Theorem 7 Let 0 < p < 1. Then every solutions of Eq. (8) is bounded and persist such that 1- p" n- 1< Yn < 1- P where C1 = p (y0 +). Proof. Let {yn} be a positive solution of Eq.(8) and p > 0. Then, we have from Eq.(8) Yo 1+p 2 > 1, Y1 Y1 1+P2 Yi-m > 1. Y2 Thus we obtain by induction yn > 1 for n > 1. Now we consider the other side. We have from Eq.(8) Yn+1 = 1+p2 Yn <1+ pyn. (12) Yn-m where According to Theorem 3, there exist a sequence yn S Un, n = 0, 1, · .., {un} satisfies (13) Un+1 = 1+ pun, n > 1, such that us = Ys,Us+1 = Ys+1, 8 E {-m, -m + 1, ·..},n > s. Therefore the solution of the difference equation (13) is 1- pr + p2-C1 1- p (14) Un =
Theorem 7 Let 0 < p < 1. Then every solutions of Eq. (8) is bounded and persist such that 1- p" n- 1< Yn < 1- P where C1 = p (y0 +). Proof. Let {yn} be a positive solution of Eq.(8) and p > 0. Then, we have from Eq.(8) Yo 1+p 2 > 1, Y1 Y1 1+P2 Yi-m > 1. Y2 Thus we obtain by induction yn > 1 for n > 1. Now we consider the other side. We have from Eq.(8) Yn+1 = 1+p2 Yn <1+ pyn. (12) Yn-m where According to Theorem 3, there exist a sequence yn S Un, n = 0, 1, · .., {un} satisfies (13) Un+1 = 1+ pun, n > 1, such that us = Ys,Us+1 = Ys+1, 8 E {-m, -m + 1, ·..},n > s. Therefore the solution of the difference equation (13) is 1- pr + p2-C1 1- p (14) Un =
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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Transcribed Image Text:Theorem 7 Let 0 < p < 1.
Then every solutions of Eq.(8) is bounded and
persist such that
1- pr
1< yn < +p"-!Cı
+p?-C1
where C1 = p (v0 +).
Proof. Let {yn} be a positive solution of Eq.(8) and p > 0. Then, we have
from Eq.(8)
Yo
> 1,
1+Pym
Y1
Y1
1+P2
Y1-m
> 1.
Y2
Thus we obtain by induction yn > 1 for n > 1.
Now we consider the other side. We have from Eq.(8)
Yn
Yn+1 = 1+p
<1+ pyn.
(12)
2
Yn-m
According to Theorem 3, there exist a sequence yn < Un, n = 0, 1, .., where
{un} satisfies
(13)
Un+1 = 1+ pun, n > 1,
such that us = Ys,Us+1 = Ys+1, 8 € {-m, -m + 1, ..:},n > s. Therefore the
solution of the difference equation (13) is
1- p"
Un
+ p?-C1
(14)
1
- p
where C1 = p( Yo+
Moreover we obtain from (12) and (14),
p-
Yn+1
- Un+1 <p(yn – Un)
where n > s and p E (0, 1). So, we get yn < Um, n > s as desired. I
![Theorem 3 (See [8]) Let n e N, and g (n, u, v) be a decreasing function in
u and v for any fixed n. Suppose that for n > no, the inequalities
Yn+1 <g (n, Yn, Yn-1) < Un+1
hold. Then
Yno-1 < Uno-1, Yno < Uno
implies that
Yn < Un, n 2 no.
Consider the scalar kth-order linear difference equation
x (n + k) + P1(n)x (n + k – 1) + ..
+ Pk(n)x (n) = 0,
(4)
%3|
where k is a positive integer and pi : Z+ → C for i = 1,... , k. Assume that
qi = lim p:(n), i = 1,
, к,
(5)
exist in C. Consider the limiting equation of (4):
x (n + k) + q1x (n + k – 1) +...
+ qkx (n) = 0.
(6)
Yn
Yn+1 = 1+p
(8)
yn-m
...hone m.
B
hendle the dieTone
uotion I0)
The unigu](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F24337687-8172-457a-8254-e047d667e397%2Fvevyxh_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 3 (See [8]) Let n e N, and g (n, u, v) be a decreasing function in
u and v for any fixed n. Suppose that for n > no, the inequalities
Yn+1 <g (n, Yn, Yn-1) < Un+1
hold. Then
Yno-1 < Uno-1, Yno < Uno
implies that
Yn < Un, n 2 no.
Consider the scalar kth-order linear difference equation
x (n + k) + P1(n)x (n + k – 1) + ..
+ Pk(n)x (n) = 0,
(4)
%3|
where k is a positive integer and pi : Z+ → C for i = 1,... , k. Assume that
qi = lim p:(n), i = 1,
, к,
(5)
exist in C. Consider the limiting equation of (4):
x (n + k) + q1x (n + k – 1) +...
+ qkx (n) = 0.
(6)
Yn
Yn+1 = 1+p
(8)
yn-m
...hone m.
B
hendle the dieTone
uotion I0)
The unigu
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