PROGRAM drsdasl5 c>> 2008-10-26 DRSDASL5 Krogh Moved Format statements up for C conv. c>> 2008-10-24 DRSDASL5 Krogh Removed in INCLUDE statement & cDEC\$... C>> 2008-08-27 DRSDASL5 solve for initial y' value, change user codes C>> 2006-12-18 DRSDASL5 add total energy as a constraint. C>> 2001-10-11 DRSDASL5 R. J. Hanson Example 5 Code, with Download C Solve a planar pendulum problem in rectangular coordinates. C The equation is transformed from "Index 3" to "Index 1" C by differentiating the length constraint twice. The system C is integrated (using SDASSFA) using these three constraints, C including total energy. C In the second integration the problem is transformed from C "Index 3" to "Index 0" by differentiating the length constraint C three times. The routine SDASSFB uses four constraints, C including total energy. C This example shows that the constraints remain satisfied, C and the integration can succeed with either approach. C It is more efficient to use the "Index 0" problem than C the "Index 1" problem. C--S replaces "?": DR?DASL5, ?DASLS, ?DASLX, ?DASSFA, ?DASSFB EXTERNAL sdassfa,sdassfb,r1mach,sdasls C Set number of equations: INTEGER neq PARAMETER (neq=5) C Set maximum number of constraints. INTEGER maxcon PARAMETER (maxcon=4) C Work space sizes: INTEGER liw,lrw,ldc PARAMETER (liw=30+neq) PARAMETER (lrw=45+(5+2*maxcon+4)*neq+2*neq**2) PARAMETER (ldc=2*neq) INTEGER ndig,kk REAL tol C++S Default NDIG = 4 C++ Default NDIG = 11 C++ Substitute for NDIG below PARAMETER (ndig=4 ) PARAMETER (tol=10.e0**(-ndig)) INTEGER i,info(16),idid,iwork(liw) REAL t,y(neq),yprime(neq),tout,rtol(neq),atol(neq), & rwork(lrw),length,drl,drv,c(ldc,neq),ftol,rnktol, & r1mach LOGICAL constraint INTEGER kr,kf,kc COMMON /counts/ kr,kf,kc 70 FORMAT (14x,'Example Results for a Constrained Pendulum Problem, I &ndex 1') 80 FORMAT (14x,'Initial Position and Derivative Values using SDASLS') 90 FORMAT (6x,'At the time value',f10.2,2x,'the integration was stopp &ed.') 100 FORMAT ('The pendulum length or variation has large weighted error &s.'/'These should remain less than 1 is magnitude:',10x,2f8.2/) 110 FORMAT (14x,'Example Results for a Constrained Pendulum Problem, I &ndex 0') 120 FORMAT (5x,'No. of Residual Evaluations',6x,'Factorizations',6x,'P &rojection Solves'/'Constraint Partials-',1x,i8,12x,i8,17x,i8/) 130 FORMAT (/8x,'T',10x,'y_1',09x,'y_2',09x,'y_3',09x,'y_4',09x,'y_5'/ &1p,6d12.4/) 140 FORMAT (/8x,'T',09x,'y''_1',08x,'y''_2',08x,'y''_3',08x,'y''_4', & 08x,'y''_5'/1p,6d12.4/) WRITE (*,70) C Tolerances: DO 10 i=1,neq atol(i)=tol rtol(i)=tol 10 CONTINUE C Setup options: DO 20 i=1,16 info(i)=0 20 CONTINUE C Use partial derivatives provided in user code: info(5)=2 C Constrain solution, forward with 3 constraints: info(10)=3 C Compute the initial value of YPRIME(*). C Base computation of the initial y' on the C report by Krogh, Hanson, (2008), "Solving C Constrained Differential-Algebraic Systems C Using Projections," (8/2008). t=0 C Tolerances for a small F and a rank tolerance: ftol=r1mach(4)**(2./3.) rnktol=ftol C Give starting values to y and y' before Newton method: do 25 i = 1, neq y(i) = 0.E0 yprime(i) = 0.E0 25 continue c (Initial value for y(1) reset when ires is 0) CALL sdasls (sdassfa, neq, t, y, yprime, info, ftol, rnktol, c, & ldc, neq, idid, rwork, lrw, iwork, liw) WRITE (*,80) C Write the computed values for initial position and derivatives. WRITE (*,130) t,y WRITE (*,140) t,yprime C Now have initial values of y' so no need to compute them. C Allow more than nominal number of steps. info(12)=50000 C This is the pendulum length length=1.1e0 C This is how long we will integrate. kk=1000 DO 30 i=1,kk,10 C Integrate from T=I-1 to TOUT=T+10. C If the solution drifts away from the constraints, stop. t=i-1 tout=t+10 CALL sdaslx (sdassfa, neq, t, y, yprime, tout, info, rtol, atol, & idid, rwork, lrw, iwork, liw) C Compute residuals on length and its variation. They C should be smaller than the tolerances used. drl=(y(1)**2+y(2)**2-length**2)/(rtol(1)*length**2+atol(1)) drv=(y(1)*y(3)+y(2)*y(4))/(rtol(1)*(abs(y(1)*y(3))+ & abs(y(1)*y(4)))+atol(1)) constraint=abs(drl).le.1.e0.and.abs(drv).le.1.e0 IF (.not.constraint) GO TO 40 30 CONTINUE 40 CONTINUE WRITE (*,130) tout,y WRITE (*,120) kr,kf,kc WRITE (*,90) tout IF (.not.constraint) THEN WRITE (*,100) drl,drv END IF C Start the integration over for the index 0 problem. info(1)=0 C Set the number of constraints to 4. info(10)=4 C Starting values to y and y' before Newton method: do 45 i = 1, neq y(i) = 0.E0 yprime(i) = 0.E0 45 continue c (Initial value for y(1) and yprime(4) reset when ires is 0) DO 50 i=1,kk,10 C Integrate from T=I-1 to TOUT=T+10. t=i-1 tout=t+10 CALL sdaslx (sdassfb, neq, t, y, yprime, tout, info, rtol, atol, & idid, rwork, lrw, iwork, liw) C Compute residuals on length and its variation. They C should be smaller than the tolerances used. drl=(y(1)**2+y(2)**2-length**2)/(rtol(1)*length**2+atol(1)) drv=(y(1)*y(3)+y(2)*y(4))/(rtol(1)*(abs(y(1)*y(3))+ & abs(y(1)*y(4)))+atol(1)) constraint=abs(drl).le.1.e0.and.abs(drv).le.1.e0 IF (.not.constraint) GO TO 60 50 CONTINUE 60 CONTINUE WRITE (*,110) WRITE (*,130) tout,y WRITE (*,120) kr,kf,kc WRITE (*,90) tout IF (.not.constraint) THEN WRITE (*,100) drl,drv END IF END SUBROUTINE sdassfa (t, y, yprime, delta, d, ldd, cj, ires, rwork, & iwork) C Routine for the swinging simple pendulum problem, C with constraints on the index 2 and 3 equations. REAL t,y(*),yprime(*),delta(*),d(ldd,5),cj,rwork(*), & lsq,mg INTEGER ires,iwork(*),ldd REAL one,two,zero,mass,length,gravity PARAMETER (one=1.e0,two=2.e0,zero=0.e0,mass=1.e0,length=1.1e0, & gravity=9.806650e0) SAVE mg,lsq INTEGER kr,kf,kc COMMON /counts/ kr,kf,kc C This is the setup call. IF (ires.eq.0) THEN mg=mass*gravity lsq=length**2 y(1)=length C This is the only nonzero value for y'_4(0) but the startup C procedure will compute it. It is written here for C reference. C YPRIME(4)=-Gravity kr=0 kf=0 kc=0 END IF mg=mass*gravity lsq=length**2 C The system residual value. IF (ires.eq.1) THEN delta(1)=y(3)-yprime(1) delta(2)=y(4)-yprime(2) delta(3)=-y(1)*y(5)-mass*yprime(3) delta(4)=-y(2)*y(5)-mass*yprime(4)-mg delta(5)=mass*(y(3)**2+y(4)**2)-mg*y(2)-lsq*y(5) C Count residual evaluations. kr=kr+1 END IF C The mixed partial derivative matrix DF/Dy + c * DF/Dy' IF (ires.eq.2) THEN d(1,1)=-cj d(3,1)=-y(5) d(2,2)=-cj d(4,2)=-y(5) d(5,2)=-mg d(1,3)=one d(3,3)=-mass*cj d(5,3)=two*mass*y(3) d(2,4)=one d(4,4)=d(3,3) d(5,4)=two*mass*y(4) d(3,5)=-y(1) d(4,5)=-y(2) d(5,5)=-lsq kf=kf+1 END IF C The constraining equations after the corrector has converged. C Both partials and residuals are required. IF (ires.eq.5) THEN d(6,1)=y(1)*two d(6,2)=y(2)*two d(6,3)=zero d(6,4)=zero d(6,5)=zero d(7,1)=y(3) d(7,2)=y(4) d(7,3)=y(1) d(7,4)=y(2) d(7,5)=zero C Constraining total energy - d(8,1)=zero d(8,2)=mg d(8,3)=mass*y(3) d(8,4)=mass*y(4) d(8,5)=zero delta(6)=y(1)**2+y(2)**2-lsq delta(7)=y(1)*y(3)+y(2)*y(4) delta(8)=0.5e0*mass*(y(3)**2+y(4)**2)+mg*y(2) kc=kc+1 END IF C What follows are Partials w.r.t. T, Y' and Y, needed for C computing the starting values of y' based on the C Krogh/Hanson starting algorithm. C These values (6,7,8) of IRES occur only for use in C the computation of initial values for y'. IF (ires.eq.6) THEN C This is the partial of F, the residual function, w.r.t. T. C There is no explicit time dependence so this is zero. C The contents of DELTA(:) are set zero by the calling program C so there is nothing to do. C DELTA(1:6)=ZERO END IF C This is the partial of F w.r.t. y'. C Values not assigned are set zero by the calling program. IF (ires.eq.7) THEN d(1,1)=-one d(2,2)=-one d(3,3)=-mass d(4,4)=-mass kf=kf+1 END IF C This is the partial of F w.r.t. y. C Values not assigned are set zero by the calling program. IF (ires.eq.8) THEN d(3,1)=-y(5) d(4,2)=-y(5) d(5,2)=-mg d(1,3)=one d(5,3)=two*mass*y(3) d(2,4)=one d(5,4)=two*mass*y(4) d(3,5)=-y(1) d(4,5)=-y(2) d(5,5)=-lsq kf=kf+1 END IF END SUBROUTINE sdassfb (t, y, yprime, delta, p, ldp, cj, ires, rwork, & iwork) C C Routine for the swinging simple pendulum problem, C with constraints on the index 1,2 amd 3 equations. C Use P(:,:) to distinguish from D(:,:) in routine SDASSFA. REAL t,y(*),yprime(*),delta(*),p(ldp,5),cj,rwork(*), & lsq,mg INTEGER ires,iwork(*),ldp REAL one,two,three,zero,mass,length,gravity PARAMETER (one=1.e0,two=2.e0,three=3.e0,zero=0.e0,mass=1.e0, & length=1.1e0,gravity=9.806650e0) SAVE mg,lsq INTEGER kr,kf,kc COMMON /counts/ kr,kf,kc C This is the setup call. IF (ires.eq.0) THEN mg=mass*gravity lsq=length**2 y(1)=length yprime(4)=-gravity kr=0 kf=0 kc=0 END IF C The sytem residual value. IF (ires.eq.1) THEN delta(1)=y(3)-yprime(1) delta(2)=y(4)-yprime(2) delta(3)=-y(1)*y(5)-mass*yprime(3) delta(4)=-y(2)*y(5)-mass*yprime(4)-mg delta(5)=-three*mg*y(4)-lsq*yprime(5) kr=kr+1 END IF C The partial derivative matrix. Entries are all zero C so only non-zero values are needed. IF (ires.eq.2) THEN p(1,1)=-cj p(3,1)=-y(5) p(2,2)=-cj p(4,2)=-y(5) p(1,3)=one p(3,3)=-mass*cj p(2,4)=one p(4,4)=p(3,3) p(5,4)=-three*mg p(3,5)=-y(1) p(4,5)=-y(2) p(5,5)=-lsq*cj kf=kf+1 END IF C The constraining equations after the corrector has converged. C Both partials and residuals are required. IF (ires.eq.5) THEN p(6,1)=y(1)*two p(6,2)=y(2)*two p(6,3)=zero p(6,4)=zero p(6,5)=zero p(7,1)=y(3) p(7,2)=y(4) p(7,3)=y(1) p(7,4)=y(2) p(7,5)=zero p(8,1)=zero p(8,2)=-mg p(8,3)=two*mass*y(3) p(8,4)=two*mass*y(4) p(8,5)=-lsq C Constraining total energy - p(9,1)=zero p(9,2)=mg p(9,3)=mass*y(3) p(9,4)=mass*y(4) p(9,5)=zero delta(6)=y(1)**2+y(2)**2-lsq delta(7)=y(1)*y(3)+y(2)*y(4) delta(8)=mass*(y(3)**2+y(4)**2)-mg*y(2)-lsq*y(5) delta(9)=0.5e0*mass*(y(3)**2+y(4)**2)+mg*y(2) kc=kc+1 END IF END