Then, the only positive equilibrium point of Eq.(1.1) is given by (2.7) (1 – A) (E Bi) Li=1 provided that A < 1. Now, let us introduce a continuous function H : (0, 00)6 → (0, 0) which is defined by 55 H(uo, ., u5) = Auo + (2.8) Therefore, it follows that H(uo,..us) duo А, H(uo.....us) H(uo....us) dug H(uo,.us) H(uo..us) Oug as E(Bu)+Bsus] - Ba E(aus)+asus] H(uo,...us) dus as (E (Bu) Consequently, we get 8H(ỹ,...) = A = - P5, (1-A)[a1 ( B.) – B (2a)] =- P4, OH (ỹ,...) (1-A)[a2 ( Bi+E-3 P) - B (ait EL, a)] =- P3, OH(ỹ....) dug (1-A)[as ( E B+E84) - Ba (E-, art Ea)] =- P2, (1-A)[a4 (Bs +E81) - BA (as+ E-s at)] = - Pl, (1-A)[as (E Bi) - As (E)] Po. (2.9) Hence, the linearized equation of Eq.(1.1) about ỹ takes the form Ym+1+ P5Ym +P4Ym-1+P3Ym-2+P2Ym-3+P1Ym-4+P0ym-5 = 0, (2.10) where po, P1, P2, P3, P4 and ps are given by (2.9). The characteristic equation associated with Eq.(2.10) is 16 + psA + paX“ + P3A3 + p212 + pid+ po = 0, (2.11)
Then, the only positive equilibrium point of Eq.(1.1) is given by (2.7) (1 – A) (E Bi) Li=1 provided that A < 1. Now, let us introduce a continuous function H : (0, 00)6 → (0, 0) which is defined by 55 H(uo, ., u5) = Auo + (2.8) Therefore, it follows that H(uo,..us) duo А, H(uo.....us) H(uo....us) dug H(uo,.us) H(uo..us) Oug as E(Bu)+Bsus] - Ba E(aus)+asus] H(uo,...us) dus as (E (Bu) Consequently, we get 8H(ỹ,...) = A = - P5, (1-A)[a1 ( B.) – B (2a)] =- P4, OH (ỹ,...) (1-A)[a2 ( Bi+E-3 P) - B (ait EL, a)] =- P3, OH(ỹ....) dug (1-A)[as ( E B+E84) - Ba (E-, art Ea)] =- P2, (1-A)[a4 (Bs +E81) - BA (as+ E-s at)] = - Pl, (1-A)[as (E Bi) - As (E)] Po. (2.9) Hence, the linearized equation of Eq.(1.1) about ỹ takes the form Ym+1+ P5Ym +P4Ym-1+P3Ym-2+P2Ym-3+P1Ym-4+P0ym-5 = 0, (2.10) where po, P1, P2, P3, P4 and ps are given by (2.9). The characteristic equation associated with Eq.(2.10) is 16 + psA + paX“ + P3A3 + p212 + pid+ po = 0, (2.11)
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
![2
The local stability of the solu tions
In this section, the local stability of the solutions of Eq.(1.1) is investigated.
The equilibrium point ỹ of Eq.(1.1) is the positive solution of the equation
ỹ = Aỹ +
(2.6)
Then, the only positive equilibrium point of Eq.(1.1) is given by
=
(2.7)
(1 – A) (E B:)
i=1
provided that A < 1. Now, let us introduce a continuous function
(0, 00)6 → (0, ∞) which is defined by
H :
H(uo, ....
,u5)
Auo +
(2.8)
Therefore, it follows that
H(uo,..us)
A,
H(uo,.,u5)
H(uo,.,us)
a2
%3D
H(u0,...,u5)
H(uo,...,us)
as E(Bu)+Bs us] - Ba E (au4)+asus
H(u0,...,us)
dus
as C (Bu)] - Bs (E (asu)]
(E (94))
Consequently, we get
8H(ỹ,...î) = A=- P5,
(1–4)[a1 ( E, ) – Bi ( E,s)]
(E-1)( Bi)
=- P4,
(1–A)[a2 ( Bi+E-3 P1) - B2 (an+ -, as)]
=- P3,
(1-A)[as ( -1 B+E, Bt) – Ba (E-, au+ E,as)]
= - P2,
(1-A)[a4 (Bs +E- 4) - Ba (as+ E, as)]
= - P1,
(1-A)[as (E Bi) – Bs ( i)]
=- Po.
(2.9)
Hence, the linearized equation of Eq.(1.1) about ỹ takes the form
Ym+1+P5Ym +P4Ym-1+P3Ym-2+P2Ym-3+P1Ym-4+Poym-5 = 0, (2.10)
where po, p1, P2, P3, P4 and p5 are given by (2.9).
The characteristic equation associated with Eq.(2.10) is
26+ ps15+ pad+ p3d3 + p2d² + pid+ po = 0,
(2.11)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F0d32c770-62e9-4807-937f-1cebf2133eb2%2F5phprhi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2
The local stability of the solu tions
In this section, the local stability of the solutions of Eq.(1.1) is investigated.
The equilibrium point ỹ of Eq.(1.1) is the positive solution of the equation
ỹ = Aỹ +
(2.6)
Then, the only positive equilibrium point of Eq.(1.1) is given by
=
(2.7)
(1 – A) (E B:)
i=1
provided that A < 1. Now, let us introduce a continuous function
(0, 00)6 → (0, ∞) which is defined by
H :
H(uo, ....
,u5)
Auo +
(2.8)
Therefore, it follows that
H(uo,..us)
A,
H(uo,.,u5)
H(uo,.,us)
a2
%3D
H(u0,...,u5)
H(uo,...,us)
as E(Bu)+Bs us] - Ba E (au4)+asus
H(u0,...,us)
dus
as C (Bu)] - Bs (E (asu)]
(E (94))
Consequently, we get
8H(ỹ,...î) = A=- P5,
(1–4)[a1 ( E, ) – Bi ( E,s)]
(E-1)( Bi)
=- P4,
(1–A)[a2 ( Bi+E-3 P1) - B2 (an+ -, as)]
=- P3,
(1-A)[as ( -1 B+E, Bt) – Ba (E-, au+ E,as)]
= - P2,
(1-A)[a4 (Bs +E- 4) - Ba (as+ E, as)]
= - P1,
(1-A)[as (E Bi) – Bs ( i)]
=- Po.
(2.9)
Hence, the linearized equation of Eq.(1.1) about ỹ takes the form
Ym+1+P5Ym +P4Ym-1+P3Ym-2+P2Ym-3+P1Ym-4+Poym-5 = 0, (2.10)
where po, p1, P2, P3, P4 and p5 are given by (2.9).
The characteristic equation associated with Eq.(2.10) is
26+ ps15+ pad+ p3d3 + p2d² + pid+ po = 0,
(2.11)
![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%2F0d32c770-62e9-4807-937f-1cebf2133eb2%2Fqg9sdms_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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