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 D = H(D,D, D,d, d, d) and d = H(d, d, d, D, D, D). Then we get a¡D+a2D+azd + a4d + azd B1D+ B2D+ B3d + Bad + Bzd ajd+ azd + azD+a4D+agD D = AD+ and d = Ad+ Bid + Bad + B3D+ B4D+ B3D’ or (a1 + a2) D+ (a3 + a4 + a5) d (B1 + B2) D+ (B3 + B4 + Bs) d (a1 + a2) d+ (a3 + a4 + as) D (B1 + B2) d+ (ß3 + B4 + B5) D D (1 – A) = and d(1 - A) = %3D 18 From which we have (a1 + a2) D+ (a3+a4 + a5) d–(1 – A) (ß1 + B2) D² = (1 – A) (ß3 + B4 + B5) Dd (5.38) and (a1 + a2) d+ (a3 + a4 + a5) D-(1- A) (ß1 + B2) d² = (1 – A) (B3 + B4 + B5) Dd (5.39) %3D From (5.38) and (5.39), we obtain (d – D) {[(a1 + a2) – (a3 + a4 + az)] – (1– A) (81 + 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 y of Eq.(1.1) is a global attractor.
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 D = H(D,D, D,d, d, d) and d = H(d, d, d, D, D, D). Then we get a¡D+a2D+azd + a4d + azd B1D+ B2D+ B3d + Bad + Bzd ajd+ azd + azD+a4D+agD D = AD+ and d = Ad+ Bid + Bad + B3D+ B4D+ B3D’ or (a1 + a2) D+ (a3 + a4 + a5) d (B1 + B2) D+ (B3 + B4 + Bs) d (a1 + a2) d+ (a3 + a4 + as) D (B1 + B2) d+ (ß3 + B4 + B5) D D (1 – A) = and d(1 - A) = %3D 18 From which we have (a1 + a2) D+ (a3+a4 + a5) d–(1 – A) (ß1 + B2) D² = (1 – A) (ß3 + B4 + B5) Dd (5.38) and (a1 + a2) d+ (a3 + a4 + a5) D-(1- A) (ß1 + B2) d² = (1 – A) (B3 + B4 + B5) Dd (5.39) %3D From (5.38) and (5.39), we obtain (d – D) {[(a1 + a2) – (a3 + a4 + az)] – (1– A) (81 + 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 y 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
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show me the steps of determine green and the inf is here
![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
D = H(D, D, D, d, d, d)
and
d = H(d, d, d, D, D, D).
Then we get
a1D+ a2D + azd + a4d+ azd
B1D+ B2D + B3d + Bad + Bzd
aid + agd + a3 D+ a4D+ agD
Bid + B2d + B3D + B4D+ BgD'
D = AD+
and d = Ad+
or
D (1 – A) =
(a1 + a2) D+ (a3+a4 + a5) d
(B1 + B2) D+ (B3 + B4 + B5) d
(a1 + a2) d+ (a3 + a4 + a5) D
(B1 + B2) d+ (B3 + B4 + B5) D
and d(1 – A) =
18
From which we have
(а1 + о2) D + (aз + a4 + as)d- (1 — А) (B1 + Bә) D? %3D (1 — A) (Вз + Ва + Bs) Dd
(5.38)
and
(a1 + a2) d+ (a3 + a4 + as) D-(1- A) (B1 + B2) d = (1– A) (B3 + B4 + B3) Dd
(5.39)
From (5.38) and (5.39), we obtain
(d- D) {[(a1 + a2) – (a3 + a4 + a5)] – (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 y 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%2Fb963c7ae-a766-4955-8a42-eaee6c4d9282%2Fpajk5n_processed.png&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
D = H(D, D, D, d, d, d)
and
d = H(d, d, d, D, D, D).
Then we get
a1D+ a2D + azd + a4d+ azd
B1D+ B2D + B3d + Bad + Bzd
aid + agd + a3 D+ a4D+ agD
Bid + B2d + B3D + B4D+ BgD'
D = AD+
and d = Ad+
or
D (1 – A) =
(a1 + a2) D+ (a3+a4 + a5) d
(B1 + B2) D+ (B3 + B4 + B5) d
(a1 + a2) d+ (a3 + a4 + a5) D
(B1 + B2) d+ (B3 + B4 + B5) D
and d(1 – A) =
18
From which we have
(а1 + о2) D + (aз + a4 + as)d- (1 — А) (B1 + Bә) D? %3D (1 — A) (Вз + Ва + Bs) Dd
(5.38)
and
(a1 + a2) d+ (a3 + a4 + as) D-(1- A) (B1 + B2) d = (1– A) (B3 + B4 + B3) Dd
(5.39)
From (5.38) and (5.39), we obtain
(d- D) {[(a1 + a2) – (a3 + a4 + a5)] – (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 y of Eq.(1.1) is a global attractor.
![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
Ут+1 — Аутt
т 3 0, 1, 2, ...,
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5
(1.1)
where the coefficients A, a;, B; E (0, 0), 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 = B3 = a4 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Fb963c7ae-a766-4955-8a42-eaee6c4d9282%2Fr4oetqg_processed.jpeg&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
Ут+1 — Аутt
т 3 0, 1, 2, ...,
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5
(1.1)
where the coefficients A, a;, B; E (0, 0), 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 = B3 = a4 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.
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