By differentiating the function H (uo,..., u5) with respect to u; (i = 0, ..., 5), we obtain %3D Huo = A, (5.26) (a1B2 – a2B1) uz+ (a1B3 – a3ß1) uz + (@1B4 – a4B1) u4 + (@1ß5 – a5ß1) u5 Hu | 2 (EL(A) (5.27) - (a1ß2– a2B1) u1 + (a2B3 – a3ß2) uz + (a2B4 – a4ß2) U4 + (a2ß5 – a5B2) u5 2 Hu2 (5.28) 16 - (a1ß3 – a3ß1) u1 – (a2B3 – azB2) u2 + (a3B4 – a483) u4 + (a3ß5 – a5ß3) ua | | | Hua (5.29) (a1ß4 – a4B1) U1 - (a2B4 – a4B2) u2 – (a3B4 – a483) uz + (a4ß5 – a5ß4) u5 - Hu4 %3D (5.30) and (a1ß5 – a5ß1) u1 - (a2B5 – a5B2) u2 – (a3B5 – asß3) uz – (a465 – a5ß4) u4 | Hus %3D (5.31)
By differentiating the function H (uo,..., u5) with respect to u; (i = 0, ..., 5), we obtain %3D Huo = A, (5.26) (a1B2 – a2B1) uz+ (a1B3 – a3ß1) uz + (@1B4 – a4B1) u4 + (@1ß5 – a5ß1) u5 Hu | 2 (EL(A) (5.27) - (a1ß2– a2B1) u1 + (a2B3 – a3ß2) uz + (a2B4 – a4ß2) U4 + (a2ß5 – a5B2) u5 2 Hu2 (5.28) 16 - (a1ß3 – a3ß1) u1 – (a2B3 – azB2) u2 + (a3B4 – a483) u4 + (a3ß5 – a5ß3) ua | | | Hua (5.29) (a1ß4 – a4B1) U1 - (a2B4 – a4B2) u2 – (a3B4 – a483) uz + (a4ß5 – a5ß4) u5 - Hu4 %3D (5.30) and (a1ß5 – a5ß1) u1 - (a2B5 – a5B2) u2 – (a3B5 – asß3) uz – (a465 – a5ß4) u4 | Hus %3D (5.31)
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 red and the inf is here
Transcribed Image Text:Theorem 7 For any values of the quotient 1 , If A < 1, then the
positive equilibrium point ỹ of Eq. (1.1) is a global attractor and the following
i=1
conditions hold
a1B2 > a2ß1, a1ß3 2 a3ß1, aiß4 2 a4B1, a1ß5 > a5B1, a2B3 2 azb2, a2b4 2 a4ß2,
a2B5 > a562, azß4 2 a4B3, a3ß5 2 a5B3, a4ß5 > a5B4 and a5 > (a1 + a2 + a3 + a4).
(5.25)
proof: Let {ym}=-5 be a positive solution of Eq.(1.1). and let H :
(0, 00)6 → (0, 0) be a continuous function which is defined by
H(uo, .., u5)
Auo +
By differentiating the function H(uo,..., u5) with respect to u; (i = 0,.., 5),
we obtain
Huo = A,
(5.26)
(a1B2 - a2B1) u2 + (a1ß3 – a3B1) uz + (@1B4 – a4ß1) u4 + (a1ß5 – a5ß1) u5
Hui
2
(5.27)
(a1B2 – a2B1) u1 +(@2B3 – azß2) u3 + (a2B4 – a4B2) u4 + (a2ß5 – a5B2) u5
Huz
i=1
(5.28)
16
(a133 – a3B1) u1 - (a23 – a382) u2 + (a3B4 – a4B3) u4 + (a3ß5 – a5B3) us
Huz
2
(5.29)
(a1B4 – 04B1) u1 - (a2B4 – a4B2) u2 - (a3B4 - a4B3) uz + (a4ß5 – asß4) u5
Hus
2
(5.30)
and
(a1ß5 – asB1) u1 - (a2B5 – a5B2) u2 - (a3ß5 – asB3) uz – (a465 – asß4) u4
Hus
i=1
(5.31)
It is observed that the function H(uo, ..., u5) is non-decreasing in uo,u1
and non-increasing in u5. Now, we consider four cases:
![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%2Fdbf66eec-f903-4108-8f56-0531f8ba33a7%2Fdsecxwe_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.
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