PENELOPE 2003 - OECD Nuclear Energy Agency

218 Appendix B. Numerical tools
To obtain the spline coefficients a i , b i , c i , d i (i = 1, . . . , N − 1) we start from the fact
that ϕ ′′ (x) is linear in [x i , x i+1 ]. Introducing the quantities
and
we can write the obvious identity
h i ≡ x i+1 − x i (i = 1, . . . , N − 1) (B.7)
σ i ≡ ϕ ′′ (x i ) (i = 1, . . . , N), (B.8)
p ′′ x i+1 − x x − x i
i (x) = σ i + σ i+1 (i = 1, . . . , N − 1). (B.9)
h i
h i
Notice that x i+1 must be larger than x i in order to have h i > 0. Integrating eq. (B.9)
twice with respect to x, gives for i = 1, . . . , N − 1
p i (x) = σ i
(x i+1 − x) 3
6h i
+ σ i+1
(x − x i ) 3
6h i
+ A i (x − x i ) + B i (x i+1 − x), (B.10)
where A i and B i are constants. These can be determined by introducing the expression
(B.10) into eqs. (B.4), this gives the pair of eqs.
σ i
h 2 i
6 + B ih i = f i and σ i+1
h 2 i
6 + A ih i = f i+1 . (B.11)
Finally, solving for A i and B i and substituting the result in (B.10), we obtain
p i (x) = σ i
6
[
(xi+1 − x) 3
]
x i+1 − x
− h i (x i+1 − x) + f i
h i h i
[ (x − xi ) 3
]
x − x i
− h i (x − x i ) + f i+1 .
h i h i
+ σ i+1
6
(B.12)
To be able to use ϕ(x) to approximate f(x), we must find the second derivatives
σ i (i = 1, . . . , N). To this end, we impose the conditions (B.5). Differentiating (B.12)
gives
p ′ i(x) = σ i
6
[
− 3(x i+1 − x) 2 ]
+ h i + σ i+1
h i 6
[ 3(x − xi ) 2
h i
− h i
]
+ δ i , (B.13)
where
Hence,
δ i = y i+1 − y i
h i
. (B.14)
p ′ h i
i(x i+1 ) = σ i
6 + σ h i
i+1
3 + δ i, (B.15a)
p ′ i(x i ) = −σ i
h i
3 − σ i+1
h i
6 + δ i (B.15b)