Hindawi Publishing Corporation Advances in Difference Equations Volume 2007, Article ID 31272, 13 pages doi:10.1155/2007/31272 Research Article On a k-Order System of Lyness-Type Difference Equations G. Papaschinopoulos, C. J. Schinas, and G. Stefanidou Received 17 January 2007; Revised 24 April 2007; Accepted 14 June 2007 Recommended by John R. Graef We consider the following system of Lyness-type difference equations: x 1 (n +1)= (a k x k (n)+b k )/x k−1 (n − 1), x 2 (n +1)= (a 1 x 1 (n)+b 1 )/x k (n − 1), x i (n +1)= (a i−1 x i−1 (n)+b i−1 )/x i−2 (n − 1), i = 3,4, ,k,wherea i , b i , i = 1,2, ,k, are positive con- stants, k ≥ 3 is an integer, and the initial values are positive real numbers. We study the existence of invariants, the boundedness, the persistence, and the periodicity of the p osi- tive solutions of this system. Copyright © 2007 G. Papaschinopoulos et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis- tribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Difference equations and systems of difference equations have many applications in bi- ology, economy, and other sciences. So there exist many papers concerning systems of difference equations (see [1–10] a nd the references cited therein). In [11], Koci ´ c and Ladas investigated the existence of invariants, the boundedness, the persistence, the periodicity, and the oscillation of the positive solutions of the Lyness difference equation x n+1 = x n + A x n−1 , n = 0,1, , (1.1) where A is a positive constant and the initial conditions x −1 , y −1 , x 0 , y 0 are positive real numbers. 2AdvancesinDifference Equations In [6–8], the authors studied the behavior of the p ositive solutions of the system of two Lyness difference equations x n+1 = by n + c x n−1 , y n+1 = dx n + e y n−1 , n = 0,1, , (1.2) where b, c, d, e are positive constants and the initial conditions x −1 , y −1 , x 0 , y 0 are positive numbers. Now in this paper, we consider the system of difference equations: x 1 (n +1)= a k x k (n)+b k x k−1 (n −1) , x 2 (n +1)= a 1 x 1 (n)+b 1 x k (n −1) , x i (n +1)= a i−1 x i−1 (n)+b i−1 x i−2 (n −1) , i = 3,4, ,k, (1.3) where a i , b i , i = 1,2, ,k, are positive constant numbers, k ≥ 3 is an integer, and the initial values x i (−1), x i (0), i = 1,2, ,k, are positive real numbers. For simplicity, system (1.3)canbewrittenasfollows: x i (n +1)= a i−1 x i−1 (n)+b i−1 x i−2 (n −1) , i = 1,2, ,k, (1.4) where a 0 = a k , b 0 = b k , x j (n) = x k+ j (n), j =−1,0, n =−1,0, (1.5) We study the existence of invariants, the boundedness, the persistence, and the periodicity of the positive solutions of the system (1.3). 2. Boundedness and persistence In this section, we study the boundedness and the persistence of the positive solutions of (1.3). For this goal, we show the following proposition in w hich we find conditions so that system (1.3) has an invariant. Proposition 2.1. Let k ≥ 3 and λ k+i = λ i , i ∈{−2,−1,0,1,2,3,4}, a k+i = a i , i ∈{−3,−2,−1,0,1}, b k+i = b i , i ∈{−2,−1,0}. (2.1) Assume that the system of 2k equations, with k unknowns λ 1 ,λ 2 , ,λ k of the form λ i+2 b i−1 + λ i+3 a i a i−1 = λ i−2 b i−3 + λ i−3 a i−4 a i−3 , i ∈{1,2, ,k}, λ i+4 a i+1 b i = λ i−1 a i−2 b i−1 , i ∈{1,2, ,k}, (2.2) G. Papaschinopoulos et al. 3 has a nontriv ial solution λ 1 ,λ 2 , λ k .Thensystem(1.3) has an invariant of the form I n = k i=1 λ i+2 x i (n)+ k i=1 λ i+2 x i (n −1) + k i=1 λ i b i−1 + λ i−1 a i−1 a i−2 1 x i (n) + k i=1 λ i b i−1 + λ i−1 a i−1 a i−2 1 x i (n −1) + k i=1 λ i−1 a i−2 b i−1 1 x i (n)x i−1 (n −1) + k i=1 λ i a i−1 x i−1 (n −1) x i (n) + k i=1 λ i+3 a i x i (n) x i−1 (n −1) . (2.3) Proof. From (1.5), (1.4), (2.1), (2.2), and (2.3), we have I n+1 = k i=1 λ i+2 a i−1 x i−1 (n) x i−2 (n −1) + k i=1 λ i+2 b i−1 1 x i−2 (n −1) + k i=1 λ i+2 x i (n) + k i=1 λ i b i−1 + λ i−1 a i−1 a i−2 + λ i−1 a i−2 b i−1 1 x i−1 (n) + λ i a i−1 x i−1 (n) × x i−2 (n −1) a i−1 x i−1 (n)+b i−1 + k i=1 λ i b i−1 + λ i−1 a i−1 a i−2 1 x i (n) + k i=1 λ i+3 a i a i−1 1 x i−2 (n −1) + k i=1 λ i+3 a i b i−1 1 x i−1 (n)x i−2 (n −1) = k i=1 λ i+2 x i (n)+ k i=1 λ i x i−2 (n −1) + k i=1 λ i b i−1 + λ i−1 a i−1 a i−2 1 x i (n) + k i=1 λ i+2 b i−1 + λ i+3 a i a i−1 1 x i−2 (n −1) + k i=1 λ i+3 a i b i−1 1 x i−1 (n)x i−2 (n −1) + k i=1 λ i−1 a i−2 x i−2 (n −1) x i−1 (n) + k i=1 λ i+2 a i−1 x i−1 (n) x i−2 (n −1) = I n . (2.4) This completes t he proof of the proposition. 4AdvancesinDifference Equations Corollary 2.2. Let k = 3.Thensystem(1.3)fork = 3 has the following invariant: I n = b 1 x 1 (n)+b 2 x 2 (n)+b 3 x 3 (n)+b 1 x 1 (n −1) + b 2 x 2 (n −1) + b 3 x 3 (n −1) + b 2 b 3 + b 1 a 2 a 3 1 x 1 (n) + b 3 b 1 + b 2 a 3 a 1 1 x 2 (n) + b 1 b 2 + b 3 a 1 a 2 1 x 3 (n) + b 2 b 3 + b 1 a 2 a 3 1 x 1 (n −1) + b 3 b 1 + b 2 a 3 a 1 1 x 2 (n −1) + b 1 b 2 + b 3 a 1 a 2 1 x 3 (n −1) + b 1 a 2 b 3 1 x 1 (n)x 3 (n −1) + b 2 a 3 b 1 1 x 2 (n)x 1 (n −1) + b 3 a 1 b 2 1 x 3 (n)x 2 (n −1) + b 2 a 3 x 3 (n −1) x 1 (n) + b 3 a 1 x 1 (n −1) x 2 (n) + b 1 a 2 x 2 (n −1) x 3 (n) + b 2 a 1 x 1 (n) x 3 (n −1) + b 3 a 2 x 2 (n) x 1 (n −1) + b 1 a 3 x 3 (n) x 2 (n −1) . (2.5) Proof. From (2.1)and(2.2), we get λ 2 b 2 = λ 1 b 3 , λ 3 b 3 = λ 2 b 1 , λ 1 b 1 = λ 3 b 2 .Wesetλ 1 = b 2 , λ 2 = b 3 , λ 3 = b 1 .Thenfrom(2.3), the proof follows immediately. Corollary 2.3. Let k = 4.Supposethat b 1 = b 2 = b 3 = b 4 = b. (2.6) Then system (1.3)fork = 4 has an invariant of the form I n = a 1 x 1 (n)+a 2 x 2 (n)+a 3 x 3 (n)+a 4 x 4 (n)+a 1 x 1 (n −1) + a 2 x 2 (n −1) + a 3 x 3 (n −1) + a 4 x 4 (n −1) + a 3 b + a 4 a 2 a 3 1 x 1 (n) + a 4 b + a 4 a 3 a 1 1 x 2 (n) + a 1 b + a 4 a 1 a 2 1 x 3 (n) + a 2 b + a 3 a 1 a 2 1 x 4 (n) + a 3 b + a 4 a 2 a 3 1 x 1 (n −1) + a 4 b + a 4 a 3 a 1 1 x 2 (n −1) + a 1 b + a 4 a 1 a 2 1 x 3 (n −1) + a 2 b + a 3 a 1 a 2 1 x 4 (n −1) + a 3 a 2 b 1 x 1 (n)x 4 (n −1) + a 3 a 4 b 1 x 2 (n)x 1 (n −1) G. Papaschinopoulos et al. 5 + a 1 a 4 b 1 x 3 (n)x 2 (n −1) + a 1 a 2 b 1 x 4 (n)x 3 (n −1) + a 1 a 4 x 1 (n −1) x 2 (n) + a 1 a 2 x 2 (n −1) x 3 (n) + a 3 a 2 x 3 (n −1) x 4 (n) + a 3 a 4 x 4 (n −1) x 1 (n) + a 1 a 4 x 4 (n) x 3 (n −1) + a 3 a 4 x 3 (n) x 2 (n −1) + a 3 a 2 x 2 (n) x 1 (n −1) + a 1 a 2 x 1 (n) x 4 (n −1) . (2.7) Proof. From (2.1), (2.2), and (2.6), we obtain λ 2 b + λ 3 a 4 a 3 = λ 2 b + λ 1 a 4 a 1 , λ 1 b + λ 2 a 2 a 3 = λ 1 b + λ 4 a 4 a 3 , λ 3 b + λ 4 a 4 a 1 = λ 3 b + λ 2 a 2 a 1 , λ 4 b + λ 1 a 2 a 1 = λ 4 b + λ 3 a 2 a 3 , λ 1 a 4 b = λ 2 a 3 b, λ 2 a 1 b = λ 3 a 4 b, λ 3 a 2 b = λ 4 a 1 b, λ 4 a 3 b = λ 1 a 2 b. (2.8) We set in (2.8) λ 1 = a 3 , λ 2 = a 4 , λ 3 = a 1 , λ 4 = a 2 .Thenfrom(2.3), the proof follows immediately. Corollary 2.4. Consider system (1.3), w here k = 5.Supposethat a 4 a 5 = b 2 , a 3 a 2 = b 5 , a 5 a 1 = b 3 , a 4 a 3 = b 1 , a 1 a 2 = b 4 . (2.9) Then system (1.3), with k = 5, has an invariant of the form I n = λ 3 x 1 (n)+λ 4 x 2 (n)+λ 5 x 3 (n)+λ 1 x 4 (n)+λ 2 x 5 (n) + λ 3 x 1 (n −1) + λ 4 x 2 (n −1) + λ 5 x 3 (n −1) + λ 1 x 4 (n −1) + λ 2 x 5 (n −1) + λ 1 a 2 a 3 + λ 5 a 4 a 5 1 x 1 (n) + λ 2 a 3 a 4 + λ 1 a 5 a 1 1 x 2 (n) 6AdvancesinDifference Equations + λ 3 a 4 a 5 + λ 2 a 1 a 2 1 x 3 (n) + λ 4 a 1 a 5 + λ 3 a 2 a 3 1 x 4 (n) + λ 5 a 1 a 2 + λ 4 a 3 a 4 1 x 5 (n) + λ 1 a 2 a 3 + λ 5 a 4 a 5 1 x 1 (n −1) + λ 2 a 3 a 4 + λ 1 a 5 a 1 1 x 2 (n −1) + λ 3 a 4 a 5 + λ 2 a 1 a 2 1 x 3 (n −1) + λ 4 a 1 a 5 + λ 3 a 2 a 3 1 x 4 (n −1) + λ 5 a 1 a 2 + λ 4 a 3 a 4 1 x 5 (n −1) + λ 5 a 4 a 3 a 2 1 x 1 (n)x 5 (n −1) + λ 1 a 5 a 4 a 3 1 x 2 (n)x 1 (n −1) + λ 2 a 1 a 4 a 5 1 x 3 (n)x 2 (n −1) + λ 3 a 2 a 5 a 1 1 x 4 (n)x 3 (n −1) + λ 4 a 3 a 1 a 2 1 x 5 (n)x 4 (n −1) + λ 1 a 5 x 5 (n −1) x 1 (n) + λ 2 a 1 x 1 (n −1) x 2 (n) + λ 3 a 2 x 2 (n −1) x 3 (n) + λ 4 a 3 x 3 (n −1) x 4 (n) + λ 5 a 4 x 4 (n −1) x 5 (n) + λ 4 a 1 x 1 (n) x 5 (n −1) + λ 5 a 2 x 2 (n) x 1 (n −1) + λ 1 a 3 x 3 (n) x 2 (n −1) + λ 2 a 4 x 4 (n) x 3 (n −1) + λ 3 a 5 x 5 (n) x 4 (n −1) , (2.10) where λ i , i = 1,2,3,4,5, are real numbers. Proof. Using (2.1), (2.2), and (2.9), we get λ 1 a 1 a 5 + λ 2 a 4 a 3 = λ 2 a 3 a 4 + λ 1 a 5 a 1 , λ 2 a 2 a 1 + λ 3 a 4 a 5 = λ 3 a 4 a 5 + λ 2 a 1 a 2 , λ 5 a 4 a 5 + λ 1 a 3 a 2 = λ 1 a 2 a 3 + λ 5 a 4 a 5 , λ 3 a 3 a 2 + λ 4 a 1 a 5 = λ 4 a 1 a 5 + λ 3 a 2 a 3 , λ 4 a 4 a 3 + λ 5 a 2 a 1 = λ 5 a 1 a 2 + λ 4 a 3 a 4 , λ 1 a 3 a 4 a 5 = λ 1 a 3 a 4 a 5 , λ 2 a 4 a 5 a 1 = λ 2 a 4 a 5 a 1 , λ 3 a 5 a 1 a 2 = λ 3 a 5 a 1 a 2 , λ 4 a 1 a 2 a 3 = λ 4 a 1 a 2 a 3 , λ 5 a 2 a 3 a 4 = λ 5 a 2 a 3 a 4 , (2.11) G. Papaschinopoulos et al. 7 which are satisfied for any real numbers λ i , i = 1, 2,3,4,5. Then from (2.3), the corollary is proved. 3. Periodicity We study the periodicity of the positive solutions of (1.3) by investigating three cases: k = 3, k = 4, and k ∈{5,6, }. For the first case, we show the following proposition. Proposition 3.1. Consider system (1.3)fork = 3.If a 1 = a 2 = a 3 = a, b 1 = b 2 = b 3 = b, a 2 = b, (3.1) then every positive solution of system (1.3) is periodic of period 15. Proof. We have x 1 (n +5)= ax 3 (n +4)+a 2 x 2 (n +3) = a ax 2 (n +3)+a 2 /x 1 (n +2) + a 2 x 2 (n +3) = a 2 x 2 (n +3)+a 3 + a 2 x 1 (n +2) x 1 (n +2)x 2 (n +3) = a 2 ax 1 (n +2)+a 2 /x 3 (n +1) + a 3 + a 2 x 1 (n +2) x 1 (n +2)((ax 1 (n +2)+a 2 )/x 3 (n +1)) = a 3 x 1 (n +2)+a 4 + a 3 x 3 (n +1)+a 2 x 1 (n +2)x 3 (n +1) x 1 (n +2) ax 1 (n +2)+a 2 = ax 1 (n +2)+a 2 ax 3 (n +1)+a 2 x 1 (n +2) ax 1 (n +2)+a 2 = ax 3 (n +1)+a 2 x 1 (n +2) = x 2 (n). (3.2) Working in a similar way, we can prove that x 2 (n +5)= x 3 (n), x 3 (n +5)= x 1 (n). (3.3) Thus, x 1 (n +15) = x 2 (n +10) = x 3 (n +5)= x 1 (n). (3.4) Similarly, x 2 (n +15) = x 2 (n), x 3 (n +15) = x 3 (n), (3.5) and the proof of the proposition is complete. 8AdvancesinDifference Equations In the sequel, we prove the following proposition which concerns the case k = 4. Proposition 3.2. Consider system (1.3)fork = 4.If a 1 = a 2 = a 3 = a 4 = a, b 1 = b 2 = b 3 = b 4 = b, a 2 = b, (3.6) then every positive solution of system (1.3) is periodic of period 20. Proof. We have x 1 (n +5)= ax 4 (n +4)+a 2 x 3 (n +3) = a (ax 3 (n +3)+a 2 )/x 2 (n +2) + a 2 x 3 (n +3) = a 2 x 3 (n +3)+a 3 + a 2 x 2 (n +2) x 2 (n +2)x 3 (n +3) = a 2 ax 2 (n +2)+a 2 /x 1 (n +1) + a 3 + a 2 x 2 (n +2) x 2 (n +2)((ax 2 (n +2)+a 2 )/x 1 (n +1)) = a 3 x 2 (n +2)+a 4 + a 3 x 1 (n +1)+a 2 x 1 (n +1)x 2 (n +2) x 2 (n +2) ax 2 (n +2)+a 2 = ax 2 (n +2)+a 2 ax 1 (n +1)+a 2 x 2 (n +2) ax 2 (n +2)+a 2 = ax 1 (n +1)+a 2 x 2 (n +2) = x 4 (n). (3.7) Arguing as above, we can show that x 2 (n +5)= x 1 (n), x 3 (n +5)= x 2 (n), x 4 (n +5)= x 3 (n). (3.8) So, x 1 (n +20) = x 4 (n +15) = x 3 (n +10) = x 2 (n +5)= x 1 (n). (3.9) Similarly, x 2 (n +20) = x 2 (n), x 3 (n +20) = x 3 (n), x 4 (n +20) = x 4 (n), (3.10) and the proof of the proposition is complete. G. Papaschinopoulos et al. 9 Finally, we study the case k ∈{5,6, }. To this end, we have at first to prove the fol- lowing lemma. Lemma 3.3. Let k ≥ 5.If a 1 = a 2 =···= a k = a, b 1 = b 2 =···= b k = b, a 2 = b (3.11) then x i (n +5)= x k−5+i (n), i ∈{1,2, ,5}, x i (n +5)= x i−5 (n), i ∈{6,7, ,k}. (3.12) Proof. From (1.3), we have x 1 (n +5)= ax k (n +4)+a 2 x k−1 (n +3) = a ax k−1 (n +3)+a 2 /x k−2 (n +2) + a 2 x k−1 (n +3) = a 2 x k−1 (n +3)+a 3 + a 2 x k−2 (n +2) x k−2 (n +2)x k−1 (n +3) = a 2 ax k−2 (n +2)+a 2 /x k−3 (n +1) + a 3 + a 2 x k−2 (n +2) x k−2 (n +2) ax k−2 (n +2)+a 2 /x k−3 (n +1) = a 3 x k−2 (n +2)+a 4 + a 3 x k−3 (n +1)+a 2 x k−2 (n +2)x k−3 (n +1) x k−2 (n +2) ax k−2 (n +2)+a 2 = ax k−2 (n +2)+a 2 ax k−3 (n +1)+a 2 x k−2 (n +2) ax k−2 (n +2)+a 2 . (3.13) Then since, from (1.3), x k−4 (n) = ax k−3 (n +1)+a 2 x k−2 (n +2) , (3.14) it follows that x 1 (n +5)= x k−4 (n). (3.15) Similarly, we can prove that x i (n +5)= x k−5+i (n), i ∈{2,3, ,5}. (3.16) 10 Advances in Difference Equations Let i ∈{6,7, ,k}.Then x i (n +5)= ax i−1 (n +4)+a 2 x i−2 (n +3) = a ax i−2 (n +3)+a 2 /x i−3 (n +2) + a 2 x i−2 (n +3) = a 2 x i−2 (n +3)+a 3 + a 2 x i−3 (n +2) x i−2 (n +3)x i−3 (n +2) = a 2 ax i−3 (n +2)+a 2 /x i−4 (n +1) + a 3 + a 2 x i−3 (n +2) x i−3 (n +2) ax i−3 (n +2)+a 2 /x i−4 (n +1) = a 3 x i−3 (n +2)+a 4 + a 3 x i−4 (n +1)+a 2 x i−3 (n +2)x i−4 (n +1) x i−3 (n +2) ax i−3 (n +2)+a 2 = ax i−3 (n +2)+a 2 ax i−4 (n +1)+a 2 x i−3 (n +2) ax i−3 (n +2)+a 2 . (3.17) Then since, from (1.3), x i−5 (n) = ax i−4 (n +1)+a 2 x i−3 (n +2) , (3.18) it follows that x i (n +5)= x i−5 (n), i ∈{6,7, ,k}. (3.19) Now we can show the following proposition. Proposition 3.4. Consider system (1.3), where k ≥ 5. Assume that relations (3.11)hold. Then the following statements are t rue. (i) Every positive solution of system (1.3) is periodic of period k if k = 5r, r = 1, 2, (ii) Every positive solution of system (1.3) is periodic of period 5k if k = 5r, r = 1, 2, Proof. Consider an arbitrary solution (x 1 (n), ,x k (n)) of (1.3). (i) Suppose that k = 5r, r = 1, 2, Thenfrom(3.12), we have x i (n +5)= x 5r−5+i (n), i ∈{1,2, ,5}, x i (n +5)= x i−5 (n), i ∈{6,7, ,5r}. (3.20) We claim that for i = 1,2, ,5, x i (n +5s) = x 5r−5s+i (n), s = 1,2, ,r. (3.21) From (3.20), it is obvious that (3.21)istruefors = 1. Suppose that for i = 1,2, ,5, relation (3.21)istruefors = 1,2, ,r − 1. Then since 6 ≤ 5r − 5s + i ≤ 5r,from(3.20) and (3.21), we get for i = 1,2, ,5, x i (n +5+5s) = x 5r−5s+i (n +5)= x 5r−5(s+1)+i (n), (3.22) [...]... 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The Netherlands, 1993 G Papaschinopoulos: School of Engineering, Democritus University of Thrace, 67100 Xanthi, Greece Email address: gpapas@ee.duth.gr C J Schinas: School of Engineering, Democritus University of Thrace, 67100 Xanthi, Greece Email address: cschinas@ee.duth.gr G Stefanidou: School of Engineering, Democritus University of Thrace, 67100 Xanthi, Greece Email address: gstefmath@yahoo.gr ... implies that x1 (n) is a periodic sequence of period 5(5r + 4) = 5k Then from relations (3.33), we can prove that the sequences xi (n), i = 1,2, ,k, are periodic of period 5k This completes the proof of the proposition Acknowledgment The authors would like to thank the referee for their helpful suggestions References [1] D Clark and M R S Kulenovi´ , A coupled system of rational difference equations,”... implies that x1 (n) is a periodic sequence of period 5k Then by relations (3.27), we can prove that the sequences xi (n), i = 2,3, ,k, are periodic of period 5k Let k = 5r + 3 Then from (3.12), we have xi (n + 5) = x5r −2+i (n), xi (n + 5) = xi−5 (n), i ∈ {1,2, ,5}, i ∈ {6,7, ,5r + 3} (3.30) 12 Advances in Difference Equations Then from (3.30) and using the same argument to show (3.21), we can prove that for...G Papaschinopoulos et al 11 and so (3.21) is true Then from (3.21) for s = r, we have xi (n + 5r) = xi (n), i = 1,2, ,5 (3.23) Therefore the sequences xi (n), i = 1,2, ,5 are periodic of period 5 Then from (3.20), all the sequences xi (n), i = 1,2, ,k, are periodic of period k (ii) Suppose that k = 5r, r = 1,2, Let k = 5r + 1, r = 1,2, Then from (3.12), we have xi (n + 5) = x5r −4+i... Then from (3.30) and (3.31) for i = 1, s = r, we get x1 (n + 25r + 15) = x4 (n + 20r + 15) = x7 (n + 15r + 15) = x2 (n + 15r + 10) = x5 (n + 10r + 10) (3.32) = x8 (n + 5r + 10) = x3 (n + 5r + 5) = x6 (n + 5) = x1 (n), which implies that x1 (n) is a periodic sequence of period 5(5r + 3) = 5k Then from relations (3.30), we can prove that the sequences xi (n), i = 1,2, ,k, are periodic of period 5k Let . λ 3 a 2 a 3 , λ 4 a 4 a 3 + λ 5 a 2 a 1 = λ 5 a 1 a 2 + λ 4 a 3 a 4 , λ 1 a 3 a 4 a 5 = λ 1 a 3 a 4 a 5 , λ 2 a 4 a 5 a 1 = λ 2 a 4 a 5 a 1 , λ 3 a 5 a 1 a 2 = λ 3 a 5 a 1 a 2 , λ 4 a 1 a 2 a 3 =. λ 2 a 4 a 3 = λ 2 a 3 a 4 + λ 1 a 5 a 1 , λ 2 a 2 a 1 + λ 3 a 4 a 5 = λ 3 a 4 a 5 + λ 2 a 1 a 2 , λ 5 a 4 a 5 + λ 1 a 3 a 2 = λ 1 a 2 a 3 + λ 5 a 4 a 5 , λ 3 a 3 a 2 + λ 4 a 1 a 5 = λ 4 a 1 a 5 +. behavior of a three-dimensional linear fractional system of difference equations,” Journal of Mathematical Analysis and Applications, vol. 310, no. 2, pp. 673–689, 2005. [5] M.R.S.Kulenovi ´ c and