# Representation of a Solution of the Cauchy Problem for an Oscillating System with Multiple Delays and Pairwise Permutable Matrices

Authors: Josef Diblík; Michal Fečkan; Michal Pospíšil
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

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.


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## Body

## 1. Introduction

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]
(1)x¨(t)=-B2x(t-τ)
to the case of two delays
(2)x¨(t)=-B12x(t-τ1)-B22x(t-τ2),
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
(3)x¨(t)=-b2x(t)
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-1BS=J=(λ00J~) where J~ is (N-1)×(N-1) matrix. On letting x=Sy, one gets
(4)y¨=-J2y(t-τ)
or rewrites as the system
(5)y¨1=-λ2y1(t-τ),y¨2=-J~2y2(t-τ),
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 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=-λ2e-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. For instance, if N=2 and B=(01-10), then (1) has the form
(6)x¨(t)=x(t-τ)
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.
Letτ1,τ2>0, τ:=max{τ1,τ2}, and φ∈C1([-τ,0],ℝN). Let B1,B2 be N×N permutable matrices; that is, B1B2=B2B1, and let f:[0,∞)→ℝN be a given function. Solution x(t) of
(7)x¨(t)=-B12x(t-τ1)-B22x(t-τ2)+f(t)
satisfying initial condition
(8)x(t)=φ(t),x˙(t)=φ˙(t),-τ≤t≤0
has the form
(9)x(t)={φ(t),-τ≤t<0,𝒳(t)φ(0)+𝒴(t)φ˙(0)-B12∫-τ10‍𝒴(t-τ1-s)φ(s)ds-B22∫-τ20‍𝒴(t-τ2-s)φ(s)ds+∫0t‍𝒴(t-s)f(s)ds,0≤t,
where
(10)𝒳(t)=𝒳τ1,τ2B12,B22(t)𝒳(t):=∑i,j≥0iτ1+jτ2≤t‍(-1)i+j𝒳(t)=×(i+ji)B12iB22j(t-iτ1-jτ2)2(i+j)(2(i+j))!,𝒴(t)=𝒴τ1,τ2B12,B22(t)𝒴(t):=∑i,j≥0iτ1+jτ2≤t‍(-1)i+j𝒴(t)=×(i+ji)B12iB22j(t-iτ1-jτ2)2(i+j)+1(2(i+j)+1)!.We will denoteΘ 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(11)x¨(t)=-B12x(t-τ1)-⋯-Bn2x(t-τn)+f(t)
satisfying the initial condition (8), where n≥3, τ1,…,τn>0, τ:=maxi=1,…,nτi, B1,…,Bn are N×N pairwise permutable matrices; that is, BiBj=BjBi 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,(12)∑i∈∅‍f(i)=0,∑i∈∅‍F(i)=Θ,∏i∈∅‍f(i)=1,∏i∈∅‍F(i)=E
for any function f and matrix function F, whether they are defined or not for indicated argument.We recall that(j1,…,jn)! is a multinomial coefficient [19] given by
(13)(j1,…,jn)!=(j1+⋯+jn)!j1!⋯jn!.
Note that if n=2, then (j1,j2)=(j1+j2j1) and (20) coincides with (10).We will need a property of multinomial coefficients described in the next lemma.Lemma 2.
Letn≥2 be fixed. Then
(14)(i1,i2,…,in)!=(i1-1,i2,…,in)!(i1,i2,…,in)!=+(i1,i2-1,i3,…,in)!+(i1,…,in-1,in-1)!
for any i1,…,in≥1.Proof.
Ifn=2, then the statement follows from the property of binomial coefficients:
(15)(i1,i2)!=(i1+i2i1)=(i1+i2-1i1-1)+(i1+i2-1i1)(i1,i2)!=(i1-1,i2)!+(i1,i2-1)!.
Let the statement be true for n-1. Next, we use the property of multinomial coefficient
(16)(i1,i2,i3,…,in)!=(i1+i2,i3,…,in)!(i1,i2)!
with inductive hypothesis to derive
(17)(i1,i2,i3,…,in)!=[(i1+i2-1,i3,…,in)!+(i1+i2,i3-1,…,in)!+⋯+(i1+i2,i3,…,in-1)!](i1,i2)!.
Clearly, from (16), we get
(18)(i1+i2,i3-1,…,in)!(i1,i2)!=(i1,i2,i3-1,…,in)!,(+i2,i3)⋮(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
(19)(i1+i2-1,i3,…,in)!(i1,i2)!=(i1+i2-1,i3,…,in)![(i1-1,i2)!+(i1,i2-1)!]=(i1-1,i2,i3,…,in)!+(i1,i2-1,i3,…,in)!.
Putting (18) and (19) in (17), we obtain that the statement holds for n and the proof is complete.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∈∅})!:=1.Define the functions𝒳τ1,…,τnB12,…,Bn2,𝒴τ1,…,τnB12,…,Bn2:ℝ→L(ℝN) as
(20)𝒳τ1,…,τnB12,…,Bn2(t):=∑j1,…,jn≥0j1τ1+⋯+jnτn≤t‍(-1)j1+⋯+jn(j1,…,jn)!×∏i=1n‍Bi2ji(t-j1τ1-⋯-jnτn)2(j1+⋯+jn)(2(j1+⋯+jn))!,𝒴τ1,…,τnB12,…,Bn2(t):=∑j1,…,jn≥0j1τ1+⋯+jnτn≤t‍(-1)j1+⋯+jn(j1,…,jn)!×∏i=1n‍Bi2ji(t-j1τ1-⋯-jnτn)2(j1+⋯+jn)+1(2(j1+⋯+jn)+1)!
for any t∈ℝ.We will need functions𝒳τB2,𝒴τB2:ℝ→L(ℝN) for τ>0 and N×N complex matrix B (cf. [16]) defined as
(21)𝒳τB2(t):=∑i≥0iτ≤t‍(-1)iB2i(t-iτ)2i(2i)!,𝒴τB2(t):=∑i≥0iτ≤t‍(-1)iB2i(t-iτ)2i+1(2i+1)!
with the properties
(22)𝒳˙τB2(t)=-B2𝒴τB2(t-τ),𝒳¨τB2(t)=-B2𝒳τB2(t-τ),𝒴˙τB2(t)=𝒳τB2(t),𝒴¨τB2(t)=-B2𝒴τB2(t-τ)
for any t∈ℝ, considering the one-sided derivatives at -τ, 0.Some of properties of functions𝒳τ1,…,τnB12,…,Bn2 and 𝒴τ1,…,τnB12,…,Bn2 are concluded in Lemma 4, but to prove it we will need the next lemma.Lemma 3.
Letn≥1 and τ1,…,τn>0. Let B1,…,Bn be N×N pairwise permutable matrices, that is, BiBj=BjBi for each i,j∈{1,…,n}. Then for any t∈ℝ,
(23)𝒳τ1,…,τnB12,…,Bn2(t)=∑M⊂{1,…,n}‍SM(t),𝒴τ1,…,τnB12,…Bn2(t)=∑M⊂{1,…,n}‍S~M(t),
where the sums are taken over all subsets of {1,…,n} including the trivial ones, and
(24)SM(t):=∑ji≥1,i∈M∑i∈Mjiτi≤t‍(-1)∑i∈Mji({ji∣i∈M})!SM(t):=×∏i∈M‍Bi2ji(t-∑i∈Mjiτi)2∑i∈Mji(2∑i∈Mji)!,(25)S~M(t):=∑ji≥1,i∈M∑i∈Mjiτi≤t‍(-1)∑i∈Mji({ji∣i∈M})!SM(t):=×∏i∈M‍Bi2ji(t-∑i∈Mjiτi)2∑i∈Mji+1(2∑i∈Mji+1)!.Proof.
Denoteℕ0, ℕ the set of all nonnegative, positive integers, respectively; that is, ℕ0={0}∪ℕ. Thus, we have the trivial identity
(26)ℕ0×⋯×ℕ0︸n=({0}×ℕ0×⋯×ℕ0︸n-1)∪(ℕ×ℕ0×⋯×ℕ0︸n-1)=⋯=⋃M1,…,Mn∈{{0},ℕ}‍M1×⋯×Mn.
Analogically, for anyt∈ℝ each n-tuple j1,…,jn≥0 such that ∑i=1n‍jiτi≤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, ∑i=1n‍jiτi=∑i∈M‍jiτi. Accordingly, we can write
(27){(j1,…,jn)∈ℕ0n∣∑i=1n‍jiτi≤t}=⋃M⊂{1,…,n}‍{∑i∈M‍(j1,…,jn)∈ℕ0n∣=⋃M⊂{1,…,n}‍ji=0∀i∉M,∑i∈M‍jiτi≤t},
where the union is taken over all subsets of {1,…,n} including the trivial ones. So, in the view of definition (20), the statement for 𝒳τ1,…,τnB12,…,Bn2 follows.
Statement for𝒴τ1,…,τnB12,…,Bn2 can be proved in a similar way.Lemma 4.
Letn≥3 and τ1,…,τn>0. Let B1,…,Bn be N×N pairwise permutable matrices; that is, BiBj=BjBi for each i,j∈{1,…,n}. Then the following holds for any t∈ℝ:(1)
ifBi=Θ for some i∈{1,…,n}, then
(28)𝒳τ1,…,τi-1,τi,τi+1,…,τnB12,…,Bi-12,Bi2,Bi+12,…,Bn2(t)=𝒳τ1,…,τi-1,τi+1,…,τnB12,…,Bi-12,Bi+12,…,Bn2(t),(2)
ifτi=τk for i<k, i,k∈{1,…,n}, then
(29)𝒳τ1,…,τi-1,τi,τi+1,…,τk-1,τk,τk+1,…,τnB12,…,Bi-12,Bi2,Bi+12,…,Bk-12,Bk2,Bk+12,…,Bn2(t)=𝒳τ1,…,τi-1,τi,τi+1,…,τk-1,τk+1,…,τnB12,…,Bi-12,Bi2+Bk2,Bi+12,…,Bk-12,Bk+12,…,Bn2(t),(3)
for any bijective mappingσ:{1,…,n}→{1,…,n} we get
(30)𝒳τ1,…,τnB12,…,Bn2(t)=𝒳τσ(1),…,τσ(n)Bσ(1)2,…,Bσ(n)2(t),(4)
taking the one-sided derivatives at0,τ1,…,τn, then
(31)𝒳¨τ1,…,τnB12,…,Bn2(t)=-B12𝒳τ1,…,τnB12,…,Bn2(t-τ1)𝒳¨τ1,…,τnB12,…,Bn2(t)=-⋯-Bn2𝒳τ1,…,τnB12,…,Bn2(t-τn),(5)
considering the one-sided derivatives at 0 (they both equalΘ), then
(32)𝒴˙τ1,…,τnB12,…,Bn2(t)=𝒳τ1,…,τnB12,…,Bn2(t).Statements (1)–(4) hold with 𝒴 instead of 𝒳.Proof.
Statement (1) follows easily from definition of𝒳τ1,…,τnB12,…,Bn2, because Θ2i=E if i=0 and Θ2i=Θ whenever i>0. Next, if τi=τk, then
(33)∑j1,…,jn≥0j1τ1+⋯+jnτn≤t‍F(j1,…,jn)=∑j1,…,ji-1,l,ji+1,…,jk-1,jk+1,…,jn≥0j1τ1+⋯+ji-1τi-1+lτi+ji+1τi+1+⋯+jk-1τk-1+jk+1τk+1+⋯+jnτn≤t‍∑ji,jk≥0ji+jk=l‍F(j1,…,jn)
for any matrix function F. Thus, using the property of multinomial coefficient (see (16))
(34)(j1,…,jn)!=(j1,…,ji-1,ji+jk,ji+1,…,jk-1,jk+1,…,jn)!(ji,jk)!,
for (2), we obtain
(35)𝒳τ1,…,τnB12,…,Bn2(t)=∑j1,…,ji-1,l,ji+1,…,jk-1,jk+1,…,jn≥0j1τ1+⋯+ji-1τi-1+lτi+ji+1τi+1+⋯+jk-1τk-1+jk+1τk+1+⋯+jnτn≤t‍(-1)∑s∈{1,…,n}s≠i,kjs+l×(j1,…,ji-1,l,ji+1,…,jk-1,jk+1,…,jn)!×(∑ji,jk≥0ji+jk=l‍(ji,jk)!Bi2jiBk2jk)×∏s∈{1,…,n}s≠i,k‍Bs2js(t-∑s∈{1,…,n}s≠i,kjsτs-lτi)2(∑s∈{1,…,n}s≠i,kjs+l)(2(∑s∈{1,…,n}s≠i,kjs+l))!=∑j1,…,ji-1,l,ji+1,…,jk-1,jk+1,…,jn≥0j1τ1+⋯+ji-1τi-1+lτi+ji+1τi+1+⋯+jk-1τk-1+jk+1τk+1+⋯+jnτn≤t‍(-1)∑s∈{1,…,n}s≠i,kjs+l×(j1,…,ji-1,l,ji+1,…,jk-1,jk+1,…,jn)!(Bi2+Bk2)l×∏s∈{1,…,n}s≠i,k‍Bs2js(t-∑s∈{1,…,n}s≠i,kjsτs-lτi)2(∑s∈{1,…,n}s≠i,kjs+l)(2(∑s∈{1,…,n}s≠i,kjs+l))!=𝒳τ1,…,τi-1,τi,τi+1,…,τk-1,τk+1,…,τnB12,…,Bi-12,Bi2+Bk2,Bi+12,…,Bk-12,Bk+12,…,Bn2(t).
Property (3) is trivial.
Now, we prove the statement (4). Ifτ:=τ1=⋯=τn, then
(36)𝒳¨τ1,…,τnB12,…,Bn2(t)=𝒳¨τB12+⋯+Bn2(t)𝒳¨τ1,…,τnB12,…,Bn2(t)=-(B12+⋯+Bn2)𝒳τB12+⋯+Bn2(t-τ)𝒳¨τ1,…,τnB12,…,Bn2(t)=-B12𝒳τ1,…,τnB12,…,Bn2(t-τ1)𝒳¨τ1,…,τnB12,…,Bn2(t)=-⋯-Bn2𝒳τ1,…,τnB12,…,Bn2(t-τn)
by (2) and from the property of 𝒳τB12+⋯+Bn2(t) (see (22)).
Hence, without any loss of generality, we assume thatτi≠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 proved in [16, Lemma 2.3.] Now, assume that 𝒳τ1,…,τn-1B12,…,Bn-12(t) solves
(37)x¨(t)=-B12x(t-τ1)-⋯-Bn-12x(t-τn-1),
that is, that the statement is fulfilled for n-1 different delays.
Letτk:=maxi=1,…,nτi. If t<τk, then t-τk<0, that is,
(38)𝒳τ1,…,τnB12,…,Bn2(t-τk)=Θ,
and from definition (20) it holds
(39)𝒳τ1,…,τnB12,…,Bn2(t)=𝒳τ1,…,τk-1,τk+1,…,τnB12,…,Bk-12,Bk+12,…,Bn2(t)
for such t. Consequently,
(40)𝒳¨τ1,…,τnB12,…,Bn2(t)=𝒳¨τ1,…,τk-1,τk+1,…,τnB12,…,Bk-12,Bk+12,…,Bn2(t)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑i=1,…,ni≠k‍Bi2𝒳τ1,…,τk-1,τk+1,…,τnB12,…,Bk-12,Bk+12,…,Bn2(t-τi)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑i=1n‍Bi2𝒳τ1,…,τnB12,…,Bn2(t-τi)
by the inductive hypothesis.
Now, lett≥maxi=1,…,nτi. Applying Lemma 3, we get
(41)𝒳τ1,…,τnB12,…,Bn2(t)=∑M⊂{1,…,n}SM(t)‍
with SM(t) given by (24) and the sum taken over all subsets of {1,…,n} including the trivial ones. Note that
(42)S∅(t)=∑ji≥1,i∈∅0≤t‍(-1)0({ji∣i∈∅})!E(t-0)00!S∅(t)=∑0≤t‍E=Eχ[0,∞)(t)
with a characteristic function χM~ of a set M~ given by
(43)χM~(t)={1,t∈M~,0,t∉M~.
Since each M⊂{1,…,n} is a finite set, Lemma 2 yields
(44)({ji∣i∈M})!=∑i∈M‍(ji-1,{jk∣k∈M∖{i}})!.
We apply this identity to derive a formula for the second derivative of SM for any ∅≠M⊂{1,…,n}:
(45)SM′′(t)=∑ji≥1,i∈M∑i∈Mjiτi≤t‍(-1)∑i∈Mji({ji∣i∈M})!×∏i∈M‍Bi2ji(t-∑i∈Mjiτi)2(∑i∈Mji-1)(2(∑i∈Mji-1))!=∑i∈M‍∑jk≥1,k∈M∑k∈Mjkτk≤t‍(-1)∑k∈Mjk(ji-1,{jk∣k∈M∖{i}})!×∏k∈M‍Bk2jk(t-τi-∑k∈M∖{i}jkτk-(ji-1)τi)2(∑k∈Mjk-1)(2(∑k∈Mjk-1))!.
Next, for any fixed i∈{1,…,n} we split the second sum to ji=1 and ji≥2, that is,
(46)∑jk≥1,k∈M∑k∈Mjkτk≤t‍F(j1,…,ji-1,ji,ji+1,…,jn)=∑jk≥1,k∈M∖{i}∑k∈M∖{i}jkτk≤t-τi‍F(j1,…,ji-1,1,ji+1,…,jn)+∑jk≥1,k∈M∖{i}ji≥2∑k∈Mjkτk≤t‍F(j1,…,ji-1,ji,ji+1,…,jn),
and use the equality
(47)∑jk≥1,k∈M∖{i}ji≥2∑k∈Mjkτk≤t‍F(j1,…,ji-1,ji,ji+1,…,jn)=∑jk≥1,k∈M∑k∈Mjkτk≤t-τi‍F(j1,…,ji-1,ji+1,ji+1,…,jn)
since
(48)∑k∈M‍jkτk≤t⟺∑k∈M∖{i}‍jkτk+(ji-1)τi≤t-τi.
So we obtain
(49)SM′′(t)=-∑i∈M‍Bi2(SM∖{i}(t-τi)+SM(t-τi))
for each ∅≠M⊂{1,…,n}. Obviously, S∅′′(t)=Θ. Consequently,
(50)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2(SM∖{i}(t-τi)+SM(t-τi))=-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2SM∖{i}(t-τi)-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2SM(t-τi).
Now, we add and subtract
(51)∑∅≠M⊂{1,…,n}‍∑i∉M‍Bi2SM(t-τi)
to the right-hand side of (50) to get
(52)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑∅≠M⊂{1,…,n}‍∑i=1n‍Bi2SM(t-τi)=+∑∅≠M⊂{1,…,n}‍∑i∉M‍Bi2SM(t-τi)=-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2SM∖{i}(t-τi)
and apply M=M∖{i} whenever i∉M:
(53)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑∅≠M⊂{1,…,n}‍∑i=1n‍Bi2SM(t-τi)=+∑∅≠M⊂{1,…,n}‍∑i∉M‍Bi2SM∖{i}(t-τi)=-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2SM∖{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
(54)∑∅≠M⊂{1,…,n}‍=∑M⊂{1,…,n}1≤#M≤n-1‍+∑M⊂{1,…,n}#M=n‍‍=∑M⊂{1,…,n}#M=1‍+∑M⊂{1,…,n}2≤#M≤n‍.
Hence, we have
(55)∑∅≠M⊂{1,…,n}‍∑i∉M‍Bi2SM∖{i}(t-τi)-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2SM∖{i}(t-τi)=∑M⊂{1,…,n}1≤#M≤n-1‍∑i∉M‍Bi2SM∖{i}(t-τi)+∑M={1,…,n}‍∑i∉M‍Bi2SM∖{i}(t-τi)-∑M∈{{1},…,{n}}‍∑i∈M‍Bi2SM∖{i}(t-τi)-∑M⊂{1,…,n}2≤#M≤n‍∑i∈M‍Bi2SM∖{i}(t-τi).
Now, we show that
(56)∑M⊂{1,…,n}1≤#M≤n-1∑i∉M‍Bi2SM∖{i}(t-τi)‍=∑M⊂{1,…,n}2≤#M≤n‍∑i∈M‍Bi2SM∖{i}(t-τi).
Let M⊂{1,…,n}, and let i∉M be arbitrary and fixed such that 1≤#M≤n-1. Then, clearly,
(57)BiSM∖{i}(t-τi)=BiS(M∪{i})∖{i}(t-τi)
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}.
On the other side, ifM⊂{1,…,n}, i∈M are arbitrary and fixed such that 2≤#M≤n, then
(58)BiSM∖{i}(t-τi)=BiS(M∖{i})∖{i}(t-τi)
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
(59)∑∅≠M⊂{1,…,n}‍∑i∉M‍Bi2SM∖{i}(t-τi)-∑∅≠M⊂{1,…,n}‍∑i∈M‍Bi2SM∖{i}(t-τi)=∑M={1,…,n}‍∑i∉M‍Bi2SM∖{i}(t-τi)-∑M∈{{1},…,{n}}‍∑i∈M‍Bi2SM{i}(t-τi).
Next, by the property of empty sum, we get
(60)∑M={1,…,n}‍∑i∉M‍Bi2SM∖{i}(t-τi)=∑M={1,…,n}‍Θ=Θ.
Moreover, it holds
(61)∑M∈{{1},…,{n}}‍∑i∈M‍Bi2SM∖{i}(t-τi)=∑i=1n‍Bi2S∅(t-τi)=∑M=∅‍∑i=1n‍Bi2SM(t-τi).
Therefore, putting (60) and (61) in (59) and the result in (53), we obtain
(62)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑∅≠M⊂{1,…,n}‍∑i=1n‍Bi2SM(t-τi)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑M=∅‍∑i=1n‍Bi2SM(t-τi)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑i=1n‍Bi2∑M⊂{1,…,n}‍SM(t-τi)𝒳¨τ1,…,τnB12,…,Bn2(t)=-∑i=1n‍Bi2𝒳τ1,…,τnB12,…,Bn2(t-τi).
Hence, 𝒳τ1,…,τnB12,…,Bn2(t) solves (31) for all t≥0. Clearly, the same is true for t<0.
For𝒴τ1,…,τnB12,…,Bn2(t), statements (1)–(3) can be proved as for 𝒳τ1,…,τnB12,…,Bn2(t). Next, if τ:=τ1=⋯=τn, we apply the point (2) of this lemma and property (22) for 𝒴τB12+⋯+Bn2(t) to see that
(63)𝒴¨τ1,…,τnB12,…,Bn2(t)=𝒴¨τB12+⋯+Bn2(t)𝒴¨τ1,…,τnB12,…,Bn2(t)=-(B12+⋯+Bn2)𝒴τB12+⋯+Bn2(t-τ)𝒴¨τ1,…,τnB12,…,Bn2(t)=-B12𝒴τ1,…,τnB12,…,Bn2(t-τ1)𝒴¨τ1,…,τnB12,…,Bn2(t)=-⋯-Bn2𝒴τ1,…,τnB12,…,Bn2(t-τn).
So, 𝒴τ1,…,τnB12,…,Bn2(t) is a solution of (31) when all delays are the same.
Again, the casen=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,…,nτi, then
(64)𝒴τ1,…,τnB12,…,Bn2(t)=𝒴τ1,…,τk-1,τk+1,…,τnB12,…,Bk-12,Bk+12,…,Bn2(t)
by definition (20), and the statement follows from the inductive hypothesis. For t≥maxi=1,…,nτi, we apply Lemma 3 to see that
(65)𝒴τ1,…,τnB12,…,Bn2(t)=∑M⊂{1,…,n}‍S~M(t)
with S~M(t) given by (25). This time
(66)S∅(t)=∑ji≥1,i∈∅0≤t‍(-1)0({ji∣i∈∅})!E(t-0)10!S∅(t)=∑0≤t‍Et=tEχ[0,∞)(t)
and S∅′′(t)=Θ. The rest proceeds analogically to 𝒳τ1,…,τnB12,…,Bn2(t).
The final statement follows directly from definition (20).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(67)𝒳τ1,…,τnB12,…,Bn2(t),𝒳τ1,…,τi-1,τi+1,…,τnB12,…,Bi-12,Bi+12,…,Bn2(t)
solve
(68)x¨(t)=-B12x(t-τ1)-⋯-Bi-12x(t-τi-1)-Bi+12x(t-τi+1)-⋯-Bn2x(t-τn)
with initial condition
(69)x(t)={Θ,-τ≤t<0,E,t=0,x˙(t)=Θ,-τ≤t≤0
and τ=maxi=1,…,nτi.We are ready to state and prove our main result.


## 3. Main Result

Here we find a solution of the initial value problem (11), (8) in the sense of the next definition.Definition 6.
Letτ1,…,τn>0, τ:=maxi=1,…,nτi, 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.
Letn≥3, τ1,…,τn>0, τ:=maxi=1,…,nτi, and φ∈C1([-τ,0],ℝN), and let B1,…,Bn be N×N pairwise permutable matrices; that is, BiBj=BjBi 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
(70)x(t)={φ(t),-τ≤t<0,𝒳(t)φ(0)+𝒴(t)φ˙(0)-∑i=1n‍Bi2∫-τi0‍𝒴(t-τi-s)φ(s)ds+∫0t‍𝒴(t-s)f(s)ds,0≤t,
where 𝒳(t)=𝒳τ1,…,τnB12,…,Bn2(t) and 𝒴(t)=𝒴τ1,…,τnB12,…,Bn2(t).Proof.
Obviously,x(t) satisfies the initial condition on [-τ,0), and, from definition (20), x(0)=φ(0). For the derivative, it holds limt→0-x˙(t)=φ˙(0). Moreover, if 0≤t<mini=1,…,nτi, then
(71)x(t)=φ(0)+tφ˙(0)-∑i=1n‍Bi2∫-τit-τi‍(t-τi-s)φ(s)ds+∫0t‍(t-s)f(s)ds
since
(72)𝒴(t-τi-s)={(t-τi-s)E,s∈[-τi,t-τi],Θ,s∈(t-τi,0]
for each i=1,…,n. Thus
(73)x˙(t)=φ˙(0)-∑i=1n‍Bi2∫-τit-τi‍φ(s)ds+∫0t‍f(s)ds
and limt→0+x˙(t)=φ˙(0). Clearly,
(74)x∈C1((-τ,∞),ℝN)∩C2((0,∞)∖{τ1,…,τn},ℝN).
We show that, although𝒳(t) is not C2 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).
Assume that0≤t<mini=1,…,nτi. Then identities (71) and (73) are valid, and by differentiating (73) for such t we get
(75)x¨(t)=-∑i=1n‍Bi2φ(t-τi)+f(t)=-∑i=1n‍Bi2x(t-τi)+f(t)
since x(t-τi)=φ(t-τi) for each i=1,…,n.
Now, let∅≠M1,2⊂{1,…,n} be such that τi≤t<τj for each i∈M1, j∈M2. Then
(76)𝒴(t-τj-s)={𝒴(t-τj-s),s∈[-τj,t-τj],Θ,s∈(t-τj,0]
whenever j∈M2, and (70) becomes
(77)x(t)=𝒳(t)φ(0)+𝒴(t)φ˙(0)-∑i∈M1‍Bi2∫-τi0‍𝒴(t-τi-s)φ(s)ds-∑j∈M2‍Bj2∫-τjt-τj‍𝒴(t-τj-s)φ(s)ds+∫0t‍𝒴(t-s)f(s)ds.
By the point (5) of Lemma 4, we get
(78)x˙(t)=𝒳˙(t)φ(0)+𝒴˙(t)φ˙(0)-∑i∈M1‍Bi2∫-τi0‍𝒴˙(t-τi-s)φ(s)ds-∑j∈M2‍Bj2∫-τjt-τj‍𝒳(t-τj-s)φ(s)ds+∫0t‍𝒳(t-s)f(s)ds,
and for the second derivative it holds
(79)x¨(t)=𝒳¨(t)φ(0)+𝒴¨(t)φ˙(0)-∑i∈M1‍Bi2∫-τi0‍𝒴¨(t-τi-s)φ(s)ds-∑j∈M2‍Bj2(φ(t-τj)+∫-τjt-τj‍𝒴¨(t-τj-s)φ(s)ds)+f(t)+∫0t‍𝒴¨(t-s)f(s)ds
since 𝒳(0)=E. Now, we apply the property (4) of Lemma 4 together with
(80)𝒳(t-τj)=𝒴(t-τj)=Θ,∀j∈M2
to see that both 𝒳 and 𝒴 are solutions of
(81)y¨(t)=-∑i∈M1‍Bi2y(t-τi).
Therefore,
(82)x¨(t)=-∑k∈M1‍Bk2(∑k∈M1‍𝒳(t-τk)φ(0)+𝒴(t-τk)φ˙(0)-∑i∈M1‍Bi2∫-τi0‍𝒴(t-τi-τk-s)φ(s)ds-∑j∈M2‍Bj2∫-τjt-τj‍𝒴(t-τj-τk-s)φ(s)ds+∫0t‍𝒴(t-τk-s)f(s)ds∑i∈M1‍)=-∑j∈M2‍Bj2φ(t-τj)+f(t)=-∑i∈M1‍Bi2x(t-τi)-∑j∈M2‍Bj2φ(t-τj)+f(t).
In fact, this is exactly formula (11) since x(t-τj)=φ(t-τj) for each j∈M2.
Finally, ifmaxi=1,…,nτi≤t, we have
(83)x(t)=𝒳(t)φ(0)+𝒴(t)φ˙(0)-∑i=1n‍Bi2∫-τi0‍𝒴(t-τi-s)φ(s)ds+∫0t‍𝒴(t-s)f(s)ds.
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,…,nτi<τi for each i=1,…,k-1,k+1,…,n, then by (75) and (82) we get
(84)limt→τk-x¨(t)=-∑i=1i≠kn‍Bi2φ(τk-τi)-Bk2φ(0)+f(τk)=-∑j∈M2‍Bj2φ(τk-τj)-Bk2x(0)+f(τk)=limt→τk+x¨(t),
where M2={1,…,n}∖{k}.It is easy to see that defining functions(85)𝒳~τ1,…,τnB1,…,Bn(t):=𝒳τ1,…,τn-B1,…,-Bn(t),𝒴~τ1,…,τnB1,…,Bn(t):=𝒴τ1,…,τn-B1,…,-Bn(t)
leads to the solution of
(86)x¨(t)=B1x(t-τ1)+⋯+Bnx(t-τn)+f(t)
with pairwise permutable matrices B1,…,Bn and initial condition (8). More precisely, we have the following corollary of Theorem 7.Corollary 8.
Letn≥3, τ1,…,τn>0, τ:=maxi=1,…,nτi, φ∈C1([-τ,0],ℝN), and let B1,…,Bn be N×N pairwise permutable matrices; that is, BiBj=BjBi 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
(87)x(t)={φ(t),-τ≤t<0,𝒳(t)φ(0)+𝒴(t)φ˙(0)+∑i=1n‍Bi∫-τi0‍𝒴(t-τi-s)φ(s)ds+∫0t‍𝒴(t-s)f(s)ds,0≤t,
where 𝒳(t)=𝒳~τ1,…,τnB1,…,Bn(t) and 𝒴(t)=𝒴~τ1,…,τnB1,…,Bn(t).Proof.
The corollary can be proved exactly in the same way as Theorem7.



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