(a1 + a2 + a3 + a4) D+a5d-(1– A) (B1 + B2 + B3 + B4) D² = (1 – A) B5 Dd (5.32) and (a1 + a2 + a3 + a4) d+a;D-(1– A) (B1 + B2 + B3 + B4) d² = (1 – A) Bz Dd (5.33) From (5.32) and (5.33), we obtain (d – D) {[(a1 + a2 + az + a4) – as] – (1 – A) (ß1 + B2 + B3 + B4) (d + D)} = 0. (5.34)

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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Explain the determine green

Case 1. Let the function H(uo, ..., u5) is non-decreasing in u0,U1,U2,U3,U4
and non-increasing in u5. Suppose that (d, D) is a solution of the system
D = H(D, D, D, D, D, d)
and
= H(d, d, d, d, d, D).
d =
Then we get
a1D+ a2D+æ3D+ a4D+a5d
B1D+ B2D+ B3D+ B4D+ Bzd
a1d + a2d + azd + a4d + a5D
D = AD+
and d = Ad+
Bid + B2d + B3d + Bad + B5D'
or
(a1 + a2 + a3+a4) D+ a5d
(В1 + В2 + Вз + Ba) D + Bsd
(a1 + a2 + a3 +a4) d + a5D
(B1 + B2 + B3 + B4) d + B3D
D (1 – A) =
and d(1 – A):
From which we have
(a1+ a2 + a3 + a4) D+a5d-(1 – A) (B1 + B2 + B3 + B4) D² = (1 – A) B5Dd
(5.32)
and
(a1 + a2 + a3 + a4) d+a5D-(1– A) (B1 + B2 + B3 + B4) d² = (1 – A) B3D.
(5.33)
From (5.32) and (5.33), we obtain
(d – D) {[(@1 + a2 + a3 + a4) – a5] – (1 – A) (ß1 + B2 + B3 + B4) (d + D)} = 0.
(5.34)
Since A < 1 and as 2 (a1 + a2 + a3+ a4), we deduce from (5.34) that
D = d. It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor.
Transcribed Image Text:Case 1. Let the function H(uo, ..., u5) is non-decreasing in u0,U1,U2,U3,U4 and non-increasing in u5. Suppose that (d, D) is a solution of the system D = H(D, D, D, D, D, d) and = H(d, d, d, d, d, D). d = Then we get a1D+ a2D+æ3D+ a4D+a5d B1D+ B2D+ B3D+ B4D+ Bzd a1d + a2d + azd + a4d + a5D D = AD+ and d = Ad+ Bid + B2d + B3d + Bad + B5D' or (a1 + a2 + a3+a4) D+ a5d (В1 + В2 + Вз + Ba) D + Bsd (a1 + a2 + a3 +a4) d + a5D (B1 + B2 + B3 + B4) d + B3D D (1 – A) = and d(1 – A): From which we have (a1+ a2 + a3 + a4) D+a5d-(1 – A) (B1 + B2 + B3 + B4) D² = (1 – A) B5Dd (5.32) and (a1 + a2 + a3 + a4) d+a5D-(1– A) (B1 + B2 + B3 + B4) d² = (1 – A) B3D. (5.33) From (5.32) and (5.33), we obtain (d – D) {[(@1 + a2 + a3 + a4) – a5] – (1 – A) (ß1 + B2 + B3 + B4) (d + D)} = 0. (5.34) Since A < 1 and as 2 (a1 + a2 + a3+ a4), we deduce from (5.34) that D = d. It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor.
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