220
BOUNDARY-VALUE PROBLEMS FOR SECOND ORDER DIFFERENCE EQUATIONS WITH A SPECTRAL PARAMETER IN THE BOUNDARY
CONDITIONS
Aytekin Eryilmaz1& Bilender P. Allahverdiev2
1Department of Mathematics, Nevsehir University, Nevsehir, Turkey
2 Department of mathematics, Süleyman Demirel University, Isparta, Turkey Email: 1eryilmazaytekin@gmail.com;2bilender@fef.sdu.edu.tr
ABSTRACT
This paper is concern with the boundary-value problem in the Hilbert space
l
w2( N )
generated by an infinite Jacobi matrix with a spectral parameter in the boundary condition. We proved theorems on the completeness of the system of eigenvalues and eigenvectors of operator generated by boundary value problem.Keywords: Second order difference equation, Jacobi matrix, Dilation, Dissipative operator, Completeness of the system of eigenvectors and associated vectors, Scattering matrix, Functional model, Characteristic function.
Subject Classification 2000: 47A20, 47A40, 47A45, 47B39, 47B44
1. INTRODUCTION
Spectral theory of operators plays a major role in mathematics and applied sciences. The studies of boundary value problems with a spectral parameter in the boundary conditions take a significant place in spectral theory of operators (see [1-5,7-9,16-18]). First general method in the spectral analysis of nonselfadjoint differential operators is the method of contour integration of resolvent which is studied by Naimark (see[13]). But this method is not effective in studying the spectral analysis of boundary value problem. This method is also called functional model method (see1-5,7,14,15). The rich and comprehensive theory has been developed since pioneering works of Brodskii, Livsiç and Nagy-Foiaş.The functional model techniques are based on the fundamental theorem of Nagy-Foia
s
. In 1960's independently from Nagy-Foias
, Lax and Phillips developed abstract scattering programme that is very important in scattering theory. This theory is based on the notion of incoming and outcoming subspaces for get information about analytical properties of scattering matrix by utilizing properties of orijinal unitary group.By combining the results of Nagy-Foia
s
and Lax-Phillips, characteristic function is expressed with scattering matrix and the dilation of dissipative operator is set up. By means of different spectral representation of dilation, given operator can be written very simply and functional models are obtained. The eigevalues, eigenvectors and spectral projection of model operator is expressed obviously by characteristic function. The problem on completeness of the system of eigenvectors is solved by writing characteristic function as factorization.In this paper, we consider to study nonselfadjoint boundary value problem, generated by infinite Jakobi matrix, with a spectral parameter in the boundary conditions. We applied this approach to the study of dissipative difference operator generated in the space
l
w2( ) N
. Our goal is to compute the characteristic function of the dissipative extensions. To accomplish this goal we constructed a selfadjoint dilation of dissipative operator and its incoming and outcoming spectral representations. Because this makes it possible to determine the scattering matrix of dilation according to the Lax and Phillips scheme [11]. Moreover, we constructed a functional model of dissipative operator by means of the incoming and outcoming spectral representations and defined its characteristic function. Finally, we proved a theorem on completeness of the system of eigenvectors and associated vectors of dissipative operators under consideration.2.CONSTRUCTION OF DIFFERENCE OPERATOR IN THE HILBERT SPACE
In for
a
n 0
andIma
n= Imb
n= 0 ( n N ),
an infinite Jacobi matrix is defined as follows221
.
. . . . . . . .
. . . . . . . .
. . . 0
0
. . . 0 0
. . . 0 0
. . . 0 0 0
=
3 3 2
2 2 1
1 1 0
0 0
a b a
a b a
a b a
a b
J
For all sequences
y = y
n( n N )
that composed of complex numbersy
0, y
1,...
ofly
which has components form of( ) ( ly
nn N )
is defined as follows: := 1 ( ) = 1 (
0 0 0 1)
00 0
0
b y a y
Jy w
ly w
1, ), 1 (
= ) 1 ( :=
)
( a
1y
1 b y a y
1n
Jy w
ly w
n n n n n nn n n n
where
w
n> 0 ( n N )
.Definition 1. For all
n N ,
Green's formula is defined as fallows
=0
( ) ( ) = , ( ).
nj
w ly z
j j j w y l z
j j j y z
nn N
(2.1)Then let us construct Hilbert space
l
w2( N ) ( := w w
nn N )
consisting of all complex sequences ( )
= y n N
y
n provided n=0w
ny
n 2<
, with the inner product=0
( , ) =
n n n ny z w y z
in order to pass fromthe matrix
J
to operators. Let us denote withD
the set ofy = y
n( n N )
sequences inl
w2( ) N
providing)
2
(
w
N l
ly
. DefineL
onD
by the equalityLy = ly
. For ally , z D
, we obtain existing and being finite ofthe limit
nn
y z
z
y , = lim ,
from (2.1). Therefore, if we pass to the limit as
n
in (2.1), it is obtained , .
= ) , ( ) ,
( Ly z y Lz y z
(2.2)Let us take into account the set of linear
D
0' consisting of finite vector having only finite many nonzero components inl
w2( N )
. We denote the restriction ofL
operator inD
0' byL
0'.
It is obtained from (2.2) thatL
'0 operator is symmetric. The clousure ofL
'0 operator is denoted byL
0.
The domain ofL
0 operator isD
0 and it consists of the vector ofy D
satisfying the condition y , z
= 0, z D .
The operatorL
0 is a closed symmetric operator with defect index(0,0)
and(1.1)
. Furthermore,L = L
0(see [1,6,8,18]). The operatorsL
0 andL
are called respectively the minimal and maximal operators generated by infinite Jakobi matrix J. The operatorL
0 is a selfadjoint operator for defect index(0,0).
NamelyL
0= L
0= L
.Suppose the solution of equation of a second order difference equation on the half-line
1,2,...)
= (
1
=
1
1
y b y a y w y n
a
n n
n n
n n
n n (2.3)satisfying initial conditions of
222
0 1
0 0
0 0 1
0
= 1 ) ( 0,
= ) ( ,
= ) ( 1,
= )
( Q Q a
a b P w
P
(2.4)be
P ( ) = P
n( )
andQ ( ) = Q
n( ) .
Here the functionP
n( )
is called the first kind polynomial of degreen
in
and the functionQ
n( )
is called the second kind polynomial of degreen 1
in
(see [1]). For 1,
n P ( )
is a solution of( Jy )
n= w
ny
n isP
n( )
. ButQ ( )
is not a solution of( JQ )
n= w
nQ
nbecause of
1 = 1 = 0 = ,
0
=
= )
(
00 0 0 1 0 0 0
0
Q
a a b Q a Q b
JQ
Forn N
and under boundary condition0,
1
=
y
the equation( Jy )
n= w
ny
n is equivalent to (2.3). For two arbitrary sequencesy = y
n and z
nz =
, The Wronskian of the solutionsy
andz
of the equation (2.3) is defined as follows:1 1
( , ) := ( ) = [ , ] , ( ).
n n n n n n n
W y z a y z
y
z y z n N
The Wronskian of the two solutions of (2.3) does not depend on
n ,
and two solutions of this equations are linearly indepent if only ifW
n( y , z ) 0
. From the condition (2.4), we getW
0( P , Q ) = 1
. Consequently, from Wronskian constacy, the solutions ofP ( )
andQ ( )
form a fundamental system of solutions (2.3).Suppose that the minimal symmetric operator
L
0 has defect index (1,1) so that the Weyl limit circle case holds for the expressionly
(see [1-4], [6-8]). As the defect index ofL
0 is (1,1) for all C
the solutions ofP ( )
and) (
Q
belong tol
w2( N ).
The solutions ofu = u
n andv = v
n of the equality (2.3) beu = P (0)
and(0)
= Q
v
satisfying the initial condition of0 1 0 0 0 1 0
= 1 0,
= ,
= 1,
= v v a
a u b
u
while
= 0.
In addition it isu , v D
and( Ju )
n= 0, ( n IN ), ( Jv )
n= 0, n 1.
Lemma 1. [2]For arbitrary vectors
y = y
n D
andz = z
n D
it is y z = y u ,
n,
n[ , ] z v
n y v ,
n[ , ] z u , (n
n N ) .
Theorem 1. [2]The domain
D
0 of the operatorL
0 consists precisely of those vectorsy D
satisfying the following boundary conditions y , u
= y , v
= 0.
Consider the boundary value problem governed by
1, , ,
= )
( ly
n y
ny D n
(2.5)subject to the boundary conditions
0
>
0,
1
=
0
hy Imh
y
(2.6) ,
2 , = (
1 ,
2 , )
1
y v
y u
'y v
'y u
(2.7)for the following difference expression
1 ( )
= ) 1 (
:=
0 0 0 10 0 0
0
b y a y
Jy w
ly w
1 ), 1 (
= ) 1 ( :=
)
( a
1y
1 b y a y
1n
Jy w
ly w
n n n n n nn n n n
223
where
is spectral parameter and
1,
2,
1',
2' R
and
is defined by0.
>
=
:=
1 2 1 22 2
1
1 ' '
' '
For the sake of brevity, let us suppose that the followings
, , , :=
)
(
1 2
y y v y u
M
, , , :=
)
(
1 2
y y v y u
M
'
'
', :=
)
(
10
1
y y
N
, :=
)
(
00
2
y y
N
, ,
:=
)
1
(
v y y N
, ,
:=
)
2
(
u y y N
).
( )
( :=
)
(
20 100
y N y hN y
M
Lemma 2. [9]For arbitrary
y , z D
suppose thatM
( z ) = M
( z ), M
'( z ) = M
'( z )
and) (
= ) ( , ) (
= )
(
10 20 200
1
z N z N z N z
N
then it isi)
1
( )
( )
( )
( )
=
, z M y M z M y M z
y
' '
(2.8)ii)
, = ( ). ( ) ( ).
20( ).
0 1 0
2 0 1
1
N y N z N z N y
z
y
(2.9)Supposing
f
(1) l
w2( ), N f
(2) C ,
we denote linear spaceH = l
w2( ) N C
with two component of elementsof
=
(2).
(1)
f
f f
Supposing2 2
1 1
:=
, if
> 0
and(1) (1)
(1) (1) (1) (1)
(2) (2)
= f , = g , = (
n), = (
n)( ),
f g H f f g g n N
f g
then the formula
1 (2) (2)
(1) 0
=
= 1
, g f g w f g
f
n n n n
(2.10)
defines an inner product in Hilbert space
H
. In terms of this inner product, linear spaceH
is a Hilbert space. Thus it is Hilbert space which is suitable for boundary value problem that has been defined. Let us define operator ofH H
A
h:
with equalities suitable for boundary value problem
( )
= ( , :
=
= )
(
(1) 0 (2) (1)(2) (1)
f M f
M D f
f H f f A
D
h (2.11)and
224
.
:=
) (
=
(1)(1)
f M
f f l
l f A
h:
(2.12)
Lemma 3.[9]In Hilbert space
H = l
w2( ) N C
forA
h operator defined with equalities (2.11) and (2.12) the equality(1) (1) (1) (1)
, , = ,
1,
h h
A f g f A g f g f g
(2.13)
( ) ( ) ( ) ( )
1
(1) (1) (1) (1)g M f M g
M f
M
is provided.
Theorem 2.[9]
A
h operator is dissipative inH
.For all
C ,
the solutions of (2.5) be ( )
and ( )
for the following conditions:1,
= ) (
= )) (
(
10
1
N
,
=
= )) (
(
00
2
y h
N
,
= )) (
(
2 21
N
1
( ( )) =
1 1N
(2.14)From (2.9) for
1( )
having Wronskian is
1
11
( ) := ( ), ( )
= ( ), ( )
)) ( ( )) ( ( )) ( ( )) ( (
= N
10 N
20 N
10 N
20 ))
( ( ))
( (
= N
20 hN
10 )).
( (
= M
0
From (2.8) for
( )
having Wronskian is
( ) := ( ), ( ) = ( ), ( )
( ( ) ( ( )) ( ( )) ( ( ))
= 1
M M M M
Therefore, in terms of the definition of
, it is)) ( ( )))
( ( ))(
( ( )))
( ( 1 [(
= )
(
1 1
2 2
1 1
2 2
N N N N
))]
( ( ))
( ( ))(
( ( ))
(
(
2 2 1 1 2 21
1
N N N N
( ( )) ( ( )) ( ( )) ( ( ))
= 1
1
2
2
1 1
2
2
1
N N N N
1( ( ))
1 1 2( ( ))
2 2
= 1
N
N
( ( )) ( ( ))
)) ( ( ))
( (
=
1N
1
2N
2
1N
1
2N
2
)).
( ( )
( (
= M
M
225
Lemma 4.[9]Boundary values problem
2.5 2.7
has eigenvalues iff it consists of zeroes of ( )
. ( ) =
1( ) =
( ) .
Definition 2.If the system of vectors of
y
0, y
1, y
2,..., y
n corresponding to the eigenvalue
0 are y
0=
0y
0,
l
y
0 0M y
0= 0, M
0= 0,
0
y
M y
s
0y
s y
s1= 0,
l
0
1= 0,
y
s M y
s M y
sM
0
s= 0, = 1, 2,..., .
M y s n
(2.15)Then the system of vectors of
y
0, y
1, y
2,..., y
n corresponding to the eigenvalue
0 is called a chain of eigenvectors and associated vectors of boundary value problem(2.5) (2.7)
.Lemma 5.[9]The eigenvalue of boundary value problem
(2.5) (2.7)
coincides with the eigenvalue of dissipativeA
h operator. Additionally each chain of eigenvectors and associated vectorsy
0, y
1, y
2,..., y
n corresponding tothe eigenvalue
0 corresponds to the chain eigenvectors and associated vectorsy y y y
n
,..., , ,
1 20
corresponding to the same eigenvalue
0 of dissipativeA
h operator. In this case, the equality y k n
M y y
k k
k
= , = 0,1,2,...,
is valid.
3.MAIN RESULT
In this section we mention about selfadjoint dilation, scattering theory dilation and functional model of the maximal dissipative operator
A
h.
One of the completeness proofs of nonselfadjoint oprators is the method of dilation and characteristic function. To do this, we first build a selfadjoint dilation of the maximal dissipative operatorA
h.
Definition 3. Let add the " incoming" and " outgoing" subspaces
D
= L
2 ,0
andD
= L
2 0,
toC N ) (
= l
w2H
. The orthogonal sum H= D
H D
is called mainHilbertspace of the dilation.
In the space
H ,
we consider the operatorL
h on the setD L
h,
its elements consisting of vectorsw = ,
,
,
y
generated by the expression
d
i d y d l
i d
y , = , ,
L ,
(3.1)satisfying the conditions:
W
21 ,0 ,
W
21 0, , y H ,
,
=
21
y
y y y
1 D , y
2= M
y
1,
0 ,
=
11
0
hy
a
y y
0 h y
1= a
1
0 ,
whereW
21 .,.
are Sobolev spaces and
2:= 2 Imh , 0.
>
226
Then we state our main result for the spectral analysis of the dissipative operator
A
h.
Theorem 3. For all the values of
h
withImh > 0,
except possibly for a single valueh = h
0,
the characteristic functionS
h
of the maximal dissipative operatorA
h is a Blaschke product. The spectrum ofA
h is purely discrete and belongs to the open upper half-plane.The operatorA
h h h
0
has a countable number of isolated eigenvalues with f nite multiplicity and limit po nts at inf nity. The system of all eigenvectors and associated vectors of the operatorA
h h h
0
is complete in the spaceH
.To prove theorem we have:
Theorem 4. The operator
L
h is selfadjoint inH
and it is a selfadjoint dilation of the operatorA
h.
Proof. We first prove that
L
h is symmetric inH
. Namely L
hf , g
H f , L
hg
H= 0.
Letf , g D L
h,
, ,
= y
f
and= , , .
z
g
Then we have
, =
, ,
,
, ,
, ,
,
, ,
, g f g y z y z
f
hh
L L L
L
H H. ,
, ,
, , ,
, , ,
,
=
d i d z d l
i d y
d z i d y d l
i d
Now, passing to the components of space
H
, it yields
d
d i d z
y l d d
i d
H
, , ,
=
0 0
d
d i d z
l y d d
i d
H
, , ,
0 0
d l y z i d
i
'H
0 0
,
=
d y l z i d
i
'H '
0 0
,
By
2.8
, 2.13
and integrating by parts, we get L
hf , g
H f , L
hg
H= y
1, z
1
1 y
1, z
1
0 0 0 0
1
(1) (1) (1) (1)
'' ' '
'
y M y M y i i
M y M
1 2 0 1 11
0 1 11
0 1 11
0 1 11
1 1
,
=
i y hy
z hz
y hy
z h z
z
y
227
1,
1
1 2
0 1 01 0 1 11 11 01 11 11
=
i y z h y z
hy
z h h y
z
z
y
01 11 01 11 11 01 11 01
2
y z hy z h y z h h y z
i
01
1 1 1
1 1 2 0
1 1 1
,
= i h h y z h h y z
z
y
01
1 1 1
1 1 2 0
1 1 1
,
= i h h y z h h y z
z
y
1 1 1 1
1 1
= y , z y , z = 0.
(3.2)Thus,
L
h is symmetric operator. To prove thatL
h is selfadjoint, we need to show thatL
h L
h.
We consider the bilinear form L
hf , g
H on elements
z D
hg = , , L
, wheref y D L
h
, ,
=
, ,
1
2
W R
0 = 0.
Integrating by parts, we get
d i d d z
i d
h
g = , ,
L
, where
W
21 R
,
H z
. Similarly, if
f = 0, y ,0 D L
h,
then integrating by parts in L
hf , g
H, we obtain
,
=
.
, ,
,
= , ,
= z
1D z
2M z
1d i d z d l
i d z
h
g
L
L
(3.3)Consequently, from
(
3.3), we have L
hf , g
H f , L
hg
H, f D L
h,
where the operatorL
is defined by 3.1 .
Therefore, the sum of the integrated terms in the bilinear form L
hf , g
H must be equal to zero:
1 1
(1) (1)
(1) (1)
11
1
1
,
, z y z M y M y M y M y
y
' '
0
0
0
0 = 0.
i
' i
'
(3.4)Then from
2.8
, we get
,
1 0 0 0 0 = 0.
1
1
z
i
'
i
'
y
(3.5)From the boundary conditions for
L
h,
we have
1
=
10 0 ,
0=
10 0 0
'
i a h
i y y a
Then
2.9
and 3.5
, we get
0 0
0 0 0 0 1
1
z
i z h
i
'
'
0 0 0 0
= i
'
i
'
(3.6)Comparing the coefficients of
0
in 3,6
we obtain 0 1 =
0 1 2