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obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 272 p- Vp? - 8pi? 212 12 = -p+Vp? + 8pr? 13 = 212

obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 272 p- Vp? - 8pi? 212 12 = -p+Vp? + 8pr? 13 = 212

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 Yn Xn+1 = A + B- 2 , Yn+1 = A + B- 2 (1) Yn-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 Zn = A Уп tn = A From this, system (1) transform into following system: tn Zn , Zn+1 = 1+p2 Zn-1 tn+1 = 1+P2 (2) %3D 'n-1 where p = > 0. From now on, we study the system (2). A2 Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+T+4p 1+ VI+4p (F, 2) = ( - 2 2 Now, we consider a transformation as follows: (tn, tn-1, Zn, Zn-1) → (f, fi, 8, 81) where f = 1+ p, fi = tn, g = 1 + p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, z): n-1 1 B (,३) = %3D 12 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (i, z) XN, where In tn-1 XN = Zn Zn-1 1 B (ī, 2 : 0 0 1 Hence, the characteristic equation of B (i, z) about the unique positive equilibrium point (7, z) is 4p2 4p2 = 0. 14 Due to t = 7, we can rearrange the characteristic equation such that ?+ 4p°, 4p2 74 14 = 0. Therefore, we obtain the four roots of characteristic equation as follows: p+Vp? – 8pi? 272 – Vp² – 8pi? p- V 12 = 272 -p+Vp? + 8pi? 212 13 = Fo o o o oo
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Yn Xn+1 = A + B- 2 , Yn+1 = A + B- 2 (1) Yn-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 Zn = A Уп tn = A From this, system (1) transform into following system: tn Zn , Zn+1 = 1+p2 Zn-1 tn+1 = 1+P2 (2) %3D 'n-1 where p = > 0. From now on, we study the system (2). A2 Lemma 1 Let p > 0. Unique positive equilibrium point of system (2) is (1+T+4p 1+ VI+4p (F, 2) = ( - 2 2 Now, we consider a transformation as follows: (tn, tn-1, Zn, Zn-1) → (f, fi, 8, 81) where f = 1+ p, fi = tn, g = 1 + p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, z): n-1 1 B (,३) = %3D 12 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (i, z) XN, where In tn-1 XN = Zn Zn-1 1 B (ī, 2 : 0 0 1 Hence, the characteristic equation of B (i, z) about the unique positive equilibrium point (7, z) is 4p2 4p2 = 0. 14 Due to t = 7, we can rearrange the characteristic equation such that ?+ 4p°, 4p2 74 14 = 0. Therefore, we obtain the four roots of characteristic equation as follows: p+Vp? – 8pi? 272 – Vp² – 8pi? p- V 12 = 272 -p+Vp? + 8pi? 212 13 = Fo o o o oo
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