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 E, 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 %3D 금 -(²+2e-3) -(e+3) 금 -(c+3) -(4µ) (e²+2e-3) e+3 e+3 1 Q = (28) 1 1 1 9.
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 E, 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 %3D 금 -(²+2e-3) -(e+3) 금 -(c+3) -(4µ) (e²+2e-3) e+3 e+3 1 Q = (28) 1 1 1 9.
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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Transcribed Image Text: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,
where Ln
(Wn, Zn, Wn-1, Zn-1, Wn-2, zn-2) and the Jacobian matrix determined at Eo is equal
-(4u)
(e+2e-3)
e+3
e+3
-(e2+2e-3)
4 µ
-(e+3)
2 u
-(e+3)
2 e
1
(28)
1
1
1
9.
Hence, the characteristic equation of (28) is
P(A) = 1° – ng A° – ng X* – n3 1³ – nz X? – n1 –
- no,
(29)
where,
(2ε+6μ-2εμ + -3)
(2μ (1-0 )
(2ε+4μ-6εμ + 2έμ-2+15)
(2μ (ε-1) )
(2e+3μ+2 εμ- μ+2-12)
(2μ (ε- 1) )
(ε+2μ-2εμ +)
(и (1 — е))
n5
n4 =
n3
n2 =
%3D
(1 – e)
n1
no = -1.
Now, let T(A)
= 1° and V(A) = -ng 15 – n4 X4 – n3 A3 – n2 X? – n1 1 – no. Consider |A|=1,
then one has
(2ε+6μ-2 εμ ε2-3)
(2μ (1 - )
(2ε+4μ-6εμ# 22 μ-
(2 µ (e – 1))
(2ε+6μ-2εμ4 ε-3 )
(2μ (1 - 0 )
(e +2µ – 2 eu + 3)
(µ (1 – e))
(2ε+3μ +2εμ- μ+ 2 2?-12 )
(2µ (€ – 1))
(ε+2μ-2εμ + 3)
(и (1 — е))
(4ε+7μ-4εμ 4 έμ+ +3)
(2μ(ε-1) )
+
e2 + 15)
(1 - €)
+1,
+1,
2
(2ε+ μ-2εμ + μ+6) (E-2μ-2εμ+3)
(2μ (ε -1) )
(-3µ+2€µ+e µ)
(2µ (e – 1))
- E
+
+1,
(и (1 — е))
u (e – 1) (1- e)
u (e – 1)
-3µ +2€µ+ e? µ – µ (e – 1)2
+1,
2
-4μ +4με
+1,
2 µ (e – 1)
+1=
2µ (e – 1)
= 3 > 1.
Then, from Rouché's Theorem 1.2, V(A) and Y(A) + V(A) = P(^) have the same number of
zeros in an open unit disk || < 1, that means there are only five roots lie inside unit disk.
Thus, from Theorem 1.1 the equilibrium point E, is not asymptotically stable. By solving (29),
we find that A
1, implies E, is nonhyperbolic point.
Transcribed Image Text: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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