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12 + U2 + PU2 U2 <12+ P2 U1 +h+PT Ui+ PUI 12 + U2 + PU2 U2 -1 — 12 - р- 12 U1 < 0, | Ui+PUI | + <0. + (U2 – 12) 12 U2 (U1 – 1) - P U1 In this here if p e (0, 5) then 1- p > 0, U1 > 0. U2 1- P 12 Thus, we get that U1-1 = 0, U2 - 12 = 0. So, U1 = l1 and U2 = 12. The proof is completed as desired. %3D %3D

12 + U2 + PU2 U2 <12+ P2 U1 +h+PT Ui+ PUI 12 + U2 + PU2 U2 -1 — 12 - р- 12 U1 < 0, | Ui+PUI | + <0. + (U2 – 12) 12 U2 (U1 – 1) - P U1 In this here if p e (0, 5) then 1- p > 0, U1 > 0. U2 1- P 12 Thus, we get that U1-1 = 0, U2 - 12 = 0. So, U1 = l1 and U2 = 12. The proof is completed as desired. %3D %3D

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 7 Assume that 0 < p < . Then the positive equilibrium point of system (2) is globally asymptotically stable. Proof We have from Theorem 5, 1 < l1 = lim inftn < M1, n-00 1 < l2 = lim infzn < M2, n-00 1 < U1 = lim supt, < M1, %3D n-00 1 < U2 = lim supzn < M2. n00 By system (2), we can write 12 U2 U1 <1+ p,h 2 1+ p- U1 U2 1+P221+ p; Hence we have U2 < Ujl2 <2+ p 12 12 U1 U2 + p- < Uzlı <li+ PT U2 Therefore we obtain that U2 <12+ p U1 +l1 + p- 12 Ui + PU, + U2 + p U2 12 U2 р 12 U1 12 12 U2 Ui + PU1 р <0, + U2 + P <0. (U1 – 1) + (U2 – 12) 12 U2 U1 In this here if p e (0, 5) then 1- p > 0, U1 1- P > 0. U2 Thus, we get that U1-1 = 0, U2 - 12 = 0. So, U1 = l1 and U2 = 12. The proof is completed as desired. %3D
Transcribed Image Text:Theorem 7 Assume that 0 < p < . Then the positive equilibrium point of system (2) is globally asymptotically stable. Proof We have from Theorem 5, 1 < l1 = lim inftn < M1, n-00 1 < l2 = lim infzn < M2, n-00 1 < U1 = lim supt, < M1, %3D n-00 1 < U2 = lim supzn < M2. n00 By system (2), we can write 12 U2 U1 <1+ p,h 2 1+ p- U1 U2 1+P221+ p; Hence we have U2 < Ujl2 <2+ p 12 12 U1 U2 + p- < Uzlı <li+ PT U2 Therefore we obtain that U2 <12+ p U1 +l1 + p- 12 Ui + PU, + U2 + p U2 12 U2 р 12 U1 12 12 U2 Ui + PU1 р <0, + U2 + P <0. (U1 – 1) + (U2 – 12) 12 U2 U1 In this here if p e (0, 5) then 1- p > 0, U1 1- P > 0. U2 Thus, we get that U1-1 = 0, U2 - 12 = 0. So, U1 = l1 and U2 = 12. The proof is completed as desired. %3D
Theorem 5 Suppose that (5) holds. If x (n) is a solution of (4), then either а (п) — 0 еventually or lim sup (la; (n)|)/" = x;. where A1,... ,dk 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 1+p (8) Yn+1 || B where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is 1+ V1+ 4p 2 Motivated by the above studies, we study the dynamics of following higher order difference equation In A+B- (2) In+1 = In-m 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. Consider the scalar kth-order linear difference equation x (n + k) + P1 (n)x (n + k – 1) + ...+ Px (n)x (n) = 0, (4) where k is a positive integer and pi : Z+ C for i = 1, ... , k. Assume that %3D •.. qi = lim p:(n), i = 1,... , k, (5) %3D exist in C. Consider the limiting equation of (4): I (n + k) + q1x (n + k – 1) + + qkx (n) = 0. (6) %3D Theorem 4 (Poincaré's Theorem) Consider (4) subject to condition (5). Let A1,.., Ak be the roots of the characteristic equation A* + 91**-1 + + 9k = 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 erists an inder je {1, ... , k} such that x (n + 1) lim n-0o x (n) Aj. %3D
Transcribed Image Text:Theorem 5 Suppose that (5) holds. If x (n) is a solution of (4), then either а (п) — 0 еventually or lim sup (la; (n)|)/" = x;. where A1,... ,dk 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 1+p (8) Yn+1 || B where p = . From now on, we handle the difference equation (8). The unique positive equilibrium point of Eq.(8) is 1+ V1+ 4p 2 Motivated by the above studies, we study the dynamics of following higher order difference equation In A+B- (2) In+1 = In-m 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. Consider the scalar kth-order linear difference equation x (n + k) + P1 (n)x (n + k – 1) + ...+ Px (n)x (n) = 0, (4) where k is a positive integer and pi : Z+ C for i = 1, ... , k. Assume that %3D •.. qi = lim p:(n), i = 1,... , k, (5) %3D exist in C. Consider the limiting equation of (4): I (n + k) + q1x (n + k – 1) + + qkx (n) = 0. (6) %3D Theorem 4 (Poincaré's Theorem) Consider (4) subject to condition (5). Let A1,.., Ak be the roots of the characteristic equation A* + 91**-1 + + 9k = 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 erists an inder je {1, ... , k} such that x (n + 1) lim n-0o x (n) Aj. %3D
Expert Solution
Step 1

GIVEN ,

=> U1+PL1/U1+U2 +PL2/ U2 ≤L2+PU2/L2 +L1 +PU1/L1

NOW ,Taking Right side All to Left Side i.e.,

=> U1+PL1/U1+U2 +PL2/ U2 -L2-PU2/L2 -L1 -PU1/L1≤0

=> U1 -L1 +PL1 /U1 -PU1 /L1 +U2 -L2 +  PL2/ U2 -PU2/L2  ≤0  (shifting some element towards other to get common)

HERE TAKING  (U1 -L1 ) AND (U2 - L2 ) COMMON IN ABOVE EQUATION :-

=> (U1 -L1 )(1 + P (L1 /U1  -  U1/L1)) + (U2 - L2 )(1+P( L2/ U2 - U2/L2)) ≤0

WE GET :

(U1 -L1 )(1 - P (1 /U1  +  1/L1)) + (U2 - L2 )(1-P( 1/ U2 + 1/L2)) ≤0 -------------(1)

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