122
Chapter 20.
Solutions: Solution of Ordinary DiFerential Equations
•
perhaps
h
=
h/
2.
•
perhaps
h
=
h/
2
p
where, since we need
δ
2
−
5
p
≈
τ
, we defne
p
= (log
δ
−
log
τ
)
/
(5 log 2).
CHALLENGE 20.10.
We know that iF our old values are correct,
y
P
n
+1
−
y
(
t
n
+1
)=
3
h
4
8
y
(4)
(
η
)
.
y
C
n
+1
−
y
(
t
n
+1
−
h
4
24
y
(4)
(
ν
)
.
Subtracting, we obtain
y
P
n
+1
−
y
C
n
+1
=
3
h
4
8
y
(4)
(
η
)
−
(
−
h
4
24
y
(4)
(
ν
))
,
where
η, ν
are in the interval containing
y
P
n
+1
,
y
C
n
+1
, and the true value. Since 3
/
8+
1
/
24 = 10
/
24, the error in the corrector can be estimated as
±
=

y
P
n
+1
−
y
C
n
+1

/
10.
Now, iF
±>τ
, we might reduce
h
by a Factor oF 2 and retake the step. IF
±<<τ
,
we might double
h
in preparation For the next step (expecting that the local error
might increase by a Factor oF 2
4
).
CHALLENGE 20.11.
No answer provided.
CHALLENGE 20.12.
y
±
(
t
D
∇
y
H
(
y
)(
t
)
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

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