Theorem 14.If 0< A+B+C+D<1 and e + d, then the equilibrium point x given by (7) of Eq.(1) is global attractor. Proof. We consider the following function by (50) (dy – ez)' F(x,y, z, w) = Ax+By+Cz+ Dw+ where dy + ez, provided that B(dy – ez)²>bez and C(dy– ez)2 + bey>0. It is easy to verify the condition (i) of Theorem 3. Let us now verify the condition (ii) of Theorem 3 as follows: b [F(x,x, x, x) – x] (x– F) = {. Kgx- -d е — b {x- [Ks – 1] (e- d) X Sx(e-d)[Kg – 1] - b e-d 1 (51) [K5 – 1]' where K5 = (A+B+C+D). Since 0 < K5 < 1 and e# d, then we deduce from (51) that [F(х, х, х, х) — х] (х— X) <0. (52) According to Theorem 3, is global attractor. Thus, the proof is now completed.O On combining the two Theorems 4 and 14, we have the result.
Theorem 14.If 0< A+B+C+D<1 and e + d, then the equilibrium point x given by (7) of Eq.(1) is global attractor. Proof. We consider the following function by (50) (dy – ez)' F(x,y, z, w) = Ax+By+Cz+ Dw+ where dy + ez, provided that B(dy – ez)²>bez and C(dy– ez)2 + bey>0. It is easy to verify the condition (i) of Theorem 3. Let us now verify the condition (ii) of Theorem 3 as follows: b [F(x,x, x, x) – x] (x– F) = {. Kgx- -d е — b {x- [Ks – 1] (e- d) X Sx(e-d)[Kg – 1] - b e-d 1 (51) [K5 – 1]' where K5 = (A+B+C+D). Since 0 < K5 < 1 and e# d, then we deduce from (51) that [F(х, х, х, х) — х] (х— X) <0. (52) According to Theorem 3, is global attractor. Thus, the proof is now completed.O On combining the two Theorems 4 and 14, we have the result.
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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Show me the steps of determine green and inf is here
![Theorem 14.If 0< A+B+ C+D<1 and e+ d, then
the equilibrium point x given by (7) of Eq.(1) is global
attractor.
Proof.We consider the following function
by
F(x,y, z, w) = Ax+By+Cz+ Dw+
(dy– ez)
(50)
where dy + ez, provided that B(dy – ez)²>bez and
C(dy– ez)? + bey>0. It is easy to verify the condition (i)
of Theorem 3. Let us now verify the condition (ii) of
Theorem 3 as follows:
[F(x,x, x, x) – x] (x- x) =
K5 x-
|
е —
d
t [Ks – 1] (e- d)
Sx(e-d) [K5 – 1]-b
{xc
e-d
1
(51)
[K5 – 1]'
where K5 = (A+B+C+D). Since 0 < K5 < 1 and e+ d,
then we deduce from (51) that
[F(x, х, х, х) — х] (х— х) <0.
(52)
According to Theorem 3, is global attractor. Thus, the
proof is now completed.O
On combining the two Theorems 4 and 14, we have the
result.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F8c6ec6ab-7bfb-4476-933e-740f22808d76%2Fx40xvvo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 14.If 0< A+B+ C+D<1 and e+ d, then
the equilibrium point x given by (7) of Eq.(1) is global
attractor.
Proof.We consider the following function
by
F(x,y, z, w) = Ax+By+Cz+ Dw+
(dy– ez)
(50)
where dy + ez, provided that B(dy – ez)²>bez and
C(dy– ez)? + bey>0. It is easy to verify the condition (i)
of Theorem 3. Let us now verify the condition (ii) of
Theorem 3 as follows:
[F(x,x, x, x) – x] (x- x) =
K5 x-
|
е —
d
t [Ks – 1] (e- d)
Sx(e-d) [K5 – 1]-b
{xc
e-d
1
(51)
[K5 – 1]'
where K5 = (A+B+C+D). Since 0 < K5 < 1 and e+ d,
then we deduce from (51) that
[F(x, х, х, х) — х] (х— х) <0.
(52)
According to Theorem 3, is global attractor. Thus, the
proof is now completed.O
On combining the two Theorems 4 and 14, we have the
result.
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxp-k- exn-i]
(1)
Xn+1
Axn+ Bxn-k+ Cxp–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-,..., X_1,., X_k, •.., X_1, Xo are arbitrary positive real
numbers such that k< 1< 0. 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 :
A=C=D=0, 1=0, b is replaced by – b.
0 and in [32] when
-
b
X=
[(A+B+C+D) – 1](e – d)
(7)
Let
us
now
introduce
continuous
function
F: (0, 00)* → (0,0) which is defined by
ai
bun
Theorem 3.[12]. Consider the difference equation (2).
Let x e I be an equilibrium point of Eq. (2). Suppose also
that
(i) F
arguments.
is a nondecreasing function in each of its
(ii) The function F satisfies the negative feedback
property
[F(x, x, x, x) – x] (x– x) < 0 for all xE I- {x},
where I is an open interval of real numbers. Then x is
global attractor for all solutions of Eq. (2).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F8c6ec6ab-7bfb-4476-933e-740f22808d76%2Fs2qt4h_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- exn-i]
(1)
Xn+1
Axn+ Bxn-k+ Cxp–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-,..., X_1,., X_k, •.., X_1, Xo are arbitrary positive real
numbers such that k< 1< 0. 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 :
A=C=D=0, 1=0, b is replaced by – b.
0 and in [32] when
-
b
X=
[(A+B+C+D) – 1](e – d)
(7)
Let
us
now
introduce
continuous
function
F: (0, 00)* → (0,0) which is defined by
ai
bun
Theorem 3.[12]. Consider the difference equation (2).
Let x e I be an equilibrium point of Eq. (2). Suppose also
that
(i) F
arguments.
is a nondecreasing function in each of its
(ii) The function F satisfies the negative feedback
property
[F(x, x, x, x) – x] (x– x) < 0 for all xE I- {x},
where I is an open interval of real numbers. Then x is
global attractor for all solutions of Eq. (2).
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