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.
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.
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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Explain the determine green and it complete
Transcribed Image Text:l stc ksa
12:14 AM
@ 1 60% 4
The objective of this article is to investigate some qualitative behavior of
the solutions of the nonlinear difference equation
bxn-1 + cæn-2+ fxn-3 + ræn-4
Xn+1 = axn +
n = 0, 1, 2, . (1)
dxn-1+ exn-2 + gæn-3 + sxn-4
where the coefficients a, b, c, d, e, f, g, r, s E (0, 00), while the initial con-
ditions a_4,x_3,x_2, x-1, xo are arbitrary positive real numbers. Note that
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![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+ cxn-2 + fxn-3 + rxn-4
Xn+1 = axn +
dxn-1 + exn-2+ gxn-3 + sxn-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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86116bc7-59c5-4d89-8ef1-07148d796d94%2F84348392-9d1b-47fe-8058-ed503dee05e6%2Fnllnslm_processed.jpeg&w=3840&q=75)
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+ cxn-2 + fxn-3 + rxn-4
Xn+1 = axn +
dxn-1 + exn-2+ gxn-3 + sxn-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)
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