The equation Yk+1 = Yk + 1/yk (2.183) %3D does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect yk to become unbounded as k → o. However, we cannot use any of the above- discussed techniques to determine its asymptotic behavior. Note that we cannot use the fact that yk > 1/yk in equation (2.183)

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Chapter2: Second-order Linear Odes
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2.8.3 Example C
The equation
Yk+1 = Yk + 1/yk
(2.183)
does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect yk
to become unbounded as k
0o.
we cannot use any of the above-
discussed techniques to determine its asymptotic behavior.
Note that we cannot use the fact that Yk > 1/yk in equation (2.183)
because the solution of the difference equation yk+1 = Yk is a constant; this
violates the condition that yk - x∞ as k → .
98
To proceed, we square both sides of equation (2.183) and obtain
= y% + 2+1/y.
(2.184)
For this equation, it is valid to neglect y? in comparison with 2 as k → o.
The resulting asymptotic difference equation
(2.185)
k+1
can be solved by making the transformation vk = y. Its solution is
%3D
Vk = 2k, k → 00,
(2.186)
and
Yk = V2k,
k → 00.
(2.187)
This is the dominant asymptotic behavior of yk as k → 0o.
The next-order correction to the result of equation (2.187) is obtained by
letting
Yk = V2k + €k,
(2.188)
where we assume that e « 2k for k → o. Substituting equation (2.188) into
equation (2.184) and neglecting €k in comparison with 2k in the denominator
of the fraction gives
99
Ek+1 – Ek = 1/2k, k → o,
(2.189)
the solution of which is
/½ log k,
(2.190)
Ek =
k → 0.
From the result
V2k + 4n = /2k (1+ )" = v2k (1 + +)
1/2
= V2k (1+
= V2k (1+
2k.
1 log k
4 V2k
+..
(2.191)
V2k
+..
we conclude that the asymptotic behavior of equation (2.183) is
1 log k
+...
(2.192)
Yk = V2k +
4 V2k
Transcribed Image Text:Adobe Scan .l 2:24 AM T @ 9 93% 4 Done 2.8.3 Example C The equation Yk+1 = Yk + 1/yk (2.183) does not cross the line yk+1 = Yk; see Figure 2.9. Therefore, we expect yk to become unbounded as k 0o. we cannot use any of the above- discussed techniques to determine its asymptotic behavior. Note that we cannot use the fact that Yk > 1/yk in equation (2.183) because the solution of the difference equation yk+1 = Yk is a constant; this violates the condition that yk - x∞ as k → . 98 To proceed, we square both sides of equation (2.183) and obtain = y% + 2+1/y. (2.184) For this equation, it is valid to neglect y? in comparison with 2 as k → o. The resulting asymptotic difference equation (2.185) k+1 can be solved by making the transformation vk = y. Its solution is %3D Vk = 2k, k → 00, (2.186) and Yk = V2k, k → 00. (2.187) This is the dominant asymptotic behavior of yk as k → 0o. The next-order correction to the result of equation (2.187) is obtained by letting Yk = V2k + €k, (2.188) where we assume that e « 2k for k → o. Substituting equation (2.188) into equation (2.184) and neglecting €k in comparison with 2k in the denominator of the fraction gives 99 Ek+1 – Ek = 1/2k, k → o, (2.189) the solution of which is /½ log k, (2.190) Ek = k → 0. From the result V2k + 4n = /2k (1+ )" = v2k (1 + +) 1/2 = V2k (1+ = V2k (1+ 2k. 1 log k 4 V2k +.. (2.191) V2k +.. we conclude that the asymptotic behavior of equation (2.183) is 1 log k +... (2.192) Yk = V2k + 4 V2k
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