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Therefore, b(1 - a) P2 +c(1- a) P = 6BPQ + cBQ + aP, (31) b(1 - a) Q? +c(1 - a) Q = 6BPQ + cßP +aQ, (32) Subtracting (32) from (31) gives P+Q = 4-c(1+3-a) b(1-a) (33) Again, adding (31) and (32) yields cB(a-c(1+8-a)) PQ = -1-a)(1+B-a)* (34)

Therefore, b(1 - a) P2 +c(1- a) P = 6BPQ + cBQ + aP, (31) b(1 - a) Q? +c(1 - a) Q = 6BPQ + cßP +aQ, (32) Subtracting (32) from (31) gives P+Q = 4-c(1+3-a) b(1-a) (33) Again, adding (31) and (32) yields cB(a-c(1+8-a)) PQ = -1-a)(1+B-a)* (34)

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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THEOREM 4.7. If l is even and k, t are odd positive integers, then Eq. (1) has no positive prime period two solution. Proof: Assume that 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 is even and k, t are odd, then xn+1 = xn-k = In-t = P and xn-l = Q. It follows Eq. (1) that P = BQ + aP+P+c and Q = BP + aQ + bQ+c° Therefore, b(1 – a) P? +c(1 - a) P = 6BPQ + cBQ + aP, (31) b(1 – a) Q? + c(1 - a) Q = 6BPQ + cBP + aQ, (32) Subtracting (32) from (31) gives P+Q = a-c(1+8-a) b(1-а) (33) Again, adding (31) and (32) yields cB(a-e(1+3-a)) PQ = --a)(1+8-a)" (34) If a < 1, a <1+B and c (1+ B- a) < a, then PQ is negative. But P, Q are both positive, and we have a contradiction. Thus, the proof is completed.
Transcribed Image Text:THEOREM 4.7. If l is even and k, t are odd positive integers, then Eq. (1) has no positive prime period two solution. Proof: Assume that 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 is even and k, t are odd, then xn+1 = xn-k = In-t = P and xn-l = Q. It follows Eq. (1) that P = BQ + aP+P+c and Q = BP + aQ + bQ+c° Therefore, b(1 – a) P? +c(1 - a) P = 6BPQ + cBQ + aP, (31) b(1 – a) Q? + c(1 - a) Q = 6BPQ + cBP + aQ, (32) Subtracting (32) from (31) gives P+Q = a-c(1+8-a) b(1-а) (33) Again, adding (31) and (32) yields cB(a-e(1+3-a)) PQ = --a)(1+8-a)" (34) If a < 1, a <1+B and c (1+ B- a) < a, then PQ is negative. But P, Q are both positive, and we have a contradiction. Thus, the proof is completed.
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).
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