Sources/misc/coefpoly.f

module coefpoly_m

  IMPLICIT NONE

contains

  SUBROUTINE coefpoly(xf1, xf2, xprim1, xprim2, xtild1, xtild2, a0, a1, a2, a3)

    ! From LMDZ4/libf/dyn3d/coefpoly.F,v 1.1.1.1 2004/05/19 12:53:05

    ! ...  Auteur :   P. Le Van  ...

    ! Calcul des coefficients a0, a1, a2, a3 du polynome de degre 3 qui
    ! satisfait aux 4 equations  suivantes :

    ! a0 + a1*xtild1 + a2*xtild1*xtild1 + a3*xtild1*xtild1*xtild1 = Xf1
    ! a0 + a1*xtild2 + a2*xtild2*xtild2 + a3*xtild2*xtild2*xtild2 = Xf2
    ! a1  +     2.*a2*xtild1 +     3.*a3*xtild1*xtild1 = Xprim1
    ! a1  +     2.*a2*xtild2 +     3.*a3*xtild2*xtild2 = Xprim2

    ! On en revient a resoudre un systeme de 4 equat.a 4 inconnues a0,a1,a2,a3

    DOUBLE PRECISION, intent(in):: xf1, xf2, xprim1, xprim2, xtild1, xtild2
    DOUBLE PRECISION, intent(out):: a0, a1, a2, a3

    ! Local:
    DOUBLE PRECISION xtil1car, xtil2car, derr, x1x2car

    !------------------------------------------------------------

    xtil1car = xtild1*xtild1
    xtil2car = xtild2*xtild2

    derr = 2.*(xf2-xf1)/(xtild1-xtild2)

    x1x2car = (xtild1-xtild2)*(xtild1-xtild2)

    a3 = (derr+xprim1+xprim2)/x1x2car
    a2 = (xprim1-xprim2+3.*a3*(xtil2car-xtil1car))/(2.*(xtild1-xtild2))

    a1 = xprim1 - 3.*a3*xtil1car - 2.*a2*xtild1
    a0 = xf1 - a3*xtild1*xtil1car - a2*xtil1car - a1*xtild1

  END SUBROUTINE coefpoly

end module coefpoly_m
1	guez	120	module coefpoly_m
2	guez	3
3	guez	81	IMPLICIT NONE
4	guez	3
5	guez	120	contains
6	guez	3
7	guez	120	SUBROUTINE coefpoly(xf1, xf2, xprim1, xprim2, xtild1, xtild2, a0, a1, a2, a3)
8	guez	3
9	guez	120	! From LMDZ4/libf/dyn3d/coefpoly.F,v 1.1.1.1 2004/05/19 12:53:05
10	guez	3
11	guez	120	! ... Auteur : P. Le Van ...
12	guez	3
13	guez	120	! Calcul des coefficients a0, a1, a2, a3 du polynome de degre 3 qui
14			! satisfait aux 4 equations suivantes :
15	guez	3
16	guez	120	! a0 + a1xtild1 + a2xtild1xtild1 + a3xtild1xtild1xtild1 = Xf1
17			! a0 + a1xtild2 + a2xtild2xtild2 + a3xtild2xtild2xtild2 = Xf2
18			! a1 + 2.a2xtild1 + 3.a3xtild1*xtild1 = Xprim1
19			! a1 + 2.a2xtild2 + 3.a3xtild2*xtild2 = Xprim2
20	guez	3
21	guez	120	! On en revient a resoudre un systeme de 4 equat.a 4 inconnues a0,a1,a2,a3
22	guez	81
23	guez	120	DOUBLE PRECISION, intent(in):: xf1, xf2, xprim1, xprim2, xtild1, xtild2
24			DOUBLE PRECISION, intent(out):: a0, a1, a2, a3
25	guez	81
26	guez	120	! Local:
27			DOUBLE PRECISION xtil1car, xtil2car, derr, x1x2car
28	guez	81
29	guez	120	!------------------------------------------------------------
30	guez	81
31	guez	120	xtil1car = xtild1*xtild1
32			xtil2car = xtild2*xtild2
33	guez	81
34	guez	120	derr = 2.*(xf2-xf1)/(xtild1-xtild2)
35
36			x1x2car = (xtild1-xtild2)*(xtild1-xtild2)
37
38			a3 = (derr+xprim1+xprim2)/x1x2car
39			a2 = (xprim1-xprim2+3.a3(xtil2car-xtil1car))/(2.*(xtild1-xtild2))
40
41			a1 = xprim1 - 3.a3xtil1car - 2.a2xtild1
42			a0 = xf1 - a3xtild1xtil1car - a2xtil1car - a1xtild1
43
44			END SUBROUTINE coefpoly
45
46			end module coefpoly_m