An Inverse problem for the Magnetic Schr\"odinger Operator on a Half Space with partial data

In this paper we prove uniqueness for an inverse boundary value problem for the magnetic Schr\"odinger equation in a half space, with partial data. We prove that the curl of the magnetic potential $A$, when $A\in W_{comp}^{1,\infty}(\ov{\R^3_{-}},\R^3)$, and the electric pontetial $q \in L_{comp}^{\infty}(\ov{\R^3_{-}},\C)$ are uniquely determined by the knowledge of the Dirichlet-to-Neumann map on parts of the boundary of the half space.


Introduction
We consider a magnetic Schrödinger operator L A,q , defined by on the half space R 3 -:= {x ∈ R 3 : x 3 < 0}. We assume that Consider the Dirichlet problem where k > 0 is fixed and f ∈ H Here we also require that the solution u should satisfy a boundary condition at infinity, which will be the Sommerfeld radiation condition Solutions satisfying this condition are called outgoing or radiating solutions. We will also occasionally use the term incoming solution. This refers to a solution of (1.3) that satisfies (1.4), when the factor −ik is replaced by ik. The existence and uniqueness of a solution u ∈ H 2 loc (R 3 -) to the problem (1.3) and (1.4) is proven in [19]. This allows us to define the so called Dirichlet to Neumann map Λ A,q , (DN-map for short), Λ A,q : H where u is the solution of the Dirichlet problem (1.3), (1.4) and f is the value of u.
Here n = (0, 0, 1) is the unit outer normal to the boundary ∂ R 3 -. The inverse problem is to investigate if the DN-map uniquely determines the potentials A and q in R 3 -. It turns out that the DN-map does not in general uniquely determine A. This is due to the gauge invariance of the DN-map, which was first noticed by [22].
Part (ii) of this Lemma shows that Λ A,q cannot uniquely determine A, since we can change a potential by a gauge transformation without changing the DN-map. The DN-map does however carry enough information to determine ∇ × A, which is the magnetic field in the context of electrodynamics.
When considering a pair of magnetic potentials A j , j = 1, 2, we use the notation A j = (A j,1 , A j,2 , A j,3 ) for the component functions. Furthermore we let N ⊂ R 3 − be a relatively open and bounded set, for which j=1,2 supp(A j ) ∪ supp(q j ) ⊂ N, and for which ∂ N piecewise C 2 and R 3 -\ N is connected. We now state the main result of this paper, which generalizes the corresponding results of [13], obtained in the case of the Schrödinger operator without a magnetic potential.
The first uniqueness result, in the context of inverse boundary value problems for the magnetic Schrödinger operator on a bounded domain, was obtained by Sun in [22], under a smallness condition on A. Nakamura, Sun and Uhlmann proved the uniqueness without any smallness condition in [17], assuming that A ∈ C 2 . Tolmasky extended this result to C 1 magnetic potentials in [23], and Panchenko to some less regular but small magnetic potentials in [18]. Salo proved uniqueness for Dini continuous magnetic potentials in [20]. The most recent result is given by Krupchyk and Uhlmann in [12], where uniqueness is proved for L ∞ magnetic potentials. In all of these works, the inverse boundary value problem with full data was considered.
In [6], Eskin and Ralston obtained a uniqueness result for the closely related inverse scattering problem, assuming the exponential decay of the potentials. The partial data problem in the magnetic case was considered by Dos Santos Ferreira, Kenig, Sjöstrand and Uhlmann in [5] and by Chung in [4].
The inverse problem for the half space geometry, without a magnetic potential was examined by Cheney and Isaacson in [2]. The uniqueness for this problem in the case of compactly supported electric potentials was proved by Lassas, Cheney and Uhlmann in [13], assuming that the supports do not come close to the boundary of the half space. The result of Theorem 1.2 is therefore already a generalization of the work [13], even in the absence of magnetic potentials. Li and Uhlmann proved uniqueness for the closely related infinite slab geometry with A = 0, in [16]. Krupchyk, Lassas and Uhlmann did this for the magnetic case in [11]. In both of these works, the reflection argument of Isakov [8] played an important role. The uniqueness problem for the magnetic potentials in the slab and half space geometries has also been studied in a recent paper by Li [15]. The half space results in [15] differ from the ones given in this work, by concerning the more general matrix valued Schrödinger equation and by assuming C 6 regularity on the magnetic potential.
The half space is perhaps the simplest example of an unbounded region with an unbounded boundary. It is of special interest in many applications, such as geophysics, ocean acoustics, and optical tomography, since it provides a simple model for semi infinite geometries. We would like to mention that the magnetic Schrödinger equation is closely related to the diffusion approximation of the photon transport equation, used in optical tomography [1].
The paper is organized as follows. Section 2 contains a review of the construction of complex geometric optics solutions for magnetic Schrödinger operators with Lipschitz continuous magnetic potentials. In section 3 we derive the central integral identity. The proof of Theorem 1.2 is contained in sections 4 and 5. The appendix contains an extension of Green's second formula and a statement of the unique continuation principle for easy reference.

Complex geometric optics solutions
Let Ω ⊂ R 3 be a bounded domain with C ∞ -boundary, and let A ∈ W 1,∞ (Ω, R 3 ), q ∈ L ∞ (Ω, C). The task of this subsection is to review the construction of complex geometric optics solutions for the magnetic Schrödinger equation, A complex geometric optics solution to (2.1) is a solution of the form where ζ ∈ C 3 , ζ · ζ = 0, a is a smooth amplitude, r is a remainder, and h > 0 is a small parameter.
In the case when A ∈ C 2 (Ω) and q ∈ L ∞ (Ω), such solutions were constructed in [5] using the method of Carleman estimates, and the construction was extended to the case of less regular potentials in [9], see also [11].
Then there is C > 0 and h 0 > 0 such that for all h ∈ (0, h 0 ], and any f ∈ L 2 (Ω), the equation Our basic strategy in constructing solutions of the form (2.2) is to write (2.1), as where L ζ := e −x·ζ/h h 2 L A,q e x·ζ/h . Then we first search for a suitable a, after which we will get r by Proposition 2.1. We must however take some care in choosing a and the way it depends on h, since we need later that r H 1 scl (Ω) → 0, sufficiently fast as h → 0. We need a also to be smooth enough. This will be handled as in [9].
We extend A ∈ W 1,∞ (Ω, R 3 ) to a Lipschitz vector field, compactly supported inΩ, whereΩ ⊂ R 3 is an open bounded set such that Ω ⊂⊂Ω. We consider the mollification A ♯ := A * ψ ǫ ∈ C ∞ 0 (Ω, R 3 ). Here ǫ > 0 is small and ψ ǫ (x) = ǫ −3 ψ(x/ǫ) is the usual mollifier with ψ ∈ C ∞ 0 (R 3 ), 0 ≤ ψ ≤ 1, and ψdx = 1. We write A ♭ = A − A ♯ . Notice that we have the following estimates for A ♭ , We shall work with a complex ζ = ζ 0 + ζ 1 depending slightly on h, for which By expanding the conjugated operator we write the right hand side of (2.4) as Now we want a to be such that this expression decays more rapidly than O(h), as h → 0.
Consider the operator in (2.7), ignoring for the time being a and its possible dependence on h. We would like to eliminate from this operator the terms that are of first order in h. Notice first that ζ 1 = O(h) and that we can control A ♭ with h, if we choose ǫ to be dependent on h. Then in an attempt to eliminate first order terms in h, it is natural to search for an a for which in Ω. (2.8) We will look for a solution of the form a = e Φ . The above equation becomes then in Ω. (2.9) Pick a γ ∈ S 2 , such that γ⊥{α, β}.
Next we consider the above equation in coordinates y, associated with the basis {α, β, γ}. Let T be the coordinate transform y = T x := (x · α, x · β, x · γ). Using the chain rule and the fact that T −1 = T * , one gets that 1 We therefore have that Equation (2.9) gives hence the∂-equation where ∂z = (∂ y1 +i ∂ y2 )/2. We will solve this using the Cauchy operator which is an inverse for the∂-operator, N := (∂ y1 + i∂ y2 )/2 (see e.g. [7] Theorem 1.2.2). We will need the following straightforward continuity result for the Cauchy operator.
Returning to (2.10) we get that Φ = 1 where T −1 (s 1 , s 2 , 0) = s 1 α+s 2 β. We have thus found a solution a = e Φ to equation (2.8). We will choose ǫ so that it depends on h, which implies that a will depend on h. In order to determine how the norm of r will depend on h and also for later estimates, we will need to see how ∂ α a L ∞ depends on h. Lemma 2.2 and estimate (2.5) imply the following result.
We can now write the L ∞ (Ω) norm of (2.7) as Using (2.5), (2.12) and the fact that Finally to solve (2.4) for r, we rewrite it as If we replace e ix·Im ζ/h r byr, then the solvability result Proposition 2.1, shows that we can find a solutionr, so that a solution r to (2.13) is given by r = e −ix·Im ζ/hr .
To get a norm estimate for r, notice that for the right hand side of (2.13) we have as h → 0. The solvability result 2.1 gives then that Thus we have obtained the following existence result for complex geometric optics solutions.
where ζ ∈ C 3 , is of the form given by (2.6), a ∈ C ∞ (Ω) solves the equation (2.8), and where a and r satisfy the estimates In the sequel, we need complex geometric optics solutions belonging to H 2 (Ω). To obtain such solutions, let Ω ′ ⊃⊃ Ω be a bounded domain with smooth boundary, and let us extend and L ∞ (Ω ′ )-functions, respectively. By elliptic regularity, the complex geometric optics solutions, constructed on Ω ′ , according to Proposition 2.4, belong to H 2 (Ω). (2.14) and satisfies Remark 2.7. We shall later use a slightly more general form for the amplitude a in the complex geometric optics solutions. Namely we suppose that a = ge Φ , where g ∈ C ∞ (Ω), is such that This means that g is holomorphic in a plane spanned by α and β. Notice also that by picking a = ge Φ , we get by (2.8) that in place of (2.9). But the Φ solving (2.9) also solves the above equation. Hence we can use the same argument to obtain the Φ for the above equation, as earlier.
We thus obtain CGO solutions of the form where Φ solves (2.9).
Notice also that setting a = ge Φ does not affect the norm estimates on a in Proposition 2.4, since g does not depend on h.

An integral identity
One central step in the ideas that are used in proving uniqueness results for inverse boundary value problems, is to derive an integral equation that expresses L 2 orthogonality between the product of two solutions u 1 and u 2 , and the difference of two potentials q 1 and q 2 , see [24]. One shows that provided that the DN-maps for q 1 and q 2 are equal.
A similar thing will be done in this subsection, for the magnetic case. The integral equation, is however more involved in the case of a magnetic potential and will not by itself be interpreted as an orthogonality relation. We will be considering the integral equation in conjunction with solutions that depend on a small positive parameter h. In the later sections we will see that in the limit h → 0, we obtain a criterion for the curl being zero.
It will be convenient to set In order to prove Theorem 1.2 we shall only use the data (3.1), which turns out to be enough to determine the magnetic field and the electric potential.
We now begin deriving the integral identity. We assume that A j , q j and Γ j are as in Theorem 1.2 so that (3.1) also applies. Let It follows from (3.1) that By Lemma 1.1 we may and shall assume that -, so that ∂ n w = 0 onΓ 1 . We also conclude from (3.2) that w satisfies the Helmholtz equation Assuming that u 2 = 0 on l, we conclude that ((L A2,q2 − k 2 )w, u 2 ) L 2 (N ) = 0. Using equation (3.2) we may write this as follows, Using again the fact that (A 2 − A 1 ) · n = 0 on ∂ N and an integration by parts, we get Thus, we obtain that The following Runge type approximation result is similar to those found in [8], [16] and [11].
By the Riesz representation theorem, there is g T ∈ L 2 (N ) that corresponds to T . Extend g T by zero to the complement of N in The existence of such a solution is proved in [19]. Now let u ∈ W 1 (R 3 -). Then because T | V = 0 and supp(g T ) ⊂ N , we get by the Green's formula of Lemma 6.2 that Since the boundary condition u|Γ 2 can be chosen arbitrarily from C ∞ 0 (Γ 2 ), we get that ∂ n U |Γ 2 = 0. Since U |Γ 2 = 0, we apply the unique continuation principle to Now applying Green's formula and doing the same computation as above for u 0 and N instead of u yields Here we have used that u 0 | l = 0. It follows that T (u 0 ) = 0. This contradiction completes the proof.
Since (A 2 − A 1 ) · n = 0 on ∂ N , we can rewrite (3.4) in the following form, Hence, an application of Lemma 3.1 implies that the integral identity (3.4) is valid for any u 1 ∈ W 1 (N ) and any u 2 ∈ W * 2 (N ). We summarize the discussion in this subsection in the following result.
Proposition 3.2. Assume that A j , q j and Γ j , j = 1, 2 are as in Theorem 1.2 and that the DN-maps satisfy for any u 1 ∈ W 1 (N ) and any u 2 ∈ W * 2 (N ).

Remark 3.3.
Notice that the proof of Proposition 3.2 only uses the assumption (3.1), which follows from (3.5). Proposition 3.2 holds therefore also under the weaker assumption (3.1).

Recovering the magnetic field
The aim of this section is to prove the first part of Theorem 1.2, by showing that the curl of the magnetic potential is determined by the DN-map. We choose an open ball B centered on ∂ R 3 -with N ⊂⊂ B. And use the notations The first step in the argument will be to construct complex geometric optics solutions u 1 ∈ W 1 (N ) and u 2 ∈ W * 2 (N ) and then to examine the limit of (3.6) as h → 0.
For u 1 ∈ W 1 (N ) and u 2 ∈ W * 2 (N ), we have that u j | l = 0, j = 1, 2. To obtain solutions that satisfy this condition, we will first choose solutions defined on the larger set B and then use a reflection argument.
We need to extend the potentials A j and q j , j = 1, 2, to B + . For the component functions A j,1 , A j,2 , and q j , we do an even extension, and for A j,3 , we do an odd extension, i.e., for j = 1, 2 we set, . By our assumptions, A j,3 | x3=0 = 0, from which it follows thatÃ j ∈ W 1,∞ (B) andq j ∈ L ∞ (B), j = 1, 2.
We can now by Proposition 2.4 and Remark 2.5 pick complex geometric optics solutionsũ 1 in H 2 (B), To obtain a function that is zero on the plane x 3 = 0, we set Then it is easy to check that the restriction u 1 | N ∈ W 1 (N ).
Finally it will also be convenient to explicitly state the following norm estimates, which follow from Proposition 2.4 where γ 1 , γ 2 and ξ satisfy (4.1) and (4.8).
Proof. We will prove the statement by multiplying the integral equation (3.6) of Proposition 3.2 by h, when u 1 and u 2 are given by (4.4) and (4.6), and then take the limit as h → 0.
To begin with notice that we may integrate over B − in (3.6), since and u j are defined in B, when j = 1, 2. We first show that for the second term in (3.6) we have as h → 0. Using the phase computations (4.7) we get that This is multiplied by an L ∞ function in (4.11). Since we restricted the choice of γ 1 to make the exponents purely imaginary, we see easily using the estimates (4.9) that (4.11) holds. Equation (3.6) multiplied by h, is thus reduced in the limit to whereζ j := (ζ j,1 , ζ j,2 , −ζ j,3 ), j = 1, 2. The terms of the product that do not contain the factor 1/h, result in integrals similar to the one in (4.11). One sees similarly using estimates (4.9) that they are zero in the limit of (4.12). The first term inside the limit in (4.12) is therefore reduced to Now we use the Riemann-Lebesgue Lemma to conclude that the terms with exponents containing ξ + and ξ − are zero in the limit. To see this, notice that by Remark 2.6, we see that Φ i L ∞ (B − ) < C, for some C > 0, when h is small enough.
The first term in (4.12) is therefore as h → 0. The terms containing r i in the products of P 1 and P 2 are, because of (4.9), zero in the limit. The above limit is thus equal to Finally we split the integral and do a change of variable in the second term and arrive at the expression for the first term of (4.12).
Returning to the second term in (4.12), containing u 1 ∇u 2 . This is of the same form as the first one. By doing the above derivation by simply exchanging the roles of u 1 and u 2 , we similarly see that the second term becomes (4.14) Now ζ 1 → (γ 1 + iγ 2 ) and ζ 2 → −(γ 1 + iγ 2 ), as h → 0. Thus by using (4.13) with (4.14), we can rewrite (4.12) as The next Proposition shows that (4.10) holds even when the exponential function depending on Φ 0 i , i = 1, 2 is removed. The argument follows [5] closely. We will give details for the convenience of the reader.
Proof. By (4.3) and (4.5) we have that Remark 2.7 furthermore implies that the amplitude e Φ1 in the definition of u 1 can be replaced by ge Φ1 , if g ∈ C ∞ (B) is a solution of (4.17) (γ 1 + iγ 2 ) · ∇g = 0 in B.
Rewriting (4.18) using this and a change of variable given by T we have T B ge iy·ξ ∂ z e Ψ dy = 0, for all g satisfying (4.17).
Notice that y · ξ = y 3 ξ 3 , since ξ is in the y-coordinates of the form (0, 0, ξ 3 ). The above integral is therefore a Fourier transform w.r.t. ξ 3 . Let g ∈ C ∞ (T B) satisfy ∂z g = 0 and be independent of y 3 . Then taking the inverse Fourier transform we write 0 = where T y3 := T B ∩ Π y3 and Π y3 = {(y 1 , y 2 , y 3 ) : (y 1 , y 2 ) ∈ R 2 }. Notice that the boundary of T y3 is smooth. Multiplying the above by 2i and using Stokes' theorem we get that 0 = 2i for all holomorphic functions g ∈ C ∞ (T y3 ).
To this end, we define F to be The function F is holomorphic away from ∂ T y3 . As e Ψ is Lipschitz, we know because of the Plemelj-Sokhotski-Privalov formula (see e.g. [10]), that Now the function ζ → (ζ − z) −1 is holomorphic on T y3 when z / ∈ T y3 . By choosing g(ζ) = (ζ − z) −1 in (4.19), we get therefore that F (z) = 0, when z / ∈ T y3 . Hence, the second limit in (4.21) vanishes, and therefore, F is holomorphic function on T y3 , such that (4.20) holds.
Next we show that F is non-vanishing in T y3 . When doing so, let ∂ T y3 be parametrized by z = γ(t), and N be the number of zeros of F in T y3 . Then by the argument principle, we get To see that the last integral is zero, notice that this the winding number of the path e Ψ•γ . And that e Ψ(γ(t)) is homotopic to the constant contour {1}, with the homotopy given by e sΨ(γ(t)) , s ∈ [0, 1]. Next, since F is a non-vanishing holomorphic function on T y3 and T y3 is simply connected, it admits a holomorphic logarithm. Hence, (4.20) implies that (log F )| ∂ Ty 3 = Ψ| ∂ Ty 3 .
Because log F = Ψ is continuous on ∂ T y3 , we have by the Cauchy theorem, where g ∈ C ∞ (T y3 ) is an arbitrary function such that ∂z g = 0. Using Stokes' formula as in (4.19) allows us to write this as Taking the Fourier transform with respect to y 3 , we get T (B) e iy·ξ g ∂z Ψdy = 0, for all ξ = (0, 0, ξ 3 ), ξ 3 ∈ R. Hence, returning back to the x variable, we obtain that Using (4.16), we finally get Setting g = 1, we obtain (4.15).

Determining the electric potential
In order to complete the proof of Theorem 1.2, we need to show that the electric potential is also determined by the DN-map. Again we assume that A j , q j and Γ j , j = 1, 2 are as in Theorem 1.2 and that the DN-maps satisfy (1.6), and hence (3.1).
Since B is simply connected, it follows from the Helmholtz decomposition of A 1 −Ã 2 and (4.25) that there exists ψ ∈ C 1,1 (B) with ψ = 0 near ∂ B such that We extend ψ to a function of class C 1,1 on all of R 3 such that ψ = 0 on R 3 \ B. ThenÃ SinceΓ j ⊂ R 3 − \ N , j = 1, 2, and R 3 − \ N is connected, we can assume that ψ = 0 on R 3 -\ N and hence, we have that ψ = 0 onΓ 1 ∪Γ 2 . It follows then from Lemma 1.1 part (i) and (3.1) that for all f with supp(f ) ⊂Γ 2 , Therefore, we can apply Proposition 3.2 with A 1 = A 2 and get for all u 1 ∈ W 1 (N ) and u 2 ∈ W * 2 (N ). Choosing in (5.1) u 1 and u 2 as the complex geometric optics solutions, given by (4.4) and (4.6), passing to B − , and letting h → 0, we have By Remark 2.7 e Φ1 in the definition (4.4) of u 1 can be replaced by ge Φ1 if g ∈ C ∞ (B) is a solution of (γ 1 + iγ 2 ) · ∇g = 0 in B.
6. Appendix 6.1. Magnetic Green's formulas. Let us first recall, following [5], the standard Green formula applied to the magnetic Schrödinger operator.
We shall also need a version of the above result where Ω is replaced by R 3 -. We shall then need to put some restrictions on v and u, because R 3 -is unbounded. To this end we assume that u and v are solutions to the Helmholtz equation outside some compact set, that obey some form of radiation condition. To be precise, let A ∈ W 1,∞ comp (R 3 -, R 3 ), q ∈ L ∞ comp (R 3 -), and let u ∈ H 2 loc (R 3 -) be such that supp(u| ∂ R 3 -) is compact, and u is outgoing. Assume also that v ∈ H 2 loc (R 3 -) satisfies is compact, and v is incoming.
Proof. Let B R := B(x 0 , R) be an open ball in R 3 of radius R, and choose R > 0 large enough so that supp(A), supp(q) ⊂ B R .
Thus, to obtain (6.1) we need to show that (6.2) Let us rewrite the left hand side of the above as follows, We show that first term goes to zero as R → ∞. The second term can be handled in the same way. Applying Cauchy-Schwarz gives Here the first integral goes to zero, since ∂ n v − ikv = ∂ n v + ikv and |∂ n v + ikv| 2 is o(1/r 2 ) as r = |x| → ∞, since v is incoming. We conclude the proof by showing that the second integral is bounded as R → ∞.
6.2. The unique continuation principle. In this work we make heavy use of the so called unique continuation principle. The unique continuation principle can be seen as an extension of the familiar property that an analytic function that is zero on some open set is identically zero.
Let Ω ⊂ R n be an open connected set, and let Here a ij ∈ C 1 (Ω) are real-valued, a ij = a ji , and there is C > 0 so that n i,j=1 a ij (x)ξ i ξ j ≥ C|ξ| 2 , x ∈ Ω, ξ ∈ R n .
Then u vanishes identically in Ω.