or M (b+c+f)+ rm M (d +e+ g) + sm m (b+c+ f)+r M m (d + e+ g) + sM° М (1— а). and m (1 – a): From which we have s(1-a) Mm M (b+ c+ f)+ rm - (1- a) (d+e+g) M² (30) and Ga-a) - a) Mm = m (b+c+f)+rM- (1- a) (d+ e+g) m2. (31) From (30) and (31), we obtain (m – M) {[(b+c+ f)- r]- (1- a) (d+ e+g) (m + M)} = 0. (32) Since a < 1 and r 2 (b+c+ f), we deduce from (32) that M = m. It follows by Theorem 2, that of Eq.(1) is a global attractor and the proof is now completed.
or M (b+c+f)+ rm M (d +e+ g) + sm m (b+c+ f)+r M m (d + e+ g) + sM° М (1— а). and m (1 – a): From which we have s(1-a) Mm M (b+ c+ f)+ rm - (1- a) (d+e+g) M² (30) and Ga-a) - a) Mm = m (b+c+f)+rM- (1- a) (d+ e+g) m2. (31) From (30) and (31), we obtain (m – M) {[(b+c+ f)- r]- (1- a) (d+ e+g) (m + M)} = 0. (32) Since a < 1 and r 2 (b+c+ f), we deduce from (32) 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
Related questions
Question
Explain the determine blue
![••.
Case 1. Assume that the function F(uo, ..., u4) is non-decreasing in uo,
u1, u2, uz and non-increasing in u4. Suppose that (m, M) is a solution of the
system
М — F(M, M, М, М, т)
and
F(m, m, т, т, М).
m =
Then we get
ьМ + сМ + fM + rm
aM+
dM + eM +дM + sm
bт + ст + fm +rM
M =
аnd
т — ат+
dm + em + gm + sM'
or
M (b+c+ f)+rm
M (d + e+g) + sm
m (b+c+f)+rM
m (d +e+ g) + sM°
М (1— а) -
аnd
т (1 — а) -
From which we have
1- a) Mm = M (b +c+f) + rm - (1 – a) (d+e+ g) M²
(30)
and
- a) Mm = m (b +c+ f) + rM - (1 - a) (d +e+ g) m².
(31)
From (30) and (31), we obtain
(m – M){[(b+c+ f) – r] – (1 – a) (d+ e + g) (m + M)} = 0.
(32)
Since a < 1 and r > (b+c+ f), we deduce from (32) that M = m. It follows
by Theorem 2, that of Eq.(1) is a global attractor and the proof is now
completed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48dbf4e2-6a0f-454b-99ed-fcbc11bc7c27%2F6d578686-1984-4f9b-a459-f506f69742b0%2F21czey_processed.jpeg&w=3840&q=75)
Transcribed Image Text:••.
Case 1. Assume that the function F(uo, ..., u4) is non-decreasing in uo,
u1, u2, uz and non-increasing in u4. Suppose that (m, M) is a solution of the
system
М — F(M, M, М, М, т)
and
F(m, m, т, т, М).
m =
Then we get
ьМ + сМ + fM + rm
aM+
dM + eM +дM + sm
bт + ст + fm +rM
M =
аnd
т — ат+
dm + em + gm + sM'
or
M (b+c+ f)+rm
M (d + e+g) + sm
m (b+c+f)+rM
m (d +e+ g) + sM°
М (1— а) -
аnd
т (1 — а) -
From which we have
1- a) Mm = M (b +c+f) + rm - (1 – a) (d+e+ g) M²
(30)
and
- a) Mm = m (b +c+ f) + rM - (1 - a) (d +e+ g) m².
(31)
From (30) and (31), we obtain
(m – M){[(b+c+ f) – r] – (1 – a) (d+ e + g) (m + M)} = 0.
(32)
Since a < 1 and r > (b+c+ f), we deduce from (32) that M = m. It follows
by Theorem 2, that of Eq.(1) is a global attractor and the proof is now
completed.
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