Then we get a1D+a2d + azD+ a4d+ azd aid+ a2D+ azd+ a4D+ azD D = AD+ and d= Ad+ B1D+ B2d + B3D+ Bad + Bzd Bid + B2D+ B3d+ B4D+ Bs D' or (a1 + a3) D+ (a2 + a4 + a5) d (B1 + B3) D + (B2 + B4 + Bs) d (a1+ a3) d+ (a2 + a4 + a5) D (B1 + B3) d+ (B2 + B4 + B5) D D (1 – A) = and d(1– A) : From which we have (a1 + a3) D + (a2 + a4 + as)d-(1 – A) (81 + B3) D² = (1 – A) (32 + B4 + B3) Dd (5.41) and (a1 + a3) d+ (a2 +a4 + a5) D-(1 – A) (31 + B3) ď² = (1 - A) (32 + B4 + B5) Dd (5.42) From (5.41) and (5.42), we obtain (d - D) {[(a1 + a3) – (a2 + a4 + az3)] – (1 – A) (B1 + B3) (d + D)} = 0. (5.43)
Then we get a1D+a2d + azD+ a4d+ azd aid+ a2D+ azd+ a4D+ azD D = AD+ and d= Ad+ B1D+ B2d + B3D+ Bad + Bzd Bid + B2D+ B3d+ B4D+ Bs D' or (a1 + a3) D+ (a2 + a4 + a5) d (B1 + B3) D + (B2 + B4 + Bs) d (a1+ a3) d+ (a2 + a4 + a5) D (B1 + B3) d+ (B2 + B4 + B5) D D (1 – A) = and d(1– A) : From which we have (a1 + a3) D + (a2 + a4 + as)d-(1 – A) (81 + B3) D² = (1 – A) (32 + B4 + B3) Dd (5.41) and (a1 + a3) d+ (a2 +a4 + a5) D-(1 – A) (31 + B3) ď² = (1 - A) (32 + B4 + B5) Dd (5.42) From (5.41) and (5.42), we obtain (d - D) {[(a1 + a3) – (a2 + a4 + az3)] – (1 – A) (B1 + B3) (d + D)} = 0. (5.43)
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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Topic Video
Question
Explain the determine red and the inf is here

Transcribed Image Text:The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+ a2Ym-2 + a3Ym-3 + a4Ym-4+ a5Ym-5
Ym+1 =
Aym+
т 3D 0, 1, 2, ...,
В1ут-1 + В2ут-2 + Взут-3 + Влут-4 + B5ут-5
(1.1)
![Case 4. Let the function H(uo, ..., u5) is non-decreasing in uo,u1,u3 and
non-increasing in u2, u4, U5.
Suppose that (d, D) is a solution of the system
D =
H(D,D,d, D, d, d)
and
d = H(d, d, D, d, D, D).
Then we get
a1D+a2d + a3D+a4d + azd
a1d + a2D + azd + a4D + a5D
D = AD+
and d= Ad+
B1D+ B2d + B3D+ B4d + Bzd
Bid + B2D+ B3d + B4D+ B5D’
or
(a1 + a3) D+ (a2 + a4 + a5) d
(B1 + B3) D + (B2 + B4 + B5) d
(a1 + a3) d + (a2+ a4 + az) D
(B1 + B3) d + (82 + B4 + B5) D
D(1 – A) =
and d(1– A) =
-
From which we have
+ a3) D+ (a2 + a4 + az) d-(1 – A) (B1 + B3) D² = (1 – A) (B2 + B4 + B5) Dd
(5.41)
and
(a1 + a3) d+ (a2 + a4 + a5) D-(1 – A) (B1 + B3) ď² = (1 – A) (82 + B4 + B5) Dd
(5.42)
From (5.41) and (5.42), we obtain
(d – D) {[(a1 + a3) – (a2 + a4 + a5)] – (1 – A) (B1 + B3) (d + D)} = 0.
(5.43)
Since A < 1 and (a2 + a4 + a5) > (a1 + a3), we deduce from (5.43) that
D = d. It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor.
It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor and the
proof is now completed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1c5a833-e8e1-4192-bf76-b19122b6605c%2Fce04125c-d2ff-44cc-8a0a-b6372a0eb8f5%2Fm1afplh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Case 4. Let the function H(uo, ..., u5) is non-decreasing in uo,u1,u3 and
non-increasing in u2, u4, U5.
Suppose that (d, D) is a solution of the system
D =
H(D,D,d, D, d, d)
and
d = H(d, d, D, d, D, D).
Then we get
a1D+a2d + a3D+a4d + azd
a1d + a2D + azd + a4D + a5D
D = AD+
and d= Ad+
B1D+ B2d + B3D+ B4d + Bzd
Bid + B2D+ B3d + B4D+ B5D’
or
(a1 + a3) D+ (a2 + a4 + a5) d
(B1 + B3) D + (B2 + B4 + B5) d
(a1 + a3) d + (a2+ a4 + az) D
(B1 + B3) d + (82 + B4 + B5) D
D(1 – A) =
and d(1– A) =
-
From which we have
+ a3) D+ (a2 + a4 + az) d-(1 – A) (B1 + B3) D² = (1 – A) (B2 + B4 + B5) Dd
(5.41)
and
(a1 + a3) d+ (a2 + a4 + a5) D-(1 – A) (B1 + B3) ď² = (1 – A) (82 + B4 + B5) Dd
(5.42)
From (5.41) and (5.42), we obtain
(d – D) {[(a1 + a3) – (a2 + a4 + a5)] – (1 – A) (B1 + B3) (d + D)} = 0.
(5.43)
Since A < 1 and (a2 + a4 + a5) > (a1 + a3), we deduce from (5.43) that
D = d. It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor.
It follows by Theorem 2, that ỹ of Eq.(1.1) is a global attractor and the
proof is now completed.
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