Information Technology Reference
In-Depth Information
When the matrix is given by (
3.57
), the inverse is
:
1
1
C
t
2
=4
1
t=2
A
1
D
(3.64)
t=2
1
Consequently, we have
F
nC1
S
nC1
D
F
n
C
t
2
1
1
C
t
2
=4
1
t=2
S
n
:
(3.65)
t
2
S
n
t
C
t=2
1
F
n
By multiplying the matrix by the vector and re-arranging terms, we get
1
t
2
=4
F
n
C
t
2
C
1
tS
n
;
1
1Ct
2
=4
F
nC1
D
(3.66)
1
t
2
=4
S
n
C
t
2
1
C
tF
n
:
1
1Ct
2
=4
S
nC1
D
In Fig.
3.7
, we have plotted the state space solution computed by this scheme for
S
0
D
0:1, F
0
D
0:9,andt
D
1=1;000. Again we note that the numerical solution
seems to form a perfect circle. In order to analyze this we define, as above,
D
.F
n
1/
2
C
.S
n
1/
2
r
n
(3.67)
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
F
Fig. 3.7
The numerical solution for (
3.53
)intheF -S coordinate system, produced by the Crank-
Nicolson scheme (
3.66
)usingt
D
1=1;000, F
0
D
0:9 and S
0
D
0:1
