(D13)      CSM$USERS:[WHEREMAN.PROGRAMS.NPAINLEVE.SINGLE]P_RAM2.OUT;1


(C14) batch("np_exec.max");


(C15) /* ************************************************************************* */
/*                           Batch file NP_EXEC.MAX                          */
/* ************************************************************************* */

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

  SUBSTITUTE  X ----> G + X0

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

                                                   3
PAINLEVE ANALYSIS OF EQUATION,  F    + 4 F F  + 2 F   = 0
                                 G G        G

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

                  ALPHA
  SUBSTITUTE  U  G       FOR f IN ORIGINAL EQUATION.
               0

  MINIMUM POWERS OF g ARE  [2 ALPHA - 1, 3 ALPHA, ALPHA - 2]

                       2 ALPHA - 1         2
    * COEFFICIENT OF  G             IS  4 U  ALPHA
                                           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 .

                       3 ALPHA         3
    * COEFFICIENT OF  G         IS  2 U
                                       0

                       ALPHA - 2
    * COEFFICIENT OF  G           IS  U  (ALPHA - 1) ALPHA
                                       0

    NOTE : THIS TERM VANISHES FOR   [ALPHA = 0, ALPHA = 1] ,

    VERIFY IF ANY OF THESE VALUES FOR ALPHA LEADS TO DOMINANT

    BEHAVIOR,

    IF IT DOES THEN RUN THE PROGRAM AGAIN WITH THIS VALUE

    AS USER SUPPLIED ALPHA, CALLED BETA.

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

  FOR EXPONENTS ( 2 ALPHA - 1 ) AND ( 3 ALPHA ) OF g,

    WITH alpha =  - 1 , POWER OF g is  - 3  ----> SOLVE FOR  U
                                                              0

                       2    1
       TERM  2 (U  - 1)  U  --  IS DOMINANT
                 0        0  3
                            G

       IN EQUATION.

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

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

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

                              R - 3
       TERM ( R (R + 1) ) U  G       IS DOMINANT
                           R

       IN EQUATION.

       THE ONLY NON-NEGATIVE INTEGRAL ROOT IS  [R = 0]

 

       EXIT FIND_RESONANCES WITHOUT CHECKING.
PART called on atom: INSTANTU
Returned to Macsyma Toplevel.
(C16) QUIT();