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Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4) , then the necessary and sufficient condition for Eq.(1.1) to have positive so- lutions of prime period two is that the inequality [(A+ 1) ((B1 + B3 + B5) – (B2 + B4)) [(@1 + a3 + a5) – (a2 + a4)]? +4[(a1+a3 +a5) – (a2 + a4)] [(ß1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13)

Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4) , then the necessary and sufficient condition for Eq.(1.1) to have positive so- lutions of prime period two is that the inequality [(A+ 1) ((B1 + B3 + B5) – (B2 + B4)) [(@1 + a3 + a5) – (a2 + a4)]? +4[(a1+a3 +a5) – (a2 + a4)] [(ß1 + B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13)

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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show me the steps of determine blue and the inf is here

The main focus of this article is to discuss some qualitative behavior of the solutions of the nonlinear difference equation a1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5 Ут+1 — Аутt т 3 0, 1, 2, ..., B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5 (1.1) where the coefficients A, a;, B; E (0, 0), i = 1, ..., 5, while the initial condi- tions y-5,y-4,Y–3,Y–2, Y–1, yo are arbitrary positive real numbers. Note that the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 = = a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case B4 when a4 = B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the special case when az = B5 = 0.
Transcribed Image Text:The main focus of this article is to discuss some qualitative behavior of the solutions of the nonlinear difference equation a1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5 Ут+1 — Аутt т 3 0, 1, 2, ..., B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5 (1.1) where the coefficients A, a;, B; E (0, 0), i = 1, ..., 5, while the initial condi- tions y-5,y-4,Y–3,Y–2, Y–1, yo are arbitrary positive real numbers. Note that the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 = = a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case B4 when a4 = B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the special case when az = B5 = 0.
Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4), then the necessary and sufficient condition for Eq. (1.1) to have positive so- lutions of prime period two is that the inequality [(A+ 1) ((B1 + B3 + B3) – (B2 + Ba))] [(a1 + a3 + a5) – (a2+ a4)]² +4[(a1 + a3 + a5) – (a2 + a4)] [(B1+ B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13) is valid. proof: Suppose there exist positive distinctive solutions of prime period two .., P, Q, P,Q, .... of Eq.(1.1). From Eq.(1.1) we have a1Ym–1+ a2Ym–2+ a3Ym-3 + a4Ym-4 + a5Ym-5 + В1Ут-1 + В2ут-2 + Bзут-з + ВАУт-4 + Bзут-5 Ym+1 = Ag (a1 + a3 + a5) P+(@2 + a4) Q (B1 + B3 + B5) P+ (B2 + B4) Q (a1 + a3 + a5)Q+(@2+a4) P (B1 + B3 + B5) Q+ (B2 + B4) P´ (4.14) P = AQ+ Q3 АР+ Consequently, we get (B1 + B3 + B3) P² + (B2 + B4) PQ A (B1 + B3 + Bs) PQ + A (B2 + B4) Q² + (a1 + a3 + a5) P+ (a2 + a4)Q, (4.15) and (81 + B3 + B5) Q² + (B2 + B4) PQ A (B1 + B3 + Bs) PQ + A (B2 + B4) P² + (a1 + a3 + as) Q + (a2 + a4) P. (4.16) %3D By subtracting (4.15) from (4.16), we obtain [(31 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q). 11 Since P+ Q, it follows that [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B3) + A (B2 + B4)] P+Q = (4.17) while, by adding (4.15) and (4.16) and by using the relation p² + Q? = (P+Q)² – 2PQ for all P,Q € R, we have [(a1 + a3 + as) – (a2 + a4)] [(ß1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 +a3 + a5)] [(B1 + B3 + B5) + A (82 + B4)]² [((82 + B4) – (B1 + B3 + B5)) (A+ 1)] (4.18) РQ — Let P and Q are two distinct real roots of the quadratic equation t² – ( P+ Q)t+ PQ = 0. [(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + a3 + a5) – (a2 + a4)]t [(a1 + a3 + a5) – (a2 + a4)] [(81 + B3 + B6) (a2 + a4) + A (B2 + B4) (@1 +a3 + as)] [(B1 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A+ 1)] = 0, (4.19) and so [(@1 + a3 + a5) - (a2 + a4)]? 4 [(a1 + a3 + a5) – (a2+ a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] [((B2 + B4) – (B1 + B3 + B5)) (A+1)] > 0, or [(a1 + a3 + as) – (a2 + a4)]? (a2 + a4)] [(B1 + B3 + B3) (a2 + a4) + A (B2 + B4) (a1+a3 + a5)] [((B1 + B3 + B5) – (B2 + Ba)) (A+1)] 4[(a1 + a3 + a5) + > 0. (4.20) From (4.20), we get [((B1 + B3 + B5) – (B2 + B4)) (A+ 1)] [(a1 + a3 + a5) – (a2 + a4)]? +4[(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. Therefore, the condition (4.13) is valid. Alteratively, if we imagine that the condition (4.13) is valid where (aı + a3 + a5) > (a2 + a4) and (B1 + B3 + Bs) > (B2 + B4). Then, we can immediately discover that the inequality stands.
Transcribed Image Text:Theorem 6 If (a1 + a3 + a5) > (a2 + a4) and (B1 + B3 + B5) > (B2 + B4), then the necessary and sufficient condition for Eq. (1.1) to have positive so- lutions of prime period two is that the inequality [(A+ 1) ((B1 + B3 + B3) – (B2 + Ba))] [(a1 + a3 + a5) – (a2+ a4)]² +4[(a1 + a3 + a5) – (a2 + a4)] [(B1+ B3 + Bs) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. (4.13) is valid. proof: Suppose there exist positive distinctive solutions of prime period two .., P, Q, P,Q, .... of Eq.(1.1). From Eq.(1.1) we have a1Ym–1+ a2Ym–2+ a3Ym-3 + a4Ym-4 + a5Ym-5 + В1Ут-1 + В2ут-2 + Bзут-з + ВАУт-4 + Bзут-5 Ym+1 = Ag (a1 + a3 + a5) P+(@2 + a4) Q (B1 + B3 + B5) P+ (B2 + B4) Q (a1 + a3 + a5)Q+(@2+a4) P (B1 + B3 + B5) Q+ (B2 + B4) P´ (4.14) P = AQ+ Q3 АР+ Consequently, we get (B1 + B3 + B3) P² + (B2 + B4) PQ A (B1 + B3 + Bs) PQ + A (B2 + B4) Q² + (a1 + a3 + a5) P+ (a2 + a4)Q, (4.15) and (81 + B3 + B5) Q² + (B2 + B4) PQ A (B1 + B3 + Bs) PQ + A (B2 + B4) P² + (a1 + a3 + as) Q + (a2 + a4) P. (4.16) %3D By subtracting (4.15) from (4.16), we obtain [(31 + B3 + Bs) + A (B2 + B4)] (P² – Q²) = [(a1 + a3 + as) – (a2 + a4)] (P – Q). 11 Since P+ Q, it follows that [(a1 + a3 + a5) – (a2 + a4)] [(B1 + B3 + B3) + A (B2 + B4)] P+Q = (4.17) while, by adding (4.15) and (4.16) and by using the relation p² + Q? = (P+Q)² – 2PQ for all P,Q € R, we have [(a1 + a3 + as) – (a2 + a4)] [(ß1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 +a3 + a5)] [(B1 + B3 + B5) + A (82 + B4)]² [((82 + B4) – (B1 + B3 + B5)) (A+ 1)] (4.18) РQ — Let P and Q are two distinct real roots of the quadratic equation t² – ( P+ Q)t+ PQ = 0. [(B1 + B3 + B5) + A (B2 + B4)] t² – [(a1 + a3 + a5) – (a2 + a4)]t [(a1 + a3 + a5) – (a2 + a4)] [(81 + B3 + B6) (a2 + a4) + A (B2 + B4) (@1 +a3 + as)] [(B1 + B3 + B5) + A (B2 + B4)] [((B2 + B4) – (B1 + B3 + B5)) (A+ 1)] = 0, (4.19) and so [(@1 + a3 + a5) - (a2 + a4)]? 4 [(a1 + a3 + a5) – (a2+ a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] [((B2 + B4) – (B1 + B3 + B5)) (A+1)] > 0, or [(a1 + a3 + as) – (a2 + a4)]? (a2 + a4)] [(B1 + B3 + B3) (a2 + a4) + A (B2 + B4) (a1+a3 + a5)] [((B1 + B3 + B5) – (B2 + Ba)) (A+1)] 4[(a1 + a3 + a5) + > 0. (4.20) From (4.20), we get [((B1 + B3 + B5) – (B2 + B4)) (A+ 1)] [(a1 + a3 + a5) – (a2 + a4)]? +4[(a1 + a3 + as) – (a2 + a4)] [(B1 + B3 + B5) (a2 + a4) + A (B2 + B4) (a1 + a3 + a5)] > 0. Therefore, the condition (4.13) is valid. Alteratively, if we imagine that the condition (4.13) is valid where (aı + a3 + a5) > (a2 + a4) and (B1 + B3 + Bs) > (B2 + B4). Then, we can immediately discover that the inequality stands.
Expert Solution
Step 1

Given, if (α1+α3+α5)>(α2+α4) and (β1+β3+β5)>(β2+β4), then the necessary and sufficient condition for following equation:       ym+1=Aym+α1ym-1+α2ym-2+α3ym-3+α4ym-4+α5ym-5α1ym-1+α2ym-2+α3ym-3+α4ym-4+α5ym-5, m=0, 1, 2, .  .  .        (1.1)Where the coefficients A, αi, βi∈0, ∞, i=1, . . . , 5.

 

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