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Theorem 5 Suppose that (5) holds. If x (n) is a solution of (4), then either ¤ (n) = 0 eventually or lim sup (a; (n)|)/" = Xj. where A1,.., Ak are the (not necessarily distinct) roots of the characteristic equation (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- (8) .2 Yn-m

Theorem 5 Suppose that (5) holds. If x (n) is a solution of (4), then either ¤ (n) = 0 eventually or lim sup (a; (n)|)/" = Xj. where A1,.., Ak are the (not necessarily distinct) roots of the characteristic equation (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- (8) .2 Yn-m

Advanced Engineering Mathematics
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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Show me the steps of deremine blue and inf is here i need evey I need all the details step by step and inf is here

Theorem 5 Suppose that (5) holds. If x (n) is a solution of (4), then either x (n) : = 0 eventually or lim sup (x; (n)|)/n = Aj. where A1, , Ak are the (not necessarily distinct) roots of the characteristic .. equation (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 +p2 Уп-т В where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is Ą² • 1+ V1+4p
Transcribed Image Text:Theorem 5 Suppose that (5) holds. If x (n) is a solution of (4), then either x (n) : = 0 eventually or lim sup (x; (n)|)/n = Aj. where A1, , Ak are the (not necessarily distinct) roots of the characteristic .. equation (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 +p2 Уп-т В where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is Ą² • 1+ V1+4p
Theorem 7 Let 0 < p < 1. Then every solutions of Eq.(8) is bounded and persist such that 1-р" 1- p 1 < Yn S where C1 = p (y0 + 4). Proof. Let {Yn} be a positive solution of Eq.(8) and p > 0. Then, we have from Eq.(8) yo 1+P2 > 1, Y1 Y-m Y1 > 1. 1+P yi-m 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 Yn-m <1+ pyn• (12) According to Theorem 3, there exist a sequence yn < Un,n = 0, 1, · . · , where {un} satisfies Un+1 = 1+ pun, n > 1, (13) such that us = Ys, Us+1 = Ys+1; 8 € {-m, –m + 1,..},n 2 s. Therefore the solution of the difference equation (13) is 1-р" + p"-'C1 1- p Un = (14) where C1 = p (yo + „). Moreover we obtain from (12) and (14), Уп+1 — ип+1<p(уn — и,n) where n > s and p E (0, 1). So, we get yn < Un, n > s as desired. . 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 2 no, the inequalities Yn+1 <g (n, Yn, Yn-1) < Un+1 hold. Then Упо -1 < uпо -1, Упо < ипо implies that Yn < Un, n 2 no. Consider the scalar kth-order linear difference equation r (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): I (n + k) + q1x (n + k – 1) + ...+ qkx (n) = 0. (6) Theorem 4 (Poincaré's Theorem) Consider (4) subject to condition (5). Let d1,., Ak be the roots of the characteristic equation + +...+ qk = 0 (7) of the limiting equation (6) and suppose that |A;| + |A;| for i # j. If x (n) is a solution of (4), then either x (n) = 0 for all large n or there exists an inder je {1,..., k} such that x (n + 1) T (n) lim From this, we obtain the following difference equation Yn Yn+1 = 1+ p Yn-m (8) where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is 1+ VI+ 4p [8] A. Bilgin and M.R.S. Kulenovic, Global asymptotic stability for discrete single species population models, Discrete Dyn. Nat. Soc., 2017 (2017) 1– 15.
Transcribed Image Text:Theorem 7 Let 0 < p < 1. Then every solutions of Eq.(8) is bounded and persist such that 1-р" 1- p 1 < Yn S where C1 = p (y0 + 4). Proof. Let {Yn} be a positive solution of Eq.(8) and p > 0. Then, we have from Eq.(8) yo 1+P2 > 1, Y1 Y-m Y1 > 1. 1+P yi-m 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 Yn-m <1+ pyn• (12) According to Theorem 3, there exist a sequence yn < Un,n = 0, 1, · . · , where {un} satisfies Un+1 = 1+ pun, n > 1, (13) such that us = Ys, Us+1 = Ys+1; 8 € {-m, –m + 1,..},n 2 s. Therefore the solution of the difference equation (13) is 1-р" + p"-'C1 1- p Un = (14) where C1 = p (yo + „). Moreover we obtain from (12) and (14), Уп+1 — ип+1<p(уn — и,n) where n > s and p E (0, 1). So, we get yn < Un, n > s as desired. . 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 2 no, the inequalities Yn+1 <g (n, Yn, Yn-1) < Un+1 hold. Then Упо -1 < uпо -1, Упо < ипо implies that Yn < Un, n 2 no. Consider the scalar kth-order linear difference equation r (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): I (n + k) + q1x (n + k – 1) + ...+ qkx (n) = 0. (6) Theorem 4 (Poincaré's Theorem) Consider (4) subject to condition (5). Let d1,., Ak be the roots of the characteristic equation + +...+ qk = 0 (7) of the limiting equation (6) and suppose that |A;| + |A;| for i # j. If x (n) is a solution of (4), then either x (n) = 0 for all large n or there exists an inder je {1,..., k} such that x (n + 1) T (n) lim From this, we obtain the following difference equation Yn Yn+1 = 1+ p Yn-m (8) where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is 1+ VI+ 4p [8] A. Bilgin and M.R.S. Kulenovic, Global asymptotic stability for discrete single species population models, Discrete Dyn. Nat. Soc., 2017 (2017) 1– 15.
Expert Solution
Step 1

As per the instruction, we will be showing the steps for the blue tick marked expression.

That is, we need to show the steps for limn→∞supxj(n)1/n=λj.

From Theorem 4, we have that either x(n)=0 for all large n or there exists an index j∈1, ... , k such that limn→∞x(n+1)x(n)=λj.

We can write the above limit also as limn→∞xj(n+1)xj(n)=λj.

Now, considering the sequence xj(n) for j∈1, ... , k.

Thus, we see that all the terms xj(n) are greater than 0.

Further, the limit limn→∞xj(n+1)xj(n)=λj exists.

Now, by Cauchy's second theorem on limits, we get that limn→∞supxj(n)1/n=λj.

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