The equilibrium point of Eq.(1) is the positive solution of the equation X= (A+B+C+D)x+ (6) (d– e)x where d + e. If [(A+B+C+D) – 1] (e – d) > 0, then the only positive equilibrium point ž of Eq. (1) is given by b (7) [(A+B+C+D) – 1] (e – d)
The equilibrium point of Eq.(1) is the positive solution of the equation X= (A+B+C+D)x+ (6) (d– e)x where d + e. If [(A+B+C+D) – 1] (e – d) > 0, then the only positive equilibrium point ž of Eq. (1) is given by b (7) [(A+B+C+D) – 1] (e – d)
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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![2 The local stability of the solutions
The equilibrium point x of Eq.(1) is the positive solution
of the equation
X= (A+B+C+D)x+
(6)
(d- e)x'
|
where d+ e. If [(A+B+C+D) – 1] (e – d) > 0, then
the only positive equilibrium point ã of Eq.(1) is given by
X=
[(A+B+C+D) – 1] (e – d) '
(7)
Let
us
now
introduce
a
continuous
function
F: (0,0)4
→ (0,0) which is defined by
6.
bu
(du1 – euz)
(8)
F(uo, U1, U2, U3) = Auo + Buj +Cu2 +Du3 +
provided du # eu2. Consequently, we get
ƏF(x,x,X,X)
= A= Po,
On e
е (А+В+С+D) -1]
(е — d)
ƏF(x,X,X,X)
= B-
= P1,
|
Ine
(9)
ƏF(x,x,x,X)
e[(A+B+C+D) –1]
(e – d)
= C+
2ne
ƏF(x,x,X,X) – D=P3,
En e
where e+ d. Thus, the linearized equation of Eq.(1) about
x takes the form
Zn+1- Pozn- P1Zn–k- P2Zn–1- P3Zn-o = 0,
(10)
where po, P1, P2 and P3 are given by (9).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F47d34d77-9a40-4ab7-92d8-57c5ab376e1a%2F3o5g9o_processed.png&w=3840&q=75)
Transcribed Image Text:2 The local stability of the solutions
The equilibrium point x of Eq.(1) is the positive solution
of the equation
X= (A+B+C+D)x+
(6)
(d- e)x'
|
where d+ e. If [(A+B+C+D) – 1] (e – d) > 0, then
the only positive equilibrium point ã of Eq.(1) is given by
X=
[(A+B+C+D) – 1] (e – d) '
(7)
Let
us
now
introduce
a
continuous
function
F: (0,0)4
→ (0,0) which is defined by
6.
bu
(du1 – euz)
(8)
F(uo, U1, U2, U3) = Auo + Buj +Cu2 +Du3 +
provided du # eu2. Consequently, we get
ƏF(x,x,X,X)
= A= Po,
On e
е (А+В+С+D) -1]
(е — d)
ƏF(x,X,X,X)
= B-
= P1,
|
Ine
(9)
ƏF(x,x,x,X)
e[(A+B+C+D) –1]
(e – d)
= C+
2ne
ƏF(x,x,X,X) – D=P3,
En e
where e+ d. Thus, the linearized equation of Eq.(1) about
x takes the form
Zn+1- Pozn- P1Zn–k- P2Zn–1- P3Zn-o = 0,
(10)
where po, P1, P2 and P3 are given by (9).
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn–k
[dxp-k– exp-1]
Xn+1 =
Axn+ Bxn–k+ Cxn-1+Dxn-o +
n= 0,1,2, ...
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X-6.……., X_1,..., X_k ….., X_1, Xo are arbitrary positive real
numbers such that k < 1< o. Note that the special cases
of Eq.(1) have been studied in [1] when B=C= D=0,
and k=0,1=1, b is replaced by – b and in [27] when
B= C= D=0, and k= 0, b is replaced by – b and in
[33] when B = C = D = 0, 1 = 0 and in [32] when
A=C=D=0, 1=0, b is replaced by – b.
(1)
%3D
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F47d34d77-9a40-4ab7-92d8-57c5ab376e1a%2Fxwlr0c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn–k
[dxp-k– exp-1]
Xn+1 =
Axn+ Bxn–k+ Cxn-1+Dxn-o +
n= 0,1,2, ...
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X-6.……., X_1,..., X_k ….., X_1, Xo are arbitrary positive real
numbers such that k < 1< o. Note that the special cases
of Eq.(1) have been studied in [1] when B=C= D=0,
and k=0,1=1, b is replaced by – b and in [27] when
B= C= D=0, and k= 0, b is replaced by – b and in
[33] when B = C = D = 0, 1 = 0 and in [32] when
A=C=D=0, 1=0, b is replaced by – b.
(1)
%3D
%3D
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Given:
......
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..........
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..........
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