50
This equation is solved for Wn+1 in terms of Wn
(1 + 2AI + 2A y)Wn+1 = Wn
(4.8)
where
1 AT
= 2AAX{-
and
iBeZi
2 AY
(4.9)
4Aj;AvWn+i is added to the left hand side of the equation and 2AvWn is added and sub
tracted to the right hand side of Eq. (4.8) so that the equation can be factored to get
(1 + 2A.)(1 + 2Av)Wn+l = (1 2Av)Wn + 2AvWn (4.10)
An intermediate value W is introduced so that the above equation can be written as
two equations
(l + 2AI)W* = (l-2A)Wn
(4.11)
(1 + 2Xv)Wn+1 = W* + 2A vWn
(4.12)
Note that by eliminating W* from the two Eqs. (4.11) and (4.12), Eq. (4.8) is recovered
with the addition of an error of 4A2AyWn+l which has been introduced.
Writing out the component equations of Eq. (4.11) gives
AT AT
v* = yn -9^Svfjn
(4.13)
(4.14)
(4.15)
(4.16)
While writing out the component equations of Eq. (4.12) gives
Vn+1 + = n* + D^6yvn
un+1 = u*
yn+1 c f¡n+l y c =n
v +9AY6yV ~v +gAYW
(4.17)
(4.18)
(4.19)