Theorem 4.1. The positive equilibrium point E, of system (4) is not asymptotically stableand nonhyperbolic point .Proof. The linearized system of (4) evaluated at Eo is written in the matrix form asLn+1Q Lnwhere Ln(Wn, Zn, Wn-1, Zn–1, Wn-2, zn-2)' and the Jacobian matrix determined at Eo is equal-2e-1-(e²+2e-3)-(4 µ)(e?+2 e-3)e+3e+3-(e+3)2 µ-(e+3)1-e2(28)1119. In this paper, we solve and study the properties of the following systemWn-p 2 Zn-hp=0Wn-ph=0E zn-hh=1p=1Wn+1+u and zn+1 =+e,(4)Zn - €Wn - 4where u and e are arbitrary positive real numbers with initial conditions w; and z; for i =-2, –1,0.Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then6+ V – 4 ko k1A- V2 – 4 ko ki1- ki + ko+w2n-2+BTy – Oy -- V2 - 4 ko ki-4 ko ki+ V2 – 4 ko kiV2 - 4 ko kiAko+ Bkow2n-1- 122•((금-1) (는)1– ko + k1+ V2 - 4 vo V1- V2 - 4 VI1- v + vo22n-2+D%3D1+ Vu2 – 4 o VIgb + V2 – 4 vo vỊ11.+D- V2 – 4 v vỊV2 – 4 vo vi22n-1- v + vo1- 0 - VIwhere A, B, C and D are constants defined aswo -H2-1 +z-2koki =0 = ko + ki + ko ki,w-1 + w-220 - e20 - ew-1 +w-2, = vo + vi + vo V1,2-1 +2-21-OmVo2 – 4 ko k1-)") (v - 4ko ki – 4) + 2 (wo - (G ) -).2 62 - 8 ko kiV2- 4 ko ki262 - 8 ko ki1- ko + ki- ko – k1(1 – kj + ko1- k1 - koB =u- w-2+2wn-- vi +uC =-2+ 222 - 8 0 vI- o - VI) [(는1- v1 + o- v0 - v11- v + o2) (V2 – 4vo vI ++22 2-- 0 - v1since ko + ki #1 and vo + vı #1 for n e N.Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the followform2 zn-hWn-pp=1Wn+1h=1Zn+1and(5)1E Wn-p1> Zn-hh=0wn -up-04Then, we assume thatWn+1 - lWn - Hkn+1 =1kn =2Wn-pp=0Wn-pp=1(6)Zn+1 -€Zn - €Vn+1 =Vn2E Zn-h> Zn-hh=1h=0Substituting (6) in (5), we haveZn-hh=11== kn-1,Vnkn+1 =(7)E Wn-pp=1Vn+1 == Vn-1.kn%3DWn -Hence, we see that1ki = -,Vowo - H20 - €kow-1+ w_2Vo =Z-1+ z-2koandat n = 1, k2 = ko andV2 = Vo,at n = 2, k3 = k1 andV3 = V1,at n = 3, k4 = ko andV4 = Vo,at n = 4, ks = k andV5 = V1,(8)at n = 2 n, k2n = ko andV2n = Vo,at n = 2n + 1, k2n+1 = k, andV2n+1 = V1.Now, from the relations in (6) we getWn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2).(9)Using (8) in (9), we haveW2n = H+ ko (w2n-1 + w2n-2),W2n+1 = µ+ k1 (w2n + w2n-1),(10)Z2n = €+ vo (z2n-1 + 2n-2),22n+1 = €+ V1 (22n + 22n-1).

Theorem 4.1. The positive equilibrium point Eo of system (4) is not asymptotically stable and nonhyperbolic point . Proof. The linearized system of (4) evaluated at Eo is written in the matrix form as Ln+1 = Q Ln,

Theorem 4.1. The positive equilibrium point Eo of system (4) is not asymptotically stable and nonhyperbolic point . Proof. The linearized system of (4) evaluated at Eo is written in the matrix form as Ln+1 = Q Ln,

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Linear Programming Concepts

Linear programming is a famous mathematical modeling tool for determining the best distribution of scarce resources among competing demands. Machinery, time, personnel, raw materials, money, and space are all resources. It is used to find the most optimal solution to a problem with given constraints. The real-life situations can be formulated using linear programming concepts into a mathematical model. Linear programming problems consist of an objective function and linear equalities subjected to constraints.

Question

Theorem 4.1. The positive equilibrium point E, of system (4) is not asymptotically stable
and nonhyperbolic point .
Proof. The linearized system of (4) evaluated at Eo is written in the matrix form as
Ln+1
Q Ln
where Ln
(Wn, Zn, Wn-1, Zn–1, Wn-2, zn-2)' and the Jacobian matrix determined at Eo is equal
-2
e-1
-(e²+2e-3)
-(4 µ)
(e?+2 e-3)
e+3
e+3
-(e+3)
2 µ
-(e+3)
1-e
2
(28)
1
1
1
9.

In this paper, we solve and study the properties of the following system
Wn-p 2 Zn-h
p=0
Wn-p
h=0
E zn-h
h=1
p=1
Wn+1
+u and zn+1 =
+e,
(4)
Zn - €
Wn - 4
where u and e are arbitrary positive real numbers with initial conditions w; and z; for i =
-2, –1,0.
Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then
6+ V – 4 ko k1
A
- V2 – 4 ko ki
1- ki + ko
+
w2n-2
+B
Ty – Oy -
- V2 - 4 ko ki
-4 ko ki
+ V2 – 4 ko ki
V2 - 4 ko ki
A
ko
+ B
ko
w2n-1
- 1
2
2
•((금-1) (는)
1– ko + k1
+ V2 - 4 vo V1
- V2 - 4 VI
1- v + vo
22n-2
+D
%3D
1
+ Vu2 – 4 o VI
gb + V2 – 4 vo vỊ
1
1.
+D
- V2 – 4 v vỊ
V2 – 4 vo vi
22n-1
- v + vo
1- 0 - VI
where A, B, C and D are constants defined as
wo -H
2-1 +z-2
ko
ki =
0 = ko + ki + ko ki,
w-1 + w-2
20 - e
20 - e
w-1 +w-2
, = vo + vi + vo V1,
2-1 +2-2
1-Om
Vo2 – 4 ko k1
-)") (v - 4ko ki – 4) + 2 (wo - (G ) -).
2 62 - 8 ko ki
V2- 4 ko ki
262 - 8 ko ki
1- ko + ki
- ko – k1
(1 – kj + ko
1- k1 - ko
B =
u- w-2
+2
wn-
- vi +u
C =
-2
+ 2
22 - 8 0 vI
- o - VI
) [(는
1- v1 + o
- v0 - v1
1- v + o
2) (V2 – 4vo vI +
+2
2 2
-- 0 - v1
since ko + ki #1 and vo + vı #1 for n e N.
Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow
form
2 zn-h
Wn-p
p=1
Wn+1
h=1
Zn+1
and
(5)
1
E Wn-p
1
> Zn-h
h=0
wn -u
p-0
4
Then, we assume that
Wn+1 - l
Wn - H
kn+1 =
1
kn =
2
Wn-p
p=0
Wn-p
p=1
(6)
Zn+1 -€
Zn - €
Vn+1 =
Vn
2
E Zn-h
> Zn-h
h=1
h=0
Substituting (6) in (5), we have
Zn-h
h=1
1
== kn-1,
Vn
kn+1 =
(7)
E Wn-p
p=1
Vn+1 =
= Vn-1.
kn
%3D
Wn -
Hence, we see that
1
ki = -,
Vo
wo - H
20 - €
ko
w-1+ w_2
Vo =
Z-1+ z-2
ko
and
at n = 1, k2 = ko and
V2 = Vo,
at n = 2, k3 = k1 and
V3 = V1,
at n = 3, k4 = ko and
V4 = Vo,
at n = 4, ks = k and
V5 = V1,
(8)
at n = 2 n, k2n = ko and
V2n = Vo,
at n = 2n + 1, k2n+1 = k, and
V2n+1 = V1.
Now, from the relations in (6) we get
Wn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2).
(9)
Using (8) in (9), we have
W2n = H+ ko (w2n-1 + w2n-2),
W2n+1 = µ+ k1 (w2n + w2n-1),
(10)
Z2n = €+ vo (z2n-1 + 2n-2),
22n+1 = €+ V1 (22n + 22n-1).

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