subroutine cubspl ( tau, c, n, ibcbeg, ibcend ) 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. integer ibcbeg,ibcend,n, i,j,l,m real c(4,n),tau(n), divdf1,divdf3,dtau,g 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) return end