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system of difference equations Yn 2Yn+1 = A + B- Yn-1 Xn 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 In = A Уп = 42. A From this, system (1) transform into following system: Zn In+1 = 1+p2 Zn+1 = n-1 In 1+P (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+ /T+4p (7, 2) = 2 2 Now, we consider a transformation as follows: (In, In-1, Zn, Zn-1) → (f, fi, 8, g1) where f = 1+ p, fi = tn, 8 = 1+ p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, 2): n-I n-1 0 0 0 0 0 1 B (, ३) = 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (i, 2) XN, where In In-1 XN = uz Zn-1 0 0 1 B (i, 2 : 号部0

system of difference equations Yn 2Yn+1 = A + B- Yn-1 Xn 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 In = A Уп = 42. A From this, system (1) transform into following system: Zn In+1 = 1+p2 Zn+1 = n-1 In 1+P (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+ /T+4p (7, 2) = 2 2 Now, we consider a transformation as follows: (In, In-1, Zn, Zn-1) → (f, fi, 8, g1) where f = 1+ p, fi = tn, 8 = 1+ p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, 2): n-I n-1 0 0 0 0 0 1 B (, ३) = 0 0 1 Thus, the linearized system of system (2) about the unique positive equilibrium point is given by XN+1 = B (i, 2) XN, where In In-1 XN = uz Zn-1 0 0 1 B (i, 2 : 号部0

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 red and inf is here i need evey I need all the details step by step and inf is here

Motivated by difference equations and their systems, we consider the following system of difference equations Ул Xn 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 tn = Уп , Zn = A A From this, system (1) transform into following system: tn Zn Zn+1 = 1+ p n-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+ /I+4p 1+ VI+4p (f,३) = 2 Now, we sider a transfo as follows: (In, In-1, Zn, Zn-1) → (f, f1, g, g1) where f = 1+ p, fi = tn, 8 = 1+ p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, z): P: 'n-1 n-I 1 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 (i, z) X N, where In tn-1 XN = uz Zn-1 0 0 울글 1 0 0 B (i, 2): -2p 1 Hence, the characteristic equation of B (î, z) about the unique positive equilibrium point (7, z) is 4p2 4p2 = 0. Due to t = 7, we can rearrange the characteristic equation such that 4p2 4p? = 0. 74 Therefore, we obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 212 Vp² – 8 pi? 272 P-VP- 12 = -p+Vp? + 8pi? 212 13 =
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Ул Xn 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 tn = Уп , Zn = A A From this, system (1) transform into following system: tn Zn Zn+1 = 1+ p n-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+ /I+4p 1+ VI+4p (f,३) = 2 Now, we sider a transfo as follows: (In, In-1, Zn, Zn-1) → (f, f1, g, g1) where f = 1+ p, fi = tn, 8 = 1+ p, 81 = Zn. Thus we get the jacobian matrix about equilibrium point (7, z): P: 'n-1 n-I 1 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 (i, z) X N, where In tn-1 XN = uz Zn-1 0 0 울글 1 0 0 B (i, 2): -2p 1 Hence, the characteristic equation of B (î, z) about the unique positive equilibrium point (7, z) is 4p2 4p2 = 0. Due to t = 7, we can rearrange the characteristic equation such that 4p2 4p? = 0. 74 Therefore, we obtain the four roots of characteristic equation as follows: p+ Vp? – 8pi? 212 Vp² – 8 pi? 272 P-VP- 12 = -p+Vp? + 8pi? 212 13 =
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