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By subtracting (23) from (22), we get ´a+c(1-a+3) While, by adding (22) and (23), we deduce PQ = c(1-a)(a+c(1-a+8)) 62B(1-a+3) If a <1 and a <1+B then from (24) and (25), we have PQ (P+Q) = - e(1-a)(a+c(1=a+8))² <0 682 (1-a+B)

By subtracting (23) from (22), we get ´a+c(1-a+3) While, by adding (22) and (23), we deduce PQ = c(1-a)(a+c(1-a+8)) 62B(1-a+3) If a <1 and a <1+B then from (24) and (25), we have PQ (P+Q) = - e(1-a)(a+c(1=a+8))² <0 682 (1-a+B)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Our goal is to obtain some qualitative behavior of the positive solutions of the difference equation aIn-t In+1 = Bxn-i+axn-k + п%3D 0, 1, (1) bxn-t+c' where the parameters B, a, a, b and c are positive real numbers and the initial conditions x-s, x-s+1,..., x-1, xo are positive real numbers where s = таxfl, k, t).
Transcribed Image Text:Our goal is to obtain some qualitative behavior of the positive solutions of the difference equation aIn-t In+1 = Bxn-i+axn-k + п%3D 0, 1, (1) bxn-t+c' where the parameters B, a, a, b and c are positive real numbers and the initial conditions x-s, x-s+1,..., x-1, xo are positive real numbers where s = таxfl, k, t).
THEOREM 4.4. If 1, t are even and k is odd positive integers then Eq. (1) has no positive prime period two solution. Proof: Let there exists distinct positive solution P and Q, such that ....Р, Q, Р, Q, .., is a prime period two solution of Eq.(1). We see from Eq. (1) when l, t are even and k is odd, then xn+1 = xn-k = P and xn-1 = xn-t =Q. It follows Eq. (1) that P = BQ + aP+ aQ bQ+c and Q = BP + aQ + Pc: bP+c Therefore, b(1 - a) PQ+c(1 – a) P = 6BQ² + (cß + a) Q, (22) b(1- a) PQ +c(1- a) Q = bB P² + (cB+a) P, (23) By subtracting (23) from (22), we get a+c(1-a+3) Q = - (24) While, by adding (22) and (23), we deduce PQ = c(1-a)(a+c(1-a+8)) 623(1-a+B) (25) If a <1 and a <1+8 then from (24) and (25), we have PQ (P+ Q) c(1-a)(a+c(1-a+3))? 682 (1-a+8) This contradicts the hypothesis that both P, Q are positive. Thus, the proof is now completed.
Transcribed Image Text:THEOREM 4.4. If 1, t are even and k is odd positive integers then Eq. (1) has no positive prime period two solution. Proof: Let there exists distinct positive solution P and Q, such that ....Р, Q, Р, Q, .., is a prime period two solution of Eq.(1). We see from Eq. (1) when l, t are even and k is odd, then xn+1 = xn-k = P and xn-1 = xn-t =Q. It follows Eq. (1) that P = BQ + aP+ aQ bQ+c and Q = BP + aQ + Pc: bP+c Therefore, b(1 - a) PQ+c(1 – a) P = 6BQ² + (cß + a) Q, (22) b(1- a) PQ +c(1- a) Q = bB P² + (cB+a) P, (23) By subtracting (23) from (22), we get a+c(1-a+3) Q = - (24) While, by adding (22) and (23), we deduce PQ = c(1-a)(a+c(1-a+8)) 623(1-a+B) (25) If a <1 and a <1+8 then from (24) and (25), we have PQ (P+ Q) c(1-a)(a+c(1-a+3))? 682 (1-a+8) This contradicts the hypothesis that both P, Q are positive. Thus, the proof is now completed.
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