(b+ f) M + (c+r)m (d+ g) M + (e+ s) m (b+ f) m+ (c+r) M (d + g) m + (e + s) M M (1 – a) = and m(1- a) = From which we have (e+s) (1- a) Mm = M (b+ f) + (c+r)m- (1- a) (d + g) M² (39) and (e+s) (1-a) Mm = m (b+f) + (c+r) M - (1- a) (d+g) m². (40) From (39) and (40), we obtain (m– M) {[(b+ f)- (c+r)]- (1– a) (d+ g) (m + M)} = 0. Since a < 1 and (c+r) 2 (b+ f), we deduce from (41) that M = m. It follows by Theorem 2, that of Eq.(1) is a global attractor and the proof is now completed. (41) %3!

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Explain the determine green

Case 4. Assume that the function F(uo, .. ug) is non-decreasing in uo,u1,U3
and non-increasing in u2,u4. Suppose that (m, M) is a solution of the system
М — F(M, M, т, М, т)
and
т%3D F(m, т, м, т, М).
Then we get
(b+ f) M + (c + r) m
(d+g) M + (e + s) m
(b+ f) m + (c+r) M
(d+ g) m + (e + s) M’
M = aM+
and
m = am+
or
(b+ f) M + (c+r) m
(d+ g) M + (e + s) m
(b+ f) m + (c+r) M
(d + g) m + (e+ s) M
М (1—а) —
and m (1 – a) =
From which we have
(39)
(e+s) (1– a) Mm = M (b+ f) + (c+r) m – (1 – a) (d + g) M²
and
(e+s) (1 – a) Mm = m (b+ f) + (c+r) M – (1 – a) (d + g) m².
(40)
From (39) and (40), we obtain
(m – M) {[(b+ f) – (c+r)] – (1 – a) (d + g) (m + M)} = 0.
(41)
Since a < 1 and (c+ r) > (b+ f), we deduce from (41) that M = m. It
follows by Theorem 2, that a of Eq.(1) is a global attractor and the proof is
now completed.
Transcribed Image Text:Case 4. Assume that the function F(uo, .. ug) is non-decreasing in uo,u1,U3 and non-increasing in u2,u4. Suppose that (m, M) is a solution of the system М — F(M, M, т, М, т) and т%3D F(m, т, м, т, М). Then we get (b+ f) M + (c + r) m (d+g) M + (e + s) m (b+ f) m + (c+r) M (d+ g) m + (e + s) M’ M = aM+ and m = am+ or (b+ f) M + (c+r) m (d+ g) M + (e + s) m (b+ f) m + (c+r) M (d + g) m + (e+ s) M М (1—а) — and m (1 – a) = From which we have (39) (e+s) (1– a) Mm = M (b+ f) + (c+r) m – (1 – a) (d + g) M² and (e+s) (1 – a) Mm = m (b+ f) + (c+r) M – (1 – a) (d + g) m². (40) From (39) and (40), we obtain (m – M) {[(b+ f) – (c+r)] – (1 – a) (d + g) (m + M)} = 0. (41) Since a < 1 and (c+ r) > (b+ f), we deduce from (41) that M = m. It follows by Theorem 2, that a of Eq.(1) is a global attractor and the proof is now completed.
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