Case 2. Assume that the function F(uo,..., u4) is non-decreasing in uo,u1 and non-increasing in u2, u3, U4. Suppose that (m, M) is a solution of the system M = F(M, M, т, т, т) and F(m, m, М, М, М). m = Then we get bM + cm + fm + rm bm + cM + fМ +rM dm + eM +gМ + sM' М —аМ+ аnd m = am+ dM + em + gm + sm or 6M + (c+ f +r) m a) = dM + (e+g + s) m bm + (c+ f + r) M dm + (e +g + s) M М (1 and т (1 - а) From which we have (е +g+s) (1- а) Мт %3DЬМ + (с +f+r)m -d(1 - а) М? (33) and (e+g+s) (1- a) Mm = bm + (c+ f +r) M – d (1– a) m2. (34) From (33) and (34), we obtain (m – M) {[b – (c +f+r)] – d(1– a) (m + M)} = 0. (35) Since a <1 and (c+ f + r) > b, we deduce from (35) 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
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Case 2. Assume that the function F(uo, ..., u4) is non-decreasing in uo,u1
and non-increasing in u2, u3, U4.
Suppose that (m, M) is a solution of the system
М — F(M, M, т, т, т)
and
F(m, т, М, М, M).
m =
Then we get
ьМ + ст + fm + rm
bm + cM + fМ + rM
dm + eM + gм + sM
М —аМ+
and
т — ат +
dM + em + gm + sm
or
ьМ + (с+f+r) m
dM + (e+ g + s) m
bm + (с+f+r) M
dm + (е + g +s) M
М (1— а)
т (1 — а)
and
From which we have
+gts) (1—а) Мт
= bM + (c+ f +r) m – d (1 – a) M²
(33)
and
(e+g+ s) (1 – a) Mm = bm + (c+ f +r) M – d (1 – a) m².
(34)
From (33) and (34), we obtain
(т — М) {[b — (с +ftr)]-d(1— а) (т + M)} -
= 0.
(35)
Since a < 1 and (c+ f + r) > b, we deduce from (35) 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 2. Assume that the function F(uo, ..., u4) is non-decreasing in uo,u1 and non-increasing in u2, u3, U4. Suppose that (m, M) is a solution of the system М — F(M, M, т, т, т) and F(m, т, М, М, M). m = Then we get ьМ + ст + fm + rm bm + cM + fМ + rM dm + eM + gм + sM М —аМ+ and т — ат + dM + em + gm + sm or ьМ + (с+f+r) m dM + (e+ g + s) m bm + (с+f+r) M dm + (е + g +s) M М (1— а) т (1 — а) and From which we have +gts) (1—а) Мт = bM + (c+ f +r) m – d (1 – a) M² (33) and (e+g+ s) (1 – a) Mm = bm + (c+ f +r) M – d (1 – a) m². (34) From (33) and (34), we obtain (т — М) {[b — (с +ftr)]-d(1— а) (т + M)} - = 0. (35) Since a < 1 and (c+ f + r) > b, we deduce from (35) 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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