1549301742-The_Theory_of_Difference_Schemes__Samarskii

Other problems^197

We introduce on the segrnent 0 ::; r < R the equidistant grid w h =

{r; = ih, i = 0, 1, ... , N, hN = R} and by analogy with (31) may attempt

the scheme in the form

(59) Ayi = 7'

1

2 (r;_ 112 a; Yr,i)r,i - di Yi= -tpi' i=l,2, ... ,N-l.

'

At the point r = 0 we impose the difference boundary condition

(60)

a1 y,. o

I ' - qo Yo = - fo '

'*

h = h

* 6 '

and for i = N set

(61)

These equations adn1it the forn1 (39) and can be solved by the elimination
method. The n1ain difference fron1 the cylindrical case lies in the selection
rules for tp and d. To obtain the formulae for tp and d, we consider the
residual

(62) 1/J; = Au; + tp, = ~r. 2 (ri-^2 112 ai ur , ;) r,i. - di 1li + tpi

z

and subtract from equality (62) the balance equation written on the segment

T'i-1/2 < T' < T'i+1;2

(^10) i+1/2 (^1) 'i+1/2
Wi+l/2 - Wi-1/2^1
J
2 1
J
')
0= h T'2 --- q u r dr + -- 2 fr~ dr,
z h r^2 h T'
' '"i-1/2 ' (^1) 'i-1/2
w = r^2 ku^1.
The outco111e of this is
1 *
1/J,: = r2^1 71·,i + 1/J; '

• ,2 1)^0

(^1) 7£ - (^1) i-1 /2 i )
O I
17; =a,: 1t,.; ' - (ku )i-1/2,
'
where
'"i+1/2 r,:+1/2
(63) 1/J7 dr).
1
J
f r^2 dr - (cl u - -
1

• J

= tp i - y;:--;; q u T'2

T'':' ' ' h r2

' ri-1/2 z ri-1/2