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О некоторых обратных спектральных задачах для произвольного возмущения бигармонического оператора с сингулярными коэффициентами.

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12. Mathews J.H., Fink K.D. Numerical Methods Using
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13. Yang W.Y., Cao W., Chung T-S., Morris J. Applied
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procedures optimization]. Moscow, PMSOFT Publ., 2006, 290 p.
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Sons Publ., 2005, 526 p.
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ON SOME INVERSE SPECTRAL PROBLEMS
FOR AN ARBITRARY PERTURBATION OF THE BI-HARMONIC OPERATOR
WITH SINGULAR COEFFICIENTS
V.S. Serov, Dr.Sc. (Physics and Mathematics), Professor
(University of Oulu, P.O. Box 3000, Fin-90014, Oulu, Finland, vserov@cc.oulu.fi)
Received 17.04.2014
The classical inverse boundary spectral problem for any perturbation of the bi-harmonic operator with singular
coefficients from some Sobolev spaces is considered. The problem is formulated as follows: Do the Dirichlet eigenvalues and
some derivatives of the corresponding normalized eigenfunctions at the boundary of smooth bounded domain uniquely
determine the coefficients of this operator?
We proved the first step of this problem, i.e. we proved that the Dirichlet eigenvalues and the derivatives up to the second
order of the normalized eigenfunctions at the boundary uniquely determine the so-called Dirichlet-to-Neumann map that
corresponds to the Friedrichs self-adjoint extension of any perturbation of the bi-harmonic operator with singular coefficients.
The main role in this proof is played by the existence of the Green?s function and its estimates up to the boundary of the
domain. These facts will allow us to prove the classical Borg-Levinson theorem for the operators of fourth order of such type.
Keywords: Green?s function, Friedrichs extension, Dirichlet-to-Neumann map.
??? 517.95
? ????????? ???????? ???????????? ???????
??? ????????????? ?????????? ???????????????? ?????????
? ???????????? ??????????????
????? ?.?., ?.?.-?.?., ?????????
(??????????? ?????????? ???? ????, P.O. Box 3000, Fin-90014, ?. ????, ?????????)
???? ?????? ??????: 17.04.2014
?????????. ??????????????? ???????????? ???????? ????????? ???????????? ?????? ??? ????????????? ?????????? ???????????????? ????????? ? ???????????? ?????????????? ?? ????????? ??????????? ????????. ??????
????????????? ????????? ???????: ?????????? ?? ?????????? ??????????? ???????? ?????? ??????? ? ?????????
??????????? ??????????????? ????????????? ??????????? ??????? ?? ??????? ??????? ???????????? ??????? ???????????? ????? ??????????
???????? (??? ?????? ??? ? ??????? ???? ????????), ??? ??????????? ???????? ?????? ??????? ? ???????????
?? ??????? ??????? ????????????? ??????????? ??????? ?? ??????? ?????????? ?????????? ??? ?????????? ???-
192
??????????? ???????? ? ???????
? 2, 2014 ?.
??????????? ?? ??????? ? ???????, ??????? ????????????? ???????????????? ?????????? ?? ????????? ????????????? ?????????? ??-?????????????? ????????? ? ???????????? ??????????????. ??????? ???? ? ???? ?????????????? ?????? ????????????? ??????? ????? ? ?? ?????? ?????? ?? ??????? ???????. ??? ????? ???????? ??? ???????? ???????????? ??????? ?????-????????? ??? ?????????? ?????????? ??????? ?????? ????.
???????? ?????: ??????? ?????, ?????????? ?? ?????????, ?????????????? ?? ??????? ? ???????.
The subject of this work concerns to the classical
inverse spectral problem. This inverse problem can be
formulated as follows: do the Dirichlet eigenvalues
and the derivatives (which order?) of the normalized
eigenfunctions at the boundary determine uniquely
the coefficients of the corresponding differential
operator? For operators of order two this type of
theorem is called Borg-Levinson theorem. In the case
of the Schr鰀inger operators the knowledge of the
Dirichlet eigenvalues and the normal derivatives of
the normalized eigenfunctions at the boundary
uniquely determine unknown potential. BorgLevinson theorem for the Schr鰀inger operators was
proved for the first time by Nachman, Sylvester and
Uhlmann [1] for the potentials from the space
C ? (?). Their proof remains however valid if one
assumes that potential just from L?(?). This problem
was reduced finally to the fact that the Dirichley-toNeumann map uniquely determines such potentials.
The same result was obtained independently by
Novikov [2]. For singular potentials from the space
Lp(?); n/2<p??, this theorem was proved by
P鋓v鋜inta and Serov [3]. For inverse boundary
spectral problems on Riemannian manifolds some
related results were proved by Kachalov, Kurylev and
Lassas [4] (see also [5]).
For the magnetic Schr鰀inger operator with
singular coefficients Borg-Levinson theorem was
proved for the first time by Serov [6]. The proof of
this result for the magnetic Schr鰀inger operators uses
the same technique as for the Schr鰀inger operators
with singular potentials. It must be mentioned here
that the magnetic Schr鰀inger operator cannot be
considered as a "small" perturbation of the
Schr鰀inger (or Laplace) operator.
For the operator of order 4 which is the first order
perturbation of the bi-harmonic operator with Navier
boundary conditions on a smooth bounded domain in
Rn, n?3, it is proved by Krupchyk, Lassas and
Uhlmann [7] that the Dirichlet-to-Neumann map
uniquely determine this first order perturbation.
The main goal of present work is to show that the
knowledge of the discrete Dirichlet spectrum and
some special derivatives up to the second order of the
normalized eigenfunctions at the boundary uniquely
determine the Dirichlet-to-Neumann map that
corresponds to an arbitrary perturbation of the biharmonic operator with singular coefficient from
some Sobolev spaces on the smooth bounded domain
in Rn, n?2. The next step might be to prove that the
knowledge of the Dirichlet-to-Neumann map uniquely
determines the coefficients of the operator H4 which is
an arbitrary perturbation of the bi-harmonic operator.
The solution of this problem will be given in future
publications.
Below we use the following notations. The space
Wpt(?), t?0,1?p??, denotes the Lp ? based Sobolev
space in the domain ? and the space
Btp,? (??), t ? R, 1 ? p, ? ? ?, denotes the Besov space
on the boundary of the domain ?. The Besov space
Btp,? (??) for negative t is defined as the adjoint space
to the corresponding Besov space with positive index
of smoothness ? t. Throughout this article we use the
trace type theorem for the functions from the Sobolev
spaces at the boundary (see, for example, [8]). More
precisely, for any function f from the Sobolev space
Wpt (?) with t ? 1 and p > 1 there is the trace on the
t?
1
boundary of ? from the Besov space Bpp p (??).
Green?s function
Let ? be a bounded domain with smooth
boundary in Rn, n?2. We consider in this domain the
following operator of order four that we will call the
"magnetic" operator of 4th order with variable
coefficients
H 4 ? H 4 ( x, ?) := ? 2 + i?(?( A( x) ??)) + i? ? ( A( x)?) ?
(1)
?? ? ( F ( x )?) ? i? ? (G ( x)) ? iG ( x) ? ? + V ( x ),
n
where ? denotes the gradient in R , ? denotes the
Laplacian in Rn and where the coefficients A( x ) ,
F(x), G ( x ) and V(x) are assumed to be real-valued.
We assume also that
A(x) ? (Wp3 (?))n , F (x) ?Wp2 (?), G( x) ? (Wp1 (?))n (2)
n
, n ? 2,
(3)
2
where it is assumed (WLG) that the value of p is the
same for all these spaces. It is well-known that under
the conditions (2) and (3) for the coefficients the
following G錼ding?s inequality holds (see, for
instance, [9]):
and V ( x) ? Lp (?),
p>
2
(H4 u, u)L2 ? c1 u W 2 (?) ? c2 u
2
2
L2 ( ? )
,
(4)
where 0<c1<1, c2>0. This inequality allows us to
define symmetric operator (1) by the method of
quadratic forms. Then H4 has a self-adjoint Friedrichs
extension denoted by (H4)F with domain
D(( H 4 ) F ) = { f ( x) ? W22 (?) : H 4 f ( x) ? L2 (?)},
where W22 (?) denotes the closure of the space
C0? (?) by the norm of Sobolev space W22 (?) . It is
193
??????????? ???????? ? ???????
? 2, 2014 ?.
possible to prove actually that under the conditions
(2) and (3)
D(( H 4 ) F ) = W22 (?) ? W24 (?).
(5)
The spectrum of this extension is purely discrete,
of finite multiplicity and has an accumulation point
only at the +?: ?1??2????k?? ?+?.
The corresponding orthonormal eigenfunctions
{?k ( x)}?k =1 form orthonormal basis in L2(?). The
G錼ding?s inequality (4) allows us also to conclude
that there is a positive constant �such that the
operator (H4)F+� is positive.
Therefore for any ? ? �the operator (H4)F+?I is
positive and its inverse
((H4)F + ?I)?1: L2(?) ? L2(?)
(6)
is compact. It is an integral operator with kernel
denoted by G(x, y, ?). If we use for this integral
operator the symbol G? (?) then we have
(( H ) + ? I )G? (? ) = I , G? (? )(( H ) + ?I ) = I ,
4 F
4 F
(7)
Definition 1. The kernel G(x, y, ?) of the integral
operator G? (?) is called the Green?s function of the
operator (H4)F + ?I.
Our first result is:
Theorem 1. Suppose that A (x), F(x), G (x) and
V(x) satisfy the conditions (2) and (3). Then for any
??�the Green?s function of the operator (H4)F+?I
exists and satisfies the following estimates:
G ( x, y , ? ) = G ( x, y, ? ).
1
| G ( x, y, ?) |? C | x ? y |
4? n
e
?? | x ? y| ? 4
, n ? 5,
?
?
| G ( x, y, ?) |? C ?1+ | log(| x ? y | ? ) | ? e
?
?
n = 4,
1
4
and | G ( x, y, ?) |?
C
4?n
4
1
4
e??| x ? y|? , n ? 3,
1
??| x ? y | ? 4
n
(( H 4 ) F + ?I )?? f
,
(9)
(10)
Rn Rn
Using this fact we can easily prove the existence
of the fundamental solution for the operator H4+?I.
Moreover, the same estimates (8)?(10) will be
fulfilled for this fundamental solution too. In order to
obtain the same estimates (8)?(10) for the Green?s
function G(x, y, ?) we refer to the paper [10] (with
L? ( ? )
? C? 8
??
f
L2 ( ? )
,
where ??�with �as in Theorem 1.
n
Corollary 2. Assume that ? > , n ? 2. Then
8
there is a constant C>0 depending only on ?, such
n
? | ? ( x) |2
? 2?
k
4
that the estimate ?
?
C
?
holds
2?
k =1 ( ? k + ? )
uniformly in x?? and ??�
Corollary 3. Assume that ? >
n
, n?2. Then the
8
following series
?
1
(11)
?
2?
k =1 (? k + � 0 )
converges.
Remark 1. It can be mentioned here that the
estimates (8)?(10) of the Green?s function of the
"magnetic" operator (1) are obtained in Theorem 1 for
quite weak conditions of the coefficients of H4. As far
as we know they never appeared in the literature.
(8)
?
where x, y?? and constants C>0 and ?>0 do not
depend on x, y?? and ?.
Proof. It is well-known that the fundamental
solution of the operator ?2+?I, ?>0, has exactly these
estimates (8)?(10). The conditions (2) and (3) for the
coefficients that are in front of the derivatives of order
one and higher of the operator H4 allow us to
conclude that they belong to the Kato space Kn?1(Rn)
(if we extend them by zero outside of the domain ?),
i.e.
K n ?1 ( Rn ) = { f ( x) : sup ? | x ? y |1? n | f ( y) |dy < ?}.
194
some changes that are connected to the operator of
order 4). Theorem 1 is proved.
We have three immediate corollaries of Theorem
1 (see again [10]).
Corollary 1. Assume that A (x), F(x), G (x) and
n
V(x) are as above and ? > , n?2. Then for any
8
function f (x)?L2(?) the following inequality holds
Dirichlet-to-Neumann map
and eigenfunctions
Lemma 1. Under the conditions (2) and (3) for the
coefficients of H4 we have that for any two functions
u and ? from Wp4 (?) the following equality holds
( H 4u, ? )
L2 ( ? )
+(? 2u, ? ? ?)
? (u, H 4 ?)
L2 ( ?? )
L2 ( ? )
? (u, ?1? )
= (?1u, ?)
L2 ( ?? )
L2 ( ?? )
+
? (? ? u, ? 2 ?)
L2 ( ?? )
,
where ?1 and ?2 are defined as
?1u(x)=?v(?u)(x)+i?v( A ??u)(x)+iv? A ? u(x)?
?F?vu(x)?iv? G u(x), x???
(12)
and ?2u(x)=??u(x)?i A ??u(x), x???,
(13)
respectively. Here v is unit outward normal vector at
the boundary of the domain ?.
Proof. The proof of this lemma is straightforward
and is based only on the divergence theorem
("integration by parts"). And the main purpose of this
lemma is to define the map ?=(?1, ?2). It must be
mentioned also here that the conditions (2) and (3) for
the coefficients of the operator H4 and the conditions
for the functions u and ? from this lemma guarantee
the existence of the traces of the corresponding terms
in ?1u(x), ?1?(x), ?2u(x) and ?2?(x) on the boundary
?? from some Besov spaces. This fact justifies the
??????????? ???????? ? ???????
? 2, 2014 ?.
correct application of the integration by parts and
proves this lemma.
Let ??�with �as in Chapter 1. Consider the
following Dirichlet problem:
((H4)F+?I)u(x)=0; x??, u(x)=f0(x), ?vu(x)=f1(x),
x???,
(14)
where the boundary functions f0(x) and f1(x) satisfy
the following conditions:
1
p
1
p
n
f0 ( x) ? Bpp (??), f1 ( x) ? Bpp (??), p > , (15)
2
where Btpp (??), t ? R, denotes Besov space on the
boundary and p is the same as in (2) and (3).
Using the technique of the multipliers from
Sobolev spaces (see, for example, [11]) it can be
proved that there exists a unique solution of the
Dirichlet boundary value problem (14)?(15) from the
spaces
(16)
u ( x ) ? W p4,loc (?) ? W p3 (?).
Thus, we may define the Dirichlet-to-Neumann
map ??.
Definition 2. The Dirichlet-to-Neumann map ??
for Dirichlet boundary problem (14)?(15) is defined
as the following two-dimensional vector
?? { f0 , f1}( x) :=
(?? (?u)( x) + i?? ( A ??u)( x) + i? ? A?u( x) ?
?F ?? f1 ( x) ? i? ? Gf 0 ( x), ? ?u( x) ? iA ??u( x)), (17)
where v is outward normal vector at the boundary ??.
Conditions (2), (3), (15) and (16) imply that the
Dirichlet-to-Neumann map (17) acts as (for fixed ?)
3?
3?
1
2?
2?
1
?
1
?
1
?? : Bpp p (??) � Bpp p (??) ? Bppp (??) � Bppp (??) (18)
with p as in (15).
The following theorem can be considered as one
of the main results of this work.
Theorem 2. Assume that A1 , A2 , F1 , F2 , G1 , G2 and
V1, V2 satisfy the conditions (2), (3) and f0, f1 satisfy
the condition (15). In addition we assume that
A1 ( x) = A2 ( x), F1 ( x) = F2 ( x), G1 ( x) = G2 ( x) on the
boundary ??. Then, for any 0 < ? < 1 ?
(1)
(2)
lim ?? { f 0 , f1} ? ? ? { f 0 , f1} B?
??+?
1
p
pp ( ?? )
(? 2 + ?I )? + (i?(?( A1??)) + i?( A1??) ??( F1??) ?
?i?(G1?) + V1?) = i?(?( A2 ? A1 )?u2 ) +
(20)
+i?(( A2 ? A1 )?u2 ) ? ?(( F2 ? F1 )?u2 ) ?
?i?((G2 ? G1 )u2 ) + (V2 ? V1 )u2
and with Dirichlet boundary conditions
?(x)=0, ?v?(x)=0, x???.
(21)
?
Denote by G0 (? ) the integral operator with the
kernel which is the Green?s function of the operator
?2+?I with Dirichlet boundary conditions (21).
Applying G? 0 (? ) to the left and to the right hand-sides
in (20) we obtain the following integral equation
(I + K)?(x)=F(x),
(22)
where the integral operator K and the function F are
given by
?
K := G0 (?)(i?(?( A1?)) + i?( A1?) ?
(23)
??( F1?) ? i?(G1 ) + V1 )
and
?
F ( x) := G0 (?)(i?(?( A2 ? A1 )?u2 ) +
+i?(( A2 ? A1 )?u2 ) ? ?(( F2 ? F1 )?u2 ) ?
(24)
?i?((G2 ? G1 )u2 ) + (V2 ? V1 )u2 )( x).
We consider this equation (22) in the space of
functions from the Sobolev space Wp4 (?) which are
vanishing with their first normal derivatives at the
boundary ??.
Due to the assumptions (2) and (3) for the
coefficients Aj , F j , G j and Vj, j=1,2, and embedding
(16) we may conclude that F belongs to this space and
K is compact there. Since the operator H 4(1) + ? I is
positive for ??�then the boundary value problem
( H 4(1) + ? I )?( x ) = 0, x ? ?, ?(x)=0, ?v?(x)=0, x???
has only trivial solution ??0. The same is true for the
homogeneous equation corresponding to (22). By the
Fredholm?s alternative the operator I+K has a
bounded inverse in the indicated Sobolev space and
therefore the solution ? of the equation (22) satisfies
the following inequality
(25)
? W 4 (?) ? C F W 4 (?) ,
p
= 0,
(19)
where ? (?j ) denotes the corresponding Dirichlet-to
Neumann map for Aj , F j , G j ,V j + ?, j = 1, 2 .
Proof. Let ? (x): =u1(x)?u2(x), where uj(x), j=1,2,
solves the problem (14) with Aj , F j , G j ,V j ,
respectively. We denote the corresponding operators
(1) by H4( j ) . Then ? (x) solves the boundary value
problem ( H 4(1) + ? I )?( x ) = ( H 4(2) ? H 4(1) )u2 ( x ), x ? ?,
?(x)=0, ?v?(x)=0, x???.
This problem can be rewritten in the domain ? as
p
where constant C is independent on ??�
Since p satisfies the inequality p >
n
then the
2
following embeddings hold
Wp3 (?) ? Wp2 (?) ? L? (?).
This fact and the conditions (2) and (3) allow us to
obtain from (24) and (25) that
(26)
? W 4 (?) ? C u2 W 3 (?) .
p
p
We apply now the result from [12] and obtain that
(27)
( ? 2 + ?I ) ? p ? C ? ? Lp ( ? ) .
L (?)
195
??????????? ???????? ? ???????
? 2, 2014 ?.
By combining the inequalities (26) and (27) we
get the following inequality
C
(28)
u2 W 3 (? ) .
? Lp ( ? ) ?
p
?
The interpolation of (26) and (28) leads us to the
inequality
C
? W s (? ) ?
u2 W 3 ( ? ) ,
(29)
s
p
p
1?
? 4
where 0?s?4. Using the definition (17) and the
equalities for the terms A1 ( x) = A2 ( x), F1 ( x) =
= F2 ( x), G1 ( x) = G2 ( x) on the boundary ?? we have
? ?(1) { f0 , f1} ? ? (2)
? { f 0 , f1 }
+ ? ? ( A1??)
+ A1??
B ?pp ( ?? )
B ? ( ?? )
pp
B ? ( ?? )
pp
+ ?A1??
? ? ? (??)
B ?pp ( ?? )
+
B ?pp ( ?? )
B ?pp ( ?? )
C
?
1?? 1
?
4 4p
u2
W p3 ( ? )
.
(30)
??
where the kernels glij , i, j = 1, 2, are defined by
( ? ( A ??? ( x)) + ?? A?? (x)) ( ? ( A ??? ( y)) + ? ? A?? ( y)) ,
?
n
and �is as in Theorem 1.
2
Proof. Let ?k be an eigenvalue and ?k(x)
corresponding orthonormal eigenfunction. Then
Corollary 2 from Theorem 1 allows us to obtain quite
n
? C (? k + �) 8 and
n
? C (? k + �) 8 ,
(32)
where 1?p?? and constant C>0 depends only on n, p
and Vol(?).
Rewriting the equation for the eigenfunctions
?k(x) in the form (?2+�+Q(x, ?))?k(x)=(?k+�?k(x),
x??, ?k(x)=0, ?v?k(x)=0, x???, where Q(x, ?) is the
rest of the operator H4, and applying the inequality
n
(25), we obtain for any p > that
2
k
?
gl(11) ( x, y, ?) = (?1)l +1 l ! ?
k
(
?
g l( 21) ( x , y , ?) = ( ?1) l +1 l ! ?
k
?
(? k + ? )
k =1
?
(31)
(34)
+ ? gl(22) ( x, y, ?) f1 ( y) d ?( y),
k
l +1
)(
i? ? ( A ? ??k ( x)) + i?? A??k ( x) ??k ( y ) ? iA ? ??k ( y)
k =1
? C (? k + � 0 ) 4 8 ,
L? ( ? )
. (33)
( x, y, ?) f1 ( y) d ?( y),
gl(12) ( x, y, ?) = (?1)l l ! ?
where 0?s?4, p >
196
(12)
l
s n
+
Lp ( ? )
n
8
(21)
?(2)
( x, y, ?) f0 ( y) d ?( y) +
?,l = ? gl
1
taking into account the
p
boundedness of the norm of u2 in ? we may conclude
from (30) that Theorem 2 is completely proved.
We are in the position now to estimate the
normalized eigenfunctions of the magnetic
Schr鰀inger operator.
Lemma 2. Under the assumptions (2) and (3) for
the coefficients A, F , G and V the orthonormal
eigenfunctions ?k(x) satisfy the estimate
?k
L (? )
? C (? k + � 0 )
??
? < 1?
easily that ?k
1+
p
??
1
?+ 3 +
p (?)
Wp
W ps ( ? )
+?g
.
?
?k
? C (? k + � 0 ) ? k
Now by interpolation of (32) and (33) we obtain
(31). Thus, Lemma 2 is proved.
The next lemma shows us the representation for
the kernel of the operator ??.
?n?
Lemma 3. For l = ? ? + 1 (here [?] denotes the
?2?
entire part of ?) and f0 and f1 as in (15) we have for
l
? d ?
(1)
(2)
the vector ?
? ( ?? { f 0 , f1}( x) ) := (? ?,l , ? ?,l ) the
? d? ?
following representations
(11)
?(1)
( x, y, ?) f 0 ( y) d ?( y) +
?,l = ? gl
+
B pp ( ?? )
Since
W p4 ( ? )
??
+ ??
Using trace theorem and (29) we can estimate the
latter terms as follows:
(2)
? (1)
?
? { f 0 , f1 } ? ? ? { f 0 , f1 } ?
?C ?
?k
(
(?k + ?)l +1
??k ( x ) + iA ? ?? k ( x ) i? ? ( A ??? k ( y )) + i? ? A??k ( y )
)(
( ?k + ?)l +1
k =1
?
( ?? ( x) + iA ??? ( x) ) ( ?? ( y ) ? iA ? ?? ( y ) )
k =1
(? k + ?) l +1
g l(22) ( x , y , ?) = ( ?1)l l ! ?
k
),
),
k
k
(35)
k
and where for ??�the right-hand sides of (35) are
converging in Lp(??�?).
Proof. Solution u of the problem (14)?(15)
definitely depends on ? and we will denote it from
now on as u(x, ?) with ??� Integration by parts for
the problem (14) with the boundary conditions f0 and
f1 from (15) leads to
(
)
u( x, ?) = ? ? ? ( ? y G( x, y, ?)) ? i? ? ( A ? ? y G( x, y, ?) ? i? y ? A? y G( x, y, ?) f0 ( y ) d ?( y ) ?
y
y
??
??
(
)
? y G ( x, y , ?) ? iA ? ? y G ( x, y, ?) f1 ( y ) d ?( y ),
(36)
??
where G(x, y, ?) is the Green?s function of (H4)F+?I
defined in (6)?(10) and vy denotes the outward normal
vector in y. In our case the Green?s function is given
by
? ? ( x )? ( y )
k
(37)
G ( x, y, ? ) = ? k
.
?k + ?
k =1
Since u solves the problem (14)?(15) then using
J. von Neumann spectral theorem it can be easily
proved by induction that
l
?l
? d ?
l
(38)
?
? u( x, ?) = (?1) l !( ( H 4 ) F + ?I ) u ( x, ?),
? d? ?
l = 1, 2,....
The operator ((H4)F + ?I)?1 is well-defined by the
spectral theorem and it is the integral operator with
kernel denoted by Gl(x, y, ?)
??????????? ???????? ? ???????
? 2, 2014 ?.
1
? k ( x )? k ( y )
.
l
k =1 ( ? k + ? )
?
?? ( A ???k )
Gl ( x, y , ? ) = ?
This fact allows us to represent ((H4)F+?I)?1u(x, ?)
as follows
( ( H 4 ) F + ?I )? l u ( x, ? ) = ?Gl ( x, y, ? )u ( y , ? ) dy =
?
(39)
? k ( x )uk (? )
,
l
k =1 (? k + ? )
?
=?
Lp ( ?? )
p
L ( ?? )
? C ?k
)
(40)
)
Combining (39) and (40) we obtain the following
equality ((H4)F+?I)?1u(x, ?)=
?
=???
(
)
? k ( x ) ? ? ( ?? k ( y )) ? i ? ? ( A ??? k ( y )) ? i ? ? A?? k ( y ) f 0 ( y )
(? k + ?) l +1
?? k =1
?
+??
(
)
?k ( x) ??k ( y ) ? iA ? ??k ( y ) f1 ( y )
(?k + ?)l +1
??k =1
d ?( y) +
(41)
d ?( y )
which coincides for l=0 with (36).
Since u solves the boundary value problem
(15)?(16) using (17) we can obtain
l
? d ?
?
? ( ?? { f 0 , f1}( x) ) =
? d? ?
? d lu ?
? ? d lu ? ?
? ? d lu ? ?
= ? ? ?? ? ? l ? ?? + i? ? ?? A ?? ? l ? ?? + i? ? A? ? l ? ?
? d? ??
? d? ?
? ? d? ??
?
l
l
?du? ?du?
?? ? l ? ? iA ?? ? l ? .
(42)
? d? ?
? d? ?
Thus, the equalities (38), (41) and (42) give us that
l
? d ?
formally we have for the vector ?
? ( ?? { f0 , f1})
? d? ?
the following relations ?(1)
?,l =
?
(
A
???
(
x
))
+
?
?
A
??
(
x
) ) ( ? ( A ??? ( y )) + ? ? A?? ( y ) ) f ( y )
(
= ( ?1) l !
d ?( y ) +
l +1
?
k
?
??
k
?
k
k
(?k + ?) l +1
i? ? ( A ???k ( x )) + i? ? A??k ( x) ?? k ( y ) ? iA ???k ( y ) f1 ( y )
0
?? k=1
?
+( ?1) l l ! ? ?
(
)(
)
( ?k + ?)l +1
?? k=1
d?( y)
and ?(2)
?,l =
( ?? ( x) + iA ??? ( x) ) (i?
?
= ( ?1)l +1 l ! ? ?
k
k
?
)
( A ??? k ( y )) + i? ? A?? k ( y ) f 0 ( y )
(?k + ?)l +1
?? k ( x) + iA ???k ( x ) ??k ( y ) ? iA ??? k ( y ) f1 ( y )
d?( y) +
?? k =1
?
+ (?1)l l ! ? ?
??k =1
(
)(
(?k + ?)l +1
)
2+
Wp
1
p
(? )
? C (? k + � 0 ) 2
d ?( y ).
Here we have used the following equalities for the
eigenfunctions ?k(x) at the boundary (see (15) and
(31)): ?v?k=0, ?v(??k)=?(?v?k), x???.
The latter equalities show that this lemma will be
proved if we show the convergence of all series (35)
in Lp(??�?). To this end, the inequality (31) from
Lemma 2, the conditions (2) and (3) for the
coefficients and Sobolev imbedding theorem allow us
to conclude that
+
1 n
+
4p 8
+
1 n
+
4p 8
,
.
?
L p ( ?? )
??
1
? ?? (??k ( y )) ? i? ? ( A ? ??k ( y )) ? i ? ? A??k ( y ) f0 ( y) d ?( y) +
? k + ? ??
(
? C (? k + � 0 ) 2
By using these estimates and taking now
?n?
l = ? ? + 1 , we have for k, m=1,2 and j=0,1
?2?
?
1
+
? ??k ( y) ? iA ???k ( y ) f1 ( y ) d ?( y ).
? k + ? ??
Wp
1
p (?)
1
??k
Integration by parts in the last equality gives us
(
2+
( km )
? gl ( x, y, ? ) f j ( y) d ?( y)
where uk(?) is given by uk (?) = ? ?k ( y )u ( y, ?) dy.
u k (? ) = ?
? C ?k
?k
?
? C?
k =1
2
1
2+
Wp p (?)
(? k + ? )
n
[ ]+ 2
2
fj
?
L p ( ?? )
1
?
? C fj
?
Btpp ( ?? )
k =1
?
n
1 n
[ ] +1?
?
2p 4
,
(? k + �) 2
1
1
for j=0 and t = 2 ? for j=1. Thus,
p
p
due to Corollary 3 of Theorem 1 (see estimate (11))
n
the latter series converges since p > , n ? 2, and
2
therefore Lemma 3 is completely proved.
where t = 3 ?
From spectral data
to Dirichlet-to-Neumann map
Now we are in the position to formulate and to
prove the second main result of this work. By symbol
?k ( A, F , G,V ) we denote the discrete Dirichlet
spectrum of the operator H4 defined in (1) with the
coefficients A, F , G, V and by ?k ( x; A, F , G,V ) the
corresponding eigenfunctions accounting their
multiplicities.
Theorem 3. Assume that Aj ( x) ? (W p3 (?))n ,
F j ( x) ? W p2 (?), G j ( x) ? (W p1 (?))n , V j ( x) ? Lp (?),
n
p > , n ? 2, for j=1, 2. Assume in addition that
2
A1 ( x) = A2 ( x) and ?? A1 ( x) = ?? A2 ( x) at the boundary
??. Assume also that for each k=1, 2, ?
?k ( A1 , F1 , G1 ,V1 ) = ?k ( A2 , F2 , G2 ,V2 )
(43)
and ??k ( x; A1 , F1 , G1 ,V1 ) = ??k ( x; A2 , F2 , G2 ,V2 ),
??k ( x; A1 , F1 , G1 ,V1 ) = ??k ( x; A2 , F2 , G2 ,V2 ),
(44)
? ? (??k ( x; A1 , F1 , G1 ,V1 )) = ? ? (??k ( x; A2 , F2 , G2 ,V2 )),
x ? ??.
Then for all ??�(2)
? (1)
? { f 0 , f1 } = ? ? { f 0 , f1 }
for any f 0 ? B
1
3?
p
pp
(??) and f1 ? B
(45)
1
2?
p
pp
(??) .
197
??????????? ???????? ? ???????
? 2, 2014 ?.
Proof. The conditions (43) and (44), Lemma 3
(see formulas (34), (35)) imply that for all ??�and
l
?n?
? d ?
(1)
for l = ? ? + 1 we have ?
? ( ? ? { f0 , f1}( x) ?
?2?
? d? ?
?? (2)
for any
? { f 0 , f1 }( x) ) = 0
2?
3?
References
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differential equations. Dekker Publ., NY, 1997.
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equations. Springer-Verlag Publ., NY, 1983, vols. 1?2.
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1
f0 ? Bpp p (??) and
1
f1 ? Bpp p (??) . This equality can be read as
l
(2)
m
?(1)
? { f 0 , f1 }( x ) ? ? ? { f 0 , f1 }( x) = ? ? Lm { f 0 , f1 }, (46)
m =0
where two-dimensional vector-valued operators Lm
are
p
bounded
from
3?
1
2?
1
Bpp p (??) � Bpp p (??)
to
p
L (??)譒 (??). But Theorem 2 (see (19)) shows us
that the polynomial in the right-hand side of (46) is
zero. Hence, the equality (45) holds. It means that
Theorem 3 is proved.
Remark 2. This theorem means that the inverse
boundary spectral problem for the operator H4 is
reduced to the problem of the reconstruction of the
unknown coefficients of this operator by the
knowledge of Dirichlet-to-Neumann map.
Acknowledgments. This work was supported by the
Academy of Finland (Application No. 250215, Finnish
Programme for Centres of Excellence in Research 2012?
2017).
?
25 ??? ?????????? ?????????? ?????????????? ??????????
12 ??? 2014 ???? ? ???????????? ?????? ??????? ?????????? ?????????, ??????????? ?????? ?????????? ?????????? ?????????????? ?????????? (????). ? ??????????? ???????????? ??????? ???????
????????? ???? ?.?.-?.?., ????????? ?.?. ??????, ????????? ??????????? ?????????? ??????????????
?????????? ?.?.?., ????????? ?.?. ????????, ????-?????????? ???? ?.?. ???????? ? ?.?. ??????, ???????????? ???????? ?????? ?.?. ????????, ????? ??????: ?.?. ???????, ?.?. ???????, ?.?. ???????,
?.?. ??????????, ?.?. ?????????, ?.?. ??????, ?.?. ???????, ?.?. ???????, ?.?. ???????????, ?????
???????????? ????????? ???? ?.?. ???????, ?.?. ???????, ????????????? ??????? ???????? ????
?.?. ????????, ?.?. ????????, ?.?. ????????, ?.?. ?????????, ?.?. ????????, ?.?. ????????,
?.?. ???????, ?.?. ???????.
? ??? 1989 ???? ? ??????? ????????? ????????????? ????? ????????? ?????????? ??????????????
??????????, ??? ????????? ????? ??? ?????? ?? ?????? ????????? ? ??????? ????. ????? ??? ??????????? ???????????? ??????? ?? ????? ? ???????????????? ?????????????? ?????????? ? ????? ??????
??????????? ???????? ??????????????? ??????????, ?? ?? ??? ?????? ?????? ??????????? ??????????, ? ????????????? ???????? ?????? ? ???????? ?? ???? ???????? ????????? ????????.
??? ? ???????? ???? ?????, ???? ?????? ?????????? ??????: ????????? ? ???????? ????? ?? ?????
??????? ? ????? ?????? ???? ?? ?????????????? ??????????, ?????????????? ??? ?.?. ??????????,
??????????? ???????????? ? ????????????? ???????????, ??????? ???????? ???????, ????????? ???????????? ????? ?????? ????.
??????????? ?????????? ?????????? ?????????????? ?????????? ? 25-??????
???????, ?????? ????? ?????????? ????????, ??????? ????????? ? ?????????? ??
??????????????!
????????
????????? ?????????? ? ??????????? ? ?? ????? ??????? www.swsys.ru.
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