stellinstall/cubsplb_8f_source.html

      subroutine cubsplb ( tau, c, n, ibcbeg, ibcend )

c

c  from  * a practical guide to splines *  by c. de boor

c     ************************  input  ***************************

c     n = number of data points. assumed to be .ge. 2.

c     (tau(i), c(1,i), i=1,...,n) = abscissae and ordinates of the

c        data points. tau is assumed to be strictly increasing.

c     ibcbeg, ibcend = boundary condition indicators, and

c     c(2,1), c(2,n) = boundary condition information. specifically,

c        ibcbeg = 0  means no boundary condition at tau(1) is given.

c           in this case, the not-a-knot condition is used, i.e. the

c           jump in the third derivative across tau(2) is forced to

c           zero, thus the first and the second cubic polynomial pieces

c           are made to coincide.)

c        ibcbeg = 1  means that the slope at tau(1) is made to equal

c           c(2,1), supplied by input.

c        ibcbeg = 2  means that the second derivative at tau(1) is

c           made to equal c(2,1), supplied by input.

c        ibcend = 0, 1, or 2 has analogous meaning concerning the

c           boundary condition at tau(n), with the additional infor-

c           mation taken from c(2,n).

c     ***********************  output  **************************

c     c(j,i), j=1,...,4; i=1,...,l (= n-1) = the polynomial coefficients

c        of the cubic interpolating spline with interior knots (or

c        joints) tau(2), ..., tau(n-1). precisely, in the interval

c        (tau(i), tau(i+1)), the spline f is given by

c           f(x) = c(1,i)+h*(c(2,i)+h*(c(3,i)+h*c(4,i)/3.)/2.)

c        where h = x - tau(i). the function program *ppvalu* may be

c        used to evaluate f or its derivatives from tau,c, l = n-1,

c        and k=4.

c

      implicit none

      integer ibcbeg,ibcend,n,   i,j,l,m

c

      real c(4,n),tau(n),   divdf1,divdf3,dtau,g

c

c****** a tridiagonal linear system for the unknown slopes s(i) of

c  f  at tau(i), i=1,...,n, is generated and then solved by gauss elim-

c  ination, with s(i) ending up in c(2,i), all i.

c     c(3,.) and c(4,.) are used initially for temporary storage.

      l = n-1

compute first differences of tau sequence and store in c(3,.). also,

compute first divided difference of data and store in c(4,.).

      do 10 m=2,n

         c(3,m) = tau(m) - tau(m-1)

   10    c(4,m) = (c(1,m) - c(1,m-1))/c(3,m)

construct first equation from the boundary condition, of the form

c             c(4,1)*s(1) + c(3,1)*s(2) = c(2,1)

      if (ibcbeg-1)                     11,15,16

   11 if (n .gt. 2)                     go to 12

c     no condition at left end and n = 2.

      c(4,1) = 1.

      c(3,1) = 1.

      c(2,1) = 2.*c(4,2)

                                        go to 25

c     not-a-knot condition at left end and n .gt. 2.

   12 c(4,1) = c(3,3)

      c(3,1) = c(3,2) + c(3,3)

      c(2,1) =((c(3,2)+2.*c(3,1))*c(4,2)*c(3,3)+

     >   c(3,2)**2*c(4,3))/c(3,1)

                                        go to 19

c     slope prescribed at left end.

   15 c(4,1) = 1.

      c(3,1) = 0.

                                        go to 18

c     second derivative prescribed at left end.

   16 c(4,1) = 2.

      c(3,1) = 1.

      c(2,1) = 3.*c(4,2) - c(3,2)/2.*c(2,1)

   18 if(n .eq. 2)                      go to 25

c  if there are interior knots, generate the corresp. equations and car-

c  ry out the forward pass of gauss elimination, after which the m-th

c  equation reads    c(4,m)*s(m) + c(3,m)*s(m+1) = c(2,m).

   19 do 20 m=2,l

         g = -c(3,m+1)/c(4,m-1)

         c(2,m) = g*c(2,m-1) + 3.*(c(3,m)*c(4,m+1)+c(3,m+1)*c(4,m))

   20    c(4,m) = g*c(3,m-1) + 2.*(c(3,m) + c(3,m+1))

construct last equation from the second boundary condition, of the form

c           (-g*c(4,n-1))*s(n-1) + c(4,n)*s(n) = c(2,n)

c     if slope is prescribed at right end, one can go directly to back-

c     substitution, since c array happens to be set up just right for it

c     at this point.

      if (ibcend-1)                     21,30,24

   21 if (n .eq. 3 .and. ibcbeg .eq. 0) go to 22

c     not-a-knot and n .ge. 3, and either n.gt.3 or  also not-a-knot at

c     left end point.

      g = c(3,n-1) + c(3,n)

      c(2,n) = ((c(3,n)+2.*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)

                                        go to 29

c     either (n=3 and not-a-knot also at left) or (n=2 and not not-a-

c     knot at left end point).

   22 c(2,n) = 2.*c(4,n)

      c(4,n) = 1.

                                        go to 28

c     second derivative prescribed at right endpoint.

   24 c(2,n) = 3.*c(4,n) + c(3,n)/2.*c(2,n)

      c(4,n) = 2.

                                        go to 28

   25 if (ibcend-1)                     26,30,24

   26 if (ibcbeg .gt. 0)                go to 22

c     not-a-knot at right endpoint and at left endpoint and n = 2.

      c(2,n) = c(4,n)

                                        go to 30

   28 g = -1./c(4,n-1)

complete forward pass of gauss elimination.

   29 c(4,n) = g*c(3,n-1) + c(4,n)

      c(2,n) = (g*c(2,n-1) + c(2,n))/c(4,n)

carry out back substitution

   30 j = l

   40    c(2,j) = (c(2,j) - c(3,j)*c(2,j+1))/c(4,j)

         j = j - 1

         if (j .gt. 0)                  go to 40

c****** generate cubic coefficients in each interval, i.e., the deriv.s

c  at its left endpoint, from value and slope at its endpoints.

      do 50 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.*divdf1

         c(3,i-1) = 2.*(divdf1 - c(2,i-1) - divdf3)/dtau

   50    c(4,i-1) = (divdf3/dtau)*(6./dtau)

c

                                        return

      end