Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 931493, 10 pages
http://dx.doi.org/10.1155/2013/931493
Research Article
Representation of a Solution of the Cauchy Problem for
an Oscillating System with Multiple Delays and Pairwise
Permutable Matrices
Josef Diblík,1 Michal FeIkan,2,3 and Michal Pospíšil4
1
Department of Mathematics, Faculty of Electrical Engineering and Communication, Brno University of Technology,
Technická 3058/10, 616 00 Brno, Czech Republic
2
Department of Mathematical Analysis and Numerical Mathematics, Comenius University, Mlynská dolina,
842 48 Bratislava, Slovakia
3
Mathematical Institute of Slovak Academy of Sciences, Štefánikova 49, 814 73 Bratislava, Slovakia
4
Centre for Research and Utilization of Renewable Energy, Faculty of Electrical Engineering and Communication,
Brno University of Technology, Technická 3058/10, 616 00 Brno, Czech Republic
Correspondence should be addressed to Michal Pospı́šil; pospisilm@feec.vutbr.cz
Received 13 January 2013; Accepted 19 April 2013
Academic Editor: Jaan Janno
Copyright © 2013 Josef Diblı́k et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Nonhomogeneous system of linear differential equations of second order with multiple different delays and pairwise permutable
matrices defining the linear parts is considered. Solution of corresponding initial value problem is represented using matrix
polynomials.
Motivated by delayed exponential representing a solution
of a system of differential or difference equations with
one or multiple fixed or variable delays [1–6], which has
many applications in theory of controllability, asymptotic
properties, boundary-value problems, and so forth [3–5, 7–
15], we extended representation of a solution of a system of
differential equations of second order with delay [1]
representing linear oscillator, to 𝑁-dimensional space with
one or multiple fixed delays. Clearly, each solution of the latter
equation is oscillating whenever 0 ≠ 𝑏 ∈ R. Analogically,
(1) with 𝑥 ∈ R𝑁 can have at least one oscillating solution
whenever 𝑁 is odd. Indeed, if 𝐵 is 𝑁 × 𝑁 matrix, 𝑁 ≥ 3 is
odd, and 𝐵 has a simple real nonzero eigenvalue 𝜆, then there
exists a regular matrix 𝑆 such that 𝑆−1 𝐵𝑆 = 𝐽 = ( 𝜆0 0𝐽̃ ) where
𝐽̃ is (𝑁 − 1) × (𝑁 − 1) matrix. On letting 𝑥 = 𝑆𝑦, one gets
𝑥̈ (𝑡) = −𝐵2 𝑥 (𝑡 − 𝜏)
𝑦̈ = −𝐽2 𝑦 (𝑡 − 𝜏)
1. Introduction
(1)
or rewrites as the system
to the case of two delays
𝑥̈ (𝑡) = −𝐵12 𝑥 (𝑡 − 𝜏1 ) − 𝐵22 𝑥 (𝑡 − 𝜏2 ) ,
(2)
where the linear parts were given by permutable matrices [16].
Equations (1), (2), and the below-stated (11) with 𝑓 ≡ 0 are
generalizations of the scalar equation
𝑥̈ (𝑡) = −𝑏2 𝑥 (𝑡)
(4)
(3)
𝑦1̈ = −𝜆2 𝑦1 (𝑡 − 𝜏) ,
𝑦2̈ = −𝐽̃2 𝑦2 (𝑡 − 𝜏) ,
(5)
where 𝑦 = (𝑦1 , 𝑦2 ) ∈ R × R𝑁−1 . Note that the first
column V of 𝑆 is the eigenvector of 𝐵 corresponding to 𝜆.
Clearly, if solution 𝑦1 of (5) is oscillating, then solution 𝑦 of
(4) is oscillating in the first coordinate whenever its initial
2
Abstract and Applied Analysis
condition satisfies {𝑦(𝑡) | 𝑡 ∈ [−𝜏, 0]} ⊂ R × {0}𝑁−1 .
Consequently, solution 𝑥 of (1) is oscillating in span{V}
whenever {𝑥(𝑡) | 𝑡 ∈ [−𝜏, 0]} ⊂ span{V}. Taking 𝑦1 (𝑡) = 𝑒𝜇𝑡 ,
one obtains characteristic equation 𝜇2 = −𝜆2 𝑒−𝜇𝜏 of (5),
which has solutions 𝜇1,2 = 𝛼 ± 𝚤𝛽 ∈ C with 𝛽 ≠ 0. Thus, 𝑦1
is oscillating.
On the other hand, there can exist a nonoscillating
solution of the system (1) whenever 𝑥 ∈ R𝑁 and 𝑁 is even.
0 1 ), then (1) has the form
For instance, if 𝑁 = 2 and 𝐵 = ( −1
0
𝑥̈ (𝑡) = 𝑥 (𝑡 − 𝜏)
(6)
with 𝑥 ∈ R2 , which, obviously, does not have an oscillating
solution satisfying nonoscillating initial condition. Similarly,
it can be shown that system with odd dimension can possess
a nonoscillating solution satisfying an appropriate initial
condition.
For simplicity, we call the generalizations (1), (2), and (11)
with 𝑓 ≡ 0, of scalar equation (3), oscillating although their
solutions do not always have to be oscillating. Nevertheless, at
the end of this paper, in Corollary 8 we state the representation of a solution of more general system (86) without squares
of matrices.
We note that the delayed matrix exponential from [1–5]
as well as the representation of a solution of second-order
differential equations derived in [1, 16] and in this paper can
lead to new results in nonlinear boundary value problems for
impulsive functional differential equations considered in [17]
or stochastic delayed differential equations from [18].
So, in the present paper, we extend our result from
[16] to three and more delays by the assumption of pairwise permutable matrices defining linear parts. By such an
assumption, we are able to construct matrix functions solving
homogeneous system of differential equations of second
order with any number of fixed delays, and, consequently,
we use these functions to represent a solution of the corresponding nonhomogeneous initial value problem. As will
be shown in the next sections, extending from two to more
delays brings many technical difficulties, for example, the
use of multinomial coefficients. Naturally, the results of the
present paper hold with one or two different delays as well.
However, these cases can by studied in a simpler way, which
was already done in [1, 16]. Thus, we focus our attention on
the case of three and more different delays.
First, we recall our result from [16].
Theorem 1. Let 𝜏1 , 𝜏2 > 0, 𝜏 := max{𝜏1 , 𝜏2 }, and 𝜑 ∈
𝐶1 ([−𝜏, 0], R𝑁). Let 𝐵1 , 𝐵2 be 𝑁×𝑁 permutable matrices; that
is, 𝐵1 𝐵2 = 𝐵2 𝐵1 , and let 𝑓 : [0, ∞) → R𝑁 be a given function.
Solution 𝑥(𝑡) of
𝑥̈ (𝑡) = −𝐵12 𝑥 (𝑡 − 𝜏1 ) − 𝐵22 𝑥 (𝑡 − 𝜏2 ) + 𝑓 (𝑡)
(7)
satisfying initial condition
𝑥 (𝑡) = 𝜑 (𝑡) ,
𝑥̇ (𝑡) = 𝜑̇ (𝑡) ,
−𝜏 ≤ 𝑡 ≤ 0
(8)
has the form
𝜑 (𝑡) ,
{
{
{
{
X (𝑡) 𝜑 (0) + Y (𝑡) 𝜑̇ (0)
{
{
{
0
{
{
{
{ −𝐵12 ∫ Y (𝑡 − 𝜏1 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
{
−𝜏1
𝑥 (𝑡) = {
0
{
{
2
{
Y (𝑡 − 𝜏2 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
−𝐵
∫
{
2
{
{
−𝜏2
{
{
𝑡
{
{
{ +∫ Y (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠,
{
0
−𝜏 ≤ 𝑡 < 0,
0 ≤ 𝑡,
(9)
where
2
2
X (𝑡) = X𝐵𝜏11,𝜏,𝐵22 (𝑡)
:=
∑ (−1)𝑖+𝑗
𝑖,𝑗≥0
𝑖𝜏1 +𝑗𝜏2 ≤𝑡
2(𝑖+𝑗)
𝑖 + 𝑗 2𝑖 2𝑗 (𝑡 − 𝑖𝜏1 − 𝑗𝜏2 )
×(
) 𝐵1 𝐵2
𝑖
(2 (𝑖 + 𝑗))!
Y (𝑡) =
:=
2 2
Y𝐵𝜏11,𝜏,𝐵22
,
(10)
(𝑡)
∑ (−1)𝑖+𝑗
𝑖,𝑗≥0
𝑖𝜏1 +𝑗𝜏2 ≤𝑡
2(𝑖+𝑗)+1
𝑖 + 𝑗 2𝑖 2𝑗 (𝑡 − 𝑖𝜏1 − 𝑗𝜏2 )
×(
) 𝐵1 𝐵2
𝑖
(2 (𝑖 + 𝑗) + 1)!
.
We will denote Θ and 𝐸 the 𝑁 × 𝑁 zero and identity
matrix, respectively.
2. Systems with Multiple Delays
In this section, we derive the representation of a solution of
𝑥̈ (𝑡) = −𝐵12 𝑥 (𝑡 − 𝜏1 ) − ⋅ ⋅ ⋅ − 𝐵𝑛2 𝑥 (𝑡 − 𝜏𝑛 ) + 𝑓 (𝑡)
(11)
satisfying the initial condition (8), where 𝑛 ≥ 3, 𝜏1 , . . . , 𝜏𝑛 > 0,
𝜏 := max𝑖=1,...,𝑛 𝜏𝑖 , 𝐵1 , . . . , 𝐵𝑛 are 𝑁 × 𝑁 pairwise permutable
matrices; that is, 𝐵𝑖 𝐵𝑗 = 𝐵𝑗 𝐵𝑖 for each 𝑖, 𝑗 ∈ {1, . . . , 𝑛}, 𝜑 ∈
𝐶1 ([−𝜏, 0], R𝑁), and 𝑓 : [0, ∞) → R𝑁 are given functions.
The solution 𝑥(𝑡) will be represented using matrix functions
analogical to (10) and will be stated in Section 3. We note that
the same problems with 𝑛 = 1, 2 were studied in [1, 16].
From now on, we assume the property of empty sum and
empty product; that is,
∑𝑓 (𝑖) = 0,
∑𝐹 (𝑖) = Θ,
𝑖∈0
𝑖∈0
∏𝑓 (𝑖) = 1,
∏𝐹 (𝑖) = 𝐸
𝑖∈0
𝑖∈0
(12)
for any function 𝑓 and matrix function 𝐹, whether they are
defined or not for indicated argument.
Abstract and Applied Analysis
3
We recall that (𝑗1 , . . . , 𝑗𝑛 )! is a multinomial coefficient [19]
given by
(𝑗1 , . . . , 𝑗𝑛 )! =
(𝑗1 + ⋅ ⋅ ⋅ + 𝑗𝑛 )!
.
𝑗1 ! ⋅ ⋅ ⋅ 𝑗𝑛 !
(13)
𝑗 +𝑗
Note that if 𝑛 = 2, then (𝑗1 , 𝑗2 ) = ( 1𝑗1 2 ) and (20) coincides
with (10).
We will need a property of multinomial coefficients
described in the next lemma.
In further work, we write ({𝑗 | 𝑗 ∈ 𝑀})! for the
multinomial coefficient of elements of the finite set 𝑀, and
(𝑖, {𝑗 | 𝑗 ∈ 𝑀})! for the multinomial coefficient of 𝑖 and
elements of the finite set 𝑀; for example, if 𝑀 = {1, 2}, then
(𝑎, {𝑗 | 𝑗 ∈ 𝑀})! = (𝑎, 1, 2)!. For the completeness, we define
({𝑗 | 𝑗 ∈ 0})! := 1.
2
2
,...,𝐵𝑛2
,...,𝐵𝑛2
, Y𝐵𝜏11,...,𝜏
: R → 𝐿(R𝑁) as
Define the functions X𝐵𝜏11,...,𝜏
𝑛
𝑛
2
2
,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡)
𝑛
:=
Lemma 2. Let 𝑛 ≥ 2 be fixed. Then
∑
𝑗1 ,...,𝑗𝑛 ≥0
𝑗1 𝜏1 +⋅⋅⋅+𝑗𝑛 𝜏𝑛 ≤𝑡
(𝑖1 , 𝑖2 , . . . , 𝑖𝑛 )! = (𝑖1 − 1, 𝑖2 , . . . , 𝑖𝑛 )!
𝑛
+ (𝑖1 , 𝑖2 − 1, 𝑖3 , . . . , 𝑖𝑛 )! + (𝑖1 , . . . , 𝑖𝑛−1 , 𝑖𝑛 − 1)!
(14)
2𝑗𝑖
× ∏𝐵𝑖
𝑖=1
2
(−1)𝑗1 +⋅⋅⋅+𝑗𝑛 (𝑗1 , . . . , 𝑗𝑛 )!
2(𝑗 +⋅⋅⋅+𝑗𝑛 )
(𝑡 − 𝑗1 𝜏1 − ⋅ ⋅ ⋅ − 𝑗𝑛 𝜏𝑛 ) 1
(2 (𝑗1 + ⋅ ⋅ ⋅ + 𝑗𝑛 ))!
,
(20)
2
,...,𝐵𝑛
Y𝐵𝜏11,...,𝜏
(𝑡)
𝑛
for any 𝑖1 , . . . , 𝑖𝑛 ≥ 1.
Proof. If 𝑛 = 2, then the statement follows from the property
of binomial coefficients:
𝑖 +𝑖
𝑖 +𝑖 −1
𝑖 +𝑖 −1
(𝑖1 , 𝑖2 )! = ( 1 2 ) = ( 1 2
)
)+(1 2
𝑖1
𝑖1 − 1
𝑖1
:=
𝑗1 ,...,𝑗𝑛 ≥0
𝑗1 𝜏1 +⋅⋅⋅+𝑗𝑛 𝜏𝑛 ≤𝑡
𝑛
2𝑗𝑖
× ∏𝐵𝑖
(15)
𝑖=1
= (𝑖1 − 1, 𝑖2 )! + (𝑖1 , 𝑖2 − 1)!.
Let the statement be true for 𝑛 − 1. Next, we use the property
of multinomial coefficient
(𝑖1 , 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 )! = (𝑖1 + 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 )! (𝑖1 , 𝑖2 )!
∑
(−1)𝑗1 +⋅⋅⋅+𝑗𝑛 (𝑗1 , . . . , 𝑗𝑛 )!
2(𝑗 +⋅⋅⋅+𝑗𝑛 )+1
(𝑡 − 𝑗1 𝜏1 − ⋅ ⋅ ⋅ − 𝑗𝑛 𝜏𝑛 ) 1
(2 (𝑗1 + ⋅ ⋅ ⋅ + 𝑗𝑛 ) + 1)!
for any 𝑡 ∈ R.
2
2
We will need functions X𝐵𝜏 , Y𝐵𝜏 : R → 𝐿(R𝑁) for 𝜏 > 0
and 𝑁 × 𝑁 complex matrix 𝐵 (cf. [16]) defined as
(16)
2
X𝐵𝜏 (𝑡) := ∑ (−1)𝑖 𝐵2𝑖
with inductive hypothesis to derive
𝑖≥0
𝑖𝜏≤𝑡
(𝑖1 , 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 )!
2
Y𝐵𝜏
= [(𝑖1 + 𝑖2 − 1, 𝑖3 , . . . , 𝑖𝑛 )! + (𝑖1 + 𝑖2 , 𝑖3 − 1, . . . , 𝑖𝑛 )!
+ ⋅ ⋅ ⋅ + (𝑖1 + 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 − 1)!] (𝑖1 , 𝑖2 )!.
(17)
(𝑡) := ∑ (−1) 𝐵
𝑖≥0
𝑖𝜏≤𝑡
with the properties
2
(𝑖1 + 𝑖2 , 𝑖3 − 1, . . . , 𝑖𝑛 )! (𝑖1 , 𝑖2 )! = (𝑖1 , 𝑖2 , 𝑖3 − 1, . . . , 𝑖𝑛 )!,
..
.
2
2
2
Ẏ 𝐵𝜏 (𝑡) = X𝐵𝜏 (𝑡) ,
2
2
Ẍ 𝐵𝜏 (𝑡) = −𝐵2 X𝐵𝜏 (𝑡 − 𝜏) ,
2
2
Ÿ 𝐵𝜏 (𝑡) = −𝐵2 Y𝐵𝜏 (𝑡 − 𝜏)
(18)
(𝑖1 + 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 − 1)! (𝑖1 , 𝑖2 )! = (𝑖1 , 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 − 1)!.
Applying the case 𝑛 = 2 (property of binomial coefficient)
and (16), we get
(𝑖1 + 𝑖2 − 1, 𝑖3 , . . . , 𝑖𝑛 )! (𝑖1 , 𝑖2 )!
= (𝑖1 + 𝑖2 − 1, 𝑖3 , . . . , 𝑖𝑛 )! [(𝑖1 − 1, 𝑖2 )! + (𝑖1 , 𝑖2 − 1)!]
Lemma 3. Let 𝑛 ≥ 1 and 𝜏1 , . . . , 𝜏𝑛 > 0. Let 𝐵1 , . . . , 𝐵𝑛 be
𝑁 × 𝑁 pairwise permutable matrices, that is, 𝐵𝑖 𝐵𝑗 = 𝐵𝑗 𝐵𝑖 for
each 𝑖, 𝑗 ∈ {1, . . . , 𝑛}. Then for any 𝑡 ∈ R,
2
2
(19)
2
Putting (18) and (19) in (17), we obtain that the statement
holds for 𝑛 and the proof is complete.
(22)
for any 𝑡 ∈ R, considering the one-sided derivatives at −𝜏, 0.
2
2
,...,𝐵𝑛2
,...,𝐵𝑛2
Some of properties of functions X𝐵𝜏11,...,𝜏
and Y𝐵𝜏11,...,𝜏
are
𝑛
𝑛
concluded in Lemma 4, but to prove it we will need the next
lemma.
,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡) =
𝑛
= (𝑖1 − 1, 𝑖2 , 𝑖3 , . . . , 𝑖𝑛 )! + (𝑖1 , 𝑖2 − 1, 𝑖3 , . . . , 𝑖𝑛 )!.
(21)
− 𝑖𝜏)2𝑖+1
(2𝑖 + 1)!
𝑖 2𝑖 (𝑡
Ẋ 𝐵𝜏 (𝑡) = −𝐵2 Y𝐵𝜏 (𝑡 − 𝜏) ,
Clearly, from (16), we get
(𝑡 − 𝑖𝜏)2𝑖
,
(2𝑖)!
2
,...𝐵𝑛
Y𝐵𝜏11,...,𝜏
(𝑡) =
𝑛
𝑆𝑀 (𝑡) ,
∑
𝑀⊂{1,...,𝑛}
∑
𝑆̃𝑀 (𝑡) ,
𝑀⊂{1,...,𝑛}
(23)
4
Abstract and Applied Analysis
where the sums are taken over all subsets of {1, . . . , 𝑛} including
the trivial ones, and
𝑆𝑀 (𝑡) :=
∑
𝑗𝑖 ≥1,𝑖∈𝑀
∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 ≤𝑡
×
𝑆̃𝑀 (𝑡) :=
2𝑗
∏ 𝐵𝑖 𝑖
𝑖∈𝑀
∑
𝑗𝑖 ≥1,𝑖∈𝑀
∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 ≤𝑡
2
(𝑡 − ∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 )
(2 ∑𝑖∈𝑀 𝑗𝑖 )!
2
(24)
2∑
2
2
2
2
2
2
,
(28)
(2) if 𝜏𝑖 = 𝜏𝑘 for 𝑖 < 𝑘, 𝑖, 𝑘 ∈ {1, . . . , 𝑛}, then
2
2
2
2
2
2
2
2
,...,𝐵𝑖−1 ,𝐵𝑖 ,𝐵𝑖+1 ,...,𝐵𝑘−1 ,𝐵𝑘 ,𝐵𝑘+1 ,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡)
𝑖−1 ,𝜏𝑖 ,𝜏𝑖+1 ,...,𝜏𝑘−1 ,𝜏𝑘 ,𝜏𝑘+1 ,...,𝜏𝑛
=
(−1)∑𝑖∈𝑀 𝑗𝑖 ({𝑗𝑖 | 𝑖 ∈ 𝑀})!
𝑖∈𝑀
2𝑗 (𝑡 − ∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 )
× ∏ 𝐵𝑖 𝑖
(2 ∑𝑖∈𝑀 𝑗𝑖 + 1)!
𝑖∈𝑀
2
,...,𝐵𝑖−1 ,𝐵𝑖 ,𝐵𝑖+1 ,...,𝐵𝑛
,...,𝐵𝑖−1 ,𝐵𝑖+1 ,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡) = X𝐵𝜏11,...,𝜏
(𝑡) ,
𝑖−1 ,𝜏𝑖 ,𝜏𝑖+1 ,...,𝜏𝑛
𝑖−1 ,𝜏𝑖+1 ,...,𝜏𝑛
(−1)∑𝑖∈𝑀 𝑗𝑖 ({𝑗𝑖 | 𝑖 ∈ 𝑀})!
2 ∑𝑖∈𝑀 𝑗𝑖
(1) if 𝐵𝑖 = Θ for some 𝑖 ∈ {1, . . . , 𝑛}, then
(25)
𝑗𝑖 +1
.
2
2
2
2
2
,...,𝐵𝑖−1
,𝐵𝑖2 +𝐵𝑘2 ,𝐵𝑖+1
,...,𝐵𝑘−1
,𝐵𝑘+1
,...,𝐵𝑛2
X𝐵𝜏11,...,𝜏
𝑖−1 ,𝜏𝑖 ,𝜏𝑖+1 ,...,𝜏𝑘−1 ,𝜏𝑘+1 ,...,𝜏𝑛
N
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
0 × ⋅ ⋅ ⋅ × N0
(𝑡) ,
(3) for any bijective mapping 𝜎 : {1, . . . , 𝑛} → {1, . . . , 𝑛}
we get
2
Proof. Denote N0 , N the set of all nonnegative, positive
integers, respectively; that is, N0 = {0} ∪ N. Thus, we have
the trivial identity
(29)
𝐵2
2
,...,𝐵2
,...,𝐵𝑛
𝜎(1)
𝜎(𝑛)
X𝐵𝜏11,...,𝜏
(𝑡) = X𝜏𝜎(1)
,...,𝜏𝜎(𝑛) (𝑡) ,
𝑛
(30)
(4) taking the one-sided derivatives at 0, 𝜏1 , . . . , 𝜏𝑛 , then
𝑛
2
2
2
2
,...,𝐵𝑛
,...,𝐵𝑛
Ẍ 𝐵𝜏11,...,𝜏
(𝑡 − 𝜏1 )
(𝑡) = −𝐵12 X𝐵𝜏11,...,𝜏
𝑛
𝑛
= ({0} × N
N0 × ⋅ ⋅ ⋅ × N 0 )
⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
0 × ⋅ ⋅ ⋅ × N0 ) ∪ (N × ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
𝑛−1
= ⋅⋅⋅ =
2
(31)
2
,...,𝐵𝑛
− ⋅ ⋅ ⋅ − 𝐵𝑛2 X𝐵𝜏11,...,𝜏
(𝑡 − 𝜏𝑛 ) ,
𝑛
𝑛−1
𝑀1 × ⋅ ⋅ ⋅ × 𝑀𝑛 .
⋃
𝑀1 ,...,𝑀𝑛 ∈{{0},N}
(26)
Analogically, for any 𝑡 ∈ R each 𝑛-tuple 𝑗1 , . . . , 𝑗𝑛 ≥ 0
such that ∑𝑛𝑖=1 𝑗𝑖 𝜏𝑖 ≤ 𝑡 can be divided in two distinct sets of
𝑖-s so that 𝑗𝑖 ≥ 1 if 𝑖 ∈ 𝑀 ⊂ {1, . . . , 𝑛} and 𝑗𝑖 = 0 if 𝑖 ∈
{1, . . . , 𝑛} \ 𝑀. That is, 𝑀 denotes the set of all indices 𝑖 such
that 𝑗𝑖 = 0. Moreover, ∑𝑛𝑖=1 𝑗𝑖 𝜏𝑖 = ∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 . Accordingly, we
can write
(5) considering the one-sided derivatives at 0 (they both
equal Θ), then
2
2
2
2
,...,𝐵𝑛
,...,𝐵𝑛
Ẏ 𝐵𝜏11,...,𝜏
(𝑡) = X𝐵𝜏11,...,𝜏
(𝑡) .
𝑛
𝑛
(32)
Statements (1)–(4) hold with Y instead of X.
2
{(𝑗1 , . . . , 𝑗𝑛 ) ∈ N𝑛0 | ∑𝑗𝑖 𝜏𝑖 ≤ 𝑡}
𝑖=1
=
⋃
𝑀⊂{1,...,𝑛}
{ (𝑗1 , . . . , 𝑗𝑛 ) ∈ N𝑛0 |
(27)
∑
𝑗1 ,...,𝑗𝑛 ≥0
𝑗1 𝜏1 +⋅⋅⋅+𝑗𝑛 𝜏𝑛 ≤𝑡
𝑗𝑖 = 0 ∀𝑖 ∉ 𝑀, ∑ 𝑗𝑖 𝜏𝑖 ≤ 𝑡} ,
𝑖∈𝑀
where the union is taken over all subsets of {1, . . . , 𝑛}
including the trivial ones. So, in the view of definition (20),
2
,...,𝐵𝑛2
follows.
the statement for X𝐵𝜏11,...,𝜏
𝑛
2
,...,𝐵𝑛2
Statement for Y𝐵𝜏11,...,𝜏
𝑛
2
,...,𝐵𝑛
Proof. Statement (1) follows easily from definition of X𝐵𝜏11,...,𝜏
,
𝑛
2𝑖
2𝑖
because Θ = 𝐸 if 𝑖 = 0 and Θ = Θ whenever 𝑖 > 0. Next, if
𝜏𝑖 = 𝜏𝑘 , then
𝑛
can be proved in a similar way.
Lemma 4. Let 𝑛 ≥ 3 and 𝜏1 , . . . , 𝜏𝑛 > 0. Let 𝐵1 , . . . , 𝐵𝑛 be
𝑁 × 𝑁 pairwise permutable matrices; that is, 𝐵𝑖 𝐵𝑗 = 𝐵𝑗 𝐵𝑖 for
each 𝑖, 𝑗 ∈ {1, . . . , 𝑛}. Then the following holds for any 𝑡 ∈ R:
=
𝐹 (𝑗1 , . . . , 𝑗𝑛 )
∑
∑ 𝐹 (𝑗1 , . . . , 𝑗𝑛 )
(33)
𝑗1 ,...,𝑗𝑖−1 ,𝑙,𝑗𝑖+1 ,...,𝑗𝑘−1 ,𝑗𝑘+1 ,...,𝑗𝑛 ≥0 𝑗𝑖 ,𝑗𝑘 ≥0
𝑗𝑖 +𝑗𝑘 =𝑙
𝑗1 𝜏1 +⋅⋅⋅+𝑗𝑖−1 𝜏𝑖−1 +𝑙𝜏𝑖 +𝑗𝑖+1 𝜏𝑖+1
+⋅⋅⋅+𝑗𝑘−1 𝜏𝑘−1 +𝑗𝑘+1 𝜏𝑘+1 +⋅⋅⋅+𝑗𝑛 𝜏𝑛 ≤𝑡
for any matrix function 𝐹. Thus, using the property of
multinomial coefficient (see (16))
(𝑗1 , . . . , 𝑗𝑛 )!
= (𝑗1 , . . . , 𝑗𝑖−1 , 𝑗𝑖 + 𝑗𝑘 , 𝑗𝑖+1 , . . . , 𝑗𝑘−1 , 𝑗𝑘+1 , . . . , 𝑗𝑛 )! (𝑗𝑖 , 𝑗𝑘 )!,
(34)
Abstract and Applied Analysis
5
Let 𝜏𝑘 := max𝑖=1,...,𝑛 𝜏𝑖 . If 𝑡 < 𝜏𝑘 , then 𝑡 − 𝜏𝑘 < 0, that is,
for (2), we obtain
𝐵12 ,...,𝐵𝑛2
𝜏1 ,...,𝜏𝑛
X
2
2
,...,𝐵𝑛
(𝑡 − 𝜏𝑘 ) = Θ,
X𝐵𝜏11,...,𝜏
𝑛
(𝑡)
=
and from definition (20) it holds
∑ 𝑠∈{1,...,𝑛} 𝑗𝑠 +𝑙
𝑠 ≠ 𝑖,𝑘
∑
(−1)
2
2
2
2
2
2
2
2
2
2𝑗
𝑛
(𝑡 − ∑ 𝑠∈{1,...,𝑛} ∑ 𝑗𝑠 𝜏𝑠 −
𝑠 ≠ 𝑖,𝑘
2(∑ 𝑠∈{1,...,𝑛} 𝑗𝑠 +𝑙)
𝑠 ≠ 𝑖,𝑘
𝑙𝜏𝑖 )
(2 (∑ 𝑠∈{1,...,𝑛} ∑ 𝑗𝑠 + 𝑙))!
× ∏ 𝐵𝑠2𝑗𝑠
(𝑡 − ∑ 𝑠∈{1,...,𝑛} ∑ 𝑗𝑠 𝜏𝑠 −
𝑠 ≠ 𝑖,𝑘
2
2
by the inductive hypothesis.
Now, let 𝑡 ≥ max𝑖=1,...,𝑛 𝜏𝑖 . Applying Lemma 3, we get
2
2
2
2
2
+
2(∑ 𝑠∈{1,...,𝑛} 𝑗𝑠 +𝑙)
𝑠 ≠ 𝑖,𝑘
𝑙𝜏𝑖 )
2
0≤𝑡
1,
𝜒𝑀
̃ (𝑡) = {
0,
̃
𝑡 ∈ 𝑀,
̃
𝑡 ∉ 𝑀.
(43)
Since each 𝑀 ⊂ {1, . . . , 𝑛} is a finite set, Lemma 2 yields
({𝑗𝑖 | 𝑖 ∈ 𝑀})! = ∑ (𝑗𝑖 − 1, {𝑗𝑘 | 𝑘 ∈ 𝑀 \ {𝑖}})!.
𝑖∈𝑀
(44)
2
,...,𝐵𝑛
= −𝐵12 X𝐵𝜏11,...,𝜏
(𝑡 − 𝜏1 )
𝑛
2
,...,𝐵𝑛2
𝐵𝑛2 X𝐵𝜏11,...,𝜏
𝑛
󸀠󸀠
𝑆𝑀
(𝑡)
2
= − (𝐵12 + ⋅ ⋅ ⋅ + 𝐵𝑛2 ) X𝐵𝜏 1 +⋅⋅⋅+𝐵𝑛 (𝑡 − 𝜏)
− ⋅⋅⋅ −
(42)
We apply this identity to derive a formula for the second
derivative of 𝑆𝑀 for any 0 ≠ 𝑀 ⊂ {1, . . . , 𝑛}:
2
,...,𝐵𝑛
Ẍ 𝐵𝜏11,...,𝜏
(𝑡) = Ẍ 𝐵𝜏 1 +⋅⋅⋅+𝐵𝑛 (𝑡)
𝑛
2
(𝑡 − 0)0
0!
= ∑𝐸 = 𝐸𝜒[0,∞) (𝑡)
2
Property (3) is trivial.
Now, we prove the statement (4). If 𝜏 := 𝜏1 = ⋅ ⋅ ⋅ = 𝜏𝑛 ,
then
2
(41)
̃
with a characteristic function 𝜒𝑀
̃ of a set 𝑀 given by
(35)
2
𝑆𝑀 (𝑡)
𝑗𝑖 ≥1,𝑖∈0
0≤𝑡
,...,𝐵𝑖−1 ,𝐵𝑖 +𝐵𝑘 ,𝐵𝑖+1 ,...,𝐵𝑘−1 ,𝐵𝑘+1 ,...,𝐵𝑛
= X𝐵𝜏11,...,𝜏
(𝑡) .
𝑖−1 ,𝜏𝑖 ,𝜏𝑖+1 ,...,𝜏𝑘−1 ,𝜏𝑘+1 ,...,𝜏𝑛
2
∑
𝑀⊂{1,...,𝑛}
𝑆0 (𝑡) = ∑ (−1)0 ({𝑗𝑖 | 𝑖 ∈ 0})!𝐸
𝑙
𝐵𝑘2 )
𝑠 ≠ 𝑖,𝑘
2
2
with 𝑆𝑀(𝑡) given by (24) and the sum taken over all subsets
of {1, . . . , 𝑛} including the trivial ones. Note that
(2 (∑ 𝑠∈{1,...,𝑛} ∑ 𝑗𝑠 + 𝑙))!
𝑠∈{1,...,𝑛}
𝑠 ≠ 𝑖,𝑘
2
𝑖=1
,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡) =
𝑛
∑ 𝑠∈{1,...,𝑛} 𝑗𝑠 +𝑙
(−1) 𝑠 ≠ 𝑖,𝑘
(𝑗1 , . . . , 𝑗𝑖−1 , 𝑙, 𝑗𝑖+1 , . . . , 𝑗𝑘−1 , 𝑗𝑘+1 , . . . , 𝑗𝑛 )!(𝐵𝑖2
2
(40)
,...,𝐵𝑛
= −∑𝐵𝑖2 X𝐵𝜏11,...,𝜏
(𝑡 − 𝜏𝑖 )
𝑛
𝑠 ≠ 𝑖,𝑘
𝑗1 ,...,𝑗𝑖−1 ,𝑙,𝑗𝑖+1 ,...,𝑗𝑘−1 ,𝑗𝑘+1 ,...,𝑗𝑛 ≥0
𝑗1 𝜏1 +⋅⋅⋅+𝑗𝑖−1 𝜏𝑖−1 +𝑙𝜏𝑖 +𝑗𝑖+1 𝜏𝑖+1
+⋅⋅⋅+𝑗𝑘−1 𝜏𝑘−1 +𝑗𝑘+1 𝜏𝑘+1 +⋅⋅⋅+𝑗𝑛 𝜏𝑛 ≤𝑡
×
2
𝑖=1,...,𝑛
𝑖 ≠ 𝑘
𝑗𝑖 ,𝑗𝑘 ≥0
𝑗𝑖 +𝑗𝑘 =𝑙
∑
2
2
,...,𝐵𝑘−1 ,𝐵𝑘+1 ,...,𝐵𝑛
= − ∑ 𝐵𝑖2 X𝐵𝜏11,...,𝜏
(𝑡 − 𝜏𝑖 )
𝑘−1 ,𝜏𝑘+1 ,...,𝜏𝑛
× ( ∑ (𝑗𝑖 , 𝑗𝑘 )!𝐵𝑖 𝑖 𝐵𝑘 𝑘 )
=
2
(39)
,...,𝐵𝑛
,...,𝐵𝑘−1 ,𝐵𝑘+1 ,...,𝐵𝑛
Ẍ 𝐵𝜏11,...,𝜏
(𝑡) = Ẍ 𝐵𝜏11,...,𝜏
(𝑡)
𝑛
𝑘−1 ,𝜏𝑘+1 ,...,𝜏𝑛
2
𝑠∈{1,...,𝑛}
𝑠 ≠ 𝑖,𝑘
2
for such 𝑡. Consequently,
× (𝑗1 , . . . , 𝑗𝑖−1 , 𝑙, 𝑗𝑖+1 , . . . , 𝑗𝑘−1 , 𝑗𝑘+1 , . . . , 𝑗𝑛 )!
× ∏ 𝐵𝑠2𝑗𝑠
2
,...,𝐵𝑛
,...,𝐵𝑘−1 ,𝐵𝑘+1 ,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡) = X𝐵𝜏11,...,𝜏
(𝑡)
𝑛
𝑘−1 ,𝜏𝑘+1 ,...,𝜏𝑛
𝑗1 ,...,𝑗𝑖−1 ,𝑙,𝑗𝑖+1 ,...,𝑗𝑘−1 ,𝑗𝑘+1 ,...,𝑗𝑛 ≥0
𝑗1 𝜏1 +⋅⋅⋅+𝑗𝑖−1 𝜏𝑖−1 +𝑙𝜏𝑖 +𝑗𝑖+1 𝜏𝑖+1
+⋅⋅⋅+𝑗𝑘−1 𝜏𝑘−1 +𝑗𝑘+1 𝜏𝑘+1 +⋅⋅⋅+𝑗𝑛 𝜏𝑛 ≤𝑡
2𝑗
(38)
(36)
∑
𝑗𝑖 ≥1,𝑖∈𝑀
∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 ≤𝑡
(𝑡 − 𝜏𝑛 )
𝐵12 +⋅⋅⋅+𝐵𝑛2
by (2) and from the property of X𝜏
(𝑡) (see (22)).
Hence, without any loss of generality, we assume that
𝜏𝑖 ≠ 𝜏𝑗 for each 𝑖 ≠ 𝑗, 𝑖, 𝑗 ∈ {1, . . . , 𝑛} (in the other case, we
collect matrices as stated in (2)). Note the case 𝑛 = 2 was
2
2
,...,𝐵𝑛−1
(𝑡)
proved in [16, Lemma 2.3.] Now, assume that X𝐵𝜏11,...,𝜏
𝑛−1
solves
2
𝑥 (𝑡 − 𝜏𝑛−1 ) ,
𝑥̈ (𝑡) = −𝐵12 𝑥 (𝑡 − 𝜏1 ) − ⋅ ⋅ ⋅ − 𝐵𝑛−1
=
(37)
that is, that the statement is fulfilled for 𝑛 − 1 different delays.
(−1)∑𝑖∈𝑀 𝑗𝑖 ({𝑗𝑖 | 𝑖 ∈ 𝑀})!
2𝑗𝑖
× ∏ 𝐵𝑖
𝑖∈𝑀
= ∑
2(∑
∑
𝑖∈𝑀 𝑗𝑘 ≥1,𝑘∈𝑀
∑𝑘∈𝑀 𝑗𝑘 𝜏𝑘 ≤𝑡
2𝑗
× ∏ 𝐵𝑘 𝑘
𝑘∈𝑀
𝑗 −1)
(𝑡 − ∑𝑖∈𝑀 𝑗𝑖 𝜏𝑖 ) 𝑖∈𝑀 𝑖
(2 (∑𝑖∈𝑀 𝑗𝑖 − 1))!
(−1)∑𝑘∈𝑀 𝑗𝑘 (𝑗𝑖 − 1, {𝑗𝑘 | 𝑘 ∈ 𝑀 \ {𝑖}})!
(𝑡 − 𝜏𝑖 − ∑𝑘∈𝑀\{𝑖} 𝑗𝑘 𝜏𝑘 − (𝑗𝑖 − 1) 𝜏𝑖 )
(2 (∑𝑘∈𝑀 𝑗𝑘 − 1))!
2(∑𝑘∈𝑀 𝑗𝑘 −1)
.
(45)
6
Abstract and Applied Analysis
Next, for any fixed 𝑖 ∈ {1, . . . , 𝑛} we split the second sum to
𝑗𝑖 = 1 and 𝑗𝑖 ≥ 2, that is,
∑
𝑗𝑘 ≥1,𝑘∈𝑀
∑𝑘∈𝑀 𝑗𝑘 𝜏𝑘 ≤𝑡
𝐹 (𝑗1 , . . . , 𝑗𝑖−1 , 𝑗𝑖 , 𝑗𝑖+1 , . . . , 𝑗𝑛 )
∑
𝑗𝑘 ≥1,𝑘∈𝑀\{𝑖}
∑𝑘∈𝑀\{𝑖} 𝑗𝑘 𝜏𝑘 ≤𝑡−𝜏𝑖
+
∑
𝑗𝑘 ≥1,𝑘∈𝑀\{𝑖}
𝑗𝑖 ≥2
∑𝑘∈𝑀 𝑗𝑘 𝜏𝑘 ≤𝑡
2
+
(46)
−
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) .
∑
Denoting #𝑀 the number of elements of the set 𝑀, we split
the last two terms of the right-hand side of the latter equality
with respect to
=
∑
∑
0 ≠ 𝑀⊂{1,...,𝑛}
𝐹 (𝑗1 , . . . , 𝑗𝑖−1 , 𝑗𝑖 + 1, 𝑗𝑖+1 , . . . , 𝑗𝑛 )
𝑘∈𝑀
=
∑ 𝑗𝑘 𝜏𝑘 + (𝑗𝑖 − 1) 𝜏𝑖 ≤ 𝑡 − 𝜏𝑖 .
(48)
𝑘∈𝑀\{𝑖}
2
(49)
+
∑
.
=
Θ.
𝑀⊂{1,...,𝑛}
2≤#𝑀≤𝑛
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
=
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑
𝑀⊂{1,...,𝑛} 𝑖∉𝑀
1≤#𝑀≤𝑛−1
2
−
∑
∑ 𝐵𝑖2
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
=−
∑
+
(𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) + 𝑆𝑀 (𝑡 − 𝜏𝑖 ))
∑
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
−
∑ 𝐵𝑖2 𝑆𝑀 (𝑡 − 𝜏𝑖 ) .
−
∑ 𝐵𝑖2 𝑆𝑀 (𝑡 − 𝜏𝑖 )
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∉𝑀
(51)
to the right-hand side of (50) to get
=−
∑
𝑛
∑ 𝐵𝑖2 𝑆𝑀 (𝑡 − 𝜏𝑖 )
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∉𝑀
−
∑
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) .
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
∑
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
𝑀⊂{1,...,𝑛} 𝑖∉𝑀
1≤#𝑀≤𝑛−1
∑
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) .
(56)
𝑀⊂{1,...,𝑛} 𝑖∈𝑀
2≤#𝑀≤𝑛
∑𝐵𝑖2 𝑆𝑀 (𝑡 − 𝜏𝑖 )
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖=1
+
∑
=
(𝑡)
∑
∑
Now, we show that
Now, we add and subtract
Ẍ
(55)
𝑀⊂{1,...,𝑛} 𝑖∈𝑀
2≤#𝑀≤𝑛
(50)
𝐵12 ,...,𝐵𝑛2
𝜏1 ,...,𝜏𝑛
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
𝑀∈{{1},...,{𝑛}} 𝑖∈𝑀
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
∑
∑
𝑀={1,...,𝑛} 𝑖∉𝑀
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
−
(54)
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
−
,...,𝐵𝑛
Ẍ 𝐵𝜏11,...,𝜏
(𝑡)
𝑛
=
∑
𝑀⊂{1,...,𝑛}
#𝑀=𝑛
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∉𝑀
𝑖∈𝑀
{1, . . . , 𝑛}. Obviously, 𝑆0󸀠󸀠 (𝑡)
+
Hence, we have
∑
󸀠󸀠
𝑆𝑀
(𝑡) = − ∑ 𝐵𝑖2 (𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) + 𝑆𝑀 (𝑡 − 𝜏𝑖 ))
⊂
∑
𝑀⊂{1,...,𝑛}
#𝑀=1
So we obtain
for each 0 ≠ 𝑀
Consequently,
∑
𝑀⊂{1,...,𝑛}
1≤#𝑀≤𝑛−1
since
∑ 𝑗𝑘 𝜏𝑘 ≤ 𝑡 ⇐⇒
(53)
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
(47)
𝑗𝑘 ≥1,𝑘∈𝑀
∑𝑘∈𝑀 𝑗𝑘 𝜏𝑘 ≤𝑡−𝜏𝑖
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∉𝑀
𝐹 (𝑗1 , . . . , 𝑗𝑖−1 , 𝑗𝑖 , 𝑗𝑖+1 , . . . , 𝑗𝑛 ) ,
𝐹 (𝑗1 , . . . , 𝑗𝑖−1 , 𝑗𝑖 , 𝑗𝑖+1 , . . . , 𝑗𝑛 )
=
𝑛
∑𝐵𝑖2 𝑆𝑀 (𝑡 − 𝜏𝑖 )
∑
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖=1
𝐹 (𝑗1 , . . . , 𝑗𝑖−1 , 1, 𝑗𝑖+1 , . . . , 𝑗𝑛 )
and use the equality
∑
2
,...,𝐵𝑛
Ẍ 𝐵𝜏11,...,𝜏
(𝑡)
𝑛
=−
=
𝑗𝑘 ≥1,𝑘∈𝑀\{𝑖}
𝑗𝑖 ≥2
∑𝑘∈𝑀 𝑗𝑘 𝜏𝑘 ≤𝑡
and apply 𝑀 = 𝑀 \ {𝑖} whenever 𝑖 ∉ 𝑀:
(52)
Let 𝑀 ⊂ {1, . . . , 𝑛}, and let 𝑖 ∉ 𝑀 be arbitrary and fixed such
that 1 ≤ #𝑀 ≤ 𝑛 − 1. Then, clearly,
𝐵𝑖 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) = 𝐵𝑖 𝑆(𝑀∪{𝑖})\{𝑖} (𝑡 − 𝜏𝑖 )
(57)
and 2 ≤ #(𝑀 ∪ {𝑖}) ≤ 𝑛, 𝑖 ∈ 𝑀 ∪ {𝑖}. Moreover, if 𝑀1 , 𝑀2 ⊂
{1, . . . , 𝑛}, 𝑖 ∉ 𝑀1,2 are such that 𝑀1 ≠ 𝑀2 , 1 ≤ #𝑀1,2 ≤ 𝑛 − 1,
then 𝑀1 ∪ {𝑖} ≠ 𝑀2 ∪ {𝑖}.
Abstract and Applied Analysis
7
On the other side, if 𝑀 ⊂ {1, . . . , 𝑛}, 𝑖 ∈ 𝑀 are arbitrary
and fixed such that 2 ≤ #𝑀 ≤ 𝑛, then
2
2
,...,𝐵𝑛
(𝑡), statements (1)–(3) can be proved as for
For Y𝐵𝜏11,...,𝜏
𝑛
2
2
,...,𝐵𝑛
(𝑡). Next, if 𝜏 := 𝜏1 = ⋅ ⋅ ⋅ = 𝜏𝑛 , we apply the point (2)
X𝐵𝜏11,...,𝜏
𝑛
2
𝐵𝑖 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) = 𝐵𝑖 𝑆(𝑀\{𝑖})\{𝑖} (𝑡 − 𝜏𝑖 )
(58)
and 1 ≤ #(𝑀 \ {𝑖}) ≤ 𝑛 − 1, 𝑖 ∉ 𝑀 \ {𝑖}. Furthermore, if
𝑀1 , 𝑀2 ⊂ {1, . . . , 𝑛}, 𝑖 ∈ 𝑀1,2 are such that 𝑀1 ≠ 𝑀2 , 2 ≤
#𝑀1,2 ≤ 𝑛, then, 𝑀1 \ {𝑖} ≠ 𝑀2 \ {𝑖}. In conclusion, there is
1 − 1 correspondence between the terms on the left-hand side
of (56) and the terms on the right-hand side. So (56) is valid.
Putting (56) in (55) we obtain
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∉𝑀
−
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖∈𝑀
=
(59)
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑
2
2
2
2
2
𝑀={1,...,𝑛} 𝑖∉𝑀
∑
2
,...,𝐵𝑛
− ⋅ ⋅ ⋅ − 𝐵𝑛2 Y𝐵𝜏11,...,𝜏
(𝑡 − 𝜏𝑛 ) .
𝑛
2
,...,𝐵𝑛
So, Y𝐵𝜏11,...,𝜏
(𝑡) is a solution of (31) when all delays are the
𝑛
same.
Again, the case 𝑛 = 2 with different delays was proved
in [16]; thus, we assume that the statement is fulfilled for 𝑛 −
1, 𝑛 ≥ 3 and that 𝜏𝑖 ≠ 𝜏𝑗 for each 𝑖 ≠ 𝑗, 𝑖, 𝑗 ∈ {1, . . . , 𝑛}. As
before, if 𝑡 < 𝜏𝑘 and 𝜏𝑘 := max𝑖=1,...,𝑛 𝜏𝑖 , then
2
2
2
2
2
2
2
,...,𝐵𝑛
Y𝐵𝜏11,...,𝜏
(𝑡) =
𝑛
Θ = Θ.
𝑀={1,...,𝑛}
(60)
𝑆̃𝑀 (𝑡)
∑
(65)
𝑀⊂{1,...,𝑛}
with 𝑆̃𝑀(𝑡) given by (25). This time
𝑆0 (𝑡) = ∑ (−1)0 ({𝑗𝑖 | 𝑖 ∈ 0})!𝐸
Moreover, it holds
𝑗𝑖 ≥1,𝑖∈0
0≤𝑡
∑ 𝐵𝑖2 𝑆𝑀\{𝑖}
∑
𝑀∈{{1},...,{𝑛}} 𝑖∈𝑀
(64)
by definition (20), and the statement follows from the
inductive hypothesis. For 𝑡 ≥ max𝑖=1,...,𝑛 𝜏𝑖 , we apply Lemma 3
to see that
Next, by the property of empty sum, we get
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 ) =
(63)
2
,...,𝐵𝑛
= −𝐵12 Y𝐵𝜏11,...,𝜏
(𝑡 − 𝜏1 )
𝑛
2
𝑀∈{{1},...,{𝑛}} 𝑖∈𝑀
∑
2
,...,𝐵𝑛
,...,𝐵𝑘−1 ,𝐵𝑘+1 ,...,𝐵𝑛
Y𝐵𝜏11,...,𝜏
(𝑡) = Y𝐵𝜏11,...,𝜏
(𝑡)
𝑛
𝑘−1 ,𝜏𝑘+1 ,...,𝜏𝑛
∑ 𝐵𝑖2 𝑆𝑀{𝑖} (𝑡 − 𝜏𝑖 ) .
∑
2
= − (𝐵12 + ⋅ ⋅ ⋅ + 𝐵𝑛2 ) Y𝐵𝜏 1 +⋅⋅⋅+𝐵𝑛 (𝑡 − 𝜏)
𝑀={1,...,𝑛} 𝑖∉𝑀
−
2
,...,𝐵𝑛
Ÿ 𝐵𝜏11,...,𝜏
(𝑡) = Ÿ 𝐵𝜏 1 +⋅⋅⋅+𝐵𝑛 (𝑡)
𝑛
2
∑ 𝐵𝑖2 𝑆𝑀\{𝑖} (𝑡 − 𝜏𝑖 )
∑
2
of this lemma and property (22) for Y𝐵𝜏 1 +⋅⋅⋅+𝐵𝑛 (𝑡) to see that
(𝑡 − 0)1
0!
(66)
= ∑𝐸𝑡 = 𝑡𝐸𝜒[0,∞) (𝑡)
(𝑡 − 𝜏𝑖 )
0≤𝑡
(61)
,...,𝐵𝑛
= Θ. The rest proceeds analogically to X𝐵𝜏11,...,𝜏
(𝑡).
and
𝑛
The final statement follows directly from definition (20).
Therefore, putting (60) and (61) in (59) and the result in (53),
we obtain
Remark 5. Another proof of statements (1)–(3) of the previous lemma can be made with the aid of statement (4)
of the same lemma and uses the uniqueness of a solution
of the corresponding initial value problem. For instance in
statement (1) of the lemma, both
=
𝑛
∑𝐵𝑖2 𝑆0
𝑖=1
2
(𝑡 − 𝜏𝑖 ) =
2
,...,𝐵𝑛
Ẍ 𝐵𝜏11,...,𝜏
(𝑡) = −
𝑛
−
𝑛
∑𝐵𝑖2 𝑆𝑀 (𝑡 − 𝜏𝑖 )
∑ ∑𝐵𝑖2 𝑆𝑀 (𝑡
𝑀=0 𝑖=1
𝑛
∑
2
,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡) ,
𝑛
− 𝜏𝑖 )
2
2
2
2
,...,𝐵𝑖−1 ,𝐵𝑖+1 ,...,𝐵𝑛
X𝐵𝜏11,...,𝜏
(𝑡)
𝑖−1 ,𝜏𝑖+1 ,...,𝜏𝑛
2
(67)
solve
(62)
𝑆𝑀 (𝑡 − 𝜏𝑖 )
𝑀⊂{1,...,𝑛}
2
,...,𝐵𝑛2
−∑𝐵𝑖2 X𝐵𝜏11,...,𝜏
𝑛
𝑖=1
2
2
𝑛
𝑖=1
2
− 𝜏𝑖 ) .
0 ≠ 𝑀⊂{1,...,𝑛} 𝑖=1
= −∑𝐵𝑖2
2
∑ ∑𝐵𝑖2 𝑆𝑀 (𝑡
𝑀=0 𝑖=1
∑
𝑛
=
𝑛
𝑆0󸀠󸀠 (𝑡)
2
𝑥 (𝑡 − 𝜏𝑖−1 )
𝑥̈ (𝑡) = − 𝐵12 𝑥 (𝑡 − 𝜏1 ) − ⋅ ⋅ ⋅ − 𝐵𝑖−1
2
𝑥 (𝑡 − 𝜏𝑖+1 ) − ⋅ ⋅ ⋅ − 𝐵𝑛2 𝑥 (𝑡 − 𝜏𝑛 )
− 𝐵𝑖+1
(68)
with initial condition
(𝑡 − 𝜏𝑖 ) .
,...,𝐵𝑛
Hence, X𝐵𝜏11,...,𝜏
(𝑡) solves (31) for all 𝑡 ≥ 0. Clearly, the same
𝑛
is true for 𝑡 < 0.
Θ, −𝜏 ≤ 𝑡 < 0,
𝑥 (𝑡) = {
𝐸, 𝑡 = 0,
𝑥̇ (𝑡) = Θ, −𝜏 ≤ 𝑡 ≤ 0
and 𝜏 = max𝑖=1,...,𝑛 𝜏𝑖 .
We are ready to state and prove our main result.
(69)
8
Abstract and Applied Analysis
3. Main Result
Here we find a solution of the initial value problem (11), (8)
in the sense of the next definition.
Definition 6. Let 𝜏1 , . . . , 𝜏𝑛 > 0, 𝜏 := max𝑖=1,...,𝑛 𝜏𝑖 , and 𝜑 ∈
𝐶1 ([−𝜏, 0], R𝑁), and let 𝐵1 , . . . , 𝐵𝑛 be 𝑁 × 𝑁 matrices, and
let 𝑓 : [0, ∞) → R𝑁 be a given function. Function 𝑥 :
[−𝜏, ∞) → R𝑁 is a solution of (11) and initial condition
(8), if 𝑥 ∈ 𝐶1 ([−𝜏, ∞), R𝑁) ∩ 𝐶2 ([0, ∞), R𝑁) (taken the
second right-hand derivative at 0) satisfies (11) on [0, ∞) and
condition (8) on [−𝜏, 0].
Theorem 7. Let 𝑛 ≥ 3, 𝜏1 , . . . , 𝜏𝑛 > 0, 𝜏 := max𝑖=1,...,𝑛 𝜏𝑖 , and
𝜑 ∈ 𝐶1 ([−𝜏, 0], R𝑁), and let 𝐵1 , . . . , 𝐵𝑛 be 𝑁 × 𝑁 pairwise
permutable matrices; that is, 𝐵𝑖 𝐵𝑗 = 𝐵𝑗 𝐵𝑖 for each 𝑖, 𝑗 ∈
{1, . . . , 𝑛}, and let 𝑓 : [0, ∞) → R𝑁 be a given function.
Solution 𝑥(𝑡) of (11) satisfying initial condition (8) has the form
Assume that 0 ≤ 𝑡 < min𝑖=1,...,𝑛 𝜏𝑖 . Then identities (71) and
(73) are valid, and by differentiating (73) for such 𝑡 we get
𝑛
𝑛
𝑖=1
𝑖=1
𝑥̈ (𝑡) = −∑𝐵𝑖2 𝜑 (𝑡 − 𝜏𝑖 ) + 𝑓 (𝑡) = −∑𝐵𝑖2 𝑥 (𝑡 − 𝜏𝑖 ) + 𝑓 (𝑡)
(75)
since 𝑥(𝑡 − 𝜏𝑖 ) = 𝜑(𝑡 − 𝜏𝑖 ) for each 𝑖 = 1, . . . , 𝑛.
Now, let 0 ≠ 𝑀1,2 ⊂ {1, . . . , 𝑛} be such that 𝜏𝑖 ≤ 𝑡 < 𝜏𝑗 for
each 𝑖 ∈ 𝑀1 , 𝑗 ∈ 𝑀2 . Then
Y (𝑡 − 𝜏𝑗 − 𝑠) , 𝑠 ∈ [−𝜏𝑗 , 𝑡 − 𝜏𝑗 ] ,
Y (𝑡 − 𝜏𝑗 − 𝑠) = {
(76)
Θ,
𝑠 ∈ (𝑡 − 𝜏𝑗 , 0]
whenever 𝑗 ∈ 𝑀2 , and (70) becomes
𝑥 (𝑡) = X (𝑡) 𝜑 (0) + Y (𝑡) 𝜑̇ (0)
0
− ∑ 𝐵𝑖2 ∫ Y (𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
𝜑 (𝑡) ,
−𝜏 ≤ 𝑡 < 0,
{
{
{
{
̇
𝜑
X
𝜑
+
Y
(𝑡)
(0)
(𝑡)
(0)
{
{
{
𝑛
0
{
𝑥 (𝑡) = { −∑𝐵𝑖2 ∫ Y (𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
{
{
−𝜏𝑖
{
𝑖=1
{
𝑡
{
{
{ +∫ Y (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠,
0 ≤ 𝑡,
{
0
(70)
2
2
2
2
𝑖∈𝑀1
− ∑ 𝐵𝑗2 ∫
𝑗∈𝑀2
Proof. Obviously, 𝑥(𝑡) satisfies the initial condition on [−𝜏, 0),
and, from definition (20), 𝑥(0) = 𝜑(0). For the derivative, it
̇ = 𝜑(0).
̇
Moreover, if 0 ≤ 𝑡 < min𝑖=1,...,𝑛 𝜏𝑖 ,
holds lim𝑡 → 0− 𝑥(𝑡)
then
𝑖=1
𝑡−𝜏𝑖
−𝜏𝑖
−𝜏𝑗
(77)
𝑡
0
𝑥̇ (𝑡) = Ẋ (𝑡) 𝜑 (0) + Ẏ (𝑡) 𝜑̇ (0)
0
− ∑ 𝐵𝑖2 ∫ Ẏ (𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
𝑖∈𝑀1
−𝜏𝑖
− ∑ 𝐵𝑗2 ∫
(𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
Y (𝑡 − 𝜏𝑗 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
+ ∫ Y (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠.
𝑥 (𝑡) = 𝜑 (0) + 𝑡𝜑̇ (0)
𝑛
𝑡−𝜏𝑗
By the point (5) of Lemma 4, we get
,...,𝐵𝑛
,...,𝐵𝑛
where X(𝑡) = X𝐵𝜏11,...,𝜏
(𝑡) and Y(𝑡) = Y𝐵𝜏11,...,𝜏
(𝑡).
𝑛
𝑛
− ∑𝐵𝑖2 ∫
−𝜏𝑖
𝑗∈𝑀2
(71)
𝑡−𝜏𝑗
−𝜏𝑗
X (𝑡 − 𝜏𝑗 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
(78)
𝑡
+ ∫ X (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠,
𝑡
0
+ ∫ (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠
and for the second derivative it holds
0
since
𝑥̈ (𝑡) = Ẍ (𝑡) 𝜑 (0) + Ÿ (𝑡) 𝜑̇ (0)
(𝑡 − 𝜏𝑖 − 𝑠) 𝐸,
Y (𝑡 − 𝜏𝑖 − 𝑠) = {
Θ,
𝑠 ∈ [−𝜏𝑖 , 𝑡 − 𝜏𝑖 ] ,
𝑠 ∈ (𝑡 − 𝜏𝑖 , 0]
(72)
𝑖∈𝑀1
for each 𝑖 = 1, . . . , 𝑛. Thus
𝑥̇ (𝑡) = 𝜑̇ (0) −
𝑛
𝑡−𝜏𝑖
∑𝐵𝑖2 ∫
−𝜏𝑖
𝑖=1
𝑡
𝜑 (𝑠) 𝑑𝑠 + ∫ 𝑓 (𝑠) 𝑑𝑠
0
0
− ∑ 𝐵𝑖2 ∫ Ÿ (𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
(73)
−𝜏𝑖
− ∑ 𝐵𝑗2 (𝜑 (𝑡 − 𝜏𝑗 ) + ∫
𝑡−𝜏𝑗
−𝜏𝑗
𝑗∈𝑀2
Ÿ (𝑡 − 𝜏𝑗 − 𝑠) 𝜑 (𝑠) 𝑑𝑠)
𝑡
+ 𝑓 (𝑡) + ∫ Ÿ (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠
̇ = 𝜑(0).
̇
and lim𝑡 → 0+ 𝑥(𝑡)
Clearly,
𝑥 ∈ 𝐶1 ((−𝜏, ∞) , R𝑁) ∩ 𝐶2 ((0, ∞) \ {𝜏1 , . . . , 𝜏𝑛 } , R𝑁) .
(74)
2
We show that, although X(𝑡) is not 𝐶 at 𝜏1 , . . . , 𝜏𝑛 ,
function 𝑥(𝑡) is 𝐶2 at these points and, therefore, in (0, ∞).
At once, we prove that 𝑥(𝑡) is a solution of (11).
0
(79)
since X(0) = 𝐸. Now, we apply the property (4) of Lemma 4
together with
X (𝑡 − 𝜏𝑗 ) = Y (𝑡 − 𝜏𝑗 ) = Θ,
∀𝑗 ∈ 𝑀2
(80)
Abstract and Applied Analysis
9
to see that both X and Y are solutions of
𝑦̈ (𝑡) = − ∑ 𝐵𝑖2 𝑦 (𝑡 − 𝜏𝑖 ) .
𝑖∈𝑀1
It is easy to see that defining functions
̃𝐵1 ,...,𝐵𝑛 (𝑡) := X−𝐵1 ,...,−𝐵𝑛 (𝑡) ,
X
𝜏1 ,...,𝜏𝑛
𝜏1 ,...,𝜏𝑛
(81)
(85)
̃ 𝐵1 ,...,𝐵𝑛 (𝑡) := Y−𝐵1 ,...,−𝐵𝑛 (𝑡)
Y
𝜏1 ,...,𝜏𝑛
𝜏1 ,...,𝜏𝑛
Therefore,
leads to the solution of
𝑥̈ (𝑡)
𝑥̈ (𝑡) = 𝐵1 𝑥 (𝑡 − 𝜏1 ) + ⋅ ⋅ ⋅ + 𝐵𝑛 𝑥 (𝑡 − 𝜏𝑛 ) + 𝑓 (𝑡)
= − ∑ 𝐵𝑘2 (X (𝑡 − 𝜏𝑘 ) 𝜑 (0) + Y (𝑡 − 𝜏𝑘 ) 𝜑̇ (0)
with pairwise permutable matrices 𝐵1 , . . . , 𝐵𝑛 and initial
condition (8). More precisely, we have the following corollary
of Theorem 7.
𝑘∈𝑀1
0
− ∑ 𝐵𝑖2 ∫ Y (𝑡 − 𝜏𝑖 − 𝜏𝑘 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
−𝜏𝑖
𝑖∈𝑀1
𝑡−𝜏𝑗
− ∑ 𝐵𝑗2 ∫
−𝜏𝑗
𝑗∈𝑀2
Y (𝑡 − 𝜏𝑗 − 𝜏𝑘 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
𝑡
+ ∫ Y (𝑡 − 𝜏𝑘 − 𝑠) 𝑓 (𝑠) 𝑑𝑠)
0
− ∑
𝑗∈𝑀2
𝐵𝑗2 𝜑 (𝑡
− 𝜏𝑗 ) + 𝑓 (𝑡)
= − ∑ 𝐵𝑖2 𝑥 (𝑡 − 𝜏𝑖 ) − ∑ 𝐵𝑗2 𝜑 (𝑡 − 𝜏𝑗 ) + 𝑓 (𝑡) .
𝑖∈𝑀1
𝑗∈𝑀2
(82)
In fact, this is exactly formula (11) since 𝑥(𝑡 − 𝜏𝑗 ) = 𝜑(𝑡 − 𝜏𝑗 )
for each 𝑗 ∈ 𝑀2 .
Finally, if max𝑖=1,...,𝑛 𝜏𝑖 ≤ 𝑡, we have
𝑥 (𝑡) = X (𝑡) 𝜑 (0) + Y (𝑡) 𝜑̇ (0)
𝑛
0
− ∑𝐵𝑖2 ∫ Y (𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
𝑖=1
−𝜏𝑖
+ ∫ Y (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠.
0
So, differentiating this formula twice and applying (4) of
Lemma 4 result in (11). Hence, one can see that function 𝑥(𝑡)
given by (70) really solves (11) and satisfies initial condition
(8) and, moreover, that 𝑥 ∈ 𝐶2 ((0, ∞), R𝑁). To see the last
one, one has to put 𝜏1 , . . . , 𝜏𝑛 into the computed derivatives,
for example, if 𝜏𝑘 := min𝑖=1,...,𝑛 𝜏𝑖 < 𝜏𝑖 for each 𝑖 = 1, . . . , 𝑘 −
1, 𝑘 + 1, . . . , 𝑛, then by (75) and (82) we get
𝑛
lim− 𝑥̈ (𝑡) = − ∑ 𝐵𝑖2 𝜑 (𝜏𝑘 − 𝜏𝑖 ) − 𝐵𝑘2 𝜑 (0) + 𝑓 (𝜏𝑘 )
𝑖=1
𝑖 ≠ 𝑘
= − ∑ 𝐵𝑗2 𝜑 (𝜏𝑘 − 𝜏𝑗 )
(84)
𝑗∈𝑀2
− 𝐵𝑘2 𝑥 (0) + 𝑓 (𝜏𝑘 ) = lim+ 𝑥̈ (𝑡) ,
𝑡 → 𝜏𝑘
where 𝑀2 = {1, . . . , 𝑛} \ {𝑘}.
Corollary 8. Let 𝑛 ≥ 3, 𝜏1 , . . . , 𝜏𝑛 > 0, 𝜏 := max𝑖=1,...,𝑛 𝜏𝑖 ,
𝜑 ∈ 𝐶1 ([−𝜏, 0], R𝑁), and let 𝐵1 , . . . , 𝐵𝑛 be 𝑁 × 𝑁 pairwise
permutable matrices; that is, 𝐵𝑖 𝐵𝑗 = 𝐵𝑗 𝐵𝑖 for each 𝑖, 𝑗 ∈
{1, . . . , 𝑛}, and let 𝑓 : [0, ∞) → R𝑁 be a given function.
Solution 𝑥(𝑡) of (86) satisfying initial condition (8) has the
form
𝜑 (𝑡) ,
{
{
{
{
{
{X (𝑡)𝑛𝜑 (0) + Y (𝑡) 𝜑̇ (0)
{
0
{
𝑥 (𝑡) = { +∑𝐵𝑖 ∫ Y (𝑡 − 𝜏𝑖 − 𝑠) 𝜑 (𝑠) 𝑑𝑠
{
{
−𝜏𝑖
{
𝑖=1
{
𝑡
{
{
{ +∫ Y (𝑡 − 𝑠) 𝑓 (𝑠) 𝑑𝑠,
{
0
−𝜏 ≤ 𝑡 < 0,
0 ≤ 𝑡,
(87)
̃𝐵1 ,...,𝐵𝑛 (𝑡) and Y(𝑡) = Y
̃ 𝐵1 ,...,𝐵𝑛 (𝑡).
where X(𝑡) = X
𝜏1 ,...,𝜏𝑛
𝜏1 ,...,𝜏𝑛
Proof. The corollary can be proved exactly in the same way as
Theorem 7.
Acknowledgments
(83)
𝑡
𝑡 → 𝜏𝑘
(86)
J. Diblı́k was supported by the Grant GAČR P201/11/0768.
M. Fečkan was supported in part by the Grants VEGAMS 1/0507/11, VEGA-SAV 2/0029/13, and APVV-0134-10. M.
Pospı́šil was supported by the Project no. CZ.1.07/2.3.00/
30.0005 funded by European Regional Development Fund.
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