On continuous spectrum of transport operator

В работе доказано, что точка Ζ=0 в отличие от других точек непрерывного спектра есть точкой разветвления логарифмического типа резольвенты транспортного оператора. В работе доказано, что точка Z=0 в отличие от других точек непрерывного спектра есть точкой разветвления логарифмического типа резольвен...

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Published in:Таврический вестник информатики и математики
Date:2010
Main Authors: Cheremnikh, E.V., Ivasyk, G.V.
Format: Article
Language:English
Published: Кримський науковий центр НАН України і МОН України 2010
Online Access:https://nasplib.isofts.kiev.ua/handle/123456789/18199
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Journal Title:Digital Library of Periodicals of National Academy of Sciences of Ukraine
Cite this:On continuous spectrum of transport operator / E.V. Cheremnikh, G.V. Ivasyk // Таврический вестник информатики и математики. — 2010. — № 2. — С. 71-77. — Бібліогр.: 5 назв. — англ.

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Digital Library of Periodicals of National Academy of Sciences of Ukraine
id nasplib_isofts_kiev_ua-123456789-18199
record_format dspace
spelling Cheremnikh, E.V.
Ivasyk, G.V.
2011-03-18T15:12:09Z
2011-03-18T15:12:09Z
2010
On continuous spectrum of transport operator / E.V. Cheremnikh, G.V. Ivasyk // Таврический вестник информатики и математики. — 2010. — № 2. — С. 71-77. — Бібліогр.: 5 назв. — англ.
1729-3901
https://nasplib.isofts.kiev.ua/handle/123456789/18199
517.983
В работе доказано, что точка Ζ=0 в отличие от других точек непрерывного спектра есть точкой разветвления логарифмического типа резольвенты транспортного оператора.
В работе доказано, что точка Z=0 в отличие от других точек непрерывного спектра есть точкой разветвления логарифмического типа резольвенты транспортного оператора.
It is proved that the point Z=0 in difference of other points of continious spectrum is point of branchement of logarithmic type of the resolvent of transport operator.
en
Кримський науковий центр НАН України і МОН України
Таврический вестник информатики и математики
On continuous spectrum of transport operator
Article
published earlier
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
title On continuous spectrum of transport operator
spellingShingle On continuous spectrum of transport operator
Cheremnikh, E.V.
Ivasyk, G.V.
title_short On continuous spectrum of transport operator
title_full On continuous spectrum of transport operator
title_fullStr On continuous spectrum of transport operator
title_full_unstemmed On continuous spectrum of transport operator
title_sort on continuous spectrum of transport operator
author Cheremnikh, E.V.
Ivasyk, G.V.
author_facet Cheremnikh, E.V.
Ivasyk, G.V.
publishDate 2010
language English
container_title Таврический вестник информатики и математики
publisher Кримський науковий центр НАН України і МОН України
format Article
description В работе доказано, что точка Ζ=0 в отличие от других точек непрерывного спектра есть точкой разветвления логарифмического типа резольвенты транспортного оператора. В работе доказано, что точка Z=0 в отличие от других точек непрерывного спектра есть точкой разветвления логарифмического типа резольвенты транспортного оператора. It is proved that the point Z=0 in difference of other points of continious spectrum is point of branchement of logarithmic type of the resolvent of transport operator.
issn 1729-3901
url https://nasplib.isofts.kiev.ua/handle/123456789/18199
citation_txt On continuous spectrum of transport operator / E.V. Cheremnikh, G.V. Ivasyk // Таврический вестник информатики и математики. — 2010. — № 2. — С. 71-77. — Бібліогр.: 5 назв. — англ.
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first_indexed 2025-11-24T11:48:40Z
last_indexed 2025-11-24T11:48:40Z
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fulltext UDC 517.983ON CONTINUOUS SPECTRUM OF TRANSPORT OPERATOR Cheremnikh E. V., Ivasyk G.V.National University "Lvivska Polite hnika"12,S.Bandery Street, 79013, Lviv, Ukrainee-mail: Ivasyk-G�yandex.ruAbstra t. It is proved that the point � = 0 in di�eren e of other points of ontinious spe trum ispoint of bran hement of logarithmi type of the resolvent of transport operator.Introdu tionWe onsider partial ase of so- alled "equation of transmission". There is mu h lit-erature on erning (during many years) di�erent problems in this dire tion. One of su hproblems, namely the problem of neutron transport, leads to the operatorLf(x; �) = �i��f�x (x; �) + (x) 1Z�1 f(x; �0)d�0 (1)in the spa e L2(D), where D = R� [�1; 1℄. In [1℄ in the ase (x) = � ; jxj < a0; jxj > a; = onstit was obtained that ontinuous spe trum of the operator L oin ides withreal axis R and that the set of eingen-values is �nite. In [2℄ in the ase 2 L1(R); supp � [�a; a℄; (x) � 0 well-known fun tional model is applied.1. Statement of the problemAmong other publi ations we mention only several of them, whi h are the losest toour problem. In [3℄ the authors use Friedri hs' model to study the operator L. In the ase of exponentialy de reasing potential the su� ient ondition of �niteness of pointspe trum was obtained. The methods of this work were used in [4℄ in more general aseof the operator Lf(x; �) = �i��f�x (x; �) + a(x) 1Z�1 b(�0)f(x; �0)d�0: (2)As it was proved in [4℄ the value � = 0 only an be the point of a umulation of pointspe trum of the operator L if the following onditions hold:a) the fun tion a(x) is lo ally integrable and satis�es the estimateja(x)j �Me�jxj; x 2 R; (3)where " > 0; M > 0 are some onstants;b) the fun tion b(�); � 2 (�1; 1) admits analyti prolongation b(z) into the ir lejzj < 1 + ". 72 Cheremnikh E. V., Ivasyk G. V.In that work it was proved that resolvent has analyti prolongation over the semi-axis (�1; 0) and (0;1): But in this work the point of spe trum � = 0 remains to beunstudied.Our aim is to prove that the point � = 0 is the point of bran hment of theresolvent.Apropos in a similar situation in the work [3℄ it was proved that the point � = 0was spe tral singularity of onsidered operator. Like [3-4℄ we use unitary equivalen e ofthe operator L to the operator of Friedri hs' model.2. PreliminaryHere we give some notations and results from [4℄.Let H be Hilbert spa e of the fun tions on two variables '(s; �); (s; �) 2 D with normjj'jj2H = ZR 1Z�1 j'(s; �)j2 1j�jdsd�and let G = L2(R). We denote by (�; �); (�; �)H s alar produ t in the spa es G and Hrespe tively. We denote by S : H ! H the operator of multipli ation by independentvariable (S')(�; �) � �'(�; �); � 2 R with maximal domain of de�nition. Using Fouriertransformation it was proved in [4℄ that the operator L : L2(D) ! L2(D) is unitaryequivalent to the operator T = S + A�B : H ! H (Friedri h's model) with boundedoperators A� : G! H; B : H ! G under the formA� (s; �) = 12� ZR a1(y) (y)e�iy s�dy; (4)and B'(x) = a2(x) ZR eix� 0� 1Z�1 b(�0)'(��0; �0)d�01A d�: (5)We use the traditional form of perturbation A�B, that's why we don't need the oper-ator A : H ! G itself. The representations (4)-(5) ontain the fa tors a1;2(x) of arbitraryfa torization su h that a(x) = a1(x)a2(x); ja1(x)j = ja2(x)j :The relation between the resolvents T� = (T � �)�1 and S� = (S� �)�1 of the operators Tand S is the following T� = S� � S�A�K(�)�1BS� , where K(�) = 1 +BS�A�.3. Estimate of the operator K(�); � ! 0It is shown in [3℄ that((K(�)� 1) )(x) = ZR k(x; y; �) (y)dy; where k(x; y; �) = 12�a2(x)a1(y)I(x� y; �) (6)¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �2' 2010 On ontinuous spe trum of transport operator. 73andI(u; �) = ZR l(�; �)eiu�d�; u = x� y; where l(�; �) = 1Z�1 b(�0)��0 � � d�0; Im(�) 6= 0: (7)Let Æ be arbitrary value su h that 0 < Æ < " ( see (3) ) and �(Æ) = f� : j�j < Æ; �Im � > 0g. By ln� we denote the bran h of logarithmi fun tionwhi h is ontinuous in the domain � =2 [0;1) and su h that ln(�1) = �i:If b(�) � 1 then we denote by I0(u; �) the expression I(u; �) (see[3℄)I0(u; �) = (�) +R0(u; �); � 2 �(Æ); (8)where (�) = ��i sign � � ln�; � = Im� (9)and the term R0(u; �) admits the estimatejR0(u; �)j �M " 1juj 1e + juj# ; p > 1; � 2 �(Æ); M = onst; (10)whi h is independent of �. Underline that (�)!1; � ! 0 and the de omposition like (8)is not unique. Let us introdu e the following notationkbkC1 = supjzj<1+" jb(z)j + supjzj<1+" jb0(z)j and N0(Æ) = sup�2 �(Æ)0� 1Z" dtjt� �jq1A 1q : (11)Lemma 1. The fun tion I(u; �), de�ned in the relations (6)-(7), an be represented inthe form I(u; �) = b(0) (�) +R(u; �); (12)where jR(u; �)j � N1(Æ)" 1juj 1e + juj# ; p > 1; � 2 �(Æ); (13)where N1(Æ) = CN0(Æ) kbkC1 and C denote some onstant, whi h is independent of Æ; �and also of the fun tion b(�).Proof. Let us denote b1(z) � b(z) � b(0). We substitute in (7) the de omposi-tion b(z) = b1(z) + b(0), separating I0(u; �) (what orrespond to with b(z) � 1) in theright part of (7) and taking into a ount the de omposition I0(u; �) itself (8), we obtainfor � 2 �(Æ)I(u; �) = 1Z0 1t� � f�!;1(t juj)dt� 1Z0 1t+ � f!;1(t juj)dt+ b(0) (�) + b(0)R0(u; �); (14)where f�!;1(�) = 1Z� 1y �b1 ��y� ei!y + b1 ���y� e�i!y� dy: (15)¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �2' 2010 74 Cheremnikh E. V., Ivasyk G. V.Integrating by parts, we get the estimatejf�!;1(�)j � � 2 kbkC1 ; � 2 (0; 1)4 kbkC1 =�; � 2 (1;1): (16)It follows from here that the interval of integrations (0;1) in (14) an be hanged bythe interval (0; ") (the value of " see in (3)) and the di�eren e between the integrals willhave the estimate like (13). In the integral (15) we put � = t juj and make the hange ofvariable yjuj = �, then in view that u = sign u � juj = ! juj we have:f�!;1(t juj) = 1Zt 1� �b1 � t�� eiu� + b1 �� t�� e�iu�� d�: (17)A ording to (14) we need the value t < ". It's easy to verify that in (17) the interval ofintegrating (t;1) an be hanged by (t; ") and therefore we an onsider the integralsg�(t; u) = "Z0 1� � b1�� t�� e�iu�d�: (18)In the right part of (14) it remains to onsider the sum I+(u; �) + I�(u; �), whereI�(u; �) = "Z0 g�(t; u)t� � dt� "Z0 g�(t;�u)t+ � dt: (19)�Theorem 1. Let Æ < ", thenK(�)� 1 = b(0)2� (�)(�; a1)a2 +Q(�); � 2 �(Æ); (20)where the elements a1, a2 are de�ned by the fa torization a(x) = a1(x)a2(x),ja1(x)j = ja2(x)j and the operator Q(�) : L2(R) ! L2(R) is ompa t with the normbounded uniformly with respe t to �, namelykQ(�)k �M kakÆ ; kak2Æ � ZR ja(x)j2 e2Æjxjdx; � 2 �(Æ): (21)Proof. A ording to (6) and (12), we havek(x; y; �) = b(0)2� (�)a2(x)a1(y) + 12�a2(x)a1(y)R(x� y; �);what proves the de omposition (20). FurtherkQ(�)k2 � 14�2 ZR ZR ja2(x)j2 ja1(y)j2 jR(x� y; �)j2 dxdy:¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �2' 2010 On ontinuous spe trum of transport operator. 75Due to the relations ja2(x)j2 = ja(x)j, ja1(y)j2 = ja(y)j, we obtainkQ(�)k2 � 14�2 ZR ZR ja(x)j ja(y)j eÆjxjeÆjyj �e�Æ(jxj+jyj) jR(x� y; �)j2� dxdy ��M20 0�ZR ZR ja(x)j2 ja(y)j2 e2Æjxje2Æjyjdxdy1A12 ==M20 0�0�ZR ja(x)j2 e2Æjxjdx1A21A 12 =M20 kak2Æ ; (22)where due to the estimate (13) under the ondition p > 2 the valueM20 = 0�ZR ZR e�2Æ(jxj+jyj) jR(x� y; �)j2 dxdy1A12is �nite. Theorem is proved. �We substitute (18) in (19), hange the order of integrating and we make the hangeof variable t� = � , then in the ase of sign ½+\I+(u; �) = 1Z0 b1(�)0� "Z�" eiu��t� � d�1A d�:By integrating by parts, we have the de omposition"Z�" eiu��� � � d� = 1� 24eiu"ln("� � �)� e�iu"ln(�"� � �)� iu "Z�" eiu�ln(�� � �)d�35 ; (22)whi h leads us to the estimate jI+(u; �)j � C kbkC1 [juj+ 1℄ ; � 2 �(Æ); C = onst:The value I�(u; �) has analogi estimate. Really (let us onsider b1(�) � 1),the value 1Z0 ln("� � �)d� is bounded for � 2 �(Æ) if the integralG(�) � 1Z0 [ln("� � �)� ln"� ℄d� = 1Z0 ln(1� �"� )d�is bounded too. If � = j�j s, thenG(�) = j�j 1j�jZ0 ln�1� �" j�j � 1s� ds¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �2' 2010 76 Cheremnikh E. V., Ivasyk G. V.and using the inequality ���ln�(1� �"j�j � 1s )���� � Ms ; s > 1; M = onst, we obtainjG(�)j � C j�j ln 1j�j or jG(�)j � C; � 2 �(Æ): Lemma is proved.Now we onsider � = 0 as the point of the spe trum of the operator L (or the opera-tor T).Statement 1. The value � = 0 is not eigen-value of the operator L.Proof. If (see(2))�i��f�x (x; �) + a(x) 1Z�1 b(�0)f(x; �0)d�0 � 0; f 2 L2(D):then integrating from 0 to x gives�i�(f(x; �))� f(�; �)) = B(x);where the fun tion B(x) = � xZ0 a(t)0� 1Z�1 b(�0)f(t; �0)d�01A dthas not limit value limx!+1B(x) = B1. Then i�f(0; �) = B1 and�i�f(x; �) +B1 = B(x): But f(x; �) = (B(x)� B1)=(�i�) =2 L2(D);what proves Statement. �Statement 2. The value � = 0 is point of bran hment of linear form of the resol-vent (T�'; ), where '; are smooth elements.Proof. We onsider the fun tions '(�); (�); whi h admit analyti prolongation in theband jIm�j < ": A ording to (20)jK+(�)j = jln j� jj+O(1); � ! 0:So, the fun tion K+(�) is not bounded if � ! 0. By the same way � = 0 is not pole ofthe fun tion K+(�) what proves the statement. �Con lusionAs a result in this work it was obtained: the point � = 0 in di�eren e of other pointsof ontinious spe trum is point of bran hement of logarithmi type of the resolvent oftransport operator. The operators whi h are more general than (1) are interesting indi�erent appli ations so the same problem will be a tual for su h operators.¾Òàâðè÷åñêèé âåñòíèê èí�îðìàòèêè è ìàòåìàòèêè¿, �2' 2010 On ontinuous spe trum of transport operator. 77Referen es1. Lehner I. The spe trum of neutron transport operator for the in�nit slab // I.Math. Me h. 11(1962),n.2, P. 173-181.2. Kuperin Yu.A., Naboko S.N., Romanov R.V. Spe tral analysis of a one speed transmission operatorand fun tional model, Fun t. anal. and its appl. (1999), v.33, n.2, P. 47-58 (Russian).3. Diaba F. Cheremnikh E.V. On the point spe trum of transport operator, Math. Fun , Anal. andTopology, v.11, n.1, 2005, P. 21-36.4. Ivasyk G.V., Cheremnikh E.V. Friedri h's model for transport operator, Journal of National University"Lvivska Polite hnika", Phys. and math. s ien es, v.643, n.643, 2009, P. 30-36 (Ukrainian).5. Latra h K. On an averaging result for transport equations. C.R.A ad. S i Paris, t.333, Serie I, 2001,P. 433-438. Ñòàòüÿ ïîñòóïèëà â ðåäàêöèþ 07.10.2010 ¾Òàâðiéñüêèé âiñíèê ií�îðìàòèêè òà ìàòåìàòèêè¿, �2' 2010