Theorem 7 Assume that 0 < p < 5. Then the positive equilibrium point of system (2) is globally asymptotically stable. Proof We have from Theorem 5, 1 < l1 = lim inftn < M1, %3D n-00 1 < l2 = lim infzn < M2, %3D n-00 1 < Uj = lim supt, < M1, n-00 1 < U2 = lim supzn < M2. %3D n-00 By system (2), we can write U1 <1+p- 12 U2 1 21+ P U2 <1+p- U1 , l2 > 1 + p- Hence we have Ui + P, U2 < Ui2 <2+ p- 12 12 U2 + PU2 U1 < Uzli sli+pT Therefore we obtain that U1 U2 +h+PT' 12 + U2 + P- U2 12 + U2 + P U1 + P- U1 U2 - 12 - - U1 < 0, Ui+ PUI U2 12 <0. (U1 – 1) + U1 + (U2 – 12) + 12 In this here if p e (0, ) then > 0, -- > 0. 1- p 12 U2 Thus, we get that U1-h =0, U2 – 12 = 0. So, U1 = l1 and U2 = l2. The proof is completed as desired.
Theorem 7 Assume that 0 < p < 5. Then the positive equilibrium point of system (2) is globally asymptotically stable. Proof We have from Theorem 5, 1 < l1 = lim inftn < M1, %3D n-00 1 < l2 = lim infzn < M2, %3D n-00 1 < Uj = lim supt, < M1, n-00 1 < U2 = lim supzn < M2. %3D n-00 By system (2), we can write U1 <1+p- 12 U2 1 21+ P U2 <1+p- U1 , l2 > 1 + p- Hence we have Ui + P, U2 < Ui2 <2+ p- 12 12 U2 + PU2 U1 < Uzli sli+pT Therefore we obtain that U1 U2 +h+PT' 12 + U2 + P- U2 12 + U2 + P U1 + P- U1 U2 - 12 - - U1 < 0, Ui+ PUI U2 12 <0. (U1 – 1) + U1 + (U2 – 12) + 12 In this here if p e (0, ) then > 0, -- > 0. 1- p 12 U2 Thus, we get that U1-h =0, U2 – 12 = 0. So, U1 = l1 and U2 = l2. The proof is completed as desired.
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 red and inf is here i need evey I need all the details step by step and inf is here
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
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following
system of difference equations
Уп
Xn
Xn+1 = A +B
, Yn+1 = A + B
(1)
2.
n-1
n-1
where A and B are positive numbers and the initial values are positive numbers. In
From this, system (1) transform into following system:
tn
In+1 = 1+P
Zn
Zn+1 =
(2)
%3D
n-1
n-1
where p = > 0. From now on, we study the system (2).
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