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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)

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)

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 her

Motivated by difference equations and their systems, we consider the following system of difference equations Xn Уп , Yn+1 = A + B- Уп-1 Xn+1 = A + B- (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: Zn , Zn+1 = 1 +p, n-1 tn+1 = 1+P7 (2) %3D 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 = (A + B (n)) En, (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)
Transcribed Image Text:Motivated by difference equations and their systems, we consider the following system of difference equations Xn Уп , Yn+1 = A + B- Уп-1 Xn+1 = A + B- (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: Zn , Zn+1 = 1 +p, n-1 tn+1 = 1+P7 (2) %3D 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 = (A + B (n)) En, (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 < such that lim tn =i and lim z, = 7. Then the error vector 5 and {(tn, Zn)} be a solution of the system (2) n-00 n-00 tn - i tn-1 - i Zn - Z en En = n-1 en n-1 Zn-1 – 2 of every solution of system (2) satisfies both of the following asymptotic relations: lim VE,| = |à1,2,3,4F1 (7, 7)|, n-00 ||En+1|| lim = |à1,2,3,4 F1(T, 2)|, n→00 || En|| where A1,2,3,4FJ (7, 7) are the characteristic roots of the Jacobian matrix Fj(i, 7). Proof To find the error terms, we set 1 1 In+l -i =a; (tn-i – i) +Bi (Zn-i – 7) , i=0 i=0 1 1 Zn+1 – ē = Eri (tn-i - i) + ôi (zn-i – 2), i=0 i=0 and e = t, – i, e? = zn – 7. Thus we have 1 1 %3D i=0 i=0 1 n-i i=0 i=0 where a0 = a1 = 0, -p (7 + Zn-1) Bo = B1 %3D Zn-1 -p(i + fn-1). Yo = , Yi = 12 'n-1 8o = 81 = 0. Now we take the limits lim ao = lim aj = 0, n-00 n-00 lim Bo = n-00 -2p lim B1 = n00 -2p lim yo = n-00 lim Yi = 12'n-00 lim 8o = lim 81 = 0. n-00 n-00 Hence -2p Bo = + an, Pi = + bn, -2p + Cn, Y1 = + dn, YO = where a, → 0, b, → 0, cn → 0, d, → 0 as n → 00. Therefore, we obtain the system of the form (14) En+1 = (A + B (n)) En where 1 A = 1 0 0 an bn 100 0 B(n) = Cn dn 0 0 0 0 1 0 io o o
Transcribed Image Text:Theorem 9 Suppose that 0 <p < such that lim tn =i and lim z, = 7. Then the error vector 5 and {(tn, Zn)} be a solution of the system (2) n-00 n-00 tn - i tn-1 - i Zn - Z en En = n-1 en n-1 Zn-1 – 2 of every solution of system (2) satisfies both of the following asymptotic relations: lim VE,| = |à1,2,3,4F1 (7, 7)|, n-00 ||En+1|| lim = |à1,2,3,4 F1(T, 2)|, n→00 || En|| where A1,2,3,4FJ (7, 7) are the characteristic roots of the Jacobian matrix Fj(i, 7). Proof To find the error terms, we set 1 1 In+l -i =a; (tn-i – i) +Bi (Zn-i – 7) , i=0 i=0 1 1 Zn+1 – ē = Eri (tn-i - i) + ôi (zn-i – 2), i=0 i=0 and e = t, – i, e? = zn – 7. Thus we have 1 1 %3D i=0 i=0 1 n-i i=0 i=0 where a0 = a1 = 0, -p (7 + Zn-1) Bo = B1 %3D Zn-1 -p(i + fn-1). Yo = , Yi = 12 'n-1 8o = 81 = 0. Now we take the limits lim ao = lim aj = 0, n-00 n-00 lim Bo = n-00 -2p lim B1 = n00 -2p lim yo = n-00 lim Yi = 12'n-00 lim 8o = lim 81 = 0. n-00 n-00 Hence -2p Bo = + an, Pi = + bn, -2p + Cn, Y1 = + dn, YO = where a, → 0, b, → 0, cn → 0, d, → 0 as n → 00. Therefore, we obtain the system of the form (14) En+1 = (A + B (n)) En where 1 A = 1 0 0 an bn 100 0 B(n) = Cn dn 0 0 0 0 1 0 io o o
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     Given,      tn+1-t¯=∑i=01αi(tn-i-t¯)+∑i=01βi(zn-i-z¯)      zn+1-z¯=∑i=01γi(tn-i-t¯)+∑i=01δi(zn-i-z¯)

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