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