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) Σ β Then, the only positive equilibrium point of Eq.(1.1) is given by (2.7) (1 – A) (E B:) ri=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) ('n'g) Therefore, it follows that H(uo,...,us) duo = A, H(uo,.,us) H(uo,...,u5) du2 duz (E(A))* H(uo,...,u5) dug a4 E- (Bru1)+Bous] - Ba E(aqu4)+a5us (EL(A4)* H(uo…...,us) dug Consequently, we get = A = - P5, (1–A)[a1 ( Ei, B ) – Bi ( E,as)] =- P4, OH(ỹ.....) (1–A)[a2 ( Bi+E-3 Ps) – B2 (ai+ E,«)] = - P3, (1-A)[as ( E-1 B+E-4 8:) – B3 (E-, ar+ E=4as)] =- P2, dug (1–4)[as (Bs +E-, Bi) – Ba (as+ Ci-gai)] %3D = - P1, OH(ỹ,....§) ðug (1–A)[as (Ei1 Bi) – As (E, a«)] (E 4)( E- Bu) =- 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 do + psdo + paXª + p3d³ + p2d² + P1d+ 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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2 The local stability of the solutions
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:)
vi=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,...,ug)
dug
a2
%3D
H(u0,...,u5)
H(uo,...us)
as (E (Bru)+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, Di) – Bi ( E)]
=- P4,
OH(ỹ...)
(1–A)[a2 ( Bi+E=3 P1) - B2 (ait - at)]
=- P3,
(1-A) [a3 ( E- 8i+E4 B:) – Ba (E=, ast Ea)]
= - P2,
(1-A)[a4 (Bs +E- 84) – B4 (as+ E, as)]
=- P1,
(1-A)[as (E Bi) – Bs ( 4)]
=- 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
18 + ps15+ paXª + p3^3 + p2d² + eid + 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.

Expert Solution

Step 1

Consider the given partial derivatives ,

$\begin{array}{rcl} \frac{\partial H (u_{0}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5})}{\partial u_{0}} & = & A \\ \frac{\partial H (u_{0}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5})}{\partial u_{1}} & = & \frac{α_{1} [\sum_{i = 2}^{5} (β_{i} u_{i})] - β_{1} [\sum_{i = 2}^{5} (α_{i} u_{i})]}{{(\sum_{i = 1}^{5} (β_{i} u_{i}))}^{2}} \\ \frac{\partial H (u_{0}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5})}{\partial u_{2}} & = & \frac{α_{2} [β_{1} u_{1} + \sum_{i = 3}^{5} (β_{i} u_{i})] - β_{2} [α_{1} u_{1} + \sum_{i = 3}^{5} (α_{i} u_{i})]}{{(\sum_{i = 1}^{5} (β_{i} u_{i}))}^{2}} \\ \frac{\partial H (u_{0}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5})}{\partial u_{3}} & = & \frac{α_{3} [\sum_{i = 1}^{2} (β_{i} u_{i}) + \sum_{i = 4}^{5} (β_{i} u_{i})] - β_{3} [\sum_{i = 1}^{2} (α_{i} u_{i}) + \sum_{i = 4}^{5} (α_{i} u_{i})]}{{(\sum_{i = 1}^{5} (β_{i} u_{i}))}^{2}} \\ \frac{\partial H (u_{0}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5})}{\partial u_{4}} & = & \frac{α_{4} [\sum_{i = 1}^{3} (β_{i} u_{i}) + (β_{5} u_{5})] - β_{4} [\sum_{i = 1}^{3} (α_{i} u_{i}) + (α_{5} u_{5})]}{{(\sum_{i = 1}^{5} (β_{i} u_{i}))}^{2}} \\ \frac{\partial H (u_{0}, u_{1}, u_{2}, u_{3}, u_{4}, u_{5})}{\partial u_{5}} & = & \frac{α_{5} [\sum_{i = 1}^{4} (β_{i} u_{i})] - β_{5} [\sum_{i = 1}^{4} (α_{i} u_{i})]}{{(\sum_{i = 1}^{5} (β_{i} u_{i}))}^{2}} \end{array}$

also the only positive equilibrium point $\tilde{y}$ is given by

$\tilde{y} = \frac{\sum_{i = 1}^{5} α_{i}}{(1 - A) (\sum_{i = 1}^{5} β_{i})}$

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