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In-Depth Information
r
N
r
0
,and
r
N
r
0
Tabl e 3. 1
The table shows t , the number of time steps N , the “relative error”
r
0
t
.
r
0
Note that the numbers
in the last column seem to tend toward a constant
t
r
N
r
0
r
N
r
0
N
r
0
r
0
t
10
1
10
2
1:7048
17.0481
10
2
10
3
10
1
1:0517
10.5165
10
3
10
4
10
2
1:0050
10.0502
10
4
10
5
1:0005
10
3
10.0050
We observe that
r
N
r
0
r
0
t
10;
and thus r
N
.1
C
10t/r
0
. Consequently, as t goes to zero, the numerical
solutions also seem to approach a perfect circle.
Let us consider this issue by analyzing the numerical scheme (
3.36
). Note that
r
nC1
D
.F
nC1
1/
2
C
.S
nC1
1/
2
;
(3.37)
and thus, by using the numerical scheme (
3.36
), we get
r
nC1
D
.F
n
1
C
t.1
S
n
//
2
C
.S
n
1
C
t.F
n
1//
2
D
.F
n
1/
2
C
2t.F
n
1/.1
S
n
/
C
t
2
.1
S
n
/
2
C
.S
n
1/
2
C
2t.F
n
1/.S
n
1/
C
t
2
.1
F
n
/
2
D
r
n
C
t
2
r
n
:
Hence
D
.1
C
t
2
/r
n
;
r
nC1
(3.38)
so by induction,
D
.1
C
t
2
/
m
r
0
:
r
m
(3.39)
Since t
D
10=N ,wehave
D
1
C
N
r
0
:
10
2
N
2
r
N
(3.40)
Now, in order to estimate r
N
r
0
, we have to study the asymptotic behavior of
1
C
10
2
N
2
N
. Note first that
1
C
N
10
2
N
2
D
e
N
ln
.
1C10
2
=N
2
/
:
(3.41)
