1549301742-The_Theory_of_Difference_Schemes__Samarskii

The summarized approximation method 615

Proof Let z(a) = zj+a/p. As usual, we 111ay atternpt a solution z(oJ =

Y(a) - uj+a/p of proble1n (25) as a sum z(o) = v(a) + T/(a) with 1)(cr) still

subject to the conditions

(38)

-----T/(a) - T/(a-1) = 1/J^0 °'

T for xEwh+i1ia, '

cx=l,2, ... ,p,

17( x) 0) = u.

From here we deduce for j = 0, 1, ... , j 0 that

0 0 0

1)j +I = 1)(p) = T)j + T ( 1/J I + 1/J 2 + · · · + 1/J p) = T)j = 0 ,

since T/o = 0. vVith regard to 17( a) it is plain to show that

0 0 0 0 0

1)(a) = T ( 1/J I + l/J 2 + ... + i,/i a) = -T (i,/i a+I + ... + 1/J p) ·

The function v(c>) is yet to be recovered from the conditions

(39) CY=l,2, ... ,p,

v(a) = -TJ(a) for x E lh,a, v(x, 0) = 0,

where i,b°' = 1/J: + AaT/(a)·
A solution of problem (39) can readily be evaluated on account of
Theore1n 1 in Section 7 with v = 0 incorporated fort = 0, stating that

(40)

j-1 p

+ 2= r 2= ll~j+a/pllc

j'=O a=I

If there exist for CY i- f3 the derivatives 84 ·u/ OX~ oa;~ continuous in the

closed cl0111ain Qy, then at all the nodes .i: E wh

0 0

Aa T/( °') = -T Aa ( 1/J a+ l + · · · + 1/J p) = 0( T) ,