668 Methods for Solving Grid Equationsoperator sk = E - TkA associated with the transition from the ( k - 1 )th
iteration to the kth iteration may be greater than 1 for negative values of
tk.
Before proceeding to further discussions, let us give t.he details of a
simple example.Example The system of equations to be solved is
v(O)=l, v(l)=O, h=l/N,
relating to the problem
u" = 0, 0 < x < 1, u(O) = 1, u(l) = 0,for which the exact solutions are known:
u(x)=l-x, v( X;) = 1 - X;.
In that case the ingredients become
4. 0 7rh 4 27rh 27rh
Ay=-Yxx> 11 = h2 sm"2, /2 = h2 cos 2' ~=tg 2·
For convenience in analysis and clarity, we take into consideration 19 equa-
tions (N=20) and set c = 10-^4. The analytical estimate (33) gives n 0 (c) =
63.2, so that n = 64 and the parameters T 1 , T 2 , ... , Tn for n = 64 are
specified by formulas (29) and (41). The final results of computational
procedures are presented below in Table 3.
Table 3 Table 4
k /::,. k k /::,. k53 0.12^1 39.6
54 1.5^2 2.6.^103
55 27 3 1.6.^105
56 6.3. 102 4 8.2. 106
57 1.9.^104 5 3.7 · 108
58 7.2.^105 6 1.2.^1010
59 3.7. 107 7 3.3. 1011
60 2.6.^109 8 7.0.^1012
61 2.5. 1011 9 1.2. 1014
62 3.3. 1013 10 1.7.^1015
63 5.0.^1015 11 1.9.^1016