From which we have (a1 + a3) D+ (a2 +a4+ a5) d-(1– A) (B1 + B3) D² = (1 – A) (B2 + B4 + B5) Dd (5.41) and (a1 + a3) d+ (a2 + a4 + a5) D-(1- A) (B1 + B3) d = (1 – A) (B2 + B4 + ß5) Dd (5.42) From (5.41) and (5.42), we obtain C(d – D) {[(a1 + a3) – (a2 + a4 + a5)] – (1 – A) (B1 + 33) (d + D)} = 0. (5.43) Since A < 1 and (a2 + a4 + a5) > (a1 + a3), we deduce from (5.43) 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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Case 4. Let the function H(uo, ..., u5) is non-decreasing in uo,u1,u3 and
non-increasing in u2, 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
aD+a2d+ a3D+a4d + a5d
B1D+ B2d + B3D+ B4d + Bzd
a1d + a2D+ 3d + a4D+ a5D
Bid + B2D + B3d + B4D+ B5D
D = AD+
аnd d — Ad+
or
D (1 – A)
(a1 + a3) D+ (a2 + a4 + a5) d
(B1 + Вз) D + (5В2 + Ba + Bs) d
(a1 + a3) d+ (a2 + a4 + a5) D
(B1 + B3) d + (B2 + Ba + B5) D
and d(1– A)
-
From which we have
(ат + аз) D + (a2 t aд t as) d-(1 — А) (B, + Вз) D? - (1 — А) (Вә + Ва + В5) Dd
(5.41)
and
(ат + аз) d + (о2t oл t as) D-(1 — А) (B1 + Вз) d? — (1 — А) (32 + Ва + B:) Dd
(5.42)
From (5.41) and (5.42), we obtain
C(d – D) {[(a1 + a3) – (a2 + a4 + a5)] – (1 – A) (B1 + B3) (d + D)} = 0.
(5.43)
Since A < 1 and (a2 + a4 + a5) > (a1 + a3), we deduce from (5.43) that
D = d. It follows by Theorem 2, that y of Eq.(1.1) is a global attractor.
Transcribed Image Text:Case 4. Let the function H(uo, ..., u5) is non-decreasing in uo,u1,u3 and non-increasing in u2, 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 aD+a2d+ a3D+a4d + a5d B1D+ B2d + B3D+ B4d + Bzd a1d + a2D+ 3d + a4D+ a5D Bid + B2D + B3d + B4D+ B5D D = AD+ аnd d — Ad+ or D (1 – A) (a1 + a3) D+ (a2 + a4 + a5) d (B1 + Вз) D + (5В2 + Ba + Bs) d (a1 + a3) d+ (a2 + a4 + a5) D (B1 + B3) d + (B2 + Ba + B5) D and d(1– A) - From which we have (ат + аз) D + (a2 t aд t as) d-(1 — А) (B, + Вз) D? - (1 — А) (Вә + Ва + В5) Dd (5.41) and (ат + аз) d + (о2t oл t as) D-(1 — А) (B1 + Вз) d? — (1 — А) (32 + Ва + B:) Dd (5.42) From (5.41) and (5.42), we obtain C(d – D) {[(a1 + a3) – (a2 + a4 + a5)] – (1 – A) (B1 + B3) (d + D)} = 0. (5.43) Since A < 1 and (a2 + a4 + a5) > (a1 + a3), we deduce from (5.43) 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
Ym+1 =
Aym+
т 3D 0, 1, 2, ...,
В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5
(1.1)
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 Ym+1 = Aym+ т 3D 0, 1, 2, ..., В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5 (1.1)
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