subroutine tautsp ( tau, gtau, ntau, gamma, s,
* break, coef, l, k, iflag )
c from * a practical guide to splines * by c. de boor
constructs cubic spline interpolant to given data
c tau(i), gtau(i), i=1,...,ntau.
c if gamma .gt. 0., additional knots are introduced where needed to
c make the interpolant more flexible locally. this avoids extraneous
c inflection points typical of cubic spline interpolation at knots to
c rapidly changing data.
c
c parameters
c input
c tau sequence of data points. must be strictly increasing.
c gtau corresponding sequence of function values.
c ntau number of data points. must be at least 4 .
c gamma indicates whether additional flexibility is desired.
c = 0., no additional knots
c in (0.,3.), under certain conditions on the given data at
c points i-1, i, i+1, and i+2, a knot is added in the
c i-th interval, i=2,...,ntau-2. see description of meth-
c od below. the interpolant gets rounded with increasing
c gamma. a value of 2.5 for gamma is typical.
c in (3.,6.), same , except that knots might also be added in
c intervals in which an inflection point would be permit-
c ted. a value of 5.5 for gamma is typical.
c output
c break, coef, l, k give the pp-representation of the interpolant.
c specifically, for break(i) .le. x .le. break(i+1), the
c interpolant has the form
c f(x) = coef(1,i) +dx(coef(2,i) +(dx/2)(coef(3,i) +(dx/3)coef(4,i)))
c with dx = x - break(i) and i=1,...,l .
c iflag = 1, ok
c = 2, input was incorrect. a printout specifying the mistake
c was made.
c workspace
c s is required, of size (ntau,6). the individual columns of this
c array contain the following quantities mentioned in the write-
c up and below.
c s(.,1) = dtau = tau(.+1) - tau
c s(.,2) = diag = diagonal in linear system
c s(.,3) = u = upper diagonal in linear system
c s(.,4) = r = right side for linear system (initially)
c = fsecnd = solution of linear system , namely the second
c derivatives of interpolant at tau
c s(.,5) = z = indicator of additional knots
c s(.,6) = 1/hsecnd(1,x) with x = z or = 1-z. see below.
c
c ------ m e t h o d ------
c on the i-th interval, (tau(i), tau(i+1)), the interpolant is of the
c form
c (*) f(u(x)) = a + b*u + c*h(u,z) + d*h(1-u,1-z) ,
c with u = u(x) = (x - tau(i))/dtau(i). here,
c z = z(i) = addg(i+1)/(addg(i) + addg(i+1))
c (= .5, in case the denominator vanishes). with
c addg(j) = abs(ddg(j)), ddg(j) = dg(j+1) - dg(j),
c dg(j) = divdif(j) = (gtau(j+1) - gtau(j))/dtau(j)
c and
c h(u,z) = alpha*u**3 + (1 - alpha)*(max(((u-zeta)/(1-zeta)),0)**3
c with
c alpha(z) = (1-gamma/3)/zeta
c zeta(z) = 1 - gamma*min((1 - z), 1/3)
c thus, for 1/3 .le. z .le. 2/3, f is just a cubic polynomial on
c the interval i. otherwise, it has one additional knot, at
c tau(i) + zeta*dtau(i) .
c as z approaches 1, h(.,z) has an increasingly sharp bend near 1,
c thus allowing f to turn rapidly near the additional knot.
c in terms of f(j) = gtau(j) and
c fsecnd(j) = 2.derivative of f at tau(j),
c the coefficients for (*) are given as
c a = f(i) - d
c b = (f(i+1) - f(i)) - (c - d)
c c = fsecnd(i+1)*dtau(i)**2/hsecnd(1,z)
c d = fsecnd(i)*dtau(i)**2/hsecnd(1,1-z)
c hence can be computed once fsecnd(i),i=1,...,ntau, is fixed.
c f is automatically continuous and has a continuous second derivat-
c ive (except when z = 0 or 1 for some i). we determine fscnd(.) from
c the requirement that also the first derivative of f be contin-
c uous. in addition, we require that the third derivative be continuous
c across tau(2) and across tau(ntau-1) . this leads to a strictly
c diagonally dominant tridiagonal linear system for the fsecnd(i)
c which we solve by gauss elimination without pivoting.
c
integer iflag,k,l,ntau, i,method,ntaum1
real break(1),coef(4,1),gamma,gtau(ntau),s(ntau,6),tau(ntau)
* ,alpha,c,d,del,denom,divdif,entry,entry3,factor,factr2,gam
* ,onemg3,onemzt,ratio,sixth,temp,x,z,zeta,zt2
alph(x) = amin1(1.,onemg3/x)
c
c there must be at least 4 interpolation points.
if (ntau .ge. 4) go to 5
print 600,ntau
600 format(8h ntau = ,i4,20h. should be .ge. 4 .)
go to 999
c
construct delta tau and first and second (divided) differences of data
c
5 ntaum1 = ntau - 1
do 6 i=1,ntaum1
s(i,1) = tau(i+1) - tau(i)
if (s(i,1) .gt. 0.) go to 6
print 610,i,tau(i),tau(i+1)
610 format(7h point ,i3,13h and the next,2e15.6,15h are disordered)
go to 999
6 s(i+1,4) = (gtau(i+1)-gtau(i))/s(i,1)
do 7 i=2,ntaum1
7 s(i,4) = s(i+1,4) - s(i,4)
c
construct system of equations for second derivatives at tau. at each
c interior data point, there is one continuity equation, at the first
c and the last interior data point there is an additional one for a
c total of ntau equations in ntau unknowns.
c
i = 2
s(2,2) = s(1,1)/3.
sixth = 1./6.
method = 2
gam = gamma
if (gam .le. 0.) method = 1
if ( gam .le. 3.) go to 9
method = 3
gam = gam - 3.
9 onemg3 = 1. - gam/3.
c ------ loop over i ------
10 continue
c construct z(i) and zeta(i)
z = .5
go to (19,11,12),method
11 if (s(i,4)*s(i+1,4) .lt. 0.) go to 19
12 temp = abs(s(i+1,4))
denom = abs(s(i,4)) + temp
if (denom .eq. 0.) go to 19
z = temp/denom
if (abs(z - .5) .le. sixth) z = .5
19 s(i,5) = z
c ******set up part of the i-th equation which depends on
c the i-th interval
if (z - .5) 21,22,23
21 zeta = gam*z
onemzt = 1. - zeta
zt2 = zeta**2
alpha = alph(onemzt)
factor = zeta/(alpha*(zt2-1.) + 1.)
s(i,6) = zeta*factor/6.
s(i,2) = s(i,2) + s(i,1)*((1.-alpha*onemzt)*factor/2. - s(i,6))
c if z = 0 and the previous z = 1, then d(i) = 0. since then
c also u(i-1) = l(i+1) = 0, its value does not matter. reset
c d(i) = 1 to insure nonzero pivot in elimination.
if (s(i,2) .le. 0.) s(i,2) = 1.
s(i,3) = s(i,1)/6.
go to 25
22 s(i,2) = s(i,2) + s(i,1)/3.
s(i,3) = s(i,1)/6.
go to 25
23 onemzt = gam*(1. - z)
zeta = 1. - onemzt
alpha = alph(zeta)
factor = onemzt/(1. - alpha*zeta*(1.+onemzt))
s(i,6) = onemzt*factor/6.
s(i,2) = s(i,2) + s(i,1)/3.
s(i,3) = s(i,6)*s(i,1)
25 if (i .gt. 2) go to 30
s(1,5) = .5
c ******the first two equations enforce continuity of the first and of
c the third derivative across tau(2).
s(1,2) = s(1,1)/6.
s(1,3) = s(2,2)
entry3 = s(2,3)
if (z - .5) 26,27,28
26 factr2 = zeta*(alpha*(zt2-1.) + 1.)/(alpha*(zeta*zt2-1.)+1.)
ratio = factr2*s(2,1)/s(1,2)
s(2,2) = factr2*s(2,1) + s(1,1)
s(2,3) = -factr2*s(1,1)
go to 29
27 ratio = s(2,1)/s(1,2)
s(2,2) = s(2,1) + s(1,1)
s(2,3) = -s(1,1)
go to 29
28 ratio = s(2,1)/s(1,2)
s(2,2) = s(2,1) + s(1,1)
s(2,3) = -s(1,1)*6.*alpha*s(2,6)
c at this point, the first two equations read
c diag(1)*x1 + u(1)*x2 + entry3*x3 = r(2)
c -ratio*diag(1)*x1 + diag(2)*x2 + u(2)*x3 = 0.
c eliminate first unknown from second equation
29 s(2,2) = ratio*s(1,3) + s(2,2)
s(2,3) = ratio*entry3 + s(2,3)
s(1,4) = s(2,4)
s(2,4) = ratio*s(1,4)
go to 35
30 continue
c ******the i-th equation enforces continuity of the first derivative
c across tau(i). it has been set up in statements 35 up to 40
c and 21 up to 25 and reads now
c -ratio*diag(i-1)*xi-1 + diag(i)*xi + u(i)*xi+1 = r(i) .
c eliminate (i-1)st unknown from this equation
s(i,2) = ratio*s(i-1,3) + s(i,2)
s(i,4) = ratio*s(i-1,4) + s(i,4)
c
c ******set up the part of the next equation which depends on the
c i-th interval.
35 if (z - .5) 36,37,38
36 ratio = -s(i,6)*s(i,1)/s(i,2)
s(i+1,2) = s(i,1)/3.
go to 40
37 ratio = -(s(i,1)/6.)/s(i,2)
s(i+1,2) = s(i,1)/3.
go to 40
38 ratio = -(s(i,1)/6.)/s(i,2)
s(i+1,2) = s(i,1)*((1.-zeta*alpha)*factor/2. - s(i,6))
c ------ end of i loop ------
40 i = i+1
if (i .lt. ntaum1) go to 10
s(i,5) = .5
c
c ------ last two equations ------
c the last two equations enforce continuity of third derivative and
c of first derivative across tau(ntau-1).
entry = ratio*s(i-1,3) + s(i,2) + s(i,1)/3.
s(i+1,2) = s(i,1)/6.
s(i+1,4) = ratio*s(i-1,4) + s(i,4)
if (z - .5) 41,42,43
41 ratio = s(i,1)*6.*s(i-1,6)*alpha/s(i-1,2)
s(i,2) = ratio*s(i-1,3) + s(i,1) + s(i-1,1)
s(i,3) = -s(i-1,1)
go to 45
42 ratio = s(i,1)/s(i-1,2)
s(i,2) = ratio*s(i-1,3) + s(i,1) + s(i-1,1)
s(i,3) = -s(i-1,1)
go to 45
43 factr2 = onemzt*(alpha*(onemzt**2-1.)+1.)/
* (alpha*(onemzt**3-1.)+1.)
ratio = factr2*s(i,1)/s(i-1,2)
s(i,2) = ratio*s(i-1,3) + factr2*s(i-1,1) + s(i,1)
s(i,3) = -factr2*s(i-1,1)
c at this point, the last two equations read
c diag(i)*xi + u(i)*xi+1 = r(i)
c -ratio*diag(i)*xi + diag(i+1)*xi+1 = r(i+1)
c eliminate xi from last equation
45 s(i,4) = ratio*s(i-1,4)
ratio = -entry/s(i,2)
s(i+1,2) = ratio*s(i,3) + s(i+1,2)
s(i+1,4) = ratio*s(i,4) + s(i+1,4)
c
c ------ back substitution ------
c
s(ntau,4) = s(ntau,4)/s(ntau,2)
50 s(i,4) = (s(i,4) - s(i,3)*s(i+1,4))/s(i,2)
i = i - 1
if (i .gt. 1) go to 50
s(1,4) = (s(1,4)-s(1,3)*s(2,4)-entry3*s(3,4))/s(1,2)
c
c ------ construct polynomial pieces ------
c
break(1) = tau(1)
l = 1
do 70 i=1,ntaum1
coef(1,l) = gtau(i)
coef(3,l) = s(i,4)
divdif = (gtau(i+1)-gtau(i))/s(i,1)
z = s(i,5)
if (z - .5) 61,62,63
61 if (z .eq. 0.) go to 65
zeta = gam*z
onemzt = 1. - zeta
c = s(i+1,4)/6.
d = s(i,4)*s(i,6)
l = l + 1
del = zeta*s(i,1)
break(l) = tau(i) + del
zt2 = zeta**2
alpha = alph(onemzt)
factor = onemzt**2*alpha
coef(1,l) = gtau(i) + divdif*del
* + s(i,1)**2*(d*onemzt*(factor-1.)+c*zeta*(zt2-1.))
coef(2,l) = divdif + s(i,1)*(d*(1.-3.*factor)+c*(3.*zt2-1.))
coef(3,l) = 6.*(d*alpha*onemzt + c*zeta)
coef(4,l) = 6.*(c - d*alpha)/s(i,1)
coef(4,l-1) = coef(4,l) - 6.*d*(1.-alpha)/(del*zt2)
coef(2,l-1) = coef(2,l) - del*(coef(3,l) -(del/2.)*coef(4,l-1))
go to 68
62 coef(2,l) = divdif - s(i,1)*(2.*s(i,4) + s(i+1,4))/6.
coef(4,l) = (s(i+1,4)-s(i,4))/s(i,1)
go to 68
63 onemzt = gam*(1. - z)
if (onemzt .eq. 0.) go to 65
zeta = 1. - onemzt
alpha = alph(zeta)
c = s(i+1,4)*s(i,6)
d = s(i,4)/6.
del = zeta*s(i,1)
break(l+1) = tau(i) + del
coef(2,l) = divdif - s(i,1)*(2.*d + c)
coef(4,l) = 6.*(c*alpha - d)/s(i,1)
l = l + 1
coef(4,l) = coef(4,l-1) + 6.*(1.-alpha)*c/(s(i,1)*onemzt**3)
coef(3,l) = coef(3,l-1) + del*coef(4,l-1)
coef(2,l) = coef(2,l-1)+del*(coef(3,l-1)+(del/2.)*coef(4,l-1))
coef(1,l) = coef(1,l-1)+del*(coef(2,l-1)+(del/2.)*(coef(3,l-1)
* +(del/3.)*coef(4,l-1)))
go to 68
65 coef(2,l) = divdif
coef(3,l) = 0.
coef(4,l) = 0.
68 l = l + 1
70 break(l) = tau(i+1)
l = l - 1
k = 4
iflag = 1
return
999 iflag = 2
return
end