Now, we consider a transformation as follows: (In, In-1, Zn, Zn-1) → (f, fi, 8, g1) where f = 1+ p, fi = tn, 8 =1+ p, 81 = zn. Thus we get the jacobian matrix about equilibrium point (7, 2): %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Show me the steps of deremine yellow and inf is here i need evey I need all the details step by step and inf is here

Motivated by difference equations and their systems, we consider the following
system of difference equations
Ул
Xn
Xn+1 = A + B-
2
, Yn+1 = A + B-
2
(1)
Yn-1
where A and B are positive numbers and the initial values are positive numbers. In
First of all, we consider the change of the variables for system (1) as follows:
Xn
tn =
Уп
, Zn =
A
A
From this, system (1) transform into following system:
tn
Zn
Zn+1 = 1+ p
n-1
tn+1 = 1+P2
(2)
n-1
where
P =
> 0. From now on, we study the system (2).
Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is
(1+ /I+4p 1+ VI+4p
(f,३) =
2
Now, we
sider a transfo
as follows:
(In, In-1, Zn, Zn-1) → (f, f1, g, g1)
where f = 1+ p, fi = tn, 8 = 1+ p, 81 = Zn. Thus we get the jacobian
matrix about equilibrium point (7, z):
n-I
'n-1
0 0
1
B (7,३) =
0 0
1
Thus, the linearized system of system (2) about the unique positive equilibrium point
is given by XN+1 =
B (i, z) X N, where
In
tn-1
XN =
uz
Zn-1
0 0 울글
1
0 0
B (i, 2):
-21
0 0
1
Hence, the characteristic equation of B (î, z) about the unique positive equilibrium
point (7, 2) is
4p?
4p2
= 0.
Due to t = 7, we can rearrange the characteristic equation such that
4p2
2²+
74
4p2
= 0.
14
Therefore, we obtain the four roots of characteristic equation as follows:
p+ Vp? – 8pi?
212
A1 =
Vp² – 8 pi?
272
P-VP-
12 =
-p+Vp? + 8pi?
212
13 =
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Ул Xn Xn+1 = A + B- 2 , Yn+1 = A + B- 2 (1) Yn-1 where A and B are positive numbers and the initial values are positive numbers. In First of all, we consider the change of the variables for system (1) as follows: Xn tn = Уп , Zn = A A From this, system (1) transform into following system: tn Zn Zn+1 = 1+ p n-1 tn+1 = 1+P2 (2) n-1 where P = > 0. From now on, we study the system (2). Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+ /I+4p 1+ VI+4p (f,३) = 2 Now, we sider a transfo as follows: (In, In-1, Zn, Zn-1) → (f, f1, g, g1) where f = 1+ p, fi = tn, 8 = 1+ p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, z): n-I 'n-1 0 0 1 B (7,३) = 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (i, z) X N, where In tn-1 XN = uz Zn-1 0 0 울글 1 0 0 B (i, 2): -21 0 0 1 Hence, the characteristic equation of B (î, z) about the unique positive equilibrium point (7, 2) is 4p? 4p2 = 0. Due to t = 7, we can rearrange the characteristic equation such that 4p2 2²+ 74 4p2 = 0. 14 Therefore, we obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 212 A1 = Vp² – 8 pi? 272 P-VP- 12 = -p+Vp? + 8pi? 212 13 =
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