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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Fb51d783c-9472-4554-9580-d4d2afbaf1bd%2Fhui80hh_processed.jpeg&w=3840&q=75)
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Fb51d783c-9472-4554-9580-d4d2afbaf1bd%2F3wektpc_processed.png&w=3840&q=75)
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.
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