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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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Hence, the characteristic equation of (28) is
P(1) = X® – n5 X5 – n4 X* – n3 X3 – n2 X² – n1 1 – no,
(29)
where,
(2ε+6μ-2 εμ+ * 3)
(2μ (1 - 9)
(2ε+4μ- 6 εμ + 2 8μ - έ + 15)
(2μ (ε - 1))
(2 e + 3µ + 2 € µ – e² µ +2 e? – 12)
(2 µ (e – 1))
( ε + 2μ-2 εμ +3)
(µ (1 – e))
n5
n4 =
n3
n2 =
n1
no = -1.
Now, let T(A) = X° and V(A) = -n5 15 – n4 X4 – n3 X3 – n2 X² – nị 1 – no. Consider |A|=1,
then one has
(2ε+3μ + 2 εμ- *μ + 22 - 12)
(2μ (ε= 1))
( ε+2μ-2εμ 3)
(µ (1 – €))
(4ε + 7μ - 4εμ + ε' μ + ?+3 )
(2μ (ε -1) )
(2ε+6μ-2εμ+-3)
(2μ (1 -6) )
(2ε+4μ-6 εμ + 2 ε' μ - 2 + 15)
(2 µ (e – 1))
(2ε+6μ-2εμ + *-3)
(2μ (1 - )
(+2μ-2 εμ +3)
(и (1 — е))
(2 e + µ – 2e u +e² µ + 6)
(2 µ (e – 1))
(-3µ+2 € µ+ e² µ), H(e – 1) (1– €)
(2μ (ε-1) )
-3µ +2 €µ+ e² µ – 1µ (e – 1)²
2 µ (e – 1)
|V(A)|
(1 – e)
+1,
2
(1 – e)
+ 1,
2
(ε+2μ-2 εμ+ 3)
(и (1 — е))
(1 – e)
+1,
2
+1,
µ(e – 1)
-4μ+4με
+1=
+1,
2μ (ε- 1)
3 > 1.
=
Then, from Rouché's Theorem 1.2, V(A) and Y(A) + ¥ (A) = P(X) have the same number of
zeros in an open unit disk JA| < 1, that means there are only five roots lie inside unit disk.
Thus, from Theorem 1.1 the equilibrium point Eo is not asymptotically stable. By solving (29),
we find that A = 1, implies E, is nonhyperbolic point.
Transcribed Image Text:Hence, the characteristic equation of (28) is P(1) = X® – n5 X5 – n4 X* – n3 X3 – n2 X² – n1 1 – no, (29) where, (2ε+6μ-2 εμ+ * 3) (2μ (1 - 9) (2ε+4μ- 6 εμ + 2 8μ - έ + 15) (2μ (ε - 1)) (2 e + 3µ + 2 € µ – e² µ +2 e? – 12) (2 µ (e – 1)) ( ε + 2μ-2 εμ +3) (µ (1 – e)) n5 n4 = n3 n2 = n1 no = -1. Now, let T(A) = X° and V(A) = -n5 15 – n4 X4 – n3 X3 – n2 X² – nị 1 – no. Consider |A|=1, then one has (2ε+3μ + 2 εμ- *μ + 22 - 12) (2μ (ε= 1)) ( ε+2μ-2εμ 3) (µ (1 – €)) (4ε + 7μ - 4εμ + ε' μ + ?+3 ) (2μ (ε -1) ) (2ε+6μ-2εμ+-3) (2μ (1 -6) ) (2ε+4μ-6 εμ + 2 ε' μ - 2 + 15) (2 µ (e – 1)) (2ε+6μ-2εμ + *-3) (2μ (1 - ) (+2μ-2 εμ +3) (и (1 — е)) (2 e + µ – 2e u +e² µ + 6) (2 µ (e – 1)) (-3µ+2 € µ+ e² µ), H(e – 1) (1– €) (2μ (ε-1) ) -3µ +2 €µ+ e² µ – 1µ (e – 1)² 2 µ (e – 1) |V(A)| (1 – e) +1, 2 (1 – e) + 1, 2 (ε+2μ-2 εμ+ 3) (и (1 — е)) (1 – e) +1, 2 +1, µ(e – 1) -4μ+4με +1= +1, 2μ (ε- 1) 3 > 1. = Then, from Rouché's Theorem 1.2, V(A) and Y(A) + ¥ (A) = P(X) have the same number of zeros in an open unit disk JA| < 1, that means there are only five roots lie inside unit disk. Thus, from Theorem 1.1 the equilibrium point Eo is not asymptotically stable. By solving (29), we find that A = 1, implies E, is nonhyperbolic point.
In this paper, we solve and study the properties of the following system
É Wn-P
E zn-h
Wn-p
Zn-h
h=1
p=0
p=1
h=0
Wn+1
and
Zn+1
+E,
(4)
Zn - €
Wn - H
where u and e are arbitrary positive real numbers with initial conditions w; and z; for i =
-2, –1,0.
The nontrivial positive equilibrium point of (4) is Eo = (w, z) = (A, 1) such that e > 1.
e+3
T
((Wn + Wn-1) (žn-1 + Zn–2)
wn-1+Wn-2) (2n + 2n-1)
Ln+1
+ µ, Wn, Wn-1,
+ €, Zn, Zn–1
Zn - E
Wn - µ
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
-(4µ)
(e2+2 €-3)
1-e
2
-2
금
e+3
e+3
-(e2+2€-3)
-(e+3)
2 µ
-(e+3)
2 µ
1-€
2
1
(28)
1
1
1
9.
Transcribed Image Text:In this paper, we solve and study the properties of the following system É Wn-P E zn-h Wn-p Zn-h h=1 p=0 p=1 h=0 Wn+1 and Zn+1 +E, (4) Zn - € Wn - H where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. The nontrivial positive equilibrium point of (4) is Eo = (w, z) = (A, 1) such that e > 1. e+3 T ((Wn + Wn-1) (žn-1 + Zn–2) wn-1+Wn-2) (2n + 2n-1) Ln+1 + µ, Wn, Wn-1, + €, Zn, Zn–1 Zn - E Wn - µ 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 -(4µ) (e2+2 €-3) 1-e 2 -2 금 e+3 e+3 -(e2+2€-3) -(e+3) 2 µ -(e+3) 2 µ 1-€ 2 1 (28) 1 1 1 9.
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