Partial data for the Neumann-to-Dirichlet-map in two dimensions

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1 Partial data for the Neumann-to-Dirichlet-map in two dimensions O. Yu. Imanuvilov, Gunther Uhlmann, M. Yamamoto Abstract For the two dimensional Schrödinger euation in a bounded domain, we prove uniueness of determination of potentials in W 1 p, p > 2 in the case where we apply all possible Neumann data supported on an arbitrarily non-empty open set Γ of the boundary and observe the corresponding Dirichlet data on Γ. An immediate conseuence is that one can determine uniuely a W 3 p p > 2 conductivity by measuring the voltage potential on an open subset of the boundary for currents supported in the same set. 1 Introduction Let R 2 be a bounded domain with smooth boundary and let ν = ν 1, ν 2 be the unit outer normal to and let ν = ν. In this domain we consider the Schrödinger euation with a potential : L x, Du = + u = 0 in. 1 Let Γ be a non-empty arbitrary fixed relatively open subset of. Consider the Neumann-to- Dirichlet map N, Γ defined by N, Γ : f u Γ, 2 where + u = 0 in, u ν \ Γ = 0, u ν Γ = f. 3 with domain DN, Γ L 2 Γ. By uniueness of the Cauchy problem for the Schödinger euation the operator N, Γ is well defined since the problem 3 has at most one solution for each f L 2 Γ. Thanks to the Fredholm alternative, we see that DN, Γ = DN, Γ and L 2 Γ \ DN, Γ is finite dimensional for any potential in W2 1. The goal of this article is to prove uniueness of the determination of the potential from the Neumann-to-Dirichlet map N, Γ given by 2 for arbitrary subboundary Γ. More precisely, we consider all Neumann data supported on an arbitrarily fixed subboundary Γ as input and we observe the Dirichlet data only on the same subboundary Γ. This map arises in electrical impedance Department of Mathematics, Colorado State University, 101 Weber Building, Fort Collins, CO , U.S.A. oleg@math.colostate.edu, Department mathematics, UC Irvine, Irvine CA Department of Mathematics, University of Washington, Seattle, WA USA gunther@math.washington.edu, Second author partly supported by NSF and a Walker Family Endowed Professorship Department of Mathematical Sciences, University of Tokyo, Komaba, Meguro, Tokyo 153, Japan myama@ms.u-tokyo.ac.jp 1

2 tomography EIT. In this problem one attempts to determine the electrical conductivity of a medium by making voltages and current measurements at the boundary. After transforming 1 to the conductivity euation, we can interpret u and u ν respectively as the voltage and the multiple of the current by a value of the surface conductivity on the boundary. In practice, we apply currents to electrodes on the boundary and observe the corresponding voltages. The current inputs are modelled by the Neumann boundary data u ν and the observation data is modelled by Dirichlet data. See e.g., [10] for applications to medical imaging of EIT. Moreover it is very desirable to restrict the supports of the current inputs as much as possible. To the author s best knowledge there are few works on the uniueness by such a Neumann-to-Dirichlet map with partial data. In Astala, Päivärinta and Lassas [3], the authors consider both the Dirichlet-to-Neumann map and Neumann-to-Dirichlet map on an arbitrarily subboundary to establish the uniueness of an anisotropic conductivity modulo the group of diffeomorphisms that is the identity on the boundary where the measurements take place. The case where the measurements are given by the Dirichlet-to-Neumann map has been extensively studied in the literature. This map is defined in the case of partial data by Λ, Γ : g u ν Γ; + u = 0 in, u \ Γ = 0, u Γ = g. We give some references but the list is not at all complete. In the case of full data Γ =, this inverse problem was formulated by Calderón [9]. In the two dimensional case given the Dirichlet-to- Neumann map with partial data, Λ, Γ, uniueness is proved under the assumption C 2+α by Imanuvilov, Uhlmann and Yamamoto [14] and for the uniueness for potentials W 1 p, p > 2 see [18]. Also see [17] for uniueness results for elliptic systems. In Guillarmou and Tzou [12], the result of [14] was extended on Riemmannian surfaces. In particular, in the two-dimensional for the case of a potential with full data, Λ,, we refer to Blasten [4], Bukhgeim [7] and Sun-Uhlmann [24] and for systems in [1]. For the case of full data, in [4] and [18], it was shown that Λ, uniuely determines in the class W 1 p with p > 2 and C α, α > 0, respectively. As for the related problem of recovery of the conductivity in EIT, [2] proved the uniueness result for conductivities in L, improving the result of [21] and [6]. Moreover for the case of dimensions n 3 with the full data Sylvester and Uhlmann [25] proved the uniueness of recovery of a conductivity in C 2, and later the regularity assumption was improved see, e.g., Brown and Torres [5], Päivärinta, Panchenko and Uhlmann [23] and Haberman and Tataru [13]. The case when voltage is applied and conductivity is measured on different subsets was studied in dimension greater than three in [8], [20] and in [19] in the two-dimensional case. Our main result is as follows Theorem 1 Let 1, 2 W 1 p for some p > 2. If DN 1, Γ DN 2, Γ and N 1, Γ f = N 2, Γ f for each f from DN 1, Γ then 1 = 2. Notice that Theorem 1 does not assume that is simply connected. An interesting inverse problem is whether one can determine the potential in a domain with holes measuring N, Γ only on some open set Γ in the outer subboundary. Let, G be bounded domains in R 2 with smooth boundaries such that G. Let Γ be an open set and W 1 p \ G with some p > 2. Consider the following Neumann-to-Dirichlet map: Ñ, Γ : f u Γ, 2

3 where u H 1 \ G, + u = 0 in \ G, u ν G \ Γ = 0, u ν Γ = f. Then we can directly derive the following from Theorem 1. Corollary 2 Let 1, 2 W 1 p \ G with some p > 2. If DÑ 1, Γ DÑ 2, Γ and Ñ 1, Γ f = Ñ 2, Γ f for each f from DÑ 1, Γ then 1 = 2 in \ G. For the case of EIT, if the conductivities are known on Γ, then we can apply our theorem to prove uniueness of the determination of conductivities in W 3 p, p > 2 from the Neumann-to- Dirichlet map. The remainder of the paper is devoted to the proof of the theorem. The main techniue is the construction of complex geometrical optics solutions whose Neumann data vanish on the complement of Γ. Throughout the article, we use the following notations. Notations. Γ 0 = \ Γ, i =, x 1, x 2 R 1, z = x 1 + ix 2, z denotes the complex conjugate of z C. We identify x = x 1, x 2 R 2 with z = x 1 + ix 2 C and ξ = ξ 1, ξ 2 with = ξ 1 + iξ 2. z = 1 2 x 1 i x2, z = 1 2 x 1 + i x2, D = 1 i x 1, 1 i x 2, = 1 2 ξ 1 i ξ2, = 1 2 ξ 1 + i ξ2. Denote by Bx, δ a ball centered at x of radius δ. For a normed space X, by o X 1 we denote κ a function f, such that f, X = o 1 as +.The tangential derivative on the κ boundary is given by = ν 2 x 1 ν 1 x 2, where ν = ν 1, ν 2 is the unit outer normal to. The operators z and z are given by z g = 1 g, π z dξ 2dξ 1, z g = z g. In the Sobolev space H 1 we introduce the following norm u H 1, = u 2 H u 2 L The Proof In this section we prove Theorem 1. Let = ϕ + iψ be a holomorphic function in such that ϕ, ψ are real-valued and C 2, Im Γ 0 = 0, Γ 0 Γ 0, 4 where Γ 0 is some open set in. Denote by H the set of the critical points of the function. Assume that H, z 2 z 0 z H, H Γ = 5 and J 1dσ = 0, J = {x; ψx = 0, x \ Γ 0}. 6 3

4 Let 1 be a bounded domain in R 2 such that 1 and C be some smooth complex-valued function in such that 2 C z = C 1x + ic 2 x in 1, 7 where C 1, C 2 are smooth real-valued functions in such that The following proposition is proved in [16]. C 1 x 1 + C 2 x 2 = 1 in 1. 8 Proposition 1 Suppose that L, the function satisfies 4, 5 and ṽ H 2. Then there exist 0 and C independent of ṽ and such that N 2 2 zṽe ϕ+nc 2 L 2 + ṽeϕ+nc 2 L 2 + ṽeϕ+nc 2 H z ṽeϕ+nc 2 L 2 L x, Dṽe ϕ+nc 2 L 2 + CN ṽeϕ+nc, ṽ ν eϕ+nc 2 H 1, L 2 9 for all > 0 N and all positive N 1. Let ṽ H 2 satisfy L x, Dṽ = f in, Using Proposition 1, we can show the following. ṽ ν Γ 0 = 0. Proposition 2 Suppose that satisfies 4, 5 and L. Then there exist 0 independent of ṽ and such that and C ṽe ϕ 2 L 2 + ṽeϕ 2 H z ṽeϕ 2 L 2 C L x, Dṽe ϕ 2 L 2 + ṽeϕ, ṽeϕ ν 2 H 1, Γ L 2 Γ 10 for all > 0 and for all ṽ H 2. Proof. Let {e j } N j=1 be a partition of unity such that e j C0 Bx j, δ where x j are some points in, N e j e j x = 1 on, ν Γ = 0 j {1,..., N} 0 j=1 and δ be a small positive number such that Bx j, δ Γ 0 implies Bx j, δ Γ 0. Denote w j = e j ṽ. Let supp w j \ Γ =. Then Proposition 1 implies N 2 2 zw j e ϕ e NC 2 L 2 + w j e ϕ+nc 2 L 2 + w je ϕ+nc 2 H z w je ϕ+nc 2 L 2 C L x, Dw j e ϕ+nc 2 L 2 + CN w je ϕ, w je ϕ 2 ν H 1, Γ L 2 Γ. 11 4

5 Next let supp w j \ Γ. We can not apply directly the Carleman estimate 10 in this case, since the function w j may not satisfy the zero Dirichlet boundary condition. To overcame this difficulty we construct an extension. Without loss of generality, using if necessary a conformal transformation, we can assume that supp w j {x 2 > 0} and supp w j {x 2 = 0}. Then using the extension w j x 1, x 2 = w j x 1, x 2, x 1, x 2 = x 1, x 2 and ϕx 1, x 2 = ϕx 1, x 2, we apply Proposition 1 to the operator L x, D in O = supp e j {x x 1, x 2 supp e j }. We have the same estimate 11. Therefore ṽ 2 := ṽe ϕ+nc 2 L 2 + ṽeϕ+nc 2 H z ṽeϕ+nc 2 L 2 N N ṽe j 2 C L x, Dw j e ϕ+nc 2 L 2 + CN w je ϕ, w je ϕ 2 ν H 1, Γ L 2 Γ j=1 C j=1 N e j ṽe ϕ+nc 2 L ze j z ṽe ϕ+nc 2 L 2 N zṽe j e ϕ+nc 2 L 2 j=1 + L x, Dṽe ϕ+nc 2 L 2 + CN ṽeϕ, ṽeϕ ν 2 H 1, Γ L 2 Γ. 12 Fixing the parameter N sufficiently large, we obtain from 12 ṽ 2 CN ṽe ϕ+nc 2 L 2 + L x, Dṽe ϕ+nc 2 L 2 + ṽeϕ, ṽeϕ ν H 1, Γ L 2 Γ The first term on the right-hand side of 13 can be absorbed into the left-hand side for all sufficiently large. Since N and C are independent of, the proof of the proposition is finished. The Carleman estimate 10 implies the existence of solutions to the following boundary value problem. Proposition 3 There exists a constant 0 such that for 0 and any f L 2, r H 1 2 Γ 0, there exists a solution to the boundary value problem such that L x, Du = fe ϕ in, u ν Γ 0 = re ϕ 14 u H 1, / C f L r L 2 Γ 0 + r 1 H Γ 0 The constants C is independent of. The proof of this proposition uses standard duality arguments, see e.g., [14]. We define two other operators: R g = 1 2 e z ge, R g = 1 2 e z ge. 16 Observe that 2 z e R g = ge, 2 z e R g = ge g L Let a C 6 be some holomorphic function on such that Im a Γ 0 = 0, lim az/ z ẑ 100 = 0 ẑ H Γ z ẑ

6 Moreover, for some x H, we assume that a x 0 and ax = 0 x H \ { x}. 19 The existence of such a function proved in Proposition 9 of [17]. Let polynomials M 1 z and M 3 z satisfy z 1 M 1 x = 0, z 1 M 3 x = The holomorphic function a 1 and the antiholomorphic function b 1 are defined by formulae a 1 z = a 1,1 z + a 1,2 z + a 1,3 z and b 1 z = b 1,1 z + b 1,2 z + b 1,3 z where a 1,1, b 1,1 C 1 are such that i ψ ν a 1,1z i ψ ν b a + a 1,1z = + i ψ ν ν a z 1 M 1 i ψ a z 1 M 3 4 z ν 4 z on Γ 0 21 and a 1,2 z,, b 1,2 z, C 1 for each are holomorphic and antiholomorphic functions such that b 1,2 z, = 1 8π ν 1 iν 2 a 1 M 1 e z dσ, a 1,2 z, = 1 8π ν 1 + iν 2 a Here the denominators of the integrands vanish in H Γ 0, but thanks to the second condition in 18 integrability is guaranteed. We represent the functions a 1,2 z,, b 1,2 z, in the form where b 1,2,1 z = 1 8π a 1,2 z, = a 1,2,1 z + a 1,2,2 z,, b 1,2 z, = b 1,2,1 z + b 1,2,2 z, Γ 0 ν 1 iν 2 a 1 M 1 dσ, a 1,2,1 z = 1 ν 1 + iν 2 a 1 M 3 dσ. z 8π Γ z 0 By 18 functions b 1,2,1, a 1,2,1 belong to C 1. By 6 1 M 3 e d z b 1,2,1 L 2 + a 1,2,1 L 2 0 as +infty Finally a 1,3 z,, b 1,3 z, W2 1 for each are holomorphic and antiholomorphic functions such that and i ψ ν a 1,3z, i ψ ν b 1,3z, = i ψ 2π ν i ψ 2π ν 1 M 3 1 M 1 e z dξ 2dξ 1 e z dξ 2dξ 1 on Γ 0, 22 a 1,3, L 2 + b 1,3, L 2 = o1 as The ineuality 23 follows from the asymptotic formula 1 M M 3 e z dξ 2dξ 1 W Γ 0 e z dξ 2dξ 1 W Γ 0 = o1 as

7 The family of functions e a 1 M 1 e z dξ 2dξ 1 C, are uniformly bounded in in C 2 and by Proposition 2.4 of [14] this function converges pointwisely to zero. Therefore 1 M 1 e e z dξ 2dξ 1 W 1 2 = o1 as Integrating by parts we obtain 1 M 1 e 1 e z dξ 2dξ 1 = 1 Thanks to 4, 18 1 e z z The functions 1 e 1 e by Proposition 2.2 of [14] functions 1 e a z z 1 M 1 e z 1 M 1 z z dξ 2dξ 1. e z 1 M 1 e z W 1 = o1 as Γ 0 a 1 M 1 e are bounded in L p uniformly in, z. So 1 e a 1 M 1 dξ 2 dξ 1 are uniformly e z bounded in Wp 1. By the trace theorem 1 1 e a 1 M 1 e dξ 2 dξ 1 z W 1 = o1 2 2 Γ 0 as By we obtain 24. We note that by 18 the function a z C2. We define the function U 1 by the formula U 1 x = e a + a 1 / + e a + b 1 / 1 2 e R {a z 1 M 1 } 1 2 e R {a z 1 M 3 }.28 Integrating by parts, we obtain the following e R {a z e R {a z We claim that 1 M 1 } = 1 2b 1,2e + e a z 1 M 1 2 z 1 M 3 } = 1 2a 1,2e + e a z 1 M 3 2 z + eiψ 2π + e 2π + e 2π 1 M 1 1 M 3 e iψ 1 M 1 e 2π z dξ 2dξ 1 L 2 1 M 3 e z dξ 2dξ 1 29 e z dξ 2dξ e z dξ 2dξ 1 L 2 0 as

8 We prove the asymptotic formula 31 for the first term. Proof of the asymptotics for the second a 1 M 1 term is the same. Denote r ξ, z = e. By 5, 19, 20 the family of these functions bounded in L p for any p < 2. Hence by Proposition 2.4 of [14] there exists a constant C independent of such that e iψ 1 M 1 e 2π z dξ 2dξ 1 L 4 C. 32 By 5, 19, 20 for any z x 1 + i x 2 function r, z belongs to L 1. Therefore by Proposition of e iψ 1 M 1 e 2π z dξ 2dξ 1 0 a.e. in. 33 From 32, 33 and Egoroff s theorem the asymptotics for the first term in 31 follows immediately. We set g = 1 e iψ a 1 / + e iψ b 1 / eiψ 2 R {a z 1 M 1 } e iψ R {a z 1 M 3 }. 2 By we have Short computations give L 1 x, DU 1 = e ϕ g in, g L 2 = O 1 as U 1 ν Γ 0 = e ϕ O 1 1 H 2 as Γ 0 Indeed, the first euation in 35 follows from 28, 19 and the factorization of the Laplace operator in the form = 4 z z. In order to prove the second euation in 35 we set U 1 ν = I 1 + I 2 where I 1 = ν a + a 1,1/e + a + b 1,1 /e = e ϕ i ψ ν a + a 1,1/ i ψ ν a + b 1,1/ + ν a + a 1,1/ + ν a + b 1,1/ = i ψ ν a z 1 M 1 i ψ 4 z ν a z 1 M 3 4 z In order to obtain the last euality, we used 18 and 21. Then e ϕ + e ϕ O C 1 Γ I 2 = ν a 1,2 + a 1,3 e + b 1,2 + b 1,3 e 1 2 ν e R {a z 1 M 1 } + e R {a z 1 M 3 } = 1 a 2 ν e z 1 M 1 a + e 1 M 1 e 2 z 2π z dξ 2dξ e a z 1 M 3 a + e 1 M 3 e 2 z 2π z dξ 2dξ 1 = i ψ ν a z 1 M 1 e a z 1 M 3 + O 1 4 z 4 z 1 H 2 Γ

9 From 36 and 37, we obtain the second euation in 35. Finally we construct the last term of the complex geometric optics solution e ϕ w. Consider the boundary value problem L 1 x, Dw e ϕ = g e ϕ in, w e ϕ Γ0 = U 1 ν ν. 38 By 34 and Proposition 3, there exists a solution to problem 38 such that Finally we set w L 2 = o 1 as u 1 = U 1 + e ϕ w. 40 By 39, 40, 23 and 28-30, we can represent the complex geometric optics solution u 1 in the form u 1 x = e a + a 1,1 + a 1,2,1 / + e a + b 1,1 + b 1,2,1 / a z 1 M 1 a z 1 M 3 e + e + e ϕ o 4 z L 4 z 2 1 as Since the Cauchy data 2 for the potentials 1 and 2 are eual, there exists a solution u 2 to the Schrödinger euation with potential 2 such that u 1 ν = u 2 ν on and u 1 = u 2 on Γ. Setting u = u 1 u 2, we obtain + 2 u = 2 1 u 1 in, u Γ = u ν = In a similar way to the construction of u 1, we construct a complex geometrical optics solution v for the Schrödinger euation with potential 2. The construction of v repeats the corresponding steps of the construction of u 1. The only difference is that instead of 1 and, we use 2 and, respectively. We skip the details of the construction and point out that similarly to 41 it can be represented in the form vx = e a + ã 1,1 + ã 1,2,1 / + e a + b 1,1 + b 1,2,1 / a z 2 M 2 a z 2 M 4 + e + e + e ϕ o 4 z L 4 z 2 1 as +, v ν Γ 0 = 0, 43 where M 2 z and M 4 z satisfy z 2 M 2 x = 0, z 2 M 4 x = 0. The functions ã 1 z = ã 1,1 z + ã 1,2 z and b 1 z = b 1,1 z + b 1,2 z are given by i ψ ψ ã1,1z+i ν ν b a + a 1,1 z = +i ψ a z 2 M 2 i ψ a z 2 M 4 ν ν 4 z ν 4 z and ã 1,2 z, b 1,2 z C 1 are holomorphic functions such that b1,2 z = 1 8π Γ 0 ν 1 iν 2 a 2 M 2 e z dσ, ã 1,2 z = 1 8π Γ 0 9 on Γ 0, ν 1 + iν 2 a ã 1,1, b 1,1 C M 4 e dσ. z

10 Denote = 1 2. Taking the scalar product of euation 42 with the function v, we have: u 1 vdx = From formulae 41 and 43 in the construction of complex geometrical optics solutions, we have u 1 vdx = 0 = a 2 + a 2 dx + 1 aa 1,1 + a 1,2,1 + b 1,1 + b 1,2,1 + aã 1,1 + ã 1,2,1 + b 1,1 + b 1,2,1 dx + aae 2iψ + aae 2iψ dx a 2 z 2 M 2 z a 2 z 1 M 1 z + a 2 z 2 M 4 z + a 2 z 1 M 3 z +o 1 = 0 dx dx as +.46 Since the potentials j are not necessarily from C 2, we can not directly use the stationary phase argument e.g., Evans [11]. Let function ˆ C0 satisfy ˆ x = x. We have Re aae 2iψ dx = ˆRe aae 2iψ dx + ˆRe aae 2iψ dx. 47 Using the stationary phase argument and 19, similarly to [14], we get ˆaae 2iψ + aae 2iψ dx = 2π a 2 xre e 2iψ x 1 + o det ψ x 1 2 For the second integral in 47 we obtain ˆaae 2iψ + aae 2iψ dx = = 1 { e 2iψ div 2i Since ψ Γ0 = 0 we have ψ, νe 2iψ aa 2i ψ 2 ψ, νe2iψ aa 2i ψ 2 ˆaa ψ ψ 2 as ψ, e2iψ ψ, e 2iψ ˆ aa 2i ψ 2 aa 2i ψ 2 dx aa ψ, νe 2iψ 2i ψ 2 e 2iψ div ψ, νe 2iψ 2i ψ 2 dσ = Γ By 4, 6 and Proposition 2.4 in [14] we conclude that ψ, νe 2iψ ψ, νe 2iψ aa 2i ψ 2 2i ψ 2 dσ = o 1 dσ ˆaa ψ ψ 2 } dx. 49 aa 2i ψ 2 ψ, νe2iψ e 2iψ dσ. as +. The last integral over in formula 49 is o 1 and therefore ˆaae 2iψ + aae 2iψ dx = o 1 as

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