From which we have (a₁ + a2) D+ (a3 + a4 + α5) d-(1 − A) (B₁ + B₂) D² = (1 − A) (B3 + B4 + B5) Dd (5.38) and (a1 + a2) d + (a3 + α4 + α5) D-(1 − A) (B₁ + B₂) d² = (1 - A) (B3 +34 +35) Dd (5.39) From (5.38) and (5.39), we obtain (d-D) {[(a₁ + a₂) - (a3+04+05)] − (1 − A) (B₁ + B₂) (d+ D)} = = 0. (5.40) Since A 1 and (a3 + a4 +a5) ≥ (a1 + a2), we deduce from (5.40) that D=d. It follows by Theorem 2. that u of Eq.(1.1) is a global attractor.
From which we have (a₁ + a2) D+ (a3 + a4 + α5) d-(1 − A) (B₁ + B₂) D² = (1 − A) (B3 + B4 + B5) Dd (5.38) and (a1 + a2) d + (a3 + α4 + α5) D-(1 − A) (B₁ + B₂) d² = (1 - A) (B3 +34 +35) Dd (5.39) From (5.38) and (5.39), we obtain (d-D) {[(a₁ + a₂) - (a3+04+05)] − (1 − A) (B₁ + B₂) (d+ D)} = = 0. (5.40) Since A 1 and (a3 + a4 +a5) ≥ (a1 + a2), we deduce from (5.40) that D=d. It follows by Theorem 2. that u of Eq.(1.1) is a global attractor.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show me the steps of determine green and the inf is here
![The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+a2Ym-2 + a3ym-3+ a4Ym-4 + a5Ym-5
Aym+
B1ym-1 + B2ym-2 + B3Ym-3 + B4Ym-4 + BsYm-5
т %3D 0, 1, 2, ...,
Ym+1 =
(1.1)
where the coefficients A, ai, Bi E (0, 00), i = 1, ..., 5, while the initial condi-
tions y-5,y-4,Y–3,Y-2, y-1, yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az =
B4
when a4 = B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.
B3 = a4 =
B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
= a5 =
Theorem 2 ([6). Let H : [a, b]k+1 → [a, b] be a continuous function, where
k is a positive integer, and where [a, b] is an interval of real numbers. Con-
sider the difference equation (1.2). Suppose that H satisfies the following
conditions:
1. For each integer i with1 < i < k+ 1; the function H(z1, z2, ..., Zk+1)
is weakly monotonic in zi for fixed z1, z2, ..., Zi-1, Zi+1, ..., Zk+1•
2. If (d, D) is a solution of the system
d = H(d1, d2, ., de+1) and D= H(D1, D2, .., Dk+1),
then d = D, where for each i = 1, 2,
..., k +1, we set
d
di = {
if F is non – decreasing in zi
if F is non – increasing in zi
D
аnd
{
(D if F is non – decreasing in z;
Di =
if
F is non – increasing in zị.
Then there exists exuctly one equilibrium y of Eq.(1.2), and every solution
of Eq. (1.2) converges to y.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F4ddcc85c-7f63-44f8-bf56-1e8822cae6db%2Fv4a955h_processed.png&w=3840&q=75)
Transcribed Image Text:The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+a2Ym-2 + a3ym-3+ a4Ym-4 + a5Ym-5
Aym+
B1ym-1 + B2ym-2 + B3Ym-3 + B4Ym-4 + BsYm-5
т %3D 0, 1, 2, ...,
Ym+1 =
(1.1)
where the coefficients A, ai, Bi E (0, 00), i = 1, ..., 5, while the initial condi-
tions y-5,y-4,Y–3,Y-2, y-1, yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az =
B4
when a4 = B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.
B3 = a4 =
B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
= a5 =
Theorem 2 ([6). Let H : [a, b]k+1 → [a, b] be a continuous function, where
k is a positive integer, and where [a, b] is an interval of real numbers. Con-
sider the difference equation (1.2). Suppose that H satisfies the following
conditions:
1. For each integer i with1 < i < k+ 1; the function H(z1, z2, ..., Zk+1)
is weakly monotonic in zi for fixed z1, z2, ..., Zi-1, Zi+1, ..., Zk+1•
2. If (d, D) is a solution of the system
d = H(d1, d2, ., de+1) and D= H(D1, D2, .., Dk+1),
then d = D, where for each i = 1, 2,
..., k +1, we set
d
di = {
if F is non – decreasing in zi
if F is non – increasing in zi
D
аnd
{
(D if F is non – decreasing in z;
Di =
if
F is non – increasing in zị.
Then there exists exuctly one equilibrium y of Eq.(1.2), and every solution
of Eq. (1.2) converges to y.
![Case 3. Let the function H(uo, ..., u5) is non-decreasing in uo,u1,u2 and
non-increasing in u3, U4, U5.
Suppose that (d, D) is a solution of the system
H(D,D, D, d, d, d)
аnd
d = H(d, d, d, D, D, D).
D =
Then we get
aD+a2D+azd + a4d + a5d
BịD+ B2D+ B3d + B4d + B5d
a1d + a2d + a3D+a4D+a5D
Bid + Bad + B3D+ B4D+ B;D
D = AD+
аnd d — Ad+
or
D (1 – A) =
(a1 + a2) D+ (a3 + a4 + a5) d
(В1 + В2) D + (Bз + Ва + Bs) d
(а1 + 02) d + (аз + а4 + as) D
(B1 + B2) d+ (B3 + B4 + B5) D
and d(1 – A) =
18
621
A. M. Alotaibi ET AL 604-627
COMPU
ONAL ANALYSIS AND APPLICATIONS, VOL. 26, NO.4, 2019, COPYRIGHT 2019 EUDOXUS PRESS, LLC
From which we have
(а1 + aэ) D + (aз + aд t as) d- (1— А) (31 + Bә) D" — (1 — А) (3з + Bа + В5) Dd
(5.38)
and
(ат + 09) d + (aз + ад + a5) D-(1 — А) (31 + B2) d? — (1 — А) (83 + Ba + B5) Dd
(5.39)
From (5.38) and (5.39), we obtain
(а — D) {(aл + a2) — (аз + од +as)] — (1 — А) (B1 + B) (d + D)} —D0.
(5.40)
Since A < 1 and (a3 + a4 + a5) > (a1 + a2), we deduce from (5.40) that
D = d. It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F4ddcc85c-7f63-44f8-bf56-1e8822cae6db%2Fwgw2l38_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Case 3. Let the function H(uo, ..., u5) is non-decreasing in uo,u1,u2 and
non-increasing in u3, U4, U5.
Suppose that (d, D) is a solution of the system
H(D,D, D, d, d, d)
аnd
d = H(d, d, d, D, D, D).
D =
Then we get
aD+a2D+azd + a4d + a5d
BịD+ B2D+ B3d + B4d + B5d
a1d + a2d + a3D+a4D+a5D
Bid + Bad + B3D+ B4D+ B;D
D = AD+
аnd d — Ad+
or
D (1 – A) =
(a1 + a2) D+ (a3 + a4 + a5) d
(В1 + В2) D + (Bз + Ва + Bs) d
(а1 + 02) d + (аз + а4 + as) D
(B1 + B2) d+ (B3 + B4 + B5) D
and d(1 – A) =
18
621
A. M. Alotaibi ET AL 604-627
COMPU
ONAL ANALYSIS AND APPLICATIONS, VOL. 26, NO.4, 2019, COPYRIGHT 2019 EUDOXUS PRESS, LLC
From which we have
(а1 + aэ) D + (aз + aд t as) d- (1— А) (31 + Bә) D" — (1 — А) (3з + Bа + В5) Dd
(5.38)
and
(ат + 09) d + (aз + ад + a5) D-(1 — А) (31 + B2) d? — (1 — А) (83 + Ba + B5) Dd
(5.39)
From (5.38) and (5.39), we obtain
(а — D) {(aл + a2) — (аз + од +as)] — (1 — А) (B1 + B) (d + D)} —D0.
(5.40)
Since A < 1 and (a3 + a4 + a5) > (a1 + a2), we deduce from (5.40) that
D = d. It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor.
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