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arXiv:math/0505554v1 [math.AP] 26 May 2005
Inverse problems for Schr¨ odinger equations
with Yang-Mills potentials in domains with
obstacles and the Aharonov-Bohm effect.
G.Eskin,Department of Mathematics, UCLA,
Los Angeles, CA 90095-1555, USA. E-mail: eskin@math.ucla.edu
February 1, 2008
Abstract
We study the inverse boundary value problems for the Schr¨ odinger
equations with Yang-Mills potentials in a bounded domain Ω0⊂ Rn
containing finite number of smooth obstacles Ωj,1 ≤ j ≤ r. We prove
that the Dirichlet-to-Neumann opeartor on ∂Ω0determines the gauge
equivalence class of the Yang-Mills potentials. We also prove that the
metric tensor can be recovered up to a diffeomorphism that is identity
on ∂Ω0.
1Introduction.
Let Ω0be a smooth bounded domain in Rn, diffeomorphic to a ball, n ≥ 2,
containing r smooth nonintersecting obstacles Ωj, 1 ≤ j ≤ r. Consider the
Schr¨ odinger equation in Ω = Ω0\ (∪r
?2
j=1Ωj) with Yang-Mills potentials
(1.1)
n
?
j=1
?
−i∂
∂xjIm+ Aj(x)
u + V (x)u − k2u = 0
with the boundary conditions
(1.2)
u??∂Ωj= 0,
1 ≤ j ≤ r,
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(1.3)
u|∂Ω0= f(x′),
where Aj(x),V (x),u(x) are m × m matrices, Im is the identity matrix in
Cm. Let G(Ω) be the gauge group of all smooth nonsingular matrices in Ω.
Potentials A(x) = (A1,...,An),V and A′(x) = (A′
gauge equivalent if there exists g(x) ∈ G(Ω) such that
A′(x) = g−1Ag − ig−1(x)∂g
1,...,A′
n),V′(x) are called
(1.4)
∂x, V′= g−1V g.
Let Λ be the Dirichlet-to-Neumann (D-to-N) operator on ∂Ω0, i.e.
Λf = (∂u
∂ν+ i(A · ν)u)??∂Ω0,
where ν = (ν1,...,νn) is the unit outward normal to ∂Ω0 and u(x) is the
solution of (1.1), (1.2), (1.3)). We assume that the Dirichlet problem (1.1),
(1.2), (1.3)) has a unique solution. We shall say that the D-to-N operators
Λ and Λ′are gauge equivalent if there exists g0∈ G(Ω) such that
Λ′= g0,∂Ω0Λg−1
0,∂Ω0,
where g0,∂Ω0is the restriction of g0 to ∂Ω0. We shall prove the following
theorem:
Theorem 1.1. Suppose that D-to-N operators Λ′and Λ corresponding to
potentials (A′,V′) and (A,V ) respectively are gauge equivalent for all k ∈
(k0− δ0,k0+ δ0), where k0> 0, δ0> 0. Then potentials (A′,V′) and (A,V )
are gauge equivalent too.
If we replace A′,V′by A(1)= g−1
Λ = Λ1where Λ1is the D-to-N operator corresponding to (A(1),V(1)). The
proof of Theorem 1.1 gives that if Λ = Λ1then (A,V ) and (A(1),V(1)) are
gauge equivalent with a gauge g ∈ G(Ω) such that g|∂Ω0= Im. We shall
denote the subgroup of G(Ω) consisting of g such that g(x)|∂Ω0= Im by
G0(Ω). In the case when Ω0contains no obstacles Theorem 1.1 was proven
in [E] for n ≥ 3 and in [E3] for n = 2. Note that the result of [E] is stronger
since it requires that Λ = Λ(1)for one value of k only. In the case n = 2 the
proof of Theorem 1.1 is simpler than that in [E3] since it does not rely on
the uniqueness of the inversion of the non-abelian Radon transform.
0A′g0− ig−1
0
∂g0
∂x, V(1)= g−1
0V g0 then
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We shall prove Theorem 1.1 in two steps. In §2 we shall prove that (A,V )
and (A(1),V(2)) are locally gauge equivalent using the reduction to the inverse
problem for the hyperbolic equations as in [B], [B1], [KKL], [KL], [E1], and
in §3 we shall prove the global gauge equivalence using the results of §2 and
of [E2]. Following Yang and Wu (see [WY]) one can describe the gauge
equivalence class of A = (A1,...,An). Fix a point x(0)∈ ∂Ω0and consider all
closed paths γ in Ω starting and ending at x(0). Let x = γ(τ), 0 ≤ τ ≤ τ0,
be a parametric equation of γ, γ(0) = γ(τ0) = x(0). Consider the Cauchy
problem for the system
∂
∂τc(τ,γ) =dγ(τ)
dτ
· A(γ(τ))c(τ,γ), c(0,γ) = Im.
By the definition the gauge phase factor c(γ,A) is c(τ0,γ). Therefore A
defines a map of the group of paths to GL(m,C). The image of this map is
a subgroup of GL(m,C) which is called the holonomy group of A (see [Va]).
It is easy to show (c.f. §3) that c(γ,A(1)) = c(γ,A(2)) for all closed paths γ iff
A(1)and A(2)are gauge equivalent in Ω. As it was shown by Aharonov and
Bohm [AB] the presence of distinct gauge equivalent classes of potentials can
be detected in an experiment and this phenomenon is called the Aharonov-
Bohm effect. In $ 4 we consider the recovery of the Riemannian metrics from
the D-to-N operator in domains with obstacles.
2Inverse problem for the hyperbolic system.
Consider two hyperbolic system:
(2.1)
L(p)u =∂2
∂t2u(p)+
n
?
j=1
(−i∂
∂xjIm+ A(p)
j(x))2u(p)+ V(p)(x)u(p)= 0, p = 1,2,
in Ω × (0,T0) with zero initial conditions
(2.2)
u(p)(x,0) = u(p)
t(x,0) = 0
and the Dirichlet boundary conditions
(2.3)
u(p)??∂Ωj×(0,T0)= 0, 1 ≤ j ≤ r, u(p)??∂Ω0×(0,T0)= f(x′,t), p = 1,2.
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Here Ω = Ω0\ (∪r
u(p)(x,t), p = 1,2, are smooth m × m matrices. As in §1 introduce D-to-N
operators Λ(p)f = (∂
j
· νj)u??∂Ω0×(0,T0), p = 1,2.
for (2.1) when T0 = ∞ determines the D-to-N operator for (1.1) for all k
except a discrete set, and vice versa.
We shall prove the following theorem:
j=1Ωj) is the same as in §1, A(p)
j(x), 1 ≤ j ≤ n, V(p)(x),
∂ν+ i?n
j=1A(p)
Making the Fourier transform in t one can show that the D-to-N operator
Theorem 2.1. Suppose Λ(1)= Λ(2)and T0> maxx∈Ωd(x,∂Ω0) where d(x,∂Ω0)
is the distance in Ω from x ∈ Ω to ∂Ω0. Then potentials A(1)
n, V1(x) and A(2)
(1.4) holds with g ∈ G0(Ω).
Note that Theorem 2.1 implies Theorem 1.1. We can consider a more
general than (2.1) equation when the Eucleadian metric is replaced by an
arbitrary Riemannian metric:
j(x),1 ≤ j ≤
j(x),1 ≤ j ≤ n, V(2)(x) are gauge equivalent in Ω, i.e.
∂2u(p)
∂t2
+
n
?
j,k=1
1
?gp(x)(−i∂
∂xjIm+ A(p)
j(x))
?
gp(x)gjk
p(x)(−i∂
∂xjIm
+A(p)
k(x))u(p)+ V(p)(x)u(p)(x,t) = 0,
(2.4)
where ?gjk
are the same as in (2.1), Ω(p)= Ω0\ Ω
subset of ∂Ω0and let 0 < T < T0be small. Denote by ∆(0,T) the intersec-
tion of the domain of influence of Γ with ∂Ω0× [0,T]. We assume that the
domain of influence of Γ does not intersect Ω′p× [0,T].
Lemma 2.1. Suppose Λ(1)= Λ(2)on ∆(0,T). There exist neighborhoods
U(p)⊂ Ω(p),p = 1,2, U(p)∩∂Ω0= Γ and the diffeomorphism ϕ : U(1)→ U(2)
such that ϕ|Γ= I and ?gjk
and ϕ ◦ A(2)
exists g(x) ∈ G0(U(1)), g(x) = I on Γ such that (1.4) holds in U(1).
The proof of Lemma 2.2 is the same as the proof of Lemma 2.1 in [E1].
One should replace only the inner products of the form?u(x,t)v(x,t)dxdt
by?Tr(uv∗)dxdt where v∗is the adjoint matrix to v(x,t). We do not assume
4
p(x)?−1are metric tensors in Ω
(p),gp(x) = det?gjk
′
p, Ω′
p?−1, A(p)
j(x),V(p)(x)
p= ∪r
j=1Ωjp. Let Γ be an open
2? = ϕ ◦ ?gjk
1?. Moreover A(1)
j, 1 ≤ j ≤ n, V(1)
j,1 ≤ j ≤ n, ϕ ◦ V(2)are gauge equivalent in U(1), i.e. there
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that matrices A(p)
obtained by the BC-method (see[B], [KKL]). Extend ϕ−1from U2to Ω(2)in
such a way that ϕ = I on ∂Ω0and ϕ is a diffeomorphism of Ω(2)and˜Ω(2)=
ϕ−1(Ω(2)). Also extend g(x) from U1to˜Ω(2)so that g(x) ∈ G0(˜Ω2), g = I
on ∂Ω0. Then we get that˜L(2)= g ◦ ϕ ◦ L(2)= L(1)in U(1).
Lemma 2.2. Let L(1)and L(2)be the operators of the form (2.4) in Ω(p)=
Ω0\ Ω′p, p = 1,2. Let B ⊂ Ω(1)∩ Ω(2)be simply-connected, ∂B ∩ ∂Ω0= Γ
be open and connected, and Ω(p)\ B be smooth. Suppose L(2)= L(1)in
B and Λ(1)= Λ(2)on ∂Ω0× (0,T0) where Λ(p)are the D-to-N operators
corresponding to L(p), p = 1,2. Then˜Λ(1)=˜Λ(2)where˜Λ(p)are the D-to-N
operators corresponding to L(p)in the domains (Ω(p)\ B) × (δ,T0− δ), δ =
maxx∈Bd(x,∂Ω0)), d(x,∂Ω0) is the distance in B between x ∈ B and ∂Ω0,
˜Λ(p)are given on ∂(Ω0\ B) × (δ,T0− δ).
Therefore Lemma 2.2 reduces the inverse problem in Ω(p)× (0,T0) to
the inverse problem in a smaller domain (Ω(p)\ B) × (δ,T0− δ). Combining
Lemmas 2.1 and 2.2 we can prove that for any x(0)∈ Ω(1)there exist a simply-
connected domain B1⊂ Ω(1), x(0)∈ B1, a diffeomorphism ϕ of˜Ω(2)onto Ω(2),
ϕ = I on ∂Ω0, such that g ∈ G0(˜Ω(2)) such that˜L(2)def
B1. To prove the global gauge equivalence and global diffeomorphism in the
case when Ω(1)is not simply-connected we shall use some additional global
quantities determined by the D-to-N operator (c.f. [E2]).
j,V(p)are self-adjoint In the latter case Lemma 2.1 can be
= g ◦ ϕ ◦ L(2)= L(1)in
3Global gauge equivalence.
In this section we shall prove Theorem 2.1. Fix arbitrary point x(0)∈ ∂Ω0.
Let γ be a path in Ω starting at x(0)and ending at x(1)∈ Ω, γ(τ) = x(τ) is
the parametric equation of γ, 0 ≤ τ ≤ τ1, x(0)= x(0), x(1)= x(τ1). Denote
by c(p)(τ,γ), p = 1,2, the solution of the system of differential equations
(3.1)
i∂c(p)(τ,γ)
∂τ
= ˙ γ(τ) · A(p)(x(τ))c(p)(τ,γ),
where
(3.2)
c(p)(0,γ) = Im, p = 1,2, 0 ≤ τ ≤ τ1,
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(3.3)˙ γ(τ) =dx
dτ.
Denote c(p)(x(1),γ) = c(p)(τ1,γ), p = 1,2.
Lemma 3.1. Suppose A(1)and A(2)are locally gauge equivalent. Then the
matrix c(2)(x(1),γ)(c(1)(x(1),γ))−1depends only on the homotopy class of the
path γ connecting x(0)and x(1).
Proof Let γ1 and γ2 be two homotopic paths connecting x(0)and x(1).
Consider the path γ0= γ1γ−1
2
that starts and ends at x(0). It follows from
(3.1) that c(2)(τ,γ)(c(1)(τ,γ))−1satisfies the following system of differential
equations:
i∂
∂τ(c(2)(τ,γ)(c(1)(τ,γ))−1) = ˙ γ(τ) · A(2)(x(τ))(c(2)(τ,γ)(c(1)(τ,γ))−1)
−(c(2)(τ,γ)(c(1)(τ,γ))−1)A(1)(x(τ)) · ˙ γ(τ).
Let b(τ,γ) = c(2)(τ,γ)(c(1)(τ,γ))−1, b(x(1),γ1) = b(τ1,γ1), b(x(1),γ2) = b(τ2,γ2),
where x = x(p)(τ) are parametric equations of γ(p), 0 ≤ τ ≤ τp, p = 1,2.
We have that b(x(1),γ1) = b(x(1),γ2) iff b(x(0),γ0) = Im, where x(0)is the
endpoint of path γ0 = γ1γ−1
2
and b(x(0),γ2) is the value at the endpoint
of the solution of (3.4) along γ0 with the initial value (3.2). If γ0 can be
contracted to a point in Ω there exists closed paths σ1,...,σN such that
γ0 = σ1...σN and each σj is contained in a neighborhood Uj ⊂ Ω where
A(1)and A(2)a gauge equivalent (see Lemma 2.1).
bj(τ,σj) is continuous on σj where bj(τ,σj) is the solution of (3.4) with γ
replaced by σj, σj(τ) = x(j)(τ), 0 ≤ τ ≤ τj, is the parametric equation of
σj, σj(0) = σj(τj). The continuity on σj means that b(0,σj) = b(τj,σj).
Since A(1)and A(2)are gauge equivalent in Ujthere exists gj(x) ∈ C∞(Uj)
such that (1.4) holds in Uj. It follows from (3.4) and (1.4) that
(3.4)
We shall show that
(3.5)
i∂
∂τ(bj(τ,σj)g−1
j(x(j)(τ)) = 0 for 0 ≤ τ ≤ τj.
Therefore bj(τ,σj)g−1
C gj(x(j)(0)). Since bj(τ,σj) is continuous on each σj, 1 ≤ j ≤ N, we get
that b(τ,γ0) is continuous on γ0, in particular, b(x(0),0,γ0) = Im. Therefore
b(x(1),γ1) = b(x(1),γ2).
j(x(j)(τ)) = C on σj. We have bj(0,σj) = bj(τj,σj) =
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Now we shall prove that b(x(1),γ1) = b(x(1),γ2) for any two paths con-
necting x(0)and x(1). As in the case of Lemma 3.1 it is enough to prove that
b(τ,γ)is continuous on γ0= γ1γ−1
2
where b(τ,γ0) is the solution of (3.4) for
γ0.
We say that ˜ γ = ˜ γ1,..., ˜ γNis a broken ray in Ω×[0,T0] with legs ˜ γj, 1 ≤
j ≤ N, if it starts at some point (x(1),t(1)) ∈ ∂Ω0×[0,T0], x = x(1)+τω, t =
t(1)+τ is the parametric equation of ˜ γ1for 0 ≤ τ ≤ τ1. Then ˜ γ makes N −1
nontangential reflections at ∂Ω
∂Ω0× [0,T0]. Denote by γ = γ1...γN the projection of ˜ γ onto the x-plane.
Let c(p)(τ,γ) be the solution of the system
′× [0,T0], where Ω′= ∪r
j=1Ωj and ends at
(3.6)
i∂
∂τc(p)(τ,γ) = A(p)(γ(τ)) · ˙ γ(τ)c(p)(τ,γ),
(3.7)
c(p)(0,γ) = Im,p = 1,2,
where γ(τ) is the parametric equation of broken ray, 0 ≤ τ ≤ τN, ˙ γ =
is the direction of the broken ray, c(p)(τ,γ) is continuous on γ(τ). Note that
dγj(τ)
dτ
= θjis constant on γj, τj−1≤ τ ≤ τj.
The following lemma is the generalization of Theorem 2.1 in [E2]:
dγ
dτ
Lemma 3.2. Let (x(N),t(N)) be the endpoint of the broken ray ˜ γ : ˜ γ(τN) =
(x(N),t(N)). Denote c(p)(x(N),γ) = c(p)(τN,γ), p = 1,2. Then c(2)(x(N),γ) =
c(1)(x(N),γ) assuming that Λ(1)= Λ(2)on ∂Ω0× (0,T0).
Assuming that Lemma 3.2 is proven we shall complete the proof of The-
orem 2.1
Let γ be a broken ray starting at x(1)and ending at x(N), x(1)∈ ∂Ω0, x(N)∈
∂Ω0.
Let c(p)(τ,γ) be the solution of (3.6), (3.7). Let α1 be a path on ∂Ω0
connecting x(0)and x(1)and let α2be a path on ∂Ω0connecting x(N)and
x(0). Therefore α = α1γα2is a closed path starting and ending at x(0). If
Uj∩ ∂Ω0 ?= 0 then the gauge gj = Im on ∂Ω0. Therefore
Uj∩ ∂Ω0 for any vector
A(2)·
c(1)(τ,α2) = c(2)(τ,α2). Therefore b(τ,α) is continuous on α = α1γα2. We
shall call α = α1γα2an extended broken ray.
∂
∂xgj·
→
l= 0 on
→
l∈ Rntangent to ∂Ω0. Then (1.4) implies that
l. It follows from (3.1), (3.2) that c(1)(τ,α1) = c(2)(τ,α1) and
→
l= A(1)·
→
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We shall assume for simplicity that extended broken rays generate the
homotopy group of Ω. Otherwise we can, as in the end of §2, construct a
simply-connected domain Ω(0)such that Ω(0)⊂ Ω, ∂Ω(0)⊃ ∂Ω0, ∂Ω(0)∩
Ω′= ∅ and Ω \ Ω(0)is ”thin”, i.e. the volume of Ω \ Ω(0)is small. Since
Ω(0)is homotopic to ∂Ω0we get, using Lemmas 2.1 and 2.2 that potentials
A(1)and A(2)are globally gauge equivalent in Ω(0). Therefore the proof of
global gauge equivalence in Ω can be reduced to the proof of the global
gauge equivalence in Ω \ Ω(0). It is clear that the extended broken rays in
Ω\Ω(0)generate the fundamental group π1(Ω\Ω(0)). Note that rays without
reflections also generated the fundamental group π1(Ω\Ω(0)). Then the closed
path γ0is homotopic to α(1)...α(N1)where α(j)are extended broken rays. Since
b(τ,α(j)) is continuous on α(j), j = 1,...,N1, we get that b(τ,γ0) is continuous
on γ0. It follows from Lemma 3.2 that b(τ,σj) = Im on ∂Ω0 and hence
b(τ,γ0) = Imon ∂Ω0. Therefore we proved that c(2)(x(1),γ)(c(1)(x(1),γ))−1
does not depend on the path γ connecting x(0)and x(1). Denote g(x(1)) =
c(2)(x(1),γ)(c(1)(x(1),γ))−1. We have that g(x) is a single-valued matrix on
Ω, g(x) = Imon ∂Ω0and g(x) is nonsingular since c(p)(x,γ) are nonsingular,
p = 1,2. We have for arbitrary x(1):
A(2)(x(τ)) · ˙ γ(τ) = i∂c(2)(τ,γ)
∂xg(x(τ)) · ˙ γ(τ)c(1)+ ig(x(τ))∂c(1)
∂xg · ˙ γg−1+ gA(1)· ˙ γ(τ)g−1= (i∂
∂τ
(c(2)(τ,γ))−1
(3.8)
= (i∂
∂τ
)(c(1)(τ,γ))−1g−1
= i∂
∂xgg−1+ gA(1)g−1) · ˙ γ(τ).
Since we can choose γ(τ) such that γ(τ1) = x(1)and ˙ γ(τ) is arbitrary at
τ = τ1, we get that
(3.9)
A(2)(x(1)) = i(∂
∂xg)g−1(x(1)) + g(x(1))A(1)(x(1))g−1(x(1)),
i.e. A(2)is gauge equivalent to A(1)in Ω. We can change A(1)to A′=
gA(1)g−1+i∂
applying Lemmas 2.1 and 2.2 we get that V(2)= V′in Ω. Therefore V(2)=
gV(1)g−1where g(x) is the same as in (3.9).
It remains to prove Lemma 3.2. In the case when the broken ray γ =
γ1...γMdoes not contain caustics points the proof of Lemma 3.2 is the same
as the proof of Theorem 2.1 in [E2]. We shall consider the case when γ has
∂xgg−1, V′= gV(1)g−1. Then we will have A(2)= A(1). Therefore
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some caustics points and we shall simplify also the proof of Theorem 2.1 in
[E2]. However in this paper we shall not use rays having caustics points.
Consider, for simplicity, the case n = 2 and x∗ ∈ γM is the only caustics
point on γ. We also assume that the caustics point is generic (see [V]). Note
that if x∗ is not generic but the broken ray γ can be approximated by a
sequence of broken rays having only generic caustics points, then Lemma 3.2
holds for such γ too. This fact suggests that Lemma 3.2 is likely true for any
broken ray.
Let χ0(y′) ∈ C∞
1, χ0(y′) = 1 for |y′| <1
χ(y′) =1
0(R2), y′= (y1,y2), χ0(y′) ≥ 0, χ0(y′) = 0 for |y′| >
2,
?
εχ0(y′
R2χ2
0(y′)dy′= 1. Denote
(3.10)
ε).
We shall heavily use the notations of [E2, §2]. The difference with [E2] is that
in this paper we consider the broken ray ˜ γ in Ω × [0,T0] and its projection
on Ω will be the broken ray γ considered in [E2].
Let Π be a plane in R2× R, (x,t) ∈ Π if x = x(0)
where ω⊥· ω = 0, x(0)?∈ Ω and the plane Π does not intersect Ω × R. We
denote by ˜ γ(y′) = ˜ γ0(y′)˜ γ1(y′)...˜ γM(y′) the broken ray starting at (x(0)+
y1ω⊥,t(0)+ y2) in the direction (ω,1), y′= (y1,y2). Then the equation
of ˜ γ0(y′) is x = x(0)+ y1ω⊥+ t0ω, t = t(0)+ y2+ t0, 0 ≤ t0 ≤ t0(y1),
where (x(0)+ y1ω⊥+ t0(y1)ω,t(0)+ y2+ t0(y1)) =˜P1is the point where ˜ γ0
hits ∂Ω′× (0,T0). As in [E2] we introduce ”ray coordinates” (sp,tp) in the
neighborhood of γp, 0 ≤ p ≤ M. Denote by Dj(x(sj,tj)) the Jacobian of
the change of coordinates x = x(j)(sj,tj). Let˜Pjbe the points of reflections
of ˜ γ(y′) at ∂Ω′, 1 ≤ j ≤ M. Denote by Pj the projection of˜Pj on the x-
plane. Note that the time coordinate of˜Pjis t(j)= t(0)+ y2+?j
where tr(y1) is the distance between Prand Pr−1. Note that t = t(j)+ tjon
˜ γj, 0 ≤ tj≤ tj(y1).
Let L(p)be the same as in (2.1), p = 1,2. We construct a solution of
L(1)u = 0 of the form (c.f. (2.1), (2.9) in [E2], see also the earlier work [I]):
0 + y1ω⊥, t = y2+ t(0),
r=0tr(y1)
(3.11)
u(x,t,ω) =
M−1
?
j=0
uj(x,t,ω) + uM1+ uM2+ uM3+ u(1),
where the principal part of ujhas a form
(3.12)
uj0= aj0(x,t,ω)eik(ψj(x,ω)−t),
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ψj(x,ω) are the same as in (2.2), (2.3), (2.4) in [E2] and
(3.13)
aj0= (aj−1,0|Dj|
1
2)??Pj
1
|Dj|
1
2cj1(x,ω),
where cj1(x,ω) is the solution of the system
(3.14)
iθj· ∇cj1= (A(1)· θj)cj1,
0 ≤ j ≤ M, ∇ =
We shall assume that
tj−1(y1) ≤ tj≤ tj(y1),cj1
??Pj= Im,
∂
∂x, θjis the direction of γj, θ0= ω.
(3.15)
u0(x,t,ω) = χ(y′)α0
on the plane Π, i.e. when t0= 0 and x = x(0)+ y1ω⊥, t = t(0)+ y2. Here α0
is an arbitrary constant matrix.
Let (x∗,t∗) ∈ ˜ γM be such that x∗ is the caustics point in the x-plane.
Note that uM1has the same form as uM−1for t < t∗− Cε, where ε is the
same as in (3.10), solution uM2is defined in a Cε-neighborhood Uεof (x∗,t∗).
We will not write the explicit form of uM2(see, for example, [V]) since we
will only need an estimate
(3.16)
|uM2| ≤
Ck
1
6
1 + k
1
6d
1
4(x),
where d(x) is the distance from x ∈ U0,εto the caustics curve. Such estimate
holds in the generic case (see [V]). Moreover,
|∇uM2| ≤
Ck
7
6
1 + k
1
6d
1
4(x).
Finally, uM3 is defined for t > t∗+ Cε and it has the same form as uM1.
The main difference is that the amplitude of uM3 has an extra factor eiβ
where β is real. The construction and the estimate of u(1)in (3.11) is similar
to [E2, Lemma 2.1] with the simplification that we consider the hyperbolic
initial-boundary value problem with the zero initial conditions when t = 0
and zero boundary conditions on ∂Ω × (0,T0) instead of (2.9) in [E2]. Since
we assumed that T0is large enough we get that the endpoint of ˜ γMbelongs
to ∂Ω0× (0,T0).
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We construct a solution v(x,t,ω) of L∗
same initial data as (3.15) for v0with α0replaced by β0where β0is an aritrary
constant matrix and with the same phase function ψj(x,ω), 0 ≤ j ≤ M, as
in (3.12): We have
2v = 0 similar to (3.11) with the
(3.17)
v =
M−1
?
j=0
vj(x,t,ω) + vM1+ vM2+ vM3+ v(1)(x,t,ω),
where the principal term of vjhas the following form:
(3.18)
vj0= bj0(x,t,ω)eik(ψj(x,ω)−t),
where bj0are the same as aj0with cj1(x,ω) replaced by c∗,j(x,ω) where c∗j
is the solution of the system
(3.19)
iθj· ∇c∗j= (θj· (A(2))∗)c∗j.
Taking the adjoint of (3.19) we get
(3.20)
−iθj· ∇c∗
∗j= c∗
∗j(θj· A(2)).
Denote
(3.21)
cj2= (c∗
∗j)−1.
Then (3.20) implies that
(3.22)
iθj· ∇cj2= (θj· A(2))cj2.
We assume that v(1)satisfies zero initial conditions when t = T, x ∈ Ω, and
zero boundary condiions on ∂Ω × (0,T0). Substitute (3.11) instead of u(1)
and (3.17) instead of v(2)in the Green’s formula. Dividing by 2k and passing
to the limit when k → ∞ we obtain (c.f. [E2]):
(3.23)0 =
M−1
?
j=0
?
Ω
?T
0
((A(1)− A(2)) · ∇(ψj− t)aj0,bj0)dxdt + IM,
where IM is the integral over a neighborhood of γM. We make a series of
changes of variables as in (2.43) in [E2].
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Note that the Jacobian DM(x(M)(sM,tM)) vanishes on the caustics set
and therefore D−1
of variables this singularity in uM1,vM1and in uM3,vM3cancels. Note also
that the estimate (3.16) implies that the integral over the neighborhood Uεis
O(√ε). Therefore taking into account that α0and β0are arbitrary matrices
we get
Mhas a singularity there. However when we make changes
(3.24)
M
?
j=0
?
R2
?
˜ γj(y′)
χ2(y′)c−1
j2(A(1)− A(2)) · θjcj1dtjdy′+ O(√ε) = 0,
where ˜ γ(y′) is the broken ray starting at (x(0)+ y1ω⊥,t(0)+ y2) and we use
in (3.24) that c∗
∗j= c−1
j2(see (3.21)). Note that
(3.25)
c−1
j2(A(1)− A(2)) · θjcj1= iθj· ∇(c−1
j2cj1),
since −c−1
(3.24) satisfy the differential equations (3.14), (3.22) but the initial conditions
are different :
j2(A(2)· θj) = iθj· ∇c−1
j2. After changes of variables cj1and cj2in
(3.26)
cji
??Pj= cj−1,i
??
Pj,
1 ≤ j ≤ M, i = 1,2.
We kept the same notation for the simplicity. Taking the limit in (3.24) when
ε → 0 we get
M
?
j=0
?
γj
θj· ∇(c−1
j2cj1)dtj=
M
?
j=0
?(c−1
j2cj1)??Pj− (c−1
j2cj1)??Pj−1
?= 0.
Since c0i|P0 = Im and (3.26) holds we get that c−1
cM1|PM= cM2|PM. Lemma 3.2 is proven.
M2cM1|PM = Im, i.e.
4Global diffeomorphism.
Let L(1)u(1)= 0 and L(2)u(2)= 0 be equations of the form (2.4) in domains
Ω(p)= Ω0\ Ω′
conditions (2.2) in Ω(p), p = 1,2 and the boundary conditions (2.3) with Ωj
replaced by Ωjp, p = 1,2, are satisfied.
p, where Ω′
p= ∪r
j=1Ωjp, p = 1,2. We assume that the initial
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Theorem 4.1. Suppose Λ(1)= Λ(2)on ∂Ω0, where Λ(p)are the D-to-N op-
erators corresponding to L(p),p = 1,2. Suppose
T0> 2min
p
max
x∈Ω(p)dp(x,∂Ω0)
where dp is the distance with respect to the metric tensor ?gjk
there exists a diffeomorphism ϕ of Ω(1)onto Ω(2)such that ϕ = I on ∂Ω0
and ?gjk
We shall sketch the proof of Theorem 4.1 assuming for the simplicity
that m = 1, A(p)
j
≡ 0, 1 ≤ j ≤ r, p = 1,2, and T0= ∞. By using Lemmas
2.1 and 2.2 we can get a simply-connected domain Ω(0)⊂ Ω(1)such that
Ω(1)\ Ω(0)has a small volume. Moreover there exists a diffeomorphism ˜ ϕ of
Ω(2)onto˜Ω(2)= ˜ ϕ−1(Ω(2)), ˜ ϕ = I on ∂Ω0such that˜L(2)def
L(1)in Ω(0). Note that Ω(0)⊂ Ω(1)∩˜Ω(2). We also get from Lemma 2.2 that
Λ(1)=˜Λ(2)on ∂Ω(0)\ ∂Ω0where˜Λ(2)is the D-to-N operator corresponding
to˜L(2). Since Ω(1)\Ω(0)is thin, there is an open subset Γ1of ∂Ω(0)such that
the endpoints of geodesics corresponding to L(1)in Ω(1)\Ω(0), orthogonal to
Γ1, form an open subset Γ2⊂ ∂Ω(0). Denote by D1⊂ Ω(1)\Ω(0)the union of
these geodesics. It follows from the proof of Lemma 2.1 (see [E1]) that Λ(1)
on Γ1 uniquely determines the metric tensor ?gjk
coordinates in D1. Denote by ψ1the map of D1on˜D1= ψ1(D1) such that
ψ1(x) are the semi-geodesic coordinates in˜D1. Analogously let D2be the
union of all geodesics of˜L(2)orthogonal to Γ1and let Γ′
its endpoints. Denote by ψ2(x) the semi-geodesic coordinates for˜L(2)and let
˜D2= ψ2(D2). By Lemma 2.1 ψ1◦L(1)= ψ2◦˜L(2)in˜D1∩˜D2. It follows from
Lemma 2.1 that ψj= I on Γ1. Note that Ω(2)\Ω(0)coincide with Ω(1)\ Ω(0)
near Γ1.
p?−1. Then
2? = ϕ ◦ ?gjk
1?.
≡ ˜ ϕ◦L2is equal to
1?−1in the semi-geodesic
2⊂ ∂Ω(0)be the set of
Lemma 4.1. The following equalities hold:
I on Γ2.
˜D1 =˜D2 and ψ = ψ−1
2ψ1 =
Proof: Since we assume that T0= ∞ we can switch to the inverse problem
for the equations of the form (1.1). Choose parameter k ∈ C such that the
boundary value problem of the form (1.1), (1.2), (1.3) has a unique solution
upfor any f ∈ H 1
Choose f nonsmooth. Denote˜Γ2= ψ−1
the unique continuation theorem that ψ1◦ u1 = ψ2◦ u2in˜D2∩˜D1 since
2(∂Ω(0)\ ∂Ω0) where f is the same for p = 1 and p = 2.
1(Γ2),˜Γ′
2= ψ−1
2(Γ′
2). It follows from
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the Cauchy data of u1and u2coincide on Γ1. Here L(p)up= 0, p = 1,2. If
˜Γ2?=˜Γ′
C∞outside˜Γ′
D1onto D2. Since˜Γ2=˜Γ′
˜D1=˜D2and u1= u2= f(x) on Γ2we get f(x) = f(ψ(x)) on Γ2. Since f is
arbitrary this implies that ψ = I on Γ2.
Therefore ψ = I on ∂D1∩ ∂Ω(0). Define ϕ(1)= ˜ ϕ on Ω(0), ϕ(1)= ψ ◦ ˜ ϕ
on D1. We get that ϕ(1)◦ L(2)= L(1)in Ω(0)∪ D1.
Applying Lemma 2.2 to Ω(p)\ (Ω(0)∪ D1) and using again Lemmas 4.1,
2.1 and 2.2 we prove Theorem 4.1.
Remark 4.1 (c.f. [E1]). We shall show now that the obstacles can be
recovered up to the diffeomorphism.
close to the obstacle Ω′
orthogonal to γ0and ending on Ω′
Ω′
of variables to the semi-geodesic coordinates and let˜∆1= ϕ1(∆1). Let ϕ2
be the change of variables to the semi-geodesic coordinates for˜L(2)in ∆2
where ∆2is the union of all geodesics of˜L(2)orthogonal to γ0and ending
on Ω′
a geometric optics solution in Ω(1)\ Ω(0)similar to constructed in §3 that
starts on γ0, reflects at ∂Ω′
solution of˜L(2)u2= 0 in Ω(2)\ Ω(0)having the same boundary data as u1.
Since ϕ1◦ L(1)= ϕ2◦˜L(2)in˜∆1∩˜∆2and since ϕ1◦ u1and ϕ2◦ u2have the
same Cauchy data on γ0we get by the uniqueness continuation theorem that
ϕ1◦u1= ϕ2◦u2in˜∆1∩˜∆2. If ˜ γ1?= ˜ γ2then we can find u1such that ϕ1◦u1
and ϕ2◦u2will have different point of reflection and this will contradict that
ϕ1◦ u1= ϕ2◦ u2in˜∆1∩˜∆2. Since ˜ γ1= ˜ γ2we get that ϕ(γ1) = γ2⊂ ∂Ω′
and ϕ(∆1) = ∆2where ϕ = ϕ−1
Remark 4.2 Note that Lemma 4.1 allows to consider the inverse prob-
lems in multi-connected domains Ω with the D-to-N operator given on a not
connected part Γ0of ∂Ω.
2we get a contradiction since ψ2◦ u1is C∞outside˜Γ2and ψ1◦ u2is
2. Therefore˜D1=˜D2and ψ = ψ−1
2we have Γ′
2ψ1is a diffeomorphism of
2= ψ(Γ2). Since ψ1◦ u1= ψ2◦ u2in
Let γ0 be an open subset of ∂Ω(0)
1. Denote by ∆1 the union of all geodesics in Ω(1)
1. Denote by γ1the intersection of ∆1and
1. Introduce semi-geodesic coordinates for L(1)in ∆1. Let ϕ1be the change
2. Let γ2= ∆2∩ ∂Ω′
2, ˜ γ2= ϕ2(γ2),˜∆2= ϕ2(∆2). Let L(1)u1= 0 be
iand leaves Ω(1)\ Ω(0)again on γ0. Let u2be the
2
2ϕ1.
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