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lim ao lim a1 = 0, %3D n-00 n-00 -2p lim Bo lim Bi n-00 -2p lim yi = lim yo = n→00 lim 8o = lim 81 == 0. n→00 n-00

lim ao lim a1 = 0, %3D n-00 n-00 -2p lim Bo lim Bi n-00 -2p lim yi = lim yo = n→00 lim 8o = lim 81 == 0. n→00 n-00

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

1 + i=0 i=0 where a0 = aj = 0, -p (z + Zn-1) B1 = B : „2 "n-1 P YO = +2 'n-1 -p (i + in-1) it-1 , Yi = 8o = 81 = 0. Now we take the limits lim α0 - lim ¤1 = 0, n→00 lim Bo -2p lim B1 = -2p lim yo lim Yi = 12' n→00 n→00 lim 80 lim 81 = 0. n→00 Hence -2p + an, Bi = Bo = bn, -2p + dn, + Cn, Y1 = YO = 0, bn system of the form (14) where an -→ → 0, cn → 0, dn → 0 as n → ∞. Therefore, we obtain the En+1 —D (А + В (п)) En where 1 A = 0 0 аn bn 1 0 0 0 В (п) — Сп dn 0 0 0 0 1 0 ||
Transcribed Image Text:1 + i=0 i=0 where a0 = aj = 0, -p (z + Zn-1) B1 = B : „2 "n-1 P YO = +2 'n-1 -p (i + in-1) it-1 , Yi = 8o = 81 = 0. Now we take the limits lim α0 - lim ¤1 = 0, n→00 lim Bo -2p lim B1 = -2p lim yo lim Yi = 12' n→00 n→00 lim 80 lim 81 = 0. n→00 Hence -2p + an, Bi = Bo = bn, -2p + dn, + Cn, Y1 = YO = 0, bn system of the form (14) where an -→ → 0, cn → 0, dn → 0 as n → ∞. Therefore, we obtain the En+1 —D (А + В (п)) En where 1 A = 0 0 аn bn 1 0 0 0 В (п) — Сп dn 0 0 0 0 1 0 ||
Motivated by difference equations and their systems, we consider the following system of difference equations Уп Xn+1 = A + B Yn-1 Xn , Yn+1 = A+B- 2 (1) Xn-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 tn+1 = 1+P Zn Zn+1 = 1+ p (2) Zn-1 'n-1 where p = > 0. From now on, we study the system (2). Now we study the rate of convergence of system (2). Hence, we consider the following system: En+1 (А + B (п)) Еп, (14) where En is a k-dimensional vector, A e Ckxk is a constant matrix, and B : Z+ → Ckxk is a matrix function satisfying || B (n)|| – → 0, (15) Theorem 9 Suppose that 0 < p < ; and {(tn, Zn)} be a solution of the system (2) such that lim tn t and lim zn = 7. Then the error vector %3D -1 n-1 – En = Zn - 2 Zn-1 n-1 of every solution of system (2) satisfies both of the following asymptotic relations: lim |E,| = |21,2.3,4F1(ī, 7)|, n→00 || En+1|| lim = |A1,2.3,4F1(ī, 2)|, n→00 ||En|| where 11,2,3,4 F,(i, 7) are the characteristic roots of the Jacobian matrix FJ (i, 7). Proof To find the error terms, we set 1 1 In+1 - i =a; (tn-i – i) +Bi (zn-i – 2), i=0 i=0 1 Zn+1 = =Evi (in-i –i) +8; (zn-i – E), | i=0 i=0 and e, = tn - i, e, = zn – z. Thus we have 1 1 n+1 n-i i=0 i=0
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Уп Xn+1 = A + B Yn-1 Xn , Yn+1 = A+B- 2 (1) Xn-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 tn+1 = 1+P Zn Zn+1 = 1+ p (2) Zn-1 'n-1 where p = > 0. From now on, we study the system (2). Now we study the rate of convergence of system (2). Hence, we consider the following system: En+1 (А + B (п)) Еп, (14) where En is a k-dimensional vector, A e Ckxk is a constant matrix, and B : Z+ → Ckxk is a matrix function satisfying || B (n)|| – → 0, (15) Theorem 9 Suppose that 0 < p < ; and {(tn, Zn)} be a solution of the system (2) such that lim tn t and lim zn = 7. Then the error vector %3D -1 n-1 – En = Zn - 2 Zn-1 n-1 of every solution of system (2) satisfies both of the following asymptotic relations: lim |E,| = |21,2.3,4F1(ī, 7)|, n→00 || En+1|| lim = |A1,2.3,4F1(ī, 2)|, n→00 ||En|| where 11,2,3,4 F,(i, 7) are the characteristic roots of the Jacobian matrix FJ (i, 7). Proof To find the error terms, we set 1 1 In+1 - i =a; (tn-i – i) +Bi (zn-i – 2), i=0 i=0 1 Zn+1 = =Evi (in-i –i) +8; (zn-i – E), | i=0 i=0 and e, = tn - i, e, = zn – z. Thus we have 1 1 n+1 n-i i=0 i=0
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Given,

                                      en+12=∑γii=11en-i1+∑δii=11en-i1where                      α0=α1=0                     β0=pzn-12, β1=-pz+zn-1zzn-12                     γ0=ptn-12, γ1=-pt+tn-1ttn-12                     δ0=δ1=0

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