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 =1 B; conditions hold a132 > a2B1, a13 2 azB1, a1ß4 2 a481, a1B5 > a5,81, a233 2 azB2, a2B4 2 a4B2, a235 > a5B2, a334 2 a43, a3B5 2 a5B3, a4B5 2 a5,B4 and as > (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, us) Auo + E (Biu;)" .... By differentiating the function H(uo,..., u5) with respect to u; (i = 0, ..., 5), we obtain Huo = A, (5.26) (@1B2 – a2B1) u2 + (@1B3 – a3ß1) uz + (@1B4 – 0481) u4 + (¤15 – a5B1) u5 Ни 2 (5.27) · (a132-a2B1) ui + (a2B3 – a3B2) uz + (a2B4 – C4B2) u4 + (a2B5 – a5,B2) u5 Huz 2 (5.28)

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How to deduce this equation from Equation 1.1 Explain to me the method. Show me the steps of determine yellow and inf is here

Theorem 7 For any values of the quotient E-1 , If A < 1, then the
positive equilibrium point ỹ of Eq.(1.1) is a global attractor and the following
conditions hold
a1 b2 2 a2B1, a1ß3 > azß1, a184 2 a4B1, a1ß5 > a5,B1, a2ß3 > a3B2, a2ß4 > a4B2,
a2B5 > a5,B2, a3ß4 > a4B3, a3ß5 > a5ß3, a4ß5 > a5,B4 and ag > (a1 + a2 + a3 + a4).
(5.25)
proof: Let {ym}=-5 be a positive solution of Eq.(1.1). and let H :
(0, 0)6 –
→ (0, 0) be a continuous function which is defined by
H(uo, ..., U5
Auo +
E1(B;u;)
By differentiating the function H(uo,..., u5) with respect to u; (i = 0,..., 5),
we obtain
А,
Huo
(5.26)
(@1b2 – a2B1) u2 + (a1B3 – a3B1) uz + (a1B4 – a4B1) u4 + (a1B5 – a5,B1) ug
Hu
2
(5.27)
(a132 – a261) u1 + (@2B3 – a3B2) uz + (a2B4 – 04B2) u4 + (a2B5 – a5B2) uz
Huz
(5.28)
16
Transcribed Image Text:Theorem 7 For any values of the quotient E-1 , If A < 1, then the positive equilibrium point ỹ of Eq.(1.1) is a global attractor and the following conditions hold a1 b2 2 a2B1, a1ß3 > azß1, a184 2 a4B1, a1ß5 > a5,B1, a2ß3 > a3B2, a2ß4 > a4B2, a2B5 > a5,B2, a3ß4 > a4B3, a3ß5 > a5ß3, a4ß5 > a5,B4 and ag > (a1 + a2 + a3 + a4). (5.25) proof: Let {ym}=-5 be a positive solution of Eq.(1.1). and let H : (0, 0)6 – → (0, 0) be a continuous function which is defined by H(uo, ..., U5 Auo + E1(B;u;) By differentiating the function H(uo,..., u5) with respect to u; (i = 0,..., 5), we obtain А, Huo (5.26) (@1b2 – a2B1) u2 + (a1B3 – a3B1) uz + (a1B4 – a4B1) u4 + (a1B5 – a5,B1) ug Hu 2 (5.27) (a132 – a261) u1 + (@2B3 – a3B2) uz + (a2B4 – 04B2) u4 + (a2B5 – a5B2) uz Huz (5.28) 16
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.
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