Information Technology Reference
In-Depth Information
@f
k
!
1
f
g
k
x
kC1
y
kC1
D
x
y
k
;
@f
k
@y
@x
(4.174)
@g
k
@x
@g
k
@y
where
f
k
D
f.x
k
;y
k
/; g
k
D
g.x
k
;y
k
/;
@f
k
@x
@f
@x
.x
k
;y
k
/
D
1;
@f
k
@y
@f
@y
.x
k
;y
k
/
D
3t y
k
D
D
;
(4.175)
@g
k
@x
@g
@x
.x
k
;y
k
/
D
3t x
k
@g
k
@y
@g
@y
.x
k
;y
k
/
D
1:
D
D
;
Note that the matrix
A
D
@f
k
!
D
@f
k
@y
t
k
13
@x
(4.176)
@g
k
@x
@g
k
@y
3t x
k
1
has a determinant given by
det.
A
/
D
1
C
9t
2
x
k
y
k
>0;
(4.177)
and so
A
1
is well defined and is given by
1
:
3t y
k
1
1
C
9t
2
x
k
A
1
D
(4.178)
3t x
k
y
k
1
In order to test the method, we consider system (
4.169
) with
u
0
D
1 and
v
0
D
0.
At each time level, we iterate using (
4.174
), until
j
f.x
k
;y
k
/
j C j
g.x
k
;y
k
/
j
<"
D
10
6
:
(4.179)
We choose t
D
1=100 and compute the numerical solution from t
D
0 to t
D
1.
The numerical approximations, as functions of t , are plotted in Figs.
4.9
and
4.10
in the .
u
;
v
/-coordinate system. In Fig.
4.11
we have plotted the number of Newton
iterations needed to reach the stopping criterion (
4.179
) at each time level. Observe
that we need no more than two iterations at all time levels.
4.7
Exercises
Exercise 4.1.
Consider the function
f.x/
D
e
x
1:
(4.180)