The solution of Equation (3.6) is
An+ = e-(At)An e(S-t+) 7z + ) ds, (3.7)
where superscript n denotes the time level tn, 7 = c/e, and 7 = oz/6 is the
derivative of a/e with respect to z. First, we approximate the integral part of
Equation (3.7), denoted as I:
I = I (s-t+l) 7Z +- ds.
The following steps are conducted.
Step 1: Both ft/Oz and V J are evaluated at time 5(tn + tn+1), which is
denoted as time level n + I. Hence we have the following approximation
1 e-Y(At) / \n+i (V. J)n+)
e ( ( ) ) + ( /J)
Define
d 1 e-y(At)7
-y
Then we can rewrite the above approximation as follows :
(/-n+ d e-Y(At) (V J)n+
I Oz 7 e
Step 2: Use (1 p)(VI/Oz)" + t(I09/Oz)"+11 to approximate the term
(ap/a8z)"n+. Hence,
[ (r ) (Y +1] 1 e-(At) (V. J)+
I -d (1 /)(-z 1+/-" ) J + -
1O ( 9Z1z 7
where 0 < p < 1.