Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Theorem 4.1. The positive equilibrium point E, of system (4) is not asymptotically stable and nonhyperbolic point . Proof. The linearized system of (4) evaluated at E, is written in the matrix form as Ln+1 Q Ln, where Ln = (wn, Zn, Wn-1, zn-1, Wn-2, Zn-2)' and the Jacobian matrix determined at Eo is equal %3D 금 -(²+2e-3) -(e+3) 금 -(c+3) -(4µ) (e²+2e-3) e+3 e+3 1 Q = (28) 1 1 1 9.

Theorem 4.1. The positive equilibrium point E, of system (4) is not asymptotically stable and nonhyperbolic point . Proof. The linearized system of (4) evaluated at E, is written in the matrix form as Ln+1 Q Ln, where Ln = (wn, Zn, Wn-1, zn-1, Wn-2, Zn-2)' and the Jacobian matrix determined at Eo is equal %3D 금 -(²+2e-3) -(e+3) 금 -(c+3) -(4µ) (e²+2e-3) e+3 e+3 1 Q = (28) 1 1 1 9.

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
See similar textbooks
icon
Related questions
Question

Show me the steps of determine green and all information is here

Theorem 4.1. The positive equilibrium point Eo of system (4) is not asymptotically stable and nonhyperbolic point . Proof. The linearized system of (4) evaluated at Eo is written in the matrix form as Ln+1 Q Ln, where Ln (Wn, Zn, Wn-1, Zn-1, Wn-2, zn-2) and the Jacobian matrix determined at Eo is equal -(4u) (e+2e-3) e+3 e+3 -(e2+2e-3) 4 µ -(e+3) 2 u -(e+3) 2 e 1 (28) 1 1 1 9. Hence, the characteristic equation of (28) is P(A) = 1° – ng A° – ng X* – n3 1³ – nz X? – n1 – - no, (29) where, (2ε+6μ-2εμ + -3) (2μ (1-0 ) (2ε+4μ-6εμ + 2έμ-2+15) (2μ (ε-1) ) (2e+3μ+2 εμ- μ+2-12) (2μ (ε- 1) ) (ε+2μ-2εμ +) (и (1 — е)) n5 n4 = n3 n2 = %3D (1 – e) n1 no = -1. Now, let T(A) = 1° and V(A) = -ng 15 – n4 X4 – n3 A3 – n2 X? – n1 1 – no. Consider |A|=1, then one has (2ε+6μ-2 εμ ε2-3) (2μ (1 - ) (2ε+4μ-6εμ# 22 μ- (2 µ (e – 1)) (2ε+6μ-2εμ4 ε-3 ) (2μ (1 - 0 ) (e +2µ – 2 eu + 3) (µ (1 – e)) (2ε+3μ +2εμ- μ+ 2 2?-12 ) (2µ (€ – 1)) (ε+2μ-2εμ + 3) (и (1 — е)) (4ε+7μ-4εμ 4 έμ+ +3) (2μ(ε-1) ) + e2 + 15) (1 - €) +1, +1, 2 (2ε+ μ-2εμ + μ+6) (E-2μ-2εμ+3) (2μ (ε -1) ) (-3µ+2€µ+e µ) (2µ (e – 1)) - E + +1, (и (1 — е)) u (e – 1) (1- e) u (e – 1) -3µ +2€µ+ e? µ – µ (e – 1)2 +1, 2 -4μ +4με +1, 2 µ (e – 1) +1= 2µ (e – 1) = 3 > 1. Then, from Rouché's Theorem 1.2, V(A) and Y(A) + V(A) = P(^) have the same number of zeros in an open unit disk || < 1, that means there are only five roots lie inside unit disk. Thus, from Theorem 1.1 the equilibrium point E, is not asymptotically stable. By solving (29), we find that A 1, implies E, is nonhyperbolic point.
Transcribed Image Text:Theorem 4.1. The positive equilibrium point Eo of system (4) is not asymptotically stable and nonhyperbolic point . Proof. The linearized system of (4) evaluated at Eo is written in the matrix form as Ln+1 Q Ln, where Ln (Wn, Zn, Wn-1, Zn-1, Wn-2, zn-2) and the Jacobian matrix determined at Eo is equal -(4u) (e+2e-3) e+3 e+3 -(e2+2e-3) 4 µ -(e+3) 2 u -(e+3) 2 e 1 (28) 1 1 1 9. Hence, the characteristic equation of (28) is P(A) = 1° – ng A° – ng X* – n3 1³ – nz X? – n1 – - no, (29) where, (2ε+6μ-2εμ + -3) (2μ (1-0 ) (2ε+4μ-6εμ + 2έμ-2+15) (2μ (ε-1) ) (2e+3μ+2 εμ- μ+2-12) (2μ (ε- 1) ) (ε+2μ-2εμ +) (и (1 — е)) n5 n4 = n3 n2 = %3D (1 – e) n1 no = -1. Now, let T(A) = 1° and V(A) = -ng 15 – n4 X4 – n3 A3 – n2 X? – n1 1 – no. Consider |A|=1, then one has (2ε+6μ-2 εμ ε2-3) (2μ (1 - ) (2ε+4μ-6εμ# 22 μ- (2 µ (e – 1)) (2ε+6μ-2εμ4 ε-3 ) (2μ (1 - 0 ) (e +2µ – 2 eu + 3) (µ (1 – e)) (2ε+3μ +2εμ- μ+ 2 2?-12 ) (2µ (€ – 1)) (ε+2μ-2εμ + 3) (и (1 — е)) (4ε+7μ-4εμ 4 έμ+ +3) (2μ(ε-1) ) + e2 + 15) (1 - €) +1, +1, 2 (2ε+ μ-2εμ + μ+6) (E-2μ-2εμ+3) (2μ (ε -1) ) (-3µ+2€µ+e µ) (2µ (e – 1)) - E + +1, (и (1 — е)) u (e – 1) (1- e) u (e – 1) -3µ +2€µ+ e? µ – µ (e – 1)2 +1, 2 -4μ +4με +1, 2 µ (e – 1) +1= 2µ (e – 1) = 3 > 1. Then, from Rouché's Theorem 1.2, V(A) and Y(A) + V(A) = P(^) have the same number of zeros in an open unit disk || < 1, that means there are only five roots lie inside unit disk. Thus, from Theorem 1.1 the equilibrium point E, is not asymptotically stable. By solving (29), we find that A 1, implies E, is nonhyperbolic point.
In this paper, we solve and study the properties of the following system Wn-p 2 Zn-h p=0 Wn-p h=0 E zn-h h=1 p=1 Wn+1 +u and zn+1 = +e, (4) Zn - € Wn - 4 where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then 6+ V – 4 ko k1 A - V2 – 4 ko ki 1- ki + ko + w2n-2 +B Ty – Oy - - V2 - 4 ko ki -4 ko ki + V2 – 4 ko ki V2 - 4 ko ki A ko + B ko w2n-1 - 1 2 2 •((금-1) (는) 1– ko + k1 + V2 - 4 vo V1 - V2 - 4 VI 1- v + vo 22n-2 +D %3D 1 + Vu2 – 4 o VI gb + V2 – 4 vo vỊ 1 1. +D - V2 – 4 v vỊ V2 – 4 vo vi 22n-1 - v + vo 1- 0 - VI where A, B, C and D are constants defined as wo -H 2-1 +z-2 ko ki = 0 = ko + ki + ko ki, w-1 + w-2 20 - e 20 - e w-1 +w-2 , = vo + vi + vo V1, 2-1 +2-2 1-Om Vo2 – 4 ko k1 -)") (v - 4ko ki – 4) + 2 (wo - (G ) -). 2 62 - 8 ko ki V2- 4 ko ki 262 - 8 ko ki 1- ko + ki - ko – k1 (1 – kj + ko 1- k1 - ko B = u- w-2 +2 wn- - vi +u C = -2 + 2 22 - 8 0 vI - o - VI ) [(는 1- v1 + o - v0 - v1 1- v + o 2) (V2 – 4vo vI + +2 2 2 -- 0 - v1 since ko + ki #1 and vo + vı #1 for n e N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form 2 zn-h Wn-p p=1 Wn+1 h=1 Zn+1 and (5) 1 E Wn-p 1 > Zn-h h=0 wn -u p-0 4 Then, we assume that Wn+1 - l Wn - H kn+1 = 1 kn = 2 Wn-p p=0 Wn-p p=1 (6) Zn+1 -€ Zn - € Vn+1 = Vn 2 E Zn-h > Zn-h h=1 h=0 Substituting (6) in (5), we have Zn-h h=1 1 == kn-1, Vn kn+1 = (7) E Wn-p p=1 Vn+1 = = Vn-1. kn %3D Wn - Hence, we see that 1 ki = -, Vo wo - H 20 - € ko w-1+ w_2 Vo = Z-1+ z-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, ks = k and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k, and V2n+1 = V1. Now, from the relations in (6) we get Wn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2). (9) Using (8) in (9), we have W2n = H+ ko (w2n-1 + w2n-2), W2n+1 = µ+ k1 (w2n + w2n-1), (10) Z2n = €+ vo (z2n-1 + 2n-2), 22n+1 = €+ V1 (22n + 22n-1).
Transcribed Image Text:In this paper, we solve and study the properties of the following system Wn-p 2 Zn-h p=0 Wn-p h=0 E zn-h h=1 p=1 Wn+1 +u and zn+1 = +e, (4) Zn - € Wn - 4 where u and e are arbitrary positive real numbers with initial conditions w; and z; for i = -2, –1,0. Theorem 2.1. Let {wn, z„}-2 be a solution of (4), then 6+ V – 4 ko k1 A - V2 – 4 ko ki 1- ki + ko + w2n-2 +B Ty – Oy - - V2 - 4 ko ki -4 ko ki + V2 – 4 ko ki V2 - 4 ko ki A ko + B ko w2n-1 - 1 2 2 •((금-1) (는) 1– ko + k1 + V2 - 4 vo V1 - V2 - 4 VI 1- v + vo 22n-2 +D %3D 1 + Vu2 – 4 o VI gb + V2 – 4 vo vỊ 1 1. +D - V2 – 4 v vỊ V2 – 4 vo vi 22n-1 - v + vo 1- 0 - VI where A, B, C and D are constants defined as wo -H 2-1 +z-2 ko ki = 0 = ko + ki + ko ki, w-1 + w-2 20 - e 20 - e w-1 +w-2 , = vo + vi + vo V1, 2-1 +2-2 1-Om Vo2 – 4 ko k1 -)") (v - 4ko ki – 4) + 2 (wo - (G ) -). 2 62 - 8 ko ki V2- 4 ko ki 262 - 8 ko ki 1- ko + ki - ko – k1 (1 – kj + ko 1- k1 - ko B = u- w-2 +2 wn- - vi +u C = -2 + 2 22 - 8 0 vI - o - VI ) [(는 1- v1 + o - v0 - v1 1- v + o 2) (V2 – 4vo vI + +2 2 2 -- 0 - v1 since ko + ki #1 and vo + vı #1 for n e N. Proof. To obtain the expressions of the general solutions for (4), we rewrite it in the follow form 2 zn-h Wn-p p=1 Wn+1 h=1 Zn+1 and (5) 1 E Wn-p 1 > Zn-h h=0 wn -u p-0 4 Then, we assume that Wn+1 - l Wn - H kn+1 = 1 kn = 2 Wn-p p=0 Wn-p p=1 (6) Zn+1 -€ Zn - € Vn+1 = Vn 2 E Zn-h > Zn-h h=1 h=0 Substituting (6) in (5), we have Zn-h h=1 1 == kn-1, Vn kn+1 = (7) E Wn-p p=1 Vn+1 = = Vn-1. kn %3D Wn - Hence, we see that 1 ki = -, Vo wo - H 20 - € ko w-1+ w_2 Vo = Z-1+ z-2 ko and at n = 1, k2 = ko and V2 = Vo, at n = 2, k3 = k1 and V3 = V1, at n = 3, k4 = ko and V4 = Vo, at n = 4, ks = k and V5 = V1, (8) at n = 2 n, k2n = ko and V2n = Vo, at n = 2n + 1, k2n+1 = k, and V2n+1 = V1. Now, from the relations in (6) we get Wn = µ+ kn (wn-1 + wn-2) and zn = e + vn (žn-1 + zn-2). (9) Using (8) in (9), we have W2n = H+ ko (w2n-1 + w2n-2), W2n+1 = µ+ k1 (w2n + w2n-1), (10) Z2n = €+ vo (z2n-1 + 2n-2), 22n+1 = €+ V1 (22n + 22n-1).
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS