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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| 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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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 |
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On continuous spectrum of transport operator |
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On continuous spectrum of transport operator Cheremnikh, E.V. Ivasyk, G.V. |
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On continuous spectrum of transport operator |
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On continuous spectrum of transport operator |
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On continuous spectrum of transport operator |
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on continuous spectrum of transport operator |
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Cheremnikh, E.V. Ivasyk, G.V. |
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Cheremnikh, E.V. Ivasyk, G.V. |
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Таврический вестник информатики и математики |
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Кримський науковий центр НАН України і МОН України |
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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.
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1729-3901 |
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https://nasplib.isofts.kiev.ua/handle/123456789/18199 |
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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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AT cheremnikhev oncontinuousspectrumoftransportoperator AT ivasykgv oncontinuousspectrumoftransportoperator |
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2025-11-24T11:48:40Z |
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2025-11-24T11:48:40Z |
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1850846174773248000 |
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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
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