On time asymptotic of the solutions of transport evolution equation
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Інститут кібернетики ім. В.М. Глушкова НАН України
2010
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| Назва видання: | Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки |
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| Цитувати: | On time asymptotic of the solutions of transport evolution equation / E.V. Cheremnikh, F. Diaba, G.V. Ivasyk // Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки: зб. наук. пр. — Кам’янець-Подільський: Кам'янець-Подільськ. нац. ун-т, 2010. — Вип. 4. — С. 208-223. — Бібліогр.: 7 назв. — англ. |
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irk-123456789-487842013-09-03T03:04:26Z On time asymptotic of the solutions of transport evolution equation Cheremnikh, E.V. Diaba, F. Ivasyk, G.V. 2010 Article On time asymptotic of the solutions of transport evolution equation / E.V. Cheremnikh, F. Diaba, G.V. Ivasyk // Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки: зб. наук. пр. — Кам’янець-Подільський: Кам'янець-Подільськ. нац. ун-т, 2010. — Вип. 4. — С. 208-223. — Бібліогр.: 7 назв. — англ. XXXX-0059 http://dspace.nbuv.gov.ua/handle/123456789/48784 517.9 en Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки Інститут кібернетики ім. В.М. Глушкова НАН України |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine |
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DSpace DC |
| language |
English |
| format |
Article |
| author |
Cheremnikh, E.V. Diaba, F. Ivasyk, G.V. |
| spellingShingle |
Cheremnikh, E.V. Diaba, F. Ivasyk, G.V. On time asymptotic of the solutions of transport evolution equation Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки |
| author_facet |
Cheremnikh, E.V. Diaba, F. Ivasyk, G.V. |
| author_sort |
Cheremnikh, E.V. |
| title |
On time asymptotic of the solutions of transport evolution equation |
| title_short |
On time asymptotic of the solutions of transport evolution equation |
| title_full |
On time asymptotic of the solutions of transport evolution equation |
| title_fullStr |
On time asymptotic of the solutions of transport evolution equation |
| title_full_unstemmed |
On time asymptotic of the solutions of transport evolution equation |
| title_sort |
on time asymptotic of the solutions of transport evolution equation |
| publisher |
Інститут кібернетики ім. В.М. Глушкова НАН України |
| publishDate |
2010 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/48784 |
| citation_txt |
On time asymptotic of the solutions of transport evolution equation / E.V. Cheremnikh, F. Diaba, G.V. Ivasyk // Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки: зб. наук. пр. — Кам’янець-Подільський: Кам'янець-Подільськ. нац. ун-т, 2010. — Вип. 4. — С. 208-223. — Бібліогр.: 7 назв. — англ. |
| series |
Математичне та комп'ютерне моделювання. Серія: Фізико-математичні науки |
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AT cheremnikhev ontimeasymptoticofthesolutionsoftransportevolutionequation AT diabaf ontimeasymptoticofthesolutionsoftransportevolutionequation AT ivasykgv ontimeasymptoticofthesolutionsoftransportevolutionequation |
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2025-07-04T09:30:02Z |
| last_indexed |
2025-07-04T09:30:02Z |
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1836708187204485120 |
| fulltext |
Математичне та комп’ютерне моделювання
208
математичного моделювання, прогнозування та оптимізації». — Кам’янець-
Подільський : Кам’янець-Подільський нац. ун-т ім. І. Огієнка, 2010. —
С. 263—264.
6. Khimka U.T., Chabanyuk Ya.M. Stochastic Optimization Procedure Convergence
with Markov Switching in the Average Scheme / U. T. Khimka, Ya. M. Chabanyuk
// Математичні студії. — 2010. — Vol. 34, № 1. — P. 101—105.
It has been established the sufficient conditions for the convergence of
the continuous stochastic optimization procedure of Kiefer-Wolfowitz in
the diffusion approximation scheme with Markov switching. The conver-
gence of the proposed procedure has been proved by using the method of
the small parameter and the solution of the singular perturbation problem
for the Markov process generator.
Key words: stochastic optimization, the markov process, solution of
the singular perturbation problem.
Отримано 16.10.2010
УДК 517.9
E. V. Cheremnikh*, Ph. D.,
F. Diaba**, Ph. D.,
G. V. Ivasyk*, assistant
*Lviv Polytechnic National University, Ukraine,
**Badji Mokhtar-Annaba, Algeria
ON TIME ASYMPTOTIC OF THE SOLUTIONS OF TRANSPORT
EVOLUTION EQUATION
Authors consider in the space 2 ( ), = [ 1,1]L D D R the
transport operator
1
1
1
= ( ) ( ) ( ) , .
f
Lf i a x b b f x d
x
To obtain the representation of the solution of the equation
=0
= , = (0)
t
u iLu u u authors introduce some integral (like
known expression the semigroup by the resolvent) and prove
directly that this integral is corresponding semigroup. To simplify
the calculus the authors reduce the operator L to some Friedrichs'
model using Fourier transformation.
Key words: spectrum, transport operator, Friedrichs' model,
semigroup.
Spectral theory for various types of transport operators is the subject
of many works (see for example [1; 2]). In the work [3] the authors begin
to use Friedrichs' model to study the spectrum of some transport operator.
Analogic result was obtained in the works [4; 5].
© E. V. Cheremnikh, F. Diaba, G. V. Ivasyk, 2010
Серія: Фізико-математичні науки. Випуск 4
209
In the present article for such operators corresponding evolution is con-
sidered. We consider in the space 2 ( )L D , = [ 1,1]D R the transport
operator
1
1
1
= ( ) ( ) ( ) ( , )
f
Lf i a x b b f x d
x
(1)
with maximum domain of definition ( )D L under the following
conditions: there exist constants > 0, > 0M such that
( ) < ,xa x Me x R (2)
and the functions 1( ), ( )b b admit analytic prolongation from interval
(-1,1) into the circle < 1z . Note thaat in [4; 5] we have 1( ) 1b and
in [3] 0 ( ) ( ) 1.b b
We will study the following transport evolution equation
=0
= , > 0
= (0), (0) ( )
t
u iLu t
u u u D L
(3)
and obtain principal term of asymptotic behaviour of the solutions of this
equation corresponding to eigen-values of L . We suppose that the
operator L has not spectral singularities.
1. Friedrichs' model of transport operator
We want to transform the operator L in Friedrichs' model. We begin
by the notations. Let H be Hilbert space of the functions
( , ), ( , ) , = [ 1,1]s s D D R with the norm
1
2 2
1
1
= ( , ) .
H
R
s dsd
We introduce the operator 2
0 : ( )F L D H , where
2
0 , = , , ( ),F u u u L D R
(4)
and the operator 2: ( )Z L R H , where
2( )( , ) = , ( ).Zc c c L R
(5)
It is not difficult to verify that 20 ( )
= ,
L DH
F u u that 0F is unitary
operator and that the operator Z is bounded, namely 2.z The
Fourier transformation is denoted by
Математичне та комп’ютерне моделювання
210
1
( )( ) = ( ) ,
2
is
R
Ff s e f d s R
in the space 2 ( )L R and in the space 2 ( )L D too.
Now we apply to the equality (1) the Fourier transformation with
respect to variable x , then
1
1
1
( )
( )( , ) = ( , ) ( ) ( ) ( , ) ,
2
i x
R
b
FLf u a x b f x d e dx
where =u Ff .
Now we apply the operator 0F (see(4)), simply it is the substitution
= ,
s
then
1
1
0
1
( )
( )( , ) = , ( ) ( ) ( , ) .
2
s
ix
R
bs
F FLf s su a x b f x d e dx
We denote ( , ) = ,
s
s u
or 0 0= = .F u F Ff Let = ,
s
then
1
0( , ) = ( , ) = ( )( , )u F . That's why
1 1
0
1
( , ) = ( )( , ) = ( , ) .
2
i x
R
f x F F x e d
The change of variable =s (cases > 0 and < 0 ) gives
1
, = , .
2
s
i x
R
ds
f x s e
Finally, we obtain
1 1
0 0 , = ( , ) ( , ),F FLF F s s s V s (6)
where
1
1
1
( )
( , ) = ( ) ( ) ( , ) .
2
s s
i ix
R R
b d
V s a x b s e ds e dx
(7)
We choose some factorization for the function ( )a x such that
1 2 1 2( ) = ( ) ( ), ( ) = ( ) .a x a x a x a x a x (8)
Let 2= ( )G L R then *=V A B (see (7)), where the operators
, :A B H G are given by the expressions
Серія: Фізико-математичні науки. Випуск 4
211
*
1 1
1
2
1
1
( , ) ( ) ( ) ( ) ,
2
1
( ) ( ) ( ) ( , ) .
2
s
ix
R
s
i
R
A c s b a x c x e dx
d
B x a x b s e ds
(9)
Obviously, the operator 2
0= : ( )U F F L D H (see(4)) is unitary.
So, the following theorem is proved.
1.1. Theorem. Let 2 2: ( ) ( )L L D L D be the operator with maximal
domain of definition, given by the expression (1). Then
1 = : ,ULU T H H
where
*= , = ,T S V V A B
( )( , ) ( , ), , ( 1,1)S R and the operators ,A B act from
H into 2= ( )G L R (see (9)).
The integral operator in the right side of (1) is bounded in the space
2 ( ).L D So, the operator V is bounded in space H . However we need
following Lemma.
1.2. Lemma. The operators , : ,A B H G namely
1
1 1
1
1
2
1
1 1
( ) = ( ) ( ) ( , )
2
1 1
( ) = ( ) ( ) ( , )
2
s
ix
R
s
ix
R
A x a x b s e d ds
B x a x b s e d ds
(10)
are bounded.
1.3. Proposition. If ( ) 0,a x x R and 1( ) = ( ), ( 1,1)b b ,
then Friedrichs' model =T S V is selfadjoint operator, * = .T T
Proof. Really, as factorization 1 2( ) = ( ) ( )a x a x a x is arbitrary one can
choose 1 2( ) = ( ) = ( )a x a x a x , then according to (10) =A B , so * =T T
in view of * = .S S Proposition is proved.
2. Spectrum of Friedrichs' model
We will consider the resolvent of the operator *= .T S A B The
equation ( ) = , ,T H R takes form *( ) = .S A B
Математичне та комп’ютерне моделювання
212
Denote 1 1= ( ) , = ( )S S T T . As R then the operator S
exists, is bounded and the equation takes form
* = .S A B S (11)
Applying the operator B , we obtain *1 =BS A B BS . Let
*( ) = 1 , ,K BS A R (12)
then for the expression B in (11) we obtain 1= ( )B K BS . Taking
into account the boundness of the operators A and B , we have the
following proposition.
2.1. Proposition. If the operator ( ),K R has bounded inverse
operator 1( )K , then the value belongs to resolvent set of the operator
T and
* 1= ( ) .T S S A K BS (13)
Substituing the expressions (9) for the operators A and B into (12)
we obtain immediately (after the change of variables =
s
) the
following Lemma.
2.2. Lemma. The operator ( )K (see (12)) admits the
representation
(( ( ) 1) )( ) = ( , , ) ( ) , ,
R
K c x k x y c y dy R (14)
where 2 1
1
( , , ) = ( ) ( ) ( , )
2
k x y a x a y I x y
and
1
1
1
( ) ( )
( , ) = ( , ) , ( , ) = .iu
R
b b
I u l e d l d
(15)
This Lemma coincides with corresponding Lemma of the work [4],
where 1( ) 1b and the function ( , )l was (compare with (15))
1
1
( )
( , ) = .
b
l d
2.3. Theorem. Operator 2 2( ) 1: ( ) ( ),K L R L R R is compact
and ( ) 1 0,K uniformly in the domain Im > 0 for every
> 0 .
In the work [4] it was shown that the proof of this Theorem
practically coincides with the corresponding proof in the work [3].
Серія: Фізико-математичні науки. Випуск 4
213
2.4. Theorem. Operator ( ) 1K admits analytic prolongation
( ) 1K over semiaxes ( ,0) and (0, ) and ( ) 1 0,K
uniformly in the domain Im < for every 1 <
2
.
We have the same proof as the proof in the work [4], instead of the
function ( )b one uses the function 0 1( ) ( ) ( )b b b . This proof is
based on the representation
0 0
1 1
( , ) = , = sign ,I u f t u dt f t u dt u
t t
where
0 0
1
( ) = , = 1.i y i yf b e b e dy
y y y
Using well known Theorem on holomorphic operator function
(see [6], Ch. 1) and Theorem 2.4, we obtain that the point spectrum
outside of R can have point of accumulation = 0 only. The condition
of finiteness of such spectrum we can give by analogy to the work [5].
Later we will suppose that point spectrum of the operator T is finite. At
the end we indicate only that continuous spectrum and spectral
singularities of the operator T belong to the axis R .
3. Construction of the semi-group exp( )itT
It is known that Cauchy problem
=0
= , = (0),
t
u Mu u u where the
operator M is such that half plane Re > > 0 belongs to its resolvent
set admits under some condition following representation of the solution
11
( ) = ( , ) (0) , ( , ) = ( ) .
2
i
t
i
u t e R M u d R M M
i
Let us consider the problem
=0
= , > 0,
= (0), (0) ( ).
t
u iTu t
u u u D T
(16)
Denote = , < < ,i then 1( ) = .iiT iT
Instead of difficult verification of some sufficient condition on the
operator T , we propose directly to choose the solution under the form
( )1
( ) = (0) .
2
i t
iu t e T u d
i
(17)
Математичне та комп’ютерне моделювання
214
We denote
( )( , ) = i th t e
and we denote the element (0) = ( , )u H simply by (0) = .u So,
we must prove that the operator
1
( ) = ( , ) , > 0
2 iU t h t T d t
i
(18)
defines semi-group corresponding to the problem (16).
3.1. Theorem. If ( )D T then the integral (18) admits the
representation
1 ( , )
( ) = , > 0,
2 i
h t
U t T T d t
i i
(19)
where the integral converges with respect to the norm of H .
Proof. To prove this theorem it is sufficient to substitute in (17) such
an expression:
1
( ) = 1.i T T
i
3.2. Lemma. If ( )D T then
( , ) ( , )
= .i i
h t h t
T T d T T d
i i
(20)
Proof. As the operator T is closed it’s sufficient to replace the integral
by the corresponding integral sum. Denote ( ) =
h
t
( ( , )h t t
( , )) /h t t .As ( , ) = exp(( ) )h t i t , then ( , ) = ( ) ( , ).h t i i h t
t
3.3. Lemma. If H , then
0
1
( ) = ( , ) .lim i i
t
h
s T d i h t T d
i t
(21)
Note, that the equality (21) in view of definition (18) signifies that
0
1
( ) 2 ( ) , ( ).lim i
t
h
s T d U t D T
i t
(22)
3.4. Theorem. Let ( ),D T then
( ) = ( ) , > 0,U t iTU t t (23)
Серія: Фізико-математичні науки. Випуск 4
215
where ( )U t signifies strong derivative.
Proof. Using the representation (19) and the equality (20) we obtain
two relations:
( ) 1 1
= ( )
2 i
U t h
T T d
t i i t
(24)
and
( ) 1 1
= ( ) .
2 i
U t h
T T d
t i i t
(25)
In view of Lemma 3.3, there exists strong limit of the integral in (24)
if 0,t so there exists strong limit
0
( )
( ) = , ( ).lim
t
U t
U t s D T
t
By analogy, there exist strong limit of the integral in (25), which is
equal to 2 ( )U t (see (22)). As the operator T is closed, we obtain the
equality (23) from the equalities (24)—(25). Theorem is proved.
3.5. Theorem. Let ( ),D T then
0
( ) = 0.lim
t
U t
(26)
Proof. Recall the representation (19)
( )1 ( , )
( ) = , ( , ) = .
2
i t
i
h t
U t T T d h t e
i i
(27)
First we prove that
1
, = .iT d T
i
(28)
Note, that the value > 0 is such that the operator function
0z iz T , =z i is holomorphic in half plane Im <z .
Let = : = , < < 2i
RL z z Re be semicircle such that R .
We obtain (28) if we prove that
1
= 0.lim z i
R LR
T dz
z i
(29)
As 1( ) , = constK z i M M and
1
Imz iS
z
, where
Im < 0z , then we have
Математичне та комп’ютерне моделювання
216
* 1
0
0 0
= ( )
, = const.
Im
z i z i z i z i
z i
T S S A K z i BS
M
M S M
z
(30)
As Im = sinz R , the estimate of the integral (29) reduces to the
following estimate
sin > sin <
1
Im sinL R R R RR
dz d
z i z R
1 1
sin > sin <
1
= , .
R R
d d
O R
R R
So, the relations (28) and (29) are proved. Now, taking into ac-
count (27)—(28) to prove the relation (26) we must prove that
0
( , ) 1
= 0.lim i
t
h t
T d
i
(31)
As 0i iT M S (see(30)), it is sufficient to estimate the
following integral
( ) 1
( )
i t
i
e
I t S d
i
1 1
2 2( ) 221
.
i t
i
e
d S d
i
The function ( ) 1 /i te i is majorable by some function
from 2 ( )L R uniformly in every finite interval 10 < <t t . In addition, the
integral
21
2
2
1
( , ) 1
=
( )
iS d d d d
i
1
2 2
1
1
( , ) = <d d
converges. So, ( ) 0, 0I t t , what proves the relation (31). Theorem is
proved.
Серія: Фізико-математичні науки. Випуск 4
217
3.6. Proposition. If
=0
= , > 0,
= 0,
t
u iTu t
u
then ( ) 0, > 0u t t .
As corrolary, we have that the operator ( ), > 0U t t (see (18)) defines
a semigroup.
Now we can formulate the main theorem of this section.
3.7. Theorem. The problem
=0
= , > 0,
= , ( ),
t
u iTu t
u D T
has unique solution ( ) = ( )u t U t , given by the semigroup
( )1
( ) = .
2
i t
iU t e T d
i
Proof. Results from (23), (26) and Proposition 3.6.
4. Asymptotic behavior of the semigroup
We study time asymptotic of the solution of evolution equation
=f iLf ,
=0
= (0)
t
f f . If the operator 2 2: ( ) ( )L L D L D is self-adjoint
then corresponding semi-group is unitary and this solution is bounded with
respect to the norm of the space. We consider nonself-adjoint operator L .
Our aim is to give simple description of increasing part of the solution
when t .
It is convinient to use Friedrichs` model T (T is unitary equivalent to L)
and study the solution of the problem
=0
= , = (0)
t
u iTu u u .
Suppose that the following conditions hold:
a) 1(0) = (0) = 0b b ;
b) the operator T has not spectral singularities, set of eigen-values is
finite.
We have need of the following known statement (see, for ex. [7]).
4.1. Lemma. If 2 ( )g L R and
1 ( )
( ) =
2 R
g s
g ds
i s
, then
2 2
2 ( )
( ) , 0.
L R
R
g i d g (32)
4.2. Lemma. Vector-functions , ,AS BS H belong to Hardy
space in half planes Im > 0 and Im < 0 .
Proof. Recall that (see(9))
Математичне та комп’ютерне моделювання
218
1
2
1
1
( ) = ( ) ( ) ( , ) .
2
s
ix
R
d
B x a x b e s ds
So,
2
1
( ) = ( ) ( , ) ,
R
BS x a x s x ds
s
(33)
where
1
1
1 ( )
( , ) = ( , ) .
2
s
ixb
s x s e d
As (0) = 0b , then
11
2 21 1 122
1 1 1
( ) ( )
2 ( , ) ( , ) ( , )
b bd
s x s d s d
and
2 2 2
2 ( )
( , ) = ( , ) , .
L R H
R
x s x ds M x R
Using (32), we have
2
2 22 2
2( )
1
( , ) 4 ( , ) 4 , .
( ) L R H
R R
s x ds d x M x R
s i
(34)
The representation (33) gives
2
2 2
2
1
= ( ) ( , ) .
( )i
R R
BS a x s x ds dx
s i
Integrating this equality with respect to and using the estimate (34),
which does not depend on x , we obtain
2 2 2 22
2 14 ( ) , 0.i H H
R R
BS d M a x dx M (35)
The vector-function AS is considered by analogy.
Lemma is proved.
Let ( ), =h i be function holomorphic in half planes > 0
and < 0 .
4.3. Lemma. Suppose that
( ) , = const, 0,
R
h i d M M (36)
then the integral ( ) = ( )
R
H h i d does not depend on in the
intervals > 0 and < 0 .
Серія: Фізико-математичні науки. Випуск 4
219
Now we come back to the solution ( , , )u t of evolution equation
(see Theorem 3.7). We will write ( )u t instead of ( , , )u t . As
* 1= ( )T S S A K BS , then
( )1
( ) = = ( , ) , > 0,
2
i t iSt
iu t e T d e I t t
i
(37)
where
( ) * 11
( , ) = ( ) .
2
i t
i iI t e S A K i BS d
i
(38)
The eigen-values k of the operator T are the poles of the operator
function 1( )K . Recall that the operator ( ) 1K is compact. Therefore,
according to the theorem about holomorphic operator function (see [6]),
the coefficients ,k jQ of principal part of Loran decomposition of
1( )K , namely
1( ) = ( ) ( ), Im 0,k
k
K Q G (39)
, , 1( ) = ... ,
( )
k m k
k km
kk
Q Q
Q
are finite dimensional operators.
Operator function ( )G is bounded and gives in (38) the term
(1), ,O t (see (35)).
It remains to calculate directly (by residues) the terms of the expression
(38), containing the operators ,k jQ then the poles =k i give the
functions exp(( ) ) = exp( )ki t i t and its derivates with respect to
give the factors pt . Using unitary equivalence of semi groups
1exp( ) = exp( )itL U itT U , we obtain following main result concerrning
initial Cauchy problem (3).
4.4. Theorem. Suppose that the operator L has finite set of eigen-
values and has not spectral singularities. Then asymptotic behaviour of the
solution of evolution equation (3) is
, ,
Im <0
( ) = ( (0), ) (1), ,
i t pk
k p H k p
pk
u t e t u f g O t
where (1)O signifies the value in the space 2 ( )L D and , ,,k p k pf g denote
some elements from 2 ( )L D .
Математичне та комп’ютерне моделювання
220
5. Case of finite-dimensional kernel
Let us consider in the space 2 ( )L D more general operator
1
1
( , ) = ( , , ) ( , ) , ( , ) ,
f
Lf x i k x f x d x D
x
(40)
where
1,
=1
( , , ) = ( ) ( ) ( ).
n
j j j
j
k x a x b b (41)
We keep the same conditions (as for the operator (1)) on the
coefficients, namely, for some constants > 0, > 0M , we suppose that
( ) exp , , = 1,...,ja x M x x R j n and the functions 1, ( ), ( ),j jb b
[ 1,1] admit holomorphic prolongation in the circle < 1z .
We want to show that the operator L may be including in previous
scheme. But we consider the spectrum of L only. We choose the
factorization 1, 2,( ) = ( ) ( )j j ja x a x a x , 1, 2,( ) = ( ) ,j ja x a x which after
Fourier transformation of the equality (40) gives Friedrichs' model
*
=1
=
n
j j
j
T S A B (42)
in the same space H . Integral operators 2, : ( )j jA B H L R are
1
1, 1,
1
1
2,
1
1 1
( ) = ( ) ( ) ( , ) , ,
2
1 1
( ) = ( ) ( ) ( , ) .
2
s
ix
j j j
R
s
ix
j j j
R
A x a x b s e d ds x R
B x a x b s e d ds
*( ) = , Im 0jm jm j mK B S A (43)
and (compare with (15))
1
1,
1
( ) ( )
( , ) = .j m
jm
b b
l d
Obviously, theo-
rem 2.3 holds for the operator (43) too. One can easily repeat the proof of
Theorem 3.2 to obtain in the neighbourhood of = 0 the decomposition
21, 1, 2,( )
1
( ) = (0) (0) ( ) , ( ),
2jm jm j m m j jmL R
K b b a a Q
(44)
where the operator function ( )Q is holomorphic in some circle <
and ( ) = signIm lni . The function ln is continuous in the
domain [0, ) and ln( 1) = i .
Серія: Фізико-математичні науки. Випуск 4
221
Let us rewrite the operator (42) under the form *= .T S A B We
introduce direct sum 1= ... ,nG G G where 2= ( )jG L R . Then we
interpret the factors in (42) as operators , :j j jA B H G and introduce
the following operators , :A B H G :
1 1
. .
. , . .
. .
n n
A B
A B G
A B
(45)
If c G then
*( , ) = ( , ) = , , .G m m G m mm
m m H
A c A c A c
So,
* *= .m m
m
A c A c (46)
Therefore * *= j j
j
A B A B and *=T S A B (see (42)). We have
(see (45)—(46))
* * *( ) = =j j j m m
m
BS A c B S A c B S A c
i.e. we have matrix form *
, =1( ) = 1 = ( ( ))n
jm j mK BS A K (see (43)).
Denote
1,1 1,1 1 2,1
2,1, 1,
(0) ( ) (0) ( )
. .
. , . .
. .
(0) ( )(0) ( ) n nn n
b a x b a x
x x G
b a xb a x
Then (see (44))
1
( ( ) 1) = ( )( , ) ( ) ,
2 GK c c Q c
where ( ( ) ) = ( ) .j jm m
m
Q c Q c According to [5]
11
22
11, 1, 1,1( ) ( ) ,jm j m m jCC
Q N b b a a
(47)
Математичне та комп’ютерне моделювання
222
where ( )N is some given function on and
2 2 2( ) .x
R
a a x e dx
Using the notation (see (41))
1
22 2 22
2,1 1( , ) = ( ) x
j j j jC C
j j R
M a b b a b a x e dx
1
22 2 22
1 11, 1, 1,( , ) = ( ) x
m m m mC C
m m R
M a b b a b a x e
and the estimates (see(47)), we obtain 1( ) ( ) ( , ) ( , ).Q N M a b M a b We
also need the value (recall that 2, 1,( ) ( ) = ( )j j ja x a x a x )
1,( , ) = (0) (0) ( ) .G j j j
j R
b b a x dx
By analogy to the work [5], we obtain the following Theorem.
5.1. Theorem. Under one of the following conditions
1) 1( , ) = 0, ( ) ( , ) ( , ) < 1G N M a b M a b
or
2) 1( , ) 0, ( ) ( , ) ( , ) < ( , )G GG G
N M a b M a b
the operator L (see(40)) has not point spectrum in the circle < .
Note, that exponential decreasing of the functions ( ),ja x x
permit (again by analogy to work [5]) to prove the analogy of
Theorem 2.3. Taking into account Theorem 5.1, we obtain that point
spectrum of the operator L is finite.
References:
1. Lehner I. The spectrum of neutron transport operator for the infinit slab,
I.Math. Mech / I. Lehner. — 1962. — №. 2. — P. 173—181.
2. Kuperin Yu. A. Spectral analysis of a one speed transmission operator and
functional model. Funct. anal. and its appl / Yu. A. Kuperin, S. N. Naboko,
R. V. Romanov. — 1999. — Vol. 33, №. 2. — P. 47—58.
3. Diaba F. On the point spectrum of transport operator. Math. Func. Anal. and
Topology / F. Diaba, E. V. Cheremnikh. — 2005. — Vol. 11, №. 1. — P. 21—36.
4. Ivasyk G. V. Friedrich's model for transport operator / G. V. Ivasyk,
E. V. Cheremnikh // Journal of National University «Lvivska Politechnika»,
Phys. and math. sciencesю. — 2009. — Vol. 643, №. 643. — P. 30—36.
5. Cheremnikh E. V. Sufficient condition of finiteness of point spectrum (in print)
/ E. V. Cheremnikh, G. V. Ivasyk.
6. Gohberg I. Z. Introduction to the theory of linear non self-adjoint operators /
I. Z. Gohberg, M. G. Krein. — 1965. — 448 p.
7. Bremermann H. Distributions, Complex Variables and Fourier Transforms /
H. Bremermann, 1965.
Серія: Фізико-математичні науки. Випуск 4
223
Автори розглядають у просторі 1,1][=),(2 RDDL
транспортний оператор
.),()()()(=
1
1
1
dxfbbxa
x
f
iLf
Щоб отримати вигляд розвязку рівняння (0)=,=
0=
uuiLuu
t
автори вводять деякий інтеграл (у вигляді відомого виразу резольвен-
ти через півгрупу) і доводять прямо, що цей інтеграл є відповідною
півгрупою. Щоб спростити обчислення, автори зводять оператор L
до деякої моделі Фрідріхса, використовуючи перетворення Фур’є.
Ключові слова: спектр ,транспортний оператор, модель
Фрідріхса, півгрупа.
Отримано 16.10.2010
УДК 519.21
В. К. Ясинский*, д-р фіз-мат. наук,
В. Ю. Береза*, канд. фіз.-мат. наук,
Е. В. Ясинский**, аналітик-програміст
*Чернівецький національний університет ім. Ю. Федьковича, м. Чернівці,
**Університет Атабаска, м. Едмонтон, Канада
СУЩЕСТВОВАНИЕ ВТОРОГО МОМЕНТА РЕШЕНИЯ
ЛИНЕЙНОГО СТОХАСТИЧЕСКОГО УРАВНЕНИЯ В ЧАСТНЫХ
ПРОИЗВОДНЫХ С МАРКОВСКИМИ ВОЗМУЩЕНИЯМИ И ЕГО
ПОВЕДЕНИЕ НА БЕСКОНЕЧНОСТИ
Для стохастической задачи Коши линейного уравнения в
частных производных с непрерывным марковским процессом
доказано существование решения в среднем квадратическом,
получены достаточные условия асимптотической устойчиво-
сти в среднем квадратическом решения этой задачи.
Ключевые слова: задача Коши, стохастические диффе-
ренциальное уравнение, уравнение в частных производных,
асимптотическая устойчивость, устойчивость в среднем
квадратическом, марковский процесс.
Введение. Доказательству существования и асимптотического
поведения решений детерминированных уравнений в частных произ-
водных посвящено достаточное число монографий и статей, которые
можно найти в монографиях [13], [15], [22].
Когда было введено понятие стохастического дифференциала и
интеграла, как функции верхнего предела, замены переменных Ито
для стохастического дифференциала, введения понятия стохастиче-
© В. К. Ясинский, В. Ю. Береза, Е. В. Ясинский, 2010
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>>
<<
/AllowImageBreaks true
/AllowTableBreaks true
/ExpandPage false
/HonorBaseURL true
/HonorRolloverEffect false
/IgnoreHTMLPageBreaks false
/IncludeHeaderFooter false
/MarginOffset [
0
0
0
0
]
/MetadataAuthor ()
/MetadataKeywords ()
/MetadataSubject ()
/MetadataTitle ()
/MetricPageSize [
0
0
]
/MetricUnit /inch
/MobileCompatible 0
/Namespace [
(Adobe)
(GoLive)
(8.0)
]
/OpenZoomToHTMLFontSize false
/PageOrientation /Portrait
/RemoveBackground false
/ShrinkContent true
/TreatColorsAs /MainMonitorColors
/UseEmbeddedProfiles false
/UseHTMLTitleAsMetadata true
>>
<<
/AddBleedMarks false
/AddColorBars false
/AddCropMarks false
/AddPageInfo false
/AddRegMarks false
/BleedOffset [
0
0
0
0
]
/ConvertColors /ConvertToRGB
/DestinationProfileName (sRGB IEC61966-2.1)
/DestinationProfileSelector /UseName
/Downsample16BitImages true
/FlattenerPreset <<
/PresetSelector /MediumResolution
>>
/FormElements true
/GenerateStructure false
/IncludeBookmarks false
/IncludeHyperlinks false
/IncludeInteractive false
/IncludeLayers false
/IncludeProfiles true
/MarksOffset 6
/MarksWeight 0.250000
/MultimediaHandling /UseObjectSettings
/Namespace [
(Adobe)
(CreativeSuite)
(2.0)
]
/PDFXOutputIntentProfileSelector /DocumentCMYK
/PageMarksFile /RomanDefault
/PreserveEditing true
/UntaggedCMYKHandling /UseDocumentProfile
/UntaggedRGBHandling /LeaveUntagged
/UseDocumentBleed false
>>
]
>> setdistillerparams
<<
/HWResolution [600 600]
/PageSize [419.528 595.276]
>> setpagedevice
|