Sx+1 = zxn + E · Xn ,2 X n-l
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
Find the determine yellow In the same way as Theorem 5 in the second picture and inf is here
Transcribed Image Text:Sa+1
+E.
x2
n--
ZXn
![Theorem 5 Suppose that (5) holds. If a (n) is a solution of (4), then either
x (n) = 0 eventually or
lim sup (a; (n)|) = x;.
where A1,...,de are the (not necessarily distinct) roots of the characteristic
еquation (7).
Firstly, we take the change of the variables for Eq.(2) as follows yn =
From this, we obtain the following difference equation
Yn
Yn+1 = 1+p
Yn-m
(8)
where
p =
2. From now on, we handle the difference equation (8). The unique
A2•
positive equilibrium point of Eq.(8) is
1+ v1+ 4p
Motivated by the above studies, we study the dynamics of following higher
order difference equation
Xn+1 = A+B
(2)
where A, B are positive real numbers and the initial conditions are positive
numbers. Additionally, we investigate the boundedness, periodicity, oscillation
behaviours, global asymptotically stability and rate of convergence of related
higher order difference equations.
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 <9 (n, Yn, Yn-1) < Un+1
hold. Then
Упо-1 ипо-1, Упо ипо
implies that
Yn S Un, n > no.
Consider the scalar kth-order linear difference equation
x (n + k) + P1 (n)x (n + k- 1) +..+ Pk (n)x (n) = 0,
(4)
where k is a positive integer and p;: Z+
C for i = 1, .. , k. Assume that
qi = lim p:(n), i = 1,..., k,
(5)
exist in C. Consider the limiting equation of (4):
x (n + k) + q1r (n+k - 1) +...+ qkx (n) = 0.
(6)
Theorem 4 (Poincaré's Theorem) Consider (4) subject to condition (5).
Let A1,., Ak be the roots of the characteristic equation
* + gk-! +...+ qk = 0
(7)
of the limiting equation (6) and suppose that A;l # A;| for i # j. If x (n) is
a solution of (4), then either r (n) = 0 for all large n or there exists an inder
je{1,..,k} such that
х (п + 1)
lim
= A;.
n-00 I (n)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff415ac57-ff92-4e25-9045-4869ff575152%2Fb9552abf-0dc5-4877-9c60-f760ce6b3ae4%2Fi3ut66s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 5 Suppose that (5) holds. If a (n) is a solution of (4), then either
x (n) = 0 eventually or
lim sup (a; (n)|) = x;.
where A1,...,de are the (not necessarily distinct) roots of the characteristic
еquation (7).
Firstly, we take the change of the variables for Eq.(2) as follows yn =
From this, we obtain the following difference equation
Yn
Yn+1 = 1+p
Yn-m
(8)
where
p =
2. From now on, we handle the difference equation (8). The unique
A2•
positive equilibrium point of Eq.(8) is
1+ v1+ 4p
Motivated by the above studies, we study the dynamics of following higher
order difference equation
Xn+1 = A+B
(2)
where A, B are positive real numbers and the initial conditions are positive
numbers. Additionally, we investigate the boundedness, periodicity, oscillation
behaviours, global asymptotically stability and rate of convergence of related
higher order difference equations.
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 <9 (n, Yn, Yn-1) < Un+1
hold. Then
Упо-1 ипо-1, Упо ипо
implies that
Yn S Un, n > no.
Consider the scalar kth-order linear difference equation
x (n + k) + P1 (n)x (n + k- 1) +..+ Pk (n)x (n) = 0,
(4)
where k is a positive integer and p;: Z+
C for i = 1, .. , k. Assume that
qi = lim p:(n), i = 1,..., k,
(5)
exist in C. Consider the limiting equation of (4):
x (n + k) + q1r (n+k - 1) +...+ qkx (n) = 0.
(6)
Theorem 4 (Poincaré's Theorem) Consider (4) subject to condition (5).
Let A1,., Ak be the roots of the characteristic equation
* + gk-! +...+ qk = 0
(7)
of the limiting equation (6) and suppose that A;l # A;| for i # j. If x (n) is
a solution of (4), then either r (n) = 0 for all large n or there exists an inder
je{1,..,k} such that
х (п + 1)
lim
= A;.
n-00 I (n)
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