(D15)     CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_DIPOLE.OUT;2


(C16) batch("np_exec.max")$


(C17) /* ************************************************************************* */
/*                           Batch file NP_EXEC.MAX                          */
/* ************************************************************************* */

exec_painleve (eq, alpha, do_resonances, max_resonance, do_simplification)$

  SUBSTITUTE  X ----> G + X0

----------------------------------------------------------------

                                   4          3     2          3
PAINLEVE ANALYSIS OF EQUATION,  6 F  M G + 3 F  (F )  M - 6 E F  + 4 A  = 0
                                                  G

----------------------------------------------------------------

                  ALPHA
  SUBSTITUTE  U  G       FOR f IN ORIGINAL EQUATION.
               0

  MINIMUM POWERS OF g ARE  [5 ALPHA - 2, 4 ALPHA + 1, 3 ALPHA, 0]

                       5 ALPHA - 2         5      2
    * COEFFICIENT OF  G             IS  3 U  ALPHA  M
                                           0

     NOTE : THIS TERM VANISHES FOR  ALPHA = 0 ,

     VERIFY IF  ALPHA = 0  LEADS TO DOMINANT BEHAVIOR,

     IF IT DOES THEN RUN THE PROGRAM AGAIN WITH THIS USER 

     SUPPLIED VALUE OF ALPHA.

     HENCE, PUT BETA =  0 .

                       4 ALPHA + 1         4
    * COEFFICIENT OF  G             IS  6 U  M
                                           0

                       3 ALPHA           3
    * COEFFICIENT OF  G         IS  - 6 U  E
                                         0

    * COEFFICIENT OF  1  IS  4 A

----------------------------------------------------------------

  FOR EXPONENTS ( 5 ALPHA - 2 ) AND ( 4 ALPHA + 1 ) OF g,

    POWER OF g IS NOT MINIMAL 
						-- SKIP THIS VALUE OF ALPHA.

----------------------------------------------------------------

  FOR EXPONENTS ( 5 ALPHA - 2 ) AND ( 3 ALPHA ) OF g,

    POWER OF g IS NOT MINIMAL 
						-- SKIP THIS VALUE OF ALPHA.

----------------------------------------------------------------

  FOR EXPONENTS ( 5 ALPHA - 2 ) AND ( 0 ) OF g,

                  2
    WITH alpha =  - , POWER OF g is  0  ----> SOLVE FOR  U
                  5                                       0

                   5
       TERM  4 (3 U  M + 25 A) 1  IS DOMINANT
                   0

       IN EQUATION.

 CANNOT SOLVE EXPLICITLY FOR  U
                               0

----------------------------------------------------------------

1 )  WITH  U  = UO  ----> FIND RESONANCES
            0

                    ALPHA         R + ALPHA
    SUBSTITUTE  U  G       +  U  G           FOR f IN EQUATION
                 0             R

                             4       R
       TERM ( 60 M (R + 1) UO  ) U  G   IS DOMINANT
                                  R

       IN EQUATION.

       THERE ARE NO NON-NEGATIVE INTEGRAL ROOTS FOR r.

----------------------------------------------------------------

  FOR EXPONENTS ( 4 ALPHA + 1 ) AND ( 3 ALPHA ) OF g,

    POWER OF g IS NOT MINIMAL 
						-- SKIP THIS VALUE OF ALPHA.

----------------------------------------------------------------

  FOR EXPONENTS ( 4 ALPHA + 1 ) AND ( 0 ) OF g,

    POWER OF g IS NOT MINIMAL 
						-- SKIP THIS VALUE OF ALPHA.

----------------------------------------------------------------

  FOR EXPONENTS ( 3 ALPHA ) AND ( 0 ) OF g,

    POWER OF g IS NOT MINIMAL 
						-- SKIP THIS VALUE OF ALPHA.

----------------------------------------------------------------


(C18) output()$

----------------------------------------------------------------

AT THE END OF THE COMPUTATIONS THE FOLLOWING ARE AVAILABLE:

* U VALUE(S)

  (type uval[j,k,l] where 1 <= j <=  1 and 0 <= k <=  [0]

  and 1 <= l <=  [1] )

  j stands for j_th alpha,k stands for u[k],l stands for

  l_th solution set for u[0] 

* ALPHA VALUE(S)

  (type alpha[j] where 1 <= j <=  1 )

  j stands for j_th alpha

* COMPATIBILITY CONDITION(S)

  (type compcond[j,k] where 1 <= j <=  1 and 1 <= k <=  [1] )

  j stands for j_th alpha,k stands for k_th solution set for u[0]

* RESONANCE(S)

  (type res[j,k] where 1 <= j <=  1 and 1 <= k <=  [1] )

  j stands for j_th alpha,k stands for k_th solution set for u[0]

----------------------------------------------------------------

TO SEE THIS MENU AGAIN JUST TYPE < output() >

----------------------------------------------------------------



(C19) /* ************************** END of NP_EXEC.MAX ************************** */



(C20) closefile();