Representation of a solution of the Cauchy problem for an oscillating system with multiple delays and pairwise permutable matrices

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,1Michal FeIkan,2,3 and Michal Pospíšil4 1Department of Mathematics, Faculty of Electrical Engineering and Communication, Brno University of Technology, Technick´ a 3058/10, 616 00 Brno, Czech Republic 2Department of Mathematical Analysis and Numerical Mathematics, Comenius University, Mlynsk´ adolina, 842 48 Bratislava, Slovakia 3Mathematical Institute of Slovak Academy of Sciences, ˇ Stef´ anikova 49, 814 73 Bratislava, Slovakia 4Centre for Research and Utilization of Renewable Energy, Faculty of Electrical Engineering and Communication, Brno University of Technology, Technick´ a 3058/10, 616 00 Brno, Czech Republic Correspondence should be addressed to Michal Posp´ ıˇ sil; [email protected]r.cz Received 13 January 2013; Accepted 19 April 2013 Academic Editor: Jaan Janno Copyright © 2013 Josef Dibl´ ık 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. 1. Introduction Motivated by delayed exponential representing a solution of a system of differential or difference equations with oneormultiplefixedorvariabledelays[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] 𝑥(𝑡)=−𝐵2𝑥(𝑡−𝜏)(1) to the case of two delays 𝑥(𝑡)=−𝐵2 1𝑥(𝑡−𝜏1)−𝐵2 2𝑥(𝑡−𝜏2), (2) where the linear parts were given by permutable matrices [16]. Equations (1), (2), and the below-stated (11)with𝑓≡0are generalizations of the scalar equation 𝑥(𝑡)=−𝑏2𝑥(𝑡)(3) representing linear oscillator, to 𝑁-dimensional space with oneormultiplefixeddelays.Clearly,eachsolutionofthelatter equation is oscillating whenever 0 =𝑏∈R. Analogically, (1)with𝑥∈R𝑁can have at least one oscillating solution whenever 𝑁is odd. Indeed, if 𝐵is 𝑁×𝑁matrix, 𝑁≥3is odd, and 𝐵has a simple real nonzero eigenvalue 𝜆, then there exists a regular matrix 𝑆such that 𝑆−1𝐵𝑆=𝐽=(𝜆0 0 𝐽)where 𝐽is (𝑁−1)×(𝑁−1)matrix. On letting 𝑥=𝑆𝑦,onegets 𝑦=−𝐽2𝑦(𝑡−𝜏)(4) or rewrites as the system 𝑦1=−𝜆2𝑦1(𝑡−𝜏), 𝑦2=−𝐽2𝑦2(𝑡−𝜏),(5) where 𝑦=(𝑦 1,𝑦2)∈R×R𝑁−1.Notethatthefirst column Vof 𝑆is the eigenvector of 𝐵corresponding to 𝜆. Clearly, if solution 𝑦1of (5)isoscillating,thensolution𝑦of (4) is oscillating in the first coordinate whenever its initial 2Abstract and Applied Analysis condition satisfies {𝑦(𝑡)|𝑡∈[−𝜏,0]}⊂R×{0}𝑁−1. Consequently, solution 𝑥of (1)isoscillatinginspan{V} whenever {𝑥(𝑡)|𝑡∈[−𝜏,0]}⊂span{V}.Taking𝑦1(𝑡)=𝑒𝜇𝑡, one obtains characteristic equation 𝜇2=−𝜆 2𝑒−𝜇𝜏 of (5), which has solutions 𝜇1,2 =𝛼±𝚤𝛽∈Cwith 𝛽 =0.Thus,𝑦1 is oscillating. On the other hand, there can exist a nonoscillating solution of the system (1)whenever𝑥∈R𝑁and 𝑁is even. For instance, if 𝑁=2and 𝐵=(01 −1 0 ),then(1)hastheform 𝑥(𝑡)=𝑥(𝑡−𝜏)(6) with 𝑥∈R2,which,obviously,doesnothaveanoscillating 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,ofscalarequation(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)withoutsquares 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]andinthispapercan 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 orderwithanynumberoffixeddelays,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 thecaseofthreeandmoredifferentdelays. First, we recall our result from [16]. Theorem 1. Let 𝜏1,𝜏2>0,𝜏:=max{𝜏1,𝜏2},and𝜑∈ 𝐶1([−𝜏,0],R𝑁).Let𝐵1,𝐵2be 𝑁×𝑁permutable matrices; that is, 𝐵1𝐵2=𝐵2𝐵1,andlet𝑓:[0,∞)→R𝑁be a given function. Solution 𝑥(𝑡)of 𝑥(𝑡)=−𝐵2 1𝑥(𝑡−𝜏1)−𝐵2 2𝑥(𝑡−𝜏2)+𝑓(𝑡)(7) satisfying initial condition 𝑥(𝑡)=𝜑(𝑡), 𝑥(𝑡)=𝜑(𝑡),−𝜏≤𝑡≤0 (8) has the form 𝑥(𝑡)={ { { { { { { { { { { { { { { { { { { { { { { { { 𝜑(𝑡),−𝜏≤𝑡<0, X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) −𝐵2 1∫0 −𝜏1 Y(𝑡−𝜏1−𝑠)𝜑(𝑠)𝑑𝑠 −𝐵2 2∫0 −𝜏2 Y(𝑡−𝜏2−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0Y(𝑡−𝑠)𝑓(𝑠)𝑑𝑠, 0≤𝑡, (9) where X(𝑡)=X𝐵2 1,𝐵2 2 𝜏1,𝜏2(𝑡) := ∑ 𝑖,𝑗≥0 𝑖𝜏1+𝑗𝜏2≤𝑡(−1)𝑖+𝑗 ×(𝑖+𝑗 𝑖)𝐵2𝑖 1𝐵2𝑗 2(𝑡−𝑖𝜏1−𝑗𝜏2)2(𝑖+𝑗) (2(𝑖+𝑗))! , Y(𝑡)=Y𝐵2 1,𝐵2 2 𝜏1,𝜏2(𝑡) := ∑ 𝑖,𝑗≥0 𝑖𝜏1+𝑗𝜏2≤𝑡(−1)𝑖+𝑗 ×(𝑖+𝑗 𝑖)𝐵2𝑖 1𝐵2𝑗 2(𝑡−𝑖𝜏1−𝑗𝜏2)2(𝑖+𝑗)+1 (2(𝑖+𝑗)+1)! . (10) We will denote Θand 𝐸the 𝑁×𝑁zero and identity matrix, respectively. 2. Systems with Multiple Delays In this section, we derive the representation of a solution of 𝑥(𝑡)=−𝐵2 1𝑥(𝑡−𝜏1)−⋅⋅⋅−𝐵2 𝑛𝑥(𝑡−𝜏𝑛)+𝑓(𝑡)(11) satisfying the initial condition (8), where 𝑛≥3,𝜏1,...,𝜏𝑛>0, 𝜏:=max𝑖=1,...,𝑛𝜏𝑖,𝐵1,...,𝐵𝑛are 𝑁×𝑁pairwise permutable matrices; that is, 𝐵𝑖𝐵𝑗=𝐵𝑗𝐵𝑖for each 𝑖,𝑗∈{1,...,𝑛},𝜑∈ 𝐶1([−𝜏,0],R𝑁),and𝑓:[0,∞)→R𝑁are given functions. The solution 𝑥(𝑡)will be represented using matrix functions analogical to (10)andwillbestatedinSection 3.Wenotethat thesameproblemswith𝑛=1,2were studied in [1,16]. From now on, we assume the property of empty sum and empty product; that is, ∑ 𝑖∈0𝑓(𝑖)=0, ∑ 𝑖∈0𝐹(𝑖)=Θ, ∏ 𝑖∈0𝑓(𝑖)=1, ∏ 𝑖∈0𝐹(𝑖)=𝐸 (12) for any function 𝑓and 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) Note that if 𝑛=2,then(𝑗1,𝑗2)=(𝑗1+𝑗2 𝑗1)and (20) coincides with (10). We will need a property of multinomial coefficients described in the next lemma. Lemma 2. Let 𝑛≥2be fixed. Then (𝑖1,𝑖2,...,𝑖𝑛)!=(𝑖1−1,𝑖2,...,𝑖𝑛)! +(𝑖1,𝑖2−1,𝑖3,...,𝑖𝑛)!+(𝑖1,...,𝑖𝑛−1,𝑖𝑛−1)! (14) for any 𝑖1,...,𝑖𝑛≥1. Proof. If 𝑛=2, then the statement follows from the property of binomial coefficients: (𝑖1,𝑖2)!=(𝑖1+𝑖2 𝑖1)=(𝑖1+𝑖2−1 𝑖1−1)+(𝑖1+𝑖2−1 𝑖1) =(𝑖1−1,𝑖2)!+(𝑖1,𝑖2−1)!. (15) 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)! (16) with inductive hypothesis to derive (𝑖1,𝑖2,𝑖3,...,𝑖𝑛)! =[(𝑖1+𝑖2−1,𝑖3,...,𝑖𝑛)!+(𝑖1+𝑖2,𝑖3−1,...,𝑖𝑛)! +⋅⋅⋅+(𝑖1+𝑖2,𝑖3,...,𝑖𝑛−1)!](𝑖1,𝑖2)!. (17) Clearly, from (16), we get (𝑖1+𝑖2,𝑖3−1,...,𝑖𝑛)!(𝑖1,𝑖2)!=(𝑖1,𝑖2,𝑖3−1,...,𝑖𝑛)!, . . . (𝑖1+𝑖2,𝑖3,...,𝑖𝑛−1)!(𝑖1,𝑖2)!=(𝑖1,𝑖2,𝑖3,...,𝑖𝑛−1)!.(18) Applying the case 𝑛=2(propertyofbinomialcoefficient) and (16), we get (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)!. (19) Putting (18)and(19)in(17), we obtain that the statement holds for 𝑛and 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 𝑖and 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. Define the functions X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛,Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛:R→𝐿(R𝑁)as X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡) := ∑ 𝑗1,...,𝑗𝑛≥0 𝑗1𝜏1+⋅⋅⋅+𝑗𝑛𝜏𝑛≤𝑡(−1)𝑗1+⋅⋅⋅+𝑗𝑛(𝑗1,...,𝑗𝑛)! ×𝑛 ∏ 𝑖=1𝐵2𝑗𝑖 𝑖(𝑡−𝑗1𝜏1−⋅⋅⋅−𝑗𝑛𝜏𝑛)2(𝑗1+⋅⋅⋅+𝑗𝑛) (2(𝑗1+⋅⋅⋅+𝑗𝑛))! , Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡) := ∑ 𝑗1,...,𝑗𝑛≥0 𝑗1𝜏1+⋅⋅⋅+𝑗𝑛𝜏𝑛≤𝑡(−1)𝑗1+⋅⋅⋅+𝑗𝑛(𝑗1,...,𝑗𝑛)! ×𝑛 ∏ 𝑖=1𝐵2𝑗𝑖 𝑖(𝑡−𝑗1𝜏1−⋅⋅⋅−𝑗𝑛𝜏𝑛)2(𝑗1+⋅⋅⋅+𝑗𝑛)+1 (2(𝑗1+⋅⋅⋅+𝑗𝑛)+1)! (20) for any 𝑡∈R. We will need functions X𝐵2 𝜏,Y𝐵2 𝜏:R→𝐿(R𝑁)for 𝜏>0 and 𝑁×𝑁complex matrix 𝐵(cf. [16]) defined as X𝐵2 𝜏(𝑡):=∑ 𝑖≥0 𝑖𝜏≤𝑡(−1)𝑖𝐵2𝑖 (𝑡−𝑖𝜏)2𝑖 (2𝑖)!, Y𝐵2 𝜏(𝑡):=∑ 𝑖≥0 𝑖𝜏≤𝑡(−1)𝑖𝐵2𝑖 (𝑡−𝑖𝜏)2𝑖+1 (2𝑖+1)!(21) with the properties X𝐵2 𝜏(𝑡)=−𝐵2Y𝐵2 𝜏(𝑡−𝜏),X𝐵2 𝜏(𝑡)=−𝐵2X𝐵2 𝜏(𝑡−𝜏), Y𝐵2 𝜏(𝑡)=X𝐵2 𝜏(𝑡),Y𝐵2 𝜏(𝑡)=−𝐵2Y𝐵2 𝜏(𝑡−𝜏)(22) for any 𝑡∈R, considering the one-sided derivatives at −𝜏,0. Some of properties of functions X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛and Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛are concluded in Lemma 4,buttoproveitwewillneedthenext lemma. Lemma 3. Let 𝑛≥1and 𝜏1,...,𝜏𝑛>0.Let𝐵1,...,𝐵𝑛be 𝑁×𝑁pairwise permutable matrices, that is, 𝐵𝑖𝐵𝑗=𝐵𝑗𝐵𝑖for each 𝑖,𝑗∈{1,...,𝑛}.Thenforany𝑡∈R, X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=∑ 𝑀⊂{1,...,𝑛}𝑆𝑀(𝑡), Y𝐵2 1,...𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=∑ 𝑀⊂{1,...,𝑛}𝑆𝑀(𝑡),(23) 4Abstract 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 𝑗𝑖𝜏𝑖≤𝑡(−1)∑𝑖∈𝑀 𝑗𝑖({𝑗𝑖|𝑖∈𝑀})! ×∏ 𝑖∈𝑀𝐵2𝑗𝑖 𝑖(𝑡−∑𝑖∈𝑀 𝑗𝑖𝜏𝑖)2∑𝑖∈𝑀 𝑗𝑖 (2∑𝑖∈𝑀 𝑗𝑖)! ,(24) 𝑆𝑀(𝑡):= ∑ 𝑗𝑖≥1,𝑖∈𝑀 ∑𝑖∈𝑀 𝑗𝑖𝜏𝑖≤𝑡(−1)∑𝑖∈𝑀 𝑗𝑖({𝑗𝑖|𝑖∈𝑀})! ×∏ 𝑖∈𝑀𝐵2𝑗𝑖 𝑖(𝑡−∑𝑖∈𝑀 𝑗𝑖𝜏𝑖)2∑𝑖∈𝑀 𝑗𝑖+1 (2∑𝑖∈𝑀 𝑗𝑖+1)! .(25) Proof. Denote N0,Nthe set of all nonnegative, positive integers, respectively; that is, N0={0}∪N.Thus,wehave the trivial identity N0×⋅⋅⋅×N0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 𝑛 =({0}×N0×⋅⋅⋅×N0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 𝑛−1 )∪(N×N0×⋅⋅⋅×N0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 𝑛−1 ) =⋅⋅⋅= ⋃ 𝑀1,...,𝑀𝑛∈{{0},N}𝑀1×⋅⋅⋅×𝑀𝑛.(26) Analogically, for any 𝑡∈Reach 𝑛-tuple 𝑗1,...,𝑗𝑛≥0 such that ∑𝑛 𝑖=1 𝑗𝑖𝜏𝑖≤𝑡canbedividedintwodistinctsetsof 𝑖-s so that 𝑗𝑖≥1if 𝑖∈𝑀⊂{1,...,𝑛}and 𝑗𝑖=0if 𝑖∈ {1,...,𝑛}\𝑀.Thatis,𝑀denotes the set of all indices 𝑖such that 𝑗𝑖=0.Moreover,∑𝑛 𝑖=1 𝑗𝑖𝜏𝑖=∑𝑖∈𝑀 𝑗𝑖𝜏𝑖. Accordingly, we can write {(𝑗1,...,𝑗𝑛)∈N𝑛 0|𝑛 ∑ 𝑖=1𝑗𝑖𝜏𝑖≤𝑡} =⋃ 𝑀⊂{1,...,𝑛}{(𝑗1,...,𝑗𝑛)∈N𝑛 0| 𝑗𝑖=0∀𝑖∉𝑀,∑ 𝑖∈𝑀 𝑗𝑖𝜏𝑖≤𝑡}, (27) 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 X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛follows. Statement for Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛canbeprovedinasimilarway. Lemma 4. Let 𝑛≥3and 𝜏1,...,𝜏𝑛>0.Let𝐵1,...,𝐵𝑛be 𝑁×𝑁pairwise permutable matrices; that is, 𝐵𝑖𝐵𝑗=𝐵𝑗𝐵𝑖for each 𝑖,𝑗∈{1,...,𝑛}. Then the following holds for any 𝑡∈R: (1) if 𝐵𝑖=Θfor some 𝑖∈{1,...,𝑛},then X𝐵2 1,...,𝐵2 𝑖−1,𝐵2 𝑖,𝐵2 𝑖+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑖−1,𝜏𝑖,𝜏𝑖+1,...,𝜏𝑛(𝑡)=X𝐵2 1,...,𝐵2 𝑖−1,𝐵2 𝑖+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑖−1,𝜏𝑖+1,...,𝜏𝑛(𝑡),(28) (2) if 𝜏𝑖=𝜏𝑘for 𝑖<𝑘,𝑖,𝑘∈{1,...,𝑛},then X𝐵2 1,...,𝐵2 𝑖−1,𝐵2 𝑖,𝐵2 𝑖+1,...,𝐵2 𝑘−1,𝐵2 𝑘,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑖−1,𝜏𝑖,𝜏𝑖+1,...,𝜏𝑘−1,𝜏𝑘,𝜏𝑘+1,...,𝜏𝑛(𝑡) =X𝐵2 1,...,𝐵2 𝑖−1,𝐵2 𝑖+𝐵2 𝑘,𝐵2 𝑖+1,...,𝐵2 𝑘−1,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑖−1,𝜏𝑖,𝜏𝑖+1,...,𝜏𝑘−1,𝜏𝑘+1,...,𝜏𝑛(𝑡),(29) (3) for any bijective mapping 𝜎:{1,...,𝑛}→{1,...,𝑛} we get X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=X𝐵2 𝜎(1),...,𝐵2 𝜎(𝑛) 𝜏𝜎(1),...,𝜏𝜎(𝑛) (𝑡),(30) (4) taking the one-sided derivatives at 0,𝜏1,...,𝜏𝑛,then X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=−𝐵2 1X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏1) −⋅⋅⋅−𝐵2 𝑛X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏𝑛), (31) (5) considering the one-sided derivatives at 0 (they both equal Θ), then Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡).(32) Statements (1)–(4) hold with Yinstead of X. Proof. Statement (1) follows easily from definition of X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛, because Θ2𝑖 =𝐸if 𝑖=0and Θ2𝑖 =Θwhenever 𝑖>0. Next, if 𝜏𝑖=𝜏𝑘,then ∑ 𝑗1,...,𝑗𝑛≥0 𝑗1𝜏1+⋅⋅⋅+𝑗𝑛𝜏𝑛≤𝑡𝐹(𝑗1,...,𝑗𝑛) =∑ 𝑗1,...,𝑗𝑖−1,𝑙,𝑗𝑖+1,...,𝑗𝑘−1,𝑗𝑘+1,...,𝑗𝑛≥0 𝑗1𝜏1+⋅⋅⋅+𝑗𝑖−1𝜏𝑖−1+𝑙𝜏𝑖+𝑗𝑖+1𝜏𝑖+1 +⋅⋅⋅+𝑗𝑘−1𝜏𝑘−1+𝑗𝑘+1𝜏𝑘+1+⋅⋅⋅+𝑗𝑛𝜏𝑛≤𝑡 ∑ 𝑗𝑖,𝑗𝑘≥0 𝑗𝑖+𝑗𝑘=𝑙𝐹(𝑗1,...,𝑗𝑛)(33) for any matrix function 𝐹.Thus,usingthepropertyof 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 for (2), we obtain X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡) =∑ 𝑗1,...,𝑗𝑖−1,𝑙,𝑗𝑖+1,...,𝑗𝑘−1,𝑗𝑘+1,...,𝑗𝑛≥0 𝑗1𝜏1+⋅⋅⋅+𝑗𝑖−1𝜏𝑖−1+𝑙𝜏𝑖+𝑗𝑖+1𝜏𝑖+1 +⋅⋅⋅+𝑗𝑘−1𝜏𝑘−1+𝑗𝑘+1𝜏𝑘+1+⋅⋅⋅+𝑗𝑛𝜏𝑛≤𝑡(−1)∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 𝑗𝑠+𝑙 ×(𝑗1,...,𝑗𝑖−1,𝑙,𝑗𝑖+1,...,𝑗𝑘−1,𝑗𝑘+1,...,𝑗𝑛)! ×(∑ 𝑗𝑖,𝑗𝑘≥0 𝑗𝑖+𝑗𝑘=𝑙 (𝑗𝑖,𝑗𝑘)!𝐵2𝑗𝑖 𝑖𝐵2𝑗𝑘 𝑘) ×∏ 𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 𝐵2𝑗𝑠 𝑠(𝑡−∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 ∑𝑗𝑠𝜏𝑠−𝑙𝜏𝑖)2(∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 𝑗𝑠+𝑙) (2(∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 ∑𝑗𝑠+𝑙))! =∑ 𝑗1,...,𝑗𝑖−1,𝑙,𝑗𝑖+1,...,𝑗𝑘−1,𝑗𝑘+1,...,𝑗𝑛≥0 𝑗1𝜏1+⋅⋅⋅+𝑗𝑖−1𝜏𝑖−1+𝑙𝜏𝑖+𝑗𝑖+1𝜏𝑖+1 +⋅⋅⋅+𝑗𝑘−1𝜏𝑘−1+𝑗𝑘+1𝜏𝑘+1+⋅⋅⋅+𝑗𝑛𝜏𝑛≤𝑡(−1)∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 𝑗𝑠+𝑙 ×(𝑗1,...,𝑗𝑖−1,𝑙,𝑗𝑖+1,...,𝑗𝑘−1,𝑗𝑘+1,...,𝑗𝑛)!(𝐵2 𝑖+𝐵2 𝑘)𝑙 ×∏ 𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 𝐵2𝑗𝑠 𝑠(𝑡−∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 ∑𝑗𝑠𝜏𝑠−𝑙𝜏𝑖)2(∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 𝑗𝑠+𝑙) (2(∑𝑠∈{1,...,𝑛} 𝑠 =𝑖,𝑘 ∑𝑗𝑠+𝑙))! =X𝐵2 1,...,𝐵2 𝑖−1,𝐵2 𝑖+𝐵2 𝑘,𝐵2 𝑖+1,...,𝐵2 𝑘−1,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑖−1,𝜏𝑖,𝜏𝑖+1,...,𝜏𝑘−1,𝜏𝑘+1,...,𝜏𝑛(𝑡).(35) Property (3) is trivial. Now, we prove the statement (4). If 𝜏:=𝜏1=⋅⋅⋅=𝜏𝑛, then X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=X𝐵2 1+⋅⋅⋅+𝐵2 𝑛 𝜏(𝑡) =−(𝐵2 1+⋅⋅⋅+𝐵2 𝑛)X𝐵2 1+⋅⋅⋅+𝐵2 𝑛 𝜏(𝑡−𝜏) =−𝐵2 1X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏1) −⋅⋅⋅−𝐵2 𝑛X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏𝑛) (36) by (2) and from the property of X𝐵2 1+⋅⋅⋅+𝐵2 𝑛 𝜏(𝑡)(see (22)). Hence, without any loss of generality, we assume that 𝜏𝑖=𝜏𝑗for each 𝑖 =𝑗,𝑖,𝑗∈{1,...,𝑛}(in the other case, we collect matrices as stated in (2)). Note the case 𝑛=2was proved in [16, Lemma 2.3.] Now, assume that X𝐵2 1,...,𝐵2 𝑛−1 𝜏1,...,𝜏𝑛−1 (𝑡) solves 𝑥(𝑡)=−𝐵2 1𝑥(𝑡−𝜏1)−⋅⋅⋅−𝐵2 𝑛−1𝑥(𝑡−𝜏𝑛−1), (37) that is, that the statement is fulfilled for 𝑛−1different delays. Let 𝜏𝑘:=max𝑖=1,...,𝑛𝜏𝑖.If𝑡<𝜏𝑘,then𝑡−𝜏𝑘<0,thatis, X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏𝑘)=Θ, (38) and from definition (20)itholds X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=X𝐵2 1,...,𝐵2 𝑘−1,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑘−1,𝜏𝑘+1,...,𝜏𝑛(𝑡)(39) for such 𝑡.Consequently, X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=X𝐵2 1,...,𝐵2 𝑘−1,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑘−1,𝜏𝑘+1,...,𝜏𝑛(𝑡) =−∑ 𝑖=1,...,𝑛 𝑖 =𝑘 𝐵2 𝑖X𝐵2 1,...,𝐵2 𝑘−1,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑘−1,𝜏𝑘+1,...,𝜏𝑛(𝑡−𝜏𝑖) =−𝑛 ∑ 𝑖=1𝐵2 𝑖X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏𝑖) (40) by the inductive hypothesis. Now, let 𝑡≥max𝑖=1,...,𝑛𝜏𝑖. Applying Lemma 3,weget X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=∑ 𝑀⊂{1,...,𝑛}𝑆𝑀(𝑡)(41) with 𝑆𝑀(𝑡)given by (24) and the sum taken over all subsets of {1,...,𝑛}including the trivial ones. Note that 𝑆0(𝑡)=∑ 𝑗𝑖≥1,𝑖∈0 0≤𝑡 (−1)0({𝑗𝑖|𝑖∈0})!𝐸(𝑡−0)0 0! =∑ 0≤𝑡𝐸=𝐸𝜒[0,∞) (𝑡)(42) with a characteristic function 𝜒 𝑀of a set  𝑀given by 𝜒 𝑀(𝑡)={1, 𝑡∈ 𝑀, 0, 𝑡∉ 𝑀. (43) Since each 𝑀⊂{1,...,𝑛}is a finite set, Lemma 2 yields ({𝑗𝑖|𝑖∈𝑀})!=∑ 𝑖∈𝑀(𝑗𝑖−1,{𝑗𝑘|𝑘∈𝑀\{𝑖}})!. (44) We apply this identity to derive a formula for the second derivative of 𝑆𝑀for any 0 =𝑀⊂{1,...,𝑛}: 𝑆󸀠󸀠 𝑀(𝑡) =∑ 𝑗𝑖≥1,𝑖∈𝑀 ∑𝑖∈𝑀 𝑗𝑖𝜏𝑖≤𝑡(−1)∑𝑖∈𝑀 𝑗𝑖({𝑗𝑖|𝑖∈𝑀})! ×∏ 𝑖∈𝑀𝐵2𝑗𝑖 𝑖(𝑡−∑𝑖∈𝑀 𝑗𝑖𝜏𝑖)2(∑𝑖∈𝑀 𝑗𝑖−1) (2(∑𝑖∈𝑀 𝑗𝑖−1))! =∑ 𝑖∈𝑀 ∑ 𝑗𝑘≥1,𝑘∈𝑀 ∑𝑘∈𝑀 𝑗𝑘𝜏𝑘≤𝑡(−1)∑𝑘∈𝑀 𝑗𝑘(𝑗𝑖−1,{𝑗𝑘|𝑘∈𝑀\{𝑖}})! ×∏ 𝑘∈𝑀𝐵2𝑗𝑘 𝑘(𝑡−𝜏𝑖−∑𝑘∈𝑀\{𝑖}𝑗𝑘𝜏𝑘−(𝑗𝑖−1)𝜏𝑖)2(∑𝑘∈𝑀 𝑗𝑘−1) (2(∑𝑘∈𝑀 𝑗𝑘−1))! . (45) 6Abstract and Applied Analysis Next, for any fixed 𝑖∈{1,...,𝑛}we split the second sum to 𝑗𝑖=1and 𝑗𝑖≥2,thatis, ∑ 𝑗𝑘≥1,𝑘∈𝑀 ∑𝑘∈𝑀 𝑗𝑘𝜏𝑘≤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∈𝑀 𝑗𝑘𝜏𝑘≤𝑡 𝐹(𝑗1,...,𝑗𝑖−1,𝑗𝑖,𝑗𝑖+1,...,𝑗𝑛), (46) andusetheequality ∑ 𝑗𝑘≥1,𝑘∈𝑀\{𝑖} 𝑗𝑖≥2 ∑𝑘∈𝑀 𝑗𝑘𝜏𝑘≤𝑡 𝐹(𝑗1,...,𝑗𝑖−1,𝑗𝑖,𝑗𝑖+1,...,𝑗𝑛) =∑ 𝑗𝑘≥1,𝑘∈𝑀 ∑𝑘∈𝑀 𝑗𝑘𝜏𝑘≤𝑡−𝜏𝑖𝐹(𝑗1,...,𝑗𝑖−1,𝑗𝑖+1,𝑗𝑖+1,...,𝑗𝑛)(47) since∑ 𝑘∈𝑀𝑗𝑘𝜏𝑘≤𝑡⇐⇒ ∑ 𝑘∈𝑀\{𝑖}𝑗𝑘𝜏𝑘+(𝑗𝑖−1)𝜏𝑖≤𝑡−𝜏𝑖.(48) So we obtain 𝑆󸀠󸀠 𝑀(𝑡)=−∑ 𝑖∈𝑀𝐵2 𝑖(𝑆𝑀\{𝑖} (𝑡−𝜏𝑖)+𝑆𝑀(𝑡−𝜏𝑖)) (49) for each 0 =𝑀 ⊂ {1,...,𝑛}.Obviously,𝑆󸀠󸀠 0(𝑡) = Θ. Consequently, X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡) =− ∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖(𝑆𝑀\{𝑖} (𝑡−𝜏𝑖)+𝑆𝑀(𝑡−𝜏𝑖)) =− ∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖). (50) Now, we add and subtract ∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖)(51) to the right-hand side of (50)toget X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡) =− ∑ 0 =𝑀⊂{1,...,𝑛} 𝑛 ∑ 𝑖=1𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖) +∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖) −∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) (52) and apply 𝑀=𝑀\{𝑖}whenever 𝑖∉𝑀: X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡) =− ∑ 0 =𝑀⊂{1,...,𝑛} 𝑛 ∑ 𝑖=1𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖) +∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖). (53) Denoting #𝑀thenumberofelementsoftheset𝑀,wesplit the last two terms of the right-hand side of the latter equality with respect to ∑ 0 =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 .(54) Hence, we have ∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) =∑ 𝑀⊂{1,...,𝑛} 1≤#𝑀≤𝑛−1 ∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) +∑ 𝑀={1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 𝑀∈{{1},...,{𝑛}} ∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 𝑀⊂{1,...,𝑛} 2≤#𝑀≤𝑛 ∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖). (55) Now, we show that ∑ 𝑀⊂{1,...,𝑛} 1≤#𝑀≤𝑛−1 ∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) =∑ 𝑀⊂{1,...,𝑛} 2≤#𝑀≤𝑛 ∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖). (56) Let 𝑀⊂{1,...,𝑛}, and let 𝑖∉𝑀be arbitrary and fixed such that 1≤#𝑀≤𝑛−1. Then, clearly, 𝐵𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖)=𝐵𝑖𝑆(𝑀∪{𝑖})\{𝑖} (𝑡−𝜏𝑖)(57) and 2≤#(𝑀∪{𝑖})≤𝑛,𝑖∈𝑀∪{𝑖}.Moreover,if𝑀1,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∈𝑀are arbitrary and fixed such that 2≤#𝑀≤𝑛,then 𝐵𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖)=𝐵𝑖𝑆(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}.Inconclusion,thereis 1−1correspondence between the terms on the left-hand side of (56) and the terms on the right-hand side. So (56)isvalid. Putting (56)in(55)weobtain ∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 0 =𝑀⊂{1,...,𝑛}∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) =∑ 𝑀={1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) −∑ 𝑀∈{{1},...,{𝑛}} ∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀{𝑖} (𝑡−𝜏𝑖). (59) Next, by the property of empty sum, we get ∑ 𝑀={1,...,𝑛}∑ 𝑖∉𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖)= ∑ 𝑀={1,...,𝑛}Θ=Θ. (60) Moreover, it holds ∑ 𝑀∈{{1},...,{𝑛}} ∑ 𝑖∈𝑀𝐵2 𝑖𝑆𝑀\{𝑖} (𝑡−𝜏𝑖) =𝑛 ∑ 𝑖=1𝐵2 𝑖𝑆0(𝑡−𝜏𝑖)=∑ 𝑀=0 𝑛 ∑ 𝑖=1𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖). (61) Therefore, putting (60)and(61)in(59)andtheresultin(53), we obtain X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=− ∑ 0 =𝑀⊂{1,...,𝑛} 𝑛 ∑ 𝑖=1𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖) −∑ 𝑀=0 𝑛 ∑ 𝑖=1𝐵2 𝑖𝑆𝑀(𝑡−𝜏𝑖) =−𝑛 ∑ 𝑖=1𝐵2 𝑖∑ 𝑀⊂{1,...,𝑛}𝑆𝑀(𝑡−𝜏𝑖) =−𝑛 ∑ 𝑖=1𝐵2 𝑖X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏𝑖). (62) Hence, X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)solves (31)forall𝑡≥0.Clearly,thesame is true for 𝑡<0. For Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡),statements(1)–(3)canbeprovedasfor X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡). Next, if 𝜏:=𝜏1=⋅⋅⋅=𝜏𝑛,weapplythepoint(2) of this lemma and property (22)forY𝐵2 1+⋅⋅⋅+𝐵2 𝑛 𝜏(𝑡)to see that Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=Y𝐵2 1+⋅⋅⋅+𝐵2 𝑛 𝜏(𝑡) =−(𝐵2 1+⋅⋅⋅+𝐵2 𝑛)Y𝐵2 1+⋅⋅⋅+𝐵2 𝑛 𝜏(𝑡−𝜏) =−𝐵2 1Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏1) −⋅⋅⋅−𝐵2 𝑛Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡−𝜏𝑛). (63) So, Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)is a solution of (31) when all delays are the same. Again, the case 𝑛=2with different delays was proved in [16]; thus, we assume that the statement is fulfilled for 𝑛− 1,𝑛≥3and that 𝜏𝑖=𝜏𝑗for each 𝑖 =𝑗,𝑖,𝑗∈{1,...,𝑛}.As before, if 𝑡<𝜏𝑘and 𝜏𝑘:=max𝑖=1,...,𝑛𝜏𝑖,then Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=Y𝐵2 1,...,𝐵2 𝑘−1,𝐵2 𝑘+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑘−1,𝜏𝑘+1,...,𝜏𝑛(𝑡)(64) by definition (20), and the statement follows from the inductive hypothesis. For 𝑡≥max𝑖=1,...,𝑛𝜏𝑖, we apply Lemma 3 to see that Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)=∑ 𝑀⊂{1,...,𝑛}𝑆𝑀(𝑡)(65) with 𝑆𝑀(𝑡)given by (25). This time 𝑆0(𝑡)=∑ 𝑗𝑖≥1,𝑖∈0 0≤𝑡 (−1)0({𝑗𝑖|𝑖∈0})!𝐸(𝑡−0)1 0! =∑ 0≤𝑡𝐸𝑡=𝑡𝐸𝜒[0,∞) (𝑡)(66) and 𝑆󸀠󸀠 0(𝑡)=Θ. The rest proceeds analogically to X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡). The final statement follows directly from definition (20). Remark 5. Another proof of statements (1)–(3) of the previouslemmacanbemadewiththeaidofstatement(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 X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡),X𝐵2 1,...,𝐵2 𝑖−1,𝐵2 𝑖+1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑖−1,𝜏𝑖+1,...,𝜏𝑛(𝑡)(67) solve 𝑥(𝑡)=−𝐵2 1𝑥(𝑡−𝜏1)−⋅⋅⋅−𝐵2 𝑖−1𝑥(𝑡−𝜏𝑖−1) −𝐵2 𝑖+1𝑥(𝑡−𝜏𝑖+1)−⋅⋅⋅−𝐵2 𝑛𝑥(𝑡−𝜏𝑛)(68) with initial condition 𝑥(𝑡)={Θ, −𝜏≤𝑡<0, 𝐸, 𝑡=0, 𝑥(𝑡)=Θ,−𝜏≤𝑡≤0 (69) and 𝜏=max𝑖=1,...,𝑛𝜏𝑖. We are ready to state and prove our main result. 8Abstract and Applied Analysis 3. Main Result Herewefindasolutionoftheinitialvalueproblem(11), (8) in the sense of the next definition. Definition 6. Let 𝜏1,...,𝜏𝑛>0,𝜏:=max𝑖=1,...,𝑛𝜏𝑖,and𝜑∈ 𝐶1([−𝜏,0],R𝑁),andlet𝐵1,...,𝐵𝑛be 𝑁×𝑁matrices, and let 𝑓:[0,∞)→R𝑁be a given function. Function 𝑥: [−𝜏,∞) → R𝑁is a solution of (11) and initial condition (8), if 𝑥∈𝐶 1([−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,...,𝑛𝜏𝑖,and 𝜑∈𝐶1([−𝜏,0],R𝑁),andlet𝐵1,...,𝐵𝑛be 𝑁×𝑁pairwise permutable matrices; that is, 𝐵𝑖𝐵𝑗=𝐵 𝑗𝐵𝑖for each 𝑖,𝑗 ∈ {1,...,𝑛},andlet𝑓:[0,∞)→R𝑁be a given function. Solution 𝑥(𝑡)of (11)satisfying initial condition (8)has the form 𝑥(𝑡)={ { { { { { { { { { { { { { { { { 𝜑(𝑡),−𝜏≤𝑡<0, X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) −𝑛 ∑ 𝑖=1𝐵2 𝑖∫0 −𝜏𝑖 Y(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0Y(𝑡−𝑠)𝑓(𝑠)𝑑𝑠, 0≤𝑡, (70) where X(𝑡)=X𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(𝑡)and Y(𝑡)=Y𝐵2 1,...,𝐵2 𝑛 𝜏1,...,𝜏𝑛(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,if0≤𝑡<min𝑖=1,...,𝑛𝜏𝑖, then 𝑥(𝑡)=𝜑(0)+𝑡𝜑(0) −𝑛 ∑ 𝑖=1𝐵2 𝑖∫𝑡−𝜏𝑖 −𝜏𝑖(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0(𝑡−𝑠)𝑓(𝑠)𝑑𝑠 (71) since Y(𝑡−𝜏𝑖−𝑠)={(𝑡−𝜏𝑖−𝑠)𝐸, 𝑠∈[−𝜏𝑖,𝑡−𝜏𝑖], Θ, 𝑠∈(𝑡−𝜏𝑖,0] (72) for each 𝑖=1,...,𝑛.Thus 𝑥(𝑡)=𝜑(0)−𝑛 ∑ 𝑖=1𝐵2 𝑖∫𝑡−𝜏𝑖 −𝜏𝑖𝜑(𝑠)𝑑𝑠+∫𝑡 0𝑓(𝑠)𝑑𝑠 (73) and lim𝑡→0+𝑥(𝑡)= 𝜑(0).Clearly, 𝑥∈𝐶1((−𝜏,∞),R𝑁)∩𝐶2((0,∞)\{𝜏1,...,𝜏𝑛},R𝑁). (74) We show that, although X(𝑡)is not 𝐶2at 𝜏1,...,𝜏𝑛, function 𝑥(𝑡)is 𝐶2at these points and, therefore, in (0,∞). At once, we prove that 𝑥(𝑡)is a solution of (11). Assume that 0≤𝑡<min𝑖=1,...,𝑛𝜏𝑖. Then identities (71)and (73) are valid, and by differentiating (73)forsuch𝑡we get 𝑥(𝑡)=−𝑛 ∑ 𝑖=1𝐵2 𝑖𝜑(𝑡−𝜏𝑖)+𝑓(𝑡)=−𝑛 ∑ 𝑖=1𝐵2 𝑖𝑥(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<𝜏𝑗for each 𝑖∈𝑀1,𝑗∈𝑀2.Then Y(𝑡−𝜏𝑗−𝑠)={Y(𝑡−𝜏𝑗−𝑠), 𝑠∈[−𝜏𝑗,𝑡−𝜏𝑗], Θ, 𝑠∈(𝑡−𝜏𝑗,0] (76) whenever 𝑗∈𝑀2,and(70)becomes 𝑥(𝑡)=X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) −∑ 𝑖∈𝑀1𝐵2 𝑖∫0 −𝜏𝑖 Y(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 −∑ 𝑗∈𝑀2𝐵2 𝑗∫𝑡−𝜏𝑗 −𝜏𝑗 Y(𝑡−𝜏𝑗−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0Y(𝑡−𝑠)𝑓(𝑠)𝑑𝑠. (77) By the point (5) of Lemma 4,weget 𝑥(𝑡)=X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) −∑ 𝑖∈𝑀1𝐵2 𝑖∫0 −𝜏𝑖Y(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 −∑ 𝑗∈𝑀2𝐵2 𝑗∫𝑡−𝜏𝑗 −𝜏𝑗 X(𝑡−𝜏𝑗−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0X(𝑡−𝑠)𝑓(𝑠)𝑑𝑠, (78) andforthesecondderivativeitholds 𝑥(𝑡)=X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) −∑ 𝑖∈𝑀1𝐵2 𝑖∫0 −𝜏𝑖Y(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 −∑ 𝑗∈𝑀2𝐵2 𝑗(𝜑(𝑡−𝜏𝑗)+∫𝑡−𝜏𝑗 −𝜏𝑗Y(𝑡−𝜏𝑗−𝑠)𝜑(𝑠)𝑑𝑠) +𝑓(𝑡)+∫𝑡 0Y(𝑡−𝑠)𝑓(𝑠)𝑑𝑠 (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 Xand Yare solutions of 𝑦(𝑡)=−∑ 𝑖∈𝑀1𝐵2 𝑖𝑦(𝑡−𝜏𝑖). (81) Therefore, 𝑥(𝑡)=−∑ 𝑘∈𝑀1𝐵2 𝑘(X(𝑡−𝜏𝑘)𝜑(0)+Y(𝑡−𝜏𝑘)𝜑(0) −∑ 𝑖∈𝑀1𝐵2 𝑖∫0 −𝜏𝑖 Y(𝑡−𝜏𝑖−𝜏𝑘−𝑠)𝜑(𝑠)𝑑𝑠 −∑ 𝑗∈𝑀2𝐵2 𝑗∫𝑡−𝜏𝑗 −𝜏𝑗 Y(𝑡−𝜏𝑗−𝜏𝑘−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0Y(𝑡−𝜏𝑘−𝑠)𝑓(𝑠)𝑑𝑠) −∑ 𝑗∈𝑀2𝐵2 𝑗𝜑(𝑡−𝜏𝑗)+𝑓(𝑡) =−∑ 𝑖∈𝑀1𝐵2 𝑖𝑥(𝑡−𝜏𝑖)−∑ 𝑗∈𝑀2𝐵2 𝑗𝜑(𝑡−𝜏𝑗)+𝑓(𝑡).(82) In fact, this is exactly formula (11)since𝑥(𝑡−𝜏𝑗)=𝜑(𝑡−𝜏𝑗) for each 𝑗∈𝑀2. Finally, if max𝑖=1,...,𝑛𝜏𝑖≤𝑡,wehave 𝑥(𝑡)=X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) −𝑛 ∑ 𝑖=1𝐵2 𝑖∫0 −𝜏𝑖 Y(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0Y(𝑡−𝑠)𝑓(𝑠)𝑑𝑠. (83) 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)reallysolves(11) and satisfies initial condition (8)and,moreover,that𝑥∈𝐶2((0,∞),R𝑁).Toseethelast one, one has to put 𝜏1,...,𝜏𝑛into the computed derivatives, for example, if 𝜏𝑘:=min𝑖=1,...,𝑛𝜏𝑖<𝜏𝑖for each 𝑖=1,...,𝑘− 1,𝑘+1,...,𝑛,thenby(75)and(82)weget lim 𝑡→𝜏− 𝑘𝑥(𝑡)=−𝑛 ∑ 𝑖=1 𝑖 =𝑘𝐵2 𝑖𝜑(𝜏𝑘−𝜏𝑖)−𝐵2 𝑘𝜑(0)+𝑓(𝜏𝑘) =−∑ 𝑗∈𝑀2𝐵2 𝑗𝜑(𝜏𝑘−𝜏𝑗) −𝐵2 𝑘𝑥(0)+𝑓(𝜏𝑘)=lim 𝑡→𝜏+ 𝑘𝑥(𝑡), (84) where 𝑀2={1,...,𝑛}\{𝑘}. It is easy to see that defining functions  X𝐵1,...,𝐵𝑛 𝜏1,...,𝜏𝑛(𝑡):=X−𝐵1,...,−𝐵𝑛 𝜏1,...,𝜏𝑛(𝑡),  Y𝐵1,...,𝐵𝑛 𝜏1,...,𝜏𝑛(𝑡):=Y−𝐵1,...,−𝐵𝑛 𝜏1,...,𝜏𝑛(𝑡)(85) leads to the solution of 𝑥(𝑡)=𝐵1𝑥(𝑡−𝜏1)+⋅⋅⋅+𝐵𝑛𝑥(𝑡−𝜏𝑛)+𝑓(𝑡)(86) with pairwise permutable matrices 𝐵1,...,𝐵𝑛and initial condition (8). More precisely, we have the following corollary of Theorem 7. Corollary 8. Let 𝑛≥3,𝜏1,...,𝜏𝑛>0,𝜏:=max𝑖=1,...,𝑛𝜏𝑖, 𝜑∈𝐶1([−𝜏,0],R𝑁),andlet𝐵1,...,𝐵𝑛be 𝑁×𝑁pairwise permutable matrices; that is, 𝐵𝑖𝐵𝑗=𝐵 𝑗𝐵𝑖for each 𝑖,𝑗 ∈ {1,...,𝑛},andlet𝑓:[0,∞)→R𝑁be a given function. Solution 𝑥(𝑡)of (86)satisfying initial condition (8)has the form 𝑥(𝑡)={ { { { { { { { { { { { { { { { { 𝜑(𝑡),−𝜏≤𝑡<0, X(𝑡)𝜑(0)+Y(𝑡)𝜑(0) +𝑛 ∑ 𝑖=1𝐵𝑖∫0 −𝜏𝑖 Y(𝑡−𝜏𝑖−𝑠)𝜑(𝑠)𝑑𝑠 +∫𝑡 0Y(𝑡−𝑠)𝑓(𝑠)𝑑𝑠, 0≤𝑡, (87) where X(𝑡)= X𝐵1,...,𝐵𝑛 𝜏1,...,𝜏𝑛(𝑡)and Y(𝑡)= Y𝐵1,...,𝐵𝑛 𝜏1,...,𝜏𝑛(𝑡). Proof. The corollary can be proved exactly in the same way as Theorem 7. Acknowledgments J. Dibl´ ık was supported by the Grant GAˇ CR P201/11/0768. M. Feˇ ckan was supported in part by the Grants VEGAMS 1/0507/11, VEGA-SAV 2/0029/13, and APVV-0134-10. M. Posp´ ıˇ sil was supported by the Project no. CZ.1.07/2.3.00/ 30.0005 funded by European Regional Development Fund. References [1] D. Y. Khusainov, J. Dibl´ ık, M. R˚ uˇ ziˇ ckov´ a, and J. Luk´ aˇ cov´ a, “Representation of a solution of the Cauchy problem for an oscillating system with pure delay,” Nonlinear Oscillations,vol. 11, no. 2, pp. 276–285, 2008. [2] D. Y. Khusainov and G. V. Shuklin, “Linear autonomous timedelay system with permutation matrices solving,” Studies of the University of ˇ Zilina,vol.17,no.1,pp.101–108,2003. [3]M.Medved’andM.Posp ´ ıˇ sil, “Sufficient conditions for the asymptotic stability of nonlinear multidelay differential equations with linear parts defined by pairwise permutable matrices,” Nonlinear Analysis. Theory, Methods & Applications,vol.75, no.7,pp.3348–3363,2012. [4]M.Medved’andM.Posp ´ ıˇ sil, “Representation and stability of solutions of systems of difference equations with multiple delays and linear parts defined by pairwise permutable matrices,”