# Periodic Solutions of a System of Nonlinear Difference Equations with Periodic Coefficients

Authors: Durhasan Turgut Tollu
Journal: Journal of Mathematics
 (2020)

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## Abstract

This paper is dealt with the following system of difference equationsxn+1=an/xn+bn/yn,yn+1=cn/xn+dn/yn, where n∈ℕ0=ℕ∪0, the initial values x0andy0 are the positive real numbers, and the sequences ann≥0, bnn≥0, cnn≥0, and dnn≥0 are two-periodic and positive. The system is an extension of a system where every positive solution is two-periodic or converges to a two-periodic solution. Here, the long-term behavior of positive solutions of the system is examined by using a new method to solve the system.


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## Body

## 1. Introduction

Studying concrete nonlinear difference equations and systems has attracted a great recent interest. Many studies have been published on this topic in the last twenty years (see, e.g., [1–15]). Particularly, there has been a renewed interest in solvable nonlinear difference equations and systems for fifteen years (see, e.g., [5, 16–21], and the related references therein). Solvable difference equations are not interesting for themselves only, but they can be also applied in other areas of mathematics, as well as other areas of science (see, e.g., [22, 23]).One of the first examples of solvable nonlinear difference equations is presented in note [24] where Brand solves the nonlinear difference equation:(1)xn+1=axn+bcxn+d,n∈ℕ0,where the initial value x0 is a real number and the parameters a, b, c, and d are the real numbers with the restrictions c≠0 and ad−bc≠0 and studies long-term behavior of solutions of the equation. The note presents a transformation which transforms the nonlinear equation into a linear one. The idea has been used many times in showing solvability of some difference equations, as well as of some systems of difference equations (see, e.g., [5, 18, 20, 21, 25, 26]). Another example of solvable nonlinear difference equations is the following system of nonlinear difference equation:(2)xn+1=axn+byn,yn+1=cxn+dyn,n∈ℕ0,where the initial values x0 and y0 are the positive real numbers and the parameters a, b, c, and d are the positive real numbers. System (2) can be transformed into an equation of form (1) by dividing the first equation of (2) by its second one. So, the results on equation (1) can be used to obtain the results on system (2). System (2) was studied for the first time in [17] by using the method described above. Also, in [17], it is shown that every positive solution of the system (2) is two-periodic or converges to a two-periodic solution. For more results on system (2), see [19, 22, 27].System (2) can be extended by interchanging the parameters a, b, c, and d with the sequences ann≥0, bnn≥0, cnn≥0, and dnn≥0. More concretely, another extension of (2) is the following system of difference equations:(3)xn+1=anxn+bnyn,yn+1=cnxn+dnyn,n∈ℕ0,where the initial values x0andy0 are the positive real numbers and ann≥0, bnn≥0, cnn≥0, and dnn≥0 are the two-periodic sequences of positive real numbers. For extensions with periodic sequences of some difference equations and systems, see [2, 28, 29].Our main purpose in this paper is to determine the long-term behavior of positive solutions of system (3). We also use a new method to solve the system without needing some other nonlinear difference equations such as (1). Throughout this paper, we assume that a2n=a0, a2n+1=a1; b2n=b0, b2n+1=b1; c2n=c0, c2n+1=c1; and d2n=d0, d2n+1=d1 with a0≠a1, b0≠b1, c0≠c1, and d0≠d1. We also adopt the conventions:(4)∑k=mm−lsk=0,∏k=mm−lsk=1,l∈ℕ,where sk is any sequence and m∈ℕ0=ℕ∪0.Definition 1.
A solutionxn,ynn≥0 of the system(5)xn+1=fxn,yn,yn+1=gxn,yn,n∈ℕ0,is eventually periodic with period p, if there is n1>0 such that xn+p,yn+p=xn,yn for n≥n1. If n1=0, then the solution is periodic with period p.
The following result is extracted from [30].Remark 1.
A product∏k=0∞1+αk with positive terms αk is convergent if and only if ∑k=0∞αk converges.


## 2. Main Results

In this section, we formulate and prove our main results.Theorem 1.
Assume thatx0,y0>0 and ann≥0, bnn≥0, cnn≥0, and dnn≥0 are the two-periodic sequences of positive real numbers. Then, system (3) can be solved in closed form.Proof.
First, it is easy to show by induction thatxn, yn>0, for all n∈ℕ0. Multiplying both equations in (3) by the following positive product:(6)∏k=0nxkyk,we obtain(7)∏k=0n+1xk∏k=0nyk=an∏k=0n−1xk∏k=0nyk+bn∏k=0nxk∏k=0n−1yk,(8)∏k=0nxk∏k=0n+1yk=cn∏k=0n−1xk∏k=0nyk+dn∏k=0nxk∏k=0n−1yk,for all n∈ℕ0. Note that the equalities (7) and (8) constitute a linear system with respect to the following products:(9)un=∏k=0nxk∏k=0n−1yk,vn=∏k=0n−1xk∏k=0nyk.
Therefore, we can write this system in the vector form:(10)un+1vn+1=bnandncnunvn,where u0=x0 and v0=y0, which is simpler, for all n∈ℕ0. Let(11)An=bnandncn.
Then, since the sequencesann≥0, bnn≥0, cnn≥0, and dnn≥0 are two-periodic, the matrix An becomes(12)An=A0=b0a0d0c0,ifnis even,A1=b1a1d1c1,ifnis odd.
Now, we decompose (10) with respect to even-subscript and odd-subscript terms as follows:(13)u2n+1v2n+1=b0a0d0c0u2nv2n,(14)u2n+2v2n+2=b1a1d1c1u2n+1v2n+1,for all n∈ℕ0. From which (13) and (14) follows that(15)u2n+2v2n+2=b1a1d1c1b0a0d0c0u2nv2n,or(16)u2n+2v2n+2=a1d0+b0b1a0b1+a1c0b0d1+c1d0a0d1+c0c1u2nv2n.
LetA1A0=A. Then, we consider two cases of the matrix A as the following:Case 1:rankA=1. In this case, the first row in the matrix A is linearly dependent on the second one. Without loss of generality, we may assume that(17)b0d1+c1d0,a0d1+c0c1=Ka1d0+b0b1,a0b1+a1c0,whereK is a positive constant such that(18)K=b0d1+c1d0a1d0+b0b1=a0d1+c0c1a0b1+a1c0.Using (17) in system (16), we have(19)u2n+2=a1d0+b0b1u2n+a0b1+a1c0v2n,v2n+2=Ka1d0+b0b1u2n+a0b1+a1c0v2n,which implies the relation(20)v2n+2=Ku2n+2,for alln∈ℕ0. By the last three relations, we have(21)u2n+2=a1d0+b0b1+Ka0b1+a1c0u2n,from which it follows that(22)u2n=a1d0+b0b1+Ka0b1+a1c0n−1u2,(23)v2n=Ka1d0+b0b1+Ka0b1+a1c0n−1u2,for alln∈ℕ. Using (22) and (23) in (13), we obtain(24)u2n+1=b0+Ka0a1d0+b0b1+Ka0b1+a1c0n−1u2,v2n+1=d0+Kc0a1d0+b0b1+Ka0b1+a1c0n−1u2,for alln∈ℕ. On the other hand, the changes of variables in (9) yield(25)xn+1=un+1vn−1unvnxn−1,(26)yn+1=vn+1un−1vnunyn−1,for alln∈ℕ. Hence, from (25) and (26), we obtain(27)x2n=x0∏k=1nu2kv2k−2u2k−1v2k−1,(28)x2n+1=x1∏k=1nu2k+1v2k−1u2kv2k,(29)y2n=y0∏k=0nv2ku2k−2v2k−1u2k−1,(30)y2n+1=y1∏k=1nv2k+1u2k−1v2ku2k.By employing (13) in (27)–(30), we have the following closed formulas:(31)x2n=x0∏k=1nu2kv2k−2b0u2k−2+a0v2k−2d0u2k−2+c0v2k−2,(32)x2n+1=x1∏k=1nb0u2k+a0v2kd0u2k−2+c0v2k−2u2kv2k,(33)y2n=y0∏k=1nv2ku2k−2d0u2k−2+c0v2k−2b0u2k−2+a0v2k−2,(34)y2n+1=y1∏k=1nd0u2k+c0v2kb0u2k−2+a0v2k−2v2ku2k,which is valid for alln∈ℕ0, respectively. Consequently, in the case rankA=1, by using the formulas (22) and (23) in (31)–(34), we have the general solution of (3) as follows:(35)x2n=x0∏k=1nKa1d0+b0b1+Ka0b1+a1c0b0+Ka0d0+Kc0,(36)x2n+1=x1∏k=1nb0+Ka0d0+Kc0Ka1d0+b0b1+Ka0b1+a1c0,(37)y2n=y0∏k=1nKa1d0+b0b1+Ka0b1+a1c0d0+Kc0b0+Ka0,(38)y2n+1=y1∏k=1nd0+Kc0b0+Ka0Ka1d0+b0b1+Ka0b1+a1c0,where(39)K=b0d1+c1d0a1d0+b0b1=a0d1+c0c1a0b1+a1c0,for alln∈ℕ0.Case 2:rankA=2. In this case, both rows in the matrix A are linearly independent of each other. This case also implies that A has two different eigenvalues denoted by λ1 and λ2. Since these eigenvalues will correspond to two linear independent eigenvectors, we may write the matrix A as follows:(40)A=PΛP−1,where(41)P=a0b1+a1c0λ1−a1d0+b0b1a0b1+a1c0λ2−a1d0+b0b111,P−1=b0d1+c1d0λ1−λ2λ1−a1d0+b0b1λ1−λ2−b0d1+c1d0λ1−λ2−λ2−a1d0+b0b1λ1−λ2,Λ=λ100λ2.
Therefore, we may write system (16) as the following:(42)Z2n+2=PΛP−1Z2n,where(43)Z2n=u2nv2n,for all n∈ℕ0. From (42), we have(44)P−1Z2n+2=ΛP−1Z2n,from which it follows that(45)P−1Z2n=ΛnP−1Z0,for all n∈ℕ0. Multiplying both sides of (45) by the matrix P, we have(46)Z2n=PΛnP−1Z0,or after some computations(47)u2nv2n=C1λ1n−C2λ2nC3λ1n−C4λ2n,where(48)C1=a0b1+a1c0λ1−λ2b0d1+c1d0λ1−a1d0+b0b1u0+v0,(49)C2=a0b1+a1c0λ1−λ2b0d1+c1d0λ2−a1d0+b0b1u0+v0,(50)C3=b0d1+c1d0λ1−λ2u0+λ1−a1d0+b0b1λ1−λ2v0,(51)C4=b0d1+c1d0λ1−λ2u0+λ2−a1d0+b0b1λ1−λ2v0,for all n∈ℕ0. From the last vectorial equality, we have(52)u2n=C1λ1n−C2λ2n,(53)v2n=C3λ1n−C4λ2n,for all n∈ℕ0. From (13), (52), and (53), we have the formulas(54)u2n+1=b0C1+a0C3λ1n−b0C2+a0C4λ2n,(55)v2n+1=d0C1+c0C3λ1n−d0C2+c0C4λ2n,for all n∈ℕ0. Also, we can write the formulas (31)–(34) as the following:(56)x2n=x0∏k=1nu2k/v2k−2b0u2k−2/v2k−2+a0d0u2k−2/v2k−2+c0,(57)x2n+1=x1∏k=1nb0u2k/v2k+a0d0u2k−2/v2k−2+c0u2k/v2k−2,(58)y2n=y0∏k=1nv2k/u2k−2d0+c0v2k−2/u2k−2b0+a0v2k−2/u2k−2,(59)y2n+1=y1∏k=1nd0+c0v2k/u2kb0+a0v2k−2/u2k−2v2k/u2k−2,for all n∈ℕ0. Finally, by employing (52)–(55) in (56)–(59), we have the general solution of (3) as the following:(60)x2n=x0∏k=1nC1λ1k−C2λ2k/C3λ1k−1−C4λ2k−1b0C1λ1k−1−C2λ2k−1/C3λ1k−1−C4λ2k−1+a0d0C1λ1k−1−C2λ2k−1/C3λ1k−1−C4λ2k−1+c0,x2n+1=x1∏k=1nb0C1λ1k−C2λ2k/C3λ1k−C4λ2k+a0d0C1λ1k−1−C2λ2k−1/C3λ1k−1−C4λ2k−1+c0C1λ1k−C2λ2k/C3λ1k−1−C4λ2k−1,y2n=y0∏k=1nC3λ1k−C4λ2k/C1λ1k−1−C2λ2k−1d0+c0C3λ1k−1−C4λ2k−1/C1λ1k−1−C2λ2k−1b0+a0C3λ1k−1−C4λ2k−1/C1λ1k−1−C2λ2k−1,y2n+1=y1∏k=1nd0+c0C3λ1k−C4λ2k/C1λ1k−C2λ2kb0+a0C3λ1k−1−C4λ2k−1/C1λ1k−1−C2λ2k−1C3λ1k−C4λ2k/C1λ1k−1−C2λ2k−1,for all n∈ℕ0.The following theorem determines and characterizes the long-term behavior of positive solutions of (3) according to the parameters in the case rankA=1.Theorem 2.
Assume thatx0, y0>0 and ann≥0, bnn≥0, cnn≥0, and dnn≥0 are the two-periodic sequences of positive real numbers. If(61)ranka1d0+b0b1a0b1+a1c0b0d1+c1d0a0d1+c0c1=1,then, for the solutions of system (3), the following is true:(i)
IfKa1d0+b0b1+Ka0b1+a1c0/b0+Ka0d0+Kc0<1, then x2n⟶0, x2n+1⟶∞, y2n⟶0, and y2n+1⟶∞ as n⟶∞(ii)
IfKa1d0+b0b1+Ka0b1+a1c0/b0+Ka0d0+Kc0>1, then x2n⟶∞, x2n+1⟶0, y2n⟶∞, and y2n+1⟶0 as n⟶∞(iii)
IfKa1d0+b0b1+Ka0b1+a1c0/b0+Ka0d0+Kc0=1, then every solution of (3) is two-periodic, where K is given by (39)Proof.
The proof follows directly from formulas (35)–(38). That is to say, it is clearly seen from these formulas that if(62)Ka1d0+b0b1+Ka0b1+a1c0b0+Ka0d0+Kc0<1,then x2n⟶0, x2n+1⟶∞, y2n⟶0, and y2n+1⟶∞ as n⟶∞. If(63)Ka1d0+b0b1+Ka0b1+a1c0b0+Ka0d0+Kc0>1,then x2n⟶∞, x2n+1⟶0, y2n⟶∞, and y2n+1⟶0 as n⟶∞. If(64)Ka1d0+b0b1+Ka0b1+a1c0b0+Ka0d0+Kc0=1,then every solution of (3) is two-periodic such that x2n=x0, x2n+1=x1, y2n=y0, and y2n+1=y1.The following theorem determines and characterizes the long-term behavior of positive solutions of (3) according to the parameters in the case rankA=2.Theorem 3.
Assume thatx0, y0>0 and ann≥0, bnn≥0, cnn≥0, and dnn≥0 are the two-periodic sequences of positive real numbers. If(65)ranka1d0+b0b1a0b1+a1c0b0d1+c1d0a0d1+c0c1=2,then, for the solutions of system (3), the following is true:(i)
IfC1λ1/C3−b0C1/C3+a0d0C1/C3+c0<0, then x2n⟶0, x2n+1⟶∞, y2n⟶0, and y2n+1⟶∞ as n⟶∞(ii)
IfC1λ1/C3−b0C1/C3+a0d0C1/C3+c0>0, then x2n⟶∞, x2n+1⟶0, y2n⟶∞, and y2n+1⟶0 as n⟶∞(iii)
IfC1λ1/C3−b0C1/C3+a0d0C1/C3+c0=0, then every solution of (3) converges to a two-periodic positive solution of the system, where C1 and C3 are given by (48) and (50)Proof.
(i)-(ii). Let(66)pk=u2kv2k−2=C1λ1k−C2λ2kC3λ1k−1−C4λ2k−1,qk=u2kv2k=C1λ1k−C2λ2kC3λ1k−C4λ2k,rk=v2ku2k−2=C3λ1k−C4λ2kC1λ1k−1−C2λ2k−1,sk=v2ku2k=C3λ1k−C4λ2kC1λ1k−C2λ2k.Then, from (56)–(59), we have(67)x2n=x0∏k=1n1+pk−b0qk−1+a0d0qk−1+c0b0qk−1+a0d0qk−1+c0,(68)x2n+1=x1∏k=1n1+pk−b0qk+a0d0qk−1+c0/b0qk+a0d0qk−1+c0,(69)y2n=y0∏k=1n1+rk−d0+c0sk−1b0+a0sk−1d0+c0sk−1b0+a0sk−1,(70)y2n+1=y1∏k=1n1+rk−d0+c0skb0+a0sk−1/d0+c0skb0+a0sk−1.We assume without loss of generality thatλ1>λ2. Then, we have the limits(71)limk⟶∞pk=C1λ1C3,limk⟶∞qk=C1C3,limk⟶∞rk=C3λ1C1,limk⟶∞sk=C3C1,and so(72)L1=limk⟶∞pk−b0qk−1+a0d0qk−1+c0b0qk−1+a0d0qk−1+c0=C1λ1/C3−b0C1/C3+a0d0C1/C3+c0b0C1/C3+a0d0C1/C3+c0,L2=limk⟶∞rk−d0+c0sk−1b0+a0sk−1d0+c0sk−1b0+a0sk−1=C3λ1/C1−b0+a0C3/C1d0+c0C3/C1b0+a0C3/C1d0+c0C3/C1,where(73)C1C3=a0b1+a1c0λ1−a1d0+b0b1.Note thatL1=L2. Hereby, convergence characters of the infinite series(74)∑k=0∞pk−b0qk−1+a0d0qk−1+c0b0qk−1+a0d0qk−1+c0,(75)∑k=0∞rk−d0+c0sk−1b0+a0sk−1d0+c0sk−1b0+a0sk−1,are the same. We can say from a well-known fundamental result about infinite series that (74) and (75) are divergent, if C1λ1/C3−b0C1/C3+a0d0C1/C3+c0≠0. So, the proofs of items (i)-(ii) follow from (67)–(70) and Remark 1.(iii)
From (74), for sufficiently large n0, we have(76)∑k=0∞pk−b0qk−1+a0d0qk−1+c0b0qk−1+a0d0qk−1+c0=S1n0+K1n0,where(77)S1n0=∑k=0n0pk−b0qk−1+a0d0qk−1+c0b0qk−1+a0d0qk−1+c0,K1n0=∑k=n0∞C1λ1/C3−b0C1/C3+a0d0C1/C3+c0b0C1/C3+a0d0C1/C3+c0.
Note that ifC1λ1/C3−b0C1/C3+a0d0C1/C3+c0=0, then K1n0⟶0 as n⟶∞. That is to say, (74) is convergent. Since L1=L2, (75) is convergent too. In this case, the proof of item (iii) follows from (67)–(70) and Remark 1.



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