ABSTRACT
The four terms in parentheses involving Xi and Xi+1 on the right-hand side of Eg. (12.93) reduce as follows:
Term I: LUi Term 2: -LUT /2 Term 3: LU; /3 Term 4: -LU; /6 (12.94)
Substituting Eg. (12.94) into Eg. (12.93) and dividing through by 2 yields
a[(i) (yo + yo) j;(i) LU Q-(i)y LU Q-(i)y. LU _ = _ I-I I + I _ I I _ 1+1 I aYi LUi 2 3 6
(12.95)
The first integral in Eg. (12.79), a[u-I) jay;, is evaluated by repeating the steps presented above. The result is
(y o -yo I) j;U-Ij LU II 1-+ 1LUi_I 2
Q-U-Ij LUYi-I i-I 6
(12.96)
