On a Solvable Symmetric and a Cyclic System of Partial Difference Equations

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1 Filomat :6 018, Published by Faculty of Sciences Mathematics, University of Niš, Serbia Available at: On a Solvable Symmetric a Cyclic System of Partial Difference Equations Stevo Stević a,b,c, Bratislav Iričanin d, Zdeněk Šmarda e a Mathematical Institute of the Serbian Academy of Sciences, Knez Mihailova 6/III, Beograd, Serbia b Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 00, Taiwan, Republic of China c Department of Healthcare Administration, Asia University, 500 Lioufeng Rd., Wufeng, Taichung 15, Taiwan, Republic of China d Faculty of Electrical Engineering, Belgrade University, Bulevar Kralja Aleksra 7, Beograd, Serbia e CEITEC - Central European Institute of Technology, Brno University of Technology, Czech Republic Abstract. It is shown that the following symmetric system of partial difference equations d m 1,n c m 1,n 1, c m 1,n d m 1,n 1, is solvable on the combinatorial domain C = { m, n N : 0 n m} \ {0, 0}, by presenting some 0 formulas for the general solution to the system on the domain in terms of the boundary values c j,j, c j,0, d j,j, d j,0, j N, the indices m n. The corresponding result for a related three-dimensional cyclic system of partial difference equations is also proved. These results can serve as a motivation for further studies of the solvability of symmetric, close-to-symmetric, cyclic, close-to-cyclic other related systems of partial difference equations. 1. Introduction By C m n, 0 n m, we denote the binomial coefficients, which are obtained, for example, as the coefficients of the polynomial P m x = 1 x m, m N 0, that is, for an m N 0. The coefficients satisfy the following equalities P m x = C m 0 Cm 1 x Cm x C m m 1 xm 1 C m mx m, C m 0 = Cm m = 1, m N 0, Mathematics Subject Classification. Primary 9A1; Secondary 9A5 Keywords. System of partial difference equations, symmetric system, cyclic system, system solvable in closed form, combinatorial domain, method of half-lines. Received: 1 November 017; Revised: 1 January 018; Accepted: 6 February 018 Communicated by Jelena Manojlović The work of Bratislav Iričanin was supported by the Serbian Ministry of Education Science projects III 105 OI The work of Stevo Stević was supported by the Serbian Ministry of Education Science projects III 105 III 006. The work of Zdeňek Šmarda was supported by the project CEITEC 00 LQ1601 with financial support from the Ministry of Education, Youth Sports of the Czech Republic under the National Sustainability Programme II. addresses: sstevic@ptt.rs Stevo Stević, iricanin@etf.rs Bratislav Iričanin, zdenek.smarda@ceitec.vutbr.cz Zdeněk Šmarda

2 S. Stević et al. / Filomat :6 018, the recurrent relation C m n = C m 1 n C m 1 n 1, for all m, n N such that n < m. By definition is taken that C m n = 0 if m < n, which is used in the paper. Many books contain some results on the coefficients see, for example, [10, 1, 15,, 5, 59]; books [10] [1], among other things, contain many basic relations which include the coefficients/numbers present basic methods for dealing with them, [15] presents many relations containing the coefficients methods for proving them, as well as many methods for calculating finite sums of various types, [] presents a list of numerous relations combinatorial identities, [5] presents, among other things, some advanced methods for dealing with combinatorial identities, while [59] contains various combinatorial applications. The notation seems a bit deceiving because it does not reveal the real character of recurrent relation. However, if it is written in the following way c m 1,n c m 1,n 1, it becomes clear that is a difference equation with two independent variables, that is, a partial difference equation, while 1 are some conditions given on the discrete half-lines m = n n = 0, m N, that is, some boundary-value conditions on the following domain C = { m, n N 0 : 0 n m} \ {0, 0}, which, we call the combinatorial domain point 0, 0 is excluded since the value c 0,0 is essentially not used in calculating the other values of the sequence c m,n. One of the basic problems for any kind of difference equations, as well as for other types of equations, is their solvability. Some old results on partial difference equations, devoted mostly to the problem of their solvability, can be found, for example, in [9, Chapter 1] [11, Chapter 8]. For some results in the area up to 00, see the monograph []. See, also [16]. The solvability of difference equations systems with one independent variable is an ancient topic see, for example, [, 6, 9 1, 1, 15]. Some recent attention on the solvability of difference equations has been, among other things, attracted by S. Stević s note [7], which explained the solvability of the following nonlinear second-order difference equation x n1 = x n 1 a bx n x n 1, n N 0, by transforming the equation to a solvable one by using a suitable change of variables for more details, as well as more general results see [, 9, 0, 50] the references therein. It has turned out that closely related methods ideas can be applied to some other classes of difference equations see [8, 51, 5] the references therein, as well as to some related systems of difference equations see, for example, [1, 8, 1, 7 9, 5, 5, 55]; [8] solves a two-dimensional close-to-symmetric relative of, while [8] studies a three-dimensional relative of of the type. For some other methods for solving difference equations systems, related topics, see also [5, 8,, 8, ]. The main feature of papers [1,,, 8 51, 5], as well as of many quoted in their lists of references, is that decisive role in their solvability plays the following difference equation x n = a n x n 1 b n, n N, which is solvable one see, for example, [, 15], where three methods for solving the equation, which essentially correspond to the three methods for solving the linear first-order differential one, are presented. We would like to mention that even behind the solvability of some product-type difference equations systems is the solvability of equation 5 see, for example, [5, 7, 6, 56 58] the references therein. Moreover, in some cases, the solvability of the product-type systems is shown by using the solvability of some special cases of the corresponding product-type equation, that is, of the following equation z n = b n z a n n 1, n N. 5

3 S. Stević et al. / Filomat :6 018, For some recent results on equation 5, see [0] [], which are partly motivated by a well-known problem in [] see, also recent paper [1]. All these facts show that equation 5 occupies one of the central roles in the area of solvability of difference equations systems. During the investigation of solvable difference equations systems Stević has noticed the fact that for the binomial coefficients there is a closed-form formula. Namely, it is well-known that C m n = m! n!m n!, 0 n m. The formula presents a solution to equation, which suggests potential solvability of equation on the combinatorial domain. That equation can be solved is a well-known fact see, for example, [11, p.9], but the general solution presented therein is not suitable for solving boundary-value problems on the combinatorial domain. These facts lead us to the natural problem of trying to find closed-form formula for solutions to equation on the domain in terms of its boundary values c 0,j, c j,j, j N. The problem was solved in [6], by using the methods ideas appearing therein, including some ideas related to solving equation 5, some related results were later obtained in [9], [] [5]. On the other h, during the 90 s Papaschinopoulos Schinas have started studying concrete symmetric related systems of difference equations [17]-[19], which have motivated experts to do research in the direction see [1, 7, 0, 1, 6, 8, 1 5, 7, 6, 8, 9, 5, 5 58] the numerous references therein. Note also that some of these papers, such as [18]-[1], study, among other things, the invariants of some systems of difference equations, which in a wider sense also belong to the area of solvability invariants of equations systems are not their solutions, but can help in getting some results on their long-term behavior. These two directions in the investigation of difference equations systems of one independent variable inspired us to study the solvability of symmetric, close to symmetric other related classes of partial difference equations. In this paper we start with the investigation by showing that the following system of partial difference equations d m 1,n c m 1,n 1, c m 1,n d m 1,n 1, 6 which is a natural two-dimensional extension to equation, is solvable on C. To do this we will use the method of half-lines described in [6], a half-constructive method which essentially uses the solvability of some special cases of equation 5 on the intersection of domain C the lines y = t k, t Z, when k N, along with an inductive argument with respect to k. More precisely, in the case of system 6 will be used the case of equation 5 when a n = 1 for every n, as it was the case in [6]. The case essentially reduces application of equation 5 to telescoping summations along with iterated summations. Nevertheless the solvability of equation 5 in the special case is crucial. It should be also mentioned that the behavior of the solutions to the equation 5 in the case can be quite complex see [1, ], but this problem is not considered here. The solvability of a closely related tree-dimensional cyclic system of partial difference equations is also studied. As far as we know, these are the first results of this type in the literature this paper initiates the study of the solvability of symmetric, close-to-symmetric, cyclic, close-to-cyclic other related systems of partial difference equations. We use the stard convention l j=k a j = 0, when k, l Z l < k.. Main results In this section we formulate prove the main results in this paper.

4 S. Stević et al. / Filomat :6 018, System 6 on the combinatorial domain The method of half-lines requests solving system 6 on the half lines which are obtained as intersections of domain C the lines y = t k, t Z, when k N, for first several k-s then based on the obtained solutions should be guessed what is the general solution to the system on C. We will find solutions to the corresponding equations for k = 1,,, then guess a formula for the general solution to the system from the mathematical point of view the case k = need not be presented, but we include it for a clearer presentation the benefit of the reader. Before this we formulate a known lemma which will be frequently used in this paper. Beside the formulation in the terms of the binomial coefficients [10, 15, ], it can be also formulated as a sum of a rational sequence can be found in many problem books books on finite differences [,, 15]. Lemma 1. Assume that k, r N 0. Then C kj r = C nk1 r1 C k r1, for every n N 0. Let m = n 1, then c n1,n = c n,n 1 d n,n, d n1,n = d n,n 1 c n,n, for n N, from which it follows that c n1,n = d j,j, d n1,n = n c j,j, 7 Let m = n, then for n N, consequently c n,n = c,0 d n,n = d,0 d j1,j, c j1,j, From 7, 8 some calculation, it follows that c n,n = c,0 = c,0 n d n,n = d,0 = d,0 n c n,n = c n1,n 1 d n1,n, d n,n = d n1,n 1 c n1,n, c i,i n i 1c i,i, 9 n i 1, 10 8

5 S. Stević et al. / Filomat :6 018, Let m = n, then c n,n = c n,n 1 d n,n, d n,n = d n,n 1 c n,n, for n N, from which it follows that c n,n = c,0 d j,j, d n,n = d,0 c j,j, 11 From 9-11 some calculation, it follows that c n,n = c,0 = c,0 nd,0 = c,0 nd,0 d n,n = d,0 = d,0 nc,0 d,0 j j i 1 nn 1 nn 1 c,0 j j i 1 j=i n i 1n i j i 1c i,i nn 1 c i,i n i 1n i One of the crucial points in this analysis is to note that the last two formulas can be written in the following, combinatorial, from forms which include some binomial coefficients: c n,n = C n 1 0 c,0 C n 1 d,0 C n1 d n,n = C n 1 0 d,0 C n 1 c,0 C n1 C n i, 1 C n i c i,i, 1

6 S. Stević et al. / Filomat :6 018, Equalities 7, 9, 10, 1, 1, suggest that the following formulas hold c nl 1,n = c nl,n = d nl 1,n = d nl,n = l 1 C nl j 1 c l j j 1,0 C nl j d l j 1 j,0 C nl j 1 l j c j,0 C nl j l j1 d j 1,0 l 1 C nl j 1 d l j j 1,0 C nl j c l j 1 j,0 C nl j 1 l j d j,0 C nl j l j1 c j 1,0 C n il l, 1 C n il 1 l 1 c i,i, 15 C n il l c i,i, 16 C n il 1 l 1, 17 for n N 0 l N. Let m = n l 1, then c nl1,n = c nl,n 1 d nl,n, d nl1,n = d nl,n 1 c nl,n, for n N. From 18, it follows that c nl1,n = c l1,0 d nl1,n = d l1,0 d sl,s, c sl,s, j=s Using the hypotheses in 19, it follows that c nl1,n = c l1,0 = c l1,0 d nl1,n = d l1,0 = d l1,0 C sl j 1 d l j j,0 d j,0 C sl j 1 l j C sl j 1 c l j j,0 for every n N 0. By Lemma 1, we have C sl j 1t l jt c j,0 C sl j 1 l j C sl j l j1 c j 1,0 c j 1,0 C sl j C sl j l j1 d j 1,0 d j 1,0 l j1 C sl j C s il 1 l 1 l j1 s=i C s il 1 l 1 c i,i c i,i s=i C s il 1 l 1, 0 C s il 1 l 1, 1 = C nl jt l j1t Cl jt l j1t = Cnl jt l j1t,

7 for every 1 j l t N 0, s=i C s il 1t l 1t = C n ilt lt S. Stević et al. / Filomat :6 018, C l 1t lt for every 1 i n t N 0. Using in 0 1, we get c nl1,n = c l1,0 d j,0 C nl j l j1 l1 = c j 1,0 C nl j1 l j d nl1,n = d l1,0 = C n ilt lt, c j 1,0 C nl j1 l j d j,0 C nl j l j1 c j,0 C nl j l j1 l1 = d j 1,0 C nl j1 l j for every n N 0. Let m = n l, then d j 1,0 C nl j1 l j c j,0 C nl j l j1 C n il l, C n il l, c i,i C n il l, c i,i C n il l, 5 c nl,n = c nl1,n 1 d nl1,n, d nl,n = d nl1,n 1 c nl1,n, 6 for n N. From 6, it follows that c nl,n = c l,0 d sl1,s, d nl,n = d l,0 Using 5, as well as in 7, it follows that c nl,n = c l,0 l1 = c l,0 l1 d j 1,0 d j 1,0 C sl j1 l j C sl j1 l j l1 = c l,0 d j 1,0 C nl j l j l1 = l1 c j,0 C nl j1 l j c sl1,s, 7 c j,0 C sl j l j1 c j,0 d j 1,0 C nl j l j C sl j c j,0 C nl j1 l j l j1 c i,i C s il l c i,i s=i c i,i C n il1 l1 C s il l c i,i C n il1 l1, 8

8 S. Stević et al. / Filomat :6 018, d nl,n = d l,0 l1 = d l,0 l1 c j 1,0 c j 1,0 C sl j1 l j C sl j1 l j l1 = d l,0 c j 1,0 C nl j l j l1 = l1 d j,0 C nl j1 l j d j,0 C sl j l j1 d j,0 c j 1,0 C nl j l j C sl j d j,0 C nl j1 l j l j1 C s il l s=i C n il1 l1 C s il l C n il1 l1, 9 for every n N 0. From 7, 9, 10,, 5, 8, 9 the induction we get that 1-17 hold for all n, l N 0. Now we are in a position to formulate prove our first main result in this paper. Theorem 1. Assume that α k k N, β k k N, γ k k N δ k k N are given sequences of complex numbers. Then the solution to system 6 with the following boundary-value conditions is given by c k,0 = α k, c k,k = β k, d k,0 = γ k, d k,k = δ k, k N, when m n 1 mod C m j 1 j α j 1 C m j 1 j γ j 1 C m j 1 j γ j C m j 1 j α j δ i, 1 β i, C m j 1 j α j C m j 1 j γ j C m j j1 γ j 1 C m j j1 α j 1 β i, δ i, when m n 0 mod, for every m, n N such that m > n. Proof. If we put l = m n 1, when m n 1 mod in 1 16, that is, l = m n, when m n 0 mod, in employ in such obtained formulas the boundary-value conditions in 0, formulas 1- are easily obtained.

9 S. Stević et al. / Filomat :6 018, Remark 1. From Theorem 1 we see that the general solution to system 6 on C is given by 1 1 when m n 1 mod C m j 1 j c j 1,0 C m j 1 j d j 1,0 C m j 1 j d j,0 C m j 1 j c j,0, c i,i, C m j 1 j c j,0 C m j 1 j d j,0 C m j j1 d j 1,0 C m j j1 c j 1,0 c i,i,, when m n 0 mod, for m, n N such that m > n. Remark. If then from Theorem 1 we see that α k = γ k β k = δ k, k N, d m,n, for every m, n C. Moreover, after some calculation from 1, we obtain C m 1 j j α j β i, which is the formula for general solution to equation on domain C obtained in [6]. From formulas 1- we see that to get closed-form formulas for solutions to equation 6 we should know some closed-form formulas for the sums of the following forms: 1 C m j 1 j c j, C m j 1 j c j, c i, when m n 1 mod, that is, of the following ones: C m j 1 j c j, C m j j1 c j, c i, when m n 0 mod, for every m, n N such that m > n. One of the cases where we can give some more compact formulas for the solution to system 6 is presented in the following corollary.

10 S. Stević et al. / Filomat :6 018, Corollary 1. Assume that β, δ C, α k k N γ k k N are given sequences of complex numbers. Then the solution to system 6 with the following boundary value conditions is given by c k,0 = α k, c k,k = β, d k,0 = γ k, d k,k = δ, k N, when m n 1 mod C m j 1 j α j 1 C m j 1 j γ j 1 C m j 1 j γ j δc m 1, 6 C m j 1 j α j βc m 1, 7 C m j 1 j α j C m j 1 j γ j C m j j1 γ j 1 βc m 1, 8 C m j j1 α j 1 δc m 1, 9 when m n 0 mod, for every m, n N such that m > n. Proof. If we put boundary-value conditions 5 in 1-, we get 1 1 when m n 1 mod C m j 1 j α j 1 C m j 1 j γ j 1 C m j 1 j γ j δ C m j 1 j α j β, 0, 1 C m j 1 j α j C m j 1 j γ j C m j j1 γ j 1 β C m j j1 α j 1 δ,, On the other h, by Lemma 1, we have = Cm 1 C = C m 1. By using equality into 0-, the formulas in 6-9 follow, finishing the proof.

11 Remark. Note that in the special case, when from 6-9 it follows that when m n 1 mod, while S. Stević et al. / Filomat :6 018, α k = γ k = 0, k N, δc m 1, βc m 1, when m n 0 mod, for every m, n N such that m > n. βc m 1, δc m 1, The case when the sequences α j 1 j N, α j j N, γ j 1 j N γ j j N, are constant is one of the interesting ones. From 6-9 we see that to find closed-form formulas for the general solutions to system 6 in the case, it is necessary to calculate the following sums: 1 C m j 1 j, C m j 1 j, when m n 1 mod, C m j 1 j, C m j j1, when m n 0 mod, for every m, n N such that m > n, which can be written in the following somewhat neater forms: when m n 1 mod, a m,n := â m,n := C nj 1 j, b m,n := C nj 1 j, bm,n := C nj j1, C nj j1, when m n 0 mod, for every m, n N such that m > n, or to avoid using the modulo function as follows: k k 1 a nk1,n := C nj 1, b j nk1,n := C nj j1, when k N 0. Note that â nk,n := k a nk1,n b nk1,n = C nj 1 = j k k C nj 1 j, bnk,n := C nj j k C nj j1, C nj 1 j 1 = C nk k, 5 k1 k1 â nk,n b nk,n = C nj 1 nj = j C j C nj 1 j 1 = C nk1 k1,

12 S. Stević et al. / Filomat :6 018, for k N 0. However, we are not able to find closed-form formulas for a nk1,n, b nk1,n, â nk,n b nk,n, at the moment. Namely, all the methods that we have used so far in trying to find the closed-form formulas only transformed the sums which define the sequences to some other ones, for which we are not able to find closed-form formulas, either. Nevertheless, we have obtained several interesting relations facts which could serve as a motivation for further investigation in the area. Now we will present some of them. First, note that a nk1,n = k k 1 = C nj 1 j = 1 C nj j1 k k C n 1j 1 j C nj j 1 C nj j = b nk1,n a nk,n 1, 6 for k N 0. From 5 6, it follows that a nk1,n = a nk,n 1 1 Cnk k, for k N 0. Using the change of variables x n := a nk1,n, n N, for a fixed k, equation 7 can be written in the form of the equation in 5 by solving it, we obtain a nk1,n = a k1,0 n 1 Ckj n j1 7 k, 8 for k N 0. From 6 8, we see that the problem of calculating the sums a nk1,n b nk1,n is equivalent to the calculation of the sum s n,k := j C kj k, for k N 0. Another method for calculating the sum a nk1,n is by using suitable polynomials such that the sum in 9 is one of their coefficients the method is used in [10, 15, 5]. One of the most natural polynomials is the following one: k P nk 1 x = 1 x n 1j x k j, whose coefficients at x k is equal to a nk1,n. By some calculation we have P nk 1 x = 1 x n 1 1 xk x k 1 x x = 1 x nk1 1 x n 1 x k 1 x 1 = 1 x nk1 1 x n 1 x k j x j, 9

13 S. Stević et al. / Filomat :6 018, where, of course, the last equality holds for x < 1/. Comparing the coefficients at x k in these two expansions of the polynomial is obtained k a nk1,n = j C nk1 k j. 50 However, the sum is also not so easy to calculate... On a three-dimensional relative of system 6 Having solved system 6 on C it is natural to ask if some of its three-dimensional cousins are also solvable. One of the most natural ones is the following cyclic system: b m,n = c m 1,n b m 1,n 1, d m 1,n c m 1,n 1, b m 1,n d m 1,n 1, 51 where m, n C. Let m = n 1, then for n N, consequently b n1,n = b n,n 1 c n,n, c n1,n = c n,n 1 d n,n, d n1,n = d n,n 1 b n,n, b n1,n = b 1,0 c j,j, c n1,n = d j,j, d n1,n = b j,j, 5 Let m = n, then for n N, consequently b n,n = b n1,n 1 c n1,n, c n,n = c n1,n 1 d n1,n, d n,n = d n1,n 1 b n1,n, b n,n = b,0 c j1,j, c n,n = c,0 d j1,j, d n,n = d,0 b j1,j, 5

14 From 5, 5 some calculation, it follows that b n,n = b,0 c n,n = c,0 d n,n = d,0 Let m = n, then b 1,0 S. Stević et al. / Filomat :6 018, = b,0 n b i,i = c,0 n c i,i = d,0 nb 1,0 n i 1, 5 n i 1b i,i, 55 n i 1c i,i, 56 b n,n = b n,n 1 c n,n, c n,n = c n,n 1 d n,n, d n,n = d n,n 1 b n,n, for n N, from which it follows that b n,n = b,0 c j,j, c n,n = c,0 d n,n = d,0 d j,j, b j,j, From 5-57 some calculation, it follows that b n,n = b,0 = b,0 nc,0 c,0 j j i 1b i,i nn 1 = C n 1 0 b,0 C n 1 c,0 C n1 b i,i j i 1 j=i 57 C n i b i,i, 58 c n,n = c,0 d,0 jb 1,0 j i 1c i,i = C n 1 0 c,0 C n 1 d,0 C n1 b 1,0 d n,n = d,0 b,0 j j i 1 = C n 1 0 d,0 C n 1 b,0 C n1 C n i c i,i, 59 C n i, 60

15 S. Stević et al. / Filomat :6 018, Let m = n, then b n,n = b n,n 1 c n,n, c n,n = c n,n 1 d n,n, d n,n = d n,n 1 b n,n, for n N, from which it follows that b n,n = b,0 c j,j, c n,n = c,0 d n,n = d,0 d j,j, b j,j, From 58-61, some calculation Lemma 1, it follows that b n,n = b,0 C j 1 0 c,0 C j 1 d,0 C j1 b 1,0 = b,0 C n 1 c,0 C n1 d,0 C n b 1,0 c n,n = c,0 C j 1 0 d,0 C j 1 b,0 C j1 = c,0 C n 1 d,0 C n1 b,0 C n d n,n = d,0 Let m = n 5, then C j 1 0 b,0 C j 1 c,0 C j1 = d,0 C n 1 b,0 C n1 c,0 C n C j i c i,i 61 C n i c i,i, 6 C j i C n i, 6 C j i b i,i C n i b i,i, 6 b n5,n = b n,n 1 c n,n, c n5,n = c n,n 1 d n,n, d n5,n = d n,n 1 b n,n, for n N, from which it follows that b n5,n = b 5,0 c j,j, c n5,n = c 5,0 d n5,n = d 5,0 d j,j, b j,j, 65

16 S. Stević et al. / Filomat :6 018, From 6-65, some calculation Lemma 1, it follows that b n5,n = b 5,0 c,0 C j 1 d,0 C j1 b,0 C j = b 5,0 C n 1 c,0 C n1 d,0 C n b,0 C n c n5,n = c 5,0 d,0 C j 1 b,0 C j1 c,0 C j = c 5,0 C n 1 d,0 C n1 b,0 C n c,0 C n d n5,n = d 5,0 Let m = n 6, then b,0 C j 1 c,0 C j1 d,0 C j b 1,0 = d 5,0 C n 1 b,0 C n1 c,0 C n d,0 C n b 1,0 C j i C n i, 66 C j i b i,i C n i b i,i, 67 C j i c i,i C n i c i,i, 68 b n6,n = b n5,n 1 c n5,n, c n6,n = c n5,n 1 d n5,n, d n6,n = d n5,n 1 b n5,n, for n N 0, from which it follows that b n6,n = b 6,0 c j5,j, c n6,n = c 6,0 d n6,n = d 6,0 d j5,j, b j5,j, From 66-69, some calculation Lemma 1, it follows that b n6,n = b 6,0 c 5,0 C j 1 d,0 C j1 b,0 C j c,0 C j = b 6,0 C n 1 c 5,0 C n1 d,0 C n b,0 C n c,0 C n 5 c n6,n = c 6,0 d 5,0 C j 1 b,0 C j1 c,0 C j d,0 C j b 1,0 = c 6,0 C n 1 d 5,0 C n1 b,0 C n c,0 C n d,0 C n 5 b 1,0 C j i b i,i 69 C n i5 5 b i,i, 70 C j i c i,i C n i5 5 c i,i, 71

17 S. Stević et al. / Filomat :6 018, d n6,n = d 6,0 b 5,0 C j 1 c,0 C j1 d,0 C j b,0 C j = d 6,0 C n 1 b 5,0 C n1 c,0 C n d,0 C n b,0 C n 5 C j i C n i5 5, 7 Equalities 5, 5-56, 58-60, 6-6, 66-68, 70-7, suggest that the following formulas hold: b nl,n = b nl 1,n = b nl,n = c nl,n = c nl 1,n = c nl,n = d nl,n = d nl 1,n = d nl,n = l 1 C nl j 1 b l j j,0 C nl j 1 l j b j 1,0 C nl j 1 l j b j,0 l 1 C nl j 1 c l j j,0 C nl j 1 l j c j 1,0 C nl j 1 l j c j,0 l 1 C nl j c l j j,0 C nl j l j 1 d j 1,0 l 1 C nl j l j1 c j,0 C nl j d l j 1 j,0 l 1 C nl j 1 d l j j,0 C nl j 1 l j d j 1,0 C nl j 1 l j d j,0 C nl j l j1 c j 1,0 l 1 C nl j d l j j,0 C nl j1 l j d j,0 C nl j l j 1 b j 1,0 l 1 C nl j l j1 d j,0 C nl j b l j 1 j,0 C nl j l j1 d j 1,0 l 1 C nl j b l j j,0 C nl j1 l j b j,0 C nl j l j 1 c j 1,0 l 1 C nl j l j1 b j,0 C nl j c l j 1 j,0 C nl j l j1 b j 1,0 C nl j1 l j c j,0 C n il l c i,i, 7 C n il l, 7 C n il 1 l 1 b i,i, 75 C n il l, 76 C n il l b i,i, 77 C n il 1 l 1 c i,i, 78 C n il l b i,i, 79 C n il l c i,i, 80 C n il 1 l 1, 81 for every n N 0 l N. Let m = n l 1, then b nl1,n = b nl,n 1 c nl,n, c nl1,n = c nl,n 1 d nl,n, d nl1,n = d nl,n 1 b nl,n, 8 for n N.

18 From 8, it follows that b nl1,n = b l1,0 c jl,j, c nl1,n = c l1,0 d nl1,n = d l1,0 d jl,j, b jl,j, S. Stević et al. / Filomat :6 018, Using the hypotheses 75, in 8, employing Lemma 1, it follows that b nl1,n = b l1,0 C sl j 1 c l j j,0 l1 = C nl j b l j j,0 C nl j l j1 c j,0 C sl j l j1 d j 1,0 C nl j1 l j d j 1,0 C sl j1 l j b j,0 C s il 1 l 1 c i,i 8 C n il l c i,i, 8 c nl1,n = c l1,0 C sl j 1 d l j j,0 l1 = C nl j c l j j,0 C nl j l j1 d j,0 C sl j l j1 b j 1,0 C nl j1 l j b j 1,0 C sl j1 l j c j,0 C s il 1 l 1 C n il l, 85 d nl1,n = d l1,0 C sl j 1 b l j j,0 l1 = C nl j d l j j,0 for every n N 0. Let m = n l, then b nl,n = b nl1,n 1 c nl1,n, c nl,n = c nl1,n 1 d nl1,n, d nl,n = d nl1,n 1 b nl1,n, for n N. From 87, it follows that b nl,n = b l,0 c jl1,j, c nl,n = c l,0 d nl,n = d l,0 d jl1,j, b jl1,j, C nl j l j1 b j,0 C sl j l j1 c j 1,0 C nl j1 l j c j 1,0 C sl j1 l j d j,0 C s il 1 l 1 b i,i C n il l b i,i,

19 S. Stević et al. / Filomat :6 018, Using 8-86 in 88, employing Lemma 1, it follows that b nl,n = b l,0 l1 = c nl,n = c l,0 l1 = l1 l1 C nl j b l j j 1,0 l1 C sl j l j c j,0 l1 C nl j c l j j 1,0 C nl j l j c j,0 C sl j l j d j,0 C nl j l j d j,0 C sl j l j1 d j,0 C sl j l j1 b j,0 C nl j1 l j d j,0 C nl j1 l j b j,0 C sl j1 l j b j 1,0 C sl j1 l j c j 1,0 C s il l C n il1 l1 89 C s il l b i,i C n il1 l1 b i,i 90 d nl,n = d l,0 l1 = l1 l1 C nl j d l j j 1,0 C sl j l j b j,0 C nl j l j b j,0 C sl j l j1 c j,0 C nl j1 l j c j,0 C sl j1 l j d j 1,0 C s il l c i,i C n il1 l1 c i,i, 91 for every n N 0. Let m = n l, then b nl,n = b nl,n 1 c nl,n, c nl,n = c nl,n 1 d nl,n, d nl,n = d nl,n 1 b nl,n, for n N. From 9, it follows that for n N 0 b nl,n = b l,0 c jl,j, c nl,n = c l,0 d nl,n = d l,0 d jl,j, b jl,j. Using in 9, employing Lemma 1, it follows that 9 9 b nl,n = b l,0 l1 = l1 C sl j c l j j 1,0 C sl j d l j j,0 l1 l1 C nl j b l j j,0 l1 C nl j c l j j 1,0 C nl j l j5 d j,0 C sl j1 l j b j,0 C s il1 l1 b i,i C n il l b i,i, 9

20 c nl,n = c l,0 l1 = l1 S. Stević et al. / Filomat :6 018, l1 C sl j d l j j 1,0 C sl j b l j j,0 l1 C nl j c l j j,0 l1 C nl j d l j j 1,0 C nl j l j5 b j,0 C sl j1 l j c j,0 C s il1 l1 c i,i C n il l c i,i, 95 d nl,n = d l,0 l1 = l1 C sl j b l j j 1,0 C sl j c l j j,0 l1 l1 C nl j d l j j,0 l1 C nl j b l j j 1,0 C nl j l j5 c j,0 C sl j1 l j d j,0 C s il1 l1 C n il l, 96 for every n N 0. From this by induction we see that formulas 7-81 hold for every n N 0 l N. Now we formulate prove the main result regarding the three-dimensional system 51. Theorem. Assume that β k k N, β k k N, γ k k N, γ k k N, δ k k N δ k k N are given sequences of complex numbers. Then the solution to system 51 with the following boundary value conditions b k,0 = β k, b k,k = β k, c k,0 = γ k, c k,k = γ k, d k,0 = δ k, d k,k = δ k, 97 k N, is given by b m,n = C m j 1 j β j 1 C m j j1 γ j 1 C m j j1 δ j 1 when m n mod, by b m,n = C m j1 j β j C m j1 j γ j C m j1 j δ j C m j1 j γ j C m j1 j δ j C m j1 j β j C m j 1 j γ j C m j 1 j δ j C m j 1 j β j C m j 1 j δ j C m j 1 j β j C m j 1 j γ j C m j j1 δ j 1 C m j j1 β j 1 C m j j1 γ j 1 δ i, 98 β i, 99 γ i, 100 γ i, 101 δ i, 10 β i, 10

21 S. Stević et al. / Filomat :6 018, when m n 1 mod, by b m,n = C m j 1 j β j C m j 1 j γ j C m j 1 j δ j C m j j1 γ j 1 C m j j1 δ j 1 C m j j1 β j 1 C m j1 j δ j C m j1 j β j C m j1 j γ j β i, 10 γ i, 105 δ i, 106 when m n 0 mod, for every m, n N 0 such that m n. Proof. If we put m = n l 1, when m n mod in 7, 77 80, put m = n l, when m n 1 mod in 7, 76 79, put l = m n, when m n 0 mod in 75, 78 81, we get b m,n = C m j 1 j b j 1,0 C m j j1 c j 1,0 C m j j1 d j 1,0 when m n mod, by b m,n = C m j1 j b j,0 C m j1 j c j,0 C m j1 j d j,0 when m n 1 mod, by C m1 j j c j,0 C m j1 j d j,0 C m j1 j b j,0 C m j 1 j c j,0 C m j 1 j d j,0 C m j 1 j b j,0 C m j 1 j d j,0 C m j 1 j b j,0 C m j 1 j c j,0 C m j j1 d j 1,0 C m j j1 b j 1,0 C m j j1 c j 1,0, 107 b i,i, 108 c i,i, 109 c i,i, 110, 111 b i,i, 11 b m,n = C m j 1 j b j,0 C m j 1 j c j,0 C m j 1 j d j,0 C m j j1 c j 1,0 C m j j1 d j 1,0 C m j j1 b j 1,0 C m j1 j d j,0 C m j1 j b j,0 C m j1 j c j,0 b i,i, 11 c i,i, 11, 115

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