Alternative cubic spline derivation for derivative primary variables

Alternative cubic spline derivation for derivative primary variables

Alternative cubic spline derivation for derivative primary variables:
yi
yi+1
Yi (x)
xi+1
xi
Let Yi (x) = ai (x − xi )3 + bi (x − xi )2 + ci (x − xi ) + di
Yi (x) = 3ai (x − xi )2 + 2bi (x − xi ) + ci
Yi (x) = 6ai (x − xi ) + 2bi
yi = Yi (xi ) = di
yi+1 = Yi (xi+1 ) = ai h3i + bi h2i + ci hi + yi where hi = xi+1 − xi .
Let si = Yi (xi ) = ci
Yi (xi )
=
si+1 = Yi (xi+1 ) = 3ai h2i + 2bi hi
2bi = Yi−1 (xi ) = 6ai−1 hi−1 + 2bi−1
3
hi
ai
yi+1 − yi − si hi
h2i
=
3h2i 2hi
bi
si+1 − si
∴ ai =
bi =
∴
∴ 6(
6 Δyi
hi hi
−
(−h4i ) − (si+1 − si )h2i (−h4i )
h3i (si+1 − si ) − 3h2i (yi+1 − yi − si hi )
(−h4i )
si+1 + si
yi+1 − yi
−2
+
h3i
h2
i
si+1 + 2si
yi+1 − yi
3
−
hi
h2i
s + si−1
(s + 2si−1 )
−12 Δyi−1
2 6 Δyi−1
si+1 + 2si =
+6 i
−2 i
+
hi
hi−1 hi−1
hi−1
hi−1 hi−1
hi−1
ai = 2hi (yi+1 − yi − si hi )
bi =
+ si
Δyi
Δyi−1
hi−1 +
hi ) = 2hi−1 si+1 + 4hi−1 si + 4hi si + 2hi si−1
hi
hi−1
∴
hi−1 si+1 + 2(hi−1 + hi )si + hi si−1 = 3(f [xi−1 , xi ]hi + f [xi , xi+1 ]hi−1 ).
23
Another spline formulation with si as primary variables.
yi−1
yi+1
yi
Yi−1 (x)
Yi (x)
Yi+1 (x)
hi
xi
xi−1
xi+1
Let Yi (x) = ai (x − xi )3 + bi (x − xi )2 + ci (x − xi ) + di
Yi (x) = 3ai (x − xi )2 + 2bi (x − xi ) + ci
Yi (x) = 6ai (x − xi ) + 2bi
at xi :
← interpolates the function at xi
yi = Yi (xi ) = di
yi+1 = Yi (xi+1 ) =
ai h3i
+
bi h2i
(1)
+ ci hi + yi ← enforce continuity
Introduce primary variable : si = Yi (xi ) = 2bi ⇒ bi = si /2
Build in continuity of Y ⇒ si+1 = Yi (xi+1 ) = 6ai hi + si ⇒ ai =
(2)
(3)
s
i+1 −si
6hi
(4)
∴ Yi (x) =
(2) ⇒
yi+1 =
ci =
=
si+1 − si
s
(x − xi )3 + (x − xi )2 + ci (x − xi ) + yi
6hi
2
si+1 − si
s
h3i + i h2i + ci hi + yi
6hi
2
si+1 − si
yi+1 − yi
s
hi − i hi
−
hi
6
2
s
+
2s
Δyi
i+1
i
hi
−
hi
6
Impose continuity of first derivatives:
Yi (xi ) = 3ai (xi − xi )2 + 2bi (xi − xi ) + ci = ci
∴ ci =
Δyi
−
hi
Yi−1
(xi ) = 3ai−1 h2i−1 + 2bi−1 hi−1 + ci−1
si+1 + 2si
si − si−1
si + 2si−1
Δyi−1
2
hi = 3
hi−1
−
hi−1 + si−1 hi−1 +
6
6hi−1
hi−1
6
hi−1 si−1 (−3 + 6 − 2) + (2hi + 2hi−1 )si + hi si+1
=6
or
hi−1 si−1 + (2hi + 2hi−1 )si + hi si+1
=6
i = 1, . . . N − 1
N − 1 EQ
2 EXTRA CONDITIONS
24
Δyi
hi
Δyi
hi
−
−
Δyi−1
hi−1
Δyi−1
hi−1