( zα + h)2 h2
f2 =
-
(26b)
2
2
h
f3 = ( zα + h) -
(26c)
2
f4 = [( zα + h) - h]
(26d)
The average and difference operators in Equations 21 to 25 are given by:
1 n+1
(φ - φn )
δt φ =
(27a)
∆t
1 n
(φ + φn+1 )
φn+1/ 2 =
(27b)
2
1
(φi+1, j + φi, j ) or 1 (φi+1/ 2, j + φi-1/ 2, j )
x
φ=
(27c)
2
2
1
(φi, j +1 + φi, j ) or 1 (φi, j +1/ 2 + φi, j -1/ 2 )
y
φ=
(27d)
2
2
1
1
(φi+1, j - φi, j ) or ∆x (φi+1/ 2, j - φi-1/ 2, j )
δ(x1)φ =
(27e)
∆x
1
1
(φi, j +1 - φi, j ) or ∆y (φi, j +1/ 2 - φi, j -1/ 2, j )
δ(y1)φ =
(27f)
∆y
1
(φi+3/ 2, j - φi-1/ 2, j )
δ(x2)φ =
(27g)
2∆x
1
(φi, j +3/ 2 - φi, j -1/ 2 )
δ(y2)φ =
(27h)
2∆y
1
(φi+3/ 2, j - 2φi+1/ 2, j + φi-1/ 2, j )
δxxφ =
(27i)
∆x2
1
(φi, j +3/ 2 - 2φi, j +1/ 2 + φi, j-1/ 2 )
δ yy φ =
(27j)
∆y 2
1
(φi, j+1 - 2φi, j + φi, j-1 )
δxy φ =
(27k)
∆x∆y
14
Chapter 3 Numerical Solution
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