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@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)
 
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