Theorem 6 If (b+f) > (c+r) and (d+ g) > (e+ s), then the necessary and sufficient condition for Eq.(1) to have positive solutions of prime period two is that the inequality [(a + 1) ((d + g) – (e+ s))] [(b+ f) – (c+r)]² +4[(b+ f) - (c+r)] [(c+r) (d+g) + a (e + s) (b+ f)] > 0. (13) is valid. Proof: Suppose that there exist positive distinctive solutions of prime period two P,Q, P, Q,.. of Eq.(1). From Eq.(1) we have bæn-1 + cxn-2 + fæn-3 + rxn-4 Xn+1 = axn + dæn-1 + exn-2+ gan-3 + sæn-4 (b+f) P+ (c+r) Q (d + g) P+ (e + s) Q' (b+ f)Q+ (c+ r) P (d + g) Q+ (e + s) P P = aQ + Q = aP+ Consequently, we obtain (d+g) P² + (e+ s) PQ = a (d+ g) PQ+a(e+s) Q² + (b+ f) P+(c+r) Q, (14) %3D

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
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Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 32EQ
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The objective of this article is to investigate some qualitative behavior of
the solutions of the nonlinear difference equation
bxn-1 + cxn-2 + fxn-3 +rxn-4
Xn+1 = axn +
n = 0, 1, 2, .. (1)
dxn-1+ exn-2+ gxn-3 + sxn-4
where the coefficients a b cde fa rs E (0 o) while the initial con-
Transcribed Image Text:The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn-1 + cxn-2 + fxn-3 +rxn-4 Xn+1 = axn + n = 0, 1, 2, .. (1) dxn-1+ exn-2+ gxn-3 + sxn-4 where the coefficients a b cde fa rs E (0 o) while the initial con-
Theorem 6 If (b+ f) > (c+r) and (d+ g) > (e + s), then the necessary
and sufficient condition for Eq.(1) to have positive solutions of prime period
two is that the inequality
[(a +1) ((d + g) – (e+ s))] [(b+ f) – (c+r)]²
+4[(b+ f) – (c+r)] [(c + r) (d+ g) + a (e+ s) (b+ f)] > 0. (13)
is valid.
Proof: Suppose that there exist positive distinctive solutions of prime period
two
.., P, Q, P, Q, ......
of Eq.(1). From Eq.(1) we have
bxn-1 + can-2 + fæn-3 +rxn-4
n+1 — ахn +
dæn-1 + exn-2 + gxn-3 + san-4
(b + f) P+ (c+ r) Q
(d + g) P + (e + s) Q'
(b + f) Q+ (c+ r) P
(d + g) Q + (e + s) P`
Р — аQ +
Q = aP+
Consequently, we obtain
d+g) P² +(e+ s) PQ = a (d + g) PQ+a(e+s)Q² +(b+ f) P+(c+r)Q,
(14)
Transcribed Image Text:Theorem 6 If (b+ f) > (c+r) and (d+ g) > (e + s), then the necessary and sufficient condition for Eq.(1) to have positive solutions of prime period two is that the inequality [(a +1) ((d + g) – (e+ s))] [(b+ f) – (c+r)]² +4[(b+ f) – (c+r)] [(c + r) (d+ g) + a (e+ s) (b+ f)] > 0. (13) is valid. Proof: Suppose that there exist positive distinctive solutions of prime period two .., P, Q, P, Q, ...... of Eq.(1). From Eq.(1) we have bxn-1 + can-2 + fæn-3 +rxn-4 n+1 — ахn + dæn-1 + exn-2 + gxn-3 + san-4 (b + f) P+ (c+ r) Q (d + g) P + (e + s) Q' (b + f) Q+ (c+ r) P (d + g) Q + (e + s) P` Р — аQ + Q = aP+ Consequently, we obtain d+g) P² +(e+ s) PQ = a (d + g) PQ+a(e+s)Q² +(b+ f) P+(c+r)Q, (14)
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