Theorem 9 Suppose that 0 < p < and {(tn, Zn)} be a solution of the system (2) such that lim t, =i and lim z, = 7. Then the error vector n-00 n-00 el En = In - i In-1 -i Zn - Z Zn-1 - i en of every solution of system (2) satisfies both of the following asymptotic relations: lim E, = |21,2,3,4F1(1, 2)|, n00 ||En+1|| |21,2.3,4F/(T, 2)|, lim n-00 ||En|| where A1,2,3,4FJ(i, 2) are the characteristic roots of the Jacobian matrix F,(i, 2). Proof To find the error terms, we set 1 In+l -i =Lai (!n-i - i) +Bi (zn-i – 2), i=0 i=0 1 1 Zn+1 - i = En (In-i - i) + 8; (zn-i – 2), i=0 i=0 and e = tn - i, e, = zn – 7. Thus we have 1 enti =Laie-i + E Bie;i i=0 i=0 1 1 %D en+1 en-i i=0 i=0

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Show me the steps of deremine blue and inf is here i need evey I need all the details step by step and inf is her

Theorem 9 Suppose that 0 < p < and {(tn, Zn)} be a solution of the system (2)
such that lim tn = i and lim zn = 7. Then the error vector
n-00
n-00
In -i
In-1 -i
Zn - 7
En =
e?
´n-1
Zn-1 – 7)
of every solution of system (2) satisfies both of the following asymptotic relations:
lim En| = |21,2,3,4F, (7, 2)|,
n-00
||En+1||
|21,2.3,4F1(, 2)|,
lim
n→∞ || En ||
where 11,2,3,4 FJ(ī, z) are the characteristic roots of the Jacobian matrix F1(ī, z).
Proof To find the error terms, we set
1
1
In+1 –i =
Eai (In-i – i) + E Bi (zn-i – 2),
i=0
i=0
1
1
Zn+1 – 7 =ri (tn-i – i) + ôi (zn-i – 2),
i=0
i=0
and e = tn - i, e, = zn – 7. Thus we have
1
1
= I+"a
i=0
i=0
1
1
ent!
i=0
i=0
where
ao = aj = 0,
-p (7 + Zn-1)
P
Bo =
2
Zn-1
, B1 =
n-1
-p (ī + tn-1)
i
Yo =
, Yi =
12
'n-1
n-1
80 = 81 = 0.
Now we take the limits
lim ¤o
n→00
lim aj = 0,
n-00
-2p
lim Bo =
lim B1
n-00
-2p
lim yo =
n-00
lim
YI
n-00
lim 8o
= lim 81 = 0.
n-00
n-00
Hence
-2p
+ an, B1 =
+ bn,
Bo =
-2p
YO =
+ Cn, Y1 =
+ dn,
where an → 0, b, → 0, cn → 0, dn → 0 as n → ∞. Therefore, we obtain the
system of the form (14)
En+1
(A + B (n)) En
%3D
where
0 0
1
A =
0 0
1
0 0 an bn
1000
B(n) =
Cn dn 0 0
Transcribed Image Text:Theorem 9 Suppose that 0 < p < and {(tn, Zn)} be a solution of the system (2) such that lim tn = i and lim zn = 7. Then the error vector n-00 n-00 In -i In-1 -i Zn - 7 En = e? ´n-1 Zn-1 – 7) of every solution of system (2) satisfies both of the following asymptotic relations: lim En| = |21,2,3,4F, (7, 2)|, n-00 ||En+1|| |21,2.3,4F1(, 2)|, lim n→∞ || En || where 11,2,3,4 FJ(ī, z) are the characteristic roots of the Jacobian matrix F1(ī, z). Proof To find the error terms, we set 1 1 In+1 –i = Eai (In-i – i) + E Bi (zn-i – 2), i=0 i=0 1 1 Zn+1 – 7 =ri (tn-i – i) + ôi (zn-i – 2), i=0 i=0 and e = tn - i, e, = zn – 7. Thus we have 1 1 = I+"a i=0 i=0 1 1 ent! i=0 i=0 where ao = aj = 0, -p (7 + Zn-1) P Bo = 2 Zn-1 , B1 = n-1 -p (ī + tn-1) i Yo = , Yi = 12 'n-1 n-1 80 = 81 = 0. Now we take the limits lim ¤o n→00 lim aj = 0, n-00 -2p lim Bo = lim B1 n-00 -2p lim yo = n-00 lim YI n-00 lim 8o = lim 81 = 0. n-00 n-00 Hence -2p + an, B1 = + bn, Bo = -2p YO = + Cn, Y1 = + dn, where an → 0, b, → 0, cn → 0, dn → 0 as n → ∞. Therefore, we obtain the system of the form (14) En+1 (A + B (n)) En %3D where 0 0 1 A = 0 0 1 0 0 an bn 1000 B(n) = Cn dn 0 0
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)
Expert Solution
Step 1

Given:

  • limntn=t¯
  • limnzn=z¯

To do:

Determine how the following equations arrived in the proof of Theorem 9.

  • tn+1-t¯=i=01αitn-i-t¯+i=01βizn-i-z¯ .......... (1)
  • zn+1-z¯=i=01γitn-i-t¯+i=01δizn-i-z¯ .......... 2
  • en+11=i=01αien-i1+i=01βien-i2 .......... 3
  • en+12=i=01γien-i1+i=01δien-i2 .......... 4
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