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Advances in Difference Equations

, 2013:237 | Cite as

Factorization of the linear differential operator

  • Klara R JanglajewEmail author
  • Kim G Valeev
Open Access
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Part of the following topical collections:
  1. Progress in Functional Differential and Difference Equations

Abstract

The paper deals with the problem of factorization of a linear differential operator with matrix-valued coefficients into a product of lower order operators of the same type. Necessary and sufficient conditions for the factorization of the considered operator are given. These conditions are obtained by using the integral manifolds approach. Some consequences of the obtained results are also considered.

MSC:34A30, 47A50, 47E05.

Keywords

linear differential equations differential operator factorization integral manifold of solutions 

1 Introduction

Factorization of differential and difference operators uses analogies between these operators and algebraic polynomials. There is a number of important papers on this subject, of which we only mention a few: [1, 2, 3, 4, 5].

A linear differential (difference) operator L admits factorization if it can be represented as a product of lower order operators of the same type (see [6, 7, 8]). Methods of factorization are exploited in analytic and algebraic approaches to the problem of integration of ordinary differential equations. Many special results are scattered over a large number of research papers; see, for instance, [9, 10, 11, 12] and the references given therein.

In this paper, we focus on an n th order linear differential operator of the form
L n ( t , D ) : = I D n + k = 0 n 1 A k ( t ) D k , D k : = d k d t k , Open image in new window
(1)

where we assume that A k ( t ) Open image in new window ( k = 0 , 1 , , n 1 Open image in new window) are m × m Open image in new window real-valued matrices with the entries being continuous and bounded functions on ℝ and that I is the m × m Open image in new window identity matrix.

Our purpose here is to give a proof that (1) can be presented as
L n ( t , D ) = L p ( t , D ) L q ( t , D ) , Open image in new window
where p + q = n Open image in new window and
L p ( t , D ) = I D p + k = 0 p 1 B k ( t ) D k , L q ( t , D ) = I D q + k = 0 q 1 C k ( t ) D k . Open image in new window
(2)

We give the necessary and sufficient conditions for factorization of the above operator L n Open image in new window into the product of lower order factors L q Open image in new window and L p Open image in new window. These conditions are connected with the existence of solutions of linear vector differential equations. The results are obtained by the usage of integral manifolds approach in the form elaborated by Valeev in the work [13].

2 Splitting equations

Let us consider the linear differential equation of order n, formed by acting the operator (1) on a vector function Z:
L n ( t , D ) Z ( t ) = ( I D n + k = 0 n 1 A k ( t ) D k ) Z ( t ) = 0 , Open image in new window
(3)

where Z ( t ) R m Open image in new window for t R Open image in new window.

The linear vector differential equation (3) can be written as an equivalent system of first-order equations of dimension mn. We let
X ( t ) : = ( Z ( t ) D Z ( t ) D q 1 Z ( t ) ) , Y ( t ) : = ( D q Z ( t ) D q + 1 Z ( t ) D n 1 Z ( t ) ) . Open image in new window
Hence,
( D Z ( t ) D 2 Z ( t ) D q Z ( t ) ) = ( 0 I 0 0 0 0 I 0 0 0 0 I 0 0 0 0 ) X ( t ) + ( 0 0 0 0 0 0 0 0 0 I 0 0 ) Y ( t ) Open image in new window
and
( D q + 1 Z ( t ) D q + 2 Z ( t ) D n Z ( t ) ) = ( 0 0 0 0 0 0 0 0 0 A 0 ( t ) A 1 ( t ) A q 1 ( t ) ) X ( t ) + ( 0 I 0 0 0 0 I 0 0 0 0 I A q ( t ) A q + 1 ( t ) A q + 2 ( t ) A n 1 ( t ) ) Y ( t ) . Open image in new window
In vector notation, we get that
d X ( t ) d t = A 11 X ( t ) + A 12 Y ( t ) , d Y ( t ) d t = A 21 ( t ) X ( t ) + A 22 ( t ) Y ( t ) , Open image in new window
(4)
where X is a q m × 1 Open image in new window vector function, Y is a p m × 1 Open image in new window vector function and
A 11 = ( 0 I 0 0 0 0 I 0 0 0 0 I 0 0 0 0 ) , A 12 = ( 0 0 0 0 0 0 0 0 0 I 0 0 ) , A 21 ( t ) = ( 0 0 0 0 0 0 0 0 0 A 0 ( t ) A 1 ( t ) A q 1 ( t ) ) , A 22 ( t ) = ( 0 I 0 0 0 0 I 0 0 0 0 I A q ( t ) A q + 1 ( t ) A q + 2 ( t ) A n 1 ( t ) ) . Open image in new window
(5)

Note that the block matrices A 11 Open image in new window, A 12 Open image in new window, A 21 ( t ) Open image in new window, A 22 ( t ) Open image in new window are q m × q m Open image in new window, q m × p m Open image in new window, p m × q m Open image in new window, p m × p m Open image in new window matrices, respectively.

We recall here the following definition (see [13]).

Definition 2.1 The connected subset M of R m n + 1 Open image in new window is called the integral manifold of system (4) if ( t 0 , X 0 , Y 0 ) M Open image in new window implies ( t , X ( t ) , Y ( t ) ) M Open image in new window for all t R Open image in new window, where X ( t ) Open image in new window and Y ( t ) Open image in new window are determined by (4) with X ( t 0 ) = X 0 Open image in new window, Y ( t 0 ) = Y 0 Open image in new window.

Suppose that the linear system of differential equations (4) has an integral manifold defined by the vector equation of the form
Y ( t ) = K ( t ) X ( t ) , Open image in new window
(6)

where K ( t ) Open image in new window is a p m × q m Open image in new window real-valued matrix and X, Y satisfy (4) on ℝ, i.e., provided that (6) is satisfied for a certain t 0 R Open image in new window, then it is valid for all t R Open image in new window.

By differentiation of (6) with respect to t, we get the vector differential equation of the form
d Y ( t ) d t = d K ( t ) d t X ( t ) + K ( t ) d X ( t ) d t . Open image in new window
(7)
From (4) and (7) it follows that
A 21 ( t ) X ( t ) + A 22 ( t ) Y ( t ) = d K ( t ) d t X ( t ) + K ( t ) [ A 11 X ( t ) + A 12 Y ( t ) ] , Open image in new window
or, equivalently,
A 21 ( t ) + A 22 ( t ) K ( t ) = d K ( t ) d t + K ( t ) [ A 11 + A 12 K ( t ) ] . Open image in new window
(8)
Let us note, that the matrix equation (8) is a Riccati-type matrix differential equation (see, for instance, [14]) and defines the integral manifold of the form (6). Assuming that the integral manifold of the form (6) does exist, the linear system (4) then reduces to
d X ( t ) d t = [ A 11 + A 12 K ( t ) ] X ( t ) , Y ( t ) = K ( t ) X ( t ) , Open image in new window
(9)

i.e., the linear subsystem splits off from the linear system of differential equations (4).

We shall seek for an integral manifold in the form (6), where K ( t ) Open image in new window is a p m × q m Open image in new window block matrix
K ( t ) = ( K 11 ( t ) K 12 ( t ) K 1 q ( t ) K 21 ( t ) K 22 ( t ) K 2 q ( t ) K p 1 ( t ) K p 2 ( t ) K p q ( t ) ) . Open image in new window
Substituting the matrices K ( t ) Open image in new window, A 11 Open image in new window and A 12 Open image in new window into (8), we obtain
d K 11 ( t ) d t + K 1 q ( t ) K 11 ( t ) = K 21 ( t ) , d K 21 ( t ) d t + K 2 q ( t ) K 11 ( t ) = K 31 ( t ) , d K 31 ( t ) d t + K 3 q ( t ) K 11 ( t ) = K 41 ( t ) , , d K p 1 , 1 ( t ) d t + K p 1 , q ( t ) K 11 ( t ) = K p 1 ( t ) , d K 12 ( t ) d t + K 11 ( t ) + K 1 q ( t ) K 12 ( t ) = K 22 ( t ) , d K 22 ( t ) d t + K 21 ( t ) + K 2 q ( t ) K 12 ( t ) = K 32 ( t ) , d K 32 ( t ) d t + K 31 ( t ) + K 3 q ( t ) K 12 ( t ) = K 42 ( t ) , , d K p 1 , 2 ( t ) d t + K p 1 , 1 ( t ) + K p 1 , q ( t ) K 12 ( t ) = K p 2 ( t ) , d K 13 ( t ) d t + K 12 ( t ) + K 1 q ( t ) K 13 ( t ) = K 23 ( t ) , d K 23 ( t ) d t + K 22 ( t ) + K 2 q ( t ) K 13 ( t ) = K 33 ( t ) , d K 33 ( t ) d t + K 32 ( t ) + K 3 q ( t ) K 13 ( t ) = K 43 ( t ) , , d K p 1 , 3 ( t ) d t + K p 1 , 2 ( t ) + K p 1 , q ( t ) K 13 ( t ) = K p 3 ( t ) , . Open image in new window
We continue in this manner to obtain
d K 1 q ( t ) d t + K 1 , q 1 ( t ) + K 1 q ( t ) K 1 q ( t ) = K 2 q ( t ) , , d K p 1 , q ( t ) d t + K p 1 , q 1 ( t ) + K p 1 , q ( t ) K 1 q ( t ) = K p q ( t ) , Open image in new window
or, in a compact form,
d K s 1 ( t ) d t + K s q ( t ) K 11 ( t ) = K s + 1 , 1 ( t ) , Open image in new window
(10)
d K s j ( t ) d t + K s q ( t ) K 1 j ( t ) + K s , j 1 ( t ) = K s + 1 , j ( t ) , Open image in new window
(11)

where K s j ( t ) Open image in new window ( s = 1 , , p 1 Open image in new window; j = 2 , , q Open image in new window) are m × m Open image in new window matrices.

Substituting A 21 ( t ) Open image in new window and A 22 ( t ) Open image in new window into (8), we have
d K p 1 ( t ) d t + K p q ( t ) K 11 ( t ) = A 0 ( t ) i = 1 p A q + i 1 ( t ) K i , 1 ( t ) , Open image in new window
(12)
d K p j ( t ) d t + K p q ( t ) K 1 j ( t ) + K p , j 1 ( t ) = A j 1 ( t ) i = 1 p A q + i 1 ( t ) K i , j ( t ) , Open image in new window
(13)

where j = 2 , , q Open image in new window.

Equations (10)-(13) are called the splitting equations (see [13]). We can use them for construction of an integral manifold of the linear vector differential equation (3).

The following theorem establishes the existence of an integral manifold of the linear differential equation (3).

Theorem 2.2 If there exists a solution of the system of splitting equations (10)-(13), then the linear vector differential equation (3) possesses the integral manifold of dimension mq, given by
D q Z ( t ) i = 0 q 1 K 1 , i + 1 ( t ) D i Z ( t ) = 0 Open image in new window
(14)

for t R Open image in new window.

Proof By virtue of formulas (5), system (9) takes the form
D q Z ( t ) = i = 0 q 1 K 1 , i + 1 ( t ) D i Z ( t ) , Open image in new window
(15)
D s + q 1 Z ( t ) = i = 0 q 1 K s , i + 1 ( t ) D i Z ( t ) , Open image in new window
(16)

where s = 2 , , p Open image in new window.

Differentiating each of the equations in (16) with respect to t, we get
D s + q Z ( t ) = i = 0 q 1 [ K s , i + 1 ( t ) D i Z ( t ) + K s , i + 1 ( t ) D i + 1 Z ( t ) ] Open image in new window
for s = 2 , , p 1 Open image in new window. Next, eliminating D q Z ( t ) Open image in new window by (15) and taking into account (10) and (11), we get that
D s + q Z ( t ) = i = 0 q 1 K s + 1 , i + 1 ( t ) D i Z ( t ) . Open image in new window
(17)
Differentiating (16) for s = p Open image in new window and eliminating D q Z ( t ) Open image in new window by (15), we get, by virtue of (12)-(13),
D n Z ( t ) = i = 0 q 1 A i ( t ) D i Z ( t ) i = 0 q 1 ( j = 1 p A q + j 1 ( t ) K j , i + 1 ( t ) ) D i Z ( t ) . Open image in new window
(18)

Substituting into (3) the derivatives D ( s + q ) Z ( t ) Open image in new window ( s = 1 , , p Open image in new window) from (17) and (18), we obtain zero. This proves the theorem. □

3 Existence of the integral manifold of solutions

We consider the linear vector differential equation of order q, formed by acting L q Open image in new window, given by (2), on a vector function Z, of the form
L q ( t , D ) Z ( t ) = ( I D q + k = 0 q 1 C k ( t ) D k ) Z ( t ) = 0 , Open image in new window
(19)

with a property such that each solution Z of (19) is also a solution of (3) on ℝ. This means that Z has n derivatives on ℝ and matrices C k ( t ) Open image in new window, k = 0 , 1 , , q 1 Open image in new window are differentiable up to order p ( p + q = n Open image in new window).

The following result may be proved in much the same way as Theorem 2.2.

Theorem 3.1 If any solution of the linear differential equation (19) with coefficients bounded together with their derivatives up to order p satisfies (3) on ℝ, then the linear system of differential equations (4) has the integral manifold given by (6), where K ( t ) Open image in new window is a p m × q m Open image in new window matrix.

Proof Let us rewrite the linear differential equation (19) in the form
D q Z ( t ) = i = 0 q 1 K 1 , i + 1 ( t ) D i Z ( t ) , Open image in new window
(20)
where
K 1 , i + 1 ( t ) = C i ( t ) ( i = 0 , , q 1 ) . Open image in new window

We substitute an arbitrary but fixed solution Z of (20) into the differential equation (3). For this end, we differentiate (20) p times with respect to t. After each differentiation, we eliminate D q Z ( t ) Open image in new window by (20) and we take into account (10)-(11). In this way we get (17). Similarly, taking into account (12)-(13), we obtain (18). Hence, the existence of solutions of the linear differential equation (19), all solutions of which are the solutions of (3), guarantees the existence of the integral manifold of the form (6) of the linear system (4) provided that all the derivatives up to the order p of matrices C k ( t ) Open image in new window ( k = 0 , , q 1 Open image in new window) are bounded. This is the desired conclusion. □

Remark 3.2 This kind of integral manifolds could be used in the investigation of stability of systems of difference equations (see [15, 16]).

Example 3.3 Let us consider the differential equation of the fifth order
L 5 ( t , D ) Z ( t ) = ( D 5 + 3 D 4 + 6 D 3 + 6 D 2 + 5 D + 3 ) Z ( t ) = 0 Open image in new window
(21)
and the differential equation of the third order
L 3 ( t , D ) Z ( t ) = ( D 3 + D 2 + D + 1 ) Z ( t ) = 0 , Open image in new window
(22)
with a property such that each solution Z of (22) is also a solution of (21). Rewrite the differential equation (22) in the form
D 3 Z ( t ) = Z ( t ) D Z ( t ) D 2 Z ( t ) . Open image in new window
(23)
Differentiating this equation two times and eliminating D 3 Z ( t ) Open image in new window, we get
D 4 Z ( t ) = Z ( t ) , D 5 Z ( t ) = D Z ( t ) . Open image in new window
(24)
Indeed, substituting D 3 Z ( t ) Open image in new window, D 4 Z ( t ) Open image in new window, and D 5 Z ( t ) Open image in new window from (23) and (24) into (21) yields
D 5 Z ( t ) + 3 D 4 Z ( t ) + 6 D 3 Z ( t ) + 6 D 2 Z ( t ) + 5 D Z ( t ) + 3 Z ( t ) = D Z ( t ) + 3 Z ( t ) + 6 ( Z ( t ) D Z ( t ) D 2 Z ( t ) ) + 6 D 2 Z ( t ) + 5 D Z ( t ) + 3 Z ( t ) 0 . Open image in new window
The linear differential equation (21) can be written as
d X ( t ) d t = A 11 X ( t ) + A 12 Y ( t ) , d Y ( t ) d t = A 21 X ( t ) + A 22 Y ( t ) , Open image in new window
(25)
where
X ( t ) : = ( Z ( t ) D Z ( t ) D 2 Z ( t ) ) , Y ( t ) : = ( D 3 Z ( t ) D 4 Z ( t ) ) , A 11 = ( 0 1 0 0 0 1 0 0 0 ) , A 12 = ( 0 0 0 0 1 0 ) , A 21 = ( 0 0 0 3 5 6 ) , A 22 = ( 0 1 6 3 ) . Open image in new window
The vector equation
Y ( t ) = ( 1 1 1 1 0 0 ) X ( t ) Open image in new window
determines the integral manifold of the system of differential equations (25). In fact, the matrix
K = ( 1 1 1 1 0 0 ) Open image in new window
is a solution of the matrix equation
A 21 + A 22 K = K ( A 11 + A 12 K ) . Open image in new window

Thus, the existence of an integral manifold of the form (6) for the linear system (4) is equivalent to the fact that any solution of (19) satisfies (3) as well.

4 Factorization of the operator L n ( t , D ) Open image in new window

Let the linear vector differential equation (3) be written in the form (4). We assume that any solution of the differential equation (19) is a solution of (3). Then the linear system (4) has the integral manifold of the form (6).

In the linear system (4), let us substitute
Y ( t ) : = K ( t ) X ( t ) + V ( t ) , Open image in new window
(26)
where
V ( t ) = ( V 1 ( t ) V p ( t ) ) Open image in new window
is a p m × 1 Open image in new window vector for t R Open image in new window. From the linear system (4) and (26), we obtain
d X ( t ) d t = ( A 11 + A 12 K ( t ) ) X ( t ) + A 12 V ( t ) , Open image in new window
(27)
d V ( t ) d t = ( A 21 ( t ) + A 22 ( t ) K ( t ) d K ( t ) d t K ( t ) A 11 K ( t ) A 12 K ( t ) ) X ( t ) + A 22 ( t ) V ( t ) K ( t ) A 12 V ( t ) . Open image in new window
(28)
Since
A 21 ( t ) + A 22 ( t ) K ( t ) d K ( t ) d t K ( t ) A 11 K ( t ) A 12 K ( t ) = 0 , Open image in new window
by (8), from (28) we get the splitting vector equation
d V ( t ) d t = A 22 ( t ) V ( t ) K ( t ) A 12 V ( t ) , Open image in new window
which, for vectors V k ( t ) Open image in new window ( k = 1 , , p Open image in new window), can be written in the form
D V 1 ( t ) = V 2 ( t ) K 1 , q ( t ) V 1 ( t ) , D V 2 ( t ) = V 3 ( t ) K 2 , q ( t ) V 1 ( t ) , , D V p 1 ( t ) = V p ( t ) K p 1 , q ( t ) V 1 ( t ) Open image in new window
(29)
and
D V p ( t ) = A q ( t ) V 1 ( t ) A q + 1 ( t ) V 2 ( t ) A q + p 1 ( t ) V p ( t ) K p , q ( t ) V 1 ( t ) . Open image in new window
(30)
From (29) we have
V 2 ( t ) = D V 1 ( t ) + K 1 , q ( t ) V 1 ( t ) , V 3 ( t ) = D ( D V 1 ( t ) + K 1 , q ( t ) V 1 ( t ) ) + K 2 , q ( t ) V 1 ( t ) , V 4 ( t ) = D [ D ( D V 1 ( t ) + K 1 , q ( t ) V 1 ( t ) ) + K 2 , q ( t ) V 1 ( t ) ] + K 3 , q ( t ) V 1 ( t ) Open image in new window
and so on. It is easily seen that
V s ( t ) = D s 1 V 1 ( t ) + k = 1 s 1 D s k 1 K k , q ( t ) V 1 ( t ) Open image in new window
for s = 2 , , p Open image in new window. Substituting V s ( t ) Open image in new window ( s = 2 , , p Open image in new window) into (30), we obtain the vector differential equation of order p
( D p + A n 1 ( t ) D p 1 + A n 2 ( t ) D p 2 + + A q ( t ) ) V 1 ( t ) + A q + 1 ( t ) ( D V 1 ( t ) + K 1 , q ( t ) V 1 ( t ) ) + A q + 2 ( t ) ( D 2 V 1 ( t ) + D ( K 1 , q ( t ) V 1 ( t ) ) + K 2 , q ( t ) V 1 ( t ) ) + + A n 1 ( t ) ( D p 1 V 1 ( t ) + D p 2 ( K 1 , q ( t ) V 1 ( t ) ) + D p 3 ( K 2 , q ( t ) V 1 ( t ) ) + + K p 1 , q ( t ) V 1 ( t ) ) + D p 1 ( K 1 , q ( t ) V 1 ( t ) ) + D p 2 ( K 2 , q ( t ) V 1 ( t ) ) + D p 3 ( K 3 , q ( t ) V 1 ( t ) ) + + K p , q ( t ) V 1 ( t ) = 0 . Open image in new window
(31)
After taking differentiation, equation (31) can be written as follows:
( I D p + k = 0 p 1 B k ( t ) D k ) V 1 ( t ) = 0 . Open image in new window
(32)
Hence, the differential equation (27) takes the form
( I D q + k = 0 q 1 C k ( t ) D k ) Z ( t ) = V 1 ( t ) , Open image in new window
where C i ( t ) = K 1 , i + 1 ( t ) Open image in new window, i = 0 , 1 , , q 1 Open image in new window. Eliminating in (32) V 1 ( t ) Open image in new window, we get, for the vector function Z, the differential equation of order n = q + p Open image in new window
( I D p + k = 0 p 1 B k ( t ) D k ) ( I D q + k = 0 q 1 C k ( t ) D k ) Z ( t ) = 0 , Open image in new window
(33)
or by (2)
L p ( t , D ) L q ( t , D ) Z ( t ) = 0 Open image in new window

for t R Open image in new window. Now our main result follows from (33) as the next theorem.

Theorem 4.1 The linear differential operator L n ( t , D ) Open image in new window defined by (1) is factorized in the form
L n ( t , D ) = L p ( t , D ) L q ( t , D ) , p + q = n , Open image in new window

where L p ( t , D ) Open image in new window and L q ( t , D ) Open image in new window are given by (2), if and only if any solution of the linear vector differential equation (19) is a solution of the linear vector differential equation (3) on ℝ.

5 The case of constant coefficients

Let us consider the case when the linear differential equation (3) has constant coefficients. This case is known to be important from the point of view of applications. In this case, from Theorem 4.1 we conclude the following.

Theorem 5.1 The polynomial matrix
L n ( s ) k = 0 n A k s k , Open image in new window
where A n Open image in new window is an m × m Open image in new window nonsingular matrix, admits a factorization in the form
L n ( s ) = L p ( s ) L q ( s ) , Open image in new window
where
L p ( s ) = k = 0 p B k s k , det B p 0 , L q ( s ) = k = 0 q C k s k , det C q 0 , Open image in new window
if and only if any solution of the linear vector differential equation
L q ( D ) Z ( t ) k = 0 q C k D k Z ( t ) = 0 Open image in new window
is a solution of the vector differential equation
L n ( D ) Z ( t ) k = 0 n A k D k Z ( t ) = 0 . Open image in new window

From this we have the following corollary.

Corollary 5.2 The polynomial matrix
L ( s ) k = 0 n A k s k , det A n 0 Open image in new window
is a multiple of s I C Open image in new window if and only if
L ( C ) k = 0 n A k C k = 0 . Open image in new window
Example 5.3 Consider the polynomial matrix
L ( s ) = I s 3 + A 2 s 2 + A 1 s + A 0 . Open image in new window
We look for a condition under which the factorization
I s 3 + A 2 s 2 + A 1 s + A 0 = ( I s 2 + B 1 s + B 0 ) ( I s C ) Open image in new window
is admissible. Taking the product, we get
A 2 = B 1 C , A 1 = B 0 B 1 C , A 0 = B 0 C . Open image in new window
Eliminating matrices B 1 Open image in new window, B 0 Open image in new window, we obtain the equation for the matrix C
C 3 + A C 2 + A 1 C + A 0 = 0 . Open image in new window
(34)
On the other hand, the vector differential equation
( I D C ) Y ( t ) = 0 Open image in new window

has the fundamental matrix Y ( t ) = e C t Open image in new window, which is a solution of the system L ( D ) Y ( t ) = 0 Open image in new window under condition (34).

Remark 5.4 The main results of this paper were announced in [17].

Notes

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© Janglajew and Valeev; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Institute of MathematicsUniversity of BiałystokBiałystokPoland
  2. 2.Kyiv National Economic UniversityKyivUkraine

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