The fixed angle scattering problem with a first order perturbation

We study the inverse scattering problem of determining a magnetic field and electric potential from scattering measurements corresponding to finitely many plane waves. The main result shows that the coefficients are uniquely determined by $2n$ measurements up to a natural gauge. We also show that one can recover the full first order term for a related equation having no gauge invariance, and that it is possible to reduce the number of measurements if the coefficients have certain symmetries. This work extends the fixed angle scattering results of Rakesh and M. Salo to Hamiltonians with first order perturbations, and it is based on wave equation methods and Carleman estimates.


Introduction and Main Theorems
In this work we study the inverse scattering problem of recovering a first order perturbation from fixed angle scattering measurements. Let λ > 0, n ≥ 2 and let ω ∈ S n−1 be a fixed unit vector. Suppose that for m ∈ N, V(x, D) is a first order differential operator with C m (R n ) coefficients having compact support in B = {x ∈ R n : |x| < 1}, the open ball of radius 1. We consider a Hamiltonian H V = −∆ + V(x, D) in R n and the problem (1.1) (H V − λ 2 )ψ V = 0 ψ V (x, λ, ω) = e iλω·x + ψ s V (x, λ, ω), where ψ s V (x, λ, ω) is known as a the scattering solution. It is well known that in order to have uniqueness for this problem one needs to put further restrictions on the function ψ s V . See e.g. [Ya10] for the following facts. The outgoing Sommerfeld Radiation Condition (SRC for short) ∂ r ψ s V − iλψ s V = o(r −(n−1)/2 ) as r → ∞, where r = |x|, selects the solutions that heuristically behave as Fourier transforms in time of spherical waves that propagate towards infinity. A function ψ s V satisfying (1.1) and the SRC is called an outgoing scattering solution. This solution is given by the so called outgoing resolvent operator so that, formally, . Notice that, assuming that such a solution ψ s V exists-which would happen if the resolvent is well defined and bounded in appropriate spaces-it must satisfy the Helmholtz equation (−∆ − λ 2 )ψ s V = 0 in R n \ B, since the coefficients of V(x, D) are compactly supported in B. It is well known that a solution of Helmholtz equation satisfying the SRC has always the asymptotic expansion (1.3) ψ s V (x, λ, ω) = e iλ|x| |x| − n−1 2 a V (λ, θ, ω) + o(|x| − n−1 2 ), as |x| → ∞, where θ = x |x| and a V (λ, θ, ω) is called the scattering amplitude or far field pattern.
In this setting, the main objective of an inverse scattering problem consists in reconstructing the coefficients of V(x, D) from partial or full knowledge of a V (λ, θ, ω). Depending on the data that is assumed to be known we can distinguish several types of inverse scattering problems: 1. Full data. Recover the coefficients of V(x, D) from the knowledge of a V (λ, θ, ω) for all (λ, θ, ω) ∈ (0, ∞) × S n−1 × S n−1 . 2. Fixed frequency (or fixed energy). Recover the coefficients of V(x, D) from the knowledge of a V (λ 0 , θ, ω) for a fixed λ 0 > 0 and all (θ, ω) ∈ S n−1 × S n−1 . 3. Backscattering. Recover the coefficients of V(x, D) from the knowledge of a V (λ, ω, −ω) for all (λ, ω) ∈ (0, ∞) × S n−1 . 4. Fixed angle (single measurement). Recover the coefficients of V(x, D) from the knowledge of a V (λ, θ, ω 0 ) for a fixed ω 0 ∈ S n−1 and all (λ, θ) ∈ (0, ∞) × S n−1 . In the case of fixed angle scattering it is also interesting to consider analogous inverse problems in which a V ( · , · , ω) is assumed to be known for each ω in a fixed subset (usually finite) of S n−1 .
Let D = −i∇. We consider the Hamiltonian where both the magnetic potential A, and the electrostatic potential q are real. Then H V is selfadjoint, and if A and q are compactly supported, the resolvent (1.2) is bounded in appropriate spaces under very general assumptions on the regularity of A and q. This implies that the problem (1.1) has a unique solution ψ s V and hence that the scattering amplitude a V = a A,q is well defined, so that the fixed angle scattering problem can be appropriately stated.
In [RS20a,RS20b] it has been proved that for H V = −∆ + q, knowledge of the fixed angle scattering data a V ( · , · , ω) in two opposite directions ω = ±ω 0 for a fixed ω 0 ∈ S n−1 , determines uniquely the potential q. The present article extends the results of [RS20a,RS20b] to the case of non-vanishing first order coefficients and proves that from 2n measurements, or just n + 1 measurements under symmetry conditions, one can determine both the first and zeroth order coefficients up to natural gauges. To prove these results, we follow the approach used in [RS20b], that is we show the equivalence of the fixed angle scattering problem with an appropriate inverse problem for the wave equation. This inverse scattering problem in time domain consists in recovering information on A and q from boundary measurements of the solution U A,q of the initial value problem If the support of A and q is contained in B, the boundary measurements of U A,q are made in the set ∂B × (−T, T ) ∩ {(x, t) : t ≥ x · ω}, where ∂B denotes the boundary of the ball. We now describe some previous results on the inverse scattering problem of recovering a potential q(x) from fixed angle measurements. As discussed above, this problem can be considered in the frequency domain, as the problem of determining q from the scattering amplitude a q ( · , · , ω) for the Schrödinger operator −∆ + q with a fixed direction ω ∈ S n−1 , or alternatively in the time domain as the problem of recovering q from boundary or scattering measurements of the solution U q of the wave equation. The equivalence of these problems is discussed in [RS20b] (see [Me95,Uh01,MU] for the odd dimensional case).
The one-dimensional case is quite classical, see [Ma11,DT79]. In dimensions n ≥ 2 uniqueness has been proved for small or generic potentials [St92,B+20], recovery of singularities results are given in [Ru01,Me18], and uniqueness of the zero potential is considered in [BM89]. Recently in [RS20a,RS20b] it was proved that measurements for two opposite fixed angles uniquely determine a potential q ∈ C ∞ c (R n ). The problem with one measurement remains open, but [RS20a,RS20b] prove uniqueness for symmetric or horizontally controlled potentials (similar to angularly controlled potentials in backscattering [RU14]), and Lipschitz stability estimates are given for the wave equation version of the problem. We also mention the recent work [MS20] which studies the fixed angle problem when the Euclidean metric is replaced by a Riemannian metric, or sound speed, satisfying certain conditions, and the upcoming work [KRS20] which studies fixed angle scattering for time-dependent coefficients also in the case of first order perturbations.
We now introduce the main results in this work. Since the metric is Euclidean, the vector potential A can equivalently be seen as a 1-form A = A j dx j . We denote by dA the exterior derivative of A. Our first result shows that the magnetic field dA and the electrostatic potential q are uniquely determined by the knowledge of the fixed angle scattering amplitude a A,q ( · , · , ω) for n orthogonal directions ω = e j , 1 ≤ j ≤ n and the n opposite ones, ω = −e j . From now on, in this paper we will fix m to be the integer (1.5) m = 3 2 n + 10 if n is even, m = 3 2 (n + 1) + 10 if n is odd.
In general we consider A ∈ C m+2 (R n ; R n ) and q ∈ C m (R n ; R). This is required in order to guarantee that the solutions of (1.4) satisfy certain regularity properties.
The condition (1.6) is a technical restriction necessary to decouple the information on q from the information on A at some point in the proof of this uniqueness result.
In Theorem 1.1 one cannot recover completely the magnetic potential A due to the phenomenon of gauge invariance. This consists simply in the observation that if H A,q u = v for some functions u and v, then H A+∇f,q (e −if u) = e −if v, for any f ∈ C 2 (R n ). Therefore if f is compactly supported, the scattering amplitude is not going to be affected by f , so that a A,q = a A+∇f,q . On the other hand if we consider HamiltoniansH where V is a fixed function, then the gauge invariance is broken and knowledge of the associated scattering amplitudeã A,V ( · , · , ω) for the 2n directions ω = ±e j determines completely A. Theorem 1.2. Let n ≥ 2 and λ 0 > 0, and let e 1 , . . . e n be any orthonormal basis in R n . Assume that A 1 , A 2 ∈ C m+2 c (R n ; R n ) and V ∈ C m c (R n ; C) have compact support in B, and that the HamiltoniansH A1,V andH A2,V are both self-adjoint operators. If for all θ ∈ S n−1 and λ ≥ λ 0 we havẽ Notice that in this statement (1.6) is not assumed. This is related to the fact that V is fixed, so it is not necessary to decouple V from A in the proof. Similarly, if A is a fixed vector potential, it would be possible to determine q from the knowledge of a A,q ( · , · , ±ω) for a fixed ω ∈ S n−1 .
In both the previous theorems we need to measure the scattering amplitude generated by a wave incoming from 2n different directions. In some cases we can avoid the need of sending a wave also from the opposite direction provided we assume there are certain symmetries in the potentials. As an example of this phenomenon we state the following result. Theorem 1.3. Let n ≥ 2 and λ 0 > 0, and let e 1 , . . . e n be any orthonormal basis in R n . Assume that the pairs of potentials A 1 , A 2 ∈ C m+2 c (R n ; R n ) and q 1 , q 2 ∈ C m c (R n ; R) are compactly supported in B. Assume also that If for all θ ∈ S n−1 and λ ≥ λ 0 we have a A1,q1 (λ, θ, e j ) = a A2,q2 (λ, θ, e j ) for all j = 1, . . . n − 1, and a A1,q1 (λ, θ, ±e n ) = a A2,q2 (λ, θ, ±e n ), then dA 1 = dA 2 and q 1 = q 2 .
We assume that a A1,q1 (λ, θ, ω) = a A2,q2 (λ, θ, ω) for ω = ±e n instead of just ω = e n since we have not considered any symmetry on the potential q (a result assuming symmetries on q to reduce further the data could also be proved modifying slightly the arguments used to prove this theorem). We also prove in time domain a more technical result analogous to Theorem 1.2 that requires just n measurements instead of 2n, and that is compatible with less restrictive symmetry conditions than (1.7) (see Theorem 5.1 below). In all the previous theorems we have considered an orthonormal basis {e 1 , e 2 , . . . e n } of R n in order to simplify the notation and computations in some parts of the arguments, but we remark that our proofs can be easily adapted to allow also non-orthonormal directions of measurements.
As already mentioned, the previous theorems follow from corresponding results for the time domain inverse problem (Theorems 2.1, 2.2, and 5.2, respectively). We now state the precise result that establishes the equivalence between the inverse scattering problem in frequency domain and the inverse scattering problem in time domain, extending the results in [RS20b] to first order perturbations.
Theorem 1.4. Let n ≥ 2, ω ∈ S n−1 , and λ 0 > 0. Assume that A 1 , A 2 ∈ C m+2 c (R n ; R n ) and q 1 , q 2 ∈ C m c (R n ; R) are supported in B. For k = 1, 2, let U A k ,q k (x, t; ω) be the unique distributional solution of the initial value problem . Then one has that a A1,q1 (λ, θ, ω) = a A2,q2 (λ, θ, ω) for λ ≥ λ 0 and θ ∈ S n−1 , if and only if We remark that the restriction of the distribution U A k ,q k to the surface ∂B × R is always well defined and vanishes in the open set (∂B × R) ∩ {t < x · ω}. This can be seen from the explicit formula for U A k ,q k that we will compute in section 2.
The proof of the main results in time domain is based on a Carleman estimate method introduced in [RS20a,RS20b], which in turn adapts the method introduced in [BK81] (see [IY01,Kl13,BY17] for more information and references on the Bukhgeim-Klibanov method). Essentially, the Carleman estimate is applied to the difference of two solutions of (1.4). The general idea is to choose an appropriate Carleman weight function for the wave operator that is large on the surface {t = x · ω} and allows one to control a source term on the right hand side of the equation. Then one needs an additional energy estimate to absorb the error coming from the source term. This will allow one to control the difference of the potentials A 1 − A 2 or q 1 − q 2 . This step is the key to get the uniqueness result. Unfortunately, after doing all this, there is a remaining boundary term in the Carleman estimate that cannot be appropriately controlled. However, this term can be canceled using an equivalent Carleman estimate for solutions of the wave equation coming from the opposite direction. This is why we require 2n measurements to recover n independent functions instead of just n measurements. Assuming symmetry properties on the coefficients like in Theorem 1.3 is essentially an alternative way to get around this difficulty.
An interesting point in the proof of the time domain results is how one decouples the information concerning A from the information on q. The method used here consists in considering the solutions of the initial value problem where H stands for the Heaviside function. Since (1.4) is essentially the time derivative of the previous IVP, it turns out that it is equivalent to formulate the inverse scattering problem in time domain using any of these initial value problems. The advantage is that the solutions of (1.8) contain information only about A at the surface {t = x · ω}. By using these ideas, we are able to estimate both A 1 − A 2 in terms of q 1 − q 2 and q 1 − q 2 in terms of A 1 − A 2 . Using these two estimates in tandem allows us to recover both the magnetic field and electric potential under the assumption (1.6). This paper is structured as follows. In Section 2 we state the time domain results, Theorem 2.1 and Theorem 2.2, from which Theorems 1.1 and 1.2 follow by Theorem 1.4. We also analyze the structure of the solutions of the initial value problems (1.4) and (1.8) and we state several of their properties that will play an essential role later on. In Section 3 we introduce the Carleman estimate and in Section 4 we combine the results of the previous two sections to prove Theorems 2.1 and 2.2. In the last section of the paper we state and prove Theorems 5.1 and 5.2 in order to illustrate how the number of measurements can be reduced in time domain by imposing symmetry assumptions on the potentials (Theorem 1.3 follows from the second result). The proof of Theorem 1.4 is given in Appendix A, and Appendix B is devoted to adapting several known results for the wave operator to our purposes.
Acknowledgements. C.M. was supported by project MTM2017-85934-C3-3-P. L.P. and M.S. were supported by the Academy of Finland (Finnish Centre of Excellence in Inverse Modelling and Imaging, grant numbers 312121 and 309963), and M.S. was also supported by the European Research Council under Horizon 2020 (ERC CoG 770924).

The inverse problem in time domain
Main results in time domain. Let ω ∈ S n−1 be fixed. In the time domain setting we consider the initial value problem where δ represents the 1-dimensional delta distribution and H V = −∆ + V(x, D) . Formally, the problem (1.1) is the Fourier transform in the time variable of (2.1). As we will show later in this section, there is a unique distributional solution of U V if the first order coefficients of V are in C m+2 c (R n ) and the zero order coefficient is C m c (R n ), for m as in (1.5). The inverse problem in the time domain consists in determining the coefficients of V from certain measurements of U V at the boundary ∂B × (−T, T ) ⊂ R n+1 for some fixed T > 0. To simply the notation we define Σ := ∂B × (−T, T ).
From now, depending on the context, it will be useful to write the Hamiltonian H V both in the forms where A, W ∈ C m+2 c (R n ; C n ) and q, V ∈ C m c (R n ; C). Since the coefficients have high regularity and are complex valued, both forms are completely equivalent, but the first notation is specially convenient in the cases where there is gauge invariance. In fact, this inverse problem has an invariance equivalent to the gauge invariance present in the frequency domain problem. A straightforward computation shows that if U is a solution of where f is any C 2 (R n ) function with compact support in B. The initial condition satisfied by U is not affected by the exponential factor e −f since for t < −1 the distribution δ(t − x · ω) is supported in {x · ω < −1}, a region where f vanishes. On the other hand we also have that U | Σ = U | Σ since the support of f is contained in B. Hence, at best one can recover the magnetic field dA from the boundary data U | Σ . We now state two uniqueness results for the inverse problem that we have just introduced.
Also, let 1 ≤ j ≤ n and consider the 2n solutions U k,±j (x, t) of . If for each 1 ≤ j ≤ n one has U 1,±j = U 2,±j on the surface Σ ∩ {t ≥ ±x j }, then dA 1 = dA 2 and q 1 = q 2 .
As in the introduction, we highlight that the restriction of the distribution U 1,±j to the surface Σ is well defined and vanishes in the open set Σ ∩ {t < ±x j }, see the comments after Proposition 2.4 for more details. Theorem 1.1 follows directly from this result and Theorem 1.4. On the other hand, if we fix the zero order term to be always the same, then the gauge invariance disappears and one can recover completely the first order term of the perturbation. To state this result we use the Hamiltonian in the form (2.3).
Theorem 2.2. Let W 1 , W 2 ∈ C m+2 c (R n ; C n ) and V ∈ C m c (R n ; C) with compact support in B. Let 1 ≤ j ≤ n and k = 1, 2, and consider the corresponding 2n solutions U k,±j satisfying As in the previous case, Theorem 1.2 follows from Theorem 2.2 with W k = −iA k and Theorem 1.4. To see this, notice that for k = 1, 2 we can write the Hamiltonians in Theorem 1.2 as And since by assumption A k is real and H A k is self-adjoint, q k must be a real function. This means that the conditions required to apply Theorem 1.4 are satisfied.
In Theorem 2.2 one needs 2n measurements to obtain the unique determination of the first order coefficient W. Therefore, one could expect to need 2(n + 1) measurements in Theorem 2.1, since one now also proves the unique determination of q. In fact, 2n measurements are always enough: Figure 1. The regions Q, Q ± and Γ the gauge invariance essentially reduces one degree of freedom by making it possible to choose a gauge in which the nth component of A 1 − A 2 vanishes (as we shall see later on, the fact that the solutions U n,±j coincide at Σ ∩ {t = ±x n } guarantees that there are no obstructions for this gauge transformation).
The direct problem. In order to prove the previous theorems we need to study the direct problem (2.1) in more detail. Let ω ∈ S n−1 . Assume W ∈ C m+2 c (R n ; C n ) and V ∈ C m c (R; C), and consider the initial value problems for the wave operator , where L W,V was defined in (2.3). As mentioned in the introduction, the reason we also consider the second equation is that the δ-wave U δ and H-wave U H contain equivalent information about W and V (see Proposition 2.5 below), but H-waves decouple the information on W from the information on V .
To study (2.6) and (2.7), it is convenient use the following coordinate system in R n . We take x = (y, z) where y ∈ R n−1 and z = x · ω. For a fixed T > 7, it will be helpful to introduce the following subsets of R n+1 (see Figure 1 We now give a heuristic motivation of the existence of solutions U δ and U H of (2.6) and (2.7). For the interested reader we give a proof in Section B.1 of the properties that we now state, by means of the progressing wave expansion method. We start by making the ansatz of looking for possible solutions of (2.6) and (2.7) in the family of functions satisfying where f (y, z, t) and g(y, z, t) are C 2 functions in {t ≥ z}. A straightforward computation shows that (2.9) In the case of equation (2.7) to satisfy the initial condition we need to have f = u where u is a function satisfying u(x, t) = 1 for all t < −1, and g(x, t) = 0 for all (x, t) ∈ R n+1 . Then (2.9) implies that u must satisfy The unique solution of the last ODE is since it has to satisfy the initial condition u(y, z, z) = 1 for z < −1. We now state this and further results about the solution of (2.7).
There is a unique distributional solution U H (x, t; ω) of (2.7), and it is supported in the region {t ≥ x · ω}. In particular, where u is a C 2 function in {t ≥ x · ω} satisfying the IVP Notice that the boundary value of u at {t = x · ω} depends only on W and not on the zero order term V . We now study the solutions of (2.6). In this case we have to consider (2.9) with g = 1 for t < −1, and f = v where v is a function satisfying v(x, t) = 0 for t < −1. Then the following conditions must be satisfied: The last two conditions are required so that [ vanishes completely in all R n+1 as a distribution. As in the case of (2.10), the third ODE and the initial condition imply that g(y, z, z) = e ψ(y,z) = e z −∞ ω·W(y,s) ds , and in fact we are going to choose g(y, z, t) = e ψ(y,z) for all (y, z, t) ∈ R n+1 . We can freely do this: g 1 (y, z, t)δ(t − z) = g 2 (y, z, t)δ(t − z) iff g 1 (y, z, z) = g 2 (y, z, z). Also, the previous choice implies that ∂ t g = 0, so that the last condition is satisfied too. Computing explicitly L W,V (g), the second equation in (2.13) becomes the ODE We state this rigorously in the following proposition.
One of the consequences of the formula (2.14) is that the restriction of U δ (x, t; ω) to the surface Σ is well defined. This essentially follows from the fact that the wave front set of the distribution δ(t − x · ω) is disjoint from the normal bundle of Σ. We emphasize that U δ satisfies the initial condition U δ | {t<−1} = δ(t − x · ω), even if it does not look that way at a first glance. This is due to the fact that when t < −1 the distribution δ(t − x · ω) is supported in {x · ω < −1}, a region in which ψ vanishes. Therefore As mentioned previously, for more details about the proofs of Propositions 2.3 and 2.4 see Section B.1. We remark that the condition (1.5) on the regularity of the coefficients appears in the proofs of these propositions in order to have C 2 solutions u and v of (2.12) and (2.15). An important fact later on is that the solutions of (2.6) and (2.7) satisfy that ∂ t U H = U δ . This is consequence of the independence of V and W from t together with the uniqueness of solutions for both equations. Of particular relevance for the scattering problem will be that this equivalence also holds for the boundary data: knowledge of U H | Σ+ gives U δ | Σ+ and vice versa.
, and let ω ∈ S n−1 be fixed. If U δ and U H are, respectively, the unique distributional solutions of (2.6) and (2.7), then one has that ∂ t U H = U δ in the sense of distributions. In fact, it also holds that v(x, t) = ∂ t u(x, t), so that (2.14) can be written as Specifically, if ψ is given by (2.11) we have that Notice that the previous identity holds in particular for every x ∈ ∂B, so we can write that Proof of Proposition 2.5. Since W and V are independent of t, we can take a time derivative of both sides of (2.7). This implies that ∂ t U H satisfies Proposition 2.4 implies there is a unique distributional solution of (2.7), and hence ∂ t U H = U δ . Then ∂ t u = v. We also get that u(x, x · ω) = e ψ , but we already knew this from Proposition 2.3. Therefore (2.16) holds true. Identity (2.17) follows directly by the fundamental theorem of calculus.
As an immediate consequence of identity (2.18) we get the following lemma.
. For k = 1, 2 and ω ∈ S n−1 , consider the solutions U H,k and U δ,k of Then Energy estimates. To finish this section we state three different estimates related to the wave operator that will be useful later on. They are analogues of the estimates given in [RS20a, Lemmas 3.3-3.5], modified in order to account for the presence of a first order perturbation not considered in the mentioned paper. For completeness we have included the proofs in Section B.2. The first two lemmas will be used to control certain boundary terms appearing in the Carleman estimate. We denote by ∇ Γ α the component of ∇α tangential to Γ.
holds for every α ∈ C ∞ (Q + ) and for every σ ≥ σ 0 . The implicit constant depends on φ C 2 (Q + ) , 1 Throughout the paper we write a b or equivalently b a, when a and b are positive constants and there exists C > 0 so that a ≤ Cb. We refer to C as the implicit constant in the estimate.
The last lemma will be used to show that the normal derivative of a function α that vanishes at the set Σ + will also vanish provided that certain conditions are met. Here ν denotes the unit vector field normal to Σ + .
Assume that on the region |(y, z)| ≥ 1 we have for any function χ ∈ C 1 (Q + ). The implicit constant depends on T .

The Carleman estimate and its consequences
In this section we are going to apply a suitable Carleman estimate in order to be able to control the difference of the potentials with the boundary data. For this purpose we adapt a Carleman estimate for general second order operators stated in [RS20b,Theorem A.7]. The trick is to choose an appropriate weight function to obtain a meaningful estimate for the wave operator.
First, take any ϑ ∈ R n such that |ϑ| = 2 and consider the following smooth function, The weight in the Carleman estimate is going to be the function φ = e λη for some λ > 0 large enough. This choice is made in order to have several properties. On the one hand we want φ to be sufficiently pseudoconvex so that the Carleman estimate holds for the wave operator. On the other hand, we want φ to decay very fast when t > z in order to deal with certain terms appearing in the Carleman estimate (see Lemma 3.2 below and its application in the proof of Lemma 3.3). And finally, since the z coordinate is going to be determined by the direction ω of the traveling wave, we require the restriction of φ to the surface {t = z} to be independent of ω (or in other words, dependent on x but independent of the choice of coordinates x = (y, z)). This is of great help since we recover A combining measurements made from waves traveling in different directions. In order to state the Carleman estimate we fix an open set D ⊂ {(x, t) ∈ R n+1 : |x| < 3/2} such that Q ⊂ D (recall the notation introduced in (2.8)).
holds for all u ∈ C 2 (Ω) and all σ ≥ σ 0 . The implicit constant depends on n, λ, and Ω. Here ν is the outward pointing unit vector normal to ∂Ω, and for real v and 0 ≤ j ≤ n, E j is given by where g is some real valued and bounded function independent of σ and v (here the index 0 corresponds to t, so that ∂ 0 φ = ∂ t φ).
Proof. Since h is real if h is a real function, the statement follows from applying Theorem A.7 of [RS20b] to the real and imaginary parts of u, and then adding the resulting estimates. However, to apply the mentioned result, one needs to verify that φ is strongly pseudoconvex in the domain D with respect the wave operator . The reader can find the precise definition of this condition in [RS20b,Appendix]), though it is not necessary for the discussion that follows.
Denote by (ξ, τ ) ∈ R n+1 the Fourier variables corresponding to (x, t), where ξ ∈ R n and τ ∈ R. It can be proved that a function φ = e λη will be strongly pseudoconvex for λ > 0 large enough provided η satisfies certain technical conditions. By Propositions A.3 and A.5 in [RS20b], these conditions are the following: one needs to verify that the level surfaces of η are pseudoconvex with respect to the wave operator , and that |(∇ x,t η)(x, t)| > 0 for all (x, t) ∈ D. In our case, the second property is immediate since |ϑ| = 2 implies that |x − ϑ| 2 has non-vanishing gradient in D. The reader can find in [RS20b, Definition A.1] a precise definition of the first property. For the purpose of this work it is enough to use that the level surfaces of a function f are pseudoconvex w.
We now consider f given by where |ϑ| = 2, b > 0, and we are using coordinates (x, t) = (y, z, t), y ∈ R n−1 . Let (ζ, ρ, τ ) ∈ R n−1 × R × R be the Fourier variables counterpart to (y, z, t). A straightforward computation with j, k = 1, . . . , n − 1 and ∂ j = ∂ yj , shows that the only non-vanishing second order derivatives of f are Note that both conditions are homogeneous in the variables (ζ, ρ, τ ), so it is enough to study the case τ = 1. Thus (3.4) becomes It is not difficult to verify that for b > 3 this inequality always holds. This proves that for b = 4 the level surfaces of f are strongly pseudoconvex with respect to , and therefore, the same holds for η = 1 2 f . To finish, we mention that precise formula (3.2) of the quadratic forms E j is computed in detail in [RS20b, Section A.2].
In the following lemma we prove a couple of properties of the weight φ that will be important later in order to show that some terms appearing in the Carleman estimate are suitably small in the parameter σ.
Lemma 3.2. Let ϑ ∈ R n with |ϑ| = 2, and η as in (3.1). Then, the following properties are satisfied for any T > 7: i) The smallest value of φ on Γ is strictly larger than the largest value of φ on Proof. Let η 0 (y, z) := |x − ϑ| 2 . For the first assertion it is enough to prove that for T large enough, since η(y, z, z) = η 0 (y, z). Observing that |(y, z)| ≤ 1 in Γ and Γ ±T , the previous inequality will hold if and since max B η 0 − min B η 0 = 8, this is true for any T > 7. This yields the first assertion. To prove the second assertion we are going to use the following inequality Since e λη0 > 1 always, we have and hence, since |z| ≤ 1, By the dominated convergence theorem, the last integral goes to zero when σ → ∞, and therefore lim σ→∞ κ(σ) = 0. This completes the proof.
We now adapt the Carleman estimate of Proposition 3.1 to our purposes. First, define for x ∈ R n and |ϑ| = 2, so that φ(y, z, z) = φ 0 (y, z). From now on it is convenient to use the notation We are interested in applying Proposition 3.1 for Ω = Q ± . The boundary of Q ± is composed of the following regions: ∂Q ± = Γ ∪ Σ ± ∪ Γ ±T . We are going to use the precise formula (3.2) only in Γ, where in fact an explicit computation yields that for any real function v ∈ C 2 (Q ± ) (for a detailed derivation of this formula see [RS20b, Section A.2]). The important thing about this identity is that it does not depend on ∇v but just on Zv and ∇ y v, which are derivatives along directions tangent to Γ.
The following lemma is the consequence of the Carleman estimate in Proposition 3.1 that is relevant for our fixed angle scattering problem. It is an analogue of [RS20b, Proposition 3.2] adapted to the case of magnetic potentials.
Lemma 3.3. Let T > 7 and ω ∈ S n−1 , and let φ 0 be as above. Assume that and that hold for some vector fields E ± , A ± ∈ C(R n ; C n ), some functions f ± , h ± , q ± ∈ C(R n ; C), and w ± ∈ C 2 (Q ± ). Then there is a constant c > 0 such that, for σ > 0 large enough, where γ is a positive function satisfying γ(σ) → 0 as σ → ∞. The implicit constant is independent of σ and ϑ and w ± .
We remark that F j depends on the function g in Proposition 3.1 which could in principle be different in the domains Q + and Q − , but as discussed in [RS20b, footnote 1 in the proof of Proposition 3.2] we can choose g to be the same both in Q + and Q − .
By (3.6) and Lemma 3.2 we have , where κ(σ) → 0 as σ → ∞. Using this in (3.14) yields where γ(σ) := κ(σ) + σ 3 e −2δσ also satisfies γ(σ) → 0 as σ → ∞. We now use that φ(y, z, z) = e λ|x−v| 2 = φ 0 (x), which means that we can write the previous estimate changing the L 2 (Γ) norms to L 2 (B) norms (this is possible since the integrands do not depend on t any more). Also, (3.5) and (3.9) imply that F j (x, σw, ∇w)ν j | Γ = F j (x, σw, ∇ Γ w)ν j | Γ , that is, F j ν j on Γ only depends on the part of the gradient of w tangential to Γ. Applying these observations and rewriting the previous estimate with w = w + and A = A + , yields Fix ν to be the downward pointing unit normal to Γ, so ν is an exterior normal for Q + . An analogous argument in Q − yields the estimate where the minus sign in the boundary term comes from the fact that the outward pointing normal at Γ seen as part of the boundary of Q − is the opposite to that of the case of Q + . Now, since F j (x, σw, ∇ Γ w)ν j is quadratic in w and ∇ Γ w (and the coefficients are bounded functions), we have that Therefore, adding (3.15) and (3.16) and applying the previous estimate gives the desired result. Lemma 3.3 is going to be used for two different purposes and, as a consequence, it will be convenient to restate the estimate in a more specific way. We do this in the following couple of lemmas.
The proof is immediate from Lemma 3.3.
Lemma 3.5. Let T > 7. Also, let A ± , q ± , and γ(σ) be as in Lemma 3.3. Suppose that for a fixed ω ∈ S n−1 there exist w ± , E ± , and f ± satisfying the assumptions of Lemma 3.3, and such that (3.6) and (3.7) hold with w + = w − on Γ and h ± = ω · A ± . Assume also that w ± | Σ± = 0 and ∂ ν w ± | Σ± = 0. Then one has that Moreover, let {e j } 1≤j≤n be an orthonormal basis of R n , and let J ⊆ {1, . . . , n} be the set of natural numbers satisfying that at least one of the components e j · A + and e j · A − does not vanish completely in R n . Suppose that for each j ∈ J, estimate (3.18) holds with ω = e j . Then one has that for σ > 0 large enough.
Proof. Estimate (3.18) follows immediately from Lemma 3.3 under the assumptions in the statement. Let us prove (3.19). By assumption for each j ∈ J we have But notice that, by definition, e j · A ± = 0 in R n if j / ∈ J, so that A ± = j∈J (e j · A ± )e j . This means that adding (3.20) for all j ∈ J gives Then, using that γ(σ) → 0 as σ → ∞, to finish the proof is enough to take σ large enough in order to absorb in the left hand side the first term on the right.
We have already seen in Section 2 that the restriction to Γ of the solutions of (2.7) and (2.6) is in a certain sense related to the coefficients of the perturbation. In the next section we use this fact to construct appropriate functions w + and w − so that estimates (3.17) and (3.19) hold simultaneously with A ± and q ± being quantities related to the differences of potentials A 1 − A 2 and q 1 − q 2 . This will yield the proofs of Theorems 2.1 and 2.2, since the function γ(σ) in the previous lemmas goes to zero when σ → ∞.

Proof of the uniqueness theorems with 2n measurements
With Lemmas 3.4 and 3.5 we can finally prove the uniqueness results, Theorem 2.1 and 2.2. Both theorems are stated in terms of pairs of measurements, one for each direction ±e j . Due to this fact, it is convenient to give an explicit expression in the same coordinate system for solutions of (2.7) and (2.6) that correspond to the opposite directions ±ω 0 , where ω 0 ∈ S n−1 is fixed.
In the case of Proposition 2.4, it is convenient to state the results in the case where the Hamiltonian is written in the form (2.2). This is easily obtained making the change W = −iA and V = A 2 + D · A + q in the previous results. Therefore, using the same coordinates as before, the solutions U δ,± (y, z, t) := U δ (y, z, t; ±ω 0 ) of (2.6) given by Proposition 2.4 satisfy Having these explicit coordinate expressions, we now prove Theorems 2.1 and 2.2. We start with the second one which is the simplest, since the zeroth order term V is fixed. The proof consists in the construction of two appropriate functions w + and w − using the solutions U k,±j of (2.7) in order to apply the results introduced in the previous section. For a fixed direction ω = e j and k = 1, 2, we use U k,+j to construct w + and, with a certain gauge change, we use U k,−j to construct w − . The gauge change is a technical requirement necessary to have w + = w − in Γ, as assumed in Lemmas 3.4 and 3.5. Notice that the reason for this assumption comes from the fact that one wants to get rid of the first term on the right hand side of the estimate (3.8) which is large when the parameter σ is large.
Proof of Theorem 2.2. By Lemma 2.6 we know that it is completely equivalent to consider that the U k,±j are solutions of the initial value problem (∂ 2 t − ∆ + 2W k · ∇ + V )U k,±j = 0, in R n+1 , U k,±j | {t<−1} = H(t − ±x j ). instead of the IVP (2.5). Here it is convenient to work with H-waves instead of δ-waves since, as mentioned previously, the former have boundary values on Γ that are independent of V (see Proposition 2.3).
To apply Lemma 3.3 and Lemma 3.5 we need also to define an appropriate function w − in Q − . To obtain a useful choice we now consider the solutions u k,−j of (4.4) and we take w − (y, z, t) = e µj (y) (u 1,−j (y, z, −t) − u 2,−j (y, z, −t)).
We now go to the remaining case, Theorem 2.1. In this result the zero order term is not fixed so that there is gauge invariance (in fact, to simplify the proof it will be convenient fix a specific gauge). This is a harder proof than the previous one since we need to decouple information about q from the information about A.
Proof of Theorem 2.1. Let k = 1, 2. Since making a change of gauge A k − ∇f k with f k compactly supported leaves invariant the measured values U k,±j | Σ+,±j we can freely choose a suitable f k in order to simplify the problem. In fact, to show that dA 1 = dA 2 it is enough to prove that A 1 = A 2 in a specific fixed gauge. Under the assumptions in the statement one can always take since (1.6) implies that f k must be compactly supported in B. With this choice, one has that e n · (A 1 − ∇f 1 ) = e n · (A 2 − ∇f 2 ) = 0, in R n . Therefore, the previous arguments show that from now on we can assume without loss of generality that we have fixed a gauge such that e n · A 1 = e n · A 2 = 0 in R n .
Fix 1 ≤ j ≤ n − 1. We again take the coordinates in R n given by x = (y, z), where y ∈ R n−1 and z = x j . Let k = 1, 2. As in the proof of Theorem 2.2, by Lemma 2.6 we can assume that U k,±j satisfies the IVP , instead of (2.4). By Proposition 2.3 we know that the U k,±j have the structure described in (4.3) for 1 ≤ j ≤ n − 1 where u k,±j satisfies (4.1) with W = −iA k , ω = ±e j , and ω 0 = e j . Writing this in detail we obtain that (4.10) By the assumption that U 1,±j = U 2,±j on Σ ± and (4.3), we have that (4.5) holds analogously in this case. Specifically, in Σ + ∩ Γ this implies that there is a function µ j such that We now define w + := u 1,j − u 2,j , so that w + ∈ C 2 (Q + ) by Proposition 2.3, and To define w − we consider the solutions u k,−j of (4.10) and we take Then w − ∈ C 2 (Q − ) as desired. Also, for t = z = e µj (y)+i ∞ z ej ·A1(y,s) ds − e µj (y)+i ∞ z ej ·A2(y,s) ds = w + (y, z, z). Hence, if we define (4.12) and w − satisfies On the other hand, u 2,±j (x, ±t) and |∇u 2,±j (x, ±t)| are also bounded in Q ± , and on Γ we have that |u 2,±j | 1 by (4.10). Therefore, if f + = V 1 , f − = V 1 , E + = −iA 1 , and E − = −iA 1 + ∇µ j (notice that e j · ∇µ j = 0), one gets so that (3.6) and (3.7) are satisfied with h ± = e j · (A 1 − A 2 ), (4.13) As mentioned previously, (4.5) holds analogously in this case, so that w + | Σ+ = w − | Σ− = 0. Again, applying Lemma 2.9 respectively with α = w + and χ = 1, or α = e −µj w − and χ = e µj yields that ∂ ν w ± | Σ± = 0. Also, (4.11) shows that w + = w − in Γ. These assertions hold for each j = 1, . . . , n − 1, and the n-th component e n · (A 1 − A 2 ) vanishes in R n . This means that the assumptions of Lemma 3.5 are satisfied with J = {1, . . . , n − 1}. As a consequence (3.18) holds for each j ∈ J and hence (3.18) also holds with the choices established in (4.13). This gives (4.14) We now use the information provided by U k,±n . We use the same coordinates as before, in this case with z = x n . Since e n · A 1 = e n · A 2 = 0 in all R n , we have that

Reducing the number of measurements
In this section we prove an analogous result to Theorem 2.2, in which the number of measurements is reduced to n. To compensate this, one needs assume that the potentials have certain symmetries (in fact, each component of W must satisfy some kind of antisymmetry property). The main change in the proof is in the definition of w − in Q − , which now is constructed by symmetry from w + , instead of using new information coming from the solution associated to the opposite direction. For each 0 ≤ j ≤ the symmetry of e j · W plays an essential role since it is necessary to show that w + = w − in Γ, in order to extract meaningful results from the Carleman estimate.
Theorem 5.1. Let W 1 , W 2 ∈ C m+2 c (R n ; C n ), and V ∈ C m c (R n ; C) with compact support in B. Also, let 1 ≤ j ≤ n and k = 1, 2, and consider the n solutions U k,j of . Assume also that for each 1 ≤ j ≤ n there exists an orthogonal transformation O j satisfying that O j (e j ) = −e j and such that Then, if for all 1 ≤ j ≤ n one has U 1,j = U 2,j on the surface Σ ∩ {t ≥ x j }, it holds that W 1 = W 2 .
The simplest example of a vector field satisfying the previous conditions is the case of an antisymmetric vector fields W k , that is, such that W k (−x) = −W k (x) (for example, the gradient of a radial function). We remark that it is possible to show that the previous theorem also holds if one substitutes condition (5.2) by e j · W k (x) = −e j · W k (O j (x)) for all k = 1, 2 and 1 ≤ j ≤ n, that is, a symmetry condition instead of an antisymmetry condition in the imaginary part of e j · W k .
Proof of Theorem 5.1. By Lemma 2.6 we know that it is completely equivalent to assume that the U k,j solutions solve . instead of (5.1). Take 1 ≤ j ≤ n. In this proof we fix x = (y, z) in R n , where z = x j and y ∈ R n−1 . Let k = 1, 2. By Proposition 2.3 and (4.1) we know that U k,j (y, z, t) = u k,j (y, z, t)H(t − z) where u k,j is the same function as u k,+j in (4.4) Under the assumption in the statement we have that u 1,j = u 2,j on the surface Σ + and hence, also in Σ + ∩ Γ = Σ ∩ {t = z}, which implies that there is a function µ j : R n−1 → C such that (4.6) holds.
We now define w + := u 1,j − u 2,j in Q + , so that In this coordinates, for each matrix O j in the statement there is by definition a n − 1 × n − 1 orthogonal matrix T j such that O j (y, z) = (T j (y), −z).
Since r(x) is a radial, and A + and A − differ in an orthogonal transformation by (5.6), a direct change of variables shows that , and therefore, taking into account that A + = W 1 − W 2 we get from (5.7) that . This estimate can be proved for any 1 ≤ j ≤ n. Adding over all directions, and using that γ(σ) → 0 as σ → 0 to absorb the resulting term on the right hand side in the left, yields L 2 (B) ≤ 0, for σ > 0 large enough. Since r(x, σ) > 0 for all x ∈ R n , and σ > 0, the previous estimate implies that W 1 = W 2 . This finishes the proof.
Combining the techniques used in this proof with the techniques used in the proof of Theorem 2.1, the reader can obtain many possible results similar to the previous one, always interchanging some measurements for symmetry assumptions on A and q. A specially simple case is the following.
Proof. We use the notation introduced in (4.12). Here is convenient to use that U k,j is the same as U k,+j in the proof of Theorem 2.1. We only give a sketch of the main ideas in the proof. One can start as in the proof of Theorem 2.1 by making a change of gauge such that e n · A 1 = e n · A 2 . Let 0 ≤ j ≤ n − 1. By Lemma 2.6 we can assume that the U k,j = U k,+j satisfy (4.9) and (4.10) with W k = −iA k and V = V k . We define w + = u 1,j − u 2,j and w − (y, z, t) = w + (−y, −z, −t) (the µ j function vanishes due to the antisymmetry condition). It follows that w + = w − in Γ, this can be verified as in (5.4). The remaining conditions to apply Lemma 3.5 with A + = A 1 − A 2 , A − = A + (−x), q + = V 1 − V 2 and q − (x) = q + (−x) are easily verified. Then Lemma 3.5 and a change of variables to transform q − in q + , and A − in A + yields the estimate . We can now repeat exactly the same arguments used in the proof of Theorem 2.1 to prove (4.19). In fact we have that (4.19) holds independently for both the weight functions φ 0 (·, ϑ) and φ 0 (·, −ϑ). Adding these two possible estimates yields . The previous inequality and (5.8) imply that q 1 = q 2 and A 1 = A 2 in the gauge fixed at the beginning of the proof.

Appendix A. Stationary scattering
In this section we prove Theorem 1.4. We have adapted the proof of [RS20b, Theorem 5.1] in order to allow for the presence of a first order perturbation, but the main ideas and the exposition are similar to the work in that paper.
We define C + := {λ ∈ C : Im(λ) > 0}, and we write R V (λ) for the resolvent operator R V (λ) = (H V − λ 2 ) −1 in case it is well defined. We also use the following nonstandard convention for the Fourier transform and its inverse for Schwartz functions on the real line: (and equally for the extension of the Fourier transform to tempered distributions).
In order to illustrate why it is reasonable to expect an equivalence between the stationary scattering data and the time domain data as stated in Theorem 1.4, we reproduce here the following heuristic argument given in [RS20b]. Let U V (x, t; ω) be the solution of Suppose for the moment that the Fourier transform of U V in the time variable is well defined. Then for each λ ∈ R the function U V (x, λ; ω) should solve the equation If we define the time domain scattering solution to be u Since these are the properties that characterize the the outgoing eigenfunctions of (A.1) one might expect that where ψ V is the solution of (1.1). Now, the condition a V1 (λ, ·, ω) = a V2 (λ, ·, ω) implies by the Rellich uniqueness theorem that the outgoing eigenfunctions for H V1 and H V2 agree outside the support of the potentials: If the map λ → ψ V (λ, x, ω) were smooth near λ = 0, then one could have (A.2) for all λ ∈ R.
The argument above is only formal since requires the regularity of the map λ → ψ V (λ, x, ω) on the real line. The regularity of this map is related to the poles of the meromorphic continuation of the resolvent R V (λ), initially defined in the resolvent set of H V . Indeed, in some cases there is a pole located at λ = 0 and thus the argument above does not work in general. To get around these difficulties we start by recalling the following property of the Fourier transform.
Lemma A.1. Suppose F (z) is analytic on {Im(z) > r} for some r ∈ R and |F (z)| ≤ C(1 + |z|) N e R Im(z) , for Im(z) > r, for some positive R, C, N independent of z. There exist an f ∈ D(R) with supp(f ) ⊂ [−R, ∞) and e −(µ−r)t f ∈ S(R) that also satisfies (e −(µ−r)t f ) ∼ (·) = F (· + iµ) for every µ > r. This is essentially a Paley-Wiener theorem that we have stated in the form given in [RS20b,Lemma 5.3]. In the following proposition we give the precise relation between the time domain and frequency measurements.
Proposition A.2. Let ω ∈ S n−1 and let V(x, D) = W · ∇ + V with W ∈ C m+2 c (R n ; C n ), and V ∈ C m c (R n ; C) compactly supported in B. Let U V be the solution of given by Proposition 2.4, and let u V (x, t; ω) = U V (x, t; ω) − δ(t − x · ω). Assume also that there exists some r ≥ 0 such that for Im(λ) ≥ r for Im(λ) > r, the following identity holds , for all µ > r, and all ϕ ∈ C ∞ c (R n ) and χ ∈ C ∞ c (R). Proof. If (A.4) holds for some λ, then z = λ 2 is, by definition, in the resolvent set ρ(H V ) ⊂ C of the operator H V . It is well known that the resolvent map z → (H V − z) −1 forms a holomorphic family of bounded L 2 operators in the open set ρ(H V ) (see for example [Te09,Theorem 2.15]). Since the map z = λ 2 is also holomorphic and (A.4) holds for Im(λ) ≥ r, then λ → R V (λ) = (H V − λ 2 ) −1 is also an holomorphic map for Im(λ) ≥ r.
Then, Lemma A.1 implies that there is a function f ϕ ∈ D ′ (R) supported on [−1, ∞) such that for all µ > r: e −(µ−r)t f ϕ , χ = F ϕ (· + iµ),χ , χ ∈ C ∞ c (R). Now, given µ > r, define the linear map K : The map K is continuous. To see this, take a sequence ϕ j → 0 in C ∞ c (R n ), then (A.5) implies that F ϕj → 0 when Im(λ) ≥ r, and hence Since K is continuous, the Schwartz kernel theorem ensures that there is a unique K ∈ D ′ (R n × R) such that Since f ϕ is supported in [−1, ∞), it follows that K is supported in {t ≥ 1}. We now define the distribution v(x, t) := e µt K(x, t) ∈ D ′ (R n × R).
We claim that v is a solution in R n+1 of (A.7) Since, by (A.3), this is also the equation satisfied by u V , then the uniqueness of distributional solutions of the wave equation supported in {t ≥ −1} (see [Hö76,Theorem 9.3.2]) implies that , ϕ(x)(e µt χ)˘(σ) R n x ×Rσ , which finishes the proof of the proposition.
To prove the previous claim we use (A.6) in the following computations. First . Also, if we denote by V * the formal adjoin of V (with respect to the distribution pairing ·, · ), we have Then putting this together we obtain that Hence v satisfies (A.7), which proves the claim.
The following proposition gives in the self-adjoint case the properties of the resolvent that we require to apply the previous Proposition. Therefore we assume that V(x, D) can be written as V(x, D) = 2A · D + D · A + q for real A and q.
For any δ > 1/2 and λ in the region C + \ i(0, r 0 ], the family is a holomorphic family of bounded operators on L 2 (R n ) that can be extended continuously in the weak operator topology to C + \ i[0, r 0 ].
Let λ ≥ 0. As mentioned in the introduction, the direct problem (1.1) will have a unique scattering solution ψ s V satisfying the SRC if the outgoing resolvent operator is bounded in appropriate spaces. Then one can take , as solution of (1.1). Under the assumptions of the previous proposition, for real λ > 0 the operator R A,q (λ) given by the continuous extension to C + \ i[0, r 0 ] of the resolvent is exactly the outgoing resolvent operator. On the other hand, if λ < 0 the resolvent R A,q (λ) is the operator known as the incoming resolvent operator. This can be seen in the following computation: for real λ = 0, taking the limits in an appropriate topology, one has where the ± is given by the sign of λ.
Proof of Proposition A.3. We have that The operator H A,q is self-adjoint with domain H 2 (R n ) and, as such, it has real spectrum: the resolvent R A,q (λ) is a bounded operator in L 2 if λ ∈ C + satisfies λ 2 / ∈ R. Also, since A and q are compactly supported-V(x, D) is a short range perturbation of −∆-it is well known that the continuous spectrum of H A,q is (0, ∞) without embedded eigenvalues (see, for example, [Hö83,Chapter 14]). We now show that the point spectrum of H A,q is contained in [−r 2 0 , 0] where r 2 0 = 2 A 2 L ∞ + q L ∞ . Indeed, assume that λ 2 ∈ R and ψ ∈ L 2 are such that H A,q ψ = λ 2 ψ, is satisfied in the sense of distributions. Then ψ ∈ H 2 (R n ) by elliptic regularity, so taking the L 2 product with ψ and integrating by parts gives us where we have used that A 2 ≥ 0 and Young's inequality with ε. Hence and thus we must necessarily have λ 2 ≥ −(2 A 2 L ∞ + q L ∞ ). With this we can conclude that the full spectrum of H A,q is contained in [−r 2 0 , ∞), so that R A,q (λ) is a bounded operator in L 2 for all λ ∈ C + \ i(0, r 0 ]. Then the theory of self-adjoint operators implies two important facts.
The first is that one has the estimate , (see, for example, [Te09, Theorem 2.15]). And the second is that R A,q (λ) : L 2 → H 2 is an holomorphic family of operators for λ ∈ C + . This last statement follows from the fact that, outside the spectrum, for all λ, λ 0 ∈ C + \ i(0, r 0 ] one has the resolvent formula see for example [Te09,p. 75]. One can take the limit m → ∞ in the L 2 → L 2 operator norm to obtain an analytic expansion of the resolvent around λ 0 , since the remainder of the series goes to zero if λ is close enough to λ 0 . Then, for λ close enough to λ 0 . Since R A,q (λ) is also bounded from L 2 to H 2 , (A.10) implies that R A,q (λ) : L 2 → H 2 is holomorphic in C + \ i(0, r 0 ], and therefore so it is since the weight x −δ and all its derivatives are bounded in R n . We now prove (A.8) using (A.9). To see this write λ = σ + iµ. Then it is enough to use that dist(λ 2 , [−r 2 0 , ∞)) ≥ | Im(λ 2 )| ≥ 2|σµ| when µ 2 ≥ σ 2 −r 0 , and that dist(λ 2 , [−r 2 0 , ∞)) ≥ | Re(λ 2 )| = |σ 2 − µ 2 − r 0 | otherwise.
The continuity of x −δ R A,q (λ) x −δ for λ ∈ C + \i[0, r 0 ] it is the well known limiting absorption principle. See for example [Ya10, Proposition 1.7.1] for the free resolvent, and [Hö83,Chapter 14] for the case of short range magnetic potentials (as in this case). A more specific statement of the limiting absorption principle (also including long range magnetic potentials) can be found in [Hö83, Theorem 30.2.10], which implies that R A,q (λ) is continuous as a function from C + \ i[0, r 0 ] to the space of bounded operators between the Hörmander spaces B and B * considered with the weak operator topology. Since B is continuously embedded in x −δ L 2 this implies the (weak) continuity of Putting together Propositions A.2 and A.3 we can now formalize the heuristic argument given at the beginning of this section in order to prove Theorem 1.4.
A straightforward computation shows that the remainder term R N must satisfy The task now is to prove the existence of the coefficients (a j ) N j=−1 and R N , satisfying the recursive identity (B.4) with the corresponding initial value conditions. One expects getting smoother remainder terms R N as N grows, or at least with better regularity than the Delta distribution and Heaviside function. This can be achieved by killing most the non-smooth terms on the right-hand side of (B.4). We now split the proof into two cases depending on the nature of the initial value conditions.
The desired claim is proved by combining above choices with (B.9). On the other hand, by (B.3) and (B.8), we deduce that U δ can be written as follows is of class C 2 in the region {t ≥ x · ω}. This shows that v satisfies all the properties stated in Proposition 2.4, and hence the proof of existence and uniqueness of solutions to (B.1) is completed.
Second case. Existence of solutions of (B.2). This case is quite similar to the previous one, and hence we only give a brief explanation of the proof. Here we have a −1 = 0 and according to identity (B.4), the function a 0 must satisfy (B.5) in place of a −1 . The remaining coefficients a k satisfy (B.6). The remainder term R N satisfies the same equation as in (B.7), and thus its existence is guaranteed by [Hö76, Theorems 9.3.1 and 9.3.2]. In this case, the solution U H to (B.2) has the form where u is of class C 2 in the region {t ≥ x · ω}. Moreover, it satisfies all the desired properties stated in Proposition 2.3.
B.2. Proof of the energy lemmas. The proofs of the following lemmas are mainly based on standard multiplier techniques. Here div m and ∇ m stand respectively for the divergence and gradient operators with respect to a set of variables m; while ∇ is reserved for the gradient operator with respect to the full spatial variables, that is ∇ = ∇ x . Let r = |x|. We define the radial and angular derivatives as ∂ r = x |x| · ∇, Ω jk = x j ∂ k − x k ∂ j , j, k = 1, 2, . . . , n.
Thanks to the domain of dependence theorem for the wave equation α = 0 in H T , see for instance [Ev97, Section 2.4, Theorem 6], we deduce that α is compactly supported for each fixed t. In particular, for each fixed t, one integral involving α on |x| ≥ 1 is actually on M ≥ |x| ≥ 1 for a large enough M > 1. This fact will be used several times throughout this proof.
We have used that the unit normal vectors ν(y, z, t) are respectively equal to (0, 0, 1), 1 √ 2 (0, 1, −1) and − 1 |x| (y, z, 0) on the regions H T ∩ {t = τ }, H T ∩ {t = z} and Σ + ∩ {t ≤ τ }. A domain of dependence argument shows that H T ∩ supp α is bounded and far away from the origin, thus 1/|x| is well defined on that region. In particular, α r is well defined on H T ∩ supp α.
Let χ ∈ C 1 (Q). The chain rule shows that it also holds for χα in place of α, where the implicit constant is proportional to χ C 1 (Q) . This finishes the proof.