Show me the steps of deremine green 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 followingsystem of difference equationsУл, Yn+1 = A + B-Yn-1Xn+1 = A + B-(1)where A and B are positive numbers and the initial values are positive numbers. InFirst of all, we consider the change of the variables for system (1) as follows:XnУпtn, Zn =AAFrom this, system (1) transform into following system:tnZn, Zn+1 1+PZn-1tn+1 = 1+P2(2)'n-1where p => 0. From now on, we study the system (2).Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is(1+T+4p 1+ VI+4p(7, 2) =( -%3DNow, we consider a transformation as follows:(tn, tn-1, Zn, Zn-1) →(f, fi, 8, 81)where f = 1+ p, fi = tn, 8 = 1+p,, g1 = Zn. Thus we get the jacobiann-1matrix about equilibrium point (7, z):1 0B (f, ३) =0 01Thus, the linearized system of system (2) about the unique positive equilibrium pointis given by XN+1= B (ī, 2) XN, where-)Intn-1XN =uzZn-10 0 물글0 01B (f, ३)0 00 01Hence, the characteristic equation of B (7, z) about the unique positive equilibriumpoint (7, 2) is4p2.4p2= 0.Due to t = 7, we can rearrange the characteristic equation such that4p2.4p2= 0.741474Therefore, we obtain the four roots of characteristic equation as follows:p+Vp²- Vp² – 8pr?212Vp² – 8pi?27212 =-p+Vp? + 8pi27213 =

From equation (2), we have tn+1=1+pznzn-12 ...(2a)and…

Answered: (1+T+4p 1+ /T+4p (F, 2) = %3D 2 2

Math

Advanced Math

(1+T+4p 1+ /T+4p (F, 2) = %3D 2 2

Algebra for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter13: Conic Sections

Section13.5: Systems Involving Nonlinear Equations

Problem 39PS

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Question

Show me the steps of deremine green 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
, Yn+1 = A + B-
Yn-1
Xn+1 = A + B-
(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
Уп
tn
, Zn =
A
A
From this, system (1) transform into following system:
tn
Zn
, Zn+1 1+P
Zn-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+T+4p 1+ VI+4p
(7, 2) =
( -
%3D
Now, we consider a transformation as follows:
(tn, tn-1, Zn, Zn-1) →
(f, fi, 8, 81)
where f = 1+ p, fi = tn, 8 = 1+p,
, g1 = Zn. Thus we get the jacobian
n-1
matrix about equilibrium point (7, z):
1 0
B (f, ३) =
0 0
1
Thus, the linearized system of system (2) about the unique positive equilibrium point
is given by XN+1
= B (ī, 2) XN, where
-)
In
tn-1
XN =
uz
Zn-1
0 0 물글
0 0
1
B (f, ३)
0 0
0 0
1
Hence, the characteristic equation of B (7, z) about the unique positive equilibrium
point (7, 2) is
4p2.
4p2
= 0.
Due to t = 7, we can rearrange the characteristic equation such that
4p2.
4p2
= 0.
74
14
74
Therefore, we obtain the four roots of characteristic equation as follows:
p+Vp²
- Vp² – 8pr?
212
Vp² – 8pi?
272
12 =
-p+Vp? + 8pi
272
13 =

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