Homec4science
Diffusion SPClibs (feature/refactoring) Browse

spclibs/pppack/cubspl.f90
8f478ec0accefeature/refactoring

Last Change
Raw File

cubspl.f90
View Options

!>
!> @file cubspl.f90
!>
!> @brief
!>
!> @copyright
!> Copyright (©) 2021 EPFL (Ecole Polytechnique Fédérale de Lausanne)
!> SPC (Swiss Plasma Center)
!>
!> SPClibs is free software: you can redistribute it and/or modify it under
!> the terms of the GNU Lesser General Public License as published by the Free
!> Software Foundation, either version 3 of the License, or (at your option)
!> any later version.
!>
!> SPClibs is distributed in the hope that it will be useful, but WITHOUT ANY
!> WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS
!> FOR A PARTICULAR PURPOSE. See the GNU General Public License for more details.
!>
!> You should have received a copy of the GNU Lesser General Public License
!> along with this program. If not, see <https://www.gnu.org/licenses/>.
!>
!> @author
!> (in alphabetical order)
!> @author Trach-Minh Tran <trach-minh.tran@epfl.ch>
!>
subroutine cubspl ( tau, c, n, ibcbeg, ibcend )
!*******************************************************************************
!
!! CUBSPL defines an interpolatory cubic spline.
!
! Discussion:
!
! A tridiagonal linear system for the unknown slopes S(I) of
! F at TAU(I), I=1,..., N, is generated and then solved by Gauss
! elimination, with S(I) ending up in C(2,I), for all I.
!
! Reference:
!
! Carl DeBoor,
! A Practical Guide to Splines,
! Springer Verlag.
!
! Parameters:
!
! Input, real ( kind = 8 ) TAU(N), the abscissas or X values of
! the data points. The entries of TAU are assumed to be
! strictly increasing.
!
! Input, integer N, the number of data points. N is
! assumed to be at least 2.
!
! Input/output, real ( kind = 8 ) C(4,N).
! On input, if IBCBEG or IBCBEG is 1 or 2, then C(2,1)
! or C(2,N) should have been set to the desired derivative
! values, as described further under IBCBEG and IBCEND.
! On output, C contains the polynomial coefficients of
! the cubic interpolating spline with interior knots
! TAU(2) through TAU(N-1).
!
! In the interval interval (TAU(I), TAU(I+1)), the spline
! F is given by
!
! F(X) =
! C(1,I) +
! C(2,I) * H +
! C(3,I) * H**2 / 2 +
! C(4,I) * H**3 / 6.
!
! where H=X-TAU(I). The routine PPVALU may be used to
! evaluate F or its derivatives from TAU, C, L=N-1,
! and K=4.
!
! Input, integer IBCBEG, IBCEND, boundary condition indicators.
!
! IBCBEG=0 means no boundary condition at TAU(1) is given.
! In this case, the "not-a-knot condition" is used. That
! is, the jump in the third derivative across TAU(2) is
! forced to zero. Thus the first and the second cubic
! polynomial pieces are made to coincide.
!
! IBCBEG=1 means the slope at TAU(1) is to equal the
! input value C(2,1).
!
! IBCBEG=2 means the second derivative at TAU(1) is
! to equal C(2,1).
!
! IBCEND=0, 1, or 2 has analogous meaning concerning the
! boundary condition at TAU(N), with the additional
! information taken from C(2,N).
!
implicit none
integer n
real ( kind = 8 ) c(4,n)
real ( kind = 8 ) divdf1
real ( kind = 8 ) divdf3
real ( kind = 8 ) dtau
real ( kind = 8 ) g
integer i
integer ibcbeg
integer ibcend
real ( kind = 8 ) tau(n)
!
! C(3,*) and C(4,*) are used initially for temporary storage.
!
! Store first differences of the TAU sequence in C(3,*).
!
! Store first divided difference of data in C(4,*).
!
do i = 2, n
c(3,i) = tau(i) - tau(i-1)
end do
do i = 2, n
c(4,i) = ( c(1,i) - c(1,i-1) ) / ( tau(i) - tau(i-1) )
end do
!
! Construct the first equation from the boundary condition
! at the left endpoint, of the form:
!
! C(4,1)*S(1) + C(3,1)*S(2) = C(2,1)
!
! IBCBEG = 0: Not-a-knot
!
if ( ibcbeg == 0 ) then
if ( n <= 2 ) then
c(4,1) = 1.0D+00
c(3,1) = 1.0D+00
c(2,1) = 2.0D+00 * c(4,2)
go to 120
end if
c(4,1) = c(3,3)
c(3,1) = c(3,2) + c(3,3)
c(2,1) = ( ( c(3,2) + 2.0D+00 * c(3,1) ) * c(4,2) * c(3,3) &
+ c(3,2)**2 * c(4,3) ) / c(3,1)
!
! IBCBEG = 1: derivative specified.
!
else if ( ibcbeg == 1 ) then
c(4,1) = 1.0D+00
c(3,1) = 0.0D+00
if ( n == 2 ) then
go to 120
end if
!
! Second derivative prescribed at left end.
!
else
c(4,1) = 2.0D+00
c(3,1) = 1.0D+00
c(2,1) = 3.0D+00 * c(4,2) - c(3,2) / 2.0D+00 * c(2,1)
if ( n == 2 ) then
go to 120
end if
end if
!
! If there are interior knots, generate the corresponding
! equations and carry out the forward pass of Gauss elimination,
! after which the I-th equation reads:
!
! C(4,I) * S(I) + C(3,I) * S(I+1) = C(2,I).
!
do i = 2, n-1
g = -c(3,i+1) / c(4,i-1)
c(2,i) = g * c(2,i-1) + 3.0D+00 * ( c(3,i) * c(4,i+1) + c(3,i+1) * c(4,i) )
c(4,i) = g * c(3,i-1) + 2.0D+00 * ( c(3,i) + c(3,i+1))
end do
!
! Construct the last equation from the second boundary condition, of
! the form
!
! -G * C(4,N-1) * S(N-1) + C(4,N) * S(N) = C(2,N)
!
! If slope is prescribed at right end, one can go directly to
! back-substitution, since the C array happens to be set up just
! right for it at this point.
!
if ( ibcend == 1 ) then
go to 160
end if
if ( 1 < ibcend ) then
go to 110
end if
90 continue
!
! Not-a-knot and 3 <= N, and either 3 < N or also not-a-knot
! at left end point.
!
if ( n /= 3 .or. ibcbeg /= 0 ) then
g = c(3,n-1) + c(3,n)
c(2,n) = ( ( c(3,n) + 2.0D+00 * g ) * c(4,n) * c(3,n-1) + c(3,n)**2 &
* ( c(1,n-1) - c(1,n-2) ) / c(3,n-1) ) / g
g = - g / c(4,n-1)
c(4,n) = c(3,n-1)
c(4,n) = c(4,n) + g * c(3,n-1)
c(2,n) = ( g * c(2,n-1) + c(2,n) ) / c(4,n)
go to 160
end if
!
! N=3 and not-a-knot also at left.
!
100 continue
c(2,n) = 2.0D+00 * c(4,n)
c(4,n) = 1.0D+00
g = -1.0D+00 / c(4,n-1)
c(4,n) = c(4,n) - c(3,n-1) / c(4,n-1)
c(2,n) = ( g * c(2,n-1) + c(2,n) ) / c(4,n)
go to 160
!
! IBCEND = 2: Second derivative prescribed at right endpoint.
!
110 continue
c(2,n) = 3.0D+00 * c(4,n) + c(3,n) / 2.0D+00 * c(2,n)
c(4,n) = 2.0D+00
g = -1.0D+00 / c(4,n-1)
c(4,n) = c(4,n) - c(3,n-1) / c(4,n-1)
c(2,n) = ( g * c(2,n-1)+c(2,n))/c(4,n)
go to 160
!
! N = 2.
!
120 continue
if ( ibcend == 2 ) then
c(2,n) = 3.0D+00 * c(4,n) + c(3,n) / 2.0D+00 * c(2,n)
c(4,n) = 2.0D+00
g = -1.0D+00 / c(4,n-1)
c(4,n) = c(4,n) - c(3,n-1) / c(4,n-1)
c(2,n) = (g*c(2,n-1)+c(2,n)) / c(4,n)
else if ( ibcend == 0 .and. ibcbeg /= 0 ) then
c(2,n) = 2.0D+00 * c(4,n)
c(4,n) = 1.0D+00
g = -1.0D+00 / c(4,n-1)
c(4,n) = c(4,n) - c(3,n-1) / c(4,n-1)
c(2,n) = (g*c(2,n-1)+c(2,n))/c(4,n)
else if ( ibcend == 0 .and. ibcbeg == 0 ) then
c(2,n) = c(4,n)
end if
!
! Back solve the upper triangular system
! C(4,I) * S(I) + C(3,I) * S(I+1) = B(I)
! for the slopes C(2,I), given that S(N) is already known.
!
160 continue
do i = n-1, 1, -1
c(2,i) = ( c(2,i) - c(3,i) * c(2,i+1) ) / c(4,i)
end do
!
! Generate cubic coefficients in each interval, that is, the
! derivatives at its left endpoint, from value and slope at its
! endpoints.
!
do i = 2, n
dtau = c(3,i)
divdf1 = ( c(1,i) - c(1,i-1) ) / dtau
divdf3 = c(2,i-1) + c(2,i) - 2.0D+00 * divdf1
c(3,i-1) = 2.0D+00 * ( divdf1 - c(2,i-1) - divdf3 ) / dtau
c(4,i-1) = 6.0D+00 * divdf3 / dtau**2
end do
return
end
c4science · Help