Then we get ьм + сM + fm + rm dM + eM + gm + sm bm + cm + f M +rM dm + em + gM + sM' М - аМ+ and m = am+ or M (b+c)+ (f+ r)m M (d + e) + (g+s) m m (b+c) + (f + r) M m (d + e) + (g + s) M М (1—а) %3D and m (1 - a) = From which we have (g+ s) (1– a) Mm = M (b+ c) + (f +r) m – (1- a) (d+ e) M² (36) and (g+s) (1 – a) Mm = m (b+c) + (f + r) M – (1– a) (d +e) m². (37) From (36) and (37), we obtain (m – M) {[(b+c) - (f +r)]- (1- a) (d+ e) (m + M)} = 0. (38) Since a < 1 and (f +r) > (b+ c), we deduce from (38) that M = m. It follows by Theorem 2, that of Eq.(1) is a global attractor and the proof is now completed.

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 3. Assume that the function F(uo,
and non-increasing in u3,u4. Suppose that (m, M) is a solution of the system
u4) is non-decreasing in uo,u1,U2
М — F(M, M, М, т, т)
and
F(m, m, m,
%3D
m =
Then we get
bM + cM + fm + rm
dM + eM + gm + sm
bm + cm + fM+rM
М -аМ+
and
m = am+
dm + em + gM + sM'
or
M (b+ c) + (f + r) m
M (d + e) + (g + s) m
m (b + c) + (f + r) M
М (1 — а)
аnd m(1 - а) %—
m (d + e) + (g + s) M
From which we have
(9+ s) (1– a) Mm = M (b+ c) + (f +r)m - (1- a) (d+ e) M²
(36)
and
(g+s) (1- a) Mm = m (b+ c) + (f+ r) M – (1– a) (d + e) m².
(37)
From (36) and (37), we obtain
(m – M) {[(b+c) – (S +r)] – (1– a) (d+ e) (m + M)} = 0.
(38)
Since a < 1 and (f +r) > (b+ c), we deduce from (38) that M = m. It
follows by Theorem 2, that ã of Eq.(1) is a global attractor and the proof is
now completed.
Transcribed Image Text:Case 3. Assume that the function F(uo, and non-increasing in u3,u4. Suppose that (m, M) is a solution of the system u4) is non-decreasing in uo,u1,U2 М — F(M, M, М, т, т) and F(m, m, m, %3D m = Then we get bM + cM + fm + rm dM + eM + gm + sm bm + cm + fM+rM М -аМ+ and m = am+ dm + em + gM + sM' or M (b+ c) + (f + r) m M (d + e) + (g + s) m m (b + c) + (f + r) M М (1 — а) аnd m(1 - а) %— m (d + e) + (g + s) M From which we have (9+ s) (1– a) Mm = M (b+ c) + (f +r)m - (1- a) (d+ e) M² (36) and (g+s) (1- a) Mm = m (b+ c) + (f+ r) M – (1– a) (d + e) m². (37) From (36) and (37), we obtain (m – M) {[(b+c) – (S +r)] – (1– a) (d+ e) (m + M)} = 0. (38) Since a < 1 and (f +r) > (b+ c), we deduce from (38) that M = m. It follows by Theorem 2, that ã of Eq.(1) is a global attractor and the proof is now completed.
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