cairy.f

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      SUBROUTINE cairy(Z, ID, KODE, AI, NZ, IERR)
C***BEGIN PROLOGUE  CAIRY
C***DATE WRITTEN   830501   (YYMMDD)
C***REVISION DATE  890801   (YYMMDD)
C***CATEGORY NO.  B5K
C***KEYWORDS  AIRY FUNCTION,BESSEL FUNCTIONS OF ORDER ONE THIRD
C***AUTHOR  AMOS, DONALD E., SANDIA NATIONAL LABORATORIES
C***PURPOSE  TO COMPUTE AIRY FUNCTIONS AI(Z) AND DAI(Z) FOR COMPLEX Z
C***DESCRIPTION
C
C         ON KODE=1, CAIRY COMPUTES THE COMPLEX AIRY FUNCTION AI(Z) OR
C         ITS DERIVATIVE DAI(Z)/DZ ON ID=0 OR ID=1 RESPECTIVELY. ON
C         KODE=2, A SCALING OPTION CEXP(ZTA)*AI(Z) OR CEXP(ZTA)*
C         DAI(Z)/DZ IS PROVIDED TO REMOVE THE EXPONENTIAL DECAY IN
C         -PI/3.LT.ARG(Z).LT.PI/3 AND THE EXPONENTIAL GROWTH IN
C         PI/3.LT.ABS(ARG(Z)).LT.PI WHERE ZTA=(2/3)*Z*CSQRT(Z)
C
C         WHILE THE AIRY FUNCTIONS AI(Z) AND DAI(Z)/DZ ARE ANALYTIC IN
C         THE WHOLE Z PLANE, THE CORRESPONDING SCALED FUNCTIONS DEFINED
C         FOR KODE=2 HAVE A CUT ALONG THE NEGATIVE REAL AXIS.
C         DEFINITIONS AND NOTATION ARE FOUND IN THE NBS HANDBOOK OF
C         MATHEMATICAL FUNCTIONS (REF. 1).
C
C         INPUT
C           Z      - Z=CMPLX(X,Y)
C           ID     - ORDER OF DERIVATIVE, ID=0 OR ID=1
C           KODE   - A PARAMETER TO INDICATE THE SCALING OPTION
C                    KODE= 1  RETURNS
C                             AI=AI(Z)                ON ID=0 OR
C                             AI=DAI(Z)/DZ            ON ID=1
C                        = 2  RETURNS
C                             AI=CEXP(ZTA)*AI(Z)       ON ID=0 OR
C                             AI=CEXP(ZTA)*DAI(Z)/DZ   ON ID=1 WHERE
C                             ZTA=(2/3)*Z*CSQRT(Z)
C
C         OUTPUT
C           AI     - COMPLEX ANSWER DEPENDING ON THE CHOICES FOR ID AND
C                    KODE
C           NZ     - UNDERFLOW INDICATOR
C                    NZ= 0   , NORMAL RETURN
C                    NZ= 1   , AI=CMPLX(0.0,0.0) DUE TO UNDERFLOW IN
C                              -PI/3.LT.ARG(Z).LT.PI/3 ON KODE=1
C           IERR   - ERROR FLAG
C                    IERR=0, NORMAL RETURN - COMPUTATION COMPLETED
C                    IERR=1, INPUT ERROR   - NO COMPUTATION
C                    IERR=2, OVERFLOW      - NO COMPUTATION, REAL(ZTA)
C                            TOO LARGE WITH KODE=1.
C                    IERR=3, CABS(Z) LARGE      - COMPUTATION COMPLETED
C                            LOSSES OF SIGNIFCANCE BY ARGUMENT REDUCTION
C                            PRODUCE LESS THAN HALF OF MACHINE ACCURACY
C                    IERR=4, CABS(Z) TOO LARGE  - NO COMPUTATION
C                            COMPLETE LOSS OF ACCURACY BY ARGUMENT
C                            REDUCTION
C                    IERR=5, ERROR              - NO COMPUTATION,
C                            ALGORITHM TERMINATION CONDITION NOT MET
C
C
C***LONG DESCRIPTION
C
C         AI AND DAI ARE COMPUTED FOR CABS(Z).GT.1.0 FROM THE K BESSEL
C         FUNCTIONS BY
C
C            AI(Z)=C*SQRT(Z)*K(1/3,ZTA) , DAI(Z)=-C*Z*K(2/3,ZTA)
C                           C=1.0/(PI*SQRT(3.0))
C                           ZTA=(2/3)*Z**(3/2)
C
C         WITH THE POWER SERIES FOR CABS(Z).LE.1.0.
C
C         IN MOST COMPLEX VARIABLE COMPUTATION, ONE MUST EVALUATE ELE-
C         MENTARY FUNCTIONS. WHEN THE MAGNITUDE OF Z IS LARGE, LOSSES
C         OF SIGNIFICANCE BY ARGUMENT REDUCTION OCCUR. CONSEQUENTLY, IF
C         THE MAGNITUDE OF ZETA=(2/3)*Z**1.5 EXCEEDS U1=SQRT(0.5/UR),
C         THEN LOSSES EXCEEDING HALF PRECISION ARE LIKELY AND AN ERROR
C         FLAG IERR=3 IS TRIGGERED WHERE UR=R1MACH(4)=UNIT ROUNDOFF.
C         ALSO, IF THE MAGNITUDE OF ZETA IS LARGER THAN U2=0.5/UR, THEN
C         ALL SIGNIFICANCE IS LOST AND IERR=4. IN ORDER TO USE THE INT
C         FUNCTION, ZETA MUST BE FURTHER RESTRICTED NOT TO EXCEED THE
C         LARGEST INTEGER, U3=I1MACH(9). THUS, THE MAGNITUDE OF ZETA
C         MUST BE RESTRICTED BY MIN(U2,U3). ON 32 BIT MACHINES, U1,U2,
C         AND U3 ARE APPROXIMATELY 2.0E+3, 4.2E+6, 2.1E+9 IN SINGLE
C         PRECISION ARITHMETIC AND 1.3E+8, 1.8E+16, 2.1E+9 IN DOUBLE
C         PRECISION ARITHMETIC RESPECTIVELY. THIS MAKES U2 AND U3 LIMIT-
C         ING IN THEIR RESPECTIVE ARITHMETICS. THIS MEANS THAT THE MAG-
C         NITUDE OF Z CANNOT EXCEED 3.1E+4 IN SINGLE AND 2.1E+6 IN
C         DOUBLE PRECISION ARITHMETIC. THIS ALSO MEANS THAT ONE CAN
C         EXPECT TO RETAIN, IN THE WORST CASES ON 32 BIT MACHINES,
C         NO DIGITS IN SINGLE PRECISION AND ONLY 7 DIGITS IN DOUBLE
C         PRECISION ARITHMETIC. SIMILAR CONSIDERATIONS HOLD FOR OTHER
C         MACHINES.
C
C         THE APPROXIMATE RELATIVE ERROR IN THE MAGNITUDE OF A COMPLEX
C         BESSEL FUNCTION CAN BE EXPRESSED BY P*10**S WHERE P=MAX(UNIT
C         ROUNDOFF,1.0E-18) IS THE NOMINAL PRECISION AND 10**S REPRE-
C         SENTS THE INCREASE IN ERROR DUE TO ARGUMENT REDUCTION IN THE
C         ELEMENTARY FUNCTIONS. HERE, S=MAX(1,ABS(LOG10(CABS(Z))),
C         ABS(LOG10(FNU))) APPROXIMATELY (I.E. S=MAX(1,ABS(EXPONENT OF
C         CABS(Z),ABS(EXPONENT OF FNU)) ). HOWEVER, THE PHASE ANGLE MAY
C         HAVE ONLY ABSOLUTE ACCURACY. THIS IS MOST LIKELY TO OCCUR WHEN
C         ONE COMPONENT (IN ABSOLUTE VALUE) IS LARGER THAN THE OTHER BY
C         SEVERAL ORDERS OF MAGNITUDE. IF ONE COMPONENT IS 10**K LARGER
C         THAN THE OTHER, THEN ONE CAN EXPECT ONLY MAX(ABS(LOG10(P))-K,
C         0) SIGNIFICANT DIGITS; OR, STATED ANOTHER WAY, WHEN K EXCEEDS
C         THE EXPONENT OF P, NO SIGNIFICANT DIGITS REMAIN IN THE SMALLER
C         COMPONENT. HOWEVER, THE PHASE ANGLE RETAINS ABSOLUTE ACCURACY
C         BECAUSE, IN COMPLEX ARITHMETIC WITH PRECISION P, THE SMALLER
C         COMPONENT WILL NOT (AS A RULE) DECREASE BELOW P TIMES THE
C         MAGNITUDE OF THE LARGER COMPONENT. IN THESE EXTREME CASES,
C         THE PRINCIPAL PHASE ANGLE IS ON THE ORDER OF +P, -P, PI/2-P,
C         OR -PI/2+P.
C
C***REFERENCES  HANDBOOK OF MATHEMATICAL FUNCTIONS BY M. ABRAMOWITZ
C                 AND I. A. STEGUN, NBS AMS SERIES 55, U.S. DEPT. OF
C                 COMMERCE, 1955.
C
C               COMPUTATION OF BESSEL FUNCTIONS OF COMPLEX ARGUMENT
C                 AND LARGE ORDER BY D. E. AMOS, SAND83-0643, MAY, 1983
C
C               A SUBROUTINE PACKAGE FOR BESSEL FUNCTIONS OF A COMPLEX
C                 ARGUMENT AND NONNEGATIVE ORDER BY D. E. AMOS, SAND85-
C                 1018, MAY, 1985
C
C               A PORTABLE PACKAGE FOR BESSEL FUNCTIONS OF A COMPLEX
C                 ARGUMENT AND NONNEGATIVE ORDER BY D. E. AMOS, TRANS.
C                 MATH. SOFTWARE, 1986
C
C***ROUTINES CALLED  CACAI,CBKNU,I1MACH,R1MACH
C***END PROLOGUE  CAIRY
      COMPLEX AI, CONE, CSQ, CY, S1, S2, TRM1, TRM2, Z, ZTA, Z3
      REAL AA, AD, AK, ALIM, ATRM, AZ, AZ3, BK, CK, COEF, C1, C2, DIG,
     * DK, D1, D2, ELIM, FID, FNU, RL, R1M5, SFAC, TOL, TTH, ZI, ZR,
     * Z3I, Z3R, R1MACH, BB, ALAZ
      INTEGER ID, IERR, IFLAG, K, KODE, K1, K2, MR, NN, NZ, I1MACH
      dimension cy(1)
      DATA tth, c1, c2, coef /6.66666666666666667e-01,
     * 3.55028053887817240e-01,2.58819403792806799e-01,
     * 1.83776298473930683e-01/
      DATA  cone / (1.0e0,0.0e0) /
C***FIRST EXECUTABLE STATEMENT  CAIRY
      ierr = 0
      nz=0
      IF (id.LT.0 .OR. id.GT.1) ierr=1
      IF (kode.LT.1 .OR. kode.GT.2) ierr=1
      IF (ierr.NE.0) RETURN
      az = cabs(z)
      tol = amax1(r1mach(4),1.0e-18)
      fid = float(id)
      IF (az.GT.1.0e0) GO TO 60
C-----------------------------------------------------------------------
C     POWER SERIES FOR CABS(Z).LE.1.
C-----------------------------------------------------------------------
      s1 = cone
      s2 = cone
      IF (az.LT.tol) GO TO 160
      aa = az*az
      IF (aa.LT.tol/az) GO TO 40
      trm1 = cone
      trm2 = cone
      atrm = 1.0e0
      z3 = z*z*z
      az3 = az*aa
      ak = 2.0e0 + fid
      bk = 3.0e0 - fid - fid
      ck = 4.0e0 - fid
      dk = 3.0e0 + fid + fid
      d1 = ak*dk
      d2 = bk*ck
      ad = amin1(d1,d2)
      ak = 24.0e0 + 9.0e0*fid
      bk = 30.0e0 - 9.0e0*fid
      z3r = real(z3)
      z3i = aimag(z3)
      DO 30 k=1,25
        trm1 = trm1*cmplx(z3r/d1,z3i/d1)
        s1 = s1 + trm1
        trm2 = trm2*cmplx(z3r/d2,z3i/d2)
        s2 = s2 + trm2
        atrm = atrm*az3/ad
        d1 = d1 + ak
        d2 = d2 + bk
        ad = amin1(d1,d2)
        IF (atrm.LT.tol*ad) GO TO 40
        ak = ak + 18.0e0
        bk = bk + 18.0e0
   30 CONTINUE
   40 CONTINUE
      IF (id.EQ.1) GO TO 50
      ai = s1*cmplx(c1,0.0e0) - z*s2*cmplx(c2,0.0e0)
      IF (kode.EQ.1) RETURN
      zta = z*csqrt(z)*cmplx(tth,0.0e0)
      ai = ai*cexp(zta)
      RETURN
   50 CONTINUE
      ai = -s2*cmplx(c2,0.0e0)
      IF (az.GT.tol) ai = ai + z*z*s1*cmplx(c1/(1.0e0+fid),0.0e0)
      IF (kode.EQ.1) RETURN
      zta = z*csqrt(z)*cmplx(tth,0.0e0)
      ai = ai*cexp(zta)
      RETURN
C-----------------------------------------------------------------------
C     CASE FOR CABS(Z).GT.1.0
C-----------------------------------------------------------------------
   60 CONTINUE
      fnu = (1.0e0+fid)/3.0e0
C-----------------------------------------------------------------------
C     SET PARAMETERS RELATED TO MACHINE CONSTANTS.
C     TOL IS THE APPROXIMATE UNIT ROUNDOFF LIMITED TO 1.0E-18.
C     ELIM IS THE APPROXIMATE EXPONENTIAL OVER- AND UNDERFLOW LIMIT.
C     EXP(-ELIM).LT.EXP(-ALIM)=EXP(-ELIM)/TOL    AND
C     EXP(ELIM).GT.EXP(ALIM)=EXP(ELIM)*TOL       ARE INTERVALS NEAR
C     UNDERFLOW AND OVERFLOW LIMITS WHERE SCALED ARITHMETIC IS DONE.
C     RL IS THE LOWER BOUNDARY OF THE ASYMPTOTIC EXPANSION FOR LARGE Z.
C     DIG = NUMBER OF BASE 10 DIGITS IN TOL = 10**(-DIG).
C-----------------------------------------------------------------------
      k1 = i1mach(12)
      k2 = i1mach(13)
      r1m5 = r1mach(5)
      k = min0(iabs(k1),iabs(k2))
      elim = 2.303e0*(float(k)*r1m5-3.0e0)
      k1 = i1mach(11) - 1
      aa = r1m5*float(k1)
      dig = amin1(aa,18.0e0)
      aa = aa*2.303e0
      alim = elim + amax1(-aa,-41.45e0)
      rl = 1.2e0*dig + 3.0e0
      alaz=alog(az)
C-----------------------------------------------------------------------
C     TEST FOR RANGE
C-----------------------------------------------------------------------
      aa=0.5e0/tol
      bb=float(i1mach(9))*0.5e0
      aa=amin1(aa,bb)
      aa=aa**tth
      IF (az.GT.aa) GO TO 260
      aa=sqrt(aa)
      IF (az.GT.aa) ierr=3
      csq=csqrt(z)
      zta=z*csq*cmplx(tth,0.0e0)
C-----------------------------------------------------------------------
C     RE(ZTA).LE.0 WHEN RE(Z).LT.0, ESPECIALLY WHEN IM(Z) IS SMALL
C-----------------------------------------------------------------------
      iflag = 0
      sfac = 1.0e0
      zi = aimag(z)
      zr = real(z)
      ak = aimag(zta)
      IF (zr.GE.0.0e0) GO TO 70
      bk = real(zta)
      ck = -abs(bk)
      zta = cmplx(ck,ak)
   70 CONTINUE
      IF (zi.NE.0.0e0) GO TO 80
      IF (zr.GT.0.0e0) GO TO 80
      zta = cmplx(0.0e0,ak)
   80 CONTINUE
      aa = real(zta)
      IF (aa.GE.0.0e0 .AND. zr.GT.0.0e0) GO TO 100
      IF (kode.EQ.2) GO TO 90
C-----------------------------------------------------------------------
C     OVERFLOW TEST
C-----------------------------------------------------------------------
      IF (aa.GT.(-alim)) GO TO 90
      aa = -aa + 0.25e0*alaz
      iflag = 1
      sfac = tol
      IF (aa.GT.elim) GO TO 240
   90 CONTINUE
C-----------------------------------------------------------------------
C     CBKNU AND CACAI RETURN EXP(ZTA)*K(FNU,ZTA) ON KODE=2
C-----------------------------------------------------------------------
      mr = 1
      IF (zi.LT.0.0e0) mr = -1
      CALL cacai(zta, fnu, kode, mr, 1, cy, nn, rl, tol, elim, alim)
      IF (nn.LT.0) GO TO 250
      nz = nz + nn
      GO TO 120
  100 CONTINUE
      IF (kode.EQ.2) GO TO 110
C-----------------------------------------------------------------------
C     UNDERFLOW TEST
C-----------------------------------------------------------------------
      IF (aa.LT.alim) GO TO 110
      aa = -aa - 0.25e0*alaz
      iflag = 2
      sfac = 1.0e0/tol
      IF (aa.LT.(-elim)) GO TO 180
  110 CONTINUE
      CALL cbknu(zta, fnu, kode, 1, cy, nz, tol, elim, alim)
  120 CONTINUE
      s1 = cy(1)*cmplx(coef,0.0e0)
      IF (iflag.NE.0) GO TO 140
      IF (id.EQ.1) GO TO 130
      ai = csq*s1
      RETURN
  130 ai = -z*s1
      RETURN
  140 CONTINUE
      s1 = s1*cmplx(sfac,0.0e0)
      IF (id.EQ.1) GO TO 150
      s1 = s1*csq
      ai = s1*cmplx(1.0e0/sfac,0.0e0)
      RETURN
  150 CONTINUE
      s1 = -s1*z
      ai = s1*cmplx(1.0e0/sfac,0.0e0)
      RETURN
  160 CONTINUE
      aa = 1.0e+3*r1mach(1)
      s1 = cmplx(0.0e0,0.0e0)
      IF (id.EQ.1) GO TO 170
      IF (az.GT.aa) s1 = cmplx(c2,0.0e0)*z
      ai = cmplx(c1,0.0e0) - s1
      RETURN
  170 CONTINUE
      ai = -cmplx(c2,0.0e0)
      aa = sqrt(aa)
      IF (az.GT.aa) s1 = z*z*cmplx(0.5e0,0.0e0)
      ai = ai + s1*cmplx(c1,0.0e0)
      RETURN
  180 CONTINUE
      nz = 1
      ai = cmplx(0.0e0,0.0e0)
      RETURN
  240 CONTINUE
      nz = 0
      ierr=2
      RETURN
  250 CONTINUE
      IF(nn.EQ.(-1)) GO TO 240
      nz=0
      ierr=5
      RETURN
  260 CONTINUE
      ierr=4
      nz=0
      RETURN
      END