PARTIAL CAUCHY DATA FOR GENERAL SECOND ORDER ELLIPTIC
OPERATORS IN TWO DIMENSIONS
OLEG YU. IMANUVILOV, GUNTHER UHLMANN, AND MASAHIRO YAMAMOTO
Abstract. We consider the problem of determining the coefficients of a first-order perturbation of the Laplacian in two dimensions by measuring the corresponding Cauchy data
on an arbitrary open subset of the boundary. From this information we obtain a coupled
system of ∂z̄ and ∂z which the coefficients satisfy. As a corollary we show that for a simply
connected domain we can determine uniquely the coefficients up to the natural obstruction. Another consequence of our result is that the magnetic field and the electric potential
are uniquely determined by measuring the partial Cauchy data associated to the magnetic
Schrödinger equation measured on an arbitrary open subset of the boundary. of the boundary. We also show that the coefficients of any real vector field perturbation of the Laplacian,
the convection terms, are uniquely determined by their partial Cauchy data.
1. Introduction
We consider the problem of determining a complex-valued potential q and complex-valued
coefficients A and B in a bounded two dimensional domain from the Cauchy data measured
on an arbitrary open subset of the boundary for the associated second order elliptic operator
∂
∂
+ 2B ∂z
+ q. Specific cases of interest are the magnetic Schrödinger operator
∆ + 2A ∂z
and the Laplacian with convection terms. We remark that general second order elliptic
operators can be reduced to this form by using isothermal coordinates (e.g., [21]). The case
of the conductivity equation has been considered in [13]. For global uniqueness results in
the two dimensional case for the conductivity equation with full data measurements under
different regularity assumptions see [1], [4], [17]. Such a problem originates in [7].
Below we more precisely formulate our inverse problem under consideration. Let Ω ⊂ R2
be a bounded domain with smooth boundary ∂Ω = ∪N
k=1 γk , where γk , 1 ≤ k ≤ N , are
connected components and smooth closed contours and let √
ν be the unit outward normal
∂u
vector to ∂Ω. We denote ∂ν = ∇u · ν. Henceforth we set i = −1, x1 , x2 ∈ R, z = x1 + ix2 ,
z denotes the complex conjugate of z ∈ C, and we identify x = (x1 , x2 ) ∈ R2 with z =
∂
∂
x1 + ix2 ∈ C. We also denote ∂z
= 12 ( ∂x∂ 1 + i ∂x∂ 2 ), ∂z
= 12 ( ∂x∂ 1 − i ∂x∂ 2 ).
From now on we assume that (A, B, q), (Aj , Bj , qj ) ∈ C 5+α (Ω) × C 5+α (Ω) × C 4+α (Ω),
j = 1, 2 for some α > 0 are complex-valued functions.
Let u ∈ H 1 (Ω) be a solution of the Dirichlet problem
(1.1)
L(x, D)u = ∆u + 2A
∂u
∂u
+ 2B
+ qu = 0 in Ω,
∂z
∂z
u∂Ω = f,
First author partly supported by NSF grant DMS 0808130.
Second author partly supported by NSF and a Walker Family Endowed Professorship.
1
2
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
1
where f ∈ H 2 (∂Ω) is a given boundary input. The Dirichlet-to-Neumann (DN) map, assuming that 0 is not a Dirichlet eigenvalue, is defined by
∂u (1.2)
ΛA,B,q (f ) =
.
∂ν Γe
More generally we define the set of Cauchy data by:
∂u
∂
∂u
1
\
+ q)u = 0 in Ω, u ∈ H (Ω) .
(1.3)
C
u|∂Ω , | (∆ + 2A + 2B
A,B,q =
∂ν ∂Ω
∂z
∂z
1
−1
\
We have C
A,B,q ⊂ H 2 (∂Ω) × H 2 (∂Ω).
e ⊂ ∂Ω be a fixed non-empty open subset of the boundary and Γ0 = ∂Ω \ Γ̃.
Let Γ
Our main theorem states that the coefficients must satisfy a system of ∂ and ∂ if the set
e for
of Cauchy data for the pairs coincide. Consider the following sets of Cauchy data on Γ
j = 1, 2:
(1.4)
∂u ∂
∂
1
C(Aj ,Bj ,qj ) =
u|Γe , | (∆ + 2Aj
+ 2Bj
+ qj )u = 0 in Ω, u|Γ0 = 0, u ∈ H (Ω) .
∂ν Γe
∂z
∂z
Our main result is:
Theorem 1.1. Assume C(A1 ,B1 ,q1 ) = C(A2 ,B2 ,q2 ) . Then
(1.5)
A 1 = A2 ,
B1 = B2
on
e
Γ,
(1.6)
∂
(A1 − A2 ) − (B1 − B2 )A1 − (A1 − A2 )B2 + (q1 − q2 ) = 0 in Ω,
−2 ∂z
(1.7)
∂
−2 ∂z
(B1 − B2 ) − (A1 − A2 )B1 − (B1 − B2 )A2 + (q1 − q2 ) = 0 in Ω.
Remark. In the case that A1 = A2 and B1 = B2 in Ω, then Theorem 1.1 yields that
q1 = q2 , which is the main result in [13]. The latter result was extended to Riemann surfaces
in [11]. The case of full data in two dimensions was settled in [5]. This case is closed related
to the inverse conductivity problem, or Calderón problem. See the articles [17], [4], [1] in
two dimensions.
We observe that cannot uniquely determine (A, B, q) from the set of partial Cauchy data.
∂η
|Γ̃ = 0. Then it is easy to check the
Consider a function η ∈ C ∞ (Ω) such that η|Γ̃ = ∂ν
η
−η
operators L(x, D) and e L(x, D)e have the same Dirichlet-to-Neumann map.
In the case of a simply connected domain Ω, we can show that from the set of partial
Cauchy data we can determine uniquely a first order perturbation of the Laplacian up to
the natural obstruction mentioned above from the set of partial Cauchy data.
Corollary 1.1. Let Ω be a simply connected domain. Then C(A1 ,B1 ,q1 ) = C(A2 ,B2 ,q2 ) if and
∂η
only if there exists a function η ∈ C 5+α (Ω̄), η|Γ̃ = ∂ν
|Γ̃ = 0 such that
(1.8)
L1 (x, D) = e−η L2 (x, D)eη .
Proof. We only prove the sufficiency since the necessity of the condition is easy to check. By
∂
∂
(A1 − A2 ) = ∂z
(B1 − B2 ). Since the domain Ω is simply connected we claim
(1.6), (1.7), ∂z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
3
that there exists a function η ∈ C 5+α (Ω̄) such that
∂η
∂η
, B1 − B2 = 2 .
∂ z̄
∂z
The existence of such an η is proved as follows. We define f = A1 − A2 and g = B1 − B2 .
Then ∂f
= ∂g
. Now we a one-form a by a = f dz + gdz. Then we have
∂z
∂z
A1 − A2 = 2
da =
∂g
∂f
∂g
∂f
dz ∧ dz + dz ∧ dz =
dz ∧ dz − dz ∧ dz = 0
∂z
∂z
∂z
∂z
Therefore in a simply connected domain Ω, we can choose η such that a = dη =
and therefore f = ∂η
and g = ∂η
, proving proving the existence of η.
∂z
∂z
Therefore by (1.6)
∂η
dz + ∂η
dz
∂z
∂z
∂η ∂η
∂η
∂η
+ 2 A2 + 2 B2 .
∂z ∂ z̄
∂z
∂ z̄
The operator L1 (x, D) given by (1.8) has the Laplace operator as the principal part, the
∂
coefficients of ∂z
is 2A2 + 4 ∂η
, the coefficient of ∂∂z̄ is 2B2 + 4 ∂η
, and the coefficient of the
∂ z̄
∂z
∂η
zero order term is given by the right-hand side of (1.9). By (1.5) we have that ∂ν
|Γ̃ = 0 and
η|Γ̃ = 0. The proof of the corollary is completed.
(1.9)
q1 = q2 + ∆η + 4
e=
We now apply our result to the case of the magnetic Schrödinger operator. Denote A
e
e
∂
A
∂
A
e1 , A
e2 ), where A
ej are real-valued, Ae = A
e1 − iA
e2 , rot A
e = 2 − 1 , Dj = 1 ∂ . Consider
(A
∂x1
∂x2
i ∂xj
the magnetic Schrödinger operator
(1.10)
LA,e
e q (x, D) =
2
X
ek )2 + qe.
(Dk + A
k=1
Let us define the following set of partial Cauchy data
∂u
1
e
CAe(j) ,eq(j) = (u|Γe , |Γe )|LAe(j) ,eq(j) (x, D)u = 0 in Ω, u|Γ0 = 0, u ∈ H (Ω) .
∂ν
For the case of full data it is known that there is a gauge invariance in the problem and we
can recover at best the magnetic field [20]. The same is valid for the case of partial Cauchy
data [10]. We prove here that the converse holds in two dimensions.
e(1) , A
e(2) ∈ C 5+α (Ω) and complexCorollary 1.2. Let α > 0, real-valued vector fields A
e e(1) (1) = C
e e(2) (2) . Then qe(1) = qe(2) and
valued potentials qe(1) , qe(2) ∈ C 4+α (Ω) be such that C
A ,e
q
A ,e
q
e(1) = rot A
e(2) .
rot A
Proof. A straightforward calculation gives
e
e
2e ∂
2e ∂
e 2 + 1 ∂ A1 + 1 ∂ A2 + qe
LA,e
+ A
+ |A|
e q (x, D) = −∆ + A1
2
i ∂x1
i ∂x2
i ∂x1
i ∂x2
2 ∂
2 ∂
2 ∂ Ae
e + |A|
e 2 + qe.
(1.11)
= −∆ + Ae + Ae +
− rot A
i ∂z
i ∂z
i ∂z
Then the operator LA,e
e q (x, D) is a particular case of (1.1). Suppose that the vector fields
(1)
(2)
e ,A
e and the potentials qe(1) , qe(2) have the same Dirichlet-to-Neumann map. Taking into
A
4
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
e(j)
e(j) + |A
e(j) |2 + qe(j) ), we see
account that Aj = − 1i Ae(j) , Bj = − 1i Ae(j) , qj = −( 2i ∂ A∂z − rot A
that (1.6) gives
e(1) − rot A
e(2) + qe(2) − qe(1) ≡ 0
rot A
and (1.7) gives
(1.12)
2 ∂ Ae(1) 2 ∂ Ae(2) 2 ∂ Ae(1) 2 ∂ Ae(2)
e(1) − rot A
e(2) + qe(2) − qe(1) ≡ 0.
−
−
+
+ rot A
i ∂z
i ∂z
i ∂z
i ∂z
Using the identity
2 ∂A
i ∂z
−
2 ∂A
i ∂z
e we transform (1.12) to the form
= −2rot A
e(1) − rot A
e(2) ) + qe(2) − qe(1) ≡ 0.
−(rot A
The proof of the corollary is completed.
Corollary 1.2 is new even in the case when the data is measured on the whole boundary. In
two dimensions, Sun proved in [20] that for measurements on the whole boundary uniqueness
holds assuming that both the magnetic potential and the electric potential are small. Kang
and Uhlmann proved global uniqueness for the case of measurements on the whole boundary
for a special case of the magnetic Schrödinger equation, namely the Pauli Hamiltonian [14].
In dimension n ≥ 3 global uniqueness was shown in [18] for the case of full data. The
regularity assumptions in the result were improved by Salo in [19]. The case of partial data
was considered in [10], based on the methods of [15] and [6], with an improvement on the
regularity of the coefficients in [16].
Our main theorem implies that the Dirichlet-to-Neumann map can uniquely determine
any two of (A, B, q). First we can prove that A and B are uniquely determined if q is known.
We discuss the uniqueness for Laplace operators with convection terms.
(1.13)
L(x, D)u = ∆u + a(x)
∂u
∂u
+ b(x)
+ q(x)u.
∂x1
∂x2
Here a, b, q are complex-valued functions. Let us define the following set of partial Cauchy
data
∂u
∂u
∂u
(j)
(j)
1
e
+ b (x)
+ q(x)u = 0 in Ω, u|Γ0 = 0, u ∈ H (Ω) .
Ca(j) ,b(j) = (u|Γe , |Γe )|∆u + a (x)
∂ν
∂x1
∂x2
We have
Corollary 1.3. Let α > 0 and two pairs of complex-valued coefficients (a(1) , b(1) ) ∈ C 5+α (Ω)×
ea(1) ,b(1) = C
ea(2) ,b(2) . Then
C 5+α (Ω) and (a(2) , b(2) ) ∈ C 5+α (Ω) × C 5+α (Ω) be such that C
(a(1) , b(1) ) ≡ (a(2) , b(2) ).
Proof. Taking into account that
operator (1.13) in the form
∂
∂x1
∂
= ( ∂z
+
L(x, D)u = ∆u + (a(x) + ib(x))
∂
)
∂z
and
∂
∂x2
∂
= i( ∂z
−
∂
),
∂z
we can rewrite the
∂u
∂u
+ (a(x) − ib(x))
+ q(x)u.
∂z
∂z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
5
The pairs (a(1) , b(1) ) and (a(2) , b(2) ) be such that corresponding operators defined by (1.13)
have the same Dirichlet-to-Neumann map. Denote 2Ak (x) = a(k) (x) + ib(k) (x) and 2Bk (x) =
a(k) (x) − ib(k) (x). By (1.6), we have
(1.14)
∂
−2 ∂z
(A1 − A2 ) − (B1 − B2 )A1 − (A1 − A2 )B2 = 0 in Ω,
(1.15)
∂
−2 ∂z
(B1 − B2 ) − (A1 − A2 )B1 − (B1 − B2 )A2 = 0 in Ω.
∂
∂
Applying to equation (1.14) the operator 2 ∂z
and to equation (1.15) the operator 2 ∂z
we
have
(1.16)
−∆(A1 − A2 ) − 2 ∂∂z̄ ((B1 − B2 )A1 + (A1 − A2 )B2 ) = 0 in Ω,
(1.17)
∂
−∆(B1 − B2 ) − 2 ∂z
((A1 − A2 )B1 + (B1 − B2 )A2 ) = 0 in Ω.
By (1.5)
(A1 − A2 )|Γe = (B1 − B2 )|Γe = 0.
Using these identities and equations (1.14), (1.15) we obtain
∂(B1 − B2 )
∂(A1 − A2 )
|Γe =
|Γe = 0.
∂ν
∂ν
The uniqueness of the Cauchy problem for the system (1.16)-(1.17) can be proved in the
standard way by using a Carleman estimate (e.g., [12]). Therefore we have A1 = A2 and
B1 = B2 in Ω.
We remark that Corollary generalizes the result of [9] who proved this result assuming that
the measurements are made on the whole boundary. In dimension n ≥ 3 global uniqueness
was shown in [8] for the case of full data.
Similarly to Corollary 1.2, we can prove that the Dirichlet-to-Neumann map can uniquely
determine a potential q and one of A and B in (1.1).
Corollary 1.4. For j = 1, 2, let (Aj , Bj , qj ) ∈ C 5+α (Ω)×C 5+α (Ω)×C 4+α (Ω) for some α > 0
and be complex-valued. We assume either A1 = A2 or B1 = B2 in Ω. Then C(A1 ,B1 ,q1 ) =
C(A2 ,B2 ,q2 ) implies (A1 , B1 , q1 ) = (A2 , B2 , q2 ).
The proof of Theorem 1.1 follows the general method of [13]. In this case we need to
prove a new Carleman estimate with degenerate harmonic weights to construct appropriate
complex geometrical optics solutions. These solutions have a different form to take into
account the first order terms. The new form of these solutions complicates considerably
the arguments, especially the asymptotic expansions needed to analyze the behavior of the
solutions. In Section 2 we prove the Carleman estimate which we need. In Section 3 we
state the estimates and asymptotics which we will use in the construction of the complex
geometrical optics solutions. This construction is done in Section 4. The proof of Theorem
1.1 is completed in Section 5. In section 6 and 7 we discuss some technical lemmas needed
in the previous sections.
6
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
2. Carleman estimate
√
Notations We use throughout the paper the following notations. i = −1, x1 , x2 , ξ1 , ξ2 ∈ R,
z = x1 + ix2 , ζ = ξ1 + iξ2 , z denotes the complex conjugate of z ∈ C, Dk = 1i ∂x∂ k . We
∂
identify x = (x1 , x2 ) ∈ R2 with z = x1 + ix2 ∈ C. We set ∂z = ∂z
= 12 ( ∂x∂ 1 − i ∂x∂ 2 ),
∂z̄ = ∂∂z̄ = 12 ( ∂x∂ 1 + i ∂x∂ 2 ), O = {x ∈ Ω|dist(x, ∂Ω) ≤ }. Throughout the paper we use both
2
∂
notations ∂z and ∂z
, etc. and for example we denote ∂z̄2 = ∂∂z̄2 . We say that a function a(x)
∂
=
is antiholomorphic if ∂z a(x) = 0. The tangential derivative on the boundary is given by ∂~
τ
∂
∂
ν2 ∂x1 − ν1 ∂x2 , with ν = (ν1 , ν2 ) the unit outer normal to ∂Ω. The ball of radius δ centered
at x
b is denoted by B(b
x, δ) = {x ∈ R2 ||x − x
b| < δ}. the corresponding sphere is denoted by
2
S(b
x, δ) = {x ∈ R ||x − x
b| = δ}.if f (x) : R2 → R1 is a function, f 00 is the Hessian matrix
2
f
with entries ∂x∂i ∂x
. k·k2H k,τ (Ω) = k·k2H k (Ω) +|τ |2k k·k2L2 (Ω) is the standard semiclassical Sobolev
j
space with inner product given by (·, ·)H k,τ (Ω) = (·, ·)H k (Ω) + |τ |2k (·, ·)L2 (Ω) . L(X, Y ) denotes
the Banach space of all bounded linear operators from a Banach space X to another Banach
space Y . By oX ( τ1κ ) we denote a function f (τ, ·) such that kf (τ, ·)kX = o( τ1κ ) as |τ | → +∞.
Let Φ(z) = ϕ(x1 , x2 ) + iψ(x1 , x2 ) ∈ C 5 (Ω) with real-valued ϕ and ψ satisfy
∂Φ
(z) = 0 in Ω, Im Φ|Γ0 = 0.
∂ z̄
Denote by H the set of all the critical points of the function Φ
(2.1)
H = {z ∈ Ω|
∂Φ
(z) = 0}.
∂z
Assume that Φ has no critical points on Γ̃, and that all critical points on the boundary are
nondegenerate:
∂2Φ
(z) 6= 0, ∀z ∈ H.
∂z 2
Then Φ has only a finite number of critical points and we can set:
(2.2)
(2.3)
H ∩ ∂Ω ⊂ Γ0 ,
H \ Γ0 = {e
x1 , ..., x
e` },
H ∩ Γ0 = {e
x`+1 , ..., x
e`+`0 }.
The following proposition was proved in [13].
Proposition 2.1. Let x
e be an arbitrary point in Ω. There exists a sequence of functions
{Φ }∈(0,1) satisfying (2.1) such that all the critical points of Φ are nondegenerate and there
exists a sequence {e
x }, ∈ (0, 1) such that
∂Φ
(z) = 0},
∂z
Moreover for any j from {1, . . . , N } we have
x
e ∈ H = {z ∈ Ω|
x
e → x
e as → +0.
H ∩ γj = ∅
if γj ∩ Γ̃ 6= ∅,
H ∩ γj ⊂ Γ0
if γj ∩ Γ̃ = ∅,
Im Φ (e
x ) ∈
/ {Im Φ (x)|x ∈ H \ {xe }} and Im Φ (e
x ) 6= 0.
In order to prove (1.5) we need the following proposition.
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
7
e be an arc with the left endpoint x− and the right endpoint
Proposition 2.2. Let Γ∗ ⊂⊂ Γ
x+ oriented clockwise. For any x
b ∈ Int Γ∗ there exists a function Φ(z) which satisfies (2.1),
(2.2), Im Φ|∂Ω\Γ∗ = 0 and
x
b ∈ G = {x ∈ Γ∗ |
(2.4)
(2.5)
(
∂Im Φ
(x) = 0},
∂~τ
cardG < ∞,
∂ 2
) Im Φ(x) 6= 0 ∀x ∈ G \ {x− , x+ },
∂~τ
Moreover
Im Φ(b
x) 6= Im Φ(x) ∀x ∈ G \ {b
x} and
(2.6)
(2.7)
(
∂
)6 Im Φ(x− ) 6= 0,
∂~τ + 0
(
Im Φ(b
x) 6= 0.
∂
)6 Im Φ(x+ ) 6= 0.
∂~τ − 0
Proof. Denote Γ∗0 = ∂Ω \ Γ∗ . Let x
b− , x
b+ ∈ ∂Ω be points such that the arc [b
x− , x
b+ ] ⊂ (x− , x+ )
and x
b ∈ (b
x− , x
b+ ) be an arbitrary point and x0 be another fixed point from the interval
(b
x, x
b+ ). We claim that there exists a pair (ϕ, ψ) ∈ C 5 (Ω) × C 5 (Ω) which solves the system
of Cauchy-Riemann equations in Ω such that
∂ψ
∂
∂ϕ
|γj \Γ∗ > 0 ifγj ∩ Γ∗ 6= ∅,
(b
x) = 0, ( )2 ψ(b
x) 6= 0,
∂~τ
∂~τ
∂~τ
∂
∂
A0 ) (
)6 ψ(x− ) 6= 0, (
)6 ψ(x+ ) 6= 0,
∂~τ + 0
∂~τ − 0
A) ψ|Γ∗0 = 0, |
B) The restriction of the function ψ to the arc [b
x− , x
b+ ] is a Morse function,
∂ψ
> 0 on (x− , x
b− ],
∂~τ
∂ψ
< 0 on [b
x+ , x+ ),
∂~τ
∂ψ
(x) = 0},
D) ψ(b
x) ∈
/ {ψ(x)|x ∈ ∂Ω \ {b
x},
∂~τ
E) if γj ∩ Γ∗ = ∅, then the restriction of the function ϕ on γj
C)
has only two nondegenerate critical points.
Such a pair of functions may be constructed in the following way. Let γ1 ∩ Γ∗ 6= ∅ and
γj ∩ Γ∗ = ∅ for all j ∈ {2, . . . N }. First by Corollary 6.1 in the Appendix, for some α ∈ (0, 1),
e ∈ C 5+α (Ω) × C 5+α (Ω) to the Cauchy-Riemann equations with
there exists a solution (ϕ,
e ψ)
the following boundary data
∂ϕ
e
|γ \[x ,bx ] < β < 0
∂~τ 0 0 +
and such that if γj ∩ Γ∗ = ∅ the function ϕ̃ has only two nondegenerate critical points
∗
>
located on the contour γj . The function ψ∗ has the following properties: ψ∗ |Γ∗0 = 0, ∂ψ
∂~
τ
∂ψ∗
0 on (x− , x
b− ], ∂~τ < 0 on [b
x+ , x+ ). The function ψ∗ on the set [b
x− , x0 ] has only one critical
point x
b and ψ∗ (x̂) 6= 0. On the set (x0 , x̂+ ) the Cauchy data is not fixed. The restriction of
e ∂Ω\[x ,bx ] = ψ∗ ,
ψ|
0 +
8
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
the function ψe on [x0 , x
b+ ] can be approximated in the space C 5+α ([x0 , x
b+ ]) by a sequence of
Morse functions {g }∈(0,1) such that
(
∂ ke
∂
) ψ(x) = ( )k g (x) x ∈ {b
x+ , x0 },
∂~τ
∂~τ
k ∈ {0, 1, . . . , 5},
and
ψ∗ (x̂) ∈
/ {g (x)|
∂g (x)
= 0}.
∂~τ
Let us consider some arc J ⊂⊂ (x− , x
b− ). On this arc we have
(2.8)
∂ ψe
> β 0 > 0 on J
∂~τ
e
∂ψ
∂~
τ
> 0, say,
for some positive β 0 .
Let (ϕ , ψ ) ∈ C 5+α (Ω) × C 5+α (Ω) be a solution to the Cauchy-Riemann equations with
boundary data ψ = 0 on ∂Ω \ (J ∪ [x0 , x
b+ ]) and ψ = g − ψe on [x0 , x
b+ ] and on J the
Cauchy data is chosen in such a way that
(2.9)
e C 5+α ([x ,bx ]) → 0.
kψ kC 5+α (∂Ω) + kϕ kC 5+α (∂Ω) → 0 as kg − ψk
0 +
By (2.8), (2.9) for all small positive , the restriction of the function ψe + ψ to ∂Ω satisfies
(ψe + ψ )|Γ∗0 = 0,
∂(ψe + ψ )
|γ0 \[x0 ,x̂+ ] < 0,
∂ν
∂(ψe + ψ )
> 0 on [x− , x̂− ],
∂~τ
∂(ψe + ψ )
< 0 on [x̂+ , x+ ], (ψe + ψ )|[x0 ,bx+ ] = g , (ψe + ψ )|[x̂− ,x0 ] = ψ∗ .
∂~τ
If j ≥ 2 then the restriction of the function ϕ + ϕ
e on γj has only two critical points located
∗
on the contour γj ⊂ Γ0 . These critical points are nondegenerate if is sufficiently small.
Therefore the restriction of the function (ψe + ψ ) on Γ∗ has a finite number of critical
points. Some of these points may be the critical points of (ψe + ψ ) considered as the function
on Ω. We change slightly the function (ψe + ψ ) such that all of its critical points are in Ω.
Suppose that function ψe + ψ has critical points on Γ∗ . Then these critical points should be
among the set of critical points of the function g , otherwise it would be the point x
b. We
b ∈ C 5+α (Ω) × C 5+α (Ω) be a solution to the
denote these points by x
b1 , . . . , x
bm . Let (ϕ,
b ψ)
Cauchy-Riemann problem (6.1) with the following boundary data
b Γ∗ = 0, ψ(b
b x) = 1, ψ|
b G\{bx} = 0,
ψ|
0
|
∂ ψb
||
> 0.
∂ν γ0 \J
For all small positive 1 the function ψe + ψ + 1 ψb does not have a critical point on ∂Ω and
e has a finite number of nondegenerate critical points.
the restriction of this function on Γ
b as the pairs of functions satisfying A) - E).
Therefore we take (ϕ
e + ϕ + 1 ϕ,
b ψe + ψ + 1 ψ)
The function ϕ+iψ with pair (ϕ, ψ) satisfying conditions A)-E) satisfies all the hypotheses
of Proposition 2.2 except that some of its critical points might possibly be degenerate.
In order to fix this problem we consider a perturbation of the function ϕ + iψ which is
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
9
constructed in the following way. By Proposition 6.2, there exists a function w holomorphic
in Ω, such that
∂w
∂2w
|H0 = 0,
|H 6= 0.
∂z
∂z 2 0
Denote Φδ = ϕ + iψ + δw. For all sufficiently small positive δ, we have
∂
H0 ⊂ Hδ ≡ {x ∈ Ω| Φδ (x) = 0}.
∂z
We now show that for all sufficiently small positive δ, all critical points of the function Φδ
are nondegenerate. Let x
e be a critical point of the function ϕ + iψ. If x
e is a nondegenerate
critical point, by the implicit function theorem, there exists a ball B(e
x, δ1 ) such that the
function Φδ in this ball has only one nondegenerate critical point for all small δ. Let x
e be a
degenerate critical point of ϕ + iψ. Without loss of generality we may assume that x
e = 0.
P∞
P
k+b
k
k
δ
=
−δ ∞
In some neighborhood of 0, we have ∂Φ
k=1 ck z
k=1 bk z for some natural
∂z
number b
k and some c1 6= 0. Moreover (2.10) implies b1 6= 0. Let (x1,δ , x2,δ ) ∈ Hδ and
zδ = x1,δ + ix2,δ → 0. Then either
(2.10)
Im w|Γ∗0 = 0, w|H0 =
zδ = 0 or zδk = δb1 /c1 + o(δ) as δ → 0.
(2.11)
Therefore
b
∂ 2 Φδ
(zδ )
∂z 2
6= 0 for all sufficiently small δ.
The following proposition was proven in [13]:
Proposition 2.3. Let Φ satisfy (2.1) and (2.2). Let fe ∈ L2 (Ω), and ve ∈ H 1 (Ω) be a solution
to
∂
∂Φ
(2.12)
2 ve − τ
ve = fe in Ω
∂z
∂z
or ve be a solution to
∂
∂Φ
(2.13)
2 ve − τ
ve = fe in Ω.
∂z
∂z
In the case (2.12) we have
Z
∂ −iτ ψ 2
k
(e
ve)kL2 (Ω) − τ
(∇ϕ, ν)|e
v |2 dσ
∂x1
∂Ω
Z
∂
∂
∂ −iτ ψ 2
i
ν2
(2.14) +Re
− ν1
ve vedσ + k
(e
ve)kL2 (Ω) = kfek2L2 (Ω).
∂x1
∂x2
∂x2
∂Ω
In the case (2.13) we have
Z
Z
∂ iτ ψ
∂
∂
2
k
(e ve)kL2 (Ω) − τ
(∇ϕ, ν)|e
v | dσ + Re
i
−ν2
+ ν1
ve vedσ
∂x1
∂x1
∂x2
∂Ω
∂Ω
∂ iτ ψ 2
(2.15)
+k
(e ve)kL2 (Ω) = kfek2L2 (Ω) .
∂x2
Let α ∈ (0, 1) and A, B ∈ C 6+α (Ω) be two complex-valued solutions to the boundary value
problem
∂A
∂B
(2.16)
2
= −A in Ω, Im A|Γ0 = 0, 2
= −B in Ω, Im B|Γ0 = 0.
∂z
∂z
10
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Consider the boundary value problem

 K(x, D)u = (4 ∂ ∂ + 2A ∂ + 2B ∂ )u = f
∂z ∂z
∂z
∂z

u|∂Ω = 0.
in Ω,
For this problem we have the following Carleman estimate with boundary terms.
Proposition 2.4. Suppose that Φ satisfies (2.1), (2.2), u ∈ H01 (Ω) and the coefficients
A, B ∈ {D ∈ C 1 (Ω)|kDkC 1 (Ω) ≤ K}. Then there exist τ0 = τ0 (K, Φ) and C = C(K, Φ)
independent of u and τ such that for all |τ | > τ0
∂u τ ϕ 2
∂Φ
e kL2 (Γ0 ) + τ 2 k| |ueτ ϕ k2L2 (Ω)
∂ν
Z ∂z
∂u
≤ C1 (k(K(x, D)u)eτ ϕ k2L2 (Ω) + |τ | | |2 e2τ ϕ dσ).
e ∂ν
Γ
|τ |kueτ ϕ k2L2 (Ω) + kueτ ϕ k2H 1 (Ω) + k
(2.17)
Proof. Denote ve = ueτ ϕ , K(x, D)u = f. Observe that ϕ(x1 , x2 ) = 12 (Φ(z) + Φ(z)). Therefore
∂
∂Φ
∂
∂Φ
∂A
− (τ
− B))(2 − (τ
− A))e
v + (−2
− AB)e
v=
∂z
∂z
∂z
∂z
∂z
∂
∂Φ
∂
∂Φ
∂B
(2 − (τ
− A))(2 − (τ
− B))e
v + (−2
− AB)e
v = f eτ ϕ .
∂z
∂z
∂z
∂z
∂z
eτ ϕ K(x, D)(e−τ ϕ ve) = (2
(2.18)
∂
∂
e
e
Denote w
e1 = Q(z)(2
−τ ∂Φ
+A)e
v, w
e2 = Q(z)(2 ∂z
−τ ∂Φ
+B)e
v , where Q(z), Q(z)
∈ C 2 (Ω)
∂z
∂z
∂z
are some holomorphic functions in Ω that will be specified below. Thanks to the zero Dirichlet
boundary condition for u we have
v
v
∂e
v
∂e
v
e ∂e
e ∂e
w
e1 |∂Ω = 2Q(z)
|∂Ω = (ν1 + iν2 )Q(z)
|∂Ω , w
e2 |∂Ω = 2Q(z) |∂Ω = (ν1 − iν2 )Q(z) |∂Ω .
∂z
∂ν
∂z
∂ν
By Proposition 2.3 we have the following integral equalities:
Z
v B2
∂ψ
∂
B 2
e 2 | ∂e
(2.19)
− iτ
)(w
e1 e )kL2 (Ω) − τ
(∇ϕ, ν)|Q|
e | dσ
k(
∂x1
∂x1
∂ν
∂Ω
Z
∂
∂
+Re
i((ν2
− ν1
)(w
e1 eB ))w
e1 eB dσ +
∂x
∂x
1
2
∂Ω
∂A
∂
∂ψ
e eτ ϕ + (2
+k(
− iτ
)(w
e1 eB )k2L2 (Ω) = kQ(f
+ AB)e
v )eB k2L2 (Ω)
∂x2
∂x2
∂z
and
(2.20)
Z
∂
∂e
v
∂ψ
A 2
k(
+ iτ
)(w
e2 e )kL2 (Ω) − τ
(∇ϕ, ν)|Q|2 | eA |2 dσ
∂x1
∂x1
∂ν
∂Ω
Z
∂
∂
+Re
i((−ν2
+ ν1
)(w
e2 eA ))w
e2 eA dσ +
∂x
∂x
1
2
∂Ω
∂B
∂
∂ψ
+k(
+ iτ
)(w
e2 eA )k2L2 (Ω) = kQ(f eτ ϕ + (2
+ AB)e
v )eA k2L2 (Ω) .
∂x2
∂x2
∂z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
11
R
We now simplify the integral Re i ∂Ω ((ν2 ∂x∂ 1 − ν1 ∂x∂ 2 )(w
e1 eB dσ. We recall that ve =
e1 eB ))w
∂e
v
∂u τ ϕ
B
e
e
e
ueτ ϕ and w
e1 = Q(z)(ν
= Q(z)(ν
e . Denote R + iP = Q(z)(ν
1 + iν2 )
1 + iν2 )
1 + iν2 )e .
∂ν
∂ν
Therefore
Z
(2.21)
∂
∂
e1 eB dσ =
− ν1
)(w
e1 eB ))w
∂x
∂x
1
2
∂Ω
Z
∂
∂u τ ϕ
∂
∂u
Re
i((ν2
− ν1
)[(R + iP ) e ])(R − iP ) eτ ϕ dσ =
∂x1
∂x2
∂ν
∂ν
∂Ω
Z
∂
∂
∂e
v
Re
i[(ν2
− ν1
)(R + iP )]| |2 (R − iP )dσ +
∂x1
∂x2
∂ν
∂Ω
Z
∂
∂ ∂e
v
i 2
(R + P 2 )(ν2
− ν1
)| |2 dσ =
Re
∂x1
∂x2 ∂ν
∂Ω 2
Z
∂R
∂P
∂e
v
(
P−
R)| |2 dσ.
τ
∂~τ
∂ν
∂Ω ∂~
Re
i((ν2
R
e2 eA ))w
e2 eA dσ. We recall that ve =
Let us simplify the integral Re ∂Ω i((−ν2 ∂x∂ 1 + ν1 ∂x∂ 2 )(w
∂e
v
ueτ ϕ and w
e2 = (ν1 − iν2 )Q(z) ∂ν
= (ν1 − iν2 )Q(z) ∂u
eτ ϕ . Denote R̃ + iP̃ = Q(z)(ν1 − iν2 )eA .
∂ν
Thus
Z
∂
∂
(2.22)
Re
i((−ν2
+ ν1
)(w
e2 eA ))w
e2 eA dσ =
∂x
∂x
1
2
∂Ω
Z
∂u
∂
∂
e + iPe) eτ ϕ ])(R
e − iPe) ∂u eτ ϕ dσ =
Re
i((−ν2
+ ν1
)[(R
∂x1
∂x2
∂ν
∂ν
∂Ω
Z
∂
∂
v 2 e
e + iPe)]| ∂e
Re
i[(−ν2
+ ν1
)(R
| (R − iPe)dσ −
∂x1
∂x2
∂ν
∂Ω
Z
∂
∂ ∂e
v
i e2 e 2
(R + P )(ν2
− ν1
)| |2 dσ =
Re
∂x1
∂x2 ∂ν
∂Ω 2
Z
e
∂R
∂ Pe e ∂e
v
(
Pe −
R)| |2 dσ.
τ
∂~τ
∂ν
∂Ω ∂~
Using the above formula we obtain
∂
∂
∂ψ
∂ψ
+ iτ
)(w
e2 eA )k2L2 (Ω) + k(
+ iτ
)(w
e2 eA )k2L2 (Ω)
∂x1
∂x1
∂x2
∂x2
Z
Z
v B2
∂e
v
2 ∂e
e
−τ
(ν, ∇ϕ)|Q| | e | dσ − τ
(ν, ∇ϕ)|Q|2 | eA |2 dσ
∂ν
∂ν
∂Ω
∂Ω
∂
∂ψ
∂
∂ψ
+k(
− iτ
)(w
e1 eB )k2L2 (Ω) + k(
− iτ
)(w
e1 eB )k2L2 (Ω)
∂x1
∂x2
∂x2
∂x2
Z
Z
e
∂R
∂P
∂e
v B2
∂R
∂ Pe e ∂e
v
Pe −
R)| eA |2 dσ =
+
(
P−
R)| e | dσ +
(
τ
∂~τ
∂ν
τ
∂~τ
∂ν
∂Ω ∂~
∂Ω ∂~
∂B
∂A
(2.23) kQ(f eτ ϕ + (2
+ AB)e
v )eB k2L2 (Ω) + kQ(f eτ ϕ + (2
+ AB)e
v )eA k2L2 (Ω) .
∂z
∂z
k(
12
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
We can rewrite (2.23) in the form
∂ iψτ
∂ iψτ
(e w
e2 eA )k2L2 (Ω) + k
(e w
e2 eA )k2L2 (Ω)
∂x1
∂x2
Z
Z
∂e
v
v B2
2 ∂e
e
(ν, ∇ϕ)|Q|2 | eA |2 dσ
−τ
(ν, ∇ϕ)|Q| | e | dσ − τ
∂ν
∂ν
∂Ω
∂Ω
∂ −iψτ
∂ −iψτ
(e
w
e1 eB )k2L2 (Ω) + k
(e
w
e1 eB )k2L2 (Ω)
+k
∂x1
∂x2
Z
Z
e
∂R
∂R
∂ Pe e ∂e
v
∂P
∂e
v B2
+
(
P−
R)| e | dσ +
(
Pe −
R)| eA |2 dσ =
τ
∂~τ
∂ν
τ
∂~τ
∂ν
∂Ω ∂~
∂Ω ∂~
∂B
e eτ ϕ + (2 ∂A + AB)e
(2.24) kQ(f
v )eB k2L2 (Ω) + kQ(f eτ ϕ + (2
+ AB)e
v )eA k2L2 (Ω) .
∂z
∂z
k
Next we show that it is possible to make a choice of functions Q̃ and Q such that
e
∂P
∂R
∂ Pe e
∂R
P−
R) > 0 on Γ0 , (
Pe −
R) > 0 on Γ0 .
∂~τ
∂~τ
∂~τ
∂~τ
Let γj be a contour from ∂Ω. We parametrize the curve γj by arc length s starting from one
fixed point on γj : x(s) : [0, `j ] → γj . Here `j denotes the total length of γj . We note that
d
∂R
= ds
R(x(s)). If there exists a holomorphic function Q̃ such that R(x(s)) = `j sin(s/`j ),
∂~
τ
P (x(s)) = `j cos(s/`j ). Then
(2.25)
(2.26)
(
∂R
∂P
P−
R = `j on γj
∂~τ
∂~τ
∀j ∈ {1, . . . , N }.
Taking into account that R + iP = Q̃(z)(ν1 + iν2 )eB we set
(2.27)
`j sin(s/`j ) − i`j cos(s/`j ) −B
`j sin(s/`j ) − i`j cos(s/`j ) −B
b1 = Re
e
, b2 = Im
e
.
(ν1 − iν2 )
(ν1 − iν2 )
e
Using Proposition 5.1 in [13] we choose Q(z)
such that on Γ0 the function Q̃(z) is close to
1
b1 +ib2 in the norm of the space C (Γ0 ). Then by (2.27) we have the first inequality in (2.25).
The choice of the function Q may be done in a similar fashion.
By (2.25) we obtain from (2.24) that
(2.28)
k
∂ṽ
kL2 (Γ0 ) ≤ C2 (kf eτ ϕ kL2 (Ω) + kṽkL2 (Ω) ).
∂ν
From now on we assume that Q̃ = Q = 1. Observe that there exists a positive constant C3 ,
independent of τ , such that
1 ∂ iτ ψ
1
1 ∂ iτ ψ
e2 k2L2 (Ω) ) ≤ k
(kw
e1 k2L2 (Ω) + kw
(e w
e2 eA )k2L2 (Ω) + k
(e w
e2 eA )k2L2 (Ω)
C3
2 ∂x1
2 ∂x2
Z
Z
∂e
v
∂e
v
e 2 | |2 |eB |2 dσ − τ
−τ
(ν, ∇ϕ)|Q|
(ν, ∇ϕ)|Q|2 | |2 |eA |2 dσ
∂ν
∂ν
∂Ω−
∂Ω−
1 ∂ −iτ ψ
1 ∂ −iτ ψ
(2.29)
(e
w
e1 eB )k2L2 (Ω) + k
(e
w
e1 eB )k2L2 (Ω) .
+ k
2 ∂x1
2 ∂x2
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
Here ∂Ω− = {x ∈ ∂Ω|(ν, ∇ϕ) < 0}. (Since
R
∂ϕ
dσ
∂Ω ∂ν
=0=
R
∂ϕ
dσ
Γ̃ ∂ν
13
the domain ∂Ω− is not
empty.) Observe that ū satisfies the equation K(x, D)u = f . Denoting w1∗ = Q̃(z)(2 ∂∂z̄ −
∂
− τ ∂Φ
+ A)ṽ and repeating the above arguments we have the
τ ∂∂Φ̄z̄ + B)ṽ, w2∗ = Q̃(z)(2 ∂z
∂z
analog of (2.29):
1 ∂ iτ ψ ∗ B 2
1 ∂ iτ ψ ∗ B 2
1
(kw1∗ k2L2 (Ω) + kw2∗ k2L2 (Ω) ) ≤ k
(e w2 e )kL2 (Ω) + k
(e w2 e )kL2 (Ω)
C4
2 ∂x1
2 ∂x2
Z
Z
∂e
v 2 A2
v
2 ∂e
e
−τ
(ν, ∇ϕ)|Q| | | |e | dσ − τ
(ν, ∇ϕ)|Q|2 | |2 |eB |2 dσ
∂ν
∂ν
∂Ω−
∂Ω−
1 ∂ −iτ ψ ∗ A 2
1 ∂ −iτ ψ ∗ A 2
(e
w1 e )kL2 (Ω) + k
(e
w1 e )kL2 (Ω) .
(2.30)
+ k
2 ∂x1
2 ∂x2
By (2.29), (2.30) and the definitions of w
e1 , w1∗ and w
e2 , w2∗ , we have
k
∂Im ve
∂Re ve
∂Φ
∂Φ
−τ
Re vek2L2 (Ω) + k2
−τ
Im vek2L2 (Ω)
∂z
∂z
∂z
∂z
≤ C5 (kw
e1 k2L2 (Ω) + kw
e2 k2L2 (Ω) + ke
v k2L2 (Ω) ).
Therefore
∂Φ 2
vek 2
∂z L (Ω)
≤ C6 (kw
e1 k2L2 (Ω) + kw
e2 k2L2 (Ω) + ke
v k2L2 (Ω) ).
k∇e
v k2L2 (Ω) + τ 2 k
(2.31)
Now since by assumption (2.2) the function Φ has zeros of at most second order, there
exists a constant C7 > 0 independent of τ such that
τ ke
v k2L2 (Ω) ≤ C7 (ke
v k2H 1 (Ω) + τ 2 k|
(2.32)
∂Φ 2
|e
v kL2 (Ω) ).
∂z
By (2.31) and (2.32)
∂Φ 2
|e
v kL2 (Ω) ≤ C8 (kw
e1 k2L2 (Ω) + kw
e2 k2L2 (Ω) + ke
v k2L2 (Ω) ).
∂z
By (2.33) and (2.28) we obtain from (2.24), (2.29)
(2.33)
(2.34)
τ ke
v k2L2 (Ω) + ke
v k2H 1 (Ω) + τ 2 k|
1
∂Φ 2
(τ ke
v k2L2 (Ω) + ke
v k2H 1 (Ω) + τ 2 k| |e
v kL2 (Ω) )
C9
∂z
Z
Z
∂e
v 2 B2
∂e
v
(ν, ∇ϕ)| | |e | dσ − τ
−τ
(ν, ∇ϕ)| |2 |eA |2 dσ
∂ν
∂ν
∂Ω
∂Ω
Z
∂e
v
≤ kf esϕ k2L2 (Ω) + τ |(ν, ∇ϕ)|| |2 (|eB |2 + |eA |2 )dσ.
∂ν
e
Γ
This concludes the proof of the proposition.
As a corollary we derive a Carleman inequality for the function u which satisfies the
integral equality
(2.35)
(u, K(x, D)∗ w)L2 (Ω) + (f, w)H 1,τ (Ω) + (geτ ϕ , e−τ ϕ w)H 12 ,τ (Γ)
e = 0
for all w ∈ X = {w ∈ H 1 (Ω)|w|Γ0 = 0, K(x, D)∗ w ∈ L2 (Ω)}. We have
14
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
1
e u ∈ L2 (Ω)
Corollary 2.1. Suppose that Φ satisfies (2.1), (2.2), f ∈ H 1 (Ω), g ∈ H 2 (Γ),
and the coefficients A, B ∈ {C ∈ C 1 (Ω)|kCkC 1 (Ω) ≤ K}. Then there exist τ0 = τ0 (K, Φ) and
C = C(K, Φ), independent of u and τ , such that for solutions of (2.35):
(2.36)
kueτ ϕ k2L2 (Ω) ≤ C10 |τ |(kf eτ ϕ k2H 1,τ (Ω) + kgeτ ϕ k2
1
e
H 2 ,τ (Γ)
) ∀|τ | ≥ τ0 .
Proof. Let be some positive number. Consider the extremal problem
(2.37)
(2.38)
1
1
1
J (w) = kwe−τ ϕ k2L2 (Ω) + kK(x, D)∗ w − ue2τ ϕ k2L2 (Ω) +
kwe−τ ϕ k2L2 (Γ)
e → inf,
2
2
2|τ |
1
w ∈ X̂ = {w ∈ H 2 (Ω)|K(x, D)∗ w ∈ L2 (Ω), w|Γ0 = 0}.
There exists a unique solution to (2.37), (2.38) which we denote by w
b . By Fermat’s theorem
J0 (w
b )[δ] = 0 ∀δ ∈ X̂ .
b − ue2τ ϕ ) this implies
Using the notation p = 1 (K(x, D)∗ w
(2.39)
K(x, D)p + w
b e−2τ ϕ = 0 in Ω,
∂p
w
b −2τ ϕ
|Γe =
e
.
∂ν
|τ |
p |∂Ω = 0,
By Proposition 2.4 we have
∂Φ
∂p τ ϕ 2
e kL2 (Γ0 ) + τ 2 k| |p eτ ϕ k2L2 (Ω)
∂ν Z
∂z
1
≤ C11 (kw
b e−τ ϕ k2L2 (Ω) +
|w
b |2 e−2τ ϕ dσ) ≤ 2C11 J (w
b ).
|τ | Γe
|τ |kp eτ ϕ k2L2 (Ω) + kp eτ ϕ k2H 1 (Ω) + k
(2.40)
Taking the scalar product of equation (2.39) with w
b we obtain
2J (w
b ) + (ue2τ ϕ , p )L2 (Ω) = 0.
Applying to the second term of the above equality estimate (2.40) we have
|τ |J (w
b ) ≤ C12 kueτ ϕ k2L2 (Ω) .
Using this estimate we pass to the limit in (2.39) as goes to zero. We obtain
(2.41)
(2.42)
K(x, D)p + we
b −2τ ϕ = 0 in Ω,
p|∂Ω = 0,
K(x, D)∗ w
b − ue2τ ϕ = 0 in Ω,
w
b −2τ ϕ
∂p
|Γe =
e
,
∂ν
|τ |
w|
b Γ0 = 0,
and
(2.43)
τϕ 2
b −τ ϕ k2L2 (Γ)
|τ |kwe
b −τ ϕ k2L2 (Ω) + kwe
e ≤ C13 kue kL2 (Ω) .
Since w
b ∈ L2 (Ω) we have p ∈ H 2 (Ω),
(2.43) we get
(2.44)
∂p
∂ν
3
1
∈ H 2 (∂Ω) and therefore w
b ∈ H 2 (∂Ω). By (2.40)1
kwe
b −τ ϕ kH 12 ,τ (∂Ω) ≤ C14 |τ | 2 kueτ ϕ kL2 (Ω) .
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
15
Taking the scalar product of (2.42) with ŵe−2τ ϕ and using the estimates (2.44), (2.43) we
get
1
1
k∇we
b −τ ϕ k2L2 (Ω) + |τ |kwe
b −τ ϕ k2L2 (Ω) +
kwe
b −τ ϕ k2 21 ,τ e ≤ C15 kueτ ϕ k2L2 (Ω) .
(Γ)
H
|τ |
|τ |
(2.45)
From this estimate and a standard duality argument, the statement of Corollary 2.1 follows
immediately.
Consider the following boundary value problem
∂d
∂a
= 0 in Ω,
= 0 in Ω, (a(z)eA + d(z̄)eB )|Γ0 = β.
∂ z̄
∂z
The existence of such functions a(z) and d(z̄) is given by the following proposition.
(2.46)
Proposition 2.5. Let α ∈ (0, 1), A and B be as in (2.16). If β ∈ C 5+α (Γ0 ) the problem
(2.46) has at least one solution (a, d) ∈ C 5+α (Ω) × C 5+α (Ω) such that
k(a, d)kC 5+α (Ω)×C 5+α (Ω) ≤ C16 kβkC 5+α (Γ0 ) .
(2.47)
1
If β ∈ H 2 (Γ0 ), then the problem (2.46) has at least one solution (a, d) ∈ H 1 (Ω) × H 1 (Ω)
such that
k(a, d)kH 1 (Ω)×H 1 (Ω) ≤ C18 kβkH 12 (Γ ) .
(2.48)
0
Proof. Let Ω̃ be a domain in R2 with smooth boundary such that Ω ⊂ Ω̃ and there exits an
open subdomain Γ̃0 ⊂ ∂ Ω̃ satisfying Γ0 ⊂ Γ̃0 . Denote Γ∗ = ∂ Ω̃ \ Γ̃0 . We extend A, B to Γ̃0
keeping the regularity and we extend β to Γ̃0 in such a way that kβkH 21 (Γ̃ ) ≤ C19 kβkH 12 (Γ )
0
0
or kβkC 5+α (Γ̃ ) ≤ C19 kβkC 5+α (Γ0 ) where the constant C19 is independent of β. By the trace
0
theorem there exist a constant C20 independent of β, and a pair (r, r̃) such that (reA +
1
r̃eB )|Γ̃0 = β and if β ∈ H 2 (Γ̃0 ) then (r, r̃) ∈ H 1 (Ω) × H 1 (Ω) and
k(r, r̃)kH 1 (Ω)×H 1 (Ω) ≤ C20 kβkH 21 (Γ ) .
0
Similarly if β ∈ C 5+α (Γ̃0 ) then (r, r̃) ∈ C 5+α (Ω) × C 5+α (Ω) and
k(r, r̃)kC 5+α (Ω)×C 5+α (Ω) ≤ C21 kβkC 5+α (Γ0 ) .
Let f =
∂r
∂z
and f˜ =
∂ r̃
.
∂ z̄
Consider the extremal problem
1 ∂p
1 ∂ p̃
J (p, p̃) = k(p, p̃)k2L2 (Ω̃) + k − f k2L2 (Ω̃) + k − f˜k2L2 (Ω̃) → inf,
∂z
∂ z̄
(p, p̃) ∈ K,
where K = {(h1 , h2 ) ∈ L2 (Ω̃) × L2 (Ω̃)|(h1 eA + h2 eB )|Γ̃0 = 0}. Denote the solution to this
extremal problem as (p , p̃ ). Then
J0 (p , p̃ )(δ, δ̃) = 0 ∀(δ, δ̃) ∈ K.
Hence
1 ∂p
∂δ
1 ∂ p̃
∂ δ̃
(2.49) ((p , p̃ ), (δ, δ̃))L2 (Ω̃) + (
− f, )L2 (Ω̃) + (
− f˜, )L2 (Ω̃) = 0 ∀(δ, δ̃) ∈ K.
∂z
∂z
∂ z̄
∂ z̄
16
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Denote P = − 1 ( ∂p
− f ), P̃ = − 1 ( ∂∂p̃z̄ − f˜). From (2.49) we obtain
∂z
∂ P̃
∂P
= p ,
= p̃ , P |Γ∗ = P̃ |Γ∗ = 0, ((ν1 + iν2 )P eB − (ν1 − iν2 )P̃ eA )|Γ̃0 = 0.
∂ z̄
∂z
We claim that there exists a constant C22 independent of such that
(2.50)
(2.51)
k(P , P̃ )kH 1 (Ω̃) ≤ C22 (k(p , p̃ )kL2 (Ω̃) + k(P , P̃ )kL2 (Ω̃) ).
It clearly suffices to prove the estimate (2.51) locally assuming that supp (p , p̃ ) is in a small
neighborhood of zero and the vector (0, 1) is orthogonal to ∂Ω on the intersection of this
neighborhood with the boundary. Using a conformal transformation we may assume that
∂Ω ∩ supp P , ∂Ω ∩ supp P̃ ⊂ {x1 = 0}. In order to prove this fact we consider the system of
equations
∂u
∂u
+ B̂
= F, supp u ⊂ B(0, δ) ∩ {x|x2 ≥ 0}.
(2.52)
∂x2
∂x1
Here u = (u1 , u2 , u3 , u4 ) = (Re P , Im P , Re P̃ , Im P̃ ), F = 2(Re p , Im p , Re p̃ , Im p̃ ), B̂ =


0 1 0 0
−1 0 0 0 


 0 0 0 −1 . The matrix B̂ has two eigenvalues ±i and four linearly independent
0 0 1 0
eigenvectors:
q3 = (0, 0, 1, i), q4 = (1, −i, 0, 0) corresponding to the eigenvalue − i,
q1 = (1, i, 0, 0), q2 = (0, 0, 1, −i)
corresponding to the eigenvalue i.
We set r1 = (ν1 eB , −ν2 eB , −ν1 eA , −ν2 eA ), r2 = (ν2 eB , ν1 eB , ν2 eA , −ν1 eA ). Consider the matrix D = {dj` } where dj` = rj · q` . We have
(ν1 − iν2 )eB −(ν1 − iν2 )eA
D=
.
(ν2 + iν1 )eB (ν2 + iν1 )eA
Since the Lopatinski determinant det D 6= 0 we obtain (2.51) (see e.g. [22]).
Suppose that for any C̃ one can find such that the estimate
k(P , P̃ )kH 1 (Ω̃) ≤ C̃k(p , p̃ )kL2 (Ω̃)
fails. That is, for all ∈ (0, 1), there exist p , p̃ , P , P̃ , C > 0 such that lim→0 C = ∞
and
1
(p , p̃ )
.
<
k(P , P̃ )kH 1 (Ω̃) 2
C
L (Ω̃)
We set (Q , Q̃ ) = (P , P̃ )/k(P , P̃ )kH 1 (Ω̃) and (q , q̃ ) = (p , p̃ )/k(P , P̃ )kH 1 (Ω̃) .
k(q , q̃ )kL2 (Ω̃) → 0 as → 0. Passing to the limit in (2.50) we have
Then
∂Q
∂ Q̃
= 0 in Ω̃,
= 0 in Ω̃, Q|Γ∗ = Q̃|Γ∗ = 0.
∂ z̄
∂z
By the uniqueness for the Cauchy problem for the operator ∂z we have Q = Q̃ = 0. On the
other hand, since k(Q , Q̃ )kH 1 (Ω̃) = 1, we can extract a subsequence, denoted the same, which
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
17
is convergent in L2 (Ω̃). Therefore the sequence (Q , Q̃ ) converges to zero in L2 (Ω̃). By (2.51),
we have 1/C23 ≤ k(q , q̃ )kL2 (Ω̃) + k(Q , Q̃ )kL2 (Ω̃) . Therefore lim inf →0 k(Q , Q̃ )kL2 (Ω̃) 6= 0,
and this is a contradiction. Hence
k(P , P̃ )kH 1 (Ω̃) ≤ C24 k(p , p̃ )kL2 (Ω̃) ,
∀ > 0.
Let us plug in (2.49) the function (p , p̃ ) instead of (δ, δ̃). Then, by the above inequality,
in view of the definitions of P and P̃ , we have
k(p , p̃ )k2L2 (Ω̃) ≤ C25 ((f, f˜), (P , P̃ ))L2 (Ω̃) ≤ C26 k(f, f˜)kL2 (Ω̃) k(P , P̃ )kL2 (Ω̃)
≤ C27 k(f, f˜)kL2 (Ω̃) k(p , p̃ )kL2 (Ω̃) .
This inequality implies that the sequence (p , p̃ ) is bounded in L2 (Ω̃) and
∂p ∂ p̃
,
) → (f, f˜) in L2 (Ω̃) × L2 (Ω̃).
∂z ∂ z̄
Then we construct a solution to (2.46) such that
(
k(p, p̃)kL2 (Ω̃) ≤ C28 k(f, f˜)kL2 (Ω̃) .
(2.53)
Observe that we can write the boundary value problem
∂p
=f
∂z
in Ω,
∂ p̃
= f˜ in Ω,
∂ z̄
(peA + p̃eB )|Γ̃0 = 0
in the form of (2.52) with u = (Re p, Im p, Re p̃, Im p̃), F = 2(Re f, Im f, Re f˜, Im f˜). We set
r1 = (eA , −eA , −eB , −eB ), r2 = (eA , eA , eB , −eB ). Consider the matrix D = {dj` } where
dj` = rj · q` . We have
B
e −eA
D= B
.
e
eA
Since the Lopatinski determinant det D 6= 0 the estimate (2.53) imply (2.47) and (2.48) (see
e.g., [22] Theorem 4.1.2.). This completes the proof of the proposition.
Consider the following problem
(2.54)
L(x, D)u = f eτ ϕ
in Ω,
u|Γ0 = geτ ϕ .
We have
Proposition 2.6. Let A, B ∈ C 5+α (Ω), q ∈ L∞ (Ω) and , α be a small positive numbers.
There exists τ0 > 0 such that for all |τ | > τ0 there exists a solution to the boundary value
problem (2.54) such that
p
1
p k∇ue−τ ϕ kL2 (Ω) + |τ |kue−τ ϕ kL2 (Ω) ≤ C29 (kf kL2 (Ω) + kgk 12 ,τ ).
(2.55)
(Γ0 )
H
|τ |
Let be a sufficiently small positive number. If suppf ⊂ G = {x ∈ Ω|dist(x, H \ Γ0 ) > }
and g = 0 then there exists τ0 > 0 such that for all |τ | > τ0 there exists a solution to the
boundary value problem (2.54) such that
(2.56)
k∇ue−τ ϕ kL2 (Ω) + |τ |kue−τ ϕ kL2 (Ω) ≤ C30 ()kf kL2 (Ω) .
18
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Proof. First we reduce the problem (2.54) to the case g = 0. Let r(z) be a holomorphic
function and re(z) be an antiholomorphic function such that (eA r + eB re)|Γ0 = g where A, B ∈
C 6+α (Ω) are defined as in (2.16). The existence of such functions r, re follows from Proposition
2.5, and these functions can be chosen in such a way that
krkH 1 (Ω) + ke
rkH 1 (Ω) ≤ C31 kgkH 21 (Γ ) .
0
We look for a solution u in the form
e,
u = (eA+τ Φ r + eB+τ Φ re) + u
where
(2.57)
L(x, D)e
u = feeτ ϕ
in Ω,
u
e|Γ0 = 0
and fe = f − (q − 2 ∂A
− AB)eA reiτ ψ − (q − 2 ∂B
− AB)eB ree−iτ ψ .
∂z
∂z
In order to prove (2.55) we consider the following extremal problem:
1
1
1
(2.58)
Ie (u) = kue−τ ϕ k2H 1,τ (Ω) + kL(x, D)u − f˜eτ ϕ k2L2 (Ω) + kue−τ ϕ k2 12 ,τ e → inf,
H
(Γ)
2
2
2
(2.59)
u ∈ Y = {w ∈ H 1 (Ω)|w|Γ0 = 0, L(x, D)w ∈ L2 (Ω)}.
There exists a unique solution to problem (2.58), (2.59) which we denote as u
b . By Fermat’s
theorem
(2.60)
Ie0 (b
u )[δ] = 0 ∀δ ∈ Y.
Let p = 1 (L(x, D)û − f˜eτ ϕ ). Applying Corollary 2.1 we obtain from (2.60)
(2.61)
1
kp eτ ϕ k2L2 (Ω) ≤ C32 (kb
u e−τ ϕ k2H 1,τ (Ω) + kb
u e−τ ϕ k2 12 ,τ e ) ≤ 2C32 Ie (b
u ).
H
(Γ)
|τ |
Substituting in (2.60) with δ = u
b we obtain
2Ie (b
u ) + (f˜eτ ϕ , p )L2 (Ω) = 0.
Applying to this equality estimate (2.61) we have
Ie (b
u ) ≤ C |τ |kf˜k2L2 (Ω) .
Using this estimate we pass to the limit as → +0. We obtain
(2.62)
L(x, D)u − f˜eτ ϕ = 0 in Ω, u|Γ0 = 0,
and
(2.63)
˜2
kue−τ ϕ k2H 1,τ (Ω) + kue−τ ϕ k2L2 (Γ)
e ≤ C33 |τ |kf kL2 (Ω) .
Since kf˜kL2 (Ω) ≤ C34 (kf kL2 (Ω) + kgkH 12 (Γ ) ), inequality (2.63) implies (2.55). In order to
0
prove (2.56) we consider the following extremal problem
(2.64)
(2.65)
1
1
1
Je (u) = kue−τ ϕ k2L2 (Ω) + kL(x, D)u − f eτ ϕ k2L2 (Ω) +
kue−τ ϕ k2L2 (Γ)
e → inf,
2
2
2|τ |
1
u ∈ Xe = {w ∈ H 2 (Ω)|w|Γ0 = 0, L(x, D)w ∈ L2 (Ω)}.
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
19
There exists a unique solution to problem (2.64), (2.65) which we denote as u
b . By Fermat’s
theorem
Je0 (b
u )[δ] = 0 ∀δ ∈ Xe.
This equality implies
L(x, D)∗ p + u
b e−2τ ϕ = 0 in Ω,
(2.66)
u
b −2τ ϕ
∂p
|Γe =
e
.
∂ν
|τ |
pb |∂Ω = 0,
By Proposition 2.4
1
∂p τ ϕ 2
∂Φ
kp eτ ϕ k2H 1,τ (Ω) + k
e kL2 (Γ0 ) + τ 2 k| |p eτ ϕ k2L2 (Ω)
|τ |
∂ν
∂z
Z
1
−τ ϕ 2
|b
u |2 e−2τ ϕ dσ) ≤ C36 Je (b
u ).
≤ C35 (kb
u e kL2 (Ω) +
|τ | Γe
(2.67)
Taking the scalar product of equation (2.66) with u
b we obtain
2Je (b
u ) + (f eτ ϕ , p )L2 (Ω) = 0.
Applying to this equality estimate (2.67) we have
|τ |2 Je (b
u ) ≤ C kf k2L2 (Ω) .
(2.68)
Using this estimate we pass to the limit in (2.66). We obtain that
L(x, D)∗ p + ue−2τ ϕ = 0 in Ω,
(2.69)
L(x, D)u − f e−τ ϕ = 0 in Ω,
(2.70)
∂p
u −2τ ϕ
|Γe =
e
,
∂ν
|τ |
p|∂Ω = 0,
u|Γ0 = 0.
Moreover (2.68) implies
2
|τ |2 kue−τ ϕ k2L2 (Ω) + kue−τ ϕ k2L2 (Γ)
e ≤ C37 kf kL2 (Ω) .
(2.71)
This finishes the proof of the proposition.
3. Estimates and Asymptotics
In this section we prove some estimates and obtain asymptotic expansions needed in the
construction of the complex geometrical optics solutions in Section 4.
Consider the operator
∂
∂
∂ ∂
+ 2A1
+ 2B1
+ q1 =
∂z ∂z
∂z
∂z
∂
∂
∂A1
(2 + B1 )(2 + A1 ) + q1 − 2
− A1 B1 =
∂z
∂z
∂z
∂
∂
∂B1
(2 + A1 )(2 + B1 ) + q1 − 2
− A1 B1 .
∂z
∂z
∂z
L1 (x, D) = 4
(3.1)
Let A1 , B1 , A2 , B2 ∈ C 6+α (Ω) with some α ∈ (0, 1) satisfy
(3.2)
2
∂A1
= −A1
∂z
in Ω, Im A1 |Γ0 = 0,
2
∂B1
= −B1
∂z
in Ω, Im B1 |Γ0 = 0
20
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
∂B2
∂A2
= A2 in Ω, Im A2 |Γ0 = 0, 2
= B 2 in Ω, Im B2 |Γ0 = 0.
∂z
∂z
Observe that
∂
∂
(2 + A1 )eA1 = 0 in Ω, (2 + B1 )eB1 = 0 in Ω.
∂z
∂z
Therefore if a(z), Φ(z) are holomorphic functions and b(z) is an antiholomorphic function,
we have
∂A1
L1 (x, D)(eA1 a(z)eτ Φ ) = (q1 − 2
− A1 B1 )eA1 aeτ Φ ,
∂z
∂B1
L1 (x, D)(eB1 b(z)eτ Φ ) = (q1 − 2
− A1 B1 )eB1 beτ Φ .
∂z
Let us introduce the operators:
Z
Z
g(ξ1 , ξ2 )
g(ξ1 , ξ2 )
1
1
−1
∂z g =
dζ ∧ dζ = −
dξ1 dξ2 ,
2πi Ω ζ − z
π Ω ζ −z
Z
Z
g(ξ1 , ξ2 )
1
g(ξ1 , ξ2 )
1
−1
dζ ∧ dζ = −
dξ1 dξ2 .
∂z g = −
2πi Ω ζ − z
π Ω ζ −z
We have (e.g., p.47, 56, 72 in [21]):
2
Proposition 3.1. A) Let m ≥ 0 be an integer number and α ∈ (0, 1). The operators
∂z−1 , ∂z−1 ∈ L(C m+α (Ω), C m+α+1 (Ω)).
2p
B) Let 1 ≤ p ≤ 2 and 1 < γ < 2−p
. Then ∂z−1 , ∂z−1 ∈ L(Lp (Ω), Lγ (Ω)).
C)Let 1 < p < ∞. Then ∂z−1 , ∂z−1 ∈ L(Lp (Ω), Wp1 (Ω)).
Assume that A, B satisfy (2.16). Setting TB g = eB ∂z−1 (e−B g) and PA g = eA ∂z−1 (e−A g), we
have
∂
∂
(2 + B)TB g = g in Ω, (2 + A)PA g = g in Ω.
∂z
∂z
We define two other operators:
1
e τ,B g = 1 eB eτ (Φ−Φ) ∂z−1 (ge−B eτ (Φ−Φ) ).
(3.3)
Rτ,A g = eA eτ (Φ−Φ) ∂z−1 (ge−A eτ (Φ−Φ) ), R
2
2
The following proposition follows from straightforward calculations.
Proposition 3.2. Let g ∈ C α (Ω) for some positive α. The function Rτ,A g is a solution to
(3.4)
2
∂
∂Φ
Rτ,A g − 2τ
Rτ,A g + ARτ,A g = g
∂z
∂z
in
Ω.
e τ,B g solves
The function R
(3.5)
2
∂Φ e
∂ e
e τ,B g = g
Rτ,B g + 2τ
Rτ,B g + B R
∂z
∂z
in Ω.
We have
Proposition 3.3. Let g ∈ C 2 (Ω), g|O = 0 and g|H = 0. Then for any 1 ≤ p < ∞
1
g g e
(3.6)
as |τ | → +∞.
Rτ,A g + 2τ ∂ Φ p + Rτ,B g − 2τ ∂z Φ p = o τ
z
L (Ω)
L (Ω)
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
21
e τ,B g. The proof for the Rτ,A g is
Proof. We give a proof of the asymptotic formula for R
−B
similar. Let g̃(ζ, ζ) = ge . Then
Z
g(ζ, ζ) τ (Φ(ζ)−Φ(ζ))
eτ (Φ−Φ)
−B
e
e Rτ,B g = −
dξ1 dξ2
π
Ω ζ −z
Z
g(ζ, ζ) τ (Φ(ζ)−Φ(ζ))
eτ (Φ−Φ)
dξ1 dξ2 .
=−
lim
e
δ→+0 Ω\B(z,δ) ζ − z
π
Let z = x1 + ix2 and (x1 , x2 ) be not a critical point of the function Φ. Then
Z
eτ (Φ−Φ)
g̃(ζ, ζ) ∂ζ eτ (Φ(ζ)−Φ(ζ))
−B
e Rτ,B g = −
lim
dξ1 dξ2
πτ δ→+0 Ω\B(z,δ) ζ − z
∂ζ Φ(ζ)
Z
eτ (Φ−Φ)
1 ∂ g̃(ζ, ζ)
lim
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2
=
δ→+0
πτ
∂ζ Φ(ζ)
Ω\B(z,δ) ζ − z ∂ζ
Z
eτ (Φ−Φ)
g̃(ζ, ζ) (e
ν1 − ie
ν2 ) τ (Φ(ζ)−Φ(ζ))
−
lim
e
dξ1 dξ2 .
πτ δ→+0 S(z,δ) ζ − z 2∂ζ Φ(ζ)
Since g̃|H = 0, we have
`
X
∂ g̃(ζ, ζ) kg̃kC 1 (Ω)
≤C
(3.7)
∈ Lp (Ω) ∀p ∈ (1, 2).
∂ζ ∂ζ Φ(ζ) |ζ
−
x
e
|
k
k=1
Hence
Z
g̃(z, z̄)
eτ (Φ−Φ)
1 ∂ g̃(ζ, ζ)
e Rτ,B g =
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 −
.
πτ
∂ζ Φ(ζ)
τ ∂z Φ(z)
Ω ζ − z ∂ζ
R 1 ∂ g̃(ζ,ζ) τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 . By the stationary phase argument,
Denote Gτ (x) = Ω ζ−z ∂ζ ∂ζ Φ(ζ) e
we see that
−B
(3.8)
Gτ (x) −→ 0 as |τ | → +∞ ∀x ∈ Ω.
Denote
∂ g̃(ζ, ζ̄) ζ
F (ξ1 , ξ2 ) = .
∂ζ Φ(ζ) Clearly
Z
(3.9)
|Gτ (x)| ≤
Ω
|F (ξ1 , ξ2 )|
dξ1 dξ2
|z − ζ|
a.e. in Ω ∀τ.
By (3.7) F belongs to Lp (Ω) for any p ∈ (1, 2). For f ∈ Lp (R2 ), we set
Z
2
Ir f (z) =
|z − ζ|− r f (ζ, ζ̄)dξ1 dξ2 .
R2
Then, by the Hardy-Littlewood-Sobolev inequality, if r > 1 and
p < q < ∞, then
kIr f kLq (R2 ) ≤ Cp,q kf kLp (R2 ) .
1
r
= 1−
1
p
−
1
q
for 1 <
22
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Set r = 2. Then we have to choose p1 − 1q = 21 , that is, we can arbitrarily choose p > 2 close
R F
to 2, so that q is arbitrarily large. Hence Ω |z−ζ|
dξ1 dξ2 belongs to Lq (Ω) with positive q. By
(3.8), (3.9) and the dominated convergence theorem
Gτ → 0 in Lq (Ω)
∀q ∈ (1, ∞).
The proof of the proposition is finished.
Using the stationary phase argument (e.g., Bleistein and Handelsman [2]), we will show
Proposition 3.4. Let g ∈ L1 (Ω) and a function Φ satisfy (2.1),(2.2). Then
Z
lim|τ |→+∞ geτ (Φ(z)−Φ(z)) dx = 0.
Ω
∞
1
Proof. Let {gk }∞
k=1 ∈ C0 (Ω) be a sequence of functions such that gk → g in L (Ω). Let > 0
be an arbitrary number. Suppose that b
j is large enough such that kg − gbj kL1 (Ω) ≤ 2 . Then
Z
Z
Z
τ (Φ(z)−Φ(z))
τ (Φ(z)−Φ(z))
dx| ≤ | (g − gbj )e
dx| + | gbj eτ (Φ(z)−Φ(z)) dx|.
| ge
Ω
Ω
Ω
The first term on the right-hand side of this inequality is less then /2 and the second goes
to zero as |τ | approaches to infinity by the stationary phase argument (see e.g. [2]).
We now consider the contribution from the critical points.
Proposition 3.5. Let Φ satisfy (2.1) and (2.2). Let g ∈ C 4+α (Ω) for some α > 0, g|O = 0
and g|H = 0. Then there exist constants pk such that
Z
τ (Φ(z)−Φ(z))
(3.10)
ge
Ω
`
1 X
1
pk e2τ iψ(exk ) + o( 2 ) as |τ | → +∞.
dx = 2
τ k=1
τ
Proof. Let δ > 0 be a sufficiently small number and ẽk ∈ C0∞ (B(e
xk , δ)), ẽk |B(exk ,δ/2) ≡ 1. By
the stationary phase argument
Z
τ (Φ−Φ)
I(τ ) =
ge
dx =
Ω
`
X
e2iτ ψ(exk )
` Z
X
k=1
Z
ẽk geτ (Φ−Φ) dx + o(
B(e
xk ,δ)
ẽk geτ (Φ−Φ)−2iτ ψ(exk ) dx + o(
B(e
xk ,δ)
k=1
1
)=
τ2
1
) as |τ | → +∞.
τ2
Since all the critical points of Φ are nondegenerate, in some neighborhood of x
ek one can take
local coordinates such that Φ − Φ − 2iτ ψ(e
xk ) = z 2 − z 2 . Therefore
I(τ ) =
`
X
k=1
2iτ ψ(e
xk )
Z
e
B(0,δ 0 )
qk eτ (z
2 −z 2 )
dx + o(
1
) as |τ | → +∞,
τ2
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
23
where qk ∈ C04 (B(0, δ 0 )) and qk (0) = 0. Hence there exist functions r1,k , r2,k ∈ C03 (B(0, δ 0 ))
such that qk = 2zr1,k + 2zr2,k . Integrating by parts, one can decompose I(τ ) as
Z
`
1 X 2iτ ψ(exk )
∂r1,k ∂r2,k τ (z2 −z2 )
1
I(τ ) = −
e
(
−
)e
dx + o( 2 ) =
τ k=1
∂z
τ
B(0,δ 0 ) ∂z
Z
`
1 X 2iτ ψ(exk )
∂r1,k ∂r2,k
1
2
2
−
e
(
−
)(0)χ(x)eτ (z −z ) dx + o( 2 )
τ k=1
∂z
τ
B(0,δ 0 ) ∂z
Z
`
1
1 X 2iτ ψ(exk )
2
2
e
qek eτ (z −z ) dx + o( 2 ) as |τ | → +∞,
−
τ k=1
τ
B(0,δ 0 )
where χ, qek ∈ C02 (B(0, δ 0 )), χ|B(0,δ0 /2) ≡ 1 and qek (0) = 0. Hence there exist functions re1,k , re2,k ∈
C01 (B(0, δ 0 )) such that qek = 2ze
r1,k + 2ze
r2,k . Integrating by parts and applying Proposition
3.3 we obtain
Z
Z
`
X
∂e
r1,k ∂e
r2,k τ (z2 −z2 )
2iτ ψ(e
xk )
τ (z 2 −z 2 )
e
lim
(
dx = −
dx = 0.
lim τ
qek e
−
)e
|τ |→+∞ B(0,δ 0 ) ∂z
|τ |→+∞
∂z
B(0,δ 0 )
k=1
Therefore (3.10) follows from a standard application of stationary phase. The proof of the
proposition is completed.
Proposition 3.6. Let 0 < 0 < , a function Φ satisfy (2.1), (2.2) and O ∩ (H \ Γ0 ) = ∅.
Suppose that g ∈ C α (Ω) for some α > 0, g|O = 0 and g|H = 0. Then
(3.11)
e τ,B gkL∞ (O 0 ) + k∇R
e τ,B gkL∞ (O 0 ) ≤ C1 (0 , α)kgk α
|τ |kR
C (Ω)∩H 1 (Ω) .
Moreover
e τ,B gkL2 (Ω) ≤ C2 (0 , α)kgk α
e τ,B gkL2 (Ω) + |τ | 12 kR
e τ,B gkL2 (Ω) + |τ |k ∂Φ R
(3.12) k∇R
C (Ω)∩H 1 (Ω) .
∂z
Proof. Denote ge = ge−B . Let x = (x1 , x2 ) be an arbitrary point from O0 and z = x1 + ix2 .
Then
Z
` Z
X
geeτ (Φ−Φ)
geeτ (Φ−Φ)
−1 τ (Φ−Φ)
−π∂z (e
dξ1 dξ2 = lim
dξ1 dξ2 .
ge) =
δ→+0
xk ,δ) ζ − z
Ω ζ −z
k=1 Ω\B(e
Integrating by parts and taking δ sufficiently small we have
Z
∂e
g
1
∂ζ
eτ (Φ−Φ) dξ1 dξ2
= − lim
τ δ→+0 Ω\∪`k=1 B(exk ,δ) (ζ − z) ∂Φ
∂ζ
Z
∂2Φ
ge ∂ζ 2
1
+ lim
eτ (Φ−Φ) dξ1 dξ2
∂Φ
2
δ→+0
τ
xk ,δ) (ζ − z)( ∂ζ )
Ω\∪`k=1 B(e
Z
ge
1
+
lim
(e
ν1 − ie
ν2 )
eτ (Φ−Φ) dσ.
∂Φ
2τ δ→+0 ∪`k=1 S(exk ,δ)
(ζ − z) ∂ζ
−π∂z−1 (eτ (Φ−Φ) ge)
(3.13)
24
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Since g|H = 0 and all the critical points of Φ are nondegenerate we have that kgkC 0 (S(exk ,δ)) ≤
δ α kgkC α (Ω) . Therefore
Z
ge
1
(e
ν1 − ie
ν2 )
lim
eτ (Φ−Φ) dσ = 0.
∂Φ
2τ δ→+0 ∪`k=1 S(exk ,δ)
(ζ − z) ∂ζ
Since |
2Φ
∂ζ 2
( ∂Φ
)2
∂ζ
g
e∂
(ζ, ζ̄)| ≤ C3 ke
g kC α (Ω)
P`
1
k=1 |ξ−e
xk |2−α
we see that
2Φ
∂ζ 2
( ∂Φ
)2
∂ζ
g
e∂
(ζ, ζ̄) ∈ L1 (Ω) and
Z
∂e
g
1
∂ζ
=−
eτ (Φ−Φ) dξ1 dξ2
∂Φ
τ Ω (ζ − z) ∂ζ
2
Z
ge ∂∂ζΦ2
1
eτ (Φ−Φ) dξ1 dξ2 .
+
2
τ Ω (ζ − z)( ∂Φ
)
∂ζ
−π∂z−1 (eτ (Φ−Φ) ge)
(3.14)
e τ,B , the estimate (3.11) follows
From this equality and definition (3.3) of the operator R
immediately. To prove (3.12) we observe
e τ,B g
∂R
∂B e
e τ,B { ∂g − ∂B g} + τ eτ (Φ−Φ)+B
=
Rτ,B g + R
∂z
∂z
∂z
∂z
2π
Z
∂Φ(ζ)
∂ζ
Ω
−
∂Φ(z)
∂z
ζ −z
g̃eτ (Φ−Φ) dξ1 dξ2 .
By Proposition 3.1
k
∂B e
e τ,B { ∂g − ∂B g}kL2 (Ω) ≤ C4 kgkH 1 (Ω) .
Rτ,B g + R
∂z
∂z
∂z
Using arguments similar to (3.13), (3.14) we obtain
τ
k
2π
Z
Ω
∂Φ(ζ)
∂ζ
−
∂Φ(z)
∂z
ζ −z
g̃eτ (Φ−Φ) dξ1 dξ2 kL2 (Ω) ≤ C5 kgkC α (Ω)∩H 1 (Ω) .
Hence
e τ,B g
∂R
kL2 (Ω) ≤ C6 kgkC α (Ω)∩H 1 (Ω) .
∂z
Combining this estimate with (3.11) we conclude
k
e τ,B gkL2 (Ω) ≤ C7 kgk α
k∇R
C (Ω)∩H 1 (Ω) .
Using this estimate and equation (3.4) we have
|τ |k
∂Φ e
Rτ,B gkL2 (Ω) ≤ C8 kgkC α (Ω)∩H 1 (Ω) ,
∂z
finishing the proof of the proposition.
Let e1 , e2 ∈ C ∞ (Ω) be functions such that
(3.15)
e1 + e2 = 1 in Ω,
e2 vanishes in some neighborhood of H \ Γ0 and e1 vanishes in a neighborhood of ∂Ω.
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
25
Proposition 3.7. Let for some α ∈ (0, 1) A, B ∈ C 5+α (Ω), and the functions A, B ∈
C 6+α (Ω) satisfy (2.16). Let e1 , e2 be defined as in (3.15). Let g ∈ Lp (Ω) for some p > 2,
supp g ⊂⊂ supp e1 . We define u by
fA
e τ,B (e1 (PA g − M
feA )) + e2 (PA g − M e ) ,
u=R
2τ ∂z Φ
f=M
f(z) is a polynomial such that (PA g − M
feA )|H = 0 and ∂ kk (PA g − M
feA )|H = 0
where M
∂z
for any k from {1, . . . , 4}. Then we have
eτ ϕ
∂
∂
hτ
(3.16)
P(x, D)(ueτ Φ ) , (2 + A)(2 + B)(ueτ Φ ) = geτ Φ +
∂z
∂z
|τ |
where
khτ kL∞ (Ω) ≤ C9 (p)kgkLp (Ω)
as |τ | → +∞,
and
(3.17)
1
|τ |
1
1
2
k∇ukL2 (Ω) + |τ | 2 kukL2 (Ω) + kukH 1,τ (O0 ) ≤ C10 kgkLp (Ω) .
Proof. By Proposition 3.1 PA g belongs to Wp1 (Ω). Since p > 2, by the Sobolev embedding
theorem there exists α > 0 such that PA g ∈ C α (Ω). By properties of elliptic operators and
the fact that supp e2 ∩ supp g = {∅} we have that PA g ∈ C 5 (supp e2 ). The estimate (3.17)
follows from Proposition 3.6. Short calculations give
!
τΦ
A
f
e
e2 (PA g − M e )
(3.18)
P(x, D)(ueτ Φ ) = geτ Φ +
.
P(x, D)
τ
2∂z Φ
feA )
M
This formula implies (3.16) with hτ = eiτ ψ P(x, D) e2 (PA2∂g−
/sign τ.
zΦ
The following proposition will play a critically important role in the construction of the
complex geometric optic solutions.
Proposition 3.8. Let f ∈ Lp (Ω) for some p > 2, 0 be a small positive number such that
O0 ∩ (H \ Γ0 ) = ∅. Then there exists τ0 such that for all |τ | > τ0 there exists a solution to
the boundary value problem
(3.19)
L(x, D)w = f eτ Φ
in Ω,
w|Γ0 = qeτ ϕ /τ
such that
p
1
|τ |kwe−τ ϕ kL2 (Ω) + p k∇we−τ ϕ kL2 (Ω) + kwe−τ ϕ kH 1,τ (O0 ) ≤ C11 (kf kLp (Ω) + kqkH 12 (Γ ) ).
0
|τ |
Proof. Let χ ∈ C0∞ (Ω) be equal to one in some neighborhood of the set H\Γ0 . By Proposition
2.6 there exists a solution to the problem (3.19) with inhomogeneous term (1 − χ)f and
boundary data q/τ such that
(3.20)
kw1 e−τ ϕ kH 1,τ (Ω) ≤ C12 (kf kL2 (Ω) + kqkH 12 (Γ ) ).
0
feA )
e τ,B (e1 (PA (χf ) − M
feA )) + e2 (PA (χf )−M
Denote w2 = R
2τ ∂z Φ
feA )|H = 0 and ∂ kk (PA g − M
feA )|H =
such that (PA g − M
∂z
f=M
f(z) is a polynomial
where M
0 for any k from {1, . . . , 4}. Let qτ
26
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
be the restriction of w2 to Γ0 . By (3.12) there exists a constant C13 independent of τ such
that
(3.21)
|τ |kqτ kC 1 (Γ0 ) ≤ C13 kf kLp (Ω) .
By Proposition 3.7 there exists a constant C14 independent of τ such that
p
1
(3.22)
|τ |kw2 e−τ ϕ kL2 (Ω) + p k∇w2 e−τ ϕ kL2 (Ω) + kw2 e−τ ϕ kH 1,τ (O0 ) ≤ C14 kf kLp (Ω) .
|τ |
Let e
aτ , ebτ ∈ H 1 (Ω) be holomorphic and an antiholomorphic functions respectively such
that (e
aτ eA + ebτ eB )|Γ0 = −qτ . By (3.21) and Proposition 2.5 there exist constants C15 , C16
independent of τ such that
kf kLp (Ω)
(3.23)
ke
aτ kH 1 (Ω) + kebτ kH 1 (Ω) ≤ C15 kqτ kC 1 (Γ0 ) ≤ C16
.
|τ |
The function W = (w2 + e
aτ eA )eτ Φ + ebτ eB+τ Φ satisfies
e
hτ
L(x, D)W = χf eτ Φ + eτ ϕp
|τ |
in Ω,
W |Γ0 = 0,
where
(3.24)
ke
hτ kL2 (Ω) ≤ C17 kf kL2 (Ω)
with some constant C17 independent of τ. By (3.22), (3.23)
p
1
|τ |kW e−τ ϕ kL2 (Ω) + p k∇W e−τ ϕ kL2 (Ω) + kW e−τ ϕ kH 1,τ (O0 ) ≤ C18 kf kLp (Ω) .
(3.25)
|τ |
f be a solution to problem (2.54) with inhomogeneous term and boundary data f =
Let W
e
−√hτ , g ≡ 0 respectively given by Proposition 2.6. The estimate (2.55) has the form
|τ |
(3.26)
fe−τ ϕ kH 1,τ (Ω) ≤ C19 ke
kW
hτ kL2 (Ω) ≤ C20 kf kL2 (Ω) .
f solves (3.19). The estimate (3.18) follows form (3.20), (3.25)
Then the function w1 + W + W
and (3.26). The proof of the proposition is completed.
4. Complex Geometrical Optics Solutions
For a complex-valued vector field (A1 , B1 ) and complex-valued potential q1 we will construct solutions to the boundary value problem
(4.1)
L1 (x, D)u1 = 0 in Ω,
u1 |Γ0 = 0
of the form
(4.2)
u1 (x) = aτ (z)eA1 +τ Φ + dτ (z)eB1 +τ Φ + u11 eτ ϕ + u12 eτ ϕ .
Here A1 and B1 are defined by (2.16) respectively for A1 and B1 , aτ (z) = a(z)+ a1τ(z) + a2,ττ 2(z) ,
dτ (z) = d(z̄) + d1τ(z) + d2,ττ 2(z) ,
(4.3)
a, d ∈ C 5+α (Ω),
∂a
∂d
= 0 in Ω,
= 0 in Ω,
∂z
∂z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
27
(a(z)eA1 + d(z̄)eB1 )|Γ0 = 0.
(4.4)
Let x̃ be some fixed point from H. Suppose in addition that
(4.5)
∂ka
∂kd
|
=
0,
|H∩∂Ω = 0 ∀k ∈ {0, . . . , 5}, a|H\{x̃} = d|H\{x̃} = 0, a(x̃) 6= 0, d(x̃) 6= 0.
H∩∂Ω
∂z k
∂ z̄ k
Such functions exists by Proposition 6.2.
Denote
∂B1
∂A1
−A1 B1 )deB1 )−M2 (z)eB1 , g2 = PA1 ((q1 −2
−A1 B1 )aeA1 )−M1 (z)eA1 ,
g1 = TB1 ((q1 −2
∂z
∂z
where M1 (z) and M2 (z) are polynomials such that
∂ k g1
∂ k g2
|
=
|H = 0 ∀k ∈ {0, . . . , 5}.
H
∂ z̄ k
∂z k
Thanks to our assumptions on the regularity of A1 , B1 and q, g1 , g2 belong to C 6+α (Ω).
Note that by (4.6), (4.5)
(4.6)
∂ k+j g1
∂ k+j g2
|
=
|H∩∂Ω = 0 if k + j ≤ 5.
H∩∂Ω
∂z k ∂ z̄ j
∂z k ∂ z̄ j
The function a1 (z) is holomorphic in Ω and d1 (z) is antiholomorphic in Ω and
g1
g2
a1 (z)eA1 + d1 (z)eB1 =
+
on Γ0 .
2∂z Φ 2∂z Φ
(4.7)
The existence of such functions is given again by Proposition 2.5. Observe that by (4.7) the
functions e∂2 gΦ1 , e∂2zgΦ2 ∈ C 4 (Ω). Let
z
ĝ1 = TB1 ((q1 −2
∂B1
−A1 B1 )d1 eB1 )−M̂2 (z)eB1 ,
∂z
ĝ2 = PA1 ((q1 −2
∂A1
−A1 B1 )a1 eA1 )−M̂1 (z)eA1 ,
∂z
where M̂1 (z) and M̂2 (z) are polynomials such that
(4.8)
The function u11
∂ k ĝ2
∂ k ĝ1
|
=
|H = 0 ∀k ∈ {0, . . . , 3}.
H
∂ z̄ k
∂z k
is given by
ĝ1
)
τ
e−iτ ψ
e2 g1
u11 = −e
R−τ,A1 {e1 (g1 + ĝ1 /τ )} − e
+ 2
L1 (x, D)
2τ ∂z Φ
4τ ∂z Φ
∂z Φ
ĝ2
iτ
ψ
e
e2 g2
iτ ψ e
iτ ψ e2 (g2 + τ )
(4.9)
−e Rτ,B1 {e1 (g2 + ĝ2 /τ )} − e
+ 2
L1 (x, D)
,
2τ ∂z Φ
4τ ∂z Φ
∂z Φ
−iτ ψ
−iτ ψ e2 (g1
+
Now let us determine the functions u12 , a2 (z) and d2 (z).
First we can obtain the following asymptotic formulae for any point on the boundary of
Ω.
(4.10)
eA1 +2iτ ψ
1
R−τ,A1 {e1 g1 } = 2
2τ |det Im Φ00 (x̃)| 12
(4.11)
1
eB1 −2iτ ψ
R̃τ,B1 {e1 g2 } = 2
2τ |det Im Φ00 (x̃)| 12
e−2iτ ψ(x̃) q1 (x̃) e−2iτ ψ(x̃) m1 (x̃)
+
(z − z̃)2
(z̃ − z)
e2iτ ψ(x̃) q̃1 (x̃) e2iτ ψ(x̃) m̃1 (x̃)
+
(z̄ − z̃)2
(z̃ − z̄)
+ Wτ,1 ,
+ Wτ,2 ,
28
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
where
∂z g̃1 (x̃)
q1 = 2
,
4∂z Φ(x̃)
q̃1 =
∂z̄ g̃2 (x̃)
4∂z2 Φ(x̃)
,
1
m1 =
8
∂z g̃1 (x̃) ∂z3 Φ(x̃) ∂z̄2 g̃1 (x̃) ∂z2 g̃1 (x̃)
− 2
+
∂z Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃) ∂z2 Φ(x̃)
!
1
m̃1 =
8
∂z̄ g̃2 (x̃) ∂z̄3 Φ(x̃) ∂z̄2 g̃2 (x̃) ∂z2 g̃2 (x̃)
+ 2
−
∂z Φ(x̃)
∂z2 Φ(x̃) ∂z̄2 Φ(x̃)
∂z2 Φ(x̃)
!
,
,
1
g̃1 = e−A1 g1 , g̃2 = e−B1 g2 and Wτ,1 , Wτ,2 ∈ H 2 (Γ0 ) satisfy
(4.12)
kWτ,1 kH 12 (Γ ) + kWτ,2 kH 21 (Γ ) = o(
0
0
1
) as |τ | → +∞.
τ2
The proof of (4.10) and (4.11) is given in Section 7.
Denote
q1 (x̃)
m1 (x̃)
A1
p+ = e
+
,
(z − z̃)2 (z̃ − z)
m̃1 (x̃)
q̃1 (x̃)
B1
.
p− = e
+
(z̄ − z̃)2 (z̃ − z̄)
Thanks to Proposition 2.5 we can define functions a2,± (z) ∈ C 2 (Ω) and d2,± (z) ∈ C 2 (Ω)
satisfying
a2,± (z)eA1 + d2,± (z)eB1 = p±
(4.13)
on Γ0 .
Straightforward computations give
L1 (x, D)((a(z) +
= (q1 −
1
2 ∂B
∂z
τΦ
d1 (z) B1 +τ Φ
)e
τ
− A1 B1 )e
τΦ
+ (d(z̄) +
e τ,B1 {e1 (g2 + ĝ2 /τ )} −
−R
+ eτ ϕ u11 )
e2 (g2 +ĝ2 /τ )
2τ ∂z Φ
e2 (g1 +ĝ1 /τ )
2τ ∂z Φ
−R−τ,A1 {e1 (g1 + ĝ1 /τ )} −
e2 g2
e2 g1
eτ Φ
1
eτ Φ
1
+ τ 2 L1 (x, D) 4∂z Φ L1 (x, D) ∂z Φ
.
+ τ 2 L1 (x, D) 4∂ Φ L1 (x, D) ∂ Φ
+(q1 −
(4.14)
1
2 ∂A
∂z
a1 (z) A1 +τ Φ
)e
τ
− A1 B1 )e
z
z
Using Proposition 3.3 we transform the right-hand side of (4.14) as follows.
L1 (x, D)((a(z) +
=
(4.15)
−(q1 −
1
2 ∂B
∂z
a1 (z) A1 +τ Φ
)e
τ
1
−(q1 − 2 ∂A
∂z
τΦ
− A1 B1 )e
d1 (z) B1 +τ Φ
)e
τ
A1 B1 )eτ Φ 2τg∂1z Φ
+ (d(z̄) +
−
g2
2τ ∂z Φ
+ u11 eτ ϕ )
+ oL4 (Ω)
1
τ
as |τ | → +∞.
We are looking for u12 in the form u12 = u0 + u−1 . The function u−1 is given by
(4.16)
u−1 =
eiτ ψ e
Rτ,B1 {e1 g5 }
τ
+
e−iτ ψ
R−τ,A1 {e1 g6 }
τ
+
e2 g5 eiτ ψ
2τ 2 ∂z Φ
+
e2 g6 e−iτ ψ
,
2τ 2 ∂z Φ
where
(4.17)
1
1
PA1 ((q1 − 2 ∂A
− A1 B1 )g1 ) − M5 (z)eA1
TB1 ((q1 − 2 ∂B
− A1 B1 )g2 ) − M6 (z)eB1
∂z
∂z
g5 =
, g6 =
.
2∂z Φ
2∂z Φ
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
29
Here M5 (z), M6 (z) are polynomials such that
g5 |H = g6 |H = ∇g5 |H = ∇g6 |H = 0.
Using Proposition 2.5 we introduce functions a2,0 , d2,0 ∈ C 2 (Ω̄) (holomorphic and antiholomorphic respectively) such that
g5
g6
a2,0 (z)eA1 + d2,0 (z)eB1 =
+
(4.18)
on Γ0 .
2∂z Φ 2∂z Φ
Next we claim that
1
e τ,B1 {e1 g5 }|Γ0 = o( 1 ) as |τ | → +∞.
(4.19) R−τ,A1 {e1 g6 }|Γ0 = o( ) as |τ | → +∞, R
τ
τ
To see this, let us introduce the function F with domain Γ0 .
F = 2e−A1 eτ (Φ−Φ) R−τ,A1 {e1 g6 }
=
TB ((q1
∂z−1 (e1 e−A1 +τ (Φ−Φ) 1
T
((q −2
∂B1
1
− A1 B1 )g2 ) − M6 eB1
− 2 ∂B
∂z
2∂z Φ
).
−A B )g )−M (z)eB1
6
Denoting r(x) = eA1 B1 1 ∂z 2∂1 Φ1 2
we have
z
Z
Z X
2
1
e1 (x)r(x)e2iτ ψ
1
∂
∂ψ e1 (x) r(x)
F(x) = −
dξ1 dξ2 =
e2iτ ψ dξ1 dξ2 .
2
π Ω
2iπτ Ω k=1 ∂xk ∂xk |∇ψ| ζ − z
ζ −z
∂ψ
P
∂xk e1 (x)r(x)
Since 2k=1 ∂x∂ k ( |∇ψ|
) ∈ L1 (Ω), we have F = o( τ1 ). This proves (4.19).
2
ζ−z
Now we finish the construction of the functions a2,τ (z) and d2,τ (z) by setting
!
1 d2,+ (z)e2iτ ψ(x̃)
d2,− (z)e−2iτ ψ(x̃)
d2,τ (z) = d2,0 (z) +
,
+
1
2 |det Im Φ00 (x̃)| 21
|det Im Φ00 (x̃)| 2
!
1 a2,+ (z)e2iτ ψ(x̃)
a2,− (z)e−2iτ ψ(x̃)
a2,τ (z) = a2,0 (z) +
+
,
1
2 |det Im Φ00 (x̃)| 12
|det Im Φ00 (x̃)| 2
where a2,± , d2,± satisfy (4.13). To complete the construction of a solution to (4.1) we define
u0 as the solution to the inhomogeneous problem
(4.20)
L1 (x, D)(u0 eτ ϕ ) = h1 eτ ϕ
(4.21)
u0 eτ ϕ = eτ ϕ m1
in Ω,
on Γ0 ,
where
h1 = −e−τ ϕ L1 (x, D)(aτ (z)eA1 +τ Φ + dτ (z)eB1 +τ Φ + u11 eτ ϕ + u−1 eτ ϕ ),
m1 = −e−τ ϕ (aτ (z)eA1 +τ Φ + dτ (z)eB1 +τ Φ + u11 eτ ϕ + u−1 eτ ϕ )|Γ0 .
Observe that by (4.15) - (4.17)
1
kh1 kL4 (Ω) = o( ) as |τ | → +∞
τ
and by (4.8), (4.12), (4.13), (4.18)
1
(4.22)
ku0 kH 12 (Γ ) = o( 2 ) as |τ | → +∞.
0
τ
30
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
By Proposition 2.6 and Proposition 3.8 there exists a solution to (4.20), (4.21) such that
p
1
1
p ku0 kH 1 (Ω) + |τ |ku0 kL2 (Ω) + ku0 kH 1,τ (O ) = o( ) as |τ | → +∞.
τ
|τ |
(4.23)
4.1. Complex geometrical optics solutions for the adjoint operator. We now construction complex geometrical optics solutions for the adjoint operator. This parallels the
previous construction since the adjoint has a similar form.
∂ ∂
∂
∂
Consider the operator L2 (x, D) = 4 ∂z
+ 2A2 ∂z
+ 2B2 ∂z
+ q2 . Its adjoint has the form
∂z
∂ ∂
∂A2 ∂B 2
∂
∂
− 2A2
− 2B2
+ q2 − 2
−
∂z ∂z
∂z
∂z
∂ z̄
∂z
∂
∂A2
∂
= (2 − A2 )(2 − B2 ) + q2 − 2
− A2 B2
∂z
∂z
∂ z̄
∂
∂
∂B2
= (2 − B2 )(2 − A2 ) + q2 − 2
− A2 B2 .
∂z
∂z
∂z
Next we construct solution to the following boundary value problem:
L2 (x, D)∗ = 4
L2 (x, D)∗ v = 0 in Ω,
(4.24)
v|Γ0 = 0.
We construct solutions to (4.24) of the form
v(x) = bτ (z)eB2 −τ Φ + cτ (z)eA2 −τ Φ + v11 e−τ ϕ + v12 e−τ ϕ ,
(4.25)
v|Γ0 = 0.
Here A2 , B2 ∈ C 6+α (Ω) satisfy
2
∂A2
= A2
∂z
and bτ (z) = b(z) +
(4.26)
(4.27)
in Ω,
b1 (z)
τ
+
Im A2 |Γ0 = 0,
b2,τ (z)
, cτ (z)
τ2
= c(z) +
b, c ∈ C 5+α (Ω),
2
∂B2
= B2
∂z
c1 (z)
τ
+
c2,τ (z)
τ2
in Ω,
Im B2 |Γ0 = 0,
and
∂b
∂c
= 0 in Ω,
= 0 in Ω,
∂z
∂z
(b(z)eB2 + c(z)eA2 )|Γ0 = 0,
(4.28)
∂kc
∂kb
|
=
0,
|H∩∂Ω = 0 k ∈ {0, . . . , 5}, b|H\{x̃} = c|H\{x̃} = 0, b(x̃) 6= 0, c(x̃) 6= 0.
H∩∂Ω
∂z k
∂ z̄ k
The existence of the functions b and c is given by Proposition 6.2 . Denote
∂A2
∂B2
−A2 B2 )beB2 )−M3 (z)eB2 , g4 = T−A2 ((q−2
−A2 B2 )ceA2 )−M4 (z̄)eA2 ,
∂ z̄
∂z
where the polynomials M3 (z), M4 (z̄) are chosen such that
g3 = P−B2 ((q 2 −2
(4.29)
∂ k g3
∂ k g4
|
=
|H = 0 ∀k ∈ {0, . . . , 5}.
H
∂z k
∂ z̄ k
By (4.29), (4.28)
(4.30)
∂ k+j g3
∂ k+j g4
|
=
|H∩∂Ω = 0 ∀k + j ≤ 5.
H∩∂Ω
∂z k ∂ z̄ j
∂z k ∂ z̄ j
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
31
Observe that by (4.30) ∂gz3Φ , ∂g4Φ ∈ C 4+α (Ω). Using Proposition 2.5 we introduce a holomorz
phic function b1 (z) ∈ C 2 (Ω) and an antiholomorphic function c1 (z̄) ∈ C 2 (Ω) such that
e2 g3
e2 g4
on Γ0 .
(4.31)
b1 eB2 + c1 eA2 =
+
2∂z Φ 2∂z Φ
Let
ĝ3 = P−B2 ((q 2 −2
∂A2
−A2 B2 )b1 eB2 )−M̂3 (z)eB2 ,
∂ z̄
ĝ4 = T−A2 ((q−2
∂B2
−A2 B2 )c1 eA2 )−M̂4 (z̄)eA2 ,
∂z
where the polynomials M̂3 (z), M̂4 (z̄) are chosen such that
(4.32)
The function v11
∂ k ĝ3
∂ k ĝ4
|
=
|H = 0 ∀k ∈ {0, . . . , 3}.
H
∂z k
∂ z̄ k
is defined by
e
v11 = −e−iτ ψ R
−τ,−A2 {e1 (g3 + ĝ3 /τ )} +
e−iτ ψ e2 (g3 +ĝ3 /τ )
2τ ∂z Φ
− eiτ ψ Rτ,−B2 {e1 (g4 + ĝ4 /τ )} +
e−iτ ψ
eiτ ψ
∗ e2 g4
∗ e2 g3
− 4τ 2 ∂z Φ L2 (x, D) ∂z Φ − 4τ 2 ∂ Φ L2 (x, D) ∂ Φ .
(4.33)
eiτ ψ e2 (g4 +ĝ4 /τ )
2τ ∂z Φ
z
z
Here we set
1
Rτ,−B2 {g} = eB2 eτ (Φ−Φ) ∂z−1 (ge−B2 eτ (Φ−Φ) )
2
1 A2 τ (Φ−Φ) −1 −A2 τ (Φ−Φ)
e
∂z (ge
)
e
R
−τ,−A2 {g} = e e
2
provided that A2 , B2 , A2 , B2 satisfy (3.2). By Proposition 7.1 the following asymptotic formulae holds.
2iτ ψ(x̃)
1
e−B2 −2τ iψ
e
r̃1 (x̃) e2iτ ψ(x̃) t̃1 (x̃)
f1,τ ,
+
(4.34)
Rτ,−B 2 {e1 g3 } |Γ0 = 2
+W
2τ |det Im Φ00 (x̃)| 12
(z − z̃)2
(z̃ − z)
e
(4.35) R
−τ,−A2 {e1 g4 } |Γ0
e−A2 +2τ iψ
1
= 2
2τ |det Im Φ00 (x̃)| 12
e−2iτ ψ(x̃) r1 (x̃) e−2iτ ψ(x̃) t1 (x̃)
+
(z̄ − z̃)2
(z̃ − z̄)
f2,τ ,
+W
where
∂z g̃4 (x̃)
r1 = − 2
,
4∂z Φ(x̃)
r̃1 = −
∂z̄ g̃3 (x̃)
4∂z2 Φ(x̃)
,
1
t1 =
8
1
t̃1 =
8
∂z g̃4 (x̃) ∂z3 Φ(x̃) ∂z̄2 g̃4 (x̃) ∂z2 g̃4 (x̃)
− 2
−
+ 2
∂z Φ(x̃) ∂z2 Φ(x̃)
∂z Φ(x̃)
∂z2 Φ(x̃)
∂z̄ g̃3 (x̃) ∂z̄3 Φ(x̃)
∂z̄2 g̃3 (x̃)
∂ 2 g̃3 (x̃)
−
+
− z2
∂z Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃) ∂z2 Φ(x̃)
!
,
!
,
fτ,1 , W
fτ,2 ∈ H 12 (Γ0 ) satisfy
g̃3 = eA2 g3 , g̃4 = eB2 g4 . Here the functions W
1
fτ,1 k 1
fτ,2 k 1
kW
+ kW
= o( 2 ) as |τ | → +∞.
H 2 (Γ0 )
H 2 (Γ0 )
τ
Using Proposition 2.5 we define the holomorphic functions b2,± (z) ∈ C 2 (Ω̄) and antiholomorphic c2,± (z) ∈ C 2 (Ω̄) such that
(4.36)
(4.37)
b2,± (z)eB2 + c2,± (z)eA2 = pe±
on Γ0 ,
32
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
where pek is defined as
−B2
pe+ (z) = e
−A2
pe− (z̄) = e
r̃1 (x̃)
t̃1 (x̃)
+
2
(z − z̃)
(z̃ − z)
r1 (x̃)
t1 (x̃)
+
2
(z̄ − z̃)
(z̃ − z̄)
,
.
Similarly to (4.15) we obtain
L2 (x, D)∗ ((b(z) +
(4.38)
=
g3 e−τ Φ
(q 2
2τ ∂z Φ
b1 (z) B2 −τ Φ(z)
)e
τ
2
− 2 ∂A
− A2 B2 ) −
∂ z̄
+ (b(z) +
g4 e−τ Φ
(q 2
2τ ∂z Φ
c1 (z) A2 −τ Φ(z)
)e
τ
2
− 2 ∂B
∂z
+ v11 e−τ ϕ )
− A2 B2 ) + oL4 (Ω) τ1 .
We are looking for v12 in the form v12 = v0 + v−1 . The function v−1 is given by
(4.39)
v−1 = −
eτ iψ
e−τ iψ e
e2 g7
e2 g8
Rτ,−B 2 {e1 g7 } −
,
R−τ,−A2 {e1 g8 } + 2
+ 2
τ
τ
2τ ∂z Φ 2τ ∂z Φ
where
(4.40)
2
2
− A2 B2 )g3 ) − M7 (z)eB2
T−A2 ((q2 − 2 ∂A
− A2 B2 )g4 ) − M8 (z̄)eA2
P−B2 ((q2 − 2 ∂B
∂z
∂ z̄
, g8 =
,
g7 =
2∂z Φ
2∂z Φ
and M7 (z), M8 (z̄) are polynomials such that
g7 |H = g8 |H = ∇g7 |H = ∇g8 |H = 0.
(4.41)
Using Proposition 2.5 we introduce functions b2,0 , c2,0 ∈ C 2 (Ω̄) such that
g7
g8
(4.42)
+
on Γ0 .
b2,0 (z)eB2 + c2,0 (z)eA2 =
2∂z Φ 2∂z Φ
Similarly to (4.19) we have
1
1e
1
( Rτ,−B 2 {e1 g7 } + R
−τ,−A2 {e1 g8 })|Γ0 = o( 2 ) as |τ | → +∞.
τ
τ
τ
Now we finish the construction of the functions b2,τ (z) and c2,τ (z) by setting
!
1 b2,+ (z)e2iτ ψ(x̃)
b2,− (z)e−2iτ ψ(x̃)
(4.43)
b2,τ (z) = b2,0 (z) +
+
1
2 |det Im Φ00 (x̃)| 12
|det Im Φ00 (x̃)| 2
and
(4.44)
1
c2,τ (z) = c2,0 (z) +
2
c2,+ (z)e2iτ ψ(x̃)
1
|det Im Φ00 (x̃)| 2
+
c2,− (z)e−2iτ ψ(x̃)
1
|det Im Φ00 (x̃)| 2
where b2,+ , c2,− are defined in (4.37).
Consider the following boundary value problem
(4.45)
L2 (x, D)∗ (e−τ ϕ v0 ) = h2 e−τ ϕ
(4.46)
e−τ ϕ v0 |Γ0 = m2 e−τ ϕ ,
in Ω,
!
,
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
33
where
h2 = −eτ ϕ L2 (x, D)∗ (bτ (z)eB2 −τ Φ + cτ (z)eA2 −τ Φ + v11 e−τ ϕ + v−1 e−τ ϕ )
and
m2 = −eτ ϕ (bτ (z)eA2 −τ Φ + cτ (z)eB2 −τ Φ + v11 e−τ ϕ + v−1 e−τ ϕ ).
By (4.38)-(4.40) we can estimate the norm of the function h2 as
(4.47)
1
kh2 kL4 (Ω) = o( ) as |τ | → +∞.
τ
By (4.43), (4.44), (4.37), (4.36), (4.26) we have
(4.48)
kv0 kH 21 (Γ ) = o(
0
1
) as |τ | → +∞.
τ2
Thanks to (4.47), (4.48), by Proposition 2.6 and Proposition 3.8 for sufficiently small positive
there exists a solution to problem (4.45), (4.46) such that
(4.49)
p
1
1
p kv0 kH 1 (Ω) + |τ |kv0 kL2 (Ω) + kv0 kH 1,τ (O ) = o( ) as |τ | → +∞.
τ
|τ |
5. End of the proof of the main theorem
Let u1 be a complex geometrical optics solution as in (4.2). Let u2 be a solution to the
following boundary value problem
(5.1)
L2 (x, D)u2 = 0 in Ω,
u2 |∂Ω = u1 |∂Ω ,
∂u1
∂u2
|Γe =
|e .
∂ν
∂ν Γ
Setting u = u1 − u2 , q = q1 − q2 we have
(5.2)
(5.3)
L2 (x, D)u + 2(A1 − A2 )
∂u1
∂u1
+ 2(B1 − B2 )
+ qu1 = 0 in Ω,
∂z
∂z
u|∂Ω = 0,
∂u
| = 0.
∂ν Γ̃
Let v be a solution to (4.24) in the form (4.25). Taking the scalar product of (5.2) with v in
L2 (Ω) we get
Z
∂u1
∂u1
(5.4)
0 = (2(A1 − A2 )
+ 2(B1 − B2 )
+ qu1 )vdx.
∂z
∂z
Ω
Our goal is to get the asymptotic formula for the right hand side of (5.4). We have
34
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Proposition 5.1. The following asymptotic formula is valid as |τ | → +∞:
Z
(qu, v)L2 (Ω) = (qac̄e(A1 +A2 ) + qdbe(B1 +B2 ) )dx
(5.5)
Ω
Z
q
q
+ ( (a1 b + ac1 )e(A1 −B2 ) + (ab1 + bd1 )e(A2 −B1 ) )dx
τ
Ω τ
!
Z
1
ag4 eA1
c̄g2 eA2
bg1 eB2 dg3 eB1
+
−
−
+
dx
τ Ω 2∂z Φ
2∂z Φ
2∂z̄ Φ
2∂z Φ
+2π
(qab)(x̃)e(A1 +B2 +2τ iImΦ)(x̃) + (qdc)(x̃)e(B1 +A2 −2iτ ImΦ)(x̃)
+
1
τ |det Im Φ00 (x̃)| 2
Z
Z
(ν, ∇ψ)
1
1
1
A1 +B2 +2τ iψ (ν, ∇ψ)
qabe
qdceB1 +A2 −2τ iψ
dσ −
dσ + o( ).
2
2
2τ i ∂Ω
|∇ψ|
2τ i ∂Ω
|∇ψ|
τ
Proof. By (4.2), (4.9) and Proposition 3.3 we have
(5.6) u1 (x) = (a(z) +
a1 (z) A1 +τ Φ
g1 eτ Φ
g2 eτ Φ
d1 (z) B1 +τ Φ
1
−
−
)e
+ (d(z̄) +
)e
+ oL2 (Ω) ( ).
τ
τ
τ
2τ ∂z Φ 2τ ∂z Φ
Using (4.25), (4.33) and Proposition 3.3 we obtain
(5.7) v(x) = (b(z) +
c1 (z) A2 −τ Φ g4 e−τ Φ g3 e−τ Φ
1
b1 (z) B2 −τ Φ
+
)e
+ (c(z̄) +
)e
+
+ oL2 (Ω) ( ).
τ
τ
2τ ∂z Φ
τ
2τ ∂z Φ
By (5.6), (5.7) we obtain
(qu, v)
L2 (Ω)
a1 A1 +τ Φ
d1 B1 +τ Φ
g1 eτ Φ
g2 eτ Φ
1
= (q((a + )e
+ (d + )e
−
+ oL2 (Ω) ( )),
−
τ
τ
τ
2τ ∂z Φ 2τ ∂z Φ
(b +
g4 e−τ Φ g3 e−τ Φ
b1 B2 −τ Φ
c1
1
)e
+ (c + )eA2 −τ Φ +
+
+ oL2 (Ω) ( ))L2 (Ω) =
τ
τ
2τ ∂z Φ
τ
2τ ∂z Φ
Z 1
1
B1 +B2
A1 +A2
q(db + (d1 b + db1 ))e
+ q(ac̄ + (ac1 + a1 c))e
dx
τ
τ
Ω
!
Z
ag4 eA1
c̄g2 eA2
1
bg1 eB2 dg3 eB1
+
−
−
+
dx
τ Ω 2∂z Φ
2∂z Φ
2∂z̄ Φ
2∂z Φ
Z 1
+ q abeA1 +B2 +τ (Φ−Φ) + dc̄eB1 +A2 +τ (Φ−Φ) dx + o( ).
τ
Ω
Applying the stationary phase argument to the last integral in the right hand side of this
formula we finish the proof of Proposition 5.1.
We set
(5.8)
U(x) = aτ (z)eA1 (x)+τ Φ(z) + dτ (z)eB1 (x)+τ Φ(z) , V(x) = bτ (z)eB2 (x)−τ Φ(z) + cτ (z)eA2 (x)−τ Φ(z) .
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
Short calculations give:
∂U
, V)L2 (Ω)
∂z
∂Φ
∂A1
∂aτ A1 +τ Φ
∂B1 B1 +τ Φ
= (2(A1 − A2 )(
+τ
)e
+ dτ
e
aτ +
,
∂z
∂z
∂z
∂z
1
bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω) + I1 (∂Ω)
τ
Z
3
X
=
τ 2−k κk − (A1 − A2 )B1 dτ (z)cτ (z)eB1 +A2 −2iτ ψ dx
I1 ≡ 2((A1 − A2 )
Ω
k=1
∂
A1 +τ Φ
B2 −τ Φ
− 2 (A1 − A2 )aτ e
, bτ e
∂z
L2 (Ω)
∂B2 B2 −τ Φ
bτ e
)L2 (Ω)
∂z
Z
1
+
(A1 − A2 )(ν1 − iν2 )aτ bτ eA1 +B2 +2iτ ψ dσ + o
τ
∂Ω
Z
3
X
=
τ 2−k κk +
−(A1 − A2 )B1 dτ (z)cτ (z)eB1 +A2 −2iτ ψ
−(2(A1 − A2 )aτ eA1 +τ Φ ,
Ω
k=1
∂
A1 +B2 +2iτ ψ
−(A1 − A2 )B2 aτ bτ (z)e
dx
− 2 (A1 − A2 )aτ bτ (z)e
∂z
Z
1
1
A1 +B2 +2iτ ψ
+
(A1 − A2 )(ν1 − iν2 )aτ bτ e
dσ + I1 (∂Ω) + o
τ
τ
∂Ω
A1 +B2 +2iτ ψ
(5.9)
35
36
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
and
∂U
I2 ≡ ((B1 − B2 ) , V)L2 (Ω)
∂z
∂A1
∂
= (2(B1 − B2 )(aτ eA1 +τ Φ
+
dτ eB1 +τ Φ ), bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
∂z
∂z
Z
3
X
∂A1
∂
2−k
A1 +B2 −2τ iψ
=
τ κ
ek + 2(B1 − B2 )
dx − 2 (B1 − B2 )dτ eB1 +τ Φ ,
aτ b τ e
∂z
∂z
Ω
k=1
B2 −τ Φ
A2 −τ Φ
B1 +τ Φ ∂A2
A2 −τ Φ
bτ e
+ cτ e
− 2(B1 − B2 )dτ e
,
cτ e
∂z
L2 (Ω)
L2 (Ω)
Z
1
1
(B1 − B2 )(ν1 + iν2 )dτ (z)cτ (z)eB1 +A2 −2iτ ψ dσ + I2 (∂Ω) + o
+
τ
τ
∂Ω
Z
3
n
X
τ 2−k κ
ek +
=
−(B1 − B2 )A1 aτ bτ eA1 +B2 +2τ iψ
k=1
Ω
∂
B1 +A2 −2iτ ψ
B1 +A2 −2iτ ψ
− 2 (B1 − B2 )dτ (z)cτ (z)e
− (B1 − B2 )dτ (z)cτ (z)A2 e
dx
∂z
Z
1
1
B1 +A2 −2iτ ψ
+
(B1 − B2 )(ν1 + iν2 )dτ (z)cτ (z)e
dσ + I2 (∂Ω) + o
(5.10)
.
τ
τ
∂Ω
Here κk , κ
ek are some constants independent of τ but may be dependent on Aj , Bj , Φ. The
terms I1 (∂Ω), I2 (∂Ω) are given by
!
−2τ iψ(x̃)
2τ iψ(x̃)
a
e
a
e
∂Φ
2,+
2,−
c(z̄)
dx
I1 (∂Ω) = (A1 − A2 )eA1 +A2
1 +
1
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω
!
Z
2τ iψ(x̃)
−2τ iψ(x̃)
∂Φ
c
e
c
e
2,+
2,−
+ (A1 − A2 )eA1 +A2
a(z)
dx =
1 +
1
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω
!
Z
∂ A1 +A2 ∂Φ
a2,+ e−2τ iψ(x̃)
a2,− e2τ iψ(x̃)
−2
e
c(z̄)
dx
1 +
1
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω ∂ z̄
!
Z
c2,− e−2τ iψ(x̃)
∂ A1 +A2 ∂Φ
c2,+ e2τ iψ(x̃)
−2
e
a(z)
dx =
1 +
1
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω ∂ z̄
!
Z
−2τ iψ(x̃)
2τ iψ(x̃)
∂Φ
a2,+ e
a2,− e
−2
(ν1 + iν2 )eA1 +A2
c(z̄)
dσ
1 +
1
00
∂z
|det Im Φ (x̃)| 2
|det Im Φ00 (x̃)| 2
∂Ω
!
Z
c2,− e−2τ iψ(x̃)
c2,+ e2τ iψ(x̃)
A1 +A2 ∂Φ
(5.11)
a(z)
dσ
−2
(ν1 + iν2 )e
1 +
1
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
∂Ω
Z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
37
and
!
2τ iψ(x̃)
−2τ iψ(x̃)
∂Φ
d
e
d
e
2,−
2,+
I2 (∂Ω) = (B1 − B2 )eB1 +B2
b(z)
dx
1 +
1
00
∂
z̄
2
|det Im Φ (x̃)|
|det Im Φ00 (x̃)| 2
Ω
!
Z
2τ iψ(x̃)
−2τ iψ(x̃)
∂Φ
b
e
b
e
2,+
2,−
+ (B1 − B2 )eB1 +B2
d(z̄)
dx =
1 +
1
00
∂
z̄
|det Im Φ (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω
!
Z
2τ iψ(x̃)
−2τ iψ(x̃)
d2,− e
∂ B1 +B2 ∂Φ
d2,+ e
b(z)
dx
e
−2
1 +
1
00
∂ z̄
|det Im Φ (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω ∂z
!
Z
∂ B1 +B2 ∂Φ
b2,+ e2τ iψ(x̃)
b2,− e−2τ iψ(x̃)
−2
e
d(z̄)
dx =
1 +
1
∂ z̄
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
Ω ∂z
!
Z
−2τ iψ(x̃)
2τ iψ(x̃)
∂Φ
d
e
d
e
2,+
2,−
b(z)
dσ
−2
(ν1 − iν2 )eB1 +B2
1 +
1
∂ z̄
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
∂Ω
!
Z
−2τ iψ(x̃)
2τ iψ(x̃)
b
e
b
e
∂Φ
2,−
2,+
(5.12)
d(z̄)
dσ.
−2
(ν1 − iν2 )eB1 +B2
1 +
1
00
∂
z̄
2
|det Im Φ (x̃)|
|det Im Φ00 (x̃)| 2
∂Ω
Z
Denote
U1 = −eτ Φ R−τ,A1 {e1 g1 },
e τ,B1 {e1 g2 }.
U2 = −eτ Φ R
A short calculation gives
(5.13)
2
∂U1
= (−e1 g1 + A1 R−τ,A1 {e1 g1 })eτ Φ
∂z
and
(5.14)
2
∂U2
e τ,B1 {e1 g2 })eτ Φ .
= (−e1 g2 + B1 R
∂z
We have
∂
∂A1
∂Φ
∂(e1 g1 )
R−τ,A1 {e1 g1 } =
R−τ,A1 {e1 g1 } + τ
R−τ,A1 {e1 g1 } + R−τ,A1 {
}
∂z
∂z
∂z
∂z
∂Φ
∂(e1 g1 )
∂A1
} − τ R−τ,A1 { e1 g1 } = R−τ,A1 {
}
−R−τ,A1 {e1 g1
∂z
∂z
∂z
Z ∂Φ (ζ) − ∂Φ (z)
eA1 −τ (Φ−Φ)
∂ζ
∂z
+τ
e
(e1 g1 e−A1 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2
2π
ζ −z
Ω
Z
∂A
1
1
(ζ, ζ̄) − ∂A
(z, z̄)
eA1 −τ (Φ−Φ)
∂ζ
∂z
(5.15) +
e
(e1 g1 e−A1 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 .
2π
ζ −z
Ω
∂
∂
Denote P(x, D) = 2(A1 − A2 ) ∂z
+ 2(B1 − B2 ) ∂z
. Let
1
G(x, g, A, τ ) = −
2π
Z (τ ∂Φ(ζ) +
∂ζ
Ω
∂A(ζ,ζ̄)
)
∂ζ
− (τ ∂Φ(z)
+
∂z
ζ −z
∂A(z,z̄)
)
∂z
e1 ge−A eτ (Φ−Φ) dξ1 dξ2 ,
We set G1 (x, τ ) = G(x, g1 , A1 , τ ), G2 (x, τ ) = G(x, ḡ2 , B1 , τ ), G3 (x, τ ) = G(x, g4 , A2 , −τ ), G4 =
G(x, g3 , B2 , −τ ).
38
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
By (3.15), (4.6), (5.15) and Proposition 3.3 we get
∂(e g )
1 1
1
∂
R−τ,A1 {e1 g1 } = ∂z − eA1 e−τ (Φ−Φ) G1 (·, τ ) + oL2 (Ω) ( ).
∂z
τ
2τ ∂z Φ
(5.16)
Simple computations provide the formula
(5.17)
∂ e
∂B1 e
∂Φ e
e τ,B1 { ∂(e1 g2 ) }
Rτ,B1 {e1 g2 } =
Rτ,B1 {e1 g2 } + τ
Rτ,B1 {e1 g2 } + R
∂z
∂z
∂z
∂z
e τ,B1 { ∂B1 e1 g2 } = R
e τ,B1 { ∂(e1 g2 ) }
e τ,B1 { ∂Φ e1 g2 } − R
−τ R
∂z
∂z
∂z
Z
∂Φ
∂Φ
(ζ) − ∂z (z)
eB1 τ (Φ−Φ)
∂ζ
+τ
e
(e1 g2 e−B1 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2
2π
ζ −z
Ω
Z
∂B
1
1
(ζ, ζ) − ∂B
(z, z)
eB1 τ (Φ−Φ)
∂z
∂ζ
e
(e1 g2 e−B1 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 .
+
2π
ζ
−
z
Ω
By (3.15), (4.6), (5.17), Proposition 3.3 we have
∂(e g )
1 2
1
∂ e
Rτ,B1 {e1 g2 } = ∂z − eB1 eτ (Φ−Φ) G2 (·, τ ) + oL2 (Ω) ( ).
∂z
2τ ∂z Φ
τ
(5.18)
Denote
V2 = −e−τ Φ Rτ,−B 2 {e1 g3 },
e
V1 = −e−τ Φ R
−τ,A2 {e1 g4 }.
The following proposition is proved in Section 7.
Proposition 5.2. There exist two numbers κ, κ0 independent of τ such that the following
asymptotic formula holds true:
(P(x, D)(U1 + U2 ), bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω) + (P(x, D)(aτ eA1 +τ Φ + dτ eB1 +τ Φ ), V1 + V2 )L2 (Ω) =
Z
Z
κ0
A1 +A2
(ν1 + iν2 )e
κ+
−2
cτ (z̄)G1 (x, τ )dσ − 2
(ν1 − iν2 )eB1 +B2 dτ (z)G4 (x, τ )dσ
τ
∂Ω
∂Ω
Z
(5.19)
−2
A1 +A2
aτ (z)G3 (x, τ )(ν1 + iν2 )e
∂Ω
Z
dσ − 2
(ν1 − iν2 )eB1 +B2 bτ (z)G2 (x, τ )dσ.
∂Ω
By (5.13), (5.14), (5.16), (5.18) and Proposition 3.4 there exists a constant C0 independent
of τ such that
(
P(x, D)(U1 + U2 ), V1 + V2 )L2 (Ω) = ((A1 − A2 )(−2
!
1
− eA1 e−τ (Φ̄−Φ) G1 + oL2 (Ω) ( ) eτ Φ
τ
2τ ∂z Φ
∂(e1 g1 )
∂z
1
B1 e1 g2
+ oL2 (Ω) ( ))eτ Φ ), V1 + V2 )L2 (Ω)
2τ ∂z Φ
τ
1
e1 g1
+ ((B1 − B2 )(−e1 g1 + A1
+ oL2 (Ω) ( ))eτ Φ , V1 + V2 )L2 (Ω)
τ
2τ ∂z Φ
+ (−e1 g2 +
− (2(B1 − B2 )(
∂(e1 g2 )
∂ z̄
1
C0
1
− eB1 eτ (Φ̄−Φ) G2 + oL2 (Ω) ( ))eτ Φ , V1 + V2 )L2 (Ω) =
+ o( ) as |τ | → +∞.
2τ ∂z Φ
τ
τ
τ
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
39
Next we claim that
1
(P(x, D)(u0 eτ ϕ ), v)L2 (Ω) = o( ) as |τ | → +∞,
τ
(5.20)
and
1
(P(x, D)u, v0 e−τ ϕ )L2 (Ω) = o( ) as |τ | → +∞.
τ
Let us first prove (5.21). By (4.23) and (4.49), we have
(5.21)
1
(P(x, D)u, v0 e−τ ϕ )L2 (Ω) = (P(x, D)U, v0 e−τ ϕ )L2 (Ω) + o( ) as |τ | → +∞.
τ
We remind that the function U and V are defined by (5.8). By (4.49) we obtain from (5.22)
(5.23)
Z
∂Φ
∂Φ
1
−τ ϕ
(P(x, D)u, v0 e )L2 (Ω) = τ
2χ( (A1 − A2 )aeA1 +iτ ψ +
(B1 − B2 )beB1 −iτ ψ )v0 dx + o( )
∂z
∂z
τ
Ω
(5.22)
as |τ | → +∞. Here χ ∈ C0∞ (Ω) is a function such that χ ≡ 1 in some neighborhood of
supp e2 , H \ ∂Ω ⊂ supp e2 .
The functions v0,+ = e−iτ ψ v0 and v0,− = eiτ ψ v0 satisfy eτ Φ L2 (x, D)∗ (e−τ Φ v0,+ ) = h2 eiτ ψ
and eτ Φ L2 (x, D)∗ (e−τ Φ v0,− ) = h2 e−iτ ψ . More explicitly there exist two first-order operators
Pk (x, D) such that
eτ Φ L2 (x, D)∗ (e−τ Φ v0,+ ) = ∆v0,+ −2τ
∂Φ ∂v0,+
1
(2
−A2 v 0,+ )+P1 (x, D)v0,+ = oL2 (Ω) ( ) as |τ | → +∞
∂z
∂z
τ
and
eτ Φ L2 (x, D)∗ (e−τ Φ v0,− ) = ∆v0,− −2τ
1
∂Φ ∂v0,−
(2
−B2 v 0,− )+P2 (x, D)v0,− = oL2 (Ω) ( ) as |τ | → +∞.
∂z
∂z
τ
Let χ1 ∈ C0∞ (Ω) be a function such that χ1 ≡ 1 on supp χ. Taking the scalar product of
the first equation with χ1 g where g ∈ C 2 (Ω) we obtain
Z
Z
∂Φ
∂Φ ∂
∂
1
τ
v0,+ χ1 (2 −A2 )gdx = o( )− (v0,+ (∆+P1 (x, D)∗ )(χ1 g)−τ v0,+
g( −A2 )χ1 )dx.
∂z
τ
∂z ∂z
Ω ∂z
Ω
By (4.49) we have
Z
(5.24)
τ
Ω
∂Φ
∂
1
v0,+ χ1 (2 − A2 )gdx = o( ) as |τ | → +∞.
∂z
∂z
τ
Taking the scalar product of the second equation with χ1 g where g ∈ C 2 (Ω) we have
Z
Z
∂Φ
∂
1
∂Φ ∂
τ
v0,− χ1 (2 −B2 )gdx = o( )− (v0,− (∆+P2 (x, D)∗ )(χ1 g)−τ v0,− g
(2 −B2 )χ1 )dx.
∂z
τ
∂z ∂z
Ω
Ω ∂z
By (4.49) we obtain
Z
∂Φ
∂
1
(5.25)
τ
v0,− χ1 (2 − B2 )gdx = o( ) as |τ | → +∞.
∂z
τ
Ω ∂z
∂
∂
Taking g such that (2 ∂z
−A2 )g = (A1 −A2 )eA1 a(z) in (5.24) and g such that (2 ∂z
−B2 )g =
(B1 − B2 )b(z)eB1 in (5.25) from (5.22) we obtain (5.21).
40
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
To prove (5.20) we observe that
1
1
(P(x, D)(u0 eτ ϕ ), v) = (P(x, D)(u0 eτ ϕ ), V)L2 (Ω) + o( ) = (P(x, D)(u0 eτ ϕ ), χV)L2 (Ω) + o( )
τ
τ
1
(5.26)
= (u0 eτ ϕ , P(x, D)∗ (χV))L2 (Ω) + o( ) as |τ | → +∞.
τ
Then we can finish the proof of (5.20) using arguments similar to (5.23)-(5.24).
Denote M1 =
1
L (x, D)( e∂2 gΦ1 ),
4∂z̄ Φ 1
z
M4 = − 4∂1 Φ L2 (x, D)∗
z
e2 g4
∂z̄ Φ
M2 =
1
L (x, D)( e∂2zgΦ2 ),
4∂z Φ 1
M3 =
− 4∂1z Φ L2 (x, D)∗
e2 g3
∂z Φ
,
. Then there exists a constant C independent of τ such that
(5.27)
(P(x, D)(eτ Φ
M4
M1 τ Φ M2
M3
C
1
+e
), v)L2 (Ω) +(P(x, D)u, e−τ Φ 2 +e−τ Φ 2 )L2 (Ω) = +o( ) as |τ | → +∞.
2
2
τ
τ
τ
τ
τ
τ
e2 g2
e2 g1
, X2 = − 2∂
, X3 =
Denote X1 = − 2∂
zΦ
z̄ Φ
argument we conclude
e2 g3
, X4
2∂z Φ
=
e2 g4
.
2∂z̄ Φ
Then, using the stationary phase
X2
X3
X1
X4
(P(x, D)(eτ Φ
+ eτ Φ ), v)L2 (Ω) + (P(x, D)u, e−τ Φ
+ e−τ Φ )L2 (Ω) =
τ
τ
τ
τ
Z
∂Φ
∂Φ
(∇ψ, ν)
C−1 1
dσ
C0 +
+
((A1 − A2 ) X2 beB2 e2τ iψ − (B1 − B2 ) X1 beA2 e−2τ iψ )
τ
τ Γe
∂z
∂z
2i|∇ψ|2
(5.28)
1
+
τ
Z
((A1 − A2 )
e
Γ
∂Φ
∂Φ
(∇ψ, ν)
1
dσ + o( ) as |τ | → +∞.
X3 aeA1 e2τ iψ − (B1 − B2 ) X4 aeB1 e−2τ iψ )
2
∂z
∂z
2i|∇ψ|
τ
Next we show that
Proposition 5.3. Under the conditions of Theorem 1.1
(5.29)
A1 = A2 ,
B1 = B2
e
on Γ
and
(5.30)
Z I(Φ, a, b, c, d) =
Γ̃
∂Φ
∂Φ
(ν1 + iν2 ) a(z)c(z̄)eA1 +A2 + (ν1 − iν2 ) d(z̄)b(z)eB1 +B2
∂z
∂ z̄
dσ = 0.
e and Γ∗ be an arc containing x
Proof. Let x
b be an arbitrary point from Int Γ
b such that
e
Γ∗ ⊂⊂ Γ. By Proposition 2.2 there exists a weight function Φ satisfying (2.4) and (2.6).
Then the boundary integrals in (5.9), (5.10) have the following asymptotic:
Z
Z
B1 +A2 −2iτ ψ
(B1 − B2 )dτ (z)cτ (z)e
dσ + (A1 − A2 )(ν1 − iν2 )aτ (z)bτ (z)eA1 +B2 +2iτ ψ dσ =
e
e
Γ
Γ


! 21
! 12

X
e−2τ iψ(x)
e2τ iψ(x) 
2π
2π
(ab(B1 − B2 ))(x) √
+
(ab(A1 − A2 ))(x) √
2
 i ∂ 2 ψ2 (x)
τ
τ 
−i ∂∂~τψ2 (x)
∂~
τ
x∈G\{x− ,x+ }
(5.31)
1
+O( ) as |τ | → +∞.
τ
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
41
We remind that the set G is introduced in (2.4). Moreover, in order to avoid the contribution
from the points x± we chose functions a, b in such a way that
(5.32)
∂ |β| a
∂ |β| b
∂ |β| c
∂ |β| d
(x
)
=
(x
)
=
(x
)
=
(x± ) = 0 ∀|β| ∈ {0, . . . , 5}.
±
±
±
∂xβ1 1 ∂xβ2 2
∂xβ1 1 ∂xβ2 2
∂xβ1 1 ∂xβ2 2
∂xβ1 1 ∂xβ2 2
Let χ̃1 ∈ C ∞ (∂Ω) be a function such that it is equal 1 near points x± and has support
located in a small neighborhood of these points. Then
Z
Z
B1 +A2 −2iτ ψ
χ̃1 (B1 − B2 )dτ cτ e
dσ +
χ̃1 (A1 − A2 )(ν1 − iν2 )aτ bτ eA1 +B2 +2iτ ψ dσ =
Γ∗
Z
Γ∗
Γ∗
χ̃1 (B1 − B2 )dτ cτ eB1 +A2 ∂e−2iτ ψ
dσ +
∂~τ
−2iτ ∂ψ
∂~
τ
Z
Γ∗
Z
−
Γ∗
∂
∂~τ
∂
∂~τ
Z
Γ∗
χ̃1 (A1 − A2 )(ν1 − iν2 )aτ bτ eA1 +B2 ∂e2iτ ψ
dσ =
∂~τ
2iτ ∂ψ
∂~
τ
χ̃1 (B1 − B2 )dτ cτ eB1 +A2
2iτ ∂ψ
∂~
τ
!
χ̃1 (A1 − A2 )(ν1 − iν2 )aτ bτ eA1 +B2
2iτ ∂ψ
∂~
τ
e−2iτ ψ dσ
!
1
e2iτ ψ dσ = O( ).
τ
In order to obtain the last equality used that by (5.32) and (2.7) the functions
!
!
∂ χ̃1 (B1 − B2 )dτ cτ eB1 +A2
∂ χ̃1 (A1 − A2 )(ν1 − iν2 )aτ bτ eA1 +B2
,
∂~τ
∂~τ
2i ∂ψ
2i ∂ψ
∂~
τ
∂~
τ
are bounded. By (5.5), (5.9)-(5.12), (5.19)-(5.21), (5.27) - (5.28) and (5.31), we can represent
the right-hand side of (5.4) as

! 21
X
e−2τ iψ(x)
1
2π

√
−
B
))(x)
O( ) = τ F1 + F0 +
(ab(B
1
2
2
τ
τ
i ∂∂~τψ2 (x)
x∈G\{x− ,x+ }
+
2π
2
−i ∂∂~τψ2 (x)
! 12
e2τ iψ(x)
(ab(A1 − A2 ))(x) √
τ
..
Taking into account that F1 is equal to the left-hand side of (5.30) we obtain the equality
(5.30). Using (2.6) and applying Bohr’s theorem (e.g., [3], p.393), we obtain (5.29).
Denote
∂
(A1 − A2 ),
∂z
∂
Q+ = −(A1 − A2 )B1 − (B1 − B2 )A2 − 2 (B1 − B2 ).
∂z
Q− = −(B1 − B2 )A1 − (A1 − A2 )B2 − 2
42
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Thanks to (5.5), (5.9)-(5.12), (5.19)-(5.21), (5.27)-(5.29), (5.31), we can write down the
right-hand side of (5.4) as
3
X
I1 + I2 =
τ 2−k (κk + κ
ek ) + κ
k=1
Z
A1 +B2
Z
(B1 − B2 )(ν1 + iν2 )dτ (z)cτ (z)eB1 +A2 dσ
(A1 − A2 )(ν1 − iν2 )aτ bτ e
dσ +
Γ0
Γ0
Z
Z
1
(∇ψ, ν)
(∇ψ, ν)
1
−
Q+ abeB1 +A2
dσ −
Q− abeA1 +B2
dσ
2
τ Γ0
|∇ψ|
τ Γ0
|∇ψ|2
(Q+ ab)(e
x)e(A1 +B2 +2τ iImΦ)(ex) + (Q− dc̄)(e
x)e(B1 +A2 −2iτ ImΦ)(ex)
+ 2π
1
τ |det ImΦ00 (e
x)| 2
1
+
(I1 (∂Ω) + I2 (∂Ω))
τZ
+
(ν1 + iν2 )eA1 +A2 cτ (z̄)G1 (x, τ )dσ
− 2
Z∂Ω
− 2
Z∂Ω
− 2
(ν1 − iν2 )eB1 +B2 dτ (z)G4 (x, τ )dσ
(ν1 + iν2 )eA1 +A2 aτ (z)G3 (x, τ )dσ
∂Ω
Z
(5.33)
−2
1
(ν1 − iν2 )eB1 +B2 bτ (z)G2 (x, τ )dσ + o( ) as |τ | → +∞.
τ
∂Ω
We note that κk and κ
ek denote generic constants which are independent of τ . In order to
transform some terms in the above equality, we need the following proposition:
1
Proposition 5.4. There exist a holomorphic function Ψ ∈ H 2 (Ω) and an antiholomorphic
1
function Ψ̃ ∈ H 2 (Ω) such that
Ψ|Γ̃ = eA1 +A2 ,
(5.34)
Ψ̃|Γ̃ = eB1 +B2
and
eB1 +B2 Ψ = eA1 +A2 Ψ̃ on Γ0 .
(5.35)
Proof. Consider the extremal problem:
J(Ψ, Ψ̃) = keA1 +A2
∂Φ
∂Φ
b̄d − Ψ̃k2L2 (Γ̃) → inf,
ac̄ − Ψk2L2 (Γ̃) + keB1 +B2
∂z
∂ z̄
∂Ψ
∂ Ψ̃
= 0 in Ω,
= 0 in Ω, ((ν1 − iν2 )Ψ(z) + (ν1 + iν2 )Ψ̃(z̄))|Γ0 = 0.
∂ z̄
∂z
Here the functions a, b, c, d satisfy (4.3), (4.4), (4.26) and (4.27). Denote the unique solution
ˆ Applying the Lagrange principle to this extremal problem
to this extremal problem as (Ψ̂, Ψ̃).
(5.36)
we obtain
(5.37)
Re(eA1 +A2
∂Φ
∂Φ
ˆ δ̃)
ac̄ − Ψ̂, δ)L2 (Γ̃) + Re(eB1 +B2
b̄d − Ψ̃,
L2 (Γ̃) = 0
∂z
∂ z̄
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
43
1
for any δ from H 2 (Ω) such that
∂ δ̃
= 0 in Ω,
∂z
∂δ
= 0 in Ω,
∂ z̄
(ν1 − iν2 )δ|Γ0 = −(ν1 + iν2 )δ̃|Γ0
1
and there exist two functions P, P̃ ∈ H 2 (Ω) such that
∂P
= 0 in Ω,
∂ z̄
(5.38)
(ν1 + iν2 )P = eA1 +A2
(5.39)
∂ P̃
= 0 in Ω,
∂z
∂Φ
ac̄ − Ψ̂ on Γ̃,
∂z
(ν1 − iν2 )P̃ = eB1 +B2
∂Φ
ˆ on Γ̃,
b̄d − Ψ̃
∂ z̄
(P − P̃ )|Γ0 = 0.
(5.40)
Denote Ψ0 (z) =
1
(P (z)
2i
− P̃ (z̄)), Φ0 (z) = 12 (P (z) + P̃ (z̄)). By (5.40)
Im Ψ|Γ0 = Im Φ|Γ0 = 0.
Hence
(5.41)
P = (Φ0 + iΨ0 ),
P̃ = (Φ0 − iΨ0 ).
From (5.37)
(5.42)
Re(eA1 +A2
∂Φ
∂Φ
ˆ
ˆ Ψ̃)
b̄d − Ψ̃,
ac̄ − Ψ̂, Ψ̂)L2 (Γ̃) + Re(eB1 +B2
L2 (Γ̃) = 0.
∂z
∂ z̄
By (5.38), (5.39) and (5.41), we have
∂Φ
∂Φ
∂Φ
ˆ eB1 +B2 ∂Φ b̄d)
ac̄ − Ψ̂, eA1 +A2
ac̄)L2 (Γ̃) + Re(eB1 +B2
b̄d − Ψ̃,
L2 (Γ̃)
∂z
∂z
∂ z̄
∂ z̄
∂Φ
∂Φ
ac̄)L2 (Γ̃) + Re((ν1 + iν2 )P̃ , eB1 +B2
b̄d)L2 (Γ̃) =
Re((ν1 − iν2 )P̄ , eA1 +A2
∂z
∂ z̄
∂Φ
∂Φ
2Re((ν1 − iν2 )(Φ0 + iΨ0 ), eA1 +A2
ac̄)L2 (Γ̃) + 2Re((ν1 + iν2 )(Φ0 − iΨ0 ), eB1 +B2
b̄d)L2 (Γ̃) .
∂z
∂ z̄
H1 = Re(eA1 +A2
We can rewrite
2Re((ν1 − iν2 )Φ0 , eA1 +A2
(5.43)
∂Φ
∂Φ
b̄d)L2 (Γ̃) =
ac̄)L2 (Γ̃) + 2Re((ν1 + iν2 )Φ0 , eB1 +B2
∂z
∂ z̄
I(Φ, Φ0 a, b, c, Φ0 d) + I(Φ, Φ0 a, b, c, Φ0 d)
44
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
and
∂Φ
∂Φ
ac̄)L2 (Γ̃) + 2Re((ν1 + iν2 )(−iΨ0 ), eB1 +B2
b̄d)L2 (Γ̃) =
∂z
∂ z̄
∂Φ
¯ 0 , eB1 +B2 ∂Φ ) 2 =
)L2 (Γ̃) − 2Im((ν1 + iν2 )bdΨ
−2Im((ν1 − iν2 )ācΨ0 , eA1 +A2
∂z
∂ z̄ L (Γ̃)
Z
∂Φ A1 −A2
∂Φ A1 +A2
1
((ν1 − iν2 )ācΨ0
−
e
− (ν1 + iν2 )ac̄Ψ0
e
)dx
i Ω
∂ z̄
∂z
Z
1
¯ 0 ∂Φ eB1 +B2 − (ν1 − iν2 )b̄dΨ0 ∂Φ eB1 +B2 )dx =
((ν1 + iν2 )bdΨ
−
i Ω
∂z
∂ z̄
1
(5.44)
(−I(Ψ, aΨ0 , b, c, dΨ0 ) + I(Ψ, aΨ0 , b, c, dΨ0 )).
i
2Re((ν1 − iν2 )(iΨ0 ), eA1 +A2
Then by (5.43), (5.44) and Proposition 5.3, H1 = 0. Taking into account (5.42) we obtain
ˆ = 0. Consequently (5.34) is proved. From (4.3), (4.4), (4.26), (4.27) and (5.36)
that J(Ψ̂, Ψ̃)
we obtain (5.35). The proof of the proposition is completed.
Thanks to Proposition 5.4, we can rewrite (5.33) as
3
X
1
(5.45)
τ 2−k Fek
o( ) =
τ
k=1
n
1
+ 2π
x)e(A1 +B2 +2τ iImΦ)(ex)
((Q+ + (q1 − q2 ))ab)(e
1
τ |det ImΦ00 (x̃)| 2
x)
(B1 +A2 −2iτ ImΦ)(e
+ ((Q− + (q1 − q2 ))dc̄)(e
x)e
−
−
−
−
−
−
2
τ
Z
e2τ iψ(x̃)
e−2τ iψ(x̃)
!
∂Φ
c2,−
c2,+
a(z)
dσ
1 +
1
00
∂z
|det Im Φ (x̃)| 2
|det Im Φ00 (x̃)| 2
Γ0
!
Z
−2τ iψ(x̃)
2τ iψ(x̃)
d
e
2
∂Φ
d
e
2,+
2,−
(ν1 − iν2 )(eB1 +B2 − Ψ̃) b(z)
dσ
1 +
1
τ Γ0
∂ z̄
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
!
Z
−2τ iψ(x̃)
2τ iψ(x̃)
a
e
2
∂Φ
a
e
2,+
2,−
(ν1 + iν2 )(eA1 +A2 − Ψ) c(z̄)
dσ
1 +
1
τ Γ0
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
!
Z
∂Φ
2
b2,− e−2τ iψ(x̃)
b2,+ e2τ iψ(x̃)
B1 +B2
(ν1 − iν2 )(e
− Ψ̃) d(z̄)
dσ
1 +
1
τ Γ0
∂ z̄
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
Z
Z
A1 +A2
2
(ν1 + iν2 )(e
− Ψ)cτ (z̄)G1 (x, τ )dσ − 2
(ν1 − iν2 )(eB1 +B2 − Ψ̃)dτ (z)G4 (x, τ )dσ
ZΓ0
Z∂Ω
2
aτ (z)(ν1 + iν2 )(eA1 +A2 − Ψ)G3 (x, τ )dσ − 2
(ν1 − iν2 )(eB1 +B2 − Ψ̃)bτ (z)G2 (x, τ )dσ.
(ν1 + iν2 )(eA1 +A2 − Ψ)
Γ0
Here Fek denote constants which are independent of τ .
∂Ω
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
45
Observe that
(5.46)
(ν1 + iν2 )
∂Φ
∂ Φ̄
= −(ν1 − iν2 )
∂z
∂ z̄
on Γ0 .
∂ϕ
−
To see this we argue as follows. We have that ∂∂z̄ = 12 ( ∂x∂ 1 + i ∂x∂ 2 )(ϕ + iψ) = 12 ( ∂x
1
∂ψ
∂ψ
∂ϕ
∂ψ
∂ϕ
∂ψ ∂ϕ
∂ψ
∂ψ
∂ϕ
i ∂ϕ
) + 2 ( ∂x2 + ∂x1 ). Hence ∂x1 = ∂x2 , ∂x2 = − ∂x1 , ∂ν = − ∂~τ and ∂ν = ∂~τ . Observe that
∂x2
∂
∂
∂
∂
∂
= 12 (ν1 ∂x∂ 1 +ν2 ∂x∂ 2 )+ 2i (ν2 ∂x∂ 1 −ν1 ∂x∂ 2 ) = 21 ( ∂ν
+i ∂~
) and (ν1 −iν2 ) ∂∂z̄ = 12 ( ∂ν
−i ∂~
).
(ν1 +iν2 ) ∂z
τ
τ
Hence
(ν1 + iν2 )
∂Φ
1 ∂
∂
1 ∂ϕ ∂ψ
i ∂ϕ ∂ψ
∂ψ
∂ϕ
= (
+ i )(ϕ + iψ) = (
−
)+ (
+
)=−
+i .
∂z
2 ∂ν
∂~τ
2 ∂ν
∂~τ
2 ∂~τ
∂ν
∂~τ
∂~τ
Therefore
(ν1 − iν2 )
∂ψ
∂ϕ
∂ψ
∂ϕ
∂Φ
=
+i
=−
−i .
∂ z̄
∂~τ
∂~τ
∂~τ
∂~τ
Taking into account that ψ|Γ0 = 0 we obtain (5.46).
Then, using (5.46), on Γ0 we have
∂Φ
−(ν1 + iν2 ) (eA1 +A2 − Ψ)c(z̄)
∂z
−(ν1 − iν2 )
∂Φ
−(ν1 + iν2 )
∂z
−(ν1 − iν2 )
A1 +A2
(e
− Ψ)c(z̄)
a2,− e2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
d2,+ e−2τ iψ(x̃)
∂Φ
−(ν1 − iν2 ) (eB1 +B2 − Ψ̃)b(z)
∂ z̄
∂Φ
−(ν1 + iν2 )
∂z
a2,+ e−2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
B1 +A2
1
|det Im Φ00 (x̃)| 2
+ (e
− Ψ̃e
1
|det Im Φ00 (x̃)| 2
!
2τ iψ(x̃)
d2,− e
1
|det Im Φ00 (x̃)| 2
)c(z̄)
(eA1 +A2 − Ψ)c(z̄)
a2,− e2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
1
|det Im Φ00 (x̃)| 2
!
d2,+ e−2τ iψ(x̃)
+ (eB1 +A2 − Ψe−A1 +B1 )c(z̄)
=
!
d2,− e2τ iψ(x̃)
∂Φ
a2,+ e−2τ iψ(x̃)
B1 +B2
b(z) (eA1 +B2 − ΨeB2 −A2 )
− Ψ̃)
1 + (e
1
∂ z̄
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
a2,+ e−2τ iψ(x̃)
∂Φ
B1 +B2
b(z) (eA1 +B2 − Ψ̃eA1 −B1 )
− Ψ̃)
1 + (e
1
00
∂ z̄
2
|det Im Φ (x̃)|
|det Im Φ00 (x̃)| 2
=
!
d2,− e2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
!
d2,+ e−2τ iψ(x̃)
−(ν1 + iν2 )
(5.47)
A2 −B2
+
+
!
a2,− e2τ iψ(x̃)
=
c(z̄)p− e2τ iψ(x̃)
∂Φ A2
(e − Ψe−A1 )
1
∂z
|det Im Φ00 (x̃)| 2
−(ν1 − iν2 )
∂Φ B2
b(z)p+ e−2τ iψ(x̃)
(e − Ψ̃e−B1 )
1
∂ z̄
|det Im Φ00 (x̃)| 2
46
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
and
−(ν1 + iν2 )(eA1 +A2
∂Φ
− Ψ) a(z)
∂z
−(ν1 − iν2 )(eB1 +B2 − Ψ̃)
∂Φ
d(z̄)
∂ z̄
c2,+ e2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
b2,+ e2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
+
+
!
c2,− e−2τ iψ(x̃)
1
|det Im Φ00 (x̃)| 2
!
b2,− e−2τ iψ(x̃) )
1
|det Im Φ00 (x̃)| 2
=
!
2τ iψ(x̃)
2τ iψ(x̃)
∂Φ
c
e
b
e
2,−
2,−
A1 +B2
−(ν1 + iν2 ) a(z) (eA1 +A2 − Ψ)
− e−B1 +A1 Ψ̃)
1 + (e
1
∂z
|det Im Φ00 (x̃)| 2
|det Im Φ00 (x̃)| 2
!
−2τ iψ(x̃)
−2τ iψ(x̃)
∂Φ
c
e
b
e
2,+
2,+
B1 +B2
−(ν1 − iν2 ) d(z̄) (eB1 +A2 − Ψe−A1 +B1 )
=
− Ψ̃)
1 + (e
1
00
∂ z̄
2
|det Im Φ (x̃)|
|det Im Φ00 (x̃)| 2
!
2τ iψ(x̃)
2τ iψ(x̃)
∂Φ
c
e
b
e
2,−
2,−
A1 +B2
−(ν1 + iν2 ) a(z) (eA1 +A2 − Ψ)
− eB2 −A2 Ψ)
1 + (e
1
00
∂z
|det Im Φ (x̃)| 2
|det Im Φ00 (x̃)| 2
!
−2τ
iψ(x̃)
−2τ
iψ(x̃)
∂Φ
c2,+ e
b2,+ e
B1 +B2
− Ψ̃)
=
−(ν1 − iν2 ) d(z̄) (eB1 +A2 − ΨeA2 −B2 )
1 + (e
1
00
∂ z̄
|det Im Φ (x̃)| 2
|det Im Φ00 (x̃)| 2
−(ν1 + iν2 )
−(ν1 − iν2 )
(5.48)
∂Φ A1
a(z)p̃− e2τ iψ(x̃)
(e − Ψe−A2 )
1
∂z
|det Im Φ00 (x̃)| 2
∂Φ B1
d(z̄)e−2τ iψ(x̃) p̃+
(e − Ψ̃e−B2 )
1 .
∂ z̄
|det Im Φ00 (x̃)| 2
Using (5.47), (5.48) and Proposition 7.2 in Section 7, we rewrite (5.45) as
3
X
1
o( ) =
τ 2−k Fek +
τ
k=1
2 π
(5.49)
1
1
τ |det ImΦ00 (e
x)| 2
n
x)e(A1 +B2 +2τ iImΦ)(ex)
((Q+ + (q1 − q2 ))ab)(e
o
+ ((Q− + (q1 − q2 ))dc̄)(e
x)e(B1 +A2 −2iτ ImΦ)(ex)
Z
e2iτ ψ(x̃)
∂z (g3 e−A2 )(x̃)
1
−
(ν1 − iν2 )(eB1 +B2 − Ψ̃)d(z)
dσ
1
4τ Γ0
z̃ − z
|det Im Φ00 (x̃)| 2
Z
e−2iτ ψ(x̃)
1
∂z (g1 e−A1 )(x̃)
−
(ν1 + iν2 )(eA1 +A2 − Ψ)c(z̄)
dσ
1
4τ Γ0
z̃ − z
|det Im Φ00 (x̃)| 2
Z
1
e−2iτ ψ(x̃)
∂z̄ (g4 e−B2 )(x̃)
−
(ν1 + iν2 )(eA1 +A2 − Ψ)a(z)
dσ
1
4τ Γ0
z̃ − z
|det Im Φ00 (x̃)| 2
Z
1
e2iτ ψ(x̃)
∂z̄ (g2 e−B1 )(x̃)
−
(ν1 − iν2 )(eB1 +B2 − Ψ̃)b(z)
dσ
1
4τ Γ0
z̃ − z
|det Im Φ00 (x̃)| 2
1
+ o( ) as |τ | → +∞.
τ
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
47
Using Proposition 5.3 we take a = a∗ , d = d∗ and b = b∗ , c = c∗ such that
Z
1
1
B1 +B2
− Ψ̃)b∗ (z)
(ν1 −iν2 )(e
dσ = 0 and
(ν1 +iν2 )(eA1 +A2 −Ψ)a∗ (z)
dσ = 0.
z̃ − z
z̃ − z
Γ0
Γ0
Z
Then we obtain from (5.49)
(5.50)
((Q+ + (q1 − q2 ))a∗ b∗ )(e
x)e(A1 +B2 +2τ iImΦ)(ex)
1
|det ImΦ00 (e
x)| 2
= 0.
On the other hand using the Proposition 5.3 again we can take a = a∗ , d = d∗ and
b = b∗ , c = c∗ such that
Z
Z
1
1
B1 +B2
∗
dσ = 0 and
(ν1 +iν2 )(eA1 +A2 −Ψ)c∗ (z̄)
(ν1 −iν2 )(e
− Ψ̃)d (z̄)
dσ = 0.
z̃ − z
z̃ − z
Γ0
Γ0
These equalities and (5.49) imply
(5.51)
((Q− + (q1 − q2 ))d∗ c¯∗ )(e
x)e(B1 +A2 −2iτ ImΦ)(ex)
1
|det ImΦ00 (e
x)| 2
= 0.
Let x̃ be an arbitrary point from Ω. Consider the sequence of the functions Φ given by
Proposition 2.1. From (5.50) and (5.51) we have
((Q− + (q1 − q2 ))d∗ c¯∗ )(e
x ) = 0,
((Q+ + (q1 − q2 ))a∗ b∗ )(e
x ) = 0.
The proof of the theorem is completed. 6. Appendix I
Consider the Cauchy problem for the Cauchy-Riemann equations
∂φ
∂ψ ∂φ
∂ψ
(6.1) L(φ, ψ) = (
−
,
+
) = 0 in Ω, (φ, ψ) |Γ0 = (b1 (x), b2 (x)),
∂x1 ∂x2 ∂x2 ∂x1
∂l
(φ + iψ)(x̂j ) = c0,j , ∀j ∈ {1, . . . N } and ∀l ∈ {0, . . . , 5}.
∂z l
Here x̂1 , . . . x̂N be an arbitrary fixed points in Ω. We consider the pair b1 , b2 and complex
~ = (c0,1 , c1,1 , c2,1 , c3,1 , c4,1 , c5,1 . . . c0,N , c1,N , c2,N , c3,N , c4,N , c5,N ) as initial data for
numbers C
(6.1). The following proposition establishes the solvability of (6.1) for a dense set of Cauchy
data.
~ ∈ O,
Proposition 6.1. There exists a set O ⊂ (C 5 (Γ0 ))2 ×C6N such that for each (b1 , b2 , C)
5
2
5
2
6N
(6.1) has at least one solution (φ, ψ) ∈ (C (Ω)) and O = (C (Γ)) × C .
Proof. Denote B = (b1 , b2 ) an arbitrary element of the space C 7 (Γ0 ) × C 7 (Γ0 ). Consider the
following extremal problem
(6.2)
3
X
∂ k (φ, ψ) 4
J (φ, ψ) = k(φ, ψ) − Bk 274
+
k
k 274 −k
B4 (Γ0 )
∂ν k
B4
(∂Ω)
k=0
4
N X
5
X
4
1
∂k
+ ∆3 L(φ, ψ)L4 (Ω) +
| k (φ + iψ)(x̂j ) − ck,j |2 → inf,
∂z
j=1 k=0
48
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
(φ, ψ) ∈ W47 (Ω) × W47 (Ω).
(6.3)
Here Bkl denotes the Besov space of the corresponding orders.
For each > 0 there exists a unique solution to (6.2), (6.3) which we denote as (φb , ψb ).
This fact can be proved by standard arguments. We fix > 0. Denote by Uad the set of
admissible elements of the problem (6.2), (6.3), namely
Uad = {(φ, ψ) ∈ W47 (Ω) × W47 (Ω)|J (φ, ψ) < ∞}.
Denote Jˆ = inf (φ,ψ)∈W47 (Ω)×W47 (Ω) J (φ, ψ). Clearly the pair (0, 0) ∈ Uad . Therefore there exists
7
7
a minimizing sequence {(φk , ψk )}∞
k=1 ⊂ W4 (Ω) × W4 (Ω) such that
Jˆ = lim J (φk , ψk ).
k→+∞
Observe that the minimizing sequence is bounded in W47 (Ω) × W47 (Ω). Indeed, since the se27
3
−k
∂
4
3
4
quence {∆3 L(φk , ψk ), L(φk , ψk )|∂Ω , . . . , ∂ν
(∂Ω)
3 L(φk , ψk )|∂Ω } is bounded in L (Ω)×Πk=0 B4
p
the standard elliptic L -estimate implies that the sequence {L(φk , ψk )} is bounded in the
space W46 (Ω)×W46 (Ω). Taking into account that the sequence traces of the functions (φk , ψk )
27
27
is bounded in the Besov space B44 (∂Ω) × B44 (∂Ω) and applying the estimates for elliptic
operators one more time we obtain that {(φk , ψk )} bounded in W47 (Ω) × W47 (Ω). By the
Sobolev imbedding theorem the sequence {(φk , ψk )} is bounded in C 6 (Ω) × C 6 (Ω). Then
taking if necessary a subsequence, (which we denote again as {(φk , ψk )} ) we obtain
(φk , ψk ) → (φb , ψb ) weakly in W47 (Ω) × W47 (Ω),
27
27
∂ j φb ∂ j ψb
∂ j φk ∂ j ψk
−j
−j
4
4
(∂Ω)
×
B
(∂Ω) ∀j ∈ {0, 1, 2, 3},
,
)
→
(
,
)
weakly
in
B
4
4
j
j
j
j
∂ν
∂ν
∂ν ∂ν
∂k
(φ + iψ)(x̂j ) − ck,j → Ck,j, , k ∈ {0, . . . , 5},
∂z k
∆3 L(φk , ψk ) → r weakly in L4 (Ω) × L4 (Ω), L(φk , ψk ) → r̃ weakly in W46 (Ω) × W46 (Ω).
Obviously, r = ∆3 L(φb , ψb ), r̃ = L(φb , ψb ). Then, since the norms in the spaces L4 (Ω) and
(
27
B44
−k
(∂Ω) are lower semicontinuous with respect to weak convergence we obtain that
J (φb , ψb ) ≤ lim J (φk , ψk ) = Jˆ .
k→+∞
Thus the pair (φb , ψb ) is a solution to the extremal problem (6.2), (6.3). Since the set of an
admissible elements is convex and the functional J is strictly convex, this solution is unique.
By Fermat’s theorem we have
J0 (φb , ψb )[δ̃] = 0,
∀δ̃ ∈ W47 (Ω) × W47 (Ω).
This equality can be written in the form
(6.4)
IΓ0 0 , 27 ((φb , ψb )
4
− B)[δ̃] + 3
X
4
k=0
+
0
I∂Ω,
(
27
−k
∂k b b
∂k
(
φ
,
ψ
))[
δ̃] + (p , ∆3 Lδ̃)L2 (Ω)
∂ν k
∂ν k
N X
5
X
∂k
∂k
∂k
∂k
( k (φb + iψb )(x̂j ) − ck,j ) k (δ̃1 + iδ̃2 )(x̂j ) + ( k (φb + iψb )(x̂j ) − ck,j ) k (δ̃1 + iδ̃2 )(x̂j ) = 0,
∂z
∂z
∂z
∂z
j=1 k=0
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
49
∂ φ̂
∂ φ̂
where p = 4 ((∆3 ( ∂x
− ∂∂xψ̂2 ))3 , (∆3 ( ∂x
+ ∂∂xψ̂1 ))3 ) and IΓ0 ∗ ,κ (ŵ) denotes the derivative of the
1
2
functional w → kwk4B κ (Γ∗ ) at ŵ.
4
P P5
2
+ N
Observe that the pair J (φb , ψb ) ≤ J (0, 0) = kBk4 274
j=1
k=0 |ck,j | . This implies
B4 (Γ0 )
27
∂k b
b
that the sequence {(φb , ψb )} is bounded in B44 (Γ0 ), the sequences { ∂z
k (φ +iψ )(x̂j )−ck,j } are
P3
k
k
∂
0
bounded in C, the sequence k=0 I∂Ω,
( ∂ (φb , ψb ))[ ∂ν
27
k δ̃] converges to zero for any δ̃ from
−k ∂ν k
27
4
4
27
4
4
4
B4 (∂Ω) × B4 (Γ0 ). Then (6.4) implies that the sequence {p } is bounded in L 3 (Ω) × L 3 (Ω).
27
27
Therefore there exist B ∈ B44 (Γ0 ) × B44 (Γ0 ), C0,j , C1,j , . . . , C5,j ∈ C and p = (p1 , p2 ) ∈
4
4
L 3 (Ω) × L 3 (Ω) such that
(6.5)
27
27
4
4
(φb , ψb ) − B * B weakly in B44 (Γ0 ) × B44 (Γ0 ), p * p weakly in L 3 (Ω) × L 3 (Ω),
∂k b
(φ + iψb )(x̂j ) − ck,j * Ck,j k ∈ {0, 1, . . . , 5}, j ∈ {1, . . . , N }.
∂z k
Passing to the limit in (6.4) we obtain
(6.7)
N X
5
X
∂k
3
0
Ck,j k (δ̃1 + iδ̃2 )(x̂j ) = 0 ∀δ̃ ∈ W47 (Ω) × W47 (Ω).
IΓ0 , 27 (B)[δ̃] + (p, ∆ Lδ̃)L2 (Ω) + 2Re
4
∂z
j=1 k=0
(6.6)
Next we claim that
∆3 p = 0 in Ω \ ∪N
j=1 {x̂}
(6.8)
in the sense of distributions. Suppose that (6.8) is already proved. This implies
(p, ∆3 Lδ̃)L2 (Ω) + 2Re
N X
5
X
Ck,j
j=1 k=0
∂k
(δ̃1 + iδ̃2 )(x̂j ) = 0 ∀δ̃1 , δ̃2 ∈ C0∞ (Ω).
∂z k
If p = (p1 , p2 ), denoting P = p1 − ip2 , we have
3
Re (∆ P, ∂z̄ (δ̃1 + iδ̃2 ))
L2 (Ω)
+ Re
N X
5
X
j=1
∂k
Ck,j k (δ̃1 + iδ̃2 )(x̂j ) = 0 ∀δ̃1 , δ̃2 ∈ C0∞ (Ω).
∂z
k=0
ˆ
Since by (6.8) supp ∆3 P ⊂ ∪N
j=1 {x̂j } there exist some constants mβ,j and `j such that
P
P
ˆ
`
β
∆3 P = N
j=1
|β|=1 mβ,j D δ(x − x̂j ). The above equality can be written in the form
ˆ
`j
X
−
|β|=1
5
X
∂k
∂
(−1)k Ck,j k δ(x − x̂j ).
mβ,j Dβ δ(x − x̂j ) =
∂ z̄
∂z
k=0
From this we obtain
(6.9)
C0,j = C1,j = · · · = C5,j = 0 j ∈ {1, . . . , N }.
Therefore
(6.10)
∆3 p = 0 in Ω.
50
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
This implies
(p, ∆3 Lδ̃)L2 (Ω) = 0 ∀δ̃ ∈ W47 (Ω) × W47 (Ω),
Lδ̃|∂Ω =
∂Lδ̃
∂ 5 Lδ̃
|∂Ω = · · · =
|∂Ω = 0.
∂ν
∂ν 5
This equality and (6.7) yield
(6.11)
IΓ0 0 , 27 (B)[δ̃] = 0 ∀δ̃ ∈ W47 (Ω) × W47 (Ω),
4
Lδ̃|∂Ω =
∂Lδ̃
∂ 5 Lδ̃
|∂Ω = . . .
|∂Ω = 0.
∂ν
∂ν 5
Then using the trace theorem we conclude that B = 0. Using this and (6.5) we obtain
27
(6.12)
27
(φbk , ψbk ) − B * 0 weakly in B44 (Γ0 ) × B44 (Γ0 ).
From (6.6) and (6.9) we obtain
∂k b
(φ + iψb )(x̂) * ck,j
∂z k
k ∈ {0, 1, . . . , 5}, j ∈ {1, . . . , N }.
27
By the Sobolev embedding theorem B44 (Γ0 ) ⊂⊂ C 5 (Γ0 ). Therefore (6.12) implies
(6.13)
(φbk , ψbk ) − B → 0 in C 5 (Γ0 ) × C 5 (Γ0 ).
Let the pair (φ̃k , ψ̃k ) be a solution to the boundary value problem
(6.14)
L(φ̃k , ψ̃k ) = L(φ̂k , ψ̂k ) in Ω,
ψ̃k |∂Ω = ψ∗k .
Here ψ∗k is a smooth function such that ψ∗k |Γ0 = 0 and the pair (L(φ̂k , ψ̂k ), ψ∗k ) is orthogonal
to all solutions of the adjoint problem (see [21]). Moreover since L(φ̃k , ψ̃k ) → 0 in W46 (Ω) ×
W46 (Ω) we may assume ψ∗k → 0 in C 6 (∂Ω) × C 6 (∂Ω). Among all possible solutions to
problem
(6.14) (clearly there is no unique solution to this problem) we choose one such that
R
φ̃ dx = 0. Thus we obtain
Ω k
(6.15)
(φ̃k , ψ̃k ) → 0 in W47 (Ω) × W47 (Ω).
Therefore the sequence {(φbk − φ̃k , ψbk − ψ̃k )} represents the desired approximation to the
solution of the Cauchy problem (6.1).
Now we prove (6.8). Let x
e be an arbitrary point in Ω \ ∪N
e be a smooth
j=1 {x̂j } and let χ
N
function such that it is zero in some neighborhood of Γ0 ∪ ∪j=1 {x̂j } and the set A = {x ∈
Ω|e
χ(x) = 1} contains an open connected subset F such that x
e ∈ F and Γ̃ ∩ F is an open
set in ∂Ω. In addition we assume that Int(supp χ) is a simply connected domain. By (6.7)
we have
(6.16) 0 = (p, ∆3 L(e
χδ̃))L2 (Ω) = (e
χp, ∆3 Lδ̃)L2 (Ω) +(p, [∆3 L, χ
e]δ̃)L2 (Ω)
∀δ̃ ∈ W47 (Ω)×W47 (Ω).
Denote Lδ̃ = δ̂. Consider the functional mapping δ̂ ∈ W42 (supp χ
e) to (p, [∆3 L, χ
e]δ̃)L2 (Ω) ,
where
Z
Lδ̃ = δ̂ in Ω, Im δ̃|S = 0,
Re δ̃dx = 0,
supp χ
e
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
51
where S denotes the boundary of supp χ
e. For each δ̂ ∈ W42 (supp χ
e) × W42 (supp χ
e), there
3
3
e) × W4 (supp χ
e). Hence the functional is wellexists a unique solution δ̃ ∈ W4 (supp χ
4
2
defined
continuous on W4 (supp χ
e). Therefore there exists q, r, q0 ∈ L 3 (supp χ
e) such
R andP
2
∂2
3
that supp χe( j,k=1 rjk ∂xj ∂xk δ̂ + (q, δ̂) + q0 δ̂)dx = (p, [∆ L, χ
e]δ̃)L2 (supp χe) .
Consider the boundary value problem
∆3 P̃ = f˜ in supp χ,
P̃ |S =
∂ 2 P̃
∂ P̃
|S =
|S = 0.
∂ν
∂ν 2
P
2
Here f˜ = 2div (∇q̃) − q0 − 2j,k=1 ∂x∂j ∂xk rjk . A solution to this problem exists and is unique,
e) × W̊ 41 (supp χ
e). On the other hand, thanks
since f˜ ∈ (W̊ 2 (supp χ
e))0 . Then P ∈ W̊ 41 (supp χ
4
3
3
to (6.16), P = χ̃p ∈ W̊ 41 (supp χ
e) × W̊ 41 (supp χ
e).
3
3
Next we take another smooth cut off function χ
e1 such that supp χ
e1 ⊂ A and Int (supp χ1 )
is a simply connected domain. A neighborhood of x
e belongs to A1 = {x|e
χ1 = 1}, the interior
of A1 is connected, and Int A1 ∩ Γ̃ contains an open subset O in ∂Ω. Similarly to (6.16) we
have
(e
χ1 p, ∆3 Lδ̃)L2 (Ω) − (p, [∆3 L, χ
e1 ]δ̃)L2 (Ω) = 0 ∀δ̃ ∈ W47 (Ω) × W47 (Ω).
This equality implies that χ
e1 p ∈ W 42 (Ω), using a similar argument to the one above.
3
Next we take another smooth cut off function χ
e2 such that supp χ
e2 ⊂ A2 and Int (supp χ2 )
is a simply connected domain. A neighborhood of x
e belongs to A3 = {x|e
χ2 = 1}, the interior
of A1 is connected, and Int A3 ∩ Γ̃ contains an open subset O in ∂Ω. Similarly to (6.16) we
have
(e
χ2 p, ∆3 Lδ̃)L2 (Ω) − (p, [∆3 L, χ
e2 ]δ̃)L2 (Ω) = 0 ∀δ̃ ∈ W47 (Ω) × W47 (Ω).
This equality implies that χ
e2 p ∈ W 43 (Ω), using a similar argument to the one above. Let
3
ω be a domain such that ω ∩ Ω = ∅, ∂ω ∩ ∂Ω ⊂ O contains an open set in ∂Ω.
We extend p on ω by zero. Then
(∆3 (e
χ2 p), Lδ̃)L2 (Ω∪ω) + (p, [∆3 L, χ
e2 ]δ̃)L2 (Ω∪ω) = 0.
Hence, since [∆3 L, χ
e2 ]|A1 = 0 we have
L∗ ∆3 (e
χ2 p) = 0 in Int A2 ∪ ω,
By Holmgren’s theorem ∆3 (e
χ2 p)|Int
A1
p|ω = 0.
= 0, that is, (∆3 p)(e
x) = 0. Thus (6.8) is proved. Consider the Cauchy problem for the Cauchy-Riemann equations
(6.17)
L(φ, ψ) = (
∂φ
∂ψ ∂φ
∂ψ
−
,
+
) = 0 in Ω,
∂x1 ∂x2 ∂x2 ∂x1
(φ, ψ) |Γ0 = (b(x), 0),
∂l
(φ + iψ)(x̂j ) = c0,j , ∀j ∈ {1, . . . N } and ∀l ∈ {0, . . . , 5}.
∂z l
Here x̂1 , . . . x̂N be an arbitrary fixed points in Ω. We consider the function b and complex
~ = (c0,1 , c1,1 , c2,1 , c3,1 , c4,1 , c5,1 . . . c0,N , c1,N , c2,N , c3,N , c4,N , c5,N ) as initial data for
numbers C
(6.1). The following proposition establishes the solvability of (6.1) for a dense set of Cauchy
data.
52
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
~ ∈ O, problem
Corollary 6.1. There exists a set O ⊂ C 5 (Γ0 ) × C6N such that for each (b, C)
5
2
5
6N
(6.17) has at least one solution (φ, ψ) ∈ (C (Ω)) and O = C (Γ) × C .
The proof of Corollary 6.1 is similar to the proof of Proposition 6.1. The only difference
is that instead of extremal problem considered there we have to use the following extremal
problem
3
X
∂ k (φ, ψ) 4
J (φ, ψ) = kφ − bk 274
+
k
k 274 −k
k
B4 (Γ0 )
∂ν
(∂Ω)
B4
k=0
4
N X
5
X
4
1
∂k
3
+ ∆ L(φ, ψ) L4 (Ω) +
| k (φ + iψ)(x̂j ) − ck,j |2 → inf,
∂z
j=1 k=0
(φ, ψ) ∈ W47 (Ω) × W47 (Ω),
ψ|Γ0 = 0.
We have
Proposition 6.2. Let α ∈ (0, 1), g1 , . . . , gp be linearly independent functions in L2 (Γ0 ),
y1 , . . . , yk̂ be some arbitrary points from Γ0 , yk̂+1 , . . . , yk̃ be some arbitrary points from Ω
and x̃ be an arbitrary point from Ω \ {yk̂+1 , . . . , yk̃ }. Then there exist a holomorphic function
a ∈ C 5+α (Ω̄) and an antiholomorphic function d ∈ C 5+α (Ω̄) such that (aeA + deB )|Γ0 = 0,
Z
∂ k+j a
agp dσ = 0 p ∈ {1, . . . , m};
(y` ) = 0 k + j ≤ 5, ∀` ∈ {1, . . . , k̃},
∂xk1 ∂xj2
Γ0
and
a(x̃) 6= 0 and d(x̂) 6= 0.
Proof. Consider the operator
Z
Z
∂5a
R(γ) = (
ag1 dσ, . . . ,
agp dσ, a(y1 ), . . . 5 (y1 ), . . .
∂z
Γ0
Γ0
∂5a
∂5d
∂5d
(y
),
d(y
),
.
.
.
(y
),
.
.
.
,
d(y
),
.
.
.
(y ), a(x̃), d(x̃)).
1
1
k̃
∂z 5 k̃
∂ z̄ 5
∂ z̄ 5 k̃
Here γ ∈ C0∞ (Γ̃) and the functions a and d are solutions to the following problem
a(yk̃ ), . . . ,
∂a
∂d
= 0, in Ω,
= 0 in Ω, (aeA + deB )|∂Ω = γ.
∂ z̄
∂z
Consider the image of the operator R. Clearly it is closed. Let us show that the point
R(0, . . . , 0, 1, 1) belongs
R to the image of the operator R. Let a holomorphic function a satisfy
ag1 dσ = · · · = Γ0 agp dσ = 0 and
Γ0
∂β a
(yj ) = 0 ∀|β| ∈ {0, . . . , 5},
∂x1 β1 ∂x2 β2
j ∈ {1, . . . , k̃},
|a(x̃)| > 2.
Consider the function −eA−B a(z) and the pair (b1 , b2 ) = (Re{−eA−B a}, Im{eA−B a}). Using
Proposition 6.1 we solve problem (6.1) with l = 0 approximately. Let (φ , ψ ) be a sequence
of functions such that
∂
(φ + iψ ) = 0 in Ω, (φ , ψ )|Γ0 → (b1 , b2 ) in C 5+α (Γ0 ), (φ + iψ )(x̃) → 1.
∂z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
53
Denote d = φ − iψ , β = aeA + d eB . Then the sequence {β } converges to zero in the
space C 5+α (Γ0 ).
By Proposition 2.5 there exists a solution to problem (2.46) with the initial data β , which
we denote as {ã , d˜ } such that the sequence {ã , d˜ } converges to zero in (C 5 (Ω))2 . Denote
by γ = (a + ã , d + d˜ )|Γ0 . Clearly R(γ ) converges to (0, . . . , 0, 1, 1). The proof of the
proposition is completed.
7. Appendix II. Asymptotic formulas
Proposition 7.1. Under the conditions of Theorem 1.1 for any point x on the boundary of
Ω we have
1
−
π
∂z g̃1 (x̃)
Z
Ω
1
e1 g̃1 e−τ (Φ(ζ)−Φ(ζ))
1 ∂ 2 Φ(x̃)
e−2iτ ψ(x̃)
dξ1 dξ2 = 2 z
1 + o( 2 ) +
2
ζ −z
2τ (z̃ − z) |det Im Φ00 (x̃)| 2
τ
(7.1)
1
4τ 2
(7.2)
1
−
π
∂z g̃1 (x̃) ∂z3 Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃)
+
∂z̄2 g̃1 (x̃)
∂z2 Φ(x̃)
−
∂z2 g̃1 (x̃)
∂z2 Φ(x̃)
Ω
|detIm Φ00 (x̃)| 2
∂z̄ g̃2 (x̃)
e2iτ ψ(x̃)
e1 g̃2 eτ (Φ(ζ)−Φ(ζ))
1 ∂ 2 Φ(x̃)
dξ1 dξ2 = 2 z
2τ (z̃ − z̄)2 |det Im Φ00 (x̃)| 12
ζ̄ − z̄
3
2
∂z̄ g̃2 (x̃)
∂z̄ g̃2 (x̃) ∂z̄ Φ(x̃)
1 − ∂z2 Φ(x̃) ∂z̄2 Φ(x̃) + ∂z2 Φ(x̃) −
− 2
4τ
z̃ − z̄
(7.3)
Ω
Z
+ o(
1
2
1
)
τ2
|det Im Φ00 (x̃)|
as |τ | → +∞.
Ω
∂z̄ g̃3 (x̃) ∂z̄3 Φ(x̃)
∂z2 Φ(x̃) ∂z̄2 Φ(x̃)
−
∂z̄2 g̃3 (x̃)
∂z2 Φ(x̃)
+
∂z2 g̃3 (x̃)
∂z2 Φ(x̃)
e−2iτ ψ(x̃)
1
2
z̃ − z̄
+ o(
1
)
τ2
|det Im Φ00 (x̃)|
as |τ | → +∞.
∂z g̃4 (x̃)
1 ∂ 2 Φ(x̃)
e1 g̃4 eτ (Φ(ζ)−Φ(ζ))
e2iτ ψ(x̃)
dξ1 dξ2 = − 2 z
ζ −z
2τ (z̃k − z)2 |det Im Φ00 (x̃)| 12
1
− 2
4τ
(7.4)
e2iτ ψ(x̃)
e1 g̃3 e−τ (Φ(ζ)−Φ(ζ))
1 ∂ 2 Φ(x̃)
e−2iτ ψ(x̃)
dξ1 dξ2 = − 2 z
2τ (z̃ − z̄)2 |det Im Φ00 (x̃)| 12
ζ̄ − z̄
1
− 2
4τ
1
−
π
∂z2 g̃2 (x̃)
∂z2 Φ(x̃)
∂z̄ g̃3 (x̃)
Z
1
−
π
as |τ | → +∞.
1
z̃ − z
Z
e−2iτ ψ(x̃)
∂z̄ g̃4 (x̃) ∂z3 Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃)
+
∂z̄2 g̃4 (x̃)
∂z2 Φ(x̃)
z̃ − z
−
∂z2 g̃4 (x̃)
∂z2 Φ(x̃)
e2iτ ψ(x̃)
1
2
+ o(
1
)
τ2
|det Im Φ00 (x̃)|
as |τ | → +∞.
54
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Proof. Let δ > 0 be a sufficiently small number and ẽ ∈ C0∞ (B(e
x, δ)), ẽ|B(ex,δ/2) ≡ 1. We
compute the asymptotic formulae of the following integral as |τ | goes to infinity.
1
−
π
e1 g̃1 e−τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
ζ −z
Z
Ω
ẽg̃1 e−τ (Φ(ζ)−Φ(ζ))
1
dξ1 dξ2 + o( 2 ) =
ζ −z
τ
B(x̃,δ)
(
Z
∂z g̃1 (x̃)(ζ − z̃) + 12 ∂z̄2 g̃1 (x̃)(ζ̄ − z̃)2
1
−
ẽ
π B(x̃,δ)
ζ −z
)
∂z̄ ∂z g̃1 (x̃)(ζ − z̃)(ζ̄ − z̃) + 12 ∂z2 g̃1 (x̃)(ζ − z̃)2
1
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =
+
ζ −z
τ
−
1
π
Z
1
−
π
+
ẽ
B(x̃,δ)
z
1 ∂z̄2 g̃1 (x̃)
∂ Φ(ζ̄
2 ∂z2 Φ(x̃) ζ̄
− z̃)

z
z
ζ −z
Z
ẽ
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o(


1 (x̃)
 ∂∂z2g̃Φ(x̃)
(∂ζ Φ −
z
1 ∂z3 Φ(x̃)
∂ Φ(ζ
2 ∂z2 Φ(x̃) ζ
− z̃)) +
ζ −z

∂z̄ ∂z g̃1 (x̃)
1 ∂z2 g̃1 (x̃)

∂
Φ(ζ
−
z̃)
(ζ
−
z̃)∂
Φ
+
ζ
2
ζ̄
2 ∂ Φ(x̃)
∂ 2 Φ(x̃)
B(x̃,δ)
1 ∂z̄2 g̃1 (x̃)
∂ Φ(ζ̄
2 ∂z2 Φ(x̃) ζ̄
1
)=
τ2
− z̃)

z
z
ζ −z
1
−
πτ
+

1 (x̃)
 ∂∂z2g̃Φ(x̃)
(∂ζ Φ − 12 ∂z3 Φ(x̃)(ζ − z̃)2 ) +
ζ −z

∂z̄ ∂z g̃1 (x̃)
1 ∂z2 g̃1 (x̃)

Φ
+
(ζ
−
z̃)∂
∂
Φ(ζ
−
z̃)
ζ
2
ζ̄
2 ∂ Φ(x̃)
∂ 2 Φ(x̃)
1
−
π
+
Z
Z
ẽ


3
1 (x̃) ∂z Φ(x̃)
 1 − ∂∂z2g̃Φ(x̃)
−
∂ 2 Φ(x̃)
z
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o(
z
∂z̄2 g̃1 (x̃)
∂z2 Φ(x̃)
−
∂z g̃1 (x̃)
(1
∂z2 Φ(x̃)
1 ∂z3 Φ(x̃)
(ζ
2 ∂z2 Φ(x̃)
(ζ − z)2
−
− z̃))
ζ −z

∂z2 g̃1 (x̃)
∂z2 g̃1 (x̃)

(ζ
−
z̃)
1 ∂z2 Φ(x̃)
1 ∂z2 Φ(x̃)
1
−
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ).
2

2 ζ −z
2 (ζ − z)
τ
B(x̃,δ)
2
(7.5)
Using stationary phase we obtain
Z
ẽ
B(x̃,δ)
(ζ − z̃) −τ (Φ(ζ)−Φ(ζ))
1
e
dξ1 dξ2 = o( ) as |τ | → +∞.
2
(ζ − z)
τ
Another asymptotic calculation is
1
)=
τ2
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
1
−
π
Z
Ω
∂z g̃1 (x̃)
Z
e1 g̃1 eτ (Φ(ζ)−Φ(ζ))
1
1
∂ 2 Φ(x̃)
dξ1 dξ2 =
ẽ z
e−τ (Φ(ζ)−Φ(ζ)) dx + o( 2 ) +
2
ζ −z
πτ B(x̃,δ) (ζ − z)
τ
2
3
2
∂z g̃1 (x̃) ∂z Φ(x̃)
1 (x̃)
1 (x̃)
Z
+ ∂∂z̄2g̃Φ(x̃)
− ∂∂z2g̃Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃)
1
z
z
ẽ
e−τ (Φ(ζ)−Φ(ζ)) dx =
2πτ B(x̃,δ)
(ζ − z)
∂z g̃1 (x̃)
1 ∂ 2 Φ(x̃)
e−2iτ ψ(x̃)
1
+ 2 z
+
o(
)+
1
2τ (z̃ − z)2 |det Im Φ00 (x̃)| 2
τ2
(7.6)
1
4τ 2
∂z g̃1 (x̃) ∂z3 Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃)
+
∂z̄2 g̃1 (x̃)
∂z2 Φ(x̃)
(z̃ − z)
−
∂z2 g̃1 (x̃)
∂z2 Φ(x̃)
e−2iτ ψ(x̃)
1
|detIm Φ00 (x̃)| 2
as |τ | → +∞.
55
56
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Next we compute the asymptotics of the following integral as |τ | goes to infinity.
1
−
π
−
−
+
−
+
−
+
−
−
Z
Ω
e1 g̃2 eτ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
ζ̄ − z̄
ẽg̃2 eτ (Φ(ζ)−Φ(ζ))
1
dξ1 dξ2 + o( 2 ) =
τ
ζ̄ − z̄
B(x̃,δ)
(
Z
∂z̄ g̃2 (x̃)(ζ − z̃) + 12 ∂z̄2 g̃2 (x̃)(ζ̄ − z̃)2
1
ẽ
π B(x̃,δ)
ζ̄ − z̄
)
∂z̄ ∂z g̃2 (x̃)(ζ − z̃)(ζ̄ − z̃) + 12 ∂z2 g̃2 (x̃)(ζ − z̃)2
1
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =
τ
ζ̄ − z̄

2
Z
 ∂z̄2g̃2 (x̃) (∂ζ̄ Φ − 21 ∂z̄3 Φ(x̃)(ζ − z̃k )2 ) + 12 ∂z̄2g̃2 (x̃) ∂ζ̄ Φ(ζ − z̃)
1
∂z Φ(x̃)
∂z Φ(x̃)
ẽ
π B(x̃,δ) 
ζ − z̄

∂z̄ ∂z g̃2 (x̃)
1 ∂z2 g̃2 (x̃)

z̃)∂
Φ
+
(
ζ̄
−
∂
Φ(x̃)(ζ
−
z̃)
ζ
1
∂z2 Φ(x̃)
2 ∂z2 Φ(x̃) ζ
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =

τ
ζ̄ − z̄

3
2
Z
 ∂z̄2g̃2 (x̃) (∂ζ̄ Φ − 21 ∂z̄2 Φ(x̃) ∂ζ̄ Φ(ζ̄ − z̃)) + 12 ∂z̄2g̃2 (x̃) ∂ζ̄ Φ(ζ̄ − z̃)
1
∂
Φ(x̃)
∂z Φ(x̃)
∂z Φ(x̃)
z̄
ẽ
π B(x̃,δ) 
ζ̄ − z̄

∂z̄ ∂z g̃2 (x̃)
1 ∂z2 g̃2 (x̃)

(
ζ̄
−
∂
Φ(ζ
−
z̃)
z̃)∂
Φ
+
ζ
1
∂z2 Φ(x̃)
2 ∂z2 Φ(x̃) ζ
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =

τ
ζ − z̄

3
∂z̄ g̃2 (x̃)
1 ∂z̄2 g̃2 (x̃)
1 ∂z̄3 Φ(x̃)
Z
 − 21 ∂z̄2g̃2 (x̃) ∂∂z̄2 Φ(x̃)
+
(1
−
(ζ̄ − z̃))
2
2
2
2
1
∂z̄2 Φ(x̃)
∂z̄ Φ(x̃) z̄ Φ(x̃)
∂z̄ Φ(x̃)
∂ Φ(x̃)
ẽ
− z
πτ B(x̃,δ) 
ζ̄ − z̄
(ζ̄ − z̄)2

∂z̄2 g2 (x̃)
∂z2 g̃2 (x̃) 
ζ̄
−
(
z̃)
1 ∂z2 Φ(x̃)
1 ∂z2 Φ(x̃)
1
−
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ).
2

2 (ζ̄ − z̄)
2 ζ̄ − z̄
τ
1
π
Z
(7.7)
Observe that by the stationary phase argument
Z
ẽ
B(x̃,δ)
(ζ − z̃) τ (Φ(ζ)−Φ(ζ))
1
e
dξ
1 dξ2 = o( ) as |τ | → +∞.
τ
(ζ − z)2
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
57
Finally we have
1
−
π
1
e1 g̃2 eτ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
πτ
ζ̄ − z̄
Z
1
−
πτ
Ω
Z
ẽ
2 (x̃)
− 12 ∂∂z̄2g̃Φ(x̃)
z
∂z̄3 Φ(x̃)
∂z̄2 Φ(x̃)
+
1
2
Z
ẽ
B(x̃,δ)
∂z̄2 g̃2 (x̃)
∂z2 Φ(x̃)
−
∂z̄ g̃2 (x̃)
∂z2 Φ(x̃)
(ζ̄ −
z̄)2
1 ∂z2 g̃2 (x̃)
2 ∂z2 Φ(x̃) τ (Φ(ζ)−Φ(ζ))
e
ζ̄ − z̄
B(x̃,δ)
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2
dξ1 dξ2 + o(
1
)=
τ2
∂z̄ g̃2 (x̃)
e2iτ ψ(x̃)
1 ∂z2 Φ(x̃)
+
2τ 2 (z̃ − z̄)2 |det Im Φ00 (x̃)| 12
3
2
∂z̄ g̃2 (x̃)
∂z̄ g̃2 (x̃) ∂z̄ Φ(x̃)
1 − ∂z2 Φ(x̃) ∂z̄2 Φ(x̃) + ∂z2 Φ(x̃) −
−
4τ 2
z̃ − z̄
∂z2 g̃2 (x̃)
∂z2 Φ(x̃)
e2iτ ψ(x̃)
|det Im Φ00 (x̃)|
1
2
+ o(
1
) as |τ | → +∞.
τ2
We compute the asymptotic of the following integral as |τ | goes to infinity.
1
−
π
−
−
+
−
+
−
+
Z
Ω
e1 g̃4 eτ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
ζ −z
ẽg̃4 eτ (Φ(ζ)−Φ(ζ))
1
dξ1 dξ2 + o( 2 ) =
ζ −z
τ
B(x̃,δ)
(
Z
∂z g̃4 (x̃)(ζ − z̃) + 21 ∂z̄2 g̃4 (x̃)(ζ̄ − z̃)2
1
ẽ
π B(x̃,δ)
ζ −z
)
∂z̄ ∂z g̃4 (x̃)(ζ − z̃)(ζ̄ − z̃) + 12 ∂z2 g̃4 (x̃)(ζ − z̃)2
1
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =
ζ −z
τ

2 (x̃)
4
4 (x̃)
Z
 ∂∂z2g̃Φ(x̃)
(∂ζ Φ − 12 ∂z3 Φ(x̃)(ζ − z̃)2 ) + 12 ∂∂z̄2g̃Φ(x̃)
∂ζ̄ Φ(ζ̄ − z̃)
1
z
z
ẽ
π B(x̃,δ) 
ζ −z

∂z̄ ∂z g̃4 (x̃)
1 ∂z2 g̃4 (x̃)

(ζ
−
z̃)∂
Φ
+
∂
Φ(ζ
−
z̃)
ζ
2
ζ̄
2 ∂z Φ(x̃)
1
∂z2 Φ(x̃)
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =

ζ −z
τ

2 (x̃)
3
4
4 (x̃)
Z
 ∂∂z2g̃Φ(x̃)
(∂ζ Φ − 12 ∂∂z2 Φ(x̃)
∂ζ Φ(ζ − z̃)) + 21 ∂∂z̄2g̃Φ(x̃)
∂ζ̄ Φ(ζ̄ − z̃)
1
z
z Φ(x̃)
z
ẽ
π B(x̃,δ) 
ζ −z

∂z̄ ∂z g̃4 (x̃)
1 ∂z2 g̃4 (x̃)

(ζ
−
z̃)∂
Φ
+
∂
Φ(ζ
−
z̃)
ζ
2
ζ̄
2 ∂z Φ(x̃)
1
∂z2 Φ(x̃)
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =

ζ −z
τ
1
π
Z
58
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Z
1
−
πτ
ẽ
B(x̃,δ)
1 ∂z2 g̃4 (x̃)
2 ∂z2 Φ(x̃)
−
ζ −z

3
4 (x̃) ∂z Φ(x̃)
 12 ∂∂z̄2g̃Φ(x̃)
+
∂z Φ(x̃)
z
ζ −z

1 ∂z2 g̃4 (x̃)
(ζ − z̃) 
2 ∂ 2 Φ(x̃)
1 ∂z̄2 g̃4 (x̃)
2 ∂z2 Φ(x̃)
+
∂z g̃4 (x̃)
(1
∂z2 Φ(x̃)

+
z
(ζ −
z)2
1 ∂z3 Φ(x̃)
(ζ
2 ∂z2 Φ(x̃)
(ζ − z)2
−
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o(

− z̃))
1
).
τ2
(7.8)
Observe that
Z
ẽ
B(x̃,δ)
(ζ − z̃) τ (Φ(ζ)−Φ(ζ))
1
e
dξ1 dξ2 = o( ) as |τ | → +∞.
2
(ζ − z)
τ
Finally we have
e1 g̃4 eτ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
ζ −z
Ω

3
1 ∂z̄2 g̃4 (x̃)
4 (x̃) ∂z Φ(x̃)
Z
 21 ∂∂z̄2g̃Φ(x̃)
−
2 Φ(x̃) + 2 2
∂
1
∂z Φ(x̃)
z
z
−
ẽ
ζ −z
πτ B(x̃,δ) 
∂z g̃4 (x̃) )
1
∂z2 Φ(x̃)
eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =
+
2
(ζ − z)
τ
1
−
π
Z
1
−
4τ 2
∂z̄ g̃4 (x̃) ∂z3 Φ(x̃)
∂z2 Φ(x̃) ∂z2 Φ(x̃)
+
∂z̄2 g̃4 (x̃)
∂z2 Φ(x̃)
z̃ − z
−
∂z2 g̃4 (x̃)
∂z2 Φ(x̃)
1 ∂z2 g̃4 (x̃)
2 ∂z2 Φ(x̃)
e2iτ ψ(x̃)
1
|det Im Φ00 (x̃)| 2
∂z g̃4 (x̃)
e2iτ ψ(x̃)
1 ∂z2 Φ(x̃)
1
−
as |τ | → +∞.
1 + o( 2 )
2
2
2τ (z̃ − z) |det Im Φ00 (x̃)| 2
τ
(7.9)
We compute the asymptotic of the following integral as |τ | goes to infinity:
1
−
π
Z
Ω
e1 g̃3 e−τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
ζ̄ − z̄
1
ẽg̃3 e−τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 + o( 2 ) =
τ
ζ̄ − z̄
B(x̃,δ)
(
Z
∂z̄ g̃3 (x̃)(ζ − z̃) + 12 ∂z̄2 g̃3 (x̃)(ζ̄ − z̃)2
1
−
ẽ
π B(x̃,δ)
ζ̄ − z̄
)
∂z̄ ∂z g̃3 (x̃)(ζ − z̃)(ζ̄ − z̃) + 12 ∂z2 g̃3 (x̃)(ζ − z̃)2
1
+
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =
τ
ζ̄ − z̄
−
1
π
Z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
−
+
1
π
ẽ
B(x̃,δ)
z
1 ∂z̄2 g̃3 (x̃)
∂ Φ(ζ
2 ∂z2 Φ(x̃) ζ̄
− z̃)

z
z
ζ̄ − z̄
Z
ẽ
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o(


 ∂z̄2g̃3 (x̃) (∂ζ̄ Φ −
∂z Φ(x̃)
1 ∂z̄3 Φ(x̃)
∂ Φ(ζ̄
2 ∂z̄2 Φ(x̃) ζ̄
− z̃)) +
ζ̄ − z̄

∂z̄ ∂z g̃3 (x̃)
1 ∂z2 g̃3 (x̃)

ζ̄
−
z̃)∂
Φ
+
(
∂
Φ(ζ
−
z̃)
ζ
∂ 2 Φ(x̃)
2 ∂ 2 Φ(x̃) ζ
B(x̃,δ)
1 ∂z̄2 g3 (x̃)
∂ Φ(ζ̄
2 ∂z2 Φ(x̃) ζ̄
1
)=
τ2
− z̃)

z
z
ζ − z̄
1
−
πτ
+

 ∂z̄2g̃3 (x̃) (∂ζ̄ Φ − 12 ∂z̄3 Φ(x̃)(ζ − z̃)2 ) +
∂ Φ(x̃)
ζ − z̄

∂z̄ ∂z g̃3 (x̃)
1 ∂z2 g̃3 (x̃)
(ζ̄ − z̃)∂ζ Φ + 2 ∂ 2 Φ(x̃) ∂ζ Φ(ζ − z̃) 
∂ 2 Φ(x̃)
1
−
π
+
Z
59
Z
ẽ


3
 21 ∂z̄2g̃3 (x̃) ∂z̄2 Φ(x̃) −
∂ Φ(x̃) ∂ Φ(x̃)
z
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o(
z̄
1 ∂z̄2 g̃3 (x̃)
2 ∂z2 Φ(x̃)
+
∂z̄ g̃3 (x̃)
(1
∂z2 Φ(x̃)
−
1 ∂z̄3 Φ(x̃)
(ζ̄
2 ∂z̄2 Φ(x̃)
1
)=
τ2
− z̃))
(ζ̄ − z̄)2
ζ̄ − z̄

∂z̄2 g3 (x̃)(ζ̄−z̃) 
1 ∂z2 g̃3 (x̃)
1
1 ∂z̄2 Φ(x̃)
2 ∂z2 Φ(x̃)
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ).
+
2
2 (ζ̄ − z̄) 
τ
ζ̄ − z̄
B(x̃,δ)

(7.10)
Using stationary phase we have
Z
ẽ
B(x̃,δ)
(ζ − z̃) −τ (Φ(ζ)−Φ(ζ))
1
e
dξ1 dξ2 = o( ) as |τ | → +∞.
2
τ
(ζ − z)
Finally we have
e1 g̃3 e−τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
ζ̄ − z̄
Ω

3
2
∂z̄ g̃3 (x̃)
Z
 12 ∂z̄2g̃3 (x̃) ∂z̄2 Φ(x̃) − 12 ∂z̄2g̃3 (x̃)
1
∂z Φ(x̃) ∂z̄ Φ(x̃)
∂z Φ(x̃)
∂ 2 Φ(x̃)
ẽ
+ z
−
πτ B(x̃,δ) 
ζ̄ − z̄
(ζ̄ − z̄)2

1 ∂z2 g̃3 (x̃) 
1
2 ∂z2 Φ(x̃)
+
e−τ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 + o( 2 ) =
τ
ζ̄ − z̄ 
1
−
π
Z
1
−
4τ 2
∂z̄ g̃3 (x̃) ∂z̄3 Φ(x̃)
∂z2 Φ(x̃) ∂z̄2 Φ(x̃)
−
∂z̄2 g̃3 (x̃)
∂z2 Φ(x̃)
z̃ − z̄
+
∂z2 g̃3 (x̃)
∂z2 Φ(x̃)
e−2iτ ψ(x̃)
1
|det Im Φ00 (x̃)| 2
∂z̄ g̃3 (x̃)
1
1 ∂z2 Φ(x̃)
e−2iτ ψ(x̃)
−
as |τ | → +∞.
1 + o( 2 )
2
2
00
2τ (z̃ − z̄) |det Im Φ (x̃)| 2
τ
60
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Proposition 7.2. For any x from the boundary of Ω, the following asymptotic formulae
holds true as |τ | goes to +∞.
!
1 ∂(g1 e−A1 )
∂Φ
∂Φ
−2iτ ψ(x̃)
q
(x̃)
+
m
1
e
1
1 ∂z
1
8
∂z
∂z
+ o( ),
(7.11)
G1 (x, τ ) = −
+
1
2
00
τ |det Im Φ (x̃)| 2
z̃ − z
(z̃ − z)
τ
(7.12)
1
e2iτ ψ(x̃)
G2 (x, τ ) = −
τ |det Im Φ00 (x̃)| 12
1 ∂(g2 e−B1 )
(x̃)
8
∂ z̄
(7.13)
1
e−2iτ ψ(x̃)
G3 (x, τ ) =
τ |det Im Φ00 (x̃)| 12
1 ∂(g4 e−B2 )
(x̃)
8
∂ z̄
(7.14)
1
e2iτ ψ(x̃)
G4 (x, τ ) = −
τ |det Im Φ00 (x̃)| 21
+
∂Φ
m̃1
∂ z̄
+
∂Φ
t
∂ z̄ 1
z̃ − z
z̃ − z
1 ∂(g3 e−A2 )
(x̃)
8
∂z
z̃ − z
+
q̃1 ∂Φ
∂ z̄
+
(z̃ − z)2
r1 ∂Φ
∂ z̄
+
(z̃ − z)2
∂Φ
t̃
∂z 1
!
!
r̃1 ∂Φ
∂z
+
(z̃ − z)2
1
+ o( ),
τ
1
+ o( ),
τ
!
1
+ o( ).
τ
Here z̃ = x̃1 + ix̃2 . Moreover the following asymptotic formula holds true
(7.15)
∂G1 (·, τ )
∂G3 (·, τ )
∂G4 (·, τ )
1
∂G2 (·, τ )
k
kC(Ω) +k
kC(Ω) +k
kC(Ω) +k
kC(Ω) = o( ) as |τ | → +∞.
∂z
∂ z̄
∂ z̄
∂ z̄
τ
Proof. Observe that the functions G1 (x, τ ), . . . , G4 (x, τ ) are given by
Z ( ∂A1 (ζ, ζ̄) + τ ∂Φ (ζ)) − ( ∂A1 (z, z̄) + τ ∂Φ (z))
1
∂ζ
∂ζ
∂z
∂z
G1 (x, τ ) = −
(e1 g1 e−A1 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 ,
2π Ω
ζ −z
Z ∂B1 (ζ, ζ) + τ ∂Φ (ζ) − ( ∂B1 (z, z) + τ ∂Φ (z))
1
∂z
∂z
∂ζ
∂ζ
(e1 g2 e−B1 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 ,
G2 (x, τ ) = −
2π Ω
ζ −z
Z (τ ∂Φ(ζ̄) − ∂A2 (ζ,ζ̄) ) − (τ ∂Φ(z̄) − ∂A2 (z,z̄) )
1
∂ z̄
∂ z̄
∂ ζ̄
∂ ζ̄
(e1 g4 e−A2 )(ξ1 , ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 ,
G3 (x, τ ) =
2π Ω
ζ̄ − z̄
Z (τ ∂Φ(ζ) − ∂B2 (ζ,ζ̄) ) − (τ ∂Φ(z) − ∂B2 (z,z̄) )
1
∂ζ
∂ζ
∂z
∂z
(e1 g3 e−B2 )(ξ1 .ξ2 )eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 .
G4 (x, τ ) =
2π Ω
ζ −z
These explicit formulae and the stationary phase argument imply (7.15). Let z = x1 + ix2
where x = (x1 , x2 ) ∈ ∂Ω. The following asymptotic is valid.
Z
Z
τ
∂
∂ζ Φ(ζ)
1
1
−A1 τ (Φ(ζ)−Φ(ζ))
(e1 g1 e
)e
(e1 g1 e−A1 ) eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 =
dξ1 dξ2 = −
2π Ω ζ − z
2π Ω ζ − z
∂ζ
Z
1
∂
1
1
e−2iτ ψ(x̃)
∂z (g1 e−A1 )(x̃)
(e1 g1 e−A1 ) eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 =
.
1
2π Ω ∂ζ ζ − z
8τ |det Im Φ00 (x̃)| 2
z̃ − z
τ
2π
Z
Ω
Z
∂ζ̄ Φ
1
∂
1
−B1 τ (Φ(ζ)−Φ(ζ))
(e1 g2 e )e
dξ1 dξ2 = −
(e1 g2 e−B1 ) eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 =
2π Ω ζ̄ − z̄
ζ̄ − z̄
∂ζ
Z
2iτ ψ(x̃)
1
∂
1
1
e
∂z̄ (g2 e−B1 )(x̃)
(
(e1 g2 e−B1 ))eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 =
2π Ω ∂ζ ζ̄ − z̄
8τ |det Im Φ00 (x̃)| 12
z̃ − z
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
τ
2π
61
Z
Z
∂ζ Φ(ζ)
1
1
−A2 τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
(e1 g3 e
)e
(e1 g3 e−A2 )∂ζ eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 =
ζ
−
z
2π
ζ
−
z
Ω
Ω
Z
1
1
∂z (g3 e−A2 )(x̃)
1
e2iτ ψ(x̃)
−
∂ζ
(e1 g3 e−A2 ) eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 = −
2π Ω
ζ −z
8τ |det Im Φ00 (x̃)| 12
z̃ − z
and
Z
Z
∂ζ̄ Φ
1
1
τ
−B2 τ (Φ(ζ)−Φ(ζ))
dξ1 dξ2 =
(e1 g4 e )e
(e1 g4 e−B2 )∂ζ eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 =
2π Ω ζ̄ − z̄
2π Ω ζ̄ − z̄
Z
1
1
1
e−2iτ ψ(x̃)
∂z̄ (g4 e−B2 )(x̃)
−B2
∂ζ
(e1 g4 e ) eτ (Φ(ζ)−Φ(ζ)) dξ1 dξ2 = −
.
−
2π Ω
8τ |det Im Φ00 (x̃)| 12
ζ̄ − z̄
z̃ − z
Taking into account Proposition 7.1, we obtain (7.11)-(7.14) for the functions G1 (x, τ ), ..., G4 (x, τ ).
Proof of Proposition 5.2. Using (5.14) and (5.16) we have
L0 ≡
∂U1
, bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
∂z
∂U1
+ (2(B1 − B2 )
, bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
∂z
∂
= −2((A1 − A2 )eτ Φ R−τ,A1 {e1 g1 }, bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
∂z
+ ((B1 − B2 )(−e1 g1 + A1 R−τ,A1 {e1 g1 })eτ Φ , bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
(2(A1 − A2 )
τΦ
= −2((A1 − A2 )e (
∂(e1 g1 )
∂z
1
− eA1 e−τ (Φ−Φ) G1 + oL2 (Ω) ( )), bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
τ
2τ ∂z Φ
(7.16) + ((B1 − B2 )(−e1 g1 + A1 R−τ,A1 {e1 g1 })eτ Φ , bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω) .
By Proposition 3.4 we have
(7.17)
2((A1 − A2 )eτ Φ
∂(e1 g1 )
∂z
1
, cτ eA2 −τ Φ )L2 (Ω) = o( ) as |τ | → +∞.
τ
2τ ∂z Φ
By (7.11) and Proposition 3.4 we obtain
(7.18)
1
−2((A1 − A2 )eτ Φ eA1 e−τ (Φ−Φ) G1 , bτ eB2 −τ Φ )L2 (Ω) = o( ) as |τ | → +∞.
τ
Integrating by parts and using (7.15) and (3.2), we have
2((A1 − A2 )eτ Φ eA1 e−τ (Φ−Φ) G1 , cτ eA2 −τ Φ )L2 (Ω) = 2((A1 − A2 )eA1 G1 , cτ eA2 )L2 (Ω)
Z
Z
∂ A1 +A2
A1 +A2
2(A1 − A2 )e
cτ (z̄)G1 (x, τ )dx = −4
e
cτ (z̄)G1 (x, τ )dx
=
Ω ∂ z̄
Ω
Z
1
(ν1 + iν2 )eA1 +A2 cτ (z̄)G1 (x, τ )dσ + o( ) as |τ | → +∞.
(7.19)
= −2
τ
∂Ω
62
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
By (7.11), (7.17)-(7.19) and Propositions 3.3 and 3.5, we conclude
L0 =
(
(B1 − B2 )(−e1 g1 + A1 R−τ,A1 {e1 g1 }), bτ eB2 )L2 (Ω)
+2 ((A1 − A2 )eτ Φ eA1 e−τ (Φ−Φ) G1 , cτ eA2 −τ Φ )L2 (Ω)
−
=
−
−
∂(e1 g1 )
∂z
1
− eA1 e−τ (Φ−Φ) G1 ), bτ eB2 )L2 (Ω) + o( )
τ
2τ ∂z Φ
A1 e1 g1
((B1 − B2 )(−e1 g1 +
), bτ eB2 )L2 (Ω)
2τ ∂z Φ
2((A1 − A2 )(
∂(e1 g1 )
∂z
2((A1 − A2 )
, bτ eB2 )L2 (Ω)
2τ ∂z Φ
Z
1
2
(ν1 + iν2 )eA1 +A2 cτ (z̄)G1 (x, τ )dσ + o( ) as |τ | → +∞.
τ
∂Ω
(7.20)
Using (5.14) and (5.18) we obtain after simple computations
L1 ≡
(
+
=
−
=
∂U2
, bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
∂z
∂U2
(2(B1 − B2 )
, bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
∂z
e τ,B1 {e1 g2 })eτ Φ , bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
((A1 − A2 )(−e1 g2 + B1 R
∂ e
τΦ
B2 −τ Φ
+ cτ eA2 −τ Φ )L2 (Ω)
2((B1 − B2 ) R
τ,B1 {e1 g2 }e , bτ e
∂z
e τ,B1 {e1 g2 })eτ Φ , bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω)
((A1 − A2 )(−e1 g2 + B1 R
2(A1 − A2 )
(7.21) − 2((B1 − B2 )(
∂(e1 g2 )
∂z
1
− eB1 eτ (Φ−Φ) G2 + oL2 (Ω) ( ))eτ Φ , bτ eB2 −τ Φ + cτ eA2 −τ Φ )L2 (Ω) .
2τ ∂z Φ
τ
Then, by (7.15) we have the asymptotic formula
∂(e1 g2 )
∂z
1
− eB1 eτ (Φ−Φ) G2 + oL2 (Ω) ( ))eτ Φ , bτ eB2 −τ Φ )L2 (Ω)
2τ ∂z Φ
τ
Z
1
= 2((B1 − B2 )eB1 G2 , bτ eB2 )L2 (Ω) + o( ) = 2 (B1 − B2 )eB1 +B2 bτ (z)G2 (x, τ )dx
τ
Ω
Z
Z
∂ B1 +B2
1
1
= −4
e
bτ (z)G2 (x, τ )dx + o( ) = −2
(ν1 − iν2 )eB1 +B2 bτ (z)G2 (x, τ )dσ + o( ).
τ
τ
Ω ∂z
∂Ω
(7.22)
−2((B1 − B2 )(
By (7.12) and Proposition 3.4 we obtain using (7.21):
−2((B1 − B2 )(
(7.23)
∂(e1 g2 )
∂z
2τ ∂z Φ
= −2((B1 − B2 )(
− eB1 eτ (Φ−Φ) G2 )eτ Φ , cτ eA2 −τ Φ )L2 (Ω)
∂(e1 g2 )
∂z
1
χ
), cτ eA2 −τ Φ )L2 (Ω) + o( ) + .
2τ ∂z Φ
τ
τ
Here χ is a constant which is independent of τ .
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
By (7.22) and (7.23) we have
L1 =
(
e τ,B1 {e1 g2 }), cτ eA2 )L2 (Ω)
(A1 − A2 )(−e1 g2 + B1 R
∂(e1 g2 )
∂z
1
− eB1 eτ (Φ−Φ) G2 ), cτ eA2 )L2 (Ω) + o( )
2τ ∂z Φ
τ
B1 e1 g2
= ((A1 − A2 )(−e1 g2 +
), cτ eA2 )L2 (Ω)
2τ ∂z Φ
− 2((B1 − B2 )(
∂(e g )
(7.24)
1 2
∂B1 e1 g2
∂z
− 2((B1 − B2 )(
+
), cτ eA2 )L2 (Ω)
∂z 2τ ∂z Φ 2τ ∂z Φ
Z
1
− 2
(ν1 − iν2 )eB1 +B2 bτ (z)G2 (x, τ )dσ + o( ) as |τ | → +∞.
τ
∂Ω
−τ Φ
e
Rτ,−B2 {e1 g3 }.
Recall V1 = −e−τ Φ R
−τ,A2 {e1 g4 } and V2 = −e
By Proposition 3.2 we conclude
(7.25)
2
∂V1
−τ Φ
e
= (−e1 g4 + A2 R
−τ,A2 {e1 g4 })e
∂z
and
(7.26)
2
∂V2
= (−e1 g3 − B2 Rτ,−B 2 {e1 g3 })e−τ Φ .
∂z
1
2
Similarly to (5.15) and (5.17) we calculate ∂V
and ∂V
:
∂z
∂z
∂V1
∂(e1 g4 )
e
(7.27)
= −e−τ Φ R
+ e−τ Φ+A2 G3 (·, τ )
−τ,A2
∂z
∂z
and
∂V2
= −e−τ Φ Rτ,−B 2
∂z
(7.28)
∂(e1 g3 )
∂z
+ e−τ Φ+B2 G4 (·, τ ).
Using (7.25) and (7.26) we obtain
L2 ≡
∂
(aτ eA1 +τ Φ + dτ eB1 +τ Φ ), V1 + V2 )L2 (Ω)
∂z
= −((A1 − A2 )dτ B1 eB1 eτ Φ , V1 + V2 )L2 (Ω)
(
2(A1 − A2 )
(A1 − A2 )(ν1 − iν2 )aτ eA1 +τ Φ , V1 + V2 )L2 (∂Ω)
∂
− ( 2 (A1 − A2 )aτ eA1 +τ Φ , V1 + V2 )L2 (Ω)
∂z
∂V1 ∂V2
− ( (A1 − A2 )aτ eA1 +τ Φ , 2(
+
))L2 (Ω) .
∂z
∂z
We observe that by (7.27), (7.15), Proposition 3.3 and Proposition 3.4
Z
∂V1
∂
1
A1 +τ Φ
−((A1 − A2 )aτ e
,2
)L2 (Ω) = −4 aτ (z) eA1 +A2 G3 (x, τ )dx + o( ) =
∂z
∂ z̄
τ
Ω
Z
1
(7.29)
−2
aτ (z)G3 (x, τ )(ν1 + iν2 )eA1 +A2 dσ + o( ) as |τ | → +∞.
τ
∂Ω
+
(
63
64
O. IMANUVILOV, G. UHLMANN, AND M. YAMAMOTO
Hence, using (7.26) we have
L2 =
(
e
(A1 − A2 )dτ B1 eB1 , R
−τ,A2 {e1 g4 })L2 (Ω)
+
(
+
(
(A1 − A2 )(ν1 − iν2 )aτ eA1 +τ Φ , V1 + V2 )L2 (∂Ω)
∂
2 (A1 − A2 )aτ eA1 , Rτ,−B 2 {e1 g3 })L2 (Ω)
∂z
1
(A1 − A2 )aτ eA1 , −e1 g3 − B2 Rτ,−B 2 {e1 g3 })L2 (Ω) + o( )
τ
Z
− 2
aτ (z)G3 (x, τ )(ν1 + iν2 )eA1 +A2 dσ.
+
(
∂Ω
By (4.34) and stationary phase we get
1
((A1 − A2 )(ν1 − iν2 )aτ eA1 +τ Φ , V1 + V2 )L2 (∂Ω) = o( ) as |τ | → +∞.
τ
Therefore, by Proposition 4.1
e1 g4
)L2 (Ω)
τ ∂z Φ
e1 g3
∂
)L2 (Ω)
+ (
(A1 − A2 )aτ eA1 ,
∂z
τ ∂z Φ
e1 g3
+ ( (A1 − A2 )aτ eA1 , −e1 g3 + B2
)L2 (Ω)
τ ∂z Φ
Z
1
− 2
aτ (z)G3 (x, τ )(ν1 + iν2 )eA1 +A2 dσ + o( ) as |τ | → +∞.
τ
∂Ω
L2 = −
(7.30)
(
(A1 − A2 )dτ B1 eB1 ,
Integrating by parts we compute
L3 ≡
(
2(B1 − B2 )
∂
(aτ eA1 +τ Φ + dτ eB1 +τ Φ ), V1 + V2 )L2 (Ω) =
∂z
∂
(B1 − B2 )dτ eB1 +τ Φ , V1 + V2 )L2 (Ω)
∂z
(B1 − B2 )A1 aτ eA1 +τ Φ , V1 + V2 )L2 (Ω)
− (2
−
(
+
(
−
(
(ν1 + iν2 )(B1 − B2 )dτ eB1 +τ Φ , V1 + V2 )L2 (∂Ω)
∂V1 ∂V2
(B1 − B2 )dτ eB1 +τ Φ , 2(
+
))L2 (Ω) .
∂z
∂z
We observe that by (7.15), (7.28), Proposition 3.3 and Proposition 3.4:
(7.31)
∂V2
,
)L2 (Ω) = −4
∂z
Z
∂ B1 +B2
−2((B1 − B2 )dτ e
e
dτ (z)G4 (x, τ )dx =
Ω ∂z
Z
1
(ν1 − iν2 )eB1 +B2 dτ (z)G4 (x, τ )dσ + o( ) as |τ | → +∞.
−2
τ
∂Ω
B1 +τ Φ
PARTIAL DIRICHLET-TO-NEUMANN MAP FOR GENERAL OPERATOR
65
Hence
−
(
∂
e
(B1 − B2 )dτ eB1 , R
τ,−A2 {e1 g4 })L2 (Ω)
∂z
(B1 − B2 )A1 aτ eA1 , Rτ,−B 2 {e1 g3 })L2 (Ω)
+
(
e
(B1 − B2 )dτ eB1 , −e1 g4 + A2 R
−τ,A2 {e1 g4 })L2 (Ω)
+
(ν1 + iν2 )(B1 − B2 )dτ eB1 +τ Φ , V1 + V2 )L2 (∂Ω)
Z
1
+ 2
(ν1 − iν2 )eB1 +B2 dτ (z)G4 (x, τ )dσ + o( ) as |τ | → +∞.
τ
∂Ω
L3 = (2
(7.32)
(
By (4.34) and the stationary phase argument
1
((ν1 + iν2 )(B1 − B2 )dτ eB1 +τ Φ , V1 + V2 )L2 (∂Ω) = o( ) as |τ | → +∞.
τ
Therefore, applying Proposition 3.3 we conclude finally that
∂
e1 g4
L3 = − 2( (B1 − B2 )dτ eB1 ,
)L2 (Ω)
∂z
τ ∂z Φ
e1 g3
− ( (B1 − B2 )A1 aτ eA1 ,
)L2 (Ω)
2τ ∂z Φ
A2 e1 g4
)L2 (Ω)
+ ( (B1 − B2 )dτ eB1 , −e1 g4 −
2τ ∂z Φ
Z
1
(7.33)
+ 2
(ν1 − iν2 )eB1 +B2 dτ (z)G4 (x, τ )dσ + o( ) as |τ | → +∞.
τ
∂Ω
P3
The sum k=0 Lk equal to the left hand side of (5.19). By (7.20), (7.24), (7.30) and (7.33),
there exist numbers κ, κ0 such that the asymptotic formula (5.19) holds true. Acknowledgements.
The first named author was partly supported by NSF grant DMS 0808130, and the second
named author was partly supported by NSF and a Walker Family Endowed Professorship.
The visits of O.Imanuvilov to Seattle were supported by NSF and the Department of Mathematics of the University of Washington. O.Imanuvilov also thanks the Global COE Program
“The Research and Training Center for New Development in Mathematics” for support of
a visit to the University of Tokyo. The authors thank Professor J. Sawon for discussions on
the proof of Corollary 1.1.
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Department of Mathematics, Colorado State University, 101 Weber Building, Fort Collins
CO, 80523 USA, e-mail: [email protected]
Department of Mathematics, University of Washington, Seattle, WA 98195 USA, e-mail:
[email protected]
Department of Mathematical Sciences, University of Tokyo, Komaba, Meguro, Tokyo
153, Japan, e-mail: [email protected]