueness of the solutions in Theorem 1.1 that z = u 1 and therefore u 1 = u 2 in Ω c. Combining Lemmas we arrive at the following: Proposition 3.1 Let γ i W 1, (Ω) conductivities with Λ γ1 =Λ γ2 and let a i = γ 1/2 i.then ( a 1 (u 1 v) a 2 (u 2 v)) dx Ω (3.5) ( (a 1 u 1 ) v (a 2 u 2 ) v) dx =0, Ω for any v H 1 (Ω) and u i H 1 (Ω) solution of div(γ i u i )=0on Ω, i =1, 2.

13 Complex geometrical optics solutions for Lipschitz conductivities 69 The next lemma is the only place where the assumption γ W 3/2, (Ω) is needed. Lemma 3.4 Let γ W 3/2, (R n ), γ strictly positive and equal to one outside a large ball. Then for ω 1 (x, ) =ω 1 (x, ε, ), as in (2.28), ε = 1 we have (3.6) e ix ξ γ 1/2 ω 1 dx Proof. The statement (3.6) is equivalent to (3.7) e ix ξ γ 1/2 A ω 1 dx Using (2.12), (2.16) and (2.19), (3.7) follows if we show (3.8) e ix ξ γ 1/2 A 1 f ε dx and (3.9) e ix ξ γ 1/2 A 1 γ 1/2 A 1 γ 1/2 f ε dx We know from Lemma 2.1 that e ϕ ε 2 γ 1/2 in Hδ 1 so that it can be easily estimated. The first two terms of f ε involve ϕ ε and ϕ ε. Recall from Proposition 2. that ϕ ε = A ε and therefore ϕ ε =diva ε. Below we will see that the choice ε 1 is the only possibility to obtain (3.6). Accordingly in (2.14) the terms A ε = ϕ ε and (A A ε ) behave both as ε 1/2 o(ε). The other two terms are bounded and cause no trouble. We will give detailed estimates for the term involving (A A ε ). The same reasoning works for the other terms, as well. To this end we notice that for (3.9) we need to show C ε, := e ix ξ γ 1/2 (A A ε ) 1 γ 1/2 A 1 γ 1/2 (A A ε ) dx and D ε, := e ix ξ γ 1/2 A ε 1 γ 1/2 A 1 γ 1/2 (A A ε ) dx Since 1 and 1 are both bounded operators from L 2 δ+1 L2 δ we have by choosing ε = 1 C ε, = εo(ε)

14 70 L. Päivärinta, A. Panchenko and G. Uhlmann On the other hand, since the integration by parts gives (e ix ξ γ 1/2 A ε ) L 2 = ε 1/2 o(ε) D ε, = ε 1/2 ε 1/2 o(ε) proving (3.9) concerning this term. The proof of (3.8) is similar but easier. Note that the above proof uses the result of Theorem 2.1. The proof of this theorem uses the choice ε = 1/4 but the statement of the theorem is independent of ε. ProofofTheorem1.2. We now extend γ i W 3/2, (R n )withγ 1 = γ 2 on Ω c, γ i = 1 outside a large ball i =1, 2andγ i strictly positive. Let ξ R n. We choose 1 C n, 1 1 =0, 1 large enough and 2 C n, 2 2 =0sothat = iξ. Let ( ) (3.10) u i = e x 1 ω (i) 0 + ω (i) 1, i =1, 2, as in Theorem 1.1 with γ replaced by γ i,i=1, 2. Let (3.11) v = e x 2. Now we substitute (3.10) and (3.11) in (3.5). We note that by Lemmas 3.2 and 3.3 we can replace the integration over Ω on (3.5) by integration over R n. By taking the limit as and ε 0 we can replace by Lemmas 2.1 and 3.3 u 1 by e x 1 γ 1/2 1 and u 2 by e x 2 γ 1/2 2 in (3.5). We obtain ( ) (3.12) γ 1/2 1 (e ix ξ γ 1/2 1 ) γ 1/2 2 (e ix ξ γ 1/2 2 ) dx =0. R n This leads to i (3.13) 2 ξ e ix ξ ( log γ 1 log γ 2 )dx 1 e ix ξ (( log γ 1 ) 2 ( log γ 2 ) 2 )dx =0. 4 Then by using the Fourier inversion formula we obtain in the sense of distributions (3.14) (log γ 1 log γ 2 )+ 1 2 ( log γ 1 + log γ 2 ) (log γ 1 log γ 2 )=0. Since γ 1 Ω = γ 2 Ω, by the uniqueness of the solutions of the Dirichlet problem we conclude that γ 1 = γ 2 ending the proof of Theorem 1.2.

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16 72 L. Päivärinta, A. Panchenko and G. Uhlmann [15] Tolmasky, C.: Exponentially growing solutions for nonsmooth first order perturbations of the Laplacian. SIAM J. Math. Anal. 29 (1998), [16] Uhlmann, G.: Developments in inverse problems since Calderón s foundational paper. Harmonic analysis and partial differential equations (Chicago, IL, 1996), (M. Christ, C. Kenig, C. Sadosky editors). Chicago Lectures in Math., Univ. Chicago Press, Recibido: 7 de mayo de 2001 Lassi Päivärinta Department of Mathematical Sciences University of Oulu Oulu, Finland lassi@tol.oulu.fi Alexander Panchenko Department of Mathematics Pennsylvania State University University Park, PA 16802, USA panchenk@math.psu.edu Gunther Uhlmann Department of Mathematics University of Washington Seattle, WA 98195, USA gunther@math.washington.edu