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(1+T+4p 1+ /T+4p (F, 2) = %3D 2 2

(1+T+4p 1+ /T+4p (F, 2) = %3D 2 2

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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Motivated by difference equations and their systems, we consider the following system of difference equations Ул , Yn+1 = A + B- Yn-1 Xn+1 = A + B- (1) where A and B are positive numbers and the initial values are positive numbers. In First of all, we consider the change of the variables for system (1) as follows: Xn Уп tn , Zn = A A From this, system (1) transform into following system: tn Zn , Zn+1 1+P Zn-1 tn+1 = 1+P2 (2) 'n-1 where p = > 0. From now on, we study the system (2). Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+T+4p 1+ VI+4p (7, 2) = ( - %3D Now, we consider a transformation as follows: (tn, tn-1, Zn, Zn-1) → (f, fi, 8, 81) where f = 1+ p, fi = tn, 8 = 1+p, , g1 = Zn. Thus we get the jacobian n-1 matrix about equilibrium point (7, z): 1 0 B (f, ३) = 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (ī, 2) XN, where -) In tn-1 XN = uz Zn-1 0 0 물글 0 0 1 B (f, ३) 0 0 0 0 1 Hence, the characteristic equation of B (7, z) about the unique positive equilibrium point (7, 2) is 4p2. 4p2 = 0. Due to t = 7, we can rearrange the characteristic equation such that 4p2. 4p2 = 0. 74 14 74 Therefore, we obtain the four roots of characteristic equation as follows: p+Vp² - Vp² – 8pr? 212 Vp² – 8pi? 272 12 = -p+Vp? + 8pi 272 13 =
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Ул , Yn+1 = A + B- Yn-1 Xn+1 = A + B- (1) where A and B are positive numbers and the initial values are positive numbers. In First of all, we consider the change of the variables for system (1) as follows: Xn Уп tn , Zn = A A From this, system (1) transform into following system: tn Zn , Zn+1 1+P Zn-1 tn+1 = 1+P2 (2) 'n-1 where p = > 0. From now on, we study the system (2). Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+T+4p 1+ VI+4p (7, 2) = ( - %3D Now, we consider a transformation as follows: (tn, tn-1, Zn, Zn-1) → (f, fi, 8, 81) where f = 1+ p, fi = tn, 8 = 1+p, , g1 = Zn. Thus we get the jacobian n-1 matrix about equilibrium point (7, z): 1 0 B (f, ३) = 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (ī, 2) XN, where -) In tn-1 XN = uz Zn-1 0 0 물글 0 0 1 B (f, ३) 0 0 0 0 1 Hence, the characteristic equation of B (7, z) about the unique positive equilibrium point (7, 2) is 4p2. 4p2 = 0. Due to t = 7, we can rearrange the characteristic equation such that 4p2. 4p2 = 0. 74 14 74 Therefore, we obtain the four roots of characteristic equation as follows: p+Vp² - Vp² – 8pr? 212 Vp² – 8pi? 272 12 = -p+Vp? + 8pi 272 13 =
Expert Solution
Given

From equation (2), we have

            tn+1=1+pznzn-12                   ...(2a)and             zn+1=1+ptntn-12                   ...(2b)

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