PROGRAM axist1 C This program solves the AXIsymmetric STraining deformation of C two spheres. It calculates coefficients for the resistance functions C C Q Q C X (s,lambda) and X (s,lambda) C 11 12 C C The main counting variable is M=N+P+Q, where N, P, Q are array indices. C If the maximum power we wish to find is S**(- MAXS), then the maximum value C of M is MMAX=MAXS+2. Dimension P,V to INDX2(2,0,MAXS). Also dimension C C (n+is) / C the combinatorial factor C0(n,is)= ( ) / (n+1) C ( n )/ C to ( (MMAX+1)/2, (MMAX+1)/2 ). C The dimensions given below are for MAXS = 15. In AXIST2.FOR there is a C table of dimensions for different MAXS. C DOUBLE PRECISION P(0:558),V(0:558),C0(9,9),XN,XS DOUBLE PRECISION FCTR1,FCTR2,FCTR3,FCTRV,FAC,PSUM,VSUM INTEGER MMAX,MAXS,NM,N,IS,M,NQ,IQ,INDX2,JS,IP LOGICAL EVEN COMMON/ARRAYS/P,V,C0 C This COMMON block reduces disk space requirements with many compilers C and helps memory management on some computers. MAXS=15 MMAX=MAXS+2 C First tabulate C0. DO 10 N=1,(MMAX+1)/2 XN=DBLE(N) C0(N,1)=1D0 DO 10 IS=2,(MMAX+1)/2 XS=DBLE(IS) C0(N,IS)=C0(N,IS-1)*(XN+XS)/XS 10 CONTINUE C We set the initial conditions. The labelling scheme needs some zero C elements at the start. DO 5 M=0,5 P(M)=0D0 5 V(M)=0D0 P(INDX2(2,0,0))=1D0 V(INDX2(2,0,0))=1D0 C We start at N=1, Q=0, P=M-1 and proceed keeping M fixed. C NM is the solution of the equation M=N+P+Q=NM+NM-2+0, except C the last time through when only N=1 and 2 are calculated. DO 500 M=3,MMAX NM=M/2+1 IF(M.EQ.MMAX)NM=2 DO 400 N=1,NM C The equation for NQ is M=N+P+Q=N+N-2+NQ NQ=M-2*(N-1) XN=DBLE(N) FCTR1=XN*(XN+.5D0) FCTR2=XN*(XN-.5D0) FCTR3=XN*(2D0*XN**2-.5D0) FCTRV=2D0*XN/(2D0*XN+3D0) DO 300 IQ=0,NQ C Now that M, N, Q are specified, I know P (denoted IP) as well. C We obtain JS from the equation JS-2=IQ-JS. IP=M-IQ-N PSUM=0D0 VSUM=0D0 C The term corresponding to IQ = NQ is INDX2(N,N-2,NQ) and C this is always 0, except when N=2. Moreover, if we plug this combination C of subscripts into our recurrence formulae, we find that we strike a C subscript INDX2(IS, NQ-IS, -1), which we have not included in our C numbering scheme, so we must implement special tests. C If IQ = NQ we skip the summation. And, since we are on the subject, C the case IQ=0 is also special. IF ( IQ.EQ.0 .OR. IQ.EQ.NQ) GO TO 250 JS=IQ/2+1 DO 200 IS=1,JS XS=DBLE(IS) FAC =C0(N,IS) C The way the loops and the index mapping are set up, the first term C in (3.9) will always be legal, but the other terms in (3.8)-(3.9) must be C tested to make certain that the INDX2 is legal (an illegal argument to C INDX2 means that the term can be shown analytically to be zero). C PSUM=PSUM+FAC*FCTR1*(2D0*XN*XS-XN-XS+2D0)* & P(INDX2(IS,IQ-IS,IP-N+1))/((XN+XS)*(2D0*XS-1D0)) IF(IP.GT.N) THEN PSUM=PSUM-FAC*FCTR2*P(INDX2(IS,IQ-IS,IP-N-1)) VSUM=VSUM-FAC*FCTRV*P(INDX2(IS,IQ-IS,IP-N-1)) ENDIF C The largest value of IS for the next term is given by IS-2 = IQ-IS-2. C Thus IS = IQ/2 = JS-1. Thus we skip this if IS = JS. IF(IS.LT.JS) THEN PSUM=PSUM-FAC*FCTR3*V(INDX2(IS,IQ-IS-2,IP-N+1))/(2D0*XS+1D0) ENDIF 200 CONTINUE 250 P(INDX2(N,IP,IQ))=PSUM V(INDX2(N,IP,IQ))=PSUM+VSUM 300 CONTINUE 400 CONTINUE 500 CONTINUE do 690 m=1,maxs EVEN= M .EQ. 2*(M/2) NM = (M*(M+1))/2 v(nm)=0d0 if (even) then do 640 iq=1,m-1 ip=m-iq-1 psum=0d0 do 630 n=1,(iq+2)/2 psum=psum+p( indx2(n,iq-n,ip) ) 630 continue 640 v( nm+iq) = 2.5d0*psum v(nm+m)=0d0 else do 660 iq=1,m ip=m-iq psum=0d0 do 650 n=1,(iq+2)/2 psum=psum+p( indx2(n,iq-n-1,ip) ) 650 continue 660 v( nm+iq) = 2.5d0*psum endif 690 continue V(0)=-13D0 C We now list the coefficients as given in the paper. Note that the C coefficients stored in any file RXMxxx.DAT by version 2 of C this program are the P(2,M-q,q) terms without the 2**M factor. WRITE(6,750) 750 FORMAT(1X, &'RESISTANCE FUNCTIONS from Jeffrey, Morris & Brady, 1993') WRITE(6,760) 760 FORMAT(1X,'Phys. Fluids A5, 2317') WRITE (6,800) 800 FORMAT(/1X, & 'NON-ZERO COEFFICIENTS FOR RXQ11 AND RXQ12 (EQUNS 47)', & //28X,'AS FLOATING POINT AND AS FRACTION.' & /1X,'FOR S**( 0)') WRITE (6,805) 805 FORMAT(10X,'LAMBDA( 0) : 0.') DO 890 M=1,15 WRITE (6,850) M 850 FORMAT (1X,'FOR S**(-',I2,')') NM=((M+1)*M)/2 DO 880 IQ=0,M XN=V(NM+IQ)*DBLE(2**M) IF(XN.NE.0D0) THEN IF (M.LT.8) THEN WRITE (6,870) IQ,XN ELSE XS=XN*5D0 WRITE (6,870) IQ,XN,XS ENDIF ENDIF 870 FORMAT(10X,'LAMBDA(',I2,') : ',F15.2,10X,F13.0,'/5') 880 CONTINUE 890 CONTINUE C C Previously the equal spheres case was written out separately using C the commented out code below. C C WRITE(6,950) C950 FORMAT(/1X,'RXQ FOR EQUAL SPHERES: COEFFICIENTS FOR (2A/R)**N') C DO 990 M=0,15 C PSUM=0D0 C NM=((M+1)*M)/2 C DO 970 IQ=0,M C PSUM=PSUM+V(NM+IQ) C970 CONTINUE C PSUM=PSUM*2D0**(-M) C WRITE(6,975)M,PSUM C975 FORMAT (1X,' FOR (2A/R)**',I2,' =',E21.14) C990 CONTINUE STOP END INTEGER FUNCTION INDX2(N,IP,IQ) C -------------------------- Arguments ---------------------- INTEGER N,IP,IQ C This function maps 3-dimensional arrays onto 1-dimensional arrays. C For each array element P(n,p,q) we define M=n+p+q, which is the primary C variable. For each value of M, we find that the array elements are C non-zero only for 1=< n = P(n,n-1,m-2n+2). C INDX2 uses these facts to set up a triangular scheme for each M to convert C it to a linear array. The array is filled as follows: C n,p,q= 1,M-1,0 1,M-2,1 ....................... 1,0,M-1 1,-1,M C ***** 2,M-2,0 2,M-3,1 ............... 2,0,M-2 C ***** ******* .................... C ***** ******* n,M-n,0 ... n,n-2,M-2n+2 C Note that the integer arithmetic relies on correct divisibility. C ------------------------ Local variables ------------------- INTEGER M,M2,LM M=IP+IQ+N M2=M/2 LM=( M2*(M2+1) )/2 LM=LM*(M2+1)+ ( LM*(M2+2) )/3 C LM is calculated as above to avoid overflow. C If MMAX=200, then LM = 681750 LM=LM+(M-2*M2)*(M2+1)*(M2+1) INDX2=LM+(N-1)*(M-N+3)+IQ RETURN END